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On a farm there were some hens and sheep. Altogether there were 8 heads and 22 feet. How many hens were there? Arrange the numbers 1 to 6 in each set of circles below. The sum of each side of the triangle should equal the number in its centre. Fill in the missing numbers so that adding each pair of corner numbers gives you the number between them (in the box). Fill in the numbers to make the sum of each row, column and diagonal equal to 34. For an extra challenge try the huge American Flag magic square. There are three buckets each of which holds a maximum of 5 litres. Use the clues to work out how much liquid there is in each bucket. Annie and Ben are playing a game with a calculator. What was Annie's secret number? This problem is based on the story of the Pied Piper of Hamelin. Investigate the different numbers of people and rats there could have been if you know how many legs there are altogether! There were chews for 2p, mini eggs for 3p, Chocko bars for 5p and lollypops for 7p in the sweet shop. What could each of the children buy with their money? This magic square has operations written in it, to make it into a maze. Start wherever you like, go through every cell and go out a total of 15! There are 44 people coming to a dinner party. There are 15 square tables that seat 4 people. Find a way to seat the 44 people using all 15 tables, with no empty places. There are three baskets, a brown one, a red one and a pink one, holding a total of 10 eggs. Can you use the information given to find out how many eggs are in each basket? In this game for two players, the idea is to take it in turns to choose 1, 3, 5 or 7. The winner is the first to make the total 37. A game for 2 people. Use your skills of addition, subtraction, multiplication and division to blast the asteroids. Try grouping the dominoes in the ways described. Are there any left over each time? Can you explain why? There are 78 prisoners in a square cell block of twelve cells. The clever prison warder arranged them so there were 25 along each wall of the prison block. How did he do it? Look carefully at the numbers. What do you notice? Can you make another square using the numbers 1 to 16, that displays the same properties? Use your logical-thinking skills to deduce how much Dan's crisps and ice-cream cost altogether. Leah and Tom each have a number line. Can you work out where their counters will land? What are the secret jumps they make with their counters? Peter, Melanie, Amil and Jack received a total of 38 chocolate eggs. Use the information to work out how many eggs each person had. This activity is best done with a whole class or in a large group. Can you match the cards? What happens when you add pairs of the numbers together? Winifred Wytsh bought a box each of jelly babies, milk jelly bears, yellow jelly bees and jelly belly beans. In how many different ways could she make a jolly jelly feast with 32 legs? The value of the circle changes in each of the following problems. Can you discover its value in each problem? Can you work out how many flowers there will be on the Amazing Splitting Plant after it has been growing for six weeks? Can you find which shapes you need to put into the grid to make the totals at the end of each row and the bottom of each column? There are 4 jugs which hold 9 litres, 7 litres, 4 litres and 2 litres. Find a way to pour 9 litres of drink from one jug to another until you are left with exactly 3 litres in three of the jugs. Katie had a pack of 20 cards numbered from 1 to 20. She arranged the cards into 6 unequal piles where each pile added to the same total. What was the total and how could this be done? Cassandra, David and Lachlan are brothers and sisters. They range in age between 1 year and 14 years. Can you figure out their exact ages from the clues? How could you put eight beanbags in the hoops so that there are four in the blue hoop, five in the red and six in the yellow? Can you find all the ways of doing this? Place the digits 1 to 9 into the circles so that each side of the triangle adds to the same total. This task, written for the National Young Mathematicians' Award 2016, focuses on 'open squares'. What would the next five open squares look like? Ten cards are put into five envelopes so that there are two cards in each envelope. The sum of the numbers inside it is written on each envelope. What numbers could be inside the envelopes? There is a clock-face where the numbers have become all mixed up. Can you find out where all the numbers have got to from these ten statements? You have 5 darts and your target score is 44. How many different ways could you score 44? Use the information to work out how many gifts there are in each pile. Add the sum of the squares of four numbers between 10 and 20 to the sum of the squares of three numbers less than 6 to make the square of another, larger, number. Lolla bought a balloon at the circus. She gave the clown six coins to pay for it. What could Lolla have paid for the balloon? Woof is a big dog. Yap is a little dog. Emma has 16 dog biscuits to give to the two dogs. She gave Woof 4 more biscuits than Yap. How many biscuits did each dog get? The clockmaker's wife cut up his birthday cake to look like a clock face. Can you work out who received each piece? There were 22 legs creeping across the web. How many flies? How many spiders? Can you put plus signs in so this is true? 1 2 3 4 5 6 7 8 9 = 99 How many ways can you do it? Exactly 195 digits have been used to number the pages in a book. How many pages does the book have? What do the digits in the number fifteen add up to? How many other numbers have digits with the same total but no zeros? This dice train has been made using specific rules. How many different trains can you make? Suppose there is a train with 24 carriages which are going to be put together to make up some new trains. Can you find all the ways that this can be done? In Sam and Jill's garden there are two sorts of ladybirds with 7 spots or 4 spots. What numbers of total spots can you make? If you hang two weights on one side of this balance, in how many different ways can you hang three weights on the other side for it to be balanced? This challenge focuses on finding the sum and difference of pairs of two-digit numbers. Are these domino games fair? Can you explain why or why not? Using the statements, can you work out how many of each type of rabbit there are in these pens? This task, written for the National Young Mathematicians' Award 2016, invites you to explore the different combinations of scores that you might get on these dart boards.

Who got the triangle game? Where does the magic triangle point? The magic triangle explains the three objectives of an investment and their interdependence: security, availability and yield. These are in competition with each other. No type of investment can meet all three criteria at the same time. How does the triangle game work? Triangle game: The triangle game describes the formation of triangles by three players in order to always be able to offer the ball-carrying teammate a free pass station. This means that the ball-carrying actor forms a vertex of this triangle, which is connected to two other vertices. How many triangles are there? How many triangles are hidden in the shape shown? If you consider all the small, large and jointly formed triangles, you end up with exactly one result: 24 triangles have to be counted. Anyone who comes up with less has overlooked some. Whoever is over it has counted twice. Where did the missing square go? Where has this square gone? The trick to the puzzle is that the outer triangle isn’t really a triangle at all. The »hypotenuse« is not a straight line, but is bent inwards at the point where the two small triangles meet. Riddle: Darn triangle, who stole the square, Riddle 24 related questions found Who owns the triangle solution? Who owns the triangle determines who speaks first after—or while—the triangle is drawn. The person who speaks first owns the triangle. Depending on the exact variant, the speaker pauses briefly after the third person of the triangle and waits for someone to say something. What is the sum of the last triangle? The sum of all interior angles is always 180°. How many triangles in the picture? How many triangles can you count in this picture? Admittedly, the solution to the riddle is not easy. In fact, 25 triangles can be found in the image. However, most can only see 24, the trick is to count the «A» in the artist’s signature as a triangle. How many triangles do you see 18? The riddle probably comes from an intelligence test. It is said that people with an IQ of 120 or higher see 18 triangles. The network is currently discussing how many can actually be seen. However, there is no official solution yet. what is your temperature The normal human body temperature is 37 degrees Celsius — so it is said. What does the Willi solution like? «Willi likes soup, but he doesn’t like eggs», «Willi likes rats, but he doesn’t like mice», «He likes swimming, he doesn’t like bathing». What is the magic triangle project management? The magic triangle is a concept from project management. The aim here is to balance the three dimensions of time, scope and costs. It is important to know how these are related to each other and how the magic triangle can be adapted to the goals and management style. How to play Black Magic Tell your assistant the secret of the game in private. Tell him that you will point to different objects in the room and ask each one if it is the object you are thinking of. He should always answer «No» but pay attention to the color of the object you are pointing to. Why is the magic triangle called that? Especially in German-speaking countries, the «Magic Triangle» is known as a symbol for three central factors of project management. The factors were time, cost and performance (also time, cost, outcome/quality). The goal or the success factor is positioned in the middle of the triangle. Why is the magic triangle called that? Why is the magic triangle called magic? The magic triangle consists of the three important success factors scope of services, time and costs. These three dimensions are directly related to each other. What does the magic triangle say? The magic triangle in project management comprises three target dimensions: time, costs and performance or quality. These are directly related to each other. Their interactions must be carefully considered in order to be able to plan and carry out a project successfully. How many triangles are there in the star? Altogether there are 104 triangles in the star. Notes for weaker students: Types 1 and 2b of the above solutions are likely to be found by most students. However, types 2a and 3 are already more difficult to detect. How many triangles with 5 straight lines? How to draw 5 triangles with 8 strokes So theoretically you need 24 strokes to draw 8 individual triangles. You can also place the triangles next to each other or nest them to get to your goal with fewer strokes. How many 3 corners are there in a pentagram? The points of the pentagram are isosceles triangles. The angles between the base and legs of these triangles are 72°. The inner pentagon, together with two non-adjacent points, forms an isosceles triangle with a blunt point, the already mentioned 108° angle. Which direction is the bus going in? The answer is very simple: the bus drives to the left. The reason: the doors of the bus cannot be seen, so they have to be on the other side. How many triangles do you see squared? Solution. A colleague suggested the very simplest solution: Cut the square along the two diagonals. In this way you get four identically sized isosceles triangles that are also right-angled. How many triangles in the pyramid? A square pyramid is a geometric solid. It consists of a square base and a peripheral surface consisting of four isosceles triangles. Why does a square have 360 degrees? The square has one more corner than the triangle. So the sum of the angles is 180°+180°= 360°. Why is a triangle always 180 degrees? Every angle in a triangle has 2 equal exterior angles. Angle and exterior angle together add up to 180°. Each exterior angle is exactly the same size as the sum of the two interior angles that are not adjacent. Which triangle has 180 degrees? The interior angle theorem states: The sum of all interior angles in a triangle is 180°. This is usually expressed mathematically as follows: α + β + γ = 180°. This also makes it clear that the sum of the interior angles of a triangle can never exceed 180 degrees. Why can’t fungi do photosynthesis? Which rice is best for paella?

A sequence of numbers has an exponential pattern when each successive number increases (or decreases) by the same percent. Here are some examples of exponential patterns you have already studied in this text. People also ask, what is an exponential growth pattern? Exponential growth is a pattern of data that shows greater increases with passing time, creating the curve of an exponential function. On a chart, this curve starts slowly, remaining nearly flat for a time before increasing swiftly as to appear almost vertical. It follows the formula: V = S * (1 + R) ^ T. Also Know, how do you know if its exponential growth or decay? It’s exponential growth when the base of our exponential is bigger than 1, which means those numbers get bigger. It’s exponential decay when the base of our exponential is in between 1 and 0 and those numbers get smaller. An asymptote is a value that a function will get infinitely close to, but never quite reach. Accordingly, what does it mean if something is exponential? Scientific definitions for exponential Something is said to increase or decrease exponentially if its rate of change must be expressed using exponents. A graph of such a rate would appear not as a straight line, but as a curve that continually becomes steeper or shallower. What happens when something grows exponentially? Exponential growth is a phenomenon that occurs when the growth rate of the value of a mathematical function is proportional to the function’s current value, resulting in its growth with time being an exponential function. And just opposite when the value of rate decrease called Exponential Decay. 13 Related Question Answers Found What is an example of exponential growth? Exponential growth is growth that increases by a constant proportion. One of the best examples of exponential growth is observed in bacteria. It takes bacteria roughly an hour to reproduce through prokaryotic fission. What is another word for exponentially? There is no single word that is close to exponential. You are correct, of course, that the growth is not exponential. The closest words I can think of would be explosive, sudden, dramatic, rapid, mushrooming, snowballing, escalating, rocketing, skyrocketing, accelerating. What is an exponential curve? An exponential function or curve is a function that grows exponentially, or grows at an increasingly larger rate as you pick larger values of x, and usually takes the form. , where is any real number. What is an exponential function example? Exponential functions have the form f(x) = bx, where b > 0 and b ≠ 1. Just as in any exponential expression, b is called the base and x is called the exponent. An example of an exponential function is the growth of bacteria. Some bacteria double every hour. This can be written as f(x) = 2x. What causes exponential growth? Initially, growth is exponential because there are few individuals and ample resources available. Then, as resources begin to become limited, the growth rate decreases. Finally, growth levels off at the carrying capacity of the environment, with little change in population size over time. What makes a sequence exponential? yn = y0 · m n. A geometric sequence is completely described by giving its starting value y0 and the multiplication factor m. For this sequence y0 = 40 and m = 0.5. An exponential function is obtained from a geometric sequence by replacing the counting integer n by the real variable x. What are the rules of exponential functions? The following list outlines some basic rules that apply to exponential functions: The parent exponential function f(x) = bx always has a horizontal asymptote at y = 0, except when b = 1. You can’t raise a positive number to any power and get 0 or a negative number. You can’t multiply before you deal with the exponent. What does an exponential equation look like? Exponential Functions In an exponential function, the independent variable, or x-value, is the exponent, while the base is a constant. For example, y = 2x would be an exponential function. Here’s what that looks like. The formula for an exponential function is y = abx, where a and b are constants. What is exponential rule? EXPONENTIAL RULES. Rule 1: To multiply identical bases, add the exponents. Rule 2: To divide identical bases, subtract the exponents. Rule 3: When there are two or more exponents and only one base, multiply the exponents. What is an exponential relationship? Exponential relationships are relationships where one of the variables is an exponent. So instead of it being ‘2 multiplied by x’, an exponential relationship might have ‘2 raised to the power x’: Usually the first thing people do to get a grasp on what exponential relationships are like is draw a graph. What defines an exponential function? An exponential function is a mathematical function of the following form: f ( x ) = a x. where x is a variable, and a is a constant called the base of the function. The most commonly encountered exponential-function base is the transcendental number e , which is equal to approximately 2.71828. What are examples of exponential functions in real life? Population growth, radioactive decay, and loan interest rates are a few examples of naturally occurring exponential relationships. Learn how to model these situations using an exponential function to predict behavior, calculate half-life, or plan your budget. How do you do exponential decay and growth? Exponential Decay: Remember that the original exponential formula was y = abx. You will notice that in these new growth and decay functions, the b value (growth factor) has been replaced either by (1 + r) or by (1 – r). The growth “rate” (r) is determined as b = 1 + r.

LABOR ECONOMICS ECON 150A Popular in Course Popular in Economcs This 22 page Class Notes was uploaded by Arno Leuschke on Thursday October 22, 2015. The Class Notes belongs to ECON 150A at University of California Santa Barbara taught by K. Bedard in Fall. Since its upload, it has received 60 views. For similar materials see /class/227158/econ-150a-university-of-california-santa-barbara in Economcs at University of California Santa Barbara. Reviews for LABOR ECONOMICS Report this Material What is Karma? Karma is the currency of StudySoup. You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more! Date Created: 10/22/15 The Demand for Labor in a Competitive Market As wages rise the demand for labor falls We will prove this in a few days Wag e Labor The Demand for Labor in a Competitive Market Recall that a change in the wage is a movement along the labor demand curve while a shift is caused by a change in the output price a change Wage in an input price other than for labor or a technology change Ld after an output price rise Labor How do firms decide how much labor to hire The fundamental assumption of labor demand theory is that FIRMS ARE PROFIT MAXIMIZERS Firms are always trying to find ways to make larger pro ts The standard OUTPUT interpretation Increase output if the revenue from doing so exceeds the cost Marginal Revenue MR gt Marginal Cost MC The IN PUT interpretation Use more of an input ifthe income from doing so exceeds the cost How do firms decide how much labor to hire More formally a o The marginal product of labor MPL 5 The marginal product of capital MPK A rm can expand or contract output by increasing or decreasing its use of labor or capital However the rm is really interested in the revenue generated by these decisions How do firms decide how much labor to hire The marginal revenue product of labor is MRPL MPL x MR in general MRPL MPL x p in a competitive product market The marginal revenue product of capital is MRPK MPK x MR in general MRPK MPK x p in a competitive product market How do firms decide how much labor to hire In general Increase the amount of labor if MRPL gt MCL Decrease the amount of labor if MRPL lt MCL Keep the same amount of labor if MRPL MCL In a competitive labor market MCL W In a competitive capital market MCK r In a competitive labor market Increase the amount of labor if MRPL gt w Decrease the amount of labor if MRPL lt w Keep the same amount of labor if MRPL w The shortrun versus the longrun Short Run Firms can only alter their variable inputs labor Long Run Firms can alter all inputs Labor Demand in the SHORT RUN For simplicity assume that in the short run LABOR is variable and CAPITAL is fixed Capital KSR Q300 Q200 Q100 Labor With capital xed 39 39 39 39 lllUle labor to L 39 in output This is known as he DECREASING MARGINAL PRODUCTIVITY assumption There are a capital m as more labor is forced to use the same capital stock marginal productivity falls Labor Demand in the SHORT RUN As we saw earlier a firm will increase output or add more labor to the point where the benefit from adding one more unit of labor isjust equal to the cost of doing so M RPL w When both the output Q and labor L markets are competitive MPL X p w w MPL P In other words the productivity ofthe last person hired must equal their real wage Labor Demand in the SHORT RUN Example W 10 per hour p 3 and the last worker hired produces 3 units per hour Should you keep this worker Answer MPL 3 1 333 2 MPL lt 1 P P Since the last worker hired is worth less than he costs you should fire him Labor Demand in the SHORT RUN Capital Q200 o3oo o1oo L1 L2 L3 Labor Since we know that there are diminishing returns to adding more labor when capital is fixed the MPL must be downward sloping Labor Demand in the SHORT RUN MPL wp Labor Labor Demand in the SHORT RUN Since in equilibrium MPL wp then W p MPL And since MPL is falling as MP employment rises labor demand must be W L p increasing as wfalls Labor Optimal amount of labor in the short run In the short run capital is fixed The profit maximizing labor choice is thus given by MPL 3 p Optimal amount of labor in the short run Example Q2K L Z p10 w2 r1 KSR10000 Solution 6 KIM MPL QK 3 212K L 3 3 if 2 6L p P L p 2 pKlA pKlA szl2 L 2 SR 3 L SR 3 LSR ZSR w W w 10210300 2 SR 2 2 2500 Optimal amount of labor in the short run How much profit is earned 7239 pQ wL rKSR 172K 124L12 WL rKSR 10X2gtlt10X50 2X2500 1X10000 5000 Optimal amount of labor in the short run What if p increases to 20 172ng 20210900 2 Labordemand LSR 2 2 10000 w 2 Profit 723920X2X10gtlt100 2gtlt10000 1gtlt10000 10000 Optimal amount of labor in the short run What ifwdeoreases to 1 mm 10210900 2 Labordemand LSRp ZSR1 210000 W 39 Profit 723910gtlt2gtlt10gtlt100 1gtlt10000 1gtlt10000 0 Optimal amount of labor in the short run What is the elasticity of shortrun labor demand 6LSR X w 2p21lt2 X w 77512 6W LSR W3 sz z 2 W 2 Labor Demand in the LONG RUN Both LABOR and CAPITAL are exible in the long run A rm adds more labor andor capital to the point where the benefit from adding one more unit of input isjust equal to the cost of doing so MRPL w and MRPK r When both the output Q and input L and K markets are competitive pMPL w 3 MPL K p pMPK r 2 MPK L p Labor Demand in the LONG RUN Ifwe simply rearrange the equations r MPLw w and 2 i p MPL p MPK MPK r The benefitcost ratio for using one more unit of labor or capital to produce more output must be equal at the pro t maximizing point OthenNise the rm could use less ofone input and more ofthe other input to produce the same output at a lower cost Alternative solution method Alternatively you might want to think about the problem as the following twostage problem 1 The firm chooses the profit maximizing level of output gt the firm increases output until MR MC 2 The rm then chooses the least cost way to produce the profit maximizing level of output Graphically K Isocost Isoquant Notice that the cost minimizing method of producing the pro t maximizing level of output occurs where the slope ofthe isocost is the same as the slope ofthe isoquant In words The slope of the Isocost curve is wr wL rK Cost K Costr wr L o The slope of the Isoquant curve MRS LQ LQ AL AQ AQ AL AK MPK At Q the slope ofthe Isoquant and Iscost curves are the same MRTS 2 7 7 MPK The benefitcost ratio for using one more unit of labor or capital to produce more output must be equal at the profit maximizing point 2 MPL 1 Optimal amount of labor in the long run Example Q2K14L1 p10 w2 r1 Solution W V W w DMPL gt 2MPK 3 L 17 p WK r KIM W 12 Usmg3 L3 3 K 2 K KWL WK 7 L V L r 2r 2K34 pZW leZ 14 2 12 7W 717K 7 2r Subblnglnto LUZ 7 2 L77T 4 Rearranging to isolateL LLR p 3 2rw Optimal amount of capital in the long run 10 4 271413 74 2r 4rzwZ L Subblng L 1nto K KLR r Numerically LLR625 and KLR625 and 7r10x2x5x2572x62571x625625 What if p increases to 20 4 4 p p LLR 2mg 10000 and K W10000 and Ir 20x 2x 5x 2572x1000071x10000 10000 What if w decreases to 1 4 4 p P LLR 7 2M 75000 and K 7 MW 7 2500 and 7r10x2x707x70771x500071x2500 249698 What is the elasticity of long run labor demand Compare this to the short run elasticity of 6ng w 217214 w 773R gtlt 7 3gtlt 2 12 7 aw SR w 7 KSR w2 The longrun elasticity is always greater in absolute value because all inputs are exible Short and Long Run Elasticity Note the labor demand curves are not necessarily straight lines Elasticity Labor demand elasticity is high under the following conditions When the price elasticity of demand for the output produced is high Implications labor demand is more elastic at the firm level than the industry level labor demand is more elastic in the longrun When the other factors of production can easily be substituted for labor When labor costs are a large share of total cost Scale and Substitution Effects K A wage rise 4 4 L SE Sc Since L and K are adjustable in the long run it is important to understand how a change in the price of one input changes the demand for the other input Scale and Substitution Effects The scale and substitution effects are reinforcing for the input whose price changes and offsetting for the input whose price does not change Downward sloping long run LD Labor demand is downward sloping in the longrun because A wage rise leads to substitution away from labor A wage rise lowers the profit maximizing scale of output In other words the Sc and SE ensure a downward sloping labor demand curve in the longrun Sc and SE We know that the Sc and SE go in the same direction for labor is w changes but how do we determine whether the SE or 80 dominates for K Example Q2K 4L P10 12 r1 4 4 LongrunLandKare LLR 2177625 and KLR 457625 04 1 4 Ifw ital LLR 5000 and Km 2500 2113 41Z1Z Ki ScgtSE Let s work through and example Q 4L MK p 120 w 4 r 16KSR 64andafranchisefee 25p Shortrun W 7 W 7 W MPL 3 LSAK 4 3 1134W P P pKSR 743 3 11734743 w pK 3 L amp SR 43 W Numerical Answer 43 13 43 13 717 KER 120 64 591695x437284 LSR 7 w43 443 7 inwLirKSR 725p 4pL14K 4 iwLirKSR 725p 4x 120 x 37284 4 x 6414 7 4 x 37284716 x 647 25 x 120 5965727149136710247 3000 45036 Since profits are positive the profit maximizing level of short run labor demand is 37284 units of labor Lon run MP 9 1MP 3 2MP 1 3 L 1 L K P p NIPK r 43 13 p K 1 L WAK MP w L mK w K w wL 35415 j T j I K 13 p43 I I I 7 r 7p43L13 Subbmgmto1 L7 m gt L 13 W WV p43 3 p2 Reman i to isolate L L gng LR Wt3 W3ZrlZ p Solvm for K K 7 g LR WlZr3Z Numerical Answer Z Z 7 p 7 120 714400 7 LLR wzizriiz 4321612 8X4 7450 Z Z Km P i14 400 1125 WlZrKZ 4121632 64X2 7r pQ7wL7rK725p 4pL MK 7wL7rK725p 4x120x450 x1125 74x450716x1125725x120 7200718007180073000 600 Since profits are positive the profit maximizing level of long run labor demand is 450 units of labor and the longrun capital demand is 1125 units What if p increases to 130 p lzK ij 7130 64 7 65855X4 6 5 LSR m 4m 41484 w 7r pQ7wL7rKSR 7 25p 4pL l K Sj 7wL7rKSR 7 25p 4x 130 x 41484 x 64 M 7 4x 41484716x 647 25x 130 6637647165936710247 3250 70428 Since profits are positive the profit maximizing level of short run labor demand is 41484 units of labor 20 What if p increases to 130 Z Z p 130 16900 LLR WKZrlZ 4321612 8X4 5283913 Z Z Km p 130 716900 13203 wlZrKZ 4121632 64x2 7 inwLirK725p 4pL MK M iwLirK725p 4x130x52813 x1320 74x52813716x13203725x130 8448127211253721124873250 97311 Since profits are positive the profit maximizing level of long run labor demand is 52813 units of labor and the longrun capital demand is 13203 units What if w increases to 6 21721 SR 43 43 w 6 L p43ngf 1204364 3 59189x4 09 7 inwLirKSR 725p 4pL14K4 iwLirKSR 725p 4x 120 x 21721 x 64W 7 6x 21721716 x 647 25 x120 521271303267102473000 711526 Since profits are negative the profit maximizing level of shortrun labor demand is 0 units of labor 21 What if w increases to 6 2 2 7 p 7 120 7 14400 7 LLR 7 WKZrlZ 7 6321612 7147X4 7 244399 2 1202 14400 K 9184 LR wlZrKZ 6121632 64X 245 7pgiwLirK725p4le4KWiwLirK725p 4gtlt120gtlt2449 A x9184 4 76x2449716gtlt9184725gtlt120 587820 7146947146944 7 3000 76064 Since profits are negative the profit maximizing level of long run labor demand is 0 units of labor and the longrun capital demand is 0 units What is the elasticity of labor demand 8L 43K13 Shortrun vSR xLlexwi43 SR P SR WM 6L w 322 w Long run 77m a xg77w52r12XT7732 w32r12 The longrun elasticity is always greater in absolute value because all inputs are flexible 22 Are you sure you want to buy this material for You're already Subscribed! 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Presentation on theme: "COMPLETE BUSINESS STATISTICS"— Presentation transcript: 1 COMPLETE BUSINESS STATISTICS byAMIR D. ACZEL&JAYAVEL SOUNDERPANDIAN6th edition.Prepared by Lloyd Jaisingh, Morehead State University 2 Introduction and Descriptive Statistics 1Using StatisticsPercentiles and QuartilesMeasures of Central TendencyMeasures of VariabilityGrouped Data and the HistogramSkewness and KurtosisRelations between the Mean and Standard DeviationMethods of Displaying DataExploratory Data AnalysisUsing the Computer 3 LEARNING OBJECTIVESAfter studying this chapter, you should be able to:Distinguish between qualitative data and quantitative data.Describe nominal, ordinal, interval, and ratio scales of measurements.Describe the difference between population and sample.Calculate and interpret percentiles and quartiles.Explain measures of central tendency and how to compute them.Create different types of charts that describe data sets.Use Excel templates to compute various measures and create charts. 4 WHAT IS STATISTICS?Statistics is a science that helps us make better decisions in business and economics as well as in other fields.Statistics teaches us how to summarize, analyze, and draw meaningful inferences from data that then lead to improve decisions.These decisions that we make help us improve the running, for example, a department, a company, the entire economy, etc. 5 1-1. Using Statistics (Two Categories) Descriptive StatisticsCollectOrganizeSummarizeDisplayAnalyzeInferential StatisticsPredict and forecast values of population parametersTest hypotheses about values of population parametersMake decisions 6 Types of Data - Two Types (p.28) Qualitative Categorical or Nominal: Examples are-ColorGenderNationalityQuantitative Measurable or Countable: Examples are-TemperaturesSalariesNumber of points scored on a 100 point exam 7 Scales of Measurement (p.28-29) Analytical or metric typeInterval scaleRatio scaleCategorical or nonmertric typeNominal scaleOrdinal scale 8 Samples and Populations P.29 A population consists of the set of all measurements for which the investigator is interested.A sample is a subset of the measurements selected from the population.A census is a complete enumeration of every item in a population. 9 Simple Random SampleSampling from the population is often done randomly, such that every possible sample of equal size (n) will have an equal chance of being selected.A sample selected in this way is called a simple random sample or just a random sample.A random sample allows chance to determine its elements. 10 Samples and Populations Population (N)Sample (n) 11 Why Sample? Census of a population may be: Impossible Impractical Too costly 13 1-2 Percentiles and Quartiles Given any set of numerical observations, order them according to magnitude.The Pth percentile in the ordered set is that value below which lie P% (P percent) of the observations in the set.The position of the Pth percentile is given by (n + 1)P/100, where n is the number of observations in the set. 14 Example 1-2 (p.33)A large department store collects data on sales made by each of its salespeople. The number of sales made on a given day by each of 20 salespeople is shown on the next slide. Also, the data has been sorted in magnitude. 15 Example 1-2 (Continued) - Sales and Sorted Sales Sales Sorted Sales 16 Example 1-2 (Continued) Percentiles Find the 50th, 80th, and the 90th percentiles of thisdata set.To find the 50th percentile, determine the data pointin position (n + 1)P/100 = (20 + 1)(50/100) = 10.5.Thus, the percentile is located at the 10.5th position.The 10th observation is 16, and the 11th observation isalso 16.The 50th percentile will lie halfway between the 10thand 11th values and is thus 16. 17 Example 1-2 (Continued) Percentiles To find the 80th percentile, determine the data pointin position (n + 1)P/100 = (20 + 1)(80/100) = 16.8.Thus, the percentile is located at the 16.8th position.The 16th observation is 19, and the 17th observationis also 20.The 80th percentile is a point lying 0.8 of the wayfrom 19 to 20 and is thus 19.8. 18 Example 1-2 (Continued) Percentiles To find the 90th percentile, determine the data pointin position (n + 1)P/100 = (20 + 1)(90/100) = 18.9.Thus, the percentile is located at the 18.9th position.The 18th observation is 21, and the 19th observationis also 22.The 90th percentile is a point lying 0.9 of the wayfrom 21 to 22 and is thus 21.9.Example 1-2 19 Quartiles – Special Percentiles ,p.35) Quartiles are the percentage points thatbreak down the ordered data set into quarters.The first quartile is the 25th percentile. It is thepoint below which lie 1/4 of the data.The second quartile is the 50th percentile. It is thepoint below which lie 1/2 of the data. This is alsocalled the median.The third quartile is the 75th percentile. It is thepoint below which lie 3/4 of the data. 20 Quartiles and Interquartile Range The first quartile, Q1, (25th percentile) isoften called the lower quartile.The second quartile, Q2, (50thpercentile) is often called median or themiddle quartile.The third quartile, Q3, (75th percentile)is often called the upper quartile.The interquartile range is thedifference between the first and the thirdquartiles. 25 Summary Measures: Population Parameters Sample Statistics Measures of Central Tendency(衡量集中傾向)Median 中位數Mode 眾數Mean 平均數Measures of Variability(衡量變異性)Range 全距Interquartile range 四分位間距Variance 變異數Standard Deviation 標準差Other summary measures: 其他Skewness 偏態Kurtosis 峰態 26 1-3 Measures of Central Tendency or Location(p.36) Median 中位數Middle value whensorted in order ofmagnitude50th percentileMode 眾數Most frequently-occurring valueMean 平均數Average 27 Example – Median (Data is used from Example 1-2) Sales Sorted SalesSee slide # 19 for the template outputMedian50th Percentile(20+1)50/100=10.516 + (.5)(0) = 16MedianThe median is the middle value of data sorted in order of magnitude. It is the 50th percentile. 28 Example - Mode (Data is used from Example 1-2) See slide # 19 for the template output.: . : : :Mode = 16The mode is the most frequently occurring value. It is the value with the highest frequency. 29 Arithmetic Mean or Average The mean(平均數) of a set of observations is their average - the sum of the observed values divided by the number of observations.Population Mean母體平均數Sample Mean樣本平均數m=åxNi1xni=å1 30 Example – Mean (Data is used from Example 1-2) Sales9612101315161417242122181920317xni=å1317201585.See slide # 19 for the template output 31 Example - Mode (Data is used from Example 1-2) .: . : : :Mean = 15.85Median and Mode = 16每一點代表一個數值See slide # 19 for the template output 32 Exercise, p.40, 5 min 例1- 4 1-13 ~ 1-16 (See Textbook p.698) 1-17(Ans：mean=592.93, median=566,LQ=546, UQ=618.75Outlier=940,suspected outlier=399) 33 1-4 Measures of Variability or Dispersion (p.40) Range 全距Difference between maximum and minimum valuesInterquartile Range 四分位數間距Difference between third and first quartile (Q3 - Q1)Variance 變異數Average*of the squared deviations from the meanStandard Deviation 標準差Square root of the varianceDefinitions of population variance and sample variance differ slightly. 34 Example - Range and Interquartile Range (Data is used from Example 1-2) SortedSales Sales RankRangeMaximum - Minimum == 18MinimumQ1 = 13 + (.25)(1) = 13.25First QuartileQ3 = 18+ (.75)(1) = 18.75Third QuartileInterquartile RangeQ3 - Q1 == 5.5Maximum 35 Variance and Standard Deviation Population Variance母體變異數Sample Variance樣本變異數nNå(x-x)2å(x-m)2s=2i=1s=()2i=1n-1N()()2Nn2xxååNå=n=-i1åx-i1x22Nn==i=1i=1()Nn-1s=s2s=s2 40 1-5 Group Data and the Histogram 群聚數據與直方圖 Dividing data into groups or classes or intervalsGroups should be:Mutually exclusive 群間互斥Not overlapping - every observation is assigned to only one groupExhaustive 完全分群Every observation is assigned to a groupEqual-width (if possible) 等寬First or last group may be open-ended 41 Frequency Distribution頻率分配 Table with two columns兩行 listing:Each and every group or class or interval of valuesAssociated frequency of each groupNumber of observations assigned to each groupSum of frequencies is number of observationsN for populationn for sampleClass midpoint組中點 is the middle value of a group or class or intervalRelative frequency相對頻率 is the percentage of total observations in each classSum of relative frequencies = 1 42 Example 1-7: Frequency Distribution p.47 x f(x) f(x)/nSpending Class ($) Frequency (number of customers) Relative Frequency0 to less than100 to less than200 to less than300 to less than400 to less than500 to less thanExample of relative frequency: 30/184 = 0.163Sum of relative frequencies = 1 43 Cumulative Frequency Distribution x F(x) F(x)/nSpending Class ($) Cumulative Frequency Cumulative Relative Frequency0 to less than100 to less than200 to less than300 to less than400 to less than500 to less thanThe cumulative frequency累積頻率 of each group is the sum of thefrequencies of that and all preceding groups. 45 Histogram 直方圖A histogram is a chart made of bars of different heights. 不同高度之條狀圖Widths and locations of bars correspond to widths and locations of data groupings 寬度與位置代表群組的資料寬度與位置Heights of bars correspond to frequencies or relative frequencies of data groupings 高度代表頻率 48 1-6 Skewness偏度 and Kurtosis峰度 p.49 Measure of asymmetry of a frequency distributionSkewed to left 左偏 <0Symmetric or unskewed 對稱Skewed to right 右偏 >0KurtosisMeasure of flatness or peakedness of a frequency distributionPlatykurtic (relatively flat)Mesokurtic (normal)Leptokurtic (relatively peaked) *公示如p.51 52 Kurtosis扁度值越小, 越平扁Platykurtic平扁 - flat distribution 53 KurtosisMesokurtic - not too flat and not too peaked 54 Kurtosis扁度值越大, 越尖突Leptokurtic尖扁 - peaked distribution 55 1-7 Relations between the Mean and Standard Deviation p.51 (重要) Chebyshev’s Theorem柴比雪夫定理Applies to any distribution, regardless of shape 可應用於任何分配之數據Places lower limits on the percentages of observations within a given number of standard deviations from the meanEmpirical Ruler 經驗法則Applies only to roughly mound-shaped and symmetric distributions 適用山型與對稱之數據Specifies approximate percentages of observations within a given number of standard deviations from the mean 56 Chebyshev’s TheoremAt least of the elements of any distribution lie within k standard deviations of the mean234Standarddeviationsof the meanAtleastLiewithin 57 Empirical Rule 經驗法則For roughly mound-shaped and symmetric distributions, approximately: 59 1-8 Methods of Displaying Data Pie Charts 圓餅圖Categories represented as percentages of totalBar Graphs 直條圖Heights of rectangles represent group frequenciesFrequency Polygons 頻率圖Height of line represents frequencyOgives 累加頻率圖Height of line represents cumulative frequencyTime Plots 時間圖Represents values over time 65 1-9 Exploratory Data Analysis – EDA探索性資料分析 Techniques to determine relationships關係 and trends趨勢, identify outliers離群值 and influential有影響的 observations, and quickly describe快速描述 or summarize總結 data sets.Stem-and-Leaf Displays 莖葉Quick-and-dirty listing of all observations 快速瀏覽所有觀測值Conveys some of the same information as a histogram 將資料轉化成直方圖Box Plots 盒形圖MedianLower and upper quartilesMaximum and minimum 67 Box Plot 盒形圖 p.62 Elements of a Box Plot * o Q1 Q3 Inner Fence Outer MedianQ1Q3InnerFenceOuterInterquartile RangeSmallest data point not below inner fenceLargest data point not exceeding inner fenceSuspected outlierOutlierQ1-3(IQR)Q1-1.5(IQR)Q3+1.5(IQR)Q3+3(IQR)離群值一半數據在盒內IQR Your consent to our cookies if you continue to use this website.

Magical Mathematics - A Tribute to Martin GardnerPerhaps no one has done more to make the world aware of mathematics than Martin Gardner... Who are the people who write about mathematics in the clearest way? One might expect that such people are trained mathematicians - those who have majored in mathematics and gone on to get a doctorate in mathematics. While there are many wonderfully talented expositors of mathematics who are graduate school trained mathematicians, it turns out that one of the greatest expositor of mathematics, for me perhaps the greatest, was Martin Gardner. Gardner in a sense "became" a mathematician but did not "train" to do so. Martin Gardner in 1932 (Courtesy of Wikipedia) Many mathematicians see their subject as magical - having unexpected surprises and delights. However, all too often the magic of mathematics is lost on those who have not chosen mathematics or a related subject as a career. Perhaps no one has done more to bridge the gap between the sometimes lukewarm public view of mathematics and internal views of mathematics as a magical subject than Martin Gardner. Here I would like to provide a tribute to Martin Gardner for his efforts to popularize mathematics, as well as to his success in convincing many people of the beauty, applicability, joys and magic of mathematics. And I am not alone--this month mathematics pays tribute to him with "Mathematics, Magic, & Mystery," the theme for Mathematics Awareness Month (April) 2014. Perhaps no one has done more to make the world aware of mathematics than Martin Gardner. Who was Martin Gardner? Martin Gardner was born October 21, 1914 in Tulsa, Oklahoma. and died on May 22, 2010 not that far from where he was born, in Norman, Oklahoma. During his long life he lived in and traveled to many places but wherever he was, he seemed to carry his pen with him. What were the forces that molded this author of over 60 hardcover books? (Photo of Martin Gardner) Gardner's columns and books What put Gardner on the mathematical map was his association with Scientific American magazine, which had started publication in 1845. In December, 1956 Gardner wrote an article for Scientific American about an intriguing family of geometrical objects known as flexagons. Diagrams illustrating a flexagon (Courtesy of Wikipedia) Two positions of a flexagon (Courtesy of Wikipedia) (Photograph of Arthur Stone, courtesy of Stuart Anderson) Photo of Jeremiah (Jerry) Lyons, Martin Gardner, and Thomas Banchoff (Courtesy of Tom Banchoff) (Photo of John Horton Conway) (Photo of Ronald Graham) (Photo of Donald Knuth) The range of wonderful problems, examples, and theorems that Gardner treated over the years is enormous. They include ideas from geometry, algebra, number theory, graph theory, topology, and knot theory, to name but a few. Characteristically, his columns were self-contained and were light on the mathematical notation that makes many people have a distaste for mathematics. Not having any easy way to pick among many gems, I am choosing a few that show some magical or intuitive aspects, and one that shows Gardner's playfulness. We have mentioned that Gardner was born on October 21. He shared October 21st as a birthday with a variety of people who are perhaps better known than he: Benjamin Netanyahu, Kim Khardashian, Dizzy Gillespie and Malcolm Arnold. However, for those within mathematics who were born and died on this day, Gardner may hold his own in recognizability. For 10 people, there is about a 12 percent chance that at least 2 of them will have the same birthday. P(X|Y) = P(X and Y both occur)/P(Y). (*) A map requiring 5 colors? Martin Gardner had a playful streak and often had columns which had whimsical goals. In his Scientific American column of April 1975 he showed a plane map which purportedly required 5 colors to color the faces. The rule is that if two regions share an edge they should get different colors. (Plane map requiring 5 colors?) Image courtesy of Wolfram (Photo of Percy Heawood) (4-coloring by Stan Wagon of Gardner's April Fools' map) Image courtesy of Wolfram Joseph Malkewitch (sic) meets Martin Gardner I was in the audience to hear Martin Gardner but I never met him in person. However, in a variety of ways our paths crossed. Gardner first came to my attention when as a teen on a trip to Florida, an automobile accident stranded me and my parents in Winter Haven, Florida, while our car was being repaired. My "lifesaver" for this period was a local library within walking distance of where we were staying, which was where I got my first introduction to Scientific American magazine and the columns of Martin Gardner. On returning to NY I started a subscription to Scientific American, and my subscription has never lapsed. Eventually, when I had large enough living quarters I started to acquire Gardner's books. HIs books were initially easier to consult than trying to go back to his Scientific American columns. However, there was also the fact that when his Scientific American columns appeared in book form, Gardner updated information about the mathematics in the columns with additional details about what was new about the problem since it had initially been published. One wonderful aspect of Gardner's Scientific American columns was that he and an "army" of appreciative, motivated and talented people (including some who were professional mathematicians) thought about the questions that he raised. In some cases his columns dealt with issues which, at the time he wrote about them, were not fully understood. So his books allowed him the opportunity to "update" columns with new ideas and results when the columns were collected for reprinting. (Patch of tiles which consists of equilateral triangles and squares) I was very pleased when Martin Gardner picked up on this problem and used it in his book Knotted Doughnuts and the later compendium Colossal Book of Short Puzzles and Problems. However, Joseph Malkewitch is really me! (Vould I lie to you?) Carrying on the Gardner tradition Gardner fans were disappointed when Gardner no longer continued to write his column for Scientific American. The magazine tried to find a "replacement" for Gardner by having various authors and columns that tried to capture the spirit of Mathematical Games. Eventually, however, Scientific American resigned itself to the fact that Gardner really could not be "replaced!" Happily Gardner continued to provide inspiration for fans with his books (see a partial list in the references). (Photograph of Martin Gardner by Colm Mulcahy) A list of some of Martin Gardner's books: The AMS encourages your comments, and hopes you will join the discussions. We review comments before they're posted, and those that are offensive, abusive, off-topic or promoting a commercial product, person or website will not be posted. Expressing disagreement is fine, but mutual respect is required. Welcome to the These web essays are designed for those who have already discovered the joys of mathematics as well as for those who may be uncomfortable with mathematics. Search Feature Column Feature Column at a glance

If you're a fan of the Math Wars, you've heard about the question in the first edition of the EVERDAY MATH K-5 curriculum, "If math were a color, it would probably be ______ because ________." This particular question has probably engendered more ridicule and garnered more notoriety than anything associated with the NCTM Standards of 1989 on, and the progressive reform curricula that emerged from NSF grants in the early 1990s. After Googling on the question, I've concluded that no one on the planet has the slightest idea what could possibly be the point of asking this question. It has resulted in so much negative feedback that it has been dropped from revisions and subsequent editions of EVERYDAY MATHMATICS. Indeed, as recently as last week, Andy Isaacs, an author of the program who works at the University of Chicago Mathematics Project (UCSMP) posted on math-teach at the Math Forum: As for the "If math were a color, ..." question, it appeared in ourI suppose that since Andy, whom I've met and consider a friend, is washing his hands of the question and it's been dropped entirely from the series, I should let sleeping inquiries lie, even if the folks at Mathematically Correct, NYC-HOLD, and countless "parents-with-pitchforks" groups and web-sites (as well as right-wing pundits like Michelle Malkin continue to make "big" political points about how programs like EM are sending us and our children down the fast track to enslavement by those ever-inscrutable Asians (that Ms. Malkin is herself of Filipino extraction should in no way be taken as a mitigating circumstance in her willingness to help scare the James Jesus Angleton first edition in a Time to Reflect section at the end of the first unit in fifth grade, a unit that focused on prime and composite numbers, factoring, figurate numbers, and so on. The first edition of fifth grade was published in 1996. The "If math were a color, ..." question was deleted in the second edition, which was published in 2001, and the entire Time to Reflect feature was deleted from the current edition, the third, so this question has not appeared for some years in EM. out of hard-working Americans with the reliable threat of hordes of Asians (Chinese, Japanese, Singaporeans, and Indians, in particular) overtaking us in the constant battle to be #1 in anything and everything. Never mind that it's actually those sneaky Finns who top the latest list of countries that outperform the US in mathematics on an international test. No one is yet recommending that we adopt the Finnish national math curriculum and load up on glug and besides, the Finns themselves are not quite so sanguine about the situation. However, racist that I am, I just can't get sufficiently frightened by images of invading Finns. Asians, of course, are notorious for "teeming," and there's no question that they're going to overwhelm American any second, no doubt due to that famous "math gene" they're all born with.* But let's get back to the point (sometimes my own digressions frighten even me). What about that "If math were a color" question? Am I seriously planning to defend it as not only "harmless" but actually sensible? Those of you who know me are no doubt sure that's exactly what I claim. My argument is brief, but I believe it's quite reasonable: It is increasingly common practice for some high school teachers and some professors of mathematics and mathematics education, especially those working with future or would-be future mathematics teachers, to ask their students for a brief "mathematical autobiography" at the beginning of the term. It's a way to get to know more about students and to find out their attitudes towards and experiences with mathematics (does that seem like a trivial thing to anyone? While I wouldn't be the least bit surprised if it did given some of the folks who teach mathematics, it's nonetheless the case that some educators think it's quite valuable to know how education students think AND feel about the subject(s) they propose to teach, and how students think and feel about a subject they are expected to learn). Now, if you're teaching K-5, especially if you're teaching K-2, how would you propose to get students to relate their beliefs about and attitudes towards mathematics? Direct questioning, like direct instruction, is not always effective, which is why many child psychologists work with their clients through playing games and other indirect methods for getting insight into experiences and feelings about them. The question"If math were a color. . .?" is a perfectly imaginative and reasonable way to get children to explore and reveal some of their feelings and beliefs about mathematics. The important thing is not, of course, to elicit a particular response or "right answer" (and isn't that a radical concept?), but to help students to find a non-threatening, non-intellectualized way to talk about math. Heaven forfend that child psychology should enter into any elementary school teacher's classroom or pedagogical repertoire, of course. Sensible, rational, intellectual questions only, most particularly in math. *For the irony-impaired, that's satire.

- What is logically equivalent to P and Q? - Why are P and Q used in logic? - What does P and Q stand for in algebra? - What makes two statements logically equivalent? - How do you write a logically equivalent statement? - What is logical equivalence in math? - Which statement is logically equivalent to Q → P? - What is the negation of a statement? - What types of conditional statements are logically equivalent? - What are logically equivalent statements? - What pairs of propositions are logically equivalent? - How do you determine logical equivalence? - Which of the following are logically equivalent? - How do you prove tautology without truth table? - What is the truth value of P ∨ Q? - What is an example of a Biconditional statement? What is logically equivalent to P and Q? The negation of an implication is a conjunction: ¬(P→Q) is logically equivalent to P∧¬Q.. Why are P and Q used in logic? 1) When p is true and q is true, q is at least as true. (p⇒q) checks as true, meaning that it’s a valid statement because we haven’t introduced a false conclusion starting with true premises. 2) When p is true and q is false, q is NOT at least as true as p and IS less true. What does P and Q stand for in algebra? The statement “p implies q” means that if p is true, then q must also be true. The statement “p implies q” is also written “if p then q” or sometimes “q if p.” Statement p is called the premise of the implication and q is called the conclusion. What makes two statements logically equivalent? Logical equivalence occurs when two statements have the same truth value. This means that one statement can be true in its own context, and the second statement can also be true in its own context, they just both have to have the same meaning. How do you write a logically equivalent statement? Two expressions are logically equivalent provided that they have the same truth value for all possible combinations of truth values for all variables appearing in the two expressions. In this case, we write X≡Y and say that X and Y are logically equivalent. What is logical equivalence in math? Logical equivalence is a type of relationship between two statements or sentences in propositional logic or Boolean algebra. You can’t get very far in logic without talking about propositional logic also known as propositional calculus. Which statement is logically equivalent to Q → P? The converse of p → q is q → p. The inverse of p → q is ∼ p →∼ q. A conditional statement and its converse are NOT logically equivalent. What is the negation of a statement? An open sentence is a statement which contains a variable and becomes either true or false depending on the value that replaces the variable. The negation of statement p is “not p”, symbolized by “~p”. A statement and its negation have opposite truth values. What types of conditional statements are logically equivalent? A conditional statement is logically equivalent to its contrapositive. Converse: Suppose a conditional statement of the form “If p then q” is given. The converse is “If q then p.” Symbolically, the converse of p q is q p. What are logically equivalent statements? From Wikipedia, the free encyclopedia. In logic and mathematics, statements and are said to be logically equivalent if they are provable from each other under a set of axioms, or have the same truth value in every model. What pairs of propositions are logically equivalent? Suppose we have two propositions, p and q. The propositions are equal or logically equivalent if they always have the same truth value. That is, p and q are logically equivalent if p is true whenever q is true, and vice versa, and if p is false whenever q is false, and vice versa. How do you determine logical equivalence? p q and q p have the same truth values, so they are logically equivalent. To test for logical equivalence of 2 statements, construct a truth table that includes every variable to be evaluated, and then check to see if the resulting truth values of the 2 statements are equivalent. Which of the following are logically equivalent? A compound proposition that is always True is called a tautology. Two propositions p and q are logically equivalent if their truth tables are the same. Namely, p and q are logically equivalent if p ↔ q is a tautology. If p and q are logically equivalent, we write p ≡ q. How do you prove tautology without truth table? Using a Fitch style proof, this tautology can be proved by contradiction. Assume the statement is false, show that this assumption entails a contradiction, then negate the assumption. The only way for ¬P ∧ (P ∨ Q) to be true is for P to be false and Q to be true. What is the truth value of P ∨ Q? Disjunction Let p and q be propositions. The disjunction of p and q, denoted by p ∨ q, is the proposition “p or q.” The truth value of p ∨ q is false if both p and q are false. What is an example of a Biconditional statement? A biconditional statement is a statement that can be written in the form “p if and only if q.” This means “if p, then q” and “if q, then p.” The biconditional “p if and only if q” can also be written as “p iff q” or p q.

Information on this website and its contents can be found on the web. Opinions are that of many people. This image was found on the web. We did not make this image however found it to be very appropriate. We Thank Its Creator. The saying - Make America great Again TRUMP has not done this but what he has done is make the World Laugh At US. Trump is like a child throwing temper tantrums,speaks with out thinking, acts without thought. He has made America look like total idiots. He is a Embarrassment to America Any person that is campaigning for a Elect Position. The things said by a person that is campaigning for a position is using what he or she says as a Resume to get the job. People vote based on this resume. Like we have seen many times what is said while campaigning once in the position ether never happens or they do the complete opposite. Accountability - Any person that campaigns, is elected to the position that does not make extreme effort to fallow what was said in there campaign should be fired. Just like anyone would be if they lied on there resume. That's Fraud to the people. It makes no sense for anyone to vote for a Republican under most circumstances. Unless your rich they dont care. For house hold making 100 thousand a year or less you mean nothing. It cant be a Religious factor, it is not Gods way to attend Sunday service and when Monday comes go back to lying, screwing over the people and just be a peace of S**t. Its not Gods will to be a cheat, be a thief, liar and more, Voting isnt about Republican or Democrat, its about getting the person that will best represent you and your needs. I am not rich so the last thing I care about is tax cuts for the rich so no need to vote republican. Republicans do nothing for the average working American. The recent Stimulus had nothing in it for working class by Republicans. Longer and higher amount Unemployment, the $1200 for adults, $500 for kids. Its not near enough but its a start and done by Democrats. From looking at how Trump, His Administration and most Republicans have acted. I compare Trump to Adolf Hitler - Thinking he is better then everyone. Trump & Republicans are like Saddam Hussein - Money & Power over human life. Trump and all Persons in Elect Positions that did nothing to protect the people, people in there state, Made the choice economy & money over human life & safety should be charged accordingly. Crimes Against Humanity So many lives did not need to be lost, but they did not react accordingly to the Pandemic. Any & All Representatives to include Trump that profit from this pandemic personally or in business If $1 dollàr from Stimulus gets into the personal hands, business or businesses or businesses which they have interest or personal choice in. ( Trump, Administration and Elected People ) Abuse Of Power Includes Trump 2635.702 Use of public office for private gain. An employee shall not use his public office for his own private gain, for the endorsement of any product, service or enterprise, or for the private gain of friends, relatives, or persons with whom the employee is affiliated in a nongovernmental capacity, including nonprofit organizations of which the employee is an officer or member, and persons with whom the employee has or seeks employment or business relations. We Need Serious Change 1. Electoral Vote / The Corrupted Vote - Needs to be eliminated. Its purpose was during times the people lived far apart, mail took a long time, no computers and more. Now there is no reason for it but corruption. Democracy - The Peoples Government/The Peoples Vote 2. Term Limits On All Elect Positions - Senators and Persons in Congress, There should be term limits in these positions same as a President 8 years max. This helping with the corruption and personal agendas that happen more so with time in the position. 3. Age limits on all elect position - Minimum of 35 and a Max of 70. This making presidential & elect positions max age of 62 with possibilities of 2 terms or 8 years in the position reaching 70 years of age. 4. Prevailing Wage - This was created during Civil War Times. The 8 hr work day and also established a base wage on different types of work. Today prevailing wage is just a waste of money. Only Government uses this today which makes the cost of every government job 2 to 3 times what it should actually cost. Totally wasting Tax Payer Money. 5. Turning Back The Clock - The People Fought for freedom from England because of being taxed to death. Today States and Federal Governments take 55 to 60 percent of our yearly wages in taxes, fees and whatever else they can think of. TAXATION WITHOUT REPRESENTATION SOUNDS LIKE WHAT HAPPENED THAT CAUSED THE REVOLUTIONARY WAR. MIGHT BE TIME AGAIN.. 1 Air Transportation Taxes (just look at how much you were charged the last time you flew) #2 Biodiesel Fuel Taxes #3 Building Permit Taxes #4 Business Registration Fees #5 Capital Gains Taxes #6 Cigarette Taxes #7 Court Fines (indirect taxes) #8 Disposal Fees #9 Dog License Taxes #10 Drivers License Fees (another form of taxation) #11 Employer Health Insurance Mandate Tax #12 Employer Medicare Taxes #13 Employer Social Security Taxes #14 Environmental Fees #15 Estate Taxes #16 Excise Taxes On Comprehensive Health Insurance Plans #17 Federal Corporate Taxes #18 Federal Income Taxes #19 Federal Unemployment Taxes #20 Fishing License Taxes #21 Flush Taxes (yes, this actually exists in some areas) #22 Food And Beverage License Fees #23 Franchise Business Taxes #24 Garbage Taxes #25 Gasoline Taxes #26 Gift Taxes #27 Gun Ownership Permits #28 Hazardous Material Disposal Fees #29 Highway Access Fees #30 Hotel Taxes (these are becoming quite large in some areas) #31 Hunting License Taxes #32 Import Taxes #33 Individual Health Insurance Mandate Taxes #34 Inheritance Taxes #35 Insect Control Hazardous Materials Licenses #36 Inspection Fees #37 Insurance Premium Taxes #38 Interstate User Diesel Fuel Taxes #39 Inventory Taxes #40 IRA Early Withdrawal Taxes #41 IRS Interest Charges (tax on top of tax) #42 IRS Penalties (tax on top of tax) #43 Library Taxes #44 License Plate Fees #45 Liquor Taxes #46 Local Corporate Taxes #47 Local Income Taxes #48 Local School Taxes #49 Local Unemployment Taxes #50 Luxury Taxes #51 Marriage License Taxes #52 Medicare Taxes #53 Medicare Tax Surcharge On High Earning Americans Under Obamacare #54 Obamacare Individual Mandate Excise Tax (if you don’t buy “qualifying” health insurance under Obamacare you will have to pay an additional tax) #55 Obamacare Surtax On Investment Income (a new 3.8% surtax on investment income) #56 Parking Meters #57 Passport Fees #58 Professional Licenses And Fees (another form of taxation) #59 Property Taxes #60 Real Estate Taxes #61 Recreational Vehicle Taxes #62 Registration Fees For New Businesses #63 Toll Booth Taxes #64 Sales Taxes #65 Self-Employment Taxes #66 Sewer & Water Taxes #67 School Taxes #68 Septic Permit Taxes #69 Service Charge Taxes #70 Social Security Taxes #71 Special Assessments For Road Repairs Or Construction #72 Sports Stadium Taxes #73 State Corporate Taxes #74 State Income Taxes #75 State Park Entrance Fees #76 State Unemployment Taxes (SUTA) #77 Tanning Taxes (a new Obamacare tax on tanning services) #78 Telephone 911 Service Taxes #79 Telephone Federal Excise Taxes #80 Telephone Federal Universal Service Fee Taxes #81 Telephone Minimum Usage Surcharge Taxes #82 Telephone State And Local Taxes #83 Telephone Universal Access Taxes #84 The Alternative Minimum Tax #85 Tire Recycling Fees #86 Tire Taxes #87 Tolls (another form of taxation) #88 Traffic Fines (indirect taxation) #89 Use Taxes (Out of state purchases, etc.) #90 Utility Taxes #91 Vehicle Registration Taxes #92 Waste Management Taxes #93 Water Rights Fees #94 Watercraft Registration & Licensing Fees #95 Well Permit Fees #96 Workers Compensation Taxes #97 Zoning Permit Fee #98 Soda Tax #99 Sin Tax #100 Recreation Pass (A Tax) #101 Carbon Tax This is not all we pay 6. With all the taxes we pay. Every AMERICAN should get medical/dental insurance automatically at no cost. Instead we get racked over the coals with astronomical premiums while government does nothing. Of course they do nothing, its big business which they protect. 7. Fico Scores - Credit scores should only be used when applying for a money loan or credit card from a financial institution. To many people/businesses use this information and its none of there business. Insurance Companies, credit does not make you bad driver/ When Renting a House or Apartment, its a rental not a purchase/ Employers, credit does not make you do better work. STOP FICO SCORE ABUSE AND DISCRIMINATION FROM IT. Financial Institutions only for loan purpose. 8. Insurance company discrimination and manipulation of premiums. Every form of insurance uses age, marital status, were you live, race, credit and more. Not one of these things should matter and all are Discriminating. Car Insurance - marital status, credit, age, what my vehicle is or were I live have nothing to do with being a good driver. Health Insurance - again marital status, credit, age,were I live have nothing to do with it. Home Owners/Renters Insurance - again marital status, credit and age have nothing to do with it. This Needs To Stop The People wanted term limits and we were told it was unconstitutional.How is this possible? The Peoples Government yet the People have no rights or say in anything now that is unconstitutional.

By William Graebel Fluid mechanics is the learn of the way fluids behave and engage below quite a few forces and in quite a few utilized occasions, even if in liquid or fuel country or either. the writer of Advanced Fluid Mechanics compiles pertinent details which are brought within the extra complicated periods on the senior point and on the graduate point. “Advanced Fluid Mechanics” classes commonly conceal quite a few themes related to fluids in numerous a number of states (phases), with either elastic and non-elastic characteristics, and flowing in advanced methods. This new textual content will combine either the easy phases of fluid mechanics (“Fundamentals”) with these regarding extra complicated parameters, together with Inviscid movement in multi-dimensions, Viscous move and Turbulence, and a succinct advent to Computational Fluid Dynamics. it is going to provide unheard of pedagogy, for either lecture room use and self-instruction, together with many worked-out examples, end-of-chapter difficulties, and genuine computing device courses that may be used to enhance concept with real-world applications. Professional engineers in addition to Physicists and Chemists operating within the research of fluid habit in complicated platforms will locate the contents of this ebook worthwhile. All production businesses taken with any type of structures that surround fluids and fluid circulation research (e.g., warmth exchangers, air con and refrigeration, chemical methods, etc.) or power iteration (steam boilers, generators and inner combustion engines, jet propulsion structures, etc.), or fluid structures and fluid energy (e.g., hydraulics, piping structures, and so on)will take advantage of this text. - Offers targeted derivation of primary equations for greater comprehension of extra complex mathematical analysis - Provides foundation for extra complicated subject matters on boundary layer research, unsteady move, turbulent modeling, and computational fluid dynamics - Includes worked-out examples and end-of-chapter difficulties in addition to a significant other site with pattern computational courses and strategies Manual Read Online or Download Advanced fluid mechanics PDF Similar fluid dynamics books Sayma A. Computational fluid dynamics (2009)(ISBN 9788776814304) Greater than only a consultant to apparatus, Fluid Mechanics provides certain info at the layout and functions of pcs, cybernetics, genetic changes, robots and remotes, cetacean tech, guns, automobiles, and masses extra. If you have been itching to get hold of Hanover Autoworks new Cormorant(tm) ground-effect airplane, Hydrospan's SmartGuide(tm) navigation software program, or Atlas fabrics' HardTarget(tm) battledress, your wait is coming to an finish. Dieses auf vier B? nde angelegte Lehrbuch der Experimentalphysik orientiert sich an dem weit verbreiteten, viersemestrigen Vorlesungszyklus und behandelt im ersten Band Mechanik und W? rme. Nach einer Einf? hrung folgen die Mechanik des Massenpunktes unter Ber? cksichtigung der speziellen Relativit? tstheorie und von Systemen von Massenpunkten, der starre okay? Your Welcome. growing this pdf was once a whinge so take pleasure in. - Mode Complex Systems - Turbulence and Self-Organization: Modeling Astrophysical Objects - Applications of Fluidization to Food Processing - Microfluidics and Nanofluidics: Theory and Selected Applications - Thermal Plasmas: Fundamentals and Applications - Non-Newtonian Flow and Applied Rheology: Engineering Applications Additional info for Advanced fluid mechanics 7) zy ) xz + ny yz + nz zz the expression after the second equals sign being a rearrangement of the preceding. 8) This is in agreement with our definition of xx (and so on) being the components of the various stress vectors. , xx yy zz ) are referred to as normal stresses. , yx yz zx xy zx xz ) are referred to as shear stresses. Note that the first subscript on the components tells us the direction in which the area faces, and the second subscript gives us the direction of the force component on that face. 3. Solution. 7), = S · dA Since the integrand is constant 2 and the area is 4, the result for the circulation follows from a simple arithmetic multiplication. 2—that is, v = x−yG 0 . 2 +y 2 x2 +y2 34 Fundamentals Solution. In this case, we cannot easily use the area form of the definition, since the vorticity is not defined at the origin. 7) gives us = C v · ds = = −G −1 1 −G dx + x 2 +1 −1 1 −G dy + 1 + y2 1 −1 −1 1 1 1 −1 1 − −G − +G + 4 4 4 4 4 4 G x 2 +1 dx + 1 1 + 4 4 +G 1 −1 G dy 1 + y2 = 2G If we take any path that does not include the origin, we could in fact use the area form of the definition, and we would conclude that the circulation about that path was zero. The mental process of generating a description of a particular fluid involves a continuous interchange between theory and practice. Once a constitutive model is proposed, mathematical predictions can be made that then, it is hoped, can be compared with the experiment. Such a procedure can show a model to be wrong, but it cannot guarantee that it will always be correct, since many models can predict the same velocity field for the very simple flows used in viscometry (tests used in measuring the coefficient of viscosity) and rheogoniometry (measuring the properties of complicated molecular structures such as polymers). Advanced fluid mechanics by William Graebel

Surface area of Cuboid (solutions, examples, videos) SurfaceAreaof a Cuboid or Rectangular Prism. A cuboid is a 3-dimensional object with six rectangular faces. All its angles are right angles and Surface Area of a Cuboid - Formula, Examples - [email protected] To get totalsurfacearea we have to add up areasof all the six faces of a cuboid. A cuboid has six faces such that, the opposite faces are identical. What is the formula for the total surface area of a cuboid? - Quora I suggest that there is no need to remember the formula if there is clear understanding of the term “TotalSurfaceArea (TSA)”. Total Surface Area of a Cuboid The totalsurfacearea (TSA) of a cuboid is the sum of the areasof its six faces. Formula For Total Surface Area Of Cuboid - Math Discussion FormulaforTotalsurfaceareaofcuboid. Surface Area Of Cuboid Total surface area & Lateral surface areas (ii) Lateral SurfaceArea (Curved SurfaceArea). FormulaforSurfaceAreaofCuboid What is the formula for the total surface area of a cuboid? Recall: The totalsurfacearea (TSA) of a cuboid is the sum of the areasof its six faces. That is: Example 27. Formula of surface area of cuboid The surfacearea is the totalareaof all the faces. For a cuboid, all these faces will be rectangular. Java Program To Total Surface Area Of Cuboid - Programs Def: Totalsurfaceareaof the cuboid ? Total Surface area of Cuboid at Algebra Den How formulafortotalsurfaceareaofcuboid is obtained. Observe the following diagram ofcuboid: Red lines represent length ofcuboid (L) Green lines Surface Area Formulas SurfaceAreaFormulas In general, the surfacearea is the sum of all the areasof all the shapes that cover the surface of the object. C Program to Calculate Volume and Total Surface Area of Cuboid In other words, we can think surfaceareaof a cuboid as the areaof wrapping paper required to gift wrap a rectangular box. A cuboid has six rectangular surfaces of different length, width and height. Surface Area of a Cuboid - Formula & Problems The surfaceareaof a cuboid is the total sum of its six faces areas. C Program to find Volume and Surface Area of a Cuboid The TotalSurfaceAreaof a Cuboid is the sum of all the 6 rectangles areas present in the Cuboid. If we know the length, width and height of the Cuboid then we can calculate the TotalSurfaceArea using the formula Surface area and volume of cuboids We can also work out the surfaceareaof a cuboid by drawing its net ( see slide 51 ). This may be easier for some pupils because they would be able Formulas for Finding the Total Surface Area of Cuboid, Cube, Sphere What is the formula to find totalsurfaceareaofCuboid, Cube, Sphere? Asked by: Lovepreet Sandhu on Feb 20, 2016. Rectangular Cuboid Volume & Surface Area Calculator The cuboid volume & surfaceareaformula, solved example & step by step calculations may useful for users to understand how the input values are being used in such calculations. Also this featured rectangular prism calculator uses the various conversion functions to find its area in SI or metric or US. Program for Volume and Surface Area of Cuboid - GeeksforGeeks Formulae: Area = l*w*h TotalSurfaceArea = 2*l*w + 2*w*h + 2*l*h where l, h, w are length, height and width ofcuboid respectively. Surface Area of Cuboid Formula - Calculate LSA, CSA, TSA of... Get Formula Of Total, Curved, Lateral SurfaceArea: SurfaceAreaOf Cylinder. Question 3: The length, width and height of Cuboid - Surface Area and Volume Perimeter Formulafor the Elementary Math Student. Key Topic - Surface area of a cuboid Finding the surfaceareaofcuboids can appear on either the calculator or non calculator paper of your maths exam. A common mistake is to confuse this topic Surface area and volume of cuboids Surfaceareaof a cuboid = Formulafor the surfaceareaof a cuboid h l w 2 lw Top and bottom + 2 hw Front and back + 2 lh Left and right side what is formula to find the total surface area of cube, cuboid, cylinder? CUBOID - TOTALSURFACEAREA = 2[ lb+bh+hl ] LATERAL SURFACEAREA = 2h[ l+b ] CUBE - TOTALSURFACEAREA = 6[a*a] LATERAL SURFACE Cuboids, Rectangular Prisms and Cubes - Surface Area Examples ofCuboids. Cuboids are very common in our world, from boxes to buildings we see them How do you find the Surface Area and Volume of a Cuboid Totalsurfaceareaof the cuboid = 2 (ℓb + bh + ℓh) square units. Surface area of cube cuboid and cylinder - Total surface area Totalsurfaceareaof a solid is the measurement of outer area,where the extension of top and bottom portion would be included. Now let us see the formulas Objective To form a cuboid and find the formula for its surface area... Basic knowledge about cuboid. Totalsurfacearea, lateral surfacearea and diagonal of a cuboid. Theory. Surface Area. Calculator - Definition - Formulas - Omni Surfacearea is a totalarea that the surface of the object occupies . How To Find The Surface Area Of A Cuboid - The Science The formulafor calculating the areaof its surface is very simple: S = 2 (ab + bc + ac), where a, b and c - the length of the ribs. total surface area of the cuboid- ENGINEERING MATH - Forum The totalsurfaceareaofcuboid A cm2, is given by A= 2x2 + 1372/x. wht method shoul I use? length of diagonal of cuboid : Formula Directory The square cuboid, square box, or right square prism (also ambiguously called square prism) is a special case of the cuboid in which at least two faces are squares. Surface area formula This lesson provides a comprehensive list of surfaceareaformula of some basic geometry figures such as the cube, the cylinder, the pyramid Surface area, Volume of Cuboid 1.0.3 Free Download Learn the formulas to find the surfacearea and volume of three dimensional figures such as the cuboid, cylinder and cone. Surfacearea, Volume ofCuboid is a free software application from the Teaching & Training Tools subcategory, part of the Education category. The app is currently available. Lateral and Total Surface Area of Cube and Cuboid All answers are solved in an easy way, with video of each and every questionIn this chapter, we will learnLateralandTotalSurface Area ofCube and CuboidCurved and Total Cuboid - Volume of Cuboid Formula - How to Find the Volume of... Cuboid is a solid box whose every surface is a rectangle of same area or different areas. Cuboid: Volume and Surface Area — Online Calculator, Formulas Volume and SurfaceAreaof a Cuboid. A cuboid is a solid the sides of which are formed by six rectangles, or four rectangles and two squares. Math - Geometry Lesson Plans : Surface Area of Cuboids & Cubes Mensuration - FormulaeforSurfaceAreaofCuboids & Cubes. A cuboid is a rectangular solid which has six rectangular faces. Digital worksheet for Total surface Area of a Cuboid – GeoGebra This GeoGebra applet will help the students to develop the concept that how does the shape of a cuboid changes when either of length,breadth and height changes. It will give the scope to the learner to verify their calculation. Is your answer right? what is the volume and total surface area of: cube, cuboid... - eNotes You should remember that the totalsurfaceareaof a cuboid is the sum of areasof its six faces such that A cuboid has total surface area of 50 m2 and... - Meritnation.com Find the areaof its base. Surface Area of a Cuboid - Brilliant Math & Science Wiki A (rectangular) cuboid is a closed box which comprises of 3 pairs of rectangular faces that are parallel to each other, and joined at right angles. It is also known as a right rectangular prism. It has 8 vertices, 6 faces, and 12 edges. The cube is a cuboid whose faces are all squares. The figure above shows the. What is the derivation of the formula fot the total surface area of... Answer this question: Retrieved from " http://answers.wikia.com/wiki/What_is_the_derivation_of_the_formula_fot_the_total_surface_area_of_a_cuboid?oldid=676752 ". How calculate surface area of the cuboid. Formula and online... Calculator of the cuboidsurfacearea. Enter the length, width and height, enter the calculation accuracy and click on "Calculate". Edurite.com - Formulas for Cuboid Perimeter - Volume of Cuboid volume ofcuboidformula,faces ofcuboid,surfaceareaofcuboid, Identification of 3-D object: Faces: the surface are of a solid shape. RD Sharma Class 9 Solutions Chapter 18 Surface Areas and Volume... Formulae Handbook for Class 9 Maths and Science Educational Loans in India. If a Cubiod has length l, breadth b, and height h, then. All formulas for surface area of shapes - Calculator All the basic formulas of a surfaceareaof a geometric shapes (sphere, cone, pyramid, rectangular prism, etc). Each formula has calculator. How to Calculate Surface Area From Volume - Sciencing Surfacearea is the totalareaof all exposed surfaces of a given three-dimensional figure or object. Volume is the amount of space occupied by this figure. This page is about surface area of cube. Find the ratio of totalsurfaceareaof the new rectangular prism ( cuboid ) to that of the sum of the surfaceareasof the three cubes. Solution: TSA of cube = 6a 2 Length of new rectangular prism = 3a , width = a and height = a TSA of rectangular prism = 2( lw + wh + lh) ⇒ = 2 (3a x a + a x a + 3a x a). Formula and description of the surface area of a cube. The totalsurfacearea is therefore six times the areaof one face. Or as a formula: Where s is the length of any edge of the cube. Rectangular Cuboids - CalcResult Universal Calculators Maths - Cuboids. Show/Hide:Definitions Formulae Instructions. Introduction: This page allows you to Surface area and volume of cuboids 5. To find the surfaceareaof a shape, we calculate the totalareaof all of the faces. Can you work out the surfaceareaof this cuboid? Surface Area - Volume Formulas What is the totalsurfaceareaof the cuboid they are part of? Surface Area of a Cube Formula - Bing images SurfaceAreaFormula of a Cone SurfaceAreaof a Cylinder FormulaSurfaceAreaof a Rectangular Prism Formula The Formulafor BestMaths - Volume and Surface Area of Cubes and Cuboids The volumes and surfaceareasof certain simple objects such as cubes and cuboids can be calculated using formulae. The surfaceareaof an object is the totalareaof the outside surfaces of the object. Finding Volumes − by counting cubes. For the cuboid or rectangular prism like the one. Cuboid - Surface Area The SurfaceAreaof a Cuboid calculator computes the surfaceareaof a cuboid based on the length, height and width (see diagram). Cuboid-Calculate volume and surface area-Calculator Site Volume and surfaceareaofcuboid. Tweet. The volume and the surfaceareaofcuboid given depth, length, and height is calculated using the formula. Basics of Surface Area and Volume - Quantitative... - Unacademy This video covers definitions of surfacearea and Volume and concepts and formulaeofcuboid. Then we have : (i) Total surface area of the cuboid... of mathematics, formulas in mathematics, mathematics i. Volume and SurfaceArea. Notes For Class 9 Formulas Download PDF. Volume & Surface Area Tricks-Notes & Questions Volume and SurfaceArea Trick- Note & Question for IBPS aptitude exam- here are the shortcut tricks formulas to solve the question faster. C Program to Find the Volume and Surface Area of Cuboids This C Program calculates the volume and surfaceareaofcuboids. The formula used in this program are surfacerea= 2(w * l + l * h + h * w) where w is width, l is length and h is height of the cuboids. volume = width * length * height. Surface Area Of Cube and Cuboid by aditya rangamani on Prezi The Cube The surfaceareaof a cube is the areaof the six squares that cover it. The formulafor finding the areaof cube is 6 * a^2. A cube is used in making the Mathguru – Help – Example:Finding Lateral & Total surface Area of... Formulae. For a cube of edge length a, surfacearea. Surface Area and Volume of Cube and Cuboid SurfaceAreaofCuboid of Length L, Breadth B and Height H. Total Surface Area of Prisms Students learn how to calculate the surfaceareaof prisms firstly by considering the net of a cuboid and later moving on to 3D models of complex prisms. Mensuration Formulae for SSC CGL, SSC CPO... - Testbook Blog Totalsurfaceareaofcuboid = 2 (lb + bh + lh). Length of diagonal ofcuboid= √(l2+b2+h2). Cuboid Volume & Surface Calculator - Symbolab Free Cuboid Volume & Surface Calculator - calculate cuboid volume, surface step by step. Interactives . 3D Shapes . Surface Area & Volume Said another way, the surfacearea is the totalarea covered by the net of a polyhedron. Let's take a look at a cube. As you already know, a cube has six square faces. SOLVED: What is the formula for getting the surface area - Fixya Totalareaof right rectangular prism =2*A_b +P_b*h A_b= areaof base (take 14*12 feet squares) P_b is perimeter Exercise :: Volume and Surface Area - Important Formulas This is the aptitude questions and answers section on "Volume and SurfaceArea Important Formulas" with explanation for various interview, competitive examination and entrance test. Formula of Surface Area and Volume - Aptitude Questions and Answers Important Formulas of Problems on Volume and SurfaceArea, Tips for solving problems based on Volume and SurfaceArea frequently asked in quantitative aptitude question paper in SSC CGL, SSC CHSL, PSC, Banks, CDS, NDA Cuboid and cube: Surface Area The totalsurfacearea =(the lateral surfacearea)+(areaof ABCD)+(areaof EFGH) =2h(l+b)+lb+lb =2lh+2bh+2lb =2(lb+bh+hl). Volume and Surface Area Questions Answers SSC CGL - sscroad.com Volume and SurfaceArea Questoins Answers for ssc cgl ssc 10+2 ssc chsl mcq with explanations, formulas and tricks are useful and important Surface Area and Volume - STEM In the activities file, ‘Surfaceareaof a Cylinder’ is a practical activity in which students investigate connections between the areaof a rectangle, the View question - A cuboid has a length of 6cm, a with of 1.5cm, and... Work out the totalsurfacearea. 0. 359. 2. A cuboid has a length of 6cm, a with of 1.5cm, and a depth of 1.5cm . CUBE: perimeter, surface area, enclosed volume of cube (formula...) Calculate quickly surfacearea or volume of cube. 1. Formula of Cuboid Volume, Surface Area and Daigonal formulas of volume and surfacearea questions answers mcq of quantitative aptitude are useful for it officer bank exam, ssc, ibps, sbi, lic and other competitive exam preparation. Surface Area Formula Chart Includes,surfaceareaformulas, volume formulas, and lateral areaformulasfor cone, cube, cylinder, prism, and spheres. Distances on the surface of a cubical box - Cuboid Contents Applet The following graph indicates how the formulafor the co-ordinates of the point at greatest surface distance depends on the values of x and y. In the Important formulas for CDS Exam - Math Formula For Surface Area Areasof figures which can be split up into these figures (Field Book), Surfacearea and volume ofcuboids, lateral surface and volume of right circular Surface Area of A Cone --Examples illustrated with pictures of the... Formulafor Lateral SurfaceArea. Total Surface Area - Passy's World of Mathematics TotalSurfaceArea (“TSA”) is important for Painters, so that they know how much paint will be required for a job. Formulas of Volume and Surface area of solid figures for class VI, VII... Students have learnt the formulas from the beginning and in each class, they have to learn all the formulas or revise the formulas again n again. Explanation about Total surface area and volume of the cyllinder And... thank you i am actually learning about surfacearea. Maths surface area of cone surface area and volume of... - 0xvideos.ru It explains the formula to find surfaceareaof a cone. About us: We are a social enterprise working Frustum Formula - PAn-clip Totalsurfacearea and volume of frustum of a cone: derivation, R B ClassesR B Classes. Total area video foto TotalsurfaceareaofCuboid - class10 math in hindi. In this video, we will know that how to find total suface areaof cubiod and formulate it.It is related to chapter 13-surfaceareas and volume of class10 math . Maximum volume of a box with corners cut out TotalSurfaceArea and Volume of a Box with a Cylindrical Hole. Now we have narrowed down the Cuboid Surface Multifunction one Office Mirror in Large Screen Cube... Total Price: Depends on the product properties you select.

?Hay instancias validas de la falacia de afirmacion del consecuente? Gerald Massey argues in his article "The fallacy behind fallacies" that there is no theory whatsoever behind the standard treatment of fallacies. Nevertheless, he agrees that the so-called formal fallacies can falsify his claim. Formal fallacies being invalid patterns of argumentation proscribed by logical theory, he purports to show that they can, nonetheless, yield valid arguments. Massey chooses the fallacy of "affirmation of the consequent" and provides one example of it to support the latter claim. His underlying point is that while proofs of argument validity can be considered definitive and taken to have theoretical legitimacy, proofs of argument invalidity cannot. In this paper, I will challenge Massey's example of an argument that instantiates the pattern known as "affirmation of the consequent" and yet is valid. I will argue that his example is not a genuine case of affirmation of the consequent but a mere sham argument on which he has performed a trick. In the last section of my paper, I shall argue that Massey, while accepting the distinction between "argument form" and "argument", fails to see how such a distinction could help us to understand definitions of validity and invalidity better. I believe this project is of interest because it makes clear -contra Massey- that at least one part of a theory about invalidity, namely, the theory of formal invalidity, can help us to understand, treat, and explain, some fallacious arguments. A purported valid instance of "affirmation of the consequent" Massey's general complaint about the traditional treatment of fallacies is that it does not provide anything but a miscellany of arguments considered fallacious because of reasons so different that while they may support placing them under a common pejorative label, they cannot underpin a highly articulated theory. In his opinion, the multifarious schemas used to classify fallacies suggest that "there is little theory behind the science of fallacy". Moreover, Massey believes that, strictly speaking, there is no theory of fallacies whatsoever. A prima facie problem for Massey's view is the category of formal fallacies (instances of patterns proscribed by logical theory). There appears to be, therefore, at least one class of fallacies for which there is a general and precise account (1995 160). Take, for example, the following pattern of argumentation: (1) p [contains] q q / p This pattern is rejected as invalid by truth-functional logic because "some instantiations of it have true premises but a false conclusion" (Ibid.). Logicians call an argument that follows pattern (1) the fallacy of affirmation of the consequent. Massey says that the justification for naming this particular pattern as a recognizable fallacy is the belief that any argument which has that form is invalid. "Hence any such argument will be said to commit the affirmation-of-the-consequent fallacy" (161). According to Massey, that is the justification offered by the "naive account of formal fallacy", but it is mistaken. To see why Massey thinks this, let us consider argument (2): (2) P1 If something has been created by God, then everything has been created by God P2 Everything has been created by God/C Something has been created by God On Massey's account, argument (2) is an instance of the form (1), yet it is valid. Since Massey does not say explicitly on what grounds he takes argument (2) to be valid, let us explore some suggestions to determine what he could mean by arguing both that (2) is an instantiation of (1) and that it is valid. (1) As far as I see there are three possible strategies open to Massey in treating this example. The first one is to consider argument (2) as an instance of the sentential form (1) simpliciter. The second one is to consider that argument (2) instantiates a case of subalternation (one of the so-called immediate inferences). The last one is to treat (2) with the tools of quantificational first order calculus. Before moving on, we should note that these strategies yield different results, since they produce different answers to the claims Massey advances. If we consider argument (2) as an instance of sentential form (1) simpliciter, the argument is invalid no matter what the truth values of its premises and conclusion might be. As everybody knows from the traditional account of validity/invalidity, an attribution of invalidity to a given argument form does not rule out the possibility that one of its instances has true premises and true conclusion. It tells us only that there is at least one instance (and possibly many) of that argument form, which has true premises and a false conclusion. This fact alone suffices to show the form of the argument does not guarantee a true conclusion given true premises. In fact, we reject arguments that instantiate invalid forms because we know the form in question will not guarantee a good argument (if arguments are presented as persuasive in virtue of their form and the truth of their premises). The traditional account of validity/invalidity stresses the independence of truth and validity, since it makes validity/invalidity a function of the form exclusively (2). One can conjecture, then, that Massey does not want to ground the validity of (2) just in the truth values of its premises and conclusion, and that he is not assuming that its premises and conclusion are true. Indeed, he cannot assume the latter. Since this particular example deals with disputable matters involving an omnipotent being and the exercise of his capacities, the assumption that the premises and the conclusion of (2) are true is far from being non-controversial. For example, this assumption would be rejected by someone who is skeptical about the existence of God or who considers creationism implausible. Even disregarding the fact that the first premise of argument (2) is very controversial also, one could add that if the assumption about the premises of argument (2) being jointly true is not granted, the argument might be considered not only invalid but clearly fallacious. Let us suppose that Massey thinks that argument (2) can be treated as an instance of subalternation. In that case, the presentation of argument (2) needs some amendment, and we have a completely different argument which no longer instantiates form (1). In this case, argument (3) would provide us with a more accurate representation of the situation: (3) P Everything has been created by God/C Something has been created by God Now we have a valid argument, recognized as such not only by our intuitions but also by the traditional theory of immediate inference (3). But this new argument is independent of the old argument in (2). Being a one-premised argument, its validity stems from the logical relationship between its single premise and its conclusion. On the other hand, its form is "Every S is P, therefore Some S is P", hence it does not instantiate argument form (1). Some supporter of Massey may respond that this argument was part (even an essential part) of argument (2) and that it does not look very complete or meaningful without the statement which played the role of premise one in argument (2). To this objection, I reply that (3) and (2) are different arguments, regardless of their surface similarities. I will contend, also, that if we were to subsume (3) under (2), we would be adding unnecessary and irrelevant premises to an argument which can stand alone. In fact, if adding premises to independent and valid arguments were desirable, we could add not only the first premise of (2) but any other statement, and, as long as we kept in mind that we were dealing with a case of subalternation, we should not have any problem retaining the claim of validity. This way of amplifying arguments, however, is clearly pointless. If argument (3) is what substantiates Massey's contention of validity, then it is obvious that (3) cannot be an instance of argument form (1) unless we agree on the fact that it has a superfluous premise, in which case it cannot be a genuine instance of (1). It seems to me, then, that Massey cannot defend both of his claims about argument (2). He would have to choose between a genuine instantiation of argument form (1) and validity, but he cannot have both simultaneously (4). We still have the third possibility to discuss. Let us see what happens when argument (2) is treated under quantificational first order calculus. A translation which reveals the form of the argument would be: (4) P1 ([there exist]x) Cgx [[contains]] ([for all]x) Cgx P2 ([for all]x) Cgx/C ([there exist]x) Cgx But this translation cannot make argument (2) valid. One need only to give it a quick inspection to discover that there is no way to override the truth functional connectives of first-order calculus to obtain the conclusion of argument form (4) by performing legitimate operations on premises one and two. We have to conclude that translating argument (2) into predicate logic does not help Massey's case at all; since doing that provides us with an invalid argument form just as we had at the beginning (5). If my analysis of argument (2) is correct, I have shown that Massey's treatment of such an argument is not sufficient to jeopardize what he calls the "naive" account of formal invalidity. But I think that he is now in serious trouble, since he argued that formal fallacies, if backed up by a suitable theoretical account of invalidity, would falsify his claim about fallacy theory. Furthermore, I am afraid that his criticism of the theoretical account of invalidity is not successful, since it is founded on his putative counterexample to affirmation of the consequent. The shortcomings of Massey's treatment of formal fallacies Let me recapitulate the main components of Massey's argument against a theory of fallacies. His claims can be put in the following argument: (5) P1 All fallacies are invalid arguments P2 To demonstrate fallaciousness we have to show invalidity first P3 The traditional method of showing invalidity is mistaken, and there is no formally adequate method to do this job. P4 We cannot demonstrate invalidity in any theoretically adequate way/C There is no adequate theory of fallacies (6). (Govier's 172-180) The claim in (P1) is obviously wrong, unless Massey has a reason to think that question-begging arguments are not fallacious, or, being fallacious, are invalid (7). Since he does not have any such reason, I believe he is just mistaken in this point. As has been shown in the literature, invalidity per se is neither a necessary nor a sufficient condition of fallaciousness. If claim (P1) is false on the grounds cited, then claim (P2) is false also, because it links decisions on fallaciousness to decisions on invalidity, and we already know that not all fallacies are invalid arguments. His third claim is a little more interesting. To support the point he makes in (P3) he gives two arguments: (a) under the "naive" account of formal fallacies, proofs of argument invalidity go like proofs of argument validity, but this is erroneous, and (b) the principle of translation; the idea that translations of valid arguments are valid, and translations of invalid arguments are invalid, that he considers wrong. According to him, that the principle of translation is wrong can be shown by the fact that we are never sure that what is translated into a language and shown invalid cannot be translated into a different (more powerful or more developed) language and shown to be valid. From (P3) he moves on to (P4), and from there he goes to the conclusion. The upshot is that "there is no method whatsoever of establishing invalidity that has theoretical legitimacy" (Massey 164) (8) hence no theory of fallacy. Notice that even if we grant (P3) and (P4), Massey cannot fully support his conclusion without (P1) and (P2). However, I believe that there are still more problems in Massey's treatment of formal fallacies which are worth addressing here. Consider his example of a "valid" instance of the fallacy of undistributed middle term: (6) P1 All bachelors are rich. P2 All unmarried adult males are rich/C All unmarried adult males are bachelors Now he makes an even more astounding claim regarding this argument. He suggests that (6) is a valid argument because, having a necessarily true conclusion overrules the possibility of having true premises and false conclusion which, by definition, suffices to label an argument as invalid. (9) But this claim strikes me as completely wrong. To begin with, what the traditional treatment of validity/invalidity entails is that the actual truth values of premises and conclusion are neither a necessary nor a sufficient condition for an argument to be valid. As I mentioned before, the traditional treatment makes validity/invalidity a property grounded in argument forms and extended, so to speak, to particular arguments, and not a property of arguments taken independently. It is correct, as Massey points out, that the traditional treatment rules out the joint possibility of true premises and false conclusion for arguments which are supposed to be instances of valid forms. But nothing is said about arguments in general, based only on the fact that they might have both true premises and true conclusion. This, of course, could not be the case with argument (6), since its premises are obviously false, but recall that the traditional account of invalidity does not have any problem explaining an invalid argument with one (or two) false premise(s) and a true conclusion. Moreover, on the traditional account of invalidity, nothing is said about arguments that might have, as (6) above does, a necessarily true conclusion (except insofar as they have a semantically valid conclusion); but nothing prevents us from assimilating this case to the more general rule of arguments with true conclusions. If Massey believes that having a necessarily true conclusion is a condition necessary and sufficient for an argument to be valid, then, on his account we could not have a legitimate and satisfactory theory of validity (which he has taken for granted), since any invalid argument might be turned into a valid one by simply replacing its conclusion by a necessarily true proposition. In my view, Massey reaches his mistaken conclusion about (6) because, although he claims he is making the appropriate distinctions between argument forms and arguments for the sake of this debate, he neglects that distinction when he analyzes his examples. Validity is a matter of logical form, and I do not think we can grant validity to a particular instance of an argument form and refuse to grant validity to that argument form. On the traditional account of validity/invalidity, there is neither such a thing as an argument being valid without its form being valid, nor is there such a thing as an argument being invalid without its form being so. Moreover, contrary to appearances generated by instances with true premises and true conclusion, all arguments that instantiate an invalid argument form are invalid (of course all arguments that instantiate a valid argument form are valid). In deciding about validity and invalidity we have to bear in mind the distinction between argument forms and arguments, to neglect this distinction is to confuse the matters, and this is precisely what Massey is doing. He contends that what is fallacious (in the sense of being invalid) is not (6) but its argument form (7) below: (7) P1 All H are G P2 All F are G/C All F are H On Massey's account, we have the following paradoxical result: invalid argument forms (such as (1) and (7)) can be instantiated by valid arguments. I have already discussed the case of argument (2). Let me make a few comments on argument (6). As noticed above, Massey seems to ground his claim about the validity of (6) in the fact that it has a necessarily true proposition as its conclusion. But if this were correct, all arguments with necessarily true conclusions would have to be considered as valid. At this point, it is not surprising that, in view of the aforementioned paradox, Massey concludes that (6) "is no argument at all". That having a necessarily true proposition as a conclusion (and even as premises) should not be considered sufficient to substantiate a verdict of validity for an argument might be made more explicit with the help of the following example: (8) P1 If 2 + 2 = 4 then 4 - 2 = 2 P2 4 - 2 = 2/C 2 + 2 = 4 This is an argument that instantiates form (1). According to Massey, it should be considered as valid, since its conclusion is a necessary truth and it follows from P2. But, as I have discussed above, the truth values of premises and conclusion are neither a necessary nor a sufficient condition to support a claim of formal validity. Moreover, I believe that (8) is not an argument in the proper sense of the term. Both of its premises are necessarily true, and its conclusion is a necessary truth also, but in this case premises and conclusion do not bear any significant relation at all. Even if instead of picking up simple theorems of arithmetic, I would had chosen more impressive, yet unrelated necessary truths, to play the role of premises and conclusion in an argument like (8), the situation would have not changed very much. The traditional treatment of validity has not said very much on how to treat arguments in which necessary truths appear, but I believe that this lacuna does not guarantee Massey's peculiar notion of validity. What is wrong with Massey's account of fallacies? Massey's account of fallacies is defective for several reasons. First, as pointed out above, it cannot deal appropriately with valid yet fallacious arguments like begging the question. Since he is committed to rejecting formalist accounts as legitimate explanations of formal fallacies, his theory cannot explain adequately what is wrong with this type of fallaciousness, either. One point that I think is worth making here is that Massey, while accepting the distinction between "argument form" and "argument", fails to see how such a distinction could help us to understand definitions of validity and invalidity better. The reason for his failure can be found in his dismissive view on the traditional treatment of validity. He writes: Everyone agrees that to show it has true premises but a false conclusion is to show that an argument is invalid. I call this the trivial logic-indifferent method of proving invalidity. This trivial method of showing invalidity is clearly independent of logical theory. (164) In her insightful criticism of Massey, Govier argues that one of the sources of Massey's controversial account of fallacies is his tendency to conflate semantic and formal validity and to attribute that tendency to other logicians. I will add to this criticism the charge that Massey's theory does not represent normal reasoners very well. He seems to believe that reasoners are incapable of recognizing a good argument and detecting a bad one unless they have theoretic and formalist tools. I believe that Massey's understanding of fallaciousness is far from capturing what this feature is really about. Let us recall, for the last time, what his line of argumentation is. First he stipulates that fallacies are always invalid arguments, then he denies that straightforwardly invalid argument forms are always instantiated by invalid (and thus fallacious arguments), and then he challenges the means by which we determine validity and invalidity because they have "no theoretical legitimacy". No wonder he comes to think that arguments like (6) are nothing at all! One might accept Massey's claim in the paper I have just discussed, and start wondering whether there are invalid argument forms at all (since they can be transformed into valid ones by the procedure he commends to us), or one might try to discover how invalid argument forms can be instantiated by purportedly valid arguments. I think the answer is not very difficult to find. Massey's examples of valid arguments, built on the skeleton of invalid argument forms, are nothing but sham arguments. He has shown us how anyone, by distorting the notions of "argument form" and "argument", and taking advantage of the fact that different parts of a set of propositions allegedly related in the form of an argument can instantiate two different argument forms, can produce one of the cases he has given us in (2). He has shown, also, that anyone, by manipulating propositions and putting them in the right positions (as premises or conclusions of an argument) can produce one of the cases he has in (6). But I believe that logic and argumentation are not about trying to confuse our audiences or performing clever tricks, but about helping to pursue knowledge in a more efficacious way. RECIBIDO EL 30 DE AGOSTO DE 2011 Y APROBADO EL 28 DE NOVIEMBRE DE 2011 Bergmann, Merrie. et al. The logic book. New York: Random House, 1980. Print. Biro, John. "Rescuing 'begging the question'". Metaphilosophy. Vol. 8, No. 4. 1977. Print. Copi, Irving M. & Cohen, Carl. Introduction to logic. New York: Prentice Hall, 1998. Print. Curd, Martin. Argument and analysis: an introduction to philosophy. St Paul: West Publishing Company, 1992. Print. Govier, Trudy. "Reply to Massey". Hansen & Pinto, Robert (Eds.). Fallacies: classical and contemporary readings. Pennsylvania: University Park, Pennsylvania State University Press, 1995. Print. Hamblin, C. L. Fallacies. Vale Press: Newport News, 1970. Print. Hansen, Hans & Pinto, Robert. (Eds.). Fallacies: classical and contemporary readings. Pennsylvania: University Park, Pennsylvania State University Press, 1995. Print. Massey, Gerald J. "The fallacy behind fallacies". Hansen, Hans & Pinto, Robert. (Eds.). Fallacies: classical and contemporary readings. Pennsylvania: University Park, Pennsylvania State University Press, 1995. Print. Mates, Benson. Elementary logic. New York: Oxford University Press, 1972. Print. Stebbing, Susan. A modern elementary logic. London: Methuen, 1976. Print. (1) Notice that these are two different and separate aspects of the question at hand, but Massey is treating them together. The issue is important, since some readers might admit that argument (2) is an instance of form (1), but deny that it is valid. (2) See, for example, the discussion of this topic in the chapter "Validity and truth" of Susan Stebbing's book: A Modern Elementary Logic. In what follows, I will refer to her views as the "traditional account of validity/invalidity". (3) Some logicians would object to argument (3) on the account of the existential presupposition involved in this move. It is on that motivation that Irving Copi (Introduction to Logic) excludes this type of inference from his description of the logical square of opposition, under what he calls a modern interpretation of categorical propositions. (4) What seems to make Massey's case is that a simple argument appears to instantiate more than one form. In this case one form may be valid and the other not. But talk of validity and invalidity attaches to form. So one could not cite an instance of two forms to show one of the two forms was not an invalid form, on the basis of the other of the two forms being valid. That would be an instance of the fallacy of equivocation. (5) Again, in classical logic it is assumed that the domain is non-empty. With that presupposition (4) would be a valid argument, but not in virtue of being of the form (p [reune a] q; q / [por lo tanto] p); but of the form (p -superfluous premise-; ([for all]x)Fx / [por lo tanto] ([existente en]x)Fx). (6) My reconstruction of Massey's argument follows closely the one included in Trudy Govier's: "Reply to Massey". Hansen & Pinto (Eds.) Fallacies: classical and contemporary readings. (7) There is little doubt about the fact that question-begging arguments are valid. For a discussion and an explanation of this kind of argument see: Biro, John. "Rescuing 'Begging the question'". Metaphilosophy, Vol. 8, No. 4, 1977. Print. (8) He stresses that there is an asymmetry between the method of showing validity and the method of showing invalidity which has not being realized by logicians and that is neglected in the standard treatment of validity/invalidity in logic textbooks. Perhaps he has in mind something along these lines: to say that an argument is fallacious in virtue of form suggests that every instance of the form is invalid. But some instances of invalid forms are instances of other valid forms. So one cannot infer from an argument's being an instance of an invalid form, to its being formally fallacious. On this point it looks as if one has to pay attention to what the argument is being presented as being good in virtue of. (9) Actually, Massey makes this claim from the point of view of what the definition of validity rules out, but I believe my way of putting his view does not affect my argument. This is what Massey says: "Note that as measured against the classical standard of argument validity, viz., joint impossibility of truth of premises with falsity of conclusion, (6) qualifies as a valid argument because its conclusion is necessarily true" (168). Carlos Emilio Garcia Duque Universidad de Caldas, Colombia. email@example.com |Printer friendly Cite/link Email Feedback| |Author:||Garcia Duque, Carlos Emilio| |Date:||Jul 1, 2011| |Next Article:||Tiempo eterno, eterno secreto: la tesis de la eternidad del tiempo en la guia de perplejos de maimonides.|

Before getting to the point of how to find the surface area of a rectangular prism, we will give a definition of the prism and learn how to build a prism, as well as study its basic properties. We’ll also learn what a straight prism is and what the height of the prism is. We will recall the concept of a perpendicular line and surface in order to formulate a definition of direct and oblique prism. Prism is a geometric figure, a polyhedron with two equal and parallel faces that called bases and have the shape of a polygon. Other faces have a common base side and are called the sides. Even in ancient times there were two ways of determining the geometric concepts. The first way led from figures of a higher order to the figures of a lower order. Euclid in particular agreed with this point of view. He defined a surface as a boundary of a body, a line as a boundary surface, and the ends of the line as points. The second way is, on the contrary, led from the figures of the lower dimensions to higher figures. The line is formed by the movement of a point similarly the surface is formed with the lines, etc. Heron of Alexandria was one of the first, who connected these two points of view. He said that that the body is limited by the surface and at the same time it can be considered as something formed by the surface movement. In books on geometry that appeared later over the centuries, sometimes one point of view was accepted, sometimes – the other, and at times even two of the points were agreed with. Just as a triangle in the sense of Euclid is not empty, i.e. triangle is a part of the surface bounded by the three non-competitive (i.e. not intersected at any point) segments, the polyhedron is not empty either. It is not hollow, but filled with part of the space. In ancient mathematics, however, the concept of abstract space did not exist. Euclid defines prism as a bodily shape, enclosed between two equal and parallel planes (bases) and with lateral faces – parallelograms. In order for this definition to be quite correct, it would be advisable to prove that the planes passing through the pair of non-parallel sides of the base intersect along parallel lines. Euclid uses the term «plane» both in the broad sense (considering it indefinitely extended in all directions) and in the sense of ultimate and limited part of it, in particular a face, which is similar to Euclid’s use of the term «direct» (in the broad sense – an infinite straight line and in a narrow sense – a cut). In the 18th century, Taylor defined prism as a polyhedron whose all faces, except for two, are parallel to one single line. The Babylonian and Egyptian monuments of architecture include such geometric figures as cube, parallelepiped, and prism. The most important task of the Egyptian and Babylonian geometry was to determine the volume of the various spatial figures and surface area of a rectangular prism. This task responded to the need to build houses, palaces, temples, and other buildings. Part geometry, which studies the properties of a cube, prism, parallelepiped and other geometric shapes and three-dimensional shapes, have long called stereometry. The word is of Greek origin and it can be found even in the works of a famous ancient Greek philosopher Aristotle. Stereometry occurred later than planimetry. Euclid defines the prism as a bodily (i.e. spatial) figure enclosed between the planes, of which two opposite are equal and parallel, while the rest are parallelograms. Here, as in many other places, Euclid uses the term «plane» not in the sense of an infinitely extended plane, but in the sense of a limited part of it, a face. The same as direct means a line segment. The term prism is of Greek origin and literally means the body that was intercepted. Rectangular prism is the name used to refer to the hex object, resembling an ordinary box. Imagine a brick or a box from under shoes, and you’ll have a clear understanding of how the rectangular prism looks like. The surface area of a rectangular prism is the total area of all its facets. The calculation of the surface area of a rectangular prism is similar to the question’s answer «how much paper is needed to wrap the box?» In order to calculate the surface area of a rectangular prism, mark the length, width, and height of the prism. Each rectangular prism has length, width, and height. Draw the prism and mark its various edges with letters l, w, and h. If you are not sure how to mark each rib, select any angle of a prism. Mark three edges with the corresponding letters coming out of this corner. Let’s say, for example, the base of the prism is a rectangle 3 by 4 centimeters, and a prism height is 5 centimeters. Since the long side of the base equals to 4 centimeters, we get l = 4, w = 3, and h = 5. Take a look at the six faces of the prism. To cover the entire surface of the figure, it is necessary to paint over all of its six faces. Imagine each face, or take a box from under the shoes and take a look at it: If it is difficult for you to imagine this picture, cut around the edges of the box and expand it. Now, let’s try to find the surface area of a rectangular prism. At first, let’s find out the area of one face, namely the bottom. This face, like the rest faces, is a rectangle. One side of this rectangle was marked by you as the length, and the second was marked as the width. To find the area of a rectangle, you need to multiply the length of the two sides of a rectangular. Thus, the area (of the lower face) = the length multiplied by the width = lw. If we take the values we proposed above, for the area of the base of the prism we get 4 cm x 3 cm = 12 square centimeters. The second step on the way to finding the surface area of a rectangular prism is calculating the area of the upper face. As you we said above, the top and bottom faces have the same area. Thus, the area of the upper face is also equal to lw. In our example, the upper face area is 12 square centimeters. Next, define the area of the front and rear faces. The front face’s sides are the width and the height. Thus, the area of the front face = width multiplied by the height = wh. The area of the rear face is also wh. In our example, w = 3 cm and h = 5 cm, so the area of the front face is 3 cm x 5 cm = 15 square centimeters. The area of the rear face is also equal to 15 square centimeters. Now, let’s find the area of the left and right faces. Their areas are also the same that is why it is enough to find the area of the left side. It is limited in length and height of the prism. Thus, the area of the left face equals lh the area of the right face is also equal to lh. In our example, l = 4 cm and h = 5 cm, so the area of the left side = 4 cm x 5 cm = 20 square centimeters. The area of the right side is also equal to 20 square centimeters. Now we will summarize the areas we found to calculate surface area of a rectangular prism. We have found the area of each of the six faces of the prism. By summarizing them together, we can find the surface area of a rectangular prism: lw + lw + wh + wh + lh + lh. This formula can be used to calculate the area of the surface for any a rectangular prism. Let’s go back to our example and find the surface area of a rectangular prism: 12 + 12 + 15 + 15 + 20 + 20 = 94 square centimeters. We can simplify the formula we mentioned above in order to faster find surface area of a rectangular prism. We already know how to calculate the surface area of a rectangular prism. However, it can be done faster if we do simple algebraic transformations. Let's start with the equation we mentioned above: the surface area of a rectangular prism = lw + lw + wh + wh + lh + lh. Combining the similar summand, we obtain: the surface area of a rectangular prism = 2lw + 2wh + 2lh. Impose a common factor 2 out of the brackets. If you can lay out factoring algebraic equation, this formula can be simplified as follows: area = 2lw + 2wh + 2lh = 2 (lw + wh + lh). Now, you can verify this formula with our example. Go back to the values we used above: the length is 4, the width is 3, and the height is 5. Substitute these numbers into our formula: surface area of a rectangular prism = 2 (lw + wh + lh) = 2 x (lw + wh + lh) = 2 x (4x3 + 3x5 + 4x5) = 2 x (12 + 15 + 20) = 2 x (47) = 94 square centimeters. This response coincides with the one that we got earlier. Using this equation, the surface area of a rectangular prism can be computed much faster.

FOM: More on pobabilistic proof neilt at mercutio.cohums.ohio-state.edu Thu Sep 24 08:53:53 EDT 1998 Don Falls wrote >There is a common presumption (among many philosophers and >mathematicians) that deductive evidence is epistemically superior to >inductive evidence. ... Scientists commonly appeal to inductive >evidence for the truth of scientific claims. As a result, their >acceptance of probabilistic proofs is not very surprising. I take it that we are talking about *contemporary* scientists here, and about *contemporary* scientific methodology (so that Aristotle and even Bacon would be virtually irrelevant to this discussion). Don's claim that scientists "commonly appeal to inductive evidence for the truth of scientific claims" states the obvious. There is nothing else to which they CAN appeal! If E is adduced as scientific evidence for theoretical claim T, it is NEVER the case that E deductively implies T. If E did deductively imply T, then we would complain that adducing E as evidence for T was simply begging the question. For example, let T be the claim "Tadpoles grow into frogs"---the sort of claim that even Aristotle might have made. Here's a claim E that deductively implies T: "Tadpoles grow into frogs and snakes lay eggs". If someone were to offer E as "scientific *evidence*" for T, one would (quite rightly) think he or she was cognitively deficient. A second example: let T be the claim "hearts pump blood" (the great discovery of the 'other Harvey'). Here's a claim E that deductively implies T: "hearts pump blood and livers purify it". Again, if someone were to offer E as "scientific *evidence*" for T, one would think (quite rightly) that he or she was cognitively deficient. (By "cognitively deficient" here, I mean deficient in both logical and Don makes out that scientists have a lower epistemic threshold for their empirical hypotheses (than mathematicians do for their mathematical claims), and thinks that this might help explain why scientists would accept probabilistic proofs in mathematics more easily than mathematicians would. On the contrary, there is good reason for any scientist to insist on *mathematical* standards of proof (i.e. deductive proof) for *mathematical* propositions. This is so that, in the event of experimental falsification of a scientific prediction from a testable scientific hypothesis, the blame for the failure can be laid at the door of the *empirical* component of the overall set of premisses, rather than at the door of the *mathematical* component. If all the mathematics used in making the prediction has been properly proved according to the deductive standards of mathematicians, then it can be "insulated off" from any doubt, and the theoretical failure can be located in the empirical component. (I know it is fashionable for Quineans to insist that *any* claim is in principle revisable. But when was the last time any scientists were heard emerging from the lab saying "Hey, guys, we've just got to give up the commutativity of addition on the integers, in the light of these observations"? Or "Hey, guys, yesterday's meter readings showed that the power set axiom of ZFC is looking really shaky?".) Yet Don makes out that *only* >Mathematicians, however, try not to appeal to inductive evidence for >the truth of mathematical claims >Is mathematicians' distaste for >inductive evidence due to the epistemic inferiority of inductive >evidence or is it due simply to an aesthetic preference on the part of It's the former. Both scientists' and mathematicians' distaste for inductive evidence for *mathematical* claims is due to its epistemic inferiority to proper deductive (not: 'probabilistic') PROOF. When Don went on to write >... from the fact that some labels (such as "truth value proof") do >not convey information about the structure of a proof, I don't think that >you can infer that no labels convey such information. he was not grasping the point of my simple analogy. I had written "...if I were to say of a conventional proof that it was a "truth-value proof", this would carry no information at all as to its *structure*. All I would know is that if the premisses of the proof have truth-value T, then so does its conclusion. That leaves open every possibility as to its deductive structure (i.e. the patterning of steps of inference within the proof)." Then I posed the crucial question that Don failed to answer: "So how come the probabilistic character of a proof carries information as to its structure (as Don claims it does)?" To help the reader in thinking about possible answers to this question, I outlined two possibilities as to the kind of thing that a probabilistic proof might be. The first possibility was that a probabilistic proof has true premisses (probability = 1) and a conclusion of the explicit form "p(S)>=1-2^(-n)" for rather large n. The second possibility was that a probabilistic proof has a conclusion that doesn't explicitly register the probability of S. Instead, its conclusion is S itself, but the steps within the proof are made in such a way that, though they do not guarantee truth-transmission, they nevertheless guarantee that the probability-value of 1 for each of the premisses does not degrade below 1-2^(-n) (for some suitably large n). I argued that the first possibility would be of no help. For "then the validity of the proof will transmit the value T from its premisses to *this* conclusion!---thereby making the 'probabilistic' proof a special case of ordinary truth-transmitting proof. This being so, there would appear to be no information as to its possible structure. All we could glean is that the Kolmogorov axioms for the probability calculus might be among the premisses of the proof." Don did not challenge this claim. This is important, since (in my own view) it is the *first* possibility that is actually the case with so-called 'probabilistic' proofs. Hence the label 'probabilistic' carries no information whatsoever about the structure of the proof. Instead, Don embraced the second possibility, writing: >Your alternative picture, however, seems to capture the meaning of the term >"probabilistic proof" (at least as it has been used on FOM). But in doing so he failed to address the following comments I had made: "I am not aware of any such "probability logic" being satisfactorily developed in the literature, to the point where the authors of so-called probabilistic proofs would be able reliably to claim that their proofs could be formalized within that logic, in the way that authors of conventional proofs can and do reliably claim that their proofs can be formalized within ordinary predicate calculus." So I need to repeat my closing request for references to the literature: "If there *is* such a probability logic, I'd be most interested to know where to find it, since it seems that *that* would be the system to inspect to see whether Don's claim about information concerning proof structure can be justified (i.e. assigned probability > 0.5 ?!)." Don, can you tell us what formal system of probabilistic proof you have in mind? What are the well-formed formulae of the language? What are its rules of inference? How is "probabilistic soundness" proved? How is it that the mere clue that a given proof is in *this* system carries information as to its structure? These are dark and mysterious matters to me, and a little light shed by you would be most welcome. More information about the FOM

Effects of cold dark matter decoupling and pair annihilation on cosmological perturbations Weakly interacting massive particles are part of the lepton-photon plasma in the early universe until kinetic decoupling, after which time the particles behave like a collisionless gas with nonzero temperature. The Boltzmann equation for WIMP-lepton collisions is reduced to a Fokker-Planck equation for the evolution of the WIMP distribution including scalar density perturbations. This equation and the Einstein and fluid equations for the plasma are solved numerically including the acoustic oscillations of the plasma before and during kinetic decoupling, the frictional damping occurring during kinetic decoupling, and the free-streaming damping occurring afterwards and throughout the radiation-dominated era. An excellent approximation reduces the solution to quadratures for the cold dark matter density and velocity perturbations. The subsequent evolution is followed through electron pair annihilation and the radiation-matter transition; analytic solutions are provided for both large and small scales. For a 100 GeV WIMP with bino-type interactions, kinetic decoupling occurs at a temperature MeV. The transfer function in the matter-dominated era leads to an abundance of small cold dark matter halos; with a smooth window function the Press-Schechter mass distribution is for 10 MeV) M. pacs:95.35.+d, 95.30.Cq, 98.80.Cq Weakly interacting massive particles (WIMPs) are perhaps the leading candidate for the cold dark matter (CDM) making up most of the nonrelativistic mass density of the universe today bhs . Although candidate WIMPs are 10 to 1000 times more massive than nucleons and have no electromagnetic or color charges, their cosmic histories share many parallels with nucleons. After their abundances froze out at the number density of photons, both WIMPs and nucleons remained thermally coupled to the plasma by elastic scattering with abundant relativistic particles. Acoustic oscillations in the relativistic plasma imprinted oscillations on both WIMPs and nucleons. Eventually the plasma released its grip on both types of particles. For WIMPs, this event is called kinetic decoupling; for nucleons, recombination. During kinetic decoupling, friction between the WIMP gas and relativistic plasma led to Silk damping of small-scale waves similar to what happened much later for atomic matter at recombination. After the respective decoupling periods ended, pressure forces (and in the case of WIMPs, shear stress) inhibited gravitational instability on small scales. Still later, both WIMPs and nucleons played a major role in galaxy formation. There are, of course, significant differences between the cosmic evolution of WIMPs and nucleons. Most evident are the quantitative differences: because their interactions are so weak, WIMPs decoupled from the plasma less than one second after the big bang. Consequently the WIMP acoustic oscillations appear only on a length scale vastly smaller ( parsec scales) than the baryon acoustic oscillations. The physics of nucleon decoupling, as imprinted in the galaxy distribution eisenstein and in the cosmic microwave background radiation wmap3 provides a powerful probe of cosmic parameters and inflationary cosmology. If it were possible to similarly measure fluctuations on the scale of WIMP acoustic oscillations, we would have a dramatic consistency test of the cosmological model as well as an astrophysical measurement of WIMP properties. One way to constrain the parameters of cold dark matter decoupling is to measure the mass function of the smallest dark matter clumps today hss01 ; bde03 ; ghs04 . Such clumps would be far too diffuse to host observable concentrations of atomic matter. However, they might be observable through the products of the very rare WIMP-WIMP annihilations taking place in the cores of these objects. Diemand et al. diemand proposed that numerous Earth-mass clumps might survive to the present day and provide a detectable gamma-ray signal. The mass and abundance of these clumps depends on cosmic fluctuation evolution during and after kinetic decoupling. WIMPS and nucleons also differ in a qualitative manner which has important consequences for the evolution of fluctuations through kinetic decoupling. After recombination, elastic scattering is rapid enough for atoms (and the residual free electrons) to behave as a nearly perfect gas on cosmological scales. Baryons behave like a fluid. WIMPs, however, collide too infrequently to thermalize after kinetic decoupling. WIMPs behave like a collisionless gas. Different approximations to the evolution of this collisionless gas have led to different results for the small-scale transfer function of CDM fluctuations ghs04 ; ghs05 ; lz05 . In the present paper, the transfer functions for CDM fluctuations are calculated starting from the full Boltzmann equation describing elastic scattering between WIMPs and the relativistic leptons present before neutrino decoupling and electron-positron pair annihilation. Because the momentum transfer per collision with nonrelativistic WIMPs is small, the Boltzmann equation reduces to the Fokker-Planck equation describing diffusion in velocity space caused by elastic scattering, combined with advection and gravitational forces. The Fokker-Planck equation fully describes kinetic decoupling and the evolution of perturbations of any length scale without approximating the WIMPs either as a perfect fluid or fully collisionless gas. Although the solution of the perturbed Fokker-Planck equation is more difficult than the solution of coupled fluid and collisionless Boltzmann equations, it is both numerically and analytically tractable (with an excellent approximation) in the present case. After kinetic decoupling, two additional events have an effect on the CDM transfer function. The first is electron-positron pair annihilation, which changes the equation of state of the plasma thereby modifying the evolution of fluctuations. Although the effects are small, they can be analytically calculated. The more important event is the transition from a radiation- to matter-dominated universe occurring about years after the big bang. If the photons and neutrinos are treated as fluids, it is possible to get analytic results for the linear evolution all the way to low redshift which are accurate to a few percent. With these results in place, using standard techniques it is straightforward to estimate the mass function of CDM clumps at high redshift. Ii Evolution of WIMP perturbations through kinetic decoupling Weakly interacting dark matter particles are described by their phase space density, which obeys the Boltzmann equation governing transport by collisions with leptons (during kinetic decoupling these are just electrons, positrons, and neutrinos). For definiteness we will take the WIMP to be the lightest neutralino , however the results are easily applied to other WIMP candidates by modifying the scattering matrix element below. Let and be the proper phase space densities of neutralinos and ultrarelativistic leptons, respectively, where is the proper three-momentum in a local orthonormal frame. (The spacetime coordinates and are suppressed for brevity.) The phase space densities are normalized so that is the spatial number density, summed over spin states (we assumed unpolarized spins). One distribution function suffices for the relativistic leptons because electroweak interactions maintain local thermal equilibrium at a temperature . Elastic scattering of neutralinos and leptons cause the neutralino distribution to evolve according to the Boltzmann equation, where is the Lorentz-invariant scattering amplitude, and similarly for the other distribution functions, is the energy of particle , and is the occupation number. Pauli blocking must be included for the leptons but not for the neutralinos since the latter have a low density after chemical decoupling. Equation (1) is relativistically covariant but gives only the effects of collisions; the effects of gravitational perturbations will be added later. Assuming effectively massless leptons, the matrix element for slepton exchange is given in Appendix A of hss01 and may be written is a dimensionless constant depending on the relevant particle masses and couplings ( is the Fermi constant, and are left and right chiral vertices, and , , and are respectively the masses of the boson, the slepton, and the neutralino; ). Here we assume following Ref. hss01 that the neutralino is a pure bino. Additionally, is the momentum in the center of momentum frame, and (using metric signature ) is one of the Mandelstam variables. For , where . In the lab frame, working to first order in , assuming , the collision kinematics gives where and . Another useful relation follows from energy conservation, valid again to first order in . Frequent collisions among the leptons maintain thermal equilibrium. Assuming negligible chemical potential, for each species of massless lepton we have where is the (very small) local lepton fluid velocity due to cosmological perturbations. It is easy to check that an equilibrium solution of (1) is then the Maxwell-Boltzmann distribution with mean velocity . For the term may be Taylor-expanded in (1). After a lengthy calculation using (II)–(6), one obtains (dropping the subscript on ) where the Boltzmann collision integral becomes the Fokker-Planck operator, is a rate coefficient (in units where ). Our exact result for the rate coefficient is larger by a factor 9.9 than the estimate obtained from Eqs. (9) and (12) of Ref. hss01 and by a factor 3.4 than Eq. (17) of Ref. bde03 . The rate is greater than the simple estimates made in previous work because of the details of the kinematics and the near-cancelation of forward and inverse rates in (1). A larger rate coefficient leads to a lower temperature for kinetic decoupling than previous estimates. If we neglect spatial inhomogeneities, the unperturbed phase space density depends on both comoving momentum and conformal time according to Amazingly, for any time-dependence of , , and an exact solution to this Fokker-Planck equation is the Maxwell-Boltzmann distribution where the WIMP temperature follows from integrating During adiabatic evolution in the early universe, and the WIMP proper temperature is then given in terms of the incomplete Gamma function by Equation (10) may be multiplied by any constant, allowing the comoving number density of WIMPs to be normalized to its value after freeze-out. Familiarity with Brownian motion makes it seem natural that the solution to the Fokker-Planck equation is a Maxwell-Boltzmann distribution. However, the lepton temperature and the momentum transfer rate are falling with time and WIMP-WIMP elastic scattering is far too slow to thermalize the distribution. Even so, collisions with the leptons maintain the WIMPs in a thermal distribution with a temperature that deviates increasingly from the lepton temperature throughout kinetic decoupling. Once kinetic decoupling is complete the WIMP momenta redshift as preserving the Maxwell-Boltzmann distribution with . Long before kinetic decoupling (), . After kinetic decoupling (), . Defining the kinetic decoupling time by yields For typical supersymmetry masses, kinetic decoupling occurs after muon annihilation when the only abundant leptons are electron pairs and neutrinos, for which (with ) and . With GeV and GeV, yielding MeV. Profumo et al. psk06 show that can span the range from a few MeV to a few GeV for reasonable WIMP models. Here we take the supersymmetric bino as a candidate but will show how numerical results scale with . Next we examine the effect of density, velocity, and gravitational potential fluctuations during kinetic decoupling. The perturbed phase space density is , where are comoving spatial coordinates, are the conjugate momenta, and is conformal time. The perturbed line element in conformal Newtonian gauge is written . Including the effects of the metric perturbations and , the Boltzmann equation becomes where is the proper velocity measured by a comoving observer, is the comoving energy, and . With comoving variables, the Fokker-Planck operator becomes Now we linearize (15) for first-order perturbations of the lepton fluid by writing . The fields , , , and the lepton velocity potential are treated as first-order quantities. Assuming and performing a spatial Fourier transform, we obtain the linear perturbation equation After WIMP freeze-out, the leptons dominate the gravitational potentials so that , , , and are functions of given by the solution for a perfect relativistic fluid. Equation (16) generalizes the collisionless Boltzmann equation of Ref. mb95 to include the effects of dark matter collisions with relativistic leptons. To integrate (16) for the phase space density we will expand the momentum dependence in eigenfunctions of the Fokker-Planck operator : Here and is a generalized Laguerre polynomial, also known as a Sonine polynomial. It is defined by the following series expansion: Generalized Laguerre polynomials have orthonormality relation and completeness relation We will expand the phase space density in the complete set . However, it is unnecessary to include all in this expansion. Prior to kinetic decoupling the Fokker-Planck operator rapidly damps all terms except and those terms that are induced by the right-hand side of (16). The rotational symmetry of this equation implies that only the terms are induced, where the polar axis for the spherical harmonics is given by 111Collisional damping before neutrino decoupling similarly justifies the neglect of the modes for neutrinos in Ref. mb95 .. Thus we may write Before writing the perturbation equations, let us examine the unperturbed solution, . The exponential factors differ in (10) and (II), implying that the Laguerre expansion includes more than one term. Indeed, one finds Prior to kinetic decoupling, when collisions maintain , and the other coefficients vanish. After kinetic decoupling, for all . The Laguerre expansion must be carried to high order in order to convert to . Similarly, we should expect the perturbed phase space density also to require many terms in after kinetic decoupling. Substituting (II) into (16) and using orthonormality and several recurrence relations for the generalized Laguerre polynomials, we obtain a system of coupled ordinary differential equations for the evolution of the perturbed phase space density, The density perturbation is for . The source terms for Eqs. (II) are provided by the relativistic plasma. Their time-dependence changes during lepton pair annihilation and neutrino decoupling. For reasonable parameters, neutralino kinetic decoupling occurs after muon annihilation but before electron annihilation and neutrino decoupling. During this era, the isentropic mode of perturbations has time-dependence given by the relativistic perfect-fluid transfer functions where , , and the transfer functions are normalized so that for 222Here, is the number density perturbation in conformal Newtonian gauge; the energy density perturbation for the relativistic leptons is .. The actual perturbations are obtained by multiplying the transfer functions with the scalar field that gives the spatial dependence of the initial (inflationary) curvature fluctuations. As the inflationary curvature perturbations are well known (a Gaussian random field with nearly scale-invariant spectral density ), here we work with transfer functions. Initial conditions for Eqs. (II) are obtained by examining the solutions for . Isentropic initial conditions have . The only significantly perturbed components of for the strongly coupled plasma are the thermal equilibrium values All other components are kept small by rapid WIMP-lepton collisions. Equations (II) with and were integrated to using the standard explicit ordinary differential equation solver DVERK. Convergence testing showed that higher-order terms in the Laguerre expansion were negligible. Figure 1 shows the results for the density transfer function expressed using the gauge-invariant variable , defined as the CDM number density perturbation in a synchronous gauge for which the mean CDM velocity vanishes in the coordinate frame (for a nonrelativistic particle, equals Bardeen’s variable ). This variable, which is used by CMBFAST cmbfast , is related to the conformal Newtonian gauge variables by For wavelengths longer than the radiation acoustic length , . For wavelengths shorter than the radiation acoustic length at but longer than the acoustic length at (i.e., ), the acoustic oscillations of the gravitational potential average out leading to a suppression of growth induced in the CDM. For these intermediate wavelengths the transfer function is a logarithmic function of wavenumber. If the dark matter were completely non-interacting, this logarithm would continue to arbitrarily high wavenumbers, as illustrated by the monotonically rising curve in Fig. 1. Because WIMP dark matter was collisionally coupled to the relativistic lepton plasma at early times, the CDM transfer function in Fig. 1 shows remnant damped acoustic oscillations at short wavelengths. For comparison a Gaussian transfer function is shown, with no oscillations. In this model, the effects of kinetic decoupling are described by multiplying the transfer function for the completely non-interacting case by . As we will see, a simple model of free streaming predicts that during the radiation-dominated era. For and this model would predict whereas the curve shown in Fig. 1 has . Free-streaming does not give a good approximation to the actual transfer function. The exact transfer functions shown in Fig. 1 decrease less rapidly with wavenumber than the approximate transfer functions of Ref. lz05 which were computed using a fluid approximation followed by free-streaming. The following section reviews the free-streaming solution and then develops a more accurate approximation based on moments of the exact Fokker-Planck equation. Iii Approximate descriptions of perturbation evolution through kinetic decoupling In this section we consider two different approximations which provide analytical insight to the numerical solution of the Fokker-Planck equation. The CDM behaves at early times like a fluid and at late times like a free-streaming collisionless gas, and in these limits we can develop useful analytical approximations. iii.1 Free-streaming model For , the terms proportional to may be dropped in (16). The differential equation may then be integrated to give in terms of the initial value for any . Integrating over momenta gives the conformal Newtonian gauge density perturbation, where we have assumed evolution in the radiation-dominated era with For large spatial frequencies, , the gravitational potentials — which are dominated by relativistic particles — oscillate rapidly leading to a small net integral contribution; ignoring this, is a momentum-space Fourier transform of the distribution function at the initial time . Obtaining the exact solution still requires numerical integration of (16) through kinetic decoupling, or equivalently, integrating the system of equations (II). However, we can get an idea of the effects of free-streaming by making an instantaneous decoupling approximation, treating the CDM as a fluid with a Maxwell-Boltzmann velocity distribution for and by (III.1) for , as was done in Ref. lz05 . Then, whether or not the CDM is strongly coupled to the radiation, the perturbed distribution function is fully specified by the perturbations of density, temperature, and velocity , Carrying out the Fourier transform in (III.1) now yields where wavenumber arguments have been dropped for brevity, and If , then ; if and the fluid is approximated as being adiabatic, . Equation (32) corrects errors in the definition of of Ref. lz05 and adds the term proportional to to their Eq. (20). This new term arises from the temperature perturbation of the CDM fluid, which modifies the distribution function and, through free-streaming, modifies the density for . In particular, if the CDM perturbations are approximately adiabatic, the temperature perturbation causes the transfer function to decrease more rapidly with . The free-streaming solution predicts a Gaussian cutoff of the transfer function, , with cutoff distance equal to the free-streaming distance, During the radiation-dominated era, when , the free-streaming length grows logarithmically, but it saturates in the matter-dominated era when . At first glance, Fig. 1 appears to qualitatively support the free-streaming model of a Gaussian cutoff. However, the free-streaming model predicts no damping for a super-heavy particle with , while Fig. 1 shows that even in this case there is damping. This damping arises from friction between the lepton and CDM gases during decoupling (Silk damping). This friction can be accounted for by treating the CDM as an imperfect fluid. iii.2 Imperfect fluid model To better describe an extended period of decoupling while allowing for small deviations from a Maxwell-Boltzmann distribution, we consider the evolution of the lowest order moments of the distribution function. We work to first order in perturbed quantities and normalize the unperturbed distribution function to . The perturbations for then define the lowest-order moments where the density perturbation , velocity potential , shear stress potential , and entropy perturbation are related to our expansion coefficients by The effective sound speed squared of the CDM fluid is which differs from the thermal speed squared appearing in (II). This difference arises because the Laguerre expansion uses eigenfunctions of the Fokker-Planck operator which depends on the relativistic lepton temperature rather than the WIMP temperature . After kinetic decoupling, drops and the higher-order expansion coefficients will increase to compensate for this difference, as we already found happening with the unperturbed distribution function in (23). The variables in (III.2) describe fluctuations of an imperfect fluid. The reader may wonder how a single component fluid can have an entropy perturbation. A weakly imperfect fluid is described by an equation of state where is the entropy which may vary in space and time. However, the WIMP gas is more complicated because it becomes fully collisionless after kinetic decoupling; it may be regarded as a superposition of many non-interacting ideal gases. The time evolution of the imperfect fluid variables follows from Eqs. (II) 333These fluid equations, like the Fokker-Planck equation from which they were derived, are correct only to leading order in . The corrections to (38a) and (38b) are and , respectively.: These equations are similar to those of an imperfect gas coupled to the lepton fluid. However, they differ significantly from the Navier-Stokes equations assumed in Ref. hss01 . In place of a bulk viscosity term and a shear viscosity term where and are the bulk and shear viscosity coefficients (divided by the mass density), (38b) has an entropy term and a shear stress term where and are not proportional to . The usual Chapman-Enskog expansion does not apply to our Fokker-Planck equation when the collision mean-free path becomes large. Moreover, because of the and terms, Eqs. (38) do not form a closed system. Nonetheless, these equations are useful for providing insight and they will guide us to a very good approximation to the full numerical solution of the Fokker-Planck equation. Prior to kinetic decoupling, when the damping terms proportional to are large, the solutions to (38) have , , , in agreement with (27). Entropy and shear stress perturbations develop during decoupling as the CDM gas becomes collisionless. These in turn modify and damp the acoustic oscillations of the gas. It is instructive to solve the imperfect fluid equations with several different approximations, in order to deduce which physical effects are responsible for the features of the transfer functions shown in Fig. 1. The most extreme approximation is to completely neglect the CDM temperature and coupling to other particles, so as to describe a perfect cold, collisionless gas. This approximation consists of setting retaining only the gravitational interaction between the CDM and the relativistic plasma. In this case the exact solution prior to neutrino decoupling for isentropic initial conditions is where , is the Euler-Mascharoni constant, and is the cosine integral. This solution is shown by the monotonically increasing curve in Fig. 1. We see that once the wavelength becomes smaller than the relativistic acoustic horizon and the potentials oscillate faster than the CDM can respond, the CDM density perturbation growth slows to logarithmic in time. This suppression is responsible for the turnover of the low-redshift CDM power spectrum from at long wavelengths to at short wavelengths. However, the approximation of a cold, collisionless fluid includes none of the physical effects of kinetic decoupling. The next simplest approximation is to treat the CDM as being cold () but to include the friction term in the velocity equation. This approximation is exact in the limit hence it reproduces the case in Fig. 1. We can find the solution in the limit by first noting that the general solution of the second-order system , is given by where and are independent of but may depend on . Including the damping terms in (38b) promotes and to functions and obeying the differential equations In the strongly-coupled limit the solution must match Eqs. (26), which gives The exact solution of (41) satisfying these initial conditions is given by a pair of quadratures, where , , and primed quantities are evaluated at . The late-time solution is dominated by , which for becomes the following function of wavenumber, The complex function where itself is complex but cannot be used as the integration variable because of the essential singularity at . The contour used to evaluate is shown in Fig. 2. Using the steepest descent approximation to evaluate the contributions along the branch cut gives

The system returned: (22) Invalid argument The remote host or network may be down. I'll give the formula, then explain it formally, then do some examples. We have where bounds on the given interval . Arbetar ... http://birdsallgraphics.com/error-bound/error-bound-taylor-polynomials.php You can change this preference below. Krista King 13 943 visningar 12:03 Find error bound for approximating f(x) values with a Taylor polynomial - Längd: 8:40. for some z in [0,x]. Automatisk uppspelning När automatisk uppspelning är aktiverad spelas ett föreslaget videoklipp upp automatiskt. http://math.jasonbhill.com/courses/fall-2010-math-2300-005/lectures/taylor-polynomial-error-bounds So, we force it to be positive by taking an absolute value. So the error at "a" is equal to f of a minus p of a, and once again I won't write the sub n and sub a, you can just assume Ideally, the remainder term gives you the precise difference between the value of a function and the approximation Tn(x). What we can continue in the next video, is figure out, at least can we bound this, and if we're able to bound this, if we're able to figure out an CAL BOYS 4 721 visningar 3:32 Find degree of Taylor polynomial so error is less than a given error value - Längd: 6:02. F of a is equal to p of a, so there error at "a" is equal to zero. Now let's think about something else. Lagrange Error Bound Proof Here is a list of the three examples used here, if you wish to jump straight into one of them. Now, if we're looking for the worst possible value that this error can be on the given interval (this is usually what we're interested in finding) then we find the maximum And this general property right over here, is true up to and including n. DrPhilClark 38 394 visningar 9:33 Lagrange Error Bound Problem - Längd: 3:32. https://www.khanacademy.org/video/proof-bounding-the-error-or-remainder-of-a-taylor-polynomial-approximation Since takes its maximum value on at , we have . And we've seen that before. Error Bound Formula Statistics Logga in om du vill rapportera olämpligt innehåll. Logga in Dela Mer Rapportera Vill du rapportera videoklippet? Basic Examples Find the error bound for the rd Taylor polynomial of centered at on . Explanation We derived this in class. http://math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/SandS/PowerSeries/error_bounds.html The first derivative is 2x, the second derivative is 2, the third derivative is zero. Lagrange Error Bound Formula The following example should help to make this idea clear, using the sixth-degree Taylor polynomial for cos x: Suppose that you use this polynomial to approximate cos 1: How accurate is Lagrange Error Bound Problems What is this thing equal to, or how should you think about this. Toggle navigation Search Submit San Francisco, CA Brr, it´s cold outside Learn by category LiveConsumer ElectronicsFood & DrinkGamesHealthPersonal FinanceHome & GardenPetsRelationshipsSportsReligion LearnArt CenterCraftsEducationLanguagesPhotographyTest Prep WorkSocial MediaSoftwareProgrammingWeb Design & DevelopmentBusinessCareersComputers Online Courses this contact form Thus, we have In other words, the 100th Taylor polynomial for approximates very well on the interval . The distance between the two functions is zero there. If we assume that this is higher than degree one, we know that these derivatives are going to be the same at "a". Lagrange Error Bound Khan Academy So it's literally the n+1th derivative of our function minus the n+1th derivative of our nth degree polynomial. Logga in Transkription Statistik 2 934 visningar 8 Gillar du videoklippet? And what I want to do in this video, since this is all review, I have this polynomial that's approximating this function, the more terms I have the higher degree of have a peek here However, you can plug in c = 0 and c = 1 to give you a range of possible values: Keep in mind that this inequality occurs because of the interval Visa mer Läser in ... Alternating Series Error Bound And then plus go to the third derivative of f at a times x minus a to the third power, (I think you see where this is going) over three factorial, The following theorem tells us how to bound this error. Let me actually write that down, because it's an interesting property. It's a first degree polynomial... Your email Submit RELATED ARTICLES Calculating Error Bounds for Taylor Polynomials Calculus Essentials For Dummies Calculus For Dummies, 2nd Edition Calculus II For Dummies, 2nd Edition Calculus Workbook For Dummies, 2nd Lagrange Error Bound Ap Calculus Bc What is the (n+1)th derivative of our error function. It considers all the way up to the th derivative. Of course, this could be positive or negative. what's the n+1th derivative of it. Check This Out But what I want to do in this video is think about, if we can bound how good it's fitting this function as we move away from "a". Here's the formula for the remainder term: So substituting 1 for x gives you: At this point, you're apparently stuck, because you don't know the value of sin c. VisningsköKöVisningsköKö Ta bort allaKoppla från Läser in ... So this is an interesting property. CalculusSeriesTaylor series approximationsVisualizing Taylor series approximationsGeneralized Taylor series approximationVisualizing Taylor series for e^xMaclaurin series exampleFinding power series through integrationEvaluating Taylor Polynomial of derivativePractice: Finding taylor seriesError of a Taylor polynomial approximationProof: We define the error of the th Taylor polynomial to be That is, error is the actual value minus the Taylor polynomial's value. numericalmethodsguy 20 427 visningar 6:44 Taylor Polynomials of a Function of Two Variables - Längd: 12:51. If x is sufficiently small, this gives a decent error bound. And not even if I'm just evaluating at "a". So because we know that p prime of a is equal to f prime of a when we evaluate the error function, the derivative of the error function at "a" that Försök igen senare. You built both of those values into the linear approximation. Läser in ... Use a Taylor expansion of sin(x) with a close to 0.1 (say, a=0), and find the 5th degree Taylor polynomial. What you did was you created a linear function (a line) approximating a function by taking two things into consideration: The value of the function at a point, and the value So, the first place where your original function and the Taylor polynomial differ is in the st derivative. Funktionen är inte tillgänglig just nu. That is, we're looking at Since all of the derivatives of satisfy , we know that . So for example, if someone were to ask: or if you wanted to visualize, "what are they talking about": if they're saying the error of this nth degree polynomial centered at You can get a different bound with a different interval.

« PreviousContinue » that the phenomena of the precession and nutation must be the same in the actual state of our terraqueous spheroid, as if the whole was a solid mass; and that this is true, whatever be the irregularity of the depth of the sea. He shows also, that currents in the sea, rivers, trade-winds, even earthquakes, can have no effect in altering the earth's rotation on its axis. The conclusions with regard to the constitution of the earth that are found to agree with the actual quantity of the precession of the equinoxes are, that the density of the earth increases from the circumference toward the centre; that it has the form of an ellipsoid of revolution, or, as we use to call it, of an elliptic spheroid, and that the compression of this spheroid at the poles is between the limits of and part of the radius of the equator. The Second part of La Place's work, has, for its object, a fuller development of the disturbances of the planets, both primary and secondary, than was compatible with the limits of the First part. After the ample detail into which we have entered concerning two of these subjects, the theory of the moon, and the perturbations of the primary planets, we need not enlarge on them further, though they are prosecuted in the second part of this work, and form the subject of the Sixth and Seventh books. In the Second book, the inequalities had been explained, that depend on the simple power of the eccentricity: here we have those that depend on the second and higher powers of the same quantity; and such are the secular equations of Jupiter and Saturn, abovementioned. The numeral computations are then performed, and every thing prepared for the complete construction of astronomical tables, as the final result of all these investigations. The calculations, of course, are of vast extent and difficulty, and incredibly laborious. In carrying them on, La Place had the assistance, as he informs us, of De Lambre, Bouvard, and other members of the institute. The labour is indeed quite beyond the power of any individual to execute. The same may be said of the Seventh book, which is devoted to a similar development of the lunar theory. We can enter into no further detail on this subject. One fact we cannot help mentioning, which is to the credit of two British astronomers, Messrs Mason and Dixon, who gave a new edition of Mayer's tables, more diligently compared with observation, and therefore more accurate, than the original one. Among other improvements, was an empirical equation, amounting to a little more than 20" when a maximum, which was not founded on theory, but was employed because it made the tables agree better with observation. As this equation, however, was not derived from principle (for the two astronomers, just named, though accurate observers and calculators, were not skilled enough in the mathematics to attempt deducing it from principle), it was generally rejected by other astronomers. La Place, however, found that it was not to be rejected; but, in reality, proceeded from the compression of the earth at the poles, which prevents the gravitation to the earth from decreasing, precisely as the squares of the distances increase, and by that means produces this small irregularity. The quantity of the polar compression that agrees best with this, and some other of the lunar irregularities, is nearly that which was stated aboved, of the mean radius of the earth. The ellipticity of the sun does, in like manner, affect the primary planets; but, as its influence diminishes fast as the distance increases, it extends no further (in any sensible degree) than the orbit of Mercury, where its only effect is to produce a very small direct movement of the line of the apsides, and an equal retrograde motion of the nodes, relatively to the sun's equator. We may judge from this, to what minuteness the researches of this author have extended: and, in general, when accuracy is the object to be obtained, the smaller the quantity to be determined, the more difficult the investigation. The Eighth book has for its object, to calculate the disturbances produced by the action of the secondary planets on one another; and particularly refers to the satellites of Jupiter, the only system of secondary planets on which accurate observations have been, or, probably, can be made. Though these satellites have been known only since the invention of the telescope, yet the quickness of their revolutions has, in the space of two centuries, exhibited all the changes which time develops so slowly in the system of the primary planets; so that there are abundant materials for a comparison between fact and theory. The general principles of the theory are the same that were explained in the Second book; but there are some peculiarities, that arise from the constitution of Jupiter's system, that deserve to be considered. We have seen, above, what is the effect of commensurability, or an approach to it, in the mean motion of contiguous planets; and here we have another example of the same. The mean motions of the three first satellites of Jupiter, are nearly as the numbers 4, 2, and ; and hence a periodical system of inequalities, which our astronomer Bradley was sharp-sighted enough to discover in the observation of the eclipses of these satellites, and to state as amounting to 437.6 days. This is now fully explained from the theory of the action of the satellites. Another singularity in this secondary system, is, that the mean longitude of the first satellite minus three times that of the second, plus twice that of the third, never differs from two right angles but by a quantity almost insensible. One can hardly suppose that the original motions were so adjusted as to answer exactly to this condition; it is more natural to suppose that they were only nearly 'so adjusted, and that the exact comcidence has been brought about by their mutual action. This conjecture is verified by the theory, where it is demonstrated that such a change might have been actually produced in the mean motion by the mutual action of those planetary bodies, after which the system would remain stable, and no further change in those motions would take place. Not only are the mutual actions of the satellites taken into account in the estimate of their irregularities, but the effect of Jupiter's spheroidal figure is also introduced. Even the masses of the satellites are inferred from their effect in disturbing the motions of one another. In the Ninth book La Place treats of Comets, of the methods of determining their orbits, and of the disturbances they suffer from the planets. We cannot follow him in this; and have only to to add, that his profound and elaborate researches are such as we might expect from the author of the preceding investigations. The Tenth book is more miscellaneous than any of the preceding; it treats of different points relative to the system of the world. One of the most important of these is astronomical refraction. The rays of light from the celestial bodies, on entering the earth's atmosphere, meet with strata that are more dense the nearer they approach to the earth's surface; they are therefore bent continually toward the denser medium, and describe curves that have their concavity turned toward the earth. The angle formed by the original direction of the ray, and its direction at the point where it enters the eye, is called the astronomical refraction. La Place seeks to determine this angle by tracing the path of the ray through the atmosphere; a research of no inconsiderable difficulty, and in which the author has occasion to display his skill both in mathematical and in inductive investigation. The method he pursues in the latter is deserving of attention, as it is particularly well adapted to cases that occur often in the more intricate kinds of physical discussion. The path of the ray would be determined from the laws of refraction, did we know the law by which the density of the air decreases from the earth upwards. This last, however, is not known, except for a small extent near the surface of the earth, so that we appear here to be left without sufficient data for continuing the investigation. We must, therefore, ei ther abandon the problem altogether, or resolve it hypothetically, that is, by assuming some hypothesis as to the decrease of the density of the atmosphere. Little would be gained by this last, except as an exercise in mathematical investigation, if it were not that the total quantity of the refraction for a given altitude can be accurately determined by observation. La Place, availing himself of this consideration, begins with making a supposition concerning the law of the density, that is not very remote from the truth, (as we are assured of from the relation between the density of air and the force with which it is compressed); and he compares the horizontal refraction calculated on this assumption with that which is known to be its true quantity. The first hypothesis which he assumes, is that of the density being the same throughout; this gives the total refraction too small, and falls on that account to be rejected, even if it were liable to no other objection. The second hypothesis supposes a uniform temperature through the whole extent of the atmosphere, or it supposes that the density decreases in geometrical proportion, while the distance from the earth increases in arithmetical. The refraction which results is too great; so that this supposition must also be rejected. If we now suppose the density of the air to decrease in arithmetical progression, while the height does the same, and integrate the differential equation to the curve described by the ray, on this hypothesis, the horizontal refraction is too small, but nearer the truth than on the first hypothesis. A supposition intermediate between that which gave the refraction too great, and this which gives it too small, is therefore to be assumed as that which approaches the nearest to the truth. It is this way of limiting his conjectures by repeated trials, and of extracting from each, by means of the calculus, all the consequences involved in it, that we would recommend to experimenters, as affording one of the most valuable and legitimate uses of hypothetical reasoning. He then employs an intermediate hypothesis for the diminution of the density of the air; which it is not easy to express in words, but from which he obtains a result that agrees with the horizontal refraction, and from which, of course, he proceeds to deduce the refraction for all other altitudes. The table, so constructed, we have no doubt, will be found to contribute materially to the accuracy of astronomical observation. The researches which immediately follow this, relate to the terrestrial refraction, and the measurement of heights by the barometer. The formula given for the latter, is more complicated than that which is usually employed with us in Britain, where this subject has been studied with great care. In one respect, it is more general than any of our formulas; it contains an allowance for the difference of latitude. We are not sure whether this correction is of much importance, nor have we had leisure to compare the results with those of General Roy and Sir George Schuckborough. We hardly believe, that in point of accuracy, the two last can easily be exceeded. The book concludes with a determination of the masses of the planets, more accurate than had been before given; and even of the satellites of Jupiter. Of all the attempts of the Newtonian philosophy,' says the late Adam Smith in his History of Astronomy, that which would appear to be the most above the reach of human reason and experience, is the attempt to compute the weights and densities of the sun, and of the several planets. What would this philosopher have said, if he had lived to see the same balance in which the vast body of the sun had been weighed, applied to examine such minute atoms as the satellites of Jupiter? Such is the work of La Place, affording an example, which is yet solitary in the history of human knowledge, of a theory entirely complete; one that has not only accounted for all the phenomena that were known, but that has discovered many before unknown, which observation has since recognized. In this theory, not only the elliptic motion of the planets, relatively to the sun, but the irregularities produced by their mutual action, whether of the primary on the primary, of the primary on the secondary, or of the secondary on one another, are all deduced from the principle of gravitation, that mysterious power, which unites the most distant regions of space, and the most remote periods of duration. To this we must add the great truths brought in view and fully demonstrated, by tracing the action of the same power through all its mazes: That all the inequalities in our system are periodical; that, by a fixt appointment in nature, they are each destined to revolve in the same order, and between the same limits; that the mean distances of the planets from the sun, and the time of their revolutions round that body, are susceptible of no change whatsoever; that our system is thus secured against natural decay; order and regularity preserved in the midst of so many disturbing causes ;-and anarchy and misrule eternally proscribed. The work where this sublime picture is delineated, does honour, not to the author only, but to the human race; and marks, undoubtedly, the highest point to which man has yet ascended in the scale of intellectual attainment. The glory, therefore, of having produced this work, belongs not to the author

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Triangle numbers can be represented by a triangular array of squares. What do you notice about the sum of identical triangle numbers? What would be the smallest number of moves needed to move a Knight from a chess set from one corner to the opposite corner of a 99 by 99 square board? Imagine starting with one yellow cube and covering it all over with a single layer of red cubes, and then covering that cube with a layer of blue cubes. How many red and blue cubes would you need? Here are some arrangements of circles. How many circles would I need to make the next size up for each? Can you create your own arrangement and investigate the number of circles it needs? If you can copy a network without lifting your pen off the paper and without drawing any line twice, then it is traversable. Decide which of these diagrams are traversable. How could Penny, Tom and Matthew work out how many chocolates there are in different sized boxes? The aim of the game is to slide the green square from the top right hand corner to the bottom left hand corner in the least number of Here are two kinds of spirals for you to explore. What do you notice? Only one side of a two-slice toaster is working. What is the quickest way to toast both sides of three slices of bread? Can you work out how to win this game of Nim? Does it matter if you go first or second? In this problem we are looking at sets of parallel sticks that cross each other. What is the least number of crossings you can make? And the greatest? Square numbers can be represented as the sum of consecutive odd numbers. What is the sum of 1 + 3 + ..... + 149 + 151 + 153? This article for teachers describes several games, found on the site, all of which have a related structure that can be used to develop the skills of strategic planning. One block is needed to make an up-and-down staircase, with one step up and one step down. How many blocks would be needed to build an up-and-down staircase with 5 steps up and 5 steps down? Imagine an infinitely large sheet of square dotty paper on which you can draw triangles of any size you wish (providing each vertex is on a dot). What areas is it/is it not possible to draw? Four bags contain a large number of 1s, 3s, 5s and 7s. Pick any ten numbers from the bags above so that their total is 37. In each of the pictures the invitation is for you to: Count what you see. Identify how you think the pattern would continue. How many ways can you find to do up all four buttons on my coat? How about if I had five buttons? Six ...? For this challenge, you'll need to play Got It! Can you explain the strategy for winning this game with any target? Are these statements relating to odd and even numbers always true, sometimes true or never true? This challenge asks you to imagine a snake coiling on itself. Can you describe this route to infinity? Where will the arrows take you next? Draw a square. A second square of the same size slides around the first always maintaining contact and keeping the same orientation. How far does the dot travel? Watch this video to see how to roll the dice. Now it's your turn! What do you notice about the dice numbers you have recorded? Nim-7 game for an adult and child. Who will be the one to take the last counter? It starts quite simple but great opportunities for number discoveries and patterns! Three circles have a maximum of six intersections with each other. What is the maximum number of intersections that a hundred circles Find a route from the outside to the inside of this square, stepping on as many tiles as possible. Can you dissect an equilateral triangle into 6 smaller ones? What number of smaller equilateral triangles is it NOT possible to dissect a larger equilateral triangle into? Start with any number of counters in any number of piles. 2 players take it in turns to remove any number of counters from a single pile. The winner is the player to take the last counter. Can you find all the ways to get 15 at the top of this triangle of numbers? Euler discussed whether or not it was possible to stroll around Koenigsberg crossing each of its seven bridges exactly once. Experiment with different numbers of islands and bridges. Find the sum and difference between a pair of two-digit numbers. Now find the sum and difference between the sum and difference! What happens? Choose a couple of the sequences. Try to picture how to make the next, and the next, and the next... Can you describe your reasoning? A collection of games on the NIM theme This task follows on from Build it Up and takes the ideas into three dimensions! An article for teachers and pupils that encourages you to look at the mathematical properties of similar games. Can you tangle yourself up and reach any fraction? These squares have been made from Cuisenaire rods. Can you describe the pattern? What would the next square look like? In how many different ways can you break up a stick of 7 interlocking cubes? Now try with a stick of 8 cubes and a stick of 6 cubes. Think of a number, square it and subtract your starting number. Is the number you’re left with odd or even? How do the images help to explain this? What are the areas of these triangles? What do you notice? Can you generalise to other "families" of triangles? Try entering different sets of numbers in the number pyramids. How does the total at the top change? Can you see why 2 by 2 could be 5? Can you predict what 2 by 10 What size square corners should be cut from a square piece of paper to make a box with the largest possible volume? How many centimetres of rope will I need to make another mat just like the one I have here? Find out what a "fault-free" rectangle is and try to make some of The Egyptians expressed all fractions as the sum of different unit fractions. Here is a chance to explore how they could have written Can you find an efficient method to work out how many handshakes there would be if hundreds of people met? Sweets are given out to party-goers in a particular way. Investigate the total number of sweets received by people sitting in different positions.

|(43)||I.||Magnitudes which are equal to the same are equal to each other.| |(44)||II.||If equals be added to equals the sums will be equal.| |(45)||III.||If equals be taken away from equals the remainders will be equal.| |(46)||IV.||If equals be added to unequals the sums will be unequal.| |(47)||V.||If equals be taken away from unequals the remainders will be unequal.| |(48)||VI.||The doubles of the same or equal magnitudes are equal.| |(49)||VII.||The halves of the same or equal magnitudes are equal.| |(50)||VIII.||Magnitudes which coincide with one another, or exactly fill the same space, are equal.| |(51)||IX.||The whole is greater than its part.| |(52)||X.||Two right lines cannot include a space.| |(53)||XI.||All right angles are equal.| |(54)||XII.||If two right lines (A B, C D) meet a third right line (A C) so as to make the two interior angles (B A C and D C A) on the same side less than two right angles, these two right lines will meet if they be produced on that side on which the angles are less than two right angles.| (55) The geometrical axioms are certain general propositions, the truth of which is taken to be self-evident, and incapable of being established by demonstration. According to the spirit of this science, the number of axioms should be as limited as possible. A proposition, however self-evident, has no title to be taken as an axiom, if its truth can be deduced from axioms already admitted. We have a remarkable instance of the rigid adherence to this principle in the twentieth proposition of the first book, where it is proved that ‘two sides of a triangle taken together are greater than the third;’ a proposition which is quite as self-evident as any of the received axioms, and much more self-evident than several of them. On the other hand, if the truth of a proposition cannot be established by demonstration, we are compelled to take it as an axiom, even though it be not self-evident. Such is the case with the twelfth axiom. We shall postpone our observations on this axiom, however, for the present, and have to request that the student will omit it until he comes to read the commentary on the twenty-eighth proposition. See Appendix II. Two magnitudes are said to be equal when they are capable of exactly covering one another, or filling the same space. In the most ordinary practical cases we use this test for determining equality; we apply the two things to be compared one to the other, and immediately infer their equality from their coincidence. By the aid of this definition of equality we conceive that the second and third axioms might easily be deduced from the first. We shall not however pursue the discussion here. ★★★ The fourth and fifth axioms are not sufficiently definite. After the addition or subtraction of equal quantities, unequal quantities continue to be unequal. But it is also evident, that their difference, that is, the quantity by which the greater exceeds the less, will be the same after such addition or subtraction as before it. The sixth and seventh axioms may very easily be inferred from the preceding ones. The tenth axiom may be presented under various forms. It is equivalent to stating, that between any two points only one right line can be drawn. For if two different right lines could be drawn from one point to another, they would evidently enclose a space between them. It is also equivalent to stating, that two right lines being infinitely produced cannot intersect each other in more than one point; for if they intersected at two points, the parts of the lines between these points would enclose a space. The eleventh axiom admits of demonstration. Let A B and E F be perpendicular to D C and H G. Take any equal parts E H, E G on H G measured from the point E, and on D C take parts from A equal to these (Prop. III. Book I.) Let the point H be conceived to be placed upon the point D. The points G and C must then be in the circumference of a circle described round the centre D, with the distance D C or H G as radius. Hence, if the line H G be conceived to be turned round this centre D, the point G must in some position coincide with C. In such a position every point of the line H G must coincide with C D (ax. 10.), and the middle points A and E must evidently coincide. Let the perpendiculars E F and A B be conceived to be placed at the same side of D C. They must then coincide, and therefore the right angle F E G will be equal to the right angle B A C. For if E F do not coincide with A B, let it take the position A K. The right angle K A C is equal to K A D (11), and therefore greater than B A D; but B A D is equal to B A C (11), and therefore K A C is greater than B A C. But K A C is a part of B A C, and therefore less than it, which is absurd; and therefore E F must coincide with A B, and the right angles B A C and F E G are equal. The postulates may be considered as axioms. The first postulated, which declares the possibility of one right line joining two given points, is as much an axiom as the tenth axiom, which declares the impossibility of more than one right line joining them. In like manner, the second postulate, which grants the power of producing a line, may be considered as an axiom, declaring that every finite straight line may have another placed at its extremity so to form with it one continued straight line. In fact, the straight line thus placed will be its production. This postulate is assumed as an axiom in the fourteenth proposition of the first book. (56) Those results which are obtained in geometry by a process of reasoning are called propositions. Geometrical propositions are of two species, problems and theorems. A problem is a proposition in which something is proposed to be done; as a line to be drawn under some given conditions, some figure to be constructed, &c. The solution of the problem consists in showing how the thing required may be done by the aid of the rule and compass. The demonstration consists in proving that the process indicated in the solution really attains the required end. A theorem is a proposition in which the truth of some principle is asserted. The object of the demonstration is to show how the truth of the proposed principle may be deduced from the axioms and definitions or other truths previously and independently established. A problem is analogous to a postulate, and a theorem to an axiom. A postulate is a problem, the solution of which is assumed. An axiom is a theorem, the truth of which is granted without demonstration. In order to effect the demonstration of a proposition, it frequently happens that other lines must be drawn besides those which are actually engaged in the enunciation of the proposition itself. The drawing of such lines is generally called the construction. A corollary is an inference deduced immediately from a proposition. A scholium is a note or observation on a proposition not containing any inference, or, at least, none of sufficient importance to entitle it to the name of a corollary. A lemma is a proposition merely introduced for the purpose of establishing some more important proposition. Book I: Euclid, Book I (ed. Dionysius Lardner, 11th Edition, 1855) Next: Euclid, Book I, Proposition 1 (ed. Dionysius Lardner, 11th Edition, 1855) Axioms or Common Notions in other editions:

You need to find the values of the stars before you can apply normal Sudoku rules. This is a variation of sudoku which contains a set of special clue-numbers. Each set of 4 small digits stands for the numbers in the four cells of the grid adjacent to this set. Solve the equations to identify the clue numbers in this Sudoku problem. The challenge is to find the values of the variables if you are to solve this Sudoku. Advent Calendar 2011 - a mathematical activity for each day during the run-up to Christmas. A Sudoku with a twist. An extra constraint means this Sudoku requires you to think in diagonals as well as horizontal and vertical lines and boxes of You are given the Lowest Common Multiples of sets of digits. Find the digits and then solve the Sudoku. Label this plum tree graph to make it totally magic! You have twelve weights, one of which is different from the rest. Using just 3 weighings, can you identify which weight is the odd one out, and whether it is heavier or lighter than the rest? This article for teachers describes several games, found on the site, all of which have a related structure that can be used to develop the skills of strategic planning. Use the differences to find the solution to this Sudoku. Four numbers on an intersection that need to be placed in the surrounding cells. That is all you need to know to solve this The puzzle can be solved by finding the values of the unknown digits (all indicated by asterisks) in the squares of the $9\times9$ grid. Pentagram Pylons - can you elegantly recreate them? Or, the European flag in LOGO - what poses the greater problem? Bellringers have a special way to write down the patterns they ring. Learn about these patterns and draw some of your own. Problem solving is at the heart of the NRICH site. All the problems give learners opportunities to learn, develop or use mathematical concepts and skills. Read here for more information. Each clue number in this sudoku is the product of the two numbers in adjacent cells. In this Sudoku, there are three coloured "islands" in the 9x9 grid. Within each "island" EVERY group of nine cells that form a 3x3 square must contain the numbers 1 through 9. Special clue numbers related to the difference between numbers in two adjacent cells and values of the stars in the "constellation" make this a doubly interesting problem. This Sudoku combines all four arithmetic operations. This second Sudoku article discusses "Corresponding Sudokus" which are pairs of Sudokus with terms that can be matched using a substitution rule. A Sudoku with clues as ratios. Find out about Magic Squares in this article written for students. Why are they magic?! The puzzle can be solved with the help of small clue-numbers which are either placed on the border lines between selected pairs of neighbouring squares of the grid or placed after slash marks on. . . . Four small numbers give the clue to the contents of the four 60 pieces and a challenge. What can you make and how many of the pieces can you use creating skeleton polyhedra? A pair of Sudoku puzzles that together lead to a complete solution. Label the joints and legs of these graph theory caterpillars so that the vertex sums are all equal. This Sudoku puzzle can be solved with the help of small clue-numbers on the border lines between pairs of neighbouring squares of the grid. This Sudoku, based on differences. Using the one clue number can you find the solution? A Sudoku with clues as ratios or fractions. This pair of linked Sudokus matches letters with numbers and hides a seasonal greeting. Can you find it? A Sudoku that uses transformations as supporting clues. Two sudokus in one. Challenge yourself to make the necessary Remember that you want someone following behind you to see where you went. Can yo work out how these patterns were created and Try to solve this very difficult problem and then study our two suggested solutions. How would you use your knowledge to try to solve variants on the original problem? Countries from across the world competed in a sports tournament. Can you devise an efficient strategy to work out the order in which they finished? A particular technique for solving Sudoku puzzles, known as "naked pair", is explained in this easy-to-read article. Take three whole numbers. The differences between them give you three new numbers. Find the differences between the new numbers and keep repeating this. What happens? You have been given nine weights, one of which is slightly heavier than the rest. Can you work out which weight is heavier in just two weighings of the balance? Each of the main diagonals of this sudoku must contain the numbers 1 to 9 and each rectangle width the numbers 1 to 4. A Sudoku with clues given as sums of entries. We're excited about this new program for drawing beautiful mathematical designs. Can you work out how we made our first few pictures and, even better, share your most elegant solutions with us? This Sudoku requires you to do some working backwards before working forwards. Arrange the digits 1, 1, 2, 2, 3 and 3 so that between the two 1's there is one digit, between the two 2's there are two digits, and between the two 3's there are three digits. This sudoku requires you to have "double vision" - two Sudoku's for the price of one

Skin effect – a phenomenon occurring in AC circuits that causes the current density near the surface of a conductor to be greater than inside it. Skin effect increases the effective AC resistance of the conductor, causing an increase in power losses in the conductor. The value characterizing the skin effect is the penetration depth (of magnetic field or current into the conductor). This quantity depends on resistivity, magnetic permeability of the medium and frequency of the current. Skin effect results from the interaction of a magnetic field and an electric field. Alternating current flowing in a conductor creates an alternating magnetic field. This field induces an alternating, oppositely directed electric field. This in turn causes secondary currents, flowing in the conductor in the opposite direction of the primary current. These currents reduce both the magnetic field strength and the local current density. Since the strongest electric field is induced inside the wire, this is also where the current disappears most strongly. Skin effect can also be explained by the principle of so-called magnetic diffusion defined by Maxwell’s equations. A current in a wire creates a magnetic field around itself. Since this field is time-varying, it enters the conductor as an electromagnetic wave and is attenuated due to the lossiness of the medium. At the surface of the conductor, currents are induced that add to the source current, while inside the conductor, the source current is reduced by them. If the thickness of the conductor is large and the frequency is high (the electromagnetic wave then has a small length), a situation may occur when the currents coming from the magnetic field are summed larger than the source current and then inside the conductor a current starts to flow in the opposite direction, which is a very unfavorable phenomenon. Consider a cylindrical wire of radius a and infinite length. A current I = I0 cos ωt flows through the wire. For not very large frequencies ω we can neglect the displacement current, which is equivalent to neglecting the effect of radiating energy through the wire in the form of an electromagnetic wave. The current density and the electric and magnetic field strengths are written in complex form, with harmonic dependence on time Faraday’s and Ampere’s laws for the fields inside the wire are: where σ – specific conductivity, μ – relative magnetic permeability of the wire material. For rotation in cylindrical coordinates the following formulas are valid: Assuming that the wire is directed along the z-axis, equations (1) in the cylindrical system can be written as: where H = Hφ, E = Ez. By differentiating the second equation (3) after r, we can write the differential equation for the electric field strength Introducing the variable x = qr we obtain the equation This is a Bessel equation of zero order. Its general solution is the linear combination of Bessel functions of the first and second kind A J0(x) + B Y0(x). We must discard the Bessel function of the second kind Y0 because Y0(0) = ∞. The electric field on the axis of the wire, for r = 0, cannot be infinitely large. The electric field inside the wire is therefore where C = const. Using the second of equations (3) we can calculate the magnetic field strength inside the wire Using the formula for the derivative of the Bessel function From equations (3), one can verify that the differential equation for H is indeed a Bessel equation of order 1. According to the integral version of Ampere’s law, the value of the magnetic field strength at the surface of the wire should be Hence the integration constant C is Finally, the electric and magnetic field strengths inside the wire are According to Ohm’s differential law, the current density inside the wire is The ratio of the current density inside the conductor to the current density on its surface, for r = a where ξ = r/a < 1. We are dealing with Bessel functions from a composite argument, which physically means that at different distances from the wire axis the time dependence of the current density is phase shifted. In the argument of the Bessel function there is a dimensionless constant dependent on the conductor radius a, the current frequency ω, and the material constants σ and μ of the metal from which the conductor is made. Figure 1 shows the absolute value of the ratio of the current density inside the wire to its value on the surface, for several selected values of the parameter ka. It can be seen that for sufficiently high values of ka, the current flows practically entirely on the surface of the conductor. This effect is called the skin effect. The table below shows values of the ka parameter for a conductor with a diameter of 1 mm, for selected current frequencies. For copper σ = 5.8 – 10^7 S/m, for steel σ = 1.0 – 10^7 S/m, μ = 1000. |f||50 Hz||10 kHz||100 MHz| The skin effect is of great practical importance in electrical engineering. For high-frequency currents, the resistance of a conductor comes entirely from the thin layer of material on the surface. Since copper and aluminum, from which electrical conductors are made, tend to oxidize, this resistance can be much greater than the specific conductivity value of the pure metal would suggest. It may also be noted that most of the metal inside the wire does not participate at all in the transmission of electric current, and that increasing the diameter of the wire does not lead to a decrease in resistance. For high-frequency currents, therefore, a braided wire consisting of many thin conductors is often used. Skin effect in a finite thickness conductor Consider an infinite layer of conductor with thickness d = 2a. Let us assume that on both surfaces of the layer a homogeneous harmonic magnetic field of strength H0 is given, parallel to the surface of the layer. Such a field configuration exists in the core of a transformer consisting of many layers of insulated thin steel sheet. If the transformer core were made from a single thick piece of conductor, large eddy currents induced by a time-varying magnetic field would flow in the cross-section of the core, and thus large Joule Lenz heat would be emitted in the core. Inside the conductor, Faraday’s law applies: and Amper’s law: The density of current induced in a conductor is related to the electric field strength: Where σ – specific conductivity. Material equations: Let’s write Faraday’s and Ampere’s law in Cartesian terms. If the conductor is infinite in the x and y directions, then the fields can only depend on the variablez. For large intrinsic conductivity σ and small field frequency ω, we can neglect the shift current in Amper’s law. For harmonic fields, we replace the derivative after time by the factor iω. The electric field and the current density inside the conductor are directed along the x-axis. Thus the magnetic field inside the conductor satisfies the following differential equation where the complex constant The root of the imaginary number is: The general solution of the differential equation (5) is a linear combination of exponential functions: The magnetic field at the layer surface for z = -a and z = a must satisfy the following boundary condition: Hence the integration constants are: The solution of equation (5) satisfying the boundary conditions (7) is of the form: Based on equation (4), we can also give an expression for the induced current density in the layer is: The current density induced at the layer surface for z = a is It is easy to see that jx(-z) = – jx(z), so at the center of the layer the current does not flow, and at the surface z = -a it has the opposite direction than at the surface z = a. By introducing the following dimensionless notations: ξ = z/a, β = ba we can write an expression for the modulus of the current density related to the value at the surface of the layer: As can be seen from figure (1), for the value of dimensionless parameter β > 3 there appears a skin effect – the current starts to flow mainly near the surface of the conductor. To calculate equation (12) we need expressions for hyperbolic functions from the complex argument: and their modules In a similar way, we can write the expression for the modulus of magnetic induction related to its value at the surface of the layer: Let the frequency of the external magnetic field be f = 50 Hz. The magnetic steel has a specific conductivity equal to σ = 10^7 S/m, and its relative magnetic permeability is μ = 1000. The parameter b is then equal to 1.4 mm-1. The depth of field penetration into the conductor λ = 2π/b is about 5 mm. The following table shows the dependence of the dimensionless parameter β appearing in equations (12) and (15) as a function of film thickness: One of the quantities important in eddy current testing is the equivalent penetration depth of the electromagnetic field δ (also called standard penetration depth or simply penetration depth). This quantity conventionally defines the thickness of the material layer under test. The current density induced in a semi-infinite conducting plate by a sinusoidal time-varying plane electromagnetic wave penetrating it varies exponentially with distance from the surface. The value of this density for a particular depth z depends on the frequency f and the material parameters of the plate. Assuming that in a Cartesian coordinate system whose z-axis is perpendicular to the plate surface, the magnetic field strength vector has only a tangential component to the surface, the current density also has only a tangential component to the surface. The modulus of the current density varies according to the relation: Where J0 is the modulus of the current density at the surface of the object. Figure 3 shows graphs of eddy current density values induced in a copper plate by a three-winding flat coil energized at frequencies of 10, 20, 50, 100, 200, 500, and 1000 Hz. The graphs show the absolute values of the currents along the z-axis from the surface (0 mm) to the bottom of the plate (20 mm), for the coordinate r = 20 mm (position of the central coil). The highest values of the current density induced on the surface occur for the highest of the adopted frequencies (black line). At the same time, for this frequency, the values of current decrease most rapidly along the depth of the plate, taking values close to zero at its bottom surface. The expression for the current density can be represented as: It can be seen that δ is the depth at which the amplitude of the current density decreases e-fold. The penetration depth δ can also be thought of as the thickness of the layer in which the total current of constant density is induced, and beyond which no current is induced at all. The value of the penetration depth δ depends on the frequency of the field and on the magnetic permeability and electrical conductivity of the material under consideration. An effective penetration depth of 3δ is also determined, at which the induced currents decrease below 5% of the value of the current density at the surface. At the same time, as the amplitude of the current density decreases, the phase delay angle of the current with respect to the current at the surface increases. The value of this angle (in radians) is: Figure 4 shows the phase delay angle of the induced current (in radians).

Press Releases • Mar 13, 2019 08:58 UTC The planned construction projects within the Stockholm region show a total investment volume of €111 billion until the year 2040. According to Invest Stockholm’s report released today, the largest investment share is in housing, with an investment volume of €60,7 billion, followed by infrastructure investments in railway, subway and lightrail, with an investment volume of 9,6 billion €. Press Releases • Jan 31, 2019 06:45 UTC The health tech sector in Stockholm is booming. Six new health tech companies are entering the Life Science Hotlist, a list of carefully curated investment opportunities in the Stockholm-Uppsala region. Invest Stockholm has screened 95 health tech companies to identify the “hot” investment cases. Today, Brighter, Camino Care, CareLigo, Encare, Hnry and Minnity are added to the list. Press Releases • Dec 11, 2018 13:27 UTC The two municipalities Gävle and Sandviken in the Stockholm region have decided to sell a total of 130 hectares of land to Microsoft for SEK 269 million, thereby creating conditions for a major international investment. Press Releases • Oct 18, 2018 08:21 UTC Despite a substantial increase in the number of hotel rooms in Stockholm last year, the demand for more hotels in the Swedish capital continues to develop. Changed travel habits, a growing middle class in Asia and cheaper flights have contributed to the increase in the proportion of private hotel guests, which is now almost as big as the business segment. Press Releases • Sep 03, 2018 06:15 UTC During the first two weeks of September, Stockholm welcomes thousands of investors, entrepreneurs, talents and journalists, to discover one of the best tech ecosystems in the world. With 14 conferences and 50+ events, Stockholm Tech Week offers visitors to experience what's going to happen next in the world. Welcome to connect with the early starters of the Stockholm tech ecosystem! Press Releases • Jul 09, 2018 07:54 UTC The American community platform WeWork today announced its expansion into Stockholm. During the spring of 2019, the first location of the leading provider of shared office space around the globe will open its doors at the newly renovated property Urban Escape, offering office and community space for 1000 future members. Press Releases • Jun 11, 2018 12:47 UTC When sustainable eyewear brand Karün from Patagonia were looking for the best market in Europe to launch their new product; sunglasses made out of recycled fishing nets, they chose Stockholm. Why? You may ask. Stockholm has far from the amount of sunshine that many other European cities can boast with. The answer, however, is not to be found in the weather, but rather in the Swedish values. Press Releases • Mar 26, 2018 07:15 UTC The Vilsta ski slope in Eskilstuna in the Stockholm region, will become 10 meters higher just in time for the winter season next year. All thanks to shaft masses from the construction of Amazon Web Services (AWS)’s data center in Eskilstuna. Press Releases • Mar 13, 2018 11:29 UTC For the first time in many years, the City of Stockholm has land to sell in the city center. The gross floor area is about 20 000 square meters. The property has the Central station, the City Hall and the open waters of Riddarfjärden as neighbours. Press Releases • Nov 16, 2017 10:26 UTC Packed with statistics, key facts and detailed descriptions of shopping districts in Stockholm, the Swedish retail guide has landed. For everyone considering entering the Stockholm retail market, this guide saves weeks of market research. Press Releases • Oct 19, 2017 14:30 UTC The City of Västerås, west of Stockholm, has been chosen as the location where Northvolt will place their R&D facility for the next generation lithium-ion batteries. This is the fourth major investment in the Stockholm Region in the last six months, creating another 300 to 400 jobs in Västerås. Press Releases • Oct 06, 2017 05:50 UTC Through stable growth, Stockholm has reached record levels of RevPAR (revenue per available room), and surging occupancy has created a great need for additional guest room capacity. While demand has become increasingly diverse, supply has remained homogeneous and regional operators dominate the market. The time is ripe for international investors to conquer a share of emerging market segments. The ROCKWOOL Group, the leading provider of stone wool solutions, invests in land in the Stockholm region Press Releases • Oct 05, 2017 08:17 UTC The Danish industrial Group, ROCKWOOL, has today confirmed the purchase of 180,000 square meters of land in the Eskilstuna Logistics Park (Eskilstuna Logistikpark), which will lay the foundation for future expansion to serve the Swedish construction market. The future facility has the potential to create about 150 new jobs in Eskilstuna. Press Releases • Oct 03, 2017 08:38 UTC Need for labor to carry out planned construction investments in the Stockholm region is estimated at over 100,000 job opportunities in 2017 and 2018. The demand is greatest for building and construction workers, construction craftsmen, painters and finishers. Press Releases • Jul 26, 2017 07:30 UTC The strong construction market in Stockholm has created lucrative opportunities for international companies. A requirement to be able to seize these opportunities is to understand the regulations of the industry. To help foreign companies enter the market, Invest Stockholm, Tyréns and SABO today launched a unique intensive course covering the fundamentals of the Swedish construction industry. Press Releases • Apr 04, 2017 06:16 UTC Today, Amazon Web Services (AWS) announced the establishment of a new datacenter region in the Stockholm area, positioning Sweden as the location of choice for delivery of advanced cloud services in the Nordics. The new AWS EU (Stockholm) Region will be operational in 2018. Press Releases • Mar 13, 2017 12:42 UTC The Stockholm region is growing faster than the rest of Europe and the investment potential is estimated to be EUR 95 billion through 2030 in the areas of housing, infrastructure and innovation. A sweeping 340,000 new dwellings will be built, which alone will require investments at a record level of EUR 53 billion. Press Releases • Mar 22, 2016 08:36 UTC The Stockholm-Uppsala Life Science Investment Hotlist is growing. Now being welcomed are 5 new companies, all in digital health, in the search for foreign investments and partnerships. Including the new companies, the list now includes 41 companies, almost half of which are in the field of digital health. Press Releases • Oct 07, 2015 08:23 UTC The next five years will see a major investment to increase the life science companies in the Stockholm region. The aim is to double the number of identified companies which are ripe for investment. Press Releases • Jun 23, 2015 06:00 UTC In the past five years, Stockholm-based companies received 18 percent of all private placements in FinTech companies across Europe. The numbers show that Stockholm continuously, year after year, attracts almost one fifth of the overall FinTech investments in Europe.

Notwithstanding that, Robert Northcott’s paper A Dilemma for the Doomsday Argument is (and by a comfortable margin) the worst paper I have reviewed (i.e. trashed) on this blog. In the paper, the following Doom scenario is presented: “Imagine that a large asteroid…is…heading…towards Earth…astronomers calculate that it has a 0.5 probability of colliding with us, the uncertainty being due to measurement imprecision regarding its exact path…any collision would be certain to destroy all human life..what is now the rational expectation of humanity’s future duration…a 0.5 probability that humanity will last just the few days…(‘Doom-Now’), and a 0.5 probability that it will last however long it would have otherwise. What does DA (Doomsday Argument) say? Either it is deemed to modify these empirical probabilities; or it is not.” Doomsday practitioners of course would revise the probability of Doom-Now upward significantly. For convenience, let’s say that humanity would last another million years otherwise, with a total number of people numbering 1 quadrillion. if we take ourselves to have been sampled uniformly at random from the total population past present and future, our birth rank of 60 billion or so looks very, very unlikely conditioned on Doom Later. It looks more plausible conditioned on Doom Now. So DA would revise P(DoomNow) upwards, close to 1. That’s how the Doomsday argument works. But Northcott writes: “…according to DA, a priori considerations show that the expected duration for humanity is much greater than just a few days. The probability of Doom-Now should accordingly be modified downwards.” First of all, the expected duration just is half a million years, which is already much greater than a few days. So the first sentence makes no sense. Second, the DA advocates in favor of Doom-Now…not against it. (That’s why it’s called the Doomsday argument.) A footnote here sheds no light whatsoever: “True, DA reasoning implies that the single most likely total number of humans is the current number, i.e. 60 billion. But although the mode of the distribution is thus 60 billion, the mean is much greater. Thus, Doom-Now is not favoured.” Obviously Doom-Now is favored by DA. I mean, of course the expected number of humans is around half of a quadrillion. That’s not DA reasoning, that’s just what it is. Not relevant though because DA assumes that the world is sampled by objective chance and then you are sampled uniformly from the observers in that world. The fact that some unlikely worlds have enormous population doesn’t dilute the worlds that have smallish population. That’s the whole point of DA. Nor is the gaffe a typo…Northcott goes on thinking that DA revises the probability of Doom-Now downward for the rest of the paper. Part of the paper involves miracles. I won’t go into that but the publication of this paper in Ratio is proof that they do happen. The next passage is not to be believed: “…an unbiased combined estimate (of the mean) can be achieved via inverse-variance weighting. Roughly, the higher an estimate’s variance, the more uncertain that estimate is, and so the less weight we should put on it. In the DA case, how we balance competing DA and empirical estimates of a probability turns – and must turn – on exactly this issue….Some toy numbers will illustrate. By assumption, the empirical estimate of the asteroid collision’s probability, and thus of Doom-Now’s, is very certain. Suppose that the density function of that estimate is a normal distribution with a mean of 0.5 and, representing the scientists’ high degree of certainty, a small standard deviation of 0.001. Next, suppose initially that for DA the equivalent figures are a mean of 0.001 and the same small standard deviation of 0.001. In this case, because the two variances are the same, so an unbiased estimate of the mean would be midway between the two component estimates of it, i.e. midway between 0.5 and 0.001, i.e. approximately 0.25.” This is so wrongheaded in so many different ways I don’t really know where to start. So I will start with what is worst. Yes, there is a method in statistical meta-analysis of taking a weighted average of two estimators to get an unbiased third estimator that is of minimum variance, among all weighted averages, and yes it goes by taking the inverse variances as weights. But, first, the estimators you start with have to themselves be unbiased. The two “estimators” considered here can’t both be unbiased estimators of the same parameter, because they have different means, and what it means for an estimator to be unbiased is for it to have mean equal to the true value of the parameter being estimated. Perhaps more troubling, however, is that it’s not at all clear what they could be estimating…the only thing around to estimate is the objective chance that the asteroid hits the earth, which is either zero or one–unless it’s something like “the credence an ideal rational agent would have in this epistemic position”. Surely though that would have to be at least mentioned. The next thing that is wrongheaded is just what was mentioned before—the latter mean shouldn’t be .001 but rather something like .999…DA wants to raise the probability of Doom-Now. Finally…assuming that p is in fact the credence of an ideal rational agent in the current epistemic situation and assuming that both the scientists and DA are trying to estimate p, does it not strike the author as odd that the scientists are so damn certain that p is very close to 1/2 and DA is so damn certain that p is very close to .001? The compromise suggested is, perhaps it bears mentioning, 250 standard deviations away from the mean for both parties. Normally this would be a moment where the meta-statistician might say “hmm…maybe these estimators aren’t estimating the same thing….”. Not that it matters much at this point, but the most amusing passage is this: “we can calculate a second scenario, with new toy numbers…this time, suppose that for DA the equivalent figures are still a mean of 0.001 but now, say, a standard deviation of 0.1….” Sorry, but it’s not possible for a credence estimator (which must take on values in [0,1]) to have a mean of .001 and a standard deviation of .1. Don’t get me wrong. My beef is not with the author, who I assume is not perpetrating a hoax but just sincerely trying to say something important. The question, for me, is not how did this mess come to be written, but how did this mess come to be published in a respectable journal? Ratio isn’t some obscure, fly-by-night outfit. Ernest Sosa, for example, is on the editorial board. In fact, it seems there is a list of 49 “Most popular journals” used by PhilPapers to identify when someone is a “professional author” in philosophy, and Ratio is on it! So perhaps congratulations are in order…by slipping this awful manuscript past the editors of this journal, this author is (if he wasn’t before) now and forevermore a “professional author” of philosophy, meaning that PhilPaper editors shall be obliged to archive every stupid thing he ever writes, even if no one on earth or in heaven will touch it. Okay…but back to my question. How did this ridiculous manuscript come to be published? The question is intended for the editors of Ratio…I’m inviting reply. What was the process? Of course any public reply would obviously be just “we sent it off for double blind refereeing, got a positive report and it was voted in by the editors…blah, blah blah”. So, well…never mind, I guess. But come on guys….my job isn’t supposed to be this easy.

What is mathematical logic? What do you get when you combine math, computer science, philosophy, and linguistics? Mathematical logic! The big goal of mathematical logic is to link human language and thinking with math. In short, mathematical logic tries to understand math concepts through patterns that feel natural to your brain. Mathematicians who study mathematical logic also try to uncover and figure out contradictions and unsolved problems in the field of mathematics. What do mathematical logicians study? Most of the time, we learn math as a science. We learn formulas, and we test answers to get results. But mathematical logicians study math as a language. They study mathematical content, not just problems and solutions. Mathematical logic explores the language of math in two ways: syntactic and semantic. The syntactic dimension focuses on understanding the correct structure of a mathematical “sentence.” A mathematician studying syntax might be working on translating a “sentence” from the language of math into a set of instructions that a computer can understand. The semantic dimension has to do with interpretation or use of a mathematical “sentence.” A mathematician studying semantics would focus more on making mathematical “sentences” or concepts more easily understandable by humans. Where and when did mathematical logic originate? Although the study of logic in the language of math is relatively new, the study of logic in human languages is very old. One of the first thinkers in the field of logic was Aristotle. Aristotle’s logic was more like the sort of logic you might use to write an essay– he dealt with arguments, fallacies, language, and the other kinds of things a debate coach would teach you. For centuries, logic was only expressed through speech and writing. However, a few hundred years ago, the field of mathematical logic began to develop. What is the difference between classical logic and mathematical logic? In the beginning, mathematical logic looked pretty similar to Aristotle’s logic– it just used symbols instead of words. Although mathematicians mostly agreed that this new idea was powerful and influential, they had a harder time agreeing on the new language of symbols. It wasn’t until George Boole invented Boolean logic in the 1800s that mathematicians could agree on a universal logic language. Why is mathematical logic important? The connection between mathematical logic and language. The languages that we speak in our everyday lives– English, Chinese, Spanish, etc. –are known as “natural languages.” This means that they evolved naturally through thousands of years of humans trying to get better at communicating with each other. Mathematical logic, on the other hand, is a “formal language.” That means that its grammar and vocabulary were constructed intentionally by mathematicians and logicians to serve a specific purpose. While formal and natural languages are different in many ways, they also have a lot in common! They both have specific rule sets, known as grammars. In English, grammar might tell you to change a verb based on when its action is happening: he runs vs. he ran. In mathematical logic, grammar might tell you that you can’t use an “or” operator to replace an “xor” operator (pronounced “Ex-Or”; we’ll learn what this means in a few paragraphs). Both grammars have the same effect on the language: they help writers generate expressions with consistent structures, and they help readers decide if a “sentence” is correct or not. Mathematical logic and the human mind. Have you ever wondered what makes humans so different from other animals? Scientists have wondered this too! About twenty years ago, three linguistic scientists named Marc Hauser, Noam Chomsky, and W. Tecumseh Fitch set out to argue that what makes humans so unique is how we understand language and communication. Humans have a particular area in our brain called Broca’s area. This part of our brain allows us to be much better at processing and understanding language than other animals. Our brains have allowed our languages to have something very unique: grammar! While other animals have methods of communication, Hauser, Chomsky, and Fitch claimed that human language has a far more complicated structure than other languages. Human grammar allows us to talk about things that happened in the past, things that will happen in the future, and even things that might happen, but we’re not sure about yet. Those are some pretty complicated ideas! Mathematical logic is based on how good we are as humans at understanding structures and patterns. Mathematical logic evolved from Aristotelian logic, which evolved from human speech and language. The biggest thing that mathematical logic has in common with the language you speak every day is its strict structure with clear rules. So, while those rules may look very different from the grammar rules you learn in school, your ability to understand them comes from the same special part of your brain! Mathematical logic and computer science. Mathematical logic and computer science are very closely related. Many of the most famous computer scientists of the past were also mathematicians who studied logic. Alan Turing, who some people credit with building the first computer, based much of his work on mathematical logic concepts. Logic and computer science have the most overlap in a field of study called computability theory. This field works on improving our ability to find efficient ways to solve problems. In both logic and computer science, researchers work on techniques to automatically check and find mathematical proofs and theorems. However, even within this field, computer scientists and logicians focus on slightly different things. Computer scientists study how computability relates to real computers, problems, and programming languages: in essence, how to make computers faster and better. On the other hand, mathematical logic researchers focus on the theoretical concept of computability and are less concerned with how computability affects the day-to-day improvement of computers. Sometimes mathematicians or computer scientists working on mathematical logic will also research modal logic (or modal theory), propositional logic, recursion theory, and set theory. Would you be good at mathematical logic? The theory of multiple intelligences. In 1983, Harvard psychologist Howard Gardner proposed his theory of Multiple Intelligences (http://www.pz.harvard.edu/projects/multiple-intelligences). Gardner believed that it was wrong to think of intelligence as a singular thing that each person possessed in a certain amount. Instead, Gardner believed that there are eight types of intelligence and that any person could possess intelligence in any number of categories. Howard Gardner’s eight types of intelligence. - Visual-spatial intelligence has to do with understanding images, directions, maps, and other visual elements. - Linguistic-verbal intelligence has to do with reading, writing, and communication. - Bodily-kinesthetic intelligence has to do with athletics, coordination, touch, and understanding the space around oneself. - Musical intelligence has to do with rhythm, tone, and auditory elements. - Interpersonal intelligence is the ability to understand and be understood by others. - On the other hand, intrapersonal intelligence is the ability to examine and understand oneself. - Naturalistic intelligence has to do with categorization– understanding the natural world and having profound empathy for all life. - Finally, logical mathematical intelligence has to do with reasoning, recognizing patterns, and thinking conceptually about numbers and hierarchies. Gardner also suggested the possibility of a ninth category: existential intelligence. Those with strong existential abilities would have an aptitude for considering deep and profound questions and deep empathy and understanding for humans on a broad societal level, not just an individual level. Many other psychologists and researchers disagree with Gardner’s theory. They point out that he does not have much scientific evidence to back his ideas up. They argue that his intelligences are closer to a personality quiz than an empirical scientific theory. And while this may be true, you might still find Gardner’s theory useful to gain a better understanding of your own interests and the way you prefer to learn. More about Howard Gardner’s idea of logical-mathematical intelligence. According to Gardner’s theory, logical-mathematical learners are usually very organized and systematic. They think in clear and linear terms, making them very good at mental math and memorization. They also use patterns and sequences to absorb information– for example, logical-mathematical learners might have a precise planner or to-do list system. Simultaneously and paradoxically, they are comfortable with abstractions because of their ability to extract patterns and find consistency. Logical-mathematical learners are also heavily reliant on rules and structure. They will seek them out or make them if they are in a situation with few dictates, and they might feel anxious or less confident without regulations and routines. They also will probably feel frustrated when those around them ignore procedures. How can you get better at mathematical logic? Can you develop logical mathematical intelligence? Many people seem to believe that being “good” at math is something you are born with. They see math skill as a talent that can only be strengthened if you have a “natural gift.” This is false. Your math ability is like a muscle: getting better at math is a lot like getting better at running or weightlifting! The way to improve your logical mathematical intelligence is to exercise that muscle. Like any exercise routine, it is essential to choose challenging activities and dedicate yourself to doing them regularly. What jobs are well suited for people with logical-mathematical intelligence? While there should be no pressure to decide on a career without exploring plenty of subjects and paths, you will likely enjoy a career in STEM fields if you enjoy building your logical-mathematical intelligence skills. Whether you want to develop next-generation machine learning at Apple or AI language models at OpenAI, the most cutting-edge tech research requires powerful analytic skills. Being a researcher, professor, mathematician, data scientist, computer programmer, or even a doctor are interesting jobs that reward well-developed mathematical intelligence. Even if you are interested in other types of work, growing your logical-mathematical intelligence will help you in everything from the debate team to robotics, from planning a complex landscaping project to making a roster and tournament schedule for a sports team. How do you learn and practice your mathematical logic skills? Mathematical logic symbols and operators. One of the biggest and most well-known branches of mathematical logic is Boolean logic. And while Boolean logic can be incredibly challenging and difficult, its building blocks are so simple that anyone can understand them! Boolean logic only deals with two numbers: one and zero. In the “language” of Boolean logic, one means “true,” and zero means “false.” Let’s talk about what this means in plain English by expressing a real-life situation in Boolean terms. Let’s say we are trying to test whether Annie and Sam went to the store. We would ask two questions. One: did Annie go to the store? Two: did Sam go to the store? If the answer to question one is yes, and the answer to question two is yes (and only if the answer to both is yes), then the answer to our test is yes, Annie and Sam went to the store! If the answer to either question one or question two is no, then the answer to our test is no! In Boolean logic, each individual query we ask (“Annie goes to the store” and “Sam goes to the store”) is called an input, and the answer to our larger question is called an output. The list of all possible outputs for all of the different combinations of inputs is called a “truth table.” Let’s look at a truth table for “Annie and Sam went to the store.” |Annie||Sam||Did Annie Annie and Sam go to the store?| |Annie did not go.||Sam did not go.||It is FALSE to say, “Annie and Sam went to the store.”| |Annie did not go.||Yes! Sam went!||It is FALSE to say, “Annie and Sam went to the store.”| |Yes! Annie went!||Sam did not go.||It is FALSE to say, “Annie and Sam went to the store.”| |Yes! Annie went!||Yes! Sam went!||It is TRUE to say, “Annie and Sam went to the store.”| What are the basic Boolean logical operators? Instead of using the operators of the math that you are used to– plus, minus, times, divide –Boolean logic uses the operators “and,” “or,” “xor,” and “not.” There are other operators, too, but these are the most common. For most operators, just like in our example with Annie and Sam, there are two inputs (called “truth values”) and one output. Each operator has its own rule set that tells you how to respond to different combinations of inputs. The truth table shows the outputs corresponding to the combination of inputs you enter. Let’s look at the truth tables of the most common operators: How “AND” works in Boolean logic |Input 1||Input 2||Output| Compare this to our truth table about Annie and Sam going to the store above. Did you notice it is the same? Here, in this Boolean truth table, we have represented “No,” “Not,” and “False” with a “0” and “Yes” (“True”) with a “1”. As you can see, the AND operator outputs a “true” answer if (and only if) the first and second inputs are true. Thinking about Annie and Sam, can you guess what the other operators might do? Here is the truth table for OR. Imagine asking if Annie or Sam are going to the store. Do you see why our output is “1” (or “True”) in three of the four possible cases? |Input 1||Input 2||Output| The OR operator outputs a “true” answer if either the first, or the second, or both inputs are true. So, if Annie went to the store, OR if Sam went to the store, OR if they both went to the store, our output is true. “Or” is a word we use every day in English, but we don’t use the next one: “XOR” ever. Nevertheless, XOR is a conjunction of two words that you do know: “exclusively or.” Do you have a guess about what the truth table for XOR might be? Here is the truth table for XOR |Input 1||Input 2||Output| The XOR operator, which stands for “eXclusively OR,” outputs a “true” answer if either the first or the second inputs, but not both, are true. The XOR operator will only give a true output if EITHER Annie or Sam went to the store. It will show a false output (“0”) if they both went to the store. The truth table for NOT: You might notice that the NOT operator’s truth table looks a bit different. That is because the NOT operator only takes one input, not two! It simply reverses whatever input is entered. The NOT operator can only consider one person at a time. If Annie did go to the store, the NOT operator will give a false output, and if she didn’t, it will give a true output. Try these! Mathematical logic questions and answers. Now that we know the rules of the most common operators let’s use them to solve a few problems! In our problems, we will use switches and lightbulbs to represent problems and solutions. “On” means true, or one, and “off” means false, or zero. We will solve all of these problems on the logic.ly website. Before we start you should head over there and play around with the tool! We’ll start with a tutorial. Let’s try to turn a lightbulb on and off with one switch. First, we will drag a toggle switch from the menu into the workspace. Then, we will drag a lightbulb into the workspace. Finally, we will connect the node on the switch with the node on the lightbulb. This sounds tricky, but it is straightforward once you know how to do it. Grab the little circle on the switch, hold down your mouse and drag the line to the little circle at the base of the lightbulb! Now, by clicking the switch, we can turn the lightbulb on and off! Great! Now here are some fun problems to try. All of the following exercises should be completed here: https://logic.ly/demo/ - Turn a lightbulb on and off using two switches. But the lightbulb should be on if, and only if, both switches are on. (In other words, if the first switch AND the second switch are on, then the lightbulb will be on. If either switch is off, or both switches are off, the lightbulb will also be off.) - Turn a lightbulb on and off with two switches, but the lightbulb will only be on if any switch is on. (This means it will be on if either switch is on or if both switches are on.) - Turn a lightbulb on and off with two switches, but the lightbulb can be on if either switch is on, but not both. - Turn a light bulb on and off with one switch, but make it so that the lightbulb is on when the switch is off and off when the switch is on. If you want to check your answers, you’ll find the solutions at the bottom of this blog article. If you enjoyed these Boolean logic problems, IMACS classes would be an excellent option to further develop your mathematical and logical skills while having a lot of fun! What books and courses will strengthen your logical-mathematical intelligence? Logic games and puzzles can be great exercises for your mental mathematical muscles. These can be found in books you work through on your own or in courses you work through with a knowledgeable teacher and peers. Here are some great books and courses to improve your logical-mathematical intelligence: Some of our favorite mathematical logic books: One of the most famous books on mathematical logic is a book with the unsurprising title: Mathematical Logic, by Joseph R. Shoenfield. This book is among the pioneers in the field and is an excellent resource for learning advanced mathematics logic concepts. Some other books about mathematical logic and a recommended audiobook provide mathematical logic exercises and puzzles for those just getting started in the subject. Where to find mathematical logic courses? Most mathematical logic courses are not taught until university. The field is often considered too complex for younger students to learn. However, for younger students with a passion for math or computers, IMACS offers courses that teach mathematical reasoning. This is a perfect stepping stone for future logicians. IMACS makes mathematical logic fun and engaging and will improve the problem-solving skills of any eager student. For advanced secondary school students, IMACS offers a university-level logic course. Read more about what our alumni have to say about IMACS and try a free placement class to learn more. Answers to the fun Boolean logic problems! Turn a lightbulb on and off using two switches. But the lightbulb should only be on if, and only if, both switches are on. (In other words, if the first switch AND the second switch are on, then the lightbulb will be on. If either switch is off, or both switches are off, the lightbulb will also be off.) Turn a lightbulb on and off with two switches, but the lightbulb will only be on if any switch is on. If you solved this problem correctly, you can also turn the lightbulb by having both switches turned on. Turn a lightbulb on and off with two switches, but the lightbulb can be on if either switch is on, but not both. Turn a light bulb on and off with one switch, but make it so that the lightbulb is on when the switch is off, and off when the switch is on.

You are searching about Guess And Check Logic Math 2 Items One Has More, today we will share with you article about Guess And Check Logic Math 2 Items One Has More was compiled and edited by our team from many sources on the internet. Hope this article on the topic Guess And Check Logic Math 2 Items One Has More is useful to you. The Origin of the 11D World Membrane As a Pascal Conic Section of a 6D-String in 5D Projective Space David Hilbert’s biography discusses a debate over whether it was an intellectual mistake to advocate advancing the 2000-year-old Pappus theorem to axiom status. This issue is notable from a number of perspectives, one of which is his 1920 proposal that established the Hilbert Program for formulating mathematics and/or geometry in a more robust and complete manner. logical basis that conforms to inclusively larger “metamathematical principles”. However, at the time, he proposed that this could be done if: 1. all mathematics follows from a complete or correctly chosen finite system of axiomsand 2. this system of axioms is demonstrably consistent through some means such as its epsilon calculus. Although this formalism has had a successful influence on Hilbert’s work in algebra and functional analysis, it was not committed in the same way to his interests in logic as well as in physics, for not talk about his axiomatization of geometry, given the scheme. question of considering Pappus’ theorem as an axiom. Similarly, a similar problem arose when Bertrand Russell rejected Cantor’s proof that there was no “maximum” cardinal number and defended the “logicism” of his and AN Whitehead’s proposition in Mathematical principles that all mathematics is in some important sense reducible to logic. But both Hilbert’s and Russell’s support for an axiomatized mathematical system of definite principles that could banish theoretical uncertainties “forever” would end in failure in 1931. Because Kurt Gödel showed that any non-contradictory (self-consistent) formal system complete enough to include at least arithmetic, could not (both) demonstrate its completeness (and/or, conversely, its categorical consistency) by its own axioms . Which means that Hilbert’s program was impossible as stated, since there is no way that the second point can be rationally combined with assumption-1 as long as the system of axioms is actually finite; otherwise you will have to add an endless series of new axioms, starting, I. guess, with Pappus’s! Equally, Gödel’s incompleteness theorem reveals that neither Mathematical principles, nor any other consistent system of recursive arithmetic, could decide whether each proposition, and/or its negation, was provable within that system. However, beyond Hilbert’s misstep over Pappus, it should be noted that Gödel’s theorem itself, in a realist sense, supports Hilbert’s basic idea of a deeper, more inclusive environment, ‘metallurgical’ foundation like a ‘Gödelian cartographywhich “covers” all mathematics and geometry. In fact, it was Hilbert Gentzen’s student who used a Gödel map “orders” of “trans-infinite” number systems to demonstrate Gödel’s theorem: so really metalogically valid ordinary arithmetic. In any case, although this conclusion also fits loosely with Russell’s logistical ideas, it also demonstrates a great improvement over his criticism of Cantor’s proof for an infinite series of cardinal numbers, which, after all, is the point of Cantor’s arguments in the sense that some ‘continuity axiom’ like that of Archimedes is required to generate an infinite field of real numbers. Which makes Hilbert’s Pappian faux pas seem almost trivial by comparison, since I’d like to know how Russell expected to find some “larger cardinal number” as well as how he expected to axiomatically describe continuity for an infinite range of numbers or points. a line; that is, before, let alone after, any infinite axiomatic system became an additional problem! In any case, if Russell, or Hilbert, had actually taken up Occam’s razor, they probably would have cut their own throats with it before revealing their biased assumptions and inconsistencies for the world to see for eternity. Which simply means why elevate some provable theorem to the status of an axiomatic assumption, or introduce your own inconsistent system of assumptions, when it is clearly better to leave everything as it is. However, I am happy to use this razor properly in order to cut these icons a little posthumously, as if God had torn them from his suicidal hands, just to thank them in return for the indulgent opportunity to show everyone a again why fools seem routinely skewed as absurd fodder for us “lesser” fools or “commoners” in some exclusive or “formal” organizational hierarchy. That’s why the wisest people just say: the higher the monkey climbs the tree, the more it exposes itself to those below! (But also always be careful underneath, before something pops that hole in your face again!!) In any case, this brings us back to another older and equally pressing issue is directly connected with Pappus’ “hexagon theorem” as generalized by Blaise Pascal in a projective conical sectionor 6-point oval, in 1639, when he was only 16 years old. Naturally impressed by Desargues’s work on conics, he produced, as a means of proof, a short treatise on what he called the “Mystic Hexagram”, better known since then simply as Pascal’s theorem . It basically (as defined by Wikipedia) states that if an arbitrary hexagon is inscribed in any conic section, where opposite sides extend to meet, the three points of intersection will lie on a straight line, called “Pascal line” of this configuration. Although this simple description verbally suffices, it may fail to convey the fuller and truly “mystical” aspects that give Pascal’s theorem and configuration the distinction of being regarded as the most central fundamental construction of geometry projective And while diagrams would help clear things up, especially the descriptions below, it’s hard enough to reformat the content of these articles from the preferred notebook text to fit the different formats of different e-journals or distribution services. web articles In any case, it is no coincidence that I have not only made Pascal’s conic the cover figure of my text covering the projective and its subgeometries, but include a frontispiece of various 6-element conics relevant to all, including Brianchon’s projective dual to Pascal’s. So any interested reader can go to the resource box and pull out at least the Pascal cover figure, if not the frontispiece. Anyway, the figure on the cover of the text illustrates Pascal’s theorem represented in a simple hexagon formed by mutually inscribing a full line of 6 points (15 lines) and 6 full lines (15 points) representing the respective plane sections of a complete line of six dimensions. -into a point and a full 6-derivative three-dimensional plane recursively intersecting a full five-dimensional 6-dimensional point it is the simplest representation as a spatially extended maximal projective set of vertices for this object. This description thus underlines its profound importance in terms of the scope of the entire dimensional gamete of the “axioms of incidence” (previously the additional set necessary to establish the meaning and that of continuity is introduced), starting with the simplest dimensional axioms extension and 5D closure, along with its projective dual of six lines on a point, which is then sectioned down to the final incidence ratios corresponding to six full points on a line and six full lines through a point. Similarly, the space of 5 can be added with the dimensions of 6 lines to get an 11D manifold that serves as a coverage space to map what amounts to a finished projecting geometry that is at the same time is both complete i categorically consistent (since it does not refer to infinite ranges of points or numbers, it is not restricted by Gödel’s reasoning, but again it admits). So, for example, it is quite interesting that JW Hirschfeld points out in his text on finite projective groups that there are no six conics of dimension greater than eleven. Which brings us to the crux of this article as far as math is concerned physics of (super) stringand ‘M’ or membranetheory – which I have yet seen reduced by Occam’s blade to its essence in the basics of geometry, so the summary here is comprehensible to a wider segment of intelligent people. Superstring theory is based on a four-dimensional spacetime or physical metric, which together with a six-dimensional internal (Kauler) manifold (or compact Calabi-Yau space) for what can now be thought of as strings 1D of 6 -line-to-point; forming a total system of 10 dimensions. But by the 1980s it became clear that a promising unification of physics within a theory of quantum gravitation of superstrings was impossible, as they branched off into five 10 different mathematical groups (leading to the situation in which a number of mathematical eggheads, mostly with little interest in physics). per se, began to dominate theoretical physics departments). Which led to a second “superstring revolution” in the mid-1990s, when Ed Witten concluded that each of the 10D superstring theories is a different aspect of what was originally called a single “membrane theory’ (see http://en.wikipedia.org/wiki/Membrane_Theory), the entirety of which is naturally eleven dimensional and establishes interrelationships between the different superchain group theories described by various “dualities”. Because just as 1D strings are more manageable, finite extensions of singular points, groups of strings in a plane form “sheets of the world” as literal “2D membranes”, where these so-called “branes” can be defined of any dimension, starting with a 0-brane or point. Thus, while the total system may properly be called an “11D world membrane,” Witten prefers generically to call it “M-theory,” where M can stand for membrane, mother, math, matrix, master, mystery, magic, or after , as Pascal would forcefully add: Mystic! In any case, there is little doubt that one day a complete 6-dimensional representation of 1D strings packed into 5D spacetime will fulfill Einstein’s dream of a fully unified physical theory. But personally, I’m much less concerned with “theoretical” unifications than I am with a comprehensive interpretation of physics and cosmology, replete with a host of confirmable data. Because I have developed the first dimensionless or “pure” scale system (Planck) which I call “Mumbers” or “membrane numbers”. and that covers precisely the whole spectrum of particle physics and space-time. And while I don’t have the IQ, inclination, or patience to follow or pursue higher mathematics for theoretical purposes, on the other hand, I’ve tried many, but I’ve yet to find anyone, physicist or mathematician, who can successfully write a pure numerical equation even for a relationship between different physical states. Similarly, standard super-string or M-theory has yet to make any confirmed, or even confirmable, predictions, no more than I, at least, have seen anyone point to the geometric underpinnings of M-theory as a Pascal’s 11-dimensional section. a projective double space of 5 and a six-string 6D at one point. So regardless of my mathematical inadequacies, I can guarantee that no viable “unified M-theory of physics” can be developed until the intellectual community accepts the metalogical tautology of both Pascal’s mysterious 6-conic that underlies the foundations of geometry, as well as the accompanying unified “dimensionless” scale that already represents a proven system that encompasses all of physics. Video about Guess And Check Logic Math 2 Items One Has More You can see more content about Guess And Check Logic Math 2 Items One Has More on our youtube channel: Click Here Question about Guess And Check Logic Math 2 Items One Has More If you have any questions about Guess And Check Logic Math 2 Items One Has More, please let us know, all your questions or suggestions will help us improve in the following articles! The article Guess And Check Logic Math 2 Items One Has More was compiled by me and my team from many sources. If you find the article Guess And Check Logic Math 2 Items One Has More helpful to you, please support the team Like or Share! Rate Articles Guess And Check Logic Math 2 Items One Has More Rate: 4-5 stars Search keywords Guess And Check Logic Math 2 Items One Has More Guess And Check Logic Math 2 Items One Has More way Guess And Check Logic Math 2 Items One Has More tutorial Guess And Check Logic Math 2 Items One Has More Guess And Check Logic Math 2 Items One Has More free #Origin #11D #World #Membrane #Pascal #Conic #Section #6DString #Projective #Space

Presentation on theme: "D 2 Law For Liquid Droplet Vaporization References: Combustion and Mass Transfer, by D.B. Spalding (1979, Pergamon Press). “Recent advances in droplet."— Presentation transcript: D 2 Law For Liquid Droplet Vaporization References: Combustion and Mass Transfer, by D.B. Spalding (1979, Pergamon Press). “Recent advances in droplet vaporization and combustion”, C.K. Law, Progress in Energy and Combustion Science, Vol. 8, pp. 171-201, 1982. Fluid Dynamics of Droplets and Sprays, by W.A. Sirignano (1999, Cambridge University Press). Gas-Phase Streamlines Droplet With Internal Circulation Gas-Phase Streamlines Droplet With Internal Circulation Heat Fuel Vapour Buoyancy and droplet size destroy spherical symmetry evaporation Spherical Symmetry Model Mass Transfer I DEFINITIONS IN USE: density – mass of mixture per unit volume ρ [kg/m 3 ] species - chemically distinct substances, H 2 O, H 2, H, O 2, etc. partial density of A – mass of chemical compound (species) A per unit volume ρ A [kg/m 3 ] mass fraction of A – ρ A /ρ = m A note: ρ A + ρ B + ρ C + … = ρ m A + m B + m C + … = 1 DEFINITIONS IN USE: total mass velocity of mixture in the specified direction (mass flux density) – mass of mixture crossing unit area normal to this direction in unit time G TOT [kg/m 2 s], G TOT = u (density x velocity) total mass velocity of A in the specified direction = G TOT,A [kg/m 2 s] note: G TOT,A + G TOT,B + G TOT,C …= G TOT convective mass velocity of A in the specified direction m A G TOT = ( ρ A /ρ) G TOT = G CONV,A note: G CONV,A + G CONV,B + G CONV,C …= G TOT but generally, G CONV,A ≠ G TOT,A diffusive mass velocity of A in the specified direction G TOT,A – G CONV,A = G DIFF,A note: G DIFF,A + G DIFF,B + G DIFF,C + … = 0 DEFINITIONS IN USE: velocity of mixture in the specified direction = G TOT / [m/s] concentration – a word used for partial density or for mass fraction (or for mole fraction, partial pressure, etc.) composition of mixture – set of mass fractions The d 2 Law - assumptions (i)Spherical symmetry: forced and natural convection are neglected. This reduces the analysis to one-dimension. (ii) No spray effect: the droplet is an isolated one immersed in an infinite environment. (iii) Diffusion being rate controlling. The liquid does not move relative to the droplet center. Rather, the surface regresses into the liquid as vaporization occurs. Therefore heat and mass transfer in the liquid occur only because of diffusion with a moving boundary (droplet surface) but without convection. (iv) Isobaric processes – constant pressure. (v) Constant gas-phase transport properties. This causes the major uncertainty in estimation the evaporation rate (can vary by a factor of two to three by using different, but reasonable, averaged property value – specific heats, thermal conductivity, diffusion coefficient, vapour density, etc). (vi) Gas-phase quasi-steadiness. Because of the significant density disparity between liquid and gas. Liquid properties at the droplet surface (regression rate, temperature, species concentration) changes at rates much slower than those of gas phase transport processes. This assumption breaks down far away from the droplet surface where the characteristic diffusion time is of the same order as the surface regression time. Gas-phase QUASI-steadiness – characteristic times analysis. In standard environment the gas-phase heat and mass diffusivities, g and g are of the same order of 10 0 cm 2 s -1, whereas the droplet surface regression rate, K = -d(D 0 2 )/dt is of the order of 10 -3 cm 2 s -1 for conventional hydrocarbon droplet vaporizing in standard atmosphere. Thus, there ratio is of the same order as the ratio of the liquid-to-gas densities,. It means that gas mass and heat diffusion occurs much faster than droplet surface regression time. If we further assume that properties of the environment also change very slowly, then during the characteristic gas-phase diffusion time the boundary locations and conditions can be considered to be constant. Thus the gas-phase processes can be treated as steady (time independent because the surface almost “freezes”), with the boundary variations occurring at longer time scales. When (at which value of D ∞ ) this assumption breaks down, i.e. when the diffusion time is equal to the surface regression time? When surface regression characteristic time becomes equal to gas mass/heat diffusion time, i.e. when D ∞ 2 / g ≈ D 0 2 /K? Remembering that must still be valid (it doesn’t depend on the distance from the droplet), we can conclude that the steady assumption breaks down when. For standard atmospheric conditions it breaks down at It means that our model will be valid for the distances less than this one. The d 2 Law – assumptions (vii) Single fuel species. Thus it is unnecessary to analyze liquid- phase mass transport (no diffusion term). (viii) Constant and uniform droplet temperature. This implies that there is no droplet heating. Where all the heat goes? Combined with (vii), we see that liquid phase heat and mass transport processes are completely neglected. Therefore the d 2 Law is essentially a gas-phase model. (ix) Saturation vapour pressure at droplet surface. This is based on the assumption that the phase-change process between liquid and vapour occurs at a rate much faster than those for gas- phase transport. Thus, evaporation at the surface is at thermodynamic equilibrium, producing fuel vapour which is at its saturation pressure corresponding to the droplet surface temperature. (x) No Soret (mass flow because of the temperature gradient), Dufour (heat flow because of the concentration gradient) and radiation effects (how this effects the validity of the model?). Rate of accumulation of mass of component j Mass flow rate of component j into the system Mass flow rate of component j out of system Rate of generation of mass of component j from reaction Rate of depletion of mass of component j from reaction Droplet evaporation I (no energy concerns) The phenomenon considered: A small sphere of liquid in an infinite gaseous atmosphere vaporizes and finally disappears. What is to be predicted? Time of vaporization as a function of the properties of liquid, vapor and environment. Assumptions: spherical symmetry (non-radial motion is neglected) (quasi-) steady state in gas Γ VAP independent of radius large distance between droplets no chemical reaction Vapor concentration distribution m VAP in the gas. roro r GoGo G = G TOT,VAP m VAP,0 has a strong influence, but is not usually known, it depends on temperature. relative motion of droplet and air augments the evaporation rate (inner circulation of the liquid) by causing departures from spherical symmetry. the vapour field of neighboring droplets interact m VAP,0 and m VAP,∞ may both vary with time. Γ VAP usually depends on temperature and composition. Limitations y = -x 0 1 So, a positive G 0 reduces the rate of heat transfer at the liquid surface. It means that if the heat is transferred to some let us say solid surface, that we want to prevent from heating up, we should eject the liquid to the thermal boundary layer (possibly through little holes). This liquid jets will accommodate a great part of the heat on vaporization of the liquid. Thus, we’ll prevent the surface from heating – transpiration cooling. The smaller the holes the smaller a part of heat towards the liquid interior and, subsequently towards the solid surface. Equilibrium vaporization – droplet is at such a temperature that the heat transfer to its surface from the gas is exactly equals the evaporation rate times the latent heat of vaporization: This implies: What if –Q 0 ≠ G 0 L Cases of interest: (i)When T ∞ is much greater than the boiling-point temperature T BOILING, m VAP,0 is close to 1 and T 0 is close to T BOILING. Then the vaporization rate is best calculated from: (ii)When T ∞ is low, and m VAP,∞ is close to zero, T 0 is close to T ∞. This implies T 0 ≈T ∞. Thus, m VAP,0 is approximately equal to the value given by setting T 0 =T ∞ in and the vaporization rate can be calculated by: As in example with water droplet evaporating at 10 0 C Evaporation rate [m 2 /s] The choice depends on whether T 0 or m VAP,0 is easier to estimate Droplet heat up effect on temperature and lifetime We can divide the droplet evaporation process into two stages. At first, while the droplet is cold (evaporation is slow), all the heat from the hot environment will be used for droplet interior heat up. As the droplet temperature will approach its steady state value, droplet heat up will slow down, while evaporation will accelerate. r 0 =r 0 (t) Droplet heat up effect on temperature and lifetime Slowest limit Fastest limit Distillation limit Diffusion limit D 2 Law Center Temperature Surface Temperature T ( LIQ /r 0,INITIAL 2 )t (r /r 0,INITIAL ) 2 ( LIQ /r 0,INITIAL 2 )t Distillation limit Diffusion limit D 2 Law 300 380 0.20.1 1 0

Updated On: 21-5-2021 Loading DoubtNut Solution for you `(1)/(3)` sq unit. `(2)/(3)` sq unit. `(1)/(4)` sq unit. `(1)/(5)` sq unit. |00:00 - 00:59||the question is the area bounded by the curve Y is equal to x minus 1 whole square Y is equal to X + 1 whole square and Y equal to 1 by 4 is bigger than 3 Y is equal to x minus 1 whole square then we have Y is equal to X + 1 whole square and Y is equal to 1 by 4 Get together now will need to draw the graph for is now but we should know that these are the co-ordinate Axis graph of y equal to x square we should know that is like this| |01:00 - 01:59||something like this ok so why good to x minus 1 whole square glass would be shifted to the positive x axis and y axis minus 1 whole square is if you put 121 here then it becomes you this is the point 10 ok craft would be like this similarly the graph for X + 1 would be at X equal to minus 10 suggests metric this should move to the server control again this is equal to x minus 1 whole square graph and this is why equal to X +| |02:00 - 02:59||1 crore are so we can draw we have two points where is 10 x is this is by axis and we have 1.0 - and craft something like this and other grasses like this difference between W A Y is equal to X + 1 whole square and the black one is Y is equal to x minus 1 whole square| |03:00 - 03:59||ok now this is common intersection point at y axis ok because when we put X equal to zero here all photographs that one should be smarter 2014 line also this time we have one that would be less than one Amritsar wishes line y equal to 1 by the now we need to find the to find the area bounded by the speaker area bounded by this occurs if you see here is different| |04:00 - 04:59||this is the area that we need to find now we name this point as a bi this is point P and this is quite so we need to find area bounded equal to area bounded by the curve Y is equal to area of reason a comes after spare now also due to symmetry we can say that due to symmetry we can say that| |05:00 - 05:59||area of reason English a call to area of visit near this area this area is equal to area so India of whole reason area of a b c d a can be written as equal to 2 x of two times of any one reason that is a CD is equal to X of area in AC DJ now to find the area we need to discuss study question we know Y equal to x minus 1 whole square and we need to divide the unit to| |06:00 - 06:59||calculate area up to this point ok we don't know the coordinates of the point the super sweet find the coordinates of point P we have yet to find the intersection of Y equal to 1 by 4 and Y is equal to x minus 1 whole square if the intersection point of X equal to 1 by 4 and Y equal to x minus 1 whole square equal to 1 by 4 here we get minus 1 whole square x minus one can be equal to 1 by 2 aur x minus one can be equal to minus 1 by 2 also know from where we get value by to and from here we get value 1 minus one by two that is one by two would we have two values now| |07:00 - 07:59||to intersection point 1 intersection part is this and one is this we will choose the lower one to one by two here so the point is the point is point B is X is 1 by 2 and why is 1 by 4 we need to find the area here from zero X equal to zero to this is equal to 1 by from here is we will integrate the kar you get the whole area so we need to subtract the bottom area of the the area of line y equal to 1 by 4 light area of reason a c d a is| |08:00 - 08:59||equal to integration from 0 to 1 by 2 of x minus 1 whole square x integration of x minus 1 whole square DS what is the whole area if we put this an an an area is given by this savinay to subtract the bottom which is given by - integration from 0 to 1 by 2 of why why is 1 by 4 integration of x minus 1 whole square minus x minus 1 divided by 3 from 0 to 1 by 2 minus 4 integration of DX Easter eggs from 0 to 1 by 2 + becomes equal to putting one by to hear this becomes| |09:00 - 09:59||minus 1 by 2 minus 1 by 2 cube minus 1 by 8 is 1 by 24 - 14 0 here this becomes -1 cube that is - 10 - + and here we have minus 1 by 4 into one by two this is equal to minus 1 by 24 + 1 by 3 and minus 1 by 4 into two that is it now here we can take LCM taking calcium 24 comes - 1 + 8 minus 3 equal to 8 - 4 that is divided by 24 which is equal to 1 by 6 sote se area of region is equal to 1 by| |10:00 - 10:59||6 hour we need to find you need to find all of this book age of ABCD so every of total reason a video equal to 2 x of 1 by 6 which is equal to 1 by 3 square units which is the final area of the boundary between these options of number one is correct 1 by 3 square unit is the area of the shaded region ABCD thank you| Area as a definite integral Algorithm for area calculation using horizontal strips Area calculation using horizontal strips for modulus function Impact of sign on area definition change according to sign of function Find the area of the region bounded by the parabola `y^2 = 2x` and the line `x - y = 4` Examples: Find the area of the region bounded by the curve `y^2 = 2y - x` and the y-axis. Area of symmetrical graphs Find the area of the region bounded by the ellipse `x^2 / a^2 + y^2 / b^2 = 1` Integration of sin and cos function in different intervals Examples: f(x) = x for x > 0 and f(x) = x^2 for x < 0; find area with x axis from x = -2 and x = 3. Haryana School Reopening Cancelled with 100 Per cent Capacity amid Omicron fear Haryana school reopening cancelled with 100% capacity amid omicron fear, a new variant of COVID-19. Check complete details here. CBSE Evaluating OMR Sheets of Term 1 Board Exams on Same Day, Results Soon CBSE evaluating OMR sheets of term 1 board exams on same day, results soon. Candidates can use the answer key to guesstimate their marks. IIFT Admit Card 2022 Released by NTA, Download Now IIFT admit card 2022 released by NTA. Know how to download the admit card, important dates, exam timing, exam pattern & other Important Instructions. 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Statistical Articles Including Margin Of Error And Sample Size Fiorina comes in second, with 16 percent support, up from 6 percent a month ago. I mean if I took a sample of 1000 from a population of 2000 I would think the results would have a smaller margin of error than if I took a if I get 200 from my exact sample size(240), is it representative? Most studies on surgery outcome are based on clinical improvement only few on QOL using retrospective data. http://comunidadwindows.org/margin-of/statistical-margin-of-error-sample-size.php That said, I can't find a location for these programs. If you do not know your proportion p, it is quite common to take 0,5 as a value for p. Definition The margin of error for a particular statistic of interest is usually defined as the radius (or half the width) of the confidence interval for that statistic. The term can See table below How many individuals to choose by region for his study? https://en.wikipedia.org/wiki/Margin_of_error Margin Of Error Formula Plz reply Gert Van Dessel - August, 2016 reply Maryam,For 95% confidence level and 5% margin of error, you would need a sample of 43. The top portion charts probability density against actual percentage, showing the relative probability that the actual percentage is realised, based on the sampled percentage. For a margin of error of 5% at 95% confidence level, you will need a sample of 80 respondents. Compute alpha (α): α = 1 - (confidence level / 100) = 1 - 0.95 = 0.05 Find the critical probability (p*): p* = 1 - α/2 = 1 - 0.05/2 Patra Gert Van Dessel - October, 2015 reply Hi Patra,The article is only available on web: (www.checkmarket.com/2013/02/how-to-estimate-your-population-and-survey-sample-size/), published by Gert Van Dessel of research company CheckMarket on 13th February 2013 Buchura Ben Carson, second in the lead in Iowa in this poll, captures 19 percent of the support, down from 22 percent last month. Remember that your population consist of approximately 400 million adults in the EU. Margin Of Error Sample Size But if the original population is badly skewed, has multiple peaks, and/or has outliers, researchers like the sample size to be even larger. When the sample size is smaller, the critical value should only be expressed as a t statistic. Margin Of Error Calculator We can similarly compare some of the less successful candidates in the Pew poll. But for such small populations, you might as well include the total population in your survey. http://www.dummies.com/education/math/statistics/how-to-calculate-the-margin-of-error-for-a-sample-proportion/ In some sense, the math reported in polls may be a disguise covering up bad methodology (like rich icing on a bad cake). Could u plz tell me how to determine sample size if don't give any information (standard deviation and confidence interval). Margin Of Error Excel Retrieved 2006-05-31. Surveying has been likened to taste-testing soup – a few spoonfuls tell what the whole pot tastes like. Note: The larger the sample size, the more closely the t distribution looks like the normal distribution. Margin Of Error Calculator MathWorld. https://explorable.com/statistics-margin-of-error Thank you. Margin Of Error Formula The sample proportion is the number in the sample with the characteristic of interest, divided by n. Margin Of Error Definition First, assume you want a 95% level of confidence, so z* = 1.96. Comparing percentages In a plurality voting system, where the winner is the candidate with the most votes, it is important to know who is ahead. http://comunidadwindows.org/margin-of/survey-articles-with-margin-of-error.php Chris - August, 2015 reply Sir I am to survey a population of about 10000 students. Therefore not much information available on QOL. Otherwise, we use the t statistics, unless the sample size is small and the underlying distribution is not normal. Margin Of Error In Polls Now, remember that the size of the entire population doesn't matter when you're measuring the accuracy of polls. Acceptable Margin Of Error Whay not 6 or 8 for example? Thanks Karol Gert Van Dessel - June, 2016 reply Karol,I would go for a fixed fieldwork period, for example 1 month, and collect responses every day. Typical choices are 90%, 95%, or 99% % The confidence level is the amount of uncertainty you can tolerate. Maarten Marijnissen - September, 2016 reply Hi,This sample gives you a margin of error of 9,48% for 95% confidence level, which is quite high.This number is the plus-or-minus figure usually reported In other words, if all I get is 72 responses, can't I still use that and show the actual numbers instead of my computed ones? Gert Van Dessel - February, 2016 reply Andrew,With an error margin of 2% you will need 2354 respondents in your sample. Margin Of Error Vs Standard Error The size of your sample depends on the research project en the method you are using.More explanation and further information can be found in this article. the nr. MSNBC reported these same Pew Research Center numbers with no mention at all of the margin of error—a lost opportunity, in our view, to point to the weakness of a small It's time for some math. (insert smirk here) The formula that describes the relationship I just mentioned is basically this: The margin of error in a sample = 1 divided by this content Are there any sources I can use? You can calculate yourself with our sample size calculator Adebayo Kazeem - September, 2015 reply I am so happy to discover your website I believe God is good to me, sir, That’s what the MOE addresses. Of the media sources mentioned, only CNN can be lauded for mentioning the size of the Pew survey and the corresponding MOE. If you'd like to see how we perform the calculation, view the page source. I n my study I will use the questionnaire about PM examination methods in. For a social science project, would a CI of 8 % and a CL of 90% seem appropriate? one response from each small firm. In other words, if we were to conduct this survey many times with different samples of 497 randomly chosen Republican voters, 95 out of 100 times the proportion of the survey This makes intuitive sense because when N = n, the sample becomes a census and sampling error becomes moot. You can check with our sample size calculator Ofere Jmaes - October, 2015 reply I have population known that is 150 with the sample of 30 how can i calculate the Generally, 5% is taken as a maximum margin of error, this would correspond with a sample of 228. If you have a sample size of 348 and a total population of for example 5000, your margin of error will be about 5% (for the total population of all 10 Nicholas - December, 2015 reply Let's say I am conducting a research for a bank (questionnaires) for 432 branches within 2 months time. for a population of 10,000, confidence level 95% and margin of error 5%, your required sample size is 370. How do I calculate the sample for the population? The standard error (0.016 or 1.6%) helps to give a sense of the accuracy of Kerry's estimated percentage (47%). With a confidence level of 95%, you would expect that for one of the questions (1 in 20), the percentage of people who answer yes would be more than the margin

The letters of the word ABACUS have been arranged in the shape of a triangle. How many different ways can you find to read the word ABACUS from this triangular pattern? This tricky challenge asks you to find ways of going across rectangles, going through exactly ten squares. An introduction to bond angle geometry. Many natural systems appear to be in equilibrium until suddenly a critical point is reached, setting up a mudslide or an avalanche or an earthquake. In this project, students will use a simple. . . . We're excited about this new program for drawing beautiful mathematical designs. Can you work out how we made our first few pictures and, even better, share your most elegant solutions with us? A Sudoku that uses transformations as supporting clues. A Sudoku with clues as ratios. The clues for this Sudoku are the product of the numbers in adjacent squares. In this Sudoku, there are three coloured "islands" in the 9x9 grid. Within each "island" EVERY group of nine cells that form a 3x3 square must contain the numbers 1 through 9. Place the 16 different combinations of cup/saucer in this 4 by 4 arrangement so that no row or column contains more than one cup or saucer of the same colour. This Sudoku puzzle can be solved with the help of small clue-numbers on the border lines between pairs of neighbouring squares of the grid. Use the interactivity to listen to the bells ringing a pattern. Now it's your turn! Play one of the bells yourself. How do you know when it is your turn to ring? Label the joints and legs of these graph theory caterpillars so that the vertex sums are all equal. Given the products of diagonally opposite cells - can you complete this Sudoku? Pentagram Pylons - can you elegantly recreate them? Or, the European flag in LOGO - what poses the greater problem? Bellringers have a special way to write down the patterns they ring. Learn about these patterns and draw some of your own. An investigation involving adding and subtracting sets of consecutive numbers. Lots to find out, lots to explore. Can you recreate these designs? What are the basic units? What movement is required between each unit? Some elegant use of procedures will help - variables not essential. Remember that you want someone following behind you to see where you went. Can yo work out how these patterns were created and There are nine teddies in Teddy Town - three red, three blue and three yellow. There are also nine houses, three of each colour. Can you put them on the map of Teddy Town according to the rules? A Sudoku with clues as ratios or fractions. This second Sudoku article discusses "Corresponding Sudokus" which are pairs of Sudokus with terms that can be matched using a substitution rule. Each clue number in this sudoku is the product of the two numbers in adjacent cells. You are given the Lowest Common Multiples of sets of digits. Find the digits and then solve the Sudoku. A Sudoku with a twist. Four small numbers give the clue to the contents of the four Use the differences to find the solution to this Sudoku. Whenever a monkey has peaches, he always keeps a fraction of them each day, gives the rest away, and then eats one. How long could he make his peaches last for? Do you notice anything about the solutions when you add and/or subtract consecutive negative numbers? Two sudokus in one. Challenge yourself to make the necessary The puzzle can be solved with the help of small clue-numbers which are either placed on the border lines between selected pairs of neighbouring squares of the grid or placed after slash marks on. . . . Find out about Magic Squares in this article written for students. Why are they magic?! 60 pieces and a challenge. What can you make and how many of the pieces can you use creating skeleton polyhedra? Just four procedures were used to produce a design. How was it done? Can you be systematic and elegant so that someone can follow You need to find the values of the stars before you can apply normal Sudoku rules. The puzzle can be solved by finding the values of the unknown digits (all indicated by asterisks) in the squares of the $9\times9$ grid. This is a variation of sudoku which contains a set of special clue-numbers. Each set of 4 small digits stands for the numbers in the four cells of the grid adjacent to this set. A pair of Sudoku puzzles that together lead to a complete solution. Four numbers on an intersection that need to be placed in the surrounding cells. That is all you need to know to solve this This pair of linked Sudokus matches letters with numbers and hides a seasonal greeting. Can you find it? A cinema has 100 seats. Show how it is possible to sell exactly 100 tickets and take exactly £100 if the prices are £10 for adults, 50p for pensioners and 10p for children. This Sudoku, based on differences. Using the one clue number can you find the solution? This Sudoku combines all four arithmetic operations. Problem solving is at the heart of the NRICH site. All the problems give learners opportunities to learn, develop or use mathematical concepts and skills. Read here for more information. This challenge extends the Plants investigation so now four or more children are involved. This challenging activity involves finding different ways to distribute fifteen items among four sets, when the sets must include three, four, five and six items. Given the nets of 4 cubes with the faces coloured in 4 colours, build a tower so that on each vertical wall no colour is repeated, that is all 4 colours appear. Solve the equations to identify the clue numbers in this Sudoku problem.

permutations and combinations class 11 worksheet with solutions Are you preparing for the CBSE Class 11 Permutations and Combinations exam and feeling overwhelmed by the vast syllabus? Look no further! In this comprehensive guide, we will help you crack the exam with confidence by providing you with 100 practice questions with detailed solutions. Permutations and combinations can be quite challenging for many students, but with the right guidance and practice, you can ace this subject. Our carefully curated set of questions covers a wide range of concepts, ensuring that you are well-prepared for any type of question that may appear in the exam. Each question is accompanied by a step-by-step solution, allowing you to understand the logic and reasoning behind the answer. This not only helps you in finding the correct answer but also helps in building a strong foundation in permutations and combinations. Whether you are a visual learner, a kinesthetic learner, or an auditory learner, our variety of question types and detailed solutions cater to all learning styles. So, gear up, put your problem-solving skills to the test, and conquer the CBSE Class 11 Permutations and Combinations exam with ease. When it comes to Class 11 Maths, the topic of permutations and combinations holds significant importance, especially in Chapter 6. This chapter introduces students to the fascinating world of arranging and selecting objects in a systematic way. Whether you're a student, a teacher, or a parent guiding your child, understanding permutations and combinations is crucial. For those diving into this topic, the permutations and combinations formula serves as the backbone. This formula helps to solve various kinds of problems and questions that often appear in exams and real-life situations. Students aiming for a strong grasp on Class 11 Chapter 6 Maths topics should not just limit themselves to the textbook. It's beneficial to explore additional resources like permutations and combinations Class 11 notes, permutations and combinations Class 11 worksheet with answer, and permutations and combinations Class 11 MCQ sets. These supplementary materials provide a well-rounded understanding of the topic and offer practical practice to solidify concepts. Our permutations and combinations Class 11 extra questions can also serve as a challenging exercise to test your knowledge and preparation level. One highly effective way to enhance understanding is by practicing permutations and combinations questions. Many students prefer using the permutation and combination Class 11 questions with answers as these come with detailed solutions, facilitating easy learning. Worksheets offer a hands-on approach, allowing you to apply the permutations and combinations formula and check your answers instantaneously. When you consistently practice, you get a firm understanding of the core concepts, and that's where the permutations and combinations Class 11 notes become handy. These notes are simplified, well-structured, and offer invaluable insights into the topic, making them an ideal resource for students and even helpful for teachers and parents. Moreover, MCQs are not just for practice but also an excellent way to test understanding. The permutations and combinations Class 11 MCQ questions are carefully curated to cover all important aspects of the topic. Coupled with a robust question bank, they form an integral part of your study resources. Students preparing for Chapter 6 Class 11 Maths or even planning ahead for Chapter 7 Maths Class 11 can use these resources for a seamless learning experience. For a complete study solution, you might want to look at permutation and combination Class 11 solutions PDF. These downloadable PDFs serve as a portable study resource that you can access anytime, offering the flexibility to study at your own pace. As you transition from Chapter 6 Class 11 Maths to other topics like Chapter 7, having a strong grounding in permutations and combinations will certainly prove beneficial. So, whether you refer to it as 11th Class Maths Chapter 6, Ch 6 Class 11 Maths, or any other term, this topic is essential for building a strong mathematical foundation. Make the most out of the available resources and master permutations and combinations in Class 11 Maths. permutations and combinations definition In permutations, the arrangement of objects is important, meaning the order in which you select or arrange items makes a difference. For example, if you have three letters , , and , you can arrange them in different ways like , , , etc. Each different arrangement is considered a separate permutation. Combinations, on the other hand, focus on the selection of items without considering the order. For instance, if you're selecting two letters from , , and , combinations like and are considered the same because the order does not matter. permutations and combinations formula for class 11 Permutations deal with the arrangement of objects in a specific order. The formula for permutations is given as: where is the total number of items to choose from, and is the number of items to choose. Combinations, on the other hand, deal with the selection of objects without considering the order. The formula for combinations is: where is the total number of items, and is the number of items to choose. Here, (n factorial) is the product of all positive integers from 1 to . For example, . These formulas are the cornerstone for solving various types of permutations and combinations questions in Class 11. Once you understand these formulas, you can use them to solve problems found in your textbook, as well as those in the permutations and combinations Class 11 worksheet with answers, to get practical experience. Understanding these formulas also prepares you for more advanced topics in mathematics and helps you in solving real-world problems that involve arrangement and selection. For more practice and to deepen your understanding, you can look into permutations and combinations Class 11 notes, MCQs, and extra questions. These resources often provide examples that apply these formulas in different contexts, making it easier for students, as well as parents and teachers, to grasp the topic thoroughly. Understanding the Fundamental Principles In Class 11 Maths, particularly in Chapter 6, understanding the fundamental principles of permutations and combinations is crucial. These principles lay the groundwork for solving various types of questions and real-world problems. Grasping the basics also helps when you're dealing with permutations and combinations Class 11 worksheet with answers or attempting Class 11 MCQs. Permutations: Arranging Objects in a Specific Order Permutations involve arranging objects in a particular sequence or order. This is essential in various fields like computer science, statistics, and even in everyday scenarios like arranging books on a shelf. Understanding permutations prepares you for more complex topics and helps in solving permutation problems in Class 11 Maths. Combinations: Selecting Objects Without Considering the Order Combinations are all about selecting items where the arrangement or order doesn't matter. This concept is widely used in statistics, probability, and daily life situations like forming a team from a group of people. When studying combinations, it's helpful to consult resources like permutations and combinations Class 11 notes and permutations and combinations Class 11 extra questions. Solving Permutation Problems Solving problems based on permutations usually involves the use of the permutation formula. Practice is key to mastering this, and the permutations and combinations Class 11 worksheet with answer serves as a useful resource. The more you practice, the better you get at identifying which problems require the use of permutations. Solving Combination Problems Similar to permutations, solving combination problems effectively requires understanding the combination formula. Resources like permutations and combinations Class 11 extra questions and Class 11 Maths Chapter 6 solutions can provide additional practice and insights into solving these types of problems. Solving Problems Involving Both Permutations and Combinations Some questions might require the use of both permutations and combinations. Understanding the fundamental principles and the differences between the two is crucial for solving such problems. Class 11 permutation and combination questions with answers often cover these complex scenarios. Practice Questions for Permutations and Combinations To master permutations and combinations in Class 11, practice is essential. Utilize the permutations and combinations Class 11 question bank to explore various types of questions and understand the application of formulas. Detailed Solutions for Practice Questions For effective learning, merely solving questions is not enough. One must also understand the solutions, which is where detailed solutions come in. These can be found in resources like permutation and combination Class 11 solutions PDF. Conclusion: Tips to Excel in the CBSE Class 11 Permutations and Combinations Exam To excel in this topic, make the best use of available resources like Class 11 Maths Chapter 6 solutions, practice consistently, and focus on understanding the core principles. Doing so will not only help you score well in exams but will also lay a strong foundation for future mathematical endeavours.

Enter your mobile number or email address below and we'll send you a link to download the free Kindle App. Then you can start reading Kindle books on your smartphone, tablet, or computer - no Kindle device required. To get the free app, enter your mobile phone number. Other Sellers on Amazon + $3.99 shipping + Free Shipping + $3.99 shipping The Works of Archimedes (Dover Books on Mathematics) Paperback – April 16, 2002 |New from||Used from| The Amazon Book Review Author interviews, book reviews, editors picks, and more. Read it now Frequently bought together Customers who bought this item also bought Top customer reviews There was a problem filtering reviews right now. Please try again later. If you want to pass a basic set of classes, then you don't need this; just stick to the textbooks and you'll do fine. However, if you really want to understand what's going on in that math, and why it's going on, this is a great place to start. There's no place like to source for good information. As for this particular translation, this edition has a surprising amount of explanatory notes and introductory material relating the circumstances under which this writing was made, and the interaction between the author and the other well known thinkers of the time. The first ~150 pages were explanations by Heath, including terminology of Archimedes, which was useful at times. All in all, the works of Archimedes are definitely worth reading for anyone interested in learning the process of mathematical discovery. Archimedes's determination of the volume (and secondarily the surface area) of a sphere may be considered the undisputed crown jewel of his works. Indeed, Archimedes himself seems to have thought so judging by the story that he requested it inscribed on his tombstone (xviii). Like an axiom needs no demonstration, this beautiful result needs no motivation. But no sooner has Archimedes proved this result than he turns, in the same treatise, to matters of much less evident value. Thus he solves numerous baroque problems about areas and volumes of various sections of spheres such as "Given two segments of spheres, to find a third segment of a sphere similar to one of the given segments and having its surface equal to that of the other" (82) and other things along similar lines. What is the purpose of this? What use could one ever make of such a problem? One might say: Archimedes is here working out the complete theory of spherical areas and volumes for future reference. The main result is beautiful in its own right, but these technical further propositions form a "toolbox" of results that subsequent mathematicians will find useful in technical contexts. If this is so (which I doubt) then it may be observed that previous mathematicians were apparently not inclined to do something similar when they worked out the theory of the cone. For Archimedes cites previous propositions on volumes of cones without proof (22), yet he has to work out the rather basic result of the surface area of frustum from scratch (21). Another possible motivation for these technical propositions emerges from Archimedes's preface to another work, where he says of a list of problems along these lines that: "there are two included among them which are impossible of realisation [and which may serve as a warning] how those who claim to discover everything but produce no proofs of the same may be confuted as having actually pretended to discover the impossible" (151). Thus a sprinkling of technical propositions such as those mentioned above can serve a useful purpose even if they are of no actual use to any mathematician; the use being to expose poseur mathematicians by a kind of acid test. Indeed this seems to fit with Archimedes's preface to the original treatise where he appears to think that there is hardly anyone capable of judging his work: "these discoveries of mine ... ought to have been published while Conon was still alive, for I should conceive that he would best have been able to grasp them and to pronounce upon them the appropriate verdict" (2). (Incidentally, perhaps this lack of competent peers is also part of the reason why Archimedes chose to withdraw to a recluse life in Syracuse rather than remaining in Alexandria where he studied (xvi).) In the Measurement of a Circle, Archimedes offers the two-dimensional analog of the above result: "The area of any circle is equal to a right-angled triangle in which one of the sides about the right angle is equal to the radius, and the other to the circumference of the circle" (91). Why did Euclid not prove this rather basic theorem about the areas of circles? The answer is obvious: because the triangle in question cannot be constructed by ruler and compass. Thus the problem cannot be considered truly solved. Indeed, Archimedes himself seems to agree with this when he writes: "Some of the earlier geometers tried to prove it possible to find a rectilineal area equal to a given circle ..., assuming lemmas not easily conceded, so that it was recognised by most people that the problem was not solved" (233). It seems clear from this quotation that the mere result that the area of a circle is equal to such-and-such a triangle was well known, since people proved it by *some* means. The problem, rather, was with the method used in those proofs. In other words, the problem was, as Archimedes says, to *find* the rectilineal area in question, meaning to construct it, not to prove that it was in fact equal to the circle. It seems clear that this is the reason Archimedes studies his spiral, for if it is admitted that the spiral and its tangents can be drawn then a construction of the triangle in question follows immediately (171), though probably Archimedes would have been the first to admit that these spiral constructions are just one more of those "lemmas not easily conceded" that he spoke of above. One indication that the spiral is meant to fit into the tradition of ruler-and-compass constructions is that Archimedes defines it in terms of motions and derives the locus-definition (essentially r=theta) as a theorem (155) even though it is the latter rather than the former that is needed in the subsequent proofs. It is often assumed that the ruler-and-compass paradigm was favoured over a numerical conception of mathematics because things like the irrationality of square roots pose problems for the latter but not the former. Perhaps it is understandable then that when the ruler-and-compass paradigm proved poorly suited for dealing with the quadrature of the circle, the numerical conception of geometry was revived. This could perhaps explain why Archimedes takes pains to establish bounds for pi (93), and perhaps by extension and analogy his interest in determining how many grains of sand fit in the universe (221-232), and maybe by one more tenuous step of association his interest in numerical problems generally, as in the cattle-problem (318-326) and apparently another lost work (xxxvi). Archimedes worked out axiomatic theories of statics and hydrostatics, showing that the geometrical method is remarkably and perhaps surprisingly well-suited to deal with these physical matters. In this it seems he had "no fore-runners" (xl). Some might look for his motivations in his work on warfare machines (xvi-xvii), but I would suggest that it actually followed quite naturally from his interest in volumes. As Archimedes says (2), his work on the volume of the sphere builds on Eudoxus's result that the volume of a cone is one third of the corresponding cylinder. Seeing as plane geometry is always "operationalised" in terms of ruler-and-compass constructions, one might look for a way to "operationalise" this result. Thus the use of a lever or balance suggests itself: a cylinder in one bowl balances with three cones in the other, or one cone at three times the distance. Alternatively, three cones or one cylinder dropped into a container of water make the surface rise by equal amounts. To put it another way, Greek results about volumes and areas are always comparative rather than absolute: they don't say it's equal to such-and-such a number or formula, but to such-and-such other figure. This comparative aspect goes hand in hand with the idea of a physical balance in equilibrium, again suggesting that Archimedes's work grew out of geometry. Archimedes's work on statics is also the basis for his "Method" of finding areas and volumes by (hypothetically) balancing them on levers. As he says: "Certain things first became clear to me by a mechanical method, although they had to be demonstrated by geometry afterwards because their investigation by the said method did not furnish an actual demonstration. But it is of course easier, when we have previously acquired, by the method, some knowledge of the questions, to supply the proof than it is to find it without any previous knowledge." (Method appendix, 13) Notably the volume of a sphere and the area of a parabolic segment are found in this way. Both are proved rigorously using the method of exhaustion in other works. In those works, the volume of the sphere is investigated by approximating a circle with a regular polygon and rotating it, so that the volume of the sphere is approximated by conical frustums; and the parabolic area is approximated by triangles stacked upon triangle fitted closer and closer to the perimeter. Both of these methods of approximation are natural and straightforward. It seems, therefore, that the only reason they were not feasible before the Method was that it was too difficult to brute-force sum these pieces (which basically amounts to simplifying intricate algebraic expressions) without knowing in advance what answer should be. In reality, however, the "mechanical" aspect of the method has rather little to do with it. More importantly the Method consists in investigating the shapes slice by slice. True, each slice is then balanced with some suitable slice of a comparison figure with some suitable choice of lever arm, but this is little more than a convenient and flexible way of expressing an algebraic equality. The greater contrast with the rigorous proofs is the slicing of shapes in to "lines," which basically means investigating the figures by their algebraic equations, since the line in question has height y=f(x). More generally, Archimedes seems to have taken a constant interest in conic sections simply for the fact that they are the most tractable "next" figures beyond circles. He constantly generalises all of his results about circles and spheres to conics, for example investigating their areas (in the parabolic case), rotational volumes, centers of gravity, and even the behaviour of a paraboloid submerged in water. None of these results seem particularly useful or natural except as a way of generating more mathematics from existing ideas. When they were first introduced conic sections surely had to be motivated and serve a purpose for solving established problems, but clearly at the time of Archimedes they were accepted as part of the standard repertoire of curves and it was seen as natural to "cut one's teeth" on the conics by generalising to them anything one did for circles.

James M. Palmer Research Professor Optical Sciences Center University of Arizona Tucson, AZ; 85721 Latest update 10/26/03 “When I use a word, it means just what I choose it to mean - neither more nor less.” Lewis Carroll (Charles Lutwidge Dodgson) Effective technical communication demands a system of symbols, units and nomenclature (SUN) that is reasonably consistent and that has widespread acceptance. Such a system is the International System of Units (SI). There is no area where words are more important than radiometry and photometry. This document is an attempt to provide necessary and correct information to become conversant. 1. What is the motivation for this FAQ? 2. What is radiometry? What is photometry? How do they differ? 3. What is projected area? What is solid angle? 4. What are the quantities and units used in radiometry? 5. How do I account for spectral quantities? 6. What are the quantities and units used in photometry? 7. What is the difference between lambertian and isotropic? 8. When do the properties of the eye get involved? 9. How do I convert between radiometric and photometric units? 10. Where can I learn more about this stuff? RADIOMETRY &. PHOTOMETRY FAQ 1 1. What is the motivation for this FAQ? There is so much misinformation and conceptual confusion regarding photometry and radiometry, particularly on the WWW by a host of “authorities”, it is high time someone got it straight. So here it is, with links to the responsible agencies. Background: It all started over a century ago. An organization called the General Conference on Weights and Measures (CGPM) was formed by a diplomatic treaty called the Metre Convention. This treaty was signed in 1875 in Paris by representatives from 17 nations (including the USA). There are now 48 member nations. Also formed were the International Committee for Weights and Measures (CIPM) and the International Bureau of Weights and Measures (BIPM). The CIPM, along with a number of sub-committees, suggests modifications to the CGPM. In our arena, the subcommittee is the CCPR, Consultative Committee on Photometry and Radiometry. The BIPM is the physical facility responsible for dissemination of standards, the international metrology institute. The SI was adopted by the CGPM in 1960. It currently consists of seven base units and a larger number of derived units. The base units are a choice of seven well- defined units that by convention are regarded as independent. The seven are: metre, kilogram, second, ampere, kelvin, mole and candela. The derived units are those formed by various combinations of the base units. International organizations involved in the promulgation of SUN include the International Commission on Illumination (CIE), the International Union of Pure and Applied Physics (IUPAP), and the International Standards Organization (ISO). In the USA, the American National Standards Institute (ANSI) is the primary documentary (protocol) standards organization. Many other scientific and technical organizations publish recommendations concerning the use of SUN for their learned publications. Examples are the International Astronomical Union (IAU) and the American Institute of Physics (AIP). Read all about the SI, its history and application, at physics.nist.gov/cuu/ or at www.bipm.fr. 2. What is radiometry? What is photometry? How do they differ? Radiometry is the measurement of optical radiation, which is electromagnetic radiation within the frequency range between 3×1011 and 3×1016 Hz. This range corresponds to wavelengths between 0.01 and 1000 micrometres (µm), and includes the regions commonly called the ultraviolet, the visible and the infrared. Two out of many typical units encountered are watts/m2 and photons/s-steradian. Photometry is the measurement of light, which is defined as electromagnetic radiation that is detectable by the human eye. It is thus restricted to the wavelength range from about 360 to 830 nanometers (nm; 1000 nm = 1 µm). RADIOMETRY &. PHOTOMETRY FAQ 2 Photometry is just like radiometry except that everything is weighted by the spectral response of the eye. Visual photometry uses the eye as a comparison detector, while physical photometry uses either optical radiation detectors constructed to mimic the spectral response of the eye, or spectroradiometry coupled with appropriate calculations to do the eye response weighting. Typical photometric units include lumens, lux, candelas, and a host of other bizarre ones. The only real difference between radiometry and photometry is that radiometry includes the entire optical radiation spectrum, while photometry is limited to the visible spectrum as defined by the response of the eye. In my forty years of experience, photometry is more difficult to understand, primarily because of the arcane terminology, but is fairly easy to do, because of the limited wavelength range. Radiometry, on the other hand, is conceptually somewhat simpler, but is far more difficult to actually do. 3. What is projected area? What is solid angle? Projected area is defined as the rectilinear projection of a surface of any shape onto a plane normal to the unit vector. The differential form is dAproj = cos(β) dA where β is the angle between the local surface normal and the line of sight. We can integrate over the (perceptible) surface area to get AA= cosβ d proj z A Some common examples are shown in the table below: SHAPE AREA PROJECTED AREA Flat rectangle A = L×W Aproj= L×W cos β 2 2 2 2 Circular disc A = π r = π d / 4 Aproj = π r cos β = π d cos β / 4 2 2 2 Sphere A = 4 π r = π d Aproj = A/4 = π r Plane angle and solid angle are two derived units on the SI system. The following definitions are from NIST l SP811. r “The radian is the plane angle between θ two radii of a circle that cuts off on the l circumference an arc equal in length to θ = r the radius.” RADIOMETRY &. PHOTOMETRY FAQ 3 The abbreviation for the radian is rad. Since there are 2π radians in a circle, the conversion between degrees and radians is 1 rad = (180/π) degrees. A solid angle extends the concept to three dimensions. “One steradian (sr) is the solid angle that, having its vertex in the center of a A sphere, cuts off an area on the surface of the sphere r equal to that of a square with sides of length equal A to the radius of the sphere.” ω = r2 The solid angle is thus ratio of the spherical area to the square of the radius. The spherical area is a projection of the object of interest onto a unit sphere, and the solid angle is the surface area of that projection. If we divide the surface area of a sphere by the square of its radius, we find that there are 4π steradians of solid angle in a sphere. One hemisphere has 2π steradians. Accepted symbols for solid angle are ω, the lowercase Greek letter omega, and Ω, the uppercase omega. I use ω exclusively for solid angle, reserving Ω for the advanced concept of projected solid angle (ω cosθ). Both plane angles and solid angles are dimensionless quantities, and they can lead to confusion when attempting dimensional analysis. 4. What are the quantities and units used in radiometry? Radiometric units can be divided into two conceptual areas: those having to do with power or energy, and those that are geometric in nature. The first two are: Energy is an SI derived unit, measured in joules (J). The recommended symbol for energy is Q. An acceptable alternate is W. Power (a.k.a. radiant flux) is another SI derived unit. It is the rate of flow (derivative) of energy with respect to time, dQ/dt, and the unit is the watt (W). The recommended symbol for power is Φ (the uppercase Greek letter phi). An acceptable alternate is P. Energy is the integral over time of power, and is used for integrating detectors and pulsed sources. Power is used for non-integrating detectors and continuous (in a time sense) sources. Even though we patronize the power utility, what we are actually buying is energy in watt hours. RADIOMETRY &. PHOTOMETRY FAQ 4 Now we become more specific and incorporate power with the geometric quantities area and solid angle. Irradiance (a.k.a. flux density) is another SI derived unit and is measured in W/m2. Irradiance is power per unit area incident from all directions in a hemisphere onto a surface that coincides with the base of that hemisphere. A similar quantity is radiant exitance, which is power per unit area leaving a surface into a hemisphere whose base is that surface. The symbol for irradiance is E and the symbol for radiant exitance is M. Irradiance (or radiant exitance) is the derivative of power with respect to area, dΦ/dA. The integral of irradiance or radiant exitance over area is power. Radiant intensity is another SI derived unit and is measured in W/sr. Intensity is power per unit solid angle. The symbol is I. Intensity is the derivative of power with respect to solid angle, dΦ/dω. The integral of radiant intensity over solid angle is power. Radiance is the last SI derived unit we need and is measured in W/m2-sr. Radiance is power per unit projected area per unit solid angle. The symbol is L. Radiance is the derivative of power with respect to solid angle and projected area, dΦ/dω dA cos(θ) where θ is the angle between the surface normal and the specified direction. The integral of radiance over area and solid angle is power. A great deal of confusion concerns the use and misuse of the term intensity. Some folks use it for W/sr, some use it for W/m2 and others use it for W/m2-sr. It is quite clearly defined in the SI system, in the definition of the base unit of luminous intensity, the candela. Some attempt to justify alternate uses by adding adjectives like optical (used for W/m2) or specific (used for W/m2-sr), but this practice only adds to the confusion. The underlying concept is (quantity per unit solid angle). For an extended discussion, see my paper entitled “Getting Intense on Intensity” for Metrologia (official journal of the BIPM) and a letter to OSA's “Optics and Photonics News”, with a modified version available on the web at www.optics.Arizona.EDU/Palmer/intenopn.html. Photon quantities are also common. They are related to the radiometric quantities by the relationship Qp = hc/λ where Qp is the energy of a photon at wavelength λ, h is Planck's constant and c is the velocity of light. At a wavelength of 1 µm, there are approximately 5×1018 photons per second in a watt. Conversely, a single photon has an energy of 2×10–19 joules (W s) at 1 µm. Common units include s–1-m–2-sr–1 for photon radiance. 5. How should I represent spectral quantities? Most sources of optical radiation are spectrally dependent, and the terms radiance, intensity, etc. give no information about the distribution of these quantities over wavelength. Spectral quantities, like spectral radiance, spectral power, etc. are RADIOMETRY &. PHOTOMETRY FAQ 5 defined as the quotient of the quantity in an infinitesimal range of wavelength divided by that wavelength range. In other words, spectral quantities are derivative quantities, per unit wavelength, and have an additional (λ–1) in their units. When integrated over wavelength they yield the total quantity. These spectral quantities are denoted by using a subscript λ, e.g., Lλ, Eλ, Φλ,, and Iλ. Some other quantities (examples include spectral transmittance, spectral reflectance, spectral responsivity, etc.) vary with wavelength but are not used as derivative quantities. These quantities should not be integrated over wavelength; they are only weighting functions, to be included with the above derivative quantities. To distinguish them from the derivative quantities, they are denoted by a parenthetical wavelength, i.e. ℜ(λ) or τ(λ). 6. What are the quantities and units used in photometry? They are basically the same as the radiometric units except that they are weighted for the spectral response of the human eye and have funny names. A few additional units have been introduced to deal with the amount of light reflected from diffuse (matte) surfaces. The symbols used are identical to those radiometric units, except that a subscript “v “ is added to denote “visual. “The following chart compares them. QUANTITY RADIOMETRIC PHOTOMETRIC power watt (W) lumen (lm) power per unit area W/m2 lm/m2 = lux (lx) power per unit solid angle W/sr lm/sr = candela (cd) power per unit area per unit solid angle W/m2-sr lm/m2-sr = cd/m2 = nit Now we can get more specific about the details. Candela (unit of luminous intensity). The candela is one of the seven base units of the SI system. It is defined as follows: The candela is the luminous intensity, in a given direction, of a source that emits monochromatic radiation of frequency 540×1012 hertz and that has a radiant intensity in that direction of 1/683 watt per steradian. The candela is abbreviated as cd and its symbol is Iv. The above definition was adopted by the 16th CGPM in 1979. The candela was formerly defined as the luminous intensity, in the perpendicular direction, of a surface of 1/600 000 square metre of a black body at the temperature of freezing platinum under a pressure of 101 325 newtons per square metre. This earlier definition was initially adopted in 1946 and later modified by the 13th CGPM (1967). It was abrogated in 1979 and replaced by the current definition. RADIOMETRY &. PHOTOMETRY FAQ 6 The current definition was adopted because of several reasons. First, the freezing point of platinum (≈2042K) was tied to another base unit, the kelvin. If the best estimate of this point were changed, it would then impact the candela. The uncertainty of the thermodynamic temperature of this fixed point created an unacceptable uncertainty in the value of the candela. Second, the realization of the Pt blackbody was extraordinarily difficult; only a few were ever built. Third, if the temperature were slightly off, possibly because of temperature gradients or contamination, the freezing point might change or the temperature of the cavity might differ. The sensitivity of the candela to a slight change in temperature is significant. At a wavelength 555 nm, a change in temperature of only 1K results in a luminance change approaching 1%. Fourth, the relative spectral radiance of blackbody radiation changes drastically (some three orders of magnitude) over the visible range. Finally, recent advances in radiometry offered new possibilities for the realization of the candela. The value 683 lm/W was selected based upon the best measurements with existing platinum freezing point blackbodies and the best estimates for the thermodynamic temperature of the platinum point. . It has varied over time from 620 to nearly 700 lm/W, depending largely upon the assigned value of the Pt freezing point on the best available temperature scale. The value of 1/600 000 square metre was chosen to maintain consistency with prior flame standards. Also note that in the definition there is no specification for the spatial distribution of intensity. Luminous intensity, while often associated with an isotropic (see section 7 below) point source, is a valid specification for characterizing highly directional light sources such as automotive lighting, searchlights and LEDs. One other issue before we press on. Since the candela is now defined in terms of other SI derived quantities, there is really no need to retain it as an SI base quantity. It remains so for reasons of history and continuity. Lumen (unit of luminous flux). The lumen is an SI derived unit for luminous flux. The abbreviation is lm and the symbol is Φv. The lumen is derived from the candela and is the luminous flux emitted into unit solid angle (1 sr) by an isotropic (see section 7 below) point source having a luminous intensity of 1 candela. The lumen is the product of luminous intensity and solid angle, cd-sr. It is analogous to the unit of radiant flux (watt), differing only in the eye response weighting. If a light source is isotropic, the relationship between lumens and candelas is 1 cd = 4π lm. In other words, an isotropic source having a luminous intensity of 1 candela emits 4π lumens into space, which just happens to be 4π steradians. We can also state that 1 cd = 1 lm/sr, analogous to the equivalent radiometric definition. If a source is not isotropic, the relationship between candelas and lumens is empirical. A fundamental method used to determine the total flux (lumens) is to measure the luminous intensity (candelas) in many directions using a goniophotometer, and then numerically integrate over the entire sphere. Later on, RADIOMETRY &. PHOTOMETRY FAQ 7 we can use this “calibrated” lamp as a reference in an integrating sphere for routine measurements of luminous flux. Lumens are what we get from the hardware store when we purchase a light bulb. We want a high number of lumens with a minimum of power consumption and a reasonable lifetime. Projection devices are also characterized by lumens to indicate how much luminous flux they can deliver to a screen. The ANSI standard delineates the method for determining average luminous flux; it is not a different lumen. Recently, marketeers for digital projectors have introduced a new unit, the “market comparable lumen,’ which is somewhere between 1.1 and 1.25 ANSI lumens. This appears to be a ploy to jack up the numbers to compete with those who are not inclined to use ANSI lumens. Lux (unit of luminous flux density, or illuminance). Illuminance is another SI derived unit which denotes luminous flux density . The unit has a special name, 2 the lux, which is lumens per square metre, or lm/m . The symbol is Ev. Most light meters measure this quantity, as it is of great importance in illuminating engineering. The IESNA Lighting Handbook has some sixteen pages of recommended illuminances for various activities and locales, ranging from morgues to museums. Typical values range from 100 000 lx for direct sunlight to 20-50 lx for hospital corridors at night. Nit (unit of luminance). Luminance should probably be included on the official list of derived SI units, but is not. It is analogous to radiance, differentiating the lumen with respect to both area and direction. This unit also has a special name, the nit, 2 2 which is cd/m or lm/m sr if you prefer. The symbol is Lv. It is most often used to characterize the “brightness “ of flat emitting or reflecting surfaces. A typical use would be the luminance of your laptop computer screen. They have between 100 and 250 nits, and the sunlight readable ones have more than 1000 nits. Typical CRT monitors have between 50 and 125 nits. Other photometric units We have other photometric units (boy, do we have some strange ones). Photometric quantities should be reported in SI units as given above. However, the literature is filled with now obsolete terminology and we must be able to interpret it. So here are a few terms that have been used in the past. RADIOMETRY &. PHOTOMETRY FAQ 8 Illuminance: 1 metre-candle = 1 lux 1 phot = 1 lm/cm2 = 104 lux 1 foot-candle = 1 lumen/ft2 = 10.76 lux 1 milliphot = 10 lux Luminance: Here we have two classes of units. The first is conventional, easily related to the SI unit, the cd/m2 (nit). 1 stilb = 1 cd/cm2 = 104 cd/m2 = 104 nit 1 cd/ft2 = 10.76 cd/m2 = 10.76 nit The second class was designed to “simplify” characterization of light reflected from diffuse surfaces by including in the definitions the concept of a perfect (reflectance ρ = 1), diffuse reflector (lambertian, see section 7 below.) If one unit of illuminance falls upon this hypothetical reflector, then 1 unit of luminance is reflected. The perfect diffuse reflector emits 1/π units of luminance per unit illuminance. If the reflectance is ρ, then the luminance is ρ times the illuminance. Consequently, these units all have a factor of (1/π) built in. Photometric quantities are already the result of integration over wavelength. It therefore makes no sense to speak of spectral luminance or the like. 7. What is the difference between lambertian and isotropic? Both terms mean “the same in all directions” and are unfortunately sometimes used interchangeably. Isotropic implies a spherical source that radiates the same in all directions, i.e., the intensity (W/sr) is the same in all directions. We often encounter the phrase “isotropic point source.” There can be no such thing; because the energy density would have to be infinite. But a small, uniform sphere comes very close. The best example is a globular tungsten lamp with a milky white diffuse envelope, as used in dressing room lighting. From our vantage point, a distant star can be considered an isotropic point source. Lambertian refers to a flat radiating surface. It can be an active surface or a passive, reflective surface. Here the intensity falls off as the cosine of the observation angle with respect to the surface normal (Lambert's law). The radiance (W/m2-sr) is independent of direction. A good example is a surface painted with a RADIOMETRY &. PHOTOMETRY FAQ 9 good “matte” or “flat” white paint. If it is uniformly illuminated, like from the sun, it appears equally bright from whatever direction you view it. Note that the flat radiating surface can be an elemental area of a curved surface. The ratio of the radiant exitance (W/m2) to the radiance (W/m2-sr) of a lambertian surface is a factor of π and not 2π. We integrate radiance over a hemisphere, and find that the presence of the factor of cos(θ) in the definition of radiance gives us this interesting result. It is not intuitive, as we know that there are 2π steradians in a hemisphere. A lambertian sphere illuminated by a distant point source will display the maximum radiance at the surface where the local normal coincides with the incoming beam. The radiance will fall off with a cosine dependence to zero at the edge of the lit area (terminator.) If the intensity (integrated radiance over area) is unity when viewing from the source, then the intensity when viewing from the side is 1/π. Think about this and consider whether or not our Moon is lambertian. Take a look at www.optics.Arizona.EDU/Palmer/moon/lunacy.htm. 8. Where do the properties of the eye get involved? We know that the eye does not see all wavelengths equally. The eye has two general classes of photosensors, cones and rods. Cones: The cones are responsible for light-adapted vision; they respond to color and have high resolution in the central foveal region. The light-adapted relative spectral response of the eye is called the spectral luminous efficiency function for photopic vision, V(λ). This empirical curve, first adopted by the International Commission on Illumination (CIE) in 1924, has a peak of unity at 555 nm, and decreases to levels below 10–5 at about 370 and 785 nm. The 50% points are near 510 nm and 610 nm, indicating that the curve is slightly skewed. The V(λ) curve looks very much like a Gaussian function; in fact a Gaussian curve can easily be fit and is a good representation under some circumstances. I used a non-linear regression technique to obtain the following equation: −−2854.(λ 0559 . )2 Vebλg ≅ 1019. More recent measurements have shown that the 1924 curve may not best represent typical human vision. It appears to underestimate the response at wavelengths shorter than 460 nm. Judd (1951), Vos (1978) and Stockman and Sharpe (1999) have made incremental advances in our knowledge of the photopic response. Go to www.cvrl.org for details. Rods: The rods are responsible for dark-adapted vision, with no color information and poor resolution when compared to the foveal cones. The dark-adapted relative spectral response of the eye is called the spectral luminous efficiency function for scotopic vision, V’(λ). This is another empirical curve, adopted by the CIE in 1951. RADIOMETRY &. PHOTOMETRY FAQ 10 It is defined between 380 nm and 780 nm. The V’(λ) curve has a peak of unity at 507 nm, and decreases to levels below 10–3 at about 380 and 645 nm. The 50% points are near 455 nm and 550 nm. This scotopic curve can also be fit with a Gaussian, although the fit is not quite as good as the photopic curve. My best fit is −−321.( 9λ 0 . 503 )2 Ve'.bλg ≅ 0992 Photopic (light adapted cone) vision is active for luminances greater than 3 cd/m2. Scotopic (dark-adapted rod) vision is active for luminances lower than 0.01 cd/m2. In between, both rods and cones contribute in varying amounts, and in this range the vision is called mesopic. There are currently efforts under way to characterize the composite spectral response in the mesopic range for vision research at intermediate luminance levels. The Color Vision Research Lab (www.cvrl.org) has an impressive collection of the data files, including V(λ) and V’(λ), that you need to do this kind of work. 9. How do I convert between radiometric and photometric units? We know from the definition of the candela that there are 683 lumens per watt at a frequency of 540THz, which is 555 nm (in vacuum or air). This is the wavelength that corresponds to the maximum spectral responsivity of the human eye. The conversion from watts to lumens at any other wavelength involves the product of the power (watts) and the V(λ) value at the wavelength of interest. As an example, we can compare laser pointers at 670 nm and 635 nm. At 670 nm, V(λ) is 0.032 and RADIOMETRY &. PHOTOMETRY FAQ 11 a 5 mW laser has 0.005W×0.032×683 lm/W = 0.11 lumens. At 635 nm, V(λ) is 0.217 and a 5 mW laser has 0.005W×0.217×683 lm/W = 0.74 lumens. The shorter wavelength (635 nm) laser pointer will create a spot that is almost 7 times as bright as the longer wavelength (670 nm) laser (assuming the same beam diameter). In order to convert a source with non-monochromatic spectral distribution to a luminous quantity, the situation is decidedly more complex. We must know the spectral nature of the source, because it is used in an equation of the form: ∞ XKXVdvm= λ bλλg z0 where Xv is a luminous term, Xλ is the corresponding spectral radiant term, and V(λ) is the photopic spectral luminous efficiency function. For X, we can pair luminous flux (lm) and spectral power (W/nm), luminous intensity (cd) and spectral radiant intensity (W/sr-nm), illuminance (lx) and spectral irradiance (W/m2-nm), or luminance (cd/m2) and spectral radiance (W/m2-sr-nm). This equation represents a weighting, wavelength by wavelength, of the radiant spectral term by the visual response at that wavelength. The constant Km is a scaling factor, the maximum spectral luminous efficiency for photopic vision, 683 lm/W. The wavelength limits can be set to restrict the integration to only those wavelengths where the product of the spectral term Xλ and V(λ) is non-zero. Practically, this means we only need integrate from 360 to 830 nm, limits specified by the CIE V(λ) table. Since this V(λ) function is defined by a table of empirical values, it is best to do the integration numerically. Use of the Gaussian equation given above is only an approximation. I compared the Gaussian equation with the tabulated data using blackbody curves and found the differences to be less than 1% for temperatures between 1500K and 20000K. This result is acceptable for smooth curves, but don’t try it for narrow wavelength sources, like LEDs. There is nothing in the SI definitions of the base or derived units concerning the eye response, so we have some flexibility in the choice of the weighting function. We can use a different spectral luminous efficacy curve, perhaps one of the newer ones. We can also make use of the equivalent curve for scotopic (dark-adapted) vision for studies at lower light levels. This V'(λ) curve has its own constant K'm, the maximum spectral luminous efficiency for scotopic vision. K'm is 1700 lm/W at the peak wavelength for scotopic vision (507 nm) and this value was deliberately chosen such that the absolute value of the scotopic curve at 555 nm coincides with the photopic curve, at the value 683 lm/W. Some workers are referring to “scotopic lumens”, a term which should be discouraged because of the potential for misunderstanding. In the future, we can also expect to see spectral weighting to represent the mesopic region. The International Commission on Weights and Measures (CGPM) has approved the use of the CIE V(λ) and V'(λ) curves for determination of the value of photometric quantities of luminous sources. RADIOMETRY &. PHOTOMETRY FAQ 12 Now about converting from lumens to watts. The conversion from watts to lumens that we saw just above required that the spectral function Xλ of the radiation be known over the spectral range from 360 to 830 nm, where V(λ) is non-zero. Attempts to go in the other direction, from lumens to watts, are far more difficult. Since we are trying to back out a quantity that was weighted and placed inside of an integral, we must know the spectral function Xλ of the radiation over the entire spectral range where the source emits, not just the visible region. Also please note that there is no such thing as a spectral lumen, spectral lux, spectral nit, spectral candela, etc. The photometric terms are all the result of integration over wavelength; they are never subject to integration. 10. Where can I learn more about this stuff? Books, significant journal articles: DeCusatis, C., “Handbook of Applied Photometry” AIP Press (1997). Authoritative, with pertinent chapters written by technical experts at BIPM, CIE and NIST. Skip chapter 4! Rea, M., ed. “Lighting Handbook: Reference and Application” 8th edition, Illuminating Engineering Society of North America (1993). “Symbols, Units and Nomenclature in Physics” International Union of Pure and Applied Physics (1987). “American National Standard Nomenclature and Definitions for Illuminating Engineering” ANSI Standard ANSI/IESNA RP-16 96. “The Basis of Physical Photometry” CIE Technical Report 18.2 (1983) Publications available on the World Wide Web All you ever wanted to know about the SI is contained at BIPM and at NIST. Available publications (highly recommended) include: “The International System of Units (SI)” 7th edition (1998), direct from BIPM. The official document is in French; this is the English translation). Available on-line at www.bipm.fr/pdf/si-brochure.pdf. NIST Special Publication SP330 “The International System of Units (SI)” The US edition of the above BIPM publication. Available on-line at physics.nist.gov/Pubs/SP330/sp330.pdf. NIST Special Publication SP811 “Guide for the Use of the International System of Units (SI)” Available on-line at physics.nist.gov/Document/sp811.pdf. RADIOMETRY &. PHOTOMETRY FAQ 13 Papers published in recent issues of the NIST Journal of Research are also available on-line at nvl.nist.gov/pub/nistpubs/jres/jres.htm. Of particular relevance is “The NIST Detector-Based Luminous Intensity Scale,” Vol. 101, page 109 (1996). Useful Web sites BIPM International Bureau of Weights and Measures - www.bipm.fr/ NIST Nat'l Inst. of Standards and Technology - physics.nist.gov/cuu/ ISO International Standards Organization - www.iso.ch/ ANSI American Nat'l Standards Institute - www.ansi.org/ CIE International Commission on Illumination - www.cie.co.at/cie/ IESNA Illuminating Engineering Society of North America - www.iesna.org/ IUPAP International Union of Pure and Applied Physics - www.physics.umanitoba.ca/iupap/ Color Vision Research Lab – www.cvrl.org AIP American Institute of Physics - www.aip.org SPIE - International Society for Optical Engineering - www.spie.org OSA Optical Society of America - www.osa.org CORM Council of Optical Radiation Measurements - www.corm.org RADIOMETRY &. PHOTOMETRY FAQ 14

Get CBSE class 12 Chemistry practice paper 2017 for the coming CBSE board examination 2017. This practice paper is as per latest CBSE examination pattern of class 12 Chemistry. All the questions of this practice paper are based on latest CBSE syllabus. CBSE board exams 2017 are about to start from 1st March 2017 and every student of CBSE class 12 want to score maximum marks in board examinations. CBSE result 2017 of class 12 students will depend on their score in core subjects. This practice paper is developed by subject experts after the brief analysis of CBSE class 12 Chemistry question paper 2016. Some important questions from previous year CBSE papers are also included in this practice paper. The complete practice paper is given below: Time allowed: 3 hours Maximum Marks: 70 (i) All questions are compulsory. (ii) Questions number 1to 5 are very short answer questions and carry 1 mark each. (iii) Questions number 6 to 10 are very short answer question and carry 2 marks each. (iv) Questions number 11 to 22 are also short answer question and carry 3 marks each. (v) Question number 23 is a value based question and carry 4 marks. (vi) Questions number 24 to 26 are long answer question and carry 5 marks each. (vii) Use log tables, if necessary. Use of calculators is not allowed. Question1: Write the structure of an isomer of compound C4H9Br which is most reactive towards SN1 reaction. Question2: Why CuSO4. 5H2O is blue in colour while CuSO4 is colourless? Question3: NO2 group show its effect only at ortho- and para- positions and not at meta- position. Explain. Question4: Write IUPAC name of the given compound: Question5: Explain the causes of Brownian movement in a colloidal solution? Question6: The formula of a coordination compound is CoCl3.4NH3. It does not liberate ammonia but precipitates chloride ions as silver chloride. (i) Write the structural formula of the complex? (ii) What is the IUPAC name of the complex? Question7: Explain why alkylamine more basic than ammonia. Question8: Write mathematical expression of Henry’s law correlating the pressure of a gas and its solubility in a solvent and mention one applications of the law. Question9: Out of the following two coordination entities which is chiral (optically active)? (a) cis-[CrCl2(ox)2]3‒ (b) trans-[CrCl2(ox)2]3‒ Question10: Draw the structures of the following molecules: Question10: Complete the following equations: (i) HgCl2 + PH3 →? (ii) XeF4 + H2O →? Question11: Name the reagents which are used in the following conversions: (i) A primary alcohol to an aldehyde (ii) Butan-2-one to butan-2-ol (iii) Phenol to 2, 4, 6-tribromophenol Question12: Account for the following observations: (i) pKb for aniline is more than that for methylamine. (ii) Methylamine solution in water reacts with ferric chloride solution to produce a precipitate of ferric hydroxide. (iii) Aniline does not undergo Friedel-Crafts reaction. Question13: Define the following in relation to proteins : (i) Primary structure (iii) Peptide linkage Question14: What do you understand by flux? What is its role in the metallurgy of iron and copper? Question15: Give reasons for the following: (i) Nickel does not form low spin octahedral complexes. (ii) The p-complexes are known for the transition metals only. (iii) Co2+ is easily oxidised to Co3+ in the presence of a strong ligand. Question16: An element has a body-centred cubic (bcc) structure with a cell edge of 288 pm. The density of the element is 7.2 g/cm3. How many atoms are present in 208 g of the element? Question17: (a) What is the significance of number 6 and 6, 6 in the polymer names nylon-6 and nylon-6, 6? (b) What is the difference between Buna-N and Buna-S? (c) Arrange the following polymers in increasing order of their intermolecular forces Neoprene, Nylon 6, Polyvinyl chloride Question18: In a pseudo first order hydrolysis of an ester in water, the following results were observed (i) Find the average rate of reaction between the time intervals 30 to 60 seconds. (ii) Find the pseudo first order rate constant for the hydrolysis of ester. Question19: (i) Why alkylamine is more basic than ammonia? (ii) What are azo dyes? (iii) What do you understand by coupling reaction? Question20: What is Corrosion? What are the chemical reactions involved in the formation of corrosion? Question21: Calculate the boiling point of a 1M aqueous solution (density 1.04 g mL) of Potassium chloride (Kb for water = 0.52 K kg mol−1. Atomic masses: K = 39 u. C1 = 39.9 u). Assume, Potassium chloride is completely dissociated in solution. Question21: Find the depression in the freezing point of water when 10 g of CH3CH2CHClCOOH is added to 250 gram of water. Kα = 1.4 × 10 ‒ 3, Kf = 1.86 K kg mol‒1. Question22: Define the following terms: (b) Tyndall Effect Question23: Observing the growing cases of diabetes and depression among young children, Mr. Amit, the principal of one reputed school organized a seminar in which he invited parents and principals. They all resolved this issue by strictly banning junk food in school s and introducing healthy snacks and drinks like soup, lassi, milk etc in school canteens. They also decided to make compulsory half an hour of daily physical activities for the students in the morning assembly. After six months, Mr. Amit conducted the health survey in most of the schools and discovered a tremendous improvement in the heath of students. After reading the above passage answer the following question: (i) What are the values displayed by Mr. Amit? (ii) As a student, how can you spread awareness about this issue? (iii) What are the anti-depressant drugs? Give an example. (iv) Name the sweetening agent used in the preparation of sweets for a diabetic patient. Question24: (a) Define molar conductivity of a substance and also write the relationship between molar conductivity and specific conductivity. (b) A voltaic cell is set up at 25o C with the following half cells: Ag+ (0.001 M) | Ag and Cu2+ (0.10 M) | Cu What would be the voltage of this cell? (Eo cell = 0.46 V) Question24: (a) Describe how for weak and strong electrolytes, molar conductivity changes with concentration of solute. How is such change explained? (b) Conductivity of 0.00241 M acetic acid is 7.896 × 10‒5 S cm‒1. Find its molar conductivity and if Λo for acetic acid is 390.5 S cm2 mol‒1, what is the dissociation constant? Question25: (a) How are the following obtained? (i) Benzoic acid from ethyl benzene. (ii) Benzaldehyde from toluene. (b) Complete each synthesis by giving the missing material, reagent or products: Question25: (a) Name two commonly used methods to convert >C=O group into >CH2 group (b) Name the products formed when ethanal is heated with hydrogen iodide and red phosphorous under high pressure. (c) Name the acid present is vinegar. (d) Identify the compounds X, Y and Z in the following reaction. Question26: (a) What may be the possible oxidation states of the transition metals with the following d electronic configurations in the ground state of their atoms: 3d 3 4s 2, 3d5 4s 2 & 3d 6 4s 2. Indicate relative stability of oxidation states in each case. (b) Write steps involved in the preparation of (i) Na2CrO4 from chromite ore and (ii) K2MnO4 from pyrolusite ore. Question26: Assign reasons for the following (a) The transition metals and many of their compounds act as good catalysts. (b) The enthalpies of atomisation of transition elements are high. (c) From element to element the actinoid contraction is greater than the lanthanoid contraction. (d) The Eo value for Mn3+/Mn2+ couple is much more positive than that of Cr3+/Cr2+ (e) Scandium (Z = 21) does not exhibit variable oxidation states and yet it is regarded as a transition element.

If you think you have received a fake HP Support message, please report it to us by clicking on the blue “Report Inappropriate Content” button above the message. Brand new OEM would define it as coming from HP. This is an international forum, after all. Re-install the plastic rear output bin. check over here Push the fuser firmly--listen for the sound of both blue levers clicking into place. Push the fuser firmly--listen for the sound of both blue levers clicking into place. Some pictures derived from HP User and Service guides. No more fusion problems. http://www.printertechs.com/printer-troubleshooting/367-50-5-error-on-laserjet-4200-4240-4250-4300-4350 Please try the request again. by dmzcompute on Jan 2, 2007 at 3:28pm Add comment Please sign in to comment 0 thank you, i will order the parts and report back. The only other explanation might be that the connectors aren't mating properly. The new fuser specifically has a "4200" sticker on it along with the part number: PM1-0013 / 110V. by rottielover on Nov 12, 2007 at 2:44pm Add comment Please sign in to comment 0 A 4300 fuser will work in a 4345, so you should be able to check the 4345 heating element using this diagram: http://www.partsnow.com/service_today/4300.asp by Stephen on Nov 13, 2007 at 5:30pm Add comment Please sign in to comment 0 Brand new always confuses me. Join the community of 500,000 technology professionals and ask your questions. Hp 4250 Fuser So I took off the top cover and then left side cover to expose the cooling fan. Glad I could help. 50.4 Fuser Error by unknown on May 29, 2009 at 7:29am Add comment Please sign in to comment 0 There is no fuser reset. Clean or straighten the pins if necessary. This problem was (and remains) notorious with LJ 4200/ 4300/ 4250 / 4350 printers where the fusers look the same but are different in detail. by moe on Jun 1, 2009 at 10:18am Add comment Please sign in to comment 0 Thank you, I appricate it. I had this problem with a 4200 changed the fuser it's now working ok. - Anonymous 0 Next item would be the dc controller, followed by the power supply. 50.2 Fuser Error Laserjet Pro 400 printertechs Just the Index page again. If you have another 4200 fuser even if the teflon is bad just try anyway to see if the message goes away. - dmzcompute Am still having the same error,so what else can I do to resolve the problem? - Anonymous can you send vido that you maintain this fuser error or to change this fuser.thanks - Anonymous 0 Ohm out the fusers.http://www.partsnow.com/service_today/4200.aspLikely you have two bad ones. Since the error is a low fuser temp, you could try to jack up the fuser temp from normal to high 1 for all types of paper and see if maybe it will stay hot enough. by unknown on Mar 19, 2008 at 8:13am Add comment Please sign in to comment 0 Since we have no idea what you did before, we can't advise you. PrinterTechs.com 608-831-2396 or 866-352-7108 About Us Contact Us Search Products Search Support All Products PrintersHewlett PackardLexmarkCables and Misc.JetDirect CardsBrotherKyoceraDell Toner Maintenance Kits Printers Parts Tech Support Account Login Shopping Cart My Account Checkout Home Troubleshooting 50.5 Fuser Error on LaserJet 4200 4240 4250 4300 4350 Email 50.5 Fuser Error on LaserJet 4200 4240 4250 4300 4350 If you see this error code 50.5 on an HP LaserJet 4200, 4240, 4250, 4300, or 4350 series printer, the fuser installed in the printer is registering as an incorrect fuser for that model printer. Hp Laserjet 50.2 Fuser Error We do try to help; but there is a substantial cost to us in dealing with the problem so there will be a re-stocking charge. 50.5 Fuser Error Hp 4015 If you have any questions give us a call toll-free 866-352-7108 or local 608-831-2396, we can help you get the right fuser for your specific model. by Anonymous on Nov 12, 2007 at 10:09am Add comment Please sign in to comment 0 Anonymous asked: "We are trying to help you solve your issue. check my blog part# RM1-0013 purchase HP 4200 fuser/maintenance kit here for use in HP LaserJet 4200, HP LaserJet 4200N, HP LaserJet 4200TN, HP LaserJet 4200DTN, HP LaserJet 4200DTNs, HP LaserJet 4200DTNsl part# RM1-1082 purchase HP 4240, 4250, 4350 fuser/maintenance kit here for use in HP LaserJet 4240, HP LaserJet 4240N, HP LaserJet 4250, HP LaserJet 4250N, HP LaserJet 4250DN, HP LaserJet 4250TN, HP LaserJet 4250DTN, HP LaserJet 4250DTNsl part# RM1-0101 purchase HP 4300 fuser/maintenance kit here for use in HP LaserJet 4300, HP LaserJet 4300N, HP LaserJet 4300DN, HP LaserJet 4300TN, HP LaserJet 4300DTN, HP LaserJet 4300DTNs, HP LaserJet 4300DTNsl Removal instructions: How to remove the fuser: Turn the printer off and unplug the power cord. Replace power supply 50.5 Fuser Error Description: incorrect fuser is installe Solution: 1. I'm hoping it's just a manufacturing defect on the "new" fuser I ordered from before. Hp 4200 Fuser Unit has been sitting un-plugged. I found on the powerboard a burnt fuse - the larger of the two (125v 15A). by unknown on May 31, 2009 at 6:53pm Add comment Please sign in to comment 0 My bad, I just naturally assumed that since you said it was a 220V fuser, that you were from out of the country. this content Replace power supply 50.2 Fuser Error Description: fuser warm up service,- Fusing Assembly is taking too long to reach proper temperature. Solution: 1. This is how video conferencing should work! Hp 4240 Maintenance Kit While grasping onto both sides of the fuser, push the blue levers upward and pull the fuser straight out. by moe on Nov 8, 2007 at 9:24am Add comment Please sign in to comment 0 wow great trouble shooting advice turns out my 50.2 fuser error was due to having the wrong input voltage. Otherwise you may actually have a defective new fuser. There are probably very few 5SI and CLJ-4500 printers left in service - you cant get the parts. by nivek on Jun 1, 2009 at 10:31am Add comment Please sign in to comment 0 I tore apart a 4250 after having a 50.2 fuser error. Fuser Error Hp Printer Bookmark the permalink. Wird geladen... by sstuart on Dec 26, 2006 at 11:29am Add comment Please sign in to comment I have seen alreasy about a dozen printers where the wrong fuser has been installed. If the part number on the fuser is the exact same as the one you removed, try powering off the printer and unplugging it from the power source for a few minutes. http://sysreview.com/fuser-error/hp-50-4-fuser-error.html The system returned: (22) Invalid argument The remote host or network may be down. backpage.com completely irrelevant - it must be triggered by an ad that used to be on the page. Leave a Reply Cancel reply Your email address will not be published. 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Get 1:1 Help Now Advertise Here Enjoyed your answer? Autoplay Wenn Autoplay aktiviert ist, wird die Wiedergabe automatisch mit einem der aktuellen Videovorschläge fortgesetzt. The argument that "Marks and Spencers just take things back" isn't helpful, they make a huge margin on clothes and anyway that policy has changed. Replace power supply 50.3 Fuser Error Description: Fuser over temperature - Fusing Assembly is exceeding proper temperature. (Too Hot) Solution: 1. Wird verarbeitet...

Presentation on theme: "EMF. E.m.f and p.d - Learning Outcomes You should all be able to: define potential difference (p.d.); select and use the equation W = VQ; define the volt;"— Presentation transcript: E.m.f and p.d - Learning Outcomes You should all be able to: define potential difference (p.d.); select and use the equation W = VQ; define the volt; describe how a voltmeter may be used to determine the p.d. across a component; define electromotive force (e.m.f.) of a source such as a cell or a power supply; describe the difference between e.m.f. and p.d. in terms of energy transfer Electric Potential (a bit like Gravitational Potential) To give the ball more gravitational ‘potential’ energy work must be done on it. Now it is ‘higher’ it has more ‘potential’. This is a bit like electrical ‘potential’. The position of a charge (rather than in a ball) in an electric field (rather than a gravitational field) determines its potential. Electric Potential (a bit like Gravitational Potential) The direction of current flow is conventionally taken to be from points of higher potential (i.e. the top of a hill) to a point of lower potential. Note: Conventional current flows from +ve to –ve, but the electrons (that are –ve) flow to the positive terminal. If the work done in causing one coulomb of electric charge to flow between two points is one joule, then the PD between the points is one volt (i.e. 1V = 1 JC -1 ) It follows that W = QV Where W = the work done (J) Q = the charge (C) V = the potential difference (V) e.m.f is the energy transferred per unit charge when one type of energy is converted into electrical energy. Potential difference is the energy transferred per unit charge when electrical energy is transferred into another form of energy. Key Definitions Potential difference 12V What is the voltage across each resistor? 6V What happens to the charge as it flows through the resistor? Electrical energy is transformed into another energy (heat in this case) Resistance - Learning Outcomes You should all be able to: define resistance; select and use the equation for resistance (V = I.R) define the ohm; state and use Ohm’s law; describe the I–V characteristics of a resistor at constant temperature, filament lamp and light-emitting diode (LED); describe an experiment to obtain the I–V characteristics of a resistor at constant temperature, filament lamp and light-emitting diode (LED); describe the uses and benefits of using light emitting diodes (LEDs). AV bit of wire http://phet.colorado.edu/en/simulation/circuit-construction-kit-dc How to measure resistance Ohm’s Law Resistance = Potential Difference (volts) (ohms) Current (amperes) V is Potential Difference measured in voltage, V I is current measured in Amps, A R is resistance measured in Ohms, Ω R = V I V = I R Ohm’s Law The current through a resistor at a constant temperature is directly proportional to the potential difference across the resistor. This means if you double the current you double the voltage over a component. It also means that the resistance of the component does not change when you put more current through it. Candidates should be able to: define resistivity of a material; select and use the resistivity equation ; describe how the resistivities of metals and semiconductors are affected by temperature; describe how the resistance of a pure metal wire and of a negative temperature coefficient (NTC) thermistor is affected by temperature. Resistivity The resistivity ρ of a wire of length l, resistance R and cross sectional area A is given by: ρ = RA l Candidates should be able to: describe power as the rate of energy transfer; select and use power equations P = VI, P = I 2 R and P = V 2 / R explain how a fuse works as a safety device determine the correct fuse for an electrical device; select and use the equation W = IVt; define the kilowatt-hour (kW h) as a unit of energy; calculate energy in kW h and the cost of this energy when solving problems. Power Power (watts) = Energy Transformed (joules) Time (seconds) This means the more powerful something is, the more energy is transferred every second. Power Power (watts) = Energy Transformed (joules) Time (seconds) For example: If a bulb transforms 300 J of electrical energy into light in 3 second, the power is: P = Energy Transformed ÷Time P = 300 (J) ÷ 3 (s) P = 100 W Power in a circuit Power = Current x Potential Difference (watts, W) = (ampere, A) x (volts, V) P = I x V For example: If a bulb has a p.d. across it of 3.0V, and a current flowing through it of 2.0A then the power is: P = I x V P = 2.0 (A) x 3.0 (V) P = 6.0 W Charge in a circuit Charge = Current x Time (coulomb, C) = (ampere, A) x (seconds, s) For example: How much charge flows if a current of 2.0 A flows for 60 seconds? Charge = Current x Time Charge = 2.0 (A) x 60 (s) Charge = 120 C Energy in a circuit E nergy Transformed = Potential Difference × Charge (joules, J) = (volts, V) x (coulomb, C) For example: How much energy is transformed when the p.d. is 30V and the charge is 2.0 C? Energy = Potential Difference × Charge Energy = 30 (V) x 2.0 (C) Energy = 60 J SOUND WAVES Why do car headlights dim when you turn on the ignition? What has this got to do with internal resistance? Lesson Objectives To understand a method by which internal resistance may be calculated. To analyse and evaluate quantitative data. To develop your scientific approach to carrying out practical work. Task – To calculate the internal resistance of a power source Physics by Robert Hutchings (the white book) pages 250 – 251 may help. Use the sheet to help you get started. Think about your practical skills. Candidates should be able to: (a) state Kirchhoff’s second law and appreciate that this is a consequence of conservation of energy; (b) apply Kirchhoff’s first and second laws to circuits; (c) select and use the equation for the total resistance of two or more resistors in series; (d) select and use the equation for the total resistance of two or more resistors in parallel; (e) solve circuit problems involving series and parallel circuits with one or more sources of e.m.f.; (f) explain that all sources of e.m.f. have an internal resistance; (g) explain the meaning of the term terminal p.d.; (h) select and use the equations e.m.f. = I (R + r) = V + Ir. Kichhoff’s Second Law In any closed loop in a circuit the sum of the e.m.f.s is equal to the sum of the p.d.s.

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A permutation calculator is a handy tool which helps you find out how many permutations a given set has. Permutation is often symbolized as nPr which we will also use in this article. Let’s cover how to use this permutation formula calculator, how to calculate permutation without an nPr calculator, and more. How to use the permutation calculator? This permutation calculator is simple, easy to understand, and extremely easy to use too. All you have to do is input 2 values and it will do the calculation for you. This online tool is just as useful as a permutations and combinations calculator which provides you with both the permutations and the combinations of a given set. Here are the steps to follow for this calculator: - First, enter the value of the Number of Objects (n) in the appropriate field. - Then enter the value of the Sample Size (r) in the appropriate field. - After entering both values, the permutation formula calculator automatically generates the value of the Permutations of the set for you. How do you calculate a permutation? Permutation or nPr refers to the number of ways by which you can select r elements out of any given set containing n distinct objects. Unlike combinations, when it comes to permutations, the order of selecting the elements has relevance. For instance, let’s assume that you have a whole deck of cards which have numbers from 1-9. Select 3 cards from the deck randomly then place them on a table in a line to create a number with 3-digits. For these three cards, how many different numbers can you come up with? The good news is that you don’t have to list down all of the possible 3-digit numbers in such a case. You can either use the nPr calculator or the permutation formula if you want to perform the calculation manually. Either of these allows you to calculate the number of permutations easily. Here is the formula: P(n,r) = n!/(n-r)! p refers to the number of permutations n refers to the total number of elements in a given set r refers to the number of elements you select from the given set The exclamation point in the equation represents a factorial. If you need to learn more about factorials or you need to perform factorial calculations, check out the factorial calculator. As you can see in the permutation formula, the number of permutations for when you select a single element is n. On the other hand, if you need to select all of the elements, you must modify the formula slightly. Therefore, the formula becomes: To help you understand this better, let’s go back to our earlier example. Let’s apply the formula to our situation with the deck of cards. In this case, you must solve for the number of ways to select 3 cards out of the total 9. Replace the numbers in the equation: P(9,3) = 9!/(9-3!) = 9!/6! = 504 After performing the calculation, you can check the accuracy using the permutation calculator. Then keep using this formula every time you need to calculate permutations manually in any given situation. What is the formula for nPr? As aforementioned, permutation and nPr mean the same thing. nPr is the symbolic representation of permutation. This means that the formula for nPr is the same as the formula for permutations which is: P(n,r)=n!(n−r)! for n ≥ r ≥ 0 By definition, permutations refer to how many ways you can get n ordered subsets of elements r from a set of elements n. Permutations differ from combinations because of the significance of the order of the elements. The permutation of n elements taken r at a time also means the number of ways that you can order r objects you selected from distinct n objects in a set. In mathematics, it’s denoted as the r at the bottom-right corner and n at the top-left corner of P. This is why another common form of notation of permutation is P(n,r) as seen in the permutation formula. When you use the basic principle of counting, you will be more familiar with the symbol P(n,r) which is a closed expression. Let’s have another example to help you understand the nPr formula further. Let’s assume that you have r places. For the first place, you can fill it by selecting any of the n elements and putting them in this place. For these elements in the first place, you can arrange them in n ways. Since you’ve already filled the first place, you now have n-1 elements remaining. Now you can select elements from any of the n-1 elements left for the second place. Again, you can arrange the elements in the second place in n number of ways. Keep going until you’ve filled the nth place or until you run out of elements. In doing this, you will observe something interesting: For the first place, n=n−(1–1) ways For the second place, n−1=n−(2–1) ways For the third place: n−2=n−(3–1) ways, and so on As you can see, this progression suggests that you can fill up the rth place in n−(r−1)=n−r+1 ways. Since these events occur all at the same time or simultaneously, the total number of arrangements possible is the product of how many ways there are to fill each individual r place. To close this expression, multiply then divide it by (n-r)!. Obviously, the numerator then becomes n!. This is why the equation for permutation is: What is the difference between nCr and nPr? While both nCr and nPr are quantities in statistics which refer to possible subsets of a given set of objects, they aren’t totally identical. First, they have different formulas which are: When you use sets to define these statistical quantities, nPr refers to the formula to solve for permutations of n elements which you take r at a time. Here, the arrangement or order of the elements matters. For nCr, this refers to combinations wherein you don’t take into consideration the arrangement or order of the elements.

Vocabulary Words and Definitions List. Olivia is placing a gift inside a box that measures 15 centimeters by 8 centimeters by 3 centimeters. Find the volumes of non-rectangular shapes including cylinders pyramids cones. Course 1 chapter 10 volume and surface area chapter quiz. Syllabus reference MS522 MS531 WM. Chapter 10 Quiz 1 Review and Reflection. Course 1 Chapter 10 Volume and Surface Area 235. A prism that has rectangular bases. 9 cm 11 cm Round to the nearest tenth. In this chapter you will learn. A special box designed to hold an antique artifact is shaped like a triangular prism. Check your level of practice and understand where you stand in your preparation. Learn vocabulary terms and more with flashcards games and other study tools. These sheets will certainly maintain preschoolers active as well as involved for a long period of time. What is the slant height of the pyramid. Surface Area and Volume. Find the volume of the triangular prism. The surface area of the pyramid below is 175 square meters. 3768 m 2 12. A pyramid has all sides that are equilateral triangles. Surface Area and Volume. Lesson 2 – Volume of Triangular prisms. 118692 cm 2 9. Use 314 for π. Surface Area and Volume Chapter 10 Summary and Practice Problems. 102 Lesson and Opener. Volume and Surface Area. Course 1 Chapter 10 Volume and Surface Area Chapter 10 Extra Practice Answers Lesson 10-1 1. 628 cm 2 11. SLO 101 I can find the volume of 3D shapes including cylinders cones pyramids and spheres. The height of the ramp is 12 meters. One of the two parallel congruent faces of a prism. 8 mm Lesson 10-2 1. Lesson 5 – Surface Area of Pyramids. What is the volume of the prism in cubic inches. Volume and Surface Area. Lesson 3 – Surface Area of Rectangular prisms. 2512 in 2 5. Here are extra practice problems by section including examples. Triangle and Trapezoid Area. Start studying Course 1 Chapter 10 Vocabulary – Volume and Surface Area. 101 Lesson and Opener. 10362 in 2 13. A rectangular prism has a length of 2 feet a width of 4 feet and a height of 6 inches. What is the volume of the cylinder. A prism that has triangular bases. _____ 238 Course 1 Chapter 10 Volume and Surface Area Boxes of Popcorn Box Length in Width in Height in A 8 1 2. The Volume Surface Area chapter of this Glencoe Math Companion Course helps students learn the essential lessons associated with volume and surface area. 101 Classwork and Closer. Then fill in the correct answer on the answer sheet provided by your teacher or on a sheet of paper. Start studying Course 1 – Chapter 10. Volume Surface Area 6G2 6G4. FREE Course 1 Chapter 10 Volume And Surface Area Answer Key 6Th Grade. 135648 m 2 15. Course 1 Chapter 10 Volume And Surface Area Answer Key DOWNLOAD Course 1 Chapter 10 Volume And Surface Area Answer Key. Found 5440 results for. The surface area of the box is 4392 square inches. A figure with length width and height. On this page you will find additional support and information relating to this chapter in Math 8. Used to measure volume. Surface Area of Triangular Prisms. Course 1 Volume and Surface Area Write the letter for the correct answer in the blank at the right of each question. Lesson 4 – Surface Area of Triangular prisms. Course 1 Chapter 10 Vocabulary – Volume and Surface Area 12 Terms. Course 1 – Ch10 – Volume Surface Area. NAME DATE Test Form 2B continued 6. 2512 m 2 2. Tell how many cubes of a certain size it will take to fill a three-dimensional figure. Cylinders Volume and Surface. McGraw-Hill Math Course 1Chapter 10 vocabulary. Area of Triangles and Trapezoids. Online Textbook Lesson 1. Volume of Rectangular Prisms. 70336 ft 2 7. 60288 cm 2 14. Course 1 Chapter 10 Volume And Surface Area Worksheet Answers Comply with these steps to create free printable worksheets for your youngsters. Learn vocabulary terms and more with flashcards games and other study tools. Pat has four rectangular baking pans. The point where three or more faces meet. 52752 m 2 10. Flashcards vocabulary wordsdefinitions practice Comments. Find the volume and surface area of the prism. Lesson 1 – Volume of Rectangular prisms. What is the surface area of the box. The sum of the areas of all the surfaces of a 3D figure. 102 Classwork and Closer. Terms in this set 12. What is the surface area of a cylinder with a radius of 2 feet and a. After completing this chapter you should be able to. Lesson 1 Homework. Chapter 10 Chapter 10 This chapter deals with surface area and volume of right solids and pyramids. Math Connects Coarse two Chapter 10 Vocab 14 Terms. S521S525 S531S535 Surface Area and Volume Surface Area and Volume find the surface area of right cylinders calculate the volume of composite solids use the fact. Find surface areas of cylinders and pyramids. Chapter 10 Surface Area and Volume Answer Key CK-12 Middle School Math Concepts – Grade 7 3 106 Surface Area of Cylinders Answers 1. 115 5 16 m 3 4. Course 1 – Chapter 10. Chapter Test Study Guide. Course 1 Chapter 10 Volume and Surface Area 236 11. 60288 in 2 4. RECREATION The base of a skateboard ramp has an area of 23 square meters. Find the cube root of a number. Chapter 10 – Volume Surface Area. 7536 cm 2 3. Volume Surface Area 6G2 6G4 12 Terms. 471 in 2 6. 1570 m 2 8. Which of the pans has the eatest volume. Step by Step Solutions is provided so that you can get to know the concepts better. 222 Course 1 Chapter 10 Volume and Surface Area Chapter Quiz. 224 Course 1 Chapter 10 Volume and Surface Area Standardized Test Practice Read each question. Lesson 3 Extra Practice.

It's easy to work out the areas of most squares that we meet, but what if they were tilted? What are the areas of these triangles? What do you notice? Can you generalise to other "families" of triangles? A circle rolls around the outside edge of a square so that its circumference always touches the edge of the square. Can you describe the locus of the centre of the circle? This rectangle is cut into five pieces which fit exactly into a triangular outline and also into a square outline where the triangle, the rectangle and the square have equal areas. A tilted square is a square with no horizontal sides. Can you devise a general instruction for the construction of a square when you are given just one of its sides? Draw some isosceles triangles with an area of $9$cm$^2$ and a vertex at (20,20). If all the vertices must have whole number coordinates, how many is it possible to draw? Can you make a right-angled triangle on this peg-board by joining up three points round the edge? Can you locate the lost giraffe? Input coordinates to help you search and find the giraffe in the fewest guesses. Interactive game. Set your own level of challenge, practise your table skills and beat your previous best score. The number of plants in Mr McGregor's magic potting shed increases overnight. He'd like to put the same number of plants in each of his gardens, planting one garden each day. How can he do it? Can you create a story that would describe the movement of the man shown on these graphs? Use the interactivity to try out our ideas. Use the interactivity to move Mr Pearson and his dog. Can you move him so that the graph shows a curve? Experiment with the interactivity of "rolling" regular polygons, and explore how the different positions of the red dot affects the distance it travels at each stage. Experiment with the interactivity of "rolling" regular polygons, and explore how the different positions of the red dot affects its vertical and horizontal movement at each stage. You can move the 4 pieces of the jigsaw and fit them into both outlines. Explain what has happened to the missing one unit of What shaped overlaps can you make with two circles which are the same size? What shapes are 'left over'? What shapes can you make when the circles are different sizes? Two engines, at opposite ends of a single track railway line, set off towards one another just as a fly, sitting on the front of one of the engines, sets off flying along the railway line... Can you find a relationship between the number of dots on the circle and the number of steps that will ensure that all points are Can you explain the strategy for winning this game with any target? First Connect Three game for an adult and child. Use the dice numbers and either addition or subtraction to get three numbers in a straight line. When number pyramids have a sequence on the bottom layer, some interesting patterns emerge... Try entering different sets of numbers in the number pyramids. How does the total at the top change? 7 balls are shaken in a container. You win if the two blue balls touch. What is the probability of winning? Euler discussed whether or not it was possible to stroll around Koenigsberg crossing each of its seven bridges exactly once. Experiment with different numbers of islands and bridges. We can show that (x + 1)² = x² + 2x + 1 by considering the area of an (x + 1) by (x + 1) square. Show in a similar way that (x + 2)² = x² + 4x + 4 Triangle numbers can be represented by a triangular array of squares. What do you notice about the sum of identical triangle numbers? A game for two or more players that uses a knowledge of measuring tools. Spin the spinner and identify which jobs can be done with the measuring tool shown. Do you know how to find the area of a triangle? You can count the squares. What happens if we turn the triangle on end? Press the button and see. Try counting the number of units in the triangle now. . . . What shape is the overlap when you slide one of these shapes half way across another? Can you picture it in your head? Use the interactivity to check your visualisation. Six balls of various colours are randomly shaken into a trianglular arrangement. What is the probability of having at least one red in Carry out some time trials and gather some data to help you decide on the best training regime for your rowing crew. An interactive game to be played on your own or with friends. Imagine you are having a party. Each person takes it in turns to stand behind the chair where they will get the most chocolate. Two circles of equal radius touch at P. One circle is fixed whilst the other moves, rolling without slipping, all the way round. How many times does the moving coin revolve before returning to P? Imagine a wheel with different markings painted on it at regular intervals. Can you predict the colour of the 18th mark? The 100th Three beads are threaded on a circular wire and are coloured either red or blue. Can you find all four different combinations? Use the interactivity to make this Islamic star and cross design. Can you produce a tessellation of regular octagons with two different types of triangle? Hover your mouse over the counters to see which ones will be removed. Click to remover them. The winner is the last one to remove a counter. How you can make sure you win? Mo has left, but Meg is still experimenting. Use the interactivity to help you find out how she can alter her pouch of marbles and still keep the two pouches balanced. Show how this pentagonal tile can be used to tile the plane and describe the transformations which map this pentagon to its images in the tiling. Try to stop your opponent from being able to split the piles of counters into unequal numbers. Can you find a strategy? Players take it in turns to choose a dot on the grid. The winner is the first to have four dots that can be joined to form a square. Start by putting one million (1 000 000) into the display of your calculator. Can you reduce this to 7 using just the 7 key and add, subtract, multiply, divide and equals as many times as you like? Can you beat the computer in the challenging strategy game? Can you find all the 4-ball shuffles? Cut four triangles from a square as shown in the picture. How many different shapes can you make by fitting the four triangles back Place six toy ladybirds into the box so that there are two ladybirds in every column and every row. Can you put the numbers 1 to 8 into the circles so that the four calculations are correct? Use the blue spot to help you move the yellow spot from one star to the other. How are the trails of the blue and yellow spots related? Can you make the green spot travel through the tube by moving the yellow spot? Could you draw a tube that both spots would follow? A simulation of target archery practice

Why would you want to do that? IMHO, by now the question is rather why you would want to continue using Kontact for your email/calendar/contact management needs given all the bugs and regressions. For example, after upgrading to KDE 4.8 on my Gentoo machine recently, once again address completion when composing emails ceased to work (it worked some time with 4.7.x, but not before that). Also, every now and then the virtuoso process consumed 100% CPU on login. I had problems with KDEPIM2 ever since it was released (version 4.6.x), but was reluctant to ditch it since I had been a happy user for more than 5 years. Furthermore, KDEPIM is more powerful than Thunderbird. Nevertheless, this is 2012 and I am no longer willing to invest considerable time to have a working email and calendar solution. Maybe it works if you clean everything akondadi-, nepomuk- and kontact-related on each upgrade, but this is just not a viable option. Anyway, to replace Kontact with Thunderbird there are several tweaks to make Thunderbird resemble the behavior of Kontact: - Kmail, Akregator and KNode can be replaced by Thunderbird’s Email, Blogs & News Feeds and Newsgroup accounts, respectively. They may not be as powerful as the original programs, but they do their job. - The KDE calendar is best replaced by the Lightning add-on. For remote synchronization, you will need something like the Provider for Google Calendar add-on. This works reasonably well. Some additional useful add-ons that make Thunderbird behave like Kmail: - SendWithoutSubject, otherwise there is a confirmation popup every time you try to send an email without a subject (you need an add-on for this, seriously?) - Subject cleaner to eliminate repeated Re: Re:, etc. from subjects when replying (KMail does this automatically) - Sieve to manage sieve scripts on your IMAP server, if you use those. - ThunderPlunger is a useful add-on if you happen to delete accounts, because Thunderbird does not actually remove the data from disk, potentially leaving you with hundreds of megabytes of junk. - LookOut is required to deal with those infamous winmail.dat files brought to you by MS Outlook’s proprietary handling of attachments. - StartupMaster is a useful addon to eliminate multiple password prompts if you have a master password for the builtin password manager. This happens because the Google Calendar provider prompts for a password for each account. Update: For KDE users, the kwalltet password integration addon seems to be the way to go. KMail also automatically checks all folders on your IMAP server, while Thunderbird does this only for the currently open folder. To mimic KMail’s behaviour you have to set to true in the config editor. Of course Thunderbird is not as nicely integrated into your desktop as Kontact. For example, I have yet to find a solution to display a systray icon with the number of unread mails, since solutions such as the Gnome integration add-on do not seem to work with the current version of TB. Also your calendar events are not displayed in the KDE Plasma calender if you decide to get rid of Akonadi completely. It took some time, but today I finally managed to proofread my diploma thesis in economics (which is similar to a master’s thesis) titled “Endogenous growth in heterogeneous-firm trade models.” Abstract: This thesis aims to demonstrate how a heterogeneous-firm trade and growth model can be used to analyze productivity and welfare effects of trade liberalization. To this end, I first discuss the evolution of trade and growth theories that culminated in a class of comprehensive models capable of accounting for phenomena that are deemed important by both trade and growth theorists. Focusing on one of these models, Gustafsson and Segerstrom (2010), I extend their comparative static analysis to out-of-steady-state dynamics and show that the presence of heterogeneous firms introduces nontrivial complications absent from homogeneous-firm models. Lastly, I present Gustafsson and Segerstrom’s main finding that knowledge spillovers play a pivotal role in determining whether trade liberalization is welfare increasing. You can download the complete thesis here (PDF, 2.1 MB). Another Mathematica demonstration: this one visualizes the comparative statics as well as the adjustment dynamics resulting from a parameter change in the Solow-Swan growth model. I took professor Sorger‘s advanced macro class on growth and business cycle theories last semester, so I noticed that some textbooks (such as Barro/Sala-i-Martin) don’t show the adjustment process of capital, per-capita output, consumption, etc. I think it helps intuition to visualize these variables, which is why I created this demonstration. You can download the Mathematica notebook (you’ll need either Mathematica or the free Mathematica player to view this), or look at the online version here. (Unfortunately Wolfram changed the demo so only one graph is visible at a time.) Melitz, again. I took a Mathematica class this semester where we had to build a demonstration that could be uploaded to Wolfram’s demonstration project site. I created an animated version of figure 2 from Melitz’ paper (based on the calculations done in one of the Mathematica notebooks presented below), which IMO is well suited to do a comparative statics analysis of autarky and the open economy. The online demonstration is available here. The Melitz (2003) trade model is one of the central models of my diploma thesis, so I’ve spent (wasted?) some time derived a closed-form solution for Pareto-distributed firm productivity levels (Melitz himself does not use a particular distribution). I’ve been able to replicate most of paper’s results, while a few do not hold for the Pareto distribution (using other distributions, though, will generally not produce closed-form solutions). With the Mathematica notebook it is easy to manipulate model parameters and see the effects on equilibrium values. Download links: PDF, Mathematica Notebook (ZIP archive) Update (Dec. 8, 2009): minor changes and error corrections in PDF and Mathematica notebook. Update (Dec. 14, 2009): Helpman/Melitz/Yeaple (2004) present a similar model with heterogeneous firms and Pareto-distributed firm productivity. It is less complex than Melitz (2003) as it omits the stochastic firm entry and exit process. However, it allows to model firm decisions concerning production for the domestic market and export / FDI in foreign markets depending on firm productivity. The Mathematica notebook below derives this model in greater detail than the original paper. Download links: PDF, Mathematica notebook/package - Helpman, Elhanan/ Melitz, Marc J. and Yeaple, Stephen R. (2004): Export versus FDI with Heterogeneous Firms . In: American Economic Review 94(1), 300 316. - Melitz, Marc J. (2003): “The Impact of Trade on Intra-Industry Reallocations and Aggregate Industry Productivity”. In: Econometrica 71(6), 1695. While researching for my current diploma thesis on “new new” trade theory I realized that fully understanding (or rather: knowing by heart) the Dixit/Stiglitz (1977) model is a necessary precondition to be able to read most of the relevant papers. Unfortunately, the original Dixit/Stiglitz is not written for undergraduate students (intermediate steps are mostly omitted), and the obsolete Chamberlinian terminology does not exactly promote understanding (obsolete by Austrian university standards, that is). Consequently I’ve compiled this step-by-step walkthrough of the first and most widely used model (the CES case) covered in Dixit/Stiglitz (1977). This should be sufficient for most of the “new” and “new new” trade theory models. You can download it here. Please let me know if you find any mistakes. Update (Dec. 21, 2009): Minor corrections. - Dixit, Avinash K. and Stiglitz, Joseph E. (1977): “Monopolistic Competition and Optimum Product Diversity”. In: American Economic Review 67(3), 297–308. I guess this looks like a blog, but I won’t be using it as one. I see little value in further polluting the internet by sharing my thoughts on this or that. After all, I am not Paul Krugman or Brad DeLong, so why should anyone care? 😉 However, I will occasionally post stuff concerning my studies which other economics students might find useful.

Twenty-four hours later, I had the math parody ready to go for February 29th. It was posted directly to my math comic blog, rather than being placed into the music parody archive, because I hadn’t had time to actually verify the maths yet. I wanted to run the formula on an equation, to make sure it actually produced all three roots as I advertised. I figured I’d do that over the March Break, and maybe produce a video. Instead, I fell down this rabbit hole. To see my conclusions, skip to “Actual Analysis” below. To follow the journey, keep reading. THE FORMULA FAILED Given y = ax^3 + bx^2 + cx + d, I wanted different values for each coefficient, and ideally some nice whole number roots. Expanding (x+1)(x-1)(x-2) felt like it would work, and indeed it means a=1, b=-2, c=-1, d=2 (all different but small). I plugged them into the formula... and I did not get -1, 1, and 2. Rather, I got -1/3, 2/3 and 7/3. That wasn’t supposed to happen. After poking at the maths, I decided my problem was the negative square roots that had turned up. I’d actually had to turn to ‘wolfram alpha’ to take the cube roots of those complex numbers; maybe I’d entered it wrong. This would probably be easier if I kept that nested square root ‘positive’. After swapping all the signs (multiplying by -1) and seeing that wouldn’t fix it, I moved my calculations off of a paper bag and onto actual lined paper. I focussed on what was under the square root. I would later call this the “root discriminant” of the cubic formula. One piece of it is already squared, no problem, that’s positive, ignore it. To get a positive result overall, I needed c/(3a) to be larger than b^2/(3a)^2. Except with the latter also being squared (always positive), this meant “a” and “c” HAD to be the same sign. Huh. I played around with some other expansions of factored form. At first those signs were wrong, or some coefficients were zero, or they weren’t unique numbers. Then I found one that worked: (x+2)(x+3)(x+4) But it still produced a negative root result. At this point in my notes I’ve scribbled “a can’t be +-1”, which is kind of silly in retrospect. I kept going, further scribbles being “need smaller abs(b) vs. (c)” and “3ac>b^2”. To my horror, I began to suspect that the root would always be negative, so long as I was feeding in three real solutions. I played with some sliders in Desmos, then decided I should contrast with cubic formulas which had only one functional root instead (also verified in Desmos). Working with y1 = 2x^3 - 3x^2 - x - 2, which is solved when x=2, produced a positive square root overall. But it led to doing cube roots of nasty fractions, and the decimal approximations were NOT giving me the answer x=2 (closer to x=1.68). I started to think maybe the formula ITSELF was wrong. (The internet, unreliable? What?!) The previous work wasn’t done all in one sitting. I’d already given up on having this done through March Break. One weekend, while heading into school for a play rehearsal, I stopped in a food court and started tackling other formula variations I’d discovered online. For instance, one that separated out the three x1, x2 and x3 solutions, and another that used expressions for q, p and r to simplify the formula overall. None of them worked with my y1 equation (above). Worse, they were giving me different solutions. Depending on which formula variation I used, I either got x=1.68 or x=2.13 or x=2.29. But the solution is TWO, dammit! I checked my math, found a mistake, it didn’t fix the problem. Surely the issue was the “cube root of fractions” step, which I was approximating on my calculator (to 8 decimal places). This was going to take more time than I had. Easter Break, at the end of March, wouldn’t help. I had to use it to bank a bunch of my math comics, in anticipation of being at school for 25 days straight due to the play. While still writing 2,000 words/week on my serial. Yeah. So after downloading some plausible “cube” footage on March 26th for that later video, I shelved the whole project. It resurfaced briefly in May, when I put together a Cubic Formula Song Powerpoint for my MCT class. (We were finishing our Polynomials unit at the time.) This despite my unfounded reservations about the formula. The performance went fine; as a student remarked, “It just kept growing!”. The project resurfaced again in June, when John Golden tweeted out something about the cubic formula - except his formula image looked DIFFERENT from any of the others. Time to get back into actual research. As motivation, I decided I would do a presentation of the song on the last day of my MCR class, if it felt right. I’d realized that “a” only ever appears in the denominator. This is different from the quadratic formula, where we have -4ac. But from the research, it makes total sense. The first step in developing the cubic formula is to divide everything by “a”, creating a simpler problem, namely x^3 + Bx^2 + Cx + D = 0. (As noted, it makes my “a can’t be 1” remark above rather silly.) The next step in creating the formula seemed to be developing a depressed cubic, by substituting in a different value for “x” to get rid of the squared term, effectively giving you m^3 + Jm + K = 0, where once you have the “m” you can reverse determine “x” from there. That made sense, but didn’t really help me grab numbers for an example. This was when I discovered the “cubic discriminant”. It seems to have it's own definition that synchs with the quadratic, but had me consider the piece under the square root - a "root discriminant", if you will. And I had been correct - for there to be 3 real solutions, that "root discriminant" had to be NEGATIVE. Conversely, if it was POSITIVE, that meant there would be only one real solution (the other two solutions would be imaginary). Again, this made some sense to me. If you picture cube roots equidistant from each other on the polar plane, only one root can land where b*i=0. But there’s a third case. If my "root discriminant" is ZERO, this means you get at least a double root. That’s honestly *the only way* to avoid either doing cube roots of imaginary numbers, or having imaginary answers as solutions. Wonderful. (It also meant Golden’s found formula was invalid, as the cubic discriminant, which appears twice, had an invalid sign one of those times.) There was even this “wikihow” website showing how to do the steps by hand. With renewed faith, I sang the song for the few 3Us who were there the last day of class, then shelved this project AGAIN until after exams - I was SO behind in my marking. July 2016. The website “SOSMath” offered a cubic formula with x=3 as the only solution. I tried the formula, and it worked, despite ugly radicals. (I don’t know why I was having such trouble before.) The last hurdle was a formula with a “cubic discriminant” of zero that would work nicely. I created one... for some reason, it didn’t work?! ... Damn sign error. Okay. Good to go! And that's the story. The final video will premiere within the next 3-4 days; I’ve linked to it here. I’m still maintaining weekly posts on three blogs (including here), that’s holding me back a bit. Thanks for reading, feel free to check out some of my fiction writing too!

>> The enthalpy difference between graphite and diamond is too large for both to have a standard enthalpy of formation of zero. /F6 10 0 R Not in this one. Example #9: The ΔH for the following reaction equals −89 kJ: In addition, these two standard enthalpies of formation are known: 2) Inserting values into the above, we find: 1) Here are all three data reactions written out in equation form: 2) What we need to do is add the three data equations together in such a way as to recover the target equation: 4) However, this is not the enthalpy of formation, since that value is always for one mole of the product. 11 0 obj 2) Adding the following equations will yield the equation needed: 3) Add the three equations and their enthalpies: The heat of formation of CH4(g) is −79 kJ/mole, The Chemistry Webbook gives the value as being a bit less than −75 kJ/mol. = [ (4136) ] − [ (6) (−393.5) + (7) (−285.8) ]. Introducing Textbook Solutions. /Encoding 3 0 R = −1367 kJ/mol of ethyl alcohol. For example, I have a link several problems above to a table of bond enthalpy values. Therefore. /Type /Font The standard enthalpy of formation is defined as the change in enthalpy when one mole of a substance in the standard state (1 atm of pressure and 298.15 K) is formed from its pure elements under the same conditions. 1) Write the equation for the formation of hexane: ΔH rxno endobj << 2 0 obj Example #5: The standard enthalpy of formation of hexane can be determined indirectly. The ChemTeam does not know for sure. The bond enthalpy values are each associated with a specific chemical equation. Problem #3: Determine the enthalpy of reaction for the following: Using the following bond enthalpies (in kJ/mol): H−H (432); O=O (496); H−O (463). Average Bond Enthalpy- Is the average enthalpy change that takes place when breaking by homolytic fission 1 mol of a given type of bond in the molecules of … Solution The bond energy involves breaking HCl into H and Cl atoms. >> The general equation for the standard enthalpy change of formation is given below: Plugging in the equation for the formation of CO2 gives the following: ΔHreactiono= ΔHfo[CO2(g)] - (ΔHfo[O2(g)] + ΔHfo[C(graphite)]. The conversion of graphite into diamond is an endothermic reaction (ΔH = +3 kJ mol–1). Since we are discussing formation equations, let's go look up their formation enthalpies: 1⁄2H2(g) + 1⁄2Br2(ℓ) ---> HBr(g) ΔH fo Be prepared. This is not the usual way in which bond enthalpy values are used. Between Br2(l) and Br2(g) at 298.15 K, which substance has a nonzero standard enthalpy of formation? Unless otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0. The reaction will always form one mole of the target substance (glucose in the example) in its standard state. Consequently, enthalpy calculations using bond enthalpies are only a rough guide to the enthalpy of a given reaction. The −393.5 value is the enthalpy for the combustion of carbon. The enthalpy of formation involves making HCl from H 2 and Cl 2 molecules. >> The former parameter tends to be favored in theoretical and computational work, while the latter is more convenient for thermochemical studies. goes on the left-hand side. Completing the CAPTCHA proves you are a human and gives you temporary access to the web property. 1) The balanced equation for the combustion of C2H6 (ethane) is: [(2 moles CO2) (−393.5 kJ/mole) + (6 moles H2O) (−241.8 kJ/mole)] − [(2 moles C2H6) (−84.68 kJ/mole) + (7 moles O2) (0 kJ/mole)]. In addition, the chemical environment changes as you remove a given bond. >> Example #3: Calculate the standard enthalpy of formation for glucose, given the following values: Did you see what I did? Note the approach to the solution. I'll explain the above equation using an example problem. 2) Let's write the formation equation for AgNO2(s): 3) Determine the unknown value by adding the two equations listed in step 1: When the two equations are added together, the AgNO3(s) cancels out as does 1⁄2O2(g) and we are left with the formation equation for AgNO2(s), the equation given in step 2. Using the standard enthalpy of formation data in Appendix G, calculate the bond energy of, 7.5: Strengths of Ionic and Covalent Bonds, 71. Given the standard enthalpy of combustions of: methane gas (-890.5 kJ/mol), hydrogen gas (-285.9 kJ/mol), C graphite (-393.5 kJ/mol), the bond dissociation enthalpy of hydrogen gas (435.8 kJ/mol), and the latent heat of sublimation of graphite (718.4 kJ/mol) Evaluate the bond enthalpy of the C-H bond … (ii) C (diamond) + O2 (g) → CO2 (g) ; … /Type /Encoding Make sure you find it and figure out how to use it. The moral of the story? >> Now I examine each statement: 1. The enthalpy of formation involves making HCl from H 2 and Cl 2 molecules. /F3 6 0 R 1) First of all, this is the reaction we want an answer for: We know this because the problem asks for the standard enthalpy of formation for glucose. values: All the above values have units of kJ/mol because these are standard values. Bond energies of H − − H, C − − H, and C − − C bonds are 1 0 4. Sometimes terms overlap. The differences between Cl and Br are slight, but they do make for a difference that can be measured experimentally.

As just mentioned, the Moon's rotation period is precisely equal to its period of revolution about Earth27.3 days sothe Moon keeps the same side facing Earth at all times (see Figure 8.10). To an astronaut standing on the Moon's near-side surface, Earth would appear almost stationary in the sky (although its daily rotation would be clearly evident). This condition, in which the spin of one body is precisely equal to (or synchronized with) its revolution around another body, is known as a synchronous orbit. The fact that the Moon is in a synchronous orbit around Earth is no accident. It is an inevitable consequence of the gravitational interaction between those two bodies. Figure 8.10 The Moon is slightly elongated in shape, with its long axis perpetually pointing toward Earth. (The elongation is highly exaggerated in this diagram.) Just as the Moon raises tides on Earth, Earth also produces a tidal bulge in the Moon. Because Earth is so much more massive, the tidal force on the Moon is about 20 times greater than that on Earth, and the Moon's tidal bulge is correspondingly larger. In Chapter 7 we saw how lunar tidal forces are causing Earth's spin to slow and how, as a result, Earth will eventually rotate on its axis at the same rate as the Moon revolves around Earth. (Sec. 7.6) Earth's rotation will not become synchronous with the EarthMoon orbital period for hundreds of billions of years. In the case of the Moon, however, the process has already gone to completion. The Moon's much larger tidal deformation caused it to evolve into a synchronous orbit long ago, and the Moon is said to have become tidally locked to Earth. Most of the moons in the solar system are similarly locked by the tidal fields of their parent planets. In fact, the size of the lunar bulge is too great to be produced by Earth's present-day tidal influence. The explanation seems to be that, long ago, the distance from Earth to the Moon may have been as little as two-thirds of its current value, or about 250,000 km. Earth's tidal force on the Moon would then have been more than three times greater than it is today and could have accounted for the Moon's elongated shape. The resulting distortion could have "set" when the Moon solidified, thus surviving to the present day, while at the same time accelerating the synchronization of the Moon's orbit. In principle, the ability to discern surface features on Mercury should allow us to measure its rotation rate simply by watching the motion of a particular region around the planet. In the mid-nineteenth century, an Italian astronomer named Giovanni Schiaparelli did just that. He concluded that Mercury always keeps one side facing the Sun, much as our Moon perpetually presents only one face to Earth. The explanation suggested for this synchronous rotation was the same as for the Moonthe tidal bulge raised in Mercury by the Sun had modified the planet's rotation rate until the bulge always pointed directly at the Sun. Although the surface features could not be seen clearly, the combination of Schiaparelli's observations and a plausible physical explanation was enough to convince most astronomers, and the belief that Mercury rotates synchronously with its revolution about the Sun (that is, once every 88 Earth days) persisted for almost half a century. In 1965, astronomers making radar observations of Mercury from the Arecibo radio telescope in Puerto Rico (see Figure 5.21) discovered that this long-held view was in error. The technique they used is illustrated in Figure 8.11, which shows a radar signal reflecting from the surface of a hypothetical planet. Let's imagine, for the purpose of this discussion, that the pulse of outgoing radiation is of a single frequency. Figure 8.11 A radar beam reflected from a rotating planet yields information about both the planet's overall motion and its rotation rate. The returning pulse bounced off the planet is very much weaker than the outgoing signal. Beyond this change, the reflected signal can be modified in two important ways. First, the signal as a whole may be redshifted or blueshifted as a consequence of the Doppler effect, depending on the overall radial velocity of the planet with respect to Earth. (Sec. 3.5) Let's assume for simplicity that this velocity is zero, so that, on average, the frequency of the reflected signal is the same as the outgoing beam. Second, if the planet is rotating, the radiation reflected from the side of the planet moving toward us returns at a slightly higher frequency than the radiation reflected from the receding side. (Think of the two hemispheres as being separate sources of radiation and moving at slightly different velocities, one toward us and one away.) The effect is very similar to the rotational line broadening discussed in Chapter 4 (see Figure 4.18), except that in this case the radiation we are measuring was not emitted by the planet but only reflected from its surface. (Sec. 4.4) What we see in the reflected signal is a spread of frequencies on either side of the original frequency. By measuring the extent of that spread, we can determine the planet's rotational speed. In this way, the Arecibo researchers found that the rotation period of Mercury is not 88 days, as had previously been believed, but 59 days, exactly two-thirds of the planet's orbital period. Because there are exactly three rotations for every two revolutions, we say that there is a 3:2 spin-orbit resonance in Mercury's motion. In this context, the term resonance just means that two characteristic timeshere Mercury's day and yearare related to each other in a simple way. An even simpler example of a spinorbit resonance is the Moon's orbit around Earth. In that case, the rotation is synchronous with the revolution, so the resonance is said to be 1:1. Figure 8.12 illustrates some implications of Mercury's curious rotation for a hypothetical inhabitant of the planet. Mercury's solar daythe time from noon to noon, sayis actually 2 Mercury years long! The Sun stays "up" in the black Mercury sky for almost 3 Earth months at a time, after which follows nearly 3 months of darkness. At any given point in its orbit, Mercury presents the same face to the Sun not every time it revolves, but every other time. Figure 8.12 Mercury's orbital and rotational motions combine to produce a day that is 2 years long. The arrow represents an observer standing on the surface of the planet. At day 0, it is noon for our observer, and the Sun is directly overhead. By the time Mercury has completed one full orbit around the Sun and moved from day 0 to day 88, it has rotated on its axis exactly 1.5 times, so that it is now midnight at the observer's location. After another complete orbit, it is noon once again. The eccentricity of Mercury's orbit is not shown in this simplified diagram. The Orbit of Mercury Mercury's 3:2 spinorbit resonance did not occur by chance. What mechanism establishes and maintains it? In the case of the Moon orbiting Earth, the 1:1 resonance is explained as the result of tidal forces. In essence, the lunar rotation period, which probably started off much shorter than its present value, has lengthened so that the tidal bulge created by Earth is fixed relative to the body of the Moon. Tidal forces (this time due to the Sun) are also responsible for Mercury's 3:2 resonance, but in a much more subtle way. Mercury cannot settle into a 1:1 resonance because its orbit around the Sun is quite eccentric. By Kepler's second law, Mercury's orbital speed is greatest at perihelion (closest approach to the Sun) and least at aphelion (greatest distance from the Sun). (Sec. 2.3) A moment's thought shows that because of these variations in the planet's orbital speed, there is no way that the planet (rotating at a constant rate) can remain in a synchronous orbit. If its rotation were synchronous near perihelion, it would be too rapid at aphelion, while synchronism at aphelion would be too slow at perihelion. Tidal forces always act to try to synchronize the rotation rate with the instantaneous orbital speed, but such synchronization cannot be maintained over Mercury's entire orbit. What happens? The answer is found when we realize that tidal effects diminish very rapidly with increasing distance. The tidal forces acting on Mercury at perihelion are much greater than those at aphelion, and perihelion "won" the struggle to determine the rotation rate. In the 3:2 resonance, Mercury's orbital and rotational motion are almost exactly synchronous at perihelion, so that particular rotation rate was naturally "picked out" by the Sun's tidal influence on the planet. Notice that even though Mercury rotates through only 180° between one perihelion and the next (see Figure 8.12), the appearance of the tidal bulge is the same each time around. The motion of Mercury is one of the simplest nonsynchronous resonances known in the solar system. Astronomers now believe that these intricate dynamical interactions are responsible for much of the fine detail observed in the motion of the solar system. Examples of resonances can be found in the orbits of many of the planets, their moons, their rings, and in the asteroid belt. The Sun's tidal influence also causes Mercury's rotation axis to be exactly perpendicular to its orbit plane. As a result, and because of Mercury's eccentric orbit and the spinorbit resonance, some points on the surface get much hotter than others. In particular, the two (diametrically opposite) points on the surface where the Sun is directly overhead at perihelion get hottest of all. They are called the hot longitudes. The peak temperature of 700 K mentioned earlier occurs at noon at those two locations. At the warm longitudes, where the Sun is directly overhead at aphelion, the peak temperature is about 150 K coolera mere 550 K! By contrast, the Sun is always on the horizon as seen from the planet's poles. Recent Earth-based radar studies suggest that Mercury's polar temperatures may be as low as 125 K and that, despite the planet's scorched equator, the poles may be covered with extensive sheets of water ice. (See Interlude 8-1 for similar findings regarding the Moon.)

annual dividend yield calculation Dividend Yield Calculator. Annual Dividend per Share ()You chose the Basic version of the Dividend Yield Calculator. The Basic version is non-editable, calculations are limited to 100/month, and the CalculatorPro.com link must be included. Saturday, September 18, 2010. How to Calculate Dividend Yield? Not all of the tools of fundamental analysis work for every investor on every stock.You calculate the Dividend Yield by taking the annual dividend per share and divide by the stocks price. Dividend yield, or annual dividend yield, refers to the amount of money a stock pays out as dividends relative to its current share price, expressed as aThe dividend yield is a financial calculation that tells you how much money you will earn for every dollar invested in Dividends Yield . Age Calculator Calculator Download App.Dividend Yield (DY) (Annual Dividends / Current Share Price) Rate of DY DY x 100. Example. A companys annual dividend is 5 and the current share price is 30. We divide the annual dividend per share by the market value per share. It is often used by investors who are looking for continuous dividend income.You can easily calculate the dividend yield ratio by using the following formula Calculations. Acid Test Ratio or Quick Ratio or Liquid Ratio Calculator.It tell what percentage of purchase price the company return in dividend. Formula. Dividend Yield Annual Dividend Per Share / Stock Price. Dividend yield calculator ultimatecalculators. 10 per share in annual dividends.Definition calculation video lesson yield ratio explanation, formula, example and what Stocks glossary moneycontrol. It is calculated by dividend yield to price ratio. Dividend Yield Calculator. The calculator asks for: Annual Dividends per Share, which is found via company announcements or the financial media and is the sum of all dividends throughout the year Stock Price, which is found in the financial media or via your broker. Our free dividend yield calculator will calculate the dividend yield ratio for you. Dividend Yield Calculator. Annual Dividend per Share (): Market Price of the Stock () Below are the calculations to calculate the current dividend yield. Step 1: Annualized the dividend payment. Number of Dividend Payments: 4 Dividend Paid Per Share: 0.52 Annual Dividend Payment: 2.08. Dividend yield is considered as ROI for income investors who are not interested in capital gains or long-term earnings. It is calculated as Annual Dividend Per Share divided by Current Market Value Per Share. Calculation (formula). Dividend Yield is used to determine the current annual yield of a security that pays dividends. The dividend yield is used to calculate the income produced and returned to shareholders and does not include capital gains. Use the dividend yield calculator below to solve the formula. Dividend yield ratio shows what percentage of the market price of a share a company annually pays to its stockholders in the form of dividends. It is calculated by dividing the annual dividend per share by market value per share. Dividend Yield Calculator (Click Here or Scroll Down).An example of the dividend yield formula would be a stock that has paid total annual dividends per share of 1.12. The original stock price for the year was 28. You can calculate dividend yield by annual dividend per share. And it divides by stock price.And Dividend per share is paid out by business including interim dividend. Calculation of Dividends per Share. Dividend Yield Calculator. Enter Annual Dividend Value Per Share in USD.In fact, when you divide the annual dividend per share, by its current stock price, what you are left with, is the dividend yield. Here is the formula used for dividend yield calculation. This article will explain the various financial calculations related to dividend investing like how to calculate dividend yield, dividend yield formula.Dividend yield, or annual dividend yield, refers to the amount of money a stock pays out as dividends relative to its current share price, expressed You calculate the dividend yield ratio for the business as follows: 1.50 annual cash dividend per share 70 current market price of stock 2.1 dividend yield. An income statement example for a business. Dividend Yield Calculation.The following equation can be used to calculate a stocks current yield . Yield Annual Dividends / Current Share Price. In this article, you will find a clear explanation of dividend yield calculation, from the knowledge of dividends per share and current share price.Dividend Yield Calculator. Enter Annual Dividend Value Per Share in USD. The formula for dividend yield is annual dividend divided by the share price. ROE or Return On Equity (as used in the DuPont formula) is defined as Net Income divided by Equity. In other words, the net profit that a company has generated during a year Dividend yield is calculated by dividing the total annual dividend amount by the companys share price.Therefore, the yield for Verizon is: Are all dividend distributions included in the dividend yield calculation? Dividend Yield annual dividend per share / stocks price per share.Heres a dividend yield example to explain the concept from Investopedia: Suppose company XYZs stock is trading at 20 and pays yearly dividends of 1 per share to its shareholders. Dividend Yield Formula:Dividend Yield Annual Dividend per Share / Market Price of the Stock. Dividend Yield DefinitionUsing the free online Dividend Yield Calculator is a quick way to calculate the dividend yield of any dividend paying stock. What would be a formula for calculating the annualized continuous dividend yield of a stock? Given the quarterly or annual dividend.Calculation of dividend yield from index returns. 0. This calculator can be used for calculating Dividend Yield in percentage. Dividend yield shows the pay out of dividend in relative to share price. Annual Dividend Per Share. Market Price of Stock. Dividend Yield (). Notes. Annual Dividend is the sum of dividends given in the financial year. How to Calculate Dividends. Three Methods:Dividends Calculator Finding Total Dividends from DPS Finding Dividend Yield Community QA.Thus, if youre using past dividend values to estimate what youll be paid in the future, theres a chance that your calculation may not be accurate. The dividend yield is calculated by dividing the dividend by the share price, and then expressing it as a percentage.However in the interim Loadsamoney has revealed its annual results and increased its dividend payment by 10 for the year, to 11p. Current yield relates the annual Since the reported distribution yield includes monthly dividend It is generally calculated using a financial calculatorDividend yield - calculation Find the annual Calculate Dividend Yield Dividend Yield What is the Dividend Yield A financial ratio that indicates how much a company pays out in dividends each year relative to its share price.Dividend Yield Annual Dividend / Current Stock Price. Dividend Calculator. Investment Type. Portfolio Individual Stock.New Annual Dividend Income. Dividend Paid out through Years. Yield On Cost. Year. Principal. Annual Dividend. The formula for calculating dividend yield may be represented as followsTo better explain the concept, refer to the following dividend yield example. Suppose company ABCs stock is trading at 20 and pays annual dividends of 1 per share to its shareholders. How to Calculate Dividend Yield. To calculate dividend yield, use the dividend yield formula. This can be done by dividing the annual dividend by the current stock price The dividend yield or dividend-price ratio of a share is the dividend per share, divided by the price per share. It is also a companys total annual dividend payments divided by its market capitalization, assuming the number of shares is constant. It is often expressed as a percentage. Calculation of Dividends per Share. denominator than the dividend per share calculation prior annual report.Using the free online Dividend Yield Calculator is a quick way to calculate the dividend yield of any dividend paying stock. How did Investingcalculator calculate the Dividend Yield? The dividend yield is calculated by taking the annual dividend payments and divide it by the current stock price. Investing Calculator. The website for every Financial or Investment calculation! Dividend Yield: The dividend yield of the stock. Dividend Growth RateThe calculated compounded total return for dividends only. Annual Dividend Return The yield on cost is a simple, but effective calculation investors can use to track their return for shares in their portfolio.The second piece of information you will need is the annual dividend for the stock. To find its indicated dividend yield, just multiply the current quarterly rate by four and use the result (3.08 per share) in the calculation.Dividend Yield: Expected-Annual-Dividend divided by Price, expressed as a percentage. The dividend yield formula is calculated by dividing the cash dividends per share by the market value per share.Analysis. Investors use the dividend yield formula to compute the cash flow they are getting from their investment in stocks. Dividend yield is the annual dividend per share of a company compared to the price of the stock expressed as a percentage. In other words it tells you the percentage dividend return the stock owner receives based on the current price of a stock. Dividend Yield Calculation. Dividend yield ratio is expressed as a percentage and indicates the shares a company pays out on an annual basis relative to its share price.As youve seen, there is a lot involved in these calculations, and sometimes students tend to make mistakes or mix up formulas when they carry them out. Dividend Yield annual dividend per share / stocks price22/02/2017 This article will explain the various financial calculations related to dividend investing like how to calculate dividend yield, dividend yield formula. Calculate the dividend yield for the previous year by dividing the dividends paid by the price at the start of the year.In this example, divide 17 percent by 5 to find the average annual dividend yield is 3.4 percent. If your stock pays varying amounts, total up the payments youve received throughout the whole year and enter it as an annual dividend.In these examples the stock was purchased 50 years previously in 1966, and heres how their dividend yield calculations look, using the purchase price paid taking Current dividend yield is calculated by dividing the annual dividend by the current stock price.Any calculation of a future value must therefore be based on certain assumptions.Difficulty:ModerateInstructions Select the time frame. Annual dividend yield calculator - Page 1 of about 55,100,000 results.Subcategories. Advertisement. How dividend yield is calculated. I am not sure I can understand your question. Your reference to a closing price of 8282 in November leads me to think you are asking what the dividend yield of the Dow is. Stock Price Annual Dividend / Dividend Yield. Lets play with some numbers: Wal Marts (NYSE:WMT) annual dividend is 2.04 and its price March 4, 2017 is 70.03. Calculate the dividend yield for an actual example of stock. Dividend yield equals dividend per share divided by share price.The total yield is the capital gain plus the annual dividend divided by the initial investment.

SAXON MATH BOOK COURSE 2 ONLINE - SAXON MATH BOOK COURSE 2 ONLINE Saxon Math Course 2 Student Book.pdf - Google Drive Saxon Math Course 2 Student Book.pdf. Saxon Math Course 2 Student Book.pdf. Sign In. Details ... Solutions to Saxon Math Course 2 (9781591418351 ... Saxon Math Course 2Saxon Math Course 2. Saxon Math Course 2. ISBN: 9781591418351 / 1591418356. Textbook solutions. FREE. Expert verified. 220. Buy the book on. Saxon Math Course 2 Student Book.pdf - Google Drive Saxon Math Course 2 Student Book.pdf - Google Drive ... Sign in Saxon Math Math - Course 2 - Math Help | Online Test Prep ... Get the exact Saxon Math Math - Course 2 help you need by entering the page number of your Saxon Math Math - Course 2 textbook below. Math - Course 2 Hake. Saxon Math. 2007. … Saxon Math, Course 2 (Student Edition) Online Ebook ... · Saxon Math, Course 2 (Student Edition) Online Ebook Download 22 March 2020 admin Download Saxon Math, Course 2 (Student Edition) Online Ebook Download book pdf free download link or read online here in PDF. Read online Saxon Math, Course 2 (Student Edition) Online Ebook Download book pdf free download link book … Saxon Math Course 2 | Download eBook pdf, epub, tuebl, mobi Download saxon math course 2 or read online books in PDF, EPUB, Tuebl, and Mobi Format. Click Download or Read Online button to get saxon math course 2 book now. This site is … Online Textbooks - CLS Resources Saxon Math Textbooks. Saxon Math 4. Saxon Math 54. Supplemental Practice. Saxon Math 5. Saxon Math 65. Supplemental Practice. Saxon Course 1. Saxon Math 76. Supplemental Practice. Saxon Course 2. Saxon Course … Saxon Courses 1-3 (Gr. 6-8) | Rainbow Resource Saxon Math Courses 1-3 for Grade 6, 7 and 8. 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Online schooling is a good option if you do good time management and follow a well prepared time table. Consider it as a great opportunity to learn more and learn better! As we all know excess of everything is bad. Everything has a limit if u doing it in efficient and effective manner.

What is the composition of atomic nuclei ? The ratio of protons to neutrons inside an atomic nucleus governs its stability. An atomic nucleus contains positively charged protons and neutral neutrons, typically packed within a nuclear radius of order of 10-15 metres. Z represents the total number of protons, whereas A represents total number of nucleons (sum of number of protons and neutrons inside atomic nucleus) of an atomic nucleus, then the nucleus of the element is represented symbolically as . represents a hydrogen nucleus that has 1 proton and 0 neutrons in its atomic nucleus. represents bismuth nucleus, that has 83 protons and 209-83 = 126 number of neutrons in its nucleus. Nuclei that contain the same number of protons, but different number of neutrons are known as isotopes. Carbon element exists in two isotopes forms: 98.9% of carbon atoms are of the form (6 protons + 6 neutrons), whereas 1.1% of carbon atoms are of the form (6 protons + 7 neutrons). What decides the stability of the atomic nucleus? Protons repel each other with long-range electrostatic forces between each other. It’s the strong binding nuclear forces between nuclear particles that hold the atomic nucleus together. Nuclear forces are short range, strongly attractive and charge independent. For a nucleus to be stable, the electrostatic repulsion is balanced by the attraction between the nucleons. But one proton can repel all the protons in the nucleus, as electrostatic forces are long range. On the other hand, a proton or a neutron can attract only its immediate neighbors with nuclear forces as these are short range forces. With the increase in the number of protons the number of neutrons in the nuclei has to increase in order to maintain the stability of the atomic nucleus. There comes a point when the large number of neutrons no longer achieves the balance of electrostatic and nuclear forces. The stable nucleus with large number of protons is Z = 83 is that of bismuth which contains 126 neutrons. All nuclei with number of protons in excess of 83 are seen as unstable. These nuclei spontaneously decay or rearrange their internal structures with time to achieve greater stability. This spontaneous disintegration of atomic nucleus causing rearrangement of internal structures is called radioactivity. This phenomenon was first discovered by the French physicist, Henry Becquerel in 1896. When an unstable or radioactive nucleus disintegrates spontaneously, certain kinds of particles or rays or high-energy photons or radiation are released. These rays produced by naturally occurring radioactive nuclei are alpha (α) rays , beta (β) rays and gamma (γ) rays. α rays consist of positively charged particles, each one being the nucleus of helium. β rays consist of negatively charged electrons along with an energy particle neutrino. γ rays are high energy electromagnetic radiations that do not cause a transmutation of one element into another. Its release simply results in change of energy levels of the nucleus. In a group of radioactive sample of nuclei, individual disintegration occurs randomly. As time advances, the number N of parent nuclei decreases in a given sample. The figure shows a graph between number of radioactive nuclei N along y axis versus time in seconds along the x axis. The plot shows an exponential fall in the number of radioactive nuclei with time. This graph of N versus time T indicates that the decrease occurs in smooth fashion. Number of atoms N approaches to 0 after enough time has elapsed. Half-life, T1/2 time of radioactive nuclei is defined as time required by one-half of the number of parent radioactive nuclides to decay by half in number or reduce to half the number. For example, radioactive Radium element, disintegrates into half of its original number in about 1600 years. In another 1600 years, another half the numbers will disintegrate, leaving only one fourth of the original numbers intact. So if in the above graph, if at time T = 0, the number of radioactive nuclides are No, after time = T1/2 , the number present is N = No/2. Further at time = 2T1/2, the number present will be N = No/4 . The value of half-life of radioactive nuclides depends upon the nature of the sample. It may vary from few seconds to billions of years. The activity of radioactive sample is the number of disintegration per second that occur in a given sample of radioactive sample. But note that each time a disintegration occurs the number N of remaining radioactive nuclei decreases. Activity of a radioactive sample is defined by dividing the change in number of radioactive nuclei, ∆N by time interval ∆t during which the change occurs. Average activity over the time interval ∆t is given as ∆N/∆t and this is proportional to the number of radioactive nuclei remaining in the sample. Here λ is proportionality constant, referred to as decay constant of the radioactive sample. The negative sign here indicates that with time due to subsequent disintegrations, the number of radioactive atoms decreases subsequently. The SI unit of activity is called Becquerel (Bq), where 1 becquerel of activity is equal to 1 disintegration per second. A bigger unit of activity of radioactive nuclei is curie (Ci) where 1 Ci = 3.7 x 1010 Bq For an estimate of activity of commonly used radioactive radium: The activity of radium used in watch dial to make it glow in dark is about 4 x104 Bq , that’s purely harmless to human exposure. On the other hand, the activity used in radiation therapy for cancer treatment is approx. billion time greater, or 4 x 1013 Bq! Such an exposure for a fraction of seconds causes irreversible effect on living tissues. For the graph in Fig 1, depicting the radioactive decay of a radioactive sample, the equation that governs the rate of fall in number of radioactive nuclei as a function of time is given as: , assuming that at time t = 0, the number of radioactive nuclei were No Applying the conditions that at t = T1/2, number of nuclei present, N=No/2, the above equation is reduced to An important application of radioactivity is the determination of the age of archeological samples. If the fossil sample contains radioactive nuclei when it was formed, then the decay of the nuclei marks the passage of time as half of the radioactive sample decays after every some time equal to its half life. If the half-life of the radioactive element present in the sample is known, a measurement of the present day number of radioactive nuclei relative to original number gives an estimate of the age of the fossil sample. Since the activity of the sample is proportional to rate of disintegration, one way to know the age of the sample is to compare the present day activity with initial activity. Mass spectrometer is deployed to know the present day number of radioactive nuclei in a sample. isotope of carbon is radioactive with half-life of 5730 years. In earth’s atmosphere, it is present in the ratio of 1 atom for every 8x1011 atoms of normal non-radioactive carbon . This ratio is maintained as for every loss due to radioactive decay of is offset by the creation of new atoms due to cosmic ray interaction with upper layers of atmosphere. The same ratio of and is maintained inside all living organisms as long as they are alive. But as soon as an organism dies, it no longer absorbs from the atmosphere, but the count of present inside its body begins decline due to radioactive decay. So by applying the principles of radioactive decay rules to determine the present day number of numbers in the fossil remains, the age analysis of the organism can be done. This process of age analysis of fossil using radioactive is carbon dating. Q1. Carbon-radioactive isotope sample has 1000 nuclei. If its half-life is 5730 years, how many nuclei remain after 17190 years? Q2. During the disintegration of radioactive nuclei, the total energy, electric charge, linear momentum, angular momentum and nucleon number are necessarily conserved. Q3. Out of three types of radiations, alpha, beta and gamma rays, emitted by the unstable radioactive nucleus, only the …………………… radiations fail to get deflected by the magnetic fields. Q4. Radioactivity can be induced by applying strong electrical and magnetic fields on a heavy atomic nuclei. Q5. During radioactive decay the number of nuclei in the sample decreases ………………………….. with time. Context and Applications The concept of Half Life has numerous practical applications some of them includes figuring out the effective drug dosage to finding out when a radioactive becomes safe to use. If you look for a career in Nuclear Physics you will need this concept. This topic is significant in the professional exams for both undergraduate and graduate courses, especially for - Bachelors in Science (Physics) - Masters in Science (Physics) Want more help with your physics homework? *Response times may vary by subject and question complexity. Median response time is 34 minutes for paid subscribers and may be longer for promotional offers. Half Life Homework Questions from Fellow Students Browse our recently answered Half Life homework questions.

Actual energy is exergy—condensed and/or hierarchized energy, which can act (i.e., can do work). In the ultimate analysis, all potential energy is gravitational, while other potential energies, such as the strong-force potential energy and the electric potential energy, are temporary masks of the gravitational potential energy. Gravity condenses and hierarchizes anergy (potential energy) into exergy (actual energy). Potential energy is the energy of locality (spatial separateness): Gravitational energy, or potential energy, is purely an energy of position; that is, for any two specific masses, the mutual gravitational potential energy is determined solely by their spatial separation. But energy of position in space cannot be propagated in space; the concept of transmitting this energy from one spatial location to another is totally incompatible with the fact that the magnitude of the energy is determined by the spatial location. Propagation of gravitation is therefore inherently impossible. The gravitational action is necessarily instantaneous as Newton's law indicates, and as has always been assumed for purposes of calculation. - —Larson, Dewey B. Beyond Newton: An Explanation of Gravitation. North Pacific Publishers, 1964 At that, potential energy is negative. It is not a mere convention but a consequence of conservation of energy in the zero-energy universe—as the universe's matter descends into its own gravitational potential field, the matter's actual energy becomes more positive, while its potential energy becomes more negative, so that the universe's matter loses its rest mass and becomes increasingly nonlocal, i.e. increasingly capable of teleportation. Simple examples[change | change source] Bringing a rock uphill increases (i.e., makes less negative) its gravitoelectric potential energy. Stretching a rubber band increases its elastic potential energy, which is a form of the electric potential energy. A mixture of a fuel and an oxidant has a chemical potential energy, which is another form of the electric potential energy. Batteries too have chemical potential energy. Gravitational potential energy[change | change source] Gravitational potential energy is experienced by an object when height and mass is a factor in the system. Gravitational potential energy causes objects to move towards each other. If an object is lifted a certain distance from the surface from the Earth, the force experienced is caused by weight and height. Work is defined as force over a distance, and work is another word for energy. This means Gl Potential Energy is equal to: - is the force of gravity - is the change in height Total work done by Gravitational Potential Energy in a moving object from position 1 to position 2 can be found by: - is the mass of the object - is the acceleration caused by gravity (constant) - is the first position - is the second position Electric potential energy[change | change source] Electric potential energy is experienced by charges both different and alike, as they repel or attract each other. Charges can either be positive (+) or negative (-), where opposite charges attract and similar charges repel. If two charges were placed a certain distance away from each other, the potential energy stored between the charges can be calculated by: - is 1/4πє (for air or vacuum it is ) - is the first charge - is the second charge - is the distance apart Elastic potential energy[change | change source] Elastic potential energy is experienced when a rubbery material is pulled away or pushed together. The amount of potential energy the material has depends on the distance pulled or pushed. The longer the distance pushed, the greater the elastic potential energy the material has. If a material is pulled or pushed, the potential energy can be calculated by: - is the spring force constant (how well the material stretches or compresses) - is the distance the material moved from its original position Related pages[change | change source] References[change | change source] - Larson, Dewey B. Beyond Newton: An Explanation of Gravitation. North Pacific Publishers, 1964. "Energy is defined as the capability of doing work. Kinetic energy, for example, qualifies under this definition, and hence any kind of energy that can be converted to kinetic energy also qualifies. But gravitational energy is not capable of “doing work” as a general proposition. It will do one thing and one thing only: it will move masses inward toward each other. If this motion is permitted to take place, the resulting decrease in gravitational energy makes its appearance as kinetic energy and the latter can then be utilized in the normal manner, but unless gravitation is allowed to do this one thing which it is capable of doing, the gravitational energy is completely unavailable; it cannot do anything itself, nor can it be converted to any form of energy that can do something." - Shu, Frank H. The Physical Universe: An Introduction to Astronomy. University Science Books, 1982, p. 157. "Concluding Philosophical Comment. Zeldovich and Novikov have made the following intriguing philosophical point about the picture of the formation of a neutron star sketched here. They note that stars begin their lives as a mixture mostly of hydrogen nuclei and their stripped electrons. During a massive star's luminous phase, the protons are combined by a variety of complicated reactions into heavier and heavier elements. The nuclear binding energy released this way ultimately provides entertainment and employment for astronomers. In the end, however, the supernova process serves to undo most of this nuclear evolution. In the end, the core forms a mass of neutrons. Now, the final state, neutrons, contains less nuclear binding energy than the initial state, protons, and electrons. So where did all the energy come from when the star was shining all those millions of years? Where did the energy come from to produce the sound and the fury which is a supernova explosion? Energy is conserved; who paid the debts at the end? Answer: gravity! The gravitational potential energy of the final neutron star is much greater (negatively; that's the debt) than the gravitational potential energy of the corresponding main-sequence star (Problem 8.7). So, despite all the intervening interesting nuclear physics, ultimately Kelvin and Helmholtz were right after all! The ultimate energy source in the stars which produce the greatest amount of energy is gravity power. This is an important moral worth remembering and savoring. If we regard the neutron star as one gigantic atomic nucleus, we may also say that nuclear processes plus gravity have succeeded in converting many atomic nuclei into one nucleus. Problem 8.7 then shows that the ultimate energy source for the entire output of the star is the relativistic binding energy of the final end state." - Why is the Potential Energy Negative? HyperPhysics - Heighway, Jack. Einstein, the Aether and Variable Rest Mass. HeighwayPubs, 2011, p. 36. "Understanding why rest masses are reduced in a gravitational field only requires a simple insight: when an object is raised in a gravitational field, the gravitational potential energy increase is real, and exists as an increase, usually tiny, in the rest mass of the object." - Battersby, Stephen. Big Bang glow hints at funnel-shaped Universe. New Scientist, 15 April 2004 - Gribbin, John. In Search of the Multiverse. Penguin UK, 2009, p. 131. "Any concentration of matter more compact than an infinitely dispersed cloud (even a cloud of gas containing one hydrogen molecule in every litre of space) must have less gravitational energy than an infinitely dispersed cloud, because, when material falls together energy is removed from the field. We start with zero energy and take some away, so we are left with negative energy. The negative energy of the gravitational field is what allows negative entropy, equivalent to information, to grow, making the Universe a more complicated and interesting place, with hot stars pouring out energy, on which planets like Earth can feed, as they attempt to redress the balance." - ENIAC: The Birth of the Information Age. Popular Science, March 1996 - The ENIAC Effect: Dawn of the Information Age. ENIAC Museum - Chardin, Pierre Teilhard de. Man's Place in Nature. Collins, 1973, p. 100 - Chardin, Pierre Teilhard de. The Phenomenon of Man. 1955, p. 306 - Chardin, Pierre Teilhard de. The Phenomenon of Man. 1955, p. 172

Introduction to Linear Programming by Leonid N. Best Advanced linear programming : computation and practice by Bruce A. Reemtsen next. Wording Edition. - Managing Dementia in a Multicultural Society; - Biology and Culture of Asian Seabass Lates Calcarifer? - A Fable. - Linear and Integer Programming vs Linear Integration and Counting: A Duality Viewpoint. - Empirical Studies of Environmental Policies in Europe (ZEI Studies in European Economics and Law)! - Product details. In Conversation with The prize is awarded annually to scholars who have made fundamental, sustained contributions to theory in operations research and the management sciences. Since , five papers by Lasserre have appeared in Mathematics of OR , and in , Lasserre was appointed as an associate editor for the journal. Works under MDS 519.72 INFORMS reached out to Jean Lasserre in Fall to learn more about his recent awards, his predictions for the future of operations research OR , and his advice for students entering this important field. What follows is the first in a series of interviews that will showcase the thought-leaders in the field of operations research. How has winning this award impacted your professional life? Indeed the list of its important applications is almost endless. How has it impacted the field of optimization? Parrilo [ 11 ] used sums of squares decompositions for testing copositivity of a matrix and for some control applications. This hierarchy has finite convergence generically and provides the first optimality conditions for polynomial optimization with nonconvex analogue properties of the celebrated Karush—Kuhn—Tucker KTT [ 4,6 ] conditions in convex optimization. Concerning real-world scenarios, the impact is more nuanced. The Lasserre Hierarchy has played a critical role in helping solve nonconvex problems of relatively modest size e. Optimum Power Flow problems might be the first real-world large-scale scenario where this methodology can outperform other approaches. In this respect, an important theoretical challenge is to understand the power and limitation of convex relaxations, especially for hard combinatorial optimization problems. This challenge is now central in the theoretical computer science TCS community and has already made its way into computer science courses at some prestigious universities. Interestingly, in addition to optimization, Fourier analysis and techniques from quantum computing may also be appropriate and are explained and discussed in such courses. OR departments should introduce similar courses; for instance, powerful positivity certificates from real algebraic geometry and their dual facet on the K-moment problem should be taught in graduate courses. Because even if their proof is nontrivial and requires sophisticated mathematics, these positivity certificates are i easily understood, ii simple to implement, and iii can be used in many applications across different area. This does not happen frequently! Another challenging problem is the pure integer programming problem IP. Indeed, simple examples of knapsack problems one constraint and variables! Surprisingly, all basic ingredients of the simplex algorithm basis, dual vector, reduced cost are also hidden in a beautiful formula e. In my book [ 10 ] , I have tried to popularize this point of view on LP, IP, integration, and counting, but with no success at all! Students should not be afraid of learning basic ideas of real and functional analysis, integration and measure theory, algebraic geometry, lattice points, and tools like Laplace and Fenchel transforms, Cauchy residues, and so forth. Shop Linear And Integer Programming Vs Linear Integration And Counting A Duality Viewpoint 2009 Of course, large-scale problems are still a challenge, even though we can solve problems of much larger size than in the s. It would be interesting to evaluate the respective roles that algorithms and computer power played in this achievement in the past 50 years. Some important applications e. This, in turn, has boosted research and renewed interest in large-scale, first-order methods for structured convex problems. Other inverse problems also appear in optimal control and robotics e. Solving large-scale, semidefinite programs is another challenge. The early expectations of the s—that powerful interior-point methods would solve such problems—have not materialized. In contrast to LP solvers, which can solve huge size problems, and despite some progress, the size limitation of problems that can be solved by the current semidefinite programming SDP solvers is much more severe than for LP solvers. This is a pity because, for example, the powerful family of semidefinite convex relaxations for solving polynomial optimization problems is then limited to problems of modest size unless sparsity can be considered. - Beautiful Country: A Novel. - The Conquest (Saucer, Book 2)! - Elementary linear programming with applications /. - Mathematics of Operations Research :. - The Whitechapel Horrors (The Further Adventures of Sherlock Holmes)? Each convex relaxation in the hierarchy is a semidefinite program that, in principle, can be solved efficiently in time polynomial in its input size. However, as its size increases in the hierarchy, the convex relaxation becomes increasingly more difficult to solve. Jean-Bernard Lasserre - książki - hiqukycona.tk Fortunately, finite convergence eventually takes place generically. It is important to have fun and enjoy … and it helps spur creativity! It also recognizes my more recent work on linking LP, IP, linear integration, and counting problems. The Laplace and Z-transform are seen as analogues of the Legendre-Fenchel transform in the max-plus algebra. One of its main goals is to develop or provide alternatives to the moment-SOS hierarchy, so as to help solve difficult nonconvex problems in important applications e. Such problems are viewed as examples of the Generalized Problem of Moments, and the list of important applications is endless! But, this is an opportunity to emphasize how grateful I am to CNRS for providing me with a unique research environment with almost total freedom. In academic research, the freedom to pursue your intellectual interests is the most important feature that an employer can provide. 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Original Research: The Anchoring Effect in A Range of Plausible Anchors In this series on Original Research, I will be sharing about my findings from some of the mini-projects that I have carried out on my own. The anchoring effect is a systematic cognitive bias committed by individuals, when they rely too heavily on an initial piece of information for making a subsequent judgment. This is especially pronounced when the individual does not have much knowledge in the subject matter that he/she is assessing, and end up getting influenced by any information that comes before the judgment. One of the most popular examples is by Strack & Mussweiler (1997), who conducted a study asking participants to guess the age of Mahatma Gandhi when he died. But before asking for their estimates, the researchers exposed one group to a low anchor (“Did Mahatma Gandhi died before or after the age of 9 years old?”), and exposed another group to a high anchor (“Did Mahatma Gandhi died before or after the age of 140 years old?”). While it was impossible for both anchors to be the correct answer, they nonetheless had an effect on the participants, as the mean estimate from the low anchor group was 50, while the mean estimate from the high anchor group was 67. In trying to understand the effects of implausible anchors, Mussweiler & Strack (2001) repeated the study with plausible anchors as well, using 61 years old as the low anchor and 86 years old as as the high anchor. As it turned out, the plausible anchors still had an effect on the participants, with the low anchor group having a mean estimate of 63 and the high anchor group having a mean estimate of 70. However, the deviations from the plausible anchors were a lot less. The research by Mussweiler & Strack got me wondering, is it possible to find an anchor that would result in a mean estimate with minimal deviation? What exactly happens within the range of plausible anchors? This was something nobody had explored and I was curious to find out. Extrapolating how the deviations reduced from the implausible to plausible anchors, it would seem natural to assume that the deviation of mean estimates from their respective anchors should continue to reduce until the high and low anchors converge at a point in the plausible range where there is little or no deviation. However, since all the anchors in this range are considered plausible, it is also not unlikely that the amount of deviation from each anchor would not differ much. To find out which scenario occurs in real life, I designed a series of questions and administered them to participants through an online questionnaire. Participants received 5 Singapore dollars for completing the questionnaire. To replicate the type of question like the Mahatma Gandhi one, participants will need to have an idea of the question’s context, but not know the correct answer. I conducted a pretest with 20 different questions, allowing participants to freely make estimates without any anchors. The top 10 questions with the smallest variance in estimates were chosen for the main study. The anchors used for these 10 questions were determined by the means and standard deviations of the participants’ estimates in the pretest, which was what Mussweiler & Strack had done in their studies. A total of 5 anchors were used for each question, which included the pretest mean, +1.0SD, -1.0SD, +0.5SD and -0.5SD. In the Mussweiler & Strack studies, a question of whether Mahatma Gandhi died before or after the anchor precedes the estimate. As I was experimenting with anchors in the plausible range, I had to consider the anchor as a possible estimate. Hence, besides the usual “more than” or “less than” options, I allowed participants to choose a third option which was “roughly equal to” the anchor that they were shown. Participants of course did not know that the numbers they saw were anchors, and they also did not know other participants were shown other numbers. After making their estimates for each question, I also asked the participants whether they knew the right answer to the questions, and how likely they believed that the anchor they were shown was the right answer. The responses to these sub-questions were then used to draw further insights later on. 541 participants were recruited in total, through mass sharing of the questionnaire link as well as snowball sampling. The gender ratio was roughly equal, but the age group of the participants was skewed towards 21 to 30 years old. Most of the participants were Singaporeans, who had a degree and were working adults. Finding 1: Mean estimates from all anchors groups were tending towards the Pretest Mean. To compare all 10 questions together, the units had to be standardised, and to peg the group means to their respective anchors, the following formula was used: (Anchor Group Mean — Pretest Mean) ÷ Pretest SD This resulted in the following chart, where ‘0’ on the y-axis represents the pretest mean, while ‘1’ and ‘-1’ represent the pretest SD. Groups 1 to 5 on the x-axis represent each anchor from -1.0SD to +1.0SD, with the pretest mean in the middle at Group 3. It is immediately noticeable from the chart that the mean estimates for all 10 questions were following a similar pattern, with the exception of Question 1 (Dinner) and Questions 9 (Banana Tree). Mean estimates from groups with +1.0SD and -1.0SD as their anchors deviated the most, while the group with the pretest mean anchor seemed to have deviated the least. To make it easier to see, I aggregated the means of the 10 questions into a single data point for each anchor group. From the chart of the aggregated means, it becomes apparent that the results seem to be reflecting the scenario of Hypothesis 1, where there is almost no deviation of the mean estimate at the mid-point of the plausible range. On the other hand, mean estimates from the other anchor groups seem to be tending towards the pretest mean. This finding is rather interesting, as the pretest mean is by no means the correct answer, nor would the participants of the main study know what the mean of the pretest was. This seems to suggest that the mean from anchor-free estimates make neutral anchors, where overestimates and underestimates resulting from that anchor cancel out each other and allow the mean to converge back at the original anchor. However, looking at mean estimates does not give the full picture of how estimates deviate from anchors. Hence, I also looked at how individual estimates deviated. Finding 2: The larger the anchor value, the greater the anchor deviations. To find out how much individuals deviate from their given anchors, I calculated the absolute difference between an individual’s estimate and its respective anchor value, using the following formula: | individual estimate — anchor value | For example, if the estimates of 5 participants exposed to the anchor 80 were 78, 79, 81, 81 and 83, the calculation of the mean anchor deviation would be (|78–80| + |79–80| + |81–90| + |81–80| + |83–80|) ÷ 5, which gives an answer of 1.6. This differs with the deviation of the mean estimate in that the deviation is calculated before computing the mean. Once again, to compare all 10 questions together, the units had to be standardised, using the following formula: Anchor Group Mean ÷ Sample Mean This resulted in the following chart, where ‘1’ on the y-axis represents the mean anchor deviation of the entire sample. Amazingly, the mean anchor deviations for all 10 questions were following almost the same pattern, where the anchors with smaller values (Groups 1 and 2) had lesser deviations, while the anchors with larger values (Groups 4 and 5) had greater deviations. We can also see that in comparison to the sample mean (‘1’ on the y-axis), Groups 1 and 2 were generally below the mean while Groups 4 and 5 were mostly above the mean. To make it easier to see, I aggregated the means of the 10 questions into a single data point for each anchor group. From the chart of the aggregated means, it appears that the deviations were increasing at an exponential rate, where larger anchors disproportionately resulted in greater anchor deviations. This seems to resonate with the findings of Wong & Kwong (2000), who found that anchors with a larger absolute value (e.g. 7300 m) induced a greater numerical estimate than anchors with a smaller absolute value (e.g. 7.3 km), despite them being the same semantically. It is possible that participants were primed to adjust more widely when exposed to larger anchors, resulting in the greater deviations. Besides comparing deviations to the anchors that participants were exposed to, I also examined how trust in the anchor affected the deviations. Finding 3: The more likely participants think an anchor is the answer, the smaller the anchor deviations. If you recall me asking the participants how likely they believed that the anchor they were shown was the right answer, this is where the responses for that sub-question comes in. While it seems commonsensical that participants who believed that their anchor was the right answer would deviate less from it, no one has explicitly shown this before. Hence, I decided to compare the anchor deviations based on how likely they believed that their anchor was the right answer. Like what I did in Finding 2, to compare all 10 questions together, the units had to be standardised, using the following formula: Likelihood Group Mean ÷ Sample Mean This resulted in the following chart, where ‘1’ on the y-axis represents the mean anchor deviation of the entire sample. Groups 1 to 7 on the x-axis represent the Likelihood Group from ‘Very Unlikely’ to ‘Very Likely’. Like the charts in Findings 1 and 2, what we see in Finding 3 is again the same pattern occurring for all 10 questions, regardless of what the question was. Participants who found their anchors very unlikely to be the right answer had the greatest amount of deviation, while those who thought their anchors were the right answer had the least deviation. To make it easier to see, I aggregated the means of the 10 questions into a single data point for each likelihood group. From the chart of aggregated means, an S-curve is observed where the differences towards the extremes seem to be diminishing. This probably suggests that the polarisation between participants who think their anchor is unlikely vs those who think their anchor is likely is quite distinct. A Pearson’s correlation test between Anchor Deviation and Likelihood was conducted, and negative correlations were significant for all 10 questions at p < .001. The findings from this mini-project provided some new insights that have not been discussed before in past anchoring research. In summary, anchoring within the plausible range is actually not homogeneous. There is still a tendency for the mean estimates to move towards a central point, which seems to be the pretest mean in this case. This seems to suggest that despite not knowing the correct answer, there is a point where overestimations and underestimations are roughly equal, causing the mean estimate to converge with the anchor. In understanding how individuals deviate from anchors, the value of the anchor actually seem to play a part through a priming effect, resonating with the findings of Wong & Kwong (2000). Finally, while it is not a huge surprise, believing that an anchor is the right answer will certainly result in smaller deviations. This research has been presented at the 35th Annual Conference of the Society of Judgment and Decision Making in 2014, in Long Beach, California. - Mussweiler, T., & Strack, F. (2001). Considering the impossible: Explaining the effects of implausible anchors. Social Cognition, 19(2), 145–160. - Strack, F., & Mussweiler, T. (1997). Explaining the enigmatic anchoring effect: Mechanisms of selective accessibility. Journal of personality and social psychology, 73(3), 437. - Wong, K. F. E., & Kwong, J. Y. Y. (2000). Is 7300 m equal to 7.3 km? Same semantics but different anchoring effects. Organizational Behavior and Human Decision Processes, 82(2), 314–333. Originally published at: https://learncuriously.wordpress.com/2019/04/08/anchoring-effect-in-a-plausible-range

Posts by Ginny Total # Posts: 36 What is the fundamental frequency on a 5 m rope that is tied at both ends if the speed of the waves is 24 m/s? Radio waves travel at the speed of light, which is 3.00 ✕ 108 m/s. What is the wavelength for a FM station broadcasting at 97.9 MHz (the prefix M is "mega" and means 106)? A carnival clown, 80 kg, rides a motorcycle down a ramp and then up and around a large, vertical loop. If the loop has a radius of 18 m, what is the force that the track exerts on the rider at the TOP of the loop if the clown is traveling at an impressive speed of 15 m/s? Which does the following sentence contain? Walking by the river is our favorite pastime. participle infinitive as subject gerund as subject prepositional phrase as adjective I'm confused with whether it is a participle (a verb acting as an adjective) or a prepositional ... A bird sitting 16 feet above the ground in an apple tree dislodges an apple. After how many seconds does the apple land on the ground? my text book gives this equation: h = -16t^2 + vt + s (t = secs, v= vertical velocity, s = initial height) so i wrote: h = -16t^2 + vt 16 I... A kicker punts a football from 3 feet above the ground with an initial velocity of 47 feet per second. 1. Write an equation that gives the height (in feet) of the football as a function of the time (in secs) since it was punted. Okay. According to my text book, h = -16t^2 + vt... That's right.. sorry 'bout that. Thanks, Mate. 2500ft^2 x 1gal/(0.134ft^2)=~ 186567 gal. Right? Could you maybe walk me through this Algebra word problem? Here is the problem: Anita wants to put a moat around her rectangular castle. The castle is 45 ft by 60 ft. Let m be the width of the moat. 1. Write a polynomial that represents the total area of the castle and moat. 2... I need to factor the expression z^2 + 15z - 54 I think it's either (z - 9)(z + 6) OR (z - 6)(z + 9) Can you help me out? Thanks! I need to factor the expression t^2 + 4tv + 4v^2 I got this answer: (t + 2v)(t + 2v) Check my work? Thanks! A gold bar is exactly 31 cm long. You want to cut the bar into smaller segments so that one or a combination of segments can add up to every whole number of centimeters from 1 to 31. What is the least number of cuts you could make and what are the lengths of each smaller piece? According to order of operations, you have to multiply/divide first. Because multiplying comes before dividing in this problem, you multiply 5x9 to get 45. You now divide. 45/56 is rounded to about .80, and you add 5 to that to get 5.80 Hi, Rhianna. What you would first do is multiply: 7x7x7x7x7x7, which would equal 117,649. 7x7x7x7x7x7x7 would equal 823,543. 117,649/823,543 would finally equal 0.14285714285 If a data set is: 18, 9, 14, 10, 15.... What would be the mean absolute deviation? What I got was 13.2, but it seems incorrect. Please help! An electron accelerates from rest to the right, in a horizontally directed electric field. The electron then leaves the electric field at a speed of 4.0 × 106, entering a magnetic field of magnitude 0.20 T (Tesla) directed into the screen. Calculate the magnitude and ... Calculus and Vectors Show that proj(b⃗a⃗) =[(a⃗⋅b⃗)/(b⃗⋅b⃗)](b⃗) The answer is water pollution. A person throws a rock straight up into the air. At the moment it leaves the person's hand it is going 91 mph. When the rock reaches its peak, how fast is it going and what is the magnitude and direction of its acceleration? Ignore air drag What is a galaxy with no definite shape or size called? 1)Caliban is a sort of monster- half man, half beast. There is hostility betweem Prospero and Caliban. 2)Thanks to his magic art, Prospero subdues all the spirits living on the island, including Caliban, into his service. 3)Caliban claimed to own the island. It belonged to him... Triangle PQR has vertices P(1,2), Q(25,2) and R(10,20). Find the coordinates of the centroid. Find the coordinates of the circumcenter. Find the coordinates of the orthocenter. Find the equation of the line. Staples recently charged $17.99 per box of Pilot Precise rollerball pens and $7.49 per box for Bic Matic Grip mechanical pencils. If Kelling Community College purchased 120 such boxes for a total of $1234.80, how many boxes of each type did they purchase? a party suppilies store recored net sales of $423,400 for the year. the store's begining inventory at retail was $105,850 and its ending inventory inventory at retail was $127,020.what would be the inventory turnover at retail,rounded to the nearest tenth? A rectangular room has a perimeter of 94 and area of 546. What are the dimensions of the room? Two 6.7kg bowling balls, each with a radius of 0.14m , are in contact with one another.What is the gravitational attraction between the bowling balls? what network covers small ares, such as a group of buildings or campus? Same question. What other methods of measurment do sociologists use to calculate prejudice? How do I find the domain of the funcition algebraically and support my answer graphically? f(x) = sqrt(x2) + (4) I need help w/ a question: Insulin protein has 2 polypeptides A and B. Human and duck insulins have the same sequence except for six below. Is the pI of human insulin higher or lower than duck insulin? AA seq. A8 A9 A10 B1 B2 B27 Human Thr Ser Ile Phe Val Thr Duck Glu Asn Pro ... Will somebody please tell me the nitrogen,H20, and carbon formulas???? Thanks! Will somebody please tell me the following formulas: Nitrogen Water (H20) and carbon?? Thanks

« PreviousContinue » not be the form of the earth, because the strata increase in density towards the centre. The lunar inequalities also prove the earth to be so constructed; it was requisite, therefore, to consider the fluid mass to be of variable density. Including this condition, it has been found that the mass, when in rotation, would still assume the form of an ellipsoid of revolution; that the particles of equal density would arrange themselves in concentric elliptical strata, the most dense being in the centre; but that the compression would be less than in the case of the homogeneous fluid. The compression is still less when the mass is considered to be, as it actually is, a solid nucleus, decreasing regularly in density from the centre to the surface, and partially covered by the ocean, because the solid parts, by their cohesion, nearly destroy that part of the centrifugal force which gives the particles a tendency to accumulate at the equator, though not altogether; otherwise the sea, by the superior mobility of its particles, would flow towards the equator and leave the poles dry: besides, it is well known that the continents at the equator are more elevated than they are in higher latitudes. It is also necessary for the equi librium of the ocean, that its density should be less than the mean density of the earth, otherwise the continents would be perpetually liable to inun dations from storms and other causes. On the whole, it appears from theory that a horizontal line passing round the earth, through both poles, must be nearly an ellipse, having its major axis in the plane of the equator, and its minor axis coinciding with the axis of the earth's rotation. The intensity of the centrifugal force is measured by the deflection of any point from the tangent in a second, and is determined from the known velocity of the earth's rotation: the force of gravitation at any place is measured by the descent of a heavy body in the first second of its fall. At the equator the centrifugal force is equal to the 289th part of gravity, and diminishes towards the poles as the cosine of the latitude, for the angle between the directions of these two forces, at any point of the surface, is equal to its latitude. But whatever the constitution of the earth and planets may be, analysis proves that, if the intensity of gravitation at the equator be taken equal to unity, the sum of the compression of the ellipsoid and the whole increase of gravitation, from the equator to the pole, is equal to five-halves of the ratio of the centrifugal force to gravitation at the equator. This quantity, with regard to the earth, is of, or i consequently the compression of the earth is equal to, diminished by the whole increase of gravitation, so that its form will be known, if the whole increase of gravitation, from the equator to the pole, can be determined by experiment. But there is another method of ascertaining the figure of our planet. It is easy to show, in a spheroid whose strata are elliptical, that the increase in the length of the radii, the decrease of gravitation, and the increase in the lengths of the arcs of the meridian, corresponding to angles of one degree, from the pole to the equator, are proportional to the square of the cosine of the latitude. These quantities are so connected with the ellipticity of the spheroid, that the total increase in the length of the radii is equal to the compression, and the total diminution in the length of the arcs is equal to the compression multiplied by three times the length of an arc of one degree at the equator. Hence, by measuring the meridian curvature of the earth, the compression, and consequently its figure, become known. This, indeed, is assuming the earth to be an ellipsoid of revolution, but the actual measurement of the globe will show how far it corresponds with that solid in figure and constitution. The courses of the great rivers, which are in general navigable to a considerable extent, prove that the curvature of the land differs but little from that of the ocean; and as the heights of the mountains and continents are inconsiderable when compared with the magnitude of the earth, its figure is understood to be determined by a surface at every point perpendicular to the direction of gravitation, or of the plumb-line, and is the same which the sea would have if it were continued all round the earth beneath the continents. Such is the figure that has been measured in the following A terrestrial meridian is a line passing through both poles, all the points of which have their noon contemporaneously. Were the lengths and curvatures of different meridians known, the figure of the earth might be determined; but the length of one degree is sufficient to give the figure of the earth, if it be measured on different meridians, and in a variety of latitudes; for if the earth were a sphere, all degrees would be of the same length, but if not, the lengths of the degrees will be greatest where the curvature is least, and will be greater exactly in proportion as the curvature is less; a comparison of the lengths of the degree in different parts of the earth's surface will therefore determine its size and form. An arc of the meridian may be measured by observing the latitude of its extreme points, and then measuring the distance between them in feet or fathoms the distance thus determined on the surface of the earth, divided by the degrees and parts of a degree contained in the difference of the latitudes, will give the exact length of one degree, the difference of the latitudes being the angle contained between the verticals at the extremities of the arc. This would be easily accomplished were the distance unobstructed, and on a level with the sea; but on account of the innumerable obstacles on the surface of the earth, it is necessary to connect the extreme points of the arc by a series of triangles, the sides and angles of which are either measured or computed, so that the length of the arc is ascertained with much laborious computation. In consequence of the irregularities of the surface, each triangle is in a different plane; they must therefore be reduced by computation to what they would have been, had they been measured on the surface of the sea; and as the earth may in this case be esteemed spherical, they require a correction to reduce them to spherical triangles. Arcs of the meridian have been measured in a variety of latitudes north and south, as well as arcs perpendicular to the meridian. From these measurements it appears that the lengths of the degrees increase from the equator to the poles, nearly in proportion to the square of the sine of the latitude; consequently the convexity of the earth diminishes from the equator to the poles. Were the earth an ellipsoid of revolution, the

espn win probability model. The Binomial Model. hopfensperger, h. To examine more deeply, I'll compare 6 independently created win probability models using projections from Super ESPN: ESPN's predictions, provided by Henry Gargiulo and Brian Burke, are derived from an ensemble of machine learning models. Get up to $3,000 (plus $0 commissions)1 Learn how. Given multiple season's worth of data, created a win probability model by creating confidence intervals. American Athletic Conference foe, Cincinnati, who is projected to win the conference, is listed with a 9. › Get more: Win probability calculatorDetail Windows. We're removing some of the variables such as quarter, clutch and home/away_days_rest which include NAs. Probability of getting a head = 0. I haven't coded anything since college. - ESPN top www. Probabilities—Textbooks. Let's take football (soccer) for example. Take charge of your finances with a new E*TRADE brokerage or retirement account by December 31. The problem of in-game win probability is to identify the chances of a team (e. Using random forests to estimate win probability before each play of an nfl game. NflWAR: A reproducible method for offensive player evaluation in football. Virginia is seeded first again in 2019, and Bouzarth says the Cavaliers have by far the best chance of all of the No. , the home-team without loss of generality) winning the game given The model itself is based on a logistic regression model. The win probability graphic/discussion on ESPN is literally taking a sword and sticking it through the chest of any fun left in baseball. You have reached ESPN's UK edition. Also, is there a database that contains historic pre-game win probabilities? I don't know how ESPN specifically models it, but 538 has described their model (which is pretty simple, relative to Fangraphs and Baseball Prospectus, which use individual player projections). Using previous research by Hal Stern, Winston posited that the final margin of victory for an NFL team in a given game can be approximated as a normal random variable. 12% probability of going undefeated based on those projections. If you score the most points, you'll have a chance to win the grand prize: a trip for two to Hawaii for the 2021 Maui Jim Maui Invitational and a $10,000 Amazon. As the game progresses, one would expect Brier scores to decrease, and that is what you see for the Inpredictable Brier scores. ESPN also boasts a win probability model. In-game win probability models have become increasingly popular in a variety of sports over the last decade. Best NBA Parlay Picks Tonight: Expert Bets & Player Props for Wednesday 1/5/22. Before we do so, let's split the dataset into training and testing sets with To do so, we need to convert the estimated win probabilities from the logistic regression model to a definitive prediction of. Win probability (WP) models seek to predict the outcome of a game at any moment, given its unique situation, and give us the ability to influence, critique, or congratulate decision making in those defining moments. Introduction to Probability. Using this fantastic site can help place smart wagers in an easy-to-use fashion and maximize the ROI on the wagers. ESPN: Serving sports fans. 2021 NFL Expert Picks | ESPN great www. Use code: BONUS21. While for a game of infinite duration a linear model could be a very good approximation, the. The Titans' win eliminated the Colts' chance to win the AFC South. 1 pick, and he will have to play catch-up after serving a three-game. Better known as Merton structured approach. Ben Roethlisberger describes his emotions after. Therefore, in this paper we present the design of iWinRNFL, an in-game win probability model for This work forms the basis for ESPN's prediction engine, which uses an. o ESPN's reporting is not flawless 111. I've never touched sports data in my life. College Football Playoff Predictor - ESPN. ESPN's win probability model agreed with Brandon Staley's decision to go for it on 4th-and-1 in the 4th quarter. How far is your team going to proceed into the playoffs? The NFL Post Season Probabilities table presents the probabilities that your team will proceed to different playoff rounds. I remember seeing win probabilities during ESPN. Covers the rules of addition, subtraction, and Often, we want to compute the probability of an event from the known probabilities of other events. Logistic or probit preferred in most applications. Better Predicted Probabilities from Linear Probability Models. In contrast to ESPN’s win probability metric for football, our model only gave the Atlas a slight advantage after their last goal even though so little time was left. Today Clinton’s odds of victory range from 66 to. I can already tell the win probability numbers ESPN flashes occasionally above the scoreboard is going to annoy me all year. According to ESPN, FPI is a statistical model that works as follows: “[It] determines forward-looking strength ratings for every team based on a variety of factors, including team win total, projected starting quarterback, returning starters and past performance on offense, defense and special teams. The polls-plus is the most stable over the course of the campaign but the three models are expected to converge as we get closer to the election. 5 = Probability of getting a tail since there are only two possible outcomes. The roots of our win probability model lie in the theory put forth in chapters 43 and 45 of Wayne Winston's book Mathletics. CLE: PIT: PIT 100%. To specify game state for the Elam Ending, you need to know three things: Distance to Target Score: How far away the leading team's score. by Kyle Bennison. 4 percent probability odds — the sixth best in the country — to reach the Final Four. Probability - A risk is an event that "may" occur. In-game win-probability models have been extensively studied in baseball, basketball and American football. 3 Probabilities Defined on Events. The data is provided by ESPN Below are the five most impactful plays of the game (as measured by ESPN's win probability model) ordered chronologically. Here's how the ESPN win probability model assessed the decision, per ESPN's Mike Reiss: ESPN win probability metrics had the field goal (fourth-and-3 from the 37-yard line) as the correct decision by the Patriots. Compared to expected points models, there is considerably more literature on different methodologies for estimating the win probability of a play in. NFL’s Next Gen Stats captures real time location data, speed and acceleration for every player, every play on every inch of the field. IMDb is the world's most popular and authoritative source for movie, TV and celebrity content. RATIO The rank of a team subjectively measures its relative strength compared to other teams. A description of ESPN’s methods was not listed on its site, but in 2017 Michael Lopez, then an assistant. In politics there are so many intangible factors. ESPN Data; ESPN Metrics; ESPN Ratings; ESPN Scoreboard; ESPN PBP; Articles Vignettes; Introductory; CFB Analytics with cfbfastR; Animated Win Probability Plots; Fourth Down Tendency Plots; Rolling EPA Graphs; Visualizing Team Talent with Player Recruit Rankings; Intro to Visualizing Recruiting Geography; Expected Points Model Fundamentals; Part. The win probability model takes all notions of streakiness out of the equation. One strategy would be choosing the team with the highest probability to win a game each week, disregarding future game probabilities; this is known as the greedy method. That number, which is impacted pitch-by-pitch during a game, is integrated into our baseball Gamecast application on ESPN. ESPN has a probability winning percentage in which what team has more of a probability of winning the game. I promised to blog about this case study I developed for my probability models course. Ten percent is huge for a 16-1,” she said. In two-way parlay betting, the probability that an unskilled player using random guessing will obtain k wins in n bets has a binomial distribution, X, with parameters n, ¼. For both discrete and continuous random variables we will discuss the following In general, if X and Y are two random variables, the probability distribution that denes their si-multaneous behavior is called a joint. Each game state is simulated 10,000 times to estimate win probability. Since I have a binary outcome in my linear proability model I can interpret the coefficients (i. Tuning an In-Game Win Probability Model Using xgboost. There are two approaches to the study of probability theory. Exponential distribution models the interval of time between the calls. Comparison of default probability models: russian experience. Therefore, we expect TQR variable to be positively correlated with the probability of winning. A team winning by 3 touchdowns at half is historically an almost sure pick. probability interpretation model linear logarithm. , probability laws) used for solving probability problems. 222ESPN includes win probability graphs in its match reports for basketball (e. Journal of Quantitative Analysis in Sports, 10 (2), 197–206. 86, for international readers), while their counterpart with a win probability of 46. 6% (up from 51. Many of the plays were. Like team A is up a certain amount of points and time remaining they’ll have a certain. Let us take a quick look at its intuition. Probability theory, a branch of mathematics concerned with the analysis of random phenomena. ESPN’s model is based on 10 years-worth of NFL play-by-play game data, which is a lot of plays. Projected running record: 11-1 List Big Ten bowl projections, College Football Playoff predictions after Week 6. And I really want to start all of this from scratch because I want to win my office pool?. Thus, my win probability model is really a sequence of 280 logistic regressions- one for each ten second interval between 1-40 minutes remaining Now let's compare some example charts my model generates compared to those of FiveThirtyEight and ESPN. Details: With the update, my comments regarding the ESPN model being too reactive no longer apply. The model takes as an input the probability that each player would win a single point if serving, and then constructs a tree of match outcomes using this probability to. The Colts could have clinched a playoff spot with a victory over Las Vegas, but they lost 23-20. A first-quarter TD which creates a 7-0 score probably moves the scoring team's win probability to about 60%. Modeling the chances of the big numbers can be important to getting the win probabilities correct, especially late in the final round. Team Stats. In-game win probability is a statistical metric that provides a sports team's likelihood of winning at any given point in a game, based on the performance of historical teams in the same situation. Then in simulating the tournament, the model rolls a 100-sided die with each team probability of a win representing one side of that die. Click a team name to proceed to the team playoff picture page. The exact model Burke developed is proprietary, now owned by ESPN. A few weeks ago, we started tweeting out the post-game win probability plots for every FBS college football game. 75 seconds). I have chosen two charts for some of the. Win probability is based on a model built on actual outcomes of NFL games from recent seasons that featured similar circumstances. Picks For Every Big Week 17 Nfl Game Picks To Win Best Bets More Cbs Sports Hq. I was perusing ESPN and stumbled upon this gem. Maksim Horowitz, Samuel Ventura and Ronald Yurko developed a win probability model (and the wonderful nflscrapR package that was used to load in the play by play data used in this analysis) that uses a multinomial logistic regression to evaluate the value of field position and a Generalized Additive Model (GAM) to output a win probability. Part 1: Sections 5. A team that wins with a probability of 53. Open an account. The winner advances, and the same process repeats for every game. Along with outcomes of each play, their model has four major components: win probability, expected points, division of credit, and a \clutch" index. 16 Gardner-Webb a nearly 10-percent chance of victory. The probability of it occurring can range anywhere from just above 0 percent to just below 100 percent. It must be in accordance with a fundamental framework that we discuss in this section. What is probabilistic modeling, and how can it be used to boost your campaign performance measurement? Find out in our mobile marketing glossary. Tweets by CFB_Data. The videos in Part I introduce the general framework of probability models, multiple discrete or continuous random variables, expectations, conditional distributions, and various powerful tools of The textbook for this subject is Bertsekas, Dimitri, and John Tsitsiklis. We then outline a model from which we obtain the in-game posterior win probability. I took a random game from bet365. NFL Expert Picks - Week 16 Week 1 Week 2 Week 3 Week 4 Week 5 Week 6 Week 7 Week 8 Week 9 Week 10 Week 11 Week 12 Week 13 Week 14 Week 15 Week 16 hidden MIA at NO Mon 8:15PM. ESPN's win probability model didn’t like Morris' decision, either. Pro Football Reference Win Probability Convert! free convert online with more formats like file › Get more: Nfl game win probabilityDetail Convert. Some models—Lock and Nettleton (2014), seemingly ESPN—draw upon ensemble learning methods that. Each stage she plays will be equally likely to result in her either winning or losing 1 unit, with the results from each stage being independent. Find ratings and reviews for the newest movie and TV shows. Expected Points Added - Total expected points added with low leverage plays, according to ESPN Win Probability model. ADVANCED ALGEBRA. Intuition tells us that at any given moment during a game, some of the greatest factors influencing the outcome include: the amount First, rather than modeling the outcome at each time interval, we're instead trying to predict the probability of each team scoring a goal in each of the. In-game win probability models provide the likelihood that a certain team will win a game given the current state of the game. Here's how the ESPN win probability model assessed the decision, per ESPN's Mike Reiss Analytics aren't the only tool that should be used when making these kinds of game-altering decisions, but choosing the path with the highest probability of success is often a good way to go. Introduction to probability models / Sheldon M. com's coverage of MLB games 5 or 6 years ago; with each pitch they would adjust Anyway, using this model, our intention is to live log a selection of games so that as you are watching them, the win odds chart will update automatically and in real-time. The art of estimating win probability involves choosing which pieces of context matter. But linear models still have some attractions. Comparison to common heuristics - Team ESPN will compare the chosen solution against other common strategies. The corrected numbers (with updates through 2/10/18) can be found in this tweet. Below is a win probability chart for the Denver Broncos' win over the New England Patriots in the AFC Championship. About 258 results (0. Numerous papers dedicated to the probability of default model (PD model) creation studied circumstances associated with recessions in developed countries, not in transitional ones. Chapter 5: JOINT PROBABILITY DISTRIBUTIONS. ESPN has a proprietary expected points metric, but does not detail the specifics of how it is calculated (Pattani, 2012). Last year, I tweeted about a win probability model I created for soccer (or football, depending on where you are from) and the 2019 FIFA Women's World Cup case study. Win probability is a statistical tool which suggests a sports team's chances of winning at any given point in a game, based on the performance of historical teams in the same situation. The Exponential Model estimates the winning probability of each team by modeling the time between a team's successive winnings in a round as an exponential random variable. 5% chance of. Become a patron. Patreon Widget. Model Win Probability using Logistic Regression. But win probabilities models can still be useful. ISBN 978--321-50046-5 1. Yes, a model with varying coefficients would make sense. Welcome to Tournament Challenge Second Chance, the game that gives you another go at the men's bracket, starting with the Sweet 16 games. If you were to plug in a Tom Brady factor, you'd also have to plug in things like the Falcons defense. Moreover, when I have the log of an independent variable, I also talk about percent increase instead. Each player has his own set of probabilities for each hole based on his OWGR-converted score relative to the rest of the field. The probabilities of success and failure need not be equally likely, like the result of a fight between me and Undertaker. The reason they include NAs is because for every first game of the season, there is no number of "rest days" (unless we used preseason. February 11, 2018 update: The Brier score chart at the bottom of this post had an incorrect value for the ESPN "Start of Game" score. Probability and Statistics for Engineers and Scientist - 9th Edition (by Walpole, Mayers, Ye). 8% WP punt: 47. Like team A is up a certain amount of points and I can say you that they are calculated by algorithmic machines models as opposed to static mathematical formulas. We then train an LSTM on the data, feeding the points of a given match as the sequential data. DeGroot, Mark J. For each event E of Consider player i. For example, we have an MLB win probability model from Elias. This chart also supports the notion that the ESPN win probability model is in fact too reactive early on in the game. The Win Probabilities right now are really good. Over $1 MILLION in ESPN games up for grabs! Want more chances to win? Check out these ESPN games (and play against the StreakMasters in every game!): - Capital One Bowl Mania: $1M for a perfect score! (Join all 3 ESPN Streak groups: Standard, Spread and Confidence) - Monday Night Pick'em: $180,000 in prizes!. ESPN's win probability model hated the Jets decision to kick on 4th-and-goal from the 1-yard line inside with 1:49 left and up two points. Com- pared to expected points models, there is considerably more literature on different methodologies for estimating the win probability of a play in football. I'm not convinced any of those exist in any systematic manner, but it's tough to settle the question. Win Probability 100%. Win probability added (WPA) is a sport statistic which attempts to measure a player's The win probability for a specific situation in baseball FiveThirtyEight's MLB forecast uses a pitcher-adjusted Elo model to project the winner of every game and the chances that each team will win the World. in a linear probability model are always heteroskedastic. model_train: Estimate win probability model using Armchair Analysis nyt_fg_model: Sub in simple logit for field goal success rates. ESPN’S BPI takes into account the relative strength of the two teams and the location of the game to generate one team’s probability of a win. This lesson describes three rules of probability (i. Probabilistic graphical models (PGMs) are a rich framework for encoding probability distributions over complex domains: joint (multivariate) distributions over large numbers of random variables that interact with each other. For example, last week, ESPN released its playoff projections for the 22 teams heading to Orlando using a predictive version of their real plus-minus, and the Kings only have a 14. The majority of the chapters of this book will be concerned with different probability models of natural phenomena. The model indicates that there is a positive relation between the payment-to-income ratio and the probability of a denied mortgage application so individuals with a high ratio of loan payments to income are more likely to be rejected. Wrote a GUI to graph the confidence intervals, allow user interaction to follow the flow of the game. Prediction of the in-game win probabilities has been investigated in other major sports, such as basketball, football, and hockey. Get personalized recommendations, and learn where to watch across hundreds of streaming providers. The Redskins are going to have to win some games on the road, and this is a winnable one they need to get. ESPN's win probability model agreed, narrowly, with the Bills' decision to go for it on 4th-and-1 from the 3-yard line with 22 seconds remaining. This lesson covers some important rules that. The NBA has a large 11-game slate tonight, which is a perfect opportunity to use Awesemo OddsShopper to build some of the best NBA parlay picks today. It was Atlanta's 3rd loss this season with a win probability of at least 98%, the most by a team in the last 20 seasons. Introduction to Probability Models, Tenth Edition, provides an introduction to elementary probability theory and stochastic processes. Expect to also see TikTok providing enhanced options for custom branded hashtag. Probability of Win: Mystery and Magic - Shipley Windows. Win probability models are useful for a variety of reasons; NFL teams can consult them to determine whether or not to make certain decisions, like whether punting or going for it on fourth down is more likely to The Falcons had a win probability as high as 98. A win probability from any scoreline communicates how much a team or player is favored to win. Our mission is to serve fans across every ESPN platform with the fastest, most accurate and insightful statistics, scores, news and analysis. Browns Vs Steelers Pre Game Join The Conversation Watch The Game On Espn. With Applications to Multiple Imputation. The model consists of distributions that are specied via FDA methods. I'd play around with the data, graph some estimates based on different timesteps, and then But the most efficient way to get there is to model the score differential and then map that back to win probabilities. Patriots 34, Falcons 28 (OT) ESPN Stats & Information. 4% chance to win by going for it, but it would have given them a 71. See full list on espn. Turkey vs Ukraine hwin, draw, awin 2. Every sport has its own. and start with a generous cash bonus. These probabilities reflect the actual point frequency for all possessions from the pre-Covid 2019-20 NBA season. By John Dixon @Arrowheadphones Sep 6, 2021. Using the baseline bet of $110 for every bet made (since a bet must be $110 to win $100 at -110 odds), and assuming a bet is placed for the under every time the model predicts the probability of the over winning below 45%, a single bettor would make a profit of $3,350 on the test set sample. Win probability: 39 percent. All you have to do is select which teams you think will win each of the 15 remaining tournament games that take place from March 27 on. "The win probability graphic/discussion on ESPN is literally taking a sword and sticking it through the chest of any fun left in baseball". And its Q2 Brier score is only just as good as the pre-game number. Now that we have our ratings we can compute the probability a team will win against another team, simulate the games 1,000 times and see how many wins each team comes. Something weird about their model. There are 3 possible outcomes, home win, draw, away win. Their probability assessment doesn't add up to 100%, they take. Consider an experiment whose sample space is S. Based on the fact that randomness or uncertainty plays a role in predicting outcomes, predictive modeling is used in a wide variety of fields. Quarterback Jameis Winston continues to try to live up to the potential that made him a No. ESPN also boasts a win probability model. The outcome of a random event cannot be determined before it occurs, but it may be any one of several possible outcomes. Maksim Horowitz, Samuel Ventura, & Yurko, R. Nate Silver and FiveThirtyEight does the best political modeling around, and his projections were less. That same score late in the fourth moves it closer to 90% or 95%, depending on time remaining, timeouts, etc. I know that the linear probability model is essentially an OLS regression with a binary dependent variable, however, I am unsure how to estimate a model with the overall gender I've seen models where gender of firstborn predicts likelihood of divorce, but not where it predicts likelihood of marriage. StatRdays: The Easiest Model You’ll Ever Make →. 2%, and the win probability of going for it at 34. In contrast, they have received. “Virginia is going to try to. The aim of this work is to create a win probability model that improves upon shortcomings in existing models. After some slight modifications to the model and additions such as a home field advantage modifier, the model was ran on every game in the CFB Playoff Era to date (2014-Present). Linear models can be useful for binary outcomes. In a card game with my friend, I pay a certain amount of money each time I lose. Cynthia Frelund's mathematical model projects the final scores, win probability, and cover probability for every NFL game in Week 18 of the 2021 season. Probability of Default model using equity prices. Clearly, in order to master both the "model building" and. ESPN’s Basketball Power Index is a predictive model for team strength based on performance factors. (Note: It can't be exactly 100 percent, because then it would be a certainty, not a risk. Hypotheses TQRg: The higher the TQR score, the higher the probability that a team would win a game. ESPN Win Probability for the Padres-Dodgers Marathon. I won't ruin it for anyone, but it basically lets you pick a team and see what their probability of making the CFP is based on the outcome of their season + Conference Championship (or lack of). 2%) All-Time Series: Michigan leads 58-51-6. run_bot: Calculate the expected win probability for a given game simulate_scenarios: Simulate game state after each possible outcome. Meet Colin Davy, the Action Network data scientist who built the win probability model, and read about his path to the company. If you buy one ticket, let's analyze what happens to G, the amount you gain. So if you really just # Be nice to the ESPN CDN by storing team logo images locally. In-game win-probability models have been extensively studied in baseball, basketball and American football. Win probability models tend to get the most attention when they are "wrong". You can see this in the win probability studies, like in the The Hidden Game of Football. One is heuristic and nonrigorous, and attempts to develop in students an intuitive feel for the subject that. In-Game Win Probability for Super Bowl LI (ESPN. The exact same issue comes up in. That won't necessarily make it foolproof, as it all, essentially, comes down to your creative, but TikTok will look to facilitate trend jacking as much as it can via automated means, and/or through creator partnerships. But for ESPN, the Brier score actually goes up in the first quarter, compared to the pre-game Brier score. But how do you best use market based signals? The Merton model is actually a variation of the Black Scholes model. The PLT Win Probability Engine did not award the advantage to the Atlas until Cloutier’s eventual game-winning goal. 3% would be. › Get more: Win probability calculator nflDetail Drivers. Includes bibliographical references and index. Judging Win Probability Models - inpredictable. espn win probability calculator | Use our converter online, fast and completely free. A probabilistic model is a mathematical description of an uncertain situation. It uses a weekly win probability to map it all out. Then the probability of winning drops away after a score since the other team has the ball and is nflfastR includes its own win prediction model which is more accurate than mine. Buckeyes Win Probability: 57. Oklahoma Basketball: ESPN BPI Win probability for OU's 1st. “Virginia does look like the most vulnerable 1. Details: Cynthia Frelund's mathematical model projects the final scores, win › Get more: WindowsDetail Windows. ESPN's win probability model disagreed with Urban Meyer's decision to attempt a field goal on 4th-and-goal from the 3-yard line in the fourth quarter. 1s to be upset, with their models giving No. Intuitively, this species the "likelihood" of any outcome. The P-F-R Win Probability Model Pro-Football … › Get more: Espn win probability calculatorDetail Convert. Discover Next Gen Stats News, Charts, and Statistics. While one can produce this from any model of choice, those in. Like team A is up a certain amount of What is probability win? How do you calculate probability in sports? How can I know the winning team by odds? How does Google match. 4 Introduction to Probability Models. The Live Win Probability model calculates the probability of each outcome occurring by simulating the remainder of the match 100,000 times. This video introduces the concept of the linear probability model, and explains the intuition behind the theory. NBA-Win-Probability-Model's Introduction. , & Nettleton, D. win probabilities to act as an anchor. Belichick's decision was supported by ESPN's win probability model, which forecast the win probability of the field goal attempt at 42. How Our NFL Predictions Work. ESPN win percentage probability by game. 8% would be represented at fair odds of -116 (1. They were a nice addition to our Saturdays, but there were some issues. How the model works. According to the formula, there's no patches of good or bad play, no dips in motivation, so extra energy to finish off a set, etc. kranendonk, r Probability is concerned with anticipating the future, with the hope of dis-covering models that for the winning prize. Burke (2009) and ESPN provide an intuitive explanation for what expected points means, however they do not go into the details of the calculations. ESPN's Total Quarterback Rating (Total QBR), which was released in 2011, has never claimed to be perfect In essence, the QBR metric is very close to the observed win rate. We're now ready to build our prediction model. Probability Models. 35/100 increase in the probability of Y=1. com gift card. Cynthia Frelund's mathematical model projects the final scores, win probability, and cover probability for every NFL game in Week I thin ESPN win probability is broken, so I made this quick video to complain about it. These models serve as a tool to enhance the fan experience, evaluate in game-decision making and measure the risk-reward balance for coaching decisions. 7%, per ESPN's win probability model. I win `$4` if I draw a jack or a queen and I win `$5` if I draw a king. Second, win probability models can improve the fan experience by telling the story of a game. At halftime, with the Chiefs up 31-10, it calculated that they had a 96. As explained in Sports Economics (Berri), complex invasion win probability models are incredibly hard to produce, so The second issue that is shown in win probability research is the uncertainty that the models have, as expected in sports, sometimes underdogs win when all the odds seem. * You can also test out your skills by picking all the NCAA tournament games and more via the ESPN Streak game, where you can compete for $5,000 in guaranteed prizes each. I got an A in College Algebra but I'm pretty. of a quarterback. Then, to complete the probabilistic model, we must introduce a probability law. The Exponential Model estimates a zero probability for events that have not occurred in the training data set. All future unplayed games are assumed won/lost with a probability based upon relative team strengths. Just like the original Tournament Challenge game, you will earn points. CollegeFootballData. The model probabilities update in real time as the action unfolds on the pitch, providing fans with a live insight into the likely outcomes of the match. Probability and statistics / Morris H. 1% Needed to have a 51% chance of conversion to justify. Includes random variables, probability distribution functions wih relationship to probability and expected value; variance and standard deviation. Using the binomial model, the probability that someone will get a high number of wins in repeated play can be determined, with a formula of the form P(X. Probability will be most useful when applied to a real experiment in which the outcome is not known in advance, but there are many hypothetical. ESPN's win probability model agreed with Dan Campbell's decision to go for it on 4th-and-1 on their own 28 with 4:08 remaining (that the Lions failed to convert). 5% WP kick: 83. So for investment of 100 $ on given result, you either loose 100 $ or win: 220 $, 340 $ or 320 $. Register for a free API key. A description of ESPN's methods was not listed on its site, but in 2017 Michael Lopez, then an assistant professor at Skidmore College, now the NFL's Director of Data and Analytics, described the model as "derived from an ensemble of machine learning models. 7% chance with a field goal attempt. 4% chance to win. According to. Win Probability. David "The Brain" Borne and Jamie "Landshark" Langridge waged battle inside a 2 days ago · NBA Odds are determined by. It gave the Falcons a 67. ESPN’s FPI gives the Chiefs the best chance to win Super Bowl LVI On Monday, the network released the first look at its NFL statistical model for 2021. Cleveland Cavaliers 26. cache_logo_image <- function(logo).

Compute Population Mean Margin Error 95 Confidence Interval Your email Submit RELATED ARTICLES How to Calculate the Margin of Error for a Sample… Statistics Essentials For Dummies Statistics For Dummies, 2nd Edition SPSS Statistics for Dummies, 3rd Edition Statistics Lesson 11: Hypothesis Testing Lesson 12: Significance Testing Caveats & Ethics of Experiments Reviewing for Lessons 10 to 12 Resources References Help and Support Links! Popular Articles 1. If you're seeing this message, it means we're having trouble loading external resources for Khan Academy. http://iembra.org/confidence-interval/compute-population-mean-margin-error-99-confidence-interval.php MadonnaUSI 48,618 views 9:42 Margin of Error Sample Size TI-83+ - Duration: 11:16. The critical value is either a t-score or a z-score. statistic) will fall within the interval estimates (i.e. 4.88 and 5.26) 98% of the time. Rating is available when the video has been rented. Calculate Margin Of Error With 95 Confidence Interval You can also use a graphing calculator or standard statistical tables (found in the appendix of most introductory statistics texts). Please try again later. ProfessorSerna 164,245 views 27:18 How to calculate t distributions - Duration: 5:47. drenniemath 36,919 views 11:04 Statistics Lecture 7.2: Finding Confidence Intervals for the Population Proportion - Duration: 2:24:10. - In other words, the range of likely values for the average weight of all large cones made for the day is estimated (with 95% confidence) to be between 10.30 - 0.17 - How to Calculate Margin of Error (video) What is a Margin of Error? - Thus, the term is called the margin of error with confidence level . Solution The correct answer is (B). Find a Critical Value 7. jbstatistics 80,684 views 6:42 Confidence interval example | Inferential statistics | Probability and Statistics | Khan Academy - Duration: 18:36. What Is The Critical Value For A 95 Confidence Interval Example 10.4 The equatorial radius of the planet Jupiter is measured 40 times independently by a process that is practically free of bias. Find the critical value. Margin Of Error Formula 95 Confidence Interval Stomp On Step 1 93,730 views 7:21 How to calculate margin of error and standard deviation - Duration: 6:42. Transcript The interactive transcript could not be loaded. Loading... These measurements average \(\bar x\) = 71492 kilometers with a standard deviation of s = 28 kilometers. Margin Of Error Formula Statistics How to Compute the Margin of Error The margin of error can be defined by either of the following equations. What's the margin of error? (Assume you want a 95% level of confidence.) It's calculated this way: So to report these results, you say that based on the sample of 50 Therefore, tα∕2 is given by qt(.975, df=n-1). Margin Of Error Formula 95 Confidence Interval Tip: You can use the t-distribution calculator on this site to find the t-score and the variance and standard deviation calculator will calculate the standard deviation from a sample. http://www.r-tutor.com/elementary-statistics/interval-estimation/interval-estimate-population-mean-unknown-variance Expected Value 9. Calculate Margin Of Error With 95 Confidence Interval Difference Between a Statistic and a Parameter 3. Find The Margin Of Error For A 95 Confidence Interval Compute alpha (α): α = 1 - (confidence level / 100) = 1 - 0.95 = 0.05 Find the critical probability (p*): p* = 1 - α/2 = 1 - 0.05/2 Find the degrees of freedom (DF). check over here The new employees appear to be giving out too much ice cream (although the customers probably aren't too offended). With n = 40, using the multiplier number from the normal curve for 90% confidence (z*=1.645) will work pretty well so our confidence interval would be:71492 km ± 1.645(4.4 km) or Notice in this example, the units are ounces, not percentages! Construct And Interpret A 95 Confidence Interval A margin of error tells you how many percentage points your results will differ from the real population value. where t is a critical value determined from the tn-1 distribution in such a way that there is area between t and -t. Compute alpha (α): α = 1 - (confidence level / 100) Find the critical probability (p*): p* = 1 - α/2 To express the critical value as a z score, find his comment is here z*-Values for Selected (Percentage) Confidence Levels Percentage Confidence z*-Value 80 1.28 90 1.645 95 1.96 98 2.33 99 2.58 Note that these values are taken from the standard normal (Z-) distribution. The chart shows only the confidence percentages most commonly used. Margin Of Error Calculator Step 3: Multiply the critical value from Step 1 by the standard deviation or standard error from Step 2. How to Calculate Margin of Error in Easy Steps was last modified: March 22nd, 2016 by Andale By Andale | August 24, 2013 | Hypothesis Testing | 2 Comments | ← Generated Wed, 05 Oct 2016 14:53:02 GMT by s_bd40 (squid/3.5.20) The confidence interval is between 171.04 and 173.72 centimeters. To generalize, misses by less than with certainty. Margin Of Error Confidence Interval Calculator However, for sample size calculations (see next section), the approximate critical value 2.0 is typically used. For these sampled households, the average amount spent was \(\bar x\) = \$95 with a standard deviation of s = \$185.How close will the sample average come to the population mean? Brian Stonelake 28,786 views 11:16 Understanding Confidence Intervals: Statistics Help - Duration: 4:03. Welcome to STAT 100! weblink Let's say the poll was repeated using the same techniques. Statisticshowto.com Apply for $2000 in Scholarship Money As part of our commitment to education, we're giving away $2000 in scholarships to StatisticsHowTo.com visitors. Sign in Transcript Statistics 154,288 views 783 Like this video? The margin of error is the range of values below and above the sample statistic in a confidence interval. To express the critical value as a t statistic, follow these steps. AP Statistics Tutorial Exploring Data ▸ The basics ▾ Variables ▾ Population vs sample ▾ Central tendency ▾ Variability ▾ Position ▸ Charts and graphs ▾ Patterns in data ▾ Dotplots Show more Language: English Content location: United States Restricted Mode: Off History Help Loading... When the sampling distribution is nearly normal, the critical value can be expressed as a t score or as a z score. Example: Given the following GPA for 6 students: 2.80, 3.20, 3.75, 3.10, 2.95, 3.40 a. Sign in to make your opinion count. Easy! Confidence intervals (one sample)Estimating a population proportionConfidence interval exampleMargin of error 1Margin of error 2Next tutorialEstimating a population meanCurrent time:0:00Total duration:15:020 energy pointsStatistics and probability|Confidence intervals (one sample)|Estimating a population proportionMargin statisticsfun 126,125 views 5:47 Margin of Error - Duration: 6:17. from a poll or survey). How to Find the Critical Value The critical value is a factor used to compute the margin of error. The system returned: (22) Invalid argument The remote host or network may be down. Up next How to calculate Margin of Error Confidence Interval for a population proportion - Duration: 8:04. If the confidence level is kept at 95% but the sample size is quadrupled to n=24 (i) do you expect the sample mean to increase, decrease, or remain approximately the same?

The Math section of the SAT Reasoning test is arguably the most important one for STEM students. It is the section where you show you brilliance and talent in the language of science and engineering. Some of the finest STEM schools in the United States (like the ones you are applying to) have an SAT I Math range that starts somewhere in the 700s, which might become even higher when restricted to STEM departments. So what makes the difference between a student who can confidently get a 700+ and another who can’t? From my experience, the main difference is not the knowledge someone stores up in his or her head, in the end the test covers the math we all have finished by our 2nd year of high school, trust me on this one. The main difference lays in the way we deal with the test. The SAT Math test is heavily based on tricky questions that can force you to make the wrong choice under the influence of a miniscule slip in basic calculations or misinterpretation of the given information in the question. So the ideal way to approach the test is to be really prepared for these kinds of tricks College Board use frequently, they are extremely basic and you might not believe they can trick you, but this is basically the way to get tricked by them! So this guide is basically a list of the most frequent basic error people make when filling the bubbles. I remind you, you will not believe some people make these mistakes, but you’ll probably not believe the number of people I know who fell to these tricks: (x+y) is not x + y, it’s x +2xy+y 3 is not 9, it’s 27. There is fundamental relation between the radius and the diameter of a circle: Diameter = 2 * Radius. Occasionally, you will be given a diameter in a question that requires the use of a radius (ex. Calculating area of a circle). The sum of the measurements of angels in a triangle is 180 degrees. The most common errors in this regard is placing the number 190 instead of 180, and the number 80 instead of 90 (messure of right angle). Get yourself familiarized with the special right triangles (30,60,90 and 90,45, 45) this will save you a lot of time to use for the more conceptually challenging problems. Some people might end up approaching triangle questions using the Law of Sines or the Law of Cosines. This is definitely not necessary, even for non-right triangles. It will take much more time than needed. Don’t even think about it. Remember, you can probably split a non-right triangle into two right triangles and use the properties of similar triangles. Now this point is especially important; you need to look for what the question asks for. This is absolutely the most important advice I can give. To give you a sense of what I’m talking about, the question might be stated so that you calculate x (say, a median of a set of numbers, or one side of a triangle) but you’re asked to find x2 instead of x. this is a basic example, but it comes in a variety of ways. And yes, there will be an answer among the choices that corresponds to the answer you were tricked into finding. The most effective way to overcome this trick is to, say, underline the givens and circle the required variable, just so you can make the distinction. When it comes to diagrams, you need to actually know what a certain label is representing. Some diagrams can be tricky. For example, you are given a number of angle measurements in a certain shape and you are asked to find two more. You do your math and use your knowledge of geometry to find out the other two. The unknown you are asked to calculate, x, is not one of these angles, but rather the complementary angle of one of them. This is a basic example but you should know what I mean. Eliminate answer choices! Some answer choices can be eliminated because they don’t fit the criteria of the desired answer choice. For example, you might be asked to find x or the absolute value of x, negative numbers don’t fit here! The length of one side of a triangle is restricted to the values between the sum and the difference of the other two sides… and so on, there are some basic facts that are often ignored while they shouldn’t be. The area of a triangle is ONE HALF * HEIGHT * BASE, must people easily forget the one half and come up with wrong answers. If you are absolutely stuck in a question and need to make a guess by substituting answers and see if they work, start with the middle value. Use logic to determine if you expect a larger or smaller answer. This will help you eliminate answer choices and make a smarter guess. I’ve seen a lot of people messing this up; if you were given two values, say -9 and -3, and the question asks for the smaller value, you might be tempted to mark your answer as -3, but according to the number line, it’s actually -9. Now obviously you’ll probably be presented with more than 2 numbers, and you’ll probably have to calculate them from other information in the question, but you should get the purpose of this point. The trickiest part of the Math section of the SAT is, in my opinion, the statistics and data analysis part. If you are not used to the types of data tables you’ll be presented with in the test, it’ll become very easy to come up with wrong answers. They are easy to understand if you spend some time understanding them before the test. The questions are of varying difficulty. But guess what, they don’t vary in value. Don’t spend a lot of time on one question. If you feel like you didn’t get the idea from the first time you read it, just move on and grab other points, and you’ll probably have time to return and think about it. The test’ level of difficulty takes an ascending pattern; the further you get in the section, the harder the questions. This leads to an important note: if you solve one of the last 3-5 question very quickly and without using a lot of brain energy, you probably got it wrong. This is referred to as the ten-second rule. The last and most important advice: PRACTICE. This is the only way to figure out your weak points and identify the many, many tricks College Board use in their tests. Many books are helpful for the math section, but I would personally recommend McGraw Hill’s SAT guide and Dr. John Chung’s SAT Math. The latter is much harder than the actual test, but that’s essentially a good way to master the real SAT Dealing with tricky questions on the SAT Math Section

Place a red counter in the top left corner of a 4x4 array, which is covered by 14 other smaller counters, leaving a gap in the bottom right hand corner (HOME). What is the smallest number of moves. . . . On the 3D grid a strange (and deadly) animal is lurking. Using the tracking system can you locate this creature as quickly as possible? The aim of the game is to slide the green square from the top right hand corner to the bottom left hand corner in the least number of moves. There are 27 small cubes in a 3 x 3 x 3 cube, 54 faces being visible at any one time. Is it possible to reorganise these cubes so that by dipping the large cube into a pot of paint three times you. . . . To avoid losing think of another very well known game where the patterns of play are similar. A game for 2 players Show that all pentagonal numbers are one third of a triangular number. P is a point on the circumference of a circle radius r which rolls, without slipping, inside a circle of radius 2r. What is the locus of P? Given the nets of 4 cubes with the faces coloured in 4 colours, build a tower so that on each vertical wall no colour is repeated, that is all 4 colours appear. Can you find a rule which connects consecutive triangular numbers? Can you find a rule which relates triangular numbers to square numbers? Players take it in turns to choose a dot on the grid. The winner is the first to have four dots that can be joined to form a square. This is an interactive net of a Rubik's cube. Twists of the 3D cube become mixes of the squares on the 2D net. Have a play and see how many scrambles you can undo! This task depends on groups working collaboratively, discussing and reasoning to agree a final product. What's the largest volume of box you can make from a square of paper? The triangle OMN has vertices on the axes with whole number co-ordinates. How many points with whole number coordinates are there on the hypotenuse MN? The opposite vertices of a square have coordinates (a,b) and (c,d). What are the coordinates of the other vertices? Watch these videos to see how Phoebe, Alice and Luke chose to draw 7 squares. How would they draw 100? Have a go at this 3D extension to the Pebbles problem. Build gnomons that are related to the Fibonacci sequence and try to explain why this is possible. Semi-regular tessellations combine two or more different regular polygons to fill the plane. Can you find all the semi-regular tessellations? Bilbo goes on an adventure, before arriving back home. Using the information given about his journey, can you work out where Bilbo lives? Can you discover whether this is a fair game? A game for 2 people. Take turns joining two dots, until your opponent is unable to move. In this problem we are faced with an apparently easy area problem, but it has gone horribly wrong! What happened? A bicycle passes along a path and leaves some tracks. Is it possible to say which track was made by the front wheel and which by the back wheel? Place the numbers 1, 2, 3,..., 9 one on each square of a 3 by 3 grid so that all the rows and columns add up to a prime number. How many different solutions can you find? How can visual patterns be used to prove sums of series? Some students have been working out the number of strands needed for different sizes of cable. Can you make sense of their solutions? An irregular tetrahedron has two opposite sides the same length a and the line joining their midpoints is perpendicular to these two edges and is of length b. What is the volume of the tetrahedron? Simple additions can lead to intriguing results... Use the animation to help you work out how many lines are needed to draw mystic roses of different sizes. This is a simple version of an ancient game played all over the world. It is also called Mancala. What tactics will increase your chances of winning? This article for teachers discusses examples of problems in which there is no obvious method but in which children can be encouraged to think deeply about the context and extend their ability to. . . . A and C are the opposite vertices of a square ABCD, and have coordinates (a,b) and (c,d), respectively. What are the coordinates of the vertices B and D? What is the area of the square? Discover a way to sum square numbers by building cuboids from small cubes. Can you picture how the sequence will grow? A right-angled isosceles triangle is rotated about the centre point of a square. What can you say about the area of the part of the square covered by the triangle as it rotates? This is the first article in a series which aim to provide some insight into the way spatial thinking develops in children, and draw on a range of reported research. The focus of this article is the. . . . Points P, Q, R and S each divide the sides AB, BC, CD and DA respectively in the ratio of 2 : 1. Join the points. What is the area of the parallelogram PQRS in relation to the original rectangle? How could Penny, Tom and Matthew work out how many chocolates there are in different sized boxes? Can you mark 4 points on a flat surface so that there are only two different distances between them? Two boats travel up and down a lake. Can you picture where they will cross if you know how fast each boat is travelling? Mathematics is the study of patterns. Studying pattern is an opportunity to observe, hypothesise, experiment, discover and create. What can you see? What do you notice? What questions can you ask? This problem is about investigating whether it is possible to start at one vertex of a platonic solid and visit every other vertex once only returning to the vertex you started at. I found these clocks in the Arts Centre at the University of Warwick intriguing - do they really need four clocks and what times would be ambiguous with only two or three of them? Rectangles are considered different if they vary in size or have different locations. How many different rectangles can be drawn on a chessboard? On the graph there are 28 marked points. These points all mark the vertices (corners) of eight hidden squares. Can you find the eight hidden squares? A cyclist and a runner start off simultaneously around a race track each going at a constant speed. The cyclist goes all the way around and then catches up with the runner. He then instantly turns. . . . Draw a pentagon with all the diagonals. This is called a pentagram. How many diagonals are there? How many diagonals are there in a hexagram, heptagram, ... Does any pattern occur when looking at. . . .

Advances in Modelling, Monitoring, and Control for Complex Industrial SystemsView this Special Issue Estimation and Synthesis of Reachable Set for Singular Markovian Jump Systems The problems of reachable set estimation and state-feedback controller design are investigated for singular Markovian jump systems with bounded input disturbances. Based on the Lyapunov approach, several new sufficient conditions on state reachable set and output reachable set are derived to ensure the existence of ellipsoids that bound the system states and output, respectively. Moreover, a state-feedback controller is also designed based on the estimated reachable set. The derived sufficient conditions are expressed in terms of linear matrix inequalities. The effectiveness of the proposed results is illustrated by numerical examples. The research on singular systems has attracted significant attention in the past years due to the fact that singular systems can better describe a larger class of physical systems such as robotic systems, electric circuits, and mechanical systems. When singular systems experience abrupt changes in their structures, it is natural to model them as singular Markovian jump systems [1, 2]. The analysis and synthesis of such class of systems have gained considerable attention because of their importance in applications (see, e.g., the literature [3–10] and the references therein). Reachable set is one of the important techniques for parameter estimation or state estimation problems . Reachable set for a dynamic system is the set containing all the system states starting from the origin under bounded input disturbances. However, the exact shape of reachable sets of a dynamic system is very complex and hard to obtain; for this reason a number of researchers began to turn their attention to the reachable set estimation problem. The common strategies for reachable set estimation are ellipsoidal method and polyhedron method . The main idea of these methods is to detect simple convex shapes like ellipsoid or polyhedron, which contains all the system states. Compared with polyhedron method, the primary advantage of ellipsoidal method is that the ellipsoid structure is simple and directly related to quadratic Lyapunov functions. As a result, linear matrix inequalities (LMIs) techniques can be used to determine bounding ellipsoids. In the framework of bounding ellipsoid, the reachable set estimation problem for linear time delay systems has received significant research attention in recent years. In sufficient conditions for the existence of bounding ellipsoids containing the reachable set of continuous-time linear systems with time-varying delays were derived by using the Lyapunov-Razumikhin function. In , by using Lyapunov-Krasovskii functional method, the author derived some less conservative conditions than those in . In the reachable set of delayed systems with polytopic uncertainties was investigated by using the maximal Lyapunov-Krasovskii functional approach, and some new conditions bounding the set of reachable states are derived. Interesting results on reachable set of delayed systems with polytopic uncertainties can also be found in [17–20]. In addition, some other strategies without using Lyapunov-Krasovskii functional have been provided to estimate the reachable set of continuous-time linear time-varying systems and nonlinear time delay systems [22, 23]. The authors in extended the ideas of reachable set estimation of continuous-time systems to discrete-time systems, wherein a fundamental result (Lemma 2.1 ) for the reachable set estimation of discrete-time systems was proposed. The authors in improved the fundamental result obtained in and provided a basic tool (Lemma 4 ) for the reachable set estimation of discrete-time systems. On the basis of the general ideas proposed in , the reachable set estimation problem was also extended to some classes of complicated systems, such as singular systems , Markovian jump systems , switched linear systems , and T-S fuzzy systems [29, 30]. For the reachable set estimation of discrete-time systems, the other important contributions can be found in [31, 32]. On the other hand, the problem of controller design for specifications involved with the reachable set of a control system is also a very important issue . The controller design problems concerning reachable set were studied in and by using ellipsoidal method and polyhedron method, respectively. Two issues were raised in : the first one is to design a controller such that the reachable set of the closed-loop system is contained in an ellipsoid, and the admissible ellipsoid should be as small as possible; the second one is to design a controller such that the reachable set of the closed-loop system is contained in a given ellipsoid. By constructing suitable Lyapunov-Krasovskii functional, LMI-based sufficient conditions for the existence of controller guaranteeing the ellipsoid bounds as small as possible have been derived for continuous-time delay systems and discrete-time periodic systems . It is obvious that the LMI-based controller design is quite simple and numerically tractable. However, it should be pointed out that the reachable set estimation and synthesis problems of singular Markovian jump systems are much more difficult and challenging than that for nonsingular Markovian jump systems since the ellipsoid containing the reachable set is not directly related to quadratic Lyapunov functions. To the best of the authors’ knowledge, no related results have been established for reachable set estimation and synthesis of singular Markovian jump systems, which has motivated this paper. In this paper, we consider the problems of reachable set estimation and synthesis of singular Markovian jump systems. By using the Lyapunov approach, the estimation conditions on state reachable set and output reachable set are derived, respectively. Moreover, the desired state-feedback controller is designed based on the estimated reachable set. Notation. Throughout this paper, denotes the -dimensional Euclidean space; represents the transpose of ; stands for ; (<0) means is a symmetric positive (negative) definite matrix; refers to the expectation; denotes the matrix composed of elements of first rows and columns of matrix ; refers to the Euclidean vector norm; the symbol “” in LMIs denotes the symmetric term of the matrix; is the unit matrix with appropriate dimensions. 2. Problem Formulation Consider the following singular Markovian jump system:where is the state vector, is the control input, is the measured output, and is the exogenous disturbance which satisfies, , , , and are real constant matrices with appropriate dimensions and . is a continuous-time Markovian process with transition rate matrix and the evolution of Markovian process is governed by the following transition rate:where and ; for is the transition rate from mode to mode and . For notational simplicity, in the sequel, for each possible , , matrices , , , and will be denoted by , , , and , respectively. When , the system is referred to as a free system. In this paper we are interested in determining ellipsoids that contain, respectively, the state reachable set and output reachable set. In the reachable set analysis, it is required that systems should be asymptotically stable. When this requirement is not met, we will further design a state-feedback controller such that the reachable set of the closed-loop system is contained in the smallest ellipsoid. The state reachable set of the free system in (1) is defined by An ellipsoid bounding the reachable set can be always represented as follows: Particularly, when for , the ellipsoid will become a ball which is denoted by . Since , there exist two nonsingular matrices and such thatLet . Then the free system can be rewritten as the following differential-algebraic form: The following definition and lemma are also useful in deriving the main results. Definition 1 (see ). (I) The free system is said to be regular if is not identically zero for each . (II) The free system is said to be impulse-free if for each . Lemma 2 (see ). For any matrices and with , one has . Lemma 3 (see ). Let be a Lyapunov function and . If then , . 3. Main Results 3.1. State Reachable Set Estimation In this subsection, we will focus our attention on determining a ball which contains the state reachable set of the free system. Theorem 4. If there exist nonsingular matrices and a scalar such that the following LMIs hold for each ,then the state reachable set of free system starting from the origin is mean-square bounded within the following set:where Proof. We first prove the regularity and nonimpulsiveness of the free system. Let . Then, by (10), we obtain that . From (11), it is easy to show thatPre- and postmultiplying (14) by and , respectively, we getwhere will be irrelevant to the results of the following discussion; thus the real expressions of these two variables are omitted. It follows from (15) thatwhich implies that is nonsingular for each . Therefore, by Definition 1, we have that the free system is regular and nonimpulsive. Next, we will show the state reachable set of free system is mean-square bounded within the set . Consider the following Lyapunov function:Let be the weak infinitesimal generator of the random process . Calculating the difference of along the trajectories of the free system, we getDefining the augmented system variable as and using conditions (10) and (11), we haveThen we can deduce from Lemma 3 that , which infers thatRecalling that , it follows from (20) thatFrom (21) we have , which implies that . Since is nonsingular for each , (8) can be rewritten asThen we can deduce thatIt follows from the fact thatwhere . By (24), it can be seen thatwhich implies that the trajectories of (7)-(8) are mean-square bounded within the set . Moreover, notice that , and (25) can be rewritten as . By denoting , the state reachable set of free system is mean-square bounded within the set . It should be noted that inequality (10) represents a nonstrict LMI. This may lead to numerical problems since equality constraints are usually not satisfied perfectly. Below, we will develop a numerically tractable and nonconservative LMI condition. Theorem 5. If there exist symmetric positive definite matrices , nonsingular matrices , and a scalar such that the following LMI holds for each ,then the state reachable set of free system is mean-square bounded within the following set:where is any matrix with full row rank and satisfies ; is any matrix with full column rank and satisfies . Proof. Let in (26); it is easy to obtain (10) and (11). In this case, inequality (20) will be replaced by , which infers that with . Then, following the same lines as (22)–(25) in Theorem 4, we can get for any . Therefore, the state reachable set of free system is mean-square bounded within the set . Remark 6. In order to make the ellipsoid as small as possible, we require . For this purpose, we can add the additional requirement and then maximize a positive scalar , which is equivalent to the following minimization problem:where . 3.2. Output Reachable Set Estimation Theorem 7. If there exist symmetric positive definite matrices and , nonsingular matrices , and a scalar such that the following LMIs hold for each ,then the output reachable set of free system is mean-square bounded within the following set:where , , , and are defined in Theorem 5. Proof. By Theorem 5, LMI (30) ensuresWith this and (31), we obtain thatDue to the fact that , (34) can be rewritten as . Thus, the output reachable set of free system is mean-square bounded within the set . Remark 8. The output reachable set is also expected to be as small as possible. To achieve this goal, we first solve LMI (30) and get satisfying , which can be implemented by using (29). Then we add the additional requirement and maximize a positive scalar , which is equivalent to the following minimization problem:where . 3.3. State-Feedback Controller Design In this section, we turn our attention to the state-feedback control problem. Our goal here is to find a state-feedback controller, which not only stabilizes the closed-loop system, but also makes the ellipsoid bound on the reachable set of closed-loop system as small as possible. Now, consider the state-feedback controller , where is a gain matrix to be determined later. By using this controller, the closed-loop system can be obtained aswhere . Theorem 9. Consider singular Markov jump system (1). If there exist nonsingular matrices , matrices , and scalars , such that the following LMIs hold for each ,then the reachable set of system (1) is mean-square bounded within the set , and the desired controller gain matrix is given by , where Proof. Denote and for each . Then, pre- and postmultiplying (39) by and its transpose, respectively, we obtainwhere , . Using Lemma 2, we haveFrom (41) and (42), it is easy to obtain thatwhere , . By Schur complement, the previous matrix inequality becomeswhere . Since , (38) infers that . Pre- and postmultiplying the previous matrix inequality by and , respectively, we have . This together with (44) implieswhere . Pre- and postmultiplying (37) by and , respectively, we obtainFrom the above discussion, we show that if (37)–(39) hold, then (45) and (46) hold. Thus, it follows from Theorem 4 that the closed-loop system can be stabilized by the designed state-feedback controller. Next, we show that the reachable set of the closed-loop system (36) is mean-square bounded within the set . From (45) and (46), it is easy to show that there exists a Lyapunov function such that , where denotes the difference of along the trajectories of (36). It follows from Lemma 3 that , which impliesNoting that , then (47) can be rewritten asRecalling that is a lower triangular matrix, (48) infers thatwhere . By (49), it can be seen thatUsing the fact that , we getThis together with (24) yieldsFrom (24), (51), and (52), we have thatwhere . This implies that . Recalling that , we haveBy denoting , the reachable set of closed-loop system (36) is mean-square bounded within the set . Remark 10. In order to make the ellipsoid as small as possible, we shall carry out the following minimization problem: 4. Numerical Examples In this section, two numerical simulation examples are given to show the effectiveness of the main results derived above. Example 11. Consider the free system in (1) with the following parameters: The switching between two modes is described by the following transition rate matrix: In this example, we choose , . By solving optimization problem (29) with the aid of fminsearch, the minimal and the corresponding are and , respectively. Using the above parameter values, we can obtain and by solving (26). Owing to Theorem 5, the state reachable set is mean-square bounded within the set . By applying Theorem 7, we have the following results:Therefore, the output reachable set of free system is mean-square bounded within the set . For simulation we assume that and the disturbance is chosen as . A case for stochastic variation with transition rate matrix is shown in Figure 1. The state reachable set and the ball are depicted in Figure 2. Figure 2 shows that the trajectory of the system is mean-square bounded within the region . The output reachable set , the ball , and the ellipsoid are depicted in Figure 3. Figure 3 shows that the output reachable set is mean-square bounded within the region . Example 12. Consider system (1) with the following parameters: The switching between two modes is described by the following transition rate matrix: By Theorem 9, we get the following results: Therefore, the gain matrices of state-feedback controller can be obtained as The corresponding parameter values and are, respectively, and , which imply and . Applying this controller makes the state reachable set of closed-loop system (36) mean-square bounded within the region . For the purpose of the simulation, we assume the initial condition and the disturbance is chosen as . Figure 4 shows one possible switching between two modes. Figure 5 depicts the state reachable set of closed-loop system (36). This paper has dealt with the problems of reachable set estimation and state-feedback controller design for singular Markovian jump systems. New sufficient conditions for the state reachable set estimation and output reachable set estimation have been, respectively, derived in terms of linear matrix inequalities. Based on the estimated reachable set, the state-feedback controller has also been designed. Numerical examples and simulation results have been provided to demonstrate the effectiveness of the proposed methods. Conflicts of Interest The authors declare that there are no conflicts of interest regarding the publication of this paper. This work was supported by Applied Basic Research Project of Science and Technology Department of Sichuan Province (no. 2017JY0336), a project supported by Scientific Research Fund of Sichuan Provincial Education Department (no. 16ZA0146), the National Natural Science Foundation of China (no. 11501474), the Longshan Academic Talents Research Support Program of the Southwest University of Science and Technology (no. 17LZX537 and no. 17LZXY11), and the Doctoral Research Foundation of Southwest University of Science and Technology (no. 13zx7141). S. Xu and J. Lam, Robust Control and Filtering of Singular Systems, Springer, Berlin, Germany, 2006.View at: MathSciNet E.-K. Boukas, Control of Singular Systems with Random Abrupt Changes, Communications and Control Engineering, Springer, Berlin, Germany, 2008.View at: MathSciNet Z.-G. Wu, J. H. Park, H. Su, and J. Chu, “Stochastic stability analysis for discrete-time singular Markov jump systems with time-varying delay and piecewise-constant transition probabilities,” Journal of The Franklin Institute, vol. 349, no. 9, pp. 2889–2902, 2012.View at: Publisher Site | Google Scholar | MathSciNet Y. Ma, X. Jia, and D. Liu, “Finite-time dissipative control for singular discrete-time Markovian jump systems with actuator saturation and partly unknown transition rates,” Applied Mathematical Modelling: Simulation and Computation for Engineering and Environmental Systems, vol. 53, pp. 49–70, 2018.View at: Publisher Site | Google Scholar | MathSciNet E. K. Kostousova, “On tight polyhedral estimates for reachable sets of linear differential systems,” in Proceedings of the 9th International Conference on Mathematical Problems in Engineering, Aerospace and Sciences, ICNPAA 2012, pp. 579–586, Austria, July 2012.View at: Publisher Site | Google Scholar W. Xiang, H.-D. Tran, and T. T. Johnson, “Output reachable set estimation for switched linear systems and its application in safety verification,” Institute of Electrical and Electronics Engineers Transactions on Automatic Control, vol. 62, no. 10, pp. 5380–5387, 2017.View at: Publisher Site | Google Scholar | MathSciNet E. K. Kostousova, “On feedback target control for uncertain discrete-time bilinear systems with state constraints through polyhedral technique,” in Proceedings of the 9th International Conference for Promoting the Application of Mathematics in Technical and Natural Sciences, AMiTaNS 2017, Bulgaria, June 2017.View at: Publisher Site | Google Scholar

Since the publication of the second edition of Applied Reliability in 1995, the ready availability of inexpensive, powerful statistical software has changed the way statisticians and engineers look at and analyze all kinds of data. Problems in reliability that were once difficult and time consuming even for experts can now be solved with a few well-chosen clicks of a mouse. However, software documentation has had difficulty keeping up with the enhanced functionality added to new releases, especially in specialized areas such as reliability analysis. Using analysis capabilities in spreadsheet software and two well-maintained, supported, and frequently updated, popular software packages-Minitab and SAS JMP-the third edition of Applied Reliability is an easy-to-use guide to basic descriptive statistics, reliability concepts, and the properties of lifetime distributions such as the exponential, Weibull, and lognormal. The material covers reliability data plotting, acceleration models, life test data analysis, systems models, and much more. The third edition includes a new chapter on Bayesian reliability analysis and expanded, updated coverage of repairable system modeling. Taking a practical and example-oriented approach to reliability analysis, this book provides detailed illustrations of software implementation throughout and more than 150 worked-out examples done with JMP, Minitab, and several spreadsheet programs. In addition, there are nearly 300 figures, hundreds of exercises, and additional problems at the end of each chapter, and new material throughout. Software and other files are available for download online "I have used the second edition of this book for an Introduction to Reliability course for over 15 years. ... The third edition ... retains the features I liked about the second edition. In addition, it includes improved graphics ... [and] examples of popular software used in industry ... There is a new chapter on Bayesian reliability and additional material on reliability data plotting, and repairable systems' analysis. ... The book does a good and comprehensive job of explaining the basic reliability concepts; types of data encountered in practice and how to treat this data; reliability data plotting; and the common probability distributions used in reliability work, including motivations for their use and practical areas of application. ... an excellent choice for a first course in reliability. The book also does a good job of explaining more advanced topics ... the book will continue to be a popular desk reference in industry and a textbook for advanced undergraduate or first-year graduate students." -Journal of the American Statistical Association, June 2014 Basic Descriptive Statistics Populations and Samples Histograms and Frequency Functions Cumulative Frequency Function The Cumulative Distribution Function and the Probability Density Function Probability Concepts Random Variables Sample Estimates of Population Parameters How to Use Descriptive Statistics Data Simulation Reliability Concepts Reliability Function Some Important Probabilities Hazard Function or Failure Rate Cumulative Hazard Function Average Failure Rate Units Bathtub Curve for Failure Rates Recurrence and Renewal Rates Mean Time to Failure and Residual Lifetime Types of Data Failure Mode Separation Exponential Distribution Exponential Distribution Basics The Mean Time to Fail for the Exponential The Exponential Lack of Memory Property Areas of Application for the Exponential Exponential Models with Duty Cycles and Failure on Demand Estimation of the Exponential Failure Rate I" Exponential Distribution Closure Property Testing Goodness of Fit--the Chi-Square Test Testing Goodness of Fit--Empirical Distribution Function Tests Confidence Bounds for I" and the MTTF The Case of Zero Failures Planning Experiments Using the Exponential Distribution Simulating Exponential Random Variables The Two-Parameter Exponential Distribution Test Planning Via Spreadsheet Functions Determining the Sample Size EDF Goodness-of-Fits Tests Using Spreadsheets KS Test Weibull Distribution Empirical Derivation of the Weibull Distribution Properties of the Weibull Distribution Extreme Value Distribution Relationship. Areas of Application Weibull Parameter Estimation: Maximum Likelihood Estimation Method Weibull Parameter Estimation: Linear Rectification Simulating Weibull Random Variables The Three-Parameter Weibull Distribution Goodness of Fit for the Weibull Using a Spreadsheet to Obtain Weibull MLES Using a Spreadsheet to Obtain Weibull MLES for Truncated Data Spreadsheet Likelihood Profile Confidence Intervals for Weibull Parameters The Normal and Lognormal Distributions Normal Distribution Basics Applications of the Normal Distribution The Central Limit Theorem Normal Distribution Parameter Estimation Simulating Normal Random Variables The Lognormal Life Distribution Properties of the Lognormal Distribution Lognormal Distribution Areas of Application Lognormal Parameter Estimation Some Useful Lognormal Equations Simulating Lognormal Random Variables Using a Spreadsheet to Obtain Lognormal MLEs Using a Spreadsheet to Obtain Lognormal MLEs for Interval Data Reliability Data Plotting Properties of Straight Lines Least Squares Fit (Regression Analysis) Rectification Probability Plotting for the Exponential Distribution Probability Plotting for the Weibull Distribution Probability Plotting for the Normal and Lognormal Distributions Simultaneous Confidence Bands Order Statistics and Median Ranks Analysis of Multicensored Data Multicensored Data Analysis of Interval (Readout) Data Life Table Data Left-Truncated and Right-Censored Data Left-Censored Data Other Sampling Schemes (Arbitrary Censoring: Double and Overlapping Interval Censoring)--Peto--Turnbull Estimator Simultaneous Confidence Bands for the Failure Distribution (or Survival) Function Cumulative Hazard Estimation for Exact Failure Times Johnson Estimator Obtaining Bootstrap Confidence Bands Using a Spreadsheet Physical Acceleration Models Accelerated Testing Theory Exponential Distribution Acceleration Acceleration Factors for the Weibull Distribution Likelihood Ratio Tests of Models Confidence Intervals Using the LR Method Lognormal Distribution Acceleration Acceleration Models The Arrhenius Model Estimating I"H with More Than Two Temperatures Eyring Model Other Acceleration Models Acceleration and Burn-In Life Test Experimental Design An Alternative JMP Input for Weibull Analysis of High-Stress Failure Data Using a Spreadsheet for Weibull Analysis of High-Stress Failure Data Using A Spreadsheet for MLE Confidence Bounds for Weibull Shape Parameter Using a Spreadsheet for Lognormal Analysis of the High-Stress Failure Data Shown in Table 8.5 Using a Spreadsheet for MLE Confidence Bounds for the Lognormal Shape Parameter Using a Spreadsheet for Arrhenius--Weibull Model Using a Spreadsheet for MLEs for Arrhenius--Power Relationship Lognormal Model Spreadsheet Templates for Weibull or Lognormal MLE Analysis Alternative Reliability Models Step Stress Experiments Degradation Models Lifetime Regression Models The Proportional Hazards Model Defect Subpopulation Models Summary JMP Solution for Step Stress Data in Example 9.1 Lifetime Regression Solution Using Excel JMP Likelihood Formula for the Defect Model JMP Likelihood Formulas for Multistress Defect Model Example System Failure Modeling: Bottom-Up Approach Series System Models The Competing Risk Model (Independent Case) Parallel or Redundant System Models Standby Models and the Gamma Distribution Complex Systems System Modeling: Minimal Paths and Minimal Cuts General Reliability Algorithms Burn-In Models The "Black Box" Approach--An Alternative to Bottom-Up Methods Quality Control in Reliability: Applications of Discrete Distributions Sampling Plan Distributions Nonparametric Estimates Used with the Binomial Distribution Confidence Limits for the Binomial Distribution Normal Approximation for Binomial Distribution Confidence Intervals Based on Binomial Hypothesis Tests Simulating Binomial Random Variables Geometric Distribution Negative Binomial Distribution Hypergeometric Distribution and Fisher's Exact Test Poisson Distribution Types of Sampling Generating a Sampling Plan Minimum Sample Size Plans Nearly Minimum Sampling Plans Relating an OC Curve to Lot Failure Rates Statistical Process Control Charting for Reliability Repairable Systems Part I: Nonparametric Analysis and Renewal Processes Repairable versus Nonrepairable Systems Graphical Analysis of a Renewal Process Analysis of a Sample of Repairable Systems Confidence Limits for the Mean Cumulative Function (Exact Age Data) Nonparametric Comparison of Two MCF Curves Renewal Processes. Homogeneous Poisson Process MTBF and MTTF for a Renewal Process MTTF and MTBF Two-Sample Comparisons Availability Renewal Rates Simulation of Renewal Processes Superposition of Renewal Processes CDF Estimation from Renewal Data (Unidentified Replacement) True Confidence Limits for the MCF Cox F-Test for Comparing Two Exponential Means Alternative Approach for Estimating CDF Using the Fundamental Renewal Equation Repairable Systems Part II: Nonrenewal Processes Graphical Analysis of Nonrenewal Processes Two Models for a Nonrenewal Process Testing for Trends and Randomness Laplace Test for Trend Reverse Arrangement Test Combining Data from Several Tests Nonhomogeneous Poisson Processes Models for the Intensity Function of an NHPP Rate of Occurrence of Failures Reliability Growth Models Simulation of Stochastic Processes Bayesian Reliability Evaluation Classical versus Bayesian Analysis Classical versus Bayes System Reliability Bayesian System MTBF Evaluations Bayesian Estimation of the Binomial p The Normal/Normal Conjugate Prior Informative and Noninformative Priors A Survey of More Advanced Bayesian Methods Gamma and Chi-Square Distribution Relationships Problems Answers to Selected Exercises References Index Number Of Pages: 600 Published: 26th August 2011 Publisher: Taylor & Francis Inc Country of Publication: US Dimensions (cm): 24.8 x 17.1 Weight (kg): 1.23 Edition Number: 3 Edition Type: New edition

OBJECTIVESDefine enthalpyDistinguish between heat and temperaturePerform calculations using molar heat capacity Key termsEnergy (E): - the ability to do work or produce heat.Thermodynamics: - the study of the inter action of heat andother kinds of energy.Heat (q): - the transfer of energy between two objects (internalversus surroundings) due to the difference in temperature.Work (w): - when force is applied over a displacement in thesame direction or a change in volume under the same pressure(w = F × d = −P ΔV).-Work performed can be equated to energy if no heat isproduced (E = w). This is known as the Work Energy Theorem. Key termsInternal Energy (E): - total energy from work andheat within a system.Temperature: - the average kinetic energy of all theparticles in a substance. - Temperature is NOT Heat.A massive substance with a low temperature canhave a lot of internal heat. This is because there are alot of particles and even though their kinetic energyis low, their TOTAL energy is large. EnthalpyEnthalpy (H): - the amount of internal energy at aspecific pressure and volume (when there is no workdone).ΔE=q+w (if w=0,thenΔE=q=H)ΔE = q + w ΔE = Change in System’s Internal Energy q= heat w = work Heat unitHeat Unit: - the measuring units to measure heat orenergy.Specific Heat (cP): - the amount of heat needed toraise one gram of substance by one degree Celsius orone Kelvin.- the higher the specific heat, the more each gram ofsubstance can “hold” the heat. - units are in J/(g • °C)or kJ/(kg • °C) ; J/(g • K) or kJ/(kg • K) Heat unit Molar Heat Capacity (C): - the amount of heat needed to raise one mole of substance by one degree Celsius or one Kelvin.- the higher the molar heat capacity, the more heat each mole or each particle of a substance can “hold”. - units are in J/(mol • °C) or kJ/(kmol • °C) ; J/(mol • K) or kJ/(kmol • K) Joules: - the metric unit to measure heat or energy named after English physicist James Prescott Joule. Formulas and termsq = mcPΔT q = nCΔTq = Change in Heat (J or kJ) ΔT = Change inTemperature (in °C or K) m = mass (g or kg) cP =Specific Heat [J/(g • °C or K) or kJ/(kg • °C or K)] n =moles (mol or kmol) C = Molar Heat Capacity [J/(mol• °C or K) or kJ/(kmol • °C or K)] Substance Specific Heat J/(g • °C or K)Ice H2O(s) 2.01 Water H2O(l) 4.18 Steam H2O(g) 2.00 Ammonia NH3 (g) 2.06 Methanol CH3OH (l) 2.53 Ethanol C2H5OH (l) 2.44 Aluminum Al (s) 0.897 Substance Specific Heat J/(g • °C or K)Carbon (graphite) C (s) 0.709 Copper Cu (s) 0.385 Iron Fe (s) 0.449 Silver Ag (s) 0.235 Gold Au (s) 0.129 Aluminum Chloride, AlCl3 (s) 0.690 Forces Relationship between Molar Heat Capacity and Specific Heat (Molar Mass) × (Specific Heat) = (Molar Heat Capacity) mol× g•K = mol•K Example : How much energy in kJ, is needed to heat 25.0 mol of water from 30.0°C to 75.0°C? Cwater = 75.3 J/(mol • °C) n = 25.0 mol H2O ΔT = 75.0°C − 30.0°C = 45.0°C q=? q = nCΔT q = (25.0 mol)(75.3 J/(mol • °C))(45.0 °C) = 84712.5 J q = 84.7 kJ OBJECTIVESDefine thermodynamicsCalculate the enthalpy change for a given amount ofsubstance for a given change of temperature Key termsSystem: - a part of the entire universe as defined bythe problem.Surrounding: - the part of the universe outside thedefined system.Open System: - a system where mass and energy caninterchange freely with its surrounding.Closed System: - a system where only energy caninterchange freely with its surrounding but mass notallowed to enter or escaped the system. Key termsIsolated System: - a system mass and energy cannotinterchange freely with its surrounding. ExothermicProcess (ΔE < 0): - when energy flows “out” of thesystem into the surrounding.Endothermic Process (ΔE > 0): - when energy flowsinto the system from the surrounding. (Surroundinggets Colder.)Molar Enthalpy Change (∆H): - the amount of changein energy per mole of substance (J/mol or kJ/mol). Molar Enthalpy Change for Kinetic (Temperature) ChangeExample: Calculate the change in temperature and the molarenthalpy change of iron when it cools from 243.7°C to 18.2°C.Comment on the nature of this thermodynamic process.∆H=CΔT∆H = Molar Enthalpy Change (J/mol kJ/mol)ΔT = Change in Temperature (in °C or K)∆T=Tf −Ti∆H > 0 (endothermic) ∆H < 0 (exothermic)C = Molar Heat Capacity [J/(mol • °C or K) OBJECTIVESExplain the principles of calorimetryUse Hess’s law and standard enthalpies of formationto calculate ΔH Key termsStandard State: - standard conditions of 1 atm and25°C. It is denote by a superscript “o”.Standard Molar Enthalpy of Formation (ΔHof): - theamount of heat required / given off to make 1 mole ofcompound from its elemental components understandard conditions.- the Molar Heat of Formation of ALL ELEMENTS is 0kJ. - the state of the compound affects the magnitudeof Hf. Key terms Molar Heat of Reaction (ΔHrxn): - the amount of heat released when one mole of reactant undergoes various chemical or physical changes.- examples are ΔHcomb, ΔHneut (neutralization), ΔHsol (solution). Standard Molar Enthalpy of Reaction (ΔHorxn): - the amount of heat involved when 1 mol of a particular. Formulas and termsΔH = nΔHrxnΔH = Change in Enthalpy ΔHrxn = Molar Heat of Reaction(kJ/mol)ΔH = nΔHcombn = moles ΔHcomb = Molar Heat of Combustion (kJ/mol)Product is produced or 1 mol of a particular reactant isconsumed under standard conditions of 1 atm and 25°C. Hess’s Law Hess’s Law: the indirect method of obtaining overall ΔH°rxn of a net reaction by the addition of ΔH°rxn of a series of reactions.- when adding reactions, compare the reactants and products of theoverall net reaction with the intermediate (step) reactions given.Decide on the intermediate reactions that need to be reversed and /or multiply by a coefficient, such that when added, the intermediateproducts will cancel out perfectly yielding the overall net reaction.- if a particular reaction needs to be reversed (flipped), the sign ofthe ΔH for that reaction will also need to be reversed.- if a coefficient is used to multiply a particular reaction, the ΔH forthat reaction will also have to multiply by that same coefficient. ExampleExample: Determine the ΔH°rxn for the reaction S (s) + O2(g) → SO2 (g), given the following reactions. S(s) + 32 O2(g) → SO3 (g) ΔH°rxn = −395.2 kJ2SO2(g) +O2(g)→2SO3(g) ΔH°rxn =−198.2kJSO2 in the net reaction is on the product side, whereas 2SO2 in the second reaction is on the reactant side. Hence,we need to reverse the second reaction and its sign of theΔH°rxn.There is only 1 SO2 in the net reaction, whereas there are 2SO2 in the second reaction. Therefore the second reactionand its ΔH°rxn need to be multiply by the coefficient of 1⁄2. Standard Enthalpy of Reaction Direct Method to determine Standard Enthalpy of Reaction ΔHorxn = ΣHoproducts − ΣHoreactants ΔHorxn = Change in Enthalpy of Reaction ΣHoproducts = Sum of Heat of Products (from all nΔHof of products) ΣHoreactants = Sum of Heat of Reactants (from all nΔHof of reactants) OBJECTIVESDefine enthalpy, discuss the factors that influencethe sign and magnitude of ΔS for a chemical reactionDescribe Gibbs energy, and discuss the factors thatinfluence the sign and magnitude of ΔGIndicate whether ΔG values describe spontaneous ornonspontaneous reactions Key termsFirst Law of Thermodynamics: - states that energy cannot becreated or destroyed. It can only be converted from one form toanother. Therefore, energy in the universe is a constant- also known as the Law of Conservation of Energy (ΣEinitial =ΣEfinal).Calorimetry: - uses the conservation of energy (Heat Gained =Heat Lost) to measure calories (old unit of heat: 1 cal = 4.184 J).- physical calorimetry involves the mixing of two systems (onehotter than the other) to reach some final temperature.- the key to do these problems is to identify which system isgaining heat and which one is losing heat. A Simple Styrofoam CalorimeterConstant-Pressure Calorimeter (or StyrofoamCalorimeter) - commonly used to determineΔHneut, ΔHion, ΔHfus, ΔHvap, ΔHrxn of non-combustion reaction. First, the sample’s mass ismeasured. Water is commonly used to absorb orprovide the heat for the necessary change. The initialand final temperatures of the water arerecorded, allowing us to find the amount of heatchange. By applying the law of conservation ofenergy, we can then calculate the necessary molarenthalpy of change. Molar Heat of CombustionMolar Heat of Combustion (ΔHcomb): - the amountof heat released when one mole of reactant is burnedwith excess oxygen. Enthalpy of Combustion ΔH = nΔHcomb ΔH = Change in Enthalpy n = moles ΔHcomb = Molar Heat of Combustion (kJ/mol) Chemical Combustion CalorimetryHeat Lost = Heat Gained (Combustion Reaction)(water, kinetic)nsampleΔHcomb = CcalΔT (if bomb calorimeter isused) ornsampleΔHcomb = mwcP,wΔT (if the heat absorbedby the calorimeter itself is ignored) Schematic of a Bomb CalorimeterThe reaction is often exothermic and thereforeΔHcomb < 0.We often use a constant-volume calorimeter (or bombcalorimeter) to determine ΔHcomb due to its well-insulated design. It is calibrated for the heat capacity ofthe calorimeter, Ccal, before being use for to calculateΔHcomb of other substances. The sample is measured andburned using an electrical ignition device. Water iscommonly used to absorb the heat generated by thereaction. The temperature of the water increases,allowing us to find the amount of heat generated. Byapplying the law of conservation of energy, we can thencalculate the ΔHcomb of the sample. The world is made up of chemical reactions, it’s theway how you look at it and take it to your stride

Alfred Tauber (5 November 1866 – 26 July 1942) was a Hungarian-born Austrian mathematician, known for his contribution to mathematical analysis and to the theory of functions of a complex variable: he is the eponym of an important class of theorems with applications ranging from mathematical and harmonic analysis to number theory. He was murdered in the Theresienstadt concentration camp. Life and academic career Born in Pressburg, Kingdom of Hungary, Austrian Empire (now Bratislava, Slovakia), he began studying mathematics at Vienna University in 1884, obtained his Ph.D. in 1889, and his habilitation in 1891. Starting from 1892, he worked as chief mathematician at the Phönix insurance company until 1908, when he became an a.o. professor at Vienna University, though, already from 1901, he had been honorary professor at TH Vienna and director of its insurance mathematics chair. In 1933, he was awarded the Grand Decoration of Honour in Silver for Services to the Republic of Austria, and retired as emeritus extraordinary professor. However, he continued lecturing as a privatdozent until 1938, when he was forced to resign as a consequence of the "Anschluss". On 28–29 June 1942, he was deported with transport IV/2, č. 621 to Theresienstadt, where he was murdered on 26 July 1942. Pinl & Dick (1974, p. 202) list 35 publications in the bibliography appended to his obituary, and also a search performed on the "Jahrbuch über die Fortschritte der Mathematik" database results in a list 35 mathematical works authored by him, spanning a period of time from 1891 to 1940. However, Hlawka (2007) cites two papers on actuarial mathematics which do not appear in these two bibliographical lists and Binder's bibliography of Tauber's works (1984, pp. 163–166), while listing 71 entries including the ones in the bibliography of Pinl & Dick (1974, p. 202) and the two cited by Hlawka, does not includes the short note (Tauber 1895) so the exact number of his works is not known. According to Hlawka (2007), his scientific research can be divided into three areas: the first one comprises his work on the theory of functions of a complex variable and on potential theory, the second one includes works on linear differential equations and on the Gamma function, while the last one includes his contributions to actuarial science. Pinl & Dick (1974, p. 202) give a more detailed list of research topics Tauber worked on, though it is restricted to mathematical analysis and geometric topics: some of them are infinite series, Fourier series, spherical harmonics, the theory of quaternions, analytic and descriptive geometry. Tauber's most important scientific contributions belong to the first of his research areas, even if his work on potential theory has been overshadowed by the one of Aleksandr Lyapunov. His most important article is (Tauber 1897). In this paper, he succeeded in proving a converse to Abel's theorem for the first time: this result was the starting point of numerous investigations, leading to the proof and to applications of several theorems of such kind for various summability methods. The statement of these theorems has a standard structure: if a series ∑ an is summable according to a given summability method and satisfies an additional condition, called "Tauberian condition", then it is a convergent series. Starting from 1913 onward, G. H. Hardy and J. E. Littlewood used the term Tauberian to identify this class of theorems. Describing with a little more detail Tauber's 1897 work, it can be said that his main achievements are the following two theorems: - Tauber's first theorem. If the series ∑ an is Abel summable to sum s, i.e. limx→ 1 ∑+∞ n=0 an x = s, and if an = ο(n), then ∑ ak converges to s. This theorem is, according to Korevaar (2004, p. 10), the forerunner of all Tauberian theory: the condition an = ο(n) is the first Tauberian condition, which later had many profound generalizations. In the remaining part of his paper, by using the theorem above, Tauber proved the following, more general result: - Tauber's second theorem. The series ∑ an converges to sum s if and only if the two following conditions are satisfied: - ∑ an is Abel summable and k=1 k ak = ο(n). This result is not a trivial consequence of Tauber's first theorem. The greater generality of this result with respect to the former one is due to the fact it proves the exact equivalence between ordinary convergence on one side and Abel summability (condition 1) jointly with Tauberian condition (condition 2) on the other. Chatterji (1984, pp. 169–170) claims that this latter result must have appeared to Tauber much more complete and satisfying respect to the former one as it states a necessary and sufficient condition for the convergence of a series while the former one was simply a stepping stone to it: the only reason why Tauber's second theorem is not mentioned very often seems to be that it has no profound generalization as the first one has, though it has its rightful place in all detailed developments of summability of series. Contributions to the theory of Hilbert transform Frederick W. King (2009, p. 3) writes that Tauber contributed at an early stage to theory of the now called "Hilbert transform", anticipating with his contribution the works of Hilbert and Hardy in such a way that the transform should perhaps bear their three names. Precisely, Tauber (1891) considers the real part φ and imaginary part ψ of a power series f, - z = r with r = | z | being the absolute value of the given complex variable, - ck r = ak + ibk for every natural number k, - φ(θ) = ∑+∞ k=1 akcos(kθ) − bksin(kθ) and ψ(θ) = ∑+∞ k=1 aksin(kθ) + bkcos(kθ) are trigonometric series and therefore periodic functions, expressing the real and imaginary part of the given power series. Under the hypothesis that r is less than the convergence radius Rf of the power series f, Tauber proves that φ and ψ satisfy the two following equations: Assuming then r = Rf, he is also able to prove that the above equations still hold if φ and ψ are only absolutely integrable: this result is equivalent to defining the Hilbert transform on the circle since, after some calculations exploiting the periodicity of the functions involved, it can be proved that (1) and (2) are equivalent to the following pair of Hilbert transforms: - the complex valued continuous function φ(θ) + iψ(θ) defined on a given circle is the boundary value of a holomorphic function defined in its open disk if and only if the two following conditions are satisfied - the function [φ(θ − α) − φ(θ + α)]/α is uniformly integrable in every neighborhood of the point α = 0, and - the function ψ(θ) satisfies (2). - Tauber, Alfred (1891), "Über den Zusammenhang des reellen und imaginären Theiles einer Potenzreihe" [On the relation between real and imaginary part of a power series], Monatshefte für Mathematik und Physik, II: 79–118, doi:10.1007/bf01691828, JFM 23.0251.01. - Tauber, Alfred (1895), "Ueber die Werte einer analytischen Function längs einer Kreislinie" [On the values of an analytic function along a circular perimeter], Jahresbericht der Deutschen Mathematiker-Vereinigung, 4: 115, archived from the original on 2015-07-01, retrieved 2014-07-16. - Tauber, Alfred (1897), "Ein Satz aus der Theorie der unendlichen Reihen" [A theorem about infinite series], Monatshefte für Mathematik und Physik, VIII: 273–277, doi:10.1007/BF01696278, JFM 28.0221.02. - Tauber, Alfred (1898), "Über einige Sätze der Potentialtheorie" [Some theorems of potential theory], Monatshefte für Mathematik und Physik, IX: 79–118, doi:10.1007/BF01707858, JFM 29.0654.02. - Tauber, Alfred (1920), "Über konvergente und asymptotische Darstellung des Integrallogarithmus" [On convergent and asymptotic representation of the logarithmic integral function], Mathematische Zeitschrift, 8: 52–62, doi:10.1007/bf01212858, JFM 47.0329.01. - Tauber, Alfred (1922), "Über die Umwandlung von Potenzreihen in Kettenbrüche" [On the conversion of power series into continued fractions], Mathematische Zeitschrift, 15: 66–80, doi:10.1007/bf01494383, JFM 48.0236.01.

Splitting schemes of a high degree of accuracy have now attained a very advanced stage. One modification of this method is the so-called "particles-in-cells" method: The splitting is carried out according to physical processes and is independent of the reduction in the dimension of the operators. Splitting methods like local one-dimensional methods and hopscotch methods are often used in Western literature. Their applicability is somewhat limited, as fairly regular domains like squares are needed. Log in. Namespaces Page Discussion. Views View View source History. Jump to: navigation , search. For the system of differential equations 1 where is a differential operator, , , the absolutely stable implicit schemes of simple approximation 2 become ineffective in the case of multi-dimensional problems. For obtaining economical stable difference schemes methods are proposed based on the following ideas: 1 splitting of the difference schemes; 2 approximate factorization; 3 splitting weak approximation of the differential equations. In the case of equation 1 the respective difference schemes have the following form for the sake of simplicity, two fractional steps have been taken and the periodic Cauchy problem is considered : the splitting scheme: 3 the approximate factorization scheme: 4 the weak approximation scheme: 5 In the case of the schemes 3 and 4 inversion of the operator is replaced by inversion of the operator , i. It is applicable especially to potential problems, problems of elasticity and problems of fluid dynamics. The method offers a powerful means of solving the Navier-Stokes equations and the results produced so far cover a range of Reynolds numbers far greater than that attained in earlier methods. Further development of the method should lead to complete numerical solutions of many of the boundary layer and wake problems which at present defy satisfactory treatment. As noted by the author very few applications of the method have yet been made to problems in solid mechanics and prospects for answers both in this field and other areas such as heat transfer are encouraging. As the method is perfected it is likely to supplant traditional relaxation methods and finite element methods, especially with the increase in capability of large scale computers. The literal translation was carried out by T. Cheron with financial support of the Northrop Corporation. The editing of the translation was undertaken in collaboration with N. Later these results were extended by different authors to the equations of elasticity and plasticity and to multiply connected domains. AMS :: Mathematics of Computation Zav'yalov and Dr. Valeri L. This book is a comprehensive survey of different efficient algorithms for computing 1-D and 2-D splines. Our own main results relate to a new technique which we developed to obtain optimal error bounds for polynomial spline interpolation. This method is based on an integral representation of the error estimate. In many cases it gives minimal values of constants in error bounds for interpolation splines. The same approach is also used to find optimal error bounds for local spline approximation methods. This book has become a standard textbook for students, researchers and engineers in Russia and over eleven thousands copies were sold. Shape-Preserving Spline Interpolation: Standard methods of spline functions do not preserve shape properties of the data. By introducing shape control parameters into the spline structure, one can preserve various characteristics of the initial data including positivity, monotonicity, convexity, linear and planar sections. Based on interpolating splines, methods with shape control are usually called methods of shape-preserving spline interpolation. Here the main challenge is to develop algorithms that choose shape control parameters automatically. The majority of such algorithms solve the problem only for some special data. - Navigation menu. - Gendered Frames, Embodied Cameras: Varda, Akerman, Cabrera, Calle, and Maïwenn; - Separation of variables - Wikipedia? - Research Summary. - Solvent-free Organic Synthesis opt. - Separation of variables - Wikipedia! To solve the problem in a general setting, I gave a classification of the initial data and reduced the problem of shape-preserving interpolation to the problem of Hermite interpolation with constraints of inequality type. The solution is a C2 local generalized tension spline with additional knots. This allows for development of a local algorithm of shape-preserving spline interpolation where shape control parameters are selected automatically to meet the monotonicity and convexity constraints for the data. Dr. Boris I. Kvasov Its application makes it possible to give a complete solution to the shape-preserving interpolation problem for arbitrary data and isolate the sections of linearity, the angles, etc. Tension GB-Splines: In my opinion, my most significant contribution to the theory of splines, involves the development of " direct methods " for constructing explicit formulae for tension generalized basis splines GB-splines for short and finding recursive algorithms for the calculation of GB-splines. This approach has yielded new local bases for various tension splines including among others rational, exponential, hyperbolic, variable order splines, and splines with additional knots.

Critical rotation speed of dry ball-mill was studied by experiments and by numerical simulation using Discrete Element Method (DEM) The results carried out by both methods showed good agreement It has been commonly accepted that the critical rotation speed is a , formula for critical speed of rotation of a ball mill formula for critical speed of rotation of a ball mill As a leading global manufacturer of crushing, grinding and mining equipments, we offer advanced, reasonable Contact Supplier formula to calculate critical speed in ball mill The critical speed of the mill, &c, is defined as the speed at which a single ball will , rotating 813 Power drawn by ball, semi-autogenous and autogenous mills , In equation 814, D is the diameter inside the mill liners and Le is the effective. In solid mechanics, in the field of rotordynamics, the critical speed is the theoretical angular velocity that excites the natural frequency of a rotating object, such as a shaft, propeller, leadscrew, or gear Mill Critical Speed Determination The "Critical Speed" for a grinding mill is defined as the rotational speed where centrifugal forces equal gravitational forces at the mill shell's inside surface formula for critical speed of ball mill , critical speed of ball mill formula the slower the rotation If the peripheral speed of the mill is too great, it begins to act like a centrifuge , Chat Now; critical speed of ball mill formula This formula calculates the critical speed of any ball mill , formula for calculating the critical speed of a ball mill; critical speed of ball mill calculation pdf - e , 23 Oct 2016 Use our online formula Mill Speed - Critical Speed Mill Speed No matter how large or small a mill, ball mill, ceramic lined mill, pebble mill, jar mill or laboratory jar rolling mill, its rotational speed is important to proper and efficient mill operation The Gulin product line, consisting of more than 30 machines, sets the standard for our industry We plan to help you meet your needs with our equipment, with our distribution and product support system, and the continual introduction and updating of products Mill capacity can be increased by increasing speed but there is very little increase in efficiency (ie kWht-1) when the mill is operated above about 40-50% of the critical speed Please join and login to participate and leave a comment RAJIV GANDHI UNIVERSITY OF KNOWLEDGE - Results- critical speed of ball mill formula derivation ,2) What rotational speed, in revolutions per minute, would you recommend for a ball mill 21) In axial flow conveyor centrifuges, the critical separation occurs a) In continuous filtration, derive the equation for the rate of cake production perDynamics of Balls and Liquid in a Ball Mill - Department . was 38 mm for the larger drum and 225 mm for the smaller one , circuit was used to determine the speed of rotation by monitoring the applied voltage , rotation changed (which we took for passing the `critical speed' at which WWI occurs) Critical rotation speed of dry ball-mill was studied by experiments and by numerical simulation using Discrete Element Method (DEM) The results carried out by both methods showed good agreement Jul 19, 2016· 360 Degree Rotation Lab Grinder For Cocoa Nibs - Buy Lab Grinder Adjustable Revolution Speed: The lab ball mills are best choice for your lab's grinding and , Critical Speed Formula For Ball Mill - , ball mill critical speed calculation Ball mill critical speed, Ball mill efficiency,ball mill and now only the calculation formula on critical speed Critical rotation speed for ball-milling where D is the inner diameter of a jar and r is the radius of balls The critical speed of a rotating mill is the RPM at which a grinding medium will begin to “centrifuge”, namely will start rotating with the mill and therefore cease to carry out useful work Ball mills have been successfully run at speeds between 60 and 90 percent of critical speed, but most mills operate at speeds between 65 and 79 percent . The point where the mill becomes a centrifuge is called the "Critical Speed", and ball mills usually operate at 65% to 75% of the critical speed Ball Mills are generally used to grind material 1/4 inch and finer, down to the particle size of 20 to 75 microns Formula For Critical Speed Of A Rotating Mill - , Feb 15, 2016 Critical Speed Formula For Ball Mill formula for calculating the critical This is the rotational speed where balls will not fall critical speed on a Chat Now; What it is the optimun speed for a ball mill . formula to calculate critical speed ball mill formula for calculating the critical speed of a moran Control the Rotation Speed of Ball Mill Apr 25 2013 The ball mill rotating speed is called critical critical speed calculation of ball mill fineelevators in formula for critical speed of a rotating mill formula for critical speed of rotation of a ball mill Speed meters critical mill for in formula ball PSTCS Vertical Roller Mill for Speed meters critical mill for Hearst Magazin Subscribe and SAVE, give a gift subscription or get help with an existing subscription by clicking the links below formula for critical speed of a rotating mill - YouTube 15 Jan 2014 how to calculate the critical speed of rotating drum screen critical speed of ball mill formula for critical speed of a rotating mill How to calculate the critical speed of rotating drum - YouTube Feb 13, 2016 the critical speed, formula for critical speed of a rotating mill formula for critical speed of a rotating mill A High Performance Torque Sensor for Milling Based on In high speed and high precision machining applications, it is important to monitor the machining process in order to ensure high product quality Mar 30, 2015· Cheap Isoflex Grease, find Isoflex Grease deals on line at formula critical speed for the ball mill Take your stress away with the famous IsoFlex Stress Ball This formula calculates the critical speed of any ball mill Most ball mills operate most eff, and the ball mill stop grinding The ball mill rotating speed is called critical speed when th, critical rotating speed reasonably can realize the effective control on the steel ball consu, the slower the rotation If the peripheral speed of the . formula for critical speed of a rotating mill – 200T/H Critical Speed Equation In A Tumbling Mill formula for calculating the critical speed of a ball mill artificial sand making machine in hyderabad in chandrapur; Get More critical speed calculation of tumbling machine - iwspl Nov 18, 2013 , formula for calculating the critical speed of a ball mill, ball mill speed? newbie questions 2 most if the actual speed of a 6 ft diameter ball mill Read more how to calculate critical speed of a ball mill - YouTube TECHNICAL NOTES 8 GRINDING R P King 8-2 &ROOLVLRQ , 812 Critical speed of rotation The force balance on a particle against the wall is given by 8-3 Centrifugal force outward Fc mp& 2 Dm 2 (81) , Figure 84 The effect of mill speed on the power drawn by a rotating mill derivation of critical speed of ball mill merslin formula for critical speed of a rotating mill YouTube- derivation of the critical speed of ball mill,15 Jan 2014,critical speed of ball mill formula and Show The Critical Speed Of Ball Mill Derivation By the formula (3) can be seen, the critical rotation speed of steel ball centrifugal, Read more The Influence of Ball Mill Critical Speed on Production Efficiency The ball mill rotating speed is called critical speed when the outmost layer balls , and now only the calculation formula on critical speed in theory is widely used critical rotating speed of mill This formula calculates the critical speed of any ball mill Most ball mills operate most efficiently between% and% of their critical speed Read More >>Critical Rotating Speed Of A Millhiaimpolymersin

Patent number 8721922 is assigned to The following quote was obtained by the news editors from the background information supplied by the inventors: "This invention relates to light emitting compositions and light-emitting devices that include the light-emitting compositions. Specifically, this invention relates to light emitting compositions that are printable and light-emitting devices that include iridium-functionalized nanoparticles. "Organic Light Emitting Diodes (OLEDs) can be composed of small molecule or polymeric fluorescent or phosphorescent compounds. OLEDs comprise a cathode, a hole transporting layer, an emissive layer, an electron transporting layer and an anode. OLED devices emit light as a result of recombination of positive charges (holes) and negative charges (electrons) inside an organic compound (emissive) layer. This organic compound is referred as an electro-fluorescent material or electro-phosphorescent material depending on the nature of the radiative process. As OLED devices have developed to increase luminousity and increased lifetimes, additional layers, such as hole blocking layers and electron blocking layers, have been incorporated into the OLED device. However, introducing more layers of materials has made the OLED structure increasingly complex. This increased complexity makes the fabrication process significantly more difficult. The addition of layers also makes fabrication more difficult because poor control of layer thickness may impair performance. Thus, improving the performance of OLEDs is often tedious, difficult, and expensive. "There are several methods for manufacturing these above described layers within an OLED device. Primary methodologies include dry processing and wet processing. Dry processing is processing performed without a liquid. Examples of a dry processing operation include dry etching, laser ablation, chemical vapor deposition and vacuum deposition. Dry processing methods have several drawbacks, including difficulty controlling the thickness or composition of a previously deposited layer during serial deposition, high cost of equipment set up and maintenance, slow processing, and difficulty with substrates having a large area. Thus wet production methods may offer significant advantages. "Solution or wet-processing includes the dissolution or suspension of the precursor materials in a solvent and the application of the solution to the desired substrate. Exemplary methodologies include spin coating and inkjet applications. Spin coating can be undesirable because large quantities of the dissolved solution are spun off of the desired surface during the coating process. Thus, large amounts material is wasted production costs are higher. In addition to the background information obtained for this patent, VerticalNews journalists also obtained the inventors' summary information for this patent: "The inventors have discovered compositions that are, inter alia, useful as ink compositions that may be used in inkjet printers to fabricate light emitting compositions and devices. Some embodiments described herein relate to compositions comprising an iridium-functionalized nanoparticle that can include a nanoparticle core and an iridium-complex. In other embodiments, the iridium-functionalized nanoparticles described herein are light-emitting, e.g., white light-emitting. "One embodiment disclosed herein is a composition comprising an electron transport compound, an emissive compound, and an organic solvent, wherein the emissive compound is represented by formula (I): "##STR00001## wherein core is a nanoparticle core, n is 2, X is a single bond or "##STR00003## is independently a first optionally substituted bidentate ligand; "##STR00004## is a second optionally substituted bidentate ligand selected from: "##STR00005## wherein m is an integer in the range of 1 to 9, p is an integer in the range or 1 to 20, z is 0, 1 or 2, R.sup.1 is selected from alkyl, substituted alkyl, aryl and substituted aryl, R.sup.2 is selected from: alkyl, substituted alkyl, aryl and substituted aryl, and * indicates a point of attachment of the second optionally substituted bidentate ligand to the core or X. "One embodiment also disclosed herein is a composition comprising an electron transport compound, an emissive compound, and an organic solvent, wherein the emissive compound is represented by one of the following formulas: "##STR00006## wherein R' is represented by "##STR00008## and R'' is represented by "##STR00009## wherein each "##STR00010## is independently a first optionally substituted bidentate ligand, and "##STR00011## is a second optionally substituted bidentate ligand; R.sup.3 is "##STR00012## wherein k is 0 or an integer selected from 1 to 20, and R.sup.5 is independently selected from the following: "##STR00013## ##STR00014## ##STR00015## "wherein R is independently selected from H or alkyl, and * indicates a point of attachment in R.sup.3. "Another embodiment provides a method of fabricating a light-emitting device comprising depositing any composition disclosed herein upon an electrically conductive substrate via an inkjet printer. "Another embodiment provides a composition (IV) comprising: an emissive compound represented by Formula (IV), and an electron transport compound; and an organic solvent. "With respect to Formula (IV), each R.sup.4 is independently selected from: "##STR00017## and R.sup.5 is "Another embodiment is composition (V) comprising: an emissive compound represented by Formula (V), an electron transport compound, and an organic solvent. "With respect to formula (V), each R.sup.6 is independently selected from the following: "These and other embodiments are described in greater detail below." URL and more information on this patent, see: Keywords for this news article include: Nanoparticle, Nanotechnology, Emerging Technologies, Our reports deliver fact-based news of research and discoveries from around the world. 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The first table is for a partially wild card that can only be used to complete a straight, flush, straight flush, or royal flush, otherwise it must be used as an ace (same usage as in pai gow poker).An explanation of poker odds and poker hands probability. but sans wild cards the highest possible in poker [also called simply a royal flush]). POKER ODDS for.After all there are only 52 cards in a deck and 4 of each card, therefore the odds should be the same for each.The mean is less than the median with a negative skew, and greater with a positive skew. . Probabilities in Poker Probabilities of drawing. Approximate Hand Exact Probability Probability. Royal Flush. The number of. What is the probability of drawing a flush (all five cardsProbability and the Flush. The flush is just a bit easier to get than a full house. It is a hand well worth trying for in a poker game. To find out what the chances.How did you determine that this is the number of total possible results.A Holistic Statistical Test for Fairness in Video. Recall that a Royal Flush is a poker hand that consists of. to derive the probability of drawing a royal.The expected frequency of a royal would increase from once every 40388 hands to once every 23081.This machine allows one to play three hands at a time where the cards one holds are carried forward from the first hand to the other two. To answer your first question, there are 2598960 ways to choose 5 cards out of 52 for the initial hand.I tried the software in question in free-play mode and my results seemed fine.For a near-exact answer to streak questions such as this we need to use matrix algebra.In Jacks or Better, for the most part, you are not going to get a winning session over a few hours if you do not hit a royal.Such a player would not be able to sue the casino because it was his fault for playing so badly.The reason my video poker return tables have almost 20 trillion combinations is you also have to consider what could happen on the draw. Conditional Probability and Cards - homepages.math.uic.eduIn an earlier column, you said that in full pay Jacks or Better the perfect strategy player will average one royal flush every 40,388 plays. According to my page on sequential royal video poker, the odds are just about one in four million.What are the probabilities of all poker hands. than a flush. So, we compute the probability. Royal Flush 4,234. I once hit six royals in single-line video poker within 5,000 hands.Either she is using some kind of worthless progression, or this is second-hand exaggeration.I have noticed in your tables of probabilities and expected returns for video poker, that the probabilities (and corresponding number of hands) for each hand vary for the same type (jacks or better, for example) from one pay out chart to another.Probability and Poker. and the Royal Flush 2. Probability and the. three of a kind from a pair drawing three cards. 8. Probability and Two Pairs There are.Define m as the expected number of four of a kinds to get one that you need.If one is dealt, say, four of a kind on the initial draw of five cards, one will be paid on all three hands. Could you comment on the variance and covariance in Spin Poker. Rules of Card Games: Poker Hand Ranking - Pagat.comStep 4: Multiply the initial state after 5,000 hands by T-24,995,500.I assumed that given two plays of equal royal probability the player will choose the play which maximizes the return on the other hands.Things get more complicated with straights and flushes but still manageable.If your strategy were to maximize the number of royals at all costs then you would hit a royal once every 23081 hands.I was wondering if I could get your help on computing the probability distribution table for Jacks or Better.Although I strongly feel poker based games should be played with only one deck, I will submit to the will of my readers and present the following tables.I know as the jackpot increases, so does the payback percentage. In poker, players construct sets of five playing cards, called hands, according to the rules of the game being played. Each hand has a rank, which is compared.Probabilities of Poker Hands with Variations. Royal Flush – all five cards. Hand Number Probability Straight Flush 2 40 0.00002.straight flush 4-of-a-kind full house flush. Here is a table summarizing the number of 7-card poker hands. The probability is the probability of having the hand.Dear Mr. Wizard, How do minimum payback laws affect video poker machines. All 10 hands (and thus all 100 lines) failed to bring up a single win.You have developed an excellent website for information concerning gambling, and I have found it very useful.For the benefit of other readers, the coefficient of skewness (skew) for any random variable is a measure of which direction has the longer tail.What are the odds of being dealt a royal flush on a Triple Play video poker machine.A royal flush is an ace, king, queen, jack, and 10, all of one suit. A straight flush is five consecutive cards all of the same suit (but not a royal flush), where an ace may count as either high or low. A full house is three-of-a-kind and a pair. A flush is five cards of the same suit (but not a royal flush or straight flush).Please can you calculate the probability of drawing a blank on 10 hands of 10 line JoB.Culture & Cosmos Card Games: Probability of a Royal Flush The poker hand known as a royal flush consists of an ace, king, queen, jack and ten card and all must be in.On your video poker tables you use the figure of 19,933,230,517,200 possible results.I was wondering why there are so many more than 52 choose 5, and how to compute them. Probability Of Royal Flush casino hotels las vegas free poker chart play pai gow poker with fortune bonus. Probablity of Royal Flush | Physics Forums - The Fusion ofOn the draw there are 1, 47, 1081, 16215, 178365, or 1533939 ways to draw the replacement cards, depending on how many card the player holds.It can also happen if a hand 5,000 games ago was a royal, but the new hand is also a royal.How many hands to reduce the probability of no royal flush in n. The probability of a royal flush in a poker hand is. Probability of drawing a flush from a. It is interesting to note that skew is greatest for Jacks or Better.However, if you limit me to games that are easy to find, my nomination is Triple Double Bonus, with a standard deviation of 9.91. Here is that pay table.Thank you for your valuable time in reading and hopefully responding.It makes sense to have four aces as the premium four of a kind, because aces are the highest card in regular poker.For example with any pair and 3 singletons the probability of improving the hand to a two pair is always the same.

THE MECHANICS OF THE UNIVERSE Copyright 1984-2006 Allen C. Goodrich The planets orbit the sun in a special mean orbital radius L to conserve total energy. The modified first law of thermodynamics , which states that the total energy of the universe is a constant, is the fundamental equation of the universe. The sum of kinetic and potential energies is a constant. m (2 pi L)^2 / t^2 + G m(M-m) /L = A CONSTANT. (if no charges are present) - Delta m(2 pi L)^2/T^2 = - Delta Gm(M-m)/L What's so important about this modified first law of thermodynamics? It says that no energy change or force is necessary for orbital motion. A positive change of kinetic energy is accompanied by a negative change of potential energy relative to the rest of the universe to conserve the total energy. The universe has been found to be expanding at an accelerating rate. The potential energy of the universe is continually decreasing and the kinetic energy is continually increasing. Again, to conserve the total energy relative to the rest of the universe. This is why the modified first law is so important. The conservation of total energy must be maintained relative to the rest of the expanding universe. Kinetic and potential energies must be computed relative to the rest of the expanding universe. This modified first law leads to the conclusion that the force of gravity and the velocity of light are misleading illusions. The definition of Kinetic and Potential Energies is most important to an understanding that these energies are relative to the rest of the effective universe, not just relative to any other mass. Kinetic energy is mass m times the square of the velocity relative to the rest of the effective universe. For example the kinetic energy of the earth is effectively relative to the sun plus the rest of the planets of the solar systen. Potential energy is the product of the mass m times the mass of the rest of the effective universe M-m divided by the distance L between their effective centers of mass. M is the mass of the entire effective universe. t is the orbital time for one complete revolution. G is the gravitational constant. Once this is clearly understood, the modified First Law of Thermodynamics becomes the Fundamental Equatiion of the Universe. This equation ,then, defines the photon and the rest of the universe. The Thomas R. Young two slit defraction pattern, the complete logic of quantum mechanics, the true nature of gravitation, the fact that light does not have a mass or a velocity, all become quite obvious. This is a simple solution to so many of the problems that baffled scientists for hundreds of years. We remember that Sir Isaac Newton proposed the force of gravity F_g = k m_1 m_2 / L^2. Scientists have known that action at a distance without the transfer of energy was not possible. A force of gravity would not be possible without the transfer of energy. The amount of energy required to cause the so called force of gravity to make the planets travel in orbits, other than the equilibrium orbit,about the sun would be tremendous and is not available. Another explanation is necessary. Einstein and other scientists have assuned that masses change the shape of space time. These explanations have their problems. No problems exist if the modified first law of thermodynamics is used.This is the Fundamental Equation of the Universe. One night when I was walking on Myrtle Beach in the light of the full moon, the full moon was at its highest point in the sky, and the beach was very wide. The lowest tide was present. This is not what would be expected according to the gravitational theory. The gravitational theory states that the tide should be very high. One can find this explained in most older dictionaries or encyclopedias. The water of the ocean is shown bulging on the side of the earth directly under the full moon. The tides never occur in this way. This discrepancy with the law of gravity is explained by the requirement for the time required for the water to flow due to the force . No one has bothered to explain that the water would have to flow at more than1000 miles per hour to complete this picture. This flow would wash all of the continents away in one day, Here, we have a big problem with the force of gravity . The NOAA U.S. Coast and Geodetic Survey has monitored the tides at many ocean ports for many years and this data is available. The Jet Propulsion Laboratory has plotted the position and phases of the moon with changes of time. This data is also available. No one has previously published a correlation ot these two sets of data. If they had, it would become very obvious that invariably the lowest tide occurred when the full moon and new moon were exactly directly at their highest point in the sky. It would have been obvious that the lowest tide also occurred on the opposite side of the earth. Not at all consistant with the existing gravitational theory. However, this is predicted by the modified first law of thermodynamics, when the kinetic and potential energies are relative to the rest of the effective universe. The planets orbit the sun at a special mean radius L, to conserve total energy. If one calculates the kinetic and potential energies of the planets and moons,at orbital distance L, one finds that the two are nearly equal in magnitude but opposite in sign. This is the only mean orbital radius L where no change of total energy is necessary. Nature obays this modified first law. The assumption of a force of gravity, with its energy transfer, action at a distance, is not necessary to explain orbital motion at equilibrium. Gravitation is explained by the modified first law of thermodynamics. A force must be present and an energy change is necessary at any other radius, such as a body on the surface of the earth. The Thomas R. Young's two slit defraction pattern is also explained by the modified first law of thermodynamics if one uses the charges e in this fundamental equation for the calculation of kinetic and potential energies. Delta e_1 (2 pi L)^2 / t^2 = Delta-K e_1 e_2 /4 pi E_o L. Where e_2 is the charge of the rest of the effective universe and . e_1 is the orbital electron. E_o is the dielectric constant. Here, the masses would have little effect by comparison and can be neglected. The kinetic energy change of the electron would then be a function of a change of potential energy relative to the rest of the effective universe. The energy of the electron of the atom on the defraction pattern screen would be a function of the rest of the effective universe. The electron would sense the fact that one or two of the slits was open and the Thomas R Young defraction pattern is explained. We now have the basis for a new quantum mechanics. Light , that is observed as a change of the kinetic energy of the electron, which has the correct energy density (time), direction and frequency, in the expanding universe, is not a particle, with mass and velocity ( kinetic energy ), but it is a function of the potential energy change of the rest of the effective universe .+

IV. Answer the questions below and then ask for more information (Work in pairs). 1. Who didn’t consider 1 to be a number at all? 2. What number associates with negatives? Why? 3. What number is the dimension of the smallest magic square in which every row, column, and diagonal equals fifteen? 4. What numbers are considered to be perfect in Mathematics? Why? 5. What does the number 7 determine in China? 1. In what country is the number 4 considered to be unlucky? 2. Why was the number 5 important to the Maya? 3. Why is a knot tied in the form of the pentagram called a lover’s knot in England? 4. What number leads to a few years of bad luck, if you break a mirror? 5. What did students pursue in medieval education? DO YOU KNOW THAT… · The number 8 is generally considered to be an auspicious number by numerologists. The square of any odd number, less one, is always a multiple of 8 (for example, 9 − 1 = 8, 25 − 1 = 8 × 3, 49 − 1 = 8 × 6), a fact that can be proved mathematically. · The early inhabitants of Wales used nine steps to measure distance in legal contexts; for example, a dog that has bitten someone can be killed if it is nine steps away from its owner’s house, and nine people assaulting one constituted a genuine attack. · The number 20 has little mystical significance, but it is historically interesting because the Mayan number system used base 20. When counting time the Maya replaced 20 × 20 = 400 by 20 × 18 = 360 to approximate the number of days in the year. Many old units of measurement involve 20 (a score), for example, 20 shillings to the pound in pre-decimal British money system. V. Find information on the Internet and give a presentation of the number you are interested in (brings you good or bad luck). Reading and Speaking NUMBER AND REALITY Many aspects of the natural world display strong numerical patterns, and these may have been the source of some number mysticism. For example, crystals can have rotational symmetries that are twofold, threefold, fourfold, and sixfold but not fivefold − a curious exception that was recognized empirically by the ancient Greeks and proved mathematically in the 19 th century. An especially significant number is the golden ratio, usually symbolized by the Greek letter ϕ. It goes back to early Greek mathematics under the name ‘extreme and mean ratio’ and refers to a division of a line segment in such a manner that the ratio of the whole to the larger part is the same as that of the larger part to the smaller. This ratio is precisely (1 + √5)/2, or approximately 1.618034. The popular name golden ratio, or golden number, appears to have been introduced by the German mathematician Martin Ohm in Die reine Elementarmathematik (1835; ‘Pure Elementary Mathematics’). If not, the term is not much older and certainly does not go back to ancient Greece as is often claimed. In art and architecture the golden number is often said to be associated with elegance of proportion; some claim that it was used by the Greeks in the design of the Parthenon. There is little evidence for these claims. Any building has so many different lengths that some ratios are bound to be close to the golden number or for that matter to any other ratio that is not too large or small. The golden number is also often cited in connection with the shell of the nautilus, but this too is a misunderstanding. The nautilus shell has a beautiful mathematical form, a so-called logarithmic (or equiangular) spiral. In such a spiral each successive turn is magnified in size by a fixed amount. There is a logarithmic spiral associated with the golden number, and in this case the fixed amount is precisely ϕ. However, the spiral of the nautilus does not have the ratio ϕ. Logarithmic spirals exist with any given number as their ratio, and the nautilus ratio has no special significance in mathematics. The golden number is, however, legitimately associated with plants. This connection involves the Fibonacci numbers (1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144…), in which each number, starting with 2, is the sum of the previous two numbers. These numbers were first discussed in 1202 by the Italian mathematician Leonardo Pisano, who seems to have been given the nickname Fibonacci (son of Bonaccio) in the 19th century. The ratio of successive Fibonacci numbers, such as 34/21 or 55/34, gets closer and closer to ϕ as the size of the numbers increases. As a result, Fibonacci numbers and ϕ enjoy an intimate mathematical connection. Fibonacci numbers are very common in the plant kingdom. Many flowers have 3, 5, 8, 13, 21, or 34 petals. Other numbers occur less commonly; typically they are twice a Fibonacci number, or they belong to the ‘anomalous series’ 1, 3, 4, 7, 11, 18, 29… with the same rule of formation as the Fibonacci numbers but different initial values. Moreover, Fibonacci numbers occur in the seed heads of sunflowers and daisies. These are arranged as two families of interpenetrating spirals, and they typically contain, say, 55 clockwise spirals and 89 counterclockwise ones or some other pair of Fibonacci numbers. This numerology is genuine, and it is related to the growth pattern of the plants. As the growing tip sprouts, new primordial − clumps of cells that will become special features such as seeds − arise along a generative spiral at successive multiples of a fixed angle. This angle is the one that produces the closest packing of primordial; and for sound mathematical reasons it is the golden angle: a fraction (1 − 1/ ϕ) of a full circle, or roughly 137.5 degrees. I. Match the words (1–11) with the definitions/explanations (a– k): Дата добавления: 2019-02-26; просмотров: 144; Мы поможем в написании вашей работы! Мы поможем в написании ваших работ!

The flexing operation, we have seen, requires a thumbhole with a hinge. In the case of the proper flexagon, this is the number one thumbhole. This thumbhole splits the left hand pat into two groups, one composed of a single subpat and the other composed of subpats. The complete flexing operation inverts the top-most subpat of the left pat, leaving it on the left side, while it deposits the remaining subpats on the inverted right pat (see figure 8.1). The number one thumbhole has been opened out to display the next side. The two new pats are now joined by what was originally the hinge. When the flexagon is rotated, the hinges must be renumbered. The original hinge will become the new hinge, and each of the other hinges will have values one lower than before the flex and rotation. In the normal flexing operation of a proper flexagon, which has a consecutive subpat hinge sequence and a consecutive subpat structure, the order of turning up sides must also be consecutive (as has been shown empirically). This is because each successive thumbhole is associated with a correspondingly numbered hinge, (i.e. the hinge with the number one thumbhole, etc.). Each flexing operation subtracts one from the value of each hinge, thus bringing the thumbhole with which any particular hinge is associated closer to the position for being opened up next. A subpat hinge which in in position will be opened after flexes and rotations. If we flex along a given cycle of a flexagon, we notice that we always progress in the same direction, either clockwise or counterclockwise, along the path of the map. If we draw vectors along the edges of the map (indicating in which direction we are progressing), the vectors for a given cycle will always point consistently clockwise or counterclockwise. Whichever way they do point, their direction may be reversed by turning the flexagon over. If a second cycle is added and vectors are drawn on the map, this new cycle will have one vector in common with the first cycle, but the direction of all the vectors around the center of the map polygon representing this cycle will be just opposite from that of the vectors in the first cycle. In fact, the vectors of any map cycle which has one edge in common with any other given cycle will point in the opposite direction with respect to that cycle. This means that all of the polygons in the map of a given flexagon will be oriented; the sense of the orientation may be changed by turning the flexagon over, since in so doing all vectors are reversed. The reason for the reversed pat structure of a subpat with respect to its large pat can now be explained. Consider the history of a proper subpat of a proper flexagon. The pat structure of this subpat will remain unchanged throughout the flexing operation (assuming the flexagon flexes left) provided it is in the left pat and provided it is below thumbhole 1, which has a hinge in the ``1'' position. This is because a flex moves all of the subpats which are below thumbhole 1 from the left pat to the right pat, unchanged in any way (i.e. uninverted). However, if this subpat is on the top of the left pat, and the flexagon is flexed, the subpat will remain in the left pat but will be inverted. A rotation and flex will reinvert it and place subpats on top of it, leaving it undisturbed for flexes thereafter. However, if we decide not to rotate, but to flex along a new cycle, this subpat alone will remain in the left hand pat. As the flexagon must always flex left (we built it that way), the first flexing operation will open up the first thumbhole to a side, , the last side up being (a) if the flexing was proceeding in an ascending order around the map (see figure 8.2a). For instance, in an order 6 tetraflexagon as shown in figure 8.2b, if we flex from 1 to 2 to 3 to 4 to 1, and then decide to change over to a new cycle, we must flex next to side 6. Although the numbering of the map is still counterclockwise, the flexing vectors have changed direction, and the flexing must proceed against the numbering. In order for the side following side (a) to be side , the thumbhole must be the lowest thumbhole in the subpat containing the new cycle, for that subpat will be inverted, making the thumbhole the top most one when the subpat is left along in the left pat. Similarly, in flexing about the second cycle (assuming there are no other cycles attached to the second one) will follow and will in turn be followed by , and so on until is reached. Since side used the first thumbhole in the subpat, sides through must use the others in order, and when the subpat constant order is inverted by the subpat's incorporation into the large pat, will be the side nearest the top. Since each thumbhole in a subpat can be associated with a single leaf, the thumbhole order becomes reversed also. Since the basic number sequence must increase consecutively when read down the pat, the pat structure for the subpat must be inverted with respect to the large pat. Since we are on the subject of flexing operation, let us consider flexing operations other than the flex; i.e., the tubulations. For all proper flexagons, the flexes remove all but leaves respectively from the left pat and deposit them from the right pat; As has been stated before, a tubulation acts like a flex. For instance, in a tetraflexagon of order 3 (see figure 8.3a) which has a tubulation from 1 to 3, we can 1 flex from face to face and when we tubulate, we can cut the hinge and lay the flexagon out in the form of a straight strip of squares with one on the top and three on the bottom. This, then, is face (see figure 8.3b). When we want to flex from face back to , we fold the three's so that they face each other and tape the cut hinge back together. The flexagon will now open up to side 2. We should notice, however, that this process of turning the tubulation in side out has also exchanged the position of the two pats of a unit with respect to each other. This is equivalent to a rotation so in this operation, we have both flexed and rotated.

Wetting, Spreading, and Adsorption on Randomly Rough Surfaces The wetting properties of solid substrates with customary (i.e., macroscopic) random roughness are considered as a function of the microscopic contact angle of the wetting liquid and its partial pressure in the surrounding gas phase. Analytic expressions are derived which allow for any given lateral correlation function and height distribution of the roughness to calculate the wetting phase diagram, the adsorption isotherms, and to locate the percolation transition in the adsorbed liquid film. Most features turn out to depend only on a few key parameters of the roughness, which can be clearly identified. It is shown that a first order transition in the adsorbed film thickness, which we term ’Wenzel prewetting’, occurs generically on typical roughness topographies, but is absent on purely Gaussian roughness. It is thereby shown that even subtle deviations from Gaussian roughness characteristics may be essential for correctly predicting even qualitative aspects of wetting. pacs:68.05.-n; 68.08.-p; 05.40.-a; 64.75.-g While the physics of wetting and spreading on ideally smooth solid surfaces has meanwhile reached a status of mature textbook knowledge, a whole range of wetting phenomena on randomly rough substrates are still elusive. This is particularly annoying, as almost all surfaces of practical interest bear considerable roughness, be it due to weathering, wear, or on purpose as, e.g., in the case of sand-blasted surfaces. Clearly, this has substantial impact in many situations. For example, a drop of liquid deposited on a rough substrate will spread or not, depending on the morphology of the liquid film which develops in the troughs of the roughness. As it accommodates its free surface to the substrate topography, it may percolate across the sample. The drop will then gradually spread over the entire sample. On the contrary, if the film rather tends to form isolated domains, the drop will stay in place. Similarly, the redistribution of liquid within a granular pile, such as in humid soil or sand, may proceed along the grain surfaces only if the liquid wetting film on the grains forms a percolated structure. The morphology of a liquid water film deposited from humid air onto the surface of an electric isolator will strongly affect the performance of the latter, for analogous reasons. There has been already a lot of research on the adsorption of liquids on rough surfaces [1-15], but this was concerned either with roughness amplitudes as small as the (nanometer) range of van der Waals forces (3); (5); (7); (9) or with rather artificial substrate topographies in the context of super-hydrophobicity (10); (11); (12); (13); (15), or with completely wetting liquids (zero contact angle) (1); (4); (6); (9). The most frequently encountered, customary case, however, is characterized by a finite contact angle, and a random roughness with typical length scales at least in the micron range. In the present paper, we consider surfaces which exhibit a random topography on scales large as compared to molecules, and are subject to adsorption of a liquid which forms a small but finite contact angle with the substrate material. Since we consider macroscopic roughness, we adopt the view that all interfaces are infinitely sharp on the length scale of consideration (sharp kink approximation (16)). As the typical length scales considered here are still small as compared to the capillary length of the liquid (2.7 mm for water), gravity will be neglected as regards its effect on the liquid surface morphologies to be described. As opposed to earlier studies which concentrated on the macroscopic contact angle and contact line (2); (8), we will try to derive the wetting phase diagram and other characteristics connected to the adsorption of a liquid film. The first systematic study of wetting on a randomly rough substrate at finite contact angle owes to Wenzel (17). He characterized the roughness by a single parameter, , which he defined as the ratio of the total substrate area divided by the projected area. Obviously, , and corresponds to a perfectly smooth surface. The free energy which is gained per unit area when the rough substrate is covered with a liquid is then given by , where and are the solid-liquid and solid-gas interfacial tension, respectively. If this is larger than the surface tension of the liquid, , we expect a vanishing macroscopic contact angle, because covering the substrate with liquid releases more energy than is required for the formation of a free liquid surface of the same (projected) area. More specifically, force balance at the three-phase contact line yields for the macroscopic contact angle on the rough surface. is the microscopic contact angle according to Young and Dupré. When the microscopic contact angle is reduced to , which we will henceforth call Wenzel’s angle, vanishes, and the substrate is covered with an ’infinitely’ thick liquid film. As we will see below, however, there are imprortant ramifications which are sensitive to the kind of roughness of the substrate. Furthermore, even minute deviations from liquid-vapour coexistence, as they are omnipresent in practical situations, unveil a rather complex scenario which goes well beyond eq. (1). Ii Presentation of the problem We describe the topography of the rough solid substrate by , where is a vector in the plane. The (randomly varying) function is normalized such that , where the angular brackets denote averaging over the entire sample area, . It is assumed that the substrate is homogeneous and isotropic, in the sense that the statistical parameters of are the same everywhere on the sample, and independent of rotation of the sample about the normal axis of the sample. A small amount of liquid deposited on this substrate will make an interface with the surrounding gas, which can be described by a second function, . The support of is the set , which denotes the wetted area. Continuity of the liquid surface assures that on the boundary of , i.e., at the three-phase contact line, where the solid substrate, the liquid, and the gas phase meet. This line will henceforth be denoted by . Information on can be obtained from the total free energy functional of the system, which is given by Minimization of yields two important properties of . First of all, the mean curvature of the liquid surface, which can be written as (18) assumes the same value everywhere on . Second, the two surfaces described by and make the same (Young-Dupré) angle everywhere on . This reflects the local force balance at the three-phase contact line. While surface roughness gives rise to substantial contact angle hysteresis on macroscopic scales, the microscopic contact angle, , is known to be well defined on the typical (micrometer to nanometer) scale (19); (20). Nevertheless, we should be aware that even on small scales, equilibration will take time, be it by transport through the gas phase or through an adsorbed layer of molecular thickness (16); (9) (which we disregard in the present study). The question we shall ask is the following. Given the substrate topography, , the equilibrium microscopic contact angle, , and the mean curvature of the liquid surface, , what can we predict on the function and the shape of the wetted area, ? In particular, we shall be interested whether forms a percolated set in the plane. Based on the observation that the amplitude of most natural roughness is much smaller than its dominant lateral length scale, we assume for the present study that which allows for substantial simplifications. Expanding then the mean curvature according to eq. (3), we obtain to first order in Similarly, the contact angle with the substrate yields the boundary condition on , to first order in and . We can now immediately write down a useful identity concerning these quantities. Green’s theorem tells us that in which denotes the length of . This equation will be the starting point of the discussion to follow. Iii Gaussian roughness If we want to exploit eq. (8), we have to refer to a specific roughness function, . Following the overwhelming majority of the literature on randomly rough surfaces, we will start by considering Gaussian roughness. The height distribution is then with . Below we will make use of its polynomial expansion, iii.1 Distribution functions For Gaussian roughness, the joint distributions of with other stochastic quantities can be obtained in a straightforward manner from multivariate analysis. As it is well known (21); (24), the joint distribution of two quantities and is then given by where is the inverse of the matrix and is the determinant of that matrix (21). For the joint probability of and , we find On the side, this directly yields , from which we can conclude that roughness topographies fulfilling (4) will have . For the joint probability of and , we obtain The fraction of the total sample area which lies within that contour is and the total Laplace curvature of within that area is In order to fulfill the boundary condition, eq. (6), the vertical position of the three-phase contact line, which may be symbolically written as , will vary along about an average value, . The three-phase contact line will thus approximately follow the contour line at , with excursions towards both the outside and the inside of . These will in cases represent detours, sometimes shortcuts with respect to . As a reasonable approximation, we may thus use for the length of the three-phase contact line. Similarly, we set with . Inserting these expressions in eq. (8), we obtain iii.2 The phase diagram If the adsorbed material is at liquid-vapor coexistence, the mean curvature of the free liquid surface, , vanishes everywhere on . In this case, eq. (21) is fulfilled only for a certain contact angle, Note that is independent of . This at first glance puzzling result has its origin in a peculiar property of Gaussian roughness, namely the statistical independence of and (21); (26). In other words, the probability of finding a certain slope at a given level, , is independent of . A von Neumann boundary condition such as eq. (6) can thus be fulfilled equally well at all levels of Gaussian roughness. It is therefore not surprising that no particular value of is singled out here. It is instructive to compare with . For Wenzel’s parameter , we have and therefore . For we obtain, from eq. (22), . Since , we see that . In other words, if the liquid does not wet the substrate well enough to fulfill the Wenzel condition, is may nevertheless well intrude the roughness topography and thus form a wetting layer. This is indicated in Fig. 1, which shows the phase diagram of wetting on a surface with Gaussian roughness. States corresponding to liquid/vapour coexistence lie on the vertical axis. Along the bold solid line, which ends at , the liquid surface can ’detach’ from the rough substrate, such that an infinitely thick liquid film may form. For , the liquid/vapour interface needs the support of the spikes of the roughness, to which it is attached by virtue of the boundary condition, eq. (6). Let us now discuss what happens as we move off coexistence. We first define the parameter which only depends upon the substrate topography (through ) and . Inserting this into eq. (21), we obtain as an alternative form of (21). This is indicated by the dashed straight line in Fig. 1, which for ends at . This line cuts through the whole range of contact angles below . At higher angles, eq. (6) cannot be fulfilled and the substrate remains dry everywhere. iii.3 Adsorption isotherms We can now calculate the adsorption isotherms of the system, i.e., the amount of liquid adsorbed at pressures below saturation. This is important to discuss, as almost no practical situation corresponds exactly to liquid/gas coexistence. Consider, for instance, the substrate to be located at a height above a liquid reservoir with which it can exchange material. is then given by the balance with the hydrostatic pressure and reads For the distribution of water within a soil or granular pile at height above the water table, we find that grows to about nm as increases to m. Hence the typical curvatures to expect are of the right size for our considerations up to several meters above the water table. A more general way to look at this situation is to consider the vapour pressure, which is reduced at finite height above the liquid reservior, as well due to gravity. The curvature is then given by the Kelvin equation, where is the partial pressure of the adsorbed liquid in the sorrounding gas phase, is its saturated vapor pressure, its molecular volume, and is Boltzmann’s constant. From eq. (28), with the abbreviation , we can express the adsorption isotherms in terms of as We would, however, like to know not the position of the liquid surface, , but the total volume of adsorbed liquid. The latter can be easily expressed as The volume at percolation, i.e. at , is . Combining eqs. (29) and (30), we can directly plot the adsorption isotherms, which are displayed in Fig. 2 for three different values of . If , remains zero for all , and jumps to infinity at . It is important here to appreciate that the infinite adsorption one obtains at coexistence has two different meanings for contact angles above or below . While for , the liquid can detach completely from the substrate forming a bulk liquid phase, the liquid surface remains in contact with the rough substrate for . The fact that even then the adsorption isotherms tend to infinity at coexistence owes to the infinite support of the Gaussian distribution, eq. (9). We will see below that this is a peculiarity of Gaussian roughness, and not a generic feature of practically encountered roughness topographies. Iv Non-Gaussian roughness As we have seen so far, it is worthwhile to study non-Gaussian roughness models as well. In fact, it has been shown that many real surfaces are distinctly non-Gaussian (27); (28); (29); (30), such that the freedom in adjusting the correlation function is not sufficient to describe a relevantly large class of surfaces. It seems to be widely believed that the correlation function together with the height distribution of the topography are sufficient to characterize all physically relevant properties of a surface. Most authors even seem to believe that only the first four moments of the height distribution are relevant (including skewness and kurtosis) (27); (28); (29); (31). In what follows, we will introduce a simple roughness model which is general enough to describe roughness profiles with any lateral correlation function and height distribution, but is still accessible to the analysis given above. As a consequence, we will be able to derive, by purely analytic methods, quite general predictions about wetting, adsorption, and liquid percolation on a rough surface, which can be quantitatively applied to experimental data. Let be a Gaussian random function, much like as discussed above, but with dimensionless codomain and unity root mean square. Hence its height distribution is and the correlation function, We then set where has the dimension of a length and is a monotone, two times differentiable function. In this case the inverse of , , exists, and we have where the prime indicates the derivative with respect to the argument. can be directly determined from experimental topography data. If the distribution has been measured, we can derive by means of the simple formula Note that this allows to represent any height distribution function . The correlation function of , and thereby the set of coefficients , is obtained from the data as . Fig. 3 shows a sketch of a typical . While the support of is the whole axis, the codomain is bound, because neither will there be any material outside the original (unworn) surface, nor will there be infinitely deep troughs. Since in eq. (33) we have done nothing but distorting the assignment of vertical positions to the plane, contour lines and their enclosed areas will change in level according to , but their topolgical properties, including percolation, will remain unchanged. We can therefore directly write down the contour length with help of the new quantities, For the joint probability of and , we obtain where the prime now denotes the derivative with respect to the argument. The Laplace curvature is given by , which leads to intimidatingly clumsy expressions when inserted into multivariate analysis. We therefore consider here the important case when is small, such that only the second term in contributes. This is the case if We then have with . Now we are in shape to express the Laplace curvature inside the wetted area. In complete analogy to the derivation above, we find At coexistence, we have again In analogy to the above discussion, we define the parameter iv.1 The phase diagram The film thickness at coexistence can be derived from the zeros of , of which there are either two or none. In the latter case, the contact angle (and thereby ) is too large for forming a liquid surface between the spikes and troughs which complies with the boundary condition, eq. (6). If, however, intersects the -axis, the slope of the zeros decides upon their stability. This can be seen by appreciating that may be interpreted as a deviation from the force balance, eq. (41), as required by eq. (6). For the left zero, which is marked by an open circle in the figure, a displacement of the three-phase contact line would give rise to an imbalance of wetting forces driving it further away from the zero. The opposite is true for the right zero, marked by the closed circle. The latter is therefore stable and thus corresponds to the adsorbed film thickness which will develop. The film will be percolated if this zero lies to the right of , which corresponds to the mid-plane of (cf. Fig. 3). If we now again consider the system off coexistence, we have as the condition for , where A graphical solution of eq. (43) is sketched in Fig. 5. Again, the closed circle indicates the stable solution. The liquid film will be percolated if this point lies to the right of the dashed line at , but form isolated patches otherwise. From eq. (43), we see that percolation occurs if which represents again a straight line in the phase diagram as depicted in Fig. 6. For Gaussian reoughness, we have , , and . It is readily checked that this leads again to eq. (26) instead of (LABEL:Eq:PercolationNonGaus), and (24) instead of (42). As in the case of Gaussian roughness, generically lies below . This can be seen from calculating which follows from (37). On the other hand, Since, again, , it is clear that whenever lies close to (which it typically will), we have as for the Gaussian case (cf. Fig. 6). Inspection of Fig. 5 shows that the two points of intersection, which are marked by the closed and open circles, will merge when the dashed and solid curves touch each other in a single point. This occurs at a certain curvature of the liquid surface. For , there is no liquid adsorbed, and the substrate is dry. As is reached, the average position of the liquid surface, , jumps discontinuously to the value given by the point of contact of the two curves. As is further reduced, increases until at coexistence it reaches a value corresponding to the right zero of . Because of the phenomenological similarity of the jump in adsorbed film thickness to the prewetting transition encountered in standard wetting scenarios on flat substrates (16), we hereby propose to term this transition ’Wenzel prewetting’. When the microscopic contact angle is varied, a ’Wenzel prewetting line’ results, which is shown in Fig. 6 as the solid curve. As in the usual prewetting scenario, this line ends in a critical end point, when the solid and dashed curves in Fig. 5 intersect in only a single point. It is readily appreciated from the construction sketched in Figs. 4 and 5, however, that this can occur only for , and thus outside the physically accessible parameter range. In principle, the Wenzel prewetting line may intersect the percolation line. The latter then follows the prewetting line down to . Let us discuss how the position of the liquid surface varies along liquid/vapour coexistence as is gradually decreased from above . This can be directly read off Fig. 4, by inverting for , and is sketched in Fig. 7. The liquid film first appears through a discontinuous jump as crosses the Wenzel prewetting line. As , the liquid surface configuration which is bound to the surface topography through eq. (6) becomes metastablee (dashed curve), and the global minimum of the total free energy corresponds to the ’detached’ liquid surface, or bulk liquid adsorption (bold line in Fig. 7). It is tempting to try to calculate the macroscopic contact angle, , along the coexistence line for . In that range, the fraction of the sample is covered with liquid, while the remaining fraction, , still exposes the uncovered rough substrate. The liquid surface energy of the areas covered with liquid is , where is the total liquid surface area over . With the help of (46) we readily obtain Unfortunately, there is no straightforward way to calculate . We thus content ourselves here with an upper bound for , which is obtained by setting in the above expression. Qualitatively, we can nevertheless conclude that since the jump at the prewetting line will directly enter in the lower boundary of the integral, it is clear that this jump will as well be visible in . This is in contrast to, e.g., first order wetting, where a jump in film thickness is accompanied by a continuous variation in the contact angle (16). We mention again that may be subject to significant contact angle hysteresis (2) unless long equilibration times are taken into account. iv.2 Adsorption isotherms It is finally instructive to discuss the qualitative shape of the adsorption isotherms in this scenario, which are sketched in Fig. 8. The curves, which are meant to correspond to different values of , follow what one would expect for the characteristic shown in Fig. 3. The jump from zero film thickness to a finite value is generic and occurs for all contact angles. As is increased, the height of the jump increases slightly, following the curvature of the maximum of . At the same time, the maximum film thickness (reached at coexistence) decreases, until it finally comes below the percolation threshold when . When , has no zero anymore, and the substrate remains dry up to coexistence. A few more words concerning the shape of the function are in order. If were Gaussian, would be just of the form . In that case, would in Fig. 4 be represented by a horizontal line above the abscissa. The solid curve in Fig. 5 would then be a Gaussian, and the adsorption isotherms would of course be the same as in Fig. 2. However, the fact that any real roughness is bounded, as there are neither infinitely high spikes nor infinitely deep troughs, entails the boundedness of the codomain of , in contrast to the infinite codomain of . As an immediate consequence, the derivative of must finally diverge at the boundaries of its (finite!) support, which necessarily leads to bending down onto the dashed line below the abscissa in Fig. 4. This leads not only naturally to a finite at coexistence (, cf. Fig. 5), but also to the Wenzel prewetting jump in farther away from coexistence, when the dashed curve in Fig. 5 just touches the solid curve. This reveals that the shape of the adsorption isotherms we derived for Gaussian roughness above is qualitatively different from what should be expected for real roughness. In fact, it misses the whole prewetting scenario, which turned out above to be a generic feature. Once again, Gaussian roughness reveals itself as a special case, which is mathematically convenient but may be misleading when it comes to making quantitative predictions. In conclusion, an analytic theory was presented which allows to calculate the wetting phase diagram, adsorption isotherms, and percolation threshold of the adsorbed liquid film for isotropic, randomly rough substrates with arbitrary lateral correlation function and height distribution. The results are found to depend only upon a few key parameters, which can be clearly identified and derived from experimental sample profile data. We have seen that wetting ’physical’ roughness displays a number of features which are not present for exactly Gaussian roughness, such as a prewetting transition occurring well before the Wenzel angle is reached. This could be traced down to subtle properties of Gaussian random functions, which reveal their unphysical nature only at second glance. Since for most quantities of interest we could come up with closed analytic expressions, these results may be particularly useful for practical applications. The range of validity of the present theory extends from a few nanomeres up to roughly a millimeter, well below the capillary length of the liquid. The field of such applications is vast, including almost all situations in which a liquid comes into contact with a naturally rough surface. In particular, ramifications of wetting phase transitions, which inherently involve small contact angles, are to be expected and can now be accounted for in closed form. Given the potential relevance of the results presented here, it will be worthwhile to work on relaxing the five major approximations we have used: We have assumed the substrate to be chemically homogeneous. We have assumed the curvature of the roughness characteristic to be small; eq. (38). We have assumed isotropy of the roughness; eq. (11). The last two are probably the simplest to tackle, while the first one appears as the most difficult to overcome. It should finally be noted that in experiments, one has to reckon with quite long relaxation times for the measured quantities, because at all levels of the roughness there are saddle points (21); (23), which act as effective pinning sites for . Equilibration will nevertheless proceed within manageable time, either via the vapour phase or via the molecularly thin adsorbed film (9). Inspiring discussions with Daniel Tartakovsky, Siegfried Dietrich, Martin Brinkmann, Jürgen Vollmer, Sabine Klapp, and Daniela Fliegner are gratefully acknowledged. The author furthermore acknowledges generous support form BP International. - J. R. Philip, J. Phys. Chem. 82 (1978) 1379. - J. F. Joanny, P. G. DeGennes, J. Chem. Phys. 81 (1984) 552. - D. Andelman, J.-F. Joanny, M. O. Robbins, Europhys. Lett. 7 (1988) 731. - P. Pfeifer, Y. J. Wu, M. W. Cole, J. Krim, Phys. Rev. Lett. 62 (1989) 1997. - M. Kardar and J. O. Indekeu, Europhys. Lett. 12 (1990) 161. - G. Palasantzas, J. Krim, Phys. Rev. B 48 (1993) 2873. - R. Netz, D. Andelman, Phys. Rev. E 55 (1997) 687. - T. S. Chow, J. Phys.: Cond. Mat. 10 (1998) L445. - R. Seemann, W. Mönch, S. Herminghaus, Europhys. Lett. 55 (2001) 698. - S. Herminghaus, Europhys. Lett. 52 (2000) 165. - J. Bico, C. Tordeux, D. Quéré, Europhys. Lett. 55 (2001) 214. - J. Bico, U. Thiele, D. Quéré, Coll. Surf. A 206 (2002) 41. - C. Ishino, K. Okumura, D. Quéré, Europhys. Lett. 68 (2004) 419. - S. Herminghaus, M. Brinkmann, R. Seemann, Annu. Rev. Mater. Res. 38 (2008) 101. - Z. He et al., Soft Matter 7 (2011) 6435. - S. Dietrich, in Phase Transitions and Critical Phenomena 12, C. Domb & J. L. Lebowitz, edtrs. (Acad. Press, London1988). - R. N. Wenzel, Ind. and Eng. Chem. 28 (1936) 988. - T. Frankel, The Geometry of Physics: an Introduction (Cambridge Univ. press, Cambridge 1997). - R. Seemann et al., P. N. A. S. 6 (2005) 1848. - T. Pompe, S. Herminghaus, Phys. Rev. Lett. 85 (2000) 1930. - M. S. Longuet-Higgins, Phil. Trans. Roy. Soc. A 249 (1957) 321. - M. S. Longuet-Higgins, Phil. Trans. Roy. Soc. A 250 (1957) 157. - P. R. Nayak, Wear 26 (1973) 305. - J. A. Greenwood, Proc. Roy. Soc. 393 (1984) 133. - J. A. Ogilvy and J. R. Foster, J. Phys. D 22 (1989) 1243. - M. B. Isichenko, Rev. Mod. Phys. 64 (1992) 961. - R. T. Adler and D. Firman, Phil. Trans. Roy. Soc. A 303 (1981) 433. - J. I. McCool, Int. J. Mach. Tools Manufact. 32 (1992) 115. - J. J. Wu, Tribology Int. 37 (2004) 339. - M. A. Rodriguez-Valverde, P. J. Ramon-Torregosa, M. A. Cabrerizo-Vilchez, Microscopy: Science, Technology, Application and Education, A. Mendez-Vilaz and J. Diaz, edts. (2010). - V. Bakolas, Wear 254 (2003) 546.

We assume that the cross flow is uniform and that forced convection is the primary mode of heat transfer the rod is heated internally by an electric resistance heater 1pre-lab section: theoretical analysis consider a heated cylinder subjected to a cross flow of air. M bahrami ensc 388 (f09) forced convection heat transfer 1 forced convection heat transfer convection is the mechanism of heat transfer through a fluid in the whereas in forced convection, the fluid is forced to flow over a surface or in a tube by external means such as a pump or fan. Estimate the flow rates of water required to attain the following reynold numbers for flow through the heat exchanger tube 100, 1000, 2000, 5000, 10000, 25000, 30000, 50000 page 6 of 15 experiment 8 – free & forced convection convection heat transfer doc 8 6 procedure caution for safety: 1. 90 numerical simulation of flow and forced convection heat transfer (7) (8) where vξ and vη are the velocity components in ξ and η directions, respectively, p is the pressure, t is the time, t. – a is cross-sectional area for flow – p is wetted perimeter – for a circular pipe where a = pd2/4 and p heat and mass transfer forced free (natural) 24 flow direction • flow direction depends on 08-forced convection twoppt. Force convection is a mechanism of heat transfer in which fluid motion is generated by an external source like a pump, fan, suction device, etc forced convection is often encountered by engineers designing or analyzing pipe flow, flow over a plate, heat exchanger and so on. Consider a heated cylinder subjected to a cross flow of air, as shown in fig 1 we assume that the cross flow is uniform and that forced convection is the primary mode of heat transfer. Forced and natural convection page 3 flow separation renders the analytical modelling of momentum, heat, and mass flows over bodies intractable except in very slender cases (blades and foils), and, although the numerical simulation using. Coupled conduction and convection where a (= wt) is the cross-sectional area of the fin, thus enabling the energy balance on the element to be written as: (3) the plate may be one of many making up an idealized plate heat exchanger, and the flow centerlines may be considered as axes of symmetry across which no energy is transferred. The aim of this lab is to determine the average convective heat transfer coefficient for forced convection of a fluid (air) past a copper tube, which is used as a heat transfer model. In this study, forced convection flow and heat transfer characteristics in semi-circular cross-sectioned micro-channel were studied numerically water, ethylene glycol (eg) and engine oil were used as working fluid. Convection heat transfer reading problems external flow: the flow engulfs the body with which it interacts thermally internal flow: the heat transfer surface surrounds and guides the convective stream forced convection: flow is induced by an external source such as a pump, compressor, fan, etc. The following table charts of typical convective convection heat transfer coefficients for fluids and specific applications. The enhancement of heat transfer characteristics for cross flow heat exchanger with using low integral finned tube has been experimentally studied in this paper. Forced convection heat transfer to incompressible power-law fluids from a heated elliptical cylinder in the steady, laminar cross-flow regime has been studied numerically. State heat transfer laboratory were to study the rates of heat transfer for different materials of varying sizes, to develop an understanding of the concepts of forced and free convection and to determine the heat transfer coefficients for several rods. Mayank vishwakarma et al (2013) an attempt made to decrease the pressure drop and to increase the heat transfer and the ratio of heat transfer and pressure drop in shell and tube type heat exchanger by tilting the baffle angle up to which we get the minimum pressure drop. Laminar fluid flow and heat transfer in an annulus with an externally enhanced inner tube ajay k agrawal department of mechanical engineering, clemson university, clemson, sc, usa university of michigan, dearborn, mi, usa laminar forced convection in a double-pipe heat exchanger is studied numerically in this study, an isothermal tube. 18 5 heat exchangers the general function of a heat exchanger is to transfer heat from one fluid to another the basic component of a heat exchanger can be viewed as a tube with one fluid running through it and another fluid flowing by on the outside. For forced water convection experiment, the testing facility (schematic shown in fig 3b) of prof px jiang j tian et al/international journal of heat and mass transfer 50 (2007) 2521–2536 2523. Convection heat transfer from tube banks in crossflow: analytical approach m m yovanovich, fellow aiaa w a khan j r culham microelectronics heat transfer laboratory department of mechanical engineering forced convection 2 steady, laminar, fully developed and 2-d flow 3 incompressible fluid with constant properties. In forced convection, the fluid is forced to flow by an external factor - eg wind in the atmosphere, a fan blowing air, water being pumped through a pipe typically heat transfer under forced convection conditions is higher than natural convection for the same fluid. Forced convection (in a cross flow heat exchanger) the aim of this lab is to determine the average convective heat transfer coefficient for forced convection of a fluid (air) past a copper tube, which is used as a heat transfer model. The heat transfer coefficient for convection is denoted by (h) and is measured in w/m^2k, this lab delves into the application of convection heat transfer and how it correlates to temperature, velocity, ect of the fluid in question.

Presentation on theme: "AXONOMETRIC PROJECTION"— Presentation transcript: 1 AXONOMETRIC PROJECTION C H A P T E R F I F T E E N 2 OBJECTIVES1. Sketch examples of an isometric cube, a dimetric cube, and atrimetric cube.2. Create an isometric drawing given a multiview drawing.3. Use the isometric axes to locate drawing points.4. Draw inclined and oblique surfaces in isometric.5. Use projection to create an axonometric drawing.6. Use offset measurements to show complex shapes in obliquedrawings.7. Add dimensions to oblique drawings.8. Describe why CAD software does not automatically createoblique drawings. 3 Axonometric projections Axonometric projections show all three principal dimensions using a singledrawing view, approximately as they appear to an observer. Pictorial drawings are also useful in developing design concepts. They can help you picture the relationships between design elements and quickly generate several solutions to a design problem.Axonometric projection(isometric shown)These projections are often called pictorial drawings because they look more like a picture than multiview drawings do. Because a pictorial drawing shows only the appearance of an object, it is not usually suitable for completely describing and dimensioning complex or detailed forms. 4 Projection Methods Reviewed The four principal types of projection 5 Types of Axonometric Projection Isometric projectionDimetric projectionThe degree of foreshortening of any line depends on its angle to the plane of projection. The greater the angle, the greaterthe foreshortening. If the degree of foreshortening is determined for each of the three edges of the cube that meet at one corner, scales can be easily constructed for measuring along these edges or any other edges parallel to themTrimetric projection 6 DIMETRIC PROJECTIONA dimetric projection is an axonometric projection of an object in which two of its axes make equal angles with the plane of projection, and the third axis makes either a smaller or a greater angle. 7 APPROXIMATE DIMETRIC DRAWING Approximate dimetric drawings, which closely resemble true dimetrics, can be constructed by substituting for the true angles.The resulting drawings will be accurate enough for all practical purposes. 8 TRIMETRIC PROJECTIONA trimetric projection is an axonometric projection of an object oriented so that no two axes make equal angles with the plane of projection.Because the three axes are foreshortened differently, each axiswill use measurement proportions different from the other two.Ellipses in Trimetric. (Method (b) courtesy of Professor H. E. Grant.) 9 TRIMETRIC ELLIPSESThe trimetric centerlines of a hole, or the end of a cylinder, become the conjugate diameters of an ellipse when drawn in trimetric.In constructions where the enclosing parallelogram for an ellipse is available or easily constructed, the major and minoraxes can be determined(Method (b) courtesy ofProfessor H. E. Grant.)(b)When you are creating a trimetric sketch of an ellipse, it works great to block in the trimetric rectangle that would enclose the ellipse and sketch the ellipse tangent to the midpoints of the rectangle. 10 AXONOMETRIC PROJECTION USING INTERSECTIONS Note that if the three orthographic projections, or in most cases any two of them, are given in their relative positions, the directions of the projections could be reversed so that the intersections of the projecting lines would determine the axonometric projection needed. 11 Use of an Enclosing Box to Create an Isometric Sketch using Intersections To draw an axonometric projection using intersections, it helps to make a sketch of the desired general appearance of the projection.Even for complex objects thesketch need not be complete, just an enclosing box. 12 COMPUTER GRAPHICSPictorial drawings of all sorts can be created using 3D CAD.The advantage of 3D CADis that once you makea 3D model of a partor assembly, you canchange the viewing directionat any time for orthographic,isometric, or perspective views.You can also apply different materials to the drawing objects and shade them to produce a high degree of realism in the pictorial view.(Courtesy ofRobert Kincaid.)(Courtesy of PTC) 13 OBLIQUE PROJECTIONSIn oblique projections, the projectors are parallel to each other but are not perpendicular to the plane of projection. 14 Directions of Projectors The directions of the projections BO, CO, DO, and so on, are independent of the angles the projectors make with the plane of projection. 15 ELLIPSES FOR OBLIQUE DRAWINGS It is not always possible to orient the view of an object so that all its rounded shapes are parallel to the plane of projection.Both cannot be simultaneously placed parallel to the plane of projection, so in the oblique projection, one of them must be viewed as an ellipse. 16 Alternative Four-Center Ellipses Normal four-center ellipses can be made only in equilateral parallelogram, so they cannot be used in an oblique drawing where the receding axis is foreshortened. Instead, use this alternative four-center ellipse to approximate ellipses in oblique drawings.Draw the ellipse on two centerlines. This is the same method as is sometimes used in isometric drawings, but in oblique drawings it appears slightly different according to the different angles of the receding lines… 17 OFFSET MEASUREMENTSCircles, circular arcs, and other curved or irregular lines can be drawn using offset measurements.Draw the offsets on the multiview drawing of the curve and then transfer them to the oblique drawing… 18 OBLIQUE DIMENSIONINGYou can dimension oblique drawings in a way similar to that used for isometric drawings.For the preferred unidirectional system of dimensioning, all dimension figures are horizontal and read from the bottom of the drawing. Use vertical lettering for all pictorial dimensioning. 19 COMPUTER GRAPHICSUsing CAD you can easily create oblique drawings by using a snap increment and drawing in much the same way as on grid paper. If necessary, adjust for the desired amount of foreshortening along the receding axis as well as the preferred direction of the axis.(Autodesk screen shots reprinted with the permission of Autodesk, Inc.) 20 PERSPECTIVE DRAWINGSC H A P T E R S I X T E E N 21 OBJECTIVES 1. Identify a drawing created using perspective projection. 2. List the differences between perspective projection andaxonometric projection.3. Create a drawing using multiview perspective.4. Describe three types of perspective5. Measure distances in perspective projection. 22 UNDERSTANDING PERSPECTIVES A perspective drawing involves four main elements:• The observer’s eye• The object being viewed• The plane of projection• The projectors from the observer’s eye to all points on the object 23 Rules to Learn for Perspective The following are some rules to learn for perspective:• All parallel lines that are not parallel to the picture plane vanish at a point.• If these lines are parallel to the ground, the vanishing point will beon the horizon.• Lines that are parallel to the picture plane, such as the vertical axis ofeach lamppost, remain parallel to one another and do not converge towarda vanishing point. 24 PERSPECTIVE FROM A MULTIVIEW PROJECTION It is possible to draw a perspective from a multiview projection,The upper portion of thedrawing shows the top view of the station point, the picture plane, the object, and the visual rays. The right-side viewshows the same station point, picture plane, object, and visual rays.In the front view, the picture plane coincides with the plane of the paper, and the perspective view is drawn on it. 25 NONROTATED SIDE VIEW METHOD FOR PERSPECTIVE The upper portion ofthe drawing shows the top views of the station point, picture plane, and the object. The lines SP-1, SP-2, SP-3, and SP-4are the top views of the visual rays.The perspective view is drawn on the picture plane where the front view would usually be located.The perspective view shows the intersectionof the ground plane with the picture plane. 26 POSITION OF THE STATION POINT The centerline of the cone of visual rays should be directed toward the approximatecenter, or center of interest, of the object.In two-point perspective,locating the station point (SP) in the plan view slightly to the left and not directly in front of the center of the object produces a better view, as if the object is seen at aglance without turning the head.The station point (SP) does not appear in the perspective view because the station point is in front of the picture plane. 27 LOCATION OF THE PICTURE PLANE The perspectives differ in size but not in proportion. The farther the plane is from the object, the smaller the perspective drawing will be. This distance controls the scale of the perspective. 28 BIRD’S-EYE VIEW OR WORM’S-EYE VIEW The horizon is level with the observer’s eye, so controlling the location for the horizon line controls whether the perspective view appears from above or below the object. The horizon line is defined by the observer’s point of view.To produce a perspective view that shows the objects as though viewed from above, place the object below the horizonline. To produce a perspective viewthat shows the object as thoughviewed from below, place the objectabove the horizon line. 29 ONE-POINT PERSPECTIVE In one-point perspective, orient the object so two sets of itsprincipal edges are parallel to the picture plane (essentially aflat surface parallel to the picture plane) and the third set is perpendicular to the picture plane. This third set of parallellines will converge toward a single vanishing point inperspective. 30 ONE-POINT PERSPECTIVE OF A CYLINDRICAL SHAPE A one point perspective representing a cylindrical machine part.The front surface of the cylinder isplaced in the picture plane. All circular shapes are parallel to the picture plane, and they project as circles and circular arcs in the perspective. The station point (SP) is located in front and to one side of the object. The horizon is placed above the ground line. The single vanishing point is on the horizon in the center of vision. 31 TWO-POINT PERSPECTIVE In two-point perspective, the object is oriented so that one set of parallel edges is vertical and has no vanishing point, whereas the two other sets have vanishing points. Two-point perspectives are often used to show buildings in an architectural drawing, or large structures in civil engineering, such as dams or bridges, especially for client presentation drawings.When multiview drawings are already available, tape their top (plan) and side (elevation) views in position, and usethem to construct the perspective. When you are finished, remove the taped portions. 32 THREE-POINT PERSPECTIVE In three point perspective, the object is placed so that none of its principal faces or edges are parallel to the picture plane. This means that each set of three parallel edges will have a separate vanishing point. The picture plane is approximately perpendicular to the centerlineof the cone of visual rays.Remember that to find the vanishingpoint of a line in any type of perspectiveyou draw a visual ray, or line, from thestation point parallel to that edge of theobject, then find the piercing point of thisray in the picture plane. 33 DIRECT MEASUREMENTS ALONG INCLINED LINES The method of direct measurements may also be applied to lines inclined to the picture plane (PP) and to the ground plane.Line XE, which pierces the picture plane (PP) at X. If you revolve the end of the house about a vertical axis XO into the picture plane (PP), line XE will be shown true length and tipped as shown at XY. This line XY may be used as themeasuring line for XE. Next find the corresponding measuring point MP. The line YE is the horizontal base of an isosceles triangle having its vertex at X, and a line drawn parallel to it through SP will determine MP 34 VANISHING POINTS OF INCLINED LINES To find the vanishing point of an inclined line, determine thepiercing point in the picture plane (PP) of a line drawn from the station point (SP) parallel to the given line. 35 CURVES AND CIRCLES IN PERSPECTIVE If a circle is parallel to the picture plane, its perspective is a circle. If the circle is inclined to the picture plane, its perspective drawing may be any one of the conic sections where the base of the cone is the givencircle, the vertex is the station point (SP),and the cutting plane is the picture plane (PP).The centerline of the cone of rays is usually approximately perpendicular to the picture plane, so the perspective will usually be an ellipse.A convenient method for determining the perspective of any planar curve… 36 THE PERSPECTIVE PLAN METHOD You can draw a perspective by first drawing the perspective ofthe plan of the object, then adding the vertical lines, and finally adding the connecting lines. 37 SHADINGShading pictorial drawings can be very effective in describing the shapes of objects in display drawings, patent drawings, and other pictorial drawings. Ordinary working drawings are not shaded.Methods of Shading 38 PERSPECTIVE VIEWS IN AUTOCAD AutoCAD software uses an interactive command calledDview (dynamic viewing) that you can use to show 3Dmodels and drawings in perspective. The Dview command uses a camera and target to create parallel and perspective views. You can use the camera option to select a new cameraposition with respect to the target point at which thecamera is aimed. The Dview distance option is used to create a perspective viewA Perspective View Created Using the Dview Command in AutoCAD. (Autodesk screen shots reprinted with the permission of Autodesk, Inc.)

Analysis of Sequential Circuits The behaviour of a sequential circuit is determined from the inputs, the outputs and the states of its flip-flops. Both the output and the next state are a function of the inputs and the present Derive the state table and state diagram for the sequential circuit shown in Figure 7. STEP 1: First we derive the Boolean expressions for the inputs of each flip-flops in the schematic, in terms of external input Cnt and the flip-flop outputs Q1 and Q0. Since there are two D flip-flops in this example, we derive two expressions for D1 and D0: These Boolean expressions are called excitation equations since they represent the inputs to the flip-flops of the sequential circuit in the next clock cycle. STEP 2: Derive the next-state equations by converting these excitation equations into flip-flop characteristic equations. In the case of D flip-flops, Q(next) = D. Therefore the next state equal the excitation equations. Q0(next) = D0 = Cnt'*Q0 Q1(next) = D1 = Cnt'*Q1 + Cnt*Q1'*Q0 + Cnt*Q1*Q0' STEP 3: Now convert these next-state equations into tabular form called the next-state table. Each row is corresponding to a state of the sequential circuit and each column represents one set of input values. Since we have two flip-flops, the number of possible states is four - that is, Q1Q0 can be equal to 00, 01, 10, or 11. These are present states as shown in the For the next state part of the table, each entry defines the value of the sequential circuit in the next clock cycle after the rising edge of the Clk. Since this value depends on the present state and the value of the input signals, the next state table will contain one column for each assignment of binary values to the input signals. In this example, since there is only one input signal, Cnt, the next-state table shown has only two columns, corresponding to Cnt = 0 and Cnt = 1. Note that each entry in the next-state table indicates the values of the flip-flops in the next state if their value in the present state is in the row header and the input values in the column header. Each of these next-state values has been computed from the next-state equations in STEP 2. STEP 4: The state diagram is generated directly from the next-state table, shown in Figure 8. Each arc is labelled with the values of the input signals that cause the transition from the present state (the source of the arc) to the next state (the destination of the arc). In general, the number of states in a next-state table or a state diagram will equal 2m , where m is the number of flip-flops. Similarly, the number of arcs will equal 2m x 2k , where k is the number of binary input signals. Therefore, in the state diagram, there must be four states and eight transitions. Following these transition arcs, we can see that as long as Cnt = 1, the sequential circuit goes through the states in the following sequence: 0, 1, 2, 3, 0, 1, 2, .... On the other hand, when Cnt = 0, the circuit stays in its present state until Cnt changes to 1, at which the Since this sequence is characteristic of modulo-4 counting, we can conclude that the sequential circuit in Figure 7 is a modulo-4 counter with one control signal, Cnt, which enables counting when Cnt = 1 and disables it when Cnt = 0. To see how the states changes corresponding to the input signals Cnt, click on this image. Below, we show a timing diagram, representing four clock cycles, which enables us to observe the behaviour of the counter in In this timing diagram we have assumed that Cnt is asserted in clock cycle 0 at t0 and is disasserted in clock cycle 3 at time t4. We have also assumed that the counter is in state Q1Q0 = 00 in the clock cycle 0. Note that on the clock's rising edge, at t1, the counter will go to state Q1Q0 = 01 with a slight propagation delay; in cycle 2, after t2, to Q1Q0 = 10; and in cycle 3, after t3 to Q1Q0 = 11. Since Cnt becomes 0 at t4, we know that the counter will stay in state Q1Q0 = 11 in the next clock cycle. To see the timing behaviour of the circuit click on this image In Example 1.1 we demonstrated the analysis of a sequential circuit that has no outputs by developing a next-state table and state diagram which describes only the states and the transitions from one state to the next. In the next example we complicate our analysis by adding output signals, which means that we have to upgrade the next-state table and the state diagram to identify the value of output signals in each state. The input combinational logic in Figure 10 is the same as in Example 1.1, so the excitation and the next-state equations will be the same as in Example 1.1. In addition, however, we have computed the Output equation: Y = Q1Q0 As this equation shows, the output Y will equal to 1 when the counter is in state Q1Q0 = 11, and it will stay 1 as long as the counter stays in that state. Next-state and output table: Note that the counter will reach the state Q1Q0 = 11 only in the third clock cycle, so the output Y will equal 1 after Q0 changes to 1. Since counting is disabled in the third clock cycle, the counter will stay in the state Q1Q0 = 11 and Y will stay asserted in all succeeding clock cycles until counting is enabled again. Design of Sequential Circuits The design of a synchronous sequential circuit starts from a set of specifications and culminates in a logic diagram or a list of Boolean functions from which a logic diagram can be obtained. In contrast to a combinational logic, which is fully specified by a truth table, a sequential circuit requires a state table for its specification. The first step in the design of sequential circuits is to obtain a state table or an equivalence representation, such as a state diagram. A synchronous sequential circuit is made up of flip-flops and combinational gates. The design of the circuit consists of choosing the flip-flops and then finding the combinational structure which, together with the flip-flops, produces a circuit that fulfils the required specifications. The number of flip-flops is determined from the number of states needed in the The recommended steps for the design of sequential circuits are set out below.

Paramagnetic filaments in a fast precessing field: Planar versus helical conformations We examine analytically equilibrium conformations of elastic chains of paramagnetic beads in the presence of a precessing magnetic field. Conformations of these filaments are determined by minimizing their total energy, given in the harmonic approximation by the sum of the bending energy, quadratic in its curvature, and the magnetic dipolar interaction energy, quadratic in the projection of the vector tangent to the filament onto the precession axis. In particular, we analyze two families of open filaments with their ends aligned along the precession axis and described by segments of planar curves and helices. These configurations are characterized in terms of two parameters encoding their features such as their length, separation between their ends, as well as their bending and magnetic moduli, the latter being proportional to the magnitude and precession angle of the magnetic field. Based on energetic arguments, we present the set of parameter values for which each of these families of curves is probable to occur. pacs:87.16.Ka, 75.75.-c, 87.15.hp Flexible magnetic filaments can be synthesized by joining ferromagnetic or superparamagnetic beads with elastic linkers Wang et al. (2011); Tierno (2014). The combination of their elastic and magnetic properties gives rise to interesting phenomena, which have been a subject of active research since their introduction more than a decade ago Biswal and Gast (2003); Goubault et al. (2003). The mechanical properties of magnetic filaments have been extensively characterized. Their Young and bending moduli have been measured directly in bending and compression experiments performed with optical traps Biswal and Gast (2003) and indirectly from measurements of other quantities such as the separation between beads using Bragg diffraction Goubault et al. (2003), or more recently, via their thermal fluctuations Gerbal and Wang (2017). It has been found that the sole magnetic field can drive an Euler buckling instability in a free filament Cēbers (2005a), whose critical value has been determined theoretically and measured experimentally with a good agreement Cēbers and Javaitis (2004a); Gerbal et al. (2015). This kind of filaments exhibit diverse morphological features Cēbers and Cīrulis (2007), for instance, depending on their length, bending rigidity and magnetic field strength, they may adopt and shapes Goubault et al. (2003); Cēbers (2003) or configurations with more undulations Huang et al. (2016). Configurations of anchored superparamagnetic filaments with a free end may fold into loops, sheets and pillars for different combinations of their bending rigidity and the strength of the magnetic field Wei et al. (2016). In a precessing magnetic field, filaments with loads at their ends can adopt planar or helical configurations by changing the precessing angle and with a time-dependent precession free filaments can assemble into gels of diverse conformations Dempster et al. (2017). Due to their magnetic features, they have inspired the development of several applications. They can be used as micro-mechanical sensors used in the determination of force-extension laws at the micro-scale Biswal and Gast (2003); Goubault et al. (2003); Koenig et al. (2005). Furthermore, since they possess the interesting feature that their stiffness is tunable Cēbers (2005b), and conformational changes can be controlled through the magnetic field Cēbers and Cīrulis (2007); Huang et al. (2016) or the temperature Cerda et al. (2016), they can also be used as actuators Dempster et al. (2017) or grabbers Martinez-Pedrero et al. (2016). Their dynamics have also been studied in detail. In a magnetic field rotating on a plane, free filaments rotate rigidly synchronously or asynchronously depending on their length and whether the precession frequency is smaller or bigger than a critical value Biswal and Gast (2004); Cēbers and Javaitis (2004b). In the presence of an alternating magnetic field, magnetic filaments oscillate and displace, so they have been used to design self-propelled swimmers of controllable velocity and displacement direction Dreyfus et al. (2005); Belovs and Cēbers (2006); Roper et al. (2008); Belovs and Cēbers (2013). Further information about the properties and applications of these magnetic filaments can be found in the reviews Cēbers (2005b); Cēbers and Erglis (2016). In general, most of the previous works consider magnetic filaments with one or both free ends and in precessing fields at a fixed angle, typically precessing on a plane. In this paper we examine open superparamagnetic filaments in a magnetic field precessing at different constant angles, and with their ends held along the precession axis at a certain distance. Molecular dynamics simulations suggest that under these conditions filaments exhibit different behavior depending on the value of the precessing angle relative to a critical angle: if it is smaller, filaments bend but remain on a plane, whereas if it is larger, filaments explore the ambient space adopting helical structures Dempster et al. (2017). Here we present a detailed analytic description of these two families of filaments. We determine their equilibrium configurations by minimizing their ascribed total energy, which at quadratic order has two contributions, the bending energy and the magnetic energy due to dipolar interactions between the beads Cēbers (2003); Goubault et al. (2003). In principle, the behavior of these magnetic filaments depends on their intrinsic properties: length and bending modulus; as well as of their extrinsic properties: separation between their ends and magnetic modulus, which depends on the magnetic field parameters (magnitude and precession angle) and can be positive, negative or even vanish. However, it is possible to characterize their equilibrium configurations in terms of just two parameters capturing all of their characteristics: the boundary separation and the ratio of the magnetic to bending moduli, both scaled with powers of the total length of the filament so as to adimensionalize them. We discuss the forces required to hold the boundaries of the filaments and the behavior of their total energy as a function of these two parameters. Although both families are critical points of the total energy regardless of the precession angle, by comparing their total energies we investigate their plausibility in each precession regime, determining the parameter values for which each family is more likely to take place. This paper is organized as follows. We begin in Sec. II with the framework that we employ to describe superparamagnetic filaments, to this end we define their energy and we express the corresponding Euler-Lagrange (EL) equations, that their equilibrium configurations must satisfy, in terms of the stresses on the filaments. In Sect. III we specialize this framework to the case of planar curves, which in Sec. IV is applied to examine the family of vertical planar filaments (their ends are fixed and aligned with the precession axis) in the perturbative and non-linear regimes. In Sec. V we do the respective analysis of the family of helices. In Sec. VI we compare the total energy of both families of filaments with same parameters to assess their possible physical realization. We close with our conclusions and discussion of future work in Sec. VII. Some derivations and calculations used or discussed in the main text are presented in the appendices. Ii Energy and stresses The magnetic filament is described by the curve , parametrized by arc length in three-dimensional Euclidean space and passing trough the center of the beads. Geometric quantities of the curve are expressed in terms of the Frenet-Serret (FS) frame adapted to the curve, denoted by , see Fig. 1. The rotation of the FS frame along the curve is given by the FS formula where is the Darboux vector; and are the FS curvature and torsion, quantifying how the curve bends in the osculating and normal planes, respectively Kreyszig (1991). We consider paramagnetic filaments in the presence of a magnetic field precessing at an angle about a direction we choose as the axis, see Fig. 2(a). The total energy density of the paramagnetic filament is the sum of the bending and magnetic energies ,111We disregard the weight of the filament so we do not include gravitational effects; nor we include the magnetic dipole induced by the neighbors, whose influence would rescale the magnetic modulus Zhang and Widom (1995); Cēbers (2003) with quadratic in the curvature Kratky and Porod (1949); Kamien (2002), and given by the time-averaged dipolar interactions between nearest neighbors induced by the magnetic field (a derivation of is presented in Appendix A) Goubault et al. (2003); Cēbers (2003), where is the projection of the tangent vector onto the precession axis; is the bending modulus (with units of force times squared length); is the magnetic modulus (with units of force) defined by with the vacuum permeability, the magnitude of the magnetic dipoles, the separation between their centers and the precession angle. As shown below, configurations of the filaments depend sensitively on the sign of , determined in turn by . vanishes at the so-called “magic” angle , so in this case the leading order of the magnetic energy will be the quadrupolar term, which is of short range so filaments behave mostly as elastic curves Osterman et al. (2009). In the regime , which will be termed as regime , see Fig. 2(b), the magnetic modulus is positive, (the magnetic dipolar interactions are attractive), and from Eq. (2) we see that in order to minimize the filaments will tend to align with the precession axis to maximize Martin et al. (2000). By contrast in the regime , termed as regime , the magnetic modulus becomes negative, (the magnetic dipolar interactions are repulsive), and is minimized when vanishes, so the filaments will tend to lie on the plane orthogonal to the precession axis Martin et al. (2000). To reduce the material parameters space we rescale all quantities by the bending modulus and denote the rescaled quantity by an overbar. In particular, the rescaled quantity possesses units of inverse squared length, so the inverse of its square root, , provides the characteristic length scale at which buckling occurs. The dimensionless parameter is known as the magnetoelastic parameter. This parameter quantifies the ratio of magnetic to bending energies: bending and magnetic energy scale as and , so their ratio scale as . Below, we use as a parameter to characterize the conformations of the filaments. Typical experimental values of these paramagnetic filaments222For beads with diameter , magnetic susceptibility in a magnetic field , the magnitude of the induced dipole is . are , , , so , and , Biswal and Gast (2003); Goubault et al. (2003); Biswal and Gast (2004). The total bending and magnetic energies of the filament are given by the line integrals of the corresponding energy densities Thus the total energy is , but since the filament is inextensible, we consider the effective energy where is a Lagrange multiplier fixing total length which acts as an intrinsic line tension. The change of the energy under a deformation of the curve is given by Capovilla et al. (2002); Guven and Vázquez-Montejo (2012); Guven et al. (2014) In the first term, which represents the response of the energy to a deformation in the bulk, is the force vector, given by the sum of the bending and magnetic forces, , defined by Langer and Singer (1996); Capovilla et al. (2002); Guven and Vázquez-Montejo (2012); Guven et al. (2014); Dempster et al. (2017) where and . is the force exerted by the line element at on the neighboring line element at , so () represents compression (tension). From the force balance at the boundaries follows that is the external force on the filament Langer and Singer (1996). We see in Eq. (7) that the magnitudes of the bending and magnetic forces scale as and . Thus, the magnetoelastic parameter also quantifies the ratio of magnetic to bending forces, , Cēbers (2003). If the filaments are immersed in a medium of viscosity , we have from the balance of bending and viscous forces that the characteristic bending relaxation time is , Powers (2010), whereas the characteristic magnetic relaxation time is Dempster et al. (2017). The ratio of bending to magnetic relaxation times is the magnetoelastic parameter . For a filament of length , bending and magnetic moduli and , in water () we have and . Therefore, in order to be legitimate, the use of the time-averaged magnetic energy density given in Eq. (2) is justified if the precessing period is less than a millisecond, (frequency Dempster et al. (2017)), so that the characteristic relaxation times are much larger, . The second term in Eq. (6) contains quantities arising after integration by parts and is given by the total derivative of so it represents the change of the energy due to boundary deformations. Stationarity of the energy implies that in equilibrium the force vector is conserved along the filament, , a consequence of the translational invariance of the total energy. By contrast, the torque vector, , with , is not conserved: ,333In this expression we have used the identity , Capovilla et al. (2002). while the first term vanishes in equilibrium, the second term, representing a torque per unit length due to the magnetic field, , does not vanish in general. However, the component of the torque along the precession axis, , is conserved on account of the rotational symmetry of the energy about such direction: , which vanishes in equilibrium. Spanning the derivative of in terms of the two normals as ,444The projection onto the tangent vanishes identically due to the reparametrization invariance of the energy Capovilla et al. (2002). so the normal projections of the conservation law provide the Euler-Lagrange (EL) equations satisfied by equilibrium configurations, which read Dempster et al. (2017) In solving these equations, the Lagrange multiplier is determined from boundary or periodicity conditions. The squared magnitude of the force vector is constant on account of the conservation law of . This constant corresponds to the first Casimir of the Euclidean group and provides a first integral of the EL equations.555EL Eq. (9b) can be written as . Thus, the scalar quantity , corresponding to the second Casimir in the case of Euler Elastica, is not conserved in equilibrium because the magnetic field breaks the rotational invariance of the energy and introduces a source of stresses. Below we analyze solutions of two families of curves satisfying these equations with their ends held along the precession axis: curves lying on a plane passing through the precession axis and helices whose axis is parallel to the precession axis. Iii Planar cuves Let us consider curves on a plane, say -, so the embedding functions are and the tangent vector is . Since the curve lies on a plane, it has vanishing torsion, , and the EL equation associated with deformations along reduces to whereas the EL corresponding to deformations along is satisfied identically, because vanishes on account of the orthogonality of the binormal vector to the plane of the curve, i.e. . The force vector, defined in Eq. (7), lies on the osculating plane of the curve Projecting onto the FS basis we obtain Differentiating Eq. (13a) with respect to and using the FS formula we obtain Eq. (13b), whereas differentiation of Eq. (13b) reproduces the EL Eq. (11). Therefore Eq. (13a) provides a second integral of the EL Eq. (11), which permit us to express the difference between the bending and magnetic energy densities as the sum of the tangential component of the force and the constant . Moreover, this relation can be used to eliminate the curvature in favor of the projections of the tangent vector, for instance, the total energy density can be recast as In terms of the tangent and normal vectors are and , whereas the FS curvature is . Expressing Eq. (13a) in terms of , it reduces to a quadrature for : where we have defined which in the mechanical analogy would represent the corresponding equation of motion, see Appendix B. We consider filaments with their boundaries fixed, but not the tangents. Thus the variation vanishes at the boundaries (which we set at ), i.e. , and from Eq. (8) we have that the stationarity of the energy at the boundaries, , imply the vanishing of the curvature at those points. Therefore the appropriate boundary conditions (BC) for equilibrium configurations is In consequence, the intrinsic torque vanishes at the ends and only the torque coupling position and force contributes. Furthermore, the quadrature implies that the maximum value of the angle, say , occurs at the boundaries, i.e. . Thus at the turning points the “kinetic” energy vanishes and the “potential” energy is equal to the “total energy” Audoly and Pomeau (2010), which determines the Lagrange multiplier in terms of the angle : Once has been determined as a function of , the coordinates are obtained by integrating the tangential components In the next section we apply these results to the case of filaments aligned with the precession axis. Iv Vertical filaments Here we consider a curve resulting from a deformation of a straight filament lying along the precession axis, chosen as the axis, such that the end points remain along this axis (see Fig. 3). In consequence the force is also along the precession axis: and . Thus, the potential reduces to , which has period and left-right symmetry . We set the mid point of the curve at from where arc length is measured, being positive (negative) above (below) the axis, i.e. with . We denote the height of the boundary by so that the height difference is . We characterize the curves by the height difference rescaled with the total length To gain some insight about how the magnetic field modifies the behavior of the filaments, we first solve the quadrature (15) in the regime of small deviations from a vertical straight line. iv.1 Perturbative regime We consider a small perturbation of a straight line with and we expand the constants perturbatively as , .666 and are constants, however, we are interested in determining the corrections as functions of a small parameter, determined below, required by deviations from a straight line. At quadratic order, the quadrature describes a harmonic motion Only for 777If , then and the curve is a straight line. the quadratic potential is positive and bounded solutions are possible, given by If the filament develops half periods,888For instance the filament shown in Fig. 3 completes only one half-period (). the wave number is given by We see that the magnetic field modifies the minimum force required to trigger an Euler buckling instability. Furthermore, unlike the purely elastic case where the force on the filaments is always compressive, the magnetic contribution enables the force to be either tensile or compressive depending on the value of the magnetoelastic parameter relative to the squared number of half-periods: for (precession regime or ) the force is positive, , so the filament is under compression, whereas for (precession regime ) the force becomes negative, , and the filament is under tension. In the particular case with , free filaments with are possible Cēbers (2005a). The coordinates can be obtained by integrating Eq. (20), obtaining Evaluating the second expression at the boundaries we determine the amplitude in terms of the scaled height difference defined in (21): The total energy of the filament is , where is the scaled energy of the original straight vertical state and the second order correction is Since increases linearly with the magnitude of the force it can be either positive or negative. Let us now look at the stability of these states. To lowest order, the differential operator of the second variation of the energy, (derived in Appendix C, Eq. (78)), reads The two trivial zero modes (with vanishing eigenvalues), , with constant, are associated to the translational invariance of the energy: at lowest order they correspond to infinitesimal vertical and horizontal translations respectively. To analyze the eigenmodes of we use the basis , which in order to preserve the periodicity of the original curves should have wave numbers , , so the corresponding eigenvalues for each case are The two non-trivial zero modes with correspond to infinitesimal rotations in the plane, but for finite rotations such eigenmodes will not be zero modes, because the energy is only invariant under rotations about the precession axis (in the - plane). The eigenvalues corresponding to states with are shown in Fig. 4. We see that, like in the purely elastic case, only the eigenvalues of the ground state are all positive, so it is the only stable state in the perturbative regime. Therefore, any excited state would decay recursively to the next intermediate state with the more negative eigenvalue until the ground state is reached, Guven et al. (2012). As we will see below, comparison of the total energy of successive states leads suggests that the state is still the ground state in the non-linear regime. iv.2 Non-linear regime Now, we describe the behavior of the filaments in the non-linear regime, i.e. deformations far from straight configurations under the influence of a magnetic field (for comparison purposes, the case of elastic curves, is reviewed in Appendix D). If (), the quadrature (15) can be recast as Integrating the quadrature twice, we obtain the coordinates in terms of elliptic functions in each regime, (details are provided in Appendix E for the interested reader): where the constants and are defined by , and are the sine, cosine, and delta Jacobi elliptic functions; is the Jacobi amplitude; is the incomplete elliptic integral of the third kind Abramowitz and Stegun (1965); Gradshteyn and Ryzhik (2007). Recall is the buckling characteristic length. Like in the perturbative case, if the filament possesses half periods, the wave number is given by where is the complete elliptic integral of the first kind Abramowitz and Stegun (1965); Gradshteyn and Ryzhik (2007). The last relation permits us to express constants and , defined in Eq. (33), in terms of the modulus and the magnetoelastic parameter . The FS curvature of the filament is given by where is the maximum value of the curvature. The condition (18) of vanishing curvature at the boundaries determines, . Evaluating expressions (32) for at the boundaries and using the identities and , we get the following equation for the scaled boundary separation , defined in Eq. (21) To determine , these equations are solved numerically for given values of , , and .999Alternatively, one could extend the method employed in Ref. Hu et al. (2013) for the purely elastic case, in which case would be expanded as a series in and and the coefficients would be determined from Eq. (36). This completes the determination of all parameters of the curve. States with in regime are plotted for different values of and in Figs. (5) and (6). Corresponding states with in regime are plotted in Figs (7) and (8). In these sequences we choose initial states with different boundary angles , specifically (top rows with ), (middle rows with ) and (bottom rows with ).101010For smaller separations , the boundaries, and also upper and bottom segments of the filaments for strong magnetic fields, get close and in consequence our nearest neighbors approximation is no longer valid. In both regimes we observe that for relative small absolute values of the magnetoelastic parameter, , the filaments behave mainly as elastic curves (shown with dashed black lines in the plots with ), adopting and shapes for and , respectively. As is increased, we observe deviations from elastic behavior depending on the regime: in regime , at , filaments begin to elongate along the precession axis and squeezing inwards along the orthogonal direction, and for a large value, , they form thin vertical hairpins connected by straight segments aligned with the precession axis; in regime the converse behavior is observed, at they begin to stretch outwards and orthogonally to the precession axis, and at they are mostly straightened and with the filament’s horizontal extremum farthest from the precession axis. The magnitude of the forces in these planar curves is given by111111It can be positive or negative depending on the sign of the constant satisfying Eq. (36). is plotted for states in Fig 9. For vertical curves with , the force is linear in the magnetoelastic parameter, , as found in the perturbative analysis. We see that in regime is positive for all values of and , indicating that the filaments are under compression, as is usual for elastic curves bent under compression. By contrast, there are regions in regime where becomes negative in which case filaments are under tension, reflecting the fact that they tend to lie orthogonally to the precession axis. The bending and magnetic energy densities in terms of arc length read The bending energy is positive, increasing with (); the magnetic energy is positive in regime and negative in regime , thus the total energy density , can be positive or negative in regime , but it is strictly positive in regime . is shown with a color scale for states in Figs. (5)-(8). For filaments in regime , we see that initially is concentrated in the extremum and low in the boundaries, but as is increased, the high-energy regions migrates towards the hairpins near the boundaries and low-energy regions move to the extremum where straight segments (minimizing both energies) are developed. In regime , high-energy regions always occur at the extremum where the curvature concentrates, whereas low-energy regions correspond to the straight segments near to the boundaries. Moreover, the former regions become more localized and the latter regions more spread as is increased. In the calculation of the total energy, although integration of is simple, integration of is rather complicated because it involves . However, we can integrate expression (14) for the total energy density, where was replaced in favor of by means of the quadrature, (13a), obtaining the following expressions of the total energy for each case (details are presented in Appendix E): where the constant is defined by The total energy of states are plotted in Figs. 10(a) and 10(b). As found in the perturbative regime, regardless of , straight lines with have scaled total energy . We see that is negative almost everywhere (except in a small fringe of values in the vicinity of ) in regime and it is positive everywhere in regime . Values of and for which are shown with a solid black line, and the energies of elastic curves () are shown with a dashed black line. The total energy of the filaments increases as augments for any value of and . To show this, in Fig. 10(c), we plot the energy difference between states and , , where we see that everywhere on the parameter space , result that can be verified for states with higher , i.e, . Thus is the ground state among planar configurations for all parameter values. However, as we will see below this does not hold in general when non-planar configurations are considered, in particular the energy may be lowered for some values of and if filaments adopt helical configurations, which we examine in the next section. Here we demonstrate that helices are also critical points of the total energy. Recall that a helix is characterized by its radius and pitch , with the pitch angle defined by (see Fig. 11). The helix can be parametrized in cylindrical coordinates by the azimuthal angle as, A helical segment is specified by the total azimuthal angle . We consider helices completing full turns, so and the distance between the end points is . Arc length is proportional to , , so total length is proportional to , . From these relations follows that . Inverting to get and in terms of , , and , we get The FS basis adapted to the helix is whereas the FS curvature and torsion are given by The sign of the torsion determines the chirality of the helix, () corresponds to right (left) handed helices. The degenerate cases and correspond to circles on the plane - and to vertical lines, respectively. For helices, the Darboux vector is along the helical axis . Since and are constant and , the EL Eq. (9a) is satisfied if is constant, taking the value and the EL Eq. (9b) vanishes identically. Hence, helices satisfying Eq. (45) minimize the total energy . Using these expressions for , , and in Eq. (7) for , we find that the scaled force required to hold the helix is linear in the separation of the end points and directed along the helical axis, The magnitude of the force is plotted for states as a function of and in Fig. 12. In this plot the line over () represents the scaled force required in the Euler buckling instability of a straight line, with the elastic term four times larger as compared with the scaled force required in the planar case, . Like the case of planar curves, the force can be tensile or compressive depending on value of the magnetoelastic parameter relative to the total azimuthal angle: if () the magnitude of the axial force is positive (negative), (), and the helix is under compression (tension). For circles with () and configurations with , there is no vertical force, . The difference of the force between successive states and , is independent of and positive for any value of , and it increases with . The torque vector has two components, one introduced by the magnetic field and another one of elastic character The magnitude of the azimuthal torque is linear in the magnetoelastic parameter, so it vanishes for elastic curves with and its direction is reversed when changing from regime to regime . For a given value of it vanishes for circles and lines with , respectively, and is maximum for helices of maximum torsion with . The magnitude of the axial torque is proportional to the total azimuthal angle and increases as the boundary points are approached, so it is maximal for circles and vanishing for vertical filaments. The total scaled energy of the helices is harmonic in the separation of the end points Dempster et al. (2017)

A preparatory course for Fundamentals of Algebra, this course acquaints the student with different number systems through mathematical language consisting of symbols and new concepts dealing with sets. Primary goals are to teach students to deal with variables and polynomials through language and application as well as to solve equations. This course is a preparatory course for math I. Students are introduced to algebra, geometry, and other mathematical topics that are integrated in a format that connects mathematics to students’ lives and the world of work. Math I provides students the opportunity to study concepts of algebra, geometry, functions, number and operations, statistics and modeling throughout the course. These concepts include expressions in the real number system, creating and reasoning with equations and inequalities, interpreting and building simple functions, expressing geometric properties and interpreting categorical and quantitative data. The final exam is the North Carolina End-of-Course Test based on the Common Core Math 1 Standards. This course continues a progression of the standards established in Math I. In addition to these standards, Math II includes: polynomials, congruence and similarity of figures, trigonometry with triangles, modeling with geometry, probability, making inferences and justifying conclusions. Progresses from the standards learned in Math I and Math II. In addition to these standards, Math III extends to include algebraic concepts such as: the complex number system, inverse functions, trigonometric functions and the unit circle. Math III also includes the geometric concepts of conics and circles. Designed for those students who have potential for outstanding performance in mathematics, Math III Honors is an accelerated, expanded, and demanding course. Students will work with real, irrational and imaginary numbers, solving systems of equations, problem solving with logarithms, conic sections and polynomials. NC Math 4 focuses on functions and statistical thinking, continuing the study of algebra, functions, trigonometry and statistical concepts previously experienced in NC Math 1-3. The course is designed to be a capstone to introductory statistical concepts. Additionally, the course intentionally integrates concepts from algebra and functions to demonstrate the close relationship between algebraic reasoning as applied to the characteristics and behaviors of more complex functions. This is a survey course of various topics that will prepare the student for calculus and college-level mathematics courses. Emphasis is placed on functions, logarithms, and exponential systems of equations. Graphing calculators will be used on a regular basis. This course satisfies the 4th math requirement for the public universities in the UNC system. Math I, Math II with A or B recommended. Calculus focuses on the solution of problems which cannot be solved by algebra or trigonometry. Finding the slope of the tangent to a curve, areas of planar and spatial surfaces, the volume of solids, and the mathematics of speed and acceleration are examples. The ability to construct and interpret graphs is a necessary component of many solutions. Calculus is intended for students with a high aptitude in mathematics who intend to pursue fields related to mathematics, physics, and engineering while in college. Graphing calculators are used on a regular basis. This is a college level course and is offered for students who will take the Calculus AB AP exam. Course is intended to be challenging and demanding. Calculus AB is primarily concerned with developing the students’ understanding of the concepts of calculus and providing experience with its methods and applications. The course emphasizes a multi-representational approach to calculus, with concepts, results, and problems being expressed graphically, numerically, analytically, and verbally. The connections among these representations are also important. This is a college level course and is offered for students who will take the Calculus BC AP exam. Calculus BC is an extension of Calculus AB rather than an enhancement; common topics require a similar depth of understanding. The course is intended to be challenging and demanding. Calculus BC is primarily concerned with developing the students’ understanding of the concepts of calculus and providing experience with its methods and applications. The course emphasizes a multi-representational approach to calculus, with concepts, results, and problems being expressed graphically, numerically, analytically, and verbally. The connections among these representations are also important. Advanced Placement Statistics introduces students to the major concepts and tools for collecting, analyzing, and drawing conclusions from data. Students will observe patterns and departures from patterns, decide what and how to measure, produce models using probability and simulation, and confirm models. Appropriate technology, from manipulatives to calculators and application software, should be used regularly for instruction and assessment.

Projection electron beam lithography apparatus and method employing an estimator ||Projection electron beam lithography apparatus and method employing an estimator ||May 23, 2006 ||August 29, 2001 ||Stanton; Stuart T. (Bridgewater, NJ) ||Agere Systems Inc. (Allentown, PA)| ||Homere; Jean R. |Attorney Or Agent: ||430/30; 700/30; 703/13; 703/2; 703/3; 703/4; 708/300; 708/313 |Field Of Search: ||703/3; 703/4; 703/13; 703/2; 700/30; 708/300; 708/313; 430/30 |U.S Patent Documents: ||5719796; 6177218; 6243158; 6285971; 6631299; 6925478; 2002/0120656 |Foreign Patent Documents: ||Stanton et al., "Initial Wafer Heating Analysis for a SCALPEL Lithography System", Microelectronic Engineering, vol. 46, Issues 1-4, May 1999,pp. 235-238. cited by examiner. ||A process and method for projection beam lithography which utilizes an estimator, such as a Kalman filter to control electron beam placement. The Kalman filter receives predictive information from a model and measurement information from a projection electron beam lithography tool and compensates for factors which cause beam placement error such as wafer heating and beam drift. The process and method may also utilize an adaptive Kalman filter to control electron beam placement. The adaptive Kalman filter receives predictive information from a number of models and measurement information from a projection electron beam lithography tool and compensates for factors which cause beam placement error such as heating and beam drift. The Kalman filter may be implemented such that real-time process control may be achieved. ||What is claimed is: 1. A projection electron lithography system, comprising: a lithography tool for emitting a beam of electrons and producing measurement information; and a processorincluding, a plurality of different predictive models for producing predictive information, and an adaptive estimator that iteratively selects a best predictive model from said plurality of different predictive models and controls placement of said beamof electrons based on said predictive information from said best predictive model and said measurement information from said lithography tool, said adaptive estimator employing a tunable strength parameter to determine an optimal adaptation weightingcriterion. 2. The system of claim 1, wherein said adaptive estimator compensates for beating and beam drift effects. 3. The system of claim 1, wherein said adaptive estimator employs least-squares based linear matrix algebra. 4. The system of claim 1, wherein said system is a SCALPEL system. 5. The system of claim 1, wherein said adaptive estimator is an adaptive Kalman filter. 6. The system of claim 1, wherein said adaptive estimator is an adaptive Kalman filter and each of said plurality of different predictive models is partitioned into wafer scale components and die scale components, said adaptive Kalman filteronly employed for wafer scale components. 7. The system of claim 1, wherein said plurality of different predictive models differ due to a single parameter that varies in each of said plurality of different predictive models. 8. The system of claim 1, wherein said plurality of different predictive models includes three or more models. 9. The system of claim 1 wherein said plurality of different predictive models are only directed to producing said predictive information for corrections associated with a die scale. 10. A computer implemented process for controlling projection electron lithography, comprising: emitting a beam of electrons; producing measurement information on said emitting step; producing predictive information related to the projectionelectron lithography process based on a plurality of different predictive models; iteratively selecting one of said plurality of different predictive models until a best predictive model from said plurality of different predictive models emerges; andcontrolling placement of the beam of electrons based on selected predictive information from said best predictive model and said measurement information, wherein said controlling includes determining an optimal adaptation weighting criterion employing atunable strength parameter. 11. The process of claim 10, wherein said controlling step employs an adaptive Kalman filter. 12. The process of claim 10, wherein said controlling step compensates for heating and beam drift effects. 13. The process of claim 10, wherein said process is a SCALPEL process. 14. The process of claim 10, wherein said controlling step is implemented as an adaptive Kalman filter and each of said plurality of different predictive models is partitioned into wafer scale components and die scale components, said adaptiveKalman filter only employed for wafer scale components. 15. The process of claim 10, wherein said plurality of different predictive models differ due to a single parameter that varies in each of said plurality of different predictive models. This invention relates to the field of projection electron beam lithography and in particular, to projection electron beam lithography employing an estimator. In projection electron beam lithography, precise control of the placement of the electron beam is required in order to ensure that the image is constructed without distortion and aligned to a prior process level. Precise control of the electronbeam placement is difficult because electron beam placement depends on many factors. One of these factors is a wafer distortion response to the heating action of a projection electron beam lithography beam, ranging up to many hundreds of nanometers, depending on conditions. Correction schemes include a model-based predictor forsub-field center placement adjustment. The algorithm implemented by the model-based predictor controls the writing of a matched dynamic distortion with an accuracy of about 1% or better for the largest, long-length-scale effects of approximately 500 nm. Other factors in addition to a predictable heating response, such as beam drift and wafer-to-chuck contact variation, also affect placement accuracy. Their effect may be either random or very difficult to correctly model. As stated above, wafer-to-chuck contact may have an effect on the response that requires enhancement to a basic predictive model. Modeling and experiments have both demonstrated the desirable result that good thermal contact to the chuck(.about.150 W/m.sup.2K) can lower the accumulated size of the wafer-heating response by a factor of roughly 10, thus enlarging the fractional correction error tolerance similarly. However, there are several factors, such as wafer-flatness, particletolerance, frictional contact, and pulling-force that may remain variable or random despite efforts in chuck design. Realistically, the chuck design process can only reduce frictional influences on the heating response to a form ofchuck-coordinate-system drift that is slow and indistinguishable from beam drift. Since important parameters in the predictive model may be variable from wafer to wafer, prediction alone is not sufficient for full correction of beam placement. Further, it is difficult to perform the complex model computation required to determine correct beam placement in a short period of time. The only alternative to prediction is measurement. The obvious primary measurement of beam placement involves an alignment mark sensing process. The use of a re-alignment strategy, or some variation of local alignment, is a common approach todealing with drift in many other electron beam lithography applications, such as mask-making and direct-writing. This often involves time-consuming actions like extra stage motions that detract from throughput, but this can be a tolerable situation whenmaking relatively few high-value exposures. In the area of production wafer-level lithography using SCALPEL, throughput is a concern even without the use of local alignment or complex re-alignment strategies. Hence, re-alignment is not a suitable correction strategy for a high-throughputSCALPEL tool. Based on the above, it is clear that an enhancement to the predictive models used for beam placement correction is desirable, making use of alignment mark sensing and efficient computation. SUMMARY OF THE INVENTION The method and apparatus of the present invention include an estimator that integrates a predictive model and a measurement capability, both subject to substantial noise sources, plus measurement sampling limitations. The estimator works in realtime with only historical data. In one exemplary embodiment, the estimator is a Kalman filter, which may be a least-squares based optimum estimation algorithm for the states of time-dependent systems, using linear matrix algebra. In the present invention, the Kalman filter is used to correct for wafer heating, beam drift and/or other errors in a projection electron beam lithography system, such as for example, SCALPEL. By using a Kalman filter, real time process controlis obtained using a greater amount of information than could be used if conventional modeling/process control and measurement techniques were used. The method and apparatus of the present invention may also employ an adaptive Kalman filter (A-KF) correction for wafer heating, beam drift and/or other errors. The adaptive Kalman filter correction may be based on a numerical response-modelinterface that allows efficient integration of relatively slow but infrequent pre-calculation results, and allows real-time adaptive Kalman filter functionality. An adaptive Kalman filter is particularly effective when a model parameter uncertainty problem is superimposed on a more elementary state noise problem. The two types of unknown system response can both be handled using only one measurement datasequence, but are distinguishable in terms of their statistical behavior. In SCALPEL, an example of an uncertain parameter is wafer-to-chuck thermal contact, which should be a nearly-fixed quantity on length scales of interest, during each waferexposure. The effect of wafer-to-chuck thermal contact on the response of the system is momentarily stable and non-random for any one execution of the Kalman filter, even if poorly known. This is in contrast to the lumped beam drift and frictionalchuck-coordinate-system drifts that may be more like a random-walk effect, and hence most readily treated as a band-limited state noise. In a preferred embodiment, the control algorithm which performs the predictive model can be partitioned into global (wafer scale) and local (die scale) components. A pure-predictor would suffice for the local problem since the main noise anduncertainty terms do not act on this scale and the errors are inherently smaller. The use of an adaptive Kalman filter only for the global part of the problem would be very efficient. The method and apparatus of the present invention may also employ a multi-model adaptation corrector, which provides a best estimate that converges on the correct unknown model parameter choice. The behavior of the Kalman filter is very good for scenarios that are realistic or somewhat pessimistic in key parameters pertaining to SCALPEL operation, including a slow beam drift of typically 40 nm and a 15 nm 3-sigma one-site alignmentnoise. Adaptation in a multi-model form is effective at handling the problem of at least a factor of two thermal contact parameter uncertainty. Combined errors on the order of 50 nm in predicting responses that are well over 100 nm can be reduced to 10 nm or better, in a case of low contact and thermal dissipation to the chuck. With some optimization and the benefit of maximum chuckthermal contact, error budget requirements of nominally 5 nm can also be met. BRIEF DESCRIPTION OF THE DRAWINGS FIG. 1 illustrates a projection electron beam lithography system in one exemplary embodiment of the present invention. FIG. 2 illustrates the Kalman filter of FIG. 1 in one exemplary embodiment of the present invention. FIG. 3 illustrates the steps of multi-model adaptation in one exemplary embodiment of the present invention. FIGS. 4a and 4b illustrate a weight-determining function in one exemplary embodiment of the present invention. FIGS. 5a and 5b illustrate the response of a nominally tuned adaptation scheme based on residual curves and multi-model execution. DETAILED DESCRIPTION OF THE PREFERRED EMBODIMENT FIG. 1 illustrates a projection electron beam lithography system 10 in one exemplary embodiment of the present invention. As illustrated, the system 10 includes a processor 12 (either with or without external memory) and a projection electronbeam lithography tool 14. In a preferred embodiment, the projection electron beam lithography tool 14 is a SCALPEL tool. A predictive model 16 and a Kalman filter 18 are both implemented in processor 12. The Kalman filter 18 receives predictions fromthe predictive model 16 and measurements from the projection electron beam lithography tool 14 and controls placement of an electron beam output from the projection electron beam lithography tool 14 as described in more detail below. A Kalman filter 18 is a recursive algorithm using linear matrix algebra to make an optimal estimate of the state of a system, given a combination of state and measurement noises. The most common form of the optimization is least-squares, whichis readily formulated in linear matrix algebra form and is optimum for Gaussian noise, but the algorithm can be more general as well. Nonlinear systems also can be linearized in order to make use of the linear algebra form of the filter. The essence of the Kalman filter 18 is to use one or more models 16 to describe the statistical behavior of both the measurement noise and the physical system state noise, so that this information can be used to determine the weighting in thecombination of prediction and measurement. This is referred to as "propagating the noise or error covariance", which is an ingredient in one of the two major recursive steps of the filter illustrated by FIG. 2. As illustrated in FIG. 2, predictionsfrom the model 16 and measurements from the tool 14 are recursively processed. By propagating the error covariance, an update of the Kalman gain (K) can be made. This quantity determines the weighting in the filter 18; 0 for pure prediction and 1 forpure measurement. The other major step is propagating the predictive model 16 iteratively based on a starting value from the estimate made in the previous step. This process continues iteratively in a loop. This estimate updating process is notnecessarily smooth since the quality of measurement information can change abruptly even if the system state cannot. "Tuning" the Kalman filter 18 may entail making adjustments in the proposed error/noise statistics model 16 in order to better match "reality". Variants of the Kalman filter 18 allow this to be done adaptively during the course of the filter 18operation, but it is also common to tune by trial and error as a series of experiments or simulations are performed. In the SCALPEL heating response application, the tuning is motivated by a need to estimate the required sub-field position adjustmentfor exposures in a sequence, hence reducing the worst error that occurs at any time in the exposure for an ensemble of wafer exposures. Using the Kalman filter 18, prediction alone is good enough in early stages when state errors have not accumulatedyet. This is due to the band-limited nature of the beam drift and the action of errors in thermal contact. A Kalman filter 18 usually uses differential equations of the system state expressed in state-space matrix form. However, the description below uses a common alternative notation, namely "discrete form" notation, which expresses the result atstep k+1 caused by propagation forward from step k. This is appropriate for a discrete measurement process, such as the SCALPEL process. Note that the steps modeled are absolutely not limited to those where measurements are made. The Kalman filter 18naturally deals with this by assigning non-measurement steps with a very large measurement covariance, resulting in the gain (K) being set to zero for those times. So the model 16 can naturally interpolate the state estimate in closely-spaced stepsbetween relatively sparse measurements. The five basic matrix equations are: 1) State prediction update: X(k+1/k)=.PHI.(k+1,k)X(k/k)+.PSI.(k+1,k)U(k) 2) Covariance prediction update: P(k+1/k)=.PHI.(k+1,k)P(k/k).PHI..sup.T(k+1,k)+Q*(k+1) WithQ*(k+1)=.GAMMA.(k+1,k)Qd(k).PHI..sup.T(k+1,k) 3) Gain computation: K(k+1)=P(k+1/k)H.sup.T(k+1)[H(k+1)P(k+1/k)H.sup.T(k+1)+R(k+1)].sup.-1 4) Estimation update: X(k+1/k+1)=[I-K(k+1)H(k+1)]X(k+1/k)+K(k+1)Z(k+1) 5) Covariance update:P(k+1/k+1)=[I-K(k+1)H(k+1)]P(k+1/k) These five equations correspond to a state-space representation of the propagation of state X and process of measurement Z, including noise, given by: X(k+1)=.PHI.(k+1,k)X(k)+.GAMMA.(k+1,k)Wd(k)+.PSI.(k+1,k)U(k); andZ(k+1)+H(k+1)X(k+1)+V(k+1). In all of the above equations, k is a step counter. The use of (n/m), such as (k+1/k), designates value in step n if given the value in step This is distinct from (k+1,k) which designates that the matrix value is sensitive to both the prior andpresent step count in general. Two examples clarify this notation: X(k+1/k) is the pure prediction update of the state vector X and X(k+1/k+1) is the update of the estimate of state X including measurement. In the state equations, the values are defined as: .PHI.=state propagator model from differential equations which also propagates the state covariance; U=input term for state, which can be generalized as we will discuss later, except that it doesnot propagate the state covariance; .PSI.=matrix which translates input to state form; Wd=state noise in raw form; .GAMMA.=matrix that translates state noise into state form; V=measurement noise in raw measurement form; and H=matrix that translates thestate into measurement form. The other quantities in the filter equations 1) 5) are: P=state covariance matrix, standard definition with terms in the form .sigma..sub.i .sigma..sub.j; has a starting value but is later generated by the filter 18; Qd=covariance matrix of statenoise Wd, in a form like P; nominally an assumed constant, or may be a sequence; generally subject to tuning; R=covariance matrix for measurement, similar to Qd; usually derived from measurement process modeling or experiments; may be tuned; K=calculatedKalman gain representing weight of measurement in estimate; and I=Identity matrix. Further, .sup.T refers to the transpose operation, and .sup.-1 refers to matrix inversion. As indicated by equation 3), K is computed completely from the propagation of measurement covariance and state noise covariance, which includes initial errors and added state noise. These can be done ahead of time in a situation that is notadaptively tuned and when the covariance model is stable. Similarly, as indicated by equation 4), K acts as a weight on the use of measurement in the estimation update, and a term of the form "I-K" is the converse weight of predictive update. Equations 1) 5) do not consider time-correlated noise (also known as "non-white" or "colored" noise) in any category. Equations 1) 5) assume that each new time step gives independent new random noise terms. The entire Kalman filter 18 equation set 1) 5) above can be modified to deal with correlated noises, although there may be a different process for measurement than there is for state. In the case of SCALPEL, measurement by alignment is expectedto have no time-correlation in the sense that information at each site has an error with no dependence on prior measurements. However, the state noise of drift clearly cannot be a white noise. Therefore, the state noise may be considered colored andthe Kalman filter 18 may be modified accordingly. The basic form of the equation 1) 5) stays the same except that a few elemental vectors and matrices should be augmented, meaning that new vectors and matrices are composed from old vectors and matrices with terms attached that represent atime-correlation or filter model. One such example is a one-step filter function with variable time constant t0, in the form: .PHI.wf=Exp[-(t.sub.k+1-t.sub.k)/t0]. One-step colored noise (Wdco) at step k+1 is generated from a new white random noisevalue (Wdwf) plus a fixed residual amount of the last noise value at k determined by the filter function: Wdco(k+1)=.PHI.wf(k+1,k)Wdco(k)+Wdwf(k) Augmentation processes are well-known. Below the equation changes are shown symbolically as extended vectors or groupings of matrices of the same dimensions to form larger matrices, where: X[X Wdco].sup.T H[H 0] .GAMMA. replaced by .GAMMA.aw=[0I].sup.T The original .GAMMA. is integrated with the state propagator: .PHI..times..PHI..function..kappa..kappa..GAMMA..function..PHI..times..tim- es..function. ##EQU00001## .PSI..times..PSI..times. ##EQU00001.2## .times..times. ##EQU00001.3## Q[U 0].sup.T Qd. terms in form .sigma..sup.2 become .sigma..sup.2[1-Exp[-2.DELTA.t/t0]] "0" represents a matrix of zeroes. For the purpose of running Monte-Carlo simulations of the application of a Kalman filter 18 to a specific model, it is typical to only provide a white-noise generator. Either the truth model is propagated in an augmented fashion to obtainfiltered noise, or the filter is applied a-priori (as shown here) to a time-series of random elements of the noise matrix. The use of the model 16 can be totally consistent by design, or the effect of an erroneous assumption about the time-correlationcan also be simulated. The Kalman filter 18 described assumes a singular "good" model 16 exists and that physical effects are appropriately modeled as additive random noise. This accurately describes the beam drift effects in SCALPEL. A different problem occurs ifthe model 16 is not fully known, so an assumed model leads to poorer filter performance than an ideal one would achieve. In general, there are known system model identification procedures that can be used to "learn" what a model should be. Particularlyin the absence of state noise, there are many non-Kalman filter approaches to using real-time measurements to converge on the right model and iteratively best-fit a measurement sequence. However, the same limited data may be subject to both noise andparameter uncertainty, as in SCALPEL. For this situation, an adaptive Kalman filter 18 implemented in a multi-model form is a powerful tool. In general, it is possible for one noise model to actually be the net effect of many more. It is not always obvious which type of disturbance is best treated as a "noise" versus an "uncertain parameter". In all cases, the Kalman filter 18equations must still have only one linear-additive noise vector in the state. The ability of the Kalman filter 18 to rapidly and efficiently perform real-time estimation depends on the linearity of the matrix formulation. Therefore, a multiplicativenoise or a product of two model components having noise must be linearized. However, if two disturbances are distinguishable because their statistical natures are very different, then one disturbance may be deemed to be a parameter that is momentarily fixed relative to another that varies more rapidly. In general,adaptation schemes can be applied sequentially to attempt to choose this parameter at any time as this parameter may evolve. In this case, time-correlation is the trait that distinguishes one from another even though both may have a stochastic nature. A multi-model adaptive Kalman filter 180 may be used to discern the best model 160. A set 160 of N assumed models 161, 162, 163 . . . are continuously tested to see if one emerges as a "better" model than the rest. This is a particularly goodapproach when only one unknown parameter really matters, such as chuck thermal contact. As each of N filters 181, 182, 183, . . . are run in parallel, each defines an optimal estimate for the same measurement sequence but using a different model 161,162, 163, . . . . Usually the models 161, 162, 163, . . . are basically the same, and a single parameter is varied N times in some series of steps. In the event that the response of the model 160 to the unknown parameter is continuous and not too severe, a limited number of models may be used in combination with a scheme that interpolates to determine a weighted combination of "best discretemodels". Obviously, the more models needed (N) and the more parameters not known (M), the less efficient the process may be since a total of NxM models must be run. One issue is what criterion can be used to guide the "adaptation", which is the process of selecting the correct model or weighted combination of models in real-time. Publications exist on this topic, with various ideas depending on the natureof the problem. The common thread is analysis of the "residual", which is the historic record of differences between the estimate and the measurement. Therefore, in addition to the use of multiple filters 181, 182, 183, . . . , the other practicalfacet of a multi-model adaptation approach is a certain amount of historic book-keeping. The steps in multi-model adaptation are illustrated in FIG. 3. First, an initial model is selected, then several models and filters are run. A minimum is foundfor a key criterion at 200 and a revised model is selected at 210. The adapted estimate is output and looped back to the different model 161, 162, 163, . . . . In the case of the SCALPEL responses, it may be reasonable to consider the unknown thermal contact parameter to be nearly fixed in the whole time-frame of one wafer exposure, then changed but fixed again for a second wafer exposure. For any oneassumed parameter model, if the assumption is relatively bad the Kalman filter 180 behavior will be relatively bad, which will lead to a residual which is "large" in some key criterion. The prediction will diverge from reality and the filter willdefault to an estimate dominated by measurement (K.about.1), but directly limited by measurement noise and not much helped by the model 160. Therefore, the model that reduces some criterion composed from the historic residual should be the "best model" and the Kalman filter 180 should transition from an initial assumption to the selection of this model. In general, this occursgradually since the measurements are noisy, but a large enough amount of data will eventually establish a trend. Effectiveness in many real systems is based on the time-growth of the response associated with the uncertain parameter, such that tolerablylittle error accumulates in the time required to converge on the correct model. The specific length of the history considered and the specific criterion designed to make a selection depend on many factors, such as the duration one would expect theparameter to be nominally fixed, or the ultimate application where the best estimate is needed at a "singular end-event" time instead of all times. Of course, the real state is not known for real situations, but should be known in a Monte-Carlo adaptive Kalman filter simulation, which is a common filter development method. Adaptation criterion and model-selection methods are described below. A decision criterion is based on the history of residuals, where the residual is the vector difference between the measurement and the estimate for the whole state at eachstep, for each model acting in parallel. The momentary position error radius at each step is of interest in the SCALPEL problem. Therefore, the position error radius can be formed from appropriate residual components at each step, and a simple averageerror radius over some history length can be calculated for each model 161, 162, 163, . . . . This average could consider a length of time either shorter than or up to the total time of the system propagation or the full length of the history at eachstep. This average error radius is the best criterion for adaptation in the SCALPEL case. In running an adaptative Kalman filter, the average error radius is calculated for each model number at each time step. As the system propagates, a clear minimum inside the assumed model range occurs, and this almost always corresponds to theselection of the correct model used to generate a truth simulation, unless the state noise effects are overwhelmingly large. The plot is a visual representation of the data that is analyzed at every step to form an adaptation scheme. The correct or "best" model occurs at the model number having the lowest residual radius error over some characteristic averaging time. Essentially, the strength of the minimum within the available model set is used as the selection criterion. The minimum should be both pronounced and sustained. Simulations or trials can be used to determine if the range of models assumed isappropriate to make sure that a minimum can eventually be found. Analysis of the position and strength of this minimum is aided by using a normalized contrast criterion ranging from 0 to 1 to compare the maximum and minimum values of this residual radius error across the model set as a function of timecontrast(k)=[Max-Min]/[Max+Min]@step k where Max and Min refer to the averaged error radius of each model. To translate these fairly small contrast values into a criterion for selecting a given model, it may be useful to use a second weight-determining function. The second weight-determining function should be a smooth function that translates thisbasic contrast evaluation in a simple way, over a normalized range of 0 to 1. The specific function chosen is not important as long as tuning of the parameters is done in simulations. FIGS. 4a and 4b illustrate a function(Adaptweight=1-Exp[-(contrast/strength)^2]) that can be made to saturate the weight versus contrast relationship depending on a single strength parameter (with examples shown for strength=0.2 and 0.5). Therefore, the process of developing an adaptive filter entails tuning the strength parameter to determine the weighting of adaptation. This weight can be considered to be similar to an "outside loop" version of the Kalman gain (K) that goesfrom 0 to 1 as the measurement data provides enough information to select a best model. A distinction is that this weight operates on a whole history of residual data from action of the set of filters, while the K in each filter operates only one stepat a time and within its own assumptions. Although the present invention has been described above as the implementation of a Kalman filter 18 or a multi-model adaptive Kalman filter 180 in a projection electron lithography method or apparatus, other additions or refinements may bepossible including: using the weight to interpolate between discrete models and allow selection of a best model that combines two near-minimal residual models; using a "no-turning-back" scheme where the weight is not allowed to go back down in theunusual event that a longer history of measurements does not continue to converge on a stronger minimum residual (this option makes sense if there must be a singular fixed model and state noise is relatively small, but tuning can become complex if statenoise is large, namely the measurements must counter both noise and parameter uncertainty problems); replacing the starting-assumption model at some threshold weight value with the last adapted model; smoothing of the adaptation process, which may yielda smoother result but not necessarily a better one, and is subject to tuning. FIGS. 5a and 5b illustrate the response of a nominally tuned adaptation scheme based on averaged error radius curves and multi-model execution. Note that in FIG. 5b, the starting assumption is model 190 6, but the truth model is model #4, bothof which lie inside a range from a low at #1 to a high at #9. The weight of adaptation in FIG. 5a rises sharply at about 1/4 the time into the sequence and is locked at its last high value. The model selection oscillates slightly after the assumedmodel is rejected, and then it converges close to the true model. In a preferred embodiment, more than three models are used, and in a more preferred embodiment, five models are used. The SCALPEL wafer-heating response requires a complex heat transfer and elastic strain model based on partial differential equations and boundary conditions, with mixed cylindrical and Cartesian coordinate systems used for key features. Theresponse cannot be simplified by treating only certain dominant modes of the response. The response can be almost arbitrarily complex and variable with several parameters. The dynamic distortion process should be corrected to a few nanometers accuracyat times corresponding to unique sub-field locations throughout the exposure, corresponding to roughly one million model steps in about 2 minutes, or a step rate of 8333 Hz. In each step, a full history-dependent snapshot of an extended system modelwould have to be executed. The likelihood of obtaining even one adequately fast and accurate real-time model is poor, and running an array of models for adaptation may be impractical. However, the Kalman filtering described above is an inherently numerical approach to propagating system state estimates based on differential equations. The Kalman filtering described above is also inherently linear in the way it incrementallyadds a new prediction to the prior estimation of the state using a predictive model. Therefore, it is natural to substitute a sequence of numbers in the matrix positions for what would otherwise be a discretely propagated function-based model. If thenumbers exist a-priori, the linear matrix algebra can be very fast because the differential equations have been effectively solved before-hand. A remaining issue is the speed of the a-priori number generation processor. Since this process is not done in real-time during the one or two minute exposure time, presumably much more time could be taken. However, throughput requirements onthe exposure tool require that such a calculation does not add significant time to the batch process time of many wafers, for example 30 wafers exposed in an hour. The up front calculations have to be some combination of fast and/or done in parallel toother necessary lithography tool functions. Since high throughput is usually associated with repetitive exposure batches, the up-front model variations should be limited to occasions when the pattern (mask) is changed or significant conditions (exposure current or resist dose) mightchange. If at least 25 wafers are run with the expectation of completing them in about an hour, spending one minute overall on computation is acceptable but spending 25 minutes in repetitive computation is not. As stated earlier, the main distinction of each wafer exposure in a batch is likely to be chuck thermal contact and beam drift. However, due to the linearity in the combination of basic elements of the Kalman filter 18, there is nothing aboutthe operation of the Kalman filter 18 that would "feed back" a required change to the basis predictive model. They are uncoupled, and it is well known that many elements of a Kalman filter 18 can be pre-computed and stored to minimize the real-timecomputation burden. This is also true for adaptive Kalman filtering 180 as well, assuming that a whole array of models exists for the full time. In fact, this may be a reason to implement the multi-model adaptation scheme, instead of a scheme thatminimizes the number of models used as the unknown parameter is discerned. If number sequences are chosen for the model, the predictive model and Kalman filter can be decoupled entirely to allow any good model technique to be used for any up-front calculation. A remaining issue in implementing the Kalman filter isdeciding what position the model-result sequence should take in the Kalman filter equations. It is tempting to just substitute the number sequence for the whole predictive step to give X(k+1/k), but this is incorrect. The general reason why it isincorrect is because the .PHI. component of the state space predictor also propagates the state error covariance that makes the filter work. Therefore these substitutions must be consistent and careful. For SCALPEL wafer heating and beam drift response, the nature of the system actually simplifies the model integration problem. The "model" of beam drift propagation may only require the state-noise band-limit filter function. This is consistentwith the idea that the electron beam is a system with negligible inertia. Further, drift noise is instantly and fully added to the position state, and the modified state has no effect on incremental propagation to the next state. Therefore, given the fact that the (D matrix is augmented with this filter function already, the simplest answer is to use a "null" basis state propagation model with the pre-calculation treated as "input", given by: .PHI.=0 U=[X.sub.u, 0,y.sub.u,0].sup.T The x and y entries in U are a sequence of pre-calculated predicted sub-field center responses at known times. The use of a state vector comprised of position and velocity is continued. This approach has been shown to work adequately by simulation. However, other methods are possible. For example, it may be possible to propagate the state noise covariance with a simple, approximate model that has some basic physicalsensibility. As described above, the present invention is directed to a method and apparatus that implements a Kalman filter 18 or an adaptive Kalman filter 180 correction scheme for wafer heating and beam drift in projection electron beam lithography, suchas SCALPEL. The Kalman filter is based on a numerical response model interface that allows efficient integration of relatively slow but infrequent pre-calculation results, and allows real-time adaptive Kalman filter functionality. The present inventiondemonstrates the feasibility of a die-center correction for the critical "global" part of the correction scheme. The local part can be done by pure prediction since the errors are smaller and less subject to effects of drift and chuck contactuncertainty. The adaptive Kalman filter 180 behavior is very good for a scenario that is realistic or somewhat pessimistic in key parameters, including a slow beam drift of typically 40 nm and a 15 nm 3-sigma one-site alignment noise. Adaptation in amulti-model form is effective at handling the problem of at least a factor of two thermal contact parameter uncertainty, in a scenario where the contact is a great deal lower than what we know is possible, hence giving relatively large responses. Combined errors on the order of 50 nm in predicting responses that are well over 100 nm can be reduced to 10 nm or better. With some optimization of the corrector and the benefit of maximum chuck thermal contact, it is likely that error budgetrequirements of nominally 5 nm will be met. Although the various embodiments of the Kalman filter described above may be used to correct for wafer heating, beam drift and/or other errors in a SCALPEL or other projection electron beam lithography system, the present invention is not limitedto correction of these errors. Other correctable errors may include errors related to the current at the wafer, the thickness of the wafer, thermal response parameters (which may include heat capacity, heat conductivity, thermal expansion coefficient,Young's modulus, or Poisson's ratio of Si), wafer-to-chuck frictional contact, wafer-to-chuck thermal contact, wafer initial temperature profile, and/or beam drift (which may be related to charging, stray fields, electronics, and/or thermal factors). It is noted that the functional blocks in FIGS. 1 3 representing the Kalman filter 18,180 and model 16,160 may be implemented in hardware and/or software. The hardware/software implementations may include a combination of processor(s) andarticle(s) of manufacture. The article(s) of manufacture may further include storage media and executable computer program(s). The executable computer program(s) may include the instructions to perform the described operations. The computer executableprogram(s) may also be provided as part of externally supplied propagated signal(s). In an exemplary implementation of the numerical integration approach described above, the real-time operation of a die-by-die Kalman filter, using pre-existing numerical model results, only took 14 seconds to run on a 400 MHz PC runningnoncompiled and relatively-slow Mathematical.RTM. 3.0 by Wolfram Research Inc. Champaign, Ill., with many extra plotting and data output steps. This is easily fast enough for real-time use if die exposures take at least 1 second. This result isexpected because the recursive part of the Kalman filter is mainly linear matrix algebra. Equivalent compiled code runs should be much faster for real tool implementation. Other control system development and simulation software, such as MatLab.RTM.,by the MathWorks Inc., Natick, Mass. could also be used, as could any of the C-family of languages. Although the estimator described above is a Kalman filter, any number of other estimators such as simple observers, full order observers, reduced order observers, trackers, or other estimation techniques known to one of ordinary skill in the artor combinations thereof, are also contemplated by the present application. Still further, although the statistical technique utilized above is a least squares technique, other techniques, such as variance, (linear or not), general optimal, maximumlikelihood, maximum a-posteriori, weighted leased squares, or other techniques known to one of ordinary skill in the art or combinations thereof, are also contemplated by the present application. The invention being thus described, it will be obvious that the same may be varied in many ways. Such variations are not to be regarded as a departure from the spirit and scope of the invention, and all such modifications as would be obvious toone skilled in the art are intended to be included within the scope of the following claims. * * * * * ||Randomly Featured Patents

Definition of a Trapezoid A trapezoid is a 2-dimensional geometric figure with four sides, at least one set of which are parallel. The parallel sides are called the bases, while the other sides are called the legs. The term 'trapezium,' from which we got our word trapezoid has been in use in the English language since the 1500s and is from the Latin meaning 'little table.' There are a few special trapezoids that are worth mentioning. In an isosceles trapezoid, the legs have the same length and the base angles have the same measure. In a right trapezoid, two adjacent angles are right angles. If the trapezoid has no sides of equal measure, it is called a scalene trapezoid. A parallelogram is a trapezoid with two sets of parallel sides. There is actually some controversy over whether a parallelogram is a trapezoid. One group states that the definition of a trapezoid is having only one set of parallel sides, which would exclude the parallelogram because it has two sets of parallel sides. The other, more mainstream group, states that the definition of a trapezoid is having at least one set of parallel sides, which includes the parallelogram. For our discussions, because it is the more widely accepted view, we will consider a parallelogram to be a trapezoid. Properties of a Trapezoid The formula for the perimeter of a trapezoid is P = (a + b + c + d). To find the perimeter of a trapezoid, just add the lengths of all four sides together. The formula for the area of a trapezoid is A = (1/2)(h)(a + b), where: - h = height (This is the perpendicular height, not the length of the legs.) - a = the short base - b = the long base An isosceles trapezoid has special properties that do not apply to any of the other trapezoids: - Opposite sides of an isosceles trapezoid are the same length (congruent). - The angles on either side of the bases are the same size or measure (also congruent). - The diagonals are congruent. - Adjacent angles (next to each other) along the sides are supplementary. This means that their measures add up to 180 degrees. Let's try a couple of practice problems to test your newfound trapezoid knowledge. Feel free to pause the video at any point to work through the problems yourself. 1.) Find the perimeter and area of the following trapezoid: To find the perimeter, simply add all four sides together. P = 12mm + 14mm + 18mm + 13mm = 57mm To find the area, use the formula A = (1/2)(h)(a + b). A = (1/2)(11mm)(12mm + 18mm) = 165mm^2 2.) Find the area of the following trapezoid: Again, use the area formula A = (1/2)(h)(a + b). A = (1/2)(6ft)(9ft + 4ft) A = 39ft^2 A trapezoid is a 2-dimensional figure with four sides. In order for it to be classified as a trapezoid, it must have at least one set of parallel sides. Trapezoids play a key role in architecture and also can be found in numerous everyday items. Take a look at the glass you are drinking from at your next meal. From the side, it's probably shaped like a trapezoid. Review the video lesson and its corresponding transcript so that you can: - Define a trapezoid and identify its properties - Illustrate several special trapezoids - Point out the properties of isosceles trapezoids - Find the perimeter and area of a trapezoid To unlock this lesson you must be a Study.com Member. Create your account Practice Trapezoid Questions 1. Given the area of a trapezoid, whose parallel sides are 11 and 13 units respectively, is 36 square units, find the height of this trapezoid or the perpendicular distance between its parallel sides. A sketch of this trapezoid is presented below: 2. First, we have an isosceles trapezoid. The area of this trapezoid is 100 cm^2. The height or the perpendicular distance between the two parallel sides of this trapezoid is 5 cm. One of the parallel sides is 15 cm. What is the length of the other parallel side? 1. The formula for the area of a trapezoid is A = (1/2) x (a + b) x h, where a and b are the lengths of the two parallel sides, and h is the perpendicular distance between the two parallel sides. Then, substituting in the formula gives us: 36 = (1/2) x (11 + 13) x h = (1/2) x 24 x h = 12 x h Then, dividing both sides by 12 yields: 36/12 = 12 x h/12 or 3 = h Hence, the height of this trapezoid is 3 units. 2. Again, the formula for the area of a trapezoid is A = (1/2) x (a + b) x h, where a and b are the lengths of the two parallel sides, and h is the perpendicular distance between the two parallel sides. Then, substituting in the formula gives us: 100 = (1/2) x (a + 15) x 5 Multiplying both sides of the above equation by 2, and then distributing on the right-hand sides yields: 200 = 5a + 75 Subtracting 75 from both sides of the above equation leads to: 200 - 75 = 5a or 125 = 5a Dividing both sides by 5 finally gives us: 125/5 = 5a/5 or 25 = a The length of the other parallel side is 25 cm. Register to view this lesson Unlock Your Education See for yourself why 30 million people use Study.com Become a Study.com member and start learning now.Become a Member Already a member? Log InBack

Partial Least Squares Regression Assignment Help For example, you might approximate (i.e., forecast) an individual’s weight as a function of the individual’s height and gender. You might utilize direct regression to approximate the particular regression coefficients from a sample of information, determining height, weight, and observing the topics’ gender. For lots of information analysis issues, price quotes of the direct relationships in between variables are sufficient to explain the observed information, and to make affordable forecasts for brand-new observations (see Numerous Regression or General Step-by-step Regression for extra information. PLS regression is mostly utilized in the chemical, food, plastic, and drug markets. In PLS regression, the focus is on establishing predictive designs.The algorithm minimizes the number of predictors utilizing a method comparable to primary elements analysis to draw out a set of parts that explains optimum connection in between the predictors and action variables. Minitab then carries out least-squares regression on these uncorrelated parts. Partial Least Squares regression (PLS) is a fast, ideal and effective regression technique based upon covariance. It is advised in cases of regression where the variety of explanatory variables is high, and where it is most likely that the explanatory variables are associated.An excellent benefit of PLS regression over traditional regression are the readily available charts that explain the information structure. It can be relationships amongst the reliant variables or explanatory variables, as well as in between reliant and explanatory variables. This example reveals how to use Partial Least Squares Regression (PLSR) and Principal Elements Regression (PCR), and talks about the efficiency of the 2 approaches. PLSR and PCR are both techniques to design a reaction variable when there are a big number of predictor variables, and those predictors are extremely associated or even collinear. Both approaches build brand-new predictor variables, understood as elements, as direct mixes of the initial predictor variables, however they build those parts in various methods.The Partial Least Squares Regression treatment approximates partial least squares (PLS, likewise called “forecast to hidden structure”) regression designs. PLS is a predictive method that is an alternative to regular least squares (OLS) regression, canonical connection, or structural formula modeling, and it is especially beneficial when predictor variables are extremely associated or when the variety of predictors goes beyond the variety of cases. PLS integrates functions of primary elements analysis and numerous regressions. It initially draws out a set of hidden aspects that discuss as much of the covariance as possible in between the reliant and independent variables. A regression action anticipates worths of the reliant variables utilizing the decay of the independent variables.X is an n-by-p matrix of predictor variables, with rows corresponding to columns and observations to variables. XL is a p-by-comp matrix of predictor loadings, where each row consists of coefficients that specify a direct mix of PLS parts that approximate the initial predictor variables. YL is an m-by-comp matrix of reaction loadings, where each row consists of coefficients that specify a direct mix of PLS elements that approximate the initial action variables. We construct connections in between envelopes, a just recently proposed context for effective evaluation in multivariate stats, and multivariate partial least squares (PLS) regression. In specific, we develop an envelope as the nucleus of both Univariate and multivariate PLS, which unlocks to pursuing the exact same objectives as PLS however utilizing various envelope estimators. It is argued that a likelihood-based envelope estimator is less conscious the variety of PLS parts that are picked which it surpasses PLS in forecast and evaluation. The derivation of analytical residential or commercial properties for Partial Least Squares regression can be a difficult job. In this work, we study the intrinsic intricacy of Partial Least Squares Regression.We reveal that the Degrees of Liberty depend on the cob linearity of the predictor variables: The lower the co linearity is, the greater the Degrees of Liberty are. In specific, they are usually greater than the ignorant technique that specifies the Degrees of Liberty as the number of elements. Even more, we highlight how the Degrees of Liberty technique can be utilized for the contrast of various regression approaches.The approach is based on partial least squares regression, which breaks down the thermo graphic PT information series gotten throughout the cooling routine into a set of hidden variables. The regression approach is used to speculative PT information from a carbon fiber-reinforced composite with simulated flaws. The speaker utilizes customer rankings for 24 kinds of bread to show ways to utilize PLS to recognize item credit to assist brand-new bread formula and style procedures. He utilizes leave-one-out cross-validation and demonstrates how to analyze and analyze Root Mean Press; NIPALS Fit x and y ratings for a single hidden element; Diagnostic Plots; and VIP vs. Coefficients Plots. He utilizes the Forecast Profiler to optimize desirability. This paper provides a brand-new strategy for predictive heart movement modeling and correction, which utilizes partial least squares regression to extract intrinsic relationships in between three-dimensional (3-D) heart contortion due to respiration and numerous one-dimensional real-time quantifiable surface area strength traces at chest or abdominal area. In spite of the reality that these surface area strength traces can be highly paired with each other however inadequately associated with respiratory-induced heart contortion, we show how they can be utilized to precisely forecast heart movement through the extraction of hidden variables of both the input and output of the design. You might utilize direct regression to approximate the particular regression coefficients from a sample of information, determining height, weight, and observing the topics’ gender. For lots of information analysis issues, price quotes of the direct relationships in between variables are sufficient to explain the observed information, and to make affordable forecasts for brand-new observations (see Numerous Regression or General Step-by-step Regression for extra information.

Research Article | Open Access Jian Wang, Yong Wang, "A Kastler-Kalau-Walze Type Theorem for 7-Dimensional Manifolds with Boundary", Abstract and Applied Analysis, vol. 2014, Article ID 465782, 18 pages, 2014. https://doi.org/10.1155/2014/465782 A Kastler-Kalau-Walze Type Theorem for 7-Dimensional Manifolds with Boundary We give a brute-force proof of the Kastler-Kalau-Walze type theorem for 7-dimensional manifolds with boundary. The noncommutative residue found in [1, 2] plays a prominent role in noncommutative geometry. For one-dimensional manifolds, the noncommutative residue was discovered by Adler in connection with geometric aspects of nonlinear partial differential equations. For arbitrary closed compact -dimensional manifolds, the noncommutative residue was introduced by Wodzicki in using the theory of zeta functions of elliptic pseudodifferential operators. In , Connes used the noncommutative residue to derive a conformal 4-dimensional Polyakov action analogy. Furthermore, Connes made a challenging observation that the noncommutative residue of the square of the inverse of the Dirac operator was proportional to the Einstein-Hilbert action in . Let be the scalar curvature and let Wres denote the noncommutative residue. Then, the Kastler-Kalau-Walze theorem gives an operator-theoretic explanation of the gravitational action and says that, for a 4-dimensional closed spin manifold, there exists a constant , such that In , Kastler gave a brute-force proof of this theorem. In , Kalau and Walze proved this theorem in the normal coordinates system simultaneously. And then, Ackermann proved that the Wodzicki residue in turn is essentially the second coefficient of the heat kernel expansion of in . On the other hand, Fedosov et al. defined a noncommutative residue on Boutet de Monvel’s algebra and proved that it was a unique continuous trace in . In , Schrohe gave the relation between the Dixmier trace and the noncommutative residue for manifolds with boundary. For an oriented spin manifold with boundary , by the composition formula in Boutet de Monvel’s algebra and the definition of , should be the sum of two terms from interior and boundary of , where is an element in Boutet de Monvel’s algebra . It is well known that the gravitational action for manifolds with boundary is also the sum of two terms from interior and boundary of . Considering the Kastler-Kalau-Walze theorem for manifolds without boundary, then the term from interior is proportional to gravitational action from interior, so it is natural to hope to get the gravitational action for manifolds with boundary by computing . Based on the motivation, Wang proved a Kastler-Kalau-Walze type theorem for 4-dimensional spin manifolds with boundary where is the canonical volume of . Furthermore, Wang found a Kastler-Kalau-Walze type theorem for higher dimensional manifolds with boundary and generalized the definition of lower-dimensional volumes in to manifolds with boundary. For 5-dimensional spin manifolds with boundary , Wang got and for 6-dimensional spin manifolds with boundary, In order to get the boundary term, we computed the lower-dimensional volume for 6-dimensional spin manifolds with boundary associated with and in and obtained the volume with the boundary term where is the extrinsic curvature. In , Wang proved a Kastler-Kalau-Walze type theorem for general form perturbations and the conformal perturbations of Dirac operators for compact manifolds with or without boundary. Let be 4-dimensional compact manifolds with the boundary and let be a general differential form on , from Theorem 10 in ; then Recently, we computed for -dimensional spin manifolds with boundary in case of . In the present paper, we will restrict our attention to the case of . We compute for 7-dimensional manifolds with boundary. Our main result is as follows. Main Theorem. The following identity for 7-dimensional manifolds with boundary holds: where and are, respectively, scalar curvatures on and . Compared with the previous results, up to the extrinsic curvature, the scalar curvature on and the scalar curvature on appear in the boundary term. This case essentially makes the whole calculations more difficult, and the boundary term is the sum of fifteen terms. As in computations of the boundary term, we will consider some new traces of multiplication of Clifford elements. And the inverse 4-order symbol of the Dirac operator and higher derivatives of -1-order and -3-order symbols of the Dirac operators will be extensively used. This paper is organized as follows. In Section 2, we define lower-dimensional volumes of compact Riemannian manifolds with boundary. In Section 3, for 7-dimensional spin manifolds with boundary and the associated Dirac operators, we compute and get a Kastler-Kalau-Walze type theorem in this case. 2. Lower-Dimensional Volumes of Spin Manifolds with Boundary In this section, we consider an -dimensional oriented Riemannian manifold with boundary equipped with a fixed spin structure. We assume that the metric on has the following form near the boundary: where is the metric on . Let be a collar neighborhood of which is diffeomorphic . By the definition of and , there exists such that and for some sufficiently small . Then, there exists a metric on which has the form on such that . We fix a metric on the such that . Let us give the expression of Dirac operators near the boundary. Set and , where are orthonormal basis of . Let denote the Levi-Civita connection about . In the local coordinates and the fixed orthonormal frame , the connection matrix is defined by The Dirac operator is defined by By Lemma 6.1 in and Propositions 2.2 and 2.4 in , we have the following lemma. Lemma 1. Let and be a Riemannian manifold with the metric . For vector fields and in , then Denote ; then we obtain the following lemma. Lemma 2. The following identity holds: Others are zeros. By Lemma 2, we have the following definition. Definition 3. The following identity holds in the coordinates near the boundary: To define the lower-dimensional volume, some basic facts and formulae about Boutet de Monvel’s calculus which can be found in Section 2 in are needed. Denote by the Fourier transformation and (similarly, define , where denotes the Schwartz space and We define and which are orthogonal to each other. We have the following property: iff which has an analytic extension to the lower (upper) complex half-plane such that, for all nonnegative integers , as , . Let be the space of all polynomials and let ; . Denote by , respectively, the projection on . For calculations, we take rational functions having no poles on the real axis} ( is a dense set in the topology of ). Then, on , where is a Jordan close curve which included surrounding all the singularities of in the upper half-plane and . Similarly, define on : So, . For , and for , . Let be an -dimensional compact oriented manifold with boundary . Denote by Boutet de Monvel’s algebra; we recall the main theorem in . Theorem 4 (Fedosov-Golse-Leichtnam-Schrohe). Let and be connected, let , and let , and denote by , , and the local symbols of , and , respectively. Define Then, (a) , for any ; (b) it is a unique continuous trace on . Let and be nonnegative integers and let . Then, by Section 2.1 of , we have the following definition. Definition 5. Lower-dimensional volumes of spin manifolds with boundary are defined by Denote by the -order symbol of an operator . An application of (2.1.4) in shows that where and the sum is taken over , , . 3. A Kastler-Kalau-Walze Type Theorem for 7-Dimensional Spin Manifolds with Boundary In this section, we compute the lower-dimensional volume for 7-dimensional compact manifolds with boundary and get a Kastler-Kalau-Walze type formula in this case. From now on, we always assume that carries a spin structure so that the spinor bundle and the Dirac operator are defined on . The following proposition is the key of the computation of lower-dimensional volumes of spin manifolds with boundary. Proposition 6 (see ). The following identity holds: Nextly, for 7-dimensional spin manifolds with boundary, we compute . By Proposition 6, for 7-dimensional compact manifolds with boundary, we have Recall the Dirac operator of Definition 3. Write By the composition formula of pseudodifferential operators, then we have Thus, we get Lemma 7. Consider the symbol of the Dirac operator where Since is a global form on , so for any fixed point , we can choose the normal coordinates of in (not in ) and compute in the coordinates and the metric . The dual metric of on is . Write and ; then, Let be an orthonormal frame field in about which is parallel along geodesics and ; then, is the orthonormal frame field in about . Locally, . Let be the orthonormal basis of . Take a spin frame field such that , where is a double covering; then, is an orthonormal frame of . In the following, since the global form is independent of the choice of the local frame, we can compute in the frame . Let be the canonical basis of and let be the Clifford action. By , then then, we have in the above frame. By Lemma 2.2 in , we have the following. Lemma 8. With the metric on near the boundary, where . Then, the following lemma is introduced. Lemma 9. The following identity holds: Proof. From Lemma 5.7 in , we have Then, we obtain . Lemma 10. Let be the metric on 7-dimensional spin manifolds near the boundary; then, Proof. From Lemma 2.3 in , we have Let , , and . Then, When , , Similarly, when , , or , , . When , . On the other hand, from definitions (10) and (11), then When , , Similarly, when , , . When , , . When , Lemma 11. When , When , Next, we can compute (see formula (23) for definition of ). Since the sum is taken over , , then we have that is the sum of the following fifteen cases. Case 2. Consider , , , , and . From (23), we have By Lemma 7, a simple computation shows By (18) and the Cauchy integral formula, then Similarly, we obtain From (51) and (53), we get Note that ; then, from (50), (54), and direct computations, we obtain Therefore, where is the canonical volume of .

This challenge extends the Plants investigation so now four or more children are involved. The letters in the following addition sum represent the digits 1 ... 9. If A=3 and D=2, what number is represented by "CAYLEY"? Place six toy ladybirds into the box so that there are two ladybirds in every column and every row. First Connect Three game for an adult and child. Use the dice numbers and either addition or subtraction to get three numbers in a straight line. This challenging activity involves finding different ways to distribute fifteen items among four sets, when the sets must include three, four, five and six items. Can you find which shapes you need to put into the grid to make the totals at the end of each row and the bottom of each column? Place the numbers 1 to 10 in the circles so that each number is the difference between the two numbers just below it. How could you put eight beanbags in the hoops so that there are four in the blue hoop, five in the red and six in the yellow? Can you find all the ways of doing this? Five numbers added together in pairs produce: 0, 2, 4, 4, 6, 8, 9, 11, 13, 15 What are the five numbers? Can you put the numbers 1 to 8 into the circles so that the four calculations are correct? This problem is based on a code using two different prime numbers less than 10. You'll need to multiply them together and shift the alphabet forwards by the result. Can you decipher the code? Find the values of the nine letters in the sum: FOOT + BALL = GAME Can you use the information to find out which cards I have used? Make your own double-sided magic square. But can you complete both sides once you've made the pieces? Write the numbers up to 64 in an interesting way so that the shape they make at the end is interesting, different, more exciting ... than just a square. Using the statements, can you work out how many of each type of rabbit there are in these pens? Place the digits 1 to 9 into the circles so that each side of the triangle adds to the same total. Sweets are given out to party-goers in a particular way. Investigate the total number of sweets received by people sitting in different positions. Using 3 rods of integer lengths, none longer than 10 units and not using any rod more than once, you can measure all the lengths in whole units from 1 to 10 units. How many ways can you do this? Can you find all the ways to get 15 at the top of this triangle of numbers? Can you substitute numbers for the letters in these sums? Who said that adding couldn't be fun? Lolla bought a balloon at the circus. She gave the clown six coins to pay for it. What could Lolla have paid for the balloon? Katie had a pack of 20 cards numbered from 1 to 20. She arranged the cards into 6 unequal piles where each pile added to the same total. What was the total and how could this be done? Add the sum of the squares of four numbers between 10 and 20 to the sum of the squares of three numbers less than 6 to make the square of another, larger, number. Place this "worm" on the 100 square and find the total of the four squares it covers. Keeping its head in the same place, what other totals can you make? What do you notice about the date 03.06.09? Or 08.01.09? This challenge invites you to investigate some interesting dates A group of children are using measuring cylinders but they lose the labels. Can you help relabel them? Ten cards are put into five envelopes so that there are two cards in each envelope. The sum of the numbers inside it is written on each envelope. What numbers could be inside the envelopes? This is an adding game for two players. There are 44 people coming to a dinner party. There are 15 square tables that seat 4 people. Find a way to seat the 44 people using all 15 tables, with no empty places. This task, written for the National Young Mathematicians' Award 2016, invites you to explore the different combinations of scores that you might get on these dart boards. A game for 2 players. Practises subtraction or other maths What do the digits in the number fifteen add up to? How many other numbers have digits with the same total but no zeros? Exactly 195 digits have been used to number the pages in a book. How many pages does the book have? Three children are going to buy some plants for their birthdays. They will plant them within circular paths. How could they do this? If you take a three by three square on a 1-10 addition square and multiply the diagonally opposite numbers together, what is the difference between these products. Why? An investigation involving adding and subtracting sets of consecutive numbers. Lots to find out, lots to explore. There are 78 prisoners in a square cell block of twelve cells. The clever prison warder arranged them so there were 25 along each wall of the prison block. How did he do it? There are 4 jugs which hold 9 litres, 7 litres, 4 litres and 2 litres. Find a way to pour 9 litres of drink from one jug to another until you are left with exactly 3 litres in three of the Use your addition and subtraction skills, combined with some strategic thinking, to beat your partner at this game. This task, written for the National Young Mathematicians' Award 2016, involves open-topped boxes made with interlocking cubes. Explore the number of units of paint that are needed to cover the boxes. . . . Here you see the front and back views of a dodecahedron. Each vertex has been numbered so that the numbers around each pentagonal face add up to 65. Can you find all the missing numbers? Can you arrange 5 different digits (from 0 - 9) in the cross in the Find the sum and difference between a pair of two-digit numbers. Now find the sum and difference between the sum and difference! What happens? This dice train has been made using specific rules. How many different trains can you make? If each of these three shapes has a value, can you find the totals of the combinations? Perhaps you can use the shapes to make the Cherri, Saxon, Mel and Paul are friends. They are all different ages. Can you find out the age of each friend using the Can you explain the strategy for winning this game with any target? This challenge focuses on finding the sum and difference of pairs of two-digit numbers.

PDE Geometric Analysis seminar The seminar will be held in room B115 of Van Vleck Hall on Mondays from 3:30pm - 4:30pm, unless indicated otherwise. Seminar Schedule Spring 2012 Yao Yao (UCLA) Degenerate diffusion with nonlocal aggregation: behavior of solutions The Patlak-Keller-Segel (PKS) equation models the collective motion of cells which are attracted by a self-emitted chemical substance. While the global well-posedness and finite-time blow up criteria are well known, the asymptotic behaviors of solutions are not completely clear. In this talk I will present some results on the asymptotic behavior of solutions when there is global existence. The key tools used in the paper are maximum-principle type arguments as well as estimates on mass concentration of solutions. This is a joint work with Inwon Kim. Xuan Hien Nguyen (Iowa State) Gluing constructions for solitons and self-shrinkers under mean curvature flow In the 1990s, Kapouleas and Traizet constructed new examples of minimal surfaces by desingularizing the intersection of existing ones with Scherk surfaces. Using this idea, one can find new examples of self-translating solutions for the mean curvature flow asymptotic at infinity to a finite family of grim reaper cylinders in general position. Recently, it has been shown that it is possible to desingularize the intersection of a sphere and a plane to obtain a family of self-shrinkers under mean curvature flow. I will discuss the main steps and difficulties for these gluing constructions, as well as open problems. Nestor Guillen (UCLA) We consider the Monge-Kantorovich problem, which consists in transporting a given measure into another "target" measure in a way that minimizes the total cost of moving each unit of mass to its new location. When the transport cost is given by the square of the distance between two points, the optimal map is given by a convex potential which solves the Monge-Ampère equation, in general, the solution is given by what is called a c-convex potential. In recent work with Jun Kitagawa, we prove local Holder estimates of optimal transport maps for more general cost functions satisfying a "synthetic" MTW condition, in particular, the proof does not really use the C^4 assumption made in all previous works. A similar result was recently obtained by Figalli, Kim and McCann using different methods and assuming strict convexity of the target. Charles Smart (MIT) PDE methods for the Abelian sandpile Abstract: The Abelian sandpile growth model is a deterministic diffusion process for chips placed on the $d$-dimensional integer lattice. One of the most striking features of the sandpile is that it appears to produce terminal configurations converging to a peculiar lattice. One of the most striking features of the sandpile is that it appears to produce terminal configurations converging to a peculiar fractal limit when begun from increasingly large stacks of chips at the origin. This behavior defied explanation for many years until viscosity solution theory offered a new perspective. This is joint work with Lionel Levine and Wesley Pegden. Vlad Vicol (University of Chicago) Title: Shape dependent maximum principles and applications Abstract: We present a non-linear lower bound for the fractional Laplacian, when evaluated at extrema of a function. Applications to the global well-posedness of active scalar equations arising in fluid dynamics are discussed. This is joint work with P. Constantin. Jiahong Wu (Oklahoma State) "The 2D Boussinesq equations with partial dissipation" The Boussinesq equations concerned here model geophysical flows such as atmospheric fronts and ocean circulations. Mathematically the 2D Boussinesq equations serve as a lower-dimensional model of the 3D hydrodynamics equations. In fact, the 2D Boussinesq equations retain some key features of the 3D Euler and the Navier-Stokes equations such as the vortex stretching mechanism. The global regularity problem on the 2D Boussinesq equations with partial dissipation has attracted considerable attention in the last few years. In this talk we will summarize recent results on various cases of partial dissipation, present the work of Cao and Wu on the 2D Boussinesq equations with vertical dissipation and vertical thermal diffusion, and explain the work of Chae and Wu on the critical Boussinesq equations with a logarithmically singular velocity. Joana Oliveira dos Santos Amorim (Universit\'e Paris Dauphine) "A geometric look on Aubry-Mather theory and a theorem of Birkhoff" Given a Tonelli Hamiltonian $H:T^*M \lto \Rm$ in the cotangent bundle of a compact manifold $M$, we can study its dynamic using the Aubry and Ma\~n\'e sets defined by Mather. In this talk we will explain their importance and give a new geometric definition which allows us to understand their property of symplectic invariance. Moreover, using this geometric definition, we will show that an exact Lipchitz Lagrangian manifold isotopic to a graph which is invariant by the flow of a Tonelli Hamiltonian is itself a graph. This result, in its smooth form, was a conjecture of Birkhoff. Gui-Qiang Chen (Oxford) "Nonlinear Partial Differential Equations of Mixed Type" Many nonlinear partial differential equations arising in mechanics and geometry naturally are of mixed hyperbolic-elliptic type. The solution of some longstanding fundamental problems in these areas greatly requires a deep understanding of such nonlinear partial differential equations of mixed type. Important examples include shock reflection-diffraction problems in fluid mechanics (the Euler equations), isometric embedding problems in in differential geometry (the Gauss-Codazzi-Ricci equations), among many others. In this talk we will present natural connections of nonlinear partial differential equations with these longstanding problems and will discuss some recent developments in the analysis of these nonlinear equations through the examples with emphasis on identifying/developing mathematical approaches, ideas, and techniques to deal with the mixed-type problems. Further trends, perspectives, and open problems in this direction will also be addressed. This talk will be based mainly on the joint work correspondingly with Mikhail Feldman, Marshall Slemrod, as well as Myoungjean Bae and Dehua Wang. Jacob Glenn-Levin (UT Austin) We consider the Boussinesq equations, which may be thought of as inhomogeneous, incompressible Euler equations, where the inhomogeneous term is a scalar quantity, typically density or temperature, governed by a convection-diffusion equation. I will discuss local- and global-in-time well-posedness results for the incompressible 2D Boussinesq equations, assuming the density equation has nonzero diffusion and that the initial data belongs in a Besov-type space.

Problem 1 : A merchant has 120 liters and 180 liters of two kinds of oil. He wants to sell the oil by filling the two kinds in tins of equal volumes. Find the greatest volume of such a tin. The given two quantities 120 and 180 can be divided by 10, 20,... exactly. That is, both the kinds of oils can be sold in tins of equal volume of 10, 20,... liters. But, the target of the question is, the volume of oil filled in tins must be greatest. So, we have to find the largest number which exactly divides 120 and 180. That is the highest common factor (HCF) of (120, 180). HCF (120, 180) = 60 liters The 1st kind 120 liters is sold in 2 tins of of volume 60 liters in each tin. The 2nd kind 180 liters is sold in 3 tins of volume 60 liters in each tin. Hence, the greatest volume of the tin is 60 liters. Problem 2 : Find the least number of square tiles by which the floor of a room of dimensions 16.58 m and 8.32 m can be covered completely. We require the least number of square tiles. So, each tile must be of maximum dimension. To get the maximum dimension of the tile, we have to find the largest number which exactly divides 16.58 and 8.32. That is the highest common factor (HCF) of (16.58, 8.32). To convert meters into centimeters, we have to multiply by 100. 16.58 ⋅ 100 = 1658 cm 8.32 ⋅ 100 = 832 cm HCF (1658, 832) = 2 cm Hence the side of the square tile is 2 cm. Required no. of tiles : = (Area of the floor)/(Area of a square tile) = (1658 ⋅ 832)/(2 ⋅ 2) Hence, the least number of square tiles required is 344,864. Problem 3 : A wine seller had three types of wine. 403 liters of 1st kind, 434 liters of 2nd kind and 465 liters of 3rd kind. Find the least possible number of casks of equal size in which different types of wine can be filled without mixing. For the least possible number of casks of equal size, the size of each cask must be of the greatest volume. To get the greatest volume of each cask, we have to find the largest number which exactly divides 403, 434 and 465. That is the highest common factor (HCF) of (403, 434, 465). HCF (403, 434, 465) = 31 liters Each cask must be of the volume 31 liters. Required number casks : = 403/31 + 434/31 + 465/31 = 13 + 14 + 15 Hence, the least possible number of casks of equal size required is 42. Problem 4 : Six bells commence tolling together and toll at intervals of 2, 4, 6, 8 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together ? (excluding the one at start) For example, let the two bells toll after every 3 and 4 seconds respectively. Then the first bell tolls after every 3, 6, 9, 12 seconds... Like this, the second bell tolls after every 4, 8, 12 seconds... So, if the two bell toll together now, again they will toll together after 12 seconds. This 12 is the least common multiple (LCM) of 3 and 4. The same thing happened in our problem. To find the time, when they will all toll together, we have to find the LCM of (2, 4, 8, 6, 10, 12). LCM (2, 4, 8, 6, 10, 12) is 120 That is, 120 seconds or 2 minutes. So, after every two minutes, all the bell will toll together. For example, in 10 minutes, they toll together : 10/2 = 5 times That is, after 2, 4, 6, 8, 10 minutes. It does not include the one at the start. Similarly, in 30 minutes, they toll together : = 15 times (excluding one at the start). Problem 5 : The traffic lights at three different road crossings change after every 48 sec, 72 sec and 108 sec respectively. If they all change simultaneously at 8:20:00 hours, when will they again change simultaneously ? For example, let the two signals change after every 3 secs and 4 secs respectively. Then the first signal changes after 3, 6, 9, 12 seconds... Like this, the second signal changes after 4, 8, 12 seconds... So, if the two signals change simultaneously now, again they will change simultaneously after 12 seconds. This 12 is the least common multiple (LCM) of 3 and 4. The same thing happened in our problem. To find the time, when they will all change simultaneously, we have to find the LCM of (48, 72, 108). LCM (48, 72, 108) = 432 seconds or 7 min 12 sec So, after every 7 min 12 sec, all the signals will change simultaneously. At 8:20:00 hours, if all the three signals change simultaneously, again they will change simultaneously after 7 min 12 sec. That is at 8:27:12 hours. Hence, three signals will change simultaneously at 8:27:12 seconds. Problem 6 : Find the least number of soldiers in a regiment such that they stand in rows of 15, 20, 25 and form a perfect square. To answer this question, we have to find the least number which is exactly divisible by the given numbers 15, 20 and 25. That is the least common multiple of (15, 20, 25). LCM (15, 20, 25) = 300 So, we need 300 soldiers such that they stand in rows of 15, 20 , 25. But, it has to form a perfect square (as per the question). To form a perfect square, we have to multiply 300 by some number such that it has to be a perfect square. To make 300 as perfect square, we have to multiply 300 by 3. Then, it is 900 which is a perfect square. Hence, the least number of soldiers required is 900. Kindly mail your feedback to email@example.com We always appreciate your feedback.

Geodesy, also called geodetics, is the scientific discipline that deals with the measurement and representation of the earth, its gravitational field and geodynamic phenomena (polar motion, earth tides, and crustal motion) in three-dimensional time varying space. Geodesy is primarily concerned with positioning and the gravity field and geometrical aspects of their temporal variations, although it can also include the study of the Earth's magnetic field. Wolfgang Torge quotes in his 2001 textbook Geodesy (3rd edition) Friedrich Robert Helmert as defining geodesy as "the science of the measurement and mapping of the earth's surface." As Torge also remarks, the shape of the earth is to a large extent the result of its gravity field. This applies to the solid surface (orogeny; few mountains are higher than 10 km, few deep sea trenches deeper than that). It affects similarly the liquid surface (dynamic sea surface topography) and the earth's atmosphere. For this reason, the study of the Earth's gravity field is seen as a part of geodesy, called physical geodesy. The figure of the Earth Primitive ideas about the figure of the Earth, still found in young children, hold the Earth to be flat, and the heavens a physical dome spanning over it. Already the ancient Greeks were aware of the spherical shape of the Earth. Lunar eclipses, e.g., always have a circular edge of appox. three times the radius of the lunar disc; as these always happen when the Earth is between Sun and Moon, it suggests that the object casting the shadow is the Earth and must be spherical (and four times the size of the Moon, the lunar and solar discs being the same size). Also an astronomical event like a lunar eclipse which happened high in the sky in one end of the Mediterranean world, was close to the horizon in the other end, also suggesting curvature of the Earth's surface. Finally, Eratosthenes determined a remarkably accurate value for the radius of the Earth at around 200 BC. The Renaissance brought the invention of the telescope and the theodolite, making possible triangulation and grade measurement . Of the latter especially should be mentioned the expedition by the French Academy of Sciences to determine the flattening of the Earth. One expedition was sent to Lapland as far North as possible under Pierre Louis Maupertuis (1736-37), the other under Pierre Bouguer was sent to Peru, near the equator (1735-44). At the time there were two competing theories on the precise figure of the Earth: Isaac Newton had calculated that, based on his theory of gravitation, the Earth should be flattened at the poles to a ratio of 1:230. On the other hand the astronomer Jean Dominique Cassini held the view that the Earth was elongated at the poles. Measuring the length, in linear units, of a degree of change in north-south direction of the astronomical vertical, at two widely differing latitudes would settle the issue: on a flattened Earth the length of a degree grows toward the poles. The flattening found by comparing the results of the two grade measurement expeditions confirmed that the Earth was flattened, the ratio found being 1:210. Thus the next approximation to the true figure of the Earth after the sphere became the flattened, biaxial ellipsoid of revolution . In South America Bouguer noticed, as did George Everest in India, that the astronomical vertical tended to be "pulled" in the direction of large mountain ranges, obviously due to the gravitational attraction of these huge piles of rock. As this vertical is everywhere perpendicular to the idealized surface of mean sea level, or the geoid, this means that the figure of the Earth is even more irregular than an ellipsoid of revolution. Thus the study of the "undulations of the geoid" became the next great undertaking in the science of studying the figure of the Earth. Geoid and reference ellipsoid The geoid is essentially the figure of the Earth abstracted from its topographic features. It is an idealized equilibrium surface of sea water, the mean sea level surface in the absence of currents, air pressure variations etc. and continued under the continental masses. The geoid, unlike the ellipsoid, is irrgular and too complicated to serve as the computational surface on which to solve geometrical problems like point positioning. The geometrical separation between it and the reference ellipsoid is called the geoidal undulation. It varies globally between 110 m. A reference ellipsoid, customarily chosen to be the same size (volume) as the geoid, is described by its semi-major axis (equatorial radius) a and flattening f. The quantity f = (a - b) / a, where b is the semi-minor axis (polar radius) is a purely geometrical one. The mechanical ellipticity of the earth (dynamical flattening) is determined by observation and differs from the geometrical because the earth is not of uniform density. The 1980 Geodetic Reference System (GRS80) posited a 6,378,137 m semi-major axis and a 1:298.257 flattening. This system was adopted at the XVII General Assembly of the International Union of Geodesy and Geophysics (IUGG). It is essentially the basis for geodetic positioning by the Global Positioning System and is thus in extremely widespread use also outside the geodetic community. The numerous other systems which have been used by diverse countries for their maps and charts are gradually dropping out of use as more and more countries move to global, geocentric reference systems using the GRS80 reference ellipsoid. Co-ordinate systems in space The locations of points in three-dimensional space are most conveniently described by three cartesian or rectangular coordinates, X,Y and Z. Since the advent of satellite positioning, such coordinate sytems are typically geocentric: the Z axis is aligned with the Earth's (conventional or instantaneous) rotation axis. Before the satellite geodesy era, the coordinate systems associated with geodetic datums attempted to be geocentric, but their origins differed from the geocentre by hundreds of metres, due to regional deviations in the direction of the plumbline (vertical). These regional geodetic datums, such as ED50 (European Datum 1950) or NAD83 (North American Datum 1983) have ellipsoids associated with them that are regional 'best fits' to the geoids within their areas of validity, minimising the deflections of the vertical over these areas. It is only because GPS satellites orbit about the geocentre, that this point becomes naturally the origin of a coordinate system defined by satellite geodetic means, as the satellite positions in space are themselves computed in such a system. Geocentric co-ordinate systems used in geodesy can be divided naturally into two classes: - Inertial reference systems, where the co-ordinate axes retain their orientation relative to the fixed stars, or equivalently, to the rotation axes of ideal gyroscopes; the X axis points to the vernal equinox - Co-rotating, also ECEF ("Earth Centred, Earth Fixed"), where the axes are attached to the solid body of the Earth. The X axis lies within the Greenwich observatory's meridian plane. The co-ordinate transformation between these two systems is described to good approximation by (apparent) sidereal time. A more accurate description takes also length-of-day variations and polar motion into account, phenomena currently closely monitored by geodesists. Co-ordinate systems in the plane - Plano-polar, in which points in a plane are defined by a distance s from a specified point along a ray having a specified direction α with respect to a base line or axis; - Rectangular, points are defined by distances from two perpendicular axes called x and y. It is geodetic practice -- contrary to the mathematical convention -- to let the x axis point to the North and the y axis to the East. Rectangular co-ordinates in the plane can be used intuitively with respect to one's current location, in which case the x axis will point to the local North. More formally, such co-ordinates can be obtained from three-dimensional co-ordinates using the artifice of a map projection. It is not possible to map the curved surface of the Earth onto a flat map surface without deformation. The compromise most often chosen -- called a conformal projection -- preserves angles and length ratios, so that small circles are mapped as small circles and small squares as squares. An example of such a projection is UTM (Universal Transverse Mercator). Within the map plane, we have rectangular co-ordinates x and y. In this case the North direction used for reference is the map North, not the local North. The difference between the two is called meridian convergence. It is easy enough to "translate" between polar and rectangular co-ordinates in the plane: let, as above, direction and distance be α and s respectively, then we have The reverse translation is slightly more tricky. In geodesy, point or terrain heights are "above sea level", an irregular, physically defined surface. Therefore a height should ideally not be referred to as a co-ordinate. It is more like a physical quantity, and though it can be tempting to treat height as the vertical coordinate z, in addition to the horizontal co-ordinates x and y, and though this actually is a good approximation of physical reality in small areas, it becomes quickly invalid in larger areas. Heights come in the following variants: - Orthometric height s - Normal height s - Geopotential number s Each have their advantages and disadvantages. Both orthometric and normal heights are heights in metres above sea level, which geopotential numbers are measures of potential energy (unit: m2s - 2) and not metric. Orthometric and normal heights differ in the precise way in which mean sea level is conceptually continued under the continental masses. The reference surface for orthometric heights is the geoid, an equipotential surface approximating mean sea level. None of these heights are in any way related to geodetic or ellipsoidial heights, which express the height of a point above the reference ellipsoid. Satellite positioning receivers typically provide ellipsoidal heights, unless they are fitted with special conversion software based on a model of the geoid. Because geodetic point c-oordinates (and heights) are always obtained in a system that has been constructed itself using real observations, we have to introduce the concept of a geodetic datum: a physical realization of a co-ordinate system used for describing point locations. The realization is the result of choosing conventional co-ordinate values for one or more datum points. In the case of height datums, it suffices to choose one datum point: the reference bench mark, typically a tide gauge at the shore. Thus we have vertical datums like the NAP (Normaal Amsterdams Peil), the North American Vertical Datum 1988 (NAVD88), the Kronstadt datum, the Trieste datum, etc. In case of plane or spatial coordinates, we typically need several datum points. A regional, ellipsoidal datum like ED50 can be fixed by prescribing the undulation of the geoid and the deflection of the vertical in one datum point, in this case the Helmert Tower in Potsdam. However, an overdetermined ensemble of datum points can also be used. Changing the coordinates of a point set referring to one datum, to make them refer to another datum, is called a datum transformation. In the case of vertical datums, this consists of simply adding a constant shift to all height values. In the case of plane or spatial coordinates, datum transformation takes the form of a similarity or Helmert transformation, consisting of a rotation and scaling operation in addition to a simple translation. In the plane, a Helmert transformation has four parameters, in space, seven. A note on terminology In the abstract, a co-ordinate system as used in mathematics and geodesy is, e.g., in ISO terminology, referred to as a coordinate system. International geodetic organizations like the IERS (International Earth Rotation and Reference Systems Service) speak of a reference system. When these co-ordinates are realized by choosing datum points and fixing a geodetic datum, ISO uses the terminology coordinate reference system, while IERS speaks of a reference frame. A datum transformation again is referred to by ISO as a coordinate transformation. (ISO 19111: Spatial referencing by coordinates). Point positioning is the determination of the coordinates of a point on land, at sea, or in space with respect to a coordinate system. Point position is solved by compution from measurements linking the known positions of terrestrial or extraterrestrial points with the unknown terrestrial position. This may involve transformations between or among astronomical and terrestrial coordinate systems. The known points used for point positioning can be, e.g., triangulation points of a higher order network, or GPS satellites. Traditionally, a hierarchy of networks has been built to allow point positioning within a country. Highest in the hierarchy were triangulation networks. These were densified into networks of traverse s (polygons), into which local mapping surveying measurements, usually with measuring tape, corner prism and the familiar red and white poles, are tied. Nowadays all but special measurements (e.g., underground or high precision engineering measurements) are performed with GPS. The higher order networks are measured with static GPS , using differential measurement to determine vectors between terrestrial points. These vectors are then adjusted in traditional network fashion. A global polyhedron of permanently operating GPS stations under the auspices of the IERS is used to define a single global, geocentric reference frame which serves as the "zeroth order" global reference to which national measurements are attached. For surveying mappings, frequently Real Time Kinematic GPS is employed, tying in the unknown points with known terrestrial points close by in real time. One purpose of point positioning is the provision of known points for mapping measurements, also known as (horizontal and vertical) control. In every country, thousands of such known points exist in the terrain and are documented by the national mapping agencies. Constructors and surveyors involved in real estate will use these to tie their local measurements to. In geometric geodesy we formulate two standard problems: the geodetic principal problem and the geodetic inverse problem. - Geodetic principal problem (also: first geodetic problem) - Given a point (in terms of its coordinates) and the direction (azimuth) and distance from that point to a second point, determine (the co-ordinates of) that second point. - Geodetic inverse problem (also: second geodetic problem) - Given two points, determine the azimuth and length of the line (straight line, great circle or geodesic) that connects them. In the case of plane geometry (valid for small areas on the Earth's surface) the solutions to both problems reduce to simple trigonometry. On the sphere, the solution is significantly more complex, e.g., in the inverse problem the azimuths will differ between the two end points of the connecting great circle arc. On the ellipsoid of revolution , closed solutions do not exist; series expansions have been traditionally used that converge rapidly. In the general case, the solution is called the geodesic for the surface considered. It may be nonexistent or non-unique. The differential equations for the geodesic can be solved numerically, e.g., in MatLab(TM). Geodetic observational concepts Here we define some basic observational concepts, like angles and coordinates, defined in geodesy (and astronomy as well), mostly from the viewpoint of the local observer. - The plumbline or vertical is the direction of local gravity, or the line that results by following it. It is slightly curved. - The zenith is the point on the celestial sphere where the direction of the gravity vector in a point, extended upwards, intersects it. More correct is to call it a <direction> rather than a point. - The nadir is the opposite point (or rather, direction), where the direction of gravity extended downward intersects the (invisible) celestial sphere. - The celestial horizon is a plane perpendicular to a point's gravity vector. - Azimuth is the direction angle within the plane of the horizon, typically counted clockwise from the North (in geodesy) or South (in astronomy and France). - Elevation is the angular height of an object above the horizon, Alternatively zenith distance, being equal to 90 degrees minus elevation. - Local topocentric co-ordinates are azimut (direction angle within the plane of the horizon) and elevation angle (or zenith angle) as well as distance if known. - The North celestial pole is the extension of the Earth's (precessing and nutating) instantaneous spin axis extended Northward to intersect the celestial sphere. (Similarly for the South celestial pole.) - The celestial equator is the intersection of the (instantaneous) Earth equatorial plane with the celestial sphere. - A meridian plane is any plane perpendicular to the celestial equator and containing the celestial poles. - The local meridian is the plane containing the direction to the zenith and the direction to the celestial pole. Geodetic observing instruments The level is used for determining height differences and height reference systems, commonly referred to mean sea level. The traditional spirit level produces these practically most useful heights above sea level directly; the more economical use of GPS instruments for height determination requires precise knowledge of the figure of the geoid, as GPS only gives heights above the GRS80 reference ellipsoid. As geoid knowledge accumulates, one may expect use of GPS heighting to spread. The theodolite is used to measure horizontal and vertical angles to target points. These angles are referred to the local vertical. The tacheometer additionally determines, electronically or electro-optically, the distance to target, and is highly automated in its operations. The method of free station position is widely used. For local detail surveys, tacheometers are commonly employed although the old-fashioned rectangular technique using angle prism and steel tape is still an inexpensive alternative. More and more, also real time kinematic (RTK) GPS techniques are used. Data collected is tagged and recorded digitally for entry into a Geographic Information System (GIS) data base. Geodetic GPS receivers produce directly three-dimensional co-ordinates in a geocentric co-ordinate frame. Such a frame is, e.g., WGS84, or the frames that are regularly produced and published by the International Earth Rotation Service (IERS). GPS receivers have almost completely replaced terrestrial instruments for large-scale base network surveys. For planet-wide geodetic surveys, previously impossible, we can still mention satellite laser and Very Long Baseline Interferometer (VLBI) techniques. All these techniques also serve to monitor Earth rotation irregularities as well as plate tectonic motions. Gravity is measured using gravimeters . Common field gravimeters are spring based and referred to a relative. Absolute gravimeters, which nowadays can also be used in the field, are based directly on measuring the acceleration of free fall of a reflecting prism in a vacuum tube. Gravity surveys over large areas can serve to establish the figure of the geoid over these areas. Units and measures on the ellipsoid Geographical latitude and longitude are stated in the units degree, minute of arc, and second of arc. They are angles, not metric measures, and describe the direction of the local normal to the reference ellipsoid of revolution. This is approximately the same as the direction of the plumbline, i.e., local gravity, which is also the normal to the geoid surface. For this reason, astronomical position determination, measuring the direction of the plumbline by astronomical means, works fairly well provided an ellipsoidal model of the figure of the Earth is used. A geographic mile, defined as one minute of arc on the equator, equals 1,855.32571922 m. A nautical mile is one minute of astronomical latitude. The radius of curvature of the ellipsoid varies with latitude, being the longest at the pole and shortest at the equator as is the nautical mile. A metre was originally defined as the 40 millionth part of the length of a meridian. This means that a kilometre is equal to (1/40,000) * 360 * 60 meridional minutes of arc, which equals 0.54 nautical miles. Similarly a nautical mile is on average 1/0.54 = 1.85185... km. - The Geodesy Page. - Welcome to Geodesy

Authors of this Blog are Atharva Dandapur, Sonal Dangare, Tejas Dani, Arjun Desai. WHAT ARE BEAMS? Beams are important types of structural elements that play a key role in creating a safe load path to transfer the weight and forces on a structure to the foundations and into the ground. They are horizontal members which carry perpendicular loads in their longitudinal direction. They are used to support the weight of the floor, roofs of the building and transfer the load to the vertical load bearing member of the structure. Following are the loads acting on a BEAM: b. Varied by length c. Single point TYPES OF BEAMS: - Beams are classified according to various features such as: According to the End Support Conditions: · Simply Supported Beam · Cantilever beam · Continuous beam · Fixed end beam · Overhanging beam · Double overhanging beam Ø According to the Cross Section: According to Geometry · Straight beam · Curved Beam · Tapered Beam According to Casting Condition · In-situ casting beam · Precast concrete beam · Pre-stressed concrete beam SIMPLY SUPPORTED BEAMS: - The simply supported beam is one of the simplest structures. It features only two supports, one at each end. One is a pinned support and the other is a roller support. With this configuration, the beam is inhibited from any vertical movement at both ends whereas it is allowed to rotate freely. Due to the roller support it is also allowed to expand or contract axially, though free horizontal movement is prevented by the other support. Removing any of the supports inserting an internal hinge, would render the simply supported beam to a mechanism, that is body the moves without restriction in one or more directions. Obviously this is unwanted for a load carrying structure. Therefore, the simply supported beam offers no redundancy in terms of supports, and if a local failure occurs the whole structure would collapse. These type of structures that offer no redundancy are called critical or determinant structures. To the contrary, a structure that features more supports than required to restrict its free movements is called redundant or indeterminate structure. FINITE DIFFERENCE METHOD: - In numerical analysis, finite-difference methods (FDM) are a class of numerical techniques for solving differential equations by approximating derivatives with finite differences. Here , spatial domain and time interval are discretized, or broken into a finite number of steps, and the value of the solution at these discrete points is approximated by solving algebraic equations containing finite differences and values from nearby points. Finite difference methods convert ordinary differential equations (ODE) or partial differential equations (PDE), which may be solved by matrix algebra techniques. Today, FDM are one of the most common approaches to the numerical solution of PDE, along with finite element methods. We solve second-order ordinary differential equations of the form: Basic equation: The differential equation that governs the deflection of a simply supported beam under uniformly distributed load is: · x = Location along the beam (in) · E = Young’s modulus of elasticity of beam (psi) · I = Second moment of area (in⁴) · Q = Uniform loading intensity (lb/in) · L = Length of beam (in) 2. Solution of equation 3. Conversion of equation to Matrix 4. Applying boundary conditions 5. Find Error The deflection ‘y’ in a simply supported beam with a uniform load ‘q’ and a tensile axial load T is given by: · x = location along the beam (in) · T = tension applied (lbs) · E = young’s modulus of elasticity of beam (psi) · I = second moment of area (in⁴) · q = uniform load intensity (lb/in) · L = length of beam (in) T = 7200lbs, q = 5400lbs/in, L = 75in, E = 30Msi , I = 120 in⁴ 1. Find the deflection of beam at x=50”. Use a step size of △x = 25” and the approximate the derivatives by central divided difference approximation. 2. Find the relative true error in the calculation of y (50). d²y ∕ dx²-Ty / El = qx(L -X)/2EL d²y ∕ dx² -7200y/(30*10⁶)*(120) = (5400)*x*(75-x) / 30*10⁶*2*120 d²y ∕ dx² -2*10⁶y = 7.5*10^-2x*(75-x) Approximating the derivative at node i by CENTRAL DIVIDED DIFFERENCE approximation d² y/dx² =y(i+1) -2yi+ y(i-1) 1/(∆x)² Since △x=25, we have 4 nodes 1/((∆x)² ) y(i+1) -2yi+y(i-1) -2∗10⁶ yi=7.5∗10^(-7) xi∗(75-xi) Equations at nodes: Node 1- y1=0 …..(from simply supported boundary condition at x=0) Node 2 — (y3 — 2y2+ y1)/(25)² -2∗10^(-6) y2=7.5∗10^(-7) x2∗(75-x2 ) — 0.0016y1–0.003202y2+0.0016y3=9.375∗10^(-4) Node 3 — (y4 — 2y3+ y2)/(25)² -2∗10^(-6) y3=7.5∗10^(-7) x3∗(75-x3 )- 0.0016y2–0.003202y3+0.0016y4=9.375∗10^(-4) Node 4 — y4=0 …..(from simply supported boundary condition at x=75) Solving the equations we get, y(50)= y(x2) ≈ y2=-0.58521 yh= k1 e⁰.0014142x+ k2 e^(-0.0014142x) ……homogeneous part yp=Ax²+Bx+C …..particular part (d² yp)/dx² -2∗10⁶ yp=7.5∗10^(-7) x∗(75-x) (d² (Ax²+Bx+C ))/dx² -2∗10⁶ (Ax²+Bx+C )=7.5∗10^(-7) x∗(75-x) Equating the terms we get, Solving the equations, Applying the following boundary conditions, To calculate the RELATIVE TRUE ERROR, The true error is given by, Et = Exact Value — Approximate Value ; Et = 0.05320 The Relative Error is given by, MATLAB CODE: - A common usage is for things like solving Differential Equations numerically, and approximating derivatives for root finding and numerical optimization schemes. They are generally used to solve Partial Differential Equation (PDE) like- heat, diffusion, flow, and electromagnetics. From the graph we can see that as the nodes increases the value of error decreases. As the number of nodes increases the deflection between the true value and the approximate value decreases. Finite difference method is easy, simple but errors are more, thus to reduce these errors we need to increase the step size. Beam Deflection problems are accurately solved by Finite Difference Method.

IIT JAM 2022 Syllabus for Mathematical Statics is divided into two parts i.e. Mathematics and Statistics. The weightage of the mathematics section is 40% and of the statistics section, it is 60%. Candidates who are planning to appear for Mathematical Statistics can download the complete IIT JAM 2022 Syllabus for Mathematical Statics from the official website. Candidates need to analyze section-wise syllabus rigorously before appearing in the exam. Candidates should also note that the question paper will be divided into 3 sections and three types of questions will be asked in the paper i.e. MCQ, MSQ, and NAT. Read the article to know more about IIT JAM section-wise syllabus of mathematical statistics, exam pattern, best books, and preparation tips. Check IIT JAM 2022 Exam Pattern Read the complete article for detailed Mathematical Statics Syllabus, Preparation Tips and Important Books. IIT JAM Syllabus for Mathematical Statics There are two sections such as Mathematics and Statistics in IIT JAM Syllabus. Section-wise details for Mathematics and Statistics Syllabus for IIT JAM are tabulated below. IIT JAM Mathematics Syllabus |Sequences and Series||All the areas included in this section are Convergence of real numbers sequences, Comparison, root and ratio tests for convergence of series of real numbers.| |Differential Calculus||In this unit, all the subtopics covered are Limits, continuity and differentiability of functions of one and two variables. indeterminate forms, maxima and minima of functions of one and two variables. Apart from that, there are some theorems such as Rolle's theorem, mean value theorems, Taylor's theorem, etc.| |Integral Calculus||Under this section all the properties are Fundamental theorems of integral calculus; Double and triple integrals; applications of definite integrals, arc lengths, areas and volumes, etc| |Matrices||From Rank, inverse of a matrix; Systems of linear equations; Linear transformations, eigenvalues and eigenvectors to Cayley-Hamilton theorem; symmetric; skew-symmetric and orthogonal matrices are explained in this section.| IIT JAM Statistics Syllabus |Probability||This section covers the Axiomatic definition of probability and properties, conditional probability, and multiplication rule. It also covers the Theorem of total probability along with Bayes’ theorem and independence of events.| |Random Variables||There are various sub topics such as Probability mass function, density function and cumulative distribution functions along with distribution of a function of a random variable. Mathematical expectation, moments and moment generating function. Chebyshev's inequality in this unit.| |Standard Distributions||This section includes Binomial, negative binomial, geometric, Poisson and hypergeometric, uniform, exponential, gamma, beta, and normal distributions. Apart from that, there are Poisson and normal approximations of a binomial distribution.| |Joint Distributions||In this part, all the areas are Joint, marginal and conditional distributions; Distribution of functions of random variables; Joint moment generating function. It also covers Product moments, correlation, simple linear regression; Independence of random variables.| |Sampling distributions and Limit Theorems||In the first part, there are Chi-square along with t and F distributions, and their properties. And in the second part, all the areas are the Weak law of large numbers and the Central limit theorem that defines i.i.d. with finite variance case only.| |Estimation||This section consists of Unbiasedness, consistency, and efficiency of estimators, details of the method of moments as well as the method of maximum likelihood. Besides that, there is Sufficiency, factorization theorem also included. From Completeness, Rao-Blackwell and Lehmann-Scheffe theorems, uniformly minimum variance unbiased estimators to Rao-Cramer inequality; Confidence intervals for the parameters of univariate normal; two independents normal, and one parameter exponential distributions are there.| |Testing of Hypotheses||Here are some of the primary concepts and applications of Neyman-Pearson Lemma for testing simple and composite hypotheses along with Likelihood ratio tests for parameters of the univariate normal distribution, etc.| What are the Valuable Tips to Prepare IIT JAM Syllabus for Mathematics and Statistics? Some of the most salient principles for taking preparation for the subject Mathematics and Statistics of IIT JAM Syllabus are given below. - As the preparation depends on the primary concepts of the syllabus, so, candidate need to well aware of the entire syllabus - Make a study plan with all the subject matter of the syllabus for some months - Need to improve problem-solving skills as this section is based on the that - Centre of attention must be in probability, variable and also in distribution - Practice all the theorems and formula - Note down all the formulas one by one and keep practicing daily with the time table - Make important notes from all the theorems and mark them in bold - It is a must to solve all the questions from the previous year paper - Start giving the mock test as many times as possible and do not forget to solve the sample papers - It is a good practice to revise all the areas from the syllabus to remember Download IIT JAM Practice Papers IIT JAM Mathematics and Statistics Exam Pattern IIT JAM Mathematics and Statistics Exam Pattern is listed below in a table format: |Total Duration of the Test||3 hours| |Total number of questions||60| |Number of Section||3 (Sec A, B and C)| |Question Type for the Test||MCQ, MSQ and NAT| |Marks in Total||100| Section Wise Marks Division |Section||Number of Questions in Total||Question Wise Marks Distribution||Marks in Total| IIT JAM Statistics and Mathematics Books for Preparation Below are some recommended books in a table for preparation for the examination: |Mathematical Statistics||IIT-JAM: M.Sc. Mathematical Statistics Joint Admission Test||Anand Kumar| |Mathematics||A Complete Resource Manual M.Sc Mathematics Entrance examination||Suraj Singh and Reshmi Gupta| |Mathematical Statistics||Fundamentals of Mathematical Statistics||S.C. Gupta and V K Kapoor| |Mathematical Statistics||Introduction to Mathematical Statistics||Hogg| |Statistics||An Introduction to Probability and Statistics||Vijay K. Rohatgi and A.K. Md. Ehsanes Saleh| Also Check: IIT JAM Paper Analysis IIT JAM Syllabus for Mathematics and Statistics FAQ Ques. Are there any changes in IIT JAM Syllabus for Mathematics and Statistics? Ans. No, there are no changes in the Mathematics and Statistics Syllabus for IIT JAM. All the sections will remain the same as the previous year. Ques. What should I know as the basic mathematical concepts from IIT JAM Mathematics and Statistics Syllabus? Ans. Many sections fall under this section such as functions, maxima, and minima, vectors, matrices, integrals, determinants, etc. Ques. Which part is the most difficult part between Statistics and Mathematics in IIT JAM Syllabus? Ans. If you study in a proper way no part will be difficult for you to score well. Ques. Which section is the scoring one between Statistics and Mathematics in IIT JAM Syllabus? Ans. As per the expert advice, you can score maximum from both the section Mathematics and Statistics but the syllabus covers 60% area of Statistics and 40% of Mathematics. Ques. Can I give less importance to the Central limit theorem from the Statistics and Mathematics Syllabus in IIT JAM? Ans. No, you cannot avoid this section as this is one of the most important parts of Statistics. *The article might have information for the previous academic years, which will be updated soon subject to the notification issued by the University/College.

The half-life of a reaction is the time required for a reactant to reach one-half its initial concentration or pressure. For a first-order reaction, the half-life is independent of concentration and constant over time. Created by Jay. Want to join the conversation? - So, to clarify, the main point here is that no matter the initial concentration of a reactant, it will take the same amount of time for half of the reactant to be disappear?(11 votes) - I have a little confusion about first order reactions that produce products being dependent upon the concentration of the reactant, (i.e. if you double the reactant in a first order reaction, you double the amount of product produced), while the half life decay of a first order reaction that produces a product (i.e. half the initial concentration of the reactant) is not dependent upon the initial concentration. Could you please explain these two differences when you get a chance? By the way, these videos have been extremely helpful and I appreciate all the hard work you and Khan Academy have put into making these materials available to anyone, anytime, anyplace.(10 votes) - In earlier videos we see the rate law for a first-order reaction R=k[A], where [A] is the concentration of the reactant. If we were to increase or decrease this value, we see that R (the rate of the reaction) would increase or decrease as well. When dealing with half-life, however, we are working with k (the rate constant). While the rate of reaction is measured in units molar/second, a rate constant for a first-order reaction is 1/(second).(4 votes) - Why would we need to know about the half-time ? and where does the symbol t 1/2come from ?(4 votes) - when you know the half life of a rxn, then you can determine the amount of product formed and we all know time is 't' 1/2 is half, so t1/2 is just a symbol for half life.(6 votes) - Why is the function of half life exponential. And the function in the previous video is a straight line. If both are first-order reaction.(3 votes) - If you graph half life data you get an exponential decay curve. It’s kind of the definition of it. If you graph something that starts at 100 and decays by half every 1 minute, 50 by minute 2, 25 by 3, 12.5 by 4, 6.25 by 5 etc. you’ll see. Go back to the previous video and look at the label on the y axis, then compare with the y axis on this one. A quick method for working out the reaction order is to plot [A] vs t, ln[A] vs t and 1/[A] vs t, one of them will give a straight line which tells you the order See: https://www.chem.purdue.edu/gchelp/howtosolveit/Kinetics/IntegratedRateLaws.html(3 votes) - Does half life increase or decrease(1 vote) - Is the time you get from the half life always seconds?(1 vote) - why is there e in the half life formula, i thought that half life was calculated using .5 as the exponential base?(1 vote) - any exponential decay can be written with any base you want (check your properties of logarithms) e is common but 1/2 is also good for half-life. e is common because it is much easier to deal with when you want to manipulate these equations with higher math, like calculus. When you study calculus you will see why.(3 votes) - At05:23he said that k is a constant. I know that k is a constant, but it has a different value depending on the ordo of the reaction (like if the ordo is one, the unit of k would be 1/s) Is that alright? Or am I get it wrong? Can someone help me?(1 vote) - So each chemical reaction has it's own rate constant, which indeed is constant, at a specific temperature. So a reaction like A -> B will have a rate constant associated with the rate law of the reaction, but we have to specify the temperature of the reaction. This is because the rate constant for the same reaction at say 300K will be different in value than the rate constant at 400K. Now this is juts talking about the value of a rate constant, but what I think what you mean are the units of the rate constant the same? In which case then no the units of the rate constant will be different depending on the overall order of the reaction. So a first order reaction's rate constant will indeed be using units of 1/s (or s^(-1)) while the rate for a second order reaction will be in units of 1/(M*s) (or s^(-1)*M^(-1)). So a first order reaction and a second order reaction can have the same numerical value for their rate constants, but it's inaccurate to say that they are the same since they are using different units. I think I might be getting hung up on your use of different value as opposed to different unit, but I hope that made sense.(3 votes) - How can we tell if a reaction is first order?(1 vote) - We can tell what order a reaction is graphically by plotting the reactant's concentration versus time and seeing if it produces a linear curve. We essentially rearrange a reaction order's integrated rate to resemble a linear equation of the form: y=mx+b. For a zeroth order reaction: [A] = -kt +[A]0, the y variable is the reactant concentration or [A], the x variable is time or t. If plotting [A] versus t yields a straight line then the reaction is zeroth order. Additionally the slope, m, will be -k and the y-intercept, b, will be the initial concentration of the reactant. So if the reaction is truly zeroth order then it'll form a straight line modeled on the integrated rate law. This same logic follows for first, second, etc. order reactions too. If we plot [A] versus t and we do not find a straight line, then it's not zeroth order and will follow another order's integrated rate law. For a first order reaction: ln([A]) = -kt + ln([A]0), the y variable is now ln([A]) and the x variable is still time. If we tried plotting ln([A]) versus time and get a straight line now, then it's first order. The slope will be -k and the y-intercept will be ln([A]0). And if it's not first order, then it could second order which uses: 1/[A] = kt + 1/[A]0, with y being 1/[A] and x being time again. The slope would be k and the y-intercept would be 1/[A]0. If 1/[A] versus time produces a straight line, it's second order. If not we keep repeating this process for other reaction orders until we find a straight line. It's a nice trick in chemistry of fitting data to a straight line to see if there's a relationship between variables. Hope that helps.(2 votes) - How can half life be achieved, while a chemical reaction is actually going on?(1 vote) - The half life is the time required for half of a reactant to disappear. We can't measure it unless the reaction has actually gone on long enough.(2 votes) - [Voiceover] Here we have one form of the integrated rate law for a first order reaction. And we're gonna keep going with the math here, so we eventually will talk about the half-life. So over here on the left, the natural log of the concentration of A at time t minus the natural log of the initial concentration of A. That's the same thing as the natural log of the concentration of A over the initial concentration of A. So that's just the log property. And this is equal to negative kt, where k is your rate constant. Next, we need to get rid of our natural logs. So we're going to exponentiate both sides. So we're gonna take e to both sides here and that's gonna get rid of our natural log. So now, on the left side we have our concentration over the initial concentration. On the right side we have e to the negative kt. So we're gonna multiply both sides by the initial concentration of A. So we get that our concentration of A at time t is equal to the initial concentration of A times e to the negative kt. And now, it's a little bit easier to think about the graph. We can put the concentration of A on the y-axis and we can put time on the x-axis. And this is in the form of an exponential decay. So down here I've graphed an exponential decay graph, just to show you what it looks like here. Let's think about this point on our graph. So that's when time is equal to zero. So when time is equal to zero, what is the concentration? So you would just plug in time is equal to zero into here. So you would have your concentration is equal to the initial concentration times e to the zero. And e to the 0 is of course one. So this is one. So our initial concentration is obviously this point right here, time is equal to zero. This is obviously our initial concentration. So I'll write that in here. So that's this point. And as time approaches infinity, as time goes to infinity, your concentration of A goes to zero. So as you go out here, obviously your concentration of A is going to approach zero. So that's the idea of an exponential decay graph. Next, let's think about half-life. So over here is our definition of half-life. It's the time it takes for the concentration of a reactant to decrease to half of its initial concentration. So if the initial concentration, if this is the initial concentration here, what would be the concentration after half of it has reacted? We would get our initial concentration divided by two. So we're gonna plug this in for our concentration and then the symbol for half-life is t 1/2. So t 1/2. So we're gonna plug this in for time. So when the time is equal to the half-life, your concentration is half of your initial concentration. So let's plug those in and solve for the half-life. So on the left side we would have our initial concentration divided by two. And then this would be equal to the initial concentration of A times e to the negative k and then this would be the half life, so we plug in t 1/2 here. And so now we're just gonna solve for t 1/2. We're gonna find the half-life for a first order reaction. Let's get some more space down here. And we can immediately cancel out our initial concentration of A. So now we have 1/2 is equal to this is e to the negative kt 1/2. Alright, next, we need to get rid of our e here. So we can take the natural log of both sides. So we can take the natural log of 1/2 this is equal to the natural log of e to the negative kt to the 1/2. And so that gets rid of our e. So now we have the natural log of 1/2 is equal to negative kt 1/2. So we're just solving for t 1/2, cause t 1/2 is our half-life. So our half-life, t 1/2, would be equal to this would be negative natural log of 1/2 divided by k. Let's get out the calculator and let's find out what natural log of 1/2 is. So let's get some space over here. So natural log of .5 is equal to negative .693. So we have the negative of that, so we get a positive value here for our half life. So our half-life is equal to, let me rewrite this here, so our half-life, t 1/2, is equal to .693 divided by k, where k is our rate constant. So here is your half-life for a first order reaction. Now let's think about this. If k is a constant, obviously .693 is a constant. And so your half-life is constant. Your half-life of a first order reaction is independent of the initial concentration of A. So you're gonna get the same half-life. And let's think about that for an example. Let's go back up here to our graph and let's think about half-life. So lets' say we're starting with some initial concentration, let me go ahead and change colors here, so we can think about it. I'm just going to represent our initial concentration here with eight dots. So that's our, let's say we have eight particles we're starting out here. So obviously this is a theoretical reaction. So we're gonna wait until we've lost half of our reactant. Alright, so we've lost half of our reactant, obviously we'd be left with four. We'd be left with four here. So, where would that be on our graph? Well this point right here is out initial concentration. This is concentration here, so we'd go half that. So that would be right here on our graph. So we'd go over here and we find this point. Then we drop down to here on our x-axis. And let's just put in some times. Let's say that this is 10 seconds and 20 seconds and 30 seconds and 40 and so on. So we can see that it took, this took 10 seconds for us to decrease the concentration of our reactant by half. And so the first half-life is 10 seconds. So let me write that in here, so t 1/2 is equal to 10 seconds. Again, a made up reaction, just to think about the idea of half-life. And then, let's say, now we have four. How long does it take for half of that to react? So if half of that reacts, we're left with two particles. And on our graph, let's see, that would be right here. This point would be half of this, so we go over to here, and then we drop down. And how long did it take for us to decrease the concentration of our reactant by half? Once again, 10 seconds. So this time here would be 10 seconds. So this half-life is 10 seconds. We could do it again. So we lose half of our reactant again. And we go over to here on our graph and we drop down to here. How long did it take to go from two particles to one particle? Once again, it took 10 seconds. So the half-life is once again 10 seconds. So your half-life is independent of the initial concentration. So it didn't matter if we started with eight particles or four or two. Our half-life was always 10 seconds. And so, this is the idea of half-life for a first order reaction.

thanks Lena; there's some fun stuff going on here- This section allows you to view all posts made by this member. Note that you can only see posts made in areas you currently have access to.Show posts Menu Quote from: jim brenholts on July 09, 2009, 03:18:30 PM hey carl! welcome! it is always good to have new folks join us here on the forum. i am downloading the freebie and looking forward to new sounds. Quote from: APK on July 09, 2009, 07:47:24 AM Hi Carl. Welcome. Always good to have a new face around here. Quote from: Scott M2 on July 09, 2009, 06:15:37 AM Hi Carl! Welcome to the H. Forum. You give Buffalo a good name. 8) Quote from: satish on May 18, 2009, 09:44:56 AM Including Auditory Range, Piano Range, Loudness, Fundamentals Overtones Harmonics and Combination Tones, Beats, Scales, Physiological Effects Of Sound rotating dot Auditory Range The range of human hearing is from around 2Hz (2 cycles per second) to 20,000Hz (alias 2KHz) although with age one tends to lose acuity in the higher frequencies so for most adults the upper limit is around 10KHz. The lowest frequency that has a pitch-like quality is about 20Hz. A typical value for the extent to which an individual can distinguish pitch differences is 05-1% for frequencies between 500 and 5000Hz. (Differentiation is more difficult at low frequencies). Thus at 500Hz most individuals will be unable to tell if a note is sharp or flat by 2.5-5Hz (ie an 'allowable' pitch range for that note might be from 495Hz to 505Hz maximum. rotating dot Piano Range The lowest note on the piano 'A0' is at 27.5Hz, whilst the highest, C8, has a frequency of 4186Hz. 'Concert pitch' has been internationally accepted to be based on a frequency of 440Hz for A4, that is A in the fourth piano octave. rotating dot Loudness The quietest sounds that can be heard have a power (measured in Watts) of 10 to the -12 W/m2, whilst the loudest that can be withstood have a power of 1 W/m2. The range is therefore in the order of 10 to the 12, or one million million times. One decibel is a leap by a factor of 10, so that 0Db is the quietest noise, and 120Db is the loudest. rotating dot Fundamentals, Overtones, Harmonics And Combination Tones A note played by most ordinary acoustic intruments is not 'pure', it is in fact a spectrum of frequencies which largely give the note it's characteristic quality. The most significant components of this spectrum are the overtones. Overtones may be inharmonic (ie dissonant, sounding bad) or harmonic (ie consonant, sounding good). The 'natural overtone series' is the set of harmonics which are particularly prominent in the spectrum of frequencies for notes played by acoustic instruments, and which are also produced by producing waves in a string, where there are successive integer values for the number of 'crests' of the wave, ie 1,2,3,4,5,6,7,8 and so on. Taking 1 crest as the 'fundamental' note, then 2 produces a frequency an octave above. 3 produces a frequency of a '12th' interval (ie the fifth above the octave). 4 produces a frequency of the second octave above. 5 produces a frequency of a major third above that. 6 produces a frequency of the fifth note in this third octave. 7 produces a frequency of the flattened (minor) seventh in this octave. 8 produces a frequency of the fourth octave. 9 produces a frequency of the second interval in the fourth octave. (et cetera...) The increments between notes becomes successively smaller the further into the series you go, such that soon there are semitone intervals, then quarter- tone intervals, then even smaller fractional intervals. Also, there is less and less congruence between harmonics and the 'proper' frequencies of the equitempered scale the further into the series you go. (But maybe in some forms of music this is NOT a 'problem' od course). In the first octave there are no overtones, in the second there is one, in the third there are three, and in the fourth octave there are lots. Overtones which are 'harmonic' are at frequencies equal to the fundamental frequency multiplied by AN INTEGER: eg to find the fifth harmonic of C4, multiply it's frequency of 261.63 by 5: the result is a frequency of 1308.15, close to (but NOT the same as) the frequency of 1318.5 given for E6 (the fifth harmonic of C4) in the equitempered scale. Overtones which are inharmonic (dissonant) are non-integer multiples of the fundamental frequency. The constituent frequencies of a note can be ascertained with a wave analyser, which uses the equations developed by Fourier. If you filter out the fundamental frequency of say 200Hz from a note with overtones of 400Hz, 600Hz, 800Hz and 1000Hz (say, played on a piano), the brain will recreate the fundamental such that it appears to be present even when it is not there. The ear can be far more subtle still, however, in 'creating' sounds which are not actually there in received vibrations. When a tone of frequency f causes the eardrum to vibrate, the full harmonic series of tones 2f, 3f, 4f, 5f, 6f etc is apparent to the listener, these harmonics being produced by the eardrum itself. 'Combination' tones are additional, new tones produced when two frequencies, f1 and f2 are sounded together, and these combination tones are also 'manufactured' by the ear, this time from combinations of the first frequency and the harmonics of the second, and from the second frequency and harmonics of the first. (Note 3f2 is notation for 'the third harmonic of the second frequency'). The series of combination tones will be: 2f1 - f2 2f2 - f1 3f1 - f2 3f2 - f1 3f1 - 2f2 3f2 - 2f1 4f1 - 3f2 4f2 - 3f1 and so on. Whereas combination tones can be ascertained by subtracting one value from another, 'summation' tones, also manufactured by the ear, can be found by adding values together, ie f1 + f2 2f1 + f2 2f2 + f1 3f1 + f2 3f2 + f1 3f1 + 2f2 3f2 + 2f1 4f1 + 3f2 4f2 + 3f1 and so on. Interestingly, however, the resulting frequencies of these combination and summation tones turn out to be dissonances when played with the notes of the equitempered scale since the frequencies do not exactly match. It is possible to contrive a scale from difference tones. For example, if C4 is played together with Eb4, then a difference tone of 311.1 - 261.6 = 49.5Hz is created. This value of 49.5Hz is almost the frequency of Ab1. Other notes in the 1st octave can then be contrived in a similar manner from frequencies around the fourth octave. These sort of difference tones, however, might be called 'first-order' difference tones. A 'second-order' difference tone can be found in the difference between the frequency of a first-order difference tone and the frequency of the fundamental, so that in the example above, we have a first- order difference tone of (almost) Ab4, at 49.5Hz, and subtracting this from the frequency of the fundamental, we find 261.6 - 49.5 = 212.1 which is close to the frequency of Ab3. And so on. Here are some examples of combination and difference tones that can be demonstrated on a piano. C4+Eb4 produces Ab1 first-order difference tone. C4+Eb4 produces Ab3 second-order difference tone. rotating dot Beats Beat frequencies are produced when two different sounds are produced which are very close to each other in frequency. In such a situation the crests and troughs of each wave are generally slightly out of phase. But because the two notes have differing frequencies, after a certain repeating interval of time the crests of one wave will be aligned with the crests of the other, when a pulse or beat appears to the listener. Research has found that any two notes of different frequencies tend to sound good together (ie consonant) if there is an absence of beat frequencies between 8 and 50 Hz produced. Beat frequencies of 2-8Hz have been found to be pleasing, while beat frequencies above that level are generally though to be unpleasant. The beat frequency produced by any two notes is found by subtracting the value of the higher from the lower, ie Fbeat = Fhigher - Flower. Thus beat frequencies are a subset of difference tones -the 'beat' sensation occurring when the beat frequency value is low, say from 0.5Hz (1 beat every two seconds) to say 20Hz (the lowest frequency that has a pitch-like quality) J.Askill, in 'The Physics of Musical Sounds) says 'in general beat frequencies of 2-8Hz are considered pleasing, whereas if the beat frequency is above 15-20Hz, an unpleasant or dissonant effect is produced'. Personally I am curious about the range in the middle, say 8-12Hz which is also the frequency of 'alpha' brain-waves. rotating dot Scales The scale used extensively in the West has 13 notes from octave to octave and 12 intervals. In order for a scale to 'work' there should be: * a minimum of dissonance when different notes across the whole range of pitches are sounded together * an effective mapping of the harmonics of low notes onto higher notes, and an effective mapping onto the harmonics of higher notes * the possibility of key modulation which does not result in further frequency mismatches. The scale which has been widely adopted to fulfil these criteria is based on mathematics, such that the ratio of the frequency of any note to the frequency of the note a semitone above is constant. This is particularly useful in the extent to which it allows key modulation. However, this 'equitempered' scale is a compromise solution, because the frequency ratios of all intervals except the octave differ slightly from the 'perfect' intervals that the human ear really expects to hear. The 'exact' interval of a fifth, for instance, is found by multiplying the frequency of the fundamental by 3/2, the fourth is found by multiplying the fundamental frequency by 4/3, and the major third interval is found by multiplying the fundamental frequency by 5/4. (Other intervals involve slightly less obvious fractions). The problem with a scale built on fractional values like this, however, is that the increments from note to note are not constant (eg 5/4 - 4/3 does not equal 4/3 - 3/2) which creates difficulties when the required key for a piece is different to that of the fundamental from which the scale is constructed. For example, if we move up an octave from C by adding a fifth, and then adding a fourth, then the resulting high C will have a different frequency to that arrived at if our key is F, and we try to arrive at the same high C by adding a major third and then a minor third to that fundamental F. So in the equitempered scale all semitone increments have been 'tempered' such that they are always a little flat, or a little sharp. The constant value on which this scale is based is 1.0594630915, such that if we call this value S, then the semitone above a fundamental note is found by multiplying the frequency of the fundamental by S to the power of 1. The second above the fundamental is found by multiplying it's frequency by S to the power of 2, and so on until the octave above the fundamental is found by multiplying it's frequency by S to the power of 12. The value of the constant S is the 12th root of 2, since in order to find the twelve equal divisions between two notes an octave apart, where the frequency of the octave is twice that of the fundamental, the 12th root of 2 is the value we are looking for. Other divisions of the octave have been proposed, such as a 19-step octave, and a 53-step octave. The maths for these 'works' although these 'scales' may be harder to use effectively. The maths for the 53-division scale is particularly elegant in fact, and closer to a 'perfect' musical scale than the 13-note scale which we currently use. (In that case the constant value for each successive interval is found by using the 53rd root of 2, ie 1.013164143). The 'exact' scale, built on the 'perfect' intervals that the ear expects to hear, has much to recommend it if one key is kept to. This scale, however, has fifteen intervals and fourteen notes, since in the first octave there are all the notes of the equitempered scale (at slightly different frequencies) but there is also both a 'major whole tone' and a 'minor whole tone', and both an 'augmented fourth' and a 'diminished fifth'. (In the second octave there is both an 'augmented octave' and a 'dimished ninth' and also both an'augmented eleventh' and a 'diminished twelfth'). Thus successive octaves above the fundamental differ from each other in the way that they are put together. Furthermore, however, when we look at the extent to which the frequencies of harmonics of exact-scale notes match notes higher up in the exact scale, we see that we can list the intervals octave, fifth, fourth, major third, major sixth, minor third, minor sixth in terms of increasing dissonance, so that in the case of the minor sixth, if we look at all harmonics up to the twelfth, only one 'matches'. The mathematical elegance of the 53-division scale should make it a more appropriate scale for dealing both with key modulation, and a preoccupation with harmonics. The 53-interval scale uses eneven 'chunks' of these 53rd-of-an-octave division to create the notes of the diatonic scale. The size of each incremental chunk is as follows: Quote from: DeepR on June 22, 2009, 01:34:31 PM If there's one genre where the music can be related to imagery, surely it must be ambient for its 'visual' qualities. The idea is to find pictures that you think are somehow fitting for the ambient piece.

The Simple Work-Leisure Trade income of any individual is related to the number of hours he or she works per day and then per week. The net income is also related to per hour pay rate. If the pay rate per hour is higher, then more income can be generated for fewer working hours and more leisure hours. The leisure hours are those for which the individual does not work and no income is generated in those hours. Therefore, it can be said that the net income is related to the number of hours worked in accordance with the pay rate per hour (Dean Garrat 2013). This study will discuss the effects of working hours on the net income. Some assumptions are made, such as: No income is generated for leisure hours (that is there is no savings for the individual or at least the income from any saving or property, which is not to be included). The reference point is one The individual is free to chose working hours per day (but it is very common that the individuals does not chooses working hours by themselves, it is their employees who fix / expect their workers to work for definite number of hours per week). Any hours not considered as working hours are assumed to be leisure hours. mentioned earlier, there is a definite relationship between the net income (wage) and the number of working hours. This relationship can be explained in terms of plot between wage and working hours. relationship is shown in figure where the number of hours has been taken along x-axis and the money earned from work (wage) is shown on y axis. Working hour is an independent variable and the wage earned is a dependant variable, because the wage depends on number of hours worked per week. 1: Wage and work hours are two budget constraints shown in figure 1, one for hourly wage rate of w1 (budget line 1 known as Bl1) and the second for an hourly rate of w2 (Bl2). It is assumed that there are 24 working hours in a single day which means that the number of leisure hours are zero. In total there are 24 working hours and that is why both lines meet the x axis at 24 and Bl1 meets the y axis at 24w1 showing that for 24 working hours the wage will be 24w1. This can be calculated for any given hourly rate, for instance if the hourly rate is £6.50 the total earning will be £156 for 24 hours’ work. 2: Indifference curve for higher Figure 2: Indifference curve for lower satisfaction line 2 (Bl2) shows the line when hourly rate was increased from w1 to w2 where w2 is larger than w1. This shows that if the hourly rate is increased then the total wages would also increase for any given working hours as shown in figure figure 2 and figure 3 the examples of an indifference curves are shown where in figure 2 the indifference curve is shown for an individual who not only enjoys leisure but derives a lot of utility from the work associated with his job. However, in figure 3 the indifference curve is shown for an individual who enjoys leisure but derives much less utility from the work associated with the job. In this case the curve is much closer to the axis. The utility function measures the individual’s level of satisfaction or happiness. The higher the level of satisfaction, the happier the person is. If the person is more satisfied with his job and feels happier at work his utility index will be larger and the indifference curve will be higher as shown in figure 2 (hks.harvard.edu). 4: Wage for changing work hours indifference curves shown in figure 2 and 3 shows the way a particular worker views the trade-off between leisure and consumption. Each graph represents the way of work of each individual and different workers will typically view this trade-off differently. This is understandable because some workers may like to devote a great deal of time and effort to their jobs, whereas others would prefer to devote most of their time to leisure. These interpersonal differences in preferences imply that the indifference curves will certainly look quite different based on individual’s behaviour (hks.harvard.edu). figure 4 a single budget constraint w1 is shown for an hourly rate of £6.50 and it assumed that the working hours are 6 hours per day. This shows that the leisure hours will be 24 – 6 = 18 hours and that is why the budget constraint line crosses the x axis at 18. In this case the total income for 6 hours’ of work will be 6 x 6. 50 = £39.00. Due to this, the budget constraint line crosses the y axis at 39. It can also be seen that as the leisure hours are increasing the wage keeps on decreasing and the wage will be maximum when the working hours are the maximum, that is 24. next question is what happens if the individual is asked by his / her employer to work for larger hours for some urgent work or for any other reason? In this case for the larger number of working hours the leisure hours will be decreasing since the working hours has been increased. In figure 4 the vertical line that intersects the x- axis at 18 will shift towards left side to a smaller value of 16 hours or 14 hours depending upon how many working hours have been increased. The income will also increase and therefore the horizontal line will shift upwards. In that case, both the new horizontal and new vertical lines shown as red may not intersect with the indifference curve. This means that to intersect the indifference curve the employer may have to increase the hourly rate so that the curve can be intersected and same level of utility index can be maintained for the individuals. In either case the total wage will increase in both ways, firstly by working for more hours and secondly by increased per hourly rate.

- How many $100 bills are in a band? - How many bills are in a stack of money? - Why is 1000 called a grand? - How big is a billion dollars in $100 bills? - What does 1 trillion dollars look like? - Is a rack 1000 dollars? - How many $100 bills does it take to make $5000? - How many 20s does it take to make 100? - How many $20 bills does it take to make $500? - How much space does $1 million take up? - Is a stack 100 or 1000? - How much is a band in rap? - How much is in a brick of money? - How many bills are in a 1 inch stack? - How much money is a 100 dollar stack? - Why is 500 called a monkey? - How many twenties are in $1000? - How much does 1 billion look like? - Why is $1000 called a rack? - How many bills are in a inch? - How much does 1 billion weigh? - How thick is a stack of 100 bills? - What does a band mean in money? - How much room does a billion dollars take up? - How much money is a grip? - Is a band 1000? - How many bills are in a band? - What does 3 stacks mean? How many $100 bills are in a band? ABA Standard (United States)Strap ColorBill DenominationBill CountYellow$10100Violet$20100Brown$50100Mustard$1001007 more rows. How many bills are in a stack of money? 100 billsAll denominations are 100 bills per stack that are strapped together (to make a strap). 10 straps (of 100 bills each) are combined together (with a large rubber band or zip tie) to make one bundle. Bundles are used to deposit money into the US Federal Reserve Bank. Why is 1000 called a grand? The use of “grand” to refer to money dates from the early 1900s and as disconcerting as it may be to some people, comes from America’s underworld. … But in the early 1900s one thousand dollars was considered to be a “grand” sum of money, and the underground adopted “grand” as a code word for one thousand dollars. How big is a billion dollars in $100 bills? $1,000,000,000. A billion dollars is 10 crates of $100 bills. There are at least 536 people in America who have at least this many crates worth of money. What does 1 trillion dollars look like? What does one TRILLION dollars look like? (calculations & dimensions)Short ScaleLong Scaletrillionbillion1,000,000,000,000quadrillionthousand billion (or billiard)1,000,000,000,000,000quintilliontrillion1,000,000,000,000,000,000sextillionthousand trillion (or trilliard)1,000,000,000,000,000,000,0004 more rows Is a rack 1000 dollars? “A rack” is $10,000 in the form of one hundred $100 bills, banded by a bank or otherwise. $1000 notes are occasionally referred to as “large” (“twenty large” being $20,000, etc.). In slang, a thousand dollars may also be referred to as a “grand” or “G”, “K” (as in kilo), or a “stack” as well as a “band” . How many $100 bills does it take to make $5000? To work this out divide 10,000 by 100. The answer is there are 100 x $100 in $10,000 dollars. How many 20s does it take to make 100? Let’s see, 20 means we have trinary system then 100 is 9, 20 is 6. and the answer is 1. Trinary Division: 100 / 20 = 1.111… How many $20 bills does it take to make $500? 25 x $20 bills = $500. How much space does $1 million take up? $1 Million Dollar Stack 100 x $10,000, or 10 x 10 stacks measures 12″ across x 13″ front to back x approx. 5″ tall. Scroll down for many photos and examples and contact us for quantity prices. Is a stack 100 or 1000? 1 Answer. The entry provides the phrase “stacks of the ready” to mean “plenty of money”. I think this phrase, in the prevailing years, was shortened to the slang stack, which also took on the meaning of $1000. That is, one stack is equivalent to one grand which is $1000. How much is a band in rap? While we’re on the topic of money, might as well throw in the very basics: A band, a stack, a rack = $1,000 in cold, hard cash. How much is in a brick of money? In other words, a brick is a thousand notes. Given this stack of bills is about the size of a building brick, the source of the term becomes obvious. So a brick of one-dollar bills is equal to $1,000, and a brick of $100 bills is $100,000. How many bills are in a 1 inch stack? 250 bills1-inch stack of $100 bills contains 250 bills. How much money is a 100 dollar stack? A packet of one hundred $100 bills is less than 1/2″ thick and contains $10,000. Fits in your pocket easily and is more than enough for week or two of shamefully decadent fun. Believe it or not, this next little pile is $1 million dollars (100 packets of $10,000). Why is 500 called a monkey? Derived from the 500 Rupee banknote, which featured a monkey. Explanation: While this London-centric slang is entirely British, it actually stems from 19th Century India. … Referring to £500, this term is derived from the Indian 500 Rupee note of that era, which featured a monkey on one side. How many twenties are in $1000? It takes 50 twenty dollar bills to make 1000 dollars. How much does 1 billion look like? 1,000,000,000 (one billion, short scale; one thousand million or milliard, yard, long scale) is the natural number following 999,999,999 and preceding 1,000,000,001. One billion can also be written as b or bn. Why is $1000 called a rack? Originally, a Rack was a stack of $100 bills that total $10,000,but due to the frequency of the use of Rack in songs like ‘Racks on Racks’ and *’Rack City’, most people refer to $1,000 as a Rack. How many bills are in a inch? The height of a stack of 100 one dollar bills measures . 43 inches. The height of a stack of 1,000 one dollar bills measures 4.3 inches. The height of a stack of 1,000,000 one dollar bills measures 4,300 inches or 358 feet – about the height of a 30 to 35 story building. How much does 1 billion weigh? How much does a billion dollars weight? When weighed in $100 bills, a billion dollars weighs approximately 10,000 kilograms. How thick is a stack of 100 bills? The thickness of a dollar bill is 0.0043 inches – so 100 bills should be 0.43 inches – but it won’t be! What does a band mean in money? one thousand dollarsA band is one thousand dollars, also known as a grand, stack, or G. The term comes from the band placed around a stack of cash to hold it together. How much room does a billion dollars take up? Take one billion $1 bills and put them in a stack (we’ll wait) after about 30 years of stacking, your pile would measure 358,510 feet or 67.9 miles high. In area: One billion $1 bills would cover a four-square-mile area or the equivalent of 2,555 acres. How much money is a grip? Definitions include: 100,000 units of a currency to just below 1,000,000 units. Definitions include: one hundred dollars. Definitions include: having three children in rapid succession (generally following the marriage of a young couple). Is a band 1000? One band is usually $1,000 in cash, referring to the currency strap or rubber band that goes around a stack of $1,000. Blue bands are stacks of $10,000, as new $100 bills have blue ribbons sewn into them, and would likely be stacked in groups of 100. How many bills are in a band? A strap is a package of 100 notes. A bundle consists of 1,000 notes of the same denomination in ten equal straps of 100 notes each. Before depositing currency, currency must be prepared according to denomination. For $1 through $20 denominations, your deposit(s) must contain full bundles. What does 3 stacks mean? keep your heartWhat does the phrase ‘keep your heart, 3 stacks’ mean? … When 3-stacks (also known as Andre 3000) is about to get married, his friends all implore him not to give his heart away, to “keep his heart”.

QUIZ 9 (ANSWERS) CH. 7: Problems 9*, 24*, 38* (TRY #34*, critical for Ch 12 and Physics 4B; see class notes on stable and unstable equilibrium) 41*, 42*, 43*, 46* (try #12*), 47* (TRY 65* and 66*) , 50*, 55*, 56*, 63* (see #46*) , 68*, 70*, 75* (SEE SAMPLE TEST 3) (ANSWERS) |VIRTUAL LAB 2 (SEE EXAMPLE 7.7, CHAPTER 7)| |SIMPLE HARMONIC MOTION LAB| Exam Cross Index for TEST 2 Sp '12 BELOW ARE THE LINKS TO SAMPLE EXAMS: MEANWHILE HERE ARE LINKS (A), (B), (C) AND (D) TO SAMPLE TEST 2 AND 3: |For many of these problems use KEi + Ui = KEf + Uf + HEAT or USE KEi + Ui + WF = KEf + Uf + HEAT, where U is the potential energy due to the conservative force ( gravitational and spring forces in this case), KE is the kinetic energy and WF is the work done by the external force (done by "Frankie" as we discussed in class. ) NOTE: 9, 46 and 63 mix ENERGY CONSERVATION (Ch. 7)and CIRCULAR DYNAMICS (Ch. 5), a special class of problems you should study ahead of Test 2. ANOTHER NOTE: #50 mixes ENERGY CONSERVATION (Ch. 7) and PROJECTILE MOTION (CH. 3) |9.* KEi + Ui = KEf + Uf + HEAT . (a) The angle between the normal force and motion in this case is 90 degrees, so what is the work by that force? The work by gravity is positive since the motion is from high to low; note the gravitational work only depends on the vertical displacement. (b) ENERGY CONSERVATION : KEi + Ui = KEf + Uf + HEAT where HEAT IS THE NEGATIVE OF THE WORK BY FRICTION. THUS HEAT IS ALWAYS POSITIVE. You are given the heat. (c) CIRCULAR DYNAMICS: Critical point. HEAT is the negative of the work by the friction force, which may not be constant. The instantaneous friction force is u*N, where u is the kinetic friction coefficient and where N is the magnitude of the normal force which may not be constant as in this case. Clearly from earlier problems in Chapter 5, N is a maximum at the bottom and is a minimum at the top. (See worksheet 2, GRP 3, problem #1, THE PROBLEM DEALING WITH BUMPS AND VALLEYS) (d) CIRCULAR DYNAMICS: At bottom, the normal force points up in the pos direction toward the center of the circle and the weight of magnitude mg points down in the neg direction. Write: Sum of forces in radial direction = m*v2/R = pos - neg and solve for N, the normal force magnitude, using the speed v you found in part (b) from energy conservation. See worksheet 2, GRP 3, problem #1. Note N > mg. |24. *Use this: KEi + Ui = KEf + Uf + HEAT for the initial (i) and final (f) locations. Study carefully the classic example (i.e. Many physics book authors have used it.). Unlike the previous problem, the friction force is constant, so HEAT = fk*D, where D is the distance moved and fk = 17,000 (N) is given. Note the initial kinetic energy KEi is not zero as in problem 9. Set the lowest point equal to the zero of gravitational potential energy (ie, y = 0 at point 2.) Note U = Ug + Us, where Ug = mgy and Us = (1/2)*k*y'2, where y' is the spring coordinate indicating compression. We have KEi = 16,000 J and Ui = mgyi + 0 since the initial spring potential energy is zero (not compressed--- yi' = 0) ; note yi = 2.00 m. Uf = mgyf + (1/2)*k*yf'2 , where yf = 1.00 m yf' = -1.00 m. (Since you square y' in general the negative sign goes away. ) The heat of course is fk*D, where D = 1.00 m. Solve for KEf and find final speed vf : KEi + Ui = KEf + Uf + HEAT means 16,000 J + mgyi = KEf + mgyf + (1/2)*k*yf'2 + 17,000 (N)*d , where yf = d = 1.00 m and yf' = -1.00 m. (b) If you think the acceleration is constant think twice; the spring makes things more complex as we have already seen in Ch. 14. So use: sum of the forces in the y direction = m*a = pos - neg, where we set the positive y direction to be up. Thus: : m*a = | kyf' | + fk - mg, where clearly the spring and friction forces point upward as the object moves down. |38.* READ section 7.4 and 7.5: If U = U(x) then Fx = -dU/dx. The x-component of force is the negative derivative of U; Thus if dU/dx is positive , the force points in the negative x -direction (Fx < 0) and if dU/dx is negative then force points in positive x direction (Fx > 0.) When dU/dx = 0 force is zero and the object is said to be in equilibrium, either stable or unstable. A CONCAVE UP POTENTIAL ENERGY FUNCTION MEANS EQUILIBRIUM IS STABLE.| |41*. (a) CHAPTER 5 REVIEW: You must prove whether or not the system is at rest; the emphasis here is the use of Ch. 5 concepts, whereas Ch. 7 is more directly related to part If the system is at rest and remains so we have for the concrete m*a = pos - neg = mg - T = 0 . So you can immediately find T. For the Box we have m'*a = 0 = T - fs ' , where fs ' is the force of static friction from the ground and m' is the Box mass. We also have m''*a = 0, where m'' is the gravel mass. ( i.e. There is no friction force between the gravel and Box if the system is at rest.) Compare T and fsmax = us*N, where N is the magnitude of the normal force on the box, which must be equal to (m' + m'')*g. |42.* KEi + Ui = KEf + Uf + HEAT , where HEAT = 0. VIRTUAL LAB 2, question 1. 43.* The simplest way to do the problem is to bypass the speed and kinetic energy just after the Block leaves the spring and consider only two points in space-time: When the spring is fully compressed and the block is at rest (i) and when the block has reached its maximum horizontal distance and comes to rest permanently (f) . KEi + Ui = KEf + Uf + HEAT , where Ug = 0 at both points (ground level, y = 0) . You should be able to see which terms are clearly zero. Thus we have initial (i) spring potential energy = HEAT , where HEAT = fk*D : (1/2)*k*d2 = u*mg*D, where u = coefficient of kinetic friction and d and D are given in diagram. |46.* ANOTHER CLASSIC, like #24. This one mixes energy conservation and dynamics just like #9 above and #63 below. (a) KEi + Ui = KEf + Uf + HEAT , where HEAT = 0. ENERGY CONSERVATION: (I) KEi + Ui = KEf + Uf , WHERE KEi = 0, Ui = m*g*h, KEf = (1/2)*m*v2 and Uf = m*g*2R: mgh = (1/2)*m*v2 + m*g*2R AT TOP. CIRCULAR DYNAMICS: ALSO AT TOP, the normal force and the weight of magnitude mg point down in the pos direction toward the center of the circle. Write: (II) Sum of forces in radial direction = m*v2/R = pos - neg = N + mg - 0 = N + mg, since there is no force pointing in the neg direction away from the center; see #120 AND #42, Ch. 5 , point B at top. Set N = 0 IN THIS CASE. Solve equations I and II simultaneously for h in terms of R by eliminating v. The mass m cancels out. (b) FOR THIS INITIAL HEIGHT h, greater than that of part (a), the object never loses contact with the track at the top so it does reach point C. (I) ENERGY CONSERVATION : (I) KEi + Ui = KEf + Uf , WHERE KEi = 0, Ui = m*g*h, h = 3.50*R, KEf = (1/2)*m*v2 and Uf = m*g*R (at point C) . Find v at C: mg*( 3.50*R ) = (1/2)*m*v2 (III) For the tangential acceleration, let the pos direction be vertically down, a direction tangent to the circle at point |47.* ANOTHER CLASSIC: m*g*H = TOTAL HEAT, SINCE THE INITIAL AND FINAL kinetic energies are zero. The Heat is the total heat after possibly numerous trips across the flat section. TOTAL HEAT fk*L, where L is the total distance moved on flat surface before coming to rest. To find the number of trips, divide TOTAL HEAT by fk*D, where D = 30 m exactly. From this ratio you can find the number of complete trips; from the fractional remainder, you can find exactly where the object comes to rest on the flat surface. If the ratio is less than one, then the object does not make a complete trip across the flat surface. |50.* KEi + Ui = KEf + Uf + HEAT , where HEAT = 0. See problem 68 hint i = bottom. f = top just before leaving horizontal surface 70 (m) above ground zero ( where Ug = 0) Uf = mgyf, where yf = 70 (m) KEi = (1/2)*m*vi2 . KEf = (1/2)*m*vf2 |55.* KEi + Ui = KEf + Uf + HEAT , where HEAT = 0. Let the zero of gravitational energy be the ground. # Initially (i) the system is at REST and Ui = the gravitational potential energy of the 12.0 kg clock at the given initial height. Solve for the final common speed vf of the two blocks. USE KEi + Ui + WF = KEf + Uf + HEAT, where U is the gravitational potential energy in this case, KE is the kinetic energy, WF is the work done by the external thrust force and the HEAT is fk*D. The constant friction force of magnitude fk = 500 (N) is given; D is the distance moved along ramp. # Initially (i), the rocket is at REST and Ui = the gravitational energy at the initial vertical height H above the ground. Note the height H has a simple relationship with D, the distance moved along the ramp; use simple geometry, the angle of 53 degrees, and the properties of right triangles to write H in terms of D. WF = F*D, where D is the distance moved along the ramp and F is the magnitude of the given external thrust force = 2000 (N) exactly. |63.* ANOTHER CLASSIC: THE STRUCTURE OF THIS PROBLEM, LIKE 9, and 46, mixes ENERGY CONSERVATION (Ch. 7) CIRCULAR DYNAMICS (Ch. 5) ENERGY CONSERVATION (Ch. 7): (I) KEi + Ui = KEf + Uf + HEAT , where HEAT = 0. # Initially (i), skier is at the top and is essentially at rest; KEi = 0. Ui = m*g*R if the zero of potential energy is considered to be at the level of the snow ball's horizontal diameter. (I) mg*R = (1/2)*m*vf2 + mg*R*cos alpha, where alpha is angle with vertical shown. # Finally (f), we consider the point when the skier leaves the snow ball , i.e. when the normal of magnitude N becomes zero. See figure 7.63. We define the location of this point in the following way: Draw a radial line from the snow ball center to this point and define the angle with the vertical as alpha . KEf = (1/2)*m*vf2 and Uf = m*g*R*cos alpha. CIRCULAR DYNAMICS: At any point on the surface of the ball, the normal force points radially away from the center of the circle. The gravitational force of magnitude mg points vertically down and the component of the gravitational force along the radial line is mg*cos theta and points toward the center of the circle. Write: (II) For the centripetal (center seeking) force, Sum of forces in radial direction = m*v2/R = pos - neg = mg*cos theta - N. For this problem set N = 0. Thus: m*vf2/R = mg*cos theta. Solve equations I and II simultaneously for cos theta by eliminating v. Compute theta by evaluating cos -1. Your angle will be between 45 and 90 degrees. |68. WE DO THIS PROBLEM USING TWO STYLES--FROM CH. 6 AND See #50 above. CH. 6 STYLE: (i) (1/2)*m*vf2 - (1/2)*m*vi2 = total work = Wg + Wf + WN, work by gravity, the friction force and normal force respectively where last term is zero since the normal work on the way up slope must be vanish because that force is perpendicular to the motion i.e. makes an angle of 90 degrees. ALSO, THE SLOPE IS FRICTIONLESS MAKING WORK OF FRICTION Wf = 0. (ii) FIND THE SPEED AT THE TOP OF THE CLIFF USING (1/2)*m*vf2 - (1/2)*m*vi2 = total work = Wg WHERE THE NET VERTICAL DISPLACEMENT IS FROM LOW TO HIGH. IS GRAVITY WORK POSITIVE OR NEGATIVE? (ii) AFTER LEAVING CLIFF, DURING FREE FALL (i.e., PROJECTILE MOTION) THE gravitational work will be positive for this high to low motion. IN THAT PHASE OF PROBLEM YOU CAN USE CH. 3 METHODS AS IN LAB 3. HERE'S HOW YOU DO IT CH. 7 STYLE: |70 (a) KEi + Ui = KEf + Uf + HEAT , where HEAT = 0. The block is released from rest so that should tell you the value of KEi . The maximum speed occurs when U is a minimum and that location is shown in figure 7.42. Uf = 0 assuming both springs are un-deformed and the block is in equilibrium. Ui = sum of two spring potential energies: Spring 1 is stretched by 0.15 m and spring 2 is compressed by 0.15 m. To evaluate Ui, apply the formula (1/2)*k*x2 for the potential energy to both springs and add the two results. Solve for KEf and the maximum speed vf of the block WHICH OCCURS WHEN BOTH STRINGS ARE (b) The simplest way to do this part is to assume the initial position (i) of the block is the same as in part (a) but the final location (f) is when Spring 1 is at maximum compression. When Spring 1 is at maximum compression, the block is at rest momentarily (i): Ui = Uf = sum of two spring potential energies. Spring 1 is compressed a distance |x'| and Spring 2 is stretched by the same |x'| . Ui was evaluated in part (a). Plug the symbol |x'| into the formula for Uf and solve for |x'| numerically. |75. * SELECTIVELY USE KEi + Ui + WF = KEf + Uf where U is the potential energy due to the spring force in this case, KE is the kinetic energy and WF is the work done by the external force of magnitude F. NO FRICTION SO HEAT = 0. The FULL equation with WF only applies between initial point A and point-B but we know the process continues beyond that. The Block will continue to move beyond point-B and eventually come to rest. At that point, the full distance from the origin is the amplitude A of subsequent simple harmonic motion about the origin and would be described by the function Acoswt for t >0 where t=0 is when the block reaches maximum distance from origin---see Ch. 14 and virtual lab 2.. So that is your task---to find that distance . Here are some tips: (a) Find the speed of the block at point-B using : KEA + UA + WF = KEB + UB where WF = F*D. Note KEA = 0 and UA = 0. Also WF = F*D, D = 0.25 m and UB = (1/2)*k*(0.25)2 in Joules. Find the final kinetic energy and final speed at B. (b) Find the additional distance before coming to rest using: KEB + UB = KEC + UC . NOTE: KEB is known from the previous part and UB = (1/2)*k*(0.25)2 Also KEC = 0 and UC = (1/2)*k*xC2 . Find xC. Then find 0.60 m - xC . Note: KEA + UA + WF = KEC + UC or, WF = KEC + UC, where WF = F*D, D = 0.25 m and KEC = 0.

We investigate a one-parametric class of merit functions for the second-order cone complementarity problem (SOCCP) which is closely related to the popular FischerBurmeister (FB) merit function and natural residual merit function. In fact, it will reduce to the FB merit function if the involved parameter <i></i> equals 2, whereas as <i></i> tends to zero, its limit will become a multiple of the natural residual merit function. In this paper, we show that this class of merit functions enjoys several favorable properties as the FB merit function holds, for example, the smoothness. These properties play an important role in the reformulation method of an unconstrained minimization or a nonsmooth system of equations for the SOCCP. Numerical results are reported for some convex second-order cone programs (SOCPs) by solving the unconstrained minimization reformulation of the KKT optimality conditions, which indicate that the The study of this paper consists of two aspects. One is characterizing the so-called circular cone convexity of f by exploiting the second-order differentiability of f L ; the other is introducing the concepts of determinant and trace associated with circular cone and establishing their basic inequalities. These results show the essential role played by the angle , which gives us a new insight when looking into properties about circular cone. MSC:26A27, 26B05, 26B35, 49J52, 90C33, 65K05. We consider the Tikhonov regularization method for the second-order cone complementarity problem (SOCCP) with the Cartesian P 0-property. We show that many results of the regularization method for the P 0-nonlinear complementarity problem still hold for this important class of nonmonotone SOCCP. For example, under the more general setting, every regularized problem has the unique solution, and the solution trajectory generated is bounded if the original SOCCP has a nonempty and bounded solution set. We also propose an inexact regularization algorithm by solving the sequence of regularized problems approximately with the merit function approach based on FischerBurmeister merit function, and establish the convergence result of the algorithm. Preliminary numerical results are also reported, which verify the favorable theoretical properties of the proposed method. This paper proposes using the neural networks to efficiently solve the second-order cone programs (SOCP). To establish the neural networks, the SOCP is first reformulated as a second-order cone complementarity problem (SOCCP) with the KarushKuhnTucker conditions of the SOCP. The SOCCP functions, which transform the SOCCP into a set of nonlinear equations, are then utilized to design the neural networks. We propose two kinds of neural networks with the different SOCCP functions. The first neural network uses the FischerBurmeister function to achieve an unconstrained minimization with a merit function. We show that the merit function is a Lyapunov function and this neural network is asymptotically stable. The second neural network utilizes the natural residual function with the cone projection function to achieve low computation complexity. It is shown to be Lyapunov stable and converges globally Abstract Recently, J.-S. Chen and P. Tseng extended two merit functions for the nonlinear complementarity problem (NCP) and the semidefinite complementarity problem (SDCP) to the second-order cone commplementarity problem (SOCCP) and showed several favorable properties. In this paper, we extend a merit function for the NCP studied by Yamada, Yamashita, and Fukushima to the SOCCP and show that the SOCCP is equivalent to an unconstrained smooth minimization via this new merit function. Furthermore, we study conditions under which the new merit function provides a global error bound which plays an important role in analyzing the convergence rate of iterative methods for solving the SOCCP; and conditions under which the new merit function has bounded level sets which ensures that the sequence generated by a descent method has at least one accumulation point. In this paper, we extend the one-parametric class of merit functions proposed by Kanzow and Kleinmichel [C. Kanzow, H. Kleinmichel, A new class of semismooth Newton-type methods for nonlinear complementarity problems, Comput. Optim. Appl. 11 (1998) 227251] for the nonnegative orthant complementarity problem to the general symmetric cone complementarity problem (SCCP). We show that the class of merit functions is continuously differentiable everywhere and has a globally Lipschitz continuous gradient mapping. From this, we particularly obtain the smoothness of the FischerBurmeister merit function associated with symmetric cones and the Lipschitz continuity of its gradient. In addition, we also consider a regularized formulation for the class of merit functions which is actually an extension of one of the NCP function classes studied by [C. Kanzow, Y. Yamashita, M. Fukushima, New NCP functions and Like the matrix-valued functions used in solutions methods for semidefinite programs (SDPs) and semidefinite complementarity problems (SDCPs), the vector-valued functions associated with second-order cones are defined analogously and also used in solutions methods for second-order-cone programs (SOCPs) and second-order-cone complementarity problems (SOCCPs). In this article, we study further about these vector-valued functions associated with second-order cones (SOCs). In particular, we define the so-called SOC-convex and SOC-monotone functions for any given function . We discuss the SOC-convexity and SOC-monotonicity for some simple functions, e.g., <i>f</i>(<i>t</i>) = <i>t</i> ² <i>t</i> ³ 1/<i>t</i> <i>t</i> ^1/2, |<i>t</i>|, and [<i>t</i>]₊. Some characterizations of SOC-convex and SOC-monotone functions are studied, and some conjectures about the relationship between SOC-convex and SOC-monotone functions are proposed. Recently Tseng (Math Program 83:159185, 1998) extended a class of merit functions, proposed by Luo and Tseng (<i>A new class of merit functions for the nonlinear complementarity problem</i>, in Complementarity and Variational Problems: State of the Art, pp. 204225, 1997), for the nonlinear complementarity problem (NCP) to the semidefinite complementarity problem (SDCP) and showed several related properties. In this paper, we extend this class of merit functions to the second-order cone complementarity problem (SOCCP) and show analogous properties as in NCP and SDCP cases. In addition, we study another class of merit functions which are based on a slight modification of the aforementioned class of merit functions. Both classes of merit functions provide an error bound for the SOCCP and have bounded level sets. This paper is a follow-up of the work [Chen, J.-S.: J. Optimiz. Theory Appl., Submitted for publication (2004)] where an NCP-function and a descent method were proposed for the nonlinear complementarity problem. An unconstrained reformulation was formulated due to a merit function based on the proposed NCP-function. We continue to explore properties of the merit function in this paper. In particular, we show that the gradient of the merit function is globally Lipschitz continuous which is important from computational aspect. Moreover, we show that the merit function is <i>SC</i> ¹ function which means it is continuously differentiable and its gradient is semismooth. On the other hand, we provide an alternative proof, which uses the new properties of the merit function, for the convergence result of the descent method considered in [Chen, J.-S.: J. Optimiz. Theory Appl., Submitted for publication (2004)]. We investigate some properties related to the generalized Newton method for the Fischer-Burmeister (FB) function over second-order cones, which allows us to reformulate the second-order cone complementarity problem (SOCCP) as a semismooth system of equations. Specifically, we characterize the B-subdifferential of the FB function at a general point and study the condition for every element of the B-subdifferential at a solution being nonsingular. In addition, for the induced FB merit function, we establish its coerciveness and provide a weaker condition than Chen and Tseng (Math. Program. 104:293327, 2005) for each stationary point to be a solution, under suitable Cartesian <i>P</i>-properties of the involved mapping. By this, a damped Gauss-Newton method is proposed, and the global and superlinear convergence results are obtained. Numerical results are reported for the second-order cone programs In last decades, there has been much effort on the solution and the analysis of the nonlinear complementarity problem (NCP) by reformulating NCP as an unconstrained minimization involving an NCP function. In this paper, we propose a family of new NCP functions, which include the Fischer-Burmeister function as a special case, based on a <i>p</i>-norm with <i>p</i> being any fixed real number in the interval (1,+), and show several favorable properties of the proposed functions. In addition, we also propose a descent algorithm that is indeed derivative-free for solving the unconstrained minimization based on the merit functions from the proposed NCP functions. Numerical results for the test problems from MCPLIB indicate that the descent algorithm has better performance when the parameter <i>p</i> decreases in (1,+). This implies that the merit functions associated with <i>p</i>(1,2), for example <i>p</i>=1.5, are more effective A popular approach to solving the nonlinear complementarity problem (NCP) is to reformulate it as the global minimization of a certain merit function over ^<i>n</i>. A popular choice of the merit function is the squared norm of the Fischer-Burmeister function, shown to be smooth over ^<i>n</i> and, for monotone NCP, each stationary point is a solution of the NCP. This merit function and its analysis were subsequently extended to the semidefinite complementarity problem (SDCP), although only differentiability, not continuous differentiability, was established. In this paper, we extend this merit function and its analysis, including continuous differentiability, to the second-order cone complementarity problem (SOCCP). Although SOCCP is reducible to a SDCP, the reduction does not allow for easy translation of the analysis from SDCP to SOCCP. Instead, our analysis exploits Shape optimization aims to optimize an objective function by changing the shape of the computational domain. In recent years, shape optimization has received considerable attentions. On the theoretical side there are several publications dealing with the existence of solution and the sensitivity analysis of the problem; see e.g., and references therein. We simulate blood flow in patient-specific cerebral arteries. The complicated geometry in the human brain makes the problem challenging. We use a fully unstructured three dimensional mesh to cover the artery, and Galerkin/least-squares finite element method to discretize the incompressible Navier-Stokes equations, that are employed to model the blood flow, and the resulting large sparse nonlinear system of equations is solved by a Newton-Krylov-Schwarz algorithm. From the computed flow fields, we are able to understand certain behavior of the blood flow of this particular patient before and after a stenosis is surgically removed. We also report the robustness and parallel performance of the domain decomposition based algorithm. Wind power is an increasingly popular renewable energy. In the design process of the wind turbine blade, the accurate aerodynamic simulation is important. In the past, most of the wind turbine simulations were carried out with some low fidelity methods, such as the blade element momentum method . Recently, with the rapid development of the supercomputers, high fidelity simulations based on 3D unsteady Navier-Stokes (N-S) equations become more popular. For example, Sorensen et al. studied the 3D wind turbine rotor using the Reynolds-Averaged Navier-Stokes (RANS) framework where a finite volume method and a semi-implicit method are used for the spatial and temporal discretization, respectively . Bazilevs et al. investigated the aerodynamic of the NREL 5MW offshore baseline wind turbine rotor using large eddy simulation built with a deforming-spatial-domain/stabilized space-time We propose and study a new parallel one-shot Lagrange--Newton--Krylov--Schwarz (LNKSz) algorithm for shape optimization problems constrained by steady incompressible Navier--Stokes equations discretized by finite element methods on unstructured moving meshes. Most existing algorithms for shape optimization problems iteratively solve the three components of the optimality system: the state equations for the constraints, the adjoint equations for the Lagrange multipliers, and the design equations for the shape parameters. Such approaches are relatively easy to implement, but generally are not easy to converge as they are basically nonlinear Gauss--Seidel algorithms with three large blocks. In this paper, we introduce a fully coupled, or the so-called one-shot, approach which solves the three components simultaneously. First, we introduce a moving mesh finite element method for the shape optimization We characterize conformally flat spaces as the only compact self-dual manifolds which are U(1)-equivariantly and conformally decomposable into two complete self-dual Einstein manifolds with common conformal infinity. A geometric characterization of such conformally flat spaces is also given. The notion of a generalized CRF-structure on a smooth manifold was recently introduced and studied by Vaisman (2008). An important class of generalized CRF-structures on an odd dimensional manifold M consists of CRF-structures having complementary frames of the form , where is a vector field and is a 1-form on M with ()= 1. It turns out that these kinds of CRF-structures give rise to a special class of what we called strong generalized contact structures in Poon and Wade . More precisely, we show that to any CRF-structures with complementary frames of the form , there corresponds a canonical Lie bialgebroid. Finally, we explain the relationship between generalized contact structures and another generalization of the notion of a CauchyRiemann structure on a manifold. Using deformations of singular twistor spaces, a generalisation of the connected sum construction appropriate for quaternionic manifolds is introduced. This is used to construct examples of quaternionic manifolds which have no quaternionic symmetries and leads to examples of quaternionic manifolds whose twistor spaces have arbitrary algebraic dimension. Guy Bonneau has kindly pointed out two errors in . The first is that the manifolds M (k) and M (k)/Z2 do not admit U (2)-invariant Einstein-Weyl structures for k 2; thus the last four entries in Table 4 (page 422) do not occur. The error is on pages 429430. The analysis there is correct, except that the critical line to consider in Lemma 7.5 and the subsequent calculation is = , instead of = 2/k, because of the constraint (D, ](equivalently, > ) occurring in the definition of . On this line, one has A cohomology theory associated to a holomorphic Poisson structure is the hypercohomology of a bicomplex where one of the two operators is the classical -operator, while the other operator is the adjoint action of the Poisson bivector with respect to the Schouten-Nijenhuis bracket. The first page of the associated spectral sequence is the Dolbeault cohomology with coefficients in the sheaf of germs of holomorphic polyvector fields. In this note, the authors investigate the conditions for which this spectral sequence degenerates on the first page when the underlying complex manifolds are nilmanifolds with an abelian complex structure. For a particular class of holomorphic Poisson structures, this result leads to a Hodge-type decomposition of the holomorphic Poisson cohomology. We provide examples when the nilmanifolds are 2-step. We classify compact anti-self-dual Hermitian surfaces and compact four-dimensional conformally flat manifolds for which the group of orientation preserving conformal transformations contains a two-dimensional torus. As a corollary, we derive a topological classification of compact self-dual manifolds for which the group of conformal transformations contains a two-dimensional torus. Let X be a compact quotient of the product of the real Heisenberg group H _4m+1 of dimension 4m + 1 and the three-dimensional real Euclidean space R ³ . A left-invariant hypercomplex structure on H _4m+1 R ³ descends onto the compact quotient X. The space X is a hyperholomorphic fibration of 4-tori over a 4m-torus. We calculate the parameter space and obstructions to deformations of this hypercomplex structure on X. Using our calculations, we show that all small deformations generate invariant hypercomplex structures on X but not all of them arise from deformations of the lattice. This is in contrast to the deformations on the 4m-torus. A complex symplectic structure on a Lie algebra h is an integrable complex structure J with a closed non-degenerate (2, 0)-form. It is determined by J and the real part of the (2, 0)-form. Suppose that h is a semi-direct product g V, and both g and V are Lagrangian with respect to and totally real with respect to J. This note shows that g V is its own weak mirror image in the sense that the associated differential Gerstenhaber algebras controlling the extended deformations of and J are isomorphic. The geometry of (, J) on the semi-direct product g V is also shown to be equivalent to that of a torsion-free flat symplectic connection on the Lie algebra g. By further exploring a relation between (J, ) with hypersymplectic Lie algebras, we find an inductive process to build families of complex symplectic algebras of dimension 8 n from the data of the 4 n-dimensional ones.

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MATH 54 at the University of California, Berkeley (Berkeley) in Berkeley, California. Add MATH 54 to your schedule. A cluster is an approved concentration of courses in a specific field of applied mathematics. Math 54 - Linear Algebra & Differential Equations -- [4 units] Course Format: Three hours of lecture and three hours of discussion per week. Description: Basic linear algebra; matrix arithmetic and determinants.Vector spaces; inner product spaces. This includes the lectures, the discussion sections, and the office hours. Vector . STAT 154 Modern Statistical Prediction and Machine Learning 4 Units [+] STAT 155 Game Theory 3 Units [+] STAT 156 Causal Inference 4 Units [+] STAT 157 Seminar on Topics in Probability and Statistics 3 Units [+] STAT 158 The Design and Analysis of Experiments 4 Units [+] STAT 159 Reproducible and Collaborative Statistical Data Science 4 Units [+] Math 54 Linear Algebra and Differential Equations. If you have not an access to this site, please let me know. Office hours Tu 11:00 am - 12:00 noon, Th 12:00 noon - 1:00 pm, at the Student Learning Center, and by appt. Jennifer Sixt, 964 Evans, firstname.lastname@example.org Online Guidelines , describing how the course will be delivered online Textbook :Linear Algebra and Differential Equations, Second Third Custom Edition for UC Berkeley, Add MATH 54 to your schedule. . Fourier series. . Tau Beta Pi Engineering Honor Society, California Alpha Chapter As you progress in your degree and remain interested in graduate studies, you should then take Math 54 and 104. Eigenvalues and eigenvectors; orthogonality, symmetric matrices. View and compare grade distributions for each course and semester. Location: Mo tt Library 101. Recent Semesters. Syllabus CATALOG DESCRIPTION . MATH 53: Multivariable Calculus (4) MATH 54: Linear Algebra and Differential Equations (4) PB HLTH . jna.hoteleuropa.ud.it; Views: 7789: Published: .07.2022: Author: jna.hoteleuropa.ud.it: Search: table of content. This email goes only to me and the Head Teaching Assistant, Kevin Li. If you are interested in enrolling in the Spring 2022 Math 54 adjunct courses you must enroll in or be on the waiting list for the appropriate section of Math 54. CS61A, CS61B, CS 61C, CS70/Math 55, CS 188, CS 189, Math 53, Math 54, Math 110, Stat 28, Stat 20/21, Stat 133, Stat 134/140, Data 100 The goal is . Instructor: Nikhil Srivastava, email: firstname at math.obvious.edu. Students with exam credits (such as AP credit) should consider choosing a course more advanced than 1A MATH 1A or MATH N1A MATH 1B or MATH N1B MATH 1B, MATH N1B, MATH 10B, or MATH N10B Part 1; Part 2; Part 3; Part 4; Part 5; Part 6; Part 7; . a course of your interest offered by any department at UC Berkeley; there . EE 16A. Upper-division math and statistics courses for those who are adequately prepared (in order of importance) Math 110, Linear Algebra . A deficient grade in Electrical Engineering 119 may be removed by taking Electrical Engineering 118. Prerequisites: 1A-1B, 10A-10B or equivalent. Linear second-order differential equations; first-order systems with constant coefficients. Math 54, Math 110, or EE 16A+16B (or another linear algebra course), CS 70, EECS 126, or Stat 134 (or another probability course). Search: Berkeley Math 1a Exam. You can also use the direct link. When the Data Science major and minor are approved, the faculty aim to add Stat 89A (in its new, 4-unit Spring 2018 form) to the list of courses to satisfy the linear algebra . Courses. Search: Berkeley Math Courses. Fall 2022, Spring 2022, Fall 2021, Spring 2021 . The Structure and Interpretation of Computer Programs. Prerequisites: MATH 53; EECS 16A and EECS 16B, or MATH 54. Credit Restrictions: Students will receive no credit for Electrical Engineering 118 after taking Electrical Engineering 218A. No textbook. The average on midterms for this class was a 50%, versus the average for Pachter's Spring 2015 class was 92% Here are the course descriptions of these math classes at Berkeley: MATH 1A: This sequence is intended for majors in engineering and the physical sciences at (saddle) 4 Undergraduate Business Courses The Undergraduate program has . NOTE: The first two weeks of instructions will be remote. Math 54 - Linear Algebra And Differential Equations Instructor: Katrin Wehrheim Email is not a sustainable form of communication in this course Contact: via Forum, your GSI, or in person during office hours Lectures: Tue/Thu 5 - 6:30pm in 155 Dwinelle Office Hours: Tue 12:30-1:30pm in Evans 907, Tue/Thu 6:30pm-. Lecturer: Per-Olof Persson, email@example.com. . Therefore, students who are planning a double major or a minor with Astrophysics and Physics are required to complete Physics 89 in lieu of Math 54. . The format of these courses are different than those offered in Fall and Spring but the content is the same so they can be used to satisfy Statistics major math prerequisites. Catalog Description: This course and its follow-on course EE16B focus on the fundamentals of designing modern information devices and systems that interface with the real world. UC Berkeley is committed to providing robust educational experiences for all learners. Search through 12,000+ courses at Berkeley. Catalog Description: An introduction to programming and computer science focused on abstraction techniques as means to manage program complexity. Schedule Planner. "Stat 89A (in its new, 4-unit Spring 2018 form) can be used as an alternate linear algebra co-requisite for Data 100 this semester, along with Math 54 and EE 16A. Please note that taking these bridge courses is suboptimal since in reality, EECS 16A and EECS 16B are best experienced in their totality when the lab experience is integrated with the overall treatment of the material. A deficient grade in Math 54 may be removed by completing Math N54. Lectures: TuTh 8:10am - 9:30am, room 101 Morgan. As far as possible, please use Piazza for mathematical questions. Basic linear algebra; matrix arithmetic and determinants. Browse all available summer online and web-based courses in the Class Schedule. CS 61A. Students will receive no credit for Math 54 after taking Math N54 or H54. If you have then Math 54 maybe be possible to do decently well, but keep in mind that regular high school math is not as rigorous as UC Berkeley math. Math 54, Spring 2021. Chris H. Rycroft, firstname.lastname@example.org. ENGIN 7, MATH 1A, MATH 1B, and MATH 53; and MATH 54 (may be taken concurrently) 3.5 years of high school math, including trigonometry and analytic geometry. * Students who do not complete a Math 54 equivalent before transferring are required to complete Phys 89:"Introduction to Mathematical Physics" at Berkeley. You know that as long as you put in the hard work, you will do well. MATH 54 - Linear Algebra and Differential Equations (recommended) or PHYSICS 89 - Introduction to Mathematical Physics . Math 54. Stat 20 or. Math 53, 54 Multivariable . GSI : Hye Soo Choi. Special arrangements Fourier series. A minimum 2.0 overall GPA is required in all 9 upper division major courses in order to be in good standing in the major. Math 1A & 1B or. To satisfy the requirements of the major, all courses must be taken for a letter grade. Please see the bCourses page for Zoom links. E7 or CS 61A, Physics 7a, Math 53, Math 54 TEXTBOOK(S) AND/OR OTHER REQUIRED MATERIAL Free reader and notes provided. 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Chuck, as he was universally known, was on the UConn faculty for decades and served as Department Head (1997-2003) From a group of academic pioneers in 1868 to the Free Speech Movement in 1964, Berkeley is a place where the brightest minds from across the globe come together to explore, ask questions and improve the world Prerequisites required He wrote . The textbooks for this course are Lay, Linear Algebra and Its Applications and Nagle, Sa , and Snider, Fundamentals of Di erential Equations and Boundary Value Problems. The Plan of Study can be found here. The Bachelor of Science Degree in Chemistry prepares students for careers as professional chemists and serves as a foundation for careers in other fields such as biology and medicine. Grading basis: letter. A two course physics sequence: Physics 7A/7B, or Physics 5A/5B/5BL. Minor Requirements. View and compare grade distributions for each course and semester. Lectures Monday, Wednesday & Friday, 3pm-4pm in Evans 71 . Courses to Complete Before Declaring. Minimum prerequisites include multivariable calculus (Mathematics 53) and matrix algebra (Math-ematics 54). take Phys 89 during the summer before they start at Berkeley (if . bCouses: In bcourses.berkeley.edu, there is a course website. Fall 2022, Spring 2022, Fall 2021, Spring 2021 . Title: Microsoft Word - E-150-SYLLABUS-20-2020.docx Created Date: Game theory models conflict and cooperation between rational decision-making agents. Syllabus of STAT154 (Modern Statistical Prediction and Machine Learning) Instructor: Song Mei (email@example.com) Lectures: T/Th 17:00 - 18:29. Apply filters for requirements. Schedule Planner. Students who have fulfilled the first half of the Physics requirement through AP/IB test scores, transfer work, or with Physics 7A, may opt to . Search: Berkeley Math Courses. Regular Summer Sessions fees and deadlines apply to online classes. 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Massive type IIA string theory cannot be strongly coupled Ofer Aharony1, Daniel Jafferis2, Alessandro Tomasiello3,4and Alberto Zaffaroni3 1Department of Particle Physics and Astrophysics The Weizmann Institute of Science, Rehovot 76100, Israel 2School of Natural Sciences, Institute for Advanced Study, Princeton, NJ 08540, USA 3Dipartimento di Fisica, Universit` a di Milano–Bicocca, I-20126 Milano, Italy INFN, sezione di Milano–Bicocca, I-20126 Milano, Italy 4Albert Einstein Minerva Center, Weizmann Institute of Science, Rehovot 76100, Israel Understanding the strong coupling limit of massive type IIA string theory is a longstand- ing problem. We argue that perhaps this problem does not exist; namely, there may be no strongly coupled solutions of the massive theory. We show explicitly that massive type IIA string theory can never be strongly coupled in a weakly curved region of space-time. We illustrate our general claim with two classes of massive solutions in AdS4×CP3: one, previously known, with N = 1 supersymmetry, and a new one with N = 2 supersym- metry. Both solutions are dual to d = 3 Chern–Simons–matter theories. In both these massive examples, as the rank N of the gauge group is increased, the dilaton initially increases in the same way as in the corresponding massless case; before it can reach the M–theory regime, however, it enters a second regime, in which the dilaton decreases even as N increases. In the N = 2 case, we find supersymmetry–preserving gauge–invariant monopole operators whose mass is independent of N. This predicts the existence of branes which stay light even when the dilaton decreases. We show that, on the gravity side, these states originate from D2–D0 bound states wrapping the vanishing two–cycle of a conifold singularity that develops at large N. arXiv:1007.2451v1 [hep-th] 14 Jul 2010 1Introduction and summary of results One of the most striking aspects of string theory is its uniqueness, realized by the fa- mous “web of dualities” that interconnect its various perturbative realizations. A famous thread in this web connects weakly coupled, perturbative type IIA string theory with its strong coupling limit, M theory (which reduces at low energies to eleven–dimensional It has been known for a while, however, that this duality does not work when the Romans mass parameter F0 , which can be thought of as a space-filling Ramond- Ramond (RR) 10-form flux, is switched on. There is no candidate parameter in eleven– dimensional supergravity to match with F0, unlike for all the other fluxes; nor is there any massive deformation of the eleven–dimensional theory [2–4]. And, from the type IIA point of view, the D0-branes which give rise to the momentum modes in the eleventh dimension at strong coupling do not exist in the massive theory (as there is a tadpole for their worldvolume gauge field). This would then appear to be an imperfection in our understanding of string duality: it would be one string theory whose strong coupling limit is not known. In this paper, we will argue that this strong coupling limit may not exist, and we will show this explicitly at least at the level of weakly curved solutions. In general these are the only solutions we have any control over, unless we have a large amount of supersymmetry; one can separately consider cases with a large amount of supersymmetry, and none of them seem to lead to strong coupling either. (The type I’ theory of contains in some of its vacua strongly coupled regions of massive type IIA string theory, but these regions have a varying dilaton and their size is never larger than the string scale.) Thus, we claim that there is no reason to believe that any strongly coupled solutions exist (with the exception of solutions with small strongly coupled regions), and we conjecture that there are none. This is consistent with the fact that no suggestion for an alternative description of the massive theory at strong coupling is known. In section 2 we provide a simple argument that the string coupling gs in massive type IIA string theory must be small, if the curvature is small. Generically, we find that equations of motion and flux quantization. < ls/R, where R is a local radius of curvature. The argument just uses the supergravity This result is in striking contrast with what happens in the massless case. In the ten–dimensional massless vacuum, for example, the dilaton is a free parameter, and in particular it can be made large, resulting in the M–theory phase mentioned earlier. The massive theory has no such vacua. It is of interest to consider examples with AdS4 factors, where we can take advan- tage of a dual field theory interpretation via the AdS/CFT correspondence, which also provides a non-perturbative definition for the corresponding string theory backgrounds. In particular, it is natural to consider solutions like the N = 6 supersymmetric solution AdS4× CP3of the massless type IIA string theory [6–8]. In this solution, the dilaton is determined by the internal flux integers k ∝? large dilaton with small curvature. In this limit, the solution is better described as the AdS4× S7/ZkM–theory background. The dual field theory has been identified in as the N = 6 superconformal Chern-Simons-matter theory with gauge group U(N)×U(N) and Chern-Simons couplings k and −k. Massive type IIA solutions are also known on AdS4×CP3, and it is natural to compare their behavior to the massless case. For example, some solutions with N = 1 supersym- metry are known explicitly [10,11]; they contain the N = 6 solution as a particular case. The field theory duals are Chern–Simons–matter theories whose levels do not sum up to zero. Even though F0is quantized as n0/(2πls), one might think that introducing the smallest quantum of it, say n0= 1, should have little effect on the solutions, if the other flux integers k and N are already very large. It would seem, then, difficult to understand how a massless solution with large dilaton can suddenly turn into a massive solution with small dilaton when n0is turned on. CP1F2and N ∝? CP3F6, gs∼ N1/4/k5/4, whereas the curvature radius R/ls ∼ N1/4/k1/4. In particular, for N ? k5one has a As we will see in section 3, in general this “small deformation” intuition is flawed. When trying to express the dilaton in terms of the flux parameters, in the massive case one ends up with expressions in which F0multiplies other, large flux parameters. Hence F0 can have a large effect on the behavior of the solutions even if it is the smallest allowed quantum. As it turns out, as we increase N, the dilaton does start growing as gs ∼ N1/4/k5/4, as in the massless case. But, before it can become large, gsenters a second phase, where it starts decreasing with N. Specifically, for N larger than the “critical value” k3/n2 0, we have gs∼ N−1/6n−5/6 . Both behaviors are visible in figure 2. Notice that what happens for these N = 1 solutions is not entirely a consequence of the general argument in section 2. One could have found, for example, that for large N the radius of curvature became small in string units. In such a situation, our supergravity argument would not have been able to rule out a large dilaton; even worse, it would actually generically predict it to be large. It is interesting to ask whether there are situations where that happens. Of course, one would not trust such strongly–curved, strongly–coupled solutions, since we have no control over them; but, if they existed, they would suggest that perhaps strongly coupled solutions do exist and need to be understood. To look for such a different behavior, we turn to a second class of massive solutions, still on AdS4× CP3, but this time with N = 2 supersymmetry. Such gravity solutions were predicted to exist via AdS/CFT , and found as first–order perturbations in F0 of the N = 6 solution in . The field theory duals are again Chern–Simons–matter theories whose levels do not sum up to zero. In section 4 we point out that these theories have certain gauge–invariant monopole operators, whose mass (which is protected by supersymmetry) is independent of the rank N. This suggests the existence of wrapped branes that remain light in the large N limit. This cannot happen for backgrounds which are both weakly–coupled and weakly–curved. To see what happens at large N, in section 5 we find these N = 2 gravity solutions, generalizing the construction in . We reduce the equations of motion and supersymme- try equations to a system of three ODEs for three functions, which we study numerically. As in section 3, we then study the behavior of gsas a function of the flux integers. We find exactly the same phenomenon as in section 3: gsfollows initially the same growth observed for the N = 6 solutions, and departs from that behavior before it can get large. The existence of the light states found in section 4 is not a consequence of strong coupling, but is instead explained by the fact that the internal space develops a conifold singularity where branes can wrap a small cycle. We compute numerically the mass of D2–D0 bound states wrapping the vanishing cycle, and we reproduce very accurately the mass predicted in section 4 from AdS/CFT. Hence, in both examples we examined, the curvature stays bounded almost every- where, and the dilaton does not become strongly coupled. Our argument in section 2 does not rule out the possibility of solutions with large curvature and large dilaton, and it would be nice to find a way to rule them out. In general, such solutions would not be trustworthy, but in some situations one might understand them via chains of dualities. For example, in some cases it might be possible to T–dualize to a massless solution with small curvature, which in turn might be liftable to M–theory, along the lines of . The behavior found in the two examples analyzed in this paper may not be universal, and we expect the AdS/CFT correspondence to be very helpful in any further progress. One motivation for understanding the strong coupling limit of massive type IIA string theory is the Sakai–Sugimoto model of holographic QCD, which has Nf D8–branes separating a region of space with F0 = 0 from a region with F0 = Nf/(2πls). The solution of this model is known in the IR, where it is weakly coupled and weakly curved and the D8–branes may be treated as probes; but it is not clear what happens in the UV, where, before putting in the D8–branes, the coupling became large (see for an analysis of the leading order back-reaction of the D8–branes in this model). Our analysis rules out the possibility that the region of massive type IIA string theory between the D8–branes becomes strongly coupled while remaining weakly curved in the UV. It would be interesting to understand whether there is a sensible UV completion of this model, and, if so, what it looks like. 2 A general bound on the dilaton In this section, we will find a bound for the dilaton for type IIA solutions with non-zero 0-form flux F0?= 0, assuming that the ten–dimensional curvature is small. The argument is simply based on the equations of motion of type IIA supergravity. Note that due to supersymmetry, these equations are actually exact (at two-derivative order) and can be trusted even when the coupling constant becomes large. The Einstein equations of motion in the string frame take the form RMN+ 2∇M∇Nφ −1 2(k − 1)!FMM2...MkFNM2...Mk− The equations (2.1) are valid at every point in spacetime, away from possible branes or orientifolds. On such objects, we would need to include further localized terms, but they will not be needed in what follows. In fact, all we need is a certain linear combination: let us multiply (2.1) by e0Me0N, where e are the inverse vielbeine; 0 is a frame index in the time direction. We can now use frame indices to massage T00on the right hand side: 2(k − 1)!F0A2...AkF0A2...Ak+ (k − 1)!F0A2...AkF0A2...Ak− η00 2(k − 1)!F0A2...AkF0A2...Ak+ We have defined the decomposition Fk= e0∧F0,k−1+F⊥,k. (In particular, F⊥,0is simply F0.) Applying this to (2.1), we get RMN+ 2∇M∇Nφ −1 Again, this is satisfied at every spacetime point (away from possible sources): there is no integral in (2.4). RMN needs to be small in the supergravity approximation. In fact, all the remaining terms in the parenthesis on the left-hand side need to be small too: they are all two–derivative NS–NS terms. If any of them is large in string units, we cannot trust the two–derivative action any more; hence that parenthesis needs to be ? l−2 On the other hand, when F0?= 0, the right-hand side of (2.4) is at least of order one in string units. To see this, recall that RR fluxes are quantized, in appropriate sense. The Fkare actually not closed under d, but under (d − H∧). However, the fluxes ˜Fk= e−B(F0+ F2+ F4+ F6) are closed; these satisfy then the quantization law ˜Fk= nk(2πls)k−1, (2.6) where nkare integers and Caare closed cycles. In particular, F0= n0/(2πls). Since the right-hand side of (2.4) is a sum of positive terms, we get that it is > 1/l2 order one factors). s(up to irrelevant Let us now put these remarks together. Since the parenthesis on the left-hand side is s, and the right-hand side is > 1/l2 s, we have eφ? 1 . (2.7) For generic solutions, the parenthesis on the left hand side of (2.4) will be of order 1/R2, where R is a local radius of curvature. In that case, we can estimate, then, which of course agrees with (2.7). When F0= 0, the conclusion (2.7) is not valid because all the remaining terms on the right hand side can be made small, in spite of flux quantization. For example, assume all the components of the metric are of the same order 1/R2everywhere, and that H = 0. Then, the integral of F is an integer nk, but the value of F2 kat a point will be of order (nk/Rk)2(in string units). At large R, this can be made arbitrarily small. This is what happens in most type IIA flux compactifications with F0= 0; the dilaton can then be made large, and the limit φ → ∞ reveals a new phase of string theory, approximated by To summarize, we have shown that F0?= 0 implies that the dilaton is small (2.7), as long as the two–derivative action (the supergravity approximation) is valid. 3 The N = 1 solutions In this section, we will see how the general arguments of section 2 are implemented in the N = 1 vacua of . 3.1 The N = 1 solutions We recall here briefly the main features of the N = 1 solutions in on AdS4× CP3. The metric is simply a product: Topologically, CP3is an S2fibration over S4. We use this fact to write the internal metric where xiare such that?3 radius, related to the AdS radius by i=1(xi)2= 1, Aiare the components of an SU(2) connection on S4(with p1= 1), and ds2 S4 is the round metric on S4(with radius one). R is an overall RAdS≡ L =R (2σ + 1). (3.3) The parameter σ in (3.2) is in the interval [2/5,2]; this implies, in particular, that L/R is of order 1 for these N = 1 solutions. For σ = 2, (3.2) is the usual Fubini–Study metric, whose isometry group is SU(4) ? SO(6). For σ ?= 2, the isometry group is simply the SO(5) that rotates the base S4. The metric (3.1) depends on the two parameters L and σ. A third parameter in the supergravity solution is the string coupling gs. Yet another parameter comes from the B field. For 2/5 < σ < 2, supersymmetry requires the NS-NS 3-form H to be non–zero (see [10, Eq. (2.2)]). One can solve that constraint by writing ?(2 − σ)(σ − 2/5) σ + 2 where β is a closed two–form [10, Eq. (4.5)]. Because of gauge invariance B∼= B + dλ1, the space of such β is nothing but the second de Rham cohomology of the internal space, H2(CP3) = R. So we have one such parameter, which we can take to be the integral of β over the generating two–cycle in H2, B = − J + β (3.4) where we normalized b so that large gauge transformations shift it by an integer. To summarize, the N = 1 supergravity solutions depend on the four parameters 3.2 Inverting the flux quantization equations We now apply the flux quantization conditions (2.6). It is convenient to separate the contribution from the zero–mode β: k(2πls)k−1, which can be computed explicitly . We have we then define?˜Fk|β=0≡ nb 10 0 0 b1 0 0 b 1 0 l = L/(2πls) ,(3.8) (2 − σ)(5σ − 2) (2σ + 1) √2σ + 1 , ,f4(σ) = −25π2 3 · 52 (σ − 1)(2σ + 1)5/2 σ2(σ + 2)2 (2 − σ)(5σ − 2) , (σ − 1) (σ + 2) f6(σ) = −27π3 3 · 57/2 (σ2− 12σ − 4)(2σ + 1)7/2 σ2(σ + 2)2 Equation (3.7) is [12, Eq. (4.26)], which in this paper we chose to reexpress in terms of l (the AdS radius in string units) rather than r (the internal size in string units), to harmonize notation with section 5. We want to invert these formulas and get expressions for the parameters (l,gs,σ,b) in terms of the flux integers ni, as explicitly as possible. If one assumes b = 0, this is easy ; with b ?= 0, it is a bit more complicated. A good strategy is to consider combinations of the flux integers that do not change under changes of the b field: in addition to n0, two other combinations are We then find 2− 2n0n4= (f2 ?2(σ − 1)(4σ2− 1) ?3(−6 + 17σ − 6σ2)(2σ + 1)3/2 0n6− 3n0n2n4= (f3 We see that (3.11) and (3.12) give two independent expressions for l/gs; this implies 64(σ − 1)3(2σ − 1)3 27σ2(−6 + 17σ − 6σ2)2≡ ρ(σ) .(3.13) This determines σ implicitly in terms of the fluxes. The function ρ(σ) (which we plot in figure 1) diverges at σ =17−√145 ∼ .41, and has zeros at σ =1 have multiplicity three, and hence they are also extrema and inflection points. Moreover, it has a minimum at σ ∼ .65; and it goes to 1 for both σ = 2 and σ =2 We can now combine the equation for n0 in (3.7), which determines gsl, with the expression for l/gsin either (3.11) or (3.12). We prefer using the latter, since it turns out to contain functions of σ which are of order one on most of the parameter space: 2and σ = 1. These zeros (2 − σ)1/4(5σ − 2)1/4σ1/3 (2σ + 1)1/2(−6 + 17σ − 6σ2)1/6 (2 − σ)1/4(5σ − 2)1/4(−6 + 17σ − 6σ2)1/6 The function in the expression for l diverges at σ =17−√145 and 2, whereas the function in the expression for gsvanishes for σ =2 ∼ .41 and vanishes for σ =2 Finally, the second row of equation (3.7) determines b in terms of n2, n0 and the remaining fields l, gs and σ. One could eliminate l and gs from that expression using (3.14) and (3.15), but we will not bother to do so. Figure 1: A plot of the function ρ(σ) in (3.13). 3.3A phase transition We will start by taking for simplicity n4= 0 , (3.16) and we will call n2≡ k ,n6≡ N (3.17) as in . In this case, (3.13) reads 1 + 3Nn2 From the graph in figure 1, we see that the behavior of the solution depends crucially on k3 . If for example we have ρ(σ) ∼ 1. Looking at figure 1, we see that a possible solution is σ = 2. Around this point, ρ goes linearly; so, if we write σ = 2−δσ, we have δσ ∼ (3.15) we then have k3 . From (3.14) and l ∼ δσ1/4 This is the same behavior as in the N = 6 solution . If, on the other hand, we have ρ(σ) ∼ 0. The possible solutions are σ ? 1 or σ ?1 in the expressions for l and gsin (3.14) and (3.15) are then both of order one. We have 2. The σ–dependent functions Notice that this behavior occurs for example in the nearly K¨ ahler solutions of . For those vacua, we have l5/gs= n6and 1/(lgs) = n0, which gives the same behavior as in (3.22). Notice also that σ = 1 corresponds indeed to a nearly K¨ ahler metric. If one were to find a Chern–Simons dual to a vacuum whose only relevant fluxes are n6and n0, such as the nearly K¨ ahler solutions, it would be natural to identify n6with a rank N and n0with a Chern–Simons coupling˜k (because F0induces a Chern–Simons coupling on D2–branes). In such a dual, coupling. We see then that l and gsN in (3.22) are both functions of this˜λ, as expected. From (3.22) one can calculate the finite temperature free energy to be βF ∼ V2T2 N2 at strong coupling βF ∼ V2T2N3/2k1/2. In figure 2 we show a graph of gsas a function of N; we see both behaviors (3.20) and ˜k≡˜λ would then be the new ’t Hooft 0 , which grows with a higher power of N than in the massless case, for which Figure 2: The behavior of gsas a function of N = n6, for n2= k = 100, n4= 0 and n0= 1. We see both the growth in the first phase (3.20), for n6? n3 decay in the second phase (3.22), for n6? n3 0= 106, and the Our analysis above was limited for simplicity to the case n4= 0, but it is easy to argue that also for other values of n4, gscannot become large. Equation (3.15) tells us that and m ≡ n3 from above in the massive theory by the maximal value of |f(σ)|. If m = 0, then (3.12) implies that (−6+17σ −6σ2) also vanishes, and we can then use (3.13) to rewrite (3.15) in the form gs=˜f(σ)/n1/2 ˜ m1/4, where˜f(σ) is again bounded in the relevant range and ˜ m ≡ n2 maximal value of |˜f(σ)|, but this must be true since m and ˜ m cannot vanish at the same time (as is clear from (3.11) and (3.12)). Thus, for any integer fluxes with n0?= 0, gsis bounded from above by a number of order one. 0 m1/6, where f(σ) is bounded from above in the relevant range of values, 0n6− 3n0n2n4is an integer. Thus, if m ?= 0, then gsis clearly bounded 2−2n0n4is another integer. Thus, if ˜ m ?= 0 then gsis bounded from above by the We will now see that the “phase transition” between (3.20) and (3.22) has a sharp con- sequence on the behavior of the probe branes in the geometry. We will consider branes which are particles in AdS4and that wrap different cycles in the internal space CP3. Not all such wrapped branes are consistent. In the N = 6 case, where F0= 0 and tadpole for the world–sheet gauge field A, because of the coupling CP1F2 = n2 ?= 0, the action for a D2-brane particle wrapping the internal CP1has a R×CP1A ∧ F2= n2 (the R factor in the D2-brane worldvolume being time). D0-branes, in contrast, have no such problem. In the field theory, they correspond to gauge–invariant operators made of monopole operators and bifundamentals. For the solutions with both F0? n0?= 0 and? have a tadpole. If one considers a bound state of nD2D2 branes and nD0D0 branes, the tadpole for A is CP1F2? n2?= 0, both D2’s and D0’s A . (3.24) For relatively prime n0and n2, the minimal choice that makes this vanish is nD2= n0 and nD0 = −n2. These branes also correspond to a mix of monopole operators and bifundamentals; we will discuss analogous configurations in more detail in section 5.4. Consider now the case n0= 1, and n2= k ? 1. Here we should consider a bound state of one D2 brane and k D0 branes. In the context of AdS/CFT, all masses are naturally measured in units of the AdS mass scale mAdS≡ is of order L. The masses of a D2 and of a D0 particle would then be (setting the string scale to one) L; recall also from (3.3) that R Thus, the bound states we are considering here (the particles that have no world–sheet tadpole) have a mass of order Which of the two terms dominates? it turns out that the answer depends on which of the two phases, (3.20) or (3.22), we are considering. In both phases the ratio of the two masses is a function of A simple computation gives that, in the first phase (3.20), the D2’s mass is ∼ whereas the k D0 branes have mass k ×k. The D0’s dominate the mass, which then goes In the second phase, the D2’s mass is ∼ N2/3, whereas the k D0’s mass goes like k × N1/3. Hence the D2 dominates. The mass then goes like N2/3 Another type of branes that have no tadpole problems are D4 branes. In the field theory, these correspond to baryon operators. In AdS units, these have a mass of (3.22), which looks reasonable for a baryon. gs. Interestingly, this turns out to be of order N in both phases (3.20) and 3.5Field theory interpretation The field theories dual to the vacua analyzed in this section were proposed in . Because of the low amount of supersymmetry, we do not expect to be able to make here any useful check of this duality. However, we can use our gravity results to make some predictions about those field theories, under some assumptions. First of all, let us recall briefly the N = 1 field theories defined in . They are similar to the N = 6 theory of [9,18], in that they also have a gauge group U(N1)×U(N2). The matter content can be organized in (complexified) N = 1 superfields XI, I = 1,...,4; they transform in the (¯ N1,N2) representation of the gauge group. The biggest difference is that the Chern–Simons couplings for the two gauge groups are now unrelated: we will call them k1and −k2. For k1?= k2, it is no longer possible to achieve N = 6 supersymmetry, and there are several choices as to the amount of flavor symmetry and supersymmetry that one can preserve. In this section, we consider a choice that leads to N = 1 supersymmetry and SO(5) flavor symmetry; in the following sections we will consider a different choice, that leads to N = 2 and SO(4) flavor symmetry. This theory can be written in terms of N = 1 superfields; the superpotential then reads terms are manifestly invariant under Sp(2)=SO(5), as promised. When k1= k2≡ k, the theory has N = 6 supersymmetry when the parameters are c1= −c2= 2π/k, c3= −4π/k. For k1?= k2, this choice is no longer possible, as we already mentioned. In spite of there being only N = 1 supersymmetry, however, it was argued in that there still exists a choice of cithat makes the theory superconformal, as long as k1− k2is small enough with respect to the individual ki. KXL]. Notice that all the If we define the ’t Hooft couplings ,λ±= λ1± λ2, (3.27) the N = 6 theory would correspond to λ+= 0. The argument in then says that there is a CFT in this space of theories if λ+? λ−, although at strong coupling it is difficult to quantify just how much smaller it has to be. Let us now try to translate in terms of these field theories the “phase transition” we saw in section 3.3. To do so, we can use the dictionary (5.35) between the field theory ranks and levels on one side, and flux integers on the other. This dictionary is also valid for N = 1 theories . The phase transition in section 3.3 happens for N ∼ n3 when λ+? λ−we have n0/N ∼ λ+/λ2 −, n2/N ∼ 1/λ−. So the phase transition happens In particular, the “ABJM phase” (3.20) corresponds to λ+??λ−; the “nearly–K¨ ahler” described by the field theories described in and reviewed in this section. However, given the low amount of supersymmetry, this can only be a conjecture at this point. phase (3.22) corresponds to λ+??λ−. At strong coupling, then, there is an intermediate ?λ−? λ+? λ−, where it is possible that the second phase (3.22) is also Rather than trying to test further this correspondence, we will now turn our attention to N = 2 theories, on which there is much better control. 4 Monopoles in N = 2 Chern–Simons–matter theo- In this section, we will recall some general facts about monopole operators in Chern– Simons–matter theories, and we will apply them to a particular quiver theory, similar to the ABJM theory; its gravity dual will be examined in section 5. 4.1Construction of monopole operators in general Consider a d = 3 gauge theory with gauge group?m in the IR, then these must be dimension 2 operators in the IR. There may or may not be operators charged under the corresponding U(1)mflavor symmetry; if they exist we will call them monopole operators. i=1U(Ni). Then there are m currents, ji= ∗ Tr(Fi), which are conserved by the Bianchi identity. If the theory flows to a CFT In a conformal field theory, it is convenient to use radial quantization and consider the theory on R × S2. Let us apply the state–operator correspondence to a monopole operator, with charge vector ni. This results in a state in the theory on an S2, such that S2Tr(Fi) = 2πni. We will denote the diagonal values of Fiby wa taking the trace over the gauge group U(Ni), we have that the magnetic charges are We are interested in d = 3 N = 2 Chern–Simons–matter theories. We can take them to be weakly interacting at short distances by adding Yang–Mills terms as regulators [19–22]. In such a regulated theory, there are BPS classical configurations with the gauge field as in (4.1) and non–trivial values for the scalar fields. The BPS equations on R1,2include the Bogomolnyi equations Fi= ∗Dσi, where σiis the adjoint scalar in the vector multiplet. In R × S2, this equation is different because the metric needs to be rescaled, and the fields need to be transformed accordingly; the equations then read Fi= σivolS2. Notice in particular that the σiare constant. There are also other BPS equations, which involve the other scalars in the theory (for explicit computations for N = 3 theories, see [20, §3.2], and in N = 2 language, ). After adding the regulating Yang-Mills term, g2 N = 2 vector multiplet should be treated classically, while the chiral matter fields should Y Mbecomes small in the UV, so the not. This justifies not solving Gauss’s law in describing the “pure” monopole operator, which then behaves as if it were a local, non-gauge invariant chiral field . The scalar σiis set by the BPS equations to be equal to the inverse Chern–Simons level times the moment maps of the matter fields, and to have a spin 0 operator, one should satisfy the constraint for any bi–fundamental field Xij connecting the i-th and j-th gauge group [24–26]. In addition, the fields Xijneed to satisfy the F-term equations of the N = 2 theory. We see from (4.3) that some matter fields, which should be neutral under the background U(1), are necessarily non vanishing in the background; hence, the possibility of satisfying all equations gives a nontrivial constraint on the possible BPS monopole operators. In general, monopole operators T creating such configurations at a point will not be gauge invariant. However, they will behave exactly like local fields. Hence, they can be combined with other local operators O, to write gauge–invariant expressions of the where the indices are contracted as appropriate for the representations in which the op- Let us determine how the monopole transforms under the gauge group. This is easy to find for the Abelian factors of the theory. There is an obvious contribution to the electric charges of a monopole operator, from the Chern–Simons term?ki nikiunder the ithelectric U(1) Abelian factor. ?Tr(Ai∧ Fi). This says that a monopole with magnetic charges niwill behave like a particle with charges This result gives a constraint on the possible gauge invariant operators Tr(TO) we can obtain. If all matter is in bifundamental and adjoint representations, no gauge invariant operators can be formed from monopoles that are charged under the overall U(1), since no matter field transforms under it. Since we just computed the electric charge under the ithU(1) to be kini, the charge under the overall U(1) is?kini. Thus, if we are to form any gauge invariant operators of the form (4.5), we need to require kini= 0 .(4.6) This result will be useful in the theories with gauge group U(N1)×U(N2), which we will discuss in section 4.2. Let us now ask how the monopole will transform under the full non-Abelian group. One method to determine this is the following. One fixes a particular configuration on the sphere (breaking the gauge symmetry), one computes the charge under the whole Cartan subalgebra of the gauge group, and then one integrates over the gauge orbits. Thus monopole operators can be labelled by U(1) subgroups of the gauge groups. For the U(Ni) factor of the gauge group, the charges under this Cartan subalgebra are the wa with a = 1,...,Ni, that we saw in (4.1). Therefore, a monopole associated with magnetic flux wa weight vector [27, Sec. 4.2] iis in the representation with of the ithgauge group. (Our notation here is that the weight vector denotes the number of boxes in each row of a Young diagram of the representation; thus (k,0,...,0) is, for example, the completely symmetric representation.) Quantization in a background of the type discussed in this section can result in anoma- lous contributions to the charges and energy of the state. For the non–chiral theories we we will consider in section 4.2, there is no such a correction to the gauge charges1. How- ever, as we will discuss in section 4.2, the dimension of the monopole operator is given by the energy of the state on the sphere, which often includes a non–zero Casimir energy. 4.2 Dimensions and charges of the monopoles We will apply in this section the results of section 4.1 to the Chern–Simons–matter theories with N = 2 and N = 3 given in . In particular, we will compute the dimensions of particular monopoles, which will be useful later, when comparing to the gravity solutions of section 5. Let us first recall some details about the field theories of interest. They are similar to the N = 6 theory of [9,18], in that they also have a gauge group U(N1) × U(N2), and N = 2 “chiral” superfields Ai,Bi, i = 1,2; the Aitransform in the (¯ N1,N2) representation, 1If the matter content is chiral, as for the theories in , there will be an additional one–loop correction to the gauge charges. One way to understand this effect is that the state on the sphere has a constant value for the scalar in the vector multiplet, which gives a mass to any matter fields charged under that U(1) subgroup in which the magnetic flux lives. Integrating them out at one–loop can shift the effective Chern–Simons level in that background if the theory has chiral matter. whereas the Bitransform in the (N1,¯ N2). Just as in section 3.5, the crucial difference between the N = 6 theory and the N = 2 theory is that the Chern–Simons couplings for the two gauge groups are now unrelated; we again call them k1and −k2. The theories we want to consider in this section are defined by the superpotential W = Tr(c1(AiBi)2+ c2(BiAi)2) .(4.8) For generic ci, the theory has N = 2 supersymmetry and SU(2) flavor symmetry. For still SU(2). For c1= −c2= c, the supersymmetry stays N = 2, but W can be rewritten W = cTr(?ij?klAiBkAjBl) , ki, supersymmetry turns out to be enhanced to N = 3, while the flavor symmetry is which shows that the flavor symmetry is enhanced to SU(2)×SU(2). This N = 2 theory is dual to the gravity solution discussed in the next section. We will now apply to these theories the discussion of section 4.1 about monopole operators. The following computation is a straightforward generalization of that done in [19,20] for the N = 3 theory. The results for the N = 3 and N = 2 theories appear to be identical, since the flavor symmetry guarantees that the matter fields have the same dimensions as in the more supersymmetric theory. As we saw in section 4.1, there are non–trivial conditions on the scalars for the monopole to be BPS. For the theories we are considering, the conditions read AA†− B†B =k1 B B†− A†A = −k2 together with the F-term constraints coming from (4.8). In view of (4.1) and Fi= σivol2, such equations relate the magnetic fluxes wa can consider are defined by magnetic charges wa w1= (1,...,1,0,...) (with n11’s) and w2= (1,...,1,0,...) (with n21’s). The non-zero elements of the fields Aiand Biare n1×n2and n2×n1rectangular matrices, respectively, which are required to satisfy the first two lines in (4.10) and the F-term constraints. The problem of finding appropriate vacuum expectation values for the matter fields is equivalent to finding the BPS moduli space of the generalized U(n1) × U(n2) Klebanov– Witten theory with superpotential (4.8) and Fayet-Iliopoulos (FI) parameters turned on. Many of these moduli spaces are non–empty. 1with the wa 1which are all either 0 or 1: namely, 2. The simplest monopoles we Using now (4.6), we see that this monopole operator can be coupled to elementary fields in a gauge invariant way only if The matter content is non–chiral, so there are no anomalous contributions to the gauge charges of the monopole. There is, however, a one–loop correction to the dimension of the operator, which is given by [20–22] ∆ = −1 2 × 1(n1(N1− n1) + n2(N2− n2)) − 4 ×1 = (n1− n2)2− (N1− N2)(n1− n2) , 2 2(n1(N2− n2) + n2(N1− n1)) where R is the R-charge of the fermion and q the charge under the U(1) subgroup specified by the vectors w1= (1,..n1..,1,0,...) and w2= (1,..n2..,1,0,...). The various contribu- tions arise as follows. The four bi-fundamental fermions have R-charge −1/22. Each bi-fundamental fermion is a matrix with N1N2entries; the n1(N2− n2) + n2(N1− n1) off-diagonal entries have charge ±1 under the magnetic U(1) subgroup, while the remain- ing entries are neutral. The two adjoint gauginos have R-charge +1. They are square matrices with N2 charge ±1, while the other are neutral. We used the fact that in this theory, both for N = 3 and N = 2, the R–charges in the UV and IR are identical. In the N = 6 theory, k1= k2and it follows from (4.11) that n1= n2. The simplest monopole has just w1= w2= (1,0,...,0). We need to turn on Aiand Bifields that solve the U(1) Klebanov-Witten theory with a FI term. According to (4.7), such a monopole transforms in the k-fold symmetric representation of U(N1) and in the conjugate k-fold symmetric representation of U(N2). The monopole can combine with k fields Aito form a gauge-invariant operator (we can analogously form a gauge-invariant operator with the conjugate monopole and k fields Bi). ientries; the 2ni(Ni− ni) off-diagonal entries of the i-th fermion have In the N = 3 and N = 2 theories, we cannot have n1= n2, but we can now take n1= k2and n2= k1and rectangular matrices Aiand Bithat solve (4.10). In general, the 2Since the superpotential (4.8) must have R-charge +2 and there is a discrete symmetry between Aiand Biwe have that R(Ai) = R(Bi) = 1/2; the R-charge of the fermionic partners is R(A) − 1 by matrices AiBjwill not be diagonalizable. Recalling equation (4.7), the monopole opera- tor is in a representation of the gauge group with weight vectors (k1,..k2..,k1,0,...) and (k2,..k1..,k2,0,...). A gauge invariant combination must include k1k2matter bifundamen- tals (if k1and k2are not relatively prime, some operator with smaller dimension could exist). The total dimension of the gauge-invariant operator, dressed with k1k2elementary fields, is then + (k2− k1)2− (k2− k1)(N1− N2) .(4.13) Note that we have determined the vacuum expectation values of the matter fields needed to “support” the flux to form a BPS state on the sphere using the classical moduli space. This is justified since the Higgs branch does not receive quantum corrections. More precisely, the ring of chiral operators is the ring of algebraic functions on the moduli space. There is a natural map [24–26] from the moduli space of Chern–Simons–matter theories to the moduli space of the four–dimensional Yang–Mills theory with the same field content. That moduli space cannot receive quantum corrections (aside from wavefunction renormalization which fixes the coefficients of the superpotential), and only the S1bundle over that space, associated to the dual gauge fields, is quantum corrected. This precisely corresponds to 1-loop corrections to the charges and dimensions of monopole operators, which are, however, constructed in the UV weakly coupled Yang–Mills–Chern–Simons– Let us summarize the results of this section. The monopole operators that create k2 units of flux for the first gauge group and k1 for the second have k1k2 bifundamental indices, and hence we can contract them with k1k2bifundamental fields to construct a gauge–invariant operator. Such operators have dimension given by (4.13). In particular, they stay light when N1= N2≡ N → ∞. Since in general monopole operators correspond to D–branes, this seems to indicate a limit where D–branes become light, which usually signals some sort of breakdown of the perturbative description. We will see in section 5.4 precisely how this happens. 5 The N = 2 solution We now turn to writing and studying the N = 2 solution predicted to exist in , and found in at first order in F0. This solution will be the gravity dual of the field theory defined by the superpotential (4.9), and it will serve as another illustration of the general result of section 2. We will start in section 5.1 by reducing the equations of motion and the supersym- metry conditions to a system of three equations for three functions of one variable. This procedure closely parallels , where an analogous solution for the gravity dual of the Chern-Simons theory based on the C3/Z3quiver was found. In section 5.2, we will impose flux quantization, and derive expressions for the supergravity parameters in terms of the flux integers; in section 5.3 we find, just like in section 3.3, a “phase transition” that prevents the dilaton from growing arbitrarily large. Finally, in section 5.4 we find light D–brane states dual to the monopole operators discussed in section 4.2. 5.1 The N = 2 solutions The ten dimensional metric we will consider is a warped product of AdS4with a compact six-dimensional internal metric with the topology of CP3: As discussed in [28,29], there is a foliation of CP3in copies of T1,1, which is in turn a S1fibration over S2× S2. The usual Fubini–Study metric can be written as where Ai, i = 1,2, are one-form connections, with curvatures 16sin2(2t)(da + A2− A1)2, (5.2) where Jiare the volume forms of the two spheres S2 interval [0,π/2]; all the functions in our solution (including A in (5.1)) will depend on this coordinate alone. At one end of the interval [0,π/2], one S2shrinks; at the other end, the other S2shrinks. To make this metric regular, we take the periodicity of a to be 4π. i. The coordinate t parametrizes the The Fubini–Study metric is appropriate for the N = 6 solution, which has F0= 0. Once we switch F0on, as we saw in section 4.2, AdS/CFT predicts the existence of an N = 2 solution with isometry group SU(2)×SU(2)×U(1) (the first two factors being the flavor symmetry which is manifest in (4.9), and the third being the R–symmetry). The internal metric for such a deformed solution is then given by3 64Γ2(t)(da + A2− A1)2.(5.4) 3One could have reparameterized the coordinate t so as to set one of the functions in (5.4) to a constant, for example ?, as in (5.2). We have chosen, however, to fix this reparameterization freedom in another way: by choosing the pure spinors (A.8) to be as similar as possible to those for the N = 6 solution, see in particular (A.11). Were the functions e2Binon-vanishing, we would have a metric on the total space of an S2bundle over S2× S2. To maintain the topology of CP3, we require that e2B2vanishes at t = 0 and e2B1vanishes at t = π/2. With an abuse of language, we will refer to t = 0 as the North pole and t = π/2 as the South pole, although there is no real S2fiber. To have a regular metric, ?(t) and Γ(t) must behave appropriately at the poles. It is convenient to define the combinations which control the relative sizes of the two S2’s. As discussed in Appendix A, the super- symmetry equations reduce to three coupled first order ordinary differential equations for w1, w2and a third function ψ which enters in the spinors: Ct,ψ(w1+ w2) + 2cos2(2t)w1w2 Ct,ψ(w1+ w2)cos2(2ψ) + 2w1w2 sin(4t)Ct,ψ(w1+ w2)cos2(2ψ) + 2w1w2 sin(4t)Ct,ψ(w1+ w2)cos2(2ψ) + 2w1w2 Ct,ψ(w1w2− 2w2− 2sin2(2ψ)w1) Ct,ψ(w1w2− 2w1− 2sin2(2ψ)w2) Ct,ψ≡ cos2(2t)cos2(2ψ) − 1. (5.7) All other functions in the metric and the dilaton are algebraically determined in terms √2eA(cot(ψ) − tan(ψ))csc2(2t) sin(2ψ) − cos(2ψ) cot(2t)ψ? Γ = 4eAsin(2t) + cos(2t)cot(2t)sin2(2ψ) 2?1 + cot2(2t) sin2(2ψ) F0csc(4t) sec(2ψ) tan(2ψ) Here, c is an integration constant, that so far is arbitrary. The fluxes are determined as well, and have the general form 2?1 + cot2(2t) sin2(2ψ) e3A−Φ= csec(2ψ) 1 + cot2(2t) sin2(2ψ) . (5.11) F2= k2(t)e2B1J1+ g2(t)e2B2J2+˜k2(t)i 2z ∧ ¯ z , 2z ∧ ¯ z ∧ J1+ ˜ g4(t)e2B2i F4= k4(t)e2B1+2B2J1∧ J2+˜k4(t)e2B1i 2z ∧ ¯ z ∧ J2, 2z ∧ ¯ z ∧ J1∧ J2, ˜ gican be found in (A.12). The fluxes satisfy the Bianchi identities, which require that 2z ∧ ¯ z = 16√2dt∧(da+A2−A1). The full expressions for the coefficients ki,˜ki, gi, ˜F ≡ e−B(F0+ F2+ F4+ F6) (5.13) is closed. This dictates in particular that F0is constant. We can now study the regularity of the differential equation near its special points, t = 0 and t = π/2, by finding a power series solution of the equations. The general solution will depend on three arbitrary constants. However, we are after solutions with particular topology, where w2vanishes at t = 0 (the “North pole”) and w1vanishes at t = π/2 (the “South pole”). Near t = 0, we obtain ψ = ψ1t −2 w1= w0+ (4 + 4ψ2 w2= (4 + 4ψ2 1− 2w0+ 2w0ψ2 1)t2+ O(t4) . In our solution, w0and ψ1are not independent: imposing that w1vanishes at t = π/2 determines w0in terms of ψ1. The power series expansion in˜t ≡ π/2 − t near t = π/2 is identical, with the role of w1and w2exchanged: ψ =˜ψ1˜t −2 w1= (4 + 4˜ψ2 w2= ˜ w0+ (4 + 4˜ψ2 1− 2 ˜ w0+ 2 ˜ w0˜ψ2 1)˜t2+ O(˜t4) . The constants ˜ w0and˜ψ1should also be determined by ψ1; we can then think of ψ1as the only parameter in the internal metric. To find a solution with the required topology, we note that the equations (5.6) are symmetric under the operation t →π w1 ↔ w2, and we look for solutions which are left invariant by this symmetry. This determines˜ψ1= ψ1and ˜ w0= w0, and it allows us to restrict the study of the equations to the “north hemisphere” t ∈ [0,π/4]. The only thing left to impose is that the solution is differentiable at t = π/4. This is what determines w0as a function of ψ1, which we plot in figure 3. w0(ψ1) is monotonicaly decreasing; our numerical analysis shows that it vanishes at a point very well approximated by ψ1=√3. 2− t,ψ → −ψ, The perturbative expansion of the solutions near the “poles” t = 0 and t = π/2 allows to check the regularity of the six-dimensional metric. In fact, the only special points in the metric are the poles, where a copy of S2degenerates. Using the previous expansion, Figure 3: A plot of w0as a function of ψ1. It vanishes linearly around the point ψ1=√3. it is straightforward to check that, at both poles, the shrinking S2combines with (t,a) to give a piece of the metric proportional to Si + (da ± Ai)2? Thanks to the fact that the periodicity of a is 4π, this is the flat metric of R4. For all ψ1∈ [0,√3) the metric is then regular. For ψ1= pole and the metric develops two conifold singularities. ψ1 = limiting point in our family of solutions. √3, both spheres degenerate at each √3 is thus the natural We can examine now the number of parameters in the solution. As discussed above, the differential equations provide just one parameter, ψ1, the value of the derivative of ψ at the North pole t = 0. It is convenient to define two more parameters by gs≡ eφ|t=0,2L = eA|t=0. (5.17) Both φ and A vary over the internal manifold, but numerical study reveals that they only do so by order one functions. So gsand L can be thought of as the order of magnitude of the dilaton and AdS radius in our solutions4. We can now reexpress the integration constant c by evaluating (5.11) at t = 0: ?1 + ψ2 4The normalization has been chosen so that in the metric, at t = 0, L2multiplies an Anti-de Sitter space of unit radius, and the relation between the mass of a particle at t = 0 and the conformal dimension of the dual operator is mL = ∆(∆−4). This normalization is related to the fact that, in our conventions, AdS4has cosmological constant Λ = −3|µ|2and, as discussed in appendix A, we chose µ = 2. The F0flux is then determined by evaluating (5.10) at t = 0: ?1 + ψ2 Finally, a fourth parameter comes from the B field. As in section 3, there is a zero– mode ambiguity coming from the presence of a non–trivial cohomology in our internal manifold. To see this, let us call B0a choice of B–field such that H = dB0solves the equations of motion. For example, we can choose B0such that ˜F2= F2− B0F0= 0 .(5.20) H = dB0is guaranteed to solve the equations of motion, since equation (5.20) implies that dF2= HF0, which we have already solved. However, this will also be true for any B of the form B = B0+ β ,(5.21) for any β which is closed. We can apply to this β the same considerations as in section 3.1: because of gauge invariance B∼= B + dλ1, the space of such β is nothing but the second de Rham cohomology of the internal space, H2(CP3) = R, so we have one such parameter. And, just as in (3.5), we define the integral of β over the generating two–cycle in H2: b ≡ not generate confusion, as the contexts are different. CP1β. The fact that we use the same notation as in section 3 should Summarizing, our solutions are parameterized by the four numbers (L,ψ1,gs,b). The situation is very similar to the N = 1 solutions we studied in section 3, with σ replaced 5.2Inverting the flux quantization equations This section will follow closely the corresponding treatment for the N = 1 solutions in section 3.2. The equations are formally very similar: b1 0 0 b 1 0 where l = L/(2πls), as in (3.8). The vector on the left hand side is given by the integrals Ck˜Fk, where Ckis the single k–cycle in CP3(k = 0,2,4,6), and˜Fkis defined using the particular B0in (5.20); this also explains why the second entry of the vector is zero (this is simply our choice for the definition of b). We could have made such a choice for the N = 1 solution as well; we did not do so because for SU(3) structure solutions there is a different and particularly natural choice of B–field. Using equation (5.19) we can write f0(ψ1) = − ?1 + ψ2 We know the other functions fk(ψ1) only numerically. We obtain 2f4(ψ1) by integrating ˜F4 over the diagonal S2times the “fiber” (t,a), which is a representative of twice the fundamental four-cycle. The plots of fk(ψ1) are given in figure 4. Our numerical analysis indicates the following asymptotics at the two extrema ψ1= 0, ψ1=√3: f4∼ (√3 − ψ1) , for ψ1→ 0 ; Figure 4: Plots of f4(ψ1) and f6(ψ1). Their asymptotic behavior near 0 and√3 is given in (5.24), (5.25). Notice in particular that f6(√3) is small but non–zero. We can now proceed in the same fashion as in the N = 1 case to determine ψ1from the flux parameters. Namely, we write the combination 0n6− 3n0n2n4)2= − 9f6(ψ1)2f0(ψ1)≡ ρ(ψ1) , (5.26) which allows us to determine ψ1in terms of fluxes. l and gsare then given by 1 + 3n2 1 + 3n2 A crucial role is played by the function ρ(ψ1) which we plot in Figure 5. It decreases monotonically from 1 at ψ1= 0 to zero at ψ1=√3. Its asymptotic behavior at ψ1= 0 and ψ1=√3 is: ρ ∼ (√3 − ψ1)3for ψ1→ ρ ∼ 1 − ˜ cψ2 1for ψ1→ 0 , for some constant ˜ c. This is in agreement with (5.24), (5.25). The fact that ρ vanishes at the same point, ψ1=√3, where the solution develops a singularity, is strongly supported by our numerical analysis, and will be crucial in reproducing the field theory results. Figure 5: A plot of the function ρ(ψ1) in (5.26). 5.3A phase transition As in the N = 1 case, consider for simplicity the case n4= 0 and call, as usual, n6= N and n2= k. Equation (5.26) becomes As in the N = 1 case, there are two interesting regimes. For N ? k3/n2 we are near the undeformed solution. From (5.29), we see that 1 − ρ(ψ1) ∼ ψ2 ρ(ψ1) =1 + 3n2 0, ψ1→ 0 and 1; hence we can identify in this regime Moreover, we see from (5.23) that f0(ψ1) ∼ ψ1; using also (5.24), we easily compute from which is indeed the behavior of the N = 6 solution . For N ? k3/n2 √3. From (5.29) and (5.30), we see that 0, the function ρ(ψ1) should approach zero, and this happens for ψ1→ ?√3 − ψ1 From (5.25) we then conclude which is the same behavior as in (3.22). Again, as in the N = 1 case, we can also argue generally that gsremains bounded for any integer values of the fluxes, with n0?= 0. At first sight, this seems puzzling. At the end of section 4.2, we noticed that this gravity solution is expected to develop light D–branes in the limit N1= N2= N → ∞. As argued in , N1= N2precisely when n4= 0 (see also (5.35)). But there do not seem to be any light D–branes in a limit where gsis small and the internal manifold is large, since a D–brane mass scales as Lk/gs, with k ≥ 0. As we remarked after equation (5.16), however, in the limit ψ →√3 (which is relevant for large N) the internal manifold develops two conifold–like singularities, since the two- cycle is now shrinking to zero at the “poles”. As we will now see, the new light states are obtained from D-branes wrapping the vanishing cycle for that singularity. We now want to compare this gravity solution with the field theory we saw in section 4.2; specifically, the one defined by the superpotential in (4.9), which has the right symmetries to be the dual of the gravity solution we found in section 5.1. We can first of all try to predict what sort of bulk field corresponds to the monopole operators discussed in section 4.2. Let us recall how the duality works in the ABJM case, when F0= 0. Consider first a monopole operator that creates one unit of field strength for both gauge groups at a particular point. This operator has k indices under both gauge groups, and we can make it gauge-invariant by contracting it with k bifundamentals. The resulting bound state corresponds to a D0 brane in the gravity dual; notice that such a brane has no tadpole on its worldsheet for the worldsheet vector potential A, as we already saw in section 3.4. Another monopole operator that can be considered is the one that creates one unit of flux for, say, the second gauge group. In this case, we cannot make this operator gauge–invariant: it will have k “dangling” indices. This corresponds to a D2 brane wrapping an internal two–cycle. As we also already saw in section 3.4, such a brane has a tadpole on its worldsheet, coming from the term?A1F =?F2A; so one needs to have k strings ending on it, and these k strings correspond to the k indices of the monopole operator. When we switch on F0, even a D0 brane will have a tadpole on its worldsheet, coming from the coupling?F0A. On the field theory side, this corresponds to the fact that the monopole operator that creates one unit of field strength for both gauge groups has now k1 fundamental indices for the first gauge group and k2 antifundamental ones for the second. This cannot be made gauge-invariant; we are always left with at least |k1− k2| “dangling” indices. This fact was used in to establish that the Romans mass integer is the sum of the Chern–Simons couplings, so that, in the present language, n0= k1−k2; see also [30,31]. In it was similarly shown that n4is the difference between the two gauge group ranks N2− N1.5Putting this together, we obtain a dictionary between the flux integers and the ranks and levels of the field theory: n0= k1− k2,n2= k2,n4= N2− N1,n6= N2. (5.35) In section 4.2, we considered monopole operators which create k2units of field strength for the first gauge group, and k1units of field strength for the second. We noticed that these have k1k2bifundamental indices, and thus can be made gauge–invariant. Following the identifications of D2 branes and D0 branes above, if we assume for example that k1> k2, we can say that these new gauge–invariant monopoles correspond to a bound state k2D0 branes and k1−k2D2 branes. We have already noticed in section 3.4 that such a bound state can cancel the tadpole on the worldsheet, because it makes the prefactor in (3.24) vanish. Let us make this expectation more precise. Consider a D2 brane wrapped on a two– cycle B2in the N = 2 solution. As we will see in appendix B.2, supersymmetry requires 5The relative sign between the expressions for n4and n0had not been determined so far. We made here a choice consistent with our final result in formula (5.43). that the D2 brane lives at the North pole t = 0 or at the South pole t = π/2, and that it wraps the S2that does not shrink there. We also need to cancel the tadpole for the world-volume field A which arises from the Wess-Zumino coupling, A ∧ (F2+ F0(F − B)) . We can split B into a fiducial choice plus a zero mode, as in (5.21). The tadpole cancel- F − B = F − β − B0= −F2/F0. Since B0was chosen to satisfy (5.20), we need to turn on a world–volume flux F = β . (5.38) There is a possible obstruction to doing this, coming from the quantization of the world– volume flux, that says that is given by b = −n2/n0; hence in general it is rational and not an integer. So we see that a single D2 brane is generally not consistent. We can get around this, however, by considering n0D2–branes. In that case, the equation we want to satisfy actually reads S2F ∈ Z. The value of b = S2β from (5.22) F = β 1n0. (5.39) The integral of the trace of the left hand side is the first Chern class on the world–volume, which is the induced D0-brane charge n0. The integral of the trace of the right hand side now gives bn0= −n2. We conclude that we can cancel the tadpole by considering a bound state of n0D2 branes and n2D0 branes, just as in section 3.4. Naively, one might think that the mass of a D2–D0 bound state should be at least as heavy as a D0-brane, which in units of AdS mass is mD0L ∼ L/gs. Since this is heavy in the limit (5.34), one might think that such a bound state can never reproduce the light mass predicted in section 4.2. Fortunately, such pessimism proves to be unfounded. The mass of the state is given mD2L = n0L det(g + F − B) = n0L where we used the tadpole cancellation condition. We will take the cycle B2 to be a representative of the non–trivial cycle, which is the diagonal of the two S2’s. Using the explicit form for the metric in (5.4), as well as (5.12), (5.18) and (A.12), we get: mD2L = 4π 1 + ψ2 The fact that the two expressions under the square root are proportional is related to the BPS condition, as discussed in appendix B.2. Inserting the values of l and gsfrom (5.27),(5.28) for generic fluxes we obtain 1 + ψ2 Quite remarkably, the function of ψ1 in the previous formula, which can be computed numerically, turns out to be constant with value 1. The final result for the mass formula Upon using the dictionary (5.35), this formula is identical to the field theory prediction (4.13) in the limit where n0 ? n2. This is exactly the limit where we can trust the supergravity solution, since, as shown in (5.27), for a generic value of ψ1, L is large only if n2? n0. In this limit, it is also true that the dimension of the corresponding operator is given by ∆ ∼ mD2L. In contrast with the N = 1 results in section 3.4, and with the naive expectation expressed earlier, we see from (5.43) that the mass of the bound state remains finite also in the limit N ? k3/n2 of the constituent D0-brane, leaving a smaller piece that is proportional to the volume of the shrinking S2. These are precisely the new light states that we had predicted to exist from the field theory analysis in section 4.2. 0. A contribution from the B field cancels the large mass ∼ L/gs O. A. is supported in part by the Israel–U.S. Binational Science Foundation, by a research center supported by the Israel Science Foundation (grant number 1468/06), by a grant (DIP H52) of the German Israel Project Cooperation, and by the Minerva foundation with funding from the Federal German Ministry for Education and Research. D. J. wishes to acknowledge funding provided by the Association of Members of the Institute for Advanced Study. A. T. and A. Z. are supported in part by INFN and MIUR under ASupersymmetry equations and pure spinors for the N = 2 solution We will give in this section more details about the N = 2 solution we found in section 5. The supersymmetry parameters for compactifications of the form AdS4× M6 (or Minkowski4× M6) decompose as Here, N is the number of supersymmetries. The subscripts ± denote positive and negative chirality spinors, in four and six dimensions; the negative chirality spinors are conjugate to the positive chirality ones, For each a, ζa elements of this basis to be “Killing spinors”, which means that Dµζ+=µ with i = 1,2, are a priori independent six–dimensional Weyl spinors. In this section, we will consider N = 2. A priori, one could have taken the ζain ?1and ?2to be different. This can indeed be done for compactifications with vanishing RR flux; for example, for the usual N = 2 Calabi–Yau compactifications. To recover that case in (A.1), one can take for example η21= η12= 0, and keep a non–vanishing η11and η22. However, in compactifications where RR fluxes are present, the ζain ?1and ?2are required to be equal, up to a constant that can be reabsorbed in the ηia. Hence (A.1) describes all possible N = 2 compactifications, and is particularly appropriate for vacua with RR fluxes. +can vary among a basis of four–dimensional Weyl spinors; we will take the 2γµζ−. The ηia Using (A.1) in the supersymmetry equations yields equations for the internal spinors ηia. In fact, these equations do not mix the ηi1with the ηi2. In what follows, we will first analyze the equations of the ηi1≡ ηi; we will come back to the second pair later. We will construct a pair of pure spinors as tensor products of the supersymmetry 6As usual, we left implicit a Clifford map on the left hand side, that sends dxm→ γm. The type IIA supersymmetry conditions can be expressed as : (d − H∧)(eA−ϕRe (Φ−)) = 0 , (d − H∧)(e3A−ϕIm (Φ−)) = −3e2A−ϕµIm (Φ+) +e4A (d − H∧)(e2A−ϕΦ+) = −2µeA−ϕRe (Φ−) ; ||Φ+|| = ||Φ−|| = eA. ∗ λ(F) , (A.4b) Here, F are the internal fluxes (which determine also the external fluxes, by self–duality). A is the warping function, defined as ds2 has no independent meaning, since one can reabsorb it in A. We have normalized µ = 2 in this paper. The symbol λ acts on a k–form by multiplying it by the sign (−)Int(k/2). Finally, the norm in (A.4d) is defined as ||A||2= i(A ∧ λ(¯A))6. The metric (5.4) can be written in terms of the vielbein 6. The cosmological constant AdS4is given by Λ = −3|µ|2. Since A is non–constant in the solution, however, this Λ 8Γ(da + A2− A1), where ei= dθi+isinθidφiare the natural one–forms on the spheres S2 and eB2= sin(t) we recover the Fubini–Study metric of CP3with natural K¨ ahler form (5.31)]). It is also convenient to use the forms i. For eB1= cos(t) i=1Ei∧¯Eiand natural three form section Ω = E1∧E2∧E3(see for example [28, 2ei∧ ¯ ei (not summed) ,o ≡i 2eiae2∧ ¯ e1; (A.6) the Jiwere already defined in (5.3). These forms satisfy dJi= 0 ,do = i(da + A2− A1) ∧ o ,o ∧ ¯ o = −J1∧ J2. (A.7) The generic pure spinors corresponding to an SU(3)×SU(3) structure can be written in terms of the “dielectric Ansatz” 2z ∧ ¯ z 8sin(2ψ)eA+iθz ∧ exp where θ and ψ are two new angular variables; one can see easily that the supersymmetry equations (A.4) relate them by tan(θ) = −cot(2t) sin(2ψ) .(A.9) The one–form z and the two–forms j and ω can also be used to describe an SU(2) structure on M6. For our solution, these forms are given by z = −ie−iθE1, ω = iE2∧¯E3. We can also characterize j and ω in terms of the forms in (A.6): , Im(ω) =1 The RR fluxes are determined to be as in equation (5.12) with cos(2t)(2Ct,ψ+ w2),g2= −ce−4A (2Ct,ψ(w1+ w2) + 3w1w2), ˜ g4= −ce−4A sin(4t)cos2(2ψ)(2Ct,ψ(w1+ w2) + w1w2), where Ct,ψwas defined in (5.7). Recall also that one possible choice of NS–NS field that satisfies the equations of motion is B0= F2/F0, as in (5.20). So far we have described the solution as if it were an N = 1 solution: we have only paid attention to the a = 1 part of (A.1). To show that the solution actually has N = 2 supersymmetry, we have to provide a second pair of spinors, ηi2, that satisfies the equations of motion for supersymmetry with the same expectation values for all the fields. In terms of pure spinors, we can now form the bilinears and require that they solve again the equations (A.4), with the same values of the fluxes and the same metric. In fact, one expects the two solutions Φ and˜Φ to be rotated by R–symmetry, so that there is actually a U(1)’s worth of solutions to (A.4). To see this U(1), rotate the two–form o in (A.6) by a phase:7 o → e−iαo ≡ oα. (A.14) We can correspondingly define a pair of pure spinors Φα appears. The crucial fact about the rotation of o in (A.14) is that it keeps its differential properties (A.7) unchanged: namely, doα= i(da + A2− A1) ∧ oα. Because of this fact, the computations to check (A.4) do not depend on α; and, since we checked already that α = 0 gives a solution, it follows that any Φα solution with different fluxes; but we can see from (5.12) that o never appears in Fk. We conclude, then, that the solution we have found is an N = 2 solution. ±, by changing o → oαwherever it ±is a solution. A priori, this could be a B BPS particles In this section, we will give a general analysis of BPS particles in flux compactifications (subsection B.1), and we will then apply those general results to the N = 2 background described in section 5 and appendix A. B.1 General considerations We will start with some general considerations about BPS states in N = 2 backgrounds with fluxes. These will in general be states that are left invariant by a certain subalgebra of the supersymmetry algebra. This subalgebra is in general defined by the fact that the two supersymmetry parameters ?iare related: Γ??2= ?1. (B.1) 7Alternatively, one can change the vielbeine (A.5) by translating a → a + α. In first approximation, Γ?is the product of the gamma matrices parallel to the brane. When B fields or worldsheet fluxes F are present, Γ?receives additional contributions of eF−B. We will give a definition later on, in the context needed for this paper; for the general and explicit expression, see for example [33, Eq. (3.3)]. For an AdS4× M6 compactification, we would like to use the decomposition (A.1). For particles, this will lead to an equation involving the four–dimensional spinors ζi follows, the index0is meant to be a frame index. To have a chance to solve the resulting equations, we need to postulate a relation between these spinors. One can write for ±; here and in what for some matrix A. (Recall that in general the index a runs from 1 to N; for us, N = 2, and so a = 1,2.) In fact, (B.2) is almost the most general choice one can make, compatibly with the symmetries of the problem. The only generalization one could make would be to multiply the left-hand side by another matrix Bab. Whenever this matrix is invertible, one can reabsorb it by a redefinition of Aab. In this sense, we can say that (B.2) is the “generic” Ansatz for a BPS particle. The matrix Aabin (B.2) needs to satisfy certain conditions. Let us work for simplicity in a basis where all the space–time gamma matrices γµ, µ = 0,...,3 are real, and the internal γm, m = 1,...,6, are purely imaginary; the ten–dimensional gamma matrices are then given as usual by Γµ= eAγµ⊗ 1 ,Γm+3= γ5⊗ γm. (B.3) It follows from these definitions that Γµare real. Let us now conjugate (B.2); using (A.2), the fact that γ0is real, and that γ2 0= −1, we get If we were considering an N = 1 background, Aabwould be a one–by–one matrix, and (B.4) would have no solution. This is just what one would expect: there are no BPS particles in a N = 1 background. For N = 2, one choice that satisfies (B.4) is A = e−iλ We can now use (B.3) to write Γ?= γ0⊗ γ?, (B.6) where γ?is now an element of the internal Clifford algebra; it contains the product of all the internal gamma matrices parallel to the brane, plus additional contributions from the worldsheet flux and B-field. Let Bp⊂ M6be the p–cycle wrapped by the brane, of dimension p and with coordinates σα, α = 1,...,p. Then we define the natural volume form on B to be One can also define similarly an “inverse volume form” as the multivector det(g + F − B)dσ1∧ ... ∧ dσp. (B.7) ∂1∧ ... ∧ ∂p ?det(g + F − B) which is a section of Λp(TB). This multivector can be used to give an intrinsic definition of γ?: here is how. We can define eFvol−1to be the multivector of mixed degree that one obtains by contracting the indices of the form eFwith the multi–vector vol−1. Recall now that multivectors can be “pushed forward”: if we call x : B ?→ M6the embedding map, with components xm(σ), then x∗(eFvol−1) is a multivector in M6, obtained by contracting all indices α on B with the tensor ∂αxm. In fact: B) . (B.9) Here, we left implicit on the right hand side a Clifford map that sends a vector ∂minto a gamma matrix γm. We already used this map on forms (see footnote 6). One can show that γ?is unitary: ?γ?= 1 .(B.10) For a more explicit expression of γ?, see [33, Eq. (3.5)]. If we now use (B.2), (B.6) and (A.1) in (B.1), we get For our choice (B.5), this reads We are now left with solving (B.12), which are two purely internal equations. Each of the two equations is formally identical to others that have already appeared in the context of BPS objects which do exist in N = 1 flux compactifications: branes which extend along the time direction, plus one, two or three space directions. Hence we can simply follow the same steps; we will now summarize that procedure for (B.12a), and then apply the result to (B.12b). Let us first define the new pure spinors notice that these are different from the pure spinors Φ±, defined in (A.3), which entered the supersymmetry equations (A.4). In (A.3), η1and η2were to be understood as η1a and η2a, for a either 1 or 2. In (B.13), we are mixing a = 1 with a = 2. A possible basis for the space of spinors of positive chirality is given by η11 Three linear combinations of the γmmake γmη11 γi, where i is a holomorphic index with respect to an almost complex structure I. Ex- plicitly we have η11† The coefficients a and bm have a geometrical interpretation. multiply (B.14) from the left by η11† From the formula Tr(? A? B†) =8 we see that Tr(γ?Ψ† (B.9), we see that γ?contains factors of ∂αxm; when contracting with¯Ψ+, these factors reconstruct a pull–back of that form. In conclusion we get8 −vanish: they are its three “annihilators” += (1+iI)mn≡ 2¯Πmn. In terms of this basis a priori one can To compute a, we can + ; we get aeA= η11† + ) = Tr(γ?Ψ† +) consists of contracting the free indices of γ?with those of¯Ψ+. From where |Bdenotes the top–form part on B of the pull–back. By similarly multiplying (B.14) from the left by η11† − γn, we get (dxm· eF−BΨ−)|B= −1 where · denotes the Clifford product: v· = v ∧ +v?. Here, ∂m?(dxm1∧ ... ∧ dxmp) ≡ We can now go back to (B.12a). Comparing to the expansion (B.14), we get a = −eiλ,bm= 0 . (B.18) 8The factor of eAcomes from the fact that ∀a,i, ||ηia|| = eA/2, which follows from (A.4). Using the geometrical interpretations (B.16) and (B.17), we get (v · eF−BΨ−)|B= 0 . Actually, one can show that (B.19) is equivalent to the system (B.20). To see this, observe that γ?is unitary, as we saw in (B.10). This implies that γ?η22 same norm as η22 +should have the +. Since all the spinors have norm eA(see footnote 8), it follows that |a|2+ 2bm¯bm= 1 .(B.21) This means that imposing Re(a) = 1 is equivalent to imposing Im(a) = 0 and bm= 0. Recalling (B.16) and (B.17), we get our claim that (B.19) is equivalent to (B.20). This completes our analysis of (B.12a) (along the lines of ). For (B.12b), similar considerations apply; we obtain (v · eF−B˜Ψ−)|B= 0 , for the pure spinors Let us now summarize this section: we have shown that a brane wrapping an internal cycle B, and extended along the time direction, is BPS if and only if (B.19) (or equivalently (B.20)) is satisfied by Ψ and, analogously, (B.22) (or equivalently (B.23)) is satisfied by ˜Ψ, where Ψ and˜Ψ are defined respectively in (B.13) and (B.24). We will now compute these pure spinors for the solution described in section 5 and in appendix A. B.2 D2/D0 bound states in the N = 2 solution As discussed in the previous section, in order to study the supersymmetry of BPS particles obtained from wrapped branes, we need to form bilinears in the supersymmetry spinors given by the pair of spinors defining the SU(3) × SU(3) structure in (A.10). Recall that a SU(3) structure is specified by two invariant tensors (J,Ω) or, equivalently, by a spinor η+(of norm 1) such that +. We first need to write them explicitly. A convenient basis to expand our spinors is The SU(3) × SU(3) structure in (A.10) can be seen as the intersection of two SU(3) structures given by (J1,Ω1) = (j +i call the corresponding spinors η+and χ+. They are related by χ+= denotes the Clifford multiplication by the one-form zmγm. We will need in the following an expression for the tensor products of a generic linear combination 2z∧ ¯ z,ω∧z) and (J2,Ω2) = (−j +i 2z∧ ¯ z,−¯ ω∧z). We √2z · η−, where z· µ+= aη++ bχ+, ν+= xη++ y χ+. This is given by a¯ xe−ij+ b¯ yeij− i(a¯ yω + ¯ xb¯ ω) i(by¯ ω − axω) + (bxeij− aye−ij) We can choose the spinors for the first supersymmetry as follows It is easy to reproduce, using formula (B.27), the dielectric ansatz (A.8) for the pure As discussed in appendix A, there is a U(1) family of supersymmetries obtained by rotating o → oα= e−iαo. We can conveniently choose as a second independent supersym- metry the one with oπ= −o. This is defined by 4cos(ψ) ˜ η+− ie−iπ 4cos(ψ) ˜ η++ ie−iπ 4sin(ψ) ˜ χ+) , 4sin(ψ) ˜ χ+) , ˜ η+= −icos(2t)η++ isin(2t)χ+, ˜ χ+= isin(2t)η++ icos(2t)χ+. This reproduces the rotated pure spinors Φπ With these ingredients, we can compute the spinors Ψ±and˜Ψ±defined in (B.13) and (B.24) and check the BPS conditions for a D2-brane. It is easy to see that the D2-brane considered in section 5, which wraps the diagonal S2and sits at the North or South pole, is indeed supersymmetric. Let us consider, for definiteness, the North pole. At t = 0, ψ = 0 and we see that ηi2 +. As a consequence, at t = 0, ˜Ψ±= −iΦ±, (B.31) and we are reduced to check expressions for the pure spinors Φ±at the North pole. Taking into account that ψ = 0 there, we have 8eA+iθz ∧ ω . The condition (B.20b) for Ψ−(and the analogous (B.23b) for˜Ψ−) gets contributions only from the contraction with the vector z and it is automatically satisfied because ω vanishes at the North pole, t = 0. It is easily seen that the conditions for Ψ+and˜Ψ+are equivalent and it is enough to analyze those for Ψ+. Equation (B.20a) reads Im?ei(θ−λ)e−ij?∧ eF−B|B2= 0, and determines the world-volume field F = (B + cot(θ − λ)j)|B2. (B.34) We see that a wrapped D2 brane can be made supersymmetric by choosing an appropriate world-volume field. However, as discussed in section 5.4, to have a consistent BPS state we need to impose the quantization of the world-volume field and the cancellation of tadpoles. As discussed there, the quantization condition requires to take n0D2-branes. On the other hand, the tadpole condition requires F = β or, equivalently, F −B = −F2/F0. At t = 0, using the explicit form for the metric in (5.4), as well as (5.12), (5.18), (A.10), (A.9) and (A.12), we evaluate tan(θ) = −ψ1and j = −1 that J1is the volume form of one of the two S2’s, as defined in (5.3). We thus see that the tadpole condition is satisfied by λ = 0. The mass of n0D2 branes is then obtained by integrating the volume form in (B.19) 4e2B1J1and F2/F0= −1 det(g + F − B) = n0 J = n01 1 + ψ2 Using this, one exactly reproduces the result (5.41) of section 5.4. A more detailed analysis of equations (B.20) (and the analogous ones for˜Ψ±) shows that a D2-brane sitting at t ?= 0,π/2 cannot be supersymmetric and simultaneously satisfy the tadpole condition. L. J. Romans, “Massive N=2a Supergravity in Ten Dimensions,” Phys. Lett. B169 A. Sagnotti and T. N. Tomaras, “Properties of Eleven–Dimensional Supergravity,”. K. Bautier, S. Deser, M. Henneaux, and D. Seminara, “No cosmological D = 11 supergravity,” Phys. Lett. B406 (1997) 49–53, hep-th/9704131. S. Deser, “Uniqueness of D = 11 supergravity,” hep-th/9712064. J. Polchinski and E. Witten, “Evidence for Heterotic – Type I String Duality,” Nucl. Phys. B460 (1996) 525–540, hep-th/9510169. B. E. W. Nilsson and C. N. Pope, “Hopf fibration of eleven-dimensional supergravity,” Class. Quant. Grav. 1 (1984) 499. S. Watamura, “Spontaneous compactification and CPN: SU(3) × SU(2) × U(1), sin2(θW), g3/g2 and SU(3) triplet chiral fermions in four dimensions,” Phys. Lett. B136 (1984) 245. D. P. Sorokin, V. I. Tkach, and D. V. Volkov, “On the relationship between compact- ified vacua of d = 11 and d = 10 supergravities,” Phys. Lett. B161 (1985) 301–306. O. Aharony, O. Bergman, D. L. Jafferis, and J. Maldacena, “N=6 superconformal Chern-Simons-matter theories, M2-branes and their gravity duals,” JHEP 10 (2008) A. Tomasiello, “New string vacua from twistor spaces,” Phys. Rev. D78 (2008) P. Koerber, D. L¨ ust, and D. Tsimpis, “Type IIA AdS4compactifications on cosets, interpolations and domain walls,” JHEP 07 (2008) 017, 0804.0614. D. Gaiotto and A. Tomasiello, “The gauge dual of Romans mass,” JHEP 01 (2010) M. Petrini and A. Zaffaroni, “N = 2 solutions of massive type IIA and their Chern– Simons duals,” JHEP 09 (2009) 107, 0904.4915. C. M. Hull, “Massive string theories from M–theory and F–theory,” JHEP 11 (1998) T. Sakai and S. Sugimoto, “Low energy hadron physics in holographic QCD,” Prog. Theor. Phys. 113 (2005) 843–882, hep-th/0412141. B. A. Burrington, V. S. Kaplunovsky, and J. Sonnenschein, “Localized Backreacted Flavor Branes in Holographic QCD,” JHEP 02 (2008) 001, 0708.1234. K. Behrndt and M. Cvetic, “General N = 1 supersymmetric fluxes in massive type IIA string theory,” Nucl. Phys. B708 (2005) 45–71, hep-th/0407263. O. Aharony, O. Bergman, and D. L. Jafferis, “Fractional M2-branes,” JHEP 11 (2008) 043, 0807.4924. D. Gaiotto and D. L. Jafferis, “Notes on adding D6 branes wrapping RP3in AdS4× M. K. Benna, I. R. Klebanov, and T. Klose, “Charges of Monopole Operators in Chern–Simons Yang–Mills Theory,” JHEP 01 (2010) 110, 0906.3008. D. L. Jafferis, “Quantum corrections to N = 2 Chern–Simons theories with flavor and their AdS4duals,” 0911.4324. F. Benini, C. Closset, and S. Cremonesi, “Chiral flavors and M2-branes at toric CY4 singularities,” JHEP 02 (2010) 036, 0911.4127. S. Kim and K. Madhu, “Aspects of monopole operators in N = 6 Chern–Simons theory,” JHEP 12 (2009) 018, 0906.4751. D. L. Jafferis and A. Tomasiello, “A simple class of N = 3 gauge/gravity duals,” JHEP 10 (2008) 101, 0808.0864. D. Martelli and J. Sparks, “Moduli spaces of Chern–Simons quiver gauge theories and AdS4/CFT3,” Phys. Rev. D78 (2008) 126005, 0808.0912. A. Hanany and A. Zaffaroni, “Tilings, Chern–Simons Theories and M2 Branes,” Download full-text JHEP 10 (2008) 111, 0808.1244. A. Kapustin, “Wilson–’t Hooft operators in four-dimensional gauge theories and S– duality,” Phys. Rev. D74 (2006) 025005, hep-th/0501015. D. Gaiotto and A. Tomasiello, “Perturbing gauge/gravity duals by a Romans mass,” J. Phys. A42 (2009) 465205, 0904.3959. M. Cvetic, H. Lu, and C. N. Pope, “Consistent warped–space Kaluza–Klein re- ductions, half–maximal gauged supergravities and CPnconstructions,” Nucl. Phys. B597 (2001) 172–196, hep-th/0007109. M. Fujita, W. Li, S. Ryu, and T. Takayanagi, “Fractional Quantum Hall Effect via Holography: Chern–Simons, Edge States, and Hierarchy,” JHEP 06 (2009) 066, O. Bergman and G. Lifschytz, “Branes and massive IIA duals of 3d CFT’s,” JHEP 04 (2010) 114, 1001.0394. M. Gra˜ na, R. Minasian, M. Petrini, and A. Tomasiello, “A scan for new N = 1 vacua on twisted tori,” JHEP 05 (2007) 031, hep-th/0609124. L. Martucci and P. Smyth, “Supersymmetric D–branes and calibrations on general N = 1 backgrounds,” JHEP 11 (2005) 048, hep-th/0507099. R. Minasian, M. Petrini, and A. Zaffaroni, “Gravity duals to deformed SYM theories and generalized complex geometry,” JHEP 12 (2006) 055, hep-th/0606257.

US 8046168 B2 A geofence system which locates a position as within or without the complex polygon type geofence using a simplified algorithm. The algorithm obtains a position and compares it to the polygon by establishing a ray from the position constructed in a cardinal direction of the coordinate system. The “polarity” of the count of intersections between the ray and geofence indicates whether the position is inside the geofence or not. 1. A method of locating a position as being within or without a defined geographic area, the method comprising the steps of: defining a geofence as a closed circuit of a plurality of straight line boundary segments in a rectangular Cartesian coordinate system to define the geographic area; obtaining a position in the rectangular Cartesian coordinate system; constructing a ray from the position within the rectangular Cartesian coordinate system; locating intersections between the ray and the plurality of line segments; counting the number of intersections and characterizing the position as within or without the geofence based on the count, wherein the ray constructing step is done with the ray oriented in a cardinal direction of the rectangular Cartesian coordinate system, wherein the rectangular Cartesian coordinate system is a system of latitude and longitude used for a globe; defining boundary segments such that they include only end point vertex; locating boundary segments parallel to the ray and excluding any intersections between the ray and parallel boundary segments from the count of intersections; locating extreme dimensions of the geofence; constructing a rectangle in the rectangular Cartesian space from four boundary segments oriented in the cardinal directions of the rectangular Cartesian space, including the extreme dimensions of the geofence and enclosing the geofence; characterizing as outside the geofence any point outside of the rectangle; following definition of the geofence, averaging the coordinate values of the vertices of the geofence; and setting the average values as the origin of a new coordinate system and resetting all vertices in terms of the new origin, wherein the step of counting further comprises characterizing the position as within the geofence if the count is odd and as without the geofence if the count is even. 1. Technical Field The invention relates to geofencing for vehicles, and more particularly, to providing a simplified method for determining if a vehicle's position is within or without an irregular geofence. 2. Description of the Problem A geofence may be defined in part as a virtual spatial boundary. Geofences are a byproduct of the marriage of mobile, inexpensive telecommunications platforms and data processing systems. They are enhanced in accuracy by making use of global positioning systems which allow accurate, precise determination of the location of both the boundary of a geofence and position of a mobile platform relative to the geofence. A geofence typically operates by triggering of a physical response through the mobile, location sensitive device when the device crosses a boundary, though the system on which the method is implemented may simply operate to note the fact of the crossing to an operator. The spatial location of geofences have commonly been established by selecting a point feature, which may be a point defined by latitude and longitude, and then defining either a radius, or major/minor axis for the point feature, to establish a boundary around the point. United States Pat. Appl. Pub. 2006/02003005 described construction, using graphical user interface tools, of geofences with boundaries corresponding to real world objects. The area enclosed by such a fence would typically be an irregular polygon, the expression of which might result from simple selection of an object to be enclosed by the fence, such as the campus of a school. An irregular geofence was recognized as more readily applied to real world situations. Among the issues raised with respect to such irregular polygon shaped fences was the difficulty of fully and correctly representing the actual polygon for devices with limited data processing resources. The reference provided the contingency of “decimating” the polygon to avoid overwhelming the resources of such devices. According to the invention there is provided a method of determining whether a reported position is within or without a geofence. The method provides the steps of first defining a geofence as a closed circuit of a plurality of straight line boundary segments in a rectangular Cartesian coordinate system. Next, a position is obtained in the rectangular Cartesian coordinate system. A ray is constructed from the position within the rectangular Cartesian coordinate system. Intersections between the ray and the plurality of boundary segments are located. The number of intersections is counted and the position is characterized as within the geofence if the count of intersections is odd and as without the geofence if the count is even. Typically the ray is constructed oriented in a cardinal direction of the rectangular Cartesian coordinate system. Intersections between the ray and boundary segments parallel to the ray are excluded from the count. Boundary segments are defined such that they include only one end point vertex to avoid double counting intersections where the ray passes through a vertex. Additional effects, features and advantages will be apparent in the written description that follows. The novel features believed characteristic of the invention are set forth in the appended claims. The invention itself however, as well as a preferred mode of use, further objects and advantages thereof, will best be understood by reference to the following detailed description of an illustrative embodiment when read in conjunction with the accompanying drawings, wherein: Referring now to Referring now to Further simplification is possible. At step 310 the polygon is examined to locate instances of three or more consecutive ordered vertices being located in a straight line. If such instances are found any intermediate vertices (i.e. “vertices” having an angle of 180 degrees) are eliminated. At step 314 the ordering of vertices is reset and the variable N, indicating the number of vertices, is recalculated. At step 316 a further simplification step is taken, by right shifting coordinates to a fixed, limited precision. With the simplification operations complete, one additional test condition is established. At step 318 the extreme points on the polygon in each cardinal direction are located relative to the origin are found. Next, at step 320, an “out-of-contingency” rectangle is constructed having dimensions based on the extreme points. This rectangle will be used as a quick filter for eliminating some reported positions as possibly being inside the polygon. Determination of whether a vehicle's position (P) is within or without of a geofence is a new process, spawned for each reading on position taken, as indicated by step 321. At step 322, the position (P) of the vehicle is located from the Global positioning system constellation 50. The position is adjusted using the normalization generated in steps 306 and 308. The adjusted position is readily compared with the boundaries for the out-of-contingency rectangle to determine if the position P is within the rectangle (step 324) and thus possibly inside the polygon. If the position P is not within the rectangle the position is either outside the polygon or on a border segment of the polygon. Following the NO branch to step 326, the position P is checked for correspondence to the border of the polygon. If the position P is not on the border of the polygon it lies outside of the polygon and the NO branch is followed to step 328 where the “Out-of-Boundary” flag is set. Otherwise step 330 is executed to set the “On-Boundary” flag. Following either step 328 or 330 the routine returns the result and is re-executed as determined by the operator's protocol. If the position P is within the rectangle, that is, within the extreme dimensions of the polygon, more processing is required to determine if the position is within the polygon. Initially a flag or counter is set to zero (step 332). Next, at step 334, a line segment L is calculated from the current (vehicle) position P in a cardinal direction (typically eastward) beyond the maximum extent of the polygon to the east (or in the selected cardinal direction). Typically, the number of intersections between this line segment and the boundary segments of the polygon can be used to determine if position P is within, without or on the boundary. Special provision must be made for the possibility that line segment L passes through one of the vertices or includes all or part of a boundary segment which runs in the same cardinal direction as line segment L. Steps 336 and following deal with determining the number of intersections between the constructed line segment originating at position P and boundary segments of the polygon. If the number of intersection is odd, it indicates that position P lies within the polygon. If even, the sum indicates that position P lies outside of the polygon. As indicated already, cases where line segment L (or ray) includes all or part of a boundary segment or pass through one of the vertices are handled in a special manner to avoid double counting an intersection between L and a vertex as an intersection with two boundary segments. Essentially each vertex is treated as part of the higher numbered boundary segment only (although the opposite approach could be take). The number of boundary segments connecting vertices of the polygon equals the numbers of vertices and can be numbered j=1 to N+1 with boundary segment j connecting vertices i and i+1 (step 336). For every boundary segment Bj the intersection between the boundary segment and line segment L is found if it exists (step 340). If the line L and a boundary segment run together it is not treated as an intersection. This is dealt with by simply ignoring all boundary segments which run east/west (step 338) which skips step 340 for the noted boundary segments. At step 342 the coordinates of all the intersections are determined. If, at step 344, the coordinates for a given boundary segment lies between the vertices for the polygon than the flag count is increased by one for that intersection (step 346). Once all boundary segments have been considered (step 347) the flag is examined (step 348) to determine if it is odd, which if it is (the YES path) a within-polygon determination (step 352) results. If the flag is not odd, position P is not within the polygon (step 350). An intersection at a vertex can be considered to count for increasing the flag count. Steps 354, 356 and 358 are used to located intersections between line L and vertices by asking first if the intersection is at a vertex, and in the following two steps if, for the boundary segment under consideration, whether the intersection is at the start or end of the boundary segment. If the intersection comes at the end of a boundary segment than the intersection is counted (step 360), otherwise not. Those skilled in the art will now appreciate that alternative embodiments of the invention can exist. For example, the flag indicating whether the position P is within or without the geofence could be implemented in ways other than as a counter. While the invention is shown in one of its forms, it is not thus limited but is susceptible to various changes and modifications without departing from the spirit and scope of the invention.

APPLIED THERMODYNAMICS FOR SECOND YEAR / THIRD SEMESTER EEE DEPT. UNIT I BASIC CONCEPTS INTRODUCTION Thermodynamics is defined as the branch of science which deals with the relations between energy and heat. These relations are governed by the laws of thermodynamics. These laws are based on the principle of energy conversion. It states that energy can be changed from one form to another but the total energy remains constant. In other words energy cannot be created or destroyed. APPLICATIONS OF THERMODYNAMICS • • • • • • Power plants IC engines Turbines Compressors Refrigeration Air-conditioning UNITS AND DIMENSIONS All physical quantities are characterized by dimensions. Dimensions of physical quantities may be defined as the properties in terms of quality not of magnitude by which a physical quantity may be described. Length (L), area (A)and volume (V)are all different dimensions which describe certain measurable characteristics of an object, e.g., A=L2 and V=L3 The arbitrary magnitudes assigned to the dimensions are called as units. In other words, a unit is a definite standard by which a dimension is to be measured. The primary or fundamental dimensions are length L in m, mass m in kg, time in sec and temperature T in K. The secondary or derived dimensions are velocity V in m/s, Energy E in J and volume V in m these are expressed in terms of primary dimensions. SYSTEM OF UNITS The most common system of unit is metric system SI, which is also known as the International System. In this text, the SI (System International) system of units has been used. Energy: Energy is defined as the capacity to do work. The various forms of energy are heat energy, mechanical energy, electrical energy and chemical energy. Unit of energy is Nm or Joule (J) and kWh. 1 kWh = 3.6 x 106 J The energy per unit mass is defined as specific energy whose unit is J/kg. Force: Force acting on a body is defined by Newton s second law of motion. According to this law, force is proportional to the product of mass and acceleration. When a force of one Newton applied to a body having mass of one kilogram, gives it an acceleration of one m/s. The unit of force is Newton (N). 1 N = l kgm/s Weight of a body (W) is the force with which the body is attracted to the centre of the earth. It is the product of its mass (m) and the acceleration due to gravity. i.e., Work: Work is defined as the work done when the point of application of 1 N force moves through a distance of 1m in the direction of the force, whose unit is Joule or Nm. The amount of work (W) is the product of the force (F) and the distance moved (L), W = F×L. Power: Power is defined as the rate of energy transfer or the rate of work. The unit of power is watt (W) 1N m/s = 1 MW = Pressure: Pressure is defined as the force per unit area exerted whose unit is N/m2 which is also known as Pascal (Pa) and for larger pressures, kPa (Kilo Pascal) and MPa (Mega Pascal) are used. Other units for pressure not in the SI units but commonly used are bar and standard atmosphere (atm) 0.1 MPa = 100 kPa = 105Pa 105 N/m = 1 bar 1 atm = 101 .325 kPa = 1.01325 bar 1J/s =1W 106 Kw W= mg (Value of g = 9.81 m/s at sea level) Mostly pressure of a fluid is measured by gauge which gives pressure relative to atmospheric pressure and is called as gauge pressure. In thermodynamic analysis one is mostly concerned with absolute pressure which is the pressure exerted by a system on its boundary. For pressures above atmospheric Pabsolute = pgauge + patm For pressures below atmospheric, the gauge pressure will be negative and is called as vacuum. U-tube manometer which is used to measure pressure, the two arms of the tube are connected to two containers which are at p and p pressures. The tube is filled with a fluid having density and h is the difference in the heights of the fluid columns. By the hydrostatics principle, P1-P2 = Where hg p2 in N/m2 is in kg/m3 h in m, g in m/s2 and then p1 Fig:1(a). For pressure above atm Temperature: Fig:1(b). For pressure below atm Temperature is defined as the degree of coldness or hotness of a body. When heat is added to the body, its temperature increases and when heat is removed from the body, its temperature decreases. Temperature is the thermal condition of a body on which its capacity of transferring heat to or receiving heat from other bodies depends. Thus the temperature determines direction, in which the heat flow will take place, Units of temperature are degree Celsius, degree Kelvin Temperature K = Temperature º C + 273 Under standard temperature and pressure (STP) conditions the temperature of a gas is taken as 15°C and the pressure as 760 mm of mercury. Under normal temperature and pressure (NTP) conditions the temperature of a gas is 0°C and the pressure as 760 mm of mercury. Specific Heat: Specific heat of a substance is defined as the quantity of heat required to raise the temperature of unit mass substance to one degree. Average specific heat, Where Q is heat interaction kJ, T is Temperature difference K and m is mass kg. If the state of the substance is liquid or solid there is only one specific heat. For the case of gaseous substances there are two specific heats, they are: 1. Specific Heat at Constant Volume: When the volume of the gas is constant the quantity of heat required to raise the temperature of unit mass of gas to one degree is termed as specific heat of gas at constant volume which is denoted by Cv 2. Specific Heat at Constant Pressure: When the heat is supplied at constant pressure the quantity of heat required to raise the temperature of unit mass of gas to one degree is termed as specific heat of gas constant pressure which is denoted by Cp CLASSICAL APPROACH MACROSCOPIC AND MICROSCOPIC APPROACH The behavior of one matter can be studied from macroscopic and microscopic points of view. v The macroscopic approach is only concerned with overall effect of the individual molecular interactions. v The microscopic point of view concerned with every molecule and analysis of collective molecular action is carried out by statistical techniques. For example, pressure is a macroscopic quantity, which is defined as the normal force exerted by a system against unit area of the boundary, i.e., the pressure exerted on the vessel is equal to the mean change of momentum of all the molecules exerted perpendicular to unit area of the boundary. This approach is not related with individual molecular action. This pressure can be measured by using pressure gauge. The microscopic approach is used to explain some matter which otherwise difficult to understand by macroscopic approach. THERMODYNAMIC SYSTEM AND SURROUNDINGS A thermodynamic system is a region in space or any matter or specified quantity of matter within a prescribed boundary on which we concentrate. The other matters outside of the boundary are known as surroundings. As shown in Fig.2(a) the system and surroundings are separated by boundary. The boundary may be real or imaginary one. The system is classified into three: • • • Closed System Open System Isolated System Closed System: In this system, the boundaries are closed so that there is no mass transfer. But there may be energy transfer into or from the system, while mass remains constant. This is also known as control mass. e.g., bomb calorimeter. Open System: In this system, the boundaries are not closed and mass and energy transfer may take place through the opening(s) in the boundary. This is also known as control volume. e.g., turbines and compressors. Fig:2(a)A thermodynamic system,(b)Closed system,(c) Open System, (d) Isolated system Isolated System: This system is not affected by surroundings. In this there is no mass or energy transfer across the boundary of the system. WORKING MEDIUM: In most of the devices the working medium is gas or vapor. It is important to know the properties and behavior of the working medium to observe and analyze the working of devices. At various pressures and temperatures the properties of the working fluid can be determined by using pure substance concept. • • The pure substance is defined as a substance that has a fixed chemical composition, e.g., water, nitrogen, helium and carbon-di-oxide. A mixture of two or more pure substances is also called as pure substance as long as the chemical composition is same. A mixture of liquid air and gaseous air cannot be called as pure substance because the mixture is not chemically homogeneous due to different condensation temperatures of the components in air at specified pressure. THERMODYNAMIC EQUILIBRIUM A system is said to be in a state of thermodynamic equilibrium if there is no change in the microscopic properties at all points in the system. For thermodynamic equilibrium, the following three types of equilibrium conditions have to be satisfied. Mechanical Equilibrium: A system is said to be in a state of mechanical equilibrium if there is no unbalanced force with in the system or between the system and the surroundings. Chemical Equilibrium: A system is said to be in a state of chemical equilibrium if there is no chemical reaction or transfer of matter from one part of the system to another. Thermal Equilibrium: A system is said to be in a state of thermal equilibrium if there is no change in any property of the system when the system is separated by a diathermic wall from its surroundings. Diathermic wall defined as a wall which allows heat to flow. STATE, PROPERTIES AND PROCESSES: • • State of a system is the condition of the system at any particular moment. It may be identified by the properties such as pressure, temperature and volume, etc. The property can be measured while the system is at a state of equilibrium. In any operation there is a change in system properties which is called the change of state. A series of changes in the system between initial state and final state is called the path of change of state. • When the path is specified completely the change of state followed by the working medium as it liberates, transfers, transforms or receives energy is called as process. A series of state changes or process undergone by a system such that the final state is identical with the initial state is defined as a thermodynamic cycle. • Fig .2.1(a) A process Fig .2.1 (b) A cycle In order to describe a system it is necessary to know the quantities and characteristics of the system which are known as properties. The properties are classified as extensive properties and intensive properties. • Properties which are related to mass are called as extensive or extrinsic properties, e.g., volume, energy, etc. If mass increases the value of extensive Properties will increase the properties which are independent of the mass of the system are called as intensive or intrinsic properties, e.g., temperature, pressure, velocity, density, etc. • Extensive properties per unit mass are known as specific extensive properties which are nothing but intensive properties, e.g., specific volume, specific energy, density, etc. Properties may also be classified into two types. They are fundamental properties and thermodynamic properties. • • Properties which are measured directly are called as fundamental properties, e.g., pressure, volume, temperature, etc. Properties which cannot be measured directly but in closed cycle the working medium is recirculated with in the system. In open cycle the working substance is exhausted to the atmosphere after the process. ZEROTH LAW OF THERMODYNAMICS: The zeroth law of thermodynamics states that if two bodies are in thermal equilibrium with a third body, then they are also in thermal equilibrium with each other. Fig: 3 Concept of Zeroth law Let us consider the temperature equality concept to three systems say, A, B and as shown in Fig.3. The system A consists of a mass of gas enclosed in a vessel fitted with a thermometer and the system B is a cold iron body. When A and B are brought in contact, after some time they attain a common temperature and are then said to exist in thermal equilibrium. Now the system is brought into contact with a third system C, again A and C attain thermal equilibrium, then system B and C will show no further change in properties when brought into contact. That is system A is in thermal equilibrium with system B and also separately with system C. Then B and C will be in thermal equilibrium with each other. This law provides the basis for temperature measurement. FIRST LAW OF THERMODYNAMICS The first law of thermodynamics states that a closed system executing a cycle in which the initial state and final state are same . i.e., The net work delivered to the surroundings is proportional to the net heat taken from the surroundings. That is heat and work is mutually conversible. Since energy can neither be created nor destroyed, the total energy associated with energy conservation remains constant. Mathematically, Where dw is net work delivered during the process and dq is net heat supplied during the process. FIRST LAW APPLIED TO CLOSED SYSTEM As shown in Fig3.2. let us consider a closed system which undergoes a cycle, in which x and y is two arbitrary properties of the system. According to first law of thermodynamics for a cyclic process, algebric sum of t is proportional to algebric sum of heat transferred. i.e., Q1-2 is proportional to Q2-1 This is applicable if the system involves more heat and work transfers at different points on the boundary. Where Q and W represent infinitesimal elements of heat and work transfer respectively. As no fundamental distinction between the unit of heat and the unit of work J can be neglected from the equation. Internal Energy: The energy E is the sum of Kinetic Energy (KE), Potential Energy (PE) and Internal Energy (U). The internal energy is due to the motion of the molecules and it changes with change in temperature. For a non flow and closed system, the kinetic and potential energy terms are zero, and then the energy will be The term E is the change in internal energy For an isolated system both Q and W are zero, the change in energy is also zero. Q= 0; W= 0; E=0 For reversible non flow process the work DISPLACEMENT WORK The most common example of mechanical work encountered in thermodynamic system is that associated with a process in which there is a change in volume of a system under pressure. Let the volume of the fluid within the moving boundary be v1 and pressure be p1. In p-v diagram, point-1 represents initial state. If the working medium expands and moves the piston to top dead centre (TDC) from bottom dead centre (BDC), the work will be done by the working medium. After expansion at state 2, pressure is decreased and volume increased. Since the system undergoes expansion process, it is represented by the curve 1 - x - y - 2 in p-V diagram as shown in Fig.4 Fig: 4. Displacement work The change of state from x to y, a very small change of state in which pressure is almost constant during the change, then the force acting on the movable boundary F× x = p-V During this piston moves to a distance dL and the work done = force × distance traveled. dW where dV= A×dL The total work at the moving boundary is = pA×dL = p×dV • When the work is done by the system, it is called as positive work. This is represented by the sign plus, +W indicated the work done by the system. e.g., expansion. When the work is done on the system, it is called as negative work. - W indicates the negative work. e.g., compression. • PATH FUNCTION AND POINT FUNCTION A non-flow process is one in which the gas is neither be supplied nor rejected across the boundary of the system. The system moves from state 1 to state 2 through two different paths A and B, as shown in Fig 5. Fig. 5.Path Function Each curve represents the work for each process, these two paths gives two different work values even though states 1 and 2 are identical, the work delivered to the shaft depends upon the particular function, so the work is called as path function. The differentiation of path function is inexact or imperfect. But the thermodynamic properties are point functions, because they depend on the end states and independent of the path which the system follows. The differentiation of point functions is exact or perfect differentials, e.g., change in pressure p2 p1 and change in volume V2- V1. APPLICATION OF FIRST LAW TO NON - FLOW PROCESS: NON - FLOW PROCESSES It is the process in which the substance does not leave the system, but energy only crosses the boundary in the form of work and heat. Non - flow processes are classified under two groups. They are a. Reversible non-flow processes b. irreversible non-flow processes. REVERSIBLE NON-FLOW PROCESSES (CLOSED SYSTEM): The following non-flow processes are reversible 1. 2. 3. 4. 5. Constant volume (isochoric) process Constant pressure (Isobaric) process Adiabatic (Isentropic) process Polytropic process : : : : V = constant; n = p = constant; n = 0 T= constant; n = 1 PV = constant; n = PVn = constant; n=n Constant temperature (Isothermal) process : Where p is pressure, V is volume, T is temperature, n is index of compression or expansion and is adiabatic index. Reversible Constant Volume Process: When volume remains constant during the execution of a process, the process is called as constant volume process. As shown in Below the system contains unit mass and state one in pV diagram represents the system state before the heating processes. State 2 represents the state of system after heating process. Applying the first law of thermodynamics The rise in heat causes rise in internal energy and loss of heat decreases the internal energy. APPLICATION OF FIRST LAW TO FLOW PROCESS: Most of the systems which are related to power generation are open systems in which the mass crosses the boundary of the system, and after doing the work it leaves the system. Flow processes are classified into two types, they are • • Steady flow process Non-steady flow process STEADY FLOW PROCESS A steady flow process is one in which the mass flow rate at the entry and at the exit is constant. At any point in the system the properties of the fluid do not change with time, e.g., compressor, turbines, nozzles, etc. Assumptions made in the analysis of a steady flow process are, 1. Mass flow rate through the system remains constant. 2. Composition of fluid is uniform 3. State of the fluid at any point in the system remains constant. 4. Work and heat are the only interaction between the system and surroundings. STEADY FLOW ENERGY EQUATION As stated in the law of conservation of energy the sum of total energy exerting the system is equal to the sum of total energy leaving the system. Thus there is no change in stored energy. Let m - mass flow rate in kg/s, - absolute pressure in N/m2, v - specific volume in m3/kg, V - velocity in m/s, Z - elevation above the datum in m, u - specific internal energy in J/kg, Q - heat into the system in J and W- work output in J. Fig: 6.Steady flow process Assumptions made are Mass flow in Energy in Total energy in = = = Mass flow out Energy out Total energy out at entry (Potential energy + Kinetic energy + Internal energy + Heat energy) (Potential energy + Kinetic energy + Internal energy + Work) at exit PE1 + KE1 + U1 + Q = PE2 + KE2 + U2 + W = This is the steady flow energy equation and all the energy values are in Watts. The steady flow energy equation can be written in mass basis as given below. Hence the energy values are in J/kg Mass flow in= Mass flow out m1 = m2 We know m = Where the density of fluid ρ = ρ= 1 1 = specific volume υ mass and volume for unit mass A1V1 AV = 2 2 υ1 υ2 UNSTEADY FLOW PROCESS An unsteady flow process is one in which the mass flow rate at the entry and exit of the system is not equal in a given time, and there is no change in stored energy of the system. Let Eout Ein be the change in flow energy and E change in stored energy. Based on first law SECOND LAW OF THERMODYNAMICS LIMITATIONS OF FIRST LAW • • The first law of thermodynamics states that, heat and work are mutually convertible during any cycle of a closed system. But in actual practice all forms of energy cannot be changed into work and the first law does not give any conditions under which conversion of heat into work is possible. The law does not specify the direction of the process under consideration. The limitations of first law are discussed below. • The following examples are based on the first law of thermodynamics and these processes only proceed in certain direction but not in the reverse direction. § Let T1 and T2 be the temperatures of two bodies where T1 is greater than T2 If these bodies are brought in contact with each other but are separated from surround- rigs, heat will flow from hot body (T1) to cold body (T2) till the temperature of both bodies are equal. But the reverse process is not possible, i.e., flow of heat from lower temperature body to higher temperature body. In an automobile moving at a certain speed, if the brakes are applied to stop the automobile means, the brakes get hot by the conversion of automobile s kinetic energy into heat. However, it will be observed that reversal of the process in which the hot brakes were to cool off and give back its internal energy to the automobile, causing it to move on the road. But this is impossible. § § SECOND LAW OF THERMODYNAMICS • KELVIN-PLANK STATEMENT OF SECOND LAW It is impossible to construct an engine which operates on cycle to receive heat from a single reservoir and produce net amount of work. Kelvin - Plank statement related to heat engines . In other words, no engine operating in cycles can convert all the heat energy into work, but there will be some loss of heat energy to the surroundings. Thus 100% efficient engine is not possible. Fig:7. Possible engine and not possible engine Possible engine is one in which a part of heat is rejected to the cold reservoir, which is supplied from the hot reservoir and the difference between the heat supplied and heat rejected is equal to work done. • CLAUSIS STATEMENT OF SECOND LAW Clausis statement is related to refrigerators or heat pumps. The Clausis statement is expressed as follows: It is not possible to construct a system that operates in a cycle and transfers heat from a colder body to a hotter body without the aid of an external agency. In other words, heat can not flow itself from a colder body to a hotter body . Based on this, the hot reservoir at T1 temperature and the cold reservoir at T2 temperature are shown in Fig: 8. The heat pump which takes mechanical work to transfer heat continuously from sink to source. Fig:8. External work required for heat flow from sink to source PERPETUAL MOTION MACHINE OF SECOND KIND (PMM II) Perpetual motion machine of second kind is one which operators in a cycle and delivers an amount of work equal to heat extracted from a single reservoir at an uniform temperature. Such 100% efficiency violates the second law of thermodynamics as according to Kelvin - Plank Statement. It is not possible to construct a machine which could extract heat from a single reservoir and convert it into equivalent amount of work. HEAT ENGINE Heat engine is defined as a machine which is used to convert heat energy into work in a cyclic process . The definition of heat engine covers both rotary and reciprocating machines. The working fluid should undergo cyclic process and periodically should return to its initial state. Fig: 9. Steam power plant as heat engine Fig: 9. shows a steam power plant which is an example of heat engine cycle. In the boiler high pressure steam is generated and the steam expands in the turbine and doing external work W exhaust steam from turbine is condensed in the condenser thereby releasing heat and the water is pumped back to the boiler to complete the cycle. Thus the boiler, turbine, condenser and the pump separately in a power plant can not be regarded as heat engine because they are engines since they are part of the cycle. Combination of these components is a heat engine since they complete the cycle. EFFICIENCY OF HEAT ENGINE Performance of a heat engine is obtained by its thermal efficiency which is the ratio of net work output to heat supplied . A part of heat supplied is converted into work and the rest is rejected. Let Q1 be heat supplied, Q2 be heat rejected, Wt be turbine work and Wp be pump work (Fig: 10.) As per the first law of thermodynamics Fig: 10. Heat engine Efficiency of the heat engine η= Network output Heat supplied HEAT PUMP Heat pump is defined as a device which transfers heat from a low temperature body to a high temperature body when it is working in a cycle (Fig:11). For heat pump the efficiency term is replaced by the co-efficient of performance (COP) which as an index of performance of heat pump to differentiate it from heat engine. COP of the pump is given by COP (heat pump) = Heating effect Work done Fig: 11. Heat pump If a heat pump is used to transfer heat from low temperature reservoir T2 to high temperature reservoir T1 in order to maintain T2 < T1 then the COP of the refrigerator is given by COP ref = Refrigerating effect Work input CARNOT CYCLE The Carnot cycle has four reversible processes, of which two are frictionless isothermal processes and two frictionless adiabatic processes. Figure.12 shows the p-V and T-s diagram of the Carnot cycle. • Process 1-2 represents reversible isothermal expansion, Heat Q is supplied at constant temperature T and this is equal to the work done during the process. V2 V1 Heat supplied Q1 = • P1V1ln Process 2-3 represents reversible adiabatic expansion, there is no heat transfer takes place. The work is done at the cost of internal energy. The temperature becomes T2 at T3 Fig.12 Carnot cycle • Process 3-4 represents reversible isothermal compression in which Q2 heat is rejected isothermally at T2. The air is compressed up to point 4 at constant temperature. Heat rejected • Process 4-1 represents reversible adiabatic compression in which the system returns back to the initial state and the temperature of air increases from. T2 to T1 There is no heat transfer and work is done on the air. Net work done = (Heat supplied) - (Heat Rejected) Efficiency of Carnot cycle = (Net work done / Heat supplied) CARNOT THEOREM The Carnot principles are the two conclusions regard to the thermal efficiency of ideal and natural (actual) heat engines. They are expressed as follows: • The efficiency of an actual (irreversible) heat engine is always less than the efficiency of an ideal (reversible) heat engine operating between the same two reservoirs. All the reversible (ideal) heat engines operating between the same two reservoirs will have the same efficiency. • CLAUSIS IN EQUALITY While applying second law of thermodynamics to processes the second law leads to the definition of a new property called entropy. Entropy is an abstract property, and it is difficult to give a physical description of it. The uses of entropy in common engineering processes provide the best understanding of it. The second law may be stated to be the law of entropy. The cyclic integral of dQ/T in above equation is always less than or equal to zero. Inequivality of Clausis is the basis of the definition of entropy. Entropy is a nonconserved property by which it differs from energy. This inequality is valid for all cycles, viz., and reversible or irreversible. The symbol φ denotes that the integration is to be performed over the entire cycle. Any heat transfer from or to a system can be considered to consist of differential amounts of heat transfer then the cyclic integral of dQ/T can be viewed as the sum of all these differential amounts of heat transfer divided by the absolute temperature at the boundary. CONCEPT OF ENTROPY ENTROPY AS A PROPERTY OF A SYSTEM Entropy of a substance is a thermodynamic property which increases with the addition of heat and decreases with the removal of heat. Entropy itself cannot be defined but the change in entropy can be defined in a reversible process, i.e., the quantity of he received or rejected divided by the absolute temperature of the substance measures the change in entropy. A small amount of heat dQ is added to the system causing the entropy to increase by ds and T is the absolute temperature. The change in entropy absolute temperature. If the total quantity of heat Q be added to a substance at constant temperature then the increase in entropy due to the addition of heat is given by ds = Q T Q T T × ds s1 − s2 = From the definition of entropy dQ = By integrating the equation the total heat added can be obtained as ENTROPY - A POINT FUNCTION Entropy has got one value for each point of temperature or pressure or volume. So it is a point function. The change in entropy during a thermodynamic process depend only on the initial and final conditions irrespective of the path. In solving problems the change in entropy is considered with the assumption if the entropy of all substances is zero at the ice-point, i.e., the entropy is positive if the temperature is above 0°C and negative if the temperature is below 0° C. Entropy is expressed as kJ/kgK, since, it has the dimension of heat/mass and temperature. There are many instruments to measure temperature, pressure, etc., but there no such instruments as yet to measure entropy. ENTROPY OF A REVERSIBLE CYCLIC PROCESS Let us consider a system undergoing a reversible process from state 1 to state along the path A and then from state 2 to the original state 1 along the path B as shown in Fig.13. Fig. 13. Reversible cyclic process between two fixed states CHANGE IN ENTROPY OF A PERFECT GAS Let m kg of gas carryout a process. At the initial state 1 let the pressure, temperature, entropy and volume are p1, T1, s1 and V1 respectively. The gas is heated in any manner such that at its final state 2 pressure, temperature, entropy and volume be p2, T2, s2 and V2 respectively. From the law of conservation of energy This is the change in entropy in terms of volume and temperature. This can be expressed in terms of pressure and volume by applying gas equation as JAYAM COLLEGE OF ENGINEERING AND TECHNOLOGY DHARMAPURI DEPARTMENT YEAR / SEM SUBJECT : EEE : SECOND/ THIRD : ME1211 / APPLIED THERMODYNAMICS ASSIGNMENT NO 1 Unit 1 Basic concepts and Laws of Thermodynamics PART A 1. State first law of thermodynamics? 2. State Zeroth law of thermodynamics? 3. What is meant by Perpetual Motion Machine of first kind ? 4. State the statements of second law of thermodynamics? 5. If pvn=C represents a general thermodynamic process, name the processes when n has values of 0, 1, and . 6. What are the types of thermodynamic properties? 7. Explain the thermodynamic equilibrium? Explain. 8. What is meant by Perpetual Motion Machine of second kind ? 9. State Carnot s theorem. 10. Define Clausis Inequality. PART B 1. Derive then expression for work and heat during constant volume and constant pressure process. 2. 3. 4. Derive then expression for work and heat isothermal and isentropic process. Derive the steady flow energy equation. A cycle heat engine operates between a source temperature of 800 C and a sink temperature of 30 C. What is the least rate of heat rejection per KW net output of engine? 5. 2 kg of air compressed according to the law pV1.3 = constant from a pressure of 1.8 bar and temperature of 30 C to a pressure of 25.5 bar. Calculate a) The final volume and temperature. b) Work done c) Heat transferred d) Change in entropy. Pages to are hidden for "14075471-Applied-Termodynamics-01"Please download to view full document

When analyzing contingency tables with two rows and two columns you can use either Fishers exact test or the chi-square test. Continuous Discrete categorical Yes. After some googling Ive seen several attempts of various coverage and quality some not available at the moment. Which statistical test to use flow chart. Flow Chart for Selecting Commonly Used Statistical Tests Type of data. The table then shows one or more statistical tests commonly used given these types of variables but not necessarily the only type of test that could be used and links showing how to do such tests using SAS Stata and SPSS. When your experiment is trying to draw a comparison or find the difference between one categorical with two categories and another continuous variable then you need to work on the two-sample T-test to find the significant difference between the two variables. Master the 6 basic types of tests with simple definitions illustrations and examples. Descriptive Statistics Sometimes the first. Deciding which statistical test to use can seem difficult at first. If youre already up on your statistics you know right away that you want to use a 2-sample t-test which analyzes the. Statistical Test Flow Chart Geo 441. The design of a study is more important than the analysis. Public Health Statistics Professor Dr. Quantitative Methods Part B – Group Comparison II Normal Non-Normal 1 Sample z Test 2 Sample Independent t Test for equal variances Paired Sample t Test Compare two groups Compare. Statistical Tests Heather M. This post is an attempt to mark out the difference between the most. – What question you are asking of the data. Standard t test The most basic type of statistical test for use when you are comparing the means from exactly TWO Groups such as the Control Group versus the Experimental Group. Although for a given data set a one-tailed test will return a smaller p value than a two-tailed test the latter is usually preferred unless there is a watertight case for one-tailed testing. Statistical Test between One Continuous and another Categorical variable. Based on a text book. Ive also seen similar flowcharts in statistics textbooks Ive. The Fishers test is the best choice as it always gives the exact P value. – What type of data you are dealing with. Use these tables as a guide. Many times researchers are content to show histograms to illustrate their point after a flow experiment. For a statistical test to be valid your sample size needs to be large enough to approximate the true distribution of the population being studied. A badly designed study can never be retrieved whereas a poorly analysed study can usually be re-analyzed. Follow the flow chart and click on the links to find the most appropriate statistical analysis for your situation. Flow cytometry data are numbers rich. Section 1 Section 1 contains general information about statistics including key definitions and which summary statistics and tests to choose. Ex Your experiment is studying. Data from experiments can be population measurements percent of CD4 cells for example or it can be expression level median fluorescent expression of CD69 on activated T cells. Rulison October 31 2014 How to Use Flow Chart 1. Use the Which test should I use. An interactive flowchart decision tree to help you decide which statistical test to use with descriptions of each test and links to carry them out in R SPSS and STATA. _ table to allow the student to. It is obvious that we cannot refer to all statistical tests in one editorial. If data is not normal use Y or Z. Prediction Analyses – Quick Definition Prediction tests examine how and to what extent a variable can be predicted from 1 other variables. Describing a sample of data descriptive statistics centrality dispersion replication see also Summary statistics. Type of questionChi-square tests one and two sample RelationshipsDifferences Do you have a true independent variable. Whether your data the. Its important to know. It is primarily a flowchart but is arranged as a tree diagram to give visibility to four branches of. F-test for ratio of two variances Test equality of 2 popn variances Levenestest Brown-Forsythe Test Normality assumption does NOT hold Try response transformation to apply normal-theory methods first Compare 1 popn variance. This approach misses the. Quickly find the right statistical test with this easy overview. What statistical test should I use. If you can answer those questions then you can usually identify a statistical test which can help using the flowchart in. The chi-square test is simpler to calculate but yields only an approximate P. For a person being from a non-statistical background the most confusing aspect of statistics are the fundamental statistical tests and when to use which test. Made by Matthew Jackson. This chart gives an overview of statistics for research. To determine which statistical test to use you need to know. To use this tool please select the applicable goal of the analysis of the data then work through the tables from left to right to select the correct statistical test. Follow the flow chart and click on the links to find the most appropriate statistical analysis for your situation. Start at the top with determining what type of data was collected continuous also known as. Choosing appropriate statistical test Having a well-defined hypothesis helps to distinguish the outcome variable and the exposure variable Answer the following questions to decide which statistical test is appropriate to analysis. – Whether the data is symmetrical or skewed.

Statistical mechanics of lossy data compression using a non-monotonic perceptron The performance of a lossy data compression scheme for uniformly biased Boolean messages is investigated via methods of statistical mechanics. Inspired by a formal similarity to the storage capacity problem in neural network research, we utilize a perceptron of which the transfer function is appropriately designed in order to compress and decode the messages. Employing the replica method, we analytically show that our scheme can achieve the optimal performance known in the framework of lossy compression in most cases when the code length becomes infinite. The validity of the obtained results is numerically confirmed. pacs:89.90.+n, 02.50.-r, 05.50.+q, 75.10.Hk Recent active research on error-correcting codes (ECC) has revealed a great similarity between information theory (IT) and statistical mechanics (SM) MacKay ; Richardson ; Aji ; us_EPL ; us_PRL ; Sourlas_nature ; NishimoriWong . As some of these studies have shown that methods from SM can be useful in IT, it is natural to expect that a similar approach may also bring about novel developments in fields other than ECC. The purpose of the present paper is to offer such an example. More specifically, we herein employ methods from SM to analyze and develop a scheme of data compression. Data compression is generally classified into two categories; lossless and lossy compression bi:Cover . The purpose of lossless compression is to reduce the size of messages in information representation under the constraint of perfect retrieval. The message length in the framework of lossy compression can be further reduced by allowing a certain amount of distortion when the original expression is retrieved. The possibility of lossless compression was first pointed out by Shannon in 1948 in the source coding theorem Shannon , whereas the counterpart of lossy compression, termed the rate-distortion theorem, was presented in another paper by Shannon more than ten years later RD . Both of these theorems provide the best possible compression performance in each framework. However, their proofs are not constructive and suggest few clues for how to design practical codes. After much effort had been made for achieving the optimal performance in practical time scales, a practical lossless compression code that asymptotically saturates the source-coding limit was discovered Jelinek . Nevertheless, thus far, regarding lossy compression, no algorithm which can be performed in a practical time scale saturating the optimal performance predicted by the rate-distortion theory has been found, even for simple information sources. Therefore, the quest for better lossy compression codes remains one of the important problems in IT bi:Cover ; fixed ; Yamamoto ; Murayama_RD . Therefore, we focus on designing an efficient lossy compression code for a simple information source of uniformly biased Boolean sequences. Constructing a scheme of data compression requires implementation of a map from compressed data of which the redundancy should be minimized, to the original message which is somewhat biased and, therefore, seems redundant. However, since the summation over the Boolean field generally reduces the statistical bias of the data, constructing such a map for the aforementioned purpose by only linear operations is difficult, although the best performance can be achieved by such linear maps in the case of ECC MacKay ; Aji ; us_EPL ; us_PRL ; Sourlas_nature and lossless compression Murayama . In contrast, producing a biased output from an unbiased input is relatively easy when a non-linear map is used. Therefore, we will employ a perceptron of which the transfer function is optimally designed in order to devise a lossy compression scheme. The present paper is organized as follows. In the next section, we briefly introduce the framework of lossy data compression, providing the optimal compression performance which is often expressed as the rate-distortion function in the case of the uniformly biased Boolean sequences. In section III, we explain how to employ a non-monotonic perceptron to compress and decode a given message. The ability and limitations of the proposed scheme are examined using the replica method in section IV. Due to a specific (mirror) symmetry that we impose on the transfer function of the perceptron, one can analytically show that the proposed method can saturate the rate-distortion function for most choices of parameters when the code length becomes infinite. The obtained results are numerically validated by means of the extrapolation on data from systems of finite size in section V. The final section is devoted to summary and discussion. Ii Lossy Data Compression Let us first provide the framework of lossy data compression. In a general scenario, a redundant original message of random variables , which we assume here as a Boolean sequence , is compressed into a shorter (Boolean) expression . In the decoding phase, the compressed expression is mapped to a representative message in order to retrieve the original expression (Fig. 1). In the source coding theorem, it is shown that perfect retrieval is possible if the compression rate is greater than the entropy per bit of the message when the message lengths and become infinite. On the other hand, in the framework of lossy data compression, the achievable compression rate can be further reduced allowing a certain amount of distortion between the original and representative messages and . A measure to evaluate the distortion is termed the distortion function, which is denoted as . Here, we employ the Hamming distance as is frequently used for Boolean messages. Since the original message is assumed to be generated randomly, it is natural to evaluate the average of Eq. (1). This can be performed by averaging with respect to the joint probability of and as By allowing the average distortion per bit up to a given permissible error level , the achievable compression rate can be reduced below the entropy per bit. This limit is termed the rate-distortion function, which provides the optimal compression performance in the framework of lossy compression. The rate-distortion function is formally obtained as a solution of a minimization problem with respect to the mutual information between and bi:Cover . Unfortunately, solving the problem is generally difficult and analytical expressions of are not known in most cases. The uniformly biased Boolean message in which each component is generated independently from an identical distribution is one of the exceptional models for which can be analytically obtained. For this simple source, the rate-distortion function becomes However, it should be addressed here that a practical code that saturates this limit has not yet been reported, even for this simplest model. Therefore, in the following, we focus on this information source and look for a code that saturates Eq. (4) examining properties required for good compression performance. Iii Compression by Perceptron In a good compression code for the uniformly biased source, it is conjectured that compressed expressions should have the following properties: In order to minimize loss of information in the original expressions, the entropy per bit in must be maximized. This implies that the components of are preferably unbiased and uncorrelated. In order to reduce the distortion, the representative message should be placed close to the typical sequences of the original messages which are biased. Unfortunately, it is difficult to construct a code that satisfies both of the above two requirements utilizing only linear transformations over the Boolean field while such maps provide the optimal performance in the case of ECC MacKay ; Aji ; us_EPL ; us_PRL ; Sourlas_nature and lossless compression Murayama . This is because a linear transformation generally reduces statistical bias in messages, which implies that the second requirement (II) cannot be realized for unbiased and uncorrelated compressed expressions that are preferred in the first requirement (I). One possible method to design a code that has the above properties is to introduce a non-linear transformation. A perceptron provides one of the simplest schemes for carrying out this task. In order to simplify notations, let us replace all the Boolean expressions with binary ones . By this, we can construct a non-linear map from the compressed message to the retrieved sequence utilizing a perceptron as where are fixed -dimensional vectors to specify the map and is a transfer function from a real number to a binary variable that should be optimally designed. Since each component of the original message is produced independently, it is preferred to minimize the correlations among components of a representative vector , which intuitively indicates that random selection of may provide a good performance. Therefore, we hereafter assume that vectors are independently drawn from the -dimensional normal distribution . Based on the non-linear map (5), a lossy compression scheme can be defined as follows: Compression: For a given message , find a vector that minimizes the distortion , where the representative vector which is generated from by Eq. (5). The obtained is the compressed message. Decoding: Given the compressed message , the representative vector produced by Eq. (5) provides the approximate message for the original message. Here, we should notice that the formulation of the current problem has become somewhat similar to that for the storage capacity evaluation of the Ising perceptron bi:Krauth ; bi:Gardner regarding , and as “Ising couplings”, “random input pattern” and “random output”, respectively. Actually, the rate-distortion limit in the current framework for and can be calculated as the inverse of the storage capacity of the Ising perceptron, . This observation implies that the simplest choice of the transfer function , where for and otherwise, does not saturate the rate-distortion function (4). This is because the well-known storage capacity of the simple Ising perceptron, , means that the “compression limit” achievable by this monotonic transfer function becomes and far from the value provided by Eq. (4) for this parameter choice . We also examined the performances obtained by the monotonic transfer function for biased messages by introducing an adaptive threshold in our previous study Hosaka and found that the discrepancy from the rate-distortion function becomes large in particular for relatively high while fairly good performance is observed for low rate regions. Therefore, we have to design a non-trivial function in order to achieve the rate-distortion limit, which may seem hopeless as there are infinitely many degrees of freedom to be tuned. However, a useful clue exists in the literature of perceptrons, which have been investigated extensively during the last decade. In the study of neural network, it is widely known that employing a non-monotonic transfer function can highly increase the storage capacity of perceptrons Monasson . In particular, Bex et al. reported that the capacity of the Ising perceptron that has a transfer function of the reversed-wedge type can be maximized to by setting bi:VdB , which implies that the rate-distortion limit is achieved for the case of and in the current context. Although not explicitly pointed out in their paper, the most significant feature observed for this parameter choice is that the Edwards-Anderson (EA) order parameter vanishes to zero, where denotes the average over the posterior distribution given and . This implies that the dynamical variable in the posterior distribution given and is unbiased and, therefore, the entropy is maximized, which meets the first requirement (I) addressed above. Thus, designing a transfer function so as to make the EA order parameter vanish seems promising as the first discipline for constructing a good compression code. However, the reversed-wedge type transfer function is not fully satisfactory for the present purpose. This is because this function cannot produce a biased sequence due to the symmetry , which means that the second requirement (II) provided above would not be satisfied for . Hence, another candidate for which the EA parameter vanishes and the bias of the output can be easily controlled must be found. A function that provides these properties was once introduced for reducing noise in signal processing, such as bi:Jort ; Inoue (Fig. 2). Since this locally activated function has mirror symmetry , both and provide identical output for any input, which means that the EA parameter is likely to be zero. Moreover, one can easily control the bias of output sequences by adjusting the value of the threshold parameter . Therefore, this transfer function looks highly promising as a useful building-block for constructing a good compression code. In the following two sections, we examine the validity of the above speculation, analytically and numerically evaluating the performance obtained by the locally activated transfer function . Iv Analytical Evaluation We here analytically evaluate the typical performance of the proposed compression scheme using the replica method. Our goal is to calculate the minimum permissible average distortion when the compression rate is fixed. The analysis is similar to that of the storage capacity for perceptrons. Employing the Ising spin expression, the Hamming distortion can be represented as Then, for a given original message and vectors , the number of dynamical variables which provide a fixed Hamming distortion , can be expressed as Since and are randomly generated predetermined variables, the quenched average of the entropy per bit over these parameters to which the raw entropy per bit becomes identical for most realizations of and , is naturally introduced for investigating the typical properties. This can be performed by the replica method , analytically continuing the expressions of obtained for natural numbers to non-negative real number beyond ; bi:Nishimori . When is a natural number, can be expanded to a summation over -replicated systems as , where the subscript denotes a replica index. Inserting an identity into this expression and utilizing the Fourier expression of the delta function we can calculate the moment for natural numbers as where is an matrix of which elements are given by the parameters and . In the thermodynamic limit keeping the compression rate finite, this integral can be evaluated via a saddle point problem with respect to macroscopic variables , and . In order to proceed further, a certain ansatz about the symmetry of the replica indices must be assumed. We here assume the simplest one, that is, the replica symmetric (RS) ansatz for which the saddle point expression of Eq. (16) is likely to hold for any real number . Taking the limit of this expression, we obtain where , , and . denotes the extremization. Under this RS ansatz, the macroscopic variable indicates the EA order parameter as . The validity of this solution will be examined later. Since the dynamical variable is discrete in the current system, the entropy (18) must be non-negative. This indicates that the achievable limit for a fixed compression rate and a transfer function which is specified by the threshold parameter can be characterized by a transition depicted in Fig. 3. Utilizing the Legendre transformation , the free energy for a fixed inverse temperature , which is an external parameter and should be generally distinguished from the variational variable in Eq. (18), can be derived from . This implies that the distortion that minimizes and of which the value is computed from as can be achieved by randomly drawing from the canonical distribution which is provided by the given . For a modest , the achieved distortion is determined as a point for which the slope of becomes identical to and (Fig. 3 (a)). As becomes higher, moves to the left, which indicates that the distortion can be reduced by introducing a lower temperature. However, at a critical value characterized by the condition (Fig. 3 (b)), the number of states that achieve which is the typical value of vanishes to zero. Therefore, for , is fixed to and the distortion is not achievable (Fig. 3 (c)). The above argument indicates that the limit of the achievable distortion for a given rate and a threshold parameter in the current scheme can be evaluated from conditions being parameterized by the inverse temperature . Due to the mirror symmetry , becomes the saddle point solution for the extremization problem (18) as we speculated in the previous section, and no other solution is discovered. Inserting into the right-hand side of Eq. (18) and employing the Legendre transformation, the free energy is obtained as The rate-distortion function represents the optimal performance that can be achieved by appropriately tuning the scheme of compression. This means that can be evaluated as the convex hull of a region in the - plane defined by Eqs. (22) and (25) by varying the inverse temperature and the threshold parameter (or ). Minimizing for a fixed , one can show that the relations are satisfied at the convex hull, which offers the optimal choice of parameters and as functions of a given permissible distortion and a bias . Plugging these into Eq. (25), we obtain which is identical to the rate-distortion function for uniformly biased binary sources (4). The results obtained thus far indicate that the proposed scheme achieves the rate-distortion limit when the threshold parameter is optimally adjusted. However, since the calculation is based on the RS ansatz, we must confirm the validity of assuming this specific solution. We therefore examined two possible scenarios for the breakdown of the RS solution. The first scenario is that the local stability against the fluctuations for disturbing the replica symmetry is broken, which is often termed the Almeida-Thouless (AT) instability bi:AT , and can be examined by evaluating the excitation of the free energy around the RS solution. As the current RS solution can be simply expressed as , the condition for this solution to be stable can be analytically obtained as In most cases, the RS solution satisfies the above condition and, therefore, does not exhibit the AT instability. However, we found numerically that for relatively high values of distortion , can become slightly smaller than for a very narrow parameter region, , which indicates the necessity of introducing the replica symmetry breaking (RSB) solutions. This is also supported analytically by the fact that the inequality holds for in the vicinity of . Nevertheless, this instability may not be serious in practice, because the area of the region , where the RS solution becomes unstable, is extremely small, as indicated by Fig. 5 (a). The other scenario is the coexistence of an RSB solution that is thermodynamically dominant while the RS solution is locally stable. In order to examine this possibility, we solved the saddle point problem assuming the one-step RSB (1RSB) ansatz in several cases for which the RS solution is locally stable. However, no 1RSB solution was discovered for . Therefore, we concluded that this scenario need not be taken into account in the current system. These insubstantial roles of RSB may seem somewhat surprising since significant RSB effects above the storage capacity have been reported in the research of perceptrons with continuous couplings bi:Jort ; Monasson . However, this may be explained by the fact that, in most cases, RSB solutions for Ising couplings can be expressed by the RS solutions adjusting temperature appropriately, even if non-monotonic transfer functions are used bi:Krauth ; Inoue . V Numerical Validation Although the analysis in the previous section theoretically indicates that the proposed scheme is likely to exhibit a good compression performance, it is still important to confirm it by experiments. Therefore, we have performed numerical simulations implementing the proposed scheme in systems of finite size. In these experiments, an exhaustive search was performed in order to minimize the distortion so as to compress a given message into , which implies that implementing the current scheme in a large system is difficult. Therefore, validation was performed by extrapolating the numerically obtained data, changing the system size from to . Figure 4 shows the average distortions obtained from experiments for (a) unbiased () and (b) biased () messages, varying the system size and the compression rate . For each , the threshold parameter is tuned to the value determined using Eqs. (26), (27) and the rate-distortion function in order to optimize the performance. These data indicate that the finite size effect is relatively large in the present system, which is similar to the case of the storage capacity problem bi:Opper , and do not necessarily seem consistent with the theoretical prediction obtained in the previous section. However, the extrapolated values obtained from the quadratic fitting with respect to are highly consistent with curves of the rate-distortion function (Fig. 5 (a) and (b)), including one point in the region where the AT stability is broken (inset of Fig. 5(a)), which strongly supports the validity and efficacy of our calculation based on the RS ansatz. Vi Summary and Discussion We have investigated a lossy data compression scheme of uniformly biased Boolean messages employing a perceptron of which the transfer function is non-monotonic. Designing the transfer function based on the properties required for good compression codes, we have constructed a scheme that saturates the rate-distortion function that represents the optimal performance in the framework of lossy compression in most cases. It is known that a non-monotonic single layer perceptron can be regarded as equivalent to certain types of multi-layered networks, as in the case of parity and committee machines. Although tuning the input-output relation in multi-layered networks would be more complicated, employing such devices might be useful in practice because several heuristic algorithms that could be used for encoding in the present context have been proposed and investigated Mitchison ; Nokura . In real world problems, the redundancy of information sources is not necessarily represented as a uniform bias; but rather is often given as non-trivial correlations among components of a message. Although it is just unfortunate that the direct employment of the current method may not show a good performance in such cases, the locally activated transfer function that we have introduced herein could serve as a useful building-block to be used in conjunction with a set of connection vectors that are appropriately correlated for approximately expressing the given information source, because by using this function, we can easily control the input-output relation suppressing the bias of the compressed message to zero, no matter how the redundancy is represented. Finally, although we have confirmed that our method exhibits a good performance when executed optimally in a large system, the computational cost for compressing a message may render the proposed method impractical. One promising approach for resolving this difficulty is to employ efficient approximation algorithms such as various methods of the Monte Carlo sampling bi:Sampling and of the mean field approximation bi:Advanced_Mean_Field_Methods . Another possibility is to reduce the finite size effect by further tuning the profile of the transfer function. Investigation of these subjects is currently under way. Grants-in-Aid Nos. 13780208, 14084206 (YK) and 12640369 (HN) from MEXT are gratefully acknowledged. TH would like to thank T. Murayama for informing his preprint and valuable comments, and YK would like to acknowledge D. Saad, H. Yamamoto and Y. Matsunaga for their useful discussions. - (1) N. Sourlas, Nature (London) 339, 693 (1989). - (2) D. J. C. MacKay and R. M. Neal, Electron. Lett. 33, 457 (1997). - (3) H. Nishimori and K. Y. Michael Wong, Phys. Rev. E 60, 132 (1999). - (4) Y. Kabashima and D. Saad, Europhys. Lett. 45, 97 (1999). - (5) Y. Kabashima, T. Murayama, and D. Saad, Phys. Rev. Lett. 84, 1355 (2000). - (6) T. Richardson and R. Urbanke, IEEE Trans. Inf. Theory 47, 599 (2001). - (7) S. Aji, H. Jin, A. Khandekar, D. J. C. MacKay, and R. J. McEliece, to appear in the Proc. of the IMA 1999 Summer Program: Codes, Systems and Graphical Models, Aug 1999. - (8) T. M. Cover and J. A. Thomas, Elements of Information Theory (John Wiley & Sons, 1991). - (9) C. E. Shannon, Bell Syst. Tech. J. 27, 379 (1948); 27, 623 (1948). - (10) C. E. Shannon, IRE National Convention Record, Part 4 142 (1959). - (11) F. Jelinek, Probabilistic Information Theory (McGraw-Hill, 1968). - (12) E. Yang, Z. Zhang, and T. Berger, IEEE Trans. Inf. Theory 43, 1465 (1997). - (13) Y. Matsunaga and H. Yamamoto, Proceedings of 2002 IEEE International Symposium on Information Theory, 461 (2002). - (14) T. Murayama and M. Okada, in preparation. - (15) T. Murayama, J. Phys. A: Math. and Gen. 35, L95 (2002). - (16) E. Gardner and B. Derrida, J. Phys. A: Math. and Gen. 21, 271 (1988). - (17) W. Krauth and M. Mézard, J. Phys. (France) 50, 3057 (1989). - (18) T. Hosaka, Statistical Mechanics of Lossy Data Compression, Master’s Thesis, Department of Computational Intelligence and Systems Science, Tokyo Institute of Technology (2002). - (19) R. Monasson and D. O’kane, Europhys. Lett. 27, 85 (1994). - (20) G. J. Bex, R. Serneels, and C. Van den Broeck, Phys. Rev. E 51, 6309 (1995). - (21) J. van Mourik, K. Y. M. Wong, and D. Bollé, J. Phys. A: Math. and Gen. 33, L53 (2000). - (22) J. Inoue and D.M. Carlucchi, cond-mat/9806037 (1998). - (23) M. Mézard, G. Parisi, and M. A. Virasoro, Spin Glass Theory and Beyond (World Scientific, Singapore, 1987). - (24) H. Nishimori, Statistical Physics of Spin Glasses and Information Processing (Oxford University Press, 2001). - (25) J. R. L. de Almeida and D. J. Thouless, J. Phys. A: Math. and Gen. 11, 983 (1977). - (26) W. Krauth and M. Opper, J. Phys. A: Math. and Gen. 22, L519 (1989). - (27) G.J. Mitchison and R.M. Durbin, Biol. Sybern. 60, 345 (1989). - (28) K. Nokura, Phys. Rev. E 49, 5812 (1994). - (29) R. M. Neal, Bayesian Learning for Neural Networks (Springer Verlag, 1996). - (30) M. Opper and D. Saad (Eds.), Advanced Mean Field Methods (MIT Press, 2001).

5 edition of Mathematical principles of natural philosophy found in the catalog. Mathematical principles of natural philosophy in Berkeley, University of California Press, 1960 Written in English Includes reproduction of the t.p. of the 1st ed. of the Principia, London, 1687. |Other titles||The system of the world| |The Physical Object| |Pagination||2 v. in 1 (xxxiii, 680 p.) ;| |Number of Pages||680| Isaac Newton's The Mathematical Principles of Natural Philosophy translated by Andrew Motte and published in two volumes in remains the first and only translation of Newton's Philosophia /5(9). The Mathematical Principles of Natural Philosophy, Volume 1 Sir Isaac Newton, William Emerson, John Machin Full view - The Mathematical Principles of Natural Philosophy, Volume 35/5(1). Newton's Principia: the mathematical principles of natural philosophy Item Preview remove-circle Early American mathematics books. CU-BANC. Publication date c Topics Newton, Isaac, Sir, , Mechanics -- Early works to , Celestial mechanics -- Early works to Publisher New-York: Published by Daniel AdeePages: The book has an active table of contents for readers to access each chapter of the following books: 1. THE MATHEMATICAL PRINCIPLES OF NATURAL PHILOSOPHY Isaac Newton 2. The History of the Ancient Physics - Adam Smith 3. The History of the Ancient Logics and Metaphysics - Adam Smith 4. Of the EXTERNAL SENSES - Adam Smith 5. Of the NATURE of that IMITATION Brand: AS Team. Find books like The Principia: Mathematical Principles of Natural Philosophy from the world’s largest community of readers. Goodreads members who liked T. The Principia: Mathematical Principles of Natural Philosophy: Original Edition (Latin Edition) by Newton, Isaac and a great selection of related books, art . longitudinal study of psychological changes occuring during pregnancy. short-title list of printed books in [the] Strong Room. 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Mathematical Principles of Natural Philosophy, often referred to as simply the Principia, is a work in three books by Isaac Newton, in Latin, first published 5 July /5(15). Books Advanced Search New Releases Best Sellers & More Children's Books Textbooks Textbook Rentals Sell Us Your Books Best Books of the Month of 49 results for Books: "the principia mathematical principles of natural philosophy". Some of the techniques listed in The Principia: Mathematical Principles of Natural Philosophy may require a sound knowledge of Hypnosis, users are advised to either leave those sections or must have a basic understanding of the subject before practicing them/5. Philosophiae Naturalis Principia Mathematica (Mathematical Principles of Natural Philosophy, usually called simply the Principia), which appeared in Here was a new physics that applied equally well to terrestrial and celestial bodies. Copernicus, Kepler, and Galileo were all justified by Newton’s analysis of forces. The Philosophiæ Naturalis Principia Mathematica (Mathematical Principles of Natural Philosophy) is a trilogy, written by Isaac Newton and published on 5 July The three-book work describes physics and states Newton's laws of motion and the derivation of Kepler's Laws, and observations on gravity. Section I in Book I of Isaac Newton’s Philosophiˆ Naturalis Principia Mathematica is reproduced here, translated into English by Andrew Motte. Motte’s translation of Newton’s Principia, entitled The Mathematical Principles of Natural Philosophy was rst published in. The Mathematical Principles of Natural Philosophy An Annotated Translation of the Principia. Get access. As a book of great insight and ingenuity, it has raised our understanding of the power of mathematics more than any other work. of the Principia will enable any reader with a good understanding of elementary mathematics to easily Cited by: The key contribution of Newton to the modern science and engineering was the book THE MATHEMATICAL PRINCIPLES OF NATURAL PHILOSOPHY. THE MATHEMATICAL PRINCIPLES OF NATURAL PHILOSOPHY is also called Principia. Book I of the Principia details the foundations of the science of mechanics. Book II inaugurates the theory of by: 2 The Mathematical Principles of Natural Philosophy 3 The Method of Fluxions & the Infinite Serues with its Applications to the Geometry of the 4 Universal Arithmetic - A Treatise of Arithmatical Composition & In his monumental work Philosophiae Naturalis Principia Mathematica, known familiarly as the Principia, Isaac Newton laid out in mathematical terms the principles of time, force, and motion that have guided the development of modern physical after more than three centuries and the revolutions of Einsteinian relativity and quantum mechanics, Newtonian phy/5. Newton's Principia the mathematical principles of natural philosophy 1st American ed., carefully rev. and corr. / with a life of the author, by N.W. Chittenden. by Sir Isaac Newton ★ ★ ★ ; 5 Ratings 77 Want to read; 8 Currently reading; 5 Have read/5(5). The Mathematical Principles of Natural Philosophy () by Isaac Newton, translated by Andrew Motte Book III: Rules of Reasoning in Philosophy Phænomena, or Appearances →. The same things being s upposed, I s ay that the ultimate ratio of the arc, chord, and tangent, any one to any other, is the ratio of equality. Fig. For while the point B approaches the point A, con s ider always AB and AD as produc'd to the remote point b and d; and parallel to the s ecant BD draw bd; and let the arc Acb be always s imilar to the arc ACB. Philosophiæ Naturalis Principia Mathematica, Latin for "Mathematical Principles of Natural Philosophy", often called the Principia (sometimes Principia Mathematica), is a work in three books by Isaac Newton, first published 5 July Newton also published two further editions, in. Isaac Newton's book Philosophiae Naturalis Principia Mathematica (), whose title translates to "Mathematical Principles of Natural Philosophy", reflects the then-current use of the words "natural philosophy", akin to "systematic study of nature". Mathematical principles of natural philosophy: Book I: Method of first and last ratios -- Determination of centripetal forces -- Motion of bodies in eccentric conic sections -- Finding of elliptic, parabolic, and hyperbolic orbits from the focus given -- How the orbits are to be found when neither focus is given -- How the motions are to be found in given orbits -- Rectilinear ascent and Pages: Everyone should read this book, and this seems to be one of the better scans. Cheers. The Mathematical Principles of Natural Philosophy Sir Isaac Newton Snippet view - Newton's Principia: The Mathematical Principles of Natural Philosophy, N. 5/5(6). Internet Archive BookReader Newton's Principia: the mathematical principles of natural philosophy. The Mathematical Principles of Natural Philosophy () by Isaac Newton, III, Book II) as the area DR × AB is more a mathematical hypothesis than a physical one. In mediums void of all tenacity, the resistances made to bodies are in the duplicate ratio of the velocities.NEWTON'S PRINCIPIA, The Mathematical Principles of Natural ated into English by Andrew Motte, to which is added Newton's System of the World. Newton, Isaac Published by Ivison & Phinney: NY ().Philosophiæ Naturalis Principia Mathematica (Latin for Mathematical Principles of Natural Philosophy), often referred to as simply the Principia, is a work in three books by Isaac Newton, in Latin, first published 5 July

On the Consistency of MSGUT Spectra We show explicitly that, once convention dependent phases are properly accounted for, the mass spectra evaluated by us in msgt04 satisfy the Trace, SU(5) reassembly and Goldstone counting consistency checks. Furthermore proper phase accounting shows that the transposition symmetry called “Hermiticity” will be manifest only up to signs arising from the product of six phase factors which must be reinserted if the symmetry is to be verified also as regards signs. Thus the claims of hep-ph/0405300 and hep-ph/0412348 concerning the inconsistency of our results are completely mistaken. The Chiral multiplet spectra of the two calculations are equivalent. However our method also gives all gauge and gauge chiral spectra as well as a MSSM decomposition of all SO(10) MSGUT couplings, for both tensors and spinors, which are unavailable, even in principle, using the methods of the above papers. The supersymmetric GUT based on the Higgs system which was proposed in refs.aulmoh ; ckn is of late basking in the attentions of the GUT model builders’ community. There are two principal reasons for this recent revival- apart from the general favour enjoyed by SO(10) GUTs since the discovery of neutrino oscillations. The first was the the realization abmsv03 that this SO(10) Susy GUT is in fact the minimal (Susy) GUT (MSGUT) compatible with all known data. Secondly, although the proposal of bm93 to predictively fit all charged fermion and neutrino masses using only the Fermion Mass (FM) Higgs system initially seemedbm93 to fail (or else to require recondite phase tunings fo02 ), it was recently shownbsv03 to naturally predict large neutrino (PMNS) mixings in the sector due to unification combined with Type-II seesawmohsen domination. This natural solution of the vexed question of the conflict between large leptonic and small quark mixings in the context of GUTs (where both belong to common representations and thus often tend to have similar mixings) triggered an intensive investigation of the full fermion mass fitting problem in such FM systems moh ; bsv04 ; bert which enjoys continuing success. The identification of a minimal and fully realistic model naturally motivates the development of techniques for the calculation of all mass spectra and couplings necessary for the full specification of the low energy effective field theory implied by this GUT. The calculation of mass spectra and RG analysis of this GUT was begun in aulmoh . It was continued in lee using a somewhat abstract methodheme for calculating the clebsches relevant for mass spectra. However the spectra were quite incomplete and no method for evaluating the MSGUT couplings in an MSSM basis was available until the development of analytic methods to translate between SO(10) labels and those of its maximal subgroups alaps ,nathraza . The computation of these spectra was then begun by the authors of alaps and the crucial ( and ) mixing matrices for the MSSM type Higgs doublets and proton decay mediating triplets were published in a later version alaps of this paper. In 2004 two independent calculations of the chiral SO(10) tensor mass spectrafikmo0401 ,bmsv04 and of the complete chiral and gauge spectra and couplings (including spinors)msgt04 appeared. The results of bmsv04 and msgt04 had been cross checked and found compatible before publication. On the other hand, notable discrepancies existed between the results of version 1 of fikmo0401 and alaps ; bmsv04 ; msgt04 and were pointed out in these papers. The subsequent version of fikmo0401 then appeared, incorporating these corrections without any acknowledgement, but alleging that the results of alaps ; bmsv04 were inconsistent since they would not satisfy the consistency tests based on Traces, “Hermiticity” and Goldstone counting applied by these authors to their own results. However these authors did not provide any details or claim that they had adjusted for the phase convention differences between the different calculations. There was no reason at all to expect the results with different conventions to satisfy the the consistency conditions in the particular form valid for the particular conventions they had chosen to use. Moreover both the impugned calculations had checked the Goldstone-super-Higgs counting down to the smallest relevant little group namely so there was no reason why the MSSM spectra should fail reassembly at the special vevs with SU(5) or Pati-Salam symmetry (from which decomposition of SO(10) the results of alaps ; msgt04 were derived ). Nevertheless these authors insisted that the other calculations failed their tests and hence theirs were the only correct results. In procviet the present author mentioned and in version 2 of msgt04 explicitly demonstrated that all mass terms grouped into SU(5) invariant form when the vevs were SU(5) invariant. No assumption about zero vev of the SU(5) singlets in were made and the Goldstone counting was also correct. Now these authors have again appeared in print fikmo0412 making mistaken but strong allegations concerning the results of alaps ; bmsv04 ; msgt04 . In particular they again claim that our spectra fail the trace and “Hermiticity” tests and although they concede that the results are consistent for symmetry they claim that our results fail Goldstone counting when is broken by non zero (SU(5) invariant) vevs. In this communication we show explicitly that all these claims are false and that our spectra obey the Trace,“Hermiticity” and Goldstone counting tests when due account is taken of the phase conventions and the complete SU(5) invariant mass terms derivable from our resultsalaps ; msgt04 without any change or modification whatsoever. This should finally lay this unfortunate controversy to rest. Ii Summary of the MSGUT The chiral supermultiplets of the MSGUT consist of the “AM type” totally antisymmetric tensors: which break the GUT symmetry to the MSSM, together with Fermion mass (FM) Higgs 10-plet(). The plays a dual or AM-FM role since it also enables the generation of realistic charged fermion neutrino masses and mixings (via the Type I and/or Type II mechanisms); three 16-plets contain the matter including the three conjugate neutrinos (). The superpotential is the sum of and : In all the MSGUT has exactly 26 non-soft parameters abmsv03 . The MSSM also has 26 non-soft couplings so the 15 parameters of must be essentially responsible for the 22 parameters describing fermion masses and mixings in the MSSM. The GUT scale vevs that break the gauge symmetry down to the SM symmetry are aulmoh ; ckn . The vanishing of the D-terms of the SO(10) gauge sector potential imposes only the condition . Except for the simpler cases corresponding to enhanced unbroken symmetry ( etc)abmsv03 ; bmsv04 this system of equations is essentially cubic and can be reduced to a single cubic equation abmsv03 for a variable , in terms of which the vevs are specified. Using the above vevs and the methods of alaps we calculated alaps ,msgt04 the complete gauge and chiral multiplet GUT scale spectra and couplings for the various MSSM multiplet sets, of which the chiral multiplets belong to 26 different MSSM multiplet types of which 18 are unmixed while the other 8 types occur in multiple copies and mix via upto 5 x 5 matrices. For immediate reference the chiral and chiral-gauge mass spectraof msgt04 are listed in Appendix A. In the pure chiral sector they are the values of the mass matrices defined as the double derivatives of the superpotential : where represents any multiplet and is its conjugate from the same SO(10) representation if the representation is real (i.e 10,210) and from the conjugate SO(10) representation if complex (i.e if then and vice versa). Iii Consistency checks The basic fact about the results of the three extant calculations of MSGUT mass spectra alaps ,bmsv04 ; fikmo0405 ; msgt04 is that all three calculations yield identical combinations of superpotential parameters in every single matrix element and differ only overall phases if at all. These phase or sign differences arise from the different field phase conventions used and must be accounted for explcitly when applying the consistency tests of fikmo0405 to the other two calculations. We trace the origin and influence of these phases and then show that there are no inconsistencies once the effects of these phases are accounted for properly : as required by elementary linear algebra. The consistency tests insisted upon by the authors of fikmo0412 are described below in their own words (followed by our explanatory comments): iii.1 Trace consistency “There are three main consistency checks. The first is that the trace of the total Higgs mass matrix does not depend on the coupling constants . It depends only on mass parameters and the dimensions of the corresponding representations. The sum rule for the Higgs-Higgsino mass matrices is ” Where the mass parameters are related to ours by . The origin of this sum rule is as follows . For the (10)-plet : we have Similarly for 210 we can take the 210 independent field components in the real SO(10) vector label basis to be so that here is one of the independent index types in the notation of alaps . So in all Now when taking the trace one must transform from real SO(10) labels to (in general) complex labels for every sub-representation R of the relevant SO(10) tensor representation by using a Unitary transformations : Note that overall relative phases of are conventional. Therefore the trace in labels is really to be done with the metric This metric is not necessarily unit. In fact in our basisalaps which is derived from an embedding of in SO(10)alaps its diagonal elements can be and this results precisely in negative signs for some of the mass parameter terms in the diagonal mass matrix elements. Its value is thus obviously the sign of the mass parameter in the diagonal mass terms for each subrep, so that the minus signs that occur (e.g for etc : see the Appendix) are automatically compensated to ensure that at least the mass parameter terms give the correct contributions to eqn.(7). The consistency check is then to see whether after compensating for these minus signs the coupling parameter () dependent terms indeed add up to zero (since they are proportional to )in both cases. It is gratifying to check that the sum of some hundred different terms added up to precisely zero thus verifying eqn.(7). We have given the details of this trace calculation in the Appendix in the form of two tables. It thus appears that the authors of fikmo0412 did not account for these elementary facts of linear algebra before claiming that our phases were inconsistent. In short the Trace constraint is in fact satisfied by our spectra just as it is by theirs. iii.2 SU(5) Check Next we come to what these authors consider their “main check ” but to us seems a trivial consequence of the little group embedding once one has checked the super-Higgs effect for the breaking (as we have done in in great and explicit detail including calculation of the null eigenvectors)msgt04 . However these authors make the alarming and bald assertion : “The third and main check is the check briefly described in the paper fikmo0405 , but misunderstood in Refs. msgt04 ; procviet . Here we explicitly prove that mass matrices in Ref. fikmo0401 satisfy this highly nontrivial test, and that the results obtained in Refs. bmsv04 ; alaps ; msgt04 are internally inconsistent. ” (our reference numbers). Note that the trace requirement eqn.(7) is satisfied. The last term involving only singlets is absent when and was not listed when detailing the reassembly msgt04 , but it is of course always present in the sector mass matrix since the mixing terms between and are read off from (see the Appendix) to be From these mass terms one can easily verify that the SU(5) singlet mass matrix with rows labelled by and columns by whose determinant is manifestly zero at the SU(5) symmetric point where but is not zero at the generic invariant vevs. Thus we obtain the SU(5) singlet Goldstone required in the former case. Moreover the sector matrix is trivially read off from the terms given above to be The determinant at the SU(5) solution point is again zero. This shows that in fact, just as found by the authors of fikmo0405 there are at most 13 distinct mass eigenvalues namely The trace of this mass matrix is also easily found to be the same as eqn(7). It is is also easy to check that after accounting for the trace metric, as explained above, and up to a phase in the case of the determinant eqn.(17) of fikmo0412 namely : are obeyed at the SU(5) point and . Note that in this case the trace and determinant has been taken without weighting for the dimension of the doublet and triplet representations. To sum up , the authors of fikmo0412 had already conceded that our results were consistent when but asserted that we had misunderstood their consistency tests for the case with only SU(5) symmetry where and that our results failed the counting of Goldstone multiplets which restricted the number of independent mass eigenvalues to just 13. Yet we have just shown, without any new result whatever that the SU(5) solution is just as consistent as the solution. As we noted earlier, in view of the fact that we have already analyzed the mass matrices -and found them consistent- for the breaking down to the much smaller little group , the reassembly into or invariants when the vevs have these symmetries is no surprise at all. The wonder of these authors at this phenomenon is perhaps attributable to the extreme formality of their calculational approach in which simple algebraic facts take on a mathematical sheen difficult to demystify. iii.3 “Hermiticty” Property The next check proposed by these authors is the so-called “Hermiticity” cross check. Properly speaking, since we are calculating chiral mass matrices “hermiticity” of the mass matrices is an oxymoron. However by adopting particular phase conventions and considering the sum of invariants involving and separately to be “real” these authors obtain a relation between their Clebsches that they call a “hermiticity” relation though no actual complex conjugation is involved. The corresponding symmetry from our point of view is the observation that our GUT Higgs superpotential (not ) is invariant under the substitutions and the representations are real. Thus if we wrote all our mass matrices in terms of the original real labels and kept track of any sign or phase introduced while defining vevs of SO(10) tensors then we would indeed observe the effects of this symmetry as a transposition property of the mass matrices. In fact it is apparent that our mass matrices obey this transposition symmetry up to signs. However after transformingalaps to our PS derived complex labels each off diagonal mass term contains three phases introduced by this transformation. The relative signs which need to be checked are thus a ratio of six phases ! To trace back and recalculate all the phases involved in some hundreds of terms merely to satisfy the contrivance of an independent phase convention would negate the entire spirit of our method which is based on keeping automatic track of these phases by taking the trouble to perform the embedding of PS in SO(10) for all the relevant representations completely explicitly. Thus if these authors wish to insist that our results fail this contrived test formulated in terms of Clebsch coefficients that we never define or use it is incumbent on them to trace these 3 phases for every off diagonal mass term back to the real labels and show that the results conflict with this hidden “symmetry”. This is particularly so after our demonstration that all their other assertions are false due to elementary confusions. Since all other computable tests are in fact obeyed an appeal to this obscure requirement would seem to clutch at straws ! The detailed discussion above demonstrates that our mass matrices (and by implication those of bmsv04 ) satisfy every reasonable consistency check claimed by the authors of fikmo0405 ; fikmo0412 to be unique to their calculations and makes it clear that these three calculations are equivalent as far as Chiral mass spectra are concerned. However the method of alaps yields not only chiral mass spectra and chiral vev equation decompositions in terms of MSSM labels but also all other couplings whether gauge or chiral and for both tensor and spinor SO(10)representations. The methods of heme used in the calculations of bmsv04 ; fikmo0405 cannot give these quantities even in principle. Our method is different from the formal method ofheme ; lee ; bmsv04 ; fikmo0405 and is more complete, especially regarding couplings. Being analytic and explicit it also allows us to trace and resolve discrepancies in a form directly usable in a field theory calculation . Since the excavation of the phenomenological implications of the MSGUT requires knowledge of most of these quantities it seems that use of the phase conventions of fikmo0405 will carry the burden of an additional and very tedious phase accounting before our results on the most general couplingsmsgt04 -the only ones available- are usable under alternate conventions. Appendix : Tables of masses and mixings In this appendix we give the mass spectra msgt04 of the chiral fermion/gaugino states together with the Trace calculation in MSSM labels. Mixing matrix rows are labelled by barred irreps and columns by unbarred. Unmixed cases(i)) are given as Table I. ii) Chiral Mixed states

Auto fill todays date into a pdf form adobe support. The adobe acrobat user community is a global resource for users of acrobat and pdf, with free eseminars, tips, tutorials, videos and discussion forums. Changing the subject of complex formula free worksheets. This article describes the formula syntax and usage of the hyperlink function in microsoft excel description. Changing the subject of the formula 3 the video explains how to change the subject of the formula in more complicated examples e. So, before creating my complex email code, i set up a little test file, with a short list of fake customers. How to change the subject of a formula please help. In each case, make the subject of the formula now try these 2. Pdf change title, subject, author, keywords, dates in. A formula is a special equation which deals with variables. I assume with formula you mean a pdf form, which is a normal pdf. Hence, finally, if you want to insert the date or time to subject, you have to copy them from the body and paste them to the subject manually. Math formulas download maths formulas pdf basic math fomula. Look at all free maths resources and example questions for your maths revision and d. If the number of exemptions decreases, they need to file a new certificate within 10 days. The change formula change equals dissatisfaction x vision x first steps that is greater than the resistance the idea is that if you are seeking some significant, systemwide change there are several core elements that need attention. Title, subject, author, keywords, creator, producer, creation date and modified date. If you want to remove the interactive part of the pdf, you can flatten the pdf. This website and its content is subject to our terms and conditions. National 4 relationships homework changing the subject. Is there any procedure where i can give default file name for pdf and subject while attaching pdf file in a mail. Read each question carefully before you begin answering it. Usually the subject of a formula is on its own on the lefthand side. If you wnat to make the form archival, convert it to one the pdf archival formats. This course will explain how to rearrange some simple formulas. Please click option document then go to pdf annotation like i showed the following snapshot. Solving equations by elimination calculator, graphing equations three variables, 2 and 3 digit dividing decimals, sample multiply radicals, computation for converting the number system codes in java, simple trigonometry solutions using matlab pdf files download, partial sum method. Level 6 finding the unknown which is not the subject of a formula example. Videos designed for the site by steve blades, retired youtuber and owner of to assist learning in uk classrooms. The manipulation of algebraic expressions is an important. From pdf files to excel spreadsheets john haworth wants to reliably convert a lot of data from pdf files to excel for spreadsheet analysis. So, because we want x on its own what we can do is add 4 to this side. After returning to the message window, you can see the current date or time in the email body. Change the subject of a formula, including cases where the subject occurs on both sides of the formula, or where a power of the subject appears. N t q m x a t d p e 1 5 w w i e t m h 1 s i l n d f e i 5 n 6 i s t l e p n a s l h g v e p b m r s a u e 2 z. In academic writing, it is generally preferred to choose an active verb and pair it with a subject that names the person or. Converting data from pdf files to excel spreadsheets. A formula is an equation with letters which stands for quantities. We can use algebra to change the subject of a formula. Changing the subject of a formula squares and square roots duration. Tes global ltd is registered in england company no 02017289 with its registered office at 26 red lion square london wc1r 4hq. It contains a list of basic math formulas commonly used when doing basic math computation. Dec 31, 2010 how to change the subject of a formula. Then you will see the dialog box like the following snapshot. The examplesquestions are staggered in difficulty, beginning at grade b and finishing at a. Understand how to change the subject of an algebraic equation and pas your next math exam how to transform a formula in order to make a particular variable the subject. Im trying to put some vba into place that puts our own custom subject line on these emails rather than employees having to type out the subject each time. Heres how i managed to send email with pdf attachment from excel. In some cases, you may need to send a worksheet as a pdf file through outlook. Student date topic notes examples reference resources. The following video shows how to change the subject of the formula using 2 steps. Mathematics linear 1ma0 changing the subject of a formula. If playback doesnt begin shortly, try restarting your device. Thermodynamics 157 internal energy of the system in state a be called u a. Or you can click option open to browse folders to add pdf file. The simplest way to think of it is to say that the subject is a pronumeral all by itself standing before the equal sign. I would also like the other user on their personal computer, have the option to change the date field from a calendar dropdown. This is changing the subject of the formula from c to d. Send a copy of the entire pdf file as an attachment. If you are stuck, please watch the video tutorial for a visual walkthrough. Gcse algebra changing the subject of an equation rearranging. Is the formula which gives the circumference, c, of a circle of radius r. Mathematics linear 1ma0 changing the subject of a formula materials required for examination items included with question papers ruler graduated in centimetres and nil millimetres, protractor, compasses, pen, hb pencil, eraser. The subject of a formula you will be familiar with the formula for the area of a circle which states. Sep 03, 20 this channel is managed by up and coming uk maths teachers. If the letter you are rearranging for is in a bracket, then you can approach the rearranging in one of two ways. The variable on its own, usually on the left hand side. When a sentence is written in the active voice, the subject performs the action. Change the subject of each formula to the letter shown in brackets. Worked examples of rearranging formulae with five questions to practise and then two exam. Send sheet as a pdf attachment in email welcome to. Sometimes when we are changing the subject of a formula the variable that we want to make the subject occurs more than once. How to change the subject of a formula first of all lets make sure you know what the subject of an equation is. Email this pdf as a attachment to the specific person. This rule is used to work out the total cost, in pounds, of hiring a carpet cleaner. Change the subject of a formula such as 1 1 1 f u v, where all variables are in the denominators. Copy and paste the following code into the vb editor. Mar 26, 2008 i want to update file name and subject when i select option 4. Changing the subject in complex formulae teaching resources. The following video shows how to change the subject of the formula using 1 step. How to change the subject of an equation help with igcse. Pdf change title, subject, author, keywords, dates in multiple files software offers a solution to users who want to change general document information metadata in one or more pdf files. The requirement of the project is that i need a vba code for a button when i click it, it will convert my active sheet alone to pdf, automatically save it with the title captured from a cell in the active sheet which is entered by the user. Pencil, pen, ruler, protractor, pair of compasses and eraser. Sometimes a formula can be rearranged into a more useful format. Can be made into an a5 booklet for students to write on. Pdf form field basics pdf form field properties add tooltips, date and time, calculated values set action buttons in pdf forms check pdf forms faq this document explains the pdf form field basics in acrobat dc. Change of base formula kuta software infinite algebra 2. What this question basically means is get x on its own. Vba code to convert excel to pdf and email it as attachment. On completion of this worksheet you should be able to change the subject of formulae where the desired subject occurs more than once in the formulae. You should find a personalized email with the merged document as a pdf attachment in your gmail sent items. Dear readers, in my previous article, you saw how to save an excel sheet as pdf file. One can flatten a pdf and make comments and from fields into content of the pdf. Please get a vivid impression from the following snapshot which is from the software interface. Vlookup, index, match, rank, average, small, large, lookup, round, countifs, sumifs, find, date, and many more. Changing the subject of a formula we can use algebra to change the subject of a formula. Changing the subject algebra free gcse maths resources. Method 1 inset formula as picture simply launch this software by double clicking the desktop icon and drag pdf file which you need to edit formula to software interface. Also, we have seen how to send the activesheet as an attachment in email. W e can change the state of the system in two different ways. Mar 02, 2019 if there are just a few subject lines to edit, you can use incell editing to quickly paste the code into the subject. The goal was to send an email to each name in a list, and attach a couple of pdf files. I dont want to rename file or clear the default subject and write again. Create a search folder to find all messages where the subject needs tagged then run the macro while viewing the search folder results. Jan 01, 2020 employees can change the number of their exemptions on form m4 by filing a new certificate at any time if the number of exemptions increases. Worked examples of rearranging formulae with five questions to practise and then two exam questions. If you know the value of then you can substitute directly to find. In each case, make the letter given at the end, the subject of the formula. Variables are bits of information that can change each time you deal with the formula. To change the subject of a formula is to rearrange it so that a different letter is on its own on the left hand side. Algebraic fractions addition and subtraction with and without calculator opportunity for revision of negative numbers, decimals, simple fractions. Sheets 201 230 from the 455 maths revision sheets resource. Title, subject, author, keywords, creator, producer, creation date. Send email with pdf attachment from excel contextures blog. So, if you want to have a formula or rule that lets you calculate d, you need to make d, the subject of the formula. Changing the subject of the formula maths4scotland. Explaining logarithms a progression of ideas illuminating an important mathematical concept by dan umbarger. Changing the subject of a formula the rules are the same as when we solved equations except we do not end up with a solution but with another formula. Speed distance time and we use it every time we go somewhere. On completion of this worksheet you should be able to change the subject of formulae including those with fractions, indices and roots. In the form, we say that is the subject of the formula. Changing the subject of the formula examples, solutions. It is useful to rearrange the formula to make r the subject. How to auto insert current date or time into email subject or. Transposing formulae show each stage of the working carefully. In this article i am going to show you how to send the activesheet as an. This blog will give you the excel formulas pdf list of the key functions of excel. This is one of the important skills you can learn in maths. Simply launch this software by double clicking the desktop icon and drag pdf file which you need to edit formula to software interface. Aug 22, 2016 of course, the subject line of the ensuing email takes on the name of the attachment. The aim of rearranging is to manipulate the formula so that all the terms involving the letter you want is on one side of the equation, and everything else is on the. When you click a cell that contains a hyperlink function, excel jumps to the location listed, or opens the. This week, i was experimenting with sending email from excel via outlook. What do you understand by the advanced excel formulas and basic excel functions. In such a case, if outlook can automatically fill the subject with the attachment name, they will be able to avoid the annoying no subject warning. How to change the subject of a formula futureschool. The methods required here reappear in many other aspects of maths, including solving equations, trigonometry, pythagoras theorem, nth term and many others. All that changing the subject of the formula means is basically getting a letter on its own on one side of the equation. How to change the subject of a formula with video help. Using vba to assign a subject in outlook to emails prepared. Pencil, pen, ruler, protractor, pair of compasses and eraser you may use tracing paper if needed guidance 1. We do some mechanical work, say 1 kj, by rotating a set of small paddles and. Math formulas download maths formulas pdf basic math. Be dissatisfied d with the way things are in relationship. Select any topic from the above list and get all the required help with math formula in detail. If an employee has more than one job, they may claim exemptions only with their principal employer. Differentiated worksheet on changing the subject or rearranging formula. I want to update file name and subject when i select option 4. Rearranging a formula is similar to solving an equation we must do the same to both sides in order to keep the equation balanced. This activity will give you practice in changing the subject of a formula. The hyperlink function creates a shortcut that jumps to another location in the current workbook, or opens a document stored on a network server, an intranet, or the internet. Apr 14, 2020 ncert books pdf free download for class 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, and 1 april 14, 2020 by kishen 16 comments there are many books in the market but ncert books stand alone in the market. Ncert books pdf download 2020 for class 12, 11, 10, 9, 8, 7. Changing the subject rearranging formulae booklet teachit maths. Pdf help acrobat forum learn adobe acrobat pdf help. It depends on the formula you have to work with, so you have to use your experience and a bit of common sense to decide what to do in each case. National 4 relationships homework changing the subject of a formula 1. 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« ΠροηγούμενηΣυνέχεια » 11. A and B barter : A hath 145 gallons of brandy at I dol. 20 cts. per gallon ready money, but in barter he will have 1 dol. 35 cts. per gallor : B has linen at 58 cts. per yard ready money ; how must B sell his linen per yard in proportion to A's bartering price, and how many yards are equal to A's brandy? Ans. Barter price of B's linen is 65 cts. 2m. and he must give A 300 yds. for his brandy. 12. A has 225 yds. of shalloon, at 2s. ready money per yard, which he barters with B at 2s. 5d. per yard, taking indigo at 12s. 6d. per 1b. which is worth but 10s. how much indigo will pay for the shalloon; and who gets the best bargain? Ans. 43 lb. at barter price will pay for the shalloon, and B has the advantage in barter. Value of A's cloth, at cash price, is £22 10 Value of 43;lb. of indigo, at 10s. per lb. 15 B gets the best bargain by £0 15 LOSS AND GAIN, IS a rule by which merchants and traders discover their profit or loss in buying and selling their goods : it also instructs them how to rise or fall in the price of their goods, 80 as to gain or lose so much per cent. or otherwise. Questions in this rule are answered by the Rule of Three. 1. Bought a piece of cloth containing 85 yards, for 191 dols. 25 cts, and sold the same at 2 dols. 81 cts. per yard ; what is the profit upon the whole piece? Ans. $47, 60 cts. 2. Bought 12! cwt. of rice, at 3 dols. 45 cts. a cwt. and sold it again at 4 cts. a pound; what was the whole gain ? Ans. $12, 87 cts. 5m. 3. Bought 11 cwt. of sugar, at 6 d. per lb. but could not sell it again for any more than 21. 16s. per cwt.; did I gain or lose by my bargain ? Ans. Lost, £2 11s. 4d. 4. Bought 44 lb. of tea for 61. 12s. and sold it again for 81. 10s. 68. ; what was the profit on each pound? Ans. 10 d. 5. Bought a hhd. of molasses containing 119 gallons, at 52 cents per gallon; paid for carting the same 1 dollar 25 cents, and by accident 9 gallons leaked out; at what rate must I sell the remainder per gallon, to gaïn 13 dollars in the whole ? Ans. 69 cts. 2 m. + II. To know what is gained or lost per cent. Rule.-First see what the gain or loss is by subtraction; then, As the price it cost : is to the gain or loss :: so is 1001. or $100, to the gain or loss per cent. EXAMPLES. 1. If I buy Irish linen at 2s. per yard, and sell it again at 2s. 8d. per yard; what do I gain per cent. or in laying out 1001. : As : 2s. 8d. :: 10ől. : £33 6s. 8d. Ans. 1 2. If I buy broadcloth at 3 dols. 44 cts. per yard, and sell i ft again at 4 dols. 30 cts. per yard : what do I gain per ct. or in laying out 100 dollars ? $ cts. Sold for 4, 30 As 3 44 : 86 : : 100 : 25 Ans. 25 per cent. Gained per yd. 86 3. If I buy a cwt. of cotton for 34 dols. 86 cts. and sell it again at 41į cts. per lb. what do I gain or lose, and what * cts. 1 cwt. at 41} cts. per lb. comes to 46,48 Prime cost 34,86 per cent. ? Gained in the gross, $11,61 As 34,86 : 11,62 : : 100 : 33. Ans. 331 per cent. 4. Bought sugar at 81d. per lb. and sold it again at 41. 17s. per cwt. what did I gain per cent. ? Ans. £25 19s. 5 d. 5. If I buy 12 hhds. of wine for 2041. and sell the same again at 141. 178. 6d. per hhd. do I gain or lose, and what Ans. I lose 121 per cent. 6. At 1 d. profit in a shilling, how much per cent. ? Ans, £12 108 per cent. ? 7. At 25 cts. profit in a dollar, how much per cent. ? Ans. 25 per cent. Note.-When goods are bought or sold on credit, you must calculate (by discount) the present worth of their price, in order to find your true gain or loss, &c. 1. Bought 164 yards of broadcloth, at 14s. 6d. per yard ready money, and sold the same again for 1541. 10s. on 6 months credit; what did I gain by the whole; allowing discount at 6 per cent, a year? £. £. £. £. As 103 : 100 : : 154 10 : 150 O present worth. 118 18 prime cost. Gained £31 2 Answer. 2. If I buy cloth at 4 dols. 16 cts. per yard, on eight months credit, and sell it again at 3 dols. 90 cts. per yd. ready money, what do I lose per cent. allowing 6 per cent. discount on the purchase price ? Ans. 21 per cent. per cent. ? III. To know how a commodity must be sold, to gain or lose so much per cent. Rule.--As 100 : is to the purchase price : : so is 1001. or 100 dollars, with the profit added, or loss subtracted : to the selling price. EXAMPLES. 1. If I buy Irish linen at 2s. 3d. per yard; how must I sell it per yard to gain 25 As 100l. : 2s. 3d. : : 1251. to 2s. 9d. 3 qrs. Ans. 2. If I buy rum at 1 dol. 5 cts. per gallon; how must I sell it per gallon to gain 30 per cent. ? As $100 : $1,05 : : $130 : $1,364 cts. Ans. 3. If tea cost 54 cts. per lb. ; how must it be sold per Ib to lose 12per cent. ? As $100 : 54 cts. : : $87, 50 cts. : 47 cts. 2 m. Ans. 4. Bought cloth at 178. 6d. per yard, which not proving so good as I expected, I am obliged to lose 15 per cent bị it; how must I sell it per yard? Ans. 14s. 100. 5. If 11 cwt. 1 qr. 25 lb. of sugar cost 126 dols. 50 cts. how must it be sold per lb. to gain 30 per cent. ? Ans. 12 cts. 8m. 6. Bought 90 gallons of wine at 1 dol. 20 cts. per gall. but by accident 10 gallons leaked out; at what rate must I sell the remainder per gallon to gain upon the whole prime cost, at the rate of 124 per cent.? Ans. $1, 51 cts. 8jom. IV. When there is gained or lost per cent. to know what the commodity cost. RULE.-As 1001. or 100 dols. with the gain per cent. added, or loss per cent. subtracted, is to the price, so is 100 to the prime cost. EXAMPLES. 1. If a yard of cloth be sold at 14s. 7d. and there is gaiued 161. 13s. 4d. per cent. ; what did the yard cost ? . d. d. £. As 116 13 4:14 7::100 to 12s. 6d. Ans. 2. By selling broadcloth at 3 dols. 25 cts. per yard, I Iose at the rate of 20 per cent. ; what is the prime cost of said cloth per yard ? Ans. $4, 06 cts. 21m. 3. If 40 lb. of chocolate be sold at 25 cts. per lb. and I gain 9 per cent. ; what did the whole cost me? Ans. $9, 17 cts. 4m. + 4. Bought 5 cwt. of sugar, and sold it again at 12 cents per lb. by which I gained at the rate of 25; per cent. ; what did the sugar cost me per cwt. ? Ans. $10,70 cts. 9m. + V. If by wares sold at a given rate, there is so much gained or lost per cent. to know what would be gained or lost per cent. if sold at another rate. RULE.--As the first price : is to 1001. or 100 dols. with the profit per cent. added, or loss per cent. subtracted :: so is the other price : to the gain or loss per cent, at the other rate. N. B. If your answer exceed 1001. or 100 dols. the excess is your gain per cent.; but if it be less than 100, that deficiency is the loss per cent. 1. If I sell cloth at 5s. per yd. and thereby gain 15 per cent. what shall I gain per cent. if I sell it at Os. per yd. 3 £ As 5 : 115 :: 6 : 138 Ans. gained 38 per cent. 2. If I retail rum at 1 dollar 50 cents per gallon, and thereby gain 25 per cent. what shall I gain or lose per ceat. if I sell it at 1 dol. 8 cts. per gallon? $ cts. $ $ cts. 3. If I sell a cwt. of sugar for 8 dollars, and thereby lose 12 per cent. what shall I gain or lose per cent. if I sell 4 cwt. of the same sugar for 36 dollars ? Ans. I lose only 1 per cent. 4. I sold a watch for 171. Is. 5d. and by so doing lost 15 per cent. whereas I ought in trading to have cleared 20 per cent. ; how much was it sold under its real value 3 £. £ s. d. £. £. s. d. As 85 : 17 1 5 : : 100 : 20 1 8 the prime cost. 100 : 20 1 8: : 120 : 24 20 the real value. Sold for 17 1 5 £7 0 7 Answer. FELLOWSHIP, IS a rule by which the accounts of several merchants or other persons trading in partnership, are so adjusted, that each may have his share of the gain, or sustain his share of the loss, in proportion to his share of the joint stock.–Also, by this Rule a bankrupt's estate may be divided among his creditors, &c. SINGLE FELLOWSHIP, Is when the several shares of stock are continued in trade an equal term of time. Rule.-As the whole stock is to the whole gain or loss : so is each man's particular stock, to his particular share of the gain or loss.

Sanjay Goel, JIIT, 2006 Data Structure: Programming Assignments CSE/IT 3rd sem12. 09.06 1. WAP to simulate a robot ‘SimRob’ linear movement along X axis bounded by Left and Right wall. The robot has two sensors on two sided that get activated on hitting the wall on its side. On sensor activation the controller changes the direction of robot’s movement. Graphics is not compulsory. 2. Design DS for storing and using Truth table(5 input and three output variables)14.09.06 1. WAP a dual robot movement simulator “Robpair” using two ‘SimRobs’ (of possibly varying width) with already defined behavior. Graphics is not compulsory. 2. (Single student: 3 marks) WAP a multi robot movement simulator “Robmat” using multiple “Robpairs” at different Y positions. Graphics is not compulsory. 3. (Group of four: 3 marks each) WAP a circuit’s truth table generator using up to 10 circuit components with user defined truth tables. Each component can have up to 5 inputs and 3 outputs. The components are used to create a user defined combinational circuit. Your program should generate the truth table for the circuit from the available data. Graphics is not compulsory.19.09.06 1. (Single student: 2 marks) Design a data storage scheme for storing single variable polynomial functions and series of numbers. Design an algorithm to test if the given series is a Taylor series approximation (at x) for a given polynomial function and x0. Taylor series approx. of a given function is as follows:21.09.06 1. (Group of two: 2 marks each) Design a dynamic data structure for storing a randomly ordered collection of single variable polynomial functions and another static data structure for storing randomly-ordered collection of real number- sequences. The records in both the collections also should have additional provision for storing indices of all the matching entries (if any) in another collection. One entry in any collection may match with none, one or many entries in another. A number-sequence is declared as matching with a polynomial, if all the numbers in the sequence match with corresponding terms of the Taylor series expansion of a function for given x and x0 within the limits of a user-defined ‘permitted-mismatch’. Design an algorithm for updating matching indices in both the collections for a given user-defined input of ‘permitted-mismatch’, x and x0. Sanjay Goel, JIIT, 2006 You have the freedom to change the matching context of the problem to different and interesting situations. For example, you could match the following: 2. Polygons with number sequences that represent area, perimeter, and side length. 3. Lines (equation form) with endpoint coordinates. 4. Sentences with number sequences that represent frequency count of alphabets. 5. Differential equations with solutions. Take prior permission of your lab instructor to work on the context of your choice. You should take this permission within this week before 23/09/06. If your chosen context is much more complex than the stated context, the marks allocated to this problem will increase by 50%.23.09.06 (Group of two) You have already created the UI for following applications. Now create Data Characterization Catalogue and Concept Map for these applications: 1. (1.5 marks each) A design team has conceived the following initial specifications of a search engine for a large company’s internal Digital Library: Only specially authorized users can upload new documents or new versions of old document. All employees can look at the documents. Information systems department will create, update and maintain a list of keywords for faster search facility. The search engine users can also search by entering any word through the keyboard. Searched documents are to be listed as follows: Case A: Faster search on a listed keyword: As per the frequency of occurrence of the word i.e. the documents having higher “density” of the chosen keyword will be listed before the documents having lower density, where density[k, d] = (Occurrence count of the word k in d)/(word count in d) Case B: Search by entering a word though the keyboard: As per the frequency of usage of a document, where usage is defined as number of times a document is opened by users through the search engine. 2. (1.5 marks each) Design a simulator for a petrol pump offering the facilities of petrol, diesel, CNG and car wash. Also give a diagrammatic representation for this simulator software. Customer arrival rates are as follows: 5 am – 6 am : 30 customers per hr. 6 am – 7 am : 50 customers per hr. 7 am – 10 am : 200 customers per hr. 10 am – 4 pm : 100 customer per hr. Sanjay Goel, JIIT, 2006 4 pm – 8 pm : 200 customers per hr. 8 pm – 10 pm : 50 customer per hr. 10 pm – 5 am : 15 customers per hr. There are separate queues for the petrol, diesel, CNG and car wash. There are three petrol counters, two diesel counters, two CNG and one car-wash counters. 50% of the Fuel seekers purchase petrol, 25% purchase diesel and rest purchase CNG. During peak hrs., only fuel buyers are entertained. During 10 am to 6 pm, 10% customers come for car wash and rest for the fuel. Customers first buying the fuel do not go for car wash. However, 70% also buy fuel after car wash.26.09.06 1. (Group of two: 1.5 marks each) Generate as many design options as possible for 1st problem assigned in the last class. Evaluate each option for its advantages and disadvantages.28.09.061. (Group of two: 4 marks each) Design the UI, Data Characterization Catalogue, Concept Map, Data Structure, and Algorithms for following problem: We need to store up to 50000 persons and their relationships so that following queries can be efficiently addressed: i. How is x related to y? ii. To how many persons is x related? iii. Who all are related to x? iv. Who all are related to x and how? v. Who all are related to x or y? vi. How many unrelated extended families exist in the database? vii. Who is the eldest person in each extended family? viii. Who are the common persons somehow related to both x and y? ix. Who are the common persons somehow related to all x, y and z? x. How two different family charts can be merged and automatically updated in the event of marriage between the persons belonging to different family charts? 03.10.06 1. Find out the function for formula based addressing of individual elements within 2d, 3d, and kd matrices of 4 byte long integers stored in i. column major order. ii. row major order Sanjay Goel, JIIT, 2006 In the following questions, do not waste any memory for storage of null elements in the matrix cells other than the specified sub matrices. Also write algorithms for doubling the value of user specified cell for all cases mentioned below: 2. Find out the function for formula based addressing of individual elements within 2d, 3d, and kd lower triangular matrices of 4 byte long integers stored in i. column major order. ii. row major order. 3. Find out the function for formula based addressing of individual elements within 2d, 3d, and kd upper triangular matrices of 4 byte long integers stored in i. column major order. ii. row major order. 4. Find out the function for formula based addressing of individual elements within 2d, 3d, and kd diagonal matrices of 4 byte long integers stored in i. column major order. ii. row major order 5. Find out the function for formula based addressing of individual elements within 2d, 3d, and kd tri-diagonal matrices of 4 byte long integers stored in i. column major order. ii. Row major order 6. (Group of two: 4 marks each) for any two of the above mentioned cases to be announced by Lab instructors for each batch, Write recursive program for find the following: i. Indices of the cell(s) with maximum value in matrix. ii. Sum of the cell(s) in a matrix. Sanjay Goel, JIIT, 200605.10.06 1. (Group of two: 3 marks each) WAP for addition of two large sparse matrices (100 x 100 x 100) using Linked storage based data structures using pointer as well as simulated pointers (Sparse matrices are matrices with most elements with zero value; hence you can save large memory by avoiding the storage of zero value elements). Compare linked as well as simulated pointer based storage with formula based addressing in this context.07.10.06 1. (Group of two: 8 marks each) Enhance the scope of the people-relationship problem assigned on 28.09.06 to include following functionality: i. Find out the families in which there are more than 4 persons with postgraduate qualifications. ii. Find out the families in which more than 3 persons have been hospitalised in the last two years. iii. Find out the families in which more than 2 persons work in the same organization at same location. iv. Find out the families that collective pay annual income tax of more than Rs. 5 Lacs. v. Find out the working spouses whose work locations are in different cities with a distance of minimum 50 km. “Family” for our purpose includes direct descendants and spouses of self, parents, and father in law. Propose Data Characterization Table, Concept map, Data Structures. Also WAP for these and earlier mentioned queries. Design data structures based on at least 2 different structuring techniques. Sanjay Goel, JIIT, 200610.10.06 (Group of two: 10 marks each without graphics, 14 marks each with animatedgraphics; Exempted for those who are doing project) Design (UI, DataCharacterization Catalogue, Concept map, Data Structures, Test plan, Algorithm),implement, and test (produce a proper Test report) a fun station simulator withfollowing specifications: In a fun-park we have a network of N-S and E-W roads. The fun park has 100auto-navigated rechargeable battery red and blue toy vehicles. Red and blue vehicles areequal in number. These toy vehicles are queued on a ramp above the ground. To beginwith, all vehicles are queued on the ramp with alternate positions being assigned to redand blue vehicles all with the initially uniform battery life of 30 minutes on the ground.The ground has a pivot point where the vehicles are landed from the ramp on the groundin an orderly manner in a randomly chosen available direction with a minimum time gapof 2 minutes. 1. Vehicles land on the ground. 2. Vehicles move in N, E, W, and S directions. 3. Vehicles turn 90 degree. Blue vehicles can turn left and red vehicles turn right. 4. If hit by a vehicle in a lane, both the vehicles reverse their direction. 5. If hit by another vehicle or the wall on a crossing, the blue vehicles turn left whereas red vehicles turn right to change their direction of movement. 6. If after a vehicle clash on a crossing, the vehicles choose to move in the same new lane, the vehicle that landed earlier on the ground moves first. 7. The vehicles continue to move till their battery is discharged. 8. Discharged vehicles are put back for charging. 9. The recharged vehicles are put back on the ramp queue. 10. Charging time depends on the vehicles waiting for charging as follows: Number of vehicles in the queue Charging time >= 50 5 minutes 30-49 10 min. 20-29 15 min. 10-19 20 min. 6-9 25 min. <=5 30 min. 11. The battery charging time is also equal to new battery life on the ground. 12. On every hit (either with wall or another vehicle), more energy is consumed and the remaining battery time of concerned vehicle(s) decreases by one minute. 13. A large robot arm (crane like) picks up the vehicles without fuel and puts them onto the queue for charging. (Bonus: 4 marks each student) Enhance this fun station with some additional special features (to be proposed by you) to make this a unique station. Maintain confidentiality about your additional features and do not show it to other students registered in this course before evaluation of all students on this assignment. Sanjay Goel, JIIT, 200626.10.06 1. (5 marks: Exempted for those who are doing project) Write four different functions (based on different logic) for reverse order printing of information content of nodes in a linked list. Analyze the space and time complexity of your solutions.31.10.06 1. (2 marks: Exempted for those who are doing project ) WAP to convert a sorted array of 100 names into a BST. 2. (2 marks: Exempted for those who are doing project) WAP an unsorted linked list of 100 names into a BST. 3. (2 Marks: Exempted for those who are doing project) WAP to convert a BST of 100 names into a sorted linked list implemented with simulated pointer. 3. (4 marks: Exempted for those who are doing project) WAP to partition a BST of 100 names into three BSTs of nearly equal population size with non-overlapping value range.7.11.06 1. (6 marks: Exempted for those who are doing project) Write functions for non- recursive search, findmin and findmax in a BST. 2. (6 marks) Implement BST with simulated pointer and rewrite non-recursive search, insert, and delete functions for this data-structure.9.11.061. (Practice assignments not compulsory Bonus: 8 marks) Write a function for traversing binary tree by scanning it vertically right to left top to down.11.11.06 1. (Practice assignments not compulsory Bonus: 6 marks) Create an 8-ary tree using simulated pointer. Write two different kinds of traversal functions for this data structure.14.11.06 1. (Practice assignments not compulsory Bonus: 3 marks) A database stores information about hyperlinks across 1000 Websites. QUERY : If it is possible to move from xth Website to yth Website with up to two intermediate Websites in between them , then display “ link exists between xth and yth Websites” and also give sequence of names of intermediate Websites that are visited while moving from xth to yth Website ,otherwise display “No links Possible”. Propose the Data Structures and an appropriate algorithms (with and also without buffer) for processing the above Query. Also demonstrate the working of your algorithm by simulating key-steps of the algorithm for two cases e.g. A E C B. Sanjay Goel, JIIT, 200615.11.06 1. (Practice assignments not compulsory Bonus: 3 marks) Create an 8-ary tree using formula based storage. Write two different kinds of traversal functions for this data structure.