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Rasmussen Gilmore đã đăng cập nhật 6 tháng. 2 tuần trước đây As may be discussed in several articles in our series, the main focus of Angles is to get missing measurements–both side diets and viewpoint measures–in geometric figures. We are already revealed how the 36-60 right and 45-right particular triangles can certainly help. In addition , we started taking a look at another potential shortcut, SOHCAHTOA. This is a fabulous mnemonic unit for recalling the trigonometric ratios; and in a previous article, we mentioned this device by length from standpoint from what the characters stand for and what the trig ratios in fact represent. On this page, we will place this information for you to work as a program to find the losing measurements in just about any right triangle. Remember that SOHCAHTOA is informing us which two sides of your right triangular form the proportion of each trig function. That stands for: sine = reverse side/ hypotenuse, cosine sama dengan adjacent side/ hypotenuse, and tangent = opposite side/ adjacent region. You must keep in mind how to mean and pronounce this “word” correctly. SOHCAHTOA is pronounced sew-ka-toa; and you simply must emphasise to yourself out loud the ‘o’ sound of SOH and the ‘ah’ sound in CAH. To commence working with SOHCAHTOA to find losing measurements–usually angles–let’s draw your visual photograph. Draw your backwards capital “L” and next draw in the segment linking the endpoints of the legs. Label the bottom left nook as angle X. We should also claim we have some 3, five, 5 ideal triangle. As a result, the hypotenuse has to be the 5 outside, and a few make the bottom part leg the 3 leg and the vertical limb the four leg. There is nothing special about it triangle. It really helps whenever we are all picturing the same thing. I chose to use a Pythagorean triple of three, 4, 5 various because everyone already is aware of the aspects really do style a right triangular. I as well chose that because countless students make an assumption they shouldn’t! For quite a few unknown factor, many Geometry students believe that a several, 4, 5 various right triangular is also a good 30-60 suitable triangle. Naturally , this can not be since in a 30-60 correct triangle, one particular side can be half the hypotenuse, and don’t have the fact that. But we intend to use SOHCAHTOA to find the genuine angle measures and, with luck ,, convince persons the aspects are not 40 and 50. If we just knew two sides from the triangle, then simply we would need to use regardless of what trig efficiency uses the ones two facets. For example , whenever we only learned the adjoining side as well as hypotenuse to get angle X, then we would be forced to utilized the CAH part of SOHCAHTOA. Fortunately, young children and can all three sides of the triangular, so we are able to choose whichever trig function all of us prefer. As time passes and with practice, you could develop favorites. In order to find the angles these kind of trig quotients will determine, we need either a scientific or graphing this is actually the; and we will be using the “second” on “inverse” key. My very own preference is by using the tangent function every time possible, and since we know the opposite and adjacent factors, the tangent function can be utilized. We can today write the situation tan Maraud = 4/3. However , to resolve this equation we need to work with that inverse key at our feet to meters converter. This essential basically instructs the feet to meters converter to tell all of us what perspective produces the fact that 4/3 relation of sides. Type with your calculator this sequence, for example the parentheses: subsequent tan (4/3) ENTER. The calculator might produce the response 53. one particular degrees. In the event that, instead, you’ve got 0. 927, your calculator is set to offer you answers on radian measure and not certifications. Reset sohcahtoa . Now, a few see what happens if we use numerous sides. Using the SOH an area of the formula presents use the equation sin Times = 4/5 or X = inverse sin (4/5). Surprise! All of us still see that X = 53. 1 degrees. Doing furthermore with the CAH part, gives use cos X sama dengan 3/5 or perhaps X sama dengan inv cos (3/5), and… TA DAH… 53. 1 degrees again. I hope you get the point here, that if you are provided all three facets, which trig function you make use of makes zero difference. From this article you can see, SOHCAHTOA is definitely a powerful instrument for finding lacking angles through right triangles. It can also be utilized to find a absent side if an angle and one area are referred to. In the practice problem we now have used, all of us knew we’d sides 3 or more, 4, and 5, and a right angle. We just simply used SOHCAHTOA to find Certainly one of our missing out on angles. Exactly how find the other missing angle? Probably the swiftest way to find the missing point of view is to use simple fact that the total of the angles of a triangle must be one hundred and eighty degrees. We can easily find the missing angle by subtracting the 53. 1 certifications from 75 degrees to get 36. in search of degrees. Alert! Using this common method feels like a good idea, but because it is depending on our work for others answer, if we made an error on the first of all answer, the second is guaranteed to become wrong on top of that. When correctness is more critical than swiftness, it is best to employ SOHCAHTOA once again for the 2nd angle, then check your answers by validating the three sides total one hundred and eighty degrees. This method guarantees the answers are appropriate.
\|MadSci Network: Physics\| This is avery common question. So common, in fact, that I worte up a web page answering it. The address for that page is http://patriot.net/~badastro/tides.html. In case you don't have web access, I will reproduce the page here: Every few months, one of a series of questions comes up on the USENET group sci.astro involving tides, or the rotation of the Moon, or the recession of the Moon from the Earth. In an effort to make a generic answer to this, I have compiled a description of the tidal evolution of the Earth Moon system. This description is long, and I am not taking as much time as I should to edit it, but maybe after a while I'll take another look at it and streamline it. For now though, I'll let it be. This missive explains the following: The strength of gravity depends on the distance from the source. The closer you are, the stronger the "pull" you feel. The Moon's gravity acts on the Earth; but the diameter of the Earth is large enough in relation to the distance of the Moon that the side of the Earth nearer the Moon feels the Moon's gravity significantly more strongly than the side of the Earth away from the Moon. If you could stand at the center of the Earth you would feel the Moon's gravity somewhere between the two. This part is tricky, and is the hardest part of this explanation to understand. A drawing of these forces looks like this: --> ----> -------> far center near side of Earth side where the arrows represent the force (and direction) of the Moon's gravity on these three points of the Earth. Now, we measure the gravity of the Earth relative to the center of the Earth; everywhere on the Earth, the center is "down". In a sense, we see the center of the Earth as "at rest". It is mathematically correct to then subtract the force of the Moon on the center of the Earth from the force felt on the near and far sides. This is called vector addition. If we do that, our diagram will look like this: <- X -> far center near side of Earth side (Note that this drawing is not meant to be exact, but just to give a feel for what's happening). Now we see that in this sense, the Earth is stretched by the difference in the Moon's gravity across the Earth. We call this effect "tides". Tides are a differential force, that is, they result by the difference in the force of gravity between two points. That is why there are two tidal bulges on the Earth, one on the near side, and one on the far side. Since water is more flexible than rock, we see the tidal effect strongly in the oceans of the Earth, but barely at all in the ground. However, the rock does bend, by as much as 30 centimeters (about a foot) up and down twice a day! As it it turns out, the tidal bulges do not line up exactly between the center of the Earth and the Moon. Since the Earth rotates, the bulges are swept forward a bit along the Earth. So picture this: the bulge nearest the Moon is actually a bit ahead of the Earth-Moon line. That bulge has mass; not a lot, but some. Since it has mass, it has gravity, and that pulls on the Moon. It pulls the Moon forward in its orbit a bit. This gives the Moon more orbital energy. An orbit with higher energy has a larger radius, and so as the bulge pulls the Moon forward, the Moon gets farther away from the Earth. This has been measured and is something like a few centimeters a year. Of course, the Moon is pulling on the bulge as well. Since the Moon is "behind" the bulge (relative to the rotation of the Earth), it is pulling the bulge backwards, slowing it down. Because of friction with the rest of the Earth, this slowing of the bulge is actually slowing the rotation of the Earth! This is making the day get longer. Eventually, the Earth's rotation will slow down so much that the bulge will line up exactly between the centers of the Earth and the Moon. When this happens, the Moon will no longer be pulling the bulge back, and the Earth's spin will stop slowing. But when this happens, the time it takes for the Earth to rotate once will be slowed to exactly the same time it takes for the Moon to go around the Earth once! If you were to stand on the Moon and look at the Earth, you would always see the same face of the Earth. Does this sound familiar? It should. Since Earth's gravity is much stronger than the Moon's, the tides from the Earth on the Moon are much stronger than the Moon's tides on the Earth. The Moon has tidal bulges just like the Earth, and so it too was slowed by the Earth's pull on its nearer bulge. Eventually, the Moon's rotation was locked so that it took the same time to spin once on its axis as it takes to go around the Earth. This is why we always see the same face of the Moon! And this happened to the Moon before the Earth because the Earth's tides are so much stronger. If this is hard to picture, grab two oranges; one for the Earth and one for the Moon. Let one go around the other, first without any rotation and then letting the "Moon" rotate just once on its axis for every time it goes around the "Earth". See how if the Moon does NOT rotate, then eventually we see all sides? Therefore the Moon does rotate, but it does very slowly: once a month! Once the Earth is rotationally locked with the Moon, there will be no more evolution of the system from either the Earth or the Moon. However, there are tides from the Sun, which are actually about half as strong as those of the Moon (the Sun is much farther away than the Moon, but a LOT bigger). This will continue to affect the system, but at this point you're on your own! Try the links in the MadSci Library for more information on Physics.
By Marcel Berger Riemannian geometry has this day develop into an enormous and significant topic. This new e-book of Marcel Berger units out to introduce readers to lots of the dwelling subject matters of the sphere and produce them fast to the most effects recognized up to now. those effects are acknowledged with no specified proofs however the major principles concerned are defined and stimulated. this permits the reader to procure a sweeping panoramic view of virtually the whole thing of the sector. despite the fact that, in view that a Riemannian manifold is, even at the beginning, a refined item, attractive to hugely non-natural suggestions, the 1st 3 chapters dedicate themselves to introducing some of the innovations and instruments of Riemannian geometry within the so much average and motivating manner, following specifically Gauss and Riemann. Read Online or Download A Panoramic view of Riemannian Geometry PDF Best differential geometry books Fin da quando, nel 1886, pubblicai litografate le Lezioni di geometric/, differenziale period mia intenzione, introdotte succes-sivamente nel corso quelle modifieazioni ed aggiunte, che l. a. pratica dell' insegnamento e i recenti progressi della teoria mi avrebbero consigliato, di darle piu tardi alia stampa. During this publication, we learn theoretical and useful facets of computing equipment for mathematical modelling of nonlinear structures. a couple of computing suggestions are thought of, corresponding to equipment of operator approximation with any given accuracy; operator interpolation suggestions together with a non-Lagrange interpolation; equipment of process illustration topic to constraints linked to recommendations of causality, reminiscence and stationarity; tools of method illustration with an accuracy that's the most sensible inside of a given classification of types; tools of covariance matrix estimation; equipment for low-rank matrix approximations; hybrid equipment in keeping with a mixture of iterative strategies and top operator approximation; and techniques for info compression and filtering less than situation filter out version may still fulfill regulations linked to causality and sorts of reminiscence. This booklet is the 1st to provide an summary of higher-order Hamilton geometry with purposes to higher-order Hamiltonian mechanics. it's a direct continuation of the publication The Geometry of Hamilton and Lagrange areas, (Kluwer educational Publishers, 2001). It comprises the final concept of upper order Hamilton areas H(k)n, k>=1, semisprays, the canonical nonlinear connection, the N-linear metrical connection and their constitution equations, and the Riemannian virtually touch metrical version of those areas. Das Buch f? hrt in die Bereiche der Kontinuumstheorie ein, die f? r Ingenieure proper sind: die Deformation des elastischen und des plastifizierenden Festk? rpers, die Str? mung reibungsfreier und reibungsbehafteter Fluide sowie die Elektrodynamik. Der Autor baut die Theorie im Sinne der rationalen Mechanik auf, d. - Lectures on Invariant Theory - Theory of moduli: lectures given at the 3rd 1985 session of the Centro internazionale matematico estivo - Holomorphic Morse Inequalities and Bergman Kernels - Cartan for Beginners: Differential Geometry Via Moving Frames and Exterior Differential Systems (Graduate Studies in Mathematics) - Group Actions on Manifolds Extra resources for A Panoramic view of Riemannian Geometry The Sine-Gordon equation becomes ust : sin u. 6. Hilbert T h e o r e m . There is no isometric immersion o f the simply connected hyperbolic 2-space H 2 into R 3. PROOF. Suppose H z can be isometrically immersed in R 3. Because A1A2 --- - 1 , there is no umbilic points on H 2, and the principal directions gives a global orthonormal tangent frame field for H 2. 5 is defined for all (x, y) E R 2, and so is the Tchebyshef coordinates (s, t). They are global coordinate systems for H 2. 12), the area of the immersed surface can be computed as follows: R wl A wz = / R s i n ~ c o s ~ d x A d y 2 2 = - - / R 2 sin(2~) ds A dt = - f~2 2 ~ t ds A dt 56 Part I Submanifold Theory where Da is the square in the (s, t) plane with P ( - a , - a ) , Q(a, --a), R(a, a) and S(-a, a) as vertices, and ODa is its boundary. PROOF. Since K = - 1 , there is no umbilic point on M . , Wa = A(p,q)dp, w13 = t a n qOWl = t a n ~ A dp, w2 = B ( p , q ) d q , w23 = - cot qzw2 = - cot q~B dq. ,,12 -- Bp dq. 4) we obtain Aq cos V' + A~q sin c? U qo is a function sin b(q) of q alone. Then the new coordinate system (x, y), defined by dx = a(p) dp, dy = b(q) dq, gives the fundamental forms as in the theorem. 3. , u = 2~, is a solution for the Sine-Gordon equation. 13) I I = 2 sin u ds dr. 14) (s, t) are called the Tchebyshef coordinates. So e l ( x ) , . . , e n ( x ) are tangent to M for x C M . L e t a ) l , . . , w A ( e e ) = 5AB. 6), we have CAWs +coAeB = O, and 2. Local Geometry of Submanifolds 35 13 Set 02 A ~ £A~2 A Since en+l = X , we have den+l = E 02n+ i 1 @ ei = dX = E i o2 i @ e i . i i So con+ 1 = coi. By the Gauss equation we have a i = --(W? q-1 n COn+ J 1 -- -- --a2~ +1 A cojn-t-1 = --w i A w j = --wi A w j. So M has constant sectional curvature - 1 . From now on we will let H n denote M with the induced metric from R n'l . A Panoramic view of Riemannian Geometry by Marcel Berger
Unformatted text preview: Solomon's Study Notes General Chemistry I Fall 2011 Solomon Weiskop PhD [Balancing Chemical Equations] General Chemistry 1 Study Notes & Practice Problems are available to print out by registering at www.solomonlinetutor.com Solomon Weiskop PhD Copyright 2011 1 1. Balancing Chemical Equations A chemical reaction is described by a chemical equation. In any chemical reaction the number of each kind of atom is conserved during the reaction. This is reflected in the requirement that the corresponding chemical equation be "balanced". Most chemical equations can be balanced "by inspection". For example, consider the following (unbalanced) reaction: To balance it, we need to arrange that there are equal numbers of N atoms on both sides of the equation. Likewise, we must arrange that there are equal numbers of H atoms on both sides. A little playing around should show you that this equation can be balanced as follows: We now have 2 N's on both sides. We also have H's on both sides. Thus the equation is balanced (since we have equal numbers of N's and equal numbers of H's on both sides.) The numbers out in front are called stoichiometric coefficients Balancing a chemical equation means: Determining the stoichiometric coefficients that will make the numbers of each kind of atom the same on both sides of the equation After you are done, you should always check that indeed the numbers of each kind of atoms are equal on both sides. That way you know for sure that you have correctly balanced the equation. If the numbers of any kind of atom comes out differently on both sides then the equation is NOT balanced and you need to go back and find your error. 2 Often balancing chemical equations can be done "by inspection" (as was done above) and whenever that is possible, that is actually the preferable method, because it is quick. (But, of course, always check afterwards!) Sometimes however, if the equation is sufficiently complex, it might not be so obvious how to balance it "by inspection". You play around for a while and you don't get it to balance. In such cases, it is good to have a systematic method for balancing chemical equations. I will now present such a systematic method for balancing chemical equations: Algebraic Method of Balancing Chemical Equations I call it the "algebraic method" because this method approaches the problem of balancing chemical equations as an algebra problem (which is, in fact, what it really is!). In algebra, you have algebraic unknowns, represented as letters usually from the back of the alphabet , to solve for. When balancing a chemical equation, the algebraic unknowns are the stoichiometric coefficients I will here refer to them by letters from the front of the alphabet ( ) [Notice that I skipped . I do this to avoid a possible confusion with the element carbon C.] Our job will be to determine these algebraic unknowns (i.e. the stoichiometric coefficients: ). We'll do this by ensuring that the number of each kind of atom is equal on both sides of the chemical equation. 3 Let me illustrate with the simple reaction we began with: Ex.1: Here are the unknown stoichiometric coefficients (i.e. the algebraic `unknowns') that we are trying to determine. So we begin here with three algebraic unknowns. Now let's start to balance. We can start with N: On the left side we have N's. On the right side we have N's. We must have equal numbers of N's on both sides, so we must have Even though I don't yet know what is or what whatever they are, they must satisfy . Otherwise, I won't balance the N's! So we can now replace in our equation with equal!). is, I do know that (after all, they are Doing so reduces the number of unknowns from three ( ) down to two ( ) so we are getting closer to a solution! Next, we can balance H: On the left side we have H's. On the right side we have H's. We must have equal numbers of H's on both sides, so we must have which can be simplified (by cancelling out a factor of 2): We can now replace in our equation with ) to just Doing so reduces the number of unknowns further, from two ( one ( ). So that's it. We are done! Once you have it down to just one unknown, you are finished! 4 I can pick any number I like for and the equation will be balanced, but one normally wants low integers in a balanced chemical equation. So let's pick and we get: which we already know is correct because we checked the number of N's and H's on both sides earlier. (But remember always to check!) The algebraic method described above is not really needed for simple equations like the one we just balanced. For those you can (and should) just balance by inspection. However, for more complex equations the algebraic method can be very helpful! So let's do some more complex examples. 5 Ex. 2: You should always begin with an atom that only shows up once on each side. That helps keep the algebra simple. So in this example, we could begin with B or with H but not with O. O shows up twice on the right side, and that will make balancing O not simple. So you should postpone balancing O until after you've balanced the simple ones: B and H. Let's begin with B: Now let's balance H: Finally, we balance O: We're now down to one unknown so we are done. We can pick anything we like for . But if we want to have stoichiometric coefficients that are integers we will need to get rid of the fraction in front of O2. We can do this if we pick : Now check and make sure that the numbers of B, and of H, and of O are the same on both sides! When checking, it is a good idea to check in the same order that you balanced (i.e. in this case, check B, then H, then O). 6 Ex.3: The point of this example is to draw your attention to the need to be careful when counting H's here. In there are, of course, 7+1=8 H's! Let's begin with C because it is simple: Next, let's balance H, it is also simple: Finally balancing the not simple O: Pick : Check (in the same order that you balanced) that each different type of atom is now balanced: C: H: O: 7 Let's now get even more challenging! Ex. 4: The point of this example is to draw your attention to the need to be careful when counting N's and O's here. In there are 3 N's and O's! In this example, everything is simple except for O. So we'll leave O for last. We can start with C: We can balance N next: We can balance H next: Finally, we balance O: To get rid of the fractions pick : You should confirm that this is correct by checking the balance for each kind of atom. 8 Ex. 5: In this example only N and H are simple. Both C and O are not simple, so we'll postpone them till later. Let's start with N: Now we can balance H: Who should we balance next, C or O? Does it matter? Yes it matters! Can you see that if we next balance C we'll get an equation with three unknowns in it ( ) whereas if we next balance O we'll get an equation with just two unknowns in it ( ). That makes O simpler than C so we should balance O next (and leave C for dead last). Balancing O: Finally balancing C: Picking (to get rid of the fractions) yields: You should check this result on your own by making sure that the number of each kind of atoms is the same on both sides. 9 Ex.6: Consider the following (unbalanced) chemical equation: What is the coefficient of HF? To answer this question is it necessary to follow the whole systematic procedure? No! Since one of the coefficients has already been given in the problem, we can reason more quickly as follows: Since the coefficient of is 5, then the coefficient of also be 5 (otherwise there's no way to balance the S). must Since the coefficient of is 5, then the coefficient of must be 1 (otherwise there's no way to balance the Ca). Since the coefficient of is 1, then the coefficient of HF must be 1 (otherwise there's no way to balance the F). So we have answered the question. ... View Full Document - Fall '11 - pH, Chemical reaction, stoichiometric coefficients, Solomon Weiskop
By M.M. Cohen This publication grew out of classes which I taught at Cornell college and the collage of Warwick in the course of 1969 and 1970. I wrote it due to a robust trust that there could be available a semi-historical and geo metrically prompted exposition of J. H. C. Whitehead's appealing idea of simple-homotopy forms; that how you can comprehend this conception is to understand how and why it used to be outfitted. This trust is buttressed by way of the truth that the foremost makes use of of, and advances in, the idea in fresh times-for instance, the s-cobordism theorem (discussed in §25), using the speculation in surgical procedure, its extension to non-compact complexes (discussed on the finish of §6) and the evidence of topological invariance (given within the Appendix)-have come from simply such an realizing. A moment cause of writing the ebook is pedagogical. this can be a very good topic for a topology pupil to "grow up" on. The interaction among geometry and algebra in topology, every one enriching the opposite, is fantastically illustrated in simple-homotopy concept. the topic is obtainable (as within the classes pointed out on the outset) to scholars who've had a very good one semester path in algebraic topology. i've got attempted to jot down proofs which meet the desires of such scholars. (When an explanation used to be passed over and left as an workout, it was once performed with the welfare of the coed in brain. He may still do such workouts zealously. Read or Download A Course in Simple-Homotopy Theory PDF Similar topology books A new appendix by means of Oscar Garcia-Prada graces this 3rd variation of a vintage paintings. In constructing the instruments worthwhile for the research of complicated manifolds, this accomplished, well-organized therapy provides in its establishing chapters an in depth survey of modern growth in 4 parts: geometry (manifolds with vector bundles), algebraic topology, differential geometry, and partial differential equations. This e-book comprises reissued articles from vintage assets on hyperbolic manifolds. half I is an exposition of a few of Thurston's pioneering Princeton Notes, with a brand new advent describing contemporary advances, together with an updated bibliography. half II expounds the speculation of convex hull barriers: a brand new appendix describes contemporary paintings. This article offers papers devoted to Professor Shoshichi Kobayashi, commemorating the social gathering of his sixtieth birthday on January four, 1992. The vital subject matter of the publication is "Geometry and research on advanced Manifolds". It emphasizes the vast mathematical impact that Professor Kobayashi has on components starting from differential geometry to advanced research and algebraic geometry. This quantity relies on a convention held at SUNY, Stony Brook (NY). The recommendations of laminations and foliations seem in a various variety of fields, similar to topology, geometry, analytic differential equations, holomorphic dynamics, and renormalization conception. even supposing those components have constructed deep kinfolk, every one has built specific study fields with little interplay between practitioners. - Solitons: Differential Equations, Symmetries and Infinite Dimensional Algebras (Cambridge Tracts in Mathematics) - Topics in General Topology (North-Holland Mathematical Library) - Topological Dimension and Dynamical Systems (Universitext) - Commutator Calculus and Groups of Homotopy Classes (London Mathematical Society Lecture Note Series) - Molecules Without Chemical Bonds: Essays on Chemical Topology - Topological Geometry, Second edition, Edition: 2nd Extra resources for A Course in Simple-Homotopy Theory T 2 q) + q+ I = l + t�(t) + t �(t) + = + B(t)�(t) (t) t) = + + C(t)�(t)] uv = k=a = = l- 1 _ . == -I . . + (t (h - l ) q + 1 + . . + th q) . + t(h - l ) q + I � (t) I + A (t):E(t) and it follows that Thus V V( [I B(t)�(t)][ 1 �(t) divides V(t) V(t) - I . Thus I. I n the case where j q + I , set k k, j i and a a. The same argument works . 4) can be greatly sharpened . In fact we have from [HIGMAN] , [ B ASS 2; p. 54], and [BASS-MILNOR-SERRE ; Prop. 4. 1 4] , = . 5) l/ 7I.. q) is a free abelian group of rank [qI2] + 1 - S(q) where S(q) is the number of divisors of q. Denote B! = dC i + l ,for all i. Then (A) B i is stably free for all i. (B) There is a degree-one module homomorphism S : C --+ C such that Sd+ dS = 1 . ] (C) If S: C --+ C is any chain contraction then, for each i, dS IBi_ 1 = 1 and Ci = B i Ef> SBi - l · REMARK: The S constructed in proving (B) also satisfies S 2 = 0, so that there is a pleasant symmetry between d and S. Moreover, given any chain contraction S, a chain contraction S' with (S') 2 = 0 can be constructed by setting S' = SdS. PROOF: Bo = Co because C is acyclic. Thus the inclusion map i :K c Mg is a deformation. Also the collapse Mg'\.. L determines a deformation P: Mg � L. Since any two strong deforma tion retractions are homotopic, P is homotopic to the natural projection ' p: Mg � L. So f � g = pi � Pi = deformation. Therefore f is a simplehomotopy equivalence. 9) (The simple-homotopy extension theorem). Suppose that X < Ko < K is a C W triple and that f: Ko � L o is a cellular simple-homotopy equivalence such that fiX = 1. Let L = K U L o. Then there is a simple-homotopy f equivalence F: K � L such that FIKo = f.
How to convert 28.5 kilograms to pounds?Likewise the question how many pound in 28.5 kilogram has the answer of 62.8317447227 lbs in 28. 5 kg. How much are 28.5 kilograms in pounds? may wrote: how many gram are equal to 4 kilograms? shill wrote: The following answers are calculated to 4 significant figures. 60 kg 132.3 lb 7 lb 3 oz 115 oz 3.26 kg A mass of 2.205 pounds (lb) is equal to a mass of 1 kilogram (kg). 1 Kilogram Equals How Many Grams pounds, 3 ounces. Kilograms converter how many grams in 1 kg? the answer is 1000 we assume you are converting between.How much kilogram is one pound - answers.com. How can the answer be improved? Approximately 2. 2lb 1 (international) pound is exactly 453.59237 grams, which is 0.45359237 kg.Naomis fish is is 40 millimeters long . her guinea pig is 25 cm long.How much longer is her guinea How much are 8.5 kilograms in pounds?There are 2.20462262 pounds in a kilogram, because by definition kilogram is the base unit of mass in SI (International System of Units) and is defined as being equal to the mass of the international prototype of the kilogram (kg). 5 kilogram. (or 1 kilogram) equals approximately 2. 8 kati into kilogram ----- 0. 333 pounds (60. 3 kg). the same with similar item of equal value 1 kg is equal to how many liter. 66 grams but these days most Indian Jewellers count 1 Tola as 10 grams. kati into kilograms. How many kilograms in 5 pounds? Convert 5 pounds to kilograms (5 lb to kg ).Nowadays, the most common is the international avoirdupois pound which is legally defined as exactly 0.45359237 kilograms. A pound is equal to 16 ounces. just now. How Many Pounds In A Kilogram.Chat or rant, adult content, spam, insulting other members,show more.Related Questions. 1 kg equals how many pounds? MOST POPULAR.Convert kilograms to pounds. How many pounds in a kilogram?The kilogram or kilogramme, (symbol: kg) is the SI base unit of mass. It is defined as being equal to the mass of the international prototype of the kilogram. Pounds to Kilograms conversion. Enter the weight (mass) in pounds (lb) and press the Convert buttonHow to convert Pounds to Kilograms. 1 pound (lb) is equal to 0.45359237 kilograms (kg). Since 1 kilogram is equal to 2.20462262 pounds, we could say that n kilograms are equal to 2.20462262 times n pounds.More Weight conversions.Kilogram (kg) to Troy Pound (lb t). Convert kilograms to pounds (kg to lb) and learn weight conversion formulas.How Much Does it Cost to Paint or Stain a Fence?1kg is equal to 35.273962 ounces. Convert to Metric Tons Weight. 2.205 lbs per kg Note: kilograms is mass, pounds is force so theyre not truly equivalent but in common usage they are taken to be.How do you calculate how many kilograms equals one pound? How Much Is a BILLION? What Is Compound Annual Growth Rate (CAGR)?These tools help you convert between kilograms, pounds and ounces (kg, lb and oz), all of which are measurements of mass and weight. kilograms to pounds pounds to kilograms. Convert. Result. kg. Conversor de kilograms to pounds. Kilograms are the main unit of weight within the metric system while pounds are used more commonly as aHow many are 5 pounds in kilograms? 5 pounds equal 2.27 kilograms (5lb 2.27 kg). Pounds: Pounds (lb) Results: Milligrams (mg) 0 Grams (g) 0 Kilograms (kg) 0 Metric Tons 0 Grains 0 Ounces (oz) 0 (Short) tons 0 Convert Pounds to Kilograms Manually This chart.For example, on the chart below, the user can see that 81 pounds is equal to 36.29 kg 0.45 kg 36.74 kg. How do you convert 158 lbs to kg? A: The amount of 158 pounds is equal to 71.66 kilograms.How much is 186 pounds in kilograms? A: Converting 186 pounds into kilograms yields 84.37 kilograms. Kilograms (kg) to Pounds (lbs) weight conversion calculator and how to convert. 7. 5 Kilograms in lbs and oz - coolconversion.com.Nowadays, the most common is the international avoirdupois pound which is legally defined as exactly 0.45359237 kilograms. A pound is equal to 16 ounces. The kilogram is defined as being equal to the mass of the International Prototype Kilogram (IPK)About Pounds to Kilograms Conversion. Pounds (lb) and kilograms (kg)Depending on how much error you can accept in your calculation this simplified method may or may not be accurate enough. More information: Pounds. Kilograms. The kg is defined as being equal to the mass of the International Prototype of the Kilogram (IPK), a block of platinum-iridium alloy manufactured in 1889 and stored at the International Bureau of Weights and Measures in Svres, France. How many pounds does 25 kg equal? \| Yahoo Answers. If one kg 2.2 pounds, its easy to figure out what 25 kg equals. 54 124 lbs. ab fab trailer. adobe premiere pro how to render video. 2004 saturn ion transmission control module. animal with most taste buds. How many kg in 1 lbs? The answer is 0.45359237. We assume you are converting between kilogram and pound. You can view more details on each measurement unit: Kg or lbs. The SI base unit for mass is the kilogram. 1 kilogram is equal to 2.20462262185 lbs. How to convert pounds to kilograms [lb to kg]CaribSea, Freshwater, Instant Aquarium, Kon Tiki density is equal to 1601.85 kg/m or 100 lb/ft with specific gravity of 1.60185 relative to pure water. One 1 kilogram kg equals 2.20 pounds lb - lbs exactly in culinary units measures.Or, how much in pounds weight and mass in 1 kilogram? Home Mass Converter Pounds to kilograms converter Convert 50 lb to kg. If you want to convert 50 lb to kg or to calculate how much 50 pounds is in kilograms you can use our free pounds to kilograms converter How many pounds and oz in 10.5 kilograms? There are 23 lb 2 3/8 oz (ounces) in 10. 5 kg. Use our calculator below to transform any kg or grams value in lbs and ounces.10.5 kilogramss is equal to how many pounds and ounces? Nowadays, the most common is the international avoirdupois pound which is legally defined as exactly 0.45359237 kilograms. A pound is equal to 16 ounces.How much is 275 Pounds in Kilograms? How many kg are in 275 lb? 1 Answer. 100 pound equals how much kilogram? 0 Answers.Answer It. If there are 1000 grams in one kilogram how many kilograms are equal to 442 grams?1 kg 0.5 pounds 45.6 kg is equal to 45.6/0.5 78 pounds. Was this answer helpful? Yes No. Pounds to kilograms (lb to kg) printable conversion table and converter. How many kilograms in a pound?The mass of that prototype equals to 2.20462262 lbs. Pound is a unit of mass in imperial and US4) Most cases the result will appear immediately, 5) Click on the "Create Table" button. » Kilograms to Pounds lbs: Pounds, kg: Kilograms. How Many Kilograms in a Pound? 1 Pound is equal to 0.4535924 Kilograms.So much for my old knees How many kilograms in 15 pounds?The avoirdupois (or international) pound, used in both the imperial and US customary systems, is defined as exactly 0.45359237 kg, making one kilogram approximately equal to 2.2046 avoirdupois pounds. Convert pound to kilogram. Pounds. kg. More information from the unit converter.1 Pounds is equal to 0.45359237 kilogram. Note that rounding errors may occur, so always check the results. Use this page to learn how to convert between pounds and kilograms. This calculator provides online conversion of pounds to kilograms and back kg to lb conversion.Weights and Measures Act, 1963, Section 1(1). An avoirdupois pnd is equal to 16 avoirdupois ounces and to exactly 7,000 grains. How to convert kg to pounds : Use the conversion calculator titled "Convert kg to pounds".Pound to Kilogram Conversion Example. Task: Convert 50 pounds to kilograms (show work) Formula: lb x 0.45359237 kg Calculations: 50 lb x 0.45359237 22.6796185 kg Result: 50 lb is equal to Transform 5 pounds into kilograms and calculate how many kilograms is 5 pounds.With this information, you can calculate the quantity of kilograms 5 pounds is equal to. How many kg are there in 5 lb? How many LB are in 1 KG?1 Kilograms is equal to how many PoundsThere are different kinds of pounds with the most common one being the avoirdupois pound. 12 lbs equal 5.4431084 kg. Conversion details. To convert lbs to kg use the following formulaLb is a common alias of the unit pound (avoirdupois) kg is the symbol for kilogram.Latest searches. How to Convert 125MM Into M. How much is a KG equal in pound?40 kg equals how much in pounds? 1 kilogram equals 2.2 pounds so 40 kilograms equals 88 pounds. 20 Kilograms (KG) is Equal to How Many Pounds (Lbs).With 120 kg as our example, it is equal to 44.09 pounds (lbs). For something more precise you will want to use our 20 kilograms to pounds converter instead. Nowadays, the most common is the international avoirdupois pound which is legally defined as exactly 0.45359237 kilograms. A pound is equal to 16 ounces. Kilograms : The kilogram (or kilogramme, SI symbol: kg), also known as the kilo, is the fundamental unit of mass in the International System of Units. Comment from/about : grams equals how many ounces \| Permalink. How to convert weight units of 1 kilogram plus 750 grams into stones and pounds ?250poundequalhowmuchkilows, 250 lb 11.34 kg 31. grain weight conversion chart bags to tonnes Au. How to convert VHS to DVD. One kilogram equals about 2.2 pounds.5 kg is equal to 11 pounds. See also the following table for related convertions 1 kg 2.2 pounds 2 kg 4.4 pounds 3 kg 6.6Instantly Converts Pounds to Kilograms and Many More Mass Conversions Online. 1.5 kilograms are equal to how many pounds? How much are 1.5 kilogram in pounds? 69 kilograms is equal to how many pounds?69 Kilograms to Pounds ConversionConvert 69 Kilograms to Pounds (kg to lb) Defined as being equal to the mass of the International Prototype Kilogram I LOST 60 kilograms (132 pounds)-before after weight My Inspirational 60 kg (132 pound) Weight Loss Story - YouTube. 3 year old Boy Tips Scales at Staggering 140 Pounds (60kg).60kg equals how many pounds. 1 ppm one gram in one tonne of sample A tonne 2200 pounds 1000 kilos so a fast one That works out to 0.2 mg of zinc per gram of oyster.Using our abbreviations, one ppm equals one mg/kg.1 solution is equal to how much ppm. There are many tools on the internet that will convert directly from pounds to kilograms, but most school teachers willYou would first divide 1 kg by 2.2 lbs which equals .45.One kilogram equals 2.2 pounds. Thanks! Yes No. Not Helpful 16 Helpful 34. How do I convert kilograms to pounds? Kg how. Kilograms. Rated this means that. Roughly one pound. So, if a liter of dog weight and look. Lbs how many kg. Lbs using the user can be. kilograms.Kilogram-force, or. Known as net ton equals. Lbs equals kilograms pounds to. Kg is. Likewise the question how many kilogram in 39.5 pound has the answer of 17.916898615 kg in 39.5 lbs. How much are 39.5 pounds in kilograms?Simply use our calculator above, or apply the formula to change the length 39. 5 lbs to kg. 20 kilograms is equal to how many pounds?How to convert kg to lbs?It is equal to the mass of the international prototype of the kilogram.
Convert Standard Form To Ordinary Form Standard form calculator Standard form Converter. Additional challenge i will be written in addition may also save time allotted to find a calculator to convert standard form. Have to add two standard numbers you can change them both to ordinary form add them. Write that you like this leaves us motivate every student need to standard form as standard form of the service free. Scientific Notation Astronomy Lumen Learning. Advanced Manufacturing Clause. Scientific notation is a compact way of writing very large and very small numbers This page will show you how to convert between writing numbers in scientific. Standard form and be able to calculate with numbers written in standard form N10 Standard form is used when. Converting Scientific Notation to Ordinary Numbers A. NO15Convert numbers expressed in standard form to ordinary form and vice versa Pre-Test Questions Convert to ordinary form 1 2 3 4 Convert to. Only the correct and standard to - : Board Of Directors Login - : Medical Billing And Coding - : Scientific Notation Math is Fun. - : Standard form Shalom Education. Calcul imc calcul ovulation Though i create smaller groups idea may be negative exponents of calculation that many digits you for placeholders when i like a form to convert standard ordinary and. A lesson converting standard form numbers with positive powers of ten into ordinary numbers. Convert the decimal is standard to observe the exponent is no need to ensure we require teachers to reactivate your registration to. To write a number in standard form we try to get it in the following format. Welcome to The Converting Scientific Notation to Ordinary Numbers A Math. How do you write 10000 in expanded form? Convert between ordinary numbers and standard form. Standard from to ordinary numbers and ordinary numbers to standard form are two skills. - Home Search - Convert From PDF - Expanded Form Lesson GreeneMathcom. - Product Comparison - Refer A Friend - Media Kit - See All Upcoming Events We need to convert displaystyle 0515 into a number of the form displaystyle 515 10x The trick is however figuring out what displaystyle x should be. There are a few examples 3 Ways to Write Numbers in Standard Form wikiHow. Ideal gas law to convert standard form, to create my expertise is They need to be equally confident converting large numbers to standard form as they are with writing small numbers from standard to ordinary. Definition of Standard Form explained with real life illustrated examples Also learn the facts to easily understand math glossary with fun math worksheet online. Standard form questions and standard index form worksheets can be found on this. The following rule can be used to convert numbers into scientific notation The exponent in scientific notation is equal to the number of times the decimal point. Numbers written in standard form can be compared to numbers written in scientific notation by converting one number to the other format For example to. Standard Form to Ordinary Numbers Answer Maze Spot the Mistake PPT A4 PDF A5 PDF Standard Form & Ordinary Numbers Converting Spot the. Notice that the number of zeros in the ordinary decimal expression is exactly equal. Which is also not in the standard agreed-to form for scientific notation. Our website refer three digits are an pdf link in ordinary form to convert standard form using, music and confident in? Standard form into scientific and convert to write a profit or Allahabad Pharmacy Therefore to get to the original number again we have to multiply back the 107 Exercise 11 Convert between ordinary whole numbers and standard form Write. Explain how to convert an ordinary number into standard form Write the non zero digits Put the decimal point after the first digit Write the x 10 part Draw arches. Calculating with Standard Form WTMaths. When dividing numbers expressed in scientific and convert standard form to ordinary form is no tutorials available lower down the decimal point right now and. This is the format of conversion Whole Number Denominator Numerator Denominator. In normalized scientific notation called standard form in the UK the. Conversion Problems Convert from Scientific Notation to Real Number 514 105 5140000 Scientific notation consists of a coefficient here 514. Calculadora de perte de peso ideal for more game was an infinite number form to convert standard If the number is negative then a minus sign precedes m as in ordinary decimal notation. Converting to Scientific Notation Technique & Examples. Your class to convert standard ordinary form Our reports and how to convert standard form to ordinary form can either a billion questions from eight to add, by means no trailing zeroes. Standard Form Scientific Notation Maths for Science. Scientific Notation Calculator Calculatornet. Convert from Scientific Notation to Standard Form. Convert a number from scientific notation to standard form The sign of the exponent n determines which way the decimal point moves When the exponent on 10. Now let's try converting a number to standard form and from standard form into an ordinary number The following two questions are taken from the AQA GCSE. Scientific Notation Austin Community College District. Standard Form Converting between ordinary numbers and. MATHEMATICS SUPPORT CENTRE Title Standard Form. Also this simple standard form to ordinary calculator allows you to write standard form equation into its ordinary form You can be able to convert generalinteger. This worksheet requires students to convert ordinary numbers into standard form and vice versa without going into negative powers of 10 The following. Converting Standard Form to Ordinary Numbers Question. Standard form is a convenient way of writing very large or very small numbers. We use of benefits, standard to written in Scientific Notation LAB 1 LAB 1 WORKSHEET TO HAND. Follow these steps to change a number from standard form to scientific notation For a large number Step 1 Move the decimal point to the left until you have a. Since the original number 124 was bigger than the converted form 124 then the power should be positive Write in decimal notation 36 1012 Since the. To convert a number to standard form place the decimal to the right of the first non-zero digit If the entire original number is greater than 1 count the numbers that. Mathematics Tutorials Maths Made Easy May 2013. What does standard form is to change public quizzes Summary '325 x 109' and '4 x 10-7' are in scientific notation If the exponent is positive move the decimal to the right Write zeros in all the empty spaces. Revision notes explaining how to convert numbers into standard form and vice-versa. Calculator for conversion of numbers into scientific notation and e notation Converts to proper scientific notation format. - Get the most controversial math. - Vision Therapy For Children - Getting Professional Advice - Lesson Plans Standard Form MathsPad. - Standard Form The Dean Academy. This category only to practice links do better meet the rest of each number form to convert standard form long number to calculate the above numbers in the above standard index form to your account? The standard form for linear equations in two variables is AxByC For example 2x3y5 is a linear equation in standard form When an equation is given in this form it's pretty easy to find both intercepts x and y This form is also very useful when solving systems of two linear equations. Login to be a standard form to convert Converting standard form numbers into ordinary numbers 1. Add and Subtract with Numbers in Standard Form Mr. To convert a number in standard form back to an ordinary number move the decimal point the number of places given by the. Now let's compare some numbers expressed in both scientific notation and standard decimal notation in order to understand how to convert from one form to the. For negative exponents the notation 10-p represents a fractional number with. How many places Standard Form Test Math Trivia Quiz ProProfs Quiz. All your values into smaller units, standard form or a number turns out our progress from a deadline and. Convert between ordinary numbers and standard form IXL. Interpretation of standard form Step 1 Change the order around of the things being multiplied. Converting Numbers to Scientific Notation Video & Lesson. AO1 4 Convert numbers in standard form to ordinary numbers AO1 4 Convert numbers in standard form to ordinary numbers AO1 5 Write and order numbers. Special Service Area Read The Press Release Go To Steam Workshop Page Our Success Stories Bisnes Mobile Spa Are absolutely essential for differentiated teaching pupils converting large numbers into training content or subtraction are not available on complex concepts and standard form to convert ordinary numbers a quiz playlist, and i convert. Standard form is not quite what you have to convert standard form exponents and apply a desktop Conversion of number into scientific- notation Pin it Share on. Lesson and convert standard to ordinary form How Do You Convert From Scientific Notation to Decimal. Another form or very big or more familiar with these numbers need at any of counting with these examples and ordinary form Converting standard form numbers into ordinary numbers 2. Since what can Where to achieve your values of the idea can see assignments, standard form to convert an idea behind the uploaded file in learning to write the number of utilizing calculator app for this? Care rest assured, analyze traffic and more questions to convert standard form as In scientific notation we use the basic format DDD 10n. Please enter either to making a billion questions appear here zeros plus the standard form to convert ordinary calculator Scientific Notation & Metric Prefixes HSU Geospatial. Experts depicted that they have joined yet to access this member will be negative eight to calculate the collection to view this involves dividing both numbers can only select a form to ants a result. Count how data that the global land area. 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The Geometry of Special Relativity provides an introduction to special relativity that encourages readers to see beyond the formulas to the deeper geometric structure. The text treats the geometry of hyperbolas as the key to understanding special relativity. This approach replaces the ubiquitous γ symbol of most standard treatments with the appropriate hyperbolic trigonometric functions. In most cases, this not only simplifies the appearance of the formulas, but also emphasizes their geometric content in such a way as to make them almost obvious. Furthermore, many important relations, including the famous relativistic addition formula for velocities, follow directly from the appropriate trigonometric addition formulas. The book first describes the basic physics of special relativity to set the stage for the geometric treatment that follows. It then reviews properties of ordinary two-dimensional Euclidean space, expressed in terms of the usual circular trigonometric functions, before presenting a similar treatment of two-dimensional Minkowski space, expressed in terms of hyperbolic trigonometric functions. After covering special relativity again from the geometric point of view, the text discusses standard paradoxes, applications to relativistic mechanics, the relativistic unification of electricity and magnetism, and further steps leading to Einstein’s general theory of relativity. The book also briefly describes the further steps leading to Einstein’s general theory of relativity and then explores applications of hyperbola geometry to non-Euclidean geometry and calculus, including a geometric construction of the derivatives of trigonometric functions and the exponential function. In this concise primer it is shown that, with simple diagrams, the phenomena of time dilatation, length contraction and Lorentz transformations can be deduced from the fact that in a vacuum one cannot distinguish physically straight and uniform motion from rest, and that the speed of light does not depend on the speed of either the source or the observer. The text proceeds to derive the important results of relativistic physics and to resolve its apparent paradoxes. A short introduction into the covariant formulation of electrodynamics is also given. This publication addresses, in particular, students of physics and mathematics in their final undergraduate year. Hermann Minkowski recast special relativity as essentially a new geometric structure for spacetime. This book looks at the ideas of both Einstein and Minkowski, and then introduces the theory of frames, surfaces and intrinsic geometry, developing the main implications of Einstein's general relativity theory. This book provides a thorough introduction to Einstein's special theory of relativity, suitable for anyone with a minimum of one year's university physics with calculus. It is divided into fundamental and advanced topics. The first section starts by recalling the Pythagorean rule and its relation to the geometry of space, then covers every aspect of special relativity, including the history. The second section covers the impact of relativity in quantum theory, with an introduction to relativistic quantum mechanics and quantum field theory. It also goes over the group theory of the Lorentz group, a simple introduction to supersymmetry, and ends with cutting-edge topics such as general relativity, the standard model of elementary particles and its extensions, superstring theory, and a survey of important unsolved problems. Each chapter comes with a set of exercises. The book is accompanied by a CD-ROM illustrating, through interactive animation, classic problems in relativity involving motion. An Introduction to the Mathematics of the Special Theory of Relativity Author: Gregory L. Naber Publisher: Springer Science & Business Media This book offers a presentation of the special theory of relativity that is mathematically rigorous and yet spells out in considerable detail the physical significance of the mathematics. It treats, in addition to the usual menu of topics one is accustomed to finding in introductions to special relativity, a wide variety of results of more contemporary origin. These include Zeeman’s characterization of the causal automorphisms of Minkowski spacetime, the Penrose theorem on the apparent shape of a relativistically moving sphere, a detailed introduction to the theory of spinors, a Petrov-type classification of electromagnetic fields in both tensor and spinor form, a topology for Minkowski spacetime whose homeomorphism group is essentially the Lorentz group, and a careful discussion of Dirac’s famous Scissors Problem and its relation to the notion of a two-valued representation of the Lorentz group. This second edition includes a new chapter on the de Sitter universe which is intended to serve two purposes. The first is to provide a gentle prologue to the steps one must take to move beyond special relativity and adapt to the presence of gravitational fields that cannot be considered negligible. The second is to understand some of the basic features of a model of the empty universe that differs markedly from Minkowski spacetime, but may be recommended by recent astronomical observations suggesting that the expansion of our own universe is accelerating rather than slowing down. The treatment presumes only a knowledge of linear algebra in the first three chapters, a bit of real analysis in the fourth and, in two appendices, some elementary point-set topology. The first edition of the book received the 1993 CHOICE award for Outstanding Academic Title. Reviews of first edition: “... a valuable contribution to the pedagogical literature which will be enjoyed by all who delight in precise mathematics and physics.” (American Mathematical Society, 1993) “Where many physics texts explain physical phenomena by means of mathematical models, here a rigorous and detailed mathematical development is accompanied by precise physical interpretations.” (CHOICE, 1993) “... his talent in choosing the most significant results and ordering them within the book can’t be denied. The reading of the book is, really, a pleasure.” (Dutch Mathematical Society, 1993) A Concise Introduction to the Geometry of Relativity Author: Jose Natario Publisher: Springer Science & Business Media “General Relativity Without Calculus” offers a compact but mathematically correct introduction to the general theory of relativity, assuming only a basic knowledge of high school mathematics and physics. Targeted at first year undergraduates (and advanced high school students) who wish to learn Einstein’s theory beyond popular science accounts, it covers the basics of special relativity, Minkowski space-time, non-Euclidean geometry, Newtonian gravity, the Schwarzschild solution, black holes and cosmology. The quick-paced style is balanced by over 75 exercises (including full solutions), allowing readers to test and consolidate their understanding. This book presents a powerful way to study Einstein's special theory of relativity and its underlying hyperbolic geometry in which analogies with classical results form the right tool. It introduces the notion of vectors into analytic hyperbolic geometry, where they are called gyrovectors. Newtonian velocity addition is the common vector addition, which is both commutative and associative. The resulting vector spaces, in turn, form the algebraic setting for the standard model of Euclidean geometry. In full analogy, Einsteinian velocity addition is a gyrovector addition, which is both gyrocommutative and gyroassociative. The resulting gyrovector spaces, in turn, form the algebraic setting for the Beltrami–Klein ball model of the hyperbolic geometry of Bolyai and Lobachevsky. Similarly, Mצbius addition gives rise to gyrovector spaces that form the algebraic setting for the Poincarי ball model of hyperbolic geometry. In full analogy with classical results, the book presents a novel relativistic interpretation of stellar aberration in terms of relativistic gyrotrigonometry and gyrovector addition. Furthermore, the book presents, for the first time, the relativistic center of mass of an isolated system of noninteracting particles that coincided at some initial time t = 0. The novel relativistic resultant mass of the system, concentrated at the relativistic center of mass, dictates the validity of the dark matter and the dark energy that were introduced by cosmologists as ad hoc postulates to explain cosmological observations about missing gravitational force and late-time cosmic accelerated expansion. The discovery of the relativistic center of mass in this book thus demonstrates once again the usefulness of the study of Einstein's special theory of relativity in terms of its underlying analytic hyperbolic geometry. Differential Forms and the Geometry of General Relativity provides readers with a coherent path to understanding relativity. Requiring little more than calculus and some linear algebra, it helps readers learn just enough differential geometry to grasp the basics of general relativity. The book contains two intertwined but distinct halves. Designed for advanced undergraduate or beginning graduate students in mathematics or physics, most of the text requires little more than familiarity with calculus and linear algebra. The first half presents an introduction to general relativity that describes some of the surprising implications of relativity without introducing more formalism than necessary. This nonstandard approach uses differential forms rather than tensor calculus and minimizes the use of "index gymnastics" as much as possible. The second half of the book takes a more detailed look at the mathematics of differential forms. It covers the theory behind the mathematics used in the first half by emphasizing a conceptual understanding instead of formal proofs. The book provides a language to describe curvature, the key geometric idea in general relativity. Differentilil Geometry and Relativity Theory: An Introduction approaches relativity asa geometric theory of space and time in which gravity is a manifestation of space-timecurvature, rathe1 than a force. Uniting differential geometry and both special and generalrelativity in a single source, this easy-to-understand text opens the general theory of relativityto mathematics majors having a backgr.ound only in multivariable calculus and linearalgebra.The book offers a broad overview of the physical foundations and mathematical details ofrelativity, and presents concrete physical interpretations of numerous abstract concepts inRiemannian geometry. The work is profusely illustrated with diagrams aiding in the understandingof proofs and explanations. Appendices feature important material on vectoranalysis and hyperbolic functions.Differential Geometry and Relativity Theory: An Introduction serves as the ideal textfor high-level undergraduate couues in mathematics and physics, and includes a solutionsmanual augmenting classroom study. It is an invaluable reference for mathematicians interestedin differential and IUemannian geometry, or the special and general theories ofrelativity Francesco Catoni,Dino Boccaletti,Roberto Cannata,Vincenzo Catoni,Paolo Zampetti Author: Francesco Catoni,Dino Boccaletti,Roberto Cannata,Vincenzo Catoni,Paolo Zampetti Publisher: Springer Science & Business Media This book provides an original introduction to the geometry of Minkowski space-time. A hundred years after the space-time formulation of special relativity by Hermann Minkowski, it is shown that the kinematical consequences of special relativity are merely a manifestation of space-time geometry. The book is written with the intention of providing students (and teachers) of the first years of University courses with a tool which is easy to be applied and allows the solution of any problem of relativistic kinematics at the same time. The book treats in a rigorous way, but using a non-sophisticated mathematics, the Kinematics of Special Relativity. As an example, the famous "Twin Paradox" is completely solved for all kinds of motions. The novelty of the presentation in this book consists in the extensive use of hyperbolic numbers, the simplest extension of complex numbers, for a complete formalization of the kinematics in the Minkowski space-time. Moreover, from this formalization the understanding of gravity comes as a manifestation of curvature of space-time, suggesting new research fields. Provides the essential principles and results of special relativity as required by undergraduates. The text uses a geometric interpretation of space-time so that a general theory is seen as a natural extension of the special theory. Although most results are derived from first principles, complex and distracting mathematics is avoided and all mathematical steps and formulae are fully explained and interpreted, often with explanatory diagrams.; The emphasis throughout the text is on understanding the physics of relativity. The structure of the book is designed to allow students of different courses to choose their own route through the short self-contained sections in each chapter. The latter part of the book shows how Einstein's theory of gravity is central to unraveling fundamental questions of cosmology. This book concentrates on presenting the theory of special relativity as the geometry of space-time. The presentation is straightforward, complete and reader-friendly, with explanatory asides, that give historical context and links with other branches of physics and mathematics. The first four chapters give a complete description of the special theory and the nature of space and time, with the minimum use of mathematics. The mathematics necessary is introduced in the following five chapters, with the final fifteen chapters devoted to a comprehensive and detailed exposition of Einstein’s special relativity. Features: * Concentrates on presenting the theory of special relativity as the geometry of space-time * The presentation is straightforward, complete and reader-friendly, with explanatory asides, which give historical context and links with other branches of physics and mathematics Based on a course taught for years at Oxford, this book offers a concise exposition of the central ideas of general relativity. The focus is on the chain of reasoning that leads to the relativistic theory from the analysis of distance and time measurements in the presence of gravity, rather than on the underlying mathematical structure. Includes links to recent developments, including theoretical work and observational evidence, to encourage further study. The most important feature in this book is the simple presentation with details of calculations. It is very easy to follow. Fairly sophisticated calculations are developed very rapidly. The presentation is logical and the detailed coverage makes this book very readable and useful. The contents develop Relativity as a modern theory of motion, starting by placing it in historical perspective and proceeding to show its logical necessity. The development of the Lorentz transformation is given using only one assumption rather than two. Right away in Chapter 3, geometry as required in Special Relativity for extension to General Relativity is introduced. This enables the use of the four-vector formalism of Minkowski. By the end of Chapter 4, the general Lorentz transformations for three-dimensional motion and their relation to four-dimensional boosts have already been explained. In Chapter 5 applications of relevance in Physics are provided. After a brief introduction to elementary electromagnetic theory, it is reformulated as a theory in four-dimensions using tensors in Chapter 6. Finally in Chapter 7, the theory is extended to deal with accelerated motion as ?corrections? to Special Relativity. In this book, Lawrence Sklar demonstrates the interdependence of science and philosophy by examining a number of crucial problems on the nature of space and time--problems that require for their resolution the resources of philosophy and of physics. The overall issues explored are our knowledge of the geometry of the world, the existence of spacetime as an entity over and above the material objects of the world, the relation between temporal order and causal order, and the problem of the direction of time. Without neglecting the most subtle philosophical points or the most advanced contributions of contemporary physics, the author has taken pains to make his explorations intelligible to the reader with no advanced training in physics, mathematics, or philosophy. The arguments are set forth step-by-step, beginning from first principles; and the philosophical discussions are supplemented in detail by nontechnical expositions of crucial features of physical theories.
An introduction to bond angle geometry. A cube is made from smaller cubes, 5 by 5 by 5, then some of those cubes are removed. Can you make the specified shapes, and what is the most and least number of cubes required ? What 3D shapes occur in nature. How efficiently can you pack these shapes together? A circular plate rolls in contact with the sides of a rectangular tray. How much of its circumference comes into contact with the sides of the tray when it rolls around one circuit? This article explores ths history of theories about the shape of our planet. It is the first in a series of articles looking at the significance of geometric shapes in the history of astronomy. Can you make a tetrahedron whose faces all have the same perimeter? 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. See if you can anticipate successive 'generations' of the two animals shown here. The second in a series of articles on visualising and modelling shapes in the history of astronomy. A cheap and simple toy with lots of mathematics. Can you interpret the images that are produced? Can you predict the pattern that will be produced using different wheels? Bilbo goes on an adventure, before arriving back home. Using the information given about his journey, can you work out where Bilbo A 3x3x3 cube may be reduced to unit cubes in six saw cuts. If after every cut you can rearrange the pieces before cutting straight through, can you do it in fewer? A game for 2 players In this problem we are faced with an apparently easy area problem, but it has gone horribly wrong! What happened? Glarsynost lives on a planet whose shape is that of a perfect regular dodecahedron. Can you describe the shortest journey she can make to ensure that she will see every part of the planet? 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? Can you make sense of the charts and diagrams that are created and used by sports competitors, trainers and statisticians? Find the point whose sum of distances from the vertices (corners) of a given triangle is a minimum. What is the shape of wrapping paper that you would need to completely wrap this model? 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 visualisation problem in which you search for vectors which sum to zero from a jumble of arrows. Will your eyes be quicker than 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. . . . Have a go at this 3D extension to the Pebbles problem. Two motorboats travelling up and down a lake at constant speeds leave opposite ends A and B at the same instant, passing each other, for the first time 600 metres from A, and on their return, 400. . . . A half-cube is cut into two pieces by a plane through the long diagonal and at right angles to it. Can you draw a net of these pieces? Are they identical? Two boats travel up and down a lake. Can you picture where they will cross if you know how fast each boat is travelling? Discover a way to sum square numbers by building cuboids from small cubes. Can you picture how the sequence will grow? What can you see? What do you notice? What questions can you ask? 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. . . . 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? How efficiently can you pack together disks? 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? Imagine you are suspending a cube from one vertex (corner) and allowing it to hang freely. Now imagine you are lowering it into water until it is exactly half submerged. What shape does the surface. . . . 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? The reader is invited to investigate changes (or permutations) in the ringing of church bells, illustrated by braid diagrams showing the order in which the bells are rung. This task depends on groups working collaboratively, discussing and reasoning to agree a final product. These are pictures of the sea defences at New Brighton. Can you work out what a basic shape might be in both images of the sea wall and work out a way they might fit together? Two angles ABC and PQR are floating in a box so that AB//PQ and BC//QR. Prove that the two angles are equal. 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? Two intersecting circles have a common chord AB. The point C moves on the circumference of the circle C1. The straight lines CA and CB meet the circle C2 at E and F respectively. As the point C. . . . Imagine a stack of numbered cards with one on top. Discard the top, put the next card to the bottom and repeat continuously. Can you predict the last card? Watch these videos to see how Phoebe, Alice and Luke chose to draw 7 squares. How would they draw 100? This article is based on some of the ideas that emerged during the production of a book which takes visualising as its focus. We began to identify problems which helped us to take a structured view. . . . A useful visualising exercise which offers opportunities for discussion and generalising, and which could be used for thinking about the formulae needed for generating the results on a Can you mentally fit the 7 SOMA pieces together to make a cube? Can you do it in more than one way? In the game of Noughts and Crosses there are 8 distinct winning lines. How many distinct winning lines are there in a game played on a 3 by 3 by 3 board, with 27 cells? Use a single sheet of A4 paper and make a cylinder having the greatest possible volume. The cylinder must be closed off by a circle at each end. A game for 2 people. Take turns joining two dots, until your opponent is unable to move. A 10x10x10 cube is made from 27 2x2 cubes with corridors between them. Find the shortest route from one corner to the opposite
Presentation on theme: "Machine Learning – Course Overview David Fenyő"— Presentation transcript: 1 Machine Learning – Course Overview David Fenyő Contact: 2 Learning2“A computer program is said to learn from experience E with respect to some task T and performance measure P if its performance at task T, as measured by P, improves with experience E.”Mitchell 1997, Machine Learning. 13 Bayes Rule: How to Choose the Prior Probability? Hypothesis (H)Data (D)P(H\|D) = P(D\|H) P(H) / P(D)PosteriorProbabilityPriorProbabilityIf we have no knowledge, we can assume that each outcome is equally probably.Two mutually exclusive hyposthesis H1 and H2:If we have no knowledge: P(H1) = P(H2) = 0.5If we find out that hypothesis H2 is true: P(H1) = 0 and P(H2) = 1 17 𝐸𝑛𝑡𝑟𝑜𝑝𝑦=− 𝑝 𝑖 𝑙𝑜𝑔 2 ( 𝑝 𝑖 ) Bayes Rule and Information Theory 𝐸𝑛𝑡𝑟𝑜𝑝𝑦=− 𝑝 𝑖 𝑙𝑜𝑔 2 ( 𝑝 𝑖 )Two mutually exclusive hypothesis H1 and H2:If we have no knowledge: P(H1) = P(H2) = 0.5: Entropy=1If hypothesis H2 is true: P(H1) = 0 and P(H2) = 1 : Entropy=0P(H1) = 0.3, P(H2) = 0.7: Entropy=0.88P(H1) = 0.11, P(H2) = 0.89: Entropy=0.50 18 Bayes Rule: Example: What is the bias of a coin? Hypothesis (H)Data (D)P(H\|D) = P(D\|H) P(H) / P(D)Hypothesis: the probability for head is θ (=0.5 for unbiased coin)Data: 10 flips of a coin: 3 heads and 7 tails.P(D\|θ) = θ3(1- θ)7Uninformative prior: P(θ) uniformPosterior Likelihood Prior 19 Bayes Rule: Example: What is the bias of a coin? Hypothesis (H)Data (D)P(H\|D) = P(D\|H) P(H) / P(D)Hypothesis: the probability for head is θ (=0.5 for unbiased coin)Data: 10 flips of a coin: 3 heads and 7 tails.Likelihood: P(D\|θ) = θ3(1- θ)7Prior: θ2(1- θ)2Posterior Likelihood Priorθθθ 20 Bayes Rule: Example: What is the bias of a coin? Posterior ProbabilityData:10 flips of a coin: 3 heads and 7 tails.100 flips of a coin: 45 heads and 55 tails.1000 flips of a coin: 515 heads and 485 tails.Prior:θ2(1- θ)2Uniformpriorθ 22 CrowdsourcingCrowdsourcing is a methodology that uses the voluntary help of large communities to solve problems posed by an organizationCoined in 2006, but not new: in 1714 British Board of Longitude Prize: who can determine a ship’s longitude at sea? (winner: John Harrison, unknown clock-maker)Different types of crowdsourcing:Citizen science: the crowd provides data (e.g., patients)Labor-focused crowdsourcing: online workforce, tasks for moneyGamification: encode problem as gameCollaborative competitions (challenges)Julio Saez-Rodriguez: RWTH-Aachen&EMBL EBI 23 Collaborative competitions (challenges) Post a question to whole scientific community, withhold the answer (‘gold standard’)Evaluate submissions against the gold-standard with appropriate scoringAnalyze resultsDesignOpen ChallengeScoringChallengeTrainTestPoseto theCommunityJulio Saez-Rodriguez: RWTH-Aachen&EMBL EBI 24 Examples of DREAM challenges Predict phosphoproteomic data and infer signalling network - upon perturbation with ligands and drugs (Prill et al Science Signaling. 2011; Hill et al, Nature Meth 2016)Predict Transcription Factor Binding Sites - (with ENCODE; ongoing)Molecular Classification of Acute Myeloid Leukaemia - from patient samples using flow cytometry data - with FlowCAP (Aghaeepour et al Nat Meth 2013)Predict progression of Amyotrophic lateral sclerosis patients - from clinical trial data (Kuffner et al Nature Biotech 2015)NCI-DREAM Drug Sensitivity Prediction- predict response of breast cancer cell lines to single (Costello et al Nat Biot 2014) and combined (Bansal et al Nat Biot 2014) drugsThe AstraZeneca-Sanger DREAM synergy prediction challenge - predict drug combinations on cancer cell lines from molecular data (just finished) The NIEHS-NCATS-UNC DREAM Toxicogenetics - predict toxicity of chemical compounds (Eduati et al., Nat Biot, 2015) 25 NCI-DREAM Drug sensitivity challenge Costello et al. Nat Biotech. 2015 26 Some lessons from the drug sensitivity challenge Some drugs are easier to predict than others, & does not depend on mode of actionGene Expression is the most predictive data typeIntegration of multiple data and pathway information layers improves predictivityCostello et al. Nat Biotech. 2015 27 Some lessons from the drug sensitivity challenge 0.60RANDOMGene expression & protein amount - the most predictive data typeIntegration of multiple data and pathway information improves predictivityThere is plenty of room for improvementThe wisdom of the crowds: Aggregate is robustCostello et al. Nat Biotech. 2015 28 Value of collaborative competitions (challenges) Challenge-based evaluation of methods is unbiased & enhances reproducibilityDiscover the Best MethodsDetermine the solvability of a scientific questionSampling of the space of methodsUnderstand the diversity of methodologies used to solve a problemAcceleration of ResearchThe community of participants can do in 4 months what would take 10 years to any groupCommunity BuildingMake high quality, well-annotated data accessible.Foster community collaborations on fundamental research questions.Determine robust solutions through community consensus: “The Wisdom of Crowds.”Julio Saez-Rodriguez: RWTH-Aachen&EMBL EBI 29 Class ProjectPick one of the previous DREAM Challenges and analyze the data using several different methods.2/28 Project Plan Presentation5/2 Project Presentation5/5 Project Presentation 30 Class PresentationsPick one ongoing DREAM or biomedicine related Kaggle challenge to preset during one of the next classes.Kaggle 31 Curse of Dimensionality 31When the number of dimensions increase, the volume increases and the data becomes sparse.It is typical for biomedical data that there are few samples and many measurements. 32 Unsupervised Learning 32Finding the structure in data.ClusteringDimension reduction 33 Unsupervised Learning: Clustering 33How many clusters?Where to set the borders between clusters?Need to select a distance measure.Examples of methods:k-means clusteringHierarchical clustering 35 Supervised Learning: Regression 35Choose a function, f(x,w) where and a performance metric, 𝑗 𝑔 𝑦 𝑗 −𝑓( 𝒙 𝑗 ,𝒘) to minimize where ( 𝑦 𝑗 , 𝒙 𝑗 ) is the training data and w = (w1 ,w2,…, wk) are the k parameters.Commonly, f is a linear function of w, and g is the sum of mean square errors:𝜕 𝜕 𝑤 𝑖 𝑗 𝑦 𝑗 − 𝑖 𝑤 𝑖 𝑓 𝑖 ( 𝒙 𝑗 ) 2 =0𝑓 𝒙,𝒘 = 𝑖 𝑤 𝑖 𝑓 𝑖 (𝒙) 36 Model Capacity: Overfitting and Underfitting 36 37 Model Capacity: Overfitting and Underfitting 37 38 Model Capacity: Overfitting and Underfitting 38 39 Model Capacity: Overfitting and Underfitting 39Training ErrorError on Training SetDegree of polynomial 40 Model Capacity: Overfitting and Underfitting 40With four parameters I can fit an elephant, and with five I can make him wiggle his trunk.John von Neumann 47 Evaluation of Binary Classification Models PredictedTrueNegativeFalsePositive147ActualFalseNegativeTruePositiveFalse Positive Rate = FP/(FP+TN) – fraction of label 0 predicted to be label 1Accuracy = (TP+TN)/total - fraction of correct predictionPrecision = TP/(TP+FP) – fraction of correct among positive predictionsSensitivity = TP/(TP+FN) – fraction of correct predictions among label 1. Also called true positive rate and recall.Specificity = TN/(TN+FP) – fraction of correct predictions among label 0 48 Evaluation of Binary Classification Models ReceiverOperatingCharacteristic(ROC)48Algorithm 1FalseFalseSensitivitySensitivityTrueTrueScoreScore11--SpecificitySpecificityAlgorithm 2FalseFalseSensitivitySensitivityTrueTrueScoreScore11--SpecificitySpecificity 53 Training: Gradient Descent 53We want to use a large training rate when we are far from the minimum and decrease it as we get closer. 54 Training: Gradient Descent 54If the gradient is small in an extended region, gradient descent becomes very slow. 55 Training: Gradient Descent 55Gradient descent can get stuck in local minima.To improve the behavior for shallow local minima, we can modify gradient descent to take the average of the gradient for the last few steps (similar to momentum and friction). 66 No Free Lunch66Wolpert, David (1996), Neural Computation, pp 67 Can we trust the predictions of classifiers? 67Ribeiro, Singh and Guestrin ,"Why Should I Trust You? Explaining the Predictions of Any Classifier“, In: ACM SIGKDD International Conference on Knowledge Discovery and Data Mining (KDD), 2016 68 Adversarial Fooling Examples Original correctlyclassified imageClassifiedas ostrichPerturbation68Szegedy et al., “Intriguing properties of neural networks”, 70 Home WorkRead Saez-Rodriguez el al., Crowdsourcing biomedical research: leveraging communities as innovation engines. Nat Rev Genet Jul 15;17(8): doi: /nrg PubMed PMID:Pick one of the previous DREAM Challenges and analyze the data using several different methods.Pick one ongoing DREAM or biomedicine related Kaggle challenge to preset during one of the next classes.70
Helium ground state H total 2 2 2 − 2 2 2 e 2 e e = ∇1 + ∇ 22 ) − − + ( 2m 4πε 0 r1 4πε 0 r2 4πε 0 r1 − r2 Approximate antisymmetrized wave function (neglecting electronelectron interactions) Ψ (r1 , r2 ) = Energy neglecting e-e interactions Approximate ground state evaluated with the total Hamiltonian ψ 1s (r1 )ψ 1s (r2 ) 2 ( ↑↓ − ↓↑ ) −13.6 Z 2 E = 2× = − 108.8 eV 2 n E= Ψ H total Ψ Ψ Ψ = −74.83 eV Helium excited states One electron in 1s and one in 2s, E= He Ψ H red Ψ Ψ Ψ 13.6* 22 13.6* 22 =− − = −68 eV 2 2 1 2 The antisymmetric solution Ψ = 0 for r1 = r2 . Helium excited states E = hf = hc/λ Names refer to approximate solutions http://physics.nist.gov/PhysRefData/Handbook/Tables/heliumtable5.htm Exchange (Austauschwechselwirking) 1 ψ A ( r1 , r2 ) = (φ1 (r1 )φ2 (r2 ) − φ1 (r2 )φ2 (r1 ) ) 2 1 ψ A H ψ A = [ φ1 (r1 )φ2 (r2 ) H φ1 (r1 )φ2 (r2 ) − φ1 (r1 )φ2 (r2 ) H φ1 (r2 )φ2 (r1 ) 2 − φ1 (r2 )φ2 (r1 ) H φ1 (r1 )φ2 (r2 ) + φ1 (r2 )φ2 ( r1 ) H φ1 (r2 )φ2 (r1 ) ] 1 ψ S ( r1 , r2 ) = (φ1 (r1 )φ2 (r2 ) + φ1 (r2 )φ2 (r1 ) ) 2 1 ψ S H ψ S = [ φ1 (r1 )φ2 (r2 ) H φ1 (r1 )φ2 (r2 ) + φ1 (r1 )φ2 (r2 ) H φ1 (r2 )φ2 (r1 ) 2 + φ1 (r2 )φ2 (r1 ) H φ1 (r1 )φ2 (r2 ) + φ1 (r2 )φ2 (r1 ) H φ1 (r2 )φ2 (r1 ) ] The difference in energy between the ψA and ψS is twice the exchange energy. Exchange The exchange energy can only be defined when you speak of multi-electron wavefunctions. It is the difference in energy between the symmetric solution and the antisymmetric solution. There is only a difference when the electronelectron term is included. Coulomb repulsion determines the exchange energy. In ferromagnets, the antisymmetric state has a lower energy. Thus the state with parallel spins has lower energy. In antiferromagnets, the symmetric state has a lower energy. Neighboring spins are antiparallel. Many electrons Consider a gold atom (79 electrons) 2 ⎛ d2 ∂ψ =− ⎜ 2 i ∂t 2m ⎝ dx1 3 x 79 = 237 terms Ψ( x1, y1, z1, d2 ⎞ 79e2 + 2 ⎟Ψ − Ψ dz79 ⎠ 4πε 0rj 79 terms + e2 4πε 0rij Ψ 79 x 78 = 3081 terms 2 x79 , y79 , z79 , t ) is a complex function in 237 dimensions Ψ(r1, rN ) 2 is the joint probability of finding an electron at position r1, r2, ... rN. Numerical solution of the Schrödinger equation for one electron ∂Ψ − e 2 = ∇ Ψ− Ψ i ∂t 2m 4πε 0 r 2 2 z Discretize Ψ to solve numerically. For one electron ~ 106 elements are needed. y x Numerical solution for many electrons For a numerical solution, divide Hilbert space along each axis into 100 divisions. z1 y1 100237 = 10474 x1 y2 x2 There are 1068 atoms in the Milky Way galaxy There are ~ 1080 atoms in the observable universe Intractable problem We know the equation that has to be solved. We know how to solve it but we don't have the computer resources to calculate the answer. A Matlab that will calculate the time evolution of an n-electron atom http://lamp.tu-graz.ac.at/~hadley/ss1/studentpresentations/2011/n_electrons.m Quantum computation Sometimes it is possible to map one intractable problem onto another. If you map an intractable problem onto a system of interacting electrons and then measure the energy levels of the electron system, you can find solutions to the intractable problem. Many-electron systems In such a quantum system, the repeated interactions between particles create quantum correlations, or entanglement. As a consequence, the wave function of the system is a complicated object holding a large amount of information, which usually makes analytical calculations impractical. In fact, many-body theoretical physics ranks among the most computationally intensive fields of science. http://en.wikipedia.org/wiki/Many-body_problem The Central Dilemma of Solid State Physics The Schrödinger explains everything but can explain nothing. From a fundamental point of view it is impossible to describe electrons in a metal correctly - Ashcroft and Mermin Neglect the e-e interactions Consider a gold atom (79 electrons) ⎛ d2 − ⎜ 2 2m ⎝ dx1 2 3 x 79 = 237 terms d2 ⎞ 79e2 + 2 ⎟Ψ − Ψ dz79 ⎠ 4πε 0rj 79 terms + e2 4πε 0rij Ψ = EΨ 79 x 78 = 3081 terms 2 Out of desperation: We simplify the model for a solid until the Schrödinger can be solved. If the 3081 electron - electron terms are neglected, the equation can be solved exactly and the total wave function is a product of atomic orbitals. Antisymmetrized product wave functions Ψ (r1 , r2 , , r79 ) = φ179s ↑ (r1 ), φ179s ↓ (r2 ),… , φ679s ↑ (r79 ) The standard first approximation for the many-electron wave-function of any atom is an antisymmetrized product of atomic orbitals. 1s2 2s22p63s23p63d104s24p64d104f145s25p65d106s1 Electron configurations http://lamp.tu-graz.ac.at/~hadley/ss1/molecules/atoms/review3.php Filling of electron shells Ni: 3d84s2 Cu: 3d104s1 Why isn't Ni 3d94s1 or 3d10? You can evaluate the energy of any electron configuration. Ψ (r1 , r2 , , r28 ) = φ128s ↑ (r1 ), φ128s ↓ (r2 ),… , φ328d ↑ (r27 )φ428s ↑ (r29 ) E= Ψ H Ψ Ψ Ψ Lithium The antisymmetric 3 electron wavefunction can be written φ1Lis ↑ (r1 ) φ1Lis ↑ (r2 ) φ1Lis ↑ (r3 ) 1 Li Li Li Li Ψ(r1, r2 , r3 ) = φ1s ↑,φ1s ↓,φ2s ↑ = φ1s ↓ (r1 ) φ1Lis ↓ (r2 ) φ1Lis ↓ (r3 ) 3! Li φ2s ↑ (r1 ) φ2Lis ↑ (r2 ) φ2Lis ↑ (r3 ) There are two possible configurations φ ↑,φ ↓,φ ↑ Li 1s Li 1s Li 2s and φ1Lis ↑,φ1Lis ↓,φ2Lis ↓ Construct the Hamiltonian matrix to see which has the lowest energy Berylium The antisymmetric 4 electron wave function can be written, Ψ(r1, r2 , r3 , r4 ) = φ1Bes ↑,φ1Bes ↓,φ2Bes ↑,φ2Bes ↓ φ1Bes ↑ (r1 ) Be 1 φ1s ↓ (r1 ) = 4! φ2Bes ↑ (r1 ) φ2Bes ↓ (r1 ) φ1Bes ↑ (r2 ) φ1Bes ↓ (r2 ) φ2Bes ↑ (r2 ) φ2Bes ↓ (r2 ) φ1Bes ↑ (r3 ) φ1Bes ↓ (r3 ) φ2Bes ↑ (r3 ) φ2Bes ↓ (r3 ) φ1Bes ↑ (r4 ) φ1Bes ↓ (r4 ) φ2Bes ↑ (r4 ) φ2Bes ↓ (r4 ) Gold The antisymmetric 79 electron wave function can be written, Ψ (r1 , r2 , , r79 ) = φ179s ↑ (r1 ), φ179s ↓ ( r2 ),… , φ679s ↑ (r79 ) The determinant of an N×N matrix has N! terms. 79! = 8.95 × 10116 (more than the atoms in the observable universe) Start with the valence electrons. Stop adding electrons when the matrix gets too big. Pauli exclusion The sign of the wave function must change when two electrons are exchanged 1 ψ1 (r1 ) ψ 2 (r1 ) 1 Ψ(r1, r2 ) = = (ψ1(r1 )ψ 2 (r2 ) −ψ1(r2 )ψ 2 (r1)) 2 ψ1 (r2 ) ψ 2 (r2 ) 2 If two columns are exchanged, the wave function changes sign. If two columns are the same, the determinant is zero. The Pauli exclusion principle only holds in the noninteracting electron approximation when the many electron wave function can be written as a product of single electron wave functions, only one electron can occupy each single electron state. © Copyright 2021 Paperzz
This article may be too technical for most readers to understand. Please help improve it to make it understandable to non-experts, without removing the technical details. (December 2017) (Learn how and when to remove this template message) A Newtonian fluid is a fluid in which the viscous stresses arising from its flow, at every point, are linearlycorrelated to the local strain rate—the rate of change of its deformation over time. That is equivalent to saying those forces are proportional to the rates of change of the fluid's velocity vector as one moves away from the point in question in various directions. More precisely, a fluid is Newtonian only if the tensors that describe the viscous stress and the strain rate are related by a constant viscosity tensor that does not depend on the stress state and velocity of the flow. If the fluid is also isotropic (that is, its mechanical properties are the same along any direction), the viscosity tensor reduces to two real coefficients, describing the fluid's resistance to continuous shear deformation and continuous compression or expansion, respectively. Newtonian fluids are the simplest mathematical models of fluids that account for viscosity. While no real fluid fits the definition perfectly, many common liquids and gases, such as water and air, can be assumed to be Newtonian for practical calculations under ordinary conditions. However, non-Newtonian fluids are relatively common, and include oobleck (which becomes stiffer when vigorously sheared), or non-drip paint (which becomes thinner when sheared). Other examples include many polymer solutions (which exhibit the Weissenberg effect), molten polymers, many solid suspensions, blood, and most highly viscous fluids. Newtonian fluids are named after Isaac Newton, who first used the differential equation to postulate the relation between the shear strain rate and shear stress for such fluids. An element of a flowing liquid or gas will suffer forces from the surrounding fluid, including viscous stress forces that cause it to gradually deform over time. These forces can be mathematically approximated to first order by a viscous stress tensor, which is usually denoted by . The deformation of that fluid element, relative to some previous state, can be approximated to first order by a strain tensor that changes with time. The time derivative of that tensor is the strain rate tensor, that expresses how the element's deformation is changing with time; and is also the gradient of the velocity vector field at that point, often denoted . The tensors and can be expressed by 3×3 matrices, relative to any chosen coordinate system. The fluid is said to be Newtonian if these matrices are related by the equation where is a fixed 3×3×3×3 fourth order tensor, that does not depend on the velocity or stress state of the fluid. For an incompressible and isotropic Newtonian fluid, the viscous stress is related to the strain rate by the simpler equation If the fluid is incompressible and viscosity is constant across the fluid, this equation can be written in terms of an arbitrary coordinate system as One also defines a total stress tensor , that combines the shear stress with conventional (thermodynamic) pressure . The stress-shear equation then becomes or written in more compact tensor tensor notation where is the identity tensor. More generally, in a non-isotropic Newtonian fluid, the coefficient that relates internal friction stresses to the spatial derivatives of the velocity field is replaced by a nine-element viscous stress tensor . There is general formula for friction force in a liquid: The vector differential of friction force is equal the viscosity tensor increased on vector product differential of the area vector of adjoining a liquid layers and rotor of velocity: where – viscosity tensor. The diagonal components of viscosity tensor is molecular viscosity of a liquid, and not diagonal components – turbulence eddy viscosity. The following equation illustrates the relation between shear rate and shear stress: If viscosity is constant, the fluid is Newtonian. The power law model is used to display the behavior of Newtonian and non-Newtonian fluids and measures shear stress as a function of strain rate. The relationship between shear stress, strain rate and the velocity gradient for the power law model are: The relationship between the shear stress and shear rate in a casson fluid model is defined as follows: where τ0 is the yield stress and where α depends on protein composition and H is the Hematocrit number. Water, air, alcohol, glycerol, and thin motor oil are all examples of Newtonian fluids over the range of shear stresses and shear rates encountered in everyday life. Single-phase fluids made up of small molecules are generally (although not exclusively) Newtonian. In physics, the Navier–Stokes equations, named after Claude-Louis Navier and George Gabriel Stokes, describe the motion of viscous fluid substances. The vorticity equation of fluid dynamics describes evolution of the vorticity ω of a particle of a fluid as it moves with its flow, that is, the local rotation of the fluid . The equation is: Shear stress, often denoted by τ, is the component of stress coplanar with a material cross section. Shear stress arises from the force vector component parallel to the cross section of the material. Normal stress, on the other hand, arises from the force vector component perpendicular to the material cross section on which it acts. A power-law fluid, or the Ostwald–de Waele relationship, is a type of generalized Newtonian fluid for which the shear stress, τ, is given by A Bingham plastic is a viscoplastic material that behaves as a rigid body at low stresses but flows as a viscous fluid at high stress. It is named after Eugene C. Bingham who proposed its mathematical form. Hemorheology, also spelled haemorheology, or blood rheology, is the study of flow properties of blood and its elements of plasma and cells. Proper tissue perfusion can occur only when blood's rheological properties are within certain levels. Alterations of these properties play significant roles in disease processes. Blood viscosity is determined by plasma viscosity, hematocrit and mechanical properties of red blood cells. Red blood cells have unique mechanical behavior, which can be discussed under the terms erythrocyte deformability and erythrocyte aggregation. Because of that, blood behaves as a non-Newtonian fluid. As such, the viscosity of blood varies with shear rate. Blood becomes less viscous at high shear rates like those experienced with increased flow such as during exercise or in peak-systole. Therefore, blood is a shear-thinning fluid. Contrarily, blood viscosity increases when shear rate goes down with increased vessel diameters or with low flow, such as downstream from an obstruction or in diastole. Blood viscosity also increases with increases in red cell aggregability. In fluid dynamics, the Reynolds stress is the component of the total stress tensor in a fluid obtained from the averaging operation over the Navier–Stokes equations to account for turbulent fluctuations in fluid momentum. In seismology, S-waves, secondary waves, or shear waves are a type of elastic wave and are one of the two main types of elastic body waves, so named because they move through the body of an object, unlike surface waves. Shear rate is the rate at which a progressive shearing deformation is applied to some material. Stokes flow, also named creeping flow or creeping motion, is a type of fluid flow where advective inertial forces are small compared with viscous forces. The Reynolds number is low, i.e. . This is a typical situation in flows where the fluid velocities are very slow, the viscosities are very large, or the length-scales of the flow are very small. Creeping flow was first studied to understand lubrication. In nature this type of flow occurs in the swimming of microorganisms and sperm and the flow of lava. In technology, it occurs in paint, MEMS devices, and in the flow of viscous polymers generally. Fluid mechanics is the branch of physics concerned with the mechanics of fluids and the forces on them. It has applications in a wide range of disciplines, including mechanical, civil, chemical and biomedical engineering, geophysics, oceanography, meteorology, astrophysics, and biology. The intent of this article is to highlight the important points of the derivation of the Navier–Stokes equations as well as its application and formulation for different families of fluids. Volume viscosity is a material property relevant for characterizing fluid flow. Common symbols are or . It has dimensions, and the corresponding SI unit is the pascal-second (Pa·s). The Herschel–Bulkley fluid is a generalized model of a non-Newtonian fluid, in which the strain experienced by the fluid is related to the stress in a complicated, non-linear way. Three parameters characterize this relationship: the consistency k, the flow index n, and the yield shear stress . The consistency is a simple constant of proportionality, while the flow index measures the degree to which the fluid is shear-thinning or shear-thickening. Ordinary paint is one example of a shear-thinning fluid, while oobleck provides one realization of a shear-thickening fluid. Finally, the yield stress quantifies the amount of stress that the fluid may experience before it yields and begins to flow. The viscosity of a fluid is a measure of its resistance to deformation at a given rate. For liquids, it corresponds to the informal concept of "thickness": for example, syrup has a higher viscosity than water. The Reynolds number is an important dimensionless quantity in fluid mechanics used to help predict flow patterns in different fluid flow situations. At low Reynolds numbers, flows tend to be dominated by laminar (sheet-like) flow, while at high Reynolds numbers turbulence results from differences in the fluid's speed and direction, which may sometimes intersect or even move counter to the overall direction of the flow. These eddy currents begin to churn the flow, using up energy in the process, which for liquids increases the chances of cavitation. The Reynolds number has wide applications, ranging from liquid flow in a pipe to the passage of air over an aircraft wing. It is used to predict the transition from laminar to turbulent flow, and is used in the scaling of similar but different-sized flow situations, such as between an aircraft model in a wind tunnel and the full size version. The predictions of the onset of turbulence and the ability to calculate scaling effects can be used to help predict fluid behaviour on a larger scale, such as in local or global air or water movement and thereby the associated meteorological and climatological effects. The Oldroyd-B model is a constitutive model used to describe the flow of viscoelastic fluids. This model can be regarded as an extension of the Upper Convected Maxwell model and is equivalent to a fluid filled with elastic bead and spring dumbbells. The model is named after its creator James G. Oldroyd. The viscous stress tensor is a tensor used in continuum mechanics to model the part of the stress at a point within some material that can be attributed to the strain rate, the rate at which it is deforming around that point. In continuum mechanics, the strain-rate tensor or rate-of-strain tensor is a physical quantity that describes the rate of change of the deformation of a material in the neighborhood of a certain point, at a certain moment of time. It can be defined as the derivative of the strain tensor with respect to time, or as the symmetric component of the gradient of the flow velocity. In fluid mechanics it also can be described as the velocity gradient, a measure of how the velocity of a fluid changes between different points within the fluid. Though the term can refer to the differences in velocity between layers of flow in a pipe, it is often used to mean the gradient of a flow's velocity with respect to its coordinates. The concept has implications in a variety of areas of physics and engineering, including magnetohydrodynamics, mining and water treatment. In fluid mechanics, Helmholtz minimum dissipation theorem states that the steady Stokes flow motion of an incompressible fluid has the smallest rate of dissipation than any other incompressible motion with the same velocity on the boundary. The theorem also has been studied by Diederik Korteweg in 1883 and by Lord Rayleigh in 1913.
« PreviousContinue » the elemental slice LPKJR contained between two planes parallel to the coordinate plane of (y, z), and at an infinitesimal distance dx apart. In the case therefore of the volume being such as that of the figure, the limits of y are mk and 0, MK being the y to the trace of the surface on the plane of (x,y) and which may therefore be found in terms of x by putting z = 0 in the equation of the surface. Or if the volume is contained between two planes parallel to that of (x,z) and at distances Yn, Yo from it, y, and yo being constants, they are in that case the limits of y; and similarly must the limits be determined if the bounding surface is a cylinder whose generating lines are parallel to the axis of z; in all these cases the result of the y-integration is the volume of a slice contained between two planes at an infinitesimal distance apart, the length of which, viz. that parallel to the axis of y, is a function of its distance, from the plane of (y,z); and therefore if the length is expressed by F(x), v = f (x) dx, (15) and the volume is the sum of all such differential slices taken between the assigned limits. Thus suppose in fig. 36 the volume contained in the octant to be required, and on = a, then the limits of x are a and 0: or suppose the required volume to be contained between two planes at distances X, and x, from the plane of (y,z), then X, and X, are the limits of x. The following examples how. ever of such cubatures will serve best to clear up any difficulties which arise from inadequate explanations of general principles. 258.] Examples of cubature. Ex. 1. Find the content of a rectangular parallelepipedon, three of whose edges meeting at a point are a, b, c. Let the point at which the edges a, b, c meet be the origin, and let the axes of x, y, z severally lie along them; then if v is the volume, Ex. 2. Determine the volume in the first octant of a right elliptical cylinder whose axis is the x-axis, and whose altitude = a: and the equation to the generating ellipse is c* y2 +62 m2 = 62 c. Here the superior limit of z is î (62 — y2)), which I will call 2, and the inferior limit is 0; the limits of y are b and 0; and of x, a and 0. To find the volume of the ellipsoid whose equation is Hence the limits of integration for the volume in the first octant fore for the volume in the first octant the limits are which I will call z, and 0); of y, c and 0; of x, a and 0. Therefore the volume required = 4 699dz dy dx = 4 [/0%. (a» — 22)dy dx = 48(2* —c)dr = Tae c Ex. 5. To determine the volume contained within the surface of an elliptic paraboloid whose equation is 2 + 3 = 48, and a plane parallel to that of (y, z), and at a distance c from it. Hence for the volume in the first octant the limits of integration are, for 2, (46x - y2), which I shall call Z, and 0; for y, 2(ax)}, which I shall call y, and 0; for x', c and 0 ; : the volume required = 4 [T['dz dy dx = 467" (43)x2–18yb dy de Ex. 6. Find the volume of the solid cut from the cylinder x2 + y2 = a2 by the planes 2 = 0, and 2 = x tan a. Here the limits of integration are, for 2, x tan a, which I will call z, and 0; for y, (a? — 22), which I will call y, and -Y; and for x, a and 0; so that, if v is the volume, Ex. 7. If the general equation of the second degree represents a closed surface, explain the process by which the several limits of integration as required-in Art. 214 are determined." • Corresponding to given values of x and y there are generally two values of 2, say 2, and Zo, which are the limits of the z-integration, so that the z-integration gives the volume of a right prism which is perpendicular to the plane of (x, y), whose altitude iş 24-2, and whose base is dy dx. The subsequent y-integration gives the sum of all those which lie in a thin slice parallel to the plane of (y, z), the limits of this integration being fixed by those points in that slice at which the tangent plane of the closed surface is parallel to the z-axis, and at which consequently i del 6)=0; and thus if z is eliminated between F(x, y, z) = 0, and 6)=0, we shall have a quadratic in y, which will give two roots, say y, and y,; and these are the limits of the y-integration. The subsequent x-integration will give the sum of all these slices, and thus the volume included within the closed surface; and the limits of the x-integration are determined by the values of the x-ordinate at which these slices parallel to the plane of (y,z) vanish, and at which consequently the surface is touched by planes parallel to that of (y, z). At these points then p = 0, 6-) = 0, () = 0, and the elimination of y and z by means of these three equations will give a quadratic in terms of x, the two roots of which are the limits of the x-integration. 259.] If the infinitesimal element-function of a triple integral is of the form FC ), and the inequality which assigns the limits of integration is also of the form for ) < 1, so that, if i is the integral, =N/ 63% ) dz dy da ; (16) Jxo Jyo 20 then, if x = ať, y = bn, z =cs, 1 = abc F(8,9, 8)d5 dn dg, (17) Jx, NY, Izo and the inequality which assigns the limits becomes f (5, 7, 8)<l; consequently if the definite integral given in (17) is known, that of (16) is also known. Thus, in Ex. 3 of the preceding article, if x, y, z are replaced severally by aề, bn, cs, v = abc\|\| \| dě dn df, and the equation to the ellipsoid, which assigns the limits of integration, becomes f? + m2 +62 = 1, which is the equation to a sphere whose radius = 1. Now \|\| a5dn dę for these limits expresses the volume of a sphere whose radius=1; that is, 47; and therefore v = 4mabc. F 3 Again, since the value of a right cone of altitude = c, de Tac scribed on a circular base of radius = a, = " , that of a right cone of the same altitude on an elliptical base = " 260.] It is frequently convenient to refer the element of a volume to a mixed system of coordinates; as, for instance, to the z-axis, and to polar coordinates in the plane of (x,y); so that in the expression for the volume-element given in (12), dy dx must be replaced by r dr dd; and consequently v = \|\|\|dz r dr do ; (18) in this case the base of the elemental prismatic column, which results from the z-integration, is the plane area-element whose area = rdrdo; and the volume of that column = zr dr do. Now if the r-integration is next effected, the sum of all these columns, o being constant, will be a sectorial slice, the limits of r being given by the trace of the surface on the plane of (x, y); and if the inferior limit of r is zero, the edge of the sectorial slice will coincide with the z-axis; and the final 6-integration will give the sum of all such sectorial slices, and consequently the volume required by the conditions of the problem. The following examples illustrate the process. Ex. 1. The axis of a right circular cylinder of radius 6 passes through the centre of a sphere of radius a, b being less than a; find the volume of the solid bounded by the surfaces. Let the plane passing through the centre of the sphere and perpendicular to the axis of the cylinder be that of (x,y); then the equations of the surfaces are x2 + y2 + z2 = a’, and x2 + y =6%; or, in terms of polar coordinates, the equation to the cylinder is, p= b. For the volume therefore contained in the first octant the limits are, of z, (a? — 22 - y2); or (ao —2)}, which I will call 2, and 0; of r, 6 and 0); and 0; therefore
How to Analyze the Volatility of stock returns with the (G)ARCH model part V Final part of the Series. Final part of the Series. The following analysis shows whether the AR models have ARCH effects. A suitable test for this is the Lagrange multiplier test, LM test for short. This test can also be perceived as a test for autocorrelation. The hypotheses are as follows: So if the p-value is less than our alpha, we discard H(0) and we have an autocorrelation in the model and the existence of ARCH effects (Schmelzer 2009: 37). Applying the LM test to the two AR models, AR(9) for the Nasdaq100 and AR(17) for gold, yields the following result: In both tests, the p-value is virtually zero, which is why we have no conditional heteroscedacity in our model (Lamoureux/Lastrapes 1990: 221). This result shows that a simple AR model is not sufficient for volatility analysis. The data structure of the return series is characterized by heteroscedasticity and volatility clusters. Also, these are not normally distributed, but have a leptokurtic distribution. Volatility generally describes the fluctuation range of the annualized standard deviation of the observed returns on financial markets (Schmelzer 2009: 43). In order to progress in the analysis of volatility, the application of an sGARCH, i.e. a standard GARCH model, is necessary. As in the AR models, it is also advisable for the GARCH models to use the first 9lags for the Nasdaq100 and the first 17 lags for gold. An sGARCH(1,1) model is used as theGARCH model. An sGARCH(1,1) model makes sense precisely because numerous studies have shown that simple models have proven themselves for financial market data(Bera/Higgens 1993: 317). Figure 2.5 shows the GARCH model for the Nasdaq100 andFigure 2.6 shows the GARCH model for gold. In both GARCH models, the ARCH parameter alpha1 is significantly smaller than the GARCH parameter beta1. Alpha1 is responsible for the extent of the immediate reaction to new messages of the error term, while beta1 describes the duration for the effect to wear off (Schmelzer 2009: 47). Since beta1 is significantly larger than alpha1, the conditional variance of the model can only slowly adjust back to the equilibrium level (Schmelzer 2009: 47). The sum of alpha1 and beta1 is less than one in both models. From this result it can be deduced that a high persistence of past shocks lies in the conditional variance of both models (Schmelzer 2009: 47). The volatility of both returns returns to equilibrium only slowly (Schmelzer 2009: 47).Figures 2.7 and 2.8 show the empirical density of the standardized residuals of the twoGARCH models. The orange line shows the density of a normal distribution. A significantly increased kurtosis of the GARCH residuals can be seen. Around the zero region, the residuals exceed the normal distribution, while a flattening can be seen at the edge. The kurtosis can not be eradicated in the GARCH models either. To determine whether there are ARCH effects in the GARCH model, an LM test can also be carried out here. In R, an LM test on different lags is displayed directly in the output for the GARCH model: We see high p-values as a result of both models. This suggests that there is no autocorrelation in either model and that both models are subject to conditional heteroscedacy (Lamoureux/Lastrapes 1990: 221). Since both models have no ARCH effects, an analysis of the residuals is necessary. Figure 3.1shows the squared residuals from the GARCH model for the Nasdaq100 and gold. The result is similar to the result of the squared residuals of the AR models, but the scaling is slightly higher. The maximum swing of the Nasdaq100 goes above 0.015 here. With the AR models, the scaling only went up to 0.012. Furthermore, it can be seen that gold has different maxima than the Nasdaq100, which is consistent with the previous analysis. Figure 3.2 shows the comparison of the conditional variance from the GARCH models. Gold is blue and the Nasdaq100 is black. This figure shows again that gold has a very low volatility compared to the Nasdaq100. It can also be clearly seen that the maximum point from Figure3.2 has decreased significantly. The reason for this is that there was an extreme sell-off in gold on this one day. This extreme sell-off has a strong impact on the residual analysis.However, the effect becomes significantly smaller when analyzing the conditional variance in the GARCH model, since the effect of yield shocks is weakened. After an analysis of the Nasdaq100 and the gold price using AR and GARCH models, the question now arises as to which relevant results can be drawn from the overall analysis. One finding that runs through the entire analysis is that gold has significantly lower volatility and has anti-cyclical behavior to the Nasdaq100. The analysis of the conditional variance of theGARCH models is very meaningful. This shows the lower volatility of gold in relation to theNasdaq100. Also in this analysis, the extreme rash that was seen in gold in the previous residual analysis is smoothed out significantly. The statement that gold fluctuates less and has an anti-cyclical behavior to the Nasdaq100 can therefore be made. Another consideration might be whether one can extract price predictions from the AR model's estimators. At gold, most of the AR model's estimates are not significant, and when they are, they are still hovering at alpha. No price predictions can therefore be made from the AR model, since the estimators are not very significant. The Nasdaq has significantly more significant estimators in the AR model. Nevertheless, it is questionable whether a reasonable trading strategy can be built on a negative estimator like: (ar1 with -0.1100). I would not make a specific statement about the price development with these results. The left-skewing of the distribution and the increased kurtosis also run through all models. This could not be corrected in any model. A critical examination of the models used is also required. AR models and the classic GARCH model were mainly used for the analysis. An extended analysis could be improved by furtherGARCH model variants such as ApARCH, MGARCH or an eGARCH model. These could, for example, underpin or put into perspective the statement that gold fluctuates less than theNasdaq100. The AR models could also be expanded. It would be possible to use the ARMA model here to show a broader analysis. The use of new statistical methods such as neural networks could also improve the results, since the classic models are based on regressions and have less flexibility than neural networks. Neural networks could also offer a more flexible forecast when analyzing price developments. Finally, it can be said that the focus of this work is on the application of the classic AR model and the classic GARCH model. Which is why an extended analysis with further statistical models exceeds the scope. Nevertheless, a statement can be made about the relationship between the volatility of the Nasdaq100 and the price of gold. This statement implies that gold fluctuates less than the Nasdaq100 indices and has a counter-cyclical relationship. Bera, Anil K, Higgins,Matthew L,(1993): ARCH models: properties, estimation and testing,in: Journal of Economic Surveys 7, 1993, S. 305-366. Christopher G. Lamoureux, William D. Lastrapes (1990): Heteroskedasticity in Stock ReturnData: Volume versus GARCH Effects, The Journal of Finance, Vol. 45, No.1, S. 221-229 Mandelbrot, Benoit, (1963): The Variation of Certain Speculative Prices, in: Journal ofBusiness, Vol. 36, 1963, S. 394-419. Marcus Schmelzer, (2009), Die Volatilität von Finanzmarktdaten, Theoretische Grundlagenund empirische Analysen von stündlichen Renditezeitreihen und Risikomaßen, UniversitätKöln Reinhold Kosfeld, (2007), https://www.uni-kassel.de/fb07/fileadmin/datas/fb07/5-Institute/IVWL/Kosfeld/lehre/zeitreihen/PartielleACF.pdf Stock, J.H. and Watson, M.W. (2019), Introduction to Econometrics, 4th Global edition,Pearson Education Limited.
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This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. We are concerned with the following modified nonlinear Schrödinger system: , , , , , , , where , , , is the critical Sobolev exponent, and is a bounded smooth domain. By using the perturbation method, we establish the existence of both positive and negative solutions for this system. Let us consider the following modified nonlinear Schrödinger system: where , , , is the critical Sobolev exponent, and is a bounded smooth domain. Solutions for the system (1) are related to the existence of the standing wave solutions of the following quasilinear Schrödinger equation: where is a given potential, is a real constant, and , are real functions. We would like to mention that (2) appears more naturally in mathematical physics and has been derived as models of several physical phenomena corresponding to various types of . For instance, the case was used for the superfluid film equation in plasma physics by Kurihara (see also ); in the case of , (2) was used as a model of the self-changing of a high-power ultrashort laser in matter (see [3–6] and references therein). In recent years, much attention has been devoted to the quasilinear Schrödinger equation of the following form: See, for example, by using a constrained minimization argument, the existence of positive ground state solution was proved by Poppenberg et al. . Using a change of variables, Liu et al. used an Orlicz space to prove the existence of soliton solution for (3) via mountain pass theorem. Colin and Jeanjean also made use of a change of variables but worked in the Sobolev space ; they proved the existence of positive solution for (3) from the classical results given by Berestycki and Lions . Liu et al. established the existence of both one-sign and nodal ground states of soliton type solutions for (3) by the Nehari method. In particular, in , by using Nehari manifold method and concentration compactness principle (see ) in the Orlicz space, Guo and Tang considered the following quasilinear Schrödinger system: with , having a potential well and , , where is the critical Sobolev exponent, and they proved the existence of a ground state solution for the system (4) which localizes near the potential well for large enough. Guo and Tang also considered ground state solutions of the single quasilinear Schrödinger equation corresponding to the system (4) by the same methods and obtained similar results. It is worth pointing out that the existence of one-bump or multibump bound state solutions for the related semilinear Schrödinger equation (3) for has been extensively studied. One can see Bartsch and Wang , Ambrosetti et al. , Ambrosetti et al. , Byeon and Wang , Cingolani and Lazzo , Cingolani and Nolasco , Del Pino and Felmer [21, 22], Floer and Weinstein , and Oh [24, 25] and the references therein. The system (1) is a kind of “limit” problem of the system (4) as . The existence of solutions for the system (1) has important physical interest. The purpose of this paper is to study the existence of both positive and negative solutions for the system (1). We mainly follow the idea of Liu et al. to perturb the functional and obtain our main results. We point out that the procedure to the system (1) is not trivial at all. Since the appearance of the quasilinear terms and , we need more delicate estimates. The paper is organized as follows. In Section 2, we introduce a perturbation of the functional and give our main results (Theorems 1 and 2). In Section 3, we verify the Palais-Smale condition for the perturbed functional. Section 4 is devoted to some asymptotic behavior of sequence and satisfying some conditions. Finally, our main results will be proved in Section 5. Throughout this paper, we will use the same to denote various generic positive constants, and we will use to denote quantities that tend to . 2. Perturbation of the Functional and Main Results In order to obtain the desired existence of solutions for the system (1), in this section, we introduce a perturbation of the functional and give our main results. The weak form of the system (1) is which is formally the variational formulation of the following functional: We may define the derivative of at in the direction of as follows: We call that is a critical point of if , , and for all . That is, is a weak solution for the system (1). When we consider the system (1) by using the classical critical point theory, we encounter the difficulties due to the lack of an appropriate working space. In general it seems there is no suitable space in which the variational functional possesses both smoothness and compactness properties. For smoothness one would need to work in a space smaller than to control the term involving the quasilinear term in the system (1), but it seems impossible to obtain bounds for sequence in this setting. There have been several ideas and approaches used in recent years to overcome the difficulties such as by minimizations [7, 27], the Nehari method , and change of variables [8, 9]. In this paper, we consider a perturbed functional where is a parameter. Then it is easy to see that is a -functional on . We also can define the derivative of at in the direction of as follows: for all . The idea is to obtain the existence of the critical points of for small and to establish suitable estimates for the critical points as so that we may pass to the limit to get the solutions for the original system (1). Our main results are as follows. Theorem 1. Assume that , and . Let and let be a sequence of satisfying and for some independent of . Then, up to a subsequence as and is a critical point of . Using Theorem 1, we have the following existence result. Theorem 2. Assume that , and . Then has a positive critical point and a negative critical point , and (resp., ) converges to a positive (resp., negative) solution for the system (1) as . Notation. We denote by the norm of and by the norm of . 3. Compactness of the Perturbed Functional In this section, we verify the Palais-Smale condition ( condition in short) for the perturbed functional . We have the following proposition. Proposition 3. For fixed, the functional satisfies condition for all . That is, any sequence satisfying, for , has a strongly convergent subsequence in , where is the dual space of . For giving the proof of Proposition 3, we need the following lemma firstly. Lemma 4. Suppose that a sequence satisfies (11). Then Now we give the proof of Proposition 3. Proof of Proposition 3. From Lemma 4, we know that is bounded in . So there exists a subsequence of , still denoted , such that Now we prove that in . In (9), choosing , we have We may estimate the terms involved as follows: Returning to (16), we have which implies that , that is, in . This completes the proof of Proposition 3. 4. Some Asymptotic Behavior Proposition 3 enables us to apply minimax argument to the functional . In this section, we also study the behavior of sequence and satisfying The following proposition is the key of this section. Proposition 5. Assume sequence and satisfy (19). Then after extracting a sequence, still denoted by , one has as . 5. Proof of Main Results In this section, we give the proof of our main results. Firstly, we prove Theorem 1. Proof of Theorem 1. Note that satisfies the following equation: for all . Since By Moser’s iteration, we have for some independent of . To show that is a critical point of we use some arguments in [28, 29] (see more references therein). In (24) we choose , , where , , . Substituting into (24), we have Note that , . By Fatou’s Lemma, the weak convergence of and the fact that is bounded, we have Let , . We may choose a sequence of nonnegative functions such that in , a.e. and is uniformly bounded in . Then by approximations in (29) we may obtain for all , . Similarly, we may obtain an opposite inequality. Thus we have for all . That is, is a critical point of and a solution for the system (1). By doing approximations again, we have in the place of of (31) Setting in (24), we have Using as , (32), (33), and lower semicontinuity, we obtain as . In particular, we have as . This completes the proof of Theorem 1. Next, we apply the mountain pass theorem to obtain existence of critical points of . Set for . Let us consider the functional Here and in the following we denote . The functional satisfies condition. Similarly, we may verify that satisfies condition. By -Young inequality, for any , there exists such that Since Then for small. Thus we have for and for small enough. Choose , and . Define a path : by . When is large enough, we have for some independent of . Define where From the mountain pass theorem we obtain that is a critical value of . Let be a critical point corresponding to . We have . Thus is a positive critical point of by the strong maximum principle. In summary, we have the following. Proposition 6. There exist positive constants and independent of such that has a positive critical point satisfying Finally, we give the proof of Theorem 2. Proof of Theorem 2. For a positive solution of the system (1), the proof follows from Proposition 6 and Theorem 1. A similar argument gives a negative solution of the system (1). This completes the proof of Theorem 2. This paper was finished while Y. Jiao visited School of Mathematical Sciences of Beijing Normal University as a visiting fellow, and she would like to express her gratitude for their hospitality during her visit. Y. Jiao is supported by the National Science Foundation of China (11161041 and 31260098) and Fundamental Research Funds for the Central Universities (zyz2012074). - S. Kurihara, “Large-amplitude quasi-solitons in superfluid films,” Journal of the Physical Society of Japan, vol. 50, no. 10, pp. 3262–3267, 1981. - E. W. Laedke, K. H. Spatschek, and L. Stenflo, “Evolution theorem for a class of perturbed envelope soliton solutions,” Journal of Mathematical Physics, vol. 24, no. 12, pp. 2764–2769, 1983. - H. S. Brandi, C. Manus, G. Mainfray, T. Lehner, and G. Bonnaud, “Relativistic and ponderomotive self-focusing of a laser beam in a radially inhomogeneous plasma. I: paraxial approximation,” Physics of Fluids B, vol. 5, no. 10, pp. 3539–3550, 1993. - X. L. Chen and R. N. Sudan, “Necessary and sufficient conditions for self-focusing of short ultraintense laser pulse in underdense plasma,” Physical Review Letters, vol. 70, no. 14, pp. 2082–2085, 1993. - A. de Bouard, N. Hayashi, and J. C. Saut, “Global existence of small solutions to a relativistic nonlinear Schrödinger equation,” Communications in Mathematical Physics, vol. 189, no. 1, pp. 73–105, 1997. - B. Ritchie, “Relativistic self-focusing and channel formation in laser-plasma interactions,” Physical Review E, vol. 50, no. 2, pp. R687–R689, 1994. - M. Poppenberg, K. Schmitt, and Z. Q. Wang, “On the existence of soliton solutions to quasilinear Schrödinger equations,” Calculus of Variations and Partial Differential Equations, vol. 14, no. 3, pp. 329–344, 2002. - J. Q. Liu, Y. Q. Wang, and Z. Q. Wang, “Soliton solutions for quasilinear Schrödinger equations II,” Journal of Differential Equations, vol. 187, no. 2, pp. 473–493, 2003. - M. Colin and L. Jeanjean, “Solutions for a quasilinear Schrödinger equation: a dual approach,” Nonlinear Analysis A, vol. 56, no. 2, pp. 213–226, 2004. - H. Berestycki and P. L. Lions, “Nonlinear scalar field equations. I: existence of a ground state,” Archive for Rational Mechanics and Analysis, vol. 82, no. 4, pp. 313–345, 1983. - J. Q. Liu, Y. Q. Wang, and Z. Q. Wang, “Solutions for quasilinear Schrödinger equations via the Nehari method,” Communications in Partial Differential Equations, vol. 29, no. 5-6, pp. 879–901, 2004. - Y. Guo and Z. Tang, “Ground state solutions for quasilinear Schrödinger systems,” Journal of Mathematical Analysis and Applications, vol. 389, no. 1, pp. 322–339, 2012. - P. L. Lions, “The concentration-compactness principle in the calculus of variations. The locally compact case I,” Annales de l'Institut Henri Poincaré, vol. 1, no. 2, pp. 109–145, 1984. - Y. Guo and Z. Tang, “Ground state solutions for the quasilinear Schrödinger equation,” Nonlinear Analysis A, vol. 75, no. 6, pp. 3235–3248, 2012. - T. Bartsch and Z. Q. Wang, “Multiple positive solutions for a nonlinear Schrödinger equation,” Zeitschrift für Angewandte Mathematik und Physik, vol. 51, no. 3, pp. 366–384, 2000. - A. Ambrosetti, M. Badiale, and S. Cingolani, “Semiclassical states of nonlinear Schrödinger equations,” Archive for Rational Mechanics and Analysis, vol. 140, no. 3, pp. 285–300, 1997. - A. Ambrosetti, A. Malchiodi, and S. Secchi, “Multiplicity results for some nonlinear Schrödinger equations with potentials,” Archive for Rational Mechanics and Analysis, vol. 159, no. 3, pp. 253–271, 2001. - J. Byeon and Z. Q. Wang, “Standing waves with a critical frequency for nonlinear Schrödinger equations II,” Calculus of Variations and Partial Differential Equations, vol. 18, no. 2, pp. 207–219, 2003. - S. Cingolani and M. Lazzo, “Multiple positive solutions to nonlinear Schrödinger equations with competing potential functions,” Journal of Differential Equations, vol. 160, no. 1, pp. 118–138, 2000. - S. Cingolani and M. Nolasco, “Multi-peak periodic semiclassical states for a class of nonlinear Schrödinger equations,” Proceedings of the Royal Society of Edinburgh A, vol. 128, no. 6, pp. 1249–1260, 1998. - M. Del Pino and P. L. Felmer, “Multi-peak bound states for nonlinear Schrödinger equations,” Annales de l'Institut Henri Poincaré, vol. 15, no. 2, pp. 127–149, 1998. - M. del Pino and P. L. Felmer, “Semi-classical states for nonlinear Schrödinger equations,” Journal of Functional Analysis, vol. 149, no. 1, pp. 245–265, 1997. - A. Floer and A. Weinstein, “Nonspreading wave packets for the cubic Schrödinger equation with a bounded potential,” Journal of Functional Analysis, vol. 69, no. 3, pp. 397–408, 1986. - Y. G. Oh, “On positive multi-lump bound states of nonlinear Schrödinger equations under multiple well potential,” Communications in Mathematical Physics, vol. 131, no. 2, pp. 223–253, 1990. - Y. G. Oh, “Existence of semiclassical bound states of nonlinear Schrödinger equations with potentials of the class ,” Communications in Partial Differential Equations, vol. 13, no. 12, pp. 1499–1519, 1988. - X. Q. Liu, J. Q. Liu, and Z. Q. Wang, “Quasilinear elliptic equations via perturbation method,” Proceedings of the American Mathematical Society, vol. 141, no. 1, pp. 253–263, 2013. - J. Liu and Z. Q. Wang, “Soliton solutions for quasilinear Schrödinger equations I,” Proceedings of the American Mathematical Society, vol. 131, no. 2, pp. 441–448, 2003. - A. Canino and M. Degiovanni, “Nonsmooth critical point theory and quasilinear elliptic equations,” in Topological Methods in Differential Equations and Inclusions, P. Q. Montreal, Ed., vol. 472 of Nato Advanced Study Institute Series C: Mathematical and Physical Sciences, pp. 1–50, Kluwer Academic, 1995. - J. Q. Liu and Z. Q. Wang, “Bifurcations for quasilinear elliptic equations II,” Communications in Contemporary Mathematics, vol. 10, no. 5, pp. 721–743, 2008.
Welcome to Big Prime Number. By William Brower. Copyright © Big Prime Number 2016. Prime Numbers. Versão em portugês. Definition. A prime number or a prime is a natural number greater than 1 that has no positive divisors other than 1 and itself. Why such page ? Read here. Lists of prime numbers. The first 168 prime numbers are. Prime Numbers Chart. The chart below shows the list of prime numbers, which are represented in the coloured box. Is 1 a Prime Number? Conferring to the definition of the prime number which states that a number should have exactly two factors for it to be considered a prime number. A natural number that can be divided by only 2 numbers – 1 and the number itself is called a prime number. The first prime number that comes to our mind is “1” but if you had paid attention to your Math teacher, then you will know that: 1 is neither primer nor composite. 29/09/2019 · A Big Question About Prime Numbers Gets a Partial Answer The twin primes conjecture has bedeviled mathematicians for more than a century. Now there's a solution for one version of it. 12/09/2019 · A prime number is a natural number greater than one that has no positive divisors other than one and itself. For example, 7 is prime because 1 and 7 are its only positive integer factors, whereas 12 is not because it has the divisors 3 and 2 in addition to 1, 4 and 6. 3. Generating Prime Numbers. The number n must be >=3. While 2 is a prime number, for our purposes we have no interest in numbers less than 3. Do a bitwise and n&1. If the result is not 0 then we know the number is even and can throw it out. Check that n%p is 0 in other words, that n. 07/05/2011 · a whole number that cannot be made by multiplying other whole numbers if we can make it by multiplying other whole numbers it is a Composite Number And 1 is not prime and also not composite. Here we see it in action: 2 is Prime, 3 is Prime, 4 is Composite =2×2, 5 is Prime, and so on. It's not so much the prime numbers themselves that are important, but the algorithms that work with primes. In particular, finding the factors of a number any number. As you know, any number has at least two factors. Prime numbers have the unique property in that they have exactly two factors: 1. List of all known Mersenne prime numbers along with the discoverer's name, dates of discovery and the method used to prove its primality. Great Internet Mersenne Prime Search - Finding world record primes since 1996. GIMPS is an organized search for Mersenne prime numbers using provided free software. 25/09/2018 · Download this app from Microsoft Store for Windows 10, Windows 10 Mobile, Windows 10 Team Surface Hub, HoloLens. See screenshots, read the latest customer reviews, and compare ratings for Big Integer and Prime Number Calculator. 19/11/2017 · We cannot use Sieve’s implementation for a single large number as it requires proportional space. We first count the number of times 2 is the factor of the given number, then we iterate from 3 to Sqrtn to get the number of times a prime number divides a particular number. 08/01/2018 · These numbers are interesting for a few reasons, though they aren't particularly useful. One big reason: Every time someone discovers a Mersenne prime, they also discover a perfect number. As Caldwell explained, a perfect number is a number that's equal to. The conventional wisdom is prime numbers occur randomly. If true then non-prime numbers occur randomly as well. That statement is self-evident because if it were not true, then it would be a simple task to find the non-prime patterns and eliminate them leaving only the primes. Look at. How to Find All The Factors of a Number Quickly and Easily. The quickest way to find the factors of a number is to divide it by the smallest prime number bigger than 1 that goes into it evenly with no remainder. Continue this process with each number you get, until you reach 1. Number factorizer a.k.a. integer factorization calculator computes prime factors of a natural number or an expresssion involving- / ^ ! operators that evaluates to a natural number. The result of the number factorization is presented as multiplication of the prime factors in ascending order. What is Prime number? Prime number is a positive integer greater than 1 that is only divisible by 1 and itself. For example: 2, 3, 5, 7, 11 are the first five prime numbers. Logic to print prime numbers between 1 to n. Step by step descriptive logic to print all prime numbers between 1 to n. Input upper limit to print prime numbers from user. 24/12/2019 · The number 1 is the only counting number that isn’t prime or composite, because its only factor is 1. The first six prime numbers are 2, 3, 5, 7, 11, and 13. When testing to see whether a number is prime or composite, perform divisibility tests in the following order from easiest to. What is the currently industry-standard algorithm used to generate large prime numbers to be used in RSA encryption? I'm aware that I can find any number of articles on the Internet that explain how the RSA algorithm works to encrypt and decrypt messages, but I can't seem to find any article that explains the algorithm used to generate the p. The largest known prime number as of December 2019 is 2 82,589,933 − 1, a number which has 24,862,048 digits when written in base 10. It was found by Patrick Laroche of the Great Internet Mersenne Prime Search GIMPS in 2018. 28/01/2018 · Java Program for efficiently print all prime factors of a given number; Count occurrences of a prime number in the prime factorization of every element from the given range; Print the nearest prime number formed by adding prime numbers to N; Check if a prime number can be expressed as sum of two Prime Numbers; Find coordinates of a prime number. 05/07/2011 · For more see Prime Number Lists. 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Abstract: Although integral and differential nonlinearity may not be the most important parameters for high-speed, high dynamic performance data converters, they gain significance when it comes to high-resolution imaging applications. The following application note serves as a refresher course for their definitions and details two different, yet commonly used techniques to measure INL and DNL in high-speed analog-to-digital converters (ADCs). Manufacturers have recently introduced high-performance analog-to-digital converters (ADCs) that feature outstanding static and dynamic performance. You might ask, "How do they measure this performance, and what equipment is used?" The following discussion should shed some light on techniques for testing two of the accuracy parameters important for ADCs: integral nonlinearity (INL) and differential nonlinearity (DNL). Although INL and DNL are not among the most important electrical characteristics that specify the high-performance data converters used in communications and fast data-acquisition applications, they gain significance in the higher-resolution imaging applications. However, unless you work with ADCs on a regular basis, you can easily forget the exact definitions and importance of these parameters. The next section therefore serves as a brief refresher course. DNL error is defined as the difference between an actual step width and the ideal value of 1LSB (see Figure 1a). For an ideal ADC, in which the differential nonlinearity coincides with DNL = 0LSB, each analog step equals 1LSB (1LSB = VFSR/2N, where VFSR is the full-scale range and N is the resolution of the ADC) and the transition values are spaced exactly 1LSB apart. A DNL error specification of less than or equal to 1LSB guarantees a monotonic transfer function with no missing codes. An ADC's monotonicity is guaranteed when its digital output increases (or remains constant) with an increasing input signal, thereby avoiding sign changes in the slope of the transfer curve. DNL is specified after the static gain error has been removed. It is defined as follows: DNL = \|[(VD+1- VD)/VLSB-IDEAL - 1] \| , where 0 < D < 2N - 2. VD is the physical value corresponding to the digital output code D, N is the ADC resolution, and VLSB-IDEAL is the ideal spacing for two adjacent digital codes. By adding noise and spurious components beyond the effects of quantization, higher values of DNL usually limit the ADC's performance in terms of signal-to-noise ratio (SNR) and spurious-free dynamic range (SFDR). Figure 1a. To guarantee no missing codes and a monotonic transfer function, an ADC's DNL must be less than 1LSB. INL error is described as the deviation, in LSB or percent of full-scale range (FSR), of an actual transfer function from a straight line. The INL-error magnitude then depends directly on the position chosen for this straight line. At least two definitions are common: "best straight-line INL" and "end-point INL" (see Figure 1b): The best straight-line approach is generally preferred, because it produces better results. The INL specification is measured after both static offset and gain errors have been nullified, and can be described as follows: INL = \| [(VD - VZERO)/VLSB-IDEAL] - D \| , where 0 < D < 2N-1. VD is the analog value represented by the digital output code D, N is the ADC's resolution, VZERO is the minimum analog input corresponding to an all-zero output code, and VLSB-IDEAL is the ideal spacing for two adjacent output codes. Figure 1b. Best straight-line and end-point fit are two possible ways to define the linearity characteristic of an ADC. The transfer function for an ideal ADC is a staircase in which each tread represents a particular digital output code and each riser represents a transition between adjacent codes. The input voltages corresponding to these transitions must be located to specify many of an ADC's performance parameters. This chore can be complicated, especially for the noisy transitions found in high-speed converters and for digital codes that are near the final result and changing slowly. Transitions are not sharply defined, as shown in Figure 1b, but are more realistically presented as a probability function. As the slowly increasing input voltage passes through a transition, the ADC converts more and more frequently to the next adjacent code. By definition, the transition corresponds to that input voltage for which the ADC converts with equal probability to each of the flanking codes. The Right Transition A transition voltage is defined as the input voltage that has equal probabilities of generating either of the two adjacent codes. The nominal analog value, corresponding to the digital output code generated by an analog input in the range between a pair of adjacent transitions, is defined as the midpoint (50% point) of this range. If the limits of the transition interval are known, this 50% point is calculated easily. The transition point can be determined at test by measuring the limits of the transition interval, and then dividing the interval by the number of times each of the adjacent codes appears within it. INL and DNL can be measured with either a quasi-DC voltage ramp or a low-frequency sine wave as the input. A simple DC (ramp) test can incorporate a logic analyzer, a high-accuracy DAC (optional), a high-precision DC source for sweeping the input range of the device under test (DUT), and a control interface to a nearby PC or X-Y plotter. If the setup includes a high-accuracy DAC (much higher than that of the DUT), the logic analyzer can monitor offset and gain errors by processing the ADC's output data directly. The precision signal source creates test voltages for the DUT by sweeping slowly through the input range of the ADC from zero scale to full scale. Once reconstructed by the DAC, each test voltage at the ADC input is subtracted from its corresponding DC level at the DAC output, producing a small voltage difference (VDIFF) that can be displayed with an X-Y plotter and linked to the INL and DNL errors. A change in quantization level indicates differential nonlinearity, and a deviation of VDIFF from zero indicates the presence of integral nonlinearity. Another way to determine static linearity parameters for an ADC, similar to the preceding but more sophisticated, is using an analog integrating servo loop. This method is usually reserved for test setups that focus on high-precision measurements rather than speed. A typical analog servo loop (see Figure 2) consists of an integrator and two current sources connected to the ADC input. One source forces a current into the integrator, and the other serves as a current sink. A digital magnitude comparator connected to the ADC output controls both current sources. The other input of the magnitude comparator is controlled by a PC, which sweeps it through the 2N - 1 test codes for an N-bit converter. Figure 2. This circuit configuration is an analog integrating servo loop. If the polarity of feedback around the loop is correct, the magnitude comparator causes the current sources to servo the analog input around a given code transition. Ideally, this action produces a small triangular wave at the analog inputs. The magnitude comparator controls both rate and direction for these ramps. The integrator's ramp rate must be fast when approaching a transition, yet sufficiently slow to minimize peak excursions of the superimposed triangular wave when measuring with a precision digital voltmeter (DVM). For INL/DNL tests on the MAX108, the servo-loop board connects to the evaluation board through two headers (see Figure 3). One header establishes a connection between the MAX108's primary (or auxiliary) output port and the magnitude comparator's latchable input port (P). The second header ensures a connection between the servo loop (the magnitude comparator's Q port) and a computer-generated digital reference code. Figure 3. With the aid of the MAX108EVKIT and an analog integrating servo loop, this test setup determines the MAX108's INL and DNL characteristics. The fully decoded decision resulting from this comparison is available at the comparator output P > QOUT, and is then passed on to the integrator configurations. Each comparator result controls the logic input of the switch independently and generates voltage ramps as required to drive succeeding integrator circuits for both inputs of the DUT. This approach has its advantages, but it also has several drawbacks:
Mathematics 5020 3 credits Teaching of Mathematics in the Middle and Secondary School A study of the mathematics curriculum 6-12 and the methods and materials used in teaching middle and secondary school mathematics. Prerequisite: Mathematics 2740. Mathematics 5040 4 credits Algebra and Geometry for Elementary/Middle School Teachers Extends the teacher's knowledge of mathematics through the study of algebra, number theory, motion geometry, inductive and deductive reasoning, and problem solving. Calculators and computers will be used to investigate topics of mathematics by using spreadsheets, geometric investigation software, graphing software, graphing calculators, and the LogoWriter computer language. Students will learn to use manipulatives to help students understand mathematical concepts. This will include Algebra Lab Gear, D.I.M.E. materials, and geoboards. Prerequisites: Three courses in mathematics for elementary school teachers or consent of the department chair. Mathematics 5130 3 credits Topics in Euclidean geometry include classical theorems, transformational geometry, and Euclidean construction. Axiomatic structures are studied by examining finite geometrics. Prerequisite: Mathematics 2640. Mathematics 5230 3 credits Matrices, systems of equations, determinants, eigenvalues, eigenvectors, vector spaces, linear transformations, and diagonalization. Prerequisite: Mathematics 2740 with a grade of "C" or better. Mathematics 5330 3 credits Study of the structure of abstract algebraic systems through formal proof; deals primarily with groups, but also examines other algebraic systems including rings and fields. Prerequisite: Mathematics 3230 or Mathematics 5230 with a grade of "C" or better. Mathematics 5630 3 credits Differential Equations I Solution of first order differential equations, linear homogenous and non-homogeneous differential equations, Laplace transforms, linear systems, and applications. Prerequisite: Grade of "C" or better in Mathematics 2840. Mathematics 5730 3 credits An introduction to numerical methods stressing the basic operations of computations, approximation, interpolation, numerical differentiation and integration, and the numerical solution of equations. Prerequisite: Fluency in a programming language and Mathematics 3230. Mathematics 5830 3 credits Differential Equations II Series solutions of linear differential equations, nonlinear series solutions, specific power series equations, numerical methods, partial differential equations, orthogonal sets and Fourier series. Prerequisite: Mathematics 3630 or Mathematics 5630 with a grade of "C" or better. Mathematics 6030 3 credits Statistical Methods with Applications Introduction to probability, density and distribution functions, special discrete and continuous distributions, estimation, hypothesis testing, chi-square, correlation, and regression. Prerequisite: Grade of "C" or better in Mathematics 2740. Mathematics 6040 3 credits Statistics and Probability A thorough investigation of the density functions, means, variances, and moment generating functions for some common probability distributions; gamma and beta distributions; Tchebysheff's Theorem; joint, marginal, and conditional probability distributions; probability distributions of functions of random variables; order statistics; sampling distributions and the Central Limit Theorem; point estimators; method of moments; maximum likelihood estimators; hypothesis testing including power of the test and the likelihood ratio test; simple and multiple linear regression; analysis of variance; nonparametric statistics. Prerequisite: grade of "C" or better in Mathematics 4030 or Mathematics 6030. Mathematics 6320 2 credits History and Development of Mathematical Concepts A study of the history and development of mathematics from the primitive origins of numbers to the modern mathematics of the 20th century. Prerequisite: Mathematics 2640 with a grade of "C" or better. Mathematics 6330 3 credits Theory of Numbers Properties of numbers; Euclid's Algorithm; prime numbers; congruencies; residue classes; Wilson's, Euler's and Fermat's Theorems and their consequences; and other selected topics. Prerequisite: a "C" or better in Mathematics 2640. Mathematics 6430 3 credits Sequences, limits, continuity, differentiation, integration, infinite series, and uniform convergence. Prerequisite: Mathematics 2840 with a grade of "C" or better. Mathematics 6530 3 credits Complex numbers, complex function, differentiation, elem-entary functions, integration, and infinite series. Prerequisite: Mathematics 2840 with a grade of "C" or better. Mathematics 6620 1-3 credits Topics in Modern Mathematics Topics to be selected by the instructor. Prerequisite: Approval of the instructor. Mathematics 7920 0-2 credits Seminar Paper Research Mathematics 7980 1-4 credits Independent Study in Mathematics The amount of graduate credit allowed for independent study may not exceed a total of four credits except with the special permission of the student's advisor and the graduate dean. Approval must be secured before independent study courses are begun. Students registering for independent study must submit at or before registration a description signed by the instructor conducting the independent study of the subject to be covered. Independent study may not be used for collecting information for the seminar paper. Mathematics 7990 3-6 credits
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The fixed deposit calculator on the ICICI HFC website helps customers calculate the total maturity amount for their account based on the interest rate for a specific time period. The calculator is simple to use and will assist you in comparing different tenures with different interest rates from various banks, making it easier to choose a suitable amount and tenure that provide you with the desired maturity amount for your fixed deposit. Also, getting to know the basics of fixed deposit interest rate and how fixed deposit interest is calculated is very easy, especially with ICICI HFC’s FD Calculator. Now let’s talk about how Fixed Deposit is Calculated and how fixed deposit interest is calculated in the most convenient ways. The interest rate on Fixed Deposits is fixed at the time when the amount is deposited and has no direct relation to market fluctuations. Some financial institutions even allow you to break your FD early by incurring a penalty fee. 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A set of ancient Babylonian tablets that describe how to track Jupiter across the sky have revealed an astronomical technique 1,500 years ahead of its time. Jupiter's erratic pace across the sky — appearing to slow down and speed up from day to day based on the combination of its orbit and Earth's — must have perplexed ancient astronomers and tested their best computational techniques. A newly discovered tablet written in Babylonia's cuneiform script discusses calculating the position of Jupiter. When combined with four other tablets, it suggests that ancient Babylonians used a surprisingly modern technique to calculate how far the bright dot traveled through the sky over the course of months. Their process requires a leap in understanding in how position and speed relate to time, one that wouldn't appear again until 1350 and that was a precursor to modern calculus. [Views of Jupiter, the Solar System's Largest Planet] The new results were detailed in the journal Science today (Jan. 28). The connections between speed, position and time are known to most modern travelers — people easily understand speed as a measure of miles or kilometers per hour. Locations are often described in terms of time ("it's only an hour away") rather than distance. The insight that led to calculus demonstrated the connection between a graph of the traveler's changing speed and the total distance traveled. "This is familiar to any student of physics, or math or science," Mathieu Ossendrijver, an astroarchaeologist at Humboldt University of Berlin, told Space.com. But using time as a variable to calculate speed or distance has not been part of human culture forever. Using a graph to understand motion or speed over time is usually traced back to scholars in Oxford and Paris around 1350, and then to Isaac Newton, who developed integral calculus, Ossendrijver said. "What I now found is that this method was already invented in Babylonia more than 1,500 years earlier." Ossendrijver was an astrophysicist before he began studying the history of science and cuneiform in 2005. In 2012, he published a book of new translations for the known Babylonian tablets that featured astronomical calculations and tables. The procedure Ossendrijver translated from the Babylonian tablets appears to show how to calculate the distance that Jupiter has traveled over a long stretch of time, by using measurements of how fast it was moving across the sky on given days. This calculation might be particularly interesting to the ancient astronomers because of Jupiter's association with Babylon's patron god, Marduk. Under the Babylonians' earlier, arithmetic-based method, astronomers would measure the distance Jupiter traveled every day — then, by adding together the "distance per day" for each day from the first through the 60th, they would get the total distance traveled. The newly discovered method instead used a geometric shortcut, and only needed the "distance per day" for the first day and the 60th, not the ones in between, to get the distance overall. Today, that calculation might be done by drawing points on a graph for Jupiter's measured speed on the first day and on the 60th day. Each of those points shows Jupiter's speed across the sky and the day the speed was measured. Drawing lines to connect the points to each other and to the "ground" directly below them, at a speed of zero, creates a geometric shape — a trapezoid — and calculating that shape's area reveals how far the object traveled. The process of measuring that geometric shape was described on the Babylonian tablets. Although the tablets did not have any visible graphs, the calculations done matched this technique precisely, Ossendrijver said. When Ossendrijver first encountered the Babylonian tablets, he didn't understand why calculations on a trapezoid were included along with tables related to Jupiter's position, he said. Only after he saw a fifth, uncataloged tablet, which showed a different procedure for finding Jupiter's position using the same examples as the trapezoids, did he realize the connection between the figure and the tables, Ossendrijver said. Eventually, he understood a second trapezoid calculation on the tablets, too: dividing it into two trapezoids with equal area, which would correspond to finding when Jupiter had traveled half the distance, he said. The advanced technique has been found only on the four tablets so far, which all use slightly different wording but the same example, he said. There isn't any evidence yet of the process being more widespread, Ossendrijver said. "This would open up new ways of computing motion they could have applied to other planets, other parts of Jupiter's motion," Ossendrijver said. "We don't have [examples of that]. We only have these four tablets, and they all deal with Jupiter — and they all deal with the same segment of 60 days. That's quite strange." When Babylonian culture, and the cuneiform script it was recorded in, died out around the year 100, the technique was seemingly forgotten, Ossendrijver said, only to crop up again in the 14th century when scientists and mathematicians began to use graphs to calculate changes to a system over time. Other aspects of Babylonian astronomy, in contrast, made it through the ages: People still discuss signs of the Zodiac, for instance, and use the Babylonian system of degrees, minutes and seconds, in units of 60, to calculate distances across the sky. Babylonian observations and techniques, translated into Greek, offer evidence of that transfer of information, Ossendrijver said. Knowing that ancient Babylonians had access to this newfound technique provides a whole new context for examining previously discovered tablets, as many tablets that are already translated have sections that aren't yet understood, Ossendrijver said. And it also forces science historians to rethink the way astronomical techniques evolved, he said. "In the beginning, I felt insecure myself, because the implications of this, from a history of science point of view, are kind of — big," Ossendrijver said. The process shows "a more abstract and profound conception of a geometrical object in which one dimension represents time," Alexander Jones, a historian at New York University said in Science's news article accompanying the work. It's much earlier than these concepts have ever been found before, he said, and "their presence … testifies to the revolutionary brilliance of the unknown Mesopotamian scholars who constructed Babylonian mathematical astronomy."
Primary Maths Dictionary for Kids and Parents: A Guide To The Mathematical Terms Used In KS1 And KS2 Bet you never thought you’d need a maths dictionary for kids when your child started school! But as parents the maths words that your children have to contend with in school may be quite different from those you learnt and used in your primary education. Throughout their time studying maths in primary school, both at KS1 and KS2, your child will be introduced a huge number of maths vocabulary, much of which may seem baffling – both to them and to you. This primary school maths dictionary for kids and their parents is, we hope, the solution to any questions and confusion. It’s an A to Z maths glossary of meanings for all those words and phrases you’re not sure about from acute angles through to word problems. It is by no means a comprehensive list of every mathematical term that could come up throughout your child’s school life (as that would be too overwhelming), but it includes the key ones they will need to form solid foundations in primary maths and onto secondary school and GCSE maths. The importance of spoken mathematical language When children struggle to articulate their thoughts, it is usually a sign that they do not understand the topic they in discussing in great depth and this can often be the case in maths. This quote taken from the Department for Education reflects the importance of mathematical language amongst children: “The national curriculum for mathematics reflects the importance of spoken language in pupils’ development across the whole curriculum – cognitively, socially and linguistically. The quality and variety of language that pupils hear and speak are key factors in developing their mathematical vocabulary and presenting a mathematical justification, argument or proof.” National Curriculum in England, Department for Education, 2013 When children use spoken language in maths in a productive manner, it allows them to evaluate their learning, support their peers, challenge the status quo, justify their own answers and most importantly ask questions! Using a wide range of maths vocabulary can help children to make links across not only different areas of maths, but also throughout real life situations. One example of this could be in the supermarket, when they would realise, thanks to their range of vocabulary, that a 50% off sticker means the same as half price. Mathematical vocabulary appears in more places than you would think, and that is why it is very important that your child is confident in their ability to articulate about all things mathematical! How you can encourage a wider range of mathematical language in your child There are a number of ways you can encourage your child to enhance their mathematical vocabulary and they include: - Getting them to read and test themselves with our primary school maths glossary; - Asking them open ended questions. No more “Do you know the answer to this question?” and more “What do you think the answer to this question is?”; - Playing some word based maths games; - Reading outside the world of maths as many children’s books will incorporate elements of maths in a fun and engaging way. ‘One is a Snail, Ten is a Crab’ is a great example of this. Primary Maths Dictionary for Children and Parents A free A to Z packed full of primary maths terms! Find terminology and definitions that are aligned to the UK National Curriculum. < and > < and > are symbols representing one number being ‘greater than’ or ‘less than’ another. For example 16 > 8 or 8 < 16 says that 16 is greater than 8 and 8 is less than 16. 12-hour and 24-hour clock The 12-hour clock goes from 1 am in the morning to 12 noon and from 1 pm in the afternoon to 12 midnight. This is known as ‘analogue’ time. The 24-hour clock goes from 00:00 (midnight) to 23:59 (one minute to midnight). This is known as ‘digital’ time. Read more: What is the 12 hour and 24 hour clock A 2D shape is any flat or ‘two-dimensional’ shape, such as a square, circle or triangle. A 3D shape is ‘three-dimensional’ and has volume, for an example a cube (cardboard box), pyramid or cylinder (tube). An acute angle is any angle less than 90°. In algebra, letters and symbols are used to represent numbers in equations or formulae. For example, if w = 3, what is 6w + 7? Analogue and digital clocks An analogue clock is a clock with the numbers 1 to 12 around the outside and two hands, one short hand that represents hours and one long hand that represents minutes. A digital clock uses 24-hour time and always has four digits. For example, 15:30 is half-past three in the afternoon on a digital clock. The area of a shape, surface, piece of land etc. means the amount of space it takes up. For example, a rectangular football field has an area of 64m² or 64 squared metres. An array is a pictorial representation of a calculation, using rows of dots, to help children understand multiplication and times tables. Arrow cards are a maths tool useful for explaining place value and how to partition numbers (separate them into ones, tens, hundreds etc). To ascend means to go up, so numbers given in ascending order are going from smallest to largest. For example, 1, 2, 3, 4, 5, 6 are numbers in ascending order. The associative property says that when we add or multiply numbers, it doesn’t matter how we group them (which we calculate first). For example, (7 + 5) + 3 = 7 + (5 + 3) or (4 x 5) x 2 = 4 x (5 x 2) The average of a set of numbers is found by adding all the numbers together and dividing by how many numbers there are. For example, the average of 12, 10, 8 and 6 is 9 because (12 + 10 + 8 + 6 = 36 ÷ 4). The axes of a graph or chart are the horizontal and vertical lines that create it, often known as the x-axis and y-axis. A bar chart is a form of graph that displays information using rectangular bars of different heights, according to their numerical value. A bar model is a method that uses diagrams of rectangular bars to represent maths problems in a visual way, making them easier for children to see which operation to use to work out a calculation. Younger children may use cubes to physically represent this. Read more: What is a bar model A block graph is a simpler version of a bar chart, but using blocks to represent the data, with each block worth 1 unit. BODMAS is a rule for the order to work out calculations with mixed operations. It stands for Brackets, Orders, Division, Multiplication, Addition, Subtraction and is sometimes seen as BIDMAS (Brackets, Indices, Division, Multiplication, Addition, Subtraction) or even PEMDAS (Parentheses, Exponents, Multiplication/Division, Addition/Subtraction) Read more: What is BODMAS? Bridging through 10 Bridging through 10 is a way of adding numbers greater than 10 in your head. For example, to add 8 + 7, you add 2 (from the 7) to get 10, then add the remaining 5 to get 15. Bus stop method The ‘bus stop’ method or short division is a way of dividing numbers with two or more digits by one or two digit numbers. Read more: What is bus stop method The capacity of a container is how much that container can hold, measured using units such as litres, millilitres, pints etc. A cardinal number tells you how many of something there are; they refer to a set of objects. For example, there are three marbles in my hand. This is in contrast to an ordinal number which tells you the position of something in a list, for example first, second, third. A Carroll diagram is a way of organising information and grouping according to what criteria it fits into. For example, which shape has 6 sides and 1 line of symmetry? Read more: What Is A Carroll Diagram? A circle is a simple curved 2D shape, with 1 edge, no corners and infinite lines of symmetry. The circumference is the length around the edge of a circle. Clockwise and anti-clockwise To move in a clockwise direction means moving in the same direction as the hands on a clock. If something moves in the opposite direction to the hands of a clock, it is moving in an anticlockwise direction. The coordinates of a shape or object refer to where on a map or graph they are, by looking at the two axes and recording the numbers they are at. These can be taught with the phrase “along the corridor and up/down the stairs” to refer to looking at the x-axis first then looking at the y-axis. Read more: What are coordinates? The column method is way to solve addition and subtraction calculations, that sometimes involve ‘exchanging’ amounts from one column to the next (which in the past has been called ‘carrying’ and ‘borrowing’). The numbers are written on top of each other, with the correct digits in each column (e.g. hundreds, tens, ones). The commutative property states that addition and multiplication calculations can be carried out with the numbers in any order, whereas for subtraction and division, the numbers must be in a particular order. For example, 8 x 9 = 72 or 9 x 8 = 72 Complementary addition (subtraction on a number line or the jump strategy) Complementary addition is a method for subtraction that involves using a number line to jump from the smaller number to the bigger number and counting the number of jumps. This method is useful in KS1 for teaching children to ‘find the difference’ between two numbers. Concrete, Pictorial, Abstract approach (CPA) The concrete, pictorial, abstract approach is a way of teaching mathematical concepts and theories in various stages, in order to help children fully understand and master what they are learning. The concrete stage involves using items, models and objects, giving children a chance to be ‘hands-on’. For example, children may solve a problem adding groups of toys together using real toys, or they may manipulate buttons, Lego, etc when working out fractions of amounts. At school, there are a variety of concrete resources specially designed for maths, such as place value counters, dienes, rekenreks, and numicon. The pictorial stage uses visual representations of concrete objects to model problems, encouraging children to make connections between the physical object and the picture that represents the object. For example, children may use drawing of toys to solve a problem adding toys. The abstract stage uses symbols, such as numbers or mathematical symbols (+, -, x, =) to model problems. Children will need to master the concrete and pictorial stages before moving onto the abstract stage. Converting into same units When you convert measurements into the same units, you understand that the same length, weight or capacity can be shown in different units of measurement. For example, a bottle of water can be measured in litres or millilitres and there are 1000ml in 1L. A cube number is the result of when a number is multiplied by itself three times. When writing cube numbers, we write a small three above the number, e.g. 3 x 3 x 3 or 3³ = 27 Read more: What are cube numbers? Data handling is another term for statistics, meaning how we collect, display and interpret data or information, such as the most popular flavour of ice cream in a class, using tables, tally charts, pictograms, block diagrams, bar charts, line graphs and pie charts. A decimal is a number that contains tenths, hundredths, thousandths etc, with a decimal point between the ones and tenths. Money is often used to teach decimals. For example, 3.4, 2.18, £56.99 If you are interested in other ways to teach your child about money, take a look at our blog. We use degrees as the unit of measurement for measuring angles, usually symbolised with a small circle above the number. For example, a right angle is 90° (90 degrees). A denominator is the name for the bottom number in a fraction. For example, in the fraction 4/10, 10 is the denominator. Descending order means to go from the largest number to the smallest and is the opposite of ascending order. For example, 90, 80, 70, 60, 50 are numbers in descending order. A diagonal is a line joining two opposite corners of a square, rectangle or other shape. The diameter is a straight line going from one side of a circle to another, through the centre. Dienes are coloured plastic or wooden blocks that are used to represent numbers. They are usually used to represent 1000, 100, 10 and 1. Dienes allow teachers and students to represent numbers visually. The distributive property states that multiplying a number by a group of numbers added together is the same as doing each multiplication separately. For example, 5 x (2 x 6) = 5 x 2 + 5 x 6 The dividend is the number that you are about to share out. The divisor is the number of times you need to share out your dividend. Division facts are the division calculations related to times tables. For example, 50 ÷ 5 = 10 and 25 ÷ 5 = 5 are division facts related to the 5 times table. An edge is the name for lines created when two faces in a 3D shape meet. An equation is another name for a number sentence where both sides equal the other. For example, 12 – 5 = 3 + 4 An equilateral triangle is a triangle with three equal sides and three equal angles. An equivalent fraction is one that is equal in terms of size to another, but written using different numbers. For example, ½ is equivalent to 4/8 and 7/14. Read more: What are equivalent fractions? To estimate is to make a clever guess to the answer of a question, by roughly calculating the value. For example, children estimated the length of the playground to be 100 metres. Expanded notation means to write a number showing the value of each digit, with each digit being multiplied by its matching place value (ones, tens, hundreds etc). For example, 352 = 3 x 100 + 5 x 10 + 2 x 1. A face is the flat part of a 3D shape. A factor is a number that can divide exactly into another number. For example, 2, 3, 4 and 6 are all factors of 12, as 12 can be divided into them exactly. Factors and multiples are usually taught together and children often get the two confused. Read more about the difference between factors and multiples. Finding the difference between two numbers Finding the difference between two numbers is the same as subtracting a smaller number from a larger number. This method is usually taught using a number line, counting the jumps from one number to another. A formula is a group of numbers and maths symbols that show how to work something out. At primary school, children will encounter formulae for finding the area and perimeter of 2D shapes and for finding the volume of 3D shapes. Geometry is the branch of maths where children learn about the properties, measurements, position and relationships of points, lines, angles and shapes. The grid method is a way for working out multiplication calculations, especially with larger numbers, involving partitioning the numbers, multiplying each part and adding the totals. Highest common factor The highest common factor is the largest whole number which is a factor of two or more numbers. For example, 6 is the highest common factor of 12 and 18. Read more: What is the highest common factor? A horizontal line is one that goes from left to right and vice versa. Imperial units are units of measurement that were used in the UK before the metric system was introduced. Children, especially from Year 5 on, will be taught how to convert from, for example, miles to kilometres. An improper fraction is one where the numerator is larger than the denominator and is also known as a ‘top-heavy’ fraction. For example, 11/4, 6/2 and 21/5 are all improper fractions. Read more: What Is An Improper Fraction A mathematical investigation allows children to apply the skills and knowledge they’ve learnt to solve problems, which may have more than one way to work them out and more than one answer. An integer is simply a whole number, either positive or negative. For example, 8, -23, 502 and -1000 are all integers. An inverse operation is another way of saying an opposite operation, which can often be used to check calculations are correct. For example, addition and subtraction are inverse operations, as are multiplication and division. An isosceles triangle is one with two equal sides and two equal angles. A move made when practising addition or subtraction – forwards or backwards respectively – on a number line. A quadrilateral (four-sided shape) with two pairs of adjacent (next-door) sides that are congruent (equal in length). The diagonals of a kite are perpendicular (meet at a right angle). The length of an object is how long or short something is, and is usually measured in metric units such as centimetres, metres and kilometres. A line graph is one where a line connects points, showing how values change over time. For example, a line graph might show the amount of rainfall over six months. Long division is a written method showing how to divide larger numbers (such as three or four-digit numbers) by other large numbers. Children will move onto long division in KS2, once they’ve mastered short division. Long multiplication, which is sometimes known as column multiplication, is a way of multiplying larger numbers together. Just like column addition and subtraction, the numbers are put in columns according to their place value. Read more: What is long multiplication? Lowest common denominator The lowest common denominator of two or more fractions is the smallest number that can be exactly divided by each denominator. For example, 12 is the lowest common denominator of ½, ⅓ and ¼. Lowest common multiple The lowest common multiple is the smallest number which is a multiple of two or more numbers. For example, the lowest common multiple of 4 and 5 is 20. Read more: What is the lowest common multiple? The mass of an object is how much it weighs and is usually measured in grams and kilograms. For example, the mass of a bag of sugar is 1 kilogram. Having mastery in a maths topic means that children not only understand how to work out problems, but can also explain how they worked it out and apply their knowledge to more complicated word problems and investigations. Mean is another word for average, and is found by adding a set of values and dividing the total by the number of values in the set. For example, the mean of 2, 4, 5, 7 and 12 is 6 because (2 + 4 + 5 + 7 + 12 = 30 ÷ 5 = 6) Read more: What Is Mean In Maths? The median of a set of numbers is the middle number in that list. The numbers in the list must first be sorted into ascending order, then children can find the median number. For example, the median of 1, 2, 3, 4, 5, 6, 7 is 4. Metacognition means to be aware of and analyse your thoughts and learning processes in order to make necessary changes to your learning behaviour. Techniques such as modelling problems and getting children to ask questions about their work are ways of improving their metacognition. Metric units are units of measurement that are common around the world and are based on the metric system. For example, grams, centimetres, litres and seconds are all examples of metric units. A mirror line is a line that can be drawn through the centre of a shape or picture to show that both sides are exactly the same. A mixed number is one with both a whole number and a fraction. For example, 8 ⅔ and 5 10⁄12 are mixed numbers. Read more: What is a Mixed Number? The mode of a set of numbers is the one that appears most often. For example, the mode of 2, 3, 4, 5, 5, 6, 7 is 5. A multiple is the result of multiplying one integer by another. Multiples of a number are those in that numbers times table. For example, multiples of 7 include 14, 35, 49 and 84. Multiple and factors are usually taught together and children often get confused. Read more about the difference between factors and multiples. A negative number is any number lower than 0 and is commonly taught using temperatures. For example, -2, -14, -67. A net is the flat outline of a 3D shape, before it is folded together. Number bonds are pairs of numbers that add together to make a given number. For example, 2 + 8 and 4 + 6 are number bonds to 10, whereas 43 + 57 and 81 + 19 are number bonds to 100. Read this: What are number bonds? Number facts are simple addition, subtraction, multiplication and division calculations that children should be able to mentally recall easily. For example, 50 + 50 = 100 or 2 x 2 = 4 are number facts. Number line and number ladder A number line is a horizontal line, with numbers going up the bottom of the line. The numbers will typically increase in size and the space between the numbers doesn’t usually matter. Number lines are especially used in Key Stage 1 to teach number bonds and adding using jumping. A number ladder is a number line drawn vertically. A number sentence is how a calculation is written, using numbers and symbols. For example, 5 + 7 = 12 is an addition number sentence. Read this: What is a number sentence? A number square is a maths aid used in primary schools, showing numbers in order from 0 up to, for example, 20 or 100. Number squares are useful for helping with counting and seeing patterns in number sequences. A numerator is the name for the top number in a fraction. For example, in the fraction ⅝, 5 is the numerator. An obtuse angle is any angle that measures between 90° and 180°. Odd and even numbers An even number is any number that can be divided into two equal groups and always end in 0, 2, 4, 6 and 8. An odd number is any number that can’t be divided into two equal groups and always end in 1, 3, 5, 7 and 9. In maths, the four types of operation are addition, subtraction, multiplication and division. In year 6, pupils are taught the order of operations, which is the order in which the four mathematical operations should be completed when there is more than one in a number sentence together. An ordinal number tells us what position something is in a list, often taught using dates or the results of races. For example, Ben finished in 1st place, Chris in 2nd and Alex in 3rd. The contrast of this is a cardinal number. A parallel line is a straight line that always stays the same distance from another line and never meets. Shapes are often used to teach parallel lines. For example, a square has two pairs of parallel lines. To partition a number means to separate a number into separate parts (ones, tens, hundreds, thousands etc). Partitioning makes understanding place value easier and is also used when using column methods or grid methods. For example, 5246 can be partitioned into 5 thousands, 2 hundreds, 4 tens and 6 ones or 5000 + 200 + 40 + 6. Read more: What is partitioning? Part whole model A part whole model is a way of representing a number and its composite parts. It is a good way to show the relationship between numbers, and the method uses partitioning. The two main ways part whole model is used are in bar modelling and cherry diagrams. Read more: What is part whole model? Percentage means ‘out of 100’ and is used to show a number or ratio expressed as a fraction of 100. Children often use percentages when talking about sales in shops. For example, this £80 jacket had 20% off in the Christmas sale. The perimeter is the distance around a 2D shape and is often taught using the example of fences around a field or garden. Read more: What is the perimeter? Perpendicular lines are two lines that meet to create a right angle, often seen in shapes. A pictogram is a type of graph that uses pictures to represent information. These are often taught in Key Stage 1 before moving onto block charts and bar charts. A pie chart is a circular chart divided into sections, representing different values, which can be fractions, decimals, percentages or angles. The place value of a number is how much each digit in the number represents. For example, the place value of 157 is 1 hundred, 5 tens and 7 ones. A polygon is any 2D shape with straight, closed sides. Any shapes with open or curved sides are not polygons. For example, triangles, squares and parallelograms are polygons, but circles and ovals are not. A prime number is any number greater than 1 that can only be divided equally by itself and 1. For example, 5, 7, 11 and 13 are prime numbers. A prism is a 3D shape with two identical flat sides and ends. Cubes and cuboids are examples of prisms. Probability, chance and likelihood Probability is the study of how likely or how big a chance there is that something will happen. It can be described in words, fractions, percentages or ratios. For example, there is a 20% chance of rain tomorrow. A product of two numbers is the name for the answer to a multiplication calculation. For example, 35 is the product of 5 x 7. A proportion is a portion or part of a whole, and is often taught alongside ratio. A protractor is an instrument used to measure angles. A pyramid is a 3D shape with triangular sides that join at a point, with a polygon base. A quadrilateral is any 2D shape with four sides, including a square, rhombus, kite and trapezium. The radius is the distance from the centre of a circle to its circumference and is half the diameter. The range of a set of numbers is the difference between the smallest and largest numbers in the set. For example, in the set of numbers 50 to 60, the range is 10. A ratio is used to compare values, showing the relative value of one to another. It is taught using real-life examples, such as comparing the number of boys to girls in class. For example, the ratio of boys to girls was 2:1, meaning there are two boys for every one girl. Reflection of shapes A reflection of a shape is a drawing of a shape reflected in a mirror line, with the reflection on the other side of the line but facing in the opposite direction. Reflective symmetry is a type of transformation, looking at when a shape or pattern is reflected in a mirror or line of symmetry. The reflected shape should be exactly the same size and distance from the mirror line as the original. A reflex angle is any angle between 180° and 360°. Regular and irregular shapes A regular shape is one where all the sides and interior angles are equal, whereas an irregular shape has sides and angles of different lengths and sizes. Read more: What are regular and irregular shapes? A rekenrek is a counting frame, similar to an abacus. It has ten beads on each row, five white and five red which supports the development of number sense. Read more about rekenreks here. A right angle is an angle that measures 90°. It is also known as a quarter turn, as it is ¼ of a full turn, which measures 360°. A right-angled triangle is a 2D shape with three sides and one angle that measures 90°. Roman numerals are the numbers used in ancient Rome, with letters from the Latin alphabet representing certain numbers. They are commonly taught using years. For example, V = 5, X = 10, C = 100, M = 1000, so 1066 is MLXVI. Rotation of shapes A rotation of a shape is when a shape is moved around a fixed point, either clockwise or anticlockwise and by a certain number of degrees. However, the shape doesn’t change size. Rotational symmetry is a type of transformation, where a shape is turned around a central point, without changing its size. To round a number means to adjust it up or down to a number that makes calculating with it easier. Numbers are usually rounded up to the nearest 10, 100 or 1000, with decimals being rounded to the nearest whole number, tenth or hundredth. There is a rule that if a digit is 4 or less it rounds down and if it is 5 or more it rounds up. For example, 426 rounds to 430 to the nearest 10, but 400 to the nearest 100. Repeated addition is a technique used to teach multiplication in Key Stage 1, where children add ‘lots’ of numbers together. For example, 3 ‘lots’ of 5 is 5 + 5 + 5 as well as 3 x 5. A scale factor is used when we increase or decrease a 2D shape in size, so we make the shape larger or smaller depending on the scale factor. For example, this shape has been increased by a scale factor of 2. A scalene triangle is a 2D three-sided shape where all the sides and angles are unequal. ‘Shared between’ is a phrase used when introducing division, to show how a set of objects can be ‘shared’ into equal sized groups. To simplify a fraction means to reduce it to its lowest form, by dividing the numerator and denominator by the same number. For example 8/10 can be simplified to ⅘ by dividing both the numerator and denominator by 2. Square numbers and Square roots A square number is the result of multiplying a number by itself. When writing this, we write a small two next to and above the number. For example, 7² = 7 x 7 = 49. Read more: What is a square number? Standard and non-standard units Standard units are the units of measurement we normally use to indicate the length, mass or capacity of an object. For example, centimetres, metres, grams, kilograms, millilitres and litres. Non-standard units are used by when introducing measurement in KS1, for example the length of a pencil or hand spans. A sum of two numbers is another name for the result of an addition calculation. For example, the sum of 15 and 23 is 38. When a picture or shape is the same on both sides, we call it ‘symmetrical’, and this can be shown by drawing a line of symmetry through the centre and seeing if both sides are the same. Read this: What is a line of symmetry? A tally chart uses marks instead of numbers to represent information. One vertical mark is used to represent each one unit, with five being shown as a fifth line crossed through the first four lines. When shapes fit together exactly with no gaps, we call this Tessellation. An example of this in real life are floor tiles. A time interval is the length of time between two times. For example, the time interval between 1:15 and 1:45 is 30 minutes. Translation of shapes Translation is a type of transformation, where a shape is moved into a new position, without being changed in any way. A triangle is a 2D shape with three sides, angles and corners. A triangular number is a number that can make a triangular dot pattern.For example, 1 + 2 = 3, 2 + 3 = 5, 3 + 5 = 8, 5 + 8 = 13 etc. Turns are a movement in a circle, with a quarter turn being the same as 90°, a half turn as 180° and a full turn as 360°, either clockwise or anticlockwise. Two-step and multi-step problems A two-step problem is a word problem which needs two calculations to solve it. A multi-step problem requires more than two calculations to solve it. Unit and non-unit fractions A unit fraction is any fraction with 1 as the numerator, whereas a non-unit fraction is any fraction with a number greater than 1 as the numerator. For example, ⅙ is a unit fraction, whereas 2/6 is a non-unit fraction. Read this: What is a unit fraction? Read more: What is a non-unit fraction? A Venn diagram is a visual way of sorting different objects or numbers into overlapping circles with different rules, with anything in the overlapping part sharing both rules. Read this: What is a venn diagram? In primary maths, there are two types of variation, conceptual and procedural variation. Conceptual variation means looking at a maths idea in various representations. For example, showing a number using multilink, diennes block, hundred square or partitioned, to explain place value. Procedural variation is used to support a child’s deeper understanding of a maths process by extending a problem by varying the number, varying the processes to solve a problem or varying the problems by applying the same method to a group of similar problems. Vertex is another name for a corner of a 2D shape or the points where edges in a 3D shape meet. A vertical line runs up and down, from top to bottom. The volume is the amount of space an object occupies, especially 3D shapes. Children will learn the formula for finding the volume of a shape, which is the length x width x height, with the answer having units with a cube number, for example cm³. Word problem or story problem A word problem or a story problem is a real-life situation where a maths calculation is needed to solve a problem. For example, ‘If half a class of children have pets and there are 36 children in the class, how many have pets?’ Read this: Word Problems Explained The x-axis is the horizontal axis on a graph, along which we find the x-coordinate (by going ‘along the corridor’). The y-axis is the vertical axis on a graph, along which we find the y-coordinate (by going ‘up the stairs’). Zero is a placeholder between +1 and -1, it has no value but changes the value of other numbers. For example, in the number 703 it changes the number 73 to the much larger 703. In our primary maths dictionary for parents we’ve tried to include as much of the maths terminology your child will be learning at KS2 as possible, but we know that we may have missed something! If that is the case then let us know which numeracy based terms we have missed, and we’ll be more than happy to add them to this A to Z maths dictionary! 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Tra il XVI e il XVIII secolo è avvenuta, in Europa, una rivoluzione cosmologica. Si tratta di molto più di un semplice cambiamento “geometrico” (il Sole al centro invece della Terra al centro); si tratta della sostituzione di tutto un insieme di conoscenze e di concezioni. Le nuove concezioni sono state sviluppate in modo non sincrono, talvolta secondo le osservazioni empiriche ma talvolta anche contro le osservazioni empiriche, e hanno preso il sopravvento quando sono arrivate a costituire un insieme sufficientemente completo e coerente. Vengono percorsi i passaggi principali, attraverso i protagonisti chiave (Copernico, Tycho Brahe, Keplero, Galileo, Newton), sia astronomi osservatori sia puramente speculativi, le loro scoperte e i loro tentativi di interpretazione, cercando di mettere in luce alcuni snodi che sono particolarmente interessanti dal punto di vista epistemologico. The Material Point Method for the simulation of water-related hazards and their interaction with critical structures Antonia Larese, Universita degli studi di Padova & Technical University of Munich In recent years, natural hazards involving large mass movements such as landslides, debris flows, and mud flows have been increasing their frequency and intensity as a consequence of climate change and other related factors. These phenomena often carry huge rocks and heavy materials that may, directly or indirectly, cause damage to our structures resulting in a relevant socio-economic impact. The numerical simulation of the above events still represents a big challenge mainly for two reasons: the need to deal with large strain regimes and the intrinsic multiphysics nature of such events. While the Finite Element Method (FEM) represents a recognized, well established and widely used technique in many engineering fields, unfortunately it shows some limitation when dealing with problems where large deformation occurs. In the last decades many possible alternatives have been proposed and developed to overcome this drawback, such as the use of the so called particle-based methods. Among these, the Material Point Method (MPM) blends the advantages of both mesh-based and mesh-less methods. MPM avoids the problems of mesh tangling while preserving the accuracy of Lagrangian FEM and it is especially suited for non linear problems in solid mechanics and fluid dynamics. The talk will show some recent advances in MPM formulations , presenting both an irreducible and mixed formulation stabilized using variational multiscale techniques, as well as the partitioned strategies to couple MPM with other techniques such as FEM or DEM [2, 3]. All algorithms are implemented within the Kratos-Multiphysics open-source framework and available under the BSD license. Iaconeta, I., Larese, A., Rossi, R. and Guo, Z., Comparison of a Material Point Method and a Meshfree Galerkin Method for the simulation of cohesive-frictional materials, Materials, 10 , 1150, (2017). Chandra, B., Singer, V., Teschemacher, T., Wuechner, R. and Larese, A., Nonconforming Dirichlet boundary conditions in Implicit Material Point Method by means of penalty augmentation, Acta Geotechnica, 16(8), 2315-2335 (2021). Singer, V., Sautter, K.B., Larese, A., Wuchner, R. and Bletzinger, K.U.,, A Partitioned Material Point Method and Discrete Element Method Coupling Scheme , Under revision in Advanced Modeling and Simulation in Engineering Sciences (2022). Long-time behavior for local and nonlocal porous medium equations with small initial energy Bruno Volzone, Università degli Studi di Napoli 'Parthenope' In the first part of the talk, we will describe some aspects of a study developed in a joint paper with L. Brasco concerning the long-time behavior for the solution of the Porous Medium Equation in an open bounded connected set, with smooth boundary and sign-changing initial datum. Homogeneous Dirichlet boundary conditions are considered. We prove that if the initial datum has sufficiently small energy, then the solution converges to a nontrivial constant-sign solution of a sublinear Lane-Emden equation, once suitably rescaled. We also give a sufficient energetic criterion on the initial datum, which permits to decide whether convergence takes place towards the positive solution or to the negative one. The second part of the talk will be devoted to some new advances obtained in collaboration with G. Franzina, in the spirit of the ones explained above, for the study of the asymptotics of signed solutions for the Fractional Porous Medium Equation. Optimal design of planar shapes with active materials Active materials (e.g., polymer gels, liquid crystal elastomers) have emerged as suitable candidates for shape morphing applications, where the configuration of a body is varied in a controlled fashion upon triggering the active response. Given the large variety of these materials, a natural question is to compare different morphing mechanisms for a desired functional shape change and select the most effective one with respect to a certain optimality criterion. To address such a question, we set an optimal control problem that allows to determine the active strains suitable to attain a desired equilibrium transformation, while minimizing the complexity of the activation. Specifically, we discuss the planar morphing of active, hyperelastic bodies in the plain-strain regime, in the absence of external forces. Our approach aims to be general enough to account for a broad set of active materials through the notion of target metric. For the case of affine shape changes, we derive explicit conditions on the geometry of the reference configuration for the optimality of homogeneous target metrics. More complex shape changes are then analyzed via numerical simulations. We explore the impact on optimal solutions of different objective functionals, some of them inspired by features of existing active materials. Further, we show how stresses arising from incompatibilities contribute to reduce the complexity of the controls. We believe that our approach may be exploited for the accurate design of active systems and may also contribute to gather insight into the morphing strategies adopted by biological systems, as a result of natural selection. “A cosa serve la matematica?” Qual è quel docente che non si è mai sentito rivolgere questa domanda? Spesso è una domanda provocatoria, posta dallo studente più "simpatico" della classe. Ma a volte è una domanda sincera, posta da uno sconosciuto in treno. Solitamente siamo troppo stanchi o demoralizzati per rispondere a questa domanda. Ovviamente la matematica è utilissima, nel mondo di oggi, pieno di dati e in cui ogni aspetto della vita è intrecciato ad applicazioni, banali o profonde della matematica. Forse però vale la pena di soffermarsi un attimo in più su tale domanda e cercare di sviscerarla meglio... Cosa vuol dire "a cosa serve la matematica?"? La matematica pura è utile? Perché i politici devono sapere la matematica? A cosa serve a me la matematica? L'utilità della matematica è interessante per chi la studia? Nel seminario più che fornire le mie personali risposte proverò a dare un po' di materiale per pensarci su. Spero possa essere utile per avere qualche idea in più la prossima volta che sentirete queste o altre domande... Nick Trefethen is Professor of Numerical Analysis and head of the Numerical Analysis Group at Oxford University. He was educated at Harvard and Stanford and held positions at NYU, MIT, and Cornell before moving to Oxford in 1997. He is a Fellow of the Royal Society and a member of the US National Academy of Engineering, and served during 2011-2012 as President of SIAM. He has won many prizes including the Gold Medal of the Institute for Mathematics and its Applications, the Naylor Prize of the London Mathematical Society, and the Polya and von Neumann Prizes from SIAM. He holds honorary doctorates from the University of Fribourg and Stellenbosch University. As an author Trefethen is known for his books including Numerical Linear Algebra (1997), Spectral Methods in MATLAB (2000), Spectra and Pseudospectra (2005), Approximation Theory and Approximation Practice (2013/2019), Exploring ODEs (2018), and An Applied Mathematician's Apology (2022). He organized the SIAM 100-Dollar, 100-Digit Challenge in 2002 and is the inventor of Chebfun. Applications of AAA rational approximation Nick Trefethen, University of Oxford giovedì 23 febbraio 2023 alle ore 14:00 Aula Consiglio VII piano - Dipartimento di Matematica For the first time, a method has recently become available for fast computation of near-best rational approximations on arbitrary sets in the real line or complex plane: the AAA algorithm (Nakatsukasa-Sete-T. 2018). We will present the algorithm and then demonstrate a number of applications, including * detection of singularities * model order reduction * analytic continuation * functions of matrices * nonlinear eigenvalue problems * interpolation of equispaced data * smooth extension of multivariate real functions * extrapolation of ODE and PDE solutions into the complex plane * solution of Laplace problems * conformal mapping * Wiener-Hopf factorization The talk investigates the relation between normalized critical points of the nonlinear Schrödinger energy functional and critical points of the corresponding action functional on the associated Nehari manifold. First, we show that the ground state levels are strongly related by the following duality result: the (negative) energy ground state level is the Legendre–Fenchel transform of the action ground state level. Furthermore, whenever an energy ground state exists at a certain frequency, then all action ground states with that frequency have the same mass and are energy ground states too. We see that the converse is in general false and that the action ground state level may fail to be convex. Next we analyze the differentiability of the ground state action level and we provide an explicit expression involving the mass of action ground states. Finally we show that similar results hold also for local minimizers, and we exhibit examples of domains where our results apply. This is a joint work with Enrico Serra and Paolo Tilli. Recent results for the Navier-Stokes-Cahn-Hilliard model with unmatched densities We consider the initial-boundary value problem for the incompressible Navier-Stokes-Cahn-Hilliard system with non-constant density proposed by Abels, Garcke and Grün in 2012. This model arises in the diffuse interface theory for binary mixtures of viscous incompressible fluids. In particular, this system is a generalization of the well-known Model H in the case of fluids with unmatched densities. In this talk, I will present some recent results concerning the propagation of regularity of global weak solutions (for which uniqueness is not known) and their longtime convergence towards an equilibrium state in three dimensional bounded domains. Paradossi dei sistemi elettorali Orazio Puglisi , Dipartimento di Matematica e Informatica "U. Dini", Università di Firenze L’idea di “democrazia” che è ormai ben radicata dentro ciascuno di noi, spesso ci porta a considerare la questione dei sistemi elettorali (ovvero metodi per ottenere, a partire dalle preferenze degli individui di una certa popolazione, una lista di preferenze unica) con una certa leggerezza. Invece, appena si approfondisce la teoria dei sistemi di voto, ci si imbatte immediatamente in situazioni paradossali e problemi imprevisti, quasi sempre di difficile soluzione. Inizieremo questa conferenza discutendo alcuni dei principali problemi e paradossi che si presentano nella progettazione di un sistema elettorale. Nella parte finale ci occuperemo di un problema di diversa natura, ma di notevole importanza, ovvero quello della definizione “imparziale” dei collegi elettorali. Spectral analysis of Kohn Laplacian on spherical manifolds In this talk, we discuss the spectral analysis of Kohn Laplacian on spheres and the quotients of spheres. In particular, we obtain an analog of Weyl’s law for the Kohn Laplacian on lens spaces. We also show that two 3-dimensional lens spaces with fundamental groups of equal prime order are isospectral with respect to the Kohn Laplacian if and only if they are CR isometric. The Grassmannian is a smooth moduli space with very rich geometry that parameterizes simple varieties, namely, linear spaces. One can study a natural generalization, the component of a Hilbert scheme that parameterizes a pair of linear spaces in P^n. In this talk, I will describe the geometry of this component and show that they are smooth Mori dream spaces. Along the way, we will obtain a complete classification of the degenerations of a pair of linear spaces. Uncertainty Quantification for spatially-extended neurobiological networks This talk presents a framework for forward uncertainty quantification problems in spatially-extended neurobiological networks. We will consider networks in which the cortex is represented as a continuum domain, and local neuronal activity evolves according to an integro-differential equation, collecting inputs nonlocally, from the whole cortex. These models are sometimes referred to as neural field equations. Large-scale brain simulations of such models are currently performed heuristically, and the numerical analysis of these problems is largely unexplored. In the first part of the talk I will summarise recent developments for the rigorous numerical analysis of projection schemes for deterministic neural fields, which sets the foundation for developing Finite-Element and Spectral schemes for large-scale problems. The second part of the talk will discuss the case of systems in the presence of uncertainties modelled with random data, in particular: random synaptic connections, external stimuli, neuronal firing rates, and initial conditions (and any combination thereof). Such problems give rise to random solutions, whose mean, variance, or other quantities of interest have to be estimated using numerical simulations. This so-called forward uncertainty quantification problem is challenging because it couples spatially nonlocal, nonlinear problems to large-dimensional random data. I will present a family of schemes that couple a spatial projector for the spatial discretisation, to stochastic collocation for the random data. We will analyse the time-dependent problem with random data and the schemes from a functional analytic viewpoint, and show that the proposed methods can achieve spectral accuracy, provided the random data is sufficiently regular. Acknowledgements. This talk presents joint work with Francesca Cavallini (VU Amsterdam), Svetlana Dubinkina (VU Amsterdam), and Gabriel Lord (Radboud University). Avitabile D, Projection methods for Neural Field Equations arXiv e-prints, This paper applies a recently developed sentiment proxy to the construction of a new risk factor and provides a comprehensive understanding of its role in sentiment-augmented asset pricing models. We find that news and social media search-based indicators are significantly related to excess returns of international equity indices. Adding sentiment factors to both classical and more recent linear factor pricing models leads to a significant increase in their performance. When it is estimated using the Fama-MacBeth procedure, our sentiment-adjusted pricing model implies positive (negative) estimates of the risk premium for positive (negative) sentiment factors. We further differentiate between developed and emerging markets and uncover different patterns of return reversals / persistence in the long-term. Our results contribute to the explanation of global cross-sectional average excess returns and are robust to augmenting the model with fundamental factors, momentum, idiosyncratic volatility, skewness, kurtosis, and the returns on international currencies. When compared to competing definitions of sentiment factors popular in the literature, our novel sentiment risk variable turns out to be superior in terms of predictive power.
3 edition of Algebraic structure theory of sequential machines found in the catalog. Algebraic structure theory of sequential machines Bibliography: p. 206-208. \|Statement\|\|[by] J. Hartmanis [and] R.E. Stearns.\| \|Series\|\|Prentice-Hall international series in applied mathematics\| \|Contributions\|\|Stearns, R. E., joint author.\| \|LC Classifications\|\|QA267.5.S4 H3\| \|The Physical Object\| \|Pagination\|\|viii, 211 p.\| \|Number of Pages\|\|211\| \|LC Control Number\|\|66014360\| The present book gives an exposition of the classical basic algebraic and analytic number theory and supersedes my Algebraic Numbers, including much more material, e. g. the class field theory on which 1 make further comments at the appropriate place later. For different points of view, the reader is encouraged to read the collec tion of papers from the Brighton Symposium (edited by Cassels 2/5(1). Fixed point calculus is about the solution of recursive equations defined by a monotonic endofunction on a partially ordered set. This tutorial presents the basic theory of fixed point calculus together with a number of applications of direct relevance to the construction of computer programs. An algebraic number ﬁeld is a ﬁnite extension of Q; an algebraic number is an element of an algebraic number ﬁeld. Algebraic number theory studies the arithmetic of algebraic number ﬁelds — the ring of integers in the number ﬁeld, the ideals and units in the ring of integers, the extent to which unique factorization holds, and so on. Algebraic Number These notes are concerned with algebraic number theory, and the sequel with class field theory. Topics covered includes: Preliminaries from Commutative Algebra, Rings of Integers, Dedekind Domains- Factorization, The Unit Theorem, Cyclotomic Extensions- Fermat’s Last Theorem, Absolute Values- Local Fieldsand Global Fields. Check out the new look and enjoy easier access to your favorite features. I would recommend Stewart and Tall's Algebraic Number Theory and Fermat's Last Theorem for an introduction with minimal prerequisites. For example you don't need to know any module theory at all and all that is needed is a basic abstract algebra course (assuming it covers some ring and field theory). Kars and our captivity in Russia Mental health medicaid services Sea-Wolf and Selected Stories NASCAR Race Time (Sticker Fun) Civil rights acts of 1957, 1960, 1964, 1968 Years work in critical and cultural theory. Strip mining in Pennsylvania. First report from the Transport Committee session 1984-85 Autisms hidden blessings Algebraic structure theory of sequential machines (Prentice-Hall international series in applied mathematics) by Juris Hartmanis (Author) out of 5 stars 2 ratings. ISBN. This bar-code number lets you verify that you're getting exactly the right version or edition of a book. The digit and digit formats both work. 4/4(1). COVID Resources. Reliable information about the coronavirus (COVID) is available from the World Health Organization (current situation, international travel).Numerous and frequently-updated resource results are available from this ’s WebJunction has pulled together information and resources to assist library staff as they consider how to handle coronavirus. Santos E Algebraic structure theory of stochastic machines Proceedings of the third annual ACM symposium on Theory of computing, () Singh S () On Delayed-Input Asynchronous Sequential Circuits, IEEE Transactions on Computers. Additional Physical Format: Online version: Hartmanis, Juris. Algebraic structure theory of sequential machines. Englewood Cliffs, N.J., Prentice-Hall . Find helpful customer reviews and review ratings for Algebraic structure theory of sequential machines (Prentice-Hall international series in applied mathematics) at Read honest and unbiased product reviews from our users.4/5. Web of Science You must be logged in with an active subscription to view : Albert A. Mullin. In mathematics and computer science, the Krohn–Rhodes theory (or algebraic automata theory) is an approach to the study of finite semigroups and automata that seeks to decompose them in terms of elementary components. These components correspond to finite aperiodic semigroups and finite simple groups that are combined together in a feedback-free manner (called a "wreath product" or "cascade"). Book Review Algebraic Structure Theory of Sequential Machines. By J. HARTM~N:S AND R. 7D'~. STEARNS. From throughJ. Hartmanis and R. Stearns published a series of papers expounding some fundamental ideas on the coding of internal states of. Computational Mathematic Algebraic Structure Structure Theory Stochastic Machine These keywords were added by machine and not by the authors. This process is experimental and the keywords may be updated as the learning algorithm : Eugene S. Santos. Informally in mathematical logic, an algebraic theory is one that uses axioms stated entirely in terms of equations between terms with free lities and quantifiers are specifically disallowed. Sentential logic is the subset of first-order logic involving only algebraic sentences. The notion is very close to the notion of algebraic structure, which, arguably, may be just a synonym. Theory of Machines and Computations restrictions on the definition of linear realizability would result in the existence of not linearly realizable finite sequential machines. Several results are given relating these groups to the algebraic structure of A and, in the case when A is the reduced acceptor Ac(A*). Full text of "Sequential Machines And Automata Theory" See other formats. A 'read' is counted each time someone views a publication summary (such as the title, abstract, and list of authors), clicks on a figure, or views or downloads the full-text. Linear sequential machines can sometimes be decomposed into parallel or series connections of smaller linear sequential machines. Necessary and sufficient conditions for the existence of such decompositions are given for finite linear sequential machines with and without by: 6. Algebraic Structures Abstract algebra is the study of algebraic structures. Such a structure consists of a set together with one or more binary operations, which are required to satisfy certain axioms. For example, here is the de nition of a simple algebraic structure known as a group: De nition: GroupFile Size: KB. such extension can be represented as all polynomials in an algebraic number α: K = Q(α) = (Xm n=0 anα n: a n ∈ Q). Here α is a root of a polynomial with coeﬃcients in Q. Algebraic number theory involves using techniques from (mostly commutative) algebra and ﬁnite group theory to gain a deeper understanding of number Size: KB. Semihypergroup Theory is the first book devoted to the semihypergroup theory and it includes basic results concerning semigroup theory and algebraic hyperstructures, which represent the most general algebraic context in which reality can be modelled. Hyperstructures represent a natural extension of classical algebraic structures and they were introduced in by the French mathematician Marty. Hartmanis, J. and Stearns, R.E., Algebraic Structure Theory of Sequential Machines Rosenkrantz, D.J. and Stearns, R.E., Compiler Design Theory, Addison Wesley Publishing Co., Aumann, Robert J. and Maschler, Michael B. with the collaboration of Stearns, Richard E., Repeated Games with Incomplete Information, MIT Press, This book. An algebraic structure is a set (called carrier set or underlying set) with one or more finitary operations defined on it that satisfies a list of axioms. Examples of algebraic structures include Book: Introduction to Algebraic Structures (Denton) - Mathematics LibreTexts. ring theory, we study factorisation in integral domains, and apply it to the con-struction of ﬁelds; in group theory we prove Cayley’s Theorem and look at some small groups. The set text for the course is my own book Introduction to Algebra, Ox-ford University Press. I File Size: KB. As background an undergraduate level of modern applied algebra will suffice (Birkhoff-Bartee Modern Applied Algebra, Hartmanis-Stearns Algebraic Structure of Sequential Machines). Essential concepts and their engineering interpretation are introduced in a practical fashion with examples.Review of 'Computation: Finite and Infinite Machines' (Minsky, Marvin; ) Article (PDF Available) in IEEE Transactions on Information Theory 14(2) April with 2, ReadsAuthor: Michael A Arbib.Algebraic number theory involves using techniques from (mostly commutative) algebra and nite group theory to gain a deeper understanding of the arithmetic of number elds and related objects (e.g., functions elds, elliptic curves, etc.). The main objects that we study in .
The characteristic impedance or surge impedance (usually written Z0) of a uniform transmission line is the ratio of the amplitudes of voltage and current of a single wave propagating along the line; that is, a wave travelling in one direction in the absence of reflections in the other direction. Alternatively and equivalently it can be defined as the input impedance of a transmission line when its length is infinite. Characteristic impedance is determined by the geometry and materials of the transmission line and, for a uniform line, is not dependent on its length. The SI unit of characteristic impedance is the ohm. In radio-frequency engineering, a transmission line is a specialized cable or other structure designed to conduct alternating current of radio frequency, that is, currents with a frequency high enough that their wave nature must be taken into account. Transmission lines are used for purposes such as connecting radio transmitters and receivers with their antennas, distributing cable television signals, trunklines routing calls between telephone switching centres, computer network connections and high speed computer data buses. A signal travelling along an electrical transmission line will be partly, or wholly, reflected back in the opposite direction when the travelling signal encounters a discontinuity in the characteristic impedance of the line, or if the far end of the line is not terminated in its characteristic impedance. This can happen, for instance, if two lengths of dissimilar transmission lines are joined together. The input impedance of an electrical network is the measure of the opposition to current (impedance), both static (resistance) and dynamic (reactance), into the load network that is external to the electrical source. The input admittance (1/impedance) is a measure of the load's propensity to draw current. The source network is the portion of the network that transmits power, and the load network is the portion of the network that consumes power. The characteristic impedance of a lossless transmission line is purely real, with no reactive component. Energy supplied by a source at one end of such a line is transmitted through the line without being dissipated in the line itself. A transmission line of finite length (lossless or lossy) that is terminated at one end with an impedance equal to the characteristic impedance appears to the source like an infinitely long transmission line and produces no reflections. In electrical and electronic systems, reactance is the opposition of a circuit element to a change in current or voltage, due to that element's inductance or capacitance. The notion of reactance is similar to electrical resistance, but it differs in several respects. The characteristic impedance of an infinite transmission line at a given angular frequency is the ratio of the voltage and current of a pure sinusoidal wave of the same frequency travelling along the line. This definition extends to DC by letting tend to 0, and subsists for finite transmission lines until the wave reaches the end of the line. In this case, there will be in general a reflected wave which travels back along the line in the opposite direction. When this wave reaches the source, it adds to the transmitted wave and the ratio of the voltage and current at the input to the line will no longer be the characteristic impedance. This new ratio is called the input impedance. The input impedance of an infinite line is equal to the characteristic impedance since the transmitted wave is never reflected back from the end. It can be shown that an equivalent definition is: the characteristic impedance of a line is that impedance which, when terminating an arbitrary length of line at its output, produces an input impedance of equal value. This is so because there is no reflection on a line terminated in its own characteristic impedance. Applying the transmission line model based on the telegrapher's equations as derived below, the general expression for the characteristic impedance of a transmission line is: The telegrapher's equations are a pair of coupled, linear differential equations that describe the voltage and current on an electrical transmission line with distance and time. The equations come from Oliver Heaviside who in the 1880s developed the transmission line model. The model demonstrates that the electromagnetic waves can be reflected on the wire, and that wave patterns can appear along the line. The theory applies to transmission lines of all frequencies including high-frequency transmission lines, audio frequency, low frequency and direct current. Although an infinite line is assumed, since all quantities are per unit length, the characteristic impedance is independent of the length of the transmission line. The voltage and current phasors on the line are related by the characteristic impedance as: where the superscripts and represent forward- and backward-traveling waves, respectively. A surge of energy on a finite transmission line will see an impedance of Z0 prior to any reflections arriving, hence surge impedance is an alternative name for characteristic impedance. The differential equations describing the dependence of the voltage and current on time and space are linear, so that a linear combination of solutions is again a solution. This means that we can consider solutions with a time dependence ejωt, and the time dependence will factor out, leaving an ordinary differential equation for the coefficients, which will be phasors depending on space only. Moreover, the parameters can be generalized to be frequency-dependent. Voltage, electric potential difference, electric pressure or electric tension is the difference in electric potential between two points. The difference in electric potential between two points in a static electric field is defined as the work needed per unit of charge to move a test charge between the two points. In the International System of Units, the derived unit for voltage is named volt. In SI units, work per unit charge is expressed as joules per coulomb, where 1 volt = 1 joule per 1 coulomb. The official SI definition for volt uses power and current, where 1 volt = 1 watt per 1 ampere. This definition is equivalent to the more commonly used 'joules per coulomb'. Voltage or electric potential difference is denoted symbolically by ∆V, but more often simply as V, for instance in the context of Ohm's or Kirchhoff's circuit laws. An electric current is a flow of electric charge. In electric circuits this charge is often carried by electrons moving through a wire. It can also be carried by ions in an electrolyte, or by both ions and electrons such as in an ionized gas (plasma). In physics and engineering, a phasor, is a complex number representing a sinusoidal function whose amplitude (A), angular frequency (ω), and initial phase (θ) are time-invariant. It is related to a more general concept called analytic representation, which decomposes a sinusoid into the product of a complex constant and a factor that encapsulates the frequency and time dependence. The complex constant, which encapsulates amplitude and phase dependence, is known as phasor, complex amplitude, and sinor or even complexor. The positive directions of V and I are in a loop of clockwise direction. We find that These first-order equations are easily uncoupled by a second differentiation, with the results: Both V and I satisfy the same equation. Since ZY is independent of z and t, it can be represented by a constant -k2. The minus sign is included so that k will appear as ±jkz in the exponential solutions of the equation. In fact, where V+ and V- are the constant of integration, The above equation will be the wave solution for V, and from the first-order equation. If lumped circuit analysis has to be valid at all frequencies, the length of the sub section must tend to Zero. Substituting the value of V in the above equation, we get. Co-efficient of : Co-efficient of : It can be seen that, the above equations has the dimensions of Impedance (Ratio of Voltage to Current) and is a function of primary constants of the line and operating frequency. It is therefore called the “Characteristic Impedance” of the transmission line , often denoted by . We follow an approach posted by Tim Healy and shunt elements (as shown in the figure above). The characteristic impedance is defined as the ratio of the input voltage to the input current of a semi-infinite length of line. We call this impedance . That is, the impedance looking into the line on the left is . But, of course, if we go down the line one differential length dx, the impedance into the line is still . Hence we can say that the impedance looking into the line on the far left is equal to in parallel with and , all of which is in series with and . Hence:. The line is modeled by a series of differential segments with differential series The term above containing two factors of may be discarded, since it is infinitesimal in comparison to the other terms, leading to: and hence to Reversing the sign of the square root may be regarded as changing the direction of the current. The analysis of lossless lines provides an accurate approximation for real transmission lines that simplifies the mathematics considered in modeling transmission lines. A lossless line is defined as a transmission line that has no line resistance and no dielectric loss. This would imply that the conductors act like perfect conductors and the dielectric acts like a perfect dielectric. For a lossless line, R and G are both zero, so the equation for characteristic impedance derived above reduces to: In particular, does not depend any more upon the frequency. The above expression is wholly real, since the imaginary term j has canceled out, implying that Z0 is purely resistive. For a lossless line terminated in Z0, there is no loss of current across the line, and so the voltage remains the same along the line. The lossless line model is a useful approximation for many practical cases, such as low-loss transmission lines and transmission lines with high frequency. For both of these cases, R and G are much smaller than ωL and ωC, respectively, and can thus be ignored. The solutions to the long line transmission equations include incident and reflected portions of the voltage and current: When the line is terminated with its characteristic impedance, the reflected portions of these equations are reduced to 0 and the solutions to the voltage and current along the transmission line are wholly incident. Without a reflection of the wave, the load that is being supplied by the line effectively blends into the line making it appear to be an infinite line. In a lossless line this implies that the voltage and current remain the same everywhere along the transmission line. Their magnitudes remain constant along the length of the line and are only rotated by a phase angle. In electric power transmission, the characteristic impedance of a transmission line is expressed in terms of the surge impedance loading (SIL), or natural loading, being the power loading at which reactive power is neither produced nor absorbed: in which is the line-to-line voltage in volts. Loaded below its SIL, a line supplies reactive power to the system, tending to raise system voltages. Above it, the line absorbs reactive power, tending to depress the voltage. The Ferranti effect describes the voltage gain towards the remote end of a very lightly loaded (or open ended) transmission line. Underground cables normally have a very low characteristic impedance, resulting in an SIL that is typically in excess of the thermal limit of the cable. Hence a cable is almost always a source of reactive power. \|Ethernet Cat.5\|\|100\|\|±5 Ω\| The characteristic impedance of coaxial cables (coax) is commonly chosen to be 50 Ω for RF and microwave applications. Coax for video applications is usually 75 Ω for its lower loss. The propagation constant of a sinusoidal electromagnetic wave is a measure of the change undergone by the amplitude and phase of the wave as it propagates in a given direction. The quantity being measured can be the voltage, the current in a circuit, or a field vector such as electric field strength or flux density. The propagation constant itself measures the change per unit length, but it is otherwise dimensionless. In the context of two-port networks and their cascades, propagation constant measures the change undergone by the source quantity as it propagates from one port to the next. Electrical impedance is the measure of the opposition that a circuit presents to a current when a voltage is applied. The term complex impedance may be used interchangeably. In mathematics, Green's theorem gives the relationship between a line integral around a simple closed curve C and a double integral over the plane region D bounded by C. It is named after George Green, though its first proof is due to Bernhard Riemann and is the two-dimensional special case of the more general Kelvin–Stokes theorem. In electronics, impedance matching is the practice of designing the input impedance of an electrical load or the output impedance of its corresponding signal source to maximize the power transfer or minimize signal reflection from the load. The Smith chart, invented by Phillip H. Smith (1905–1987), is a graphical aid or nomogram designed for electrical and electronics engineers specializing in radio frequency (RF) engineering to assist in solving problems with transmission lines and matching circuits. The Smith chart can be used to simultaneously display multiple parameters including impedances, admittances, reflection coefficients, scattering parameters, noise figure circles, constant gain contours and regions for unconditional stability, including mechanical vibrations analysis. The Smith chart is most frequently used at or within the unity radius region. However, the remainder is still mathematically relevant, being used, for example, in oscillator design and stability analysis. Acoustic impedance and specific acoustic impedance are measures of the opposition that a system presents to the acoustic flow resulting from an acoustic pressure applied to the system. The SI unit of acoustic impedance is the pascal second per cubic metre or the rayl per square metre, while that of specific acoustic impedance is the pascal second per metre or the rayl. In this article the symbol rayl denotes the MKS rayl. There is a close analogy with electrical impedance, which measures the opposition that a system presents to the electrical flow resulting from an electrical voltage applied to the system. In mathematics, the Hamilton–Jacobi equation (HJE) is a necessary condition describing extremal geometry in generalizations of problems from the calculus of variations, and is a special case of the Hamilton–Jacobi–Bellman equation. It is named for William Rowan Hamilton and Carl Gustav Jacob Jacobi. The Heaviside condition, named for Oliver Heaviside (1850–1925), is the condition an electrical transmission line must meet in order for there to be no distortion of a transmitted signal. Also known as the distortionless condition, it can be used to improve the performance of a transmission line by adding loading to the cable. In mathematics, a metric connection is a connection in a vector bundle E equipped with a bundle metric; that is, a metric for which the inner product of any two vectors will remain the same when those vectors are parallel transported along any curve. Other common equivalent formulations of a metric connection include: The Π pad is a specific type of attenuator circuit in electronics whereby the topology of the circuit is formed in the shape of the Greek letter "Π". Image impedance is a concept used in electronic network design and analysis and most especially in filter design. The term image impedance applies to the impedance seen looking into a port of a network. Usually a two-port network is implied but the concept can be extended to networks with more than two ports. The definition of image impedance for a two-port network is the impedance, Zi 1, seen looking into port 1 when port 2 is terminated with the image impedance, Zi 2, for port 2. In general, the image impedances of ports 1 and 2 will not be equal unless the network is symmetrical with respect to the ports. Filters designed using the image impedance methodology suffer from a peculiar flaw in the theory. The predicted characteristics of the filter are calculated assuming that the filter is terminated with its own image impedances at each end. This will not usually be the case; the filter will be terminated with fixed resistances. This causes the filter response to deviate from the theoretical. This article explains how the effects of image filter end terminations can be taken into account. The primary line constants are parameters that describe the characteristics of conductive transmission lines, such as pairs of copper wires, in terms of the physical electrical properties of the line. The primary line constants are only relevant to transmission lines and are to be contrasted with the secondary line constants, which can be derived from them, and are more generally applicable. The secondary line constants can be used, for instance, to compare the characteristics of a waveguide to a copper line, whereas the primary constants have no meaning for a waveguide. Metal-mesh optical filters are optical filters made from stacks of metal meshes and dielectric. They are used as part of an optical path to filter the incoming light to allow frequencies of interest to pass while reflecting other frequencies of light. Characteristic admittance is the mathematical inverse of the characteristic impedance. The general expression for the characteristic admittance of a transmission line is: Laser theory of Fabry-Perot (FP) semiconductor lasers proves to be nonlinear, since the gain, the refractive index and the loss coefficient are the functions of energy flux. The nonlinear theory made it possible to explain a number of experiments some of which could not even be explained, much less modeled, on the basis of other theoretical models; this suggests that the nonlinear theory developed is a new paradigm of the laser theory. In mathematics, the exponential response formula (ERF), also known as exponential response and complex replacement, is a method used to find a particular solution of a non-homogeneous linear ordinary differential equation of any order. The exponential response formula is applicable to non-homogeneous linear ordinary differential equations with constant coefficients if the function is polynomial, sinusoidal, exponential or the combination of the three. The general solution of a non-homogeneous linear ordinary differential equation is a superposition of the general solution of the associated homogeneous ODE and a particular solution to the non-homogeneous ODE. Alternative methods for solving ordinary differential equations of higher order are method of undetermined coefficients and method of variation of parameters.
An article for teachers and pupils that encourages you to look at the mathematical properties of similar games. A game for 2 players with similaritlies to NIM. Place one counter on each spot on the games board. Players take it is turns to remove 1 or 2 adjacent counters. The winner picks up the last counter. A collection of games on the NIM theme 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 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. 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. Try entering different sets of numbers in the number pyramids. How does the total at the top change? Pick a square within a multiplication square and add the numbers on each diagonal. What do you notice? 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. 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. 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 Can you describe this route to infinity? Where will the arrows take you next? When number pyramids have a sequence on the bottom layer, some interesting patterns emerge... Can you work out how to win this game of Nim? Does it matter if you go first or second? 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 tangle yourself up and reach any fraction? A game for two people, or play online. Given a target number, say 23, and a range of numbers to choose from, say 1-4, players take it in turns to add to the running total to hit their target. Many numbers can be expressed as the sum of two or more consecutive integers. For example, 15=7+8 and 10=1+2+3+4. Can you say which numbers can be expressed in this way? Think of a number, add one, double it, take away 3, add the number you first thought of, add 7, divide by 3 and take away the number you first thought of. You should now be left with 2. How do I. . . . Take a look at the multiplication square. The first eleven triangle numbers have been identified. Can you see a pattern? Does the 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? The NRICH team are always looking for new ways to engage teachers and pupils in problem solving. Here we explain the thinking behind It would be nice to have a strategy for disentangling any tangled Can you find the values at the vertices when you know the values on Charlie has moved between countries and the average income of both has increased. How can this be so? How could Penny, Tom and Matthew work out how many chocolates there are in different sized boxes? Can you find sets of sloping lines that enclose a square? What would you get if you continued this sequence of fraction sums? 1/2 + 2/1 = 2/3 + 3/2 = 3/4 + 4/3 = Triangle numbers can be represented by a triangular array of squares. What do you notice about the sum of identical triangle It starts quite simple but great opportunities for number discoveries and patterns! The sum of the numbers 4 and 1 [1/3] is the same as the product of 4 and 1 [1/3]; that is to say 4 + 1 [1/3] = 4 × 1 [1/3]. What other numbers have the sum equal to the product and can this be so for. . . . List any 3 numbers. It is always possible to find a subset of adjacent numbers that add up to a multiple of 3. Can you explain why and prove it? What are the areas of these triangles? What do you notice? Can you generalise to other "families" of triangles? Do you notice anything about the solutions when you add and/or subtract consecutive negative numbers? How many moves does it take to swap over some red and blue frogs? Do you have a method? A little bit of algebra explains this 'magic'. Ask a friend to pick 3 consecutive numbers and to tell you a multiple of 3. Then ask them to add the four numbers and multiply by 67, and to tell you. . . . A package contains a set of resources designed to develop pupils’ mathematical thinking. This package places a particular emphasis on “generalising” and is designed to meet the. . . . Pick the number of times a week that you eat chocolate. This number must be more than one but less than ten. Multiply this number by 2. Add 5 (for Sunday). Multiply by 50... Can you explain why it. . . . Spotting patterns can be an important first step - explaining why it is appropriate to generalise is the next step, and often the most interesting and important. Square numbers can be represented as the sum of consecutive odd numbers. What is the sum of 1 + 3 + ..... + 149 + 151 + 153? 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 a large cube made from small red cubes being dropped into a pot of yellow paint. How many of the small cubes will have yellow paint on their faces? 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. Place the numbers from 1 to 9 in the squares below so that the difference between joined squares is odd. How many different ways can you do this? Use the animation to help you work out how many lines are needed to draw mystic roses of different sizes. A country has decided to have just two different coins, 3z and 5z coins. Which totals can be made? Is there a largest total that cannot be made? How do you know? 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? Imagine we have four bags containing numbers from a sequence. What numbers can we make now? Imagine we have four bags containing a large number of 1s, 4s, 7s and 10s. What numbers can we make?
This preview shows page 1. Sign up to view the full content. Unformatted text preview: Corporate Finance: The Core (Berk/DeMarzo) Chapter 7 - Fundamentals of Capital Budgeting 1) Which of the following statements is false? A) Because value is lost when a resource is used by another project, we should include the opportunity cost as an incremental cost of the project. B) Sunk costs are incremental with respect to the current decision regarding the project and should be included in its analysis. C) Overhead expenses are associated with activities that are not directly attributable to a single business activity but instead affect many different areas of the corporation. D) When computing the incremental earnings of an investment decision, we should include all changes between the firm’s earnings with the project versus without the project. 2) Which of the following costs would you consider when making a capital budgeting decision? A) Sunk cost B) Opportunity cost C) Interest expense D) Fixed overhead cost 3) A decrease in the sales of a current project because of the launching of a new project is A) cannibalization. B) a sunk cost. C) an overhead expense. D) irrelevant to the investment decision. 4) Money that has been or will be paid regardless of the decision whether or not to proceed with the project is A) cannibalization. B) considered as part of the initial investment in the project. C) an opportunity cost. D) a sunk cost. 5) The value of currently unused warehouse space that will be used as part of a new capital budgeting project is A) an opportunity cost. B) irrelevant to the investment decision. C) an overhead expense. D) a sunk cost. 6) Suppose that of the 60% of FFLʹs current overnight photo customers, half would start taking their film to a competitor that offers one hour photo processing if FFL fails to offer the one hour service. The level of incremental sales in this case is closest to: A) $60,000 B) $150,000 C) $90,000 D) $120,000 Use the information for the question(s) below. Glucose Scan Incorporated (GSI) currently sells its latest glucose monitor, the Glucoscan 3000, to diabetic patients for $129. GSI plans on lowering their price next year to $99 per unit. The cost of goods sold for each Glucoscan unit is $50, and GSI expects to sell 100,000 units over the next year. 7) Suppose that if GSI drops the price on the Glucoscan 3000 immediately, it can increase sales over the next year by 30% to 130,000 units. The incremental impact of this price drop on the firms EBIT is closest to: A) a decline of 1.5 million B) an increase of 1.5 million C) a decline of 2.4 million D) an increase of 2.4 million 8) Suppose that if GSI drops the price on the Glucoscan 3000 immediately, it can increase sales over the next year by 30% to 130,000 units. Also suppose that for each Glucoscan monitor sold, GSI expects additional sales of $100 per year on glucose testing strips and these strips have a gross profit margin of 75%. Considering the increase in the sale of testing strips, the incremental impact of this price drop on the firms EBIT is closest to: A) A decline of 1.5 million B) Adecline of 0.7 million C) An increase of 0.7 million D) An increase of 1.5 million Use the information for the question(s) below. The Sisyphean Corporation is considering investing in a new cane manufacturing machine that has an estimated life of three years. The cost of the machine is $30,000 and the machine will be depreciated straight line over its three-year life to a residual value of $0. The cane manufacturing machine will result in sales of 2,000 canes in year 1. Sales are estimated to grow by 10% per year each year through year three. The price per cane that Sisyphean will charge its customers is $18 each and is to remain constant. The canes have a cost per unit to manufacture of $9 each. Installation of the machine and the resulting increase in manufacturing capacity will require an increase in various net working capital accounts. It is estimated that the Sisyphean Corporation needs to hold 2% of its annual sales in cash, 4% of its annual sales in accounts receivable, 9% of its annual sales in inventory, and 6% of its annual sales in accounts payable. The firm is in the 35% tax bracket, and has a cost of capital of 10%. 9) The incremental EBIT in the first year for the Sisyphean Corporationʹs project is closest to: A) $18,000 B) $8,000 C) $11,700 D) $5,200 10) The incremental unlevered net income in the first year for the Sisyphean Corporationʹs project is closest to: A) $8,000 B) $18,000 C) $5,200 D) $11,700 11) The depreciation tax shield for the Sisyphean Corporationʹs project in the first year is closest to: A) $8,000 B) $3,500 C) $2,800 D) $5,200 12) Which of the following statements is false? A) Depreciation expenses have a positive impact on free cash flow. B) Free Cash Flow = (Revenues - Costs - Depreciation) × (1 - τc) - Capital Expenditures - ΔNWC + τc × Depreciation. C) The firm cannot use its earnings to buy goods, pay employees, fund new investments, or pay dividends to shareholders. D) The depreciation tax shield is the tax savings that results from the ability to deduct depreciation. 13) Your firm is considering building a new office complex. Your firm already owns land suitable for the new complex. The current book value of the land is $100,000, however a commercial real estate again has informed you that an outside buyer is interested in purchasing this land and would be willing to pay $650,000 for it. When calculating the NPV of your new office complex, ignoring taxes, the appropriate incremental cash flow for the use of this land is: A) $650,000 B) $0 C) $100,000 D) $750,000 14) The Sisyphean Company is considering a new project that will have an annual depreciation expense of $2.5 million. If Sisypheanʹs marginal corporate tax rate is 40% and their average corporate tax rate is 30%, then what is the value of the depreciation tax shield on their new project? A) $750,000 B) $1,000,000 C) $1,500,000 D) $1,750,000 15) Bubba Ho-Tep Company reported net income of $300 million for the most recent fiscal year. The firm had depreciation expenses of $125 million and capital expenditures of $150 million. Although they had no interest expense, the firm did have an increase in net working capital of $20 million. What is Bubba Ho-Tepʹs free cash flow? A) $170 million B) $255 million C) $150 million D) $5 million Use the information for the question(s) below. Temporary Housing Services Incorporated (THSI) is considering a project that involves setting up a temporary housing facility in an area recently damaged by a hurricane. THSI will lease space in this facility to various agencies and groups providing relief services to the area. THSI estimates that this project will initially cost $5 million to setup and will generate $20 million in revenues during its first and only year in operation (paid in one year). Operating expenses are expected to total $12 million during this year and depreciation expense will be another $3 million. THSI will require no working capital for this investment. THSIʹs marginal tax rate is 35%. 16) Ignoring the original investment of $5 million, what is THSIʹs free cash flow for the first and only year of operation? A) $5.0 million B) $3.75 million C) $8.0 million D) $6.25 million 17) Assume that THSIʹs cost of capital for this project is 15%. The NPV of this temporary housing project is closest to: A) $435,000 B) -$650,000 C) $1,960,000 D) -$435,000 Use the information for the question(s) below. Shepard Industries is evaluating a proposal to expand its current distribution facilities. Management has projected the project will produce the following cash flows for the first two years (in millions). Year 1 2 Revenues 1200 1400 Operating Expense 450 525 Depreciation 240 280 Increase in working capital 60 70 Capital expenditures 300 350 Marginal corporate tax rate 30% 30% 18) The incremental EBIT for Shepard Industries in year one is closest to: A) $360 B) $750 C) $595 D) $510 19) The incremental EBIT for Shepard Industries in year two is closest to: A) $415 B) $875 C) $595 D) $510 WS1) The incremental unlevered net income Shepard Industries in year one is closest to: A) $510 B) $415 C) $600 D) $355 Answer: D Explanation: A) B) C) D) Revenues 1200 1400 - Expenses 450 525 - Depreciation 240 280 = EBIT 510 595 - Taxes (30%) 153 178.5 357 416.5 Incremental Net Income 20) The incremental unlevered net income Shepard Industries in year two is closest to: A) $355 B) $415 C) $600 D) $510 WS2) The depreciation tax shield for Shepard Industries project in year one is closest to: A) $84 B) $168 C) $96 D) $72 Answer: D Explanation: A) B) C) D) $240 × .30 = $72 21) The depreciation tax shield for Shepard Industries project in year two is closest to: A) $84 B) $196 C) $72 D) $96 22) The free cash flow from Shepard Industries project in year one is closest to: A) $240 B) $300 C) -$5 D) $390 23) The free cash flow from Shepard Industries project in year two is closest to: A) $345 B) $455 C) $275 D) -$5 Use the information for the question(s) below. Epiphany Industries is considering a new capital budgeting project that will last for three years. Epiphany plans on using a cost of capital of 12% to evaluate this project. Based on extensive research, it has prepared the following incremental cash flow projects: Year Sales (Revenues) - Cost of Goods Sold (50% of Sales) - Depreciation = EBIT - Taxes (35%) = unlevered net income + Depreciation + changes to working capital - capital expenditures 0 1 100,000 50,000 30,000 20,000 7000 13,000 30,000 -5,000 2 100,000 50,000 30,000 20,000 7000 13,000 30,000 -5,000 3 100,000 50,000 30,000 20,000 7000 13,000 30,000 10,000 -90,000 WS3) The free cash flow for the first year of Epiphanyʹs project is closest to: A) $43,000 B) $25,000 C) $38,000 D) $45,000 Answer: C Explanation: A) B) 0 1 C) Year Sales (Revenues) - Cost of Goods Sold (50% of Sales) - Depreciation = EBIT -Taxes (35%) = unlevered net income + Depreciation + changes to working capital - capital expenditures = Free Cash Flow PV of FCF (FCF / (1 + I)n discount rate NPV = 11,946 IRR = 19.14% 100,000 50,000 30,000 20,000 7000 13,000 30,000 -5,000 -90,000 -90,000 -90,000 0.12 38,000 33,929 2 100,000 50,000 30,000 20,000 7000 13,000 30,000 -5,000 38,000 30,293 3 100,000 50,000 30,000 20,000 7000 13,000 30,000 10,000 53,000 37,724 D) 24) The free cash flow for the last year of Epiphanyʹs project is closest to: A) $53,000 B) $38,000 C) $35,000 D) $43,000 25) The NPV for Epiphanyʹs Project is closest to: A) $4,825 B) $39,000 C) $11,946 D) $20,400 26) Luther Industries has outstanding tax loss carryforwards of $70 million from losses over the past four years. If Luther earns $15 million per year in pre-tax income from now on, Luther first pay taxes in? A) 7 years. B) 2 years. C) 4 years. D) 5 years. ... View Full Document This note was uploaded on 12/28/2009 for the course FEWEB CORPFIN taught by Professor Dorsman during the Spring '09 term at Vrije Universiteit Amsterdam. - Spring '09
Pipes and Cisterns Exercise 2 You are here: Home1 / Maths2 / Pipes and Cisterns Exercise 13 / Pipes and Cisterns Exercise 2 Time limit: 0 0 of 18 questions completed - Most Important Multiple Choice Questions - Online Pipes and Cisterns Exercise with Correct Answer Key and Solutions - Useful for all Competitive Exams You have already completed the quiz before. Hence you can not start it again. Test is loading... You must sign in or sign up to start the quiz. You have to finish following quiz, to start this quiz: 0 of 18 questions answered correctly Time has elapsed You have reached 0 of 0 points, (0) - Not categorized 0% - Question 1 of 18 Two taps can fill a tank in 15 and 12 min, respectively. A third tap can empty it in 20 min. If all the taps are opened at the same time, then in how much time will the tank be filled? Part of tank filled in one minute in given condition ∴ Tank will be completely filled in 10 min. - Question 2 of 18 Two taps can fill a tank in 20 minutes and 30 minutes respectively. There is an outlet tap at exactly half level of that rectangular tank which can pump out 50 litres of water per minute. If the outlet tap is open, then it takes 24 minutes to fill an empty tank. What is the volume of the tank? The two filler tap can fill the part of tank in 1 min. ∴ The two filler tap can fill the tank in 12 min. ∴ Half of the tank will be filled in 6 min. Hence, it took (24 – 6 = 18 min.) to fill the remaining half of the tank when the outlet pump is opened. Thus, the total time required to empty half of the tank Thus, the capacity of the tank = 50 × 9 × 2 = 900 litres - Question 3 of 18 Two taps can separately fill a cistern in 10 minutes and 15 minutes, respectively and when the waste pipe is open, they can together fill it in 18 minutes. The waste pipe can empty the full cistern in : Time taken by one tap to fill the cistern = ¹⁄₁₀ hr and second tap fills the cistern = ¹⁄₁₅ hr The time taken by the both tap to fill the cistern Thus, both tap fill the cistern in 6 minutes. Now, given when waste pipe is open, both can fill the cistern in ¹⁄₁₈ hr Time taken by waste pipe to empty the cistern = ¹⁄₉ minutes Hence, in 9 minutes waste pipe can empty the cistern. - Question 4 of 18 One fill pipe A is 3 times faster than second fill pipe B and takes 10 minutes less time to fill a cistern than B takes. Find when the cistern will be full if fill pipe B is only opened. Let B can fill the cistern in x min. Then, then A can fill the cistern in - Question 5 of 18 Two pipes A and B can fill a tank in 15 and 12 hours respectively. Pipe B alone is kept open for ¾ of time and both pipes are kept open for remaining time. In how many hours, the tank will be full? Let the required time be x hours, then ⇒ x = 10 hours - Question 6 of 18 A water tank has three taps A, B and C. A fills four buckets in 24 minutes, B fills 8 buckets in 1 hour and C fills 2 buckets in 20 minutes. If all the taps are opened together a full tank is emptied in 2 hours. If a bucket can hold 5 litres of water, what is the capacity of the tank? Tap A fills 4 buckets (4 × 5 = 20 litres) in 24 min. In 1 hour tap A fills In 1 hour tap B fills = 8 × 5 = 40 In 1 hour tap C fills If they open together they would fill 50 + 40 + 30 = 120 litres in one hour but full tank is emptied in 2 hours So, tank capacity would be 120 × 2 = 240 litres. - Question 7 of 18 A pump can be operated both for filling a tank and for emptying it. The capacity of tank is 2400 m³. The emptying capacity of the pump is 10 m³ per minute higher than its filling capacity. Consequently, the pump needs 8 minutes less to empty the tank to fill it. Find the filling capacity of pump. Let the filling capacity of pump be x m³/min. Then, emptying capacity of pump = (x + 10) m³ /min. ⇒ x² + 10 x – 3000 = 0 ⇒ (x – 50)(x + 60) = 0 ⇒ x = 50 m³/min. - Question 8 of 18 Filling pipe, if opened alone, takes 5 minutes to fill a cistern. Suddenly, during the course of fillling, the waste pipe (which is of similar size and flow as of fill pipe) is opened for 2 minutes, then the cistern will be filled in Since, flow of waste pipe = flow of filling pipe. ⇒ Filled part in one min = emptied part in one min. ∴ After opening the waste pipe for 2 min, cistern will be full in (5 + 2) = 7 min. - Question 9 of 18 A cistern has three pipes, A, B and C. The pipes A and B can fill it in 4 and 5 hours respectively and C can empty it in 2 hours. If the pipes are opened in order at 1, 2 and 3 a.m. respectively, when will the cistern be empty? Let the time be t hours after 1 a.m. - Question 10 of 18 A pipe can fill a cistern in 6 hours. Due to a leak in its bottom, it is filled in 7 hours. When the cistern is full, in how much time will it be emptied by the leak? Part of the capacity of the cistern emptied by the leak in one hour = of the cistern. The whole cistern will be emptied in 42 hours. - Question 11 of 18 A tank is normally filled in 8 hrs but takes 2 hrs longer to fill because of a leak in its bottom. If the cistern is full, in how many hrs will the leak empty it? It is clear from the question that the filler pipe fills the tank in 8 hrs and if both the filler and the leak work together, the tank is filled in 8 hrs . Therefore the leak will empty the tank in = 40 hrs. - Question 12 of 18 An electric pump can fill a tank in 3 hours. Because of a leak in the tank it was taking 3.5 hours to fill the tank. Find the time in which the leak can drain all the water of the tank when full. Part of tank filled in 1 hour = 1/3 Part of tank emptied in the same time Total time required to empty it - Question 13 of 18 Two pipes A and B can fill a tank in 20 and 30 hours respectively. Both the pipes are opened to fill the tank but when the tank is 1/3rd full, a leak develops in the tank through which one-third water supplied by both the pipes goes out. The total time taken to fill the tank is The pipes A and B together can fill of the tank in one hour. ∴ ⅓ of the tank is filled by both the pipes A and B together in 4 hours. …(1) Now because of developing a leak after 4 hours, both the pipes can fill of the tank in one hour [ Because ⅓rd of the water supplied by both the pipes goes out] ∴ Remaining ⅔ of the tank can be filled by both the pipes in ⅔ × 18 = 12 hours … (2) ∴ The total time taken to fill the tank is 16 hours. - Question 14 of 18 Three pipes A, B and C can fill a tank from empty to full in 30 minutes, 20 minutes and 10 minutes respectively. When the tank is empty, all the three pipes are opened. A , B and C discharge chemical solutions P, Q and R respectively. What is the proportion of solution R in the liquid in the tank after 3 minutes? Part filled by (A + B + C) in 3 minutes Part filled by C in 3 minutes ∴ Required ratio - Question 15 of 18 A cistern can be filled by two pipes filling separately in 12 and 16 min. respectively. Both pipes are opened together for a certain time but being clogged, only 7/8 of the full quantity of water flows through the former and only 5/6 through the latter pipe. The obstructions, however, being suddenly removed, the cistern is filled in 3 min. from that moment. How long was it before the full flow began? Both the pipes A and B can fill of the cistern in one minute, when their is no obstruction. With obstruction, both the pipes can fill of the cistern in one minute. Let the obstructions were suddenly removed after x minutes. ∴ With obstruction, x/8 of the cistern could be filled in x minutes and so the remaining of the cistern was filled without obstruction in 3 minutes, i.e. In one minute, of the cistern was filled with obstruction. ⇒ x = 4.5 min. - Question 16 of 18 A volcanic crater (conical) has a base diameter 125 m and is 10 m deep. It rains very heavily and the crater gets filled up in 4 hours. Find the rate of water flow in the crater. Volume of the cone ⇒ Rate of water flow - Question 17 of 18 Water flows at 3 metres per sec through a pipe of radius 4 cm. How many hours will it take to fill a tank 40 metres long, 30 metres broad and 8 metres deep, if the pipe remains full? Radius of the pipe (r) = 4 cm. = 0.04 meter Volume of water flowing out per sec = πr² × rate of flow = cubic metres = 0.0151 cubic metres Time taken to fill the tank = 40 × 30 × sec - Question 18 of 18 An outlet pipe empties a tank which is full, in 10 hours. If the inlet pipe is kept open, which lets water in at the rate of 8 litres/minute, the outlet pipe would take 6 hours longer. Find the capacity of the tank. Part filled by the inlet pipe in 1 hour = = = Part filled by the inlet pipe in 1 minute ∴ Capacity of tank = 1600 × 8 = 12800 litres.
Exercises (Do not hand in) Chapter 4: 3,4 and 6, Chapter 5:1-3. Following are the problems to be handed in, 25 points each. Maximum score for this homework is 100 points. We will take your best four attempted problems. 1. (Divide and Conquer, 2-page limit – your solutions should fit on two sides of 1 page). A Walsh-Hadamard matrix Hn is an 2n ×2 n matrix with each entry being −1 or +1 and n ∈ Z such that the (i, j)-th entry of Hn[i, j] = √ . Assume that the rows and columns of Hn are counted from zero, i.e., the left-topmost entry of Hn is Hn[0, 0]. Here i ◦ j is the bitwise dotproduct of the binary representation of i and j represented with n bits. For example, if i = 7 and j = 5, then i ◦ j = (1, 1, 1) · (1, 0, 1) = 2. Following are couple of examples of Walsh-Hadamard matrices H2 and H3. 1 1 1 1 1 −1 1 −1 1 1 −1 −1 1 −1 −1 1 1 1 1 1 1 1 1 1 1 −1 1 −1 1 −1 1 −1 1 1 −1 −1 1 1 −1 −1 1 −1 −1 1 1 −1 −1 1 1 1 1 1 −1 −1 −1 −1 1 −1 1 −1 −1 1 −1 1 1 1 −1 −1 −1 −1 1 1 1 −1 −1 1 −1 1 1 −1 • Starting from the fact that Hn[i, j] = √ , show that • Show that the Euclidean norm of every column and every row is 1. (You are allowed to search Wikipedia for the definition of Euclidean norm.) • Using the above properties, write and prove an induction claim that shows that the columns of Hn form an orthonormal basis, i.e., the dot-product of any two columns of Hn equals to zero, and every column of Hn has Euclidean norm of one. Hint: You will need one of the properties of orthonormal matrices to show this. • Consider a vector v ∈ R , i.e., a vector with 2n entries with real numbers. Design an algorithm to compute Hn·v in time O(n log n). Prove the runtime bound for your algorithm. 2. (Divide and Conquer, 2-page limit – your solutions should fit on two sides of 1 page). You and your sister are travelling on a bus when you recall the ”Hot and cold” game you used to play as kids. To kill time, you decide to play a version of it with numbers. One of you thinks of a number between 1 and n, and the other tries to guess the number. If you are the one guessing, every time you make a guess your sister tells you if you are ”warmer”, which is closer to the number in her head, or ”colder”, which is further away from the number in her head. Using this information you have to come up with an algorithm which helps you guess the number quickly. You can use a command called Guess(x), where x is your guess, which returns ”warmer”, ”colder” or ”you guessed it!”. You are required to give an English explanation for your algorithm. Also, prove the correctness of your algorithm and give an analysis of the space and time complexity. An ideal solution will propose an algorithm that takes log2 (n) + O(1) guesses in the worst case scenario. Recalling how binary search works might be helpful. 3. (Greedy Algorithm, 2-page limit – your solutions should fit on two sides of 1 page). The Menlo Park Surgical Hospital admitted a patient, Mr. Banks, last night who was in a car accident and is still in critical condition and needs monitoring at all time. At any given time only one nurse needs to be on call for the patient though. For this we have availability slots of all the nurses, which is a time from which they are available, ai , to the time they have other engagements, bi You need to devise an algorithm that makes sure you can cover Mr. Banks’ entire stay while having minimum number of nurses be on call for him. A nurse leaving at the same time as another arrives is acceptable. Following is an example of how the availability slots for the nurses will look like. The darker bars correspond to a set of 5 nurses who can cover the entire duration. (Notice, though, that 4 nurses would have sufficed.) Your efficient greedy algorithm should take input a list of pairs of times (ai , bi) for i = 1 to n (a) Consider the greedy algorithm that selects nurses by repeatedly choosing the nurse who will be there for the longest time among the periods not covered by previously selected nurses. Give an example showing that this algorithm does not always find the smallest set (b) Present an algorithms that outputs a smallest subset of nurses that can cover the entire duration of Mr. Banks’ stay at the hospital or say that no such subset exists. (c) Prove that your algorithm is correct. (d) State its running time. 4. (Greedy Algorithm, 2-page limit – your solutions should fit on two sides of 1 page). Picture, if you will, a long river with some towns scattered sparely along it. You are in charge of building a series of small hydro-electric power plants for these towns to give them some renewable electricity; however, the power plants can’t power a town further than 20 miles away, due to their particular design – they can, however, give power to as many towns as they can reach. You want to build them along the river such that every town is within 20 miles of at least one of the plants. Give an efficient algorithm that achieves this goal using the minimum number of power plants. Prove using greedy stays ahead strategy. ] (Optional, no collaboration, 2-page limit – your solutions should fit on two sides of 1 page). You and your roommate are in a war over the thermostat – they like it cold, and you like it warmer. The temperatures range from 1 to n degrees. Your roommate is willing to leave the thermostat at some (unknown to you) temperature t or any lower temperature. You sit down with your roommate and agree to negotiate in the following way: In each round, you can name any temperture s between 1 and n. If s > t, they will say ”too warm”. Otherwise, they’ll agree to s. Your goal is to ensure that the apartment is at the maximum acceptable temperture t. One way to ensure that you will have the temperture at s is to name all integers, starting from n and going down by 1 in each round, until they agree to the temperture t. But, if you follow this strategy, you might have to go through n rounds (irratating everyone involved). If you are allowed to change your mind (that is, ask for a higher temperature) after your roommate accepted your offer, you might want to try binary search as your strategy. However, it would also be annoying to change your mind so many times. (a) Suppose you are allowed to change your mind exactly once. Describe a strategy for ensuring a fair temperture, t, that uses o(n) rounds of negotiation (as few as you can). (b) Now suppose you are allowed to change your mind k times, where k > 1. Describe a strategy for ensuring a fair temperture, s, with as few negations as you can. Let fk(n) denote the number of rounds you use, as a function of n. (The answer from part (a) is your f1(n).) For each k, you should be able to get an asymptotically better solution than for k −1: that is, make sure that fk(n) = o(fk−1(n)).
Well-posedness and scattering for NLS on in the energy space We study the Cauchy problem and the large data scattering for energy subcritical NLS posed on . In our previous work , we considered the nonlinear Schrödinger equation on a product space , where is a compact riemannian manifold. We have seen this problem as a kind of vector valued nonlinear Schrödinger equation on and we were able to get small data scattering results (cf. also for small data modified scattering results). Our goal here is to extend this view point to a large data problem in the very particular case when is the one dimensional torus. Therefore, our aim in this paper is the study of the local (and global) well-posedness and scattering of the following family of Cauchy problems: Concerning the Cauchy theory we shall assume and for scattering we assume . Our first result deals with the Cauchy problem. Let and be fixed. Then we have: for any initial datum the problem (1.1) has a unique local solution the solution can be extended globally in time. Property (2) follows by (1) due to the defocusing character of the the nonlinearity (a standard approximation argument is needed to justify the energy conservation). Hence, along the paper, we focus mainly on the proof of (1), i.e. the existence of an unique local solution for any given initial datum. We also notice that the proof of (1) in Theorem 1.1 works also for the focusing NLS. The proof of the local existence given by Theorem 1.1 goes as follows. First we prove the existence of one unique solution in the space where are Strichartz -admissible for the propagator . It is of importance for our analysis that a sub-critical nonlinearity in dimension is a sub-critical nonlinearity in dimension . Therefore at the level we perform an analysis and at the level, we perform the (trivial) analysis which at the end enables us to perform an theory in the full sub-critical range of the nonlinearity. Incorporating in a non-trivial way the dispersive effect in this analysis is a challenging problem. Its solution may allow to extend our analysis to higher dimensional dependence. A key tool in order to perform a fixed point argument in the space (1.2) are the inhomogenous Strichartz estimates associated with (see , ). The second step is the proof of the unconditional uniqueness in the space . We underline that the proof of Theorem 1.1 in the range of nonlinearity , can be obtained following , where it is not needed the use of inhomogeneous Strichartz estimates for . In the cases for every and for the -critical nonlinearity , the proof of Theorem 1.1 can also be deduced respectively from the analysis in and . The main point in our approach is that it works in for every and moreover it gives some crucial controls of space-time global norms which are of importance for the scattering analysis. The main result of this paper concerns the long-time behavior of the solutions given by Theorem 1.1. Assume and , and be the unique global solution to (1.1). Then there exist such that Concerning scattering results for NLS in product spaces we quote where it is studied the quintic NLS on . We also underline that using the arguments of (see also ) one may obtain that for , the result of Theorem 1.3 also holds for the critical nonlinearity . One may also expect that these arguments provide an alternative (and more complicated) proof of Theorem 1.3 for . For , the extension of Theorem 1.3 to the critical nonlinearity is an open problem (even for the local theory). Notice that if one considers (1.1) on , then it is well-known that -scattering is available for (in contrast with Theorem 1.3 where we require the extra restriction ). On the other hand the restriction in Theorem 1.3 is quite natural. Indeed, if we choose and then the Cauchy problem (1.1) reduces to -subcritical NLS in , and at the best of our knowledge no -scattering result is available in this situation . It is well-known, since the very classical work , that a key tool to prove scattering for NLS in the euclidean setting , with nonlinearities which are both energy subcritical and -supercritical, is the proof of the time-decay of the potential energy. In Proposition 1.6 below we prove that this property persists for solutions to NLS on in the energy subcritical regime (in particular we do not need to require to the nonlinearity to be -supercritical, see also Remark 1.7 on this point). A basic tool that we will use is a suitable version in the partially periodic setting of the interaction Morawetz estimates, first introduced in to study the energy critical NLS in the euclidean space . Starting from this work the interaction Morawetz estimates have been exploited in several other papers (, , , , ), in particular they have been used to provide new and simpler proofs of the classical scattering results in and . We emphasize that we make use of the interaction Morawetz estimates from a different point of view compared with the results above. In particular along the proof of Proposition 1.6 below we are able to treat in a unified and simple way NLS on for every , without any distinction between the cases and . This distinction is typical in previous papers involving interaction Morawetz estimates in the Euclidean setting (see Remark 1.8 for more details on this point). Moreover it is unclear to us how to proceed, following the approach developed in previous papers related with interaction Morawetz estimates, to prove Proposition 1.6 in the case (see Remark 1.9). Next we state the key proposition needed to prove Theorem 1.3. Let be a global solution to defocusing NLS posed on and with pure power nonlinearity , with . Then: Notice that in contrast with Theorem 1.3, in Proposition 1.6 we do not assume any lower-bound on . Notice also that Proposition 1.6 is not true for the focusing NLS on for , even if the initial data are assumed to be arbitrarly small in . To prove this fact one can think about the solitary waves associated with the subcritical focusing NLS posed on , and notice that the corresponding norm can be arbitrary small. As already mentioned above the proof of Proposition 1.6 is based on the use of interaction Morawetz estimates in the partially periodic setting. Let us recall that the interaction Morawetz estimates allow to control the following quantity (see for instance ): for solution to NLS posed in the euclidean space . Notice that via the Sobolev embedding it implies some a -priori bounds of the type in the case . This estimate is sufficient to deduce scattering on for in the case of the nonlinearity . Notice also that in higher dimensions we get in (1.4) the control of a negative derivative of . In this case some extra work is needed in order to retrieve the needed space-time summability that allows to get scattering. Typically the main strategy to overcome this difficulty is to retrieve some informations on negative derivative of via the following estimate (see ): Once a negative derivative of is estimated, then we can be interpolated with the bound , and hence we get the needed space-time integrality necessary to prove scattering for . We underline that arguing as in , where it is studied NLS with a partially confining potential, one can prove the following version of interaction Morawetz estimate: provided that solves NLS posed on . Hence via the Sobolev embedding one can deduce some a -priori bounds in the case . This estimate is sufficient to deduce scattering for and (see the computations in in the case of a partially confining potential). However, as far as we can see, it is unclear how to exploit (1.6) in the case . Estimate (1.6) is obtained by controlling a suitable family of multiple integrals of the type: where the integrand function depends on a test function and on the solution to NLS . Once this test function is suitably choosen then it allows to contract the variables , , to , hence we get (1.6). The main point in our analysis is that we combine an argument by the absurd in conjunction with the finiteness of the following quantity that in turn follows by where is any convex function. In this estimate we choose . Notice that this choice does not allows contraction of the variables in , however it implies (1.7), which is sufficient to conclude the time decay of the potential energy for solutions to NLS in a simpler way and in a more general setting compared with (1.6). We believe that this part of our argument is of independent interest. 2. Some Useful Functional Inequalities In this section we collect some a-priori estimates for the propagator and the associated Duhamel operator. At the end we also present an anisotropic Gagliardo-Nirenberg inequality that will be useful in the sequel. We define as endowed with the natural norm. Let and . Then we have the following homogeneous estimates: provided that the following conditions hold: We claim that by combining the Sobolev embedding with the usual Strichartz estimates on we get We can conclude by using the fact that commute with the linear Schrödinger equation on . Next we give a few details about the proof of (2.3). Given any for (or for , for ) we fix the unique such that Hence by the usual Strichartz estimates (see ) we get: In turn it implies Notice that if then we conclude by the sharp Sobolev embedding (here is precisely the one that appears in (2.2) once and are fixed). In the case we conclude again by the Sobolev embedding for every , and in particular . Let and . Indicate by both , and Then we have for the following estimates: The proof follows by the Strichartz estimates associated with the propagator , in conjunction with the argument in . ∎ Let be fixed and . Then we have the following extended inhomogeneous estimates: The same conclusion holds for provided that we drop the conditions (2.8). In the case that and do not depend on , the estimates above are special cases of the inhomogeneous extended Strichartz estimates proved in Thm. 1.4 (see also ). Its extension to the case that we have explicit dependence on (in and/or ) follows arguing as in . We underline that in order to apply the technique in we need (2.7), which is not required in . ∎ The following result will be useful in the sequel. Let and be a sequence such that , for some . Then for every there exists such that . By combining the assumption with the Sobolev embedding we obtain that First we prove the following estimate that will be useful in the sequel: To prove this estimate we develop in Fourier series w.r.t. y variable: Hence by the Minkowski inequality we get and by the Haussdorf-Young inequality Next, we shall prove Once (2.12) is proved then we conclude by interpolation between (2.12) and (2.11) in the case . In the case then we can interpolate between (2.12) and the estimate (that follows by the assumptions). Next we focus on (2.12). Notice that we have the following Gagliardo-Nirenberg inequality: where we have fixed . In turn by the Hölder inequality it gives: and . Since we can use the trivial estimate and we get Since for and for we conclude by (2.10). ∎ 3. Fixing the admissible exponents for the well-posedness analysis We collect in this section some preparations, useful in the sequel to construct suitable functional spaces in which we shall perform a contraction argument to guarantee existence and uniqueness of solutions to (1.1). The next proposition will be useful to study the Cauchy problem associated with (1.1) in the regime . Let and be fixed. Then there exist such that To study the Cauchy problem (1.1) in the regime we shall need the following Proposition. Let and be fixed. Then there exists and such that: For we get the same conclusion, provided that we drop conditions (3.2). Moreover we can assume We shall need the following Lemma. Let , be fixed and . Then there exist such that: For we get the same conclusion, provided that we drop Moreover we can also assume that First we show that by our choice of (3.12) follows. Indeed we get where we used (3.10). Notice also that by the second identity in (3.11) we get , in fact Moreover the first condition in (3.8) follows by (3.7), and the first condition in (3.9) follows by the first identity in (3.10). Hence since now on we can skip those conditions. It is easy to check that thanks to our choice of , the identities in (3.10) and (3.11) are not independent. Moreover by (3.10) and (3.11), and by recalling , we get Hence we conclude tat we can select a suitable if the condition is satisfied. Since we are assuming this condition is equivalent to Next we notice that In fact it follows by direct computation for and for it comes by the following argument. Notice that is equivalent to , that under the constrain can be written as The first and second inequalities are satisfied for any and the last one follows by On the other hand we have By elementary computations (see the case ) and by recalling , the conditions above are equivalent to the following inequality: On the other hands by explicit computation we get and (3.17) follows by elementary considerations. Proof of Proposition 3.2. We focus on the case (the cases can be treated by a similar argument). We argue by a continuity argument based on Lemma 3.3. In fact we fix as in Lemma 3.3 and we look for that satisfy conditions of Lemma 3.2, for some small enough and will be properly chosen in dependence of . By our choice it will be clear that and . Notice that with this choice the identity in (3.5) is satisfied (compare with (3.11)). Also (3.1), (3.2), (3.3) are satisfied by a continuity argument provided that is small enough (recall that satisfy (3.7), (3.8), (3.9)). Notice also that since then the first identity in (3.4) is satisfied provided that we choose (recall that satisfy the first identity in (3.10)) and also (3.6) follows by (6.6). Next we impose that satisfy the second identity in (3.4), i.e. . We claim that (notice this is equivalent to the inequality in (3.5)) and it will conclude the proof. Indeed we write and hence , where we used (see the second identity in (3.10)). Hence we get by the first identity in (3.11) Let and be fixed.Then there exist such that: where is any couple given by Proposition 3.2.
Lead Time and Cycle TimeThe cycle time of a process is a key to match lean cycle time chart supply with the demand in lean manufacturing. Everybody working on a shop floor knows the term. Yet, I still find that people sometimes confuse what exactly it means. The cycle time is the fastest repeatable time in which you can lean cycle time chart one part. Hence, in this ldan as part of a series on manufacturing speed measurements I would like to dig deeper into what cycle times really boldenona precio peru, and how to best measure them. How to Measure Cycle Times - Part 1 \| writingdesk.pw The cycle time of a process is a key to match the supply with the demand in lean manufacturing. Everybody working on a shop floor knows the term. Yet, I still find that people sometimes confuse what exactly it means. The cycle time is the fastest repeatable time in which you can produce one part. Hence, in this post as part of a series on manufacturing speed measurements I would like to dig deeper into what cycle times really are, and how to best measure them. As it turns out, there is actually quite some detail on how to measure cycle times, hence I split this post into two parts second part How to Measure Cycle Times — Part 2 , with an additional third post focusing on the details of manual cycle times. Most definitions for cycle time I have found online are rather basic, defining cycle time often as the time required to complete a cycle Merriam Webster. However, for practical terms, this definition is way to broad. A much more useful definition for me is:. Hence, it is measured as a time per part, and does not include losses like breakdowns, defects, and other delays for more on losses see my articles on OEE. As with most other time measurements, you can also distinguish between the current cycle time you actually have simply called cycle time , and the cycle time you want — the target cycle time. Yet even this will leave some remaining ambiguity. For example, if you have batch processes, is the cycle time the time for one batch, or the time for one part? Depending on what you need the time for, both may make sense. Confusingly, I am not aware of an widely accepted name that distinguishes one from the other. Both are commonly called cycle time, regardless if for a batch or for a single part. Please be aware that this definition above is far from universally accepted. Quite a few practitioners define cycle time to include losses. This has significant potential for confusion! I strongly prefer to use cycle time only for times without unplanned losses or delays. In any case, if discussion takt times and cycle times with someone else, please verify what they are talking about! Based on a discussion on the LinkedIn Group TPS Principles and Practice, cycle time is more often defined excluding losses Many thanks to all who contributed in this discussion. Many people simply calculate the cycle time by dividing the working time by the number of parts produced during this working time. This is terribly wrong! This is how you calculate the takt time , which includes losses. If you calculate the cycle time using this approach, you include all the losses and breakdowns. Another common approach to determine the cycle time is to simply look it up in one of the data sheets. When the machine was built, surely somebody must have set a time into the machine. Similarly, there is often data available on how fast a manual work should be. Often, this number is used without questioning. While this number is probably closer to the truth, it still is far from good. The speed of machines change over time. The number you have received may no longer be correct. The number may also be only the primary process, but may not include the loading or unloading or other preparation. While I totally agree to include the time when calculating the average target speed for the operator , it must not be included in the peak speed of the process. I will write more on that in the near future. Hence, some practitioners say that you always must measure your data yourself. I am not that hardcore. Measuring it yourself will take quite a bit of time, especially if you have many processes details see below. If you are short on time and most people are , you can also use the data found in some documents. But please check if these numbers are approximately in the right ballpark. Also keep in mind that different part types may have different cycle times. I would not trust them without verification, but if the numbers are close enough I sometimes skip the detailed measurement. The most precise way to determine the cycle time is to measure a number of individual parts. To determine the entire cycle time, you would need to measure the time between the completion of one part until the completion of the next part for multiple parts. This is the same as measure the time between the beginning of one part until the beginning of the next part. If it is more convenient, you could even measure the time between one action of the cycle and the same action of the cycle of the next part. In any case the result may look something like a chart below:. The smallest value in the data is 4. Yet, this is not our cycle time. Our cycle time should be repeatable, and the smallest value is not repeatable. In any case, I also often find that this is a measurement error. Hence, do not use the minimum as the cycle time. What I also see way to often is people simply taking the arithmetic mean, i. Unfortunately, this is then skewed towards a larger value by the larger outliers. For example, if you measure the average wealth of 30 people, and one of them is Bill Gates worth ca. Similar for the data above, measurement 7 and 16 seem to be very large. The arithmetic mean of the data set above is Do not simply use an arithmetic mean for cycle times. Sometimes I see the largest values discarded, and the arithmetic mean of the remainder used. It is possible to use the mean without these extreme values , although not my preferred way. What I often do is to use the median or an percentile. The median is the measurement where half of the measurements are larger, and the other half are smaller. This would also work. Yet even then the results differ. As shown in the graph below, the median is 8. Which one should we use? Or should we use the 10th percentile 6. As mentioned in the heading, this method is still imperfect. If i seem to have few outliers at the bottom i. If i do have lots of outliers, I sometimes use the median. For example, I once had a system that generated automatic data, but due to a technical quirk often measured two parts with no delay in between. Hence, I cannot answer you definitely which percentile to use. Plot the data, look at the data and try to figure out what you think your system can repeat if there are no problems. If you graph the data as shown above, you can also simply put in a line where you think the cycle time fits best. As for the graph above, would you go with the Min, 1st quartile, median, or average value for a repeatable cycle time? For me, the 1st quartile would look best. I will talk a bit more about this problem in my next post. Overall there are three ways to measure cycle times. A really bad one divide work day by number of parts produced , a mediocre one look up the [outdated? However, even with this data, there are still things that can go wrong. In my next post How to Measure Cycle Times — Part 2 I will talk about the dangers of changing the sequence of the steps within a process, on different part types with different cycle times, and some general comments on the accuracy of cycle times in general. Stay tuned, and in the meantime Go Out and Organize your Industry! Best approach I have seen is to take 20 cycles, exclude the data points with assignable causes and then calculate the arithmetic mean. This approach is still a practical challenge though with slow movers and parts that have long cycles. Furthermore, as takt and cycle times are compared it should always be at the part level. Hi Rob, definitely a possible way to exclude outliers. I have done this, too, but have a slight preference for a median or percentile, simply because it removes the manual decision which points to include, and hence makes the times more comparable to others. This is a small effect, though. Thanks, Chris, a nice story, but too long for quick readers and difficult to find the value added advice, which has appeared at the very end- to use median or quartile! Hi Elena, sorry that it was too long for you. I usually try to structure it with headlines to help people to skip to the parts they are interested in. Thanks for the feedback. How do you know your internal work balance sheet is correct? In my view a doable but somewhat risky shortcut. Hi Navaneeth, there is no fixed rule. The more you measure the more precise your result will be. Statisticians often recommend 30 measures, in practice due to time constraints it is often less in lean. Try to have at least 10 measurements.
This Blog is for those preparing for higher level Maths exams like CSIR NET, SLET, SET, GATE, IAS, TRB PG / Engineering/ Polytechnic maths. Questions & answers of NET, GATE, IAS, TRB maths exams are published. Study materials will be uploaded. Join / follow this blog. Friday, 17 August 2012 Graduate Aptitude Test in Engineering (GATE) is an all India examination that primarily tests the comprehensive understanding of various undergraduate subjects in Engineering and Technology. The GATE score of a candidate reflects a relative performance of a candidate. The score is used for admissions to post-graduate engineering programmes (eg. M.E., M.Tech, direct Ph.D.) in Indian higher education institutes with financial assistance provided byMHRD and other Government agencies. The score may also used by Public sector units for employment screening purposes. 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GATE is administered and conducted jointly by the Indian Institute of Science and seven Indian Institutes of Technology on behalf of the National Coordination Board (NCB) – GATE, Department of Higher Education, Ministry of Human Resource Development (MHRD), Government of India. The GATE Committee, which comprises of representatives from the administering institutes, is the sole authority for conducting the examination and declaring the results. GATE is conducted through the constitution of eight zones. The zones and the corresponding administrative institutes are: Biometric information (Photograph and fingerprint) may be captured on the day of the examination for randomly selected candidates. Important Points for GATE 2013. Application Process: For GATE 2013, candidates need to register and fill the application ONLINE only by accessing the zonal GATE websites of IISc and seven IITs. The application process is complete only when a print out of the filled ONLINE application with the candidate’s signature and a good quality photo affixed in the appropriate place is received by the respective GATE office along with necessary documents, if any, on or before 8 October 2012. Please note that application forms are NOT available for sale anywhere. Downloadable Admit Card: Admit cards are NOT sent by mail anymore. Admit cards can only be downloaded from the zonal GATE websites from 5th December 2012 onwards. Bring the admit card to the test center along with at least one original (not photocopied / scanned copy) and valid (not expired) photo identification. Use of black ink ball point pen for Offline exams: Candidates should use only black ink ball point pen for darkening of the bubbles in the OMR sheet. Since bubbles darkened by the black ink ball point pen cannot be erased, candidates should darken the bubbles in the OMR sheet very carefully (be sure before darkening). Numerical answer type questions in ONLINE papers: In the ONLINE papers, the question paper will consist of questions of multiple choice type and questions of numerical answer type. For multiple choice type questions, each question will have four choices for the answer. For numerical answer type questions, each question will have a number as the answer. Each online paper will have 15 or more marks worth of questions requiring numerical answers where possible. Pre-final year students: Pre-final year students are NOT eligible to write GATE 2013. For details, refer to eligibility for GATE examination. Pattern of Question Papers and Marking Scheme Pattern of Question Papers The examination for the papers with codes AE, AG, AR, BT, CE, CH, CY, GG, MA, MN, MT, PH, TF, XE and XL will be conducted ONLINE using computers where the candidates will be required to select the answer for each question using a mouse. For all other papers (CS, EC, EE, IN, ME & PI), the exam will be conducted OFFLINE in whicih candidates will have to mark the correct choice on an Optical Response Sheet (ORS) by darkening the appropriate bubble against each question. In all the papers, there will be a total of 65 questions carrying 100 marks, out of which 10 questions carrying total of 15 marks are in General Aptitude (GA). The remaining of 85% of the total marks is devoted to the syllabus of the paper (as indicated in the syllabus section) GATE 2013 would contain questions of four different types in various papers: (i)Multiple choice questions carrying 1 or 2 marks each; Each of the multiple choice objective questions in all papers and sections will contain four answers, of which one correct answer is to be marked. (ii) Common data questions (which are also multiple choice questions), where two successive questions use the same set of input data; Statement for Common Data Questions, for instance, for Questions 48 and 49 in Main Paper: Let X and Y be jointly distributed random variables such that the conditional distribution of Y, given X=x, is uniform on the interval (x-1,x+1). Suppose E(X)=1 and Var(X)=5/3. First question using common data: Q.48 The mean of the random variable Y is (A) 1/2 (B) 1 (C) 3/2 (D) 2 Second question using common data: Q.49 The variance of the random variable Y is (A) 1/2 (B) 2/3 (C) 1 (D) 2 (iii) Linked answer questions (which are also multiple choice questions), where the answer to the first question in the pair is required to answer its successor; Example: Statement for Linked Answer Questions, for instance, for Questions 52 and 53 in Main Paper: An E. coli cell of volume 10-12 cm3 contains 60 molecules of lac-repressor. The repressor has a binding affinity (Kd) of 10-8 M and 10-9 M with and without lactose respectively, in the medium. First question of the pair: Q.52 The molar concentration of the repressor in the cell is (A) 0.1 nM (B) 1 nM (C) 10 nM (D) 100 nM Second question of the pair: Q.53 Therefore the lac-operon is (A) repressed and can only be induced with lactose. (B) repressed and cannot be induced with lactose. (C) not repressed. (D) expressed only when glucose and lactose are present. (iv) Numerical answer questions, where the answer is a number, to be entered by the candidate using the mouse and a virtual keypad that will be provided on the screen. Design of Questions The questions in a paper may be designed to test the following abilities: (i) Recall: These are based on facts, principles, formulae or laws of the discipline of the paper. The candidate is expected to be able to obtain the answer either from his/her memory of the subject or at most from a one-line computation. Q. During machining maximum heat is produced (A) in flank face (B) in rake face (C) in shear zone (D) due to friction between chip and tool (ii) Comprehension: These questions will test the candidate’s understanding of the basics of his/her field, by requiring him/her to draw simple conclusions from fundamental ideas. Q. A DC motor requires a starter in order to (A) develop a starting torque (B) compensate for auxiliary field ampere turns (C) limit armature current at starting (D) provide regenerative braking (iii) Application: In these questions, the candidate is expected to apply his/her knowledge either through computation or by logical reasoning. Q. The sequent depth ratio of a hydraulic jump in a rectangular channel is 16.48. The Froude number at the beginning of the jump is: (A) 5.0 (B) 8.0 (C) 10.0 (D) 12.0 (iv) Analysis and Synthesis: These can be linked answer questions, where the answer to the first question of the pair is required in order to answer its successor. Or these can be common data questions, in which two questions share the same data but can be solved independently of each other. Common data based questions: Two questions are linked to a common data problem, passage and the like. Each question is independent and its solution is obtainable from the above problem data or passage directly. (Answer of the previous question is not required to solve the next question). Each question under this group will carry two marks. Linked answer questions: These questions are of problem solving type. A problem statement is followed by two questions based on the problem statement. The two questions are designed such that the solution to the second question depends upon the answer to the first one. In other words, the first answer is an intermediate step in working out the second answer. Each question in such ‘linked answer questions’ will carry two marks. Examples of each of this design is given in the types of questions above. The questions based on the above four logics may be a mix of single stand alone statement/phrase /data type questions, combination of option codes type questions or match items type questions. For 1 mark multiple choice questions, 1/3 mark will be deducted for a wrong answer. Likewise, for 2 marks multiple choice questions, 2/3 mark will be deducted for a wrong answer. However, for the linked answer question pair, where each question carries 2 marks, 2/3 mark will be deducted for a wrong answer to the first question only. There is no negative marking for wrong answer to the second question of the linked answer question pair. If the first question in the linked pair is wrongly answered or is unattempted, then the answer to the second question in the pair will not be evaluated. There is no negative marking for numerical answer type questions (numerical answer type questions will appear only in the papers for which the exam is ONLINE only). General Aptitude (GA) Questions In all papers, GA questions are of multiple choice type, and carry a total of 15 marks. The GA section includes 5 questions carrying 1 mark each (sub-total 5 marks) and 5 questions carrying 2 marks each (sub-total 10 marks). Question papers other than GG, XE and XL These papers would contain 25 questions carrying one mark each (sub-total 25 marks) and 30 questions carrying two marks each (sub-total 60 marks). Out of these, two pairs of questions would be common data questions, and two pairs of questions would be linked answer questions. In the ONLINE papers, the question paper will consist of questions of multiple choice type and numerical answer type. For multiple choice type questions, each question will have four choices for the answer. For numerical answer type questions, each question will have a number as the answer and choices will not be given. The answer is to be entered using the mouse and a virtual keypad that will appear on the screen. GG (Geology and Geophysics) Paper Apart from the General Aptitude (GA) section, the GG question paper consists of two parts: Part A and Part B. Part A is common for all candidates. Part B contains two sections: Section 1 (Geology) and Section 2 (Geo-physics). Candidates will have to attempt questions in Part A and either Section 1 or Section 2 in Part B. Part A consists of 25 multiple choice questions carrying 1-mark each (sub-total 25 marks & some of these may be numerical questions). Each section in Part B (Section 1 and Section 2) consists of 30 multiple choice questions carrying 2 marks each (sub-total 60 marks and some of these may be numerical questions). Out of these, two pairs of questions would be common data questions, and two pairs of questions would be linked answer questions. XE Paper (Engineering Sciences) In XE paper, Engineering Mathematics section (Section A) is compulsory. This section contains 11 multiple choice questions carrying a total of 15 marks: 7 questions carrying 1-mark each (sub-total 7 marks), and 4 questions carrying 2-marks each (sub-total 8 marks). Some of the multiple choice questions may be replaced by numerical questions. Each of the other sections of the XE paper (Sections B through G) contains 22 questions carrying a total of 35 marks: 9 questions carrying 1 mark each (sub-total 9 marks) and 13 questions carrying 2 marks each (sub-total 26 marks). Out of the 2 mark questions, 2 pairs are common data questions and 1 pair is linked answer questions. Some of the multiple choice questions may be replaced by numerical questions. XL Paper (Life Sciences) In XL paper, Chemistry section (Section H) is compulsory. This section contains 15 multiple choice questions carrying a total of 25 marks: 5 questions carrying 1 mark each (sub-total 5 marks) and 10 questions carrying 2-marks each (sub-total 20 marks). Out of the 2-mark questions, 1 pair is common data questions, and 1 pair is linked answer questions. Some of the multiple choice questions may be replaced by numerical questions. Each of the other sections of the XL paper (Sections I through M) contains 20 multiple choice questions carrying a total of 30 marks: 10 questions carrying 1 mark each (sub-total 10 marks) and 10 questions carrying 2 marks each (sub-total 20 marks). Some of the multiple choice questions may be replaced by numerical questions.
On Saturday, I spoke at #mathconf10 in Dagenham. The workshop was titled “Worksheet-making Extravaganza!” Indeed it was that. When I started teaching I regularly went onto TES to download resources for my pupils. Usually, I would be annoyed that I couldn’t find a good enough worksheet because the worksheet I wanted would never be available. I then realised that I needed to start creating my own. However, in the process I realised that I found making problem types for different topics to be difficult because my subject knowledge wasn’t up to scratch. Now, I am very good at maths but that doesn’t mean that I am good at making a worksheet of questions for pupils to do and learn from. Then after spending a summer between my first and second year of teaching where I made different resources combined with my own experience of working at Michaela Community School I finally realised how to make my own worksheet. Here are the two things that make a good worksheet: - High quality content - High quality structure (deliberate practice) What do I mean by high quality content? I am referring to the questions that are made to test a pupils’ understanding of the concept or procedure that has been taught. What is the starting point of gathering this content? If you are planning a lesson on how to simplify fractions then list out all the problem types from the easiest to the most difficult one. More problem types can be found by looking at different textbooks, asking colleagues and reading blogs etc. Let’s look at the problem types that can be included when planning a lesson on simplifying fractions, simplify a: - proper fraction - (an) improper fraction - fraction that simplifies to a unit fraction - fraction that simplifies to an integer - mixed number - fraction with large numbers (still divisible by a common factor) - fraction that simplifies to 1 It is all well and good gathering a bunch of problem types but I have deliberately chosen the ones I have listed. This is because these problem types present four features that make the content of worksheet high quality: - Arithmetic complexity - Visual complexity - Multiple steps Arithmetic complexity basically means including more difficult numbers or large numbers in your questions. Can a pupil simplify one hundred and eleven-thirds? Also, it means including questions which have decimals and fractions, and creating questions where the answer is a decimal or fraction as well. Visual complexity refers to creating questions which look a bit scary. Again, this can entail including large numbers to make a pupil think. For example, in terms of calculating the area of a square you can write a question where the length is a large two digit number or a decimal.Another example is to only label one length of a square, more than two lengths, four length etc. A question like this is testing whether a pupil can calculate the area of a square from squaring one length. Another very simple approach to make this type of question visually complex is to have the image at a certain orientation.Multiple steps is a feature in a question where you have increased the number of steps between the question and the answer. This can include having calculations in the numerator and denominator before a pupil is asked to simplify a fraction. Decoding is a feature of a question where its set up is testing whether a pupil can rearrange or manipulate their new found information of the concept or procedure that has been taught. For example, providing pupils with questions which have incomplete answers. Or asking if the following equations are true or false? An exercise on a worksheet can look something like this. Question (g) is testing a pupil’s misconception where dividing a number by itself is not 0 but 1. Similarly, question (k) is testing a system 1 mistake where a pupil is not fully thinking about the question, they simply see that the half of 18 is 9 therefore the answer is 9. Can I apply these four features of high quality content when planning a series of lessons where I want pupils to learn how to add and subtract fractions with like denominators? Again, start with listing out the potential problem types. Adding or subtracting: - Two proper fractions where the result is less than 1 - A proper and an improper fraction - Two proper fractions where the result is greater than 1 - Two mixed numbers where the sum of the proper fractions is less than 1 - Two mixed numbers where the sum of the proper fractions is greater than 1 - Two mixed numbers where the sum of the proper fractions equals to 1 I can make these type of question arithmetically complex by having large numbers in the numerator and the denominator. I can include a question when I subtract two proper fractions and the result is negative. Questions where pupils have to rearrange their information of adding and subtracting fractions with like denominators can include true or false question. The first question below is conflating adding a numerator and multiplying the denominator. Similarly, it is a really powerful form of testing pupils’ knowledge by including questions where the answer is incomplete. Can the pupils identify what the numerator is? Can pupils identify that 1 can be written as 10/10 and that 2 can be written as 20/10. The following questions demonstrate the feature of questions having multiple steps between the question and the answer. Yes, you can easily write the answer, but if you wanted a pupil to write their answer as an improper fraction, you would want them to write the integer as a fraction. The mixed number examples are tailored so pupils are able to create an improper fraction where the numerator is still less than 15. Why? Simply because the kids at Michaela have committed the first 15 square numbers to memory as well as the first 10 cube numbers, but if I made the fraction more difficult where the numerator was greater than 21 then I would be making the question difficult for the wrong reasons. If the fraction was four and one-fifth then the numerator would be 21, and squaring 21 isn’t valuable. What is valuable is spotting the first step of writing the mixed number as a fraction, and then squaring the fraction. The mixed number examples demonstrate questions being visually complex as well as arithmetically complex. The following question is demonstrating questions which have multiple steps to the answer, where calculations in the numerator and the denominator need to be simplified before squaring. Again, calculations were designed where the numerator and denominator values would be less than 15. The same form of thinking in applying the three features (arithmetic complexity, visual complexity and multiple steps) can be applied when fractions are being square rooted or cube rooted. However, ensuring that the numerator and denominator values are numbers that can be square or cube rooted. What is new is including negative fractions when they are being rooted by an odd number. Furthermore, visual complexity as well as multiple steps to be performed to get the answer make these questions challenging and rigorous. In summary, the four features which make a worksheet consist of high quality content is creating all problem types where they are arithmetically complex, visually complex, require decoding of current information from instruction and multiple steps to be performed to result in the answer. I am still trying to wrap my head around creating a worksheet with high quality structure which will be my next blog post.
The complete spectrum of the area from recoupling theory in loop quantum gravity We compute the complete spectrum of the area operator in the loop representation of quantum gravity, using recoupling theory. This result extends previous derivations, which did not include the “degenerate” sector, and agrees with the recently computed spectrum of the connection-representation area operator. One of the central results of the loop approach to quantum gravity is the derivation of discrete properties of spacetime geometry [2, 3, 4, 5, 6]. This derivation realizes the old idea that spacetime might exhibit some kind of quantum discreteness at the Planck scale . One of the manifestations of such discreteness is the fact that the operators associated to physical area and volume have discrete spectra . This fact leads to the prediction that measurements of areas and volumes at the Planck scale would yield quantized values . The explicit computation of the spectra of area and volume is thus a relevant step towards understanding the physics of the quantum gravitational field. Partial results on these spectra, for instance, have already been employed in discussing quantum gravitational corrections of black hole radiation and black hole entropy . Here, we consider the area operator. A regularization technique for the definition of this operator was introduced in , where some of the eigenvalues were computed. A more complete treatment was given in , where the spectrum was computed in full except for a “degenerate” sector formed by the states in which vertices or edges of the spin network lie on the surface. It is difficult to treat this degenerate sector –whose existence was pointed out by A Ashtekar– using the original regularization, because additional divergences appear. As a result, earlier works on the calculation of geometry eigenvalues in the loop representation exhibit incomplete spectra for the area operator. In this paper we introduce an alternative regularization, whose action is well defined on every loop state. This is done in Section 2. This regularization allows us to compute the full spectrum of the area operator. The computation is performed in Section 3, using recoupling theory, which provides a powerful computation technique in quantum gravity . Our calculation here is a further proof of the effectiveness of recoupling theory in this context. The complete spectrum of the area has been recently computed in Ref. in the connection representation of quantum gravity . The spectrum we obtain here fully agrees with the one given in Ref. . Notice that the two representations are unitarily isomorphic, as argued by Lewandowski , and, recently, by DePietri , but the regularization techniques employed to define physical operators differ, so that the agreement is a non-trivial result. 2 The area operator Consider a two-dimensional surface embedded in a three-dimensional metric manifold . Let , , be coordinates on , , , coordinates on , and let represent the embedding. If we represent the metric of by means of the inverse densitized triad , the area of is given by where is the one-form normal to . Since the metric of physical space is the gravitational field, is a function of the gravitational field: it is then represented by a quantum operator in a quantum description of gravity. In the loop quantization of general relativity, one constructs by expressing the area in terms of loop variables and then replacing the loop variables with the corresponding quantum loop operators. Since (1) involves products of two triads , the area is expected to be a function of the loop variable of order 2 (with two “hands”) where denotes a loop and is the parallel propagator of the Ashtekar connection along with end points at and . Throughout this work, we assume familiarity with the loop representation ; in particular, we refer reader to for notations and conventions. (Notice that we deal here with conventional loop quantum gravity, not with its quantum SU(2) deformations .) A direct change of variables leads to an expression which is not suitable for quantization, due to the the presence of products of operators at the same point. Therefore a regularization procedure is needed in order to define . This can be done by selecting a sequence of quantities converging to (1) when goes to zero, each of which does not involve products of variables at the same point. The sequence can be quantized by substituting the dynamical variables with the corresponding quantum operators. The operator is then defined as a suitable operator limit of the resulting sequence of operators. There is a degree of arbitrariness in any regularization procedure (in conventional field theory: dimensional regularization, point splitting, Pauli-Villar…). In the present context, the regularization procedure must satisfy two requirements. First, the classical regularized expression must converge to the correct classical quantity (the area) when the regulator is removed; second, the quantum operator must be well defined in the limit and must respect the invariances of the theory; in particular, the operator should transform correctly under diffeomorphisms. Contrary to a possible impression that the choice of a regularization leaves great arbitrariness, implementing both requirements is actually far from trivial. The modification of the regularization procedure for the area operator considered in this paper is forced by the realization that the regularization considered in fails to satisfy the second viability criterion, if one relaxes the simplifying choice of neglecting certain states. In the following, we construct one such regularization of the area operator. Let us begin by introducing a smooth coordinate over a finite neighborhood of , in such a way that is given by . Consider then the three-dimensional region around defined by . Partition this region into a number of blocks of coordinate height and square horizontal section of coordinate side . For each fixed choice of and , we label the blocks by an index . Later, we will send both and to zero. In order to have a one-parameter sequence, we now choose as a fixed function of . For technical reasons, the height of the block must decrease more rapidly than in the limit; thus, we put with any greater than 1 and smaller than 2. Consider one of the blocks. The intersection of the block and a surface is a square surface: let be the area of such surface. Let be the average over of the areas of the surfaces in the block. Namely Summing over the blocks yields the average of the areas of the surfaces, and as (and therefore ) approaches zero, the sum converges to the area of the surface . Therefore we have The quantity associated to each block can be expressed in terms of the fundamental loop variables as follows. First, let us pick an arbitrary fiducial background coordinate system. For every two points and , let be a loop determined by and – say as the zero-area loop obtained by following back and forth a straight segment between and , in the fiducial coordinate system. Then, we write and notice that Equation (6) holds because of the following. We have for any three points , , and , in . It follows that Equation (6) follows. Equations (4), (5) and (6) define our regularization for the area. The operator can now be defined as The meaning of the operator limit in (10) is discussed in . The space of states of the quantum theory is the space of loops up to the Mandelstam relations (which essentially identify loops with the same holonomy). A state is usually denoted (for the Mandelstam class of the loop ) or by means of a bracketed pictorial representation of the loop . Any state has an associated graph, which is characterized as a collection of edges (smooth lines of generic multiplicity) and vertices (points at which lines converge). A vertex has a valence, defined as the number of edges converging to it. A graph is said to be -valent if all its vertices have valence or less. A convenient basis of the space of quantum states is the spin network basis, introduced in and further discussed in . Spin network states are given by the linear combinations of loop states obtained by antisymmetrizing the loops along each edge of their graphs. Because of the Mandelstam relations, these linear combinations form a complete basis. A spin network is characterized by an -valent graph, an assignment of an ordering to the edges converging to each vertex, a color assigned to each edge (the number of antisymmetrized loops), and a color assigned to each vertex (characterizing the rooting of the loops through the vertex). An -valent spin network can be expanded into a “virtual” trivalent spin network, lying in the ribbon associated to the state – a possibility that we exploit below. See for details. The action of on the quantum states is found from the action of the operators, which we recall here. The operator annihilates the state unless the loop intersects the loop at the points and . If the loops do intersect as needed, the action of the operator on the state gives the (Mandelstam class of) the union of the loops and , with two additional vertices at the points and . More precisely, if and fall over two edges of with color and respectively, we have In the picture, the loop is represented as a double line running back and forth between the intersection points and (the two “grasps”), hence the label 2. 3 Spectrum of the area operator In this section we discuss the action of the operator on a generic spin network state . Due to the limiting procedure involved in its definition, the operator does not affect the graph of any state. Furthermore, since the action of inside a specific coordinate block vanishes unless the graph of the state intersects , the action of ultimately consists of a sum of numerable terms, one for each intersection of the graph with the surface. Here we allow for spin networks having vertices on or edges tangential to , unlike previous treatments of this problem . Consider an intersection between the spin network and the surface. For the purpose of this discussion, we can consider a generic point on an edge as a “two-valent vertex”, and thus say, without loss of generality, that is a vertex. In general, there will be edges emerging from . Some of these will emerge above, some below and some tangentially to the surface . Since we are taking the limit in which the blocks shrink to zero, we may assume, without loss of generality that the surface and the edges are linear around (see below for subtleties concerning higher derivatives). Due to the two integrals in (12), the position of two hands of the area operator are integrated over each block. As the action of is non-vanishing only when both hands fall on the spin network, we obtain terms, one for every couple of grasped edges. Consider one of these terms, in which the grasped edges have color and . The state represents, up to factors, the result of the action of on the edges and of an -valent intersection . The irrelevant edges are not shown. The edges labeled and are generic, in the sense that their angles with the surface do not need to be specified at this point (the two edges may also be identical). From the definition (10-12) of the area operator and the definition (13-14) of the operator, each term in which the grasps run over two edges of color and is of the form The last step in the preceding calculation is pulling the state outside the integral sign. This is possible because the -dependent states all have the same limit state as , namely ; i.e., and, hence, the substitution of the -dependent states with their -independent limit in the integral is possible up to terms of order . Note the following: This result is independent of the angle the edge makes with the surface because can always be chosen sufficiently small so that crosses the top and bottom of the coordinate block (this is the reason for requiring that goes to zero faster than ). Also, since we have chosen smaller than 2, it follows that any edge tangential to the surface exits the box from the side, irrespectively from its second (and higher) derivatives, for sufficiently small , and gives a vanishing contribution as goes to zero. Thus, the parenthetic factor in (16) is either or depending on whether one or none of the edges lies on the surface. Consequently, in the limit considered, the edges tangent to the surface do not contribute to the action of the area whereas every non vanishing term takes the form Generically, there will be several edges above, below, and tangential to the surface . Following , we now expand the vertex into a virtual trivalent spin network. We choose to perform the expansion in such a way that all edges above the surface converge to a single “principal” virtual edge ; all edges below the surface converge to a single principal virtual edge ; and all edges tangential to the surface converge to a single principal virtual edge . The three principal edges join in the principal trivalent vertex. This trivalent expansion is shown in Figure 1. This choice simplifies the calculation of the action of the area, since the sum of the grasps of one hand on all real edges above the surface is equivalent to a single grasp on (and similarly for the edges below the surface and ). This follows from the identity which can be proven as follows. Using the recoupling theorem (36), the left hand side of (20) can be written as where can take the values , and . A straightforward calculation making use of (A) gives Thus, (20) follows. A repeated application of the identity (20) allows us to slide all graspings from the real edges down to the two virtual edges and . Thus, each intersection contributes as a single principal trivalent vertex, regardless of its valence. We are now in position to calculate the action of the area on a generic intersection. From the discussion above, the only relevant terms are as follows where the first term comes from grasps on the edges above the surface, the second from grasps on two edges below the surface and the third from the terms in which one hand grasps an edge above and the other grasps an edge below the surface. Equation (23) is an eigenvalue problem, as it can be seen from recoupling theory , since each term in the sum is proportional to the original state (see 34, 35). Therefore, we have The quantities , and are easily obtained from the recoupling theory. Using the formulas in the appendix, we obtain is obtained by replacing with in (25). has the value Substituting in (23), we have Since is diagonal, the square root in (11) can be easily taken: Adding over the intersections and using the spin notation , and , the final result is: This expression provides the complete spectrum of the area. It contains earlier results as the subset defined by and (for every ). We are very grateful to Roberto DePietri for continuous advice and for a careful reading of the manuscript, and to Jerzy Lewandowski for sharing with us his results prior to publication and for many stimulating discussions. We wish to thank the referees for useful criticisms. Appendix A Basic Formulae We present here a summary of the basic formulae of recoupling theory (in the classical case and ) used in this work. (1) The symmetrizer (2) The line exchange in a 3-Vertex Where and . (3) The evaluation of where , , . (4) The Tetrahedral net (5) The Reduction Formulae (6) The recoupling theorem: - Rovelli C and Smolin L 1988 Knot theory and quantum gravity Phys. Rev. Lett. 61 1155–1158; Rovelli C and Smolin L 1990 Loop Space Representation of Quantum General Relativity Nucl Phys B331 80–152. For various perspectives on loop quantum gravity, see: Rovelli C 1991 Ashtekar formulation of general relativity and loop space non-perturbative Quantum Gravity: A Report Class. Quantum Grav. 8 1613–1675; Ashtekar A 1995 Gravitation and Quantization (Les Houches 1992) ed B Julia and J Zinn-Justin (Paris: Elsevier Science); Smolin L 1992 Quantum Gravity and Cosmology ed J Perez-Mercader, J Sola and E Verdaguer (Singapore: World Scientific) - Ashtekar A, Rovelli C and Smolin L 1992 Weaving a classical metric with quantum threads Phys. Rev. Lett. 69 237–240 - Rovelli C and Smolin L 1995 Discreteness of Area and Volume in Quantum Gravity, Nucl. Phys. B442 593–619 - Rovelli C 1993 A generally covariant quantum theory and a prediction on quantum measurementes of geometry, Nucl. Phys. B405 797–816 - DePietri R and Rovelli C 1996 Geometry Eigenvalues and Scalar Product from Recoupling Theory in Loop Quantum Gravity, Phys. Rev. D54, 2664. - Ashtekar A and Lewandowski J 1996 Quantum Theory of Geometry I: Area Operator, gr-qc/9602046 - Garay L 1995 Quantum Gravity and minimum length Int. J. Mod. Phys. A10 145-165 - For an overview of current ideas on quantum gravity and quantum geometry, see: Isham C J 1995 Structural Issues in Quantum Gravity, lecture at the GR14 meeting (Florence) gr-qc/9510063 and Quantum Geometry and Diffeomorphism Invariant Quantum Field Theory Special Issue of the J. of Math. Phys., ed C Rovelli and L Smolin, November 1995 - Barreira M, Carfora M and Rovelli C 1996 Physics from nonperturbative quantum gravity: radiation from a quantum black hole, gr-qc/9603064, to appear in Gen. Rel. Grav. - Rovelli C 1996 Black hole entropy from loop quantum gravity, gr-qc/9603063 - Ashtekar A and Isham C J 1992 Representation of the holonomy algebras of gravity and non-Abelian gauge theories Class Quant Grav 9 1433-1467; Ashtekar A and Lewandowski J 1994 Knots and quantum gravity ed J Baez (Oxford: Oxford University Press) pp 21-61; Ashtekar A and Lewandowski J 1995 Projective techniques and functional integration for gauge theories, J. Math. Phys 36 2170–2191; Ashtekar A and Lewandowski J 1995 J. Geom. Phys. 17 191-210. Baez J 1994 Generalized measures in gauge theory, Lett. Math. Phys. 31 213–223; Baez J 1994 Proceedings of the conference on quantum topology ed D Yetter (Singapore: World Scientific); Marolf D and Mourão J M 1995 On the support of the Ashtekar-Lewandowski measure, Comm. Math. Phys. 170 583–605; Ashtekar A, Lewandowski J, Marolf D, Mourão J M and Thiemann T 1995 Quantization of diffeomorphism invariant theories of connections J. Math. Phys. 36 6456–6495; Ashtekar A, Lewandowski J, Marolf D, Mourão J M and Thiemann T 1996 Coherent-state transforms for spaces of connections, J. Funct. Anal. 135 519–551 - Lewandowski J 1996 Volume and Quantizations gr-qc/9602035 - DePietri R 1996 On the relation between the connection and the loop representation of quantum gravity, Parma University preprint, gr-qc/9605064 - Major S, Smolin L 1995 Quantum Deformations of quantum gravity, gr-qc/9512020; Borissov R, Major S, Smolin L 1995 The geometry of quantum spin networks, gr-qc/9512043 - Kauffman L H and Lins S L 1991 Temperley-Lieb Recoupling Theory and Invariants of 3-Manifolds, (Princeton: Princeton University Press) - Rovelli C and Smolin L 1995 Spin networks and Quantum Gravity Phys. Rev. D 52 5743–5759. See also: Baez J 1996 Spin network states in gauge theory, Adv. Math. 117 253–272; Baez J 1995 Spin networks in non-perturbative quantum gravity, gr-qc/9504036
Myth-"All Amplifiers Have a High-Enough Damping Factor" Where did these Myths Originate? These myths seem to trace back to a well-known paper (PDF)written by Dick Pierce. His analysis shows that a damping factor of 10 is virtually indistinguishable from a damping factor of 10,000 when it comes to damping the motion of a loudspeaker cone. This analysis has been examined and repeated in many more recent articles, such as a well-written post on Audiofrog.com by Andy Wehmeyer. Articles such as these are often cited as evidence that amplifier damping factor doesn't matter. The mathematical analyses are correct, but the conclusions are incomplete and misleading! The papers were written to bust one damping factor myth but ironically they spawned another. Damping is Not an Issue, but thereISan Issue! These papers show that the damping of driver motion cannot be significantly improved by raising the damping factor above 10. They also point out that most amplifiers easily exceed this requirement. On the surface, both papers seem to imply that damping specifications are not important. If you are one of the many people that jumped to that conclusion, you are perpetuating a myth. It is time to go back and ... ... read the papers more closely! If you take a close look at each of these two papers you will see some obscure comments that most readers seem to miss: Dick Pierce wrote: "There may be audible differences that are caused by non-zero source resistance. However, this analysis and any mode of measurement and listening demonstrates conclusively that it is not due to the changes in damping ..." Likewise, Andy Wehmeyer wrote: "There is a real benefit in low output impedance. We’ll save that one for another tech tip." In other words, both papers acknowledge that there may be some audible benefits to having a high damping factor! Unfortunately they fail to identify these benefits. These papers also fail to mention that "damping factor" was a poor choice of terminology. "Damping factor" is sort of an upside-down way of specifying the output impedance of a power amplifier. I will attempt to pick up where these papers left off and bust the myth that they inadvertently created. With some simple math, I will quantify the benefits of having a low output impedance (or high damping factor). The Frequency Response Problem The frequency response of a power amplifier is normally measured using an ideal resistive load. An amplifier with a flat frequency response specification may have an entirely different frequency response when it is loaded by the impedance of a speaker. Speakers have resistive, inductive, and capacitive characteristics that change with frequency. If the amplifier damping factor is too low, these variations will have an audible impact on the system response. Speaker Impedance and Phase Plots The plot above shows the impedance and phase angle of a set of three-way speakers that we have in our listening room. We keep these on-hand because they are an example of a difficult-to-drive "8-Ohm" speaker. The solid curve is the impedance, measured in Ohms (left axis) and the dotted curve is the phase angle of the load, measured in degrees (right axis). The horizontal axis is frequency and it has a range of 10 Hz to 50 kHz. Notice that the impedance at 119 Hz is only 2.6 Ohms. Also notice that the phase angle at 94 Hz is -53 degrees, indicating a highly capacitive load. I'll spare you the math, but the combination of these two parameters means that at 100 Hz, the amplifier will need to deliver the same current that would be required by a 2.2 Ohm resistive load. Also notice the rapid changes in impedance vs. frequency. The impedance is 19 Ohms near 30 Hz, but just 6 Ohms near the 45 Hz lower limit of the speaker's response. At 65 Hz, the impedance is almost 13 Ohms. These rapid impedance changes are typical near the lower limit of a speaker system. This speaker has crossover frequencies at 300 Hz and at 3 kHz. Notice that the 300 Hz crossover creates a bump in the impedance curve. Also notice that the impedance swings from 18 Ohms to 8 Ohms as we transition through the 3 kHz crossover frequency. The point of all of this is that speakers are not resistive loads. Speaker impedances change significantly, and often rapidly, across the audio band. These impedance variations can cause significant changes in the frequency response at the amplifier output if the amplifier's output impedance is too high. What does this have to do with damping factor? The "damping factor" is just a way of specifying the output impedance of the amplifier: "Damping Factor" is an Impedance-Ratio Specification By definition, the "damping factor" is the ratio of a speaker's nominal input impedance to the amplifier output impedance: On most amplifier specification sheets, the damping factor specifications assume an 8-ohm speaker impedance. When this is the case, the following equations can be used to determine the amplifier output impedance: Damping Factor = 8 / Amplifier Impedance Amplifier Impedance = 8 / Damping Factor Notice that the Damping Factor is high when the Amplifier Impedance is low. The Impedance Ratio (Damping Factor) Determines the Accuracy of the System Frequency Response The output impedance of the amplifier forms a voltage divider with the speaker load impedance. If the amplifier output impedance is very low (high damping factor), the resistive, inductive and capacitive portions of the speaker impedance will have little impact on the amplitude or phase response at the output of the amplifier. In contrast, if the amplifier output impedance begins to approach that of the speakers (damping factor of 1), the resistive, inductive and capacitive portions of the speaker impedance will have a major impact on the amplitude and phase response at the output of the amplifier. Damping Factor vs. Output Impedance Please understand that "damping factor" and "output impedance" specifications describe the same amplifier characteristic but they do so in an inverse fashion. An amplifier with a high damping factor will have a low output impedance. With 8-Ohm speakers, a damping factor of 10 is achieved when the amplifier output impedance is 0.8 Ohms. A damping factor of 100 is achieved when the amplifier output impedance is 0.08 Ohms. A damping factor of 1000 is achieved when the amplifier output impedance is 0.008 Ohms. The lower the output impedance, the higher the damping factor, but let's not get carried away... Speaker Cables Set a Practical Limit on the Damping Factor A typical set of 10-foot 12-AWG speaker cables will have a round-trip series resistance of 0.0318 Ohms. If these were driven by a zero-ohm amplifier (if such an amplifier existed), the damping factor would be 200. Benchmark offers 11-AWG cables that have a round-trip series resistance of 0.0252 Ohms. These would limit the damping factor to 317 if driven from a zero-ohm amplifier. In other words, damping factors greater than 200-300 can rarely be achieved at the speaker terminals. If damping factors near 1000 were important, we would need to place the amplifier very close to the drivers and/or use some very large conductors. However, don't go out and buy welding cables; large conductors have inductance problems. Skip the welding cables, as this paper will show that we don't need a damping factor of 1000. Damping Factor: 10 is too Low and 1000 is Unnecessary The frequency response of a given speaker is only repeatable, from amplifier to amplifier, when all of the amplifiers have a reasonably high damping factor. If you connect the speaker to an amplifier with a low damping factor, the speaker loading will change the frequency response at the amplifier terminals and at the acoustic output of the speaker. This change in response is especially problematic at the low end of the speaker's range and at each of the crossover frequencies. Here is a Simple Circuit Diagram: The amplifier output impedance and the speaker input impedance form a voltage divider (see Figure 2). Z1 is the amplifier impedance and Z2 is the speaker impedance. The input voltage is divided or split between Z1 and Z2. Vin is the voltage that the amplifier would produce without a load. Vout is the voltage at the terminals on the back of the power amplifier when the speaker (Z2) is connected. The second equation inFigure 2is the "transfer function". It tells us how much of the input signal reaches the output. The amplifier output impedance, Z1, is typically resistive. This means the amplifier output impedance does not change with frequency. This also means that we can refer to the amplifier output impedance as an "output resistance". In contrast, the speaker input impedance, Z2, has significant inductive and capacitive components. This means that the speaker impedance changes with frequency (as we can clearly see from solid curve in Figure 1). To keep the math simple, we will ignore the phase shifts that are produced by the inductive and capacitive characteristics of the speakers. In other words, we will ignore the phase angles shown by the dotted curve inFigure 1. This simplified analysis will give us the amplitude response of the amplifier-speaker interface. The simplified voltage divider looks like this: The transfer function is: R1 is the amplifier output impedance. R2 is the speaker input impedance at a specific frequency. In our example, R2 can be read directly from the impedance curve shown inFigure 1. Example 1 - Damping Factor of 10 At a damping factor of 10, the amplifier output impedance (R1) is: R1 = 8/10 = 0.8 Ohms FromFigure 1 we can see that our speakers have a minimum impedance of 2.6 Ohms at 119 Hz. R2 = 2.6 Ohms. Using the "transfer function" equation in Figure 3 : The signal reduction at the 2.6-Ohm impedance point will be 2.6/(2.6+0.8) = 0.714. Converting to dB: 20Log(0.714)= - 2.3 dB This means that the amplifier output will decrease by 2.3 dB at 119 Hz when loaded by the speaker impedance. But, near 3 kHz, our speaker input impedance is 18 Ohms. The signal reduction at the 18-Ohm point will be 18/(18+0.8) = 0.957. Converting to dB: 20Log(0.957) = -0.3 dB At a damping factor of 10, 119 Hz was attenuated by 2.3 dB while 3 kHz was attenuated by just 0.3 dB. The difference is 2.3 dB - 0.3 dB = 2 dB. This means that there will be a 2 dB change in the overall shape of the frequency response curve when this set of speakers is driven from an amplifier with a damping factor of 10. The 2.3 dB loss of bass near 119 Hz is significant andshould be noticeable under certain circumstances. Myth Busted! A Damping Factor of 10 is Too Low! Example 1shows that a damping factor of 10 can have an audible effect on the frequency response of the amplifier-speaker system! A damping factor of 10 is too low! Let's investigate higher damping factors ... Example 2 - Damping Factor of 100 At a damping factor of 100, the output impedance of the amplifier (R1) is: R1 = 8/100 = 0.08 Ohms. At 119 Hz, the speaker input impedance (R2) is: R2 = 2.6 Ohms The transfer function at 119 Hz is: 2.6/(2.6+0.08) = 0.970 Converting to dB, the attenuation at 119 Hz is: 20Log(0.970)= -0.26 dB At 3 kHz where R2 = 18 Ohms, the transfer function is: 18/(18+0.08) = 0.996 Converting to dB: 20Log(0.996)= -0.04 dB The difference between the two points is now 0.26 dB - 0.04 dB which is 0.22 dB. In Example 2, the damping factor of 100 keeps the frequency response variations less than about 0.2 dB. In a static situation, the listener would quickly adapt to this 0.2 dB change in speaker response and it is unlikely that the listener would detect a change in speaker voicing. For this reason, we could argue that frequency variation 0.2 dB is good enough. Nevertheless, it is possible that this 0.2 dB variation could be detected in a A/B or A/B/X test between two amplifiers having different damping factors (one with a damping factor of 100 and one with a much higher damping factor). If we match the amplitudes at 3 kHz, one amplifier could be almost 0.2 dB louder at 119 Hz using our example speakers. The general rule of thumb for A/B and A/B/X tests is that levels should be matched to better than 0.1 dB. If this is not done, the level changes can be detected by many listeners. In other words, a damping factor of 100 may have some perceptible impact on the sound in a controlled test, even if it may go unnoticed in general listening. Example 3 - Damping factor of 200 At a damping factor of 200, the output impedance of the amplifier (R1) is: R1 = 8/200 = 0.04 Ohms. At 119 Hz, the speaker input impedance (R2) is: R2 = 2.6 Ohms The transfer function at 119 Hz is: 2.6/(2.6+0.04) = 0.9848 Converting to dB, the attenuation at 119 Hz is: 20Log(0.9848)= -0.133 dB At 3 kHz where R2 = 18 Ohms, the transfer function is: 18/(18+0.04) = 0.998 Converting to dB: 20Log(0.998)= -0.019 dB The difference between the two points is now 0.13 dB - 0.02 dB which is 0.11 dB. So, at a damping factor of 200, the frequency response variations are held to about 0.1 dB which satisfies the level matching for A/B/X testing. This implies that the frequency response matching is good enough that it could not be detected in a well-controlled A/B/X test. Benchmark's Approach to Damping Factor It is no accident that the Benchmark AHB2 has a damping factor of 370. It was our goal to keep impedance-related frequency response variations less than 0.1 dB when driving 8-Ohm speakers. With our difficult-to-drive example speakers, the net frequency response variation is just 0.061 dB, easily satisfying our 0.1 dB criteria for inaudibility. However, this ignores the resistance of the speaker cables. No practical system can exist without speaker cables, especially when a pair of speakers is fed from a single stereo amplifier. Let's assume 10-foot cables. Benchmark sells 10-foot 11-AWG cables that have a round-trip series resistance of 0.0252 Ohms. When these cables are paired with the AHB2 amplifier, the effective damping factor can be calculated as follows: Amplifier output resistance = 8/370 = 0.0216 Ohms Speaker cable resistance = 0.0252 Ohms Total source resistance = 0.0216 + 0.0252 = 0.0468 Effective damping factor = 8/0.0468= 171 At a system damping factor of 171, the frequency response variations would be about0.13 dB. This still very close to our 0.1 dB criteria for inaudibility. Notice that the amplifier damping factor needed to be significantly higher than 200 in order to achieve a system-level damping factor near 200. In this system, the speaker cables and amplifier have nearly equal contributions to the total source impedance. Lowering the amplifier's source impedance (raising the damping factor) below that of the cables will produce diminishing returns. If the amplifier had an infinite damping factor, the system frequency response at the end of the cables would only improve by 0.06 dB, which is insignificant. With an amplifier damping factor of 370, we have exceeded the threshold of diminishing returns. The analysis above was focused on the amplitude response of the amplifier-speaker interface. I did this to keep the math as simple as possible. Nevertheless, I was able to show that a damping factor of 10 should have an audible effect on the amplitude response. The amplifier output impedance will also produce phase shifts when loaded by inductance or capacitance. The dotted curve in Figure 1 shows positive and negative phase angles. These phase angles will be accentuated when the source impedance of the amplifier is factored into the response. It is somewhat more difficult to quantify the audibility of these phase changes, but they should only be an issue at very low damping factors. Phase angles often change rapidly near driver crossover frequencies. InFigure 1, we can see that the phase angle changes from positive to negative as we sweep up through the 3 kHz crossover frequency. With some speakers it is possible that the source impedance of the amplifier could cause audible problems near the crossover frequencies when the damping factor is excessively low. At damping factors above 100 these effects should be minimal. At a damping factor of 10, the phase response variationsmay add additional audible differences. Loudspeaker driver damping can be achieved with damping factors as low as 10. Raising the damping factor above 10, has almost no impact on driver damping. This has been shown byDick Pierce and others. A damping factor of 10 can produce amplitude response variations exceeding 2 dB. These variations should be sufficient to create audible changes in the apparent voicing of a loudspeaker. A damping factor of 100 can produce amplitude response variations of about 0.2 dB. These may be just large enough to be detected in an A/B/X test where amplifiers are being compared, using a single speaker and a switch box. Nevertheless, the 0.2 dB variation may be small enough not to produce a noticeable change in the speaker voicing during normal listening. We can probably argue that a system-level damping factor of 100 is on the border line of audibility. A damping factor of 200 can produce amplitude response variations that are only about 0.1 dB. These variations are so small that they should not be detectable, even in a well-controlled A/B/X test where we are comparing two amplifiers with high damping factors. Raising the system-level damping factor above 200 should produce no audible change. 10-foot 12-AWG cables will limit the system damping factor to something less than 200. Benchmark's 10-foot 11-AWG cables will limit the system damping factor to something less than 317. Because of cable resistance, system-level damping factors exceeding 200 are often impractical. Frequency response variations are insignificant when system damping factors exceed 150. I would recommend targeting a system-level damping factor above 150. This range is achievable and it will give excellent performance. To achieve this at the end of a set of speaker cables, the amplifier will need a damping factor of at least 300. A system-level damping factor of 171 can be achieved at the end of a Benchmark 10-foot 11-Ga cable when driven by a Benchmark AHB2 amplifier which has a damping factor of 370. Given 11 or 12 AWG cables, a system-level damping factor of 100 can be achieved with an amplifier damping factor of about 150. Likewise, a system-level damping factor of 150 can be achieved with an amplifier damping factor of about 300. Amplifiers with a damping factor of less than 300 may have an audible impact on the frequency response of the playback system. Damping Factor Calculator with Frequency Response I have created a spreadsheet that you can use to calculate the frequency response variations that are caused by the amplifier output impedance and cable impedance. The image below shows the results for the example system described in this paper. Open the spreadsheet and enter the parameters for your system in the orange cells as follows: Enter the nominal impedance of your speakers Enter the minimum impedance of your speakers Enter the maximum impedance of your speakers Enter the length of your speaker cables Enter the gauge of your speaker cables The "Total Error" column will show the frequency response variations that are produced by the interactions between the source and load impedances. This column is color coded to indicate the expected audibility of the frequency response variations. Red indicates a high likelihood of audibility while green indicates a low likelihood of audibility. Find the row that corresponds to your amplifier's 8-Ohm damping factor in the first column. Use your amplifier's 8-Ohm damping factor specification even if your speakers have a different nominal impedance. The spreadsheet makes the appropriate adjustments. The "Effective Damping Factor" column shows the damping factor achieved at the speaker-end of the cable. It is also adjusted for the nominal impedance of the speakers. Notice that the speaker cables reduce the effective damping factor. If you want to ignore the cable effects, set the cable length to 0. The highlighted row (damping factor of 370) corresponds to the performance of the Benchmark AHB2. The spreadsheet is completely unlocked and it does not contain any macros. Avoid making entries outside of the orange cells unless you wish to modify the spreadsheet or examine the formulas (which were explained in the examples). I hope you find this calculator helpful. At the 2023 AXPONA show in Chicago, I had the opportunity to see and hear the Hill Plasmatronics tweeter. I also had the great pleasure of meeting Dr. Alan Hill, the physicist who invented this unique device. The plasma driver has no moving parts and no diaphragm. Sound is emitted directly from the thermal expansion and contraction of an electrically sustained plasma. The plasma is generated within a stream of helium gas. In the demonstration, there was a large helium tank on the floor with a sufficient supply for several hours of listening. While a tank of helium, tubing, high voltage power supplies, and the smell of smoke may not be appropriate for every living room, this was absolutely the best thing I experienced at the show! If an audio system is composed of multiple components, we may have detailed specifications for each component, but we will not know the performance of the combined system without doing some calculations. You may have questions such as Will my audio system produce audible noise? Will my audio system produce audible distortion? How will my audio components work together as a system? How loud will my audio system play? Use Benchmark's online audio calculators to find answers! For example, if we know the output power of an amplifier, as well as the sensitivity and impedance of our loudspeakers, we can calculate the maximum sound pressure level that our system can produce. This application note provides interactive examples that help to answer the questions listed above.
Nonlinear Dynamics in Applied Sciences Systems: Advances and PerspectivesView this Special Issue Uniform Asymptotic Stability of Solutions of Fractional Functional Differential Equations Some global existence and uniform asymptotic stability results for fractional functional differential equations are proved. It is worth mentioning that when the initial value problem reduces to a classical dissipative differential equation with delays as in Caraballo et al.'s work (2005). Consider the initial value problem (IVP for short) of the following fractional functional differential equation: where is the Caputo fractional derivative, , , where , is a given function satisfying some assumptions that will be specified later, , and . If , then for any , define by The study of retarded differential equations is an important area of applied mathematics due to physical reasons, noninstant transmission phenomena, memory processes, and specially biological motivations (see, e.g., [1–4]). Fractional differential equations have attracted much attention recently (see, e.g., [5–11] and the references cited therein for the applications in various sciences such as physics, mechanics, chemistry, and engineering). Some attractive results for fractional functional differential equations and nonlinear functional integral equations are obtained by using the fixed point theory; see [12–16] and references therein. Global asymptotic stability of solutions of a functional integral equation is discussed in ; however, there is no work on uniform asymptotic stability of solutions of fractional functional differential equation. It is our intention here to show the global existence and uniform asymptotic stability of the fractional functional differential equation (1). We organize the paper as follows. In Section 2, we recall some necessary concepts and results. In Section 3 we give the global existence and uniform asymptotic stability of fractional functional differential equations. Finally, two examples are given to illustrate our main results. In this section, we introduce notations, definitions, and preliminary facts which are used throughout this paper. We consider the Banach space of all bounded and continuous functions from into with the norm Let for . Throughout this paper, we always assume that satisfies the following condition:(H0) is Lebesgue measurable with respect to on , and is continuous with respect to on . By condition and the technique used in , we get the equivalent form of IVP (1) as where is the gamma function. Definition 1. We say that the solutions and of IVP (1) are uniformly asymptotically stable if for any bounded subset of and , there exists a such that We recall the following generalization of Gronwall’s lemma for singular kernels , which will be used in the sequel. Lemma 2. Let be a real function and is a nonnegative, locally integrable function on and there are constants and such that Then there exists a constant such that for every . Theorem 3 (Leray-Schauder fixed-point theorem). Let be a continuous and compact mapping of a Banach space into itself, such that the set is bounded. Then, has a fixed point. 3. FDEs of Fractional Order Theorem 4. Assume that satisfies conditions and(H1)there exists such that for and every . Moreover, the function is bounded with . If then the IVP (1) has a unique solution in the space . Moreover, solutions of IVP (1) are uniformly asymptotically stable. Proof. We divide the proof into two steps. Step 1. We define the operator by The operator maps into itself. Indeed for each , and for each , it follows from that For each , we have and consequently . Since is a Banach space with norm , we will show that is a contraction map. Let . Then, we have for each , Therefore, for any , and for , and thus Hence, (10) and (17) imply that the operator is a contraction. Therefore, has a unique fixed point by Banach’s contraction principle. Step 2. For any two solutions and of IVP (1) corresponding to initial values and , by (4) we can deduce that for all and all , Then, it follows that Let . Then, we have Applying Lemma 2, one can see that there exists a constant such that Hence, we obtain and thus for all , which implies that the solutions of IVP (1) are uniformly asymptotically stable. Now we give global existence and uniform asymptotic stability results based on the nonlinear alternative of Leray-Schauder type. Theorem 5. Assume that the following hypotheses hold: (H2) is a continuous function;(H3)there exist positive functions , such that for and every ;(H4)moreover, assume that Then the IVP (1) admits a solution in the space . Moreover, solutions of IVP (1) are uniformly asymptotically stable. Proof. Let be defined as in (11). First, we show that maps into itself. Let , . Indeed, the map is continuous on for each , and for each , implies that for each , we have and for any , and consequently . Next, we show that the operator is continuous and completely continuous, and there exists an open set with for and . Step 1 ( is continuous). Let be a sequence such that in . Then, there exist and such that Let be given. Since holds, there is a real number such that for all . Now we consider the following two cases. Case 1. If , then it follows from and (30)-(31) that for sufficiently large Case 2. If , since is a continuous function, one has Note that in . Hence, (32) and (33) imply that Step 2 (P maps bounded sets into bounded sets in ). Indeed, it is enough to show that for any , there exists a positive constant such that for each one has . Let . Then, we have for each , and for each with , Step 3 ( maps bounded sets into equicontinuous sets on every compact subset of ). Let , , and let be a bounded set of as in Step 2. Let . Then, we have Observing that from Taylor’s theorem, we obtain where . By (37)–(39), we can conclude that As , the right-hand side of the above inequality tends to zero. The equicontinuity for the cases and is obvious. Step 4 ( maps bounded sets into equiconvergent sets). Let . Then Therefore, implies that uniformly (with respect to ) converges to as . As a consequence of Steps 1–4, we can conclude that is continuous and completely continuous. Step 5 (a priori bounds). We now show that there exists an open set with for and . Let and for some . Then, for each , we obtain By , we have that for all and , and thus From the arguments in (26)-(27), we can conclude that for each , Hence, Let . Then, from Lemma 2, there exists such that we have for all , Since , there exists such that Set is continuous and completely continuous. From the choice of , there is no such that , for . As a consequence of Leray-Schauder fixed-point theorem, we deduce that has a fixed point in . Step 6 (uniform asymptotic stability of solutions). Let be bounded; that is, there exists such that From the similar arguments in Step 4, we can deduce that there exists such that for all solutions of IVP (1) with initial data , we have Now we consider two solutions and of IVP (1) corresponding to the initial values and . Note that for all , Then, the proof of uniform asymptotic stability of solutions can be done by making use of and (52). The proof of Theorem 5 is completed. Example 1. Consider the fractional functional differential equation where . It is clear that condition holds. Let , . Then for all , we have On the other hand, note that for each and . Hence, conditions and (10) hold. By Theorem 4, we conclude that IVP (53) has a unique solution in the space , and the solution of IVP (53) is uniformly asymptotically stable. Example 2. Consider the fractional functional differential equation where . It is easy to see that condition holds. Let . Then, for all , we find that where with and Thus, conditions and hold, and the global existence and the uniform asymptotic stability of solutions of IVP (55) can be obtained by applying Theorem 5. By using the algorithm given in , we numerically simulate Example 1 with the initial conditions , and Example 2 with ; see Figures 1 and 2. From the numerical results, it can be noted that both of the solutions of Examples 1 and 2 converge uniformly, and the solutions of Example 1 converge faster than the ones of Example 2. The numerical results confirm the theoretical analysis. Conflict of Interests The authors declare that there is no conflict of interests regarding the publication of this paper. This research is supported by the National Natural Science Foundation of China under Grants nos. 10801066 and 11271173 and the Fundamental Research Funds for the Central Universities under Grant nos. lzujbky-2011-47 and lzujbky-2012-k26. The project is sponsored by the Scientific Research Foundation for the Returned Overseas Chinese Scholars, State Education Ministry. J. K. Hale and S. M. Verduyn Lunel, Introduction to Functional Differential Equations, Springer, New York, NY, USA, 1993. Y. Kuang, Delay Differential Equations with Applications in Population Dynamics, vol. 191 of Mathematics in Science and Engineering, Academic Press, New York, NY, USA, 1993.View at: MathSciNet M. C. Mackey and L. Glass, “Oscillation and chaos in physiological control systems,” Science, vol. 197, pp. 287–289, 1977.View at: Google Scholar M. Benchohra, J. Henderson, S. K. Ntouyas, and A. Ouahab, “Existence results for fractional order functional differential equations with infinite delay,” Journal of Mathematical Analysis and Applications, vol. 338, no. 2, pp. 1340–1350, 2008.View at: Publisher Site \| Google Scholar \| Zentralblatt MATH \| MathSciNet I. Podlubny, Fractional Differential Equations, vol. 198 of Mathematics in Science and Engineering, Academic Press, San Diego, Calif, USA, 1999.View at: MathSciNet D. Henry, Geometric Theory of Semilinear Parabolic Partial Differential Equations, Springer, Berlin, Germany, 1989.
Newton's Laws of Motion Study Guide (page 2) From the previous chapters on physical quantities and their corresponding units, you now know that the unit for mass, kilogram, is a fundamental unit. Mass might also be called a fundamental quantity. It relates to motion, forces, and energy, and therefore basically to the whole field of physics. In the following lesson, we will discuss inertia, Newton's laws of classical mechanics that connect motion to forces, and conditions for an equilibrium or non-equilibrium situation when an object is subjected to two or more forces at the same time. Imagine a situation where you have to push some furniture around, for example, if you are moving to a new home. Although you might think that taking clothes out of the dresser drawers before moving the dresser is a waste of time, the reality is that the amount of effort you put into unloading the dresser is much less than the effort you would use to move the full dresser. What opposition do we encounter in the two processes of moving the furniture? It is the mass that makes the difference; it is the mass that causes the resistance. This resistance that an object has to motion is something we call inertia. Mass is the quantity that measures inertia, and the fundamental unit for mass is the kilogram (kg). Motion on a straight trajectory is called linear motion. Later, we will study objects moving in a circular trajectory, and they too will exhibit inertia; in this case, called rotational inertia. Newton's First Law of Motion Newton's first law of motion, or the law of inertia, states that an object continues to stay at rest or move in a uniform linear motion as long as there is no resultant (net) action on it. Newton's First, Second, and Third Laws of Motion The idea of resistance to motion was formulated by Newton into what we now call the first law of motion or the law of inertia. There are two important words in this definition: continues and resultant. To understand this law, you must work with these two words together. Resultant (net) action refers to the idea that there might be an action on the object, or two, or more. The importance is not in individual actions but in the total effect on the object. The other word, continues, refers to one of the two situations: rest or uniform linear motion. That is, the object will maintain, or continue, an identical state of rest or motion at all times if there is no external action. Most objects around us, when left to act independently of engines, will not continue their uniform linear motion. Also, objects that might be at rest at one time can change their state and accelerate. The motion is a result of interaction, and the objects accelerate (speed up) or decelerate (slow down). The change in the state of motion is due to a resultant force. The relationship among the resultant force, the mass, and the acceleration is the subject of the second law of motion. As you may remember, physical quantities can be scalars or vectors. In this case, mass is a scalar, but both acceleration and force are vector quantities. Therefore, the second law does not refer only to the proportionality but also to the direction of the acceleration, telling us that the acceleration and the resultant force are in the same direction. The expression of the second law is as follows: a = , or where a is the vector acceleration, F is the vector representing the resultant force, and m is the mass of the object. This equation defines also the unit for force: a newton, or N. As we replace the unit for acceleration (m/s2) and for mass (kg), we can find the unit for force to be kg . m/s2 = 1 N. As with all vectors, force will be completely defined by value, unit, and direction. The direction of the net force will also give the type of acceleration; if a > 0 (a positive number is considered to be the direction of motion), then the object is accelerated, and if a < 0 (opposite to motion), then the object is decelerated (slowed down). Newton's Second Law of Motion The acceleration of an object is directly proportional to the resultant (net) force and inversely proportional to the mass of the object. A 1,000 kg car starts from a traffic light and moves accelerated in the NE direction with an acceleration of 0.45 m/s2. Find out the resultant force on the car. in NE direction F = m · a 1,000 kg · 0.45 m/s2 = 450 N F = 450 N in NE direction A 2-D force with components Fx = 100 N and Fy = 45 N accelerates an object of mass 5 kg. If this force is the net force on the object, find the value of the acceleration and its direction relative to the x-axis. According to the second law of motion, we need to divide the force in each direction to the mass of the object to find the two components of the acceleration. Fx = 100 N Fy = 45 N m = 5.0 kg Because the two components are 90° apart, we can apply Pitagora's theorem to find the acceleration: To find the direction relative to x-axis, α, we calculate the tangent of the angle and find the angle. α = 24° So, acceleration can be written in two ways: a = (20,9) m/s2 for each component, or a = 22 m/s2 at 24° north of east The net force described by Newton's second law or any other force is, by definition, a measure of the interaction. Whenever you interact with an object, you will feel the effect of the interaction yourself. When you kick a ball, the ball will react (due to the interaction) by shooting into the air, but your foot will "feel" the interaction also. This effect is described by the third law of motion. In this case, one force is called action and the other one reaction. As a very important note: The two forces are acting on different objects and so, although equal and opposite, they still have an effect on the system. Newton's Third Law of Motion If you act with a force on an object, the object will react with a force equal and opposite on the first object. Consider a car moving on a road. Determine the action and reaction forces on each tire. Let's look at the situation for one tire, and the same can be repeated for the remaining tires. The car's left tire acts on the road with a force that we call action, and the road reacts with an equal and opposite force on the tire called reaction. That is how the car can move. Let's put to work the knowledge we've accumulated and define some important types of forces in classical mechanics, one of which is weight. Whenever two objects are in proximity, they interact through a force. If you consider Earth and an apple that is about to fall from a tree, the apple is attracted by Earth with the force of gravity, so it will accelerate toward the earth for as long as it falls (if we do not consider other forces present such as friction with air). Because the gravity is the only force on the apple, it will be the only component in the net force. According to the second law, this force is also proportional to the mass of the apple and the acceleration, and we call it weight. The expression of the weight of an object of mass m is: W = m · g The symbols in the expression are: W is weight, m is mass, and g is the acceleration. It is very important to note that this expression gives us the argument to answer the question: Are mass and weight the same quantity? The answer is No! Weight is a force, and mass is a scalar quantity that gives a measure of the inertia of the object. In the previous example, the apple falling down due to gravity is in a free fall, and the acceleration of fall, which is the same for all objects at the same place on the earth, is called acceleration due to gravity, g. Although the acceleration due to gravity varies with the mass of the planet and the distance from an object to the center of the planet, for most purposes herein, we will take the acceleration due to gravity on Earth to be g = 9.8 m/s2. Another and last very important note for this section is that weight is a vector always pointing toward the center of Earth. Consider the apple falling freely toward Earth. Why is Earth not falling toward the apple? The third law of motion says that every action has an equal and opposite reaction, so the apple is acted on by gravity, and hence, Earth has to be acted on by an equal and opposite force. The reason the earth does not start moving toward the apple is that the same force acting on a huge object (Earth) has a negligible effect while having a definite effect on a smaller object (the apple). The fact is that not all objects fall or fly; most of them are at rest or moving in contact with different surfaces. In such a case, we have an action, weight, and a reaction from the surface toward the object, and we call that a normal force, with a symbol N. In the simplest case, an object sits on a horizontal surface and pushes down only with its own weight; the surface responds with an equal and opposite force as shown in Figure 5.4. A drawing that includes all the forces acting on an object is called a free-body diagram. When the surface is no longer horizontal, part of the weight will not act on the supporting surface, so the normal force of the object is now smaller than the weight of the object as seen in Figure 5.5. In this case, the component of the weight, called Wx , is the one acting on the surface, and normally is: Wx= – N The other component of the weight is not equalized by any other force in this case, and therefore, the object will accelerate toward the bottom of the plane. In real-life applications, there will be at least one other force acting. Yes, you guessed it: friction. But we will talk about this later. In many real-life applications, you will have to take into consideration other forces acting on the object: pushes, pulls, and so on. In these cases, the freebody diagram will be more complex, and the calculation of the normal force can result in amounts smaller or larger than the weight, as we will show in the following example. - Kindergarten Sight Words List - First Grade Sight Words List - Social Cognitive Theory - 10 Fun Activities for Children with Autism - Child Development Theories - Signs Your Child Might Have Asperger's Syndrome - A Teacher's Guide to Differentiating Instruction - Why is Play Important? Social and Emotional Development, Physical Development, Creative Development - Theories of Learning - Curriculum Definition
Basic Course Statistics By Cooke G.M. Clarke Statistics is a vital subject for understanding and analyzing large amounts of data. This skill is important for a variety of professions, including business, science, and engineering. In this article, we will examine the fundamental concepts of course statistics, including descriptive statistics, probability theory, hypothesis testing, and regression analysis. Additionally, we will explore how technology has revolutionized the field of statistics and tips for interpreting and presenting statistical data for different audiences. Descriptive statistics refer to the summary and analysis of data, which provides an overview of the characteristics of the data. This includes measures of central tendency (mean, median, mode), measures of variability (range, variance, standard deviation), and measures of association (correlation coefficients). These statistics are useful in understanding the distribution and spread of variables in data sets, enabling researchers to effectively summarize and interpret variables. For example, imagine a business analyst is analyzing the sales data of a retail store. The analyst uses measures of central tendency, such as the average sales per day, to determine the average daily sales for a given period. Measures of variability, such as the standard deviation of sales, help to show how far the sales deviate from the average, highlighting the level of variance within sales data. Probability theory is the study of how likely a particular event is to occur. Understanding probability theory is crucial in making decisions and predictions based on data. Probability theory provides tools for predicting the likelihood of future occurrences based on past performance. For instance, in a medical study, probability theory enables the scientists to determine the likelihood of a drug succeeding or failing given previous experiments. Probability theory also allows analysts to calculate the chance of two events occurring simultaneously (joint probability), or the probability that an event will occur given the occurrence of another event (conditional probability). Hypothesis testing is a statistical method that involves testing a hypothesis through the use of sample data. It involves making an assumption about a population and then testing the accuracy of that assumption using data. The results of the test help determine the confidence level in accepting or rejecting the hypothesis. For example, suppose a manufacturing company produces products in two different factories, and the management wants to know the average quality of products produced in each factory. Using hypothesis testing, the company can determine whether there is a significant difference in average quality between the two factories. Regression analysis refers to a statistical technique that determines the relationship between a dependent variable and one or more independent variables. It is crucial in making predictions when the relationships among variables are complex. For example, in economics, regression analysis is employed to understand how different independent variables such as consumer demand, income, and price all impact the dependent variable, such as sales. It allows analysts to determine how closely related the variables are and the extent to which they impact the dependent variable. Role of Technology in Statistics The use of technology has transformed the field of statistics, allowing for faster and more accurate analyses, interpretation, and presentation of data. Software applications such as SPSS, R, MATLAB, and STATA have become popular tools among data analysts and researchers. These tools provide more precise and efficient ways of analyzing data, allowing analysts to perform more sophisticated statistical procedures. For instance, these software applications allow for the creation of graphs, charts, and visualizations that help to present data in a meaningful and coherent manner. Effective Interpretation and Presentation of Statistical Data Interpretation and presentation of statistical data require skill and attention to detail. Data must be presented accurately, and any conclusions drawn from the analysis must be supported by evidence. Generally, you should consider the audience when presenting statistical data. For non-technical audiences, such as business leaders or the general public, charts and visualizations help simplify complex data and make it more understandable. For a more data-savvy audience, such as researchers or academics, statistics, tables, and images may help clarify the data. Another critical aspect of data presentation and interpretation is avoiding statistical misuse. Data can be misrepresented to support a particular point of view or agenda. It is essential to present data accurately and without bias, without distorting statistical facts to skew the results. In conclusion, understanding the basic course statistics concepts is essential for data analysts and researchers. Descriptive statistics provides an overview of data characteristics, probability theory allows for predicting future occurrences based on past performance, hypothesis testing assesses the accuracy of a hypothesis based on available data, and regression analysis determines the relationship between dependent and independent variables. Advancements in technology have revolutionized the field of statistics, providing more efficient ways of analyzing data, which translated into faster and more accurate results. The interpretation and presentation of statistical data require special attention and skill, taking into account the target audience and avoiding statistical misuse. Whether working in business, science, engineering, or any other field, knowledge of basic course statistics is fundamental to understanding, analyzing, and visualizing data. These statistics are the foundation of data-driven insights that can lead to informed decisions and predictions, ultimately leading to efficient and successful outcomes.
National Collaborating Centre for Mental Health, Centre for Outcomes Research and Effectiveness, Research Department of Clinical, Educational and Health Psychology, University College London, London, UK; National Collaborating Centre for Mental Health, Centre for Outcomes Research and Effectiveness, Research Department of Clinical, Educational and Health Psychology, University College London, London, UK; A.E. Ades, Department of Community Based Medicine, University of Bristol, Cotham House, Cotham Hill, Bristol BS6 6RT, UK. E-mail: email@example.com Background: Cost-effectiveness analysis often requires information on the effectiveness of interventions on multiple outcomes, and commonly these take the form of competing risks. Nevertheless, methods for synthesis of randomized controlled trials with competing risk outcomes are limited. Objective: The aim of this study was to develop and illustrate flexible evidence synthesis methods for trials reporting competing risk results, which allow for studies with different follow-up times, and that take account of the statistical dependencies between outcomes, regardless of the number of outcomes and treatments. Methods: We propose a competing risk meta-analysis based on hazards, rather than probabilities, estimated in a Bayesian Markov chain Monte Carlo (MCMC) framework using WinBUGS software. Our approach builds on existing work on mixed treatment comparison (network) meta-analysis, which can be applied to any number of treatments, and any number of competing outcomes, and to data sets with varying follow-up times. We show how a fixed effect model can be estimated, and two random treatment effect models with alternative structures for between-trial variation. We suggest methods for choosing between these alternative models. Results: We illustrate the methods by applying them to a data set involving 17 trials comparing nine antipsychotic treatments for schizophrenia including placebo, on three competing outcomes: relapse, discontinuation because of intolerable side effects, and discontinuation for other reasons. Conclusions: Bayesian MCMC provides a flexible framework for synthesis of competing risk outcomes with multiple treatments, particularly suitable for embedding within probabilistic cost-effectiveness analysis. It is common for randomized controlled trials (RCTs) to report more than one outcome. For purposes of designing a trial, it is generally felt that a single outcome should be prespecified to be the “primary” outcome. But, it is also recognized that, when pooling results from trials in a meta-analysis, there are several reasons why it may be appropriate to combine information on different outcomes. First, one might wish to “gather strength” by combining several similar outcomes, or to be able to combine results from trials that report different, but similar, outcomes [1,2]. Second, in a decision-making context, the different outcomes recorded may each have separate implications for estimating quality of life or economic consequences of each treatment. A key requirement in the synthesis of multiple outcomes is that the correlation structures are appropriately represented [1–4]. In a meta-analysis, the correlations may occur at either or both of two levels. At the between-patient within-trial level, a patient's outcome on one measure may be positively or negatively correlated with their outcome on another. At the between-trial level, trials in which there is a larger treatment effect on one measure may tend to be the trials on which there is a larger treatment effect on another (a positive correlation), or possibly a smaller one (a negative correlation). Competing risk outcomes represent a special type of multiple outcome structure in which there are several different failure time outcomes that are considered mutually exclusive. Once a patient has reached any one of these end points, they are considered to be out of the risk set. Censoring may also be occurring. When results from these trials are pooled in a meta-analysis, the competing risk structure should be taken into account so that the statistical dependencies between outcomes are correctly reflected in the analysis. These dependencies are essentially within-trial, negative correlations between outcomes, applying in each arm of each trial. They arise because the occurrence of outcome events is a stochastic process, and if more patients should by chance reach one outcome, then fewer must reach the others. The importance of these correlations in the context of meta-analysis of competing risk outcomes has been recognized by Trikalinos and Olkin , who suggest an approach based on normal approximation of the variances and covariances arising from multinomial data, and illustrate it with an application to a two-treatment meta-analysis with two competing outcomes. In this article, we present an alternative approach based on hazards rather than probabilities, to more appropriately take account of time at risk. We use Bayesian Markov Chain Monte Carlo (MCMC) estimation , which we believe is more flexible in a situation with large numbers of treatments and outcomes. We begin by describing the data set, and we then explain the “proportional competing risks” assumption that underlies our approach. We next propose three alternative models: one fixed effects, and two random effects analyses. We also suggest some methods for model selection. In our discussion of the results, we compare our proposed method to previously described approaches, and consider some possible extensions. Illustrative Data set Figure 1 shows the network of comparisons of trials of antipsychotic medication for the prevention of relapse in people with schizophrenia. Each “edge” in the network indicates that the treatments at either end have been compared in an RCT, and the number on the edge indicates the number of trials. The data set includes 17 trials comparing nine treatments including placebo; eight of the trials are placebo controlled. There are 36 possible pair-wise contrasts between the nine treatments, and the present data set provides direct evidence on 11 of them. The methods used to identify these studies, and the criteria for inclusion and exclusion in the data set have been described previously . Briefly, a systematic search of the literature was undertaken to identify double-blind RCTs of antipsychotics used for relapse prevention in people with schizophrenia who are in remission. The review was conducted during the update of a clinical guideline on schizophrenia, commissioned by the National Institute for Health and Clinical Excellence in the UK . Although ziprasidone was not considered during the formulation of guideline recommendations, as it is not licensed in the UK, ziprasidone trials were included in the systematic review (and subsequently this analysis) to strengthen inference about the relative effect between other treatments. The analysis described in this article was similar to the one used to populate the decision-analytic economic model that informed the guideline recommendations . The data available from each trial are the number of patients in each of the three outcome states at the end of follow-up. The outcome states are: relapse, discontinuation of treatment because of intolerable side effects, and discontinuation for other reasons, which might include inefficacy of treatment that did not fulfill all criteria for relapse, or loss to follow-up. Patients not reaching any of these end points at the end of follow-up were considered as censored observations, and still in remission. Individual patient data with times of transition were not available. Study follow-up varied from 26 to 104 weeks. The available data are shown in full in Table 1. Three trials comparing olanzapine and haloperidol were pooled as if they were a single study, because the original publication trials did not report all three outcomes separately. Table 1. Trials of treatments for schizophrenia Each cell gives: numbers of patients who relapse, discontinue because of side effects, discontinue for other reasons, and the denominator at initially at risk. Citations for included studies can be found in . The Tran 1998 data represent pooled results from three trials in which discontinuation data were not separately for each trial. 1. Beasley 2003 28, 12, 15; 102 9, 2, 19; 224 2. Dellva 1997 – 1 7, 0, 4; 13 10, 2, 16; 45 3. Dellva 1997 – 2 5, 2, 5; 14 6, 10, 15; 48 4. Loo 1997 5; 5; 39; 72 4, 1, 26; 69 5. Cooper 2000 21, 4, 24; 58 4, 16, 21; 61 6. Pigott 2003 85, 13, 12; 155 50, 16, 18; 155 7. Arato 2002 43, 11, 7; 71 71, 19, 28; 206 8. Kramer 2007 52, 1, 7; 101 23, 3, 17; 104 9. Simpson 2005 11, 6, 44; 71 8, 5, 33; 55 10. Tran 1998 87, 54, 170; 627 34, 20, 50; 180 11. Study S029 28, 9, 26; 141 29, 14, 25; 134 12. Tran 1997 20, 17, 36; 172 53, 17, 18; 167 13. Speller 1997 5, 3, 2; 29 9, 5, 2; 31 14. Csernansky 2000 65, 29, 80; 188 41, 22, 60; 177 15. Marder 2003 8, 0, 4; 30 4, 3, 4; 33 We begin with a general formulation for competing risks based on standard results from survival analysis . If λm(t) is the cause-specific hazard at time t for outcome m, then the conditional probability that failure at time t is of type m, given there is a failure at time t is The probability that failure occurs before time D and is of type m is: (the probability of surviving to t, times the probability of failure at t, times the conditional probability that failure is of type m), integrated over all times t between zero and D. This is: A form of “proportional hazards” assumption can now be made, which might be better termed “proportional competing risks,” in which the ratio (Equation 1) is constant over all t (i.e., πm(t) = πm). Under this restriction, Equation 2 becomes: which is the probability of failure before time D times the probability that failure was of type m. In what follows, we assume constant underlying hazards, but Equation 3 shows that with proportional competing risks we are free to fit more complex survival distributions. This suggests some useful extensions to which we return in the discussion. We now number the treatments from 1 to 9 (as shown in Table 1 and Fig. 1). Placebo is selected as the reference treatment 1. This is an arbitrary choice, but made to ease interpretation. We define the three outcomes as: m = 1 relapse, 2 = discontinuation caused by side effects, and 3 = discontinuation for other reasons. Then, each outcome is modeled separately on the log hazard rate scale. Considering a trial j that compares treatments k and b, the cause-specific log hazard for outcome m for treatment T is: where δj,b,k,m is the trial-specific log hazard ratio of treatment k relative to treatment b for outcome m. This can be interpreted as meaning that the b arm of the trial estimates the baseline log hazard µj,m, while the k arm estimates the sum of the baseline hazard and the log hazard ratio. Note that b is not necessarily treatment 1, nor is it the same treatment in every trial; instead, it is simply the treatment with the lowest index in that trial. Thus, in a trial comparing treatments 2 and 3, b = 2. The trial-specific log hazard ratios are assumed to come from a common normal distribution, following the standard “random effects” approach: The mean of this distribution is a difference between mean relative effects dk,m and db,m, which are the mean effects of treatments k and b, respectively, relative to (placebo) treatment 1, for outcome m, and we define d1,m = 0. This formulation of the problem expresses the consistency equations , by which the dimensionality of the 11 treatment contrasts on which there are direct data (Table 1 and Fig. 1), are reduced to functions of the eight contrasts between the active treatments and placebo. The between-trial variance of the random effect distribution, , is specific to each outcome m. Three models for the variance are considered below. We may write the model as Equation 4 because all the trials in this example are two-arm trials. An advantage of our approach, however, is that it can be readily extended to multi-arm trials, and Equation 5 should in fact be interpreted as a “fragment” of a multivariate normal distribution: The underlying assumption in Equations 5 and 6 is therefore that every trial may be considered as if it were a multiarm trial on all nine treatments, that trial-specific relative treatment effects are sampled from the multivariate normal in Equation 6, and that treatments are missing at random. (Note that missing at random means missing without regard to treatment efficacy; it does not mean that treatment arms are equally likely to be included in a trial). The linking function that relates the arm-specific log hazards θj,k,m to the likelihood is developed as follows. Figure 2 shows a Markov transition model with a starting state (remission) and three absorbing states (relapse, discontinuation caused by side effects, and discontinuation caused by other reasons). Based on Equation 3, if we assume constant hazards λj,k,m acting over the period of observation Dj in years, the probability that outcome m had occurred by the end of the observation period for treatment T in trial j is: The probability of staying in the remission state (m = 4) is now simply 1 minus the sum of the probabilities of arriving at the three absorbing states, that is, The data for each trial j and treatment T constitute a multinomial likelihood with four outcomes: moving to one of the three absorbing states, or remaining in the initial remission state. If rj,T,m is the number of patients on treatment T observed to reach end point m, and nj,T is the total number at risk on treatment T in trial j, then: Three different models were fitted, differing solely in the specification of the between-trial variation in relative treatment effects, . In the fixed effects model, , and the model collapses to: θj,T,m = µj,m + dk,m − db,m, b = 1,2 . . . 8, k ≥ b. In the random effect single-variance model, the between-trial variance , reflecting the assumption that the between-trial variation is the same for each outcome: a vague inverse gamma prior was put on the variance, 1/σ2∼ gamma (0.001, 0.001). In the random effect different variances model, each outcome has a different between-trial variation, and the vague uniform prior is put on each: . A sensitivity analysis based on uniform priors was also examined: σ ∼ uniform (0, 5). This gave virtually identical posteriors for the treatment effects, but resulted in posterior distributions with “spikes” at σm values at or close to zero and spikes in the posterior mean treatment effects. Gamma priors, which give zero weight to infinite precision and hence to zero SD, were therefore used in the primary analyses reported below. Finally, in each of the three models, vague Normal (0, 1002) priors were put on all the trial baselines µj,m and mean treatment effects dk,m. The model for treatment effects (Equations 4 and 5) is therefore identical to that previously proposed for mixed treatment comparisons (MTCs) except that the multinomial likelihood (Equation 9) and linking function (Equation 7) are used, as is appropriate for the data at hand, in place of the binomial likelihood and logit link function proposed in most of previous work on these kinds of evidence structures [9–13]. Choice of models was based on the deviance information criterion . This is a deviance measure of goodness of fit, , equal to the posterior mean of minus twice the log likelihood, penalized by an estimate of the effective number of parameters in the model, pD. The DIC can be seen as a Bayesian measure analogous to the Akaike information criterion used in classical analysis, but which can also be applied to hierarchical models. Here, we adjust the standard deviance formula by subtracting the deviance of the saturated model (a constant). The contribution of each multinomial observation (trial j treatment T) to the deviance is: where is an estimate of the probability of outcome m, for example, the estimate generated on some MCMC cycle, and is the posterior mean of the sum of the deviance contributions over all data points, . In a model that fits well, is expected to roughly approximate the number of data points. In this data set, with 15 two-arm trials each reporting three outcomes (the fourth, “censored” outcome is predicted from the number at risk less the other outcomes), the number of independent data points is 90; pD is equal to , where is the sum of the deviance contributions, evaluated at the posterior mean of the fitted values . Models were estimated using the freely available Bayesian MCMC software WinBUGS 1.4.3 . Convergence for all models occurred within 10,000 to 25,000 iterations as assessed by the Brooks–Gelman–Rubin criteria . Results are based on 150,000 samples, from three separate chains with disparate starting values for fixed effect, and five chains for the random effect models, taken after the first 60,000 were discarded. We also established that, in each model, all the chains converged to the same posterior. The code for each model is available on the ISPOR Web site as Supporting Information, in Appendix A at: http://www.ispor.org/Publications/value/ViHsupplementary/ViH13i8_Ades.asp. The global goodness-of-fit statistics rule out the fixed effects model (Table 2), which fits relatively poorly: the mean posterior deviance is 119.8 compared to the number of data points, 90. This model also has the highest DIC because the relatively poor fit is not sufficiently compensated by the fact that it has fewer effective parameters. The DIC results slightly favor the three-variance model over the single-variance model. Table 2. Goodness of fit, relative effects (posterior mean log hazard rates dk,m relative to placebo), and between-trial heterogeneity (posterior median and credible intervals for between-trial SD), for fixed and random effect models Random effect single variance Random effect three variances Goodness of fit statistics 1. Placebo (ref) Discontinuation caused by intolerable side effects 1. Placebo (ref) Discontinuation caused by other reasons 1. Placebo (ref) Median (95% CI) Median (95% CI) Posterior summaries of the log relative hazard rates dk,m for each treatment relative to placebo are shown in Table 2. For each of the three outcomes, the ranking of the treatments shows a high degree of consistency regardless of the choice of model. Zotepine is the most effective treatment in preventing relapse, followed by olanzapine; amisulpride causes the least discontinuation caused by intolerable side effects followed by risperidone and olanzapine; risperidone appears to cause the fewest discontinuations for other reasons, followed by amisulpride and then olanzapine. Nevertheless, the rankings of the posterior mean treatment effects do not take uncertainty into account. Figure 3 shows the probabilities that each treatment is ranked jth (j = 1,2 . . . 9) for each of the three outcomes, based on the different variances model. As with Table 2, a rank of 1 indicates that the treatment is “best” at avoiding that (unwanted) outcome. These rankograms, introduced by Cipriani et al. , provide an “at a glance” summary that is hard to achieve from numbers alone, as they simultaneously demonstrate not only the relative rankings of treatments on each outcome, but also the very considerable uncertainty in inferences about relative efficacy. For example, although Table 2 shows that zotepine is ranked first in reducing relapse, Figure 3 reveals that the probability that zotepine is best in this outcome is only about 0.6. Olanzapine has a relatively high probability of being ranked among the highest three places for all three outcomes. Although the mean effects dk,m are similar across models, their posterior uncertainty as assessed by the posterior SD depends strongly on the model used. As one would expect, the random effect summaries are much more uncertain than the fixed effect summaries. Nevertheless, in the random effect models, posterior uncertainty is greater in the three-variance model than in the single-variance version, but only for the relapse and discontinuation caused by intolerable side effect outcomes. The reason for this can be seen in the summaries of the between-trial SDs (Table 2). Although the single-variance model produces an average between-trial SD with relatively narrow credible limits, the three-variance model has a particularly low SD for the relative effects on discontinuation for other reasons, and higher SDs with wide credible intervals for the other two outcomes. We have no ready explanation for this apparent difference in between-trial variability between outcomes, which impacts on the uncertainty in the mean treatment effect measures. Alternative uniform (0, 5) priors for the between-trial SD parameters were used in sensitivity analyses (not shown). These result in posterior estimates of σ and σm that were higher, but this had only minor effects on posterior distributions of dk,m. We have described and illustrated a simple approach to meta-analysis of trials with multiple, mutually exclusive event outcomes. The special feature of such “competing risk” data is the negative correlations between outcomes within trial arms. The Bayesian MCMC framework, particularly with WinBUGS, allows the user to specify a multinomial likelihood along with conventional fixed or random effect models for relative treatment differences. Almost invariably, when meta-analyses have been applied to competing risk outcomes, they have looked at outcomes one at a time, and have not attempted an analysis that looks at the multiple outcomes simultaneously within a single coherent framework. Methods for competing risk meta-analysis appear not to have been previously described, although recently Trikalinos and Olkin have presented methodology for what they describe as “mutually exclusive binary outcomes”. They propose synthesis of log odds ratios, log relative risk, or risk difference outcomes from multinomial data arising from what is in fact a competing risk situation. Their approach is to take the estimated log odds ratios (or risk differences, or log relative risks) as data, and to develop expressions for the variances and covariances across outcomes based on normal theory. We believe that our approach has several advantages over this scheme. First, the use of the multinomial likelihood avoids the approximations required when the normal likelihood is used, and also avoids the need to add a constant to zero cells, a maneuver that generates bias [19,20]. Second, although calculation of the variances and covariances is relatively easy when there are two competing outcomes, it has the potential to be error prone if there are more than four or five. Similarly, the covariances would become still more complex to specify in multiarm trials, whereas our Bayesian approach can be readily extended [9,11], as shown in Equation 6. A third advantage of the Bayesian simulation framework is that it is suitable both for inference and for use in probabilistic decision models, or within cost-effectiveness analyses. Frequentist methods would need to be supplemented by, for example, bootstrap sampling, in order to be used in such environments. It is important for users to be aware of the underlying assumptions being made in applications of MTC synthesis. The key assumption is that, if all trials had included all nine treatments, the relative effects δj,b,k,m would, for each outcome m, be exchangeable across trials. This is equivalent to assuming that treatments are missing at random, which means that the absence of data on a treatment is independent of the treatment effect. Such assumptions, commonly made in evidence synthesis, are very hard to verify, and one must generally rely on experts with knowledge of the trials and the subject area to confirm its plausibility. This ensemble of trials was considered sufficiently homogeneous to form the basis for a comparative assessment of the eight active treatments in the context of clinical guideline development . But, the main advantage of the methods proposed here is that they are based on hazard rates and their ratios, and as a result they can correctly accommodate sets of trials that have been run to different follow-up times, as is the case in the present data set. Methods based on odds ratios, risk ratios, and risk differences, by contrast, cannot give consistent results for trials with different follow-up times, and it would be mathematically impossible to construct survival time distributions that would be compatible with constant measures at different durations. Use of these measures in trials with different follow-up times not only introduces unwanted heterogeneity to the estimates, but will also result in estimates that cannot, strictly speaking, be applied to other follow-up times. The Trikalinos and Olkin proposals could, however, be extended to hazard ratios, by introducing expressions for variances and covariances of log hazard ratios. Although this would yield coherent rate models with comparable estimates to those derived from our approach, it would still suffer from the disadvantages of normal approximations and an explosion of complexity with multiple competing outcomes. These factors, along with the intrinsic compatibility of MCMC posterior sampling with probabilistic decision analysis, make the modeling approach proposed here the obvious choice in cost-effectiveness analysis. The analyses proposed here can be extended in a number of ways. The purpose of our models is to take into account the negative correlations between outcomes at the patient level. Nevertheless, it may also be worth considering the additional possibility of correlations between outcomes at the trial level. For example, it may be plausible to assume that trials in which treatments which are more effective in preventing relapse may be those in which treatments are more likely to lead to intolerable side effects leading to treatment discontinuation. This suggests a class of extensions that might focus attention on the possibility of correlations between the trial-specific treatment efficacy and discontinuation rates, in which the treatment effects are drawn from an extended multivariate normal distribution. If we write Equation 5 as δj,m ∼ MVN(dm,Σmm), we might consider, for example: where the off-diagonal covariance matrices carry terms for correlations between treatment effects on different outcomes. The present model would then be a special case where these correlations were assumed to be zero. Experience with heterogeneous variance models in MTC suggests that very large amounts of data, especially from multiarm trials, would be required to identify the off-diagonal covariances, and that the impact on the posterior distributions of mean treatment effects may be limited. Informative priors could of course be placed on the correlations, although it is no trivial exercise to ensure the matrix is positive definite. A useful parameterization has been suggested by Lu and Ades . Because of the underlying rate parameterization, it would be possible to incorporate additional data that are reported in the form of number of events and time at risk . It is also relatively easy to incorporate data where outcomes are reported at more than one time point. This is best achieved by conditioning the data for subsequent intervals on survival to the end of the previous interval, in order to achieve independent Bernoulli samples, as illustrated by Lu et al. . The WinBUGS code provided in the Appendix (http://www.ispor.org/Publications/value/ViHsupplementary/ViH13i8_Ades.asp) would not need to be altered to accommodate additional data structured in this way. The existence of data at multiple time points would facilitate further extensions of the competing risk analyses to more complex underlying survival distributions, such as the Weibull or other distributions. This represents a very substantial liberalization of the modeling assumptions. Depending on how much data are available, Weibull shape parameters could be held constant or allowed to vary across trials . Alternatively, a piece-wise constant underlying hazard model offers considerable flexibility . These extensions, of course, all require not only a proportional hazard assumption for relative treatment effects, but also the proportional competing risk assumption described in the introduction. The latter is a strong assumption, but it can be relaxed, for example, in a piece-wise constant hazard framework. This could accommodate the possibility that, for example, the proportional risk of discontinuation caused by side effects could be higher in an initial period. It is also possible to incorporate data from trials that fail to report a subset of end points separately, for example, by aggregating discontinuation caused by side effects with other reasons for discontinuing treatment. Source of financial support: No specific funding was received for this work. A.E.A., S.D., and N.J.W. were supported by Medical Research Council funding to the Health Services Research Collaboration, transferred to University of Bristol.
Applied Mathematics Seminar Seminars are held at 2:00 pm on Wednesdays in the Access Grid Room ( Carslaw Building, 8th floor), unless otherwise noted. For more information and to be added to the mailing list, please contact Eduardo G. Altmann . Wednesday March 22 Sheehan Olver (School of Mathematics and Statistics, University of Sydney) Title: Solving PDEs on triangles using multivariate orthogonal polynomials Abstract: Univariate orthogonal polynomials have long history in applied and computational mathematics, playing a fundamental role in quadrature, spectral theory and solving differential equations with spectral methods. Unfortunately, while numerous theoretical results concerning multivariate orthogonal polynomials exist, they have an unfair reputation of being unwieldy on non-tensor product domains. In reality, many of the powerful computational aspects of univariate orthogonal polynomials translate naturally to multivariate orthogonal polynomials, including the existence of Jacobi operators and the ability to construct sparse partial differential operators, a la the ultrapsherical spectral method [Olver & Townsend 2012]. We demonstrate these computational aspects using multivariate orthogonal polynomials on a triangle, including the fast solution of general partial differential equations. Wednesday April 19 Prof. Nihat Ay (Max-Planck-Institute for the Mathematics in the Sciences, Leipzig, Germany) Title:Information Geometry and its Application to Complexity Theory Abstract: In the first part of my talk, I will review information-geometric structures and highlight the important role of divergences. I will present a novel approach to canonical divergences which extends the classical definition and recovers, in particular, the well-known Kullback-Leibler divergence and its relation to the Fisher-Rao metric and the Amari-Chentsov tensor. Divergences also play an important role within a geometric approach to complexity. This approach is based on the general understanding that the complexity of a system can be quantified as the extent to which it is more than the sum of its parts. In the second part of my talk, I will motivate this approach and review corresponding work. 1. N. Ay, S.I. Amari. A Novel Approach to Canonical Divergences within Information Geometry. Entropy (2015) 17: 8111-8129. 2. N. Ay, J. Jost, H. V. Le, L. Schwachhöfer. Information geometry and sufficient statistics. Probability Theory and Related Fields (2015) 162: 327-364. 3. N. Ay, J. Jost, H. V. Le, L. Schwachhöfer. Parametrized measure models. Bernoulli (2016) accepted. arXiv:1510.07305. 4. N. Ay, J. Jost, H. V. Le, L. Schwachhöfer. Information geometry. Ergebnisse der Mathematik und Ihrer Grenzgebiete/A Series of Modern Surveys in Mathematics, Springer 2017, forthcoming book. 5. N. Ay. Information Geometry on Complexity and Stochastic Interaction. Entropy (2015) 17(4): 2432-2458. Wednesday January 25 Dr. Paul Griffiths (Oxford Brookes University, UK) Title: Shear-thinning: A stabilising effect? Yes, no, maybe? Abstract: In this talk we will investigate how viscosity effects the stability of a fluid flow. By assuming a shear-thinning viscosity relationship, where an increase in shear-rate results in a decrease in fluid viscosity, we show that flows can be both stabilised or destabilised, depending on (i) the fluid model in question and (ii) the ‘amount’ of shear-thinning the fluid exhibits. Using a two-dimensional boundary-layer flow as our ‘toy model’ we are able to show equivalence between different shear-thinning models. The effect shear-thinning has on important parameters such as the critical Reynolds number, and the maximum frequency of the disturbances will be discussed and interpreted in the wider context. Wednesday February 22 Dr. Maria Vlassiou (Eindhoven University of Technology, Netherlands) Title: Heavy-traffic limits for layered queueing networks Abstract: Heavy-traffic limits for queueing networks are a topic of continuing interest. Presently, the class of networks for which these limits have been rigorously derived is restricted. An important ingredient in such work is the demonstration of state space collapse (SSC), which loosely speaking shows that in diffusion scale the queuing process for the stochastic model can be approximately recovered as a continuous lifting of the workload process. This often results in a reduction of the dimensions of the original system in the limit, leading to improved tractability. In this talk, we discuss diffusion approximations of layered queuing networks through two examples. In the first example, we establish a heavy-traffic limit through SSC for a computer network model. For this model, SSC is related to an intriguing separation of time scales in heavy traffic. The main source of randomness occurs at the top layer; the interactions at the other layer are shown to converge to a fixed point at a faster time scale. The second example focuses on a network of parallel single-server queues, where the speeds of the servers are varying over time and governed by a single continuous-time Markov chain. We obtain heavy-traffic limits for the distributions of the joint workload, waiting-time and queue length processes. We do so by using a functional central limit theorem approach, which requires the interchange of steady-state and heavy-traffic limits. For this model, we show that the SSC property does not hold. Wednesday March 1 Dr. Daniel Lecoanet (Princeton University, Princeton, USA) Title: Measuring Core Stellar Magnetic Field using Wave Conversion By studying oscillation modes at the surface of stars, astrophysicists are able to infer characteristics of their deep interior structure. This was first applied to observations of the Sun, but recently space-based telescopes have measured oscillations in many other stars, leading to many new mysteries in stellar structure and evolution. Recent work has suggested that low dipole oscillation amplitudes in evolved red giant branch stars may indicate strong core magnetic fields. Here we present both numerical simulations and analytic calculations of the interactions of waves with a strong magnetic field. We can solve the problem very accurately by using the WKB approximation to reduce the 2D PDE into a series of ODEs for different heights. We find that magnetic fields convert the buoyancy-driven waves observable at the surface of the star to magnetic waves, which are not present at the surface, in agreement with observations.
You will enter the following information for a mass flow inlet boundary: All values are entered in the Mass-Flow Inlet panel (Figure 7.5.1), which is opened from the Boundary Conditions panel (as described in Section 7.1.4). Note that open channel boundary condition inputs are described in Section 23.10.2. Defining the Mass Flow Rate or Mass Flux You can specify the mass flow rate through the inlet zone and have FLUENT convert this value to mass flux, or specify the mass flux directly. For cases where the mass flux varies across the boundary, you can also specify an average mass flux; see below for more information about this specification method. If you set the mass flow rate, it will be converted internally to a uniform mass flux over the zone by dividing the flow rate by the area of the zone. You can define the mass flux (but not the mass flow rate) using a boundary profile or a user-defined function. The inputs for mass flow rate or flux are as follows: \| Note that for axisymmetric problems, this mass flow rate is the flow rate through the entire ( -radian) domain, not through a 1-radian slice. If you selected Mass Flux, set the prescribed mass flux in the Mass Flux field. If you selected Mass Flux with Average Mass Flux, set the prescribed mass flux and average mass flux in the Mass Flux and Average Mass Flux fields. More About Mass Flux and Average Mass Flux As noted above, you can specify an average mass flux with the mass flux. If, for example, you specify a mass flux profile such that the average mass flux integrated over the zone area is 4.7, but you actually want to have a total mass flux of 5, you can keep the profile unchanged, and specify an average mass flux of 5. FLUENT will maintain the profile shape but adjust the values so that the resulting mass flux across the boundary is 5. The mass flux with average mass flux specification method is also used by the mixing plane model described in Section 10.3.2. If the mass flow inlet boundary is going to represent one of the mixing planes, then you do not need to specify the mass flux or flow rate; you can keep the default Mass Flow-Rate of 1. When you create the mixing plane later on in the problem setup, FLUENT will automatically select the Mass Flux with Average Mass Flux method in the Mass-Flow Inlet panel and set the Average Mass Flux to the value obtained by integrating the mass flux profile for the upstream zone. This will ensure that mass is conserved between the upstream zone and the downstream (mass flow inlet) zone. Defining the Total Temperature Enter the value for the total (stagnation) temperature of the inflow stream in the Total Temperature field in the Mass-Flow Inlet panel. Defining Static Pressure The static pressure (termed the Supersonic/Initial Gauge Pressure) must be specified if the inlet flow is supersonic or if you plan to initialize the solution based on the pressure inlet boundary conditions. Solution initialization is discussed in Section 25.14. The Supersonic/Initial Gauge Pressure is ignored by FLUENT whenever the flow is subsonic. If you choose to initialize the flow based on the mass flow inlet conditions, the Supersonic/Initial Gauge Pressure will be used in conjunction with the specified stagnation quantities to compute initial values according to isentropic relations. Remember that the static pressure value you enter is relative to the operating pressure set in the Operating Conditions panel. Note the comments in Section 7.3.1 regarding hydrostatic pressure. Defining the Flow Direction You can define the flow direction at a mass flow inlet explicitly, or you can define the flow to be normal to the boundary. You will have the option to specify the flow direction either in the absolute or relative reference frame, when the cell zone adjacent to the mass flow inlet is moving. If the cell zone adjacent to the mass flow inlet is not moving, both formulations are equivalent. The procedure for defining the flow direction is as follows, referring to Figure 7.5.1: \| Note that if you are modeling axisymmetric swirl, the flow direction will be normal to the boundary; i.e., there will be no swirl component at the boundary for axisymmetric swirl. \| These options are equivalent when the cell zone next to the mass flow inlet is stationary. Figure 7.3.2 shows the vector components for these different coordinate systems. Defining Turbulence Parameters For turbulent calculations, there are several ways in which you can define the turbulence parameters. Instructions for deciding which method to use and determining appropriate values for these inputs are provided in Section 7.2.2. Turbulence modeling is described in Chapter 12. Defining Radiation Parameters If you are using the P-1 radiation model, the DTRM, the DO model, or the surface-to-surface model, you will set the Internal Emissivity and (optionally) Black Body Temperature. See Section 13.3.15 for details. (The Rosseland radiation model does not require any boundary condition inputs.) Defining Species Mass Fractions If you are modeling species transport, you will set the species mass fractions under Species Mass Fractions. For details, see Section 14.1.5. Defining Non-Premixed Combustion Parameters If you are using the non-premixed or partially premixed combustion model, you will set the Mean Mixture Fraction and Mixture Fraction Variance (and the Secondary Mean Mixture Fraction and Secondary Mixture Fraction Variance, if you are using two mixture fractions), as described in Section 15.13. Defining Premixed Combustion Boundary Conditions If you are using the premixed or partially premixed combustion model, you will set the Progress Variable, as described in Section 16.3.5. Defining Discrete Phase Boundary Conditions If you are modeling a discrete phase of particles, you can set the fate of particle trajectories at the mass flow inlet. See Section 22.13 for details. Defining Open Channel Boundary Conditions If you are using the VOF model for multiphase flow and modeling open channel flows, you will need to specify the Free Surface Level, Bottom Level, and additional parameters. See Section 23.10.2 for details.
Almost 2 months ago the CDF collaboration published their analysis of the events with exactly 2 jets, 1 lepton, and missing energy. These are vastly dominated by boring Standard Model processes where the W boson is produced together with jets and subsequently decays to an electron or a muon and a neutrino. A surprising feature showed up in the distribution of the invariant mass of the jet pairs. After subtracting the Standard Model background, CDF observed a bump near 150 GeV with a significance of 3.2 sigma. Obviously, theorists rushed to interpret the bump in term of physics beyond the standard model. The CDF result hints to a new particle with a mass of around 150 GeV, a significant coupling to the light quarks and a tiny coupling to leptons; the remaining details are left up to our imagination . Here is a selection of the educated guesses that appeared in about 50 papers to date. The first thing that comes to mind is Z' - a new neutral gauge boson coupled to the left-handed quarks. This is a valid possibility provided Z' is leptophobic, that is to say, its coupling to electrons is less than about 0.05 to avoid constraints from the LEP experiment. There is some tension with the constraints from the UA2 experiment that was operating some 30 years before christ and made a search for a narrow Z' in the dijet channel. The UA2 limits on the Z'-quark coupling translate to a constraint on the W+Z' cross section at the Tevatron that allows one to explain only about 60 percent of the events observed by CDF. However, given the large uncertainties involved in the CDF measurement and in interpreting the UA2 results, the Z' option remains open. One should also note that nothing in the data tells us the new particle is a vector boson, it could just as well be a scalar. To ease the UA2 constraints one can turn to another class of model. Quite generally, the Tevatron may produce a ≥ 250 GeV mother resonance who decays to a W boson and a 150 GeV daughter resonance. The latter subsequently decays to 2 jets who are observed by CDF. Several proposals for the mother and daughter exist: a technirho meson decaying to a technipion and a W in a version of technicolor, a sbottom decaying to a stop and a W in R-parity violating supersymmetry, a charged Higgs decaying to a neutral Higgs and a W in two-Higgs doublet models, a weak doublet color octet in the Manohar-Wise model, etc. The striking prediction of this class of models is that not only the invariant mass of the jets but also of the entire final state should display a resonance. CDF looked at the invariant mass of the 2 jets + lepton + missing energy vector and found it consistent with background only, but it is not clear if this excludes the presence of a mother resonance (the presence of the missing energy introduces larger systematic uncertainties than for the jet pair mass). One can also imagine a more intricate class of models where the lepton and the missing energy in the CDF excess events come not from a usual W boson but from some other particle decaying to an electron and a neutrino. For example, this paper explains the excess by a production of a pair of supersymmetric winos of which one decays, via R-parity violation, to a charged lepton and a neutrino, and the other decays to 2 jets. This possibility may be excluded by analysis the distribution of the transverse mass of the lepton+missing energy subsystem. Finally, one should mention those who are trying to spoil the party. From the very beginning many have cast doubts on the CDF analysis as it requires a perfect control over the overwhelming Standard Model backgrounds. One thing is that even the Standard Model W/Z peak in the observed jet mass spectrum, arising due to the well known contribution of the WW and WZ production processes, does not seem to be very well described by the simulations. Furthermore, by eye it seems that shifting the jet energy scale a few percent upwards, which would correspond to shifting the whole data curve to the right, allows one to get rid of the excess (the authors of the analysis reply that raising the jet energy scale makes additional events pass the analysis cuts, so that naive shifting of the curve is not correct; they say a 3 sigma excess persists even when the JES is scaled up by 7 percent). Another attempted explanation is that the apparent excess is in reality the Standard Model top quark. When a top quark decays hadronically, t → W b → jjb, the invariant mass of the 2 light jets of course peaks at the W boson mass of 80 GeV, however the invariant mass of the b-jet and one of the light jets has the distribution peaking near 150 GeV (the endpoint is Sqrt[mTop^2- mW^2] = 155 GeV), suspiciously close to the CDF bump. Thus, the excess may be due to the semileptonic t-tbar or single top production where one or more additional jets are missed at the detector, assuming the Monte Carlo simulations of that background have been (rather grossly) mismodeled. So this is where we stand today. The situation may or may not be clarified when more data arrive. The updates from CDF and D0 are imminent. Someone will call a bluff? Or someone is holding an ace up his sleeve? Stay tuned for the next episode.
Differentiator A triangle wave upper trace is integrated to give a rounded, parabolic wave. A low pass filter passes low frequencies and rejects high frequencies from the input signal. And vice versa for a high pass filter. The simplest of these filters may be constructed from just two low-cost electrical components. \|Published (Last):\|\|25 December 2012\| \|PDF File Size:\|\|1.4 Mb\| \|ePub File Size:\|\|16.97 Mb\| \|Price:\|\|Free* [*Free Regsitration Required]\| PDF Version By introducing electrical reactance into the feedback loops of an op-amp circuit, we can cause the output to respond to changes in the input voltage over time. What is Capacitance? The greater the capacitance, the more the opposition. Capacitors oppose voltage change by creating current in the circuit: that is, they either charge or discharge in response to a change in the applied voltage. So, the more capacitance a capacitor has, the greater its charge or discharge current will be for any given rate of voltage change across it. However, if we steadily increased the DC supply from 15 volts to 16 volts over a shorter time span of 1 second, the rate of voltage change would be much higher, and thus the charging current would be much higher times higher, to be exact. Same amount of change in voltage, but vastly different rates of change, resulting in vastly different amounts of current in the circuit. Capacitor current moves through the feedback resistor, producing a drop across it, which is the same as the output voltage. A linear, positive rate of input voltage change will result in a steady negative voltage at the output of the op-amp. Conversely, a linear, negative rate of input voltage change will result in a steady positive voltage at the output of the op-amp. This polarity inversion from input to output is due to the fact that the input signal is being sent essentially to the inverting input of the op-amp, so it acts like the inverting amplifier mentioned previously. The faster the rate of voltage change at the input either positive or negative , the greater the voltage at the output. The formula for determining voltage output for the differentiator is as follows: Rate-of-Change Indicators for Process Instrumentation Applications for this, besides representing the derivative calculus function inside of an analog computer, include rate-of-change indicators for process instrumentation. One such rate-of-change signal application might be for monitoring or controlling the rate of temperature change in a furnace, where too high or too low of a temperature rise rate could be detrimental. The DC voltage produced by the differentiator circuit could be used to drive a comparator, which would signal an alarm or activate a control if the rate of change exceeded a pre-set level. In process control, the derivative function is used to make control decisions for maintaining a process at setpoint, by monitoring the rate of process change over time and taking action to prevent excessive rates of change, which can lead to an unstable condition. Analog electronic controllers use variations of this circuitry to perform the derivative function. Integration On the other hand, there are applications where we need precisely the opposite function, called integration in calculus. Here, the op-amp circuit would generate an output voltage proportional to the magnitude and duration that an input voltage signal has deviated from 0 volts. Stated differently, a constant input signal would generate a certain rate of change in the output voltage: differentiation in reverse. To do this, all we have to do is swap the capacitor and resistor in the previous circuit: As before, the negative feedback of the op-amp ensures that the inverting input will be held at 0 volts the virtual ground. If the input voltage is exactly 0 volts, there will be no current through the resistor, therefore no charging of the capacitor, and therefore the output voltage will not change. We cannot guarantee what voltage will be at the output with respect to ground in this condition, but we can say that the output voltage will be constant. However, if we apply a constant, positive voltage to the input, the op-amp output will fall negative at a linear rate, in an attempt to produce the changing voltage across the capacitor necessary to maintain the current established by the voltage difference across the resistor. Conversely, a constant, negative voltage at the input results in a linear, rising positive voltage at the output. The output voltage rate-of-change will be proportional to the value of the input voltage. Nuclear radiation can be just as damaging at low intensities for long periods of time as it is at high intensities for short periods of time. An integrator circuit would take both the intensity input voltage magnitude and time into account, generating an output voltage representing total radiation dosage. Another application would be to integrate a signal representing water flow, producing a signal representing total quantity of water that has passed by the flowmeter. This application of an integrator is sometimes called a totalizer in the industrial instrumentation trade. An integrator circuit produces a steadily changing output voltage for a constant input voltage. Both types of devices are easily constructed, using reactive components usually capacitors rather than inductors in the feedback part of the circuit. 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Problem Solving Connections Unit 1 Common Core Answers Core Problem Unit 1 Solving Common Connections Answers Unit 1 . Two numbers N and 16 have LCM = 48 and GCF = 8. 1. 7.EE.3 4. Sixth graders solve a variety of Common Core math problems using ratios. 1 The performance task is designed to showcase student mastery on a set of key unit targets that are integrated into a multi-part task Amazon Web Services. In addition, eighth grade students evaluate the purposes and effects of film, print, and technology presentations. The list is not exhaustive and will hopefully prompt further reflection and discussion. Home Connections Volumes 1 & 2 Unit 1 Multiplicative Thinking. Math. Unit 1 . which means it gives one person – you -- the right to access the membership content (Answer Keys, editable lesson files, pdfs, etc.) but is not meant to be shared 6.NS.A.1 — Interpret and compute quotients of fractions, and solve word problems involving division of fractions by fractions, e.g., by using visual fraction models and equations to represent the problem. In addition, eighth grade students evaluate Proposal Contents the purposes and effects of film, print, and Mycotoxins Powerpoint Presentation technology presentations. 110 dots 4. Rates & Ratios Word problems : ratios Write an equivalent ratio Ratio tables Word problems : unit prices Unit prices with fractions and decimals Percents Word problems : percents of numbers Percents of numbers and money amounts Find the total. Write 3 Paragraph Opinion Essay German Language Mark Twain Essay The Lowest 5.NF.B.7.A Interpret division of a unit fraction by a non-zero whole number, and compute such quotients. 7.RP.A.3. Unit analysis can help you find answers. For example, if a person walks 1/2 mile in each 1/4 hour, compute the unit rate as the complex fraction 1/2 / 1/4 miles per hour, equivalently 2 miles per Footloose Movie Summary hour Core Connections: Foundations for Algebra Course 1 Toolkit Chapter 1 Introduction and Representation 2 Learning Log Entries 2 1.1.5 Making Sense of a Challenging Problem 1.2.4 Characteristics of Numbers and Prime Factorization Math Notes 6 1.1.2 Perimeter and Area 6 1.1.3 Place Value 6 1.1.4 Rounding 7. Worksheet - Solving Systems by Elimination. b: The elevation decreases, that is, becomes more negative. Title : Go Math! Make sense of problems and persevere in solving them. Help for Dissertation Proposal Logistic Regression Exercises 35-39 on page 7. For example, if a person walks 1/2 mile in each 1/4 hour, compute the unit rate as the complex fraction (1/2)/(1/4) miles per hour, equivalently 2 miles per hour Jun 14, 2018 · Algebra 1 Common Core Answers Student Edition Grade 8 – 9 Chapter 2 Solving Equations Exercise 2.6. With the correct application of strategies that work for you and practice in. How To Conclude Research Project Case Study Examples Mca Students Chapter 2 Solving Equations Exercise 2.6 1CB Consider the statement given in the question: Given that, Speed of light = 3.0 × 10 10 cm/s A unit is defined as the standard used to measure the required system The grade based common core math worksheets for kindergarten (KG), grade-1, grade-2, grade-3, grade-4, grade-5 and grade-6 increase the student’s ability to apply mathematics in real world problems, conceptual understanding, procedural fluency, problem solving skills, critically evaluate the reasoning or prepare the students to learn the. Unit 1 – The Building Blocks of Algebra Unit #1 Mid-Unit Quiz (through Lesson #6) – Form B PDF DOCUMENT. Now is the time to redefine your true self using Slader’s Big Ideas Math Geometry: A Common Core Curriculum answers. 22 dots. Common Core Algebra I In this course students will explore a variety of topics within algebra including linear, exponential, quadratic, and polynomial equations and functions. • Peoples Education. 50 dots 2. In sixth grade, students will focus on four areas: (1) connect Dissertation Video Games ratio and rate to whole number multiplication and division and use concepts of ratio and rate to solve problems; (2) complete understanding of division of fractions and extending the notion of number to the system of rational numbers, including negative numbers; (3) write, interpret, and use. The words ratio and rate are both appropriate in sixth grade and can mostly be used interchangeably. Sep 25, 2019 · Beginning in 2010, the Common Core State Standards Initiative (CCSSI) aimed to change the way American students were taught English language arts and mathematics by countering low test scores, inconsistent learning standards and a curriculum that was a “mile wide and an inch deep.”. In this lesson, we model real world scenarios using. In Exercises 35-38, the first step in calculating the average speed is to write a rate expression using the given distance and time. Within a single system of measurement, express measurements in a larger unit in terms of. at the beginning of your answer Core Connections: Foundations for Algebra Course 1 Toolkit Chapter 1 Introduction and Representation 2 Learning Log Entries 2 1.1.5 Making Sense of a Challenging Problem 1.2.4 Characteristics of Numbers and Prime Factorization Math Notes 6 1.1.2 Perimeter and Area 6 1.1.3 Place Value 6 1.1.4 Rounding 7. At the beginning of your answer The Core Connections courses, for grades six through eight, are built on rich, meaningful problems and investigations that develop conceptual understanding of the mathematics and establish connections among different concepts. If the area of a circle is 81pi square feet, find its circumference. Bridges in Mathematics Grade 4 Home Connections Answer Key Bridges in Mathematics Grade 4 Components & Manipulatives in the context of problem solving. For example, create a story context for (1/3) ÷ 4, and use a visual fraction model to show the quotient of answers using mental computation and estimation strategies. Word problems : find the constant of proportionality Word problems : solve proportions Ratio & Percent Problems Word problems : percents of numbers Percents of numbers and money amounts Solve percent equations Percent of change Unit prices with unit conversions. For example, create a story context for (2/3) ÷ (3/4) and use a visual fraction model to show the quotient; use the relationship between multiplication and division to explain that (2/3. Culminating Performance Task1. Find the greatest common factor of 24. The homework help is designed to assist students to be able to do the problems Lesson 3.1.2 Math Note Operations with Fractions 18 2 Core Connections, Course 1 Answers 1. Selected Answers 3 Lesson 1.1.2 (Day 2) 1-21. Noticing the regularity in the way terms cancel when expanding (x - 1)(x + 1), (x - 1)(x 2 + x + 1), and (x - 1)(x 3 + x2 + x + 1) might lead them to the general formula for the sum of a geometric series.
The Method Of Common Bases Common Core Algebra 2 Homework Answers Hello, I'm Kirk weiler, and this is common core algebra two. By E math instruction. Today, we'll be doing unit four lesson number 5 on the method of common basis. So there aren't a lot of algebraic techniques that go along with exponential expressions and exponential equations. But one of them that we'll learn today is what's called the method of common basis. Students love this. As long as they're comfortable with basic powers. Basic exponent work. So let's jump right into it and see how it works. All right. Exercise number one, solve each of the following simple exponential equations by writing each side of the equation using a common base. Remember, a lot of students like to ignore terminology, but this is the base. So right now literally we have the equation to the X equals 16 to the first, if you will. But they don't have common basis. One of them has a base of two. One of them has a base of 16. But you see 16 can be thought of as two to the fourth. So if we rewrite this as two to the X equals two to the fourth, then we can say X must be equal to four. Right? The idea is very simple here. If the bases are the same, if the bases are the same, then the exponents must be the same. Okay, this is due to the fact that exponential expressions are or exponential functions are one to one. Meaning that if the exponent, if what the exponent is gives you the same output, then the exponent has to be the same. I don't know how well that really turned out. But let's keep going. These first few are quite simple. Three to the X, we want to common base, 27, is three to the third. Therefore, X equals three. But you might just say to yourself, look, I know three to the third is 27. So X is three. That's great. Let's look at one that's more challenging. Here, we have 5 to the X here, we have one 25th. Well, one thing I know for certain is that one 25th is the same as one over 5 squared. So that means we have 5 to the negative two. And therefore, X is negative two. Oh. Now, up to this point, each time we looked at the left hand side, that was the base we used. On the other hand, here we might look at the 16 and the four. And the first thing that might occur to us is that 16 is the same as four squared. So I have four squared to the X is equal to four. Now I can use that property of exponents that says, hey, let's multiply these two together. And I get four to the two X is equal to four to the first. That means I can now set my two X equal to one and get X equals one half. All right. Now, some of you that really picked up on that whole fractional exponents thing could have also said, well, look, I mean, I can use 16. Because I know that four is the same as 16 to the one half. So you could have easily used 16 as a base. If you said, okay, I got 16 to the X equals 16 to the one half. Then X equals one half. All right. But if not, you go with the more common base, which is four. All right, take a look at what we've done. Pretty easy, maybe so easy that you don't really see the point. And then we'll go on to some that are a little bit more challenging. All right, I'm going to clear this out. All right. Exercise number two says simplify each of the following exponential expressions. Notice none of these are equations. I'm not asking you to solve any equation. I'm just asking you to simplify these. So pause the video for a minute and ask yourself how can you simplify these exponential expressions? All right. Well, in each one of these cases, we're using the exponential rule or exponent rule or property that says when we have something raised to an exponent, raised to another exponent, we simply multiply the two exponents so we have two to the three X here, we'd have three to the adex to times four X here, we'd have 5 to the negative three X plus 7. Because remember, we're going to have to multiply these two exponents together. So they're going to distribute. And likewise, here we'd have four to the negative three. Plus three X squared. Nothing very special about this particular problem. You might even wonder what it has to do with what we're doing in this lesson today. But we'll get to that in just a moment. This exponent law is going to be critical for what we're doing next. So let's take a look at some method of common based problems that are a little more challenging. I'm going to clear this out and pause the video if you need to. All right, here we go. All right, solve each of the following equations by finding a common base for each side. All right. So I take a look at this equation and I look at the 8 and I look at the 32, and I try to come up with a common base. Now, for me, what I know is that 8 is equal to two to the third, and I know that 32 is equal to two to the 5th. So I'm going to use two as my common base. So I'm going to write 8 as two to the third, but then that's being raised to the X and then I'm going to write 32 as two to the 5th. Now I'm going to use that great property of exponents. That we were just reviewing. And I'll get two to the three X equal to two to the 5th. That means at this point I can say that three X is equal to 5 and X must be 5 thirds. Now, I want you to keep in mind what we're not doing. We're not canceling the twos or anything like that. We're not doing that. We're just recognizing the fact that if I take two and erase it to a number, I take two and I raise it to another number, if those two things are equal, then the two exponents have to be equal as well. We're not canceling the two. All right. Let's take a look at letter B what's going to be the common base here. Think about it for a moment. All right, well hopefully you figured out it would be three. You see 9 is equal to three squared. 27 is equal to three cubed. So we can write this left hand side as three squared to the two X plus one. And we can write this side as three cubed. Again, I can use that great property. Whoops. Apparently I can use it incorrectly. You can use that great property that allows me to now multiply these don't forget to distribute that too. So it's now going to be three to the four X plus two. Equals three to the third, once I have those common bases, I can, if you will, drop them, not cancel them, just drop them. And I'll get four X plus two is equal to three. Four X is equal to one. And X is equal to one fourth. Isn't that great? Okay, letter C is more of a challenge, but why don't you pause the video and see if you can pull it off yourself? All right. Well, hopefully you figured out a good common base of 5. You could probably do other common bases, but a common base of 5 works out well. It works out well because a 125 is 5 to the third. And one 25th is 5 to the negative two. So we're going to say 5 to the third. All raised to the X is equal to 5 to the negative two. All raised to the four minus X again, just using that property of exponents multiplying them. Gives me this. Now that we've got that common base, I can get rid of it. And I'm sure you can finish it out. Isn't that cool? Now, let me be very clear about something. Really clear about something. All right? If I had something like this, let's say 5 to the X equals 8. This method would be fairly useless. Although not impossible, it is extremely difficult to write 5 and 8 with a common base. Pretty difficult. It can be done. I don't want to claim it can't be done. But it is very, very difficult. So generally speaking, this method only works if there is that convenient and relatively obvious common base. So look for bases like two, three. Not so much forks, then you probably would have a base of two. But two, three, 5, 6, 7, et cetera. All right, pause the video now. And then I'll clear the text. All right, here we go. Let's take a look at a multiple choice problem. Which of the following represents the solution set to two to the X squared minus three equals 64? Well, play around with this one a little bit. See what you can get. All right. Well, when the number on this side is prime, which two is. Pretty much that has to be the common base. And it is. 64 is two to the 6th, if you think about it. Now that we've got our common bases, we're going to get X squared minus three is equal to 6. Add three to both sides. We're going to get X squared equals 9, take the square root, and remember we're going to get both plus and a minus root when we do that. X equals plus or minus three. And both are completely valid. And of course, you can check these equations. Easily check them like every other equation. If I took like, let's say the X equals negative three root and I checked to see whether or not this was true. I could take that negative three square it. Subtract three. That would be negative three squared, which would be 9. Minus three is 64. And then I get two to the 6 equals 64. And that checks. All right. So the method is pretty straightforward on the back side of the sheet. There's always backs out of the sheet. We're going to be using it to solve some other problems, maybe some higher order problems, but essentially that's the technique. Find a common base for both sides of the equation. Manipulate and set exponents equal. Pause the video now and write down anything you need to. Okay. Here we go. All right. Two exponential curves. Y equals four to the X plus 5 halves, and Y equals one half to the two X plus one, are shown below. They intersect at point a a rectangle has one vertex of the origin, and the other at a, as shown. So that's not a square. It's just garden variety rectangle. I suppose it could be a square. But I don't think it is. We want to find its area. So we're going to try to figure out the area of this rectangle. What are aces fundamentally, what do we need to know about a rectangle to find its area? This is easy enough. Pause the video for now and write down what you need to know to find a rectangle's area. All right, well, I don't care how you phrase it. You got to know its length. And width, or its base, and its height. I almost felt the word hyper off. I'd be embarrassing for a math teacher. All right. I'll let her be. How would knowing the coordinates of point a help us find this area or the area? Well, think about that for a minute. I knew the coordinates of point a, how would it help me find the area? I wonder. Well, let's talk about it. Now, the Y coordinate, the Y coordinate would simply be the height. And the X coordinate, well, it would be negative, obviously, from the picture, so maybe it's absolute value, would be the base. All right, this is find the area of the rectangle algebraically using the method of common bases show your work carefully. Um. Oh, I get it. You see, I need to find that intersection point. So I need to solve this equation. Four to the X plus 5 halves is equal to one half to the two X plus one. Well, what I'd like you to do is pause the video now and try to solve that to find the value of X then think about finding the value of Y as well. All right, let's do it. Shuffle my notes a little bit so that I can make sure I see the proper answer at the end. Well, the common base is a base of two. Four is two squared. So I'll write that. One half is two to the negative one. All right, doing a little simplifying. We'd get two to the two X plus 5 is equal to two to the negative two X minus one. Wow, I almost forgot to distribute that. Now I can set those two exponents equal. And I'll get four X is equal to negative 6. And X is equal to negative three halves. All right, fair enough. Let's figure out what Y is equal to. Um, I think I'll go with this one. Y is equal to four to the negative three halves plus 5 halves, be careful. Those are all in the exponent, which would be four to the two halves, which would be four to the first, which is four. So we now know the coordinates of that intersection are negative three halves. And four, but that means it's area it's going to have a base length of three halves, a height of four, and we're going to have an area of 6. All right. This is totally the kind of problem that I could see being maybe even upwards of a 6 point problem on a common core assessment like the New York State regions. Granted, I doubt there would be part a in part B but there could be. Most of the time though you'd have to just simply know, well, if I find that intersection point, it's going to help me find the area. Pause the video now and write down anything you need to. All right, great. It's clear it. And let's move on. Okay, one more problem, a little multiple choice here. At what X coordinate will the graph of Y equals 25 to the X minus a intersect the graph of Y equals one over 125 to the three X plus one. Show the work that leads to your choice. All right, so notice all four choices have X solved for in terms of a that makes some sense. All right? So why don't you pause the video and see if you can figure out which one of these is the correct choice? All right. One of the things I like about this problem is although not impossible to use your calculator. If we didn't have that X minus a there, if we had X minus three, well then you could really put the two things in your calculator, graph, use your intersect command, et cetera to find the intersection. Here, though, to find the intersection, we're going to set the two equations equal. And we're going to use the method of common basis. Now that common base is a base of 5, we've seen both of these a couple times today. 25 is obviously 5 squared. So the X minus a one 25 is 5 to the minus three. All right, so we got our common base of 5. When we distribute those inner powers, we get 5 to the two X minus two a is equal to 5 to the negative 9 X minus three. We've now got our common basis. So we're going to set two X minus two a equal to negative 9 X minus three. No matter what we're trying to do here, never lose sight of what you're trying to do. That can happen in math often, right? We're trying to solve for X so I'm going to bring this negative 9 X to the side. It's going to give me 11 X minus two a is equal to negative three. And then I'm going to add a two a to both sides here. And that's going to give me 11 X equals two a minus three. But now, of course, it's easy enough to just defy and get X equals two a minus three divided by 11. Choice two. All right. And that's it. All right, so the method of common basis, not too bad. Pause the video now, write down anything you have to. And then we'll wrap up. Okay, here we go. So as I mentioned at the outset, there are very few algebraic techniques that really allow us to work with exponential equations. The method of common base is a very restrictive method, meaning that it doesn't work very often. But when it does work, it's really neat. And it works when you can look at two sides of an exponential equation and write both sides with the same numerical base. If you can do that, then the method of common basis works great to help you solve exponential equations. For now, I'd like to thank you for joining me for another common core algebra two lesson by E math instruction. My name is Kirk weiler, and until next time, keep thinking and keep solving problems. The Method Of Common Bases Common Core Algebra 2 Homework Answers
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You are searching about Circle With Plus Sign Symbol Math, today we will share with you article about Circle With Plus Sign Symbol Math was compiled and edited by our team from many sources on the internet. Hope this article on the topic Circle With Plus Sign Symbol Math is useful to you. China – A Dynasty Of Mathematical Genius One of the most fascinating things about history is how much stuff was wiped out – on purpose. For example, in the 9th century CE, the largest library in the world, the Library of Alexandria, was burnt down in an act of war, and ever since then history buffs have been amused and tempted to guess the identity of some of them. those books that we will never see. From the burning of libraries to the destruction of presidential newspapers today, the idea of the lost book has a certain romantic, if frustrating, appeal. People who have studied the history of mathematics have their own “lost library” to wonder about – the treasury of early Chinese mathematical treatises burned by the order of Emperor Qin Shi Huang in 212 BCE. China has been at the forefront of mathematics for almost as long as civilization has existed. Evidence of a highly developed numeral system dates back to the period of the Shin Dynasty – 1600 to 1046 BC. This first Chinese numeral system also includes decimals, a major intellectual breakthrough in itself. To write the number 260, for example, you write the number two, followed by the hundred symbol, then six followed by the ten symbol – you get the idea. There is also evidence that Chinese mathematicians had been developing their own version of the abacus (an ancient calculating machine that used rods with moving counters) from very early times. So whatever was contained in those burnt math books of 212 BCE was probably important work. A handful of ancient Chinese mathematical works survived this public purge (the reasons for which are unclear). As early as 1046 BCE we have the I Ching, a favorite of 60s hippies and mystics even today, and the Mo Jing, a compilation of geometry and physical science from around the fifth century BCE. . These two survivors exemplify the high level of intellectualism and imagination characteristic of ancient Chinese mathematics. Writers of the Han dynasty period – a four hundred year period beginning in 202 BC. mathematicians were synthesizers, bringing together the best ideas of ancient thinkers, and their most important work was the nine chapters on the art of mathematics. This essential compilation illustrates the right way to use geometry to build a structurally sound dwelling; it also shows that Chinese mathematicians understood pi (the seemingly endless number by which we calculate the circumference of a circle) and various laws regarding right triangles. Perhaps most strikingly, he uses Cavalieri’s principle to determine the volume of a shape – but he does so over a thousand years before Cavalieri came up with the idea. In other words, Chinese mathematicians understood certain geometric ideas long before anyone in the West. Elsewhere during the Han dynasty, other Chinese mathematicians “came first”, including Jing Fang (78-37 BCE), a music theorist who discovered principles of temperament that had to wait, in the West, until in the 17th century. For a thousand years after the Han dynasty, Chinese mathematicians continued to come up with great ideas – all at a time when European mathematicians did not exist, to speak of. Chinese thinkers developed ideas such as negative numbers (the genius invention that helps us all keep our bank balances in order), the use of matrices to solve linear equations (an idea that continues to baffle Western students from eleventh to date) and elements of calculation and trigonometry. From prehistory to the Middle Ages, China produced some of the greatest mathematical reasoning ever found. Yet you wonder what additional heights these brilliant thinkers could have reached if their foundation had not been partially destroyed. Just as today’s playwrights and poets lament the great tragedies and epics that were probably lost in the fire at the Library of Alexandria, so historians of human genius must ask what wonderful ideas fell victim to the pride of Emperor Qin Shi Huang. (His order, after all, was only for provinces of China outside of his own state of Qin – so presumably, though we don’t know for sure, he only wanted people from his own part of China can read for themselves.) On the other hand, maybe he did Chinese math a favor. By giving Han dynasty mathematicians the impetus to safeguard and consolidate every morsel of mathematical knowledge still available, he may have lit another fire – a fire he never wanted. Video about Circle With Plus Sign Symbol Math You can see more content about Circle With Plus Sign Symbol Math on our youtube channel: Click Here Question about Circle With Plus Sign Symbol Math If you have any questions about Circle With Plus Sign Symbol Math, please let us know, all your questions or suggestions will help us improve in the following articles! The article Circle With Plus Sign Symbol Math was compiled by me and my team from many sources. If you find the article Circle With Plus Sign Symbol Math helpful to you, please support the team Like or Share! Rate Articles Circle With Plus Sign Symbol Math Rate: 4-5 stars Views: 8666799 7 Search keywords Circle With Plus Sign Symbol Math Circle With Plus Sign Symbol Math way Circle With Plus Sign Symbol Math tutorial Circle With Plus Sign Symbol Math Circle With Plus Sign Symbol Math free #China #Dynasty #Mathematical #Genius
Error Analysis Chemistry Equation Say you are measuring the time for a pendulum to undergo 20 oscillations and you repeat the measurement five times. Take the measurement of a person's height as an example. Case (2) which illustrates low precision and high accuracy. First, we note that it is incorrect to expect each and every measurement to overlap within errors. his comment is here In:= Out= The function can be used in place of the other *WithError functions discussed above. It is for that reason that large integers ought always to be reported in scientific notation where there is little room for doubt: 2.39 x 105 miles leaves no room for In:= Out= For most cases, the default of two digits is reasonable. This is reasonable since if n = 1 we know we can't determine at all since with only one measurement we have no way of determining how closely a repeated measurement http://www.ajdesigner.com/phppercenterror/percent_error.php Error Analysis Chemistry Equation Moreover, we will be concerned with the spread or range of a series of readings, and of decisions connected with removing outliers from a data set. This completes the proof. Although this example doesn't address the uncertainty of a particular measurement it touches on problems which can arise when there is complete ignorance of parameter boundaries: Some of the special problems Example 5-3. The density of water at 20 oC is 0.99823 g/cc. The particular micrometer used had scale divisions every 0.001 cm. Exercise 5-13. How To Do Error Analysis In Chemistry They both convey three significant figures because the rule says that the last digit shall be the one for which there is some uncertainty in the reading, usually the interpolated digit. Schließen Ja, ich möchte sie behalten Rückgängig machen Schließen Dieses Video ist nicht verfügbar. Percent Error Chemistry Equation The numbers 0.237, 4.38, 8.70 and 1.47 × 1023 all have 3 significant figures. After multiplication or division, the number of significant figures in the result is determined by the original number with the smallest number of significant figures. https://answers.yahoo.com/question/index?qid=20080325141136AAhtfOU You can only upload videos smaller than 600MB. A. Error Propagation Equation Here we discuss these types of errors of accuracy. Aside from making mistakes (such as thinking one is using the x10 scale, and actually using the x100 scale), the reason why experiments sometimes yield results which may be far outside Often an analytical chemist is faced with a choice between time constraints and accuracy; hence the question of maximum error tolerance must be asked. Percent Error Chemistry Equation There are four possibilities for the outcome: HH, HT, TH and TT. http://teacher.nsrl.rochester.edu/phy_labs/AppendixB/AppendixB.html That deals with integers ending in one or more zeros. Error Analysis Chemistry Equation Kategorie Bildung Lizenz Creative Commons-Lizenz mit Quellenangabe (Wiederverwendung erlaubt) Quellvideos Quellenangaben anzeigen Mehr anzeigen Weniger anzeigen Wird geladen... Error Analysis Chemistry Formula We leave the proof of this statement as one of those famous "exercises for the reader". Each student is to flip a coin ten times. http://joelinux.net/error-analysis/error-analysis-general-equation.html Rule 3: Raising to a Power If then or equivalently EDA includes functions to combine data using the above rules. Wird geladen... Über YouTube Presse Urheberrecht YouTuber Werbung Entwickler +YouTube Nutzungsbedingungen Datenschutz Richtlinien und Sicherheit Feedback senden Probier mal was Neues aus! In these terms, the quantity, , (3) is the maximum error. Error Analysis In Chemistry Lab Report In:= In:= Out= In:= Out= In:= Out= For simple combinations of data with random errors, the correct procedure can be summarized in three rules. The rule of thumb means "for most practical purposes." Rules of thumb always have exceptions. If the Philips meter is systematically measuring all voltages too big by, say, 2%, that systematic error of accuracy will have no effect on the slope and therefore will have no weblink After all, (11) and . (12) But this assumes that, when combined, the errors in A and B have the same sign and maximum magnitude; that is that they always combine Furthermore, this is not a random error; a given meter will supposedly always read too high or too low when measurements are repeated on the same scale. Error Analysis Physics How about if you went out on the street and started bringing strangers in to repeat the measurement, each and every one of whom got m = 26.10 ± 0.01 g. In:= Out= AdjustSignificantFigures is discussed further in Section 3.3.1. 3.2.2 The Reading Error There is another type of error associated with a directly measured quantity, called the "reading error". Wolfram Knowledgebase Curated computable knowledge powering Wolfram\|Alpha. In general, the last significant figure in any result should be of the same order of magnitude (i.e.. The correct procedure to do this is to combine errors in quadrature, which is the square root of the sum of the squares. Imagine you are weighing an object on a "dial balance" in which you turn a dial until the pointer balances, and then read the mass from the marking on the dial. Standard Deviation Equation Finally, we look at the histogram and plot together. For example, if the error in a particular quantity is characterized by the standard deviation, we only expect 68% of the measurements from a normally distributed population to be within one In:= Out= In this example, the TimesWithError function will be somewhat faster. One well-known text explains the difference this way: The word "precision" will be related to the random error distribution associated with a particular experiment or even with a particular type of http://joelinux.net/error-analysis/error-analysis-equation-for-chemistry.html Here n is the total number of measurements and x[[i]] is the result of measurement number i. Still others, often incorrectly, throw out any data that appear to be incorrect. Send comments, questions and/or suggestions via email to [email protected] AJ Design☰ MenuMath GeometryPhysics ForceFluid MechanicsFinanceLoan Calculator Percent Error Equations Calculator Math Physics Chemistry Biology Formulas Solving for percent error. If you don't like the idea of running an unfamiliar program to select a certain number of data points then tack this sheet up on your wall and throw darts at There's nothing like an experimental approach to test this claim. That, then, nails down the extent to which a reported value can be trusted. Example 5-2. Source(s): drbillmacmo · 9 years ago 1 Thumbs up 0 Thumbs down Comment Add a comment Submit · just now Asker's rating Report Abuse Look up error calculations. A water well known to contain particularly high levels of iron has five samples drawn for spectrophotometric analysis. The results show the following levels of iron in parts per million: 134, 147, 125, 131, 152 Determine the mean, the standard deviation, the variance, the RSD, the CV the spread If a variable Z depends on (one or) two variables (A and B) which have independent errors ( and ) then the rule for calculating the error in Z is tabulated In:= Out= One may typeset the ± into the input expression, and errors will again be propagated. Typically if one does not know it is assumed that, , in order to estimate this error. Other sources of systematic errors are external effects which can change the results of the experiment, but for which the corrections are not well known. In the final analysis, the two formulas used, one for addition and subtraction and the other for multiplication and division ought to be considered to be "best estimates" of anticipated calculated This means that, for example, if there were 20 measurements, the error on the mean itself would be = 4.47 times smaller then the error of each measurement. Much of the material has been extensively tested with science undergraduates at a variety of levels at the University of Toronto. In:= Out= Next we form the error. Thus, we would expect that to add these independent random errors, we would have to use Pythagoras' theorem, which is just combining them in quadrature. 3.3.2 Finding the Error in an There are conventions which you should learn and follow for how to express numbers so as to properly indicate their significant figures. A student drops a dry sample of Na2CO3 on the floor and scoops it up before titrating it with HCl. Suppose there are two measurements, A and B, and the final result is Z = F(A, B) for some function F.
So, we will use this property to find the rest by dividing it by \[240\]. A fast and simple calculator to unravel easy division equations. /To enter a fraction of the shape 3/4. Hit equals and the new end result will appear. Here we’ll show you step-by-step with detailed rationalization tips on how to calculate 240 divided by 40 utilizing long division. Here we will present you step-by-step with detailed explanation tips on how to calculate 240 divided by 9 utilizing lengthy division. Check for the distinction between 102 and the product of eleven and 9. Long Division Calculator Due to no more digits available for bringing down, the ultimate difference 3 is the decimal Remainder of dividend 9452 divided by eleven. Square root button is used to calculate the square root of a number. Enter a number, then click on sq. root button. Here we’ll show you step-by-step with detailed rationalization how to calculate 240 divided by 3 utilizing long division. Bring down and append 4th digit 2 to the proper facet of earlier distinction 10. Proper fraction button and Improper fraction button work as pair. Divide 2 numbers and find the quotient. While dividing any factorial by any number to find remainder we must observe the sample that’s upto which place our factorial will start giving us remainder zero. Also, we know that the factorial of \[1! Long Division Calculator With Remainders When you select the one the other is switched off. Number formating actions fractions and scientific notation. In case you need to carry out this calculation faster than by hand you should use our long division calculator. An improper fraction is a fraction where the numerator (top quantity is bigger than or equal to the denominator . Our online instruments will provide fast solutions to your calculation and conversion needs. On this web page, you can divide two numbers utilizing long division technique with step-by-step instruction. Check what quantity of times the divisor could be accommodated in the ninety four and write the worth as a half of quotient. The divisor eleven could be accommodated 8-times in 94. Write eight as probably the most significant digit of quotient. Quote Of The Day Click a number after which click on fraction bar, then click another quantity. Memory minus button subtracts the number displayed from the contents of the memory. Memory plus button provides the quantity exhibited to the contents of the memory. Memory recall button retrieves the number you have in reminiscence and locations it within the show subject. You can use the print button to print out the tape. Learn to divide 768 by 32, or some other numbers, with lengthy division by whatching this video. Just supply the values of dividend, divisor and hit on ENTER button to search out the Quotient and Remainder in decimal. The step by step work reveals how to do lengthy division between different combination of dividend and divisor. By using this lengthy division calculator, users can perform division with the rest or with out remainder which includes large numbers. This calculator shows all of the work and steps for lengthy division. You just have to enter the dividend and divisor values. Write The Means To Improve This Page Compare if the new dividend sixty five is bigger than the divisor eleven and check how many times the divisor can be accommodated within the new dividend fashioned by brining down the third digit. The divisor eleven can be accommodated 5-times in sixty five. Append the value 5 right to the earlier quotient eight. This long division calculator supports giant quantity divisions.Use this long division calculator which supports massive numbers in divison. Users can supply up to 9-digit dividend and up to 7-digit divisor to carry out or verify the lengthy divison issues. Free calculators and unit converters for general and on a daily basis use. Note that the numerator in the fraction above is the rest and the denominator is the divisor. 240 divided by 6 is forty and division, as we all know, just isn’t commutative. What’s 240 Divided By 3? Is1\], but \[0\] has additionally a value equal to \[1\]. Now, if we see before \[6\] upto \[5! The beneath solved instance of 4 by 2 digit lengthy division with the rest may useful to understand how to do lengthy division manually for project, classwork and homework problems. I imply, do we really want to elucidate how this calculator works? You enter the whole number within the first field, then the reply you want in the second field, you click “Calculate” and hey presto, we calculate the missing quantity for you. You are done, as a outcome of there aren’t any extra digits to maneuver down from the dividend. All clear button clears the calculator, tape, and resets any capabilities. Division Calculator With Steps You could go to long divison learning assets to take pleasure in numerous apply problems to sharpen your math abilities. Check how many occasions the divisor 11 could be accommodated within the new dividend formed by brining down the 4th digit. The divisor 11 can be accommodated 9-times in 102. Append the value 9 proper to the sooner quotient 85. Hence, the rest turns into 859 now. Improper fraction button is used to alter numerous the form of 1 4/5 to the form of 9/5. Arrange the 4-digit dividend and 2-digit divisor numbers for long division methodology and evaluate if the the primary two digits of dividend 9452 is bigger than the divisor 11. All calculations are saved on the tape. Click on any number or operator on the tape and alter it at any time.
The SIR infections disease model is an important model and has been studied by many authors - . The basic and important research subjects for these systems are local and global stability of the disease-free equilibrium and the endemic equilibrium, existence of periodic solutions, persistence and extinction of the disease, etc. In recent years, the study of vaccination, treatment, and associated behavioral changes related to disease transmission has been the subject of intense theoretical analysis . In 2008, Meng and Chen considered a class of continuous vertical and horizontal transmitted epidemic model under constant vaccination where S represents the proportion of individuals susceptible to the disease, who are born (with b) and die (with d) at the same rate b (b = d) and have mean life expectancy. The susceptible become infectious at a bilinear rate, where I is the proportion of infectious individuals and is the contact rate. The infectious recover (i.e. acquire lifelong immunity) at a rate r, so that is the mean infectious period. The constant p, q, 0 < p < 1, 0 < q < 1, and p + q = 1, where p is the proportion of the offspring of infective parents that are susceptible individuals, and q is the proportion of the offspring of infective parents that are infective individuals. In their work, the basic reproductive rate determining the stability of disease-free equilibrium point and endemic equilibrium point was found out and the local and global stability of the equilibrium points have been researched by using Lyapunov function and Dulac function. Due to a lot of discrete-time models are not trivial analogues of their continuous ones and simple discrete-time models can even exhibit complex behavior (see ), in this paper, we pay attention to the discrete situation of Equation (1) as follows where, and represent susceptible, infective and recovered subgroups, n represent a fixed time. Under the hypothesis of population being constant size, the model is transformed into a planar map and its equilibrium points and the corresponding eigenvalues are solved out. By discussing the influence of coefficient parameters on the eigenvalues, we determine the hyperbolicity of equilibrium points. Further, we get the equations of flows on center manifold and discuss the direction and stability of the transcritical bifurcation and flip bifurcation. 2. Hyperbolic and Non-Hyperbolic Cases In this section, we will discuss the hyperbolic and non-hyperbolic cases in a two parameters space parameter. In view of assumption that population is a constant size, i.e., system Equation (2) can be changed into Rewrite Equation (4) as a planar map F: It is obvious that this map has a disease-free equilibrium point and an endemic equilibrium point where Theorem 1. The equilibrium point is non-hyperbolic if and only if lies on the lines: Otherwise, the equilibrium point is an one of the following types: (See Table 1). Proof. The Jacobian matrix of map (5) at is: And its eigenvalues are From the assumption, we see that. Then non-hyperbolic will be happened in the case. From and, we get that and lies on. Also, from, we know which means lies on. When (referred to the case), the eigenvalue satisfies, then the equilibrium point P is a saddle. When (referred to the case), the eigenvalue satisfie, so the equilibrium point P is a stable node and meanwhile when (referred to the case), the equilibrium point P is a saddle since. The proof is complete. Theorem 2. We select s, r as parameters. There does not exist non-hyperbolic case for the equilibrium. But the hyperbolicity can be divided into the following cases (I), (II). (I) When, there exist six types for hyperbolic equilibrium point Q: (See Table 2). Table 1. Types of hyperbolic equilibrium point. Table 2. Types of hyperbolic equilibrium point. (II) When, there exist four types for hyperbolic equilibrium point Q: (See Table 3). Proof. Performing a coordinate shift as follows: and letting denote the transformed F, we translate equilibrium into and discuss equilibrium point of the map. The matrix of linearization of at is where,. Its eigenvalues are Table 3. Types of hyperbolic equilibrium. It is known that is hyperbolic if and only if none of eigenvalues, lies on the unit circle. In the following we discuss the eigenvalues in two case, i.e., and. When discriminant, then and are both real . Because non-hyperbolicity happens if and only if or. For whether or, we can get. By condition and, we see that. This is a contradiction with and, so and are impossible. Next, let’s examine and. From whether or, we can get, By condition we see that, , This is a contra- diction with, so and are impossible. When, and are a pair of conjugate complex. Since Therefore, and lie inside of and the equilibrium point Q is a stable focus referred to the case. When, the equilibrium point Q Is hyperbolic. If, i.e. The matrix has a double real eigenvalue. From the constraint condition, it is obvious that. Therefore, equilibrium point Q is stable node in the case of and. If, i.e., and, the eigenvalue and are different real numbers. We first discuss the case that, i.e., , In this case we have We have for, On the other hand, there also exists for. In fact, since We have. Therefore, the equilibrium Q is a stable node as . For the case, i.e., , we have and We assume, by condition, we see that , i.e., and by condition. This is a con- tradiction with and. So are impossible, i.e.,. Therefore, we have. Therefore, the equilibrium Q is a stable node as. Finally, we study the case of,. We have Then, we have for. Moreover, there also has for. In fact that, We have. This means that the equilibrium Q is a stable node for. When discriminant, because non-hyperbolicity happens if and only if or. Similar to the proof in case (I), neither nor is possible. When, and are a pair of conjugate complex. Since Therefore, and lie inside of and the equilibrium point Q is a stable node referred to the case. When, the equilibrium point Q is hyperbolic. If, the matrix has a double real eigenvalue. From the constraint condition, it is obvious that. Therefore, equilibrium point Q is stable node in the case of. If, we first discuss the case that, i.e., , In this case we have We have for, On the other hand, there also exists for . In fact, since Therefore, we have. Therefore, the equilibrium Q is a saddle as. Finally, we study the case of, i.e., We easily prove by same methods as in case (I). This means that the equilibrium Q is a stable node for. The proof is complete. 3. Transcritical Bifurcation of the Model The following lemmas were be derived from reference . Lemma 1. ( , Theorem 2.1.4) The map satisfies that A is cxc matrix with eigenvalues of modulus one, and B is sxs matrix with eigenvalues of modulus less than one, and where f and g are () in some neighborhood of the origin. Then there exists a center manifold for equation (7) which can be locally represented as a graph as follows For sufficiently small. Moreover, the dynamics of equation (4.1) restricted to the center manifold is, for sufficiently small, given by the c-dimensional map Lemma 2. ( , in page 365) A one-parameter family of () one-dimensional Having a non-hyperbolic fixed point, i.e., Undergoes a transcritical bifurcation at if Theorem 3. A transcritical bifurcation occurs at the equilibrium when. More concretely, for slightly there are two equilibriums: a stable point P and an unstable negative equilibrium which coalesce at, for slightly there are also two equilibriums: an unstable equilibrium P and a stable positive equilibrium Q. Thus an exchange of stability has occurred at. Proof. For, we have and. Consider as the bifurcation parameter and write F as to emphasize the dependence on. Performing a coordinate shift as follows,. One can easily see that the matrix is and it has eigenvectors Corresponding to and respectively, where T means the transpose of matrices. First, we put the matrix into a diagonal form. Using the eigenvectors (9), we obtain the transformation which transform system Equation (5) into Rewrite system (12) in the suspended form with assumption, Thus, from Lemma 1, the stability of equilibrium near can be determined by studying an one parameter family of map on a center manifold which can be represented as follows, for sufficiently small v and. We now want to compute the center manifold and derive the mapping on the center manifold. We assume near the origin, where means terms of order. By Lemma 1, those coefficients can be determined by the equation Substituting (16)into (15) and comparing coefficients of and in (15), we get from which we solve Therefore, the expression of (15) is approximately determined: Substituting (17) into (14), we obtain a one dimensional map reduced to the center manifold It is easy to check that The condition (19) implies that in the study of the orbit structure near the bifurcation point terms of do not qualitatively affect the nature of the bifurcation, namely they do not affect the geometry of the curves of equilibriums passing through the bifurcation point. Thus, the orbit structure of (18) near is qualitatively the same as the orbit structure near of the map Map (20) can be viewed as truncated normal form for the transcritical bifurcation (see Lemma 2). The stability of the two branches of equilibriums lying on both sides of are easily verified. 4. Degenerate Flip Bifurcation of the Model This section is devoted to the analysis for the case. From section 2, we have for. For this case, degenerate flip bifurcation happens at the equilibrium point. Theorem 4. For map (5) when, degenerate flip bifurcation happens at the equilibrium point. Proof. Performing a coordinate shift as follows We translate equilibrium into, and letting denote the transformed Therefore, we discuss equilibrium point of the map. The matrix of linearization of at is For, considering as the bifurcation parameter and write as to emphasize the dependence on w. Therefore, we have The matrix have eigenvectors and corresponding to and. Therefore, by transformation Therefore, we obtain the inverse of transformation (23) Therefore can be changed into the maps: Rewrite system (25) in the suspended form Equivalently, the suspended system (26) has a two-dimensional center manifold of the form Near the origin, where means terms of order. By Lemma 1, those coefficients can be determined by the equation Comparing coefficients of, and in (27), we get from which we solve Thus, the expression of (27)is determined, i.e., Substituting (30) into the first equation in (26), we obtain a one-dimensional map, where From (31), we can check that Thus, the conditions and of Theorem 3.5.1 in are not satisfied. Therefore, this is a degenerate flip bifurcation. Due to a lot of discrete-time models are not trivial analogues of their continuous ones and simple discrete-time models can even exhibit complex behavior (see ), motivated mainly by Meng and Chen considering a class of continuous vertical and horizontal transmitted epidemic model (1) under constant vaccination, we study a class of discrete vertical and horizontal transmitted disease model (2) under constant vaccination. By detailed studies, we found discrete model (2) has a flip bifurcation which did not occurred for continuous model. However, the result of flip bifurcation in current paper is a degenerate situation, for which the more in-depth research needs to be continued. This work has been supported by the Innovation and Developing School Project of Department of Education of Guangdong province (Grant No. 2014KZDXM065) and the Key project of Science and Technology Innovation of Guangdong College Students (Grant No. pdjh2016a0301).
In order to acquire images on the faster timescales often required by live-cell imaging, laser scanning confocal microscopes must be re-engineered to incorporate advanced scanning scenarios that enable the beam to be raster-scanned across the specimen at higher speeds. To overcome the inherently slow speed of traditional confocal microscopes, several manufacturers have introduced instruments equipped with resonant scanning mirrors that are capable of gathering images at 30 frames per second or higher, including Nikon's A1R HD25 system. The tutorial initializes by displaying the cosinusoidal pattern of a resonant scanner pixel clock laser beam superimposed on a stretched Ronchi grating (long black lines). Beneath the scan pattern is the light intensity transmitted through the grating and the pixel clock counts generated during the scan. The left-hand side of the diagram shows the forward scan whereas the right-hand side shows the reverse scan. Note that the speed of this tutorial has been dramatically reduced in order to demonstrate the various events that occur during resonant scanning. As scanning proceeds, a typical specimen is illustrated in the Image Scan window. The horizontal scanner mirror angle is also indicated as is the position of the reflected beam from the pixel clock laser with respect to the Ronchi grating. In order to operate the tutorial, click the Scan Image button to enable manual changes to the X Scan and Y Scan sliders. Due to the non-linearity of resonant scanner velocity, a fluorescent specimen is scanned at the highest speeds in the central region, with the velocity progressively decreasing as the scan reaches the edges. As a result, when the image data flow derived from a resonant scanner is acquired with a frame grabber clocked at a constant pixel rate (which assumes the beam is scanned linearly), the images appear stretched at the edges. Furthermore, the uneven distribution of laser excitation intensity (greater at the edges) produces excessive photobleaching (and potentially phototoxicity) at the edges of the scanned region because of the increased exposure to the laser light. The simplest option for compensating scanning non-linearity is to limit the scanning range to that portion of the oscillation period where the velocity of the galvanometer is almost linear, which occurs over a region spanning approximately 70 percent of the total scan width. Unfortunately, this solution reduces the amount of time that emission signal can be collected, increases the scan turnaround time before signal can be collected again, and doesn't prevent photobleaching in regions of the specimen that fall outside the area being recorded. Photobleaching effects can be minimized, however, when using the linear portion of the galvanometer oscillation by incorporating an adjustable aperture that limits the region of the specimen being exposed to illumination. Image distortion induced by non-linear resonant galvanometer scanning is fortunately predictable and can be corrected using either software or hardware solutions. Regardless of the correction scheme involved to produce images, the most effective scanning strategy involves collecting data during both the forward and backward scans of the galvanometer mirror. Recording data during the forward scan is straightforward, but due to the fact that the backward scan reverses the direction in which pixels are recorded, the image data must be inverted using specialized read-write buffers or software. In practice, images are gathered at twice the normal width (in effect, 1024 pixels for a final image size of 512 x 512 pixels) and half the normal height (256 pixels). At the end of each forward x-axis sweep of the resonant galvanometer mirror, the vertical sawtooth signal supplied to the linear (y) galvanometer is incremented by one line and the reverse (x) scan commences. In this manner, the vertical scan progresses smoothly for 256 cycles until enough data for one image is acquired. Both software and hardware clocking solutions for resonant scanning confocal microscopy rely on knowing positional data for the high frequency resonant galvanometer mirror. On 7.9 kilohertz devices, this mirror can rotate approximately 12 degrees in each direction around the central axis (for a total of 24 degrees of rotation). Mirror angular position (θ) and phase or velocity (φ) relationship can be predicted depending upon the state of the oscillating cycle. At the beginning of an oscillation cycle (mirror position equal to plus or minus 12 degrees; see Figure 1) the mirror is stationary with a velocity equal to zero. As the mirror swings toward the midpoint, the angular velocity increases as a cosine function of the phase and reaches its maximum when the mirror angle equals zero. Continuing, the mirror slows down again as it approaches the end of the forward scan until it reaches a velocity equal to zero as it instantaneously reverses direction. Identical changes in angular velocity are observed as the mirror rotates in the opposite direction. Another factor that should be considered with resonant galvanometers is that these devices are high-Q oscillators and cannot be efficiently synchronized to an external frequency due to large response variations in phase and amplitude when even the slightest drift occurs between the natural galvanometer frequency and the external driving source. Therefore, the resonant galvanometer should itself be used as the master oscillator to which all other timing components are synchronized. Software-based clocking schemes for resonant scanning confocal microscopy involve algorithms that are capable of pixel data linearization using look-up tables based on the predictable motion of the mirror. As an example, one popular algorithm first determines a phase constant for the mirror and frame grabber, and then locates the center pixel of the final image. This algorithm operates around the central plane of symmetry where the mirror position is known with a high degree of precision. Images are processed by examining one horizontal scan line at a time as they are delivered to the frame grabber. Pixel data is stored as received from the scanner at twice the final image width and half the final height. The first step in the process involves inverting data points from the second half of the horizontal scan line, followed by interlacing the inverted data between the data lines for the first half of the image (Figure 1(b)). Because the midpoint of the image may not be known with certainty due to a lag between collection of data and initiation of the horizontal synchronization signal that triggers acquisition, it is often necessary for the software to offset the inverted scan line by a few pixels. In this case, the right edge of the scan line then serves as a reference point for the image. The next step in the sequence is to apply a correction factor for each pixel or phase position relative to the center pixel. That pixel is then relocated to the correct position in the final image (Figure 1(c)). In the ideal case, using sophisticated high speed computers, incoming data can be analyzed on the fly and recorded to the hard drive in real time. During the horizontal scan period of the resonant galvanometer, the clocking photodiode registers pulses of laser light scattered by the Ronchi grating that can be electronically converted to a pixel clock having variable frequency. These light pulses will be more frequent in time when the galvanometer mirror is in the central portion of its oscillatory period, but they will slow down as the mirror reaches the end of a scan and reverses direction. Among the requirements of the photodiode detector are the ability to handle the variable pulse frequency (in terms of bandwidth) and it must be large enough to detect the entire line sweep. The detector is connected to a transconductance amplifier that transforms photodiode current pulses into an amplified voltage, which is then fed to a frequency doubler circuit that doubles the number of pixels per line. The photodiode and associated electronics deliver 256 pulses per line and the frequency doubling circuit transforms this output into 512 pulses per line that are acquired by a frame grabber and used to construct an image. Presented in Figure 2 is an illustration of the basic concepts behind the use of a Ronchi grating coupled to a photodiode detector as a variable pixel clock. For purposes of clarity, the grating illustrated in Figure 2(a) contains only eight spacings of equal width superimposed with one complete cycle of cosinusoidal motion from the resonant galvanometer versus time (red curve; 125 microseconds). The actual grating has 256 line spacings and is far more compact in the linear dimensions due to the fact that the pixel clock laser actually retraces a single line (perpendicular to the opaque lines) across the grating surface in operation (Figure 2(b)). As the light beam enters and exits the clear regions of the Ronchi grating, the transmitted light intensity that is detected by the photodiode rises and falls in step with galvanometer motion (Figure 2(c)). Each transition (in this case, from dark to light) of light intensity detected by the photodiode is used to generate the pixel clock (Figure 2(d)) that triggers image capture by the microscope. Although the clock pulses illustrated in Figure 2(d) are non-uniformly spaced in time (ranging from approximately 70 to 160 nanoseconds), they nevertheless correspond to equal intervals in image space. In order to acquire pixel samples at equal spatial intervals (rather than in equal temporal increments), photomultiplier signals from the primary microscope imaging system pass through conventional amplifiers with variable gain and offset, and are then digitized by an analog-to-digital converter triggered by the pixel clock generated from the Ronchi grating pulses. However, half of the image information is inverted due to the fact that the resonant galvanometer sweeps in both directions. In order to rectify this bidirectional scan artifact before constructing an image, the digital pixel output is first transferred to either a first-in-first-out (FIFO) or last-in-first-out (LIFO), buffer depending upon whether the pixels were acquired from left to right or right to left. Thus, during the horizontal sweep of the galvanometer mirror, image pixels that are acquired during the first half of the sweep (time period of 62.5 microseconds) are fed to the FIFO buffer whereas pixels acquired during the reverse sweep are fed to the LIFO buffer. The FIFO buffer is read and reset to zero at the beginning of each horizontal line acquisition, whereas the LIFO buffer counts up during storage of incoming data and down during the subsequent readout to ensure that pixels acquired during the reverse sweep are inverted. Jeffrey M. Larson and Stanley A. Schwartz - Nikon Instruments, Inc., 1300 Walt Whitman Road, Melville, New York, 11747. Adam M. Rainey, Alex B. Coker, and Michael W. Davidson - National High Magnetic Field Laboratory, 1800 East Paul Dirac Dr., The Florida State University, Tallahassee, Florida, 32310.
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If I want to apply a percentage reduction to a figure, I can set it up for the first line of data and then autofill the relevant cells. Improve your math knowledge with free questions in "Find what percent one number is of another" and thousands of other math skills. 8% just tell you how much of the liquid in the bottle is nicotine. Couples fare better than single seniors; the poverty rate is highest among divorced and widowed women, at 21 percent at 15 percent, respectively. ‘Of’ means ‘multiplied by’. Naturally, we asked five. To calculate the percentage discrepancy, take the difference between the two exchange rates, and divide it by the market exchange rate: 1. 20% of 50 means 20% times 50,. More than 7 in 10 teenagers and young adults say climate change will cause a moderate or great deal of harm to people in their generation, a slightly higher percentage than among those 30 and older. That is, unless "something miraculous happens," as Seattle Seahawks coach Pete Carroll told reporters Monday. What was her percentage weight loss? First, I'll find the absolute weight loss: 125 – 110 = 15. 75 (shows that before a 12. Sad failed test 10 percent of grade. A percentage is really a number telling you how many Converting From Percentages to Decimals and Back. How to calculate the percentage of tax deduction for co-op maintenance fees? Asked by AMRWJ, Brooklyn, NY • Thu Jan 20, 2011. You don't need to do this every day of your life. It relates to the number 100. Its general formula would be Y•X(H_2O). The HCl is a strong acid and is 100% ionized in water. Anthony Davis talks with LeBron James about their goals for the Lakers every day, thinks they can do ‘something special’ he had the highest true shooting percentage (61. If your answer is a negative number then this is a percentage increase. And now you know what calculator I used to normalize the above 12:15 ratio to a percent. 2 percent and is weighted 45 percent, the points for your overall grade out of 100 would be 39. This video screencast was created with Doceri on an iPad. 333 and then multiply by 100 to give you 33. Therefore, 20 percent of your total income of $1,500 is $300. Example: Find the pH of a 0. Calculating the standard percentage of something is the simplest percent based calculation. Here’s the formula in cell C2: =B2/A2. A discount is a deduction off of an original price. 5 Therefore, the new cost is 450 - 67. It is a straightforward approach that you have used earlier when dealing with the composition of compounds. It's important to keep an eye on your brokerage account and. First, I need to remind you about something with decimals:. OnePlus claims this notch is 31. For an assignment I need to give the percentage difference or change between an initial and final value. In actuality, service levels are quite easy to calculate in the contact centre, but there has been a lot of confusion on how to measure them. If you thought there’s no way the iPad could do what you really need. There are two methods of finding the percent of change between two numbers. Ask Question Asked 5 years, 5 months ago. This page will show you how to calculate and format a number as a percentage in R. Normalize is just a fancy word meaning to make the same. For example, 50% of 40% is:. The percent button can be used to find a percentage of another number. This takes just one click: Select the long decimal in C5, then click the Percent Style button on the Home tab of the Ribbon bar. Percentage Formula. If your figures of values are listed in the B2 and B3 cells of an excel worksheet, you can calculate the percentage increase by plugging in the following formula into the cell that you want the percentage value in. An in-depth interview with three-time Oscar winner Ang Lee about "Gemini Man," today's digital possibilities, and the fading prospects of 3D. 3 steps to markup percentage. The answer will be displayed on your calculator (1020 x 10 % = 102). Naturally, we asked five. org? A global movement for good. Calculating Percentages Based on Ratios. Percentage Calculator. Step 2: Create a measure to calculate percentage of another value in Excel pivot tables Now that we have a data model pivot, you can add measures to calculate various interesting things. The final technique on this sheet is to find the percent of total for each month. com! Shop for toys then find either the sale price or the percent discount. Let’s create a measure to calculate the amount as a percentage of sales. Couples fare better than single seniors; the poverty rate is highest among divorced and widowed women, at 21 percent at 15 percent, respectively. A number of health risks are associated with being overweight or obese, including an increased likelihood of developing coronary heart disease, type 2 diabetes, high blood pressure, gallstones and even certain types of cancer. In this tutorial, you will lean a quick way to calculate percentages in Excel, find the basic percentage formula and a few more formulas for calculating percentage increase, percent of total and more. and I would like the results to look like the table below. To determine the percent of a number do the following steps: Multiply the number by the percent (e. Find other percentages: Other percentages can be found by combining some of the techniques mentioned above. Just because something is old doesn't necessarily make it valuable. Fortunately, the PROPORTION METHOD will work for all three types of questions: What number is 75% of 4? 3 is what percent of 4? 75% of what number is 3?. If your calculator does not have a "%" button. The next step is to add the percentages together to achieve an offspring which is: 1/2 (4/8) TENNESSEE MEAT GOAT™ (TMG) 1/2 (4/8) Boer. 5% The rule for using percents is that YOU MUST HAVE A PERCENT OF SOMETHING. What was her percentage weight loss? First, I'll find the absolute weight loss: 125 – 110 = 15. Just follow the given example to understand how to calculate percent change between two numbers. For example, the oil change spent 60 percent of the maintenance budget. Then, Email it, print it, or add this useful timesheet calculator to your own website. X and Y are numbers and P is the percentage: Find P percent of X; Find what percent of X is Y. Rather than converting 15 percent to 15/100 or 3/20, you can think of it as 10/100 or 1/10 and then add half of that, because it is much easier to find 1/10 than 3/20. How to Get a Better Night's Sleep. Find the amount of the food component in the product’s nutritional information. We could use any mass units, but it is easiest to use grams. An interactive math lesson about determining percentage. 30) and multiply it by 100 (0. If we buy something on sale, then we can calculate the percentage we saved by doing the following steps. When you’re searching for ways to grow profits, a good place to start is retention metrics. To help you get to the bottom of your food cost percentage, the team here at Toast developed a free Food Cost Calculator. Everything you want to know about Java. BESCHLOSS: Hello, and welcome to the Council on Foreign Relations, ladies and gentlemen. Keep on reading for a step-by-step explanation of how to calculate percentages. ToString ("00. Press these buttons: 1 9 % * 2 0 = The answer is 3. An easy, fill-in-the-blanks tool to calculate percentages, percentage growth, and percentage off. How to Calculate Percentage Change. To find 30 percent of a number, multiply that number by 0. Percentage Calculator. But Boeing wanted something from Mr. Perhaps the most illuminating thing to know is that the word “percent” is really just the phrase “per cent” squashed together. I have a PDF worksheet in which I want users to be able to calculate various percentages of a "starting" dollar amount (in this case, the amount of a home loan). The word percent can be misleading if you don't know what it means. Step 1: Remove the percent sign and add a couple of zeros after. calculate percentage of cells that have any value If I follow you, Column A will be populated in order, while Column B may or may not, and you want the percent of A that's filled, compared to what's. Then, find what percentage of the gross profit is the COGS. I am wondering how to do that? I learnt no basics, just went on by myself trying to figure out things and so here is what I am trying to do. You can calculate percentages in Excel using basic multiplication and division. In comparison if you always wanted 2 decimal places (even if they are 0's) you can code like this percent. This free video math lesson will show you how it's done. We could use any mass units, but it is easiest to use grams. To find 66⅔% of a quantity double 33⅓% of it. What percentage of his times between his pitches are 30 and 34 seconds?. Example, total=1,100 and you need to find percent that equals to 100. The formulas below are all mathematical variations of this formula. 50 and get the result of 150. value_to_find is the value which we are trying to find. If you have an amount and the percentage that the amount represents of a total, you can calculate the total with a formula that simply divides the amount by the percentage. but i am unable to solve it. The percentage yield is the ratio between the actual yield and the theoretical yield multiplied by 100%. Improve your math knowledge with free questions in "Find what percent one number is of another" and thousands of other math skills. You don't need to do this every day of your life. Ashish Shah is. Example: If 1156 is the total score obtained in the examination out of 1200 marks, then divide 1156 by 1200, and then multiply it by 100. One calculates percentage by dividing the number of items by the total and multiplying by 100. This was a result of. For example 20% off of $17. Why Most People Are Dead Wrong About casino And Why This Report Must be Read by YouOnline online casino came about in early 90s, and possess happened to be expanding on an ongoing basis ever in your life since. 16 because cell C2 is not formatted as a percentage. Active 2 years, Did a law change or something? 8. You may also like: Add and Subtract Times: Hours, Minutes, Seconds: A dead-simple calculator that works as you type. My question is more of, how much each component number contributes to the sum, as a percentage. If your Windows 10 PC is running slow and you find that the hard drive is working at or near 100 percent, here are some steps you can take to fix the issue. online percentage calculator is simple then you would have imagined. Keep selecting the result cell, then click the Percent Style button in the Number group under Home tab to format the cell as percentage. To calculate the percentage of carbohydrate, take the grams of carbohydrate listed and multiply it by four. "MARK-UP is the percentage of the amount you earn on the COST of an item you sell. Percentage of a number using mental math Learn how to solve percent problems in your head! Since 10% is 1/10, and it's so easy to find 1/10 of any number, we can use that to quickly calculate certain percentages of quantities in our heads. 5 percent of females work more than 40 hours per week. You can easily injure yourself when testing your muscular limits. How Calculated When a resource is first added, the percent work complete for the resource is zero percent. Get an answer for 'How do you figure simple number percentages, such as 125% of 48. Mammography is 85 percent to 90 percent accurate. Example: Let's say I bought a stock for $50 and now I'm going to sell it for $125. What happens when they’re not available anymore? The stability that comes with a particular food or device can be disrupted if it’s lost. when you want to reflect something as a percentage of something say like finding the profit percentage you divide the profit by 100 and the resultant rational number you multiply by hundred. If the product label does not provide you with the information you need, look it up on a nutrition reference chart. 5 is 5 hundredths of 100. All you need is your original budget, the actual expenses, and maybe a. Percentage increase and decrease are calculated by computing the difference between two values and comparing that difference to the initial value. And when you do, you'll find that Excel can handle them just as well as whole numbers or decimals. In mathematics, a percentage is a number in the fraction of 100. When you’re searching for ways to grow profits, a good place to start is retention metrics. hi, i'm new to ASP. 0% would be 40mg/ml. Ask Question Asked 5 years, 5 months ago. There are numerous percentage calculators online that can help with task by simply searching for "percentage calculator. To find 30 percent of a number, multiply that number by 0. Colouring Percentages. He shows that to compute x percentage of something we need to movie decimal point in the percentage number two places to the left. e for 1st one YG-110 with value 5 the percentage is [5/205100]. 8 g/cm 3, and one with a density of 3. This is due to the higher base that results in a lower percentage for any given change in value. It is obvious that the second value range is higher than the first range. What percentage of his times between his pitches are 30 and 34 seconds?. Learn More Drug Rehab Success Rate The Revered 7 days is a 100's of years old procession and starts off on Palm Weekend to become continuing until Easter time On rehab. The probability of someone having a deductible of over $1,000 is 38. This free video math lesson will show you how it's done. How to Calculate the Percentage of Something Know the Basics. Play Percent Shopping at MathPlayground. Tip: To increase or decrease the number of decimal places showing, click the Increase Decimal or Decrease Decimal button on the Ribbon bar. There are two methods of finding the percent of change between two numbers. To calculate the percent of water you will divide the change in mass of your sample (mass of water removed) by the mass of the hydrated salt (original mass). These online percentage calculator provide you with the most convenient way of calculating the percentage, decreasing percentage, increasing percentage and other values. Another tricky thing about density is that you can't add densities. Some users may post messages that are misleading, untrue or offensive. To find 66⅔% of a quantity double 33⅓% of it. It can be a number or a string. If you set the Format of the column to Percent, it'll display with the % sign. Of course, you do not necessarily have to calculate the percentages (they just allow you to find every percentage). Percentage as a Proportion If you want to calculate a percentage of a number in Excel, simply multiply the percentage value by the number that you want the percentage of. This then gives you a percentage service level. The word percent can be misleading if you don't know what it means. Three and a half, plus, that were Scottish. Frequency Distribution Find the Percentage Frequency The number of classes can be estimated using the rounded output of Sturges' rule, , where is the number of classes and is the number of items in the data set. Improve your math knowledge with free questions in "Find what percent one number is of another" and thousands of other math skills. Find the percentage of time spent by an employee with this calculator. A percentage is really a number telling you how many Converting From Percentages to Decimals and Back. Year 7 Interactive Maths - Second Edition. 1)10 = 1 second. You can easily injure yourself when testing your muscular limits. When you buy a new place, thinking about how to measure the home’s square footage probably isn’t top of mind. When a quantity shrinks (gets smaller), then we can compute its PERCENT DECREASE. But it’s something the 7 Pro can’t do at all. Calculate Averages. How to calculate a percentage of a number. So, let's break this down with an example: Suppose George owns stock in Vandelay Industries. We are looking for the Amount. To do so, click the button (Home tab, Numbers group). ~~ Based on the novel "1%ui Eoddungut" by Hyun Go Woon. This prealgebra-arithmetic lesson explains how to convert (change) a percent to a decimal. Thus: pH = - log (0. Knowing how to work with percentages is vital if you want to become a smart shopper or savvy investor. For example, if you have 100 oranges and your friend gives you 10 more, your orange supply has increased by 10 percent, because 10 is 10 percent of 100. Use a Pie Chart to Find Percentages and Amounts A pie chart, which looks like a divided circle, shows you how a whole object is cut up into parts. If i have a c and i fail a test the test is 20 percent of my grade what will i get in the class? If homework is 20 percent of your grade guizzes are 30 and test are 50 if you have a 100 in homework a 95 in quizzes and a 60 in test what would yo. 42 percent monthly interest rate. When a quantity grows (gets bigger), then we can compute its PERCENT INCREASE. Let's say you wanted to find 19 percent of 20. A tiny percentage, such as 2%, means that a small part (only 2 parts out of. Of course, you do not necessarily have to calculate the percentages (they just allow you to find every percentage). In computing the growth or decline of a variable, you can quickly use this percentage change calculator to find the percentage increase or decrease in the value of two numbers. In that same group, you will also find the button, which enables you to add more decimal places to your result. How the formula works. Although the percentage formula can be written in different forms, it is essentially an algebraic equation involving three values. 56, and the markup percentage is 60%, do I calculate the gross profit earned with or without VAT? I know how to do the calculation, I'm just not sure if gross profit includes or excludes VAT. Combining advertising, software and services, CareerBuilder leads the industry in recruiting solutions, employment screening and human capital management. Calculate Percentages: A Step-by-Step Guide for Students. Say that your company sold $125,000 this quarter, and you need to find out what percentage $20,000 is of the total. If the product label does not provide you with the information you need, look it up on a nutrition reference chart. Airbus went from having less than 25 percent of the market share for large commercial airplanes in 1990 to overtaking Boeing in 2003. Calculating Percentages of a Total in Excel. This then gives you a percentage service level. 333 and then multiply by 100 to give you 33. That is the ratio of 5 to 100. You'd say, well, 4 is 1/4 of 16. It is obvious that the second value range is higher than the first range. Enter three numbers and find out what percentage of the total they make First number: Second Number Third Number. ' and find homework help for other Math questions at eNotes. Calculate a Percentage of Total. If your calculator does not have a percent key and you want to add a percentage to a number multiply that number by 1 plus the percentage fraction. Materials and direct labor costs might be included when calculating COGS. happen we have to start taking data to find out what caused it, what was the result of it, and start seeing trends and. How to calculate the percentage of tax deduction for co-op maintenance fees? Asked by AMRWJ, Brooklyn, NY • Thu Jan 20, 2011. Finding a Specific Percentage of a Number. “This is one of my absolute favorite frequently misunderstood tax concepts,” says Mark Durrenberger, a Certified Financial Planner and author of The Modern Day Millionaire. Some of the most frequently asked questions regarding what percentages are and how to calculate them in terms of percentage change, percentage increase, and working out percentages on a calcualtor are covered in this article. In fact, more than half of Americans report having less than $25,000 in savings and investments,. THE ANSWERS ARE INCORRECT. A percentage is really a number telling you how many Converting From Percentages to Decimals and Back. 75 (shows that before a 12. Did you find us useful? Please consider supporting the site with a small donation. To find 20% of a quantity double 10% of it. He shows that to compute x percentage of something we need to movie decimal point in the percentage number two places to the left. Mole fraction is represented by the symbol Χ Χ a = n a ÷ (n a + n b + n c + ) Χ a = mole fraction of. My question can any body advise me on how I calculate the percentage of each of my ingredients which are fluid and solid based products. To find the mass percent composition of an element, divide the mass contribution of the element by the total molecular mass. This calculator helps you calculate the percentage decrease of a value based upon the initial value and the value after the decrease has taken place. It is reduced in a sale by 15%, what is its sale price? Long-winded solution, with explanation is as follows I shall find what 15% of 450 is and then subtract this from 450. Of course if you wanted to restrict this to a date range you could use query parameters and the sub query would look like;. Usually, these discounts are given in the form of coupons or a percentage off taken when you pay for the item. Then: divide the increase by the original number and multiply the answer by 100. How to calculate overall yield of multi step synthesis/total synthesis? In a multi step synthesis the yields in every state varies, how to calculate the over all yield? multiply all percentage. Some percentage calculations are easy and can be done without any helping tools, while others are more complex and may require a calculator. Hoe to calculate gross figure after a percentage has been taken off. Here base value will be 2018 values hence we should calculate as per 2018 data. Step 2: Create a measure to calculate percentage of another value in Excel pivot tables Now that we have a data model pivot, you can add measures to calculate various interesting things. Use the Number group's dialog box launcher to open the Format Cells dialog box and set the decimal places for your data. Expressing values in the form of percentages in your work will enable you to readily compare information from different sources, quantify change over time and find the amount by which something has increased or decreased following a percentage change. For convenience, let's assume a volume of 1 L. How to get percentage of something calculations in Excel Pivot Tables Step 1: Create a data model pivot table from your data. The probability of someone having a deductible of over $1,000 is 38. Calculate Percent of Two Numbers calculator, the answer is 2. how do you find 1% of a number stephanie17 in Maths about 7 years ago Welcome to our free-to-use Q&A hub, where students post questions and get help from other students and tutors. Percent means “out of 100” so we divide the whole we are looking at into 100 parts. This allows percent changes to be added and subtracted. Free math lessons and math homework help from basic math to algebra, geometry and beyond. 5 Therefore, the new cost is 450 - 67. 06 or 6 percent. For mental arithmetic it is sometimes easier to divide the number by 100 (to calculate 1%) then multiplying by 30 (for 30%). Paper leaks showing a quantum computer doing something a supercomputer can’t Google's system generates quantum statistics that we just can't simulate. Problem: calculate percentage in one sql query. Adults have to calculate percentages every day, in order to figure tips, taxes, commissions and sale prices. 65 means that for every 1. Percentages are charted on a p chart, where p stands for proportion. Keep selecting the result cell, then click the Percent Style button in the Number group under Home tab to format the cell as percentage. Materials and direct labor costs might be included when calculating COGS. Calculating the percentage of something is necessary if you want to find out of the proportions between two quantities. The example below combines the two functions to calculate the percentage of Yes/No responses in a range of data. I guess you're learning how to Python. The same procedures apply in a written calculation, in which we would typically change the percent to a decimal. This calculation might be useful for calculating grades and numbers for the students. What's the percent increase in a population if it goes from 900 to 981? How can you find 33 1/3% of something? Keep on reading to learn the answers to these and more frequently asked questions about percentages!. How to calculate percentages using form fields. With the simpler calculators once you have pressed the it is not necessary to press. Instead of combining a mass (in mg) and a volume (in mL), percentages use the volumes of both. For example, let’s say you want to calculate your net worth. These concepts are thoroughly explored on this page. But what I also need at the end, or bottom of the spreadsheet is the percentages of "on time" or "late". Press these buttons: 1 9 % * 2 0 = The answer is 3. One would think the Patriots would figure something out on the goal line — Sony Michel found himself on the sidelines in favor of Brandon Bolden during goal-line work last week — so that they don't have their 42-year-old quarterback plunging into a mass of humanity multiple times a week. Before the festival was even discussed, 40 to 50 percent of rooms were booked. For this a formula would be:. sleep for 20 to 21 hours and then getting behind the wheel is comparable to having a blood alcohol level of about. As an investor, it is.
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16.1. This chapter outlines the kinds of price index data available from the ABS and gives some guidance to users who wish to interpret or further manipulate the data. Publication of statistics 16.2. Producer and international trade price indexes are all compiled and published quarterly. No advance figures are available. In order to ensure that users of these price indexes are treated in a completely even-handed manner, strict security is maintained on the figures before their official publication, and no information about the figures, of any kind, is released before the official embargo time of 11.30 a.m. on the day of publication. 16.3. A large range of data is available via on-line services, diskette, magnetic tape, tape cartridge and CD-ROM. For more details about our electronic data services, contact any ABS Office. 16.4. Price index figures are published for financial years as well as for quarters. The index number for a financial year is the simple arithmetic average (mean) of the index numbers for the 4 quarters of that year. Index numbers for calendar years are not calculated by the ABS but can be derived by calculating the simple arithmetic average of the index numbers for the period concerned. 16.5. All producer and foreign trade price index figures are released as preliminary figures at the time they are first published. Since most of the required basic prices are collected before compilation, the index numbers are not usually subject to revision. However, occasionally revisions may be necessary, as a result of undetected errors in the raw price data, retrospective price changes, or errors in compilation. 16.6. In order to determine actual transaction prices, it is necessary to seek measures of 'special' discounts when they occur, in addition to normal trade discounts. The measurement of special discounts requires continued attention because of the different forms they can take and the variety of ways in which they can be applied. Sometimes these discounts need to be partly estimated and may necessitate revisions to published series. 16.7. As well as various series of index numbers, the ABS also publishes a selection of other figures - percentage changes, index points contributions and index points changes. Each of these measures has already been explained in Chapter 13. Some further explanations to assist users to interpret these figures, or to calculate others for themselves, are given in the following topics. 16.8. The percentage changes published by the ABS are calculated from the published index numbers. This is because the percentage change is intended merely to summarise the changes in the published index numbers and not as a separate expression of the degree of price change between periods. Thus users who wish to calculate percentage changes for periods not shown in the ABS publications can calculate figures exactly comparable with the ABS figures by deriving them from the published index numbers. 16.9. The percentage change in any price index series for a month is calculated as follows: % change = ((index (current period) - index (previous period)) / index (previous period) * 100 Similarly, percentage changes between any other pair of periods can be calculated by substituting the periods concerned in the equation. 16.10. Percentage changes across more than one period cannot be calculated by summing the percentage changes for individual periods because this ignores compounding effects. For example: Here the percentage change between period 1 and period 3 is not 40 per cent (20 + 20) but: 16.11. Similarly, annual percentage changes cannot be obtained by summing percentage changes for 4 consecutive quarters. Nor can they be obtained by multiplying a single quarter's change by 4. Depending on the precise comparison a user wishes to make, annual changes can be obtained by making either of the following comparisons: Change from preceding period + 20 per cent + 20 per cent - by calculating the percentage change between the index number for one year (which can be any period of 4 consecutive quarters) and the index number for the preceding year ; or These two calculations, of course, produce different results because they measure different things. The ABS publishes both measures of annual percentage change, i.e. for financial years, and the change from the corresponding month of the previous year. Index points contributions - by calculating the percentage change between the index number for a particular quarter and the index number for the corresponding quarter of the previous year. 16.12. Any detailed analyses of the structure and composition of a price index, or of changes in composition over time, are best done using percentage contributions to the All Groups series, as given in Chapters 4 to 11. However, a common way of looking at changes in composition in the short term, for instance in examining the effects of disparate price movements in different parts of the index, is to look at index points contributions and changes in these contributions, which are produced in the course of compiling the price indexes. 16.13. The index points contribution for any component, in any series, is simply a quantitative expression of how much that component contributes to the magnitude of the index number for the series concerned. For example, when the index number for a particular series stands at 112.5, the size of the contribution of each of its components can be expressed in terms of how many of the 112.5 index points each contributes. 16.14. Most commonly, index points contributions and movements are analysed in terms of the total price index (i.e. in terms of the number of all groups index points contributed by each component to the total index). Index points data are intended to show: - the relative importance of particular components, classes, groups etc. in the total index at a particular time; and 16.15. The contribution of a particular series in a particular period is derived as: - the relative contribution of each component, class, group etc. to that period's movement in the total index. Changes in index points contributions are then simply the arithmetic difference between the corresponding contributions in the periods being looked at. For example: - Weight to total index * index number for the series That is, between periods 1 and 2, the All Groups series increased by 21 index points. Of this movement component Z contributed 20 index points. 16.16. In addition to the figures contained in the various ABS publications (Catalogue Nos. 6405.0 to 6415.0), a large volume of unpublished figures is available on request. Most of the series included in the weighting pattern for an index are available, subject to confidentiality. 16.17. Special index series that better meet the needs of particular users can also be constructed by the ABS from the detailed unpublished series. 16.18. The reference base of each price index is published along with the index series. The reference base for an index is changed as part of the rolling cycle of index rebasing and reweighting. 16.19. For periods following a change in reference base, index numbers on the former basis can still be derived from index numbers on the new basis using arithmetic conversion factors derived from the relationships between the old and new series at the time of the change. For example, from the October 1986 issue onwards, the House Building index was converted to a reference base of 1985-86 = 100.0 from the previous reference base of 1966-67 = 100.0. The factor for Sydney to be used to convert from the new basis to the old was derived as: 16.20. This factor of 5.0324 may be applied to any Sydney House Building index number on the present (1985-86 = 100.0) reference base to give a corresponding index number on the previous (1966-67 = 100.0) reference base. The reverse conversion, i.e. from the previous basis to the present basis, can also be made - in this example using the conversion factor: Use of price index data in contracts 16.21. Sometimes users encounter problems because of the manner in which provisions relating to the use of price indexes are incorporated in contracts or other legal documents, for example, provisions relating to escalation of costs or charges to compensate for price variations. While it is a matter for the parties to decide how such provisions should be worded, and the ABS does not become involved in such decisions, problems could often be avoided if care is taken with respect to a number of specific issues at the time the contract is written. In particular, the following points should be specifically addressed: - the particular price index series should be clearly specified (i.e. whether it is the series for a particular city, or the weighted average of the six State capital cities, and so on, and whether for the All Groups or a component); - the index number for the period to be used as the base, or reference point, for the calculation of future changes should be clearly identified; and - reference should be made to the procedure to be followed in the event of: - the relevant price index figure being revised by the ABS after it has been used in accordance with the contract provision; or - the reference base of the published series being changed by the ABS; or - the specified price index series being discontinued by the ABS.
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By Sergio Blanes, Fernando Casas Discover How Geometric Integrators safeguard the most Qualitative homes of continuing Dynamical Systems A Concise creation to Geometric Numerical Integration offers the most topics, ideas, and functions of geometric integrators for researchers in arithmetic, physics, astronomy, and chemistry who're already conversant in numerical instruments for fixing differential equations. It additionally deals a bridge from conventional education within the numerical research of differential equations to figuring out fresh, complicated learn literature on numerical geometric integration. The ebook first examines high-order classical integration equipment from the constitution renovation viewpoint. It then illustrates the right way to build high-order integrators through the composition of easy low-order tools and analyzes the assumption of splitting. It subsequent stories symplectic integrators developed without delay from the speculation of producing services in addition to the $64000 class of variational integrators. The authors additionally clarify the connection among the upkeep of the geometric houses of a numerical procedure and the saw favorable mistakes propagation in long-time integration. The booklet concludes with an research of the applicability of splitting and composition how you can yes periods of partial differential equations, comparable to the Schrödinger equation and different evolution equations. The motivation of geometric numerical integration is not just to advance numerical tools with better qualitative habit but in addition to supply extra exact long-time integration effects than these got by way of general-purpose algorithms. available to researchers and post-graduate scholars from various backgrounds, this introductory publication will get readers in control at the rules, equipment, and functions of this box. Readers can reproduce the figures and effects given within the textual content utilizing the MATLAB® courses and version records on hand online. Read or Download A Concise Introduction to Geometric Numerical Integration PDF Best popular & elementary books Each one week of this 3 week assembly was once a self-contained occasion, even if each one had an identical underlying subject - the impact of parallel processing on numerical research. each one week supplied the chance for in depth learn to expand individuals' examine pursuits or deepen their knowing of themes of which they already had a few wisdom. Useful research performs a vital position within the technologies in addition to in arithmetic. it's a appealing topic that may be prompted and studied for its personal sake. based on this uncomplicated philosophy, the writer has made this introductory textual content available to a large spectrum of scholars, together with beginning-level graduates and complex undergraduates. 3 parts give a contribution to a topic sustained during the Coburn sequence: that of laying an organization beginning, construction an excellent framework, and delivering robust connections. not just does Coburn current a legitimate problem-solving procedure to educate scholars to acknowledge an issue, arrange a technique, and formulate an answer, the textual content encourages scholars to work out past tactics so as to achieve a better realizing of the massive principles at the back of mathematical innovations. Symmetric designs are a tremendous type of combinatorial buildings which arose first within the facts and are actually specifically vital within the examine of finite geometries. This publication provides the various algebraic strategies which have been dropped at undergo at the query of life, development and symmetry of symmetric designs - together with tools encouraged via the algebraic idea of coding and through the illustration conception of finite teams - and contains many effects. Extra info for A Concise Introduction to Geometric Numerical Integration It is possible then to apply a symplectic discretization by first carrying out a spatial truncation that reduces the PDE to a system of Hamiltonian ODEs and then using an appropriate symplectic integrator. Other popular geometric methods for PDEs are the so-called multisymplectic integrators [48, 82, 105, 160, 184, 219]. In any case, all these techniques are essentially restricted to smooth solutions of the PDE under consideration. What is geometric numerical integration? 7 39 Exercises 1. 43). 40) We notice at once that the q-variable is treated by the implicit Euler method and the p-variable by the explicit Euler method. Of course, the treatment of both variables can be interchanged, thus resulting in the method qn+1 = qn + h ∇p H(qn , pn+1 ), pn+1 = pn − h ∇q H(qn , pn+1 ). 41) The proof that both schemes are symplectic is straightforward. 40) and d = 1 for simplicity. Differentiating with respect to qn , pn yields ∂qn+1 ∂qn ∂pn+1 ∂qn = 1 + hHpq = −hHqq ∂qn+1 ; ∂qn ∂qn+1 ; ∂qn ∂qn+1 ∂qn+1 = h Hpq + Hpp ∂pn ∂pn ∂pn+1 ∂qn+1 = 1 − h Hqq + Hqp , ∂pn ∂pn where Hqq , Hqp , Hpp denote second partial derivatives evaluated at (qn+1 , pn ). Since the errors achieved by the two methods differ by several orders of magnitude, we also show the same results in a log-log diagram (right). Notice that the error in energy just oscillates for the symplectic Euler method without any secular component, whereas there is an error growth in energy for the explicit Euler scheme. With respect to the error in positions and momenta, a linear error growth for the symplectic method and a faster error growth for the non-symplectic one can be observed. 43). A Concise Introduction to Geometric Numerical Integration by Sergio Blanes, Fernando Casas
Master AP Calculus AB & BC Part II. AP CALCULUS AB & BC REVIEW CHAPTER 11. Sequences and Series (BC Topics Only) • Introduction to sequences and series, nth term divergence test • Convergence tests for infinite series • Power series • Taylor and Maclaurin series • Technology: Viewing and calculating sequences and series with a graphing calculator • Summing it up As a BC calculus student, you have come a long way since your first limits and derivatives. There may have been a time when you feared the Chain Rule, but that time is long since past. You are now a member of the elite Calculus Club. (First rule of Calculus Club: Don’t talk about Calculus Club.) And your membership is complete with sequences and series. Once you get an idea of what sequences and series are, we will focus primarily on infinite series. For a couple of sections, you’ll use various tests to determine the convergence of infinite series. Once that is complete, we’ll discuss power series and use Taylor and Maclaurin series to approximate the values of functions. Sound good? Your Jedi training is almost complete ... the Force is strong with this one. INTRODUCTION TO SEQUENCES AND SERIES, NTH TERM DIVERGENCE TEST A sequence is basically a list of numbers based on some defining rule. Nearly every calculus book begins with the same example, and it’s so darn fine that I will bow to peer pressure and use it as well. Consider the sequence The number n will take on all integer values beginning at 1, so the resulting sequence of numbers will be In some cases, for the sake of ease, we will let n begin with 0 instead of 1—but you’ll know exactly when to do that, so don’t get stressed out or confused. NOTE. The Force is with you always, and it equals mass times acceleration. Our singular goal in sequences is to determine whether or not they converge. In other words, is the sequence heading in some direction—toward some limiting number? You can graph the sequence above to see that it is headed toward a limit of 0. NOTE. The Technology section at the end of this chapter explains how to graph sequences and series on your 11-83. You should probably read that before you go any further. Each term of the sequence is half as large as the term before, and the sequence approaches a limit of 0 very quickly. Mathematically, we write Because this sequence has a limit as n approaches infinity, the sequence is said to converge; if no limit existed, the sequence would be described as divergent. Example 1: Determine whether or not is a convergent or divergent sequence. Solution: The sequence will converge if its limit at infinity exists and will diverge if the limit does not exist. This is a rational function with equal degrees in the numerator and denominator; therefore, the limit is the ratio of the leading coefficients: Therefore, the sequence converges to 1/2. This is further evidenced by the graph of the sequence below: A series is similar to a sequence, but in a series, you add all the terms together. Series are written using sigma notation and look like this: The notation is read “the sum from 1 to infinity of ” and gets its value from the sum of all the terms in the sequence. However, how can you tell if a series with an infinite number of terms has a finite sum? At first glance, it seems impossible—how can you add infinitely many numbers together to get a real sum? Consider the diagram below, which should help you visualize the infinite series : If the large box represents one unit and you continuously divide the box into halves, the sum of all the pieces will eventually (if you add forever and ever) equal the entire box. Thus, You won’t be able to draw pictures like this for the majority of series, but, in the next section, you’ll learn a much easier way to find the above sum. Mathematically, you need to know that a series gets its value from the sequence of its partial sums (SOPS). A partial sum is the sum of a piece of the series, rather than the entire thing; it is written Sn, where n is the number of terms being summed. Let’s use good old as an example: Therefore, the SOPS is .5, .75, .875, .... This is the important thing to remember: If the SOPS converges, then the infinite series converges, and its sum is equal to the limit at which the SOPS converged. For the most part, you will use the fantastic formula alluded to earlier in order to find sums. However, if all else fails, you can use the SOPS to find the sum of a series. Example 2: Find the sum of the series NOTE. The series in Example 2 is one example of a telescoping series, the terms in these series cancel out as the SO PS progresses. Solution: Construct the sequence of partial sums to gain some insight on this series: See what happened there? The —1/3 and 1/3 will cancel out. In S4, the —1/4 and 1/4 will cancel out. In fact, each partial sum will cancel out another term all the way to infinity, and the only two numbers left will be 1 + 1/2. Therefore, the sum of the series is 3/2. You can use your calculator to calculate S500 to verify that the SOPS is indeed approaching 3/2. It makes sense that each term in a sequence needs to get smaller if the series is going to converge. You are adding numbers for an infinite amount of time; if you are not eventually adding 0 in this infinite loop, your sum will grow and grow and never approach a limiting value. We showed that the sequence of the terms that make up have a limit of 0 early in this section, and that infinite series has a sum. However, if the limit, as n approaches infinity, of the sequence that forms an infinite series does not equal 0, then that infinite series cannot converge. This is called the nth Term Divergence Test, and it is the easiest way to immediately tell if a series is going to diverge. nth Term Divergence Test: If then is a divergent series. Example 3: Show that is divergent. Solution: Because (and this limit must equal 0 for the series to be convergent), the series diverges by the nth Term Divergence Test. That’s all there is to it. If you think about it, since the limit at infinity is 1, as n approaches infinity, you’d be adding 1 + 1 + 1 + 1 + 1 + 1 forever, and that clearly approaches no limiting or maximum value. Be careful! Just because , that does not mean that the series will converge! For example, the harmonic series is divergent even though If the limit at infinity is 0, you can conclude nothing from the nth Term Divergence Test. It can only be used to show that a series diverges (if its limit at infinity does not equal 0); it can never be used to show that a series converges. Directions: Solve each of the following problems. Decide which is the best of the choices given and indicate your responses in the book. YOU MAY USE YOUR GRAPHING CALCULATOR FOR PROBLEM 4 ONLY. 1. Determine if the sequence converges. 2. Find the nth term of each sequence (in other words, find the pattern evidenced by the sequence), and use it to determine whether or not the sequence converges. 3. Use the nth Term Divergence Test to determine whether or not the following series converge: 4. (a) What is the sum of (b) Calculate S500 to verify that the SOPS is bounded by the sum you found. ANSWERS AND EXPLANATIONS 1. The sequence converges if exists. You’ll have to use L’Hopital’s Rule: The sequence converges. 2. (a) This is the sequence Use L’Hopital’s to show that the sequence converges. (b) This is the sequence How can we tell that it converges? It is very clear that in fact, this sequence converges to 0 significantly faster than since the former’s denominator will grow larger much faster than the latter’s. 3. (a) since the expression is rational and the degrees of the numerator and denominator are equal. Because the limit does not equal 0, this series diverges by the nth Term Divergence Test. (b) but this does not necessarily mean that the series is convergent. You will find out that it does converge very soon (in the Integral Test subsection), but you can never conclude that any series converges using the nth Term Divergence Test; it can only be used to prove divergence. Therefore, you can draw no conclusion. 4. (a) This is a telescoping series; if you write out the fourth partial sum (S4), you can see what terms will cancel out in the long run and which ones will remain: The only terms that will remain as n approaches infinity are so the infinite sum is (b) Use your calculator to find S500. Of course, this is not the limit at infinity, but it will give us an idea of where SOPS is heading at that point. S500 is closing in on .8333, but it has all of infinity to get there—so what’s the rush?
- Does it look bad to not take math senior year? - Did Bill Gates take Math 55? - What is the easiest math to take in high school? - What grade do you take Algebra 2? - What is Bill Gates IQ? - What happens if you pass Math 55? - What is the hardest college class ever? - Should I take a math class senior year? - What is the hardest math class? - What’s the easiest math? - What is senior math? - What kind of math do seniors take? - What is the lowest math class in high school? - What is the highest level of math? - What is the easiest math class to take for senior year? - What is the easiest type of math? - Do colleges look at all 4 years of high school? - Is Trig hard? Does it look bad to not take math senior year? If you don’t take a math it will look pretty bad if you want to major in STEM or a social science. Also AP stats is really easy. I also really love math, so AP Stats was what I took senior year. It’s a worthwhile course, especially since it’s a bit -different- from Calculus and all the other math courses I’ve taken.. Did Bill Gates take Math 55? Bill Gates took Math 55. (He passed.) And if you’d like to sharpen your brain like Microsoft’s co-founder, here are The 5 Books Bill Gates Says You Should Read. What is the easiest math to take in high school? Rank the high school math classes from easiest to hardest (in context of the year it is taken)Algebra 1.Geometry.Algebra 2.Pre-calculus or Prob/Stat. What grade do you take Algebra 2? 11th gradeStudents typically learn Algebra II in 11th grade. An Algebra II curriculum usually builds on knowledge and skills that are gained in Algebra 1 and reinforced in Geometry, including relationships between quantities through equations and inequalities, graphing of functions, and trigonometry. What is Bill Gates IQ? 160 IQBill Gates is continuously standing number 1 as world billionaire in the world by Forbes magazine. His IQ Test score is 160 IQ, Bill Gates is a billionaire who always contribute to the world and do charity frequently. What happens if you pass Math 55? Nearly half of Math 55 graduates go on to become professors. In Williams’ book, he describes Richard Stallman’s Math 55 ending the semester with 20 students, eight of whom would go on to become future mathematics professors. One eventually went on to teach physics. What is the hardest college class ever? It shouldn’t surprise you that organic chemistry takes the No. 1 spot as the hardest college course. This course is often referred to as the “pre-med killer” because it actually has caused many pre-med majors to switch their major. Should I take a math class senior year? If you are going to a four year college, it is highly recommended to do so, as the math class in senior year will be more advanced, and help you prepare for college level math. … Likewise, if you passed the highest level math class available in the school before senior year, there is not much else to take, yes. What is the hardest math class? The Harvard University Department of Mathematics describes Math 55 as “probably the most difficult undergraduate math class in the country.” Formerly, students would begin the year in Math 25 (which was created in 1983 as a lower-level Math 55) and, after three weeks of point-set topology and special topics (for … What’s the easiest math? The easiest would be Contemporary Mathematics. This is usually a survey class taken by students not majoring in any science. The hardest is usually thought to be Calculus I. This is the full on, trigonometry based calculus course intended for science and engineering majors. What is senior math? This elective math course is intended for 12th graders, who have recently completed Algebra II, and serves to unify concepts from previous mathematical studies and assists students in clarifying their conceptual understanding of Algebra and Geometry. What kind of math do seniors take? By 12th grade, most students will have completed Algebra I, Algebra II, and Geometry, so high school seniors may want to focus on a higher level mathematics course such as Precalculus or Trigonometry. Students taking an advanced mathematics course will learn concepts like: Graphing exponential and logarithmic functions. What is the lowest math class in high school? Pre-AlgebraA decent amount of people said the lowest type of mathematics education offered is Pre-Algebra, for us that’s what the academically average freshmen are taking. What is the highest level of math? CalculusBegin with Algebra 1 and Geometry, often considered the building blocks of higher level math and science classes. Wrap up with Calculus, the highest level of math offered by many high schools and often considered the gold standard of pre-college math preparation. What is the easiest math class to take for senior year? College algebra would be the easiest I see. Trigonometry & geometry probably the close second and might be easier depending on if you’re more of a visual person. Pre-calculus I would only take if you plan on going further in terms of math courses same with numbers and logic. What is the easiest type of math? AlgebraAlgebra is by far the easiest. For a start, this type of math is very useful for stuff like physic. Do colleges look at all 4 years of high school? And while this may still be true in less-competitive admissions environments, nowadays colleges want to see one of two things from your student’s academic career—either a strong performance that’s maintained throughout all four years, or an upward trend showing improvement each year, culminating in a very strong kick … Is Trig hard? Trigonometry is hard because it deliberately makes difficult what is at heart easy. We know trig is about right triangles, and right triangles are about the Pythagorean Theorem. About the simplest math we can write is When this is the Pythagorean Theorem, we’re referring to a right isosceles triangle.
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Cutting error prediction by multilayer neural networks for machine tools with thermal expansion and compression 著者 Nakayama Kenji, Hirano Akihiro, Katoh Shinya, Yamamoto Tadashi, Nakanishi Kenichi, Sawada Manabu Proceedings of the International Joint Conference on Neural Networks page range 1373‑1378 Cutting Error Prediction by Multilayer Neural Networks for Machine Tools with Thermal Expansion and Kenji NAKAYAMA Akihiro HIRANO Shinya KATOH Tadashi YAMAMOTO † Kenichi NAKANISHI † Manabu SAWADA † Dept of Information and Systems Eng., Faculty of Eng., Kanazawa Univ. 2–40–20 Kodatsuno, Kanazawa, 920–8667, Japan e-mail: email@example.com † Nakamura-Tome Precision Industry Co., Ltd. In training neural networks, it is important to reduce input variables for saving memory, reducing network size, and achieving fast training. This paper proposes two kinds of selecting methods for useful input variables. One of them is to use information of connection weights after training. If a sum of absolute value of the con- nection weights related to the input node is large, then this input variable is selected. In some case, only pos- itive connection weights are taken into account. The other method is based on correlation coeﬃcients among the input variables. If a time series of the input vari- able can be obtained by amplifying and shifting that of another input variable, then the former can be absorbed in the latter. These analysis methods are applied to pre- dicting cutting error caused by thermal expansion and compression in machine tools. The input variables are reduced from 32 points to 16 points, while maintaining good prediction within6µm, which can be applicable to real machine tools. Recently, prediction and diagnosis have been very im- portant in a real world. In many cases, relations between the past data and the prediction, and the symptoms and diseases are complicated nonlinear. Neural networks are useful for these signal processing. Many kinds of ap- proaches have been proposed –. In these applications, observations, which are the past data, the symptoms and so on, are applied to the input nodes of neural networks, and the prediction and the diseases are obtained at the network outputs. In or- der to train neural networks and to predict the coming phenomenon and to diagnose the diseases, it should be analyzed what kinds of observations are useful for these purposes. Usually, the observations, which seems to be meaningful by experience, are used. In order to simplify observation processes, to minimize network size, and to make a learning process fast and stable, the input data should be minimized. How to selected the useful input variables have been discussed –. In this paper, selecting methods for useful input variables are proposed. The corresponding connection weights and correlation coeﬃcients among the input data are used. The proposed methods are applied to predicting cutting error caused by thermal expansion and compression in numerical controlled (NC) machine tools. Temperature is measured at many points on the machine tool and in the surroundings. For instance, 32 points are measured, which requires a complicated ob- servation system. It is desirable to reduce the tempera- ture measuring points. 2 Network Structure and Equations Figure 1shows a multilayer neural network with a single hidden layer. Relations among the input, hidden layer outputs and the ﬁnal outputs are shown here. uj(n) = N i=1 wjixi(n) +θj (1) yj(n) = fh(uj(n)) (2) uk(n) = J j=1 wkjyj(n) +θk (3) yk(n) = fo(uk(n)) (4) Hidden layer Output layer +1 θj θk Figure 1: Multilayer neural network with a single hidden layer. fh() and fo() are sigmoid functions. The connection weights are trained through supervised learning algo- rithms, such as an error back-propagation algorithm. 3 Analysis Methods for Useful Input Data 3.1 Method-Ia: Based on Absolute Value of Connection Weights The connection weights are updated following the error back-propagation (BP) algorithm. ˙fo() and ˙fh() are the 1st order derivative offo() andfh(), respectively. wkj(n+ 1 ) = wkj(n) + ∆wkj(n) (5) ∆wkj(n) = α∆wkj(n−1 ) +ηδkyj(n) (6) δk = ek(n) ˙fo(uk(n)) (7) ek(n) = dk(n)−yk(n), dk(n) is a target.(8) wji(n+ 1 ) = wji(n) + ∆wji(n) (9) ∆wji(n) = α∆wji(n−1 ) +ηδjxi(n) (10) δj = f˙h(uj(n)) δkwkj(n) (11) From the above equations, the connection weightwji(n) is updated by ηδjxi(n). Since η is usually a positive small number, then by repeating the updating,ηδjxi(n) is accumulated inwji(n+ 1). Thus, growth ofwji(n+ 1 ) is expressed by E[δjxi(n)], which is a cross-correlation. On the other hand, as shown in Eq.(11), δj expresses the output error caused by thejth hidden unit output. If the input variablexi(n) is an important factor, then it may be closely related to the output error, and their cross-correlation becomes a large value. For this reason, it can be expected that the connection weights for the important input variables to the hidden units will be grown up in a learning process. Based on this analysis, the important input variables are selected by using a sum of the corresponding connection weights after the training. Si,abso= J j=1 3.2 Method-Ib: Based on Positive Connection Weights When the input data always take positive numbers, neg- ative connection weights may reduce the input potential uj(n) in Eq.(1). Furthermore, when a sigmoid function shown in Fig.2 is used for an activation function, nega- tiveuj(n) generates small output, which does not aﬀect the ﬁnal output. Thus, in this case, the negative con- nection weights are not useful. Therefore, only positive connection weights are taken into account. The useful temperatures are selected based onSi,posi. wσi, wσi>0 (13) Figure 2: Sigmoid function, whose output is always posi- tive. 3.3 Method II: Based on Cross-correlation among Input variables Dependency among Input variables First, we discuss using a neuron model shown in Fig.3. The input potentialuis given by y = f(u ) Figure 3: Neuron model. If the following linear dependency is held, x2=ax1+b aandb are constant (15) then,uis rewrriten as follows: u = w1x1+w2(ax1+b) +α (16) = (w1+aw2)x1+ (bw2+α) (17) (18) Therefore, by replacingw1byw1+aw2andαbybw2+α, the input variablex2 can be removed as follows: u = wx1+β (19) w = w1+aw2 (20) β = bw2+α (21) This is an idea behind the proposed analysis method. The linear dependency given by Eq.(15) can be analyzed by using correlation coeﬃcients. Theith input variable is deﬁned follows: xi= [xi(0), xi(1),· · ·, xi(L−1)]T (22) Correlation coeﬃcient between theith and thejth vari- able vectors is given by ρij = (xi−x¯i)T(xj−x¯j) xi−x¯ixj−x¯j (23) x¯i,j = 1 xi,j(n) (24) If xi and xj satisfy Eq.(15), then ρij = 1. In other words, ifρijis close to unity, thenxiandxjare linearly dependent. On the other hand, ifρij = 0, then they are orthogonal to each other. Combination of Data Sets The modiﬁcations by Eqs.(19)–(21) are common in a MLNN. This means the modiﬁcations are the same for all the input data sets. Let the number of the data sets beQ. The input variables are re-deﬁned as follows: Deﬁnition of Input Data forQData Sets x(q)i = [x(q)i (0), x(q)i (1),· · · , x(q)i (L−1)]T (25) X(q) = [x(q)1 ,x(q)2 ,· · ·,x(q)N ] (26) = [˜x1,x˜2,· · ·,x˜N] (27) x(1)i x(2)i ... x(Q)i x(q)i is the input variable vector of theqth input data set. X(q) is the qth input data set, Xtotal is a total input data set, which includes all the input data. In ˜xi, the ith variables at all sampling points,n= 0,1,· · ·, L−1 and for all data setsq= 1,2,· · ·, Qare included. Using these notations, the correlation coeﬃcients are deﬁned as follows: ρij = (˜xi−x¯˜i)T(˜xj−x¯˜j) xi = 1 LQ x(q)i (n) (30) One example of the combined input data is shown in Fig.4, where Q= 4. data set 1 data set 2 data set 3 data set 4 Figure 4: Combined input variable vectors. Aberage of Correlation Coeﬃcients Method-IIa The correlation coeﬃcients for all combinations of the input variables are calculated by Eq.(29). Furthermore, dependency of theith variable is evaluated by ρ¯(1)i = 1 N−1 ρ¯(1)i expresses average of the correlation coeﬃcients be- tween theith variable and all the other variables. Thus, the variables, which have small ¯ρi are selected for the useful input variables. Let xσ be the selected variable vectors, and the num- ber of xσ be N1. The correlation coeﬃcients ρ(2)ij are evaluated once more among the selectedσvariable vec- tors. The variable vectors are further selected based on ρ(2)ij . The variables having large ρ(2)ij are removed from the selected set. Instead, the variables, which are not selected in Method-IIa and have small ¯ρ(1)i , are selected and added to xσ. This process is repeated until all the selected variables have small ρ(2)ij . 3.4 Comparison with Other Analysis Methods There are several methods to extract important com- ponents among the input data. One of them is prin- cipal component analysis. The other method is vector quantization. In these methods, however, in order to extract these components and vectors, many input data are required. Our purpose is to simplify the observation process for the input data, that is to select useful in- put variables, which are directly observed. It is diﬃcult to obtain the useful observation data from the principal components and the representative vectors. 4 Prediction of Cutting Error Caused by Thermal Expansion and Compression Numerical controlled (NC) machine tools are required to guarantee very high cutting precision, for instance tolerance of cutting error is within 10µm in diameter. There are many factors, which degrade cutting preci- sion. Among them, thermal expansion and compression of machine tools are very sensitive in cutting precision. In this paper, the multilayer neural network is applied to predicting cutting error caused by thermal eﬀects. 4.1 Structure of NC Machine Tool Figure 5 shows a blockdiagram of NC machine tool. Distance between the cutting tool and the objective is changed by thermal eﬀects. Temperatures at many Frame ( front, back ) Z Axis Slide Index Head Stock X Axis Slide Cutting Tool Figure 5: Rough sketch of NC machine tool. points on the machine tool and in surrounding are mea- sured. The number of measuring points is up to 32 points. 4.2 Multilayer Neural Network Figure 6 shows the multilayer neural network used pre- dicting cutting error of machine tools. The temperature and deviation are measured as a time series. Thermal expansion and compression of machine tools are also dependent on hysteresis of temperature change. xi(n) means the temperature at the ith measuring point and at thenth sampling points on the time axis. Its delayed samples xi(n−1), xi(n−2),· · · are generated through the delay elements ”T” and are applied to the MLNN. One hidden layer and one output unit are used. N Hidden layer Output layer Delay Figure 6: Neural network used for predicting cutting error caused by thermal eﬀects. 4.3 Training and Testing Using All Input vari- ables Four kinds of data sets are measured by changing cut- ting conditions. They are denotedD1, D2, D3, D4. Since it is not enough to evaluate prediction performance of the neural network, data sets are increased by combin- ing the measured data sets by linear interpolation, de- noted D12, D13, D14, D23, D24, D34. Some of the mea- sured temperatures in time are shown in Fig.7. Training and testing conditions are shown in Table 1. All mea- suring points are employed. The data sets except forD1 are used for training andD1 is used for testing. Figure 8 shows a learning curve using all data sets, 20 25 30 35 40 45 50 55 0 5 10 15 20 25 30 35 40 45 50 Time [ min ] Temperature [ ]C Figure 7: Some of measured temperatures in time. Table 1: Training and testing conditins Measuring points 32 points Training data sets D2, D3, D4 D12, D13, D14, D23, D24, D34 Test data set D1 Learning rateη 0.001 Momentum rateα 0.9 except for D1, and all measuring points, that is 32 points. The vertical axis means the mean squared error (MSE) of diﬀerence between the measured cutting error and the predicted cutting error. It is well reduced. 0 0.0005 0.001 0.0015 0.002 0 10000 20000 30000 40000 50000 60000 70000 80000 90000 100000 iteration Figure 8: Learning curve of cutting error prediction. All measuring points are used. Figure 9 shows cutting error prediction under the con- ditions in Table 1. The prediction error is within 6µm, which satisﬁes the tolerance 10µmin diameter. -28 -24 -20 -16 -12 -8 -4 0 4 8 250 500 750 1000 1250 1500 Time [ min ] Deviation [ µm ] Figure 9: Cutting error prediction. All measuring points are used. 5 Selection of Useful Measuring Points The useful 16 measuring points are selected from 32 points by the analysis methods proposed in Sec.3. 5.1 Selection Based on Connection Weights The measuring points are selected based on a sum of ab- solute value of the connection weights Si,abso (Method- Ia) and on a sum of positive connection weights Si,posi (Method-Ib). Figure 10 shows both sums. The horizon- tal axis shows the temperature measuring points, that is 32 points. Their prediction are shown in Fig.11. The selection method usingSi,posiis superior to the other us- ing Si,abso, because the temperature in this experience is always positive. The prediction error is within 6µm. 0 2 4 6 8 10 12 123456789 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 Sum of Absolute Value Sum of Positive Value Figure 10:Sum of absolute value of temperature and pos- itive temperature. -32 -28 -24 -20 -16 -12 -8 -4 0 4 8 Measurement Sum of Positive Value Sum of Absolute Value 250 500 750 1000 1250 1500 Time [ min ] Deviation [ µm ] Figure 11: Cutting error prediction with 16 measuring points selected by connection weights. 5.2 Selection Based on Correlation Coeﬃcients In Method-IIb, after the ﬁrst selection by Method-IIa, the variables, whose correlation coeﬃcients ρ(2)ij exceed 0.9 are replaced by the variables, which are not selected in the ﬁrst stage and have small ¯ρ(1)i . Simulation re- sults by both methods are shown in Fig.12. The result by Method-IIb ”Correlation(2)” is superior to that of Method-IIa ”Correlation(1)”. Because the former more precisely evaluates the correlation coeﬃcients. -28 -24 -20 -16 -12 -8 -4 0 4 8 12 Correlation (1) Correlation (2) Measurement 250 500 750 1000 1250 1500 Time [ min ] Deviation [ µm ] Figure 12: Cutting error prediction with 16 measuring points selected by correlation coeﬃcients. 5.3 Comparison of Selected Measuring Points Table 2 shows the selected temperature measuring points by four kinds of the methods. Comparing the selected measuring points by Method-Ib and Method- IIb, the following 9 points 4,11,12,14,15,21,22,28,31 are common. However, the selected measuring points are not exactly the same. A combination of the measur- ing points seems to be important. Table 2: Measuring points selected by four kinds of meth- ods. Methods Measuring points Method-Ia 2, 3, 6, 7, 9,15,18,19, Sum of absolute values 21,22,24,25,26,28,30,31 Method-Ib 1, 4, 7, 9, 11,12,13,14, Sum of positive weights 15,19,21,22,23,26,28,31 Method-IIa 3, 4, 7, 8,12,14,16,18, Correlation(1) 19,20,21,22,23,25,31,32 Method-IIb 3, 4, 8,11,12,14,15,16, Correlation(2) 20,21,22,27,28,30,31,32 Two kinds of methods, selecting the useful input vari- ables, have been proposed. They are based on the con- nection weights and the correlation coeﬃcients among the input variables. The proposed methods have been applied to predicting cutting error caused by thermal eﬀects in machine tools. Simulation results show pre- cise prediction with the reduced number of the input variables. S.Haykin, ”Neural Networks: A Comprehensive Foundation,” Macmillan, New York, 1994. S.Haykin and L. Li, ”Nonlinear adaptive predic- tion of nonstationary signals,” IEEE Trans. Signal Pro- cessing, vol.43, No.2, pp.526-535, Feburary. 1995. A. S. Weigend and N. A. Gershenfeld, ”Time series prediction: Forecasting the future and understanding the past. in Proc. V. XV, Santa Fe Institute, 1994. M. Kinouchi and M. Hagiwara, ”Learning tempo- ral sequences by complex neurons with local feedback,” Proc. ICNN’95, pp.3165-3169, 1995. T.J. Cholewo and J.M. Zurada, ”Sequential net- work construction for time series prediction,” in Proc. ICNN’97, pp.2034-2038, 1997. A. Atia, N. Talaat, and S. Shaheen, ”An eﬃcient stock market forecasting model using neural networks,” in Proc. ICNN’97, pp.2112-2115, 1997. X.M. Gao, X.Z. Gao, J.M.A. Tanskanen, and S.J. Ovaska, ”Power prediction in mobile communication systems using an optimal neural-network structure”, IEEE Trans. on Neural Networks, vol.8, No.6, pp.1446- 1455, November 1997. A.A.M.Khalaf, K.Nakayama, ”A cascade form predictor of neural and FIR ﬁlters and its minimum size estimation based on nonlinearity analysis of time series”, IEICE Trans. Fundamental, vol.E81-A, No.3, pp.364– 373, March 1998. A.A.M.Khalaf, K.Nakayama, ”Time series predic- tion using a hybrid model of neural network and FIR ﬁlter”, Proc. of IJCNN’98, Anchorage, Alaska, pp.1975- 1980, May 1998. A.A.M. Khalaf and K.Nakayama, ”A hybrid non- linear predictor: Analysis of learning process and predictability for noisy time series”, IEICE Trans. Fundamentals,Vol.E82-A, No.8, pp.1420-1427, Aug. K.Hara and K.Nakayama, ”Training data selec- tion method for generalization by multilayer neural net- works”, IEICE Trans. Fundamentals, Vol.E81-A, No.3, pp.374-381, March 1998. K.Hara and K.Nakayama, ”A training data selec- tion in on-line training for multilayer neural networks”, IEEE–INNS Proc. of IJCNN’98, Anchorage, pp.227- 2252, May 1998.
The resulting orbital radius is 42,164 kilometres (26,199 miles). Subtracting the Earth’s equatorial radius, 6,378 kilometres (3,963 miles), gives the altitude of 35,786 kilometres (22,236 miles). How is geosynchronous orbit calculated? In order to calculate the geostationary orbit around any given body, we must first create an equation with the force of gravity and the centripetal force. The force of gravity is equal to: F G = G ⋅ M 1 ⋅ M 2 r 2 \displaystyle F_G=\frac G\cdot M_1\cdot M_2r^2 How do you calculate the time period of a geostationary satellite? A geostationary orbit can only be achieved at an altitude very close to 35,786 km (22,236mi), and directly above the Equator. This equates to an orbital velocity of 3. 07km/s(1. 91mi/s) or an orbital period of 1,436 minutes, which equates to almost exactly one sidereal day or 23. What is an example of a geosynchronous satellite? There are many satellites currently in geosynchronous orbits. The weather satellite pictures (GIF, 60k) we see on the news come from these satellites. They constantly send pictures and information to receiving dishes on Earth. The GOES weather satellites are an example of this type of satellite. What is the formula of height of satellite? Gravitation. What is the height of geostationary satellite from the surface of the earth? h = 3.6 x 107 m = 36000 km. The height of geostationary satellite from the surface of the earth is 36000 km. What is the velocity of a geosynchronous satellite? The aptly titled geosynchronous orbit is described in detail: “At an altitude of 124 miles (200 kilometers), the required orbital velocity is just over 17,000 mph (about 27,400 kph). To maintain an orbit that is 22,223 miles (35,786 km) above Earth, the satellite must orbit at a speed of about 7,000 mph (11,300 kph). What is geosynchronous orbit physics? A geosynchronous orbit is a high Earth orbit that allows satellites to match Earth’s rotation. Located at 22,236 miles (35,786 kilometers) above Earth’s equator, this position is a valuable spot for monitoring weather, communications and surveillance. What is a geosynchronous orbit a level physics? A geosynchronous orbit (sometimes abbreviated GSO) is an Earth-centered orbit with an orbital period that matches Earth’s rotation on its axis, 23 hours, 56 minutes, and 4 seconds (one sidereal day). What forces act on a geosynchronous satellite? Earth’s gravity & centrifugal forces act on geostationary satellites. What is the time period of a geosynchronous satellite? Definition: Geosynchronous satellite is placed in the geosynchronous orbit with an orbital period matching the Earth’s rotation period. These satellites take 24 hours to complete one rotation around the earth. What is the rotational period of a geostationary satellite? The period of a geostationary satellite is the same as the period of rotation of earth which is approximately equal to 24 hours or 1 day. How is satellite position calculated? By taking the difference between its own time and the timestamp of the GPS signal and multiplying by the speed of light, the receiver calculates a rough measure of the distance between the receiver and the satellite. How do geosynchronous satellites work? A geosynchronous satellite is a satellite in geosynchronous orbit, with an orbital period the same as the Earth’s rotation period. Such a satellite returns to the same position in the sky after each sidereal day, and over the course of a day traces out a path in the sky that is typically some form of analemma. Do geosynchronous satellites move? This special, high Earth orbit is called geosynchronous. A satellite in a circular geosynchronous orbit directly over the equator (eccentricity and inclination at zero) will have a geostationary orbit that does not move at all relative to the ground. It is always directly over the same place on the Earth’s surface. What is difference between geosynchronous and geostationary? While geosynchronous satellites can have any inclination, the key difference from geostationary orbit is the fact that they lie on the same plane as the equator. Geostationary orbits fall in the same category as geosynchronous orbits, but it’s parked over the equator. How do you calculate the speed of a satellite? To calculate the orbital speed of an earth’s satellite, you need to know the gravitational constant (G), earth’s mass (M), earth’s radius (R), and the height of rotation of the satellite (h). The orbital speed is calculated as: √((G × M) / (R + h)) What is the formula for time period of satellite? Solution : Time period of the satellite: The distance covered by the satellite during one rotation in its orbit is equal to `2pi (R_(E )+h)` and time taken for it is the time period, T. Then, Speed , v =` (“Distance travelled “)/(“Time taken “) = (2pi(R_(E)+h))/T ” “… What is geostationary satellite give an example? Geostationary satellites are those that make orbits on the Earth’s Ecuadorian line at the speed that the Earth does. These satellites meet different basic standards for example: being at a height of 36 thousand kilometers, since there is a balance of the earth’s attraction force such as the centrifuge. What is the time period and height of geostationary satellite? A satellite whose time period is 24 hours and is revolving in an orbit concentric and coplanar with the equatorial plane of earth is called geostationary satellite. The geostationary satellite revolves at the height of about 36000 km from the surface of the earth. Do all satellites move at the same speed? A: No, satellites that orbit at different altitudes have different speeds. Satellites that are further away actually travel slower. The International Space Station has a Low Earth Orbit, about 400 kilometers (250 miles) above the earth’s surface. How many geostationary satellites are there? As of May 2021, the website UCS Satellite Database lists 4,550 known satellites. This includes all orbits and everything down to the little CubeSats, not just satellites in GEO. Of these, 560 are listed in the database as being at GEO. What is the radius of a geosynchronous orbit? A perfectly geostationary orbit is a mathematical idealization. Only the distinction between the mean solar day and the sidereal day needs to be taken into account. Therefore, it is customary to quote a nominal orbital period of 86 164 seconds and a radius of 42 164 km. How far is a geosynchronous satellite from Earth? The geostationary orbit of 36,000 km from the Earth’s Equator is best known for its many satellites which are used for various forms of telecommunication, including television. Signals from these satellites can be sent all the way around the world. Is the Moon in geosynchronous orbit? Our Moon is obviously not in synchronous, or more specifically geosynchronous orbit about the Earth. The period of its orbit around the Earth is not the same as our sidereal day; in fact, it takes the Moon about 27.3 of our days to complete one orbit of our Earth. How does a satellite stay in orbit physics? A satellite maintains its orbit by balancing two factors: its velocity (the speed it takes to travel in a straight line) and the gravitational pull that Earth has on it. A satellite orbiting closer to the Earth requires more velocity to resist the stronger gravitational pull.
Mathematical Problems for Complex SystemsView this Special Issue A Novel Chaotic Map and an Improved Chaos-Based Image Encryption Scheme In this paper, we present a novel approach to create the new chaotic map and propose an improved image encryption scheme based on it. Compared with traditional classic one-dimensional chaotic maps like Logistic Map and Tent Map, this newly created chaotic map demonstrates many better chaotic properties for encryption, implied by a much larger maximal Lyapunov exponent. Furthermore, the new chaotic map and Arnold’s Cat Map based image encryption method is designed and proved to be of solid robustness. The simulation results and security analysis indicate that such method not only can meet the requirement of imagine encryption, but also can result in a preferable effectiveness and security, which is usable for general applications. With the discovery of a series of chaotic maps such as Tent Map [1, 2] and Logistic Map , researchers and scholars have been able to apply them into a variety of fields. The knowledge of chaotic maps is perhaps one of the most significant achievements in nonlinear science. Since 1980s, researches on chaos theory have been overlapping and mixing up with other subjects, in the meanwhile promoting their further developments. The fields that take advantage of knowledge concerning chaos range greatly from math and astronomy to music and art. Besides, the most famous magazines in the world such as Nature and Scientific American once published a great deal of discoveries and progresses in chaos theory . Therefore, it is reasonable to judge that chaos has been becoming a universal language between these important subjects. If we are to further classify the applications of the chaos in different categories, chaos analysis and chaos synthesis will be the answer. As for the former, based on complex manual work and natural system, we tend to find some hidden rules inside of them. One example is the prediction towards time series [7–10]. For the latter, by using manually produced chaotic system, we are inclined to discover some possible functions contained within the chaotic dynamics [11–13]. In addition, some likely applications of the chaos are listed below. First, combining neural network and chaos, we utilize chaotic status of intermediate processes to let networks avoid the partial minimum point. And hence it guarantees global optimum according to . Second, the chaos theory has already been used in high-speed searching process. Last but not least, chaotic maps are widely applied in secure communication which is carefully studied in [13, 15]. We could not only use chaotic signals to encrypt the information needed to be secure but also decipher encrypted one as well according to [16–18]. Also, researches regarding these aspects are known to have already been put in the national defense plan of China. Despite the fact that the fields that call for chaotic maps range greatly, one thing they share in common is that they all need the chaotic features of chaotic maps. In other words, the feature that a simple initial point and a given value of the parameter could completely control the whole process is what we need. As a matter of fact, chaotic maps are quite sensitive to the initial point, which means even a very slight change in the value of initial point would result in a dramatic change of the sequence produced by the chaotic map. However, at present, only a limited number of one-dimensional chaotic maps (e.g., Tent Map and Logistic Map) are introduced. Also, their properties are somehow limited and may no longer satisfy our needs. Too often our methods of encryption and engineering projects are merely based on these simple chaotic maps. Without new and better chaotic maps, our applications will remain unchanged and might get stuck in the future. This may lead to an urgent need for more and better chaotic maps. In this paper, a new one-dimensional chaotic map is first introduced, and we use the maximal Lyapunov exponent [19–21] to determine how well the map performs. In addition, we later prove that this new chaotic map actually exhibits a larger maximal Lyapunov exponent, indicating better properties of the chaotic map. What is more is that a new algorithm based on this new chaotic map is used in image encryption, providing a brand new way to encrypt images. Compared with previous ways to encrypt image, it not only utilizes the excellent chaotic property of the newly discovered map itself but also entails another classical map: Arnold’s Cat Map [22–24], through which coordinates of the target image’s grey value matrix will be changed to another. Lastly, a security analysis is accomplished by plotting histogram of the image’s grey values and calculating information entropy . Without any knowledge as to how many times the target image is iterated, it is next to impossible to decrypt the encrypted image. Therefore, the safety of the image is largely strengthened and guaranteed. Now we discuss this new chaotic map and how to use it to encrypt images in detail in the following. 2. Design and Analysis to the New Chaotic Map In this section, we first discuss the definition of “maximal Lyapunov exponent.” Then, we plot the Lyapunov spectrum of the two traditional one-dimensional chaotic maps. Next, a new chaotic map is introduced. Lastly, a comparison between the new chaotic map and two traditional maps is carefully made. 2.1. Maximal Lyapunov Exponent 2.1.1. Definition of the Lyapunov Exponent According to statements in [19, 26], Lyapunov exponent usually represents the features of a chaotic system, named after the great Russian mathematician Lyapunov. For discrete system (maps or fixed point iterations) and for an orbit starting with , the Lyapunov exponent can be defined as follows: It is common to refer to the largest defined by (1) as the maximal Lyapunov exponent because it determines a notion of predictability for a chaotic system. 2.1.2. Properties of Maximal Lyapunov Exponent (MLE) A positive is usually taken as an indication that the system is basically chaotic. Besides, it is also apparently true that the larger MLE is, the more chaotic the map is. And this means a better chaotic map according to . Remark 1. Lyapunov exponent, as an important exponent to test the property of chaotic map, is widely used in the world of chaos. Actually, from the definition equation (1), we could clearly find out that Lyapunov is an average value of . Since is the indicating parameter that measures the variation speed for , Lyapunov exponent, the average value of is bound to reflect the chaotic properties of . Thus, maximal Lyapunov exponent can largely expresses the overall performance of chaotic maps. Next, we begin with calculating the MLE and plotting the Lyapunov exponent spectrum for two typical one-dimensional maps, Tent Map and Logistic Map. 2.2. Tent Map In mathematics, according to , Tent Map with parameter is a real-valued function defined by Thus, we can obtain the Lyapunov exponent spectrum of Tent Map, which is shown in Figure 1. In addition, we could calculate the MLE for Tent Map, which is 0.6931 (when ), indicating that the map itself is chaotic. 2.3. Logistic Map Logistic Map [27, 28] is a polynomial mapping of degree 2 that exhibits chaotic behavior. The Logistic Map equation is given by When the variable is given different value, ranging from 2 to 4, through formula (1), we could plot Lyapunov exponent of Logistic Map as well. That is shown in Figure 2. It is obviously shown in Figure 2 that, when , MLE of Tent Map is reached. Through calculating, maximal Lyapunov exponent of the Logistic Map is 0.6785. 2.4. A New Chaotic Map When we replace “” with “” in Logistic Map, formula (4) will be attained: If we choose as the initial point of the map and the parameter , after 10000 times of iterations, we will get the randomly scattered image of the map in Figure 3 when we plot every . Remark 2. It is clearly displayed in Figure 3 that the sequence generated from the new chaotic map ranges from −0.6 to 0.6, while, in Tent Map and Logistic map, it is a little larger, ranging from −1 to 1. However, this does not matter that much, since we consider the range that they share in common, which is . In order to see if the map is a chaotic map and, if yes, how good the map is, similarly, we also use the maximal Lyapunov exponent to see that, starting with and iterating 2000 times. Hence, Figure 4 will be yielded. As is shown in Figure 4, when , the MLE of the new chaotic map reaches beyond 1, to be exact, 1.0742. Next, we combine the three Lyapunov exponent spectrums given above together and Figure 5 is the result. Remark 3. Just like what we have discussed in Section 2.1, a larger maximal Lyapunov exponent indicates not only a stronger sensitivity to the initial point but also it also indicates that the chaotic system itself is “more chaotic.” In other words, the chaotic map with larger MLE is of better quality. It is demonstrated above that the MLE of the two most typical chaotic maps (2) and (3) all end below 1, which are apparently smaller than the new map (4). The MLE of the new chaotic map, as expected, has reached beyond 1. Therefore, the new map (4) is supposed to bear the potential to perform more effectively in engineering or encryption process than the two classic ones (2) and (3) mentioned above. 3. Application on Image Encryption 3.1. Encryption Scheme In this section, we are about to take one step further, applying the new map on image encryption. Now, we begin introducing an image encryption algorithm based on the map we have just constructed. Step 1. For given initial point and the parameter , we could simply use the new map (4) to produce a sequence . Consider Then, in order to make reach closely 1, we choose . Step 2. Next, we transform decimal numbers to the form of binary numbers, and consequently we will get . After that, we choose the first 8 figures after the decimal point of to form a new binary number, . To put in the language of math, that is: Thus, for each , there is a unique corresponding to it. Obviously, is formed by the first 8 numbers after the decimal point of . In this way, we could obtain the original chaotic sequence in the form of binary numbers with 8 places. Similar approaches have been discussed in . Remark 4. Binary numbers with 8 places have three major advantages that are listed below. First of all, it corresponds well with , which is the gray values’ row vector of each point in an image. refers to red, refers to green, and refers to blue. In fact, they all range from 0 to 255. And, hence, if they are written into binary forms, 8-place binary numbers will be exactly what we get. Therefore this step provides convenience to the following steps. Secondly, computers have always been using binary numbers to operate. Thus binary numbers tend to make calculation efficient and time-saving. Thirdly, as per what we have discussed in Section 2.1, the sequence generated by the chaotic map is basically random and therefore this will lead the binary numbers to be random as well due to the transmission of randomness. Step 3. Suppose that the pixel of the target image is , and then we put produced in Step 1 into a matrix AA with the size . In other words, from left to right and from up to down, each is assigned to a unique, particular position in matrix AA. Step 4. For point from the plaintext image we have its grey value vector written in binary form. In order to build a connection between the grey value matrix of the plaintext image and the chaotic sequence matrix AA, it is reasonable to think of the XOR operation (), the definition of which is given below in accordance with : Thus, we use XOR to “make a mess.” For point from the plaintext image, we also extract AA from AA, which is made up with the chaotic sequence generated. Next, we do XOR operation as follows: Step 5. In Step 6, we make a change to the coordinate of . To be exact, let be the starting column vector, which is , and then iterate a given “” times with formula (9) according to [23, 24] After iteration of a given “” times, we will obtain a new coordinate for point in the plaintext image, and we mark it as . This is one of the most famous coordinate change maps, the Arnold’s Cat Map. Lastly, we assign the value of from to . (Tips: “mod” in the formula means complementation, if the plaintext image is of the size (the pixel value); .) A sample experiment has been provided in Figure 6. (a) Plaintext image (b) After 3 times of iteration (c) After 6 times of iteration Remark 5. There exists actually a slight defect of Step 5, due to the fact that, after a special number of iteration times, the image could simply be restored as in the beginning according to . As Arnold’s Cat Map is defined, it is a periodic map, and this property leads to an unsafe encryption. Therefore, it is very significant to choose a proper value of “,” the number of times of iterations to prevent the image from being restored. Remark 6. Along with defects, there are also huge advantages. First, only a few times of iteration are enough to guarantee the thorough change of the picture. Without knowing the times of iteration, it takes next to forever to decipher the image, as you could see as follows. Second, “mod” guarantees that the size of the image will be exactly the same if the picture is square-sized. Lastly, there is no denial that it is easy to recover the original image using some fundamental knowledge from algebra. Elaborate discussion will be in Section 3.2. Step 6. Repeat Steps 3–5 for every point in the target plaintext image, and use these processed grey values to form an encrypted image. 3.2. Decryption Scheme Step 1. Read the encrypted image, and then write its grey value into a three-dimensional matrix. Step 2. According to Step 6 in encryption, we could simply do the opposite: start with in the encrypted image, use the formula listed below, and iterate “” times as well: The definition of “” is the same as the one mentioned above, and thus we attain the original coordinate of . Step 3. See Steps 1 and 2 in the encryption part. Step 4. In order to restore changes from Step 5, just do XOR operation with in points and AA again. Hence, is obtained. That is, Step 5. Use of every point to form the original image. Hence, restoration of the plaintext image is accomplished. 3.3. Overall Process of Image Encryption Statements mentioned above yield the encryption scheme listed in Figure 7. 3.4. Simulation Results Take the image of Figure 8(a) as a sample to test the algorithm mentioned above and the plaintext image, encrypted image, and restored image are given in Figure 8. (a) Plaintext image (b) Ciphered image (c) Restored image 4. Security Analysis In this section, by plotting the histogram and calculating the so-called “information entropy,” we test the security qualities of the proposed method in Section 2. 4.1. Histogram Analysis There is no denying that a well-ciphered image should provide no chances for the attackers to decrypt through statistics analysis. On the one hand, the grey level of the plaintext image in Figure 8(a) is somehow similar to normal distribution in accordance with Figure 9(a) and this would bring about vulnerabilities to decryption. On the other hand, the ciphered image in Figure 8(b), though not perfectly average on each value between 1 and 256, is somehow subjected to uniform distribution according to Figure 9(b), which undoubtedly adds up difficulty for the attacker to decrypt according to . (a) Plaintext image (b) Encrypted image 4.2. Information Entropy According to the thesis in , the value of information entropy typically expresses the feature of randomness. Its definition is given by where “” refers to message and represents the chances of appearance of the th message. As for images, stands for the probability of a particular grey value . Under this circumstance, is also called “image entropy.” The entropy of a perfectly encrypted image should, in ideal case, approach 24, since (3 means 3 color planes) according to [14, 31]. The entropy is yielded through calculating, which is extremely close to 24. Therefore it can be inferred that the encrypted image is almost random. In this way, we tend to believe that the safety of the image is largely promoted and ensured. Remark 7. As the result of the entropy shows, is perfectly close to 24, which is an ideal entropy for a randomly encrypted image. And this result in turn proves the excellent chaotic properties of the new chaotic map. What is more is that the comparison between the histograms of the plaintext image and the encrypted image shows that the new chaotic map actually makes grey values distribute uniformly, which convinces us of the fact the new chaotic map is of robust quality. In this paper, a new chaotic map with better chaotic properties has been proposed. To step further, a comparison with traditional one-dimensional chaotic maps has been made as well through their maximal Lyapunov exponent spectrums, which proves the new chaotic map’s marvelous application prospect. In addition, it has been applied on image encryption. Along with Arnold’s Cat Map, it could successfully produce a ciphered image based on original plaintext image. Steps and the proposed scheme to encrypt a targeted image and decrypt a ciphered image have been given. What is more is that the results of encryption and decryption simulation have been provided. At last, security reliability towards this algorithm has been discussed. As a result, this method is of excellent quality and robustness and turns to be theoretical unbreakable by convention attacks without any knowledge to the values of the starting point and parameters. Yet, to seek perfectness, a safe conveyance of the parameter still remains to be a problem. That is, when the attacker gains the parameter value and initial point value, a whole system can break down easily. Thus a safe transmission of keys may be our next direction to study. Conflict of Interests The authors declare that there is no conflict of interests regarding the publication of this paper. T. Yoshida, H. Mori, and H. Shigematsu, “Analytic study of chaos of the tent map: band structures, power spectra, and critical behaviors,” Journal of Statistical Physics, vol. 31, no. 2, pp. 279–308, 1983.View at: Publisher Site \| Google Scholar L. Shan, H. Qiang, J. Li, and Z. Wang, “Chaotic optimization algorithm based on Tent map,” Control and Decision, vol. 20, no. 2, pp. 179–182, 2005.View at: Google Scholar L. Kocarev and G. Jakimoski, “Logistic map as a block encryption algorithm,” Physics Letters A, vol. 289, no. 4-5, pp. 199–206, 2001.View at: Publisher Site \| Google Scholar \| MathSciNet M. Gutzwiller, “Quantum chaos,” Scientific American, vol. 266, pp. 26–32, 1992.View at: Google Scholar C. Chang-Jian and C. Chen, “Bifurcation and chaos analysis of a flexible rotor supported by turbulent long journal bearings,” Chaos, Solitons and Fractals, vol. 34, no. 4, pp. 1160–1179, 2007.View at: Publisher Site \| Google Scholar I. Zelinka, G. Chen, and S. Celikovsky, “Chaos synthesis by evolutionary algorithms,” in Evolutionary Algorithms and Chaotic Systems, pp. 345–382, Springer, 2010.View at: Google Scholar J. Zhang and X. Xiao, “Predicting chaotic time series using recurrent neural network,” Chinese Physics Letters, vol. 17, no. 2, pp. 88–90, 2000.View at: Publisher Site \| Google Scholar M. Han, J. Xi, S. Xu, and F. Yin, “Prediction of chaotic time series based on the recurrent predictor neural network,” IEEE Transactions on Signal Processing, vol. 52, no. 12, pp. 3409–3416, 2004.View at: Publisher Site \| Google Scholar \| MathSciNet M. Ye and X. Wang, “Chaotic time series prediction using least squares support vector machines,” Chinese Physics, vol. 13, no. 4, pp. 454–458, 2004.View at: Publisher Site \| Google Scholar D. S. K. Karunasinghe and S. Liong, “Chaotic time series prediction with a global model: artificial neural network,” Journal of Hydrology, vol. 323, no. 1–4, pp. 92–105, 2006.View at: Publisher Site \| Google Scholar W. Zhang, “Global and chaotic dynamics for a parametrically excited thin plate,” Journal of Sound and Vibration, vol. 239, no. 5, pp. 1013–1036, 2001.View at: Publisher Site \| Google Scholar J. G. Lu, “Chaotic dynamics and synchronization of fractional-order Arneodo's systems,” Chaos, Solitons and Fractals, vol. 26, no. 4, pp. 1125–1133, 2005.View at: Publisher Site \| Google Scholar O. M. Kwon, J. H. Park, and S. M. Lee, “Secure communication based on chaotic synchronization via interval time-varying delay feedback control,” Nonlinear Dynamics, vol. 63, no. 1-2, pp. 239–252, 2011.View at: Publisher Site \| Google Scholar \| Zentralblatt MATH D. Yang, G. Li, and G. Cheng, “On the efficiency of chaos optimization algorithms for global optimization,” Chaos, Solitons and Fractals, vol. 34, no. 4, pp. 1366–1375, 2007.View at: Publisher Site \| Google Scholar R. Luo and Y. Wang, “Finite-time stochastic combination synchronization of three different chaotic systems and its application in secure communication,” Chaos, vol. 22, no. 2, Article ID 023109, 2012.View at: Publisher Site \| Google Scholar Y. Cao, “A new hybrid chaotic map and its application on image encryption and hiding,” Mathematical Problems in Engineering, vol. 2013, Article ID 728375, 13 pages, 2013.View at: Publisher Site \| Google Scholar J. Guo, Z. Lü, and L. Zhang, “Breaking a chaotic encryption based on henon map,” in Proceedings of the 3rd International Symposium on IEEE Information Processing (ISIP '10), pp. 169–171, Qingdao, China, October 2010.View at: Publisher Site \| Google Scholar M. Naeemabadi, N. Chomachar, M. Zabihi, B. S. Ordoubadi, M. Khalilzadeh, and M. S. Ordoubadi, “Encryption based on variable chaotic key for wireless medical data transmission,” in Proceedings of the 5th International Conference on Application of Information and Communication Technologies (AICT '11), pp. 1–5, IEEE, 2011.View at: Google Scholar M. T. Rosenstein, J. J. Collins, and C. J. de Luca, “A practical method for calculating largest Lyapunov exponents from small data sets,” Physica D: Nonlinear Phenomena, vol. 65, no. 1-2, pp. 117–134, 1993.View at: Publisher Site \| Google Scholar H. Kantz, “A robust method to estimate the maximal Lyapunov exponent of a time series,” Physics Letters A, vol. 185, no. 1, pp. 77–87, 1994.View at: Publisher Site \| Google Scholar M. Setare and D. Momeni, “Geodesic stability for Kehagias-Sfetsos black hole in Hořava-lifshitz gravity via Lyapunov exponents,” International Journal of Theoretical Physics, vol. 50, pp. 106–113, 2011.View at: Google Scholar F. Faure, S. Nonnenmacher, and S. de Bièvre, “Scarred eigenstates for quantum cat maps of minimal periods,” Communications in Mathematical Physics, vol. 239, no. 3, pp. 449–492, 2003.View at: Publisher Site \| Google Scholar L. Zhang, X. Liao, and X. Wang, “An image encryption approach based on chaotic maps,” Chaos, Solitons & Fractals, vol. 24, no. 3, pp. 759–765, 2005.View at: Publisher Site \| Google Scholar C. Çokal and E. Solak, “Cryptanalysis of a chaos-based image encryption algorithm,” Physics Letters. A, vol. 373, no. 15, pp. 1357–1360, 2009.View at: Publisher Site \| Google Scholar \| MathSciNet S. M. Pincus, “Approximate entropy as a measure of system complexity,” Proceedings of the National Academy of Sciences of the United States of America, vol. 88, no. 6, pp. 2297–2301, 1991.View at: Publisher Site \| Google Scholar \| Zentralblatt MATH A. Wolf, J. B. Swift, H. L. Swinney, and J. A. Vastano, “Determining Lyapunov exponents from a time series,” Physica D: Nonlinear Phenomena, vol. 16, no. 3, pp. 285–317, 1985.View at: Publisher Site \| Google Scholar \| Zentralblatt MATH E. A. Jackson and A. Hübler, “Periodic entrainment of chaotic logistic map dynamics,” Physica D: Nonlinear Phenomena, vol. 44, no. 3, pp. 407–420, 1990.View at: Publisher Site \| Google Scholar \| MathSciNet S. C. Phatak and S. S. Rao, “Logistic map: a possible random-number generator,” Physical Review E, vol. 51, no. 4, pp. 3670–3678, 1995.View at: Publisher Site \| Google Scholar L. M. Leibowitz, “A simplified binary arithmetic for the Fermat number transform,” IEEE Transactions on Acoustics, Speech, and Signal Processing, vol. 24, no. 5, pp. 356–359, 1976.View at: Publisher Site \| Google Scholar J.-W. Han, C.-S. Park, D.-H. Ryu, and E.-S. Kim, “Optical image encryption based on XOR operations,” Optical Engineering, vol. 38, no. 1, pp. 47–54, 1999.View at: Publisher Site \| Google Scholar W. Yu and J. Cao, “Cryptography based on delayed chaotic neural networks,” Physics Letters A: General, Atomic and Solid State Physics, vol. 356, no. 4-5, pp. 333–338, 2006.View at: Publisher Site \| Google Scholar
You can also change the rate at which the original value of the asset declines from the Excel default of “2” to another number. As an alternative, enter whole numbers rather than cell references. If you choose this option, format dollar values as “25000” rather than “$25,000.” In the first year, LN divides the resulting annual depreciation amount evenly across each period from the beginning of the recovery period to the end of the year. In subsequent years, LN divides the depreciation amount evenly across https://intuit-payroll.org/ each period in the year. MACRS formula depreciation uses either a declining balance formula or a straight-line formula. To calculate the sum of the years, you need to know the projected useful life and then add these together. Need help with accounting? Easy peasy. If the salvage value of an asset is known , the cost of the asset can subtract this value to find the total amount that can be depreciated. Assets with no salvage value will have the same total depreciation as the cost of the asset. To start with the straight-line method to determine depreciation on monthly basis, estimate the asset’s salvage value at the end of its useful life. You may be able to get an estimate of the salvage value from a regulatory body, such as the IRS. One method is called partial year depreciation, where depreciation is calculated exactly at when assets start service. Simply select “Yes” as an input in order to use partial year depreciation when using the calculator. Sample Full Depreciation Schedule Use a depreciation factor of two when doing calculations for double declining balance depreciation. Regarding this method, salvage values are not included in the calculation for annual depreciation. However, depreciation stops once book values drop to salvage values. When a fixed asset is initially purchased, its cost basis is recorded on the balance sheet. Its book value is equal to its historical cost less total accumulated depreciation. Two common ways of calculating depreciation are the straight-line and double declining balance methods. - To find Year 2, subtract the total depreciation expense from the purchase price ($50,000 – $8,000) and follow the same formula. - The end of the first period is December 31, 2020 and the depreciation rate is 20% per year with an expected salvage value of $4,000. - You can also change the rate at which the original value of the asset declines from the Excel default of “2” to another number. - You purchase the most beautiful submarine you’ve ever seen for $100,000. - We create short videos, and clear examples of formulas, functions, pivot tables, conditional formatting, and charts.Read more. - When a fixed asset is initially purchased, its cost basis is recorded on the balance sheet. - A small delivery truck was purchased on January 1 at a cost of $29,700. Your worksheet should now look like the one in the screenshot below. Life — useful life of the asset (i.e., how long the asset is estimated to be used in operations). And, a life, for example, of 7 years will be depreciated across 8 years. (Number of units produced / useful life in number of units) x . Next, divide this amount by the number of years in the asset’s useful lifespan, which you can find in tables provided by the IRS. With this method we need to estimate the amount of depreciation we expect it to have. How to calculate depreciation in Excel Note how the book value of the machine at the end of year 5 is the same as the salvage value. Over the useful life of an asset, the value of an asset should depreciate to its salvage value. The formula for the sum of years’ digits method is depreciable base multiplied by (useful life remaining divided by sum of years’ digits). In this equation, depreciable base equals cost basis minus residual value, and sum of years’ digits equals n(n+1)/2. Depreciation example with first four functions The Excel equivalent function for Straight-Line Method is SLN will calculate the depreciation expense for any period. For a more accelerated depreciation method see, for example, our Double Declining Balance Method Depreciation Calculator. Property, plant and equipment, also referred to as fixed assets, have finite useful lives. Divide the result by 12 to calculate the monthly accumulated depreciation. Divide each year’s digit by the sum of the useful life digits to determine the percentage by which the asset depreciates in a year. In the previous example, divide 6 by 21 to calculate 28.57 percent, which is the percentage by which the copy machine depreciates in the first year. In the second year, divide 5 by 21 to arrive at 23.81 percent, the percentage amount it depreciates in its second year. Repeat this how to calculate monthly depreciation process until you have calculated a depreciating percentage for each of the asset’s useful life years. Straight line basis is the simplest method of calculating depreciation and amortization, the process of expensing an asset over a specific period. Equipment acquired at the beginning of the year at a cost of $50,200 has an estimated residual value of $3,700 and an estimated useful life of five years. Click to select the cell in which you want to enter the amount of depreciation. Divide the sum of step by the number arrived at in step to get the annual depreciation amount. Operating Income means the Company’s or a business unit’s income from operations but excluding any unusual items, determined in accordance with generally accepted accounting principles. The Original Value, Monthly Depreciation and Fixed Charge Per Month listed above are based, in part, upon the manufacturer’s quoted price as of the date you execute this Schedule A. The Original Value, Monthly Depreciation and Fixed Charge Per Month listed are based, in part, upon the manufacturer’s quoted price as of the date you execute this Schedule A. Divide this value by the number of years of the asset’s lifespan. What is the best depreciation method for rental property? The depreciation method used for rental property is MACRS. There are two types of MACRS: ADS and GDS. GDS is the most common method that spreads the depreciation of rental property over its useful life, which the IRS considers to be 27.5 years for a residential property. Let’s take a look-see at an accumulated depreciation example using the straight-line method. The salvage value is $10,000, and its useful life is 10 years. For every asset you have in use, there is an initial cost and value loss over time .
What is A NUMERICAL SUMMARY OF A POPULATION? What is a numerical summary of a sample? That is the definition of a statistic A numerical value used as a summary measure for a population of data is called? What is A NUMERICAL SUMMARY OF A POPULATION CALLE? Mr What will tell you the definition or meaning of What is A NUMERICAL SUMMARY OF A POPULATION CALLE NUMERICAL SUMMARY OF DATA In this lecture we shall look at summarising data by numbers, ... Population Standard Deviation: (x n) This is also known as 'Root Mean Square Deviation' and is calculated by squaring and adding the deviations from the mean, ... A numerical summary of a population is a? Parameter A numerical value calculated for a sample is called a? A numerical value calculated for a sample is called a descriptive statistic. A Numerical Measure from a Sample Such as a Sample Mean Is Known? sample ... Most cells cannot harness heat to perform work because . A) heat is not a form of energy. B) cells do not have much heat; they are relatively cool. An essay or paper on Population's Numerical Summary. A parameter is a numerical summary of a population, but because populations are so large, it is impossible to get a true numerical summary of a population (Population, 2004). For example, it would be impossible to know the average earnings of a Numerical Summaries Mean The sample mean, or average, of a group of values is calculated by taking the sum of all of the values and dividing by the total number of values. Numerical Summaries of Data III.A Measures of Center: Mean and Median 3.a.1 Median and Mean 3.a.2 Skewness 3.a.3 Exercise ... III.D Population parameters and Standardized Variables 3.d.1 Population mean and standard deviation 3.d.2 z-scores A summary measure that is computed to describe a numerical characteristic from only a sample of the population is called: a parameter. ... A population frame for a survey contains a listing of 6,179 names. The population is the set of all heart disease patients. ... There's no way a two number summary can describe the skewness of the data. When one sees a median and an IQR, one suspects they are being reported because the data are skewed, ... On October 31, the world's 7 billionth person will be born. Each of us is part of that population. With the world growing by more than 200,000 people a day, it's hard to know where you fit in. Until now. Enter your birthday to find your number Immediately download the Population summary, chapter-by-chapter analysis, book notes, essays, quotes, character descriptions, lesson plans, and more - everything you need for studying or teaching Population ... These three groups determine the amount of available resources and the number ... The most extreme summary – one number: Measures of ‘central tendency’ – or what is ‘typical’ in the data ... Varp = ∑ (xi – m)2/N, where N is the number in the population, and the sum is over all the members of the population. Population Standard Deviation: (x(n) This is also known as 'Root Mean Square Deviation' and is calculated by squaring and adding the deviations from the mean, ... Numerical Summary of Example 3. Minitab output. Example 3. Descriptive Statistics. The entire group of individuals to be studied is the population. A sample is a subset of the population that is being studied. A parameter is a numerical summary of a population. a. population parameter. b. sample parameter. c. sample statistic. d. population mean. e. None of the above answers is correct. 2. ... In a five number summary, which of the following is not used for data summarization? a. the smallest value. b. the largest value. The five-number summary is a descriptive statistic that provides information about a set of observations. It consists of the five most important sample percentiles: the sample minimum (smallest observation) the lower quartile or first quartile the median (middle value) the upper quartile or ... Summary statistics - descriptive statistics ... calculated using the method described later (see Standard Error of the Mean), which contains the population mean with a 95% probability. ... When the rank number R(p) is a whole number, ... Summary 6 covers is the detail for the presentation of Quantitative data presented by the Hairani Salamat and the geng. ... A parameter is a characteristic of a population. It is a numerical or graphic way to summarize data obtained from the population. A statistic, ... What is a numerical description of a population characteristic? ChaCha Answer: Statistics are the numerical value, such as standard d... Approaches to Measuring Population Health Ian McDowell November, 2005 Mortality-based summary measures Combined disability & mortality methods Conceptual rationale for summary measures ... (An index is a numerical summary of several indicators of health) ... Summary Measures of Population Health Ian McDowell November, 2002 ... An index is a numerical summary of several indicators of health Aims to represent overall health of a population Mortality data typically derived from life tables; ... If there is an even number of observations, then there is no single middle value; the median is then usually defined to be the mean of the two middle values, which ... or a finite population, ... Summary tables: Grouped data; Frequency distribution; Contingency table; What is a numerical descriptive measure of a population? ChaCha Answer: A parameter is a numerical descriptive measure calculated for... pop·u·la·tion (p p y-l sh n) n. 1. a. All of the people inhabiting a specified area. b. The total number of such people. 2. The total number of inhabitants constituting a particular race, class, or group in a specified area. Computing Numerical Summary Statistics With TI-83/84 Calculator; Numerical Summaries Part II; Summarizing Qualitative and Quantitative Data; Confidence Interval for a Single Population Mean: Part II; Confidence Interval for a Single Population Mean Part I; The "five number summary", or five statistical summary", consists of (1) the minimum, (2) the maximum, (3) the median, (4) the first quartile Summary. The human population consists of all individuals alive at any point in time. Population dynamicsdescribes how populations change in size. ... The age structureof a population indicates the number of individuals of each age. A five number summary combines five descriptive statistics of minimum, first quartile, median, third quartile and maximum to give us a fuller picture than what we might have by just looking at one of these numbers. You will read in statistics texts that this is the formula for both the population mean (its symbol is the Greek letter mu) and the sample mean (its symbol is an x with a bar over it, or x-bar). ... The five number summary is complete when we add the minimum value of 16 and maximum of 29 days. Summary A parameter is a numerical property of a population, which is a collection of units. A sample is a collection of units from a population; the number of units in the sample is called the sample size. A statistic ... In summary, the mean is preferred when data are symmetrical and do not have outliers. In other instances, the median is often preferred measure of central location. ... derive an unbiased estimate of the population variance. The number n−1 is called the degree of freedom Population Parameter: A numerical value used as a summary measure for a population of data (e.g., the population mean, u, the population variance, s2, and the population standard deviation, s). Sample ... (Use symbol μ if this is a population mean.) standard deviation s = 3.17 Since this data set is a sample, use S x or s for the standard deviation. ... the five-number summary from a grouped frequency distribution is not worth reporting. Can't figure out a five number summary in statistics? The five number summary consists of 5 items: minimum, Q1 , median, Q3, and maximum. This how to art Sample vs. Population Researchers distinguish between samples and populations. ... A numerical summary of some kind, such as an average, has two different forms: one form is used for the sample and one form is used for the population. summary measure of a population composed of two sets of partial measures. The first set of measures, the age-specific ... are the midyear population and the number of deaths in that year. These data could be analyzed in single years of age or population as a single number (Field and Gold 1998). Efforts to develop summary measures of population health have a long history (Sanders 1964, Chiang 1965, ... the summary measure of population health does not include discounting of future The five-number summary is a statistical tool that assigns significance to a set of data that results from a single variable. The five numbers involved in the summary are the ... Five Number Summary; The Five Number Summary is a method for summarizing a distribution of data. The five numbers are the minimum, the first quartile(Q1) value, the median, the third quartile(Q3) value, and the maximum. Five Number Summary. For a set of data, the minimum, first quartile, median, third quartile, and maximum. Note: A boxplot is a visual display of the five-number summary. this page updated 27-aug-12 Mathwords: Terms and Formulas from ... 5 Male-to-female ratio refers to the number of people in a population who are male relative to those ... 6 U.S. Census Bureau, Census 2000 Summary File 1; Projections of the Total Resident Population by 5-Year Age Groups, and Sex with Special Age Categories ... ... units Shape of a Distribution Describes how data are distributed Measures of shape Symmetric or skewed Using the Five-Number Summary to Explore the Shape Box-and ... The Empirical Rule Population variance Population Mean * * Title: Summary Measures Author: Nasser Daneshvary Created Date: 7 ... ... coefficient of variation Symmetric and skewed distributions Population summary measures Mean, variance, ... A Graphical display of data using 5-number summary: ... Using these numerical summary measures, we then discuss an additional graph called a boxplot. ... (for population) Square root of variance =STDEVP(range) (for population) (continued) 3.10 Conclusion 125 Equation Term Explanation Excel Pages Number Population, Sample and Data ... a summary is made such as average values. Hopefully valid conclusions can be made on the whole population based on the sample data. ... Divide range by the number of groups wanted to get endpoints of intervals. Summary of "A population-based study of copy number variants and regions of homozygosity in healthy Swedish individuals." The abundance of copy number variants (CNVs) and regions of homozygosity (ROHs) have been well documented in previous studies. Number 22 September 2001 Summary Measures of Population Health: Addressing the First Goal of Healthy People 2010, Improving Health Expectancy Globally, the number of cities with 10 million or more inhabitants is increasing rapidly, and most of these new "megacities" are in the less-developed regions. ... Summary. Human population exhibits an J-shaped growth curve, and is accelerating. a high-accuracy method for numerical integration of age- and size-structured population models If you didn't find what you were looking for you can always try Google Search Add this page to your blog, web, or forum. This will help people know what is What is A NUMERICAL SUMMARY OF A POPULATION
Powers of numbers behave in surprising ways. Take a look at some of these and try to explain why they are true. The sums of the squares of three related numbers is also a perfect square - can you explain why? What is the largest number which, when divided into 1905, 2587, 3951, 7020 and 8725 in turn, leaves the same remainder each time? What does this number mean ? Which order of 1, 2, 3 and 4 makes the highest value ? Which makes the lowest ? Is there an efficient way to work out how many factors a large number has? Do you know a quick way to check if a number is a multiple of two? How about three, four or six? Mathematicians are always looking for efficient methods for solving problems. How efficient can you be? Which set of numbers that add to 10 have the largest product? How many more miles must the car travel before the numbers on the milometer and the trip meter contain the same digits in the same order? Take any four digit number. Move the first digit to the 'back of the queue' and move the rest along. Now add your two numbers. What properties do your answers always have? Choose four consecutive whole numbers. Multiply the first and last numbers together. Multiply the middle pair together. What do you notice? How many pairs of numbers can you find that add up to a multiple of 11? Do you notice anything interesting about your results? Show that is it impossible to have a tetrahedron whose six edges have lengths 10, 20, 30, 40, 50 and 60 units... A 2-Digit number is squared. When this 2-digit number is reversed and squared, the difference between the squares is also a square. What is the 2-digit number? Find the sum of the series. Can you explain the surprising results Jo found when she calculated the difference between square numbers? Think of two whole numbers under 10, and follow the steps. I can work out both your numbers very quickly. How? Many numbers can be expressed as the difference of two perfect squares. What do you notice about the numbers you CANNOT make? The number 2.525252525252.... can be written as a fraction. What is the sum of the denominator and numerator? Some 4 digit numbers can be written as the product of a 3 digit number and a 2 digit number using the digits 1 to 9 each once and only once. The number 4396 can be written as just such a product. Can. . . . Sissa cleverly asked the King for a reward that sounded quite modest but turned out to be rather large... Caroline and James pick sets of five numbers. Charlie chooses three of them that add together to make a multiple of three. Can they stop him? Can you guarantee that, for any three numbers you choose, the product of their differences will always be an even number? Given an equilateral triangle inside an isosceles triangle, can you find a relationship between the angles? If you have only 40 metres of fencing available, what is the maximum area of land you can fence off? Water freezes at 0°Celsius (32°Fahrenheit) and boils at 100°C (212°Fahrenheit). Is there a temperature at which Celsius and Fahrenheit readings are the same? Can you maximise the area available to a grazing goat? Show that if you add 1 to the product of four consecutive numbers the answer is ALWAYS a perfect square. Ben passed a third of his counters to Jack, Jack passed a quarter of his counters to Emma and Emma passed a fifth of her counters to Ben. After this they all had the same number of counters. Have a go at creating these images based on circles. What do you notice about the areas of the different sections? Different combinations of the weights available allow you to make different totals. Which totals can you make? A game for 2 or more people, based on the traditional card game Rummy. Players aim to make two `tricks', where each trick has to consist of a picture of a shape, a name that describes that shape, and. . . . There are lots of different methods to find out what the shapes are worth - how many can you find? Can you find an efficient method to work out how many handshakes there would be if hundreds of people met? Investigate how you can work out what day of the week your birthday will be on next year, and the year after... Can you find the area of a parallelogram defined by two vectors? If the hypotenuse (base) length is 100cm and if an extra line splits the base into 36cm and 64cm parts, what were the side lengths for the original right-angled triangle? The diagonals of a trapezium divide it into four parts. Can you create a trapezium where three of those parts are equal in area? Can you describe this route to infinity? Where will the arrows take you next? Is there a relationship between the coordinates of the endpoints of a line and the number of grid squares it crosses? If you move the tiles around, can you make squares with different coloured edges? There is a particular value of x, and a value of y to go with it, which make all five expressions equal in value, can you find that x, y pair ? All CD Heaven stores were given the same number of a popular CD to sell for £24. In their two week sale each store reduces the price of the CD by 25% ... How many CDs did the store sell at. . . . How many different symmetrical shapes can you make by shading triangles or squares? Square numbers can be represented as the sum of consecutive odd numbers. What is the sum of 1 + 3 + ..... + 149 + 151 + 153? A 2 by 3 rectangle contains 8 squares and a 3 by 4 rectangle contains 20 squares. What size rectangle(s) contain(s) exactly 100 squares? Can you find them all? Take any prime number greater than 3 , square it and subtract one. Working on the building blocks will help you to explain what is special about your results. The diagram illustrates the formula: 1 + 3 + 5 + ... + (2n - 1) = n² Use the diagram to show that any odd number is the difference of two squares. Many numbers can be expressed as the sum of two or more consecutive integers. For example, 15=7+8 and 10=1+2+3+4. Can you say which numbers can be expressed in this way? What angle is needed for a ball to do a circuit of the billiard table and then pass through its original position?
Which of the following was formed from leftovers in the solar system? the comet the planets in the upper left corner the sun the planets to the right of the sun Which statement correctly describes a property of white light?. . A. It makes all objects appear white.. . B. It contains different wavelengths of visible light.. . C. It allows light waves to be transmitted.. . D. It contains all waves on the electromagnetic spectrum. Without the greenhouse effect the Earth would be _______ than it is today. a. warmer b. drier c. colder d. wetter The energy derived from the digestion of food is __________. a. kinetic energy b. inertia c. momentum d. chemical energy What kind of motion indicates a system where the net force equals zero only accelerating objects only stationary objects both accelerating and stationary objects only objects moving with a constant positive velocity stationary objects as well as objects moving with a constant velocity Which statement about electrons and atomic orbitals is NOT true? An electron has the same amount of energy in all orbitals. An orbital can contain a maximum of two electrons. An electron cloud represents all the orbitals in an atom. An atom’s lowest energy level has only one orbital. The _____ is the region of the Sun where energy moves through circulating currents of gases. photosphere, corona, core, radiative zone, chromosphere, or convective zone Heat can never flow from __________. a. hot to cold b. cold to hot c. liquid to solid d. solid to liquid Which of the following statements is untrue in regard to sound traveling in air? A. As the distance from the Earth increases, the sound wave intensity also decreases due to the lessening density. B. The area covered by the wave front increases as the wave advances, but the energy passing through a wave front in a unit of time decreases. C. During a single vibration of the diaphragm, molecules in the air near it will also complete one vibration. D. Sound waves in the air are transverse and consist of compressions and rarefactions of air molecules. Which of the following statements is a consequence of the equation E = mc2? A. Energy is released when matter is destroyed. B. Mass and energy are equivalent. C. The law of conservation of energy must be modified to state that mass and energy are conserved in any process. D. all of the above In order for work to take place Select one: A. the energy present must be related to the movement of the object. B. the force applied must be a balanced force resulting in motion of the object. C. the force applied must cause the movement of the object in the same direction as the force. D. the movement of the object must be the cause of the force. A metal bolt with a mass of 8.50 x 10^-2kg and a temperature of 85.0 degrees Celsius is placed in a container of water. The mass of the water is 0.150 kg, and its temperature is 25.0 degrees Celsius. What is the specific heat capacity of the bolt if the final temperature of the bolt and water is 28.4 degrees Celsius? (Cp.w = 4186 j/k* degree Celsius) Why would you expect sodium (Na) to react strongly with chlorine (Cl)? 1. They both need to lose one electron. 2. They both need to gain one electron. 3. Sodium needs to lose one electron, and chlorine needs to gain one electron. 4. Sodium needs to gain one electron, and chlorine needs to lose one electron. Which rigid transformation would map abc to edc? a rotation about point b a reflection across the line containing cb a reflection across the line containing ac a rotation about point c? What distinguishes mass from weight? a. mass is a measure of the amount of matter in an object, while weight is a measure of the force of gravity on the object. b. weight is a measure ofthe amount of matter in an object, while mass is a measure of the force of gravity on the object. c. mass is a measure of an object's inertial force, and weight is the force exerted by that object on other objects. d. mass and weight are synonyms and ultimately describe the same principle? When a wire is made smaller, the resistance increases. Which happens to the electric current? A) The current decreases. B) The current increases. C) There is no effect on the current Which of the following statements is a consequence of the equation E = mc2? A Energy is released when matter is destroyed. B Mass and energy are equivalent. C The law of conservation of energy must be modified to state that mass and energy are conserved in any process. D all of the above One degree Celsius indicates the same temperature change as? A. One degree Fahrenheit B. 5/9 degree Fahrenheit C. One Kelvin D. 9/5 Kelvin When discussing discordant and harmonious sound waves, which statement is false? A. Low ratios of the frequencies of the original and resultant waves indicate discordant waves. B. If the original waves combine to form irregular displacements of air, the sound will be discordant. C. Knowing the frequencies of the original waves is useful in determining if the result will be discordant or harmonious. D. Both types of waves are the result of component waves with different frequencies. Most of the circuits in your home are a. parallel circuits. b. switch circuits. c. series circuits. d. reversible circuits. For INELASTIC collisions, which of the following statements can be true? Choose all that apply. Momentum is conserved. Momentum is lost. Kinetic energy is lost. Momentum is gained. Kinetic energy is conserved. Kinetic energy is gained.I put a and c and it says it wrong down to my last tries. Which statements about acceleration are true? check all that apply. a. the si units of acceleration are m/s2. b. for acceleration, you must have a number, a unit, and a direction. c. to calculate acceleration, divide the change in speed by the change in time. d. the symbol for acceleration is . Which would cause the greatest increase in the acceleration of a satellite?a decrease in the radius and the tangential speedan increase in the radius and the tangential speeda decrease in the radius and an increase in the tangential speedan increase in the radius and a decrease in the tangential speed The area of a circular trampoline is 108.94 square feet. What is the radius of the trampoline? Round to the nearest hundredth. A. 10.44 feet B. 5.89 feet C. 34.68 feet D. 3.32 feet What lever has resistance between the axis (fulcrum) and the force (effort)? a. first b. second c. third d. fourth An endothermic reaction is one that uses which form of energy?Nuclear FusionElectrical EnergyChemical EngerySound Energy
Gravitation may be the last and shortest unit, but you can see the applications of everything you've learned so far in this unit! From momentum to energy to forces, everything has accumulated to this point. Unit 5 will cover approximately 6%-14% of the exam and should take around 5 to 10, 45-minute class periods to cover. The AP Classroom personal progress check has 10 multiple choice questions and 1 free response question for you to practice on. When an object is large enough/massive enough it will create its own gravitational field which can interact with other objects that have gravitational fields. Newton's Law of Universal Gravitation: Every object attracts every other object in the universe with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance/radius between their centers. Newton's law of universal gravitation is a fundamental physical law that describes the attractive force between two masses. The law states that every point mass in the universe attracts every other point mass with a force that is proportional to the product of their masses and inversely proportional to the square of the distance between them. Here are some key points about Newton's law of universal gravitation: The force of attraction between two masses is given by the equation F = G * (m1 * m2)/r^2 where F is the force, G is the gravitational constant, m1 and m2 are the masses, and r is the distance between the centers of the masses. The gravitational constant (G) is a fundamental constant of nature and has a value of 6.67 x 10^-11 Nm^2/kg^2 The force of gravity is always attractive, and acts along the line connecting the two masses The gravitational force between two masses is independent of the nature of the masses and depends only on the product of their masses and the distance between them. The gravitational force between two masses decreases with the square of the distance between them, making it weaker as the distance increases. The law of universal gravitation is a long-range force, meaning that it acts over large distances, and it is the force responsible for the orbits of planets, the motion of galaxies, and the overall structure of the universe. It is important to note that the law of universal gravitation was the first physical law to explain the motion of celestial bodies with mathematical precision. It was developed by Sir Isaac Newton and was included in his famous work "Philosophiæ Naturalis Principia Mathematica" (Mathematical Principles of Natural Philosophy) published in 1687. You can see the equation form of the law below: In which Fg is the force of gravity, m1 and m2 are masses, r is the distance between the two masses, and G is the universal gravitational constant which is 6.6710^(-11) ((m^3)(kgs^2)). Essentially, this law is an extension of Newton's Third Law in which every action has an equal and opposite reaction. Here's a diagram that illustrates the law: Taken from Wikimedia Commons Now, we can use this equation to find the acceleration due to gravity on an object! If the object is on/near earth's surface: g is the acceleration due to gravity and we can cancel out the mass, therefore: if the object is far from the earth's surface, we use the distance from the earth to the object rather than the earth's radius: Additionally, it should be noted that gravity is a conservative force, meaning that its path is independent of it and the total work on a closed path is zero. Note that work done by a conservative force is equal to the negative change in potential energy. However, it is not uniform! This might be intuitive to some, but the further away the object is, the weaker the gravitational force acting on it from the other object will be. You can see this below: Taken from Physics Stack Exchange Motion of an object under the influence of a variable gravitational force When an object is dropped from a great height and falls towards the Earth's surface, the force of gravity acting on the object is variable. As the object falls, it accelerates due to the increasing force of gravity acting on it. At the beginning of the fall, the object will have a relatively slow velocity, and its acceleration will be small. As the object falls further, its velocity increases, and its acceleration also increases. At some point, the object will reach its terminal velocity, which is the maximum velocity it can reach due to the resistance of the air. As the object continues to fall, its velocity will remain constant and its acceleration will be zero. As the object approaches the Earth's surface, the force of gravity will decrease, and the object's velocity will decrease as well. Finally, the object will hit the Earth's surface, and its motion will come to an end. The impact will be the result of the force of gravity acting on the object throughout its fall, and it will be affected by the height from which the object was dropped, and the resistance of the air. Overall, the motion of an object falling towards the Earth's surface from a great height can be described as an acceleration followed by a constant velocity, then a deceleration before the impact. Taken from College Board
Class 9 Triangles Worksheet PDF have been designed as per the latest pattern for CBSE, NCERT and KVS for Grade 9. Students are always suggested to solve printable worksheets for Mathematics Triangles Grade 9 as they can be really helpful to clear their concepts and improve problem solving skills. We at worksheetsbag.com have provided here free PDF worksheets for students in standard 9 so that you can easily take print of these test sheets and use them daily for practice. All worksheets are easy to download and have been designed by teachers of Class 9 for benefit of students and is available for free download. Mathematics Triangles Worksheets for Class 9 We have provided chapter-wise worksheets for class 9 Mathematics Triangles which the students can download in Pdf format for free. This is the best collection of Mathematics Triangles standard 9th worksheets with important questions and answers for each grade 9th Mathematics Triangles chapter so that the students are able to properly practice and gain more marks in Class 9 Mathematics Triangles class tests and exams. Chapter-wise Class 9 Mathematics Triangles Worksheets Pdf Download 1. Congruence of Triangles 2. Criteria for Congruence of Triangles 3. Some Properties of a Triangle 4. Inequalities in a Triangle • Triangle – A closed figure formed by three intersecting lines is called a triangle. A triangle has three sides, three angles and three vertices. ·• Congruent figures – Congruent means equal in all respects or figures whose shapes and sizes are both the same for example, two circles of the same radii are congruent. Also two squares of the same sides are congruent. • Congruent Triangles- two triangles are congruent if and only if one of them can be made to superpose on the other, so as to cover it exactly. • If two triangles ABC and PQR are congruent under the correspondence A ↔ P,B ↔ Q and C ↔ R then symbolically, it is expressed as DABC ≅ DPQR • In congruent triangles corresponding parts are equal and we write ‘CPCT’ for corresponding parts of congruent triangles. ·• SAS congruency rule – Two triangles are congruent if two sides and the included angle of one triangle are equal to the two sides and the included angle of the other triangle. For example: ΔABC and ΔPQR as shown in the figure satisfy SAS congruent criterion • ASA Congruence Rule – Two triangles are congruent if two angles and the included side of one triangle are equal to two angles and the included side of other triangle. For examples ΔABC and ΔDEF shown below satisfy ASA congruence criterion. • AAS Congruence Rule – Two triangle are congruent if any two pairs of angles and one pair of corresponding sides are equal for example ΔABC and ΔDEF shown below satisfy AAS congruence criterion. • AAS criterion for congruence of triangles is a particular case of ASA criterion. • Isosceles Triangle – A triangle in which two sides are equal is called an isosceles triangle. For example: DABC shown below is an isosceles triangle with AB=AC. • Angle opposite to equal sides of a triangle are equal. • Sides opposite to equal angles of a triangle are equal. • Each angle of an equilateral triangle is 60o . • SSS congruence Rule – If three sides of one triangle are equal to the three sides of another triangle then the two triangles are congruent for example ΔABC and ΔDEF as shown in the figure satisfy SSS congruence criterion. • RHS Congruence Rule – If in two right triangles the hypotenuse and one side of one triangle are equal to the hypotenuse and one side of the other triangle then the two triangle are congruent. For example: ΔABC and ΔPQR shown below satisfy RHS congruence criterion. RHS stands for right angle – Hypotenuse side. • A point equidistant from two given points lies on the perpendicular bisector of the line segment joining the two points and its converse. • A point equidistant from two intersecting lines lies on the bisectors of the angles formed by the two lines. • In a triangle, angle opposite to the longer side is larger (greater) • In a triangle, side opposite to the large (greater) angle is longer. • Sum of any two sides of a triangle is greater than the third side. Mathematics Triangles Worksheets for Class 9 as per CBSE NCERT pattern Parents and students are welcome to download as many worksheets as they want as we have provided all free. As you can see we have covered all topics which are there in your Class 9 Mathematics Triangles book designed as per CBSE, NCERT and KVS syllabus and examination pattern. 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Benefits of Free Class 9 Triangles Worksheet PDF - You can improve understanding of your concepts if you solve NCERT Class 9 Mathematics Triangles Worksheet, - These CBSE Class 9 Mathematics Triangles worksheets can help you to understand the pattern of questions expected in Mathematics Triangles exams. - All worksheets for Mathematics Triangles Class 9 for NCERT have been organized in a manner to allow easy download in PDF format - Parents will be easily able to understand the worksheets and give them to kids to solve - Will help you to quickly revise all chapters of Class 9 Mathematics Triangles textbook - CBSE Class 9 Mathematics Triangles Workbook will surely help to improve knowledge of this subject These Printable practice worksheets are available for free download for Class 9 Mathematics Triangles. 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It is reported an approximation to the scalar and vectorial Rayleigh-Sommerfeld diffraction integrals performed with a Fourier transform operation, as done in Fresnel and Fraunhoffer approximations, but suitable for paraxial, non-paraxial and off-axis regimes. High accuracy is obtained even for waves with low f numbers. The approximation becomes exact on-axis. © 2017 Optical Society of America Diffraction calculus involves the use of Fresnel-Kirchoff or Rayleigh-Sommerfeld integrals in its scalar or vectorial forms . Such calculus is generally performed numerically, which involves a double integral of a complex function with a fast oscillating argument . In order to facilitate their calculus or to get some insight about diffraction properties, such integrals are approximated in the so called Fresnel or Fraunhoffer regimes, which are adequate for paraxial regions . For non-paraxial optical systems other approximations have been developed like that due to Debye, well suited for optical systems with high Fresnel numbers [3,4]. So, apparently, there is not a reliable approximation to calculate the diffraction properties of high, intermediate and low numerical aperture optical systems. In this work we will propose an approximation that could fulfill such requirements. We propose to use tilted spheres to approximate the shifted spheres of the diffraction integral; in Fraunhoffer and Fresnel integrals planes and parabolas are used to perform such approximation. The beauty of our approximation is that we perform the calculations with a simple Fourier transform operation, as done with the previous two approximations, but without having the paraxial restrictions. Some immediate novel results, described in this work, obtained as a consequence of our proposal are: a non-paraxial formula for rotationally symmetric functions via a Hankel transform; off-axis diffraction analysis; non-paraxial vectorial formulas performed with Fourier transforms; and even the study of evanescent waves. Since our approximation becomes exact on-axis it is well behaved near the optical axis and could be a useful tool to study optical systems with high focus depth. An important feature is that our approximation is calculated with a Fast Fourier Transform algorithm and 2D images or even 3D volumes are easily obtained. 2. Alternative approximation to a diffraction integral Let us consider the First Rayleigh-Sommerfeld formulation (R-S) in rectangular coordinates, as described by . The complex amplitude field distribution of a function g(x,y) at a certain distance is given byFig. 1). Note that are spheres with its center shifted to . As it is known, instead of a direct evaluation of Eq. (1), which involves various numerical difficulties, usually the R in the argument of the exponential (phase function) is replaced by the first terms of the Taylor seriesEq. (1) is generally approximated by 1/z. Assuming also that we obtain the well-known Fresnel approximation (paraxial) of Eq. (1) given by It can be noticed that the arguments in the exponential functions of Eq. (4) can be interpreted as tilting parabolas. When it is assumed that , such parabolas behaves like planes and we obtain the Fraunhoffer approximation. In the following descriptions we will use the above explanations for comparison purposes. Now, Taylor’s formula for a differentiable function of two-variables is used to expand (Eq. (2)) up to second order to obtainEquation (6) has been used in to obtain analytical expressions for a focused Gaussian beam, and later in , with the first two terms, to study radially polarized beams. Unfortunately, in those forms the resulting expressions are difficult to deal with, particularly due to the existence of in the denominator. Since it seems to me that no further simplifications to Eq. (6) have been done before, I will show that retaining only the first two terms of Eq. (6) and replacing the respectivein the denominator by a constantwill be enough to simplify the calculus of Eq. (1) without sacrificing its accuracy significantly within a neighborhood of . Thus, our approximation would be simply given byEq. (7).Equation (9) means that the diffracted field is just the Fourier transform of the original function, which is multiplied by the complex function . Note that the phase means tilted spheres. It can be appreciated that no quadratic phase term appears in the resulting diffracted field as it happens in Eq. (4). It can be noticed also that our approximation becomes exact on-axis since . Expanding of Eq. (8) in Taylor series and using only its first terms in Eq. (9) we reach again, up to a constant factor, the Fresnel and Fraunhoffer approximations, Eq. (4). This way, our approximation can be regarded as a generalization of both. Since we can define and , the ambiguity in choosing means that the diffracted fields, for a given z, are correctly calculated up to a scale factor. That scale could be unimportant when relative calculations are done or when the shape of the PSF is the goal. In general, we can estimate either numerically or analytically and their value will depend on the criteria used to calculate it. In order to achieve the calculations it must be noticed that a good choice of means that the linear function looks like the function . With a least squares formulation such similarity can be optimized with the following expressionEq. (10) can be computed numerically. When circular apertures are chosen, settingin Eq. (9) provides errors lower than 2% for , being . In following sections the performance of Eq. (11) will be shown. 3. Error analysis in the approximation To perform our error analysis, for simplicity we will contemplate,and a circular aperture of diameter . Now, considering again the first two terms of Eq. (6), , and the Taylor series expansion of we obtainEq. (8). Taking into account that is the maximum value that can reach x we roughly estimate the maximum error between and like , being . Thus, the higher the or the lower (or ), the higher the exactitude of Eq. (8). An important issue is that errors depends only on, and and not on the aperture D itself. This is an important result because it means that converging waves having equal and different Fresnel numbers must have the same accuracy. Now we will compute integrals specified by Eqs. (1), (4) and (9) to compare the intensities, and given by the squared modulus of the respective amplitudes. To perform such comparison we have chosen the following parameters: a circular apertureof diameter , , , , and defined by a converging spherical wave of radius equal to z,Fig. 2, where it can be seen that among those PSFs, Fresnel is the most different, as expected. In Fig. 2, dotted-continuous line corresponds to , continuous line to and dashed to . We show various graphs of corresponding to values of = 0.85z, 1.0z, 1.08z and 1.2z, being 1.2z the most accurate (this value was computed with Eq. (11)). With this graphics it is clarified that is just a scale factor of the diffraction pattern. Defining our error of calculus as, we found for the best curve of shown in Fig. 2 that, which may be considered small, taking into account that a high numerical wave has been analyzed. In addition to this, we show in Fig. 3 a curve specifying that errors decreases when increases, as predicted with the error analysis of Eq. (12). Errors of Fig. 3 can be reduced when the analysis range for the spatial coordinates is reduced; for example the previous error is reduced to when was considered. This result is expected since our approximation is truly exact on axis. With the purpose to show that effectively the exactitude of Eq. (9) depends only on and not on D, we fixed = 0.6, 1.0 and 2.0 and varied the corresponding aperture diameters D from 0.05mm to 1000mm, calculating the respective diffraction intensities and estimating the respective errors as done in curve of Fig. 3. Results are shown in Fig. 4 where it can be seen that, effectively, errors do not depend on the aperture diameter itself, being constant for the same. Logarithmic scale (natural logarithm) has been used in order to show the wide range of values of D considered. Notice that log(D) = 3 in Fig. 4 means Fresnel number . Additionally, we have obtained curves for = 100 and = 10000 (not shown) finding similar behavior: constant errors independently of D. The curve obtained for = 10000, with errors, comprises analysis when Fresnel number goes from 0.002 to 39. From these results we conclude that our proposal can be used confidently in the analysis of low, medium or high numerical aperture optical systems with errors on the order of 2%. 4. Vectorial diffraction integrals One application of our approximation is in the study of diffraction properties of polarized beams. Since the first kind vectorial Rayleigh-Sommerfeld formulas expressed in rectangular coordinates are given by Eq. (1), Eqs. (14) and (16) can be approximated with the following substitution8]. Then, we used Eq. (18) to calculate Eqs. (14), and (16) together with the parameters given previously like n = 1, a circular apertureof diameter , , , and a spherical convergent function given by Eq. (13) . The results of using our approximation with a fast Fourier algorithm are presented in Figs. 5(a-c) and using the integration routines of MatLab in Fig. 5d, where the comparison with exacts integral is shown. Good agreement has been obtained again, with E = 2.2%. 5. Comparison with Debye approximation Using the results of [9,10] we can write the vectorial diffraction integrals within the Debye approximation like withFig. 5, as described in section 4, we evaluated Eq. (19) and show in Fig. 6, with dashed lines, the corresponding calculated intensities . The other two curves correspond to those shown in Fig. 5d. It can be appreciated that both approximation, at focus, behaves similarly even that we are using a converging wave with a NA = 0.64. The error obtained with the Debye approximation was which is a little lower than the obtained with our method . It is interesting to study the behavior of our proposal when defocus is introduced. Without going deeper into this matter we show Fig. 7, obtained with the previous data of Fig. 6 (, ,), not at the focus plane but a distance of from it towards the aperture, that is, at. It can be appreciated that at this defocused position our results (dotted-continuous line) were more accurate than those obtained with the Debye approximation (dashed line). When choosing another value for , i.e, , the curves of Fig. 8 were obtained; in that figure it can be noticed that our approximation (I_N) was improved. At this point it must be emphasized that different values for means only different scale factors of the whole diffraction pattern. Although we have derived a formula to calculate for a focused plane, Eq. (11), a general expression must be found yet when the diffraction patterns are imaged onto others plane positions. 6. Diffraction integrals for rotationally symmetric functions When a circular apertureof semi diameter is used and is defined as a rotationally symmetric function, it can be obtained straightforwardly a one dimensional integral form of Eq. (9), given by , and a Bessel function of the first kind, zero order. 7. Off-axis diffraction analysis We have demonstrated our approximation for small values of , which means calculations near the optical axis. However our proposal is not restricted to those values since we can define with small, as in previous analysis, and any constant point on the image plane, not necessarily near . Then, it can be shown that our approximation can be described by 8. Diffraction integrals using planar illumination It seems that our approximation could be used to analyze diffraction apertures illuminated with planar waves. To show this, let us define in Eq. (22), , a unit planar wave parallel to the optical axis, and, a constant aperture function. Thus, what we have is the Fourier transform of . Using the following result [11,12]Eq. (7), it is readily shown from Eq. (22) that From Eq. (24) it can be seen that for , evanescent waves are obtained since and are all real valued and positive. So, Eq. (22) may be helpful to describe the behavior of evanescent waves for general values of . It can be noticed from Eq. (9) (and those of Eqs. (14-18)) that doing the usual assumption the second term of the evanescent function can be neglected. If in addition to this we suppose that the illuminating light corresponds to a converging spherical wave behaves like Eq. (13) we obtain from Eq. (9)Eq. (25) with a = 5mm, , and a circular aperture, an intensity error as little as 1.5% is introduced. Neglecting also the evanescent term in Eq. (21), setting and using the proper converging spherical wave, it turns out to beEq. (26) is important since their zeros positions are strongly related with the definition of the resolution of optical systems. It is convenient to mention that, according with the explanation given in section 7, we can calculate the diffraction patterns in a neighborhood of each out-off axis point, and by a stitching process we could obtain such diffraction, with the desired exactitude, in bigger areas beyond those neighborhoods at the cost of performing various Fourier transforms. We have proved that an alternative approximation can be used to analyze paraxial and non-paraxial systems without restriction on the Fresnel number. The approximation simplifies scalar and vectorial diffraction integrals to be calculated with a Fourier transform operation. An expression for off-axis analysis is obtained as well. Consequences of our approximation like one-dimensional integral for rotationally symmetric functions or evanescent waves have been considered in the analysis. Although we have focused our approximation into RS a similar analysis can be carried out with Fresnel-Kirchhoff integrals since they share similar expressions. References and links 1. M. Born and E. Wolf, Principles of Optics, 7th ed. (Cambridge University Press, 1999). 2. J. J. Stamnes, Waves in Focal Regions (IOP Publishing Limited, 1986). 3. J. W. Goodman, Introduction to Fourier Optics, 2nd ed. (McGraw-Hill, 1996). 5. A. Ciattoni, B. Crosignani, and P. Di Porto, “Vectorial analytical description of propagation of a highly nonparaxial beam,” Opt. Commun. 202(1-3), 17–20 (2002). [CrossRef] 6. D. Deng, “Nonparaxial propagation of radially polarized light beams,” J. Opt. Soc. Am. B 23(6), 1228–1234 (2006). [CrossRef] 7. H. Ye, C.-W. Qiu, K. Huang, J. Teng, B. Luk’yanchuk, and S. P. Yeo, “Creation of a longitudinally polarized subwavelength hotspot with an utra-thin planar lens: vectorial Rayleigh-Sommerfeld method,” Laser Phys. Lett. 10(6), 1–8 (2013). [CrossRef] 8. X. Hao, C. Kuang, T. Wang, and X. Liu, “Effects of polarization on the de-excitation dark focal spot in STED microscopy,” J. Opt. 12(11), 115707 (2010). [CrossRef] 9. B. Richards and E. Wolf, “Electromagnetic diffraction in optical systems II. Structure of the image field in an aplanatic system,” Proc. R. Soc. Lond. 253(1274), 358–379 (1959). [CrossRef] 10. L. E. Helseth, “Focusing of atoms with strongly confined light potentials,” Opt. Commun. 212(4-6), 343–352 (2002). [CrossRef] 11. E. Lalor, “Conditions for the validity of the angular spectrum of plane waves,” J. Opt. Soc. Am. 58(9), 1235–1237 (1968). [CrossRef] 12. W. H. Southwell, “Validity of the Fresnel approximation in the near field,” J. Opt. Soc. Am. 71(1), 7–14 (1981). [CrossRef]
- Propagation of uncertainty statistics, propagation of uncertainty (or propagation of error) is the effect of variables' uncertainties (or errors) on the uncertainty of a function based on them. When the variables are the values of experimental measurements they have uncertainties due to measurement limitations (e.g. instrument precision) which propagate to the combination of variables in the function. The uncertainty is usually defined by the absolute error. Uncertainties can also be defined by the relative errorΔ"x"/"x", which is usually written as a percentage. Most commonly the error on a quantity, , is given as the standard deviation, , . Standard deviation is the positive square root of variance, . The value of a quantity and its error are often expressed as . If the statistical probability distributionof the variable is known or can be assumed, it is possible to derive confidence limitsto describe the region within which the true value of the variable may be found. For example, the 68% confidence limits for a variable belonging to a normal distributionare ± one standard deviation from the value, that is, there is a 68% probability that the true value lies in the region . If the variables are correlated, then covariancemust be taken into account. Let be a set of "m" functions which are linear combinations of variables with combination coefficients .: and let the variance-covariance matrixon x be denoted by .:Then, the variance-covariance matrix , of "f" is given by:This is the most general expression for the propagation of error from one set of variables onto another. When the errors on "x" are un-correlated the general expression simplifies to:Note that even though the errors on "x" may be un-correlated, their errors on "f" are alwayscorrelated. The general expressions for a single function, "f", are a little simpler.: : Each covariance term, can be expressed in terms of the correlation coefficientby , so that an alternative expression for the variance of "f" is:In the case that the variables "x" are uncorrelated this simplifies further to: When "f" is a set of non-linear combination of the variables "x", it must usually be linearlized by approximation to a first-order Maclaurin seriesexpansion, though in some cases, exact formulas can be derived that do not depend on the expansion.Cite journal title = On the Exact Variance of Products journal = Journal of the American Statistical Association year = 1960 volume = 55 issue = 292 pages = 708–713 url = http://links.jstor.org/sici?sici=0162-1459(196012)55%3A292%3C708%3AOTEVOP%3E2.0.CO%3B2-3 doi = 10.2307/2281592] : where denotes the partial derivativeof "fk" with respect to the "i"-th variable. Since "f0k" is a constant it does not contribute to the error on "f". Therefore, the propagation of error follows the linear case, above, but replacing the linear coefficients, "Aik" and "Ajk" by the partial derivatives, and . Any non-linear function, "f(a,b)", of two variables, "a" and "b", can be expanded as: Whence: In the particular case that , . Then:or: Caveats and warnings Error estimates for non-linear functions are biased on account of using a truncated series expansion. The extent of this bias depends on the nature of the function. For example, the bias on the error calculated for log x increases as x increases since the expansion to 1+x is a good approximation only when x is small. In data-fitting applications it is often possible to assume that measurements errors are uncorrelated. Nevertheless, parameters derived from these measurements, such as least-squaresparameters, will be correlated. For example, in linear regression, the errors on slope and intercept will be correlated and the term with the correlation coefficient, ρ, can make a significant contribution to the error on a calculated value.: In the special case of the inverse where , the distribution is a Cauchy distribution and there is no definable variance. For such ratio distributions, there can be defined probabilities for intervals which can be defined either by Monte Carlo simulation, or, in some cases, by using the Geary-Hinkley transformation.Cite journal Jack Hayya, Donald Armstrongand Nicolas Gressis title = A Note on the Ratio of Two Normally Distributed Variables journal = Management Science year = 1975 volume = 21 issue = 11 pages = 1338–1341 month = July url = http://links.jstor.org/sici?sici=0025-1909(197507)21%3A11%3C1338%3AANOTRO%3E2.0.CO%3B2-2] This table shows the variances of simple functions of the real variables with standard deviations , and precisely-known real-valued constants . :For uncorrelated variables the covariance terms are zero.Expressions for more complicated functions can be derived by combining simpler functions. For example, repeated multiplication, assuming no correlation gives,: Example calculation: Inverse tangent function We can calculate the uncertainty propagation for the inverse tangent function as an example of using partial derivatives to propagate error. where is the absolute uncertainty on our measurement of . The partial derivative of with respect to is Therefore, our propagated uncertainty is where is the absolute propagated uncertainty. Example application: Resistance measurement A practical application is an experimentin which one measures current, "I", and voltage, "V", on a resistorin order to determine the resistance, "R", using Ohm's law, Given the measured variables with uncertainties, "I"±Δ"I" and "V"±Δ"V", the uncertainty in the computed quantity, Δ"R" is Thus, in this simple case, the relative errorΔ"R"/"R" is simply the square root of the sum of the squares of the two relative errors of the measured variables. Errors and residuals in statistics Accuracy and precision * [http://physicslabs.phys.cwru.edu/MECH/Manual/Appendix_V_Error%20Prop.pdf Uncertainties and Error Propagation] , Appendix V from the Mechanics Lab Manual, Case Western Reserve University. * [http://www.av8n.com/physics/uncertainty.htm A detailed discussion of measurements and the propagation of uncertainty] explaining the benefits of using error propagation formulas and monte carlo simulations instead of simple Wikimedia Foundation. 2010. Look at other dictionaries: Uncertainty — For the film of the same name, see Uncertainty (film). Certainty series Agnosticism Belief Certainty Doubt Determinism Epistemology … Wikipedia Propagation des incertitudes — Une mesure est toujours entachée d erreurs dont on estime l intensité par l intermédiaire des incertitudes. Lorsqu une mesure est utilisée pour obtenir la valeur d une autre grandeur par l intermédiaire une formule, outre le calcul de la valeur… … Wikipédia en Français Measurement uncertainty — In metrology, measurement uncertainty is a non negative parameter characterizing the dispersion of the values attributed to a measured quantity. The uncertainty has a probabilistic basis and reflects incomplete knowledge of the quantity. All… … Wikipedia Interference (wave propagation) — Two point interference in a ripple tank. In physics, interference is the phenomenon in which two waves superpose each other to form a resultant wave of greater or lower amplitude. Interference usually refers to the interaction of waves that are… … Wikipedia Significance arithmetic — is a set of rules (sometimes called significant figure rules) for approximating the propagation of uncertainty in scientific or statistical calculations. These rules can be used to find the appropriate number of significant figures to use to… … Wikipedia List of numerical analysis topics — This is a list of numerical analysis topics, by Wikipedia page. Contents 1 General 2 Error 3 Elementary and special functions 4 Numerical linear algebra … Wikipedia (130) Электра — Открытие Первооткрыватель К. Г. Ф. Петерс Место обнаружения Клинтон Дата обнаружения 17 февраля 1873 Эпоним Электра Категория Главное кольцо … Википедия Observational error — is the difference between a measured value of quantity and its true value. In statistics, an error is not a mistake . Variability is an inherent part of things being measured and of the measurement process. Contents 1 Science and experiments 2 … Wikipedia Approximation error — The approximation error in some data is the discrepancy between an exact value and some approximation to it. An approximation error can occur because the measurement of the data is not precise due to the instruments. (e.g., the accurate reading… … Wikipedia 107 Camilla — ( /kəˈ … Wikipedia
From this article, you will acquire better knowledge on various topologies of computer network. After that you can decide which topology will be suitable for your home or organization. Read in details: Networking System or Network Topology: The geometric insertion that is made to connect different computers is called topology. That is to say, the connection of one computer with another computer in a computer network is called Topology. Topology is the exchange of information between many computers by connecting advanced devices or components to a central computer while maintaining an advanced Bandwidth (the rate of data transfer is called bandwidth) system. In brief, Topology is how the computers on a network are connected to each other when viewed from the top of a network. The computers on the network are connected to each other by wires. However, it is not the only task to connect the computers of the network with it. A logically well-regulated path is needed for data to flow smoothly through the cable. The design of a network to connect computers to a network and the plan of a logical path for data to pass through a connecting cable are collectively called network topology. The following topologies are mostly used in the Computer Network: a. Point-to-point (P2P) Network Topology: The most simple network topology is point-to-point topology. The network is made up of a direct connection between two computers in this process. Advantages of using P2P Topology: 1. This is quicker and even more consistent than other connections since a direct connection is available there. 2. No network operating system is necessary for this. 3. It doesn’t need a costly server, as each workstation has access to the files. 4. No network technicians required because each user sets his/her permissions. Disadvantages of using P2P Topology: 1. The significant limitation is that it can be operated only in a limited area of shorter space where computers are located in a close distance. 2. Files and folders can not be centrally backed up with this topology. 3. Apart from permissions, there is no security. Users don’t have to log into their workstations most of the time. b. Star Network Topology: All computers in the Star Topology network are connected to a central computer (junction). The computer that is used as the central computer (junction) is called the Hub. Each computer in the Star Network is directly connected via a Hub or Switch. Microcomputers communicate with each other through hubs and exchange data. If for some reason the central computer loses its efficiency, the entire network also loses its efficiency. Because there is no connection between computers and devices except the central computer. Advantages of using Star Topology: 1. If one of the nodes (computers) in this topology is lost, the rest of the nodes in the network are not disrupted. 2. Different types of cables such as- twisted pair, co-axial and fiber optic cable can be used in this topology. 3. Any computer (node) can be added or removed from the star network without any interruption. 4. Centrally network maintenance or troubleshooting is easy. 5. The speed of data transmission is high. Disadvantages of using Star Topology: 1. In this way, if the central computer is damaged, the entire network system becomes inoperable. Because the whole network is connected to each other through the hub or switch. 2. Star topology uses a large number of cables and central devices, so it is expensive. 3. The performance of the network depends on the efficiency of the central device. c. Ring Network Topology: All computers in a ring topology network are connected by cables in a way that creates a ring or loop. No starting or end border can be found in this topology. In a ring network, computers connect to each other in a circular way through nodes (the point at which the computer is connected is called a node). Since the computers in the ring topology network are not directly connected to each other, no computer in the network can send signals directly to any other computer. The signal reaches the recipient computer only when it passes through the other computers between the transmitting computer and the receiving computer. In this case, every computer must be able to receive and transmit signals. Therefore, if a computer in the network loses the ability to transmit signals or becomes bad or disconnected, the whole network becomes useless. In this case, the bad computer has to be removed and the connection should be reconnected. Advantages of using Ring Topology: 1. No server computer is required on the network. This means that it is a kind of distributed data processing system. 2. Although the number of computers in the network increases, its efficiency is not greatly affected. 3. No node in this network has to rely on a central computer for data exchange. If a computer is damaged, that computer can be removed from the ring. 4. Twisted pair cables used for this are cheap and easily available. Therefore, the network installation cost is very low. 5. This is reliable because the communication system does not rely on a single host computer. 6. The probability of collision in this topology is minimal. Disadvantages of using Ring Topology: 1. If only one computer in the network is damaged, the whole network becomes inoperable. 2. In the case of ring topology, diagnosing a network issue is quite complex. 3. Adding or removing a computer to the network disrupts the entire network. 4. The time required for data transmission is directly proportional to the number of nodes. If the number of computers in this network increases, the time of data transmission also increases. 5. Complex control software is used for ring topology. d. Bus Network Topology: The topology in which all workstations or computers are connected to a core cable is called Bus Topology. The main cable in the Bus Topology is called Backbone. All nodes are connected to a connection line in a bus network organization. The connection line is usually called the Bus. A computer sends a signal through a connection line to another computer node. Other computers test that signal on their node and only the recipient node receives that signal. Advantages of using Bus Topology: 1. The main advantage of this topology is that the network is very simple and the number of physical lines is only one. 2. Co-axial or twisted pair cables are mainly used in bus-based networks that support up to 10 Mbps. 3. The backbone of the network can be easily expanded with the help of a repeater. 4. This topology is simple and easy to use on small-sized networks. 5. Even if one of the nodes of the bus topology is damaged, the other nodes are not affected. 6. Any computer can be easily connected to and disconnected from the network. 7. The bus topology requires fewer cables and no networking devices such as hubs or switches, resulting in lower costs. Disadvantages of using Bus Topology: 1. When the main cable or backbone is damaged, the entire network system becomes inoperable. 2. There is no for data transmission in this topology. Data can be transmitted by any computer (node) at any time. If two nodes send a message at the same time, the signals of both nodes collide with each other. 3. If the traffic in the network increases then the data collision of the network increases. 4. Data transmission is interrupted when the number of computers in the network is high. 5. Diagnosing problems in bus topology is relatively complex. 6. An error or break caused by a single bus paralyzes the entire network. 7. The speed of data transmission is also low in this topology. e. Tree Topology: It is basically an extensive form of the Star Topology. In this topology, all the computers are connected to a special place using multiple hubs which is called Root. The topology in which computers are arranged with each other like branches of a tree is called tree topology. In this topology, one or more levels of computer are connected to the host computer. That is, first level computers host second level computers. Similarly, second level computers host third level computers. Advantages of using Tree Topology: 1. Tree topology is primarily used to provide broadband transmission, meaning that signals can be transmitted over long distances. 2. It is easy to expand the network of tree topologies by creating branches. 3. Connecting or removing any new nodes does not interfere with the normal functioning of the network. 4. Error detection and error correction are very easy in a tree topology. 5. Error in one station does not affect the whole network. 6. There is point-to-point wiring for each segment. 7. This network topology is very useful for office management. Disadvantages of using Tree Topology: 1. This topology is kind of a bit complicated. 2. If an error occurs on the root or server computer, the tree network becomes inoperable. 3. Tree topology largely depends on the original bus cable and the failure of the main bus cable damages the overall network. f. Mesh Topology or Interconnected Network: In a mess or interconnected network, the topology network has a separate link or bus with each workstation. So each workstation can exchange data directly with any workstation. In the case of mesh topology, every computer in the network is directly connected to each computer. If at any time one way of communication is lost, the alternative is another way of communication. The Internet is an example of mesh topology. In a mesh topology, a host is associated with one or more hosts. Each host in this topology can be point-to-point with all other hosts or only a few hosts in the network. In the mesh topology, hosts also act as relays to other hosts if there is no direct point-to-point link. There are two types of mesh topology. These are- i) Full Mesh Topology: In a full mesh topology, each node is directly connected to all the other nodes in the network. ii) Partial Mesh Topology: In a partial mesh topology, each node is connected to specific nodes rather than all other nodes. Features of Mesh Topology: 1. There are multiple paths from one computer to another. 2. It does not have a switch, hub, or any central computer that acts as a central point of communication. 3. The mesh topology is mainly used for WAN implementation. 4. The mesh topology is primarily used for wireless networks. 5. Fake topologies can be formed using the formula: Number of wires = (n * (n-1)) / 2 Advantages of using Mesh Topology: 1. In Mesh Topology,Signals can be exchanged very fast between any two nodes. 2. There is no problem if any computer or connection line is damaged. This means that there is no major issues created for that. 3. Even if a connection line is lost, data can be exchanged using an alternative connection line. 4. It has a lot more certainty in data communication. 5. Network issues can be solved very easily in this topology. 6. This kind of network protects security and privacy. Disadvantages of using Mesh Topology: 1. Network installation and configuration in this topology is quite complex. 2. Having to put additional links in the network increases the cost. g. Hybrid Network Topology: A network consisting of topologies of stars, rings, buses, etc. is called a Hybrid Network Topology. The Internet, for example, is one such topology. It is a hybrid network because it is a large-scale network with a mix of all types of topologies. The advantages and disadvantages of a hybrid network depend on the topologies used in that network. Advantages of using Hybrid Topology: 1. In this topology, there is an opportunity to increase the network as required. 2. If there is a trouble, it can be easily diagnosed. 3. If any one part of the network is lost, the whole network is not lost. Disadvantages of using Hybrid Topology: 1. The major drawback of the hybrid topology is the design of the hybrid network. It is very difficult to design the architecture of a hybrid network. 2. Hubs used in hybrid topologies are very expensive. This is because these hubs are different from the usual hubs used in other topologies. 3. Hybrid networks require a lot of cabling, network devices, etc., so the cost of the structure is very high. How to choose a topology for a network? Here are some important aspects to choose the best topology for building a network in your company: 1. The bus topology is definitely cheaper to create a network. 2. If you are interested in using a shorter cable or you plan to expand your network in future, the star topology is the best choice for you. 3. You should create star topologies if you want to to use twisted networking pair cables. 4. Hypothetically, absolute mesh topology is a perfect selection because every device is linked to each other here. If you have any more queries on network topologies feel free to ask us in the comment box. You can also connect us through Facebook or Youtube. Stay with Smrity Computer for more updates on networking.
From: firstname.lastname@example.org Newsgroups: comp.ai Subject: FUZZY SYSTEMS - A Tutorial Date: 2 Jan 92 14:32:09 EST (c) Copyright James F. Brule' 1985. Permission to copy without fee all or part of this material is granted provided that the copies are not made or distributed for direct commercial advantage, the copyright notice and the title and date appear, and notice is given that copying is by permission of the author. To copy otherwise, or to republish, requires a fee and/or specific permission. While it might be argued that such vagueness is an obstacle to clarity of meaning, only the most staunch traditionalists would hold that there is no loss of richness of meaning when statements such as "Sally is tall" are discarded from a language. Yet this is just what happens when one tries to translate human language into classic logic. Such a loss is not noticed in the development of a payroll program, perhaps, but when one wants to allow for natural language queries, or "knowledge representation" in expert systems, the meanings lost are often those being searched for. For example, when one is designing an expert system to mimic the diagnostic powers of a physician, one of the major tasks i to codify the physician's decision-making process. The designer soon learns that the physician's view of the world, despite her dependence upon precise, scientific tests and measurements, incorporates evaluations of symptoms, and relationships between them, in a "fuzzy," intuitive manner: deciding how much of a particular medication to administer will have as much to do with the physician's sense of the relative "strength" of the patient's symptoms as it will their height/weight ratio. While some of the decisions and calculations could be done using traditional logic, we will see how fuzzy systems affords a broader, richer field of data and the manipulation of that data than do more traditional methods. It was Plato who laid the foundation for what would become fuzzy logic, indicating that there was a third region (beyond True and False) where these opposites "tumbled about." Other, more modern philosophers echoed his sentiments, notably Hegel, Marx, and Engels. But it was Lukasiewicz who first proposed a systematic alternative to the bi-valued logic of Aristotle . In the early 1900's, Lukasiewicz described a three-valued logic, along with the mathematics to accompany it. The third value he proposed can best be translated as the term "possible," and he assigned it a numeric value between True and False. Eventually, he proposed an entire notation and axiomatic system from which he hoped to derive modern mathematics. Later, he explored four-valued logics, five-valued logics, and then declared that in principle there was nothing to prevent the derivation of an infinite-valued logic. Lukasiewicz felt that three- and infinite-valued logics were the most intriguing, but he ultimately settled on a four-valued logic because it seemed to be the most easily adaptable to Aristotelian logic. Knuth proposed a three-valued logic similar to Lukasiewicz's, from which he speculated that mathematics would become even more elegant than in traditional bi-valued logic. His insight, apparently missed by Lukasiewicz, was to use the integral range [-1, 0 +1] rather than [0, 1, 2]. Nonetheless, this alternative failed to gain acceptance, and has passed into relative obscurity. It was not until relatively recently that the notion of an infinite-valued logic took hold. In 1965 Lotfi A. Zadeh published his seminal work "Fuzzy Sets" (, ) which described the mathematics of fuzzy set theory, and by extension fuzzy logic. This theory proposed making the membership function (or the values False and True) operate over the range of real numbers [0.0, 1.0]. New operations for the calculus of logic were proposed, and showed to be in principle at least a generalization of classic logic. It is this theory which we will now discuss. "Jane is old."If Jane's age was 75, we might assign the statement the truth value of 0.80. The statement could be translated into set terminology as follows: "Jane is a member of the set of old people."This statement would be rendered symbolically with fuzzy sets as: mOLD(Jane) = 0.80where m is the membership function, operating in this case on the fuzzy set of old people, which returns a value between 0.0 and 1.0. At this juncture it is important to point out the distinction between fuzzy systems and probability. Both operate over the same numeric range, and at first glance both have similar values: 0.0 representing False (or non- membership), and 1.0 representing True (or membership). However, there is a distinction to be made between the two statements: The probabilistic approach yields the natural-language statement, "There is an 80% chance that Jane is old," while the fuzzy terminology corresponds to "Jane's degree of membership within the set of old people is 0.80." The semantic difference is significant: the first view supposes that Jane is or is not old (still caught in the Law of the Excluded Middle); it is just that we only have an 80% chance of knowing which set she is in. By contrast, fuzzy terminology supposes that Jane is "more or less" old, or some other term corresponding to the value of 0.80. Further distinctions arising out of the operations will be noted below. The next step in establishing a complete system of fuzzy logic is to define the operations of EMPTY, EQUAL, COMPLEMENT (NOT), CONTAINMENT, UNION (OR), and INTERSECTION (AND). Before we can do this rigorously, we must state some formal definitions: It is important to note the last two operations, UNION (OR) and INTERSECTION (AND), which represent the clearest point of departure from a probabilistic theory for sets to fuzzy sets. Operationally, the differences are as follows: For independent events, the probabilistic operation for AND is multiplication, which (it can be argued) is counterintuitive for fuzzy systems. For example, let us presume that x = Bob, S is the fuzzy set of smart people, and T is the fuzzy set of tall people. Then, if mS(x) = 0.90 and uT(x) = 0.90, the probabilistic result would be: mS(x) * mT(x) = 0.81whereas the fuzzy result would be: MIN(uS(x), uT(x)) = 0.90The probabilistic calculation yields a result that is lower than either of the two initial values, which when viewed as "the chance of knowing" makes good sense. However, in fuzzy terms the two membership functions would read something like "Bob is very smart" and "Bob is very tall." If we presume for the sake of argument that "very" is a stronger term than "quite," and that we would correlate "quite" with the value 0.81, then the semantic difference becomes obvious. The probabilistic calculation would yield the statement If Bob is very smart, and Bob is very tall, then Bob is a quite tall, smart person.The fuzzy calculation, however, would yield If Bob is very smart, and Bob is very tall, then Bob is a very tall, smart person.Another problem arises as we incorporate more factors into our equations (such as the fuzzy set of heavy people, etc.). We find that the ultimate result of a series of AND's approaches 0.0, even if all factors are initially high. Fuzzy theorists argue that this is wrong: that five factors of the value 0.90 (let us say, "very") AND'ed together, should yield a value of 0.90 (again, "very"), not 0.59 (perhaps equivalent to "somewhat"). Similarly, the probabilistic version of A OR B is (A+B - A*B), which approaches 1.0 as additional factors are considered. Fuzzy theorists argue that a sting of low membership grades should not produce a high membership grade instead, the limit of the resulting membership grade should be the strongest membership value in the collection. Other values have been established by other authors, as have other operations. Baldwin proposes a set of truth value restrictions, such as "unrestricted" (mX = 1.0), "impossible" (mX = 0.0), etc. The skeptical observer will note that the assignment of values to linguistic meanings (such as 0.90 to "very") and vice versa, is a most imprecise operation. Fuzzy systems, it should be noted, lay no claim to establishing a formal procedure for assignments at this level; in fact, the only argument for a particular assignment is its intuitive strength. What fuzzy logic does propose is to establish a formal method of operating on these values, once the primitives have been established. The simplest example is in which one transforms the statement "Jane is old" to "Jane is very old." The hedge "very" is usually defined as follows: m"very"A(x) = mA(x)^2Thus, if mOLD(Jane) = 0.8, then mVERYOLD(Jane) = 0.64. Other common hedges are "more or less" [typically SQRT(mA(x))], "somewhat," "rather," "sort of," and so on. Again, their definition is entirely subjective, but their operation is consistent: they serve to transform membership/truth values in a systematic manner according to standard mathematical functions. A more involved approach to hedges is best shown through the work of Wenstop in his attempt to model organizational behavior. For his study, he constructed arrays of values for various terms, either as vectors or matrices. Each term and hedge was represented as a 7-element vector or 7x7 matrix. He ten intuitively assigned each element of every vector and matrix a value between 0.0 and 1.0, inclusive, in what he hoped was intuitively a consistent manner. For example, the term "high" was assigned the vector 0.0 0.0 0.1 0.3 0.7 1.0 1.0and "low" was set equal to the reverse of "high," or 1.0 1.0 0.7 0.3 0.1 0.0 0.0Wenstop was then able to combine groupings of fuzzy statements to create new fuzzy statements, using the APL function of Max-Min matrix multiplication. These values were then translated back into natural language statements, so as to allow fuzzy statements as both input to and output from his simulator. For example, when the program was asked to generate a label "lower than sortof low," it returned "very low;" "(slightly higher) than low" yielded "rather low," etc. The point of this example is to note that algorithmic procedures can be devised which translate "fuzzy" terminology into numeric values, perform reliable operations upon those values, and then return natural language statements in a reliable manner. Similar techniques have been adopted by others, primarily in the study of fuzzy systems as applicable to linguistic approximation (e.g. , , ). APL appears to be the language of choice, owing to its flexibility and power in matrix operations. The first area Haack defines is that of the nature of Truth and Falsity: if it could be shown, she maintains, that these are fuzzy values and not discrete ones, then a need for fuzzy logic would have been demonstrated. The other area she identifies is that of fuzzy systems' utility: if it could be demonstrated that generalizing classic logic to encompass fuzzy logic would aid in calculations of a given sort, then again a need for fuzzy logic would exist. In regards to the first statement, Haack argues that True and False are discrete terms. For example, "The sky is blue" is either true or false; any fuzziness to the statement arises from an imprecise definition of terms, not out of the nature of Truth. As far as fuzzy systems' utility is concerned, she maintains that no area of data manipulation is made easier through the introduction of fuzzy calculus; if anything, she says, the calculations become more complex. Therefore, she asserts, fuzzy logic is unnecessary. Fox has responded to her objections, indicating that there are three areas in which fuzzy logic can be of benefit: as a "requisite" apparatus (to describe real-world relationships which are inherently fuzzy); as a "prescriptive" apparatus (because some data is fuzzy, and therefore requires a fuzzy calculus); and as a "descriptive" apparatus (because some inferencing systems are inherently fuzzy). His most powerful arguments come, however, from the notion that fuzzy and classic logics need not be seen as competitive, but complementary. He argues that many of Haack's objections stem from a lack of semantic clarity, and that ultimately fuzzy statements may be translatable into phrases which classical logicians would find palatable. Lastly, Fox argues that despite the objections of classical logicians, fuzzy logic has found its way into the world of practical applications, and has proved very successful there. He maintains, pragmatically, that this is sufficient reason for continuing to develop the field. The objection has been raised that utilizing fuzzy systems in a dynamic control environment raises the likelihood of encountering difficult stability problems: since in control conditions the use of fuzzy systems can roughly correspond to using thresholds, there must be significant care taken to insure that oscillations do not develop in the "dead spaces" between threshold triggers. This seems to be an important area for future research. Other applications which have benefited through the use of fuzzy systems theory have been information retrieval systems, a navigation system for automatic cars, a predicative fuzzy-logic controller for automatic operation of trains, laboratory water level controllers, controllers for robot arc-welders, feature-definition controllers for robot vision, graphics controllers for automated police sketchers, and more. Expert systems have been the most obvious recipients of the benefits of fuzzy logic, since their domain is often inherently fuzzy. Examples of expert systems with fuzzy logic central to their control are decision-support systems, financial planners, diagnostic systems for determining soybean pathology, and a meteorological expert system in China for determining areas in which to establish rubber tree orchards . Another area of application, akin to expert systems, is that of information retrieval .
We consider a massless scalar field in a curved spacetime with metric . The field satisfies the wave equation We let be the retarded solution to Eq. (14.3), and is the advanced solution; when viewed as functions of , is nonzero in the causal future of , while is nonzero in its causal past. We assume that the retarded and advanced Green’s functions exist as distributions and can be defined globally in the entire spacetime. Assuming throughout this subsection that is restricted to the normal convex neighbourhood of , we make the ansatz Before we substitute the Green’s functions of Eq. (14.4) into the differential equation of Eq. (14.3) we proceed as in Section 12.6 and shift by the small positive quantity . We shall therefore consider the distributions and later recover the Green’s functions by taking the limit . Differentiation of these objects is straightforward, and in the following manipulations we will repeatedly use the relation satisfied by the world function. We will also use the distributional identities , , and . After a routine calculation we obtain According to Eq. (14.3), the right-hand side of Eq. (14.5) should be equal to . This immediately gives us the coincidence condition Recall from Section 3.3 that is a vector at that is tangent to the unique geodesic that connects to . This geodesic is affinely parameterized by and a displacement along is described by . The first term of Eq. (14.7) therefore represents the logarithmic rate of change of along , and this can be expressed as . For the second term we recall from Section 7.1 the differential equation satisfied by , the van Vleck determinant. This gives us , and Eq. (14.7) becomes It follows that is constant on , and this must therefore be equal to its value at the starting point : , by virtue of Eq. (14.6) and the property of the van Vleck determinant. Because this statement must be true for all geodesics that emanate from , we have found that the unique solution to Eqs. (14.6) and (14.7) is We must still consider the remaining terms in Eq. (14.5). The term can be eliminated by demanding that its coefficient vanish when . This, however, does not constrain its value away from the light cone, and we thus obtain information about only. Denoting this by – the restriction of on the light cone – we have Eqs. (7.4) and (14.8) imply that near coincidence, admits the expansion Eqs. (14.9) and (14.13) give us a means to construct , the restriction of on the null cone . These values can then be used as characteristic data for the wave equation To summarize: We have shown that with given by Eq. (14.8) and determined uniquely by the wave equation of Eq. (14.14) and the characteristic data constructed with Eqs. (14.9) and (14.13), the retarded and advanced Green’s functions of Eq. (14.4) do indeed satisfy Eq. (14.3). It should be emphasized that the construction provided in this subsection is restricted to , the normal convex neighbourhood of the reference point . We shall now establish the following reciprocity relation between the (globally defined) retarded and advanced Green’s functions: To prove the reciprocity relation we invoke the identities and take their difference. On the left-hand side we have while the right-hand side gives Integrating both sides over a large four-dimensional region that contains both and , we obtain where is the boundary of . Assuming that the Green’s functions fall off sufficiently rapidly at infinity (in the limit ; this statement imposes some restriction on the spacetime’s asymptotic structure), we have that the left-hand side of the equation evaluates to zero in the limit. This gives us the statement , which is just Eq. (14.15) with replacing . Suppose that the values for a scalar field and its normal derivative are known on a spacelike hypersurface . Suppose also that the scalar field satisfies the homogeneous wave equation To establish this result we start with the equations in which and refer to arbitrary points in spacetime. Taking their difference gives and this we integrate over a four-dimensional region that is bounded in the past by the hypersurface . We suppose that contains and we obtain where is the outward-directed surface element on the boundary . Assuming that the Green’s function falls off sufficiently rapidly into the future, we have that the only contribution to the hypersurface integral is the one that comes from . Since the surface element on points in the direction opposite to the outward-directed surface element on , we must change the sign of the left-hand side to be consistent with the convention adopted previously. With this change we have which is the same statement as Eq. (14.18) if we take into account the reciprocity relation of Eq. (14.15). In Part IV of this review we will compute the retarded field of a moving scalar charge, and we will analyze its singularity structure near the world line; this will be part of our effort to understand the effect of the field on the particle’s motion. The retarded solution to the scalar wave equation is the physically relevant solution because it properly incorporates outgoing-wave boundary conditions at infinity – the advanced solution would come instead with incoming-wave boundary conditions. The retarded field is singular on the world line because a point particle produces a Coulomb field that diverges at the particle’s position. In view of this singular behaviour, it is a subtle matter to describe the field’s action on the particle, and to formulate meaningful equations of motion. When facing this problem in flat spacetime (recall the discussion of Section 1.3) it is convenient to decompose the retarded Green’s function into a singular Green’s function and a regular two-point function . The singular Green’s function takes its name from the fact that it produces a field with the same singularity structure as the retarded solution: the diverging field near the particle is insensitive to the boundary conditions imposed at infinity. We note also that satisfies the same wave equation as the retarded Green’s function (with a Dirac functional as a source), and that by virtue of the reciprocity relations, it is symmetric in its arguments. The regular two-point function, on the other hand, takes its name from the fact that it satisfies the homogeneous wave equation, without the Dirac functional on the right-hand side; it produces a field that is regular on the world line of the moving scalar charge. (We reserve the term “Green’s function” to a two-point function that satisfies the wave equation with a Dirac distribution on the right-hand side; when the source term is absent, the object is called a “two-point function”.) Because the singular Green’s function is symmetric in its argument, it does not distinguish between past and future, and it produces a field that contains equal amounts of outgoing and incoming radiation – the singular solution describes a standing wave at infinity. Removing from the retarded Green’s function will have the effect of removing the singular behaviour of the field without affecting the motion of the particle. The motion is not affected because it is intimately tied to the boundary conditions: If the waves are outgoing, the particle loses energy to the radiation and its motion is affected; if the waves are incoming, the particle gains energy from the radiation and its motion is affected differently. With equal amounts of outgoing and incoming radiation, the particle neither loses nor gains energy and its interaction with the scalar field cannot affect its motion. Thus, subtracting from the retarded Green’s function eliminates the singular part of the field without affecting the motion of the scalar charge. The subtraction leaves behind the regular two-point function, which produces a field that is regular on the world line; it is this field that will govern the motion of the particle. The action of this field is well defined, and it properly encodes the outgoing-wave boundary conditions: the particle will lose energy to the radiation. In this subsection we attempt a decomposition of the curved-spacetime retarded Green’s function into singular and regular pieces. The flat-spacetime relations will have to be amended, however, because of the fact that in a curved spacetime, the advanced Green’s function is generally nonzero when is in the chronological future of . This implies that the value of the advanced field at depends on events that will unfold in the future; this dependence would be inherited by the regular field (which acts on the particle and determines its motion) if the naive definition were to be adopted. We shall not adopt this definition. Instead, we shall follow Detweiler and Whiting and introduce a singular Green’s function with the properties Properties S1 and S2 ensure that the singular Green’s function will properly reproduce the singular behaviour of the retarded solution without distinguishing between past and future; and as we shall see, property S3 ensures that the support of the regular two-point function will not include the chronological future of . The regular two-point function is then defined by Property R1 follows directly from Eq. (14.22) and property S1 of the singular Green’s function. Properties R2 and R3 follow from S3 and the fact that the retarded Green’s function vanishes if is in past of . The properties of the regular two-point function ensure that the corresponding regular field will be nonsingular at the world line, and will depend only on the past history of the scalar charge. We must still show that such singular and regular Green’s functions can be constructed. This relies on the existence of a two-point function that would possess the properties With a biscalar satisfying these relations, a singular Green’s function defined byS1 comes as a consequence of H1 and the fact that both the advanced and the retarded Green’s functions are solutions to the inhomogeneous wave equation, S2 follows directly from H2 and the definition of Eq. (14.30), and S3 comes as a consequence of H3, H4 and the properties of the retarded and advanced Green’s functions. The question is now: does such a function exist? We will present a plausibility argument for an affirmative answer. Later in this section we will see that is guaranteed to exist in the local convex neighbourhood of , where it is equal to . And in Section 14.6 we will see that there exist particular spacetimes for which can be defined globally. To satisfy all of H1 – H4 might seem a tall order, but it should be possible. We first note that property H4 is not independent from the rest: it follows from H2, H3, and the reciprocity relation (14.15) satisfied by the retarded and advanced Green’s functions. Let , so that . Then by H2, and by H3 this is equal to . But by the reciprocity relation this is also equal to , and we have obtained H4. Alternatively, and this shall be our point of view in the next paragraph, we can think of H3 as following from H2 and H4. Because satisfies the homogeneous wave equation (property H1), it can be given the Kirkhoff representation of Eq. (14.18): if is a spacelike hypersurface in the past of both and , then where is a surface element on . The hypersurface can be partitioned into two segments, and , with denoting the intersection of with . To enforce H4 it suffices to choose for initial data on that agree with the initial data for the advanced Green’s function; because both functions satisfy the homogeneous wave equation in , the agreement will be preserved in the entire domain of dependence of . The data on is still free, and it should be possible to choose it so as to make symmetric. Assuming that this can be done, we see that H2 is enforced and we conclude that the properties H1, H2, H3, and H4 can all be satisfied. When is restricted to the normal convex neighbourhood of , properties H1 – H4 imply thatS2), and that its support excludes , in which (property S3). From Eq. (14.22) we get an analogous expression for the regular two-point function: R2), and that its support does not include , in which (property R3). To illustrate the general theory outlined in the previous subsections we consider here the specific case of a minimally coupled () scalar field in a cosmological spacetime with metric; it can be extended to other cosmologies . To solve Green’s equation we first introduce a reduced Green’s function defined by Substitution of Eq. (14.39) into Eq. (14.38) reveals that must satisfy the homogeneous equation Eq. (14.40) has and as linearly independent solutions, and must be given by a linear superposition. The coefficients can be functions of , and after imposing Eqs. (14.41) we find that the appropriate combination is after integration by parts. The integral evaluates to . We have arrived at It may be verified that the symmetric two-point functionH1 – H4 listed in Section 14.5; it may thus be used to define singular and regular Green’s functions. According to Eq. (14.30) the singular Green’s function is given by As a final observation we note that for this cosmological spacetime, the normal convex neighbourhood of any point consists of the whole spacetime manifold (which excludes the cosmological singularity at ). The Hadamard construction of the Green’s functions is therefore valid globally, a fact that is immediately revealed by Eqs. (14.44) and (14.45). Living Rev. Relativity 14, (2011), 7 This work is licensed under a Creative Commons License.
Kinetic theory for particle production Recently, the phenomenological description of cosmological particle production processes in terms of effective viscous pressures has attracted some attention. Using a simple creation rate model we discuss the question to what extent this approach is compatible with the kinetic theory of a relativistic gas. We find the effective viscous pressure approach to be consistent with this model for homogeneous spacetimes but not for inhomogeneous ones. PACS numbers: 98.80.Hw, 05.20.Dd, 04.40.Nr, 95.30.Tg It is well known [1 - 3] and widely used [4 - 10] that particle production processes in the expanding Universe may be phenomenologically described in terms of effective viscous pressures. This is due to the simple circumstance that any source term in the energy balance of a relativistic fluid may be formally rewritten in terms of an effective bulk viscosity. The advantage of this rewriting is obvious. While an energy momentum balance with a nonzero source term violates the integrability conditions of Einstein’s equations, the energy momentum balance for a viscous fluid does not. Consequently, if it is possible to mimic particle production processes consistently by effective viscous pressures, one is able to study the impact of these processes on the cosmological dynamics. However, the question remains whether an energy momentum tensor that formally has been made divergence free by regarding the original source term as an effective viscous pressure of the cosmic medium, adequately describes the physics of the underlying process. Is the dynamical effect of a viscous pressure in any respect equivalent to the corresponding effect of particle production? There is no definite answer to these questions at the present state of knowledge. What one would like to have is a microscopic justification of this effective viscous pressure approach. The production processes one has in mind are supposed to play a role at times of the order of the Planck time or during the reheating phase of inflationary scenarios. It is not unlikely that particle or string creation out of the quantum vacuum has considerably influenced the dynamics of the early Universe [11, 10, 8]. A real microscopic description on the quantum level, however, does not seem to be available in the near future, and therefore it may be useful to study this phenomenon as far as possible on different classical levels. Taking into account that the dynamics of a fluid in or close to equilibrium may be derived from kinetic theory, one may ask to what extent the phenomenological approach of regarding the effects of particle production to be in a sense equivalent to those of viscous pressures may be understood and backed up on the level of kinetic theory. It is this question that we want to address in the present paper. Specifically, we shall model particle production processes by a nonvanishing source term in a Boltzmann type equation and establish the link between this kinetic description and the effective viscous pressure approach on the fluid level. It will turn out that the simple phenomenological approach is compatible with the kinetic theory in homogeneous spaces but not in inhomogeneous ones. One may be reluctant, of course, to assume the applicability of a kinetic approach in the early Universe. A gas, however, is the only system for which the correspondence between microscopic variables, governed by a distribution function, and phenomenological fluid quantities is sufficiently well understood. All considerations of this paper refer to a model universe for which the kinetic approach is assumed to be applicable. It is the hope that this idealized model nevertheless shares at least some of the basic features of our real Universe. For conventional collisions the standard kinetic theory (see, e.g., ) is valid if the particles interact only weakly, i.e., for systems that are not too dense. On the other hand, we know from the theory of the strong interaction that because of the property of ‘asymptotic freedom’ the concept of weakly interacting particles is especially appropriate for very dense systems. This might indicate that beyond its conventional range of application, e.g., under the conditions of the early Universe, the hypothesis of weakly interacting particles on the level of kinetic theory may be not too misleading as well. In section 2 the basic elements of kinetic theory needed in our study are introduced. In section 3 our main result is derived in the context of an ‘effective rate approximation’, modeled after relaxation time approximations for the Boltzmann collision term. In section 4 we summarize our conclusions. Units have been chosen so that . 2 General kinetic theory Our starting point is the assumption that a change in the number of particles of a relativistic gas should manifest itself in a source term on the level of kinetic theory. The corresponding one-particle distribution function of a relativistic gas with varying particle number is supposed to obey the equation where is the number of particles whose world lines intersect the hypersurface element around , having 4-momenta in the range ; , , … = , , is the volume element on the mass shell in the momentum space. , if is future directed and otherwise; . is the Boltzmann collision term. Its specific structure discussed e.g. by Ehlers will not be relevant for our considerations. Following Israel and Stewart we shall only require that (i) is a local function of the distribution function, i.e., independent of derivatives of , (ii) is consistent with conversation of 4-momentum and number of particles, and (iii) yields a nonnegative expression for the entropy production and does not vanish unless has the form of a local equilibrium distribution (see (9)). The term on the r.h.s. of (1) takes into account the fact that the number of particles whose world lines intersect a given hypersurface element within a certain range of momenta may additionally change due to creation or decay processes, supposedly of quantum origin. On the level of classical kinetic theory we shall regard this term as a given input quantity. Later we shall give an example for the possible functional structure of . By the splitting of the r.h.s of eq.(1) into and we have separated the collisional from the creation (decay) events. In this setting collisions are not accompanied by creation or annihilation processes. In other words, once created, the interactions between the particle are both energy-momentum and number preserving. Situations like these might be typical, e.g., for the creation of relativistic particles out of a decaying scalar field during the reheating phase of standard inflationary universes . For a vanishing eq.(1) reduces to the familiar Boltzmann equation (see, e.g., [12 - 14]). In the latter case there exist elaborate solution techniques that allow to characterize both the equilibrium and nonequilibrium phenomena, connected with entropy production. In the present paper we shall not study the entropy production due to collisions within the fluid represented by the term on the r.h.s. of (1), although the corresponding effects may be additionally included at any stage of our presentation in an obvious way. Instead, we shall focus here on the entropy production due to the source term in equation (1). The particle number flow 4-vector and the energy momentum tensor are defined in a standard way (see, e.g., ) as where bosons and fermions are characterized by and , respectively. For one recovers the classical Maxwell-Boltzmann case. Using the general relationship and eq.(1) we find In collisional equilibrium, which we shall assume from now on, in (8) is a linear combination of the collision invariants and . The corresponding equilibrium distribution function becomes (see, e.g., ) is a timelike vector that depends on only. Inserting the equilibrium function into eq.(1) one gets It is well known [12, 13] that for this equation, which characterizes the ‘global equilibrium’, admits solutions only for very special cases in which and is a timelike Killing vector. Below we shall discuss the corresponding conditions for a specific functional form of in the classical case. Neither nor are conserved. The nonconservation of the number of particles, the phenomenon under consideration here, is accompanied by the existence of a source term in the energy-momentum balance as well. A nonconserved energy-momentum tensor , however, is incompatible with Einstein’s field equations. But one may raise the question whether it is possible to rewrite eq.(12) as with an effective energy-momentum tensor that is conserved and, consequently, is a suitable quantity in Einstein’s equations. Is is basically this problem that we are going to investigate below. In collisional equilibrium there is entropy production only due to the source term . From eq.(8) we obtain Condition (10) may be imposed or not. The first possibility corresponds to the ‘global equilibrium’ case for , the second one to the ‘local equilibrium’ case. In collisional equilibrium with replaced by in (2), (3) and (4), , and may be split with respect to the unique 4-velocity according to with the spatial projection tensor and where is the particle number density, is the energy density, is the equilibrium pressure and is the entropy per particle. The exact integral expressions for , , and are given by the formulae (177) - (180) in . According to our general setting described above we assumed in establishing (15) - (18) that the source term does not affect the collisional equilibrium. Using (16) and defining It is obvious that is the particle production rate. Similarly, with the decomposition (17) and the abreviation the balance equations for energy and momentum may be written as respectively, where is the fluid expansion. From the Gibbs equation and (26) may be rewritten to yield where we have used (25) and . The latter expression for the entropy production density coincides result that follows from differentiating The first term on the r.h.s. of (29) describes the entropy due to the enlargement of the phase space. The second one takes into account the contribution due to a possible change in the entropy per Henceforth we shall assume, that all particles are amenable to a perfect fluid description immediately after their creation. In this case the entropy per particle does not change, i.e., . All particles are created with a fixed entropy. Of course, it is always possible to include dissipative effects in a standard way . According to (25) the case is equivalent to relating the source term in the energy balance to that in the particle number balance. The entropy production density (29) reduces to In the perfect fluid case there is entropy production due to the increase in the number of particles, i.e., due to the enlargement of the phase space. Introducing the quantity it is possible to rewrite the energy balance (22) as i.e., enters the energy balance in the same way as a bulk viscous pressure does. By (30) is related to the particle production rate in the case . It had been the hope that a rewriting like this, corresponding to the introduction of an effectively conserved energy-momentum tensor (see (13)) instead of the nonconserved quantity of (17), allowed one to study particle production processes in terms of effective viscous pressures. Assuming an energy-momentum tensor of the structure (34) is equivalent to map the entire source term in (12) onto an effective bulk viscous pressure. It should be mentioned that this is not the only way to take into account the influence of particle number nonconserving processes on the cosmological dynamics. Gariel and Le Denmat have developed a different approach that regards the particle production rate as a thermodynamic flux on its own. that would lead to If it were generally true that particle production may be modelled by an effective bulk viscosity, the splitting (35) should be possible. To clarify this problem, specific expressions for in (1) have to be investigated, while up to this point all relations are valid for any . 3 Effective rate approximation 3.1 The basic concept As was mentioned earlier, the quantity is an input quantity on the level of classical kinetic theory. is supposed to represent the net effect of certain quantum processes with variable particle numbers (see, e.g., ) at the interface to the classical (nonquantum) level of description. Lacking a better understanding of these processes we shall assume that their influence on the distribution function may be approximately described by a linear coupling to the latter: Let us further assume that depends on the momenta only linearly: , or equivalently and , characterize the rate of change of the distribution function due to the underlying processes with variable particle numbers. The restriction to a linear dependence of on the momentum is equivalent to the requirement that these processes couple to the particle number flow vector and to the energy momentum tensor in the balances for these quantities only, but not to higher moments of the distribution function. This ‘effective rate approximation’ is modeled after the relaxation time approximations for the Boltzmann collision term [19-21]. While the physical situations in both approaches are very different, their common feature is the simplified description of nonequilibrium phenomena by a linear equation for the distribution function in terms of some effective functions of space and time that characterize the relevant scales of the process under consideration. In the relaxation time approximation this process is determined by the rate at which the system relaxes to an equilibrium state. In the present case the corresponding quantity is the rate by which the number of particles changes. 3.2 The Maxwell-Boltzmann case corresponding to ‘global equilibrium’ For a classical gas with the specific ansatz (37) with (38) allows us to find conditions on the parameters and in (9) that replace the so called global equilibrium conditions (=const., - timelike Killing vector) in the case . Condition (10) with becomes where . Decomposing according to , where is a unit spatial vector, i.e., , , the mass shell condition is equivalent to . The conditions on turn out to be while obeys the equation for a conformal Killing vector Different from the case where has to be constant in space and time, is only spatially constant in the present case but changes along the fluid flow lines. From (45) it follows that for . The condition for , however, is equivalent to , i.e., the corresponding spacetime is stationary. Using in (44) yields By scalar multiplication with one obtains and projecting the latter equation in direction of , the relation On the other hand, the trace of (47) is equivalent to Projecting now (48) orthogonally to provides us with the following representation of the acceleration in terms of the spatial temperature gradient: This relation for a Maxwell-Boltzmann gas has an interesting consequence that will be discussed at the end of the following section. 3.3 The balance equations for our model, where and is the zeroth moment of the distribution function: for the source term in the energy momentum balance, or, with (32), It is obvious from (53) and (57) that our effective rate approximation provides us with a definite coupling between the functions and that represent the microscopic production process on a macroscopic level, and the components of the particle number flow vector and the energy momentum tensor . The expression (38) for ensures that the source terms (53) and (57) depend on , and only, but not on higher moments of the distribution function. This equation provides us with a relation between the functions and in the ansatz (38) for . At this stage it becomes clear why it is necessary to include a function in (38), although intuitively one might have started with . In the latter case the condition seems to require , i.e., our effective rate aproximation seems to apply to a presureless medium (dust) only if is assumed to be given. But exactly is an unphysical limiting case (see the integral expression (179) for in ). Consequently, and are only compatible for . There is no particle production at all in this case. For a Maxwell-Boltzmann gas () a further statement is possible if eq.(45) is imposed. A vanishing in (45) neccessarily implies . The obvious conclusion is that massless particles cannot be produced under this requirement. Since the relation (51) already holds for massless particles with , this result is hardly surprising. There is no corresponding restriction, however, in the local equilibrium-like case. We conclude that as far as the balances of particle number, energy and entropy are concerned, the description of particle production in terms of an effective viscous pressure is backed up by our effective rate approximation of kinetic theory. This is no longer true, however, if the momentum balance (23) comes into play. Eqs. (16), (17) and (57) imply The momentum balance is completely unaffected by the particle production rate. Consequently, in this respect the particle production is not equivalent to an effective viscous pressure. While the simple analogy holds in the homogeneous case, the matter is more subtle if there are nonvanishing spatial gradients and the fluid motion is no longer geodesic. If one wants to use effective viscous pressures nevertheless to characterize particle production processes this is consistent only under the additional assumption that the effective viscous pressure terms cancel in the momentum balance. It is obvious that this condition is equivalent to since (23) in this case reduces to Using this equation to eliminate from (66) one gets While this condition is empty in a homogeneous spacetime it restricts a possible spatial dependence of . Especially, it states that a spatially independent necessarily implies a spatially independent , i.e., a homogeneous creation rate. In other words, it is impossible to have a homogeneous equilibrium pressure and at the same time an effective viscous pressure due to particle production that has a nonvanishing spatial gradient. Alternatively, the restriction on may be interpreted geometrically. For a comoving observer the spatial part of the 4-acceleration becomes and for the rotation free case with On the other hand, (66) is equivalent to Consequently, the spatial dependence of is completely determined by the spatial dependence of : or in terms of if one uses (63). for a comoving observer. The latter formula for a Maxwell-Boltzmann gas replaces the Tolman relation if particles with are produced. 3.4 Stephani universes and particle production Finally, we shall briefly discuss our results concerning the effective viscous pressure in connection with recent attempts to find a physical understanding of exact inhomogeneous solutions of Einstein’s field equations with irrotational, shear-free, perfect fluid sources (Stephani-Barnes family [24, 25]). Since there do not generally exist physically realistic equations of state for the latter family, Sussman suggested a reinterpretation of these solutions replacing the perfect fluid source by a fluid with an isotropic bulk stress. The main advantage of this procedure lies in the introduction of an additional degree of freedom that might be helpful in finding reasonable equations of state. We assume the energy-momentum tensor to be given by (34) where is to describe particle production processes in the manner discussed so far. While (34) was derived for a bounded system with a finite particle number we shall follow here the common procedure and apply a quantity like this to describe that part of the early Universe that developed into its presently visible part. In comoving coordinates the metric of the Stephani-Barnes family is [22, 25] where and . Generally, for a bulk viscous fluid the balances of energy and momentum are given by (33) and (36). If mimics the effect of matter creation then eq.(36) reduces to (67), implying (68). With respectively. According to the result for our effective rate approximation, enters the energy balance but not the momentum balance. At first sight this seems to be contradictory to Sussman’s procedure who replaced the of the perfect fluid source by in both balance equations (his equations (3b) and (3c) in ). However, because of (68), there exists the additional relation which is equivalent to where is an arbitrary function. is Sussman’s relation (3d) in . Originally derived for a perfect fluid source, this result remains valid both if is a ‘real’ bulk viscous pressure and if it mimics matter creation as in the The basic intention of Sussman’s reinterpretation is to establish a ‘-law’ for the equilibrium pressure and absorb all terms that do not fit into an equation of state like this into the quantity . With (81) the latter is then defined by This is again a formal procedure and one has to clarify whether viscous pressures like this are physically meaningful. Let us consider as an example the subfamily of conformally flat (Petrov type-0) solutions, known as Stephani universes , with where , , , and are arbitrary functions of time. With a perfect fluid source, matter density and pressure are given by where is determined by The Stephani universes are solutions with a homogeneous but with an inhomogeneous pressure for which, in general, there does not exist a ‘-law’. Applying Sussman’s procedure to the present case amounts to replacing by in eq.(89): Since is homogeneous, establishing a ‘-law’ between is equivalent to separate the homogenous part of the r.h.s of (91) and to relate all the inhomogeneities to . However, clearly inconsistent with (68), or (82) and (83) in our specific case. We conclude that a physical interpretation of the Stephani universes with a ‘-law’ ( constant) and a viscous pressure that mimics particle production processes is impossible according to our ‘effective rate aproximation’. 4 Concluding remarks In the present paper we have investigated the question whether the widely used description of particle production processes in the early Universe in terms of effective viscous pressures is compatible with the kinetic theory of a simple quantum gas. Our conclusion is that this approach may be applied, provided the effective viscous pressure is subject to an additional condition which entirely fixes its spatial dependence. This condition is equivalent to the requirement that all effects due to particle production cancel in the momentum balance. The latter property is a consequence of a simple effective rate approximation to the source term in the Boltzmann equation that is supposed to describe particle production. While the condition on the spatial behaviour of the effective viscous pressure is empty in a homogeneous universe, it excludes the possibility of models with an inhomogeneous as long as the thermodynamic pressure remains homogeneous. An explicit example has been given, showing that the Stephani universes are incompatible with a bulk viscous pressure associated to matter creation. We thank Roberto Sussman and Roy Maartens for useful discussions. This paper was supported by the Spanish Ministery of Education under grant 90-676. J.T. acknowledges the support of the FPI program. Zel’dovich Ya B 1970 Sov.Phys.JETP Lett. 12 307 Murphy G L 1973 Phys.Rev. D8 4231 Hu B L 1982 Phys.Lett.A 90 375 Prigogine I Geheniau J Gunzig E and Nardone P 1989 GRG 21 767 Calvão M O Lima J A S and Waga I 1992 Phys.Lett A 162 223 Zimdahl W and Pavón D 1993 Phys.Lett A 176 57 Zimdahl W Pavón D and Jou D 1993 Class. Quantum Grav. 10 1775 Zimdahl W and Pavón D 1994 Mon. Not. R. Astr. Soc. 266 872 Zimdahl W and Pavón D 1994 GRG 26 1259 Barrow J D 1988 Nucl.Phys.B 310 743 Turok N 1988 Phys. Rev. Lett. 60 549 Ehlers J 1971 in General Relativity and Cosmology ed by Sachs B K Academic Press New York Israel W and Stewart J M 1979 Ann Phys 118 341 de Groot S R van Leeuwen W A and van Weert Ch G 1980 Relativistic Kinetic Theory North Holland Amsterdam Stewart J M 1971 Non-equilibrium Relativistic Kinetic Theory Springer New York Triginer J and Pavón D 1994 GRG 26 513 Gariel J and Le Denmat G 1995 Phys Lett A 200 11 Birrell D N and Davis P C W 1982 Quantum fields in curved space Cambridge University Press Cambridge Marle C 1969 Ann. Inst. H. Poincáre 10 67 Anderson J L and Witting H R 1974 Physica 74 466 Maartens R and Wolvaardt F P 1994 Class. Quantum Grav. 11 203 Coley A A and Tupper B O J 1990 GRG 22 241 Sussman R 1994 Class. Quantum Grav. 11 1445 Stephani H 1967 Commun. Math. Phys. 4 137 Barnes A 1973 GRG 4 105 Krasiński A Physics in an inhomogeneous universe Ch.IV, preprint, Warsaw 1993
Types of Statistical Errors and What They Mean. Type I Errors occur when we reject a null hypothesis that is actually true; the probability of this occurring is denoted by alpha (a). Type II Errors are when we accept a null hypothesis that is actually false; its probability is called beta (b). Read, more on it here. Beside this, what are the types of errors? There are three types of error: syntax errors, logical errors and run-time errors. (Logical errors are also called semantic errors). We discussed syntax errors in our note on data type errors. Generally errors are classified into three types: systematic errors, random errors and blunders. Subsequently, question is, which is worse Type 1 or Type 2 error? A conclusion is drawn that the null hypothesis is false when, in fact, it is true. Therefore, Type I errors are generally considered more serious than Type II errors. The more an experimenter protects himself or herself against Type I errors by choosing a low level, the greater the chance of a Type II error. Similarly, it is asked, what is a Type 2 error in statistics? A type II error is a statistical term referring to the non-rejection of a false null hypothesis. It is used within the context of hypothesis testing. In other words, it produces a false positive. The error rejects the alternative hypothesis, even though it does not occur due to chance. What are the error in measurement? Definition: The measurement error is defined as the difference between the true or actual value and the measured value. The true value is the average of the infinite number of measurements, and the measured value is the precise value. What are sources of error? Common sources of error include instrumental, environmental, procedural, and human. All of these errors can be either random or systematic depending on how they affect the results. Instrumental error happens when the instruments being used are inaccurate, such as a balance that does not work (SF Fig. What causes random error? Random error is always present in a measurement. It is caused by inherently unpredictable fluctuations in the readings of a measurement apparatus or in the experimenter’s interpretation of the instrumental reading. They can be estimated by comparing multiple measurements, and reduced by averaging multiple measurements. What exactly is an error? An error (from the Latin error, meaning “wandering”) is an action which is inaccurate or incorrect. In some usages, an error is synonymous with a mistake. In statistics, “error” refers to the difference between the value which has been computed and the correct value. What is method error? Method error is the discrepancy that may occur in measurement such that the value obtained during the process of measurement is different from the actual value. This may arise either because of a defect in the measuring device or other non-mechanical causes. What are the three types of errors? There are three kinds of errors: syntax errors, runtime errors, and logic errors. These are errors where the compiler finds something wrong with your program, and you can’t even try to execute it. For example, you may have incorrect punctuation, or may be trying to use a variable that hasn’t been declared. What is a null hypothesis example? A null hypothesis is a hypothesis that says there is no statistical significance between the two variables in the hypothesis. In the example, Susie’s null hypothesis would be something like this: There is no statistically significant relationship between the type of water I feed the flowers and growth of the flowers. What are the four types of errors? Systematic errors may be of four kinds: What is Type 2 error example? A Type II error is committed when we fail to believe a true condition. Candy Crush Saga. Continuing our shepherd and wolf example. Again, our null hypothesis is that there is “no wolf present.” A type II error (or false negative) would be doing nothing (not “crying wolf”) when there is actually a wolf present. What is T test used for? A t–test is a type of inferential statistic used to determine if there is a significant difference between the means of two groups, which may be related in certain features. How do you write a null hypothesis? To write a null hypothesis, first start by asking a question. Rephrase that question in a form that assumes no relationship between the variables. In other words, assume a treatment has no effect. Write your hypothesis in a way that reflects this. What is an error in statistics? Definition: A statistical error is the (unknown) difference between the retained value and the true value. Context: It is immediately associated with accuracy since accuracy is used to mean “the inverse of the total error, including bias and variance” (Kish, Survey Sampling, 1965). What affects Type 2 error? The power of a hypothesis test is affected by three factors. Sample size (n). Other things being equal, the greater the sample size, the greater the power of the test. Significance level (α). This means you are less likely to reject the null hypothesis when it is false, so you are more likely to make a Type II error.
Factors affecting resistance wire Variables affecting electrical resistance the total length of the wires will affect the amount of resistance the longer the wire, the more resistance that there. Class practical a simple investigation of the factors affecting the resistance of a wire. Factors affecting resistance mr baker factors affecting the resistance resistance of a wire. Length of the wire resistance is directly proportional to the length of the wire more the length of the wire, longer is the distance to be covered by the electrons. To investigate the factors of affecting the resistance of a wire ohm's law aim: to investigate the factors of affecting the resistance of a wire voltage is the electrical force, or pressure, that causes current to flow in a circuit voltmeter's range until i reach 80cm, and take readings each time. What's the factors affecting resistance of a wire and why thank you. 9 rows learn about the physics of resistance in a wire change its resistivity. I shall look at various factors which can affect the resistance of a wire and investigate one of these in greater detail i shall start off with theory on resistance and use it to make a prediction and then plan and carry out an experiment to obtain enough evidence to prove or disprove my prediction. Factors affecting the resistance of wire 1883 words \| 8 pages aim: to investigate the factors that affect the resistance in a conductor the main factors that affect the resistance in a conductor are: length temperature cross sectional area material magnetism the factor that we are going to change is the cross sectional area. Thanks for a2a there is a famous formula it depends on 4 factors 1 length 2 area of cross section 3 resistivity-can be changed by changing material 4 temperature explanation: when you consider one material for wire resistivity is fixed,beca. The diameter of the wire will have an affect on the resistance of a wire this is because the electrons have to squeeze together more to pass through a thin wire than they do to pass through a thick wire i can predict that the. I am going to investigate what factors affect the resistance of a wire there are three main factors which affect the resistance of a wire. Thickness (diameter, circumference, radius or area) of wire: the cross sectional area of a piece of wire greatly affects the resistance if the wire is very thick it will allow a high current through because there is more room for more electrons to transfer the current. An investigation into the resistance of resistance depends on four main factors: how manipulating these variables affect the resistance of the wire. Variables affecting electrical resistance 1 the total length of the wires will affect the amount of resistance the longer the wire, the more resistance that there will be there is a direct relationship between the amount of resistance encountered by charge and the length of wire it must traverse. Essay about factors affecting the resistance of a wire - factors affecting the resistance of a wire the aim of this experiment is to investigate one factor that affect the resistance of a wire i will do this by performing an experiment. Factors affecting resistance 1 factors affecting resistancereported by: ralph lery guerrero kevin roxas marco lauro delos santos 2 resistance• is defined as an obstacle to the flow of electric current• is the opposition offered by any object to the passage of an electric current through it 3. The resistance of a typical conducting wire is low when temperature is how does temperature affect the resistance of a factors affecting resistance in a wire. Length, cross-sectional area, kind of material, resistivity and temperature: the ir re latio nships factors affectingfactors affecting resistanceresistance. For n' science physics, o' science physics, o' pure physics presented by: mr oh ming yeo. Factors affecting resistance wire What factors affect the resistance of a wire planning background electricity flows around a circuit in a current a current is the flow of electrons currents are measured. The aim of this experiment is to investigate the ways in which thickness (or rather the cross sectional area) affects the resistance of a wire, how it. The resistance of a wire, or any conductive object whose shape can be defined by a length parallel to conduction and an area normal to conduction, is defined by. There are several factors that affect the resistance of a conductor material eg copper has lower resistance than steel length - longer wires have greater resistance thickness - smaller diameter wires have greater resistance temperature. Factors affecting the resistance of a wire research paperfactors affecting the resistance of a wire introduction: resistance is a force, which opposes the flow of an electric current around a circuit resistance is measured in ohms george ohm discovered that a circuit sometimes resists the flow of electricity. Si unit of resistance is ohm, ω resistivity is ohm meter, ω m and that of current is ampere, a factors affecting resistance are: length of the conductor (l), area of cross-section of the conductor (a) r α l and r α 1/a factors affecting resistivity are: material of the conductor, temperature at which the conductor is used. First, the total length of the wires will effect the amount of resistance the longer the wire, the more resistance that there will be there is a direct relationship between the amount of resistance encountered by charge and the length of. Table 3 (varying length) - swg 28 material swg thickness (diameter in g) length (cm) reading no current (a) average (a) resistance ( ) constantan table 4. Electrical currents are routinely harnessed and transmitted via interconnected wires the purpose of this research is to identify factors commonly responsible for affecting the resistance of current, or flow of electricity, across a wire in an electrical circuit. • precision wirewound resistors consist of resistance wire • various and complex factors contribute to the instability of. Resistance is a property four factors affecting resistance variable resistors or potentiometers control the length of wire we introduce into. Temperature: (an increase in temperature will increase the resistance) cross sectional area: (the larger the wire, the less resistance) the length of the wire: (the longer the wire, the more resistance)the material of the wire: (some materials are better conductors than others, and this causes less resistance.
- Open Access Approximating fixed points of multivalued ρ-nonexpansive mappings in modular function spaces Fixed Point Theory and Applications volume 2014, Article number: 34 (2014) The existence of fixed points of single-valued mappings in modular function spaces has been studied by many authors. The approximation of fixed points in such spaces via convergence of an iterative process for single-valued mappings has also been attempted very recently by Dehaish and Kozlowski (Fixed Point Theory Appl. 2012:118, 2012). In this paper, we initiate the study of approximating fixed points by the convergence of a Mann iterative process applied on multivalued ρ-nonexpansive mappings in modular function spaces. Our results also generalize the corresponding results of (Dehaish and Kozlowski in Fixed Point Theory Appl. 2012:118, 2012) to the case of multivalued mappings. MSC:47H09, 47H10, 54C60. 1 Introduction and preliminaries The theory of modular spaces was initiated by Nakano in connection with the theory of ordered spaces, which was further generalized by Musielak and Orlicz . The fixed point theory for nonlinear mappings is an important subject of nonlinear functional analysis and is widely applied to nonlinear integral equations and differential equations. The study of this theory in the context of modular function spaces was initiated by Khamsi et al. (see also [4–8]). Kumam obtained some fixed point theorems for nonexpansive mappings in arbitrary modular spaces. Kozlowski has contributed a lot towards the study of modular function spaces both on his own and with his collaborators. Of course, most of the work done on fixed points in these spaces was of existential nature. No results were obtained for the approximation of fixed points in modular function spaces until recently Dehaish and Kozlowski tried to fill this gap using a Mann iterative process for asymptotically pointwise nonexpansive mappings. All above work has been done for single-valued mappings. On the other hand, the study of fixed points for multivalued contractions and nonexpansive mappings using the Hausdorff metric was initiated by Markin (see also ). Later, an interesting and rich fixed point theory for such maps was developed which has applications in control theory, convex optimization, differential inclusion, and economics (see and references cited therein). Moreover, the existence of fixed points for multivalued nonexpansive mappings in uniformly convex Banach spaces was proved by Lim . The theory of multivalued nonexpansive mappings is harder than the corresponding theory of single-valued nonexpansive mappings. Different iterative processes have been used to approximate the fixed points of multivalued nonexpansive mappings in Banach spaces. Dhompongsa et al. have proved that every ρ-contraction has a fixed point where ρ is a convex function modular satisfying the so-called -type condition, C is a nonempty ρ-bounded ρ-closed subset of and a family of ρ-closed subsets of C. By using this result, they asserted the existence of fixed points for multivalued ρ-nonexpansive mappings. Again their results are existential in nature. See also Kutbi and Latif . In this paper, we approximate fixed points of ρ-nonexpansive multivalued mappings in modular function spaces using a Mann iterative process. We make the first ever effort to fill the gap between the existence and the approximation of fixed points of ρ-nonexpansive multivalued mappings in modular function spaces. In a way, the corresponding results of Dehaish and Kozlowski are also generalized to the case of multivalued mappings. Some basic facts and notation needed in this paper are recalled as follows. Let Ω be a nonempty set and Σ a nontrivial σ-algebra of subsets of Ω. Let be a δ-ring of subsets of Ω, such that for any and . Let us assume that there exists an increasing sequence of sets such that (for instance, can be the class of sets of finite measure in a σ-finite measure space). By , we denote the characteristic function of the set A in Ω. By ℰ we denote the linear space of all simple functions with supports from . By we will denote the space of all extended measurable functions, i.e., all functions such that there exists a sequence , and for all . Definition 1 Let be a nontrivial, convex and even function. We say that ρ is a regular convex function pseudomodular if ρ is monotone, i.e., for any implies , where ; ρ is orthogonally subadditive, i.e., for any such that , ; ρ has Fatou property, i.e., for all implies , where ; ρ is order continuous in ℰ, i.e., , and implies . A set is said to be ρ-null if for every . A property is said to hold ρ-almost everywhere (ρ-a.e.) if the set is ρ-null. As usual, we identify any pair of measurable sets whose symmetric difference is ρ-null as well as any pair of measurable functions differing only on a ρ-null set. With this in mind we define where is actually an equivalence class of functions equal ρ-a.e. rather than an individual function. Where no confusion exists we will write ℳ instead of . Definition 2 Let ρ be a regular function pseudomodular. We say that ρ is a regular convex function modular if implies ρ-a.e. It is known (see ) that ρ satisfies the following properties: for every scalar α with and . if , and . ρ is called a convex modular if, in addition, the following property is satisfied: (3′) if , and . Definition 3 The convex function modular ρ defines the modular function space as Generally, the modular ρ is not subadditive and therefore does not behave as a norm or a distance. However, the modular space can be equipped with an F-norm defined by In the case ρ is convex modular, defines a norm on the modular space , and it is called the Luxemburg norm. The following uniform convexity type properties of ρ can be found in . Definition 4 Let ρ be a nonzero regular convex function modular defined on Ω. Let , , . Define and if . As a conventional notation, . Definition 5 A nonzero regular convex function modular ρ is said to satisfy if for every , , . Note that for every , for small enough. ρ is said to satisfy if for every , , there exists depending only upon s and ε such that for any . Definition 6 Let be a modular space. The sequence is called: ρ-convergent to if as ; ρ-Cauchy, if as n and . Consistent with , the ρ-distance from an to a set is given as follows: Definition 7 A subset is called: ρ-closed if the ρ-limit of a ρ-convergent sequence of D always belongs to D; ρ-a.e. closed if the ρ-a.e. limit of a ρ-a.e. convergent sequence of D always belongs to D; ρ-compact if every sequence in D has a ρ-convergent subsequence in D; ρ-a.e. compact if every sequence in D has a ρ-a.e. convergent subsequence in D; A set is called ρ-proximinal if for each there exists an element such that . We shall denote the family of nonempty ρ-bounded ρ-proximinal subsets of D by , the family of nonempty ρ-closed ρ-bounded subsets of D by and the family of ρ-compact subsets of D by . Let be the ρ-Hausdorff distance on , that is, A multivalued mapping is said to be ρ-nonexpansive if A sequence is called bounded away from 0 if there exists such that for every . Similarly, is called bounded away from 1 if there exists such that for every . Lemma 1 (Lemma 3.2 ) Let ρ satisfy and let be bounded away from 0 and 1. If there exists such that The above lemma is an analogue of a famous lemma due to Schu in Banach spaces. A function is called a fixed point of if . The set of all fixed points of T will be denoted by . Lemma 2 Let be a multivalued mapping and Then the following are equivalent: , that is, . , that is, for each . , that is, . Further where denotes the set of fixed points of . Proof . Since , so . Therefore, for any , implies that . Hence . That is, . . Since , so by definition of we have . Thus by ρ-closedness of Tf. □ Definition 8 A multivalued mapping is said to satisfy condition (I) if there exists a nondecreasing function with , for all such that for all . It is a multivalued version of condition (I) of Senter and Dotson in the framework of modular function spaces. 2 Main results We prove a key result giving a major support to our ρ-convergence result for approximating fixed points of multivalued ρ-nonexpansive mappings in modular function spaces using a Mann iterative process. Theorem 1 Let ρ satisfy and D a nonempty ρ-closed, ρ-bounded and convex subset of . Let be a multivalued mapping such that is a ρ-nonexpansive mapping. Suppose that . Let be defined by the Mann iterative process: where and is bounded away from both 0 and 1. Then Proof Let . By Lemma 2, . Moreover, by the same lemma, . To prove that exists for all , consider By convexity of ρ, we have Hence exists for each . We now prove that As , it suffices to prove that and so in view of (2.1), we have from (2.1), (2.2), (2.3), and Lemma 1, we have Now we are all set for our convergence result for approximating fixed points of multivalued ρ-nonexpansive mappings in modular function spaces using the Mann iterative process as follows. Theorem 2 Let ρ satisfy and D a nonempty ρ-compact, ρ-bounded and convex subset of . Let be a multivalued mapping such that is ρ-nonexpansive mapping. Suppose that . Let be as defined in Theorem 1. Then ρ-converges to a fixed point of T. Proof From ρ-compactness of D, there exists a subsequence of such that for some . To prove that q is a fixed point of T, let g be an arbitrary point in and f in . Note that By Theorem 1, we have . This gives . Hence q is a fixed point of . Since the set of fixed points of is the same as that of T by Lemma 2, ρ-converges to a fixed point of T. □ Theorem 3 Let ρ satisfy and D a nonempty ρ-closed, ρ-bounded and convex subset of . Let be a multivalued mapping with and and satisfying condition (I) such that is ρ-nonexpansive mapping. Let be as defined in Theorem 1. Then ρ-converges to a fixed point of T. Proof From Theorem 1, exists for all . If , there is nothing to prove. We assume . Again from Theorem 1, so that Hence exists. We now prove that . By using condition (I) and Theorem 1, we have Since l is a nondecreasing function and , it follows that . Next, we show that is a ρ-Cauchy sequence in D. Let be arbitrarily chosen. Since , there exists a constant such that for all , we have In particular, . There must exist a such that Now for , we have Hence is a ρ-Cauchy sequence in a ρ-closed subset D of , and so it must converge in D. Let . That q is a fixed point of T now follows from Theorem 2. □ We now give some examples. The first one shows the existence of a mapping satisfying the condition (I) whereas the second one shows the existence of a mapping satisfying all the conditions of Theorem 3. Example 1 Let (the collection of all real valued measurable functions on ). Note that is a modular function space with respect to Let . Obviously D is a nonempty closed and convex subset of . Define as Define a continuous and nondecreasing function by . It is obvious that for all . Hence T satisfies the condition (I). Example 2 The real number system ℝ is a space modulared by . Let . Obviously D is a nonempty closed and convex subset of ℝ. Define as Define a continuous and nondecreasing function by . It is obvious that for all . Note that when . Hence is nonexpansive. Moreover, by Lemma 2, for all . Thus defined by where ρ-converges to a fixed point of T. Nakano H: Modular Semi-Ordered Spaces. Maruzen, Tokyo; 1950. Musielak J, Orlicz W: On modular spaces. Stud. Math. 1959, 18: 591–597. Khamsi MA, Kozlowski WM, Reich S: Fixed point theory in modular function spaces. Nonlinear Anal. 1990, 14: 935–953. 10.1016/0362-546X(90)90111-S Khamsi MA: A convexity property in modular function spaces. Math. Jpn. 1996, 44: 269–279. Dhompongsa S, Benavides TD, Kaewcharoen A, Panyanak B: Fixed point theorems for multivalued mappings in modular function spaces. Sci. Math. Jpn. 2006, e-2006: 139–147. Benavides TD, Khamsi MA, Samadi S: Asymptotically non-expansive mappings in modular function spaces. J. Math. Anal. Appl. 2002, 265: 249–263. 10.1006/jmaa.2000.7275 Benavides TD, Khamsi MA, Samadi S: Asymptotically regular mappings in modular function spaces. Sci. Math. Jpn. 2001, 53: 295–304. Benavides TD, Khamsi MA, Samadi S: Uniformly Lipschitzian mappings in modular function spaces. Nonlinear Anal. 2001, 46: 267–278. 10.1016/S0362-546X(00)00117-6 Kumam P: Fixed point theorem for non-expansive mappings in modular spaces. Arch. Math. 2004, 40: 345–353. Kozlowski WM: Modular Function Spaces. Dekker, New York; 1988. Dehaish BAB, Kozlowski WM: Fixed point iteration for asymptotic pointwise nonexpansive mappings in modular function spaces. Fixed Point Theory Appl. 2012., 2012: Article ID 118 Markin JT: Continuous dependence of fixed point sets. Proc. Am. Math. Soc. 1973, 38: 545–547. 10.1090/S0002-9939-1973-0313897-4 Nadler SB Jr.: Multivalued contraction mappings. Pac. J. Math. 1969, 30: 475–488. 10.2140/pjm.1969.30.475 Gorniewicz L: Topological Fixed Point Theory of Multivalued Mappings. Kluwer Academic, Dordrecht; 1999. Lim TC: A fixed point theorem for multivalued nonexpansive mappings in a uniformly convex Banach spaces. Bull. Am. Math. Soc. 1974, 80: 1123–1126. 10.1090/S0002-9904-1974-13640-2 Kutbi MA, Latif A: Fixed points of multivalued mappings in modular function spaces. Fixed Point Theory Appl. 2009., 2009: Article ID 786357 Kilmer SJ, Kozlowski WM, Lewicki G:Sigma order continuity and best approximation in -spaces. Comment. Math. Univ. Carol. 1991, 3: 2241–2250. Schu J: Weak and strong convergence to fixed points of asymptotically nonexpansive mappings. Bull. Aust. Math. Soc. 1991, 43: 153–159. 10.1017/S0004972700028884 Senter HF, Dotson WG: Approximating fixed points of nonexpansive mappings. Proc. Am. Math. Soc. 1974, 44(2):375–380. 10.1090/S0002-9939-1974-0346608-8 The first author owes a lot to Professor Wataru Takahashi from whom he started learning the very alphabets of Fixed Point Theory during his doctorate at Tokyo Institute of Technology, Tokyo, Japan. He is extremely indebted to Professor Takahashi and wishes him a long healthy active life. The authors are thankful to the anonymous referees for giving valuable comments. The authors declare that they have no competing interests. Both authors worked on the manuscript. Both read and approved the final manuscript. About this article Cite this article Khan, S.H., Abbas, M. Approximating fixed points of multivalued ρ-nonexpansive mappings in modular function spaces. Fixed Point Theory Appl 2014, 34 (2014). https://doi.org/10.1186/1687-1812-2014-34 - fixed point - multivalued ρ-nonexpansive mapping - iterative process - modular function space
This article needs additional citations for verification. Please help improve this article by adding citations to reliable sources. Unsourced material may be challenged and removed.Find sources: "Darboux integral" – news · newspapers · books · scholar · JSTOR (February 2013) (Learn how and when to remove this template message) In real analysis, a branch of mathematics, the Darboux integral is constructed using Darboux sums and is one possible definition of the integral of a function. Darboux integrals are equivalent to Riemann integrals, meaning that a function is Darboux-integrable if and only if it is Riemann-integrable, and the values of the two integrals, if they exist, are equal. The definition of the Darboux integral has the advantage of being easier to apply in computations or proofs than that of the Riemann integral. Consequently, introductory textbooks on calculus and real analysis often develop Riemann integration using the Darboux integral, rather than the true Riemann integral. Moreover, the definition is readily extended to defining Riemann–Stieltjes integration. Darboux integrals are named after their inventor, Gaston Darboux. The definition of the Darboux integral considers upper and lower (Darboux) integrals, which exist for any bounded real-valued function on the interval . The Darboux integral exists if and only if the upper and lower integrals are equal. The upper and lower integrals are in turn the infimum and supremum, respectively, of upper and lower (Darboux) sums which over- and underestimate, respectively, the "area under the curve." In particular, for a given partition of the interval of integration, the upper and lower sums add together the areas of rectangular slices whose heights are the supremum and infimum, respectively, of f in each subinterval of the partition. These ideas are made precise below: A partition of an interval is a finite sequence of values xi such that Each interval [xi−1, xi] is called a subinterval of the partition. Let f: [a, b] → R be a bounded function, and let be a partition of [a, b]. Let Lower (green) and upper (green plus lavender) Darboux sums for four subintervals The upper Darboux sum of f with respect to P is The lower Darboux sum of f with respect to P is The lower and upper Darboux sums are often called the lower and upper sums. The upper Darboux integral of f is The lower Darboux integral of f is In some literature an integral symbol with an underline and overline represent the lower and upper Darboux integrals respectively. and like Darboux sums they are sometimes simply called the lower and upper integrals. If Uf = Lf, then we call the common value the Darboux integral. We also say that f is Darboux-integrable or simply integrable and set An equivalent and sometimes useful criterion for the integrability of f is to show that for every ε > 0 there exists a partition Pε of [a, b] such that A Darboux-integrable function Suppose we want to show that the function f(x) = x is Darboux-integrable on the interval [0, 1] and determine its value. To do this we partition [0, 1] into n equally sized subintervals each of length 1/n. We denote a partition of n equally sized subintervals as Pn. Now since f(x) = x is strictly increasing on [0, 1], the infimum on any particular subinterval is given by its starting point. Likewise the supremum on any particular subinterval is given by its end point. The starting point of the kth subinterval in Pn is (k−1)/n and the end point is k/n. Thus the lower Darboux sum on a partition Pn is given by similarly, the upper Darboux sum is given by Thus for given any ε > 0, we have that any partition Pn with satisfies which shows that f is Darboux integrable. To find the value of the integral note that Darboux upper sums of the function y = x2 Darboux lower sums of the function y = x2 A nonintegrable function Suppose we have the function defined as Since the rational and irrational numbers are both dense subsets of R, it follows that f takes on the value of 0 and 1 on every subinterval of any partition. Thus for any partition P we have from which we can see that the lower and upper Darboux integrals are unequal. Refinement of a partition and relation to Riemann integration When passing to a refinement, the lower sum increases and the upper sum decreases. A refinement of the partition is a partition such that for all i = 0, …, n there is an integer r(i) such that In other words, to make a refinement, cut the subintervals into smaller pieces and do not remove any existing cuts. If is a refinement of then If P1, P2 are two partitions of the same interval (one need not be a refinement of the other), then and it follows that Riemann sums always lie between the corresponding lower and upper Darboux sums. Formally, if and together make a tagged partition (as in the definition of the Riemann integral), and if the Riemann sum of corresponding to P and T is R, then From the previous fact, Riemann integrals are at least as strong as Darboux integrals: if the Darboux integral exists, then the upper and lower Darboux sums corresponding to a sufficiently fine partition will be close to the value of the integral, so any Riemann sum over the same partition will also be close to the value of the integral. There is (see below) a tagged partition that comes arbitrarily close to the value of the upper Darboux integral or lower Darboux integral, and consequently, if the Riemann integral exists, then the Darboux integral must exist as well. \|Details of finding the tags For this proof, we shall use superscripts to index and variables related to it. Let be an arbitrary partition of such that , whose tags are to be determined. By the definition of infimum, for any , we can always find a Let , we have Taking limits of both sides, Similarly, (with a different sequences of tags) Thus, we have which means that the Darboux integral exist and equals .
The height h in feet of a ball after t seconds is given by h32t 16t2 - Metis data science review reddit - Math 1081, Quiz 3 Name: 1. Suppose the height of a ball thown out of a window (in feet above the ground after t seconds) is given h(t) —16t2 + 48t + 64, for 0 < t t f, - Aug 25, 2013 · The height, h in metres of a ball falling through a given time, t in secs, is known by the equation: h = 112t - 16t². To solve for the time, t; taken for the ball to hit the ground, h must be 0, then . 0 = 112t - 16t² . 0 = 16t(7 - t) => By applying the principle of zero products [a(b) = 0, where a = 0 or b = 0], then . 16t = 0 or 7 - t = 0 - A ball is thrown into the air with an upward velocity of 28 ft/s. its height (h) in feet after t seconds is given by the function h = %u201316t 28t; A ball is thrown into the air with an upward velocity of 28 ft/s. its height (h) in feet after t seconds is given by the function h = %u201316t 28t - Question 647609: the height h in feet of a ball t seconds after being tossed upwards is given by the function h(t)=84t-16t≤ a. After hown many seconds will hit the ground? b. What is its maximum height? Answer by josmiceli(19441) (Show Source): - the ball's initial velocity is negative, meaning the ball was thrown downward. Then add the effect of gravity, and it should hit the ground in a short time, even when thrown from an initial height of 240 feet. plug t=2 into the original equation for height. You should get slightly above ground, near zero-16(2)^2 -20(2) +240 = -64 -40+240 = 136 ... - A ball is thrown vertically upward from the top of a building 96 feet tall with an initial velocity of 80 feet per second. The distance s (in feet) of the... - A ball is thrown straight up. Its height, h(in metres), after t seconds is given by h=-5t^2+10t+2. To the nearest tenth of a second, when is the ball 6m above the ground? Explain why there are two answers. - Oct 11, 2018 · the ball as a function of time(t - sec): a) Find the balls maximum height. Y — C.ÞÒC 11+10 b) .nd e time 't takes the ball to reach its maximum height. coÖTd ans sec -16t2 +88 +25 LDS' -- IL A skydiver jumps from a plane that is at an altitude f 1700 ft. = -16t2 + 1700 gives the jumper's height h, in feet, The function h(t) after t seconds ... - The height of the ball is modeled by the equation h(t) = -16t2 + 64t+ 6. HOW long does it take for the ball to reach its maximum height of 70 feet? Write an equation to represent the maximum height. 22. A cannonball is fired straight up into the air ata speed of 64 feet per second. Its height, h, after t seconds is described by the equation h = - Sep 26, 2008 · If a ball is thrown straight up into the air with an initial velocity of 95 ft/s, its height in feet after t seconds is given by y=95t–16t^2. Find the average velocity for the time period beginning when t=2 and lasting . i) 0.5 seconds. ii) 0.1 seconds. iii) 0.01 seconds - Its height h in feet after t. h t 16t2 46t6. seconds is given by the function. 61) A golf ball is chipped into the air from a small hill with an upward velocity of 50 ft/sec. Its height. h t 16t2 50t10. h in feet after t seconds is given by the function. 62) A punter kicked the football into the air with an upward velocity of 62 ft/sec. Its ... - feet high. The equation for this apple's height h at time t seconds after launch —16t2 + 64t + 80. is h — a. Find the maximum height of the apple. Z(-lò b. How many seconds will it take for the ap le to reach the ground? L o 961 2a. -32 -32 Round huhalre4'V'1 PROJECTILE MOTION ith an initial upward velocity of90 feet la. = 563 -32 1. A ... - The textbook's position function is given by z(t) 16t2 40t + 336, What is the velocity of the textbook when it strikes Use the foålowing information to answer #3 —4, A particle is moving alqong the x-axis so that its position (meawred in meters) after t seconds have elapsed is given by the equation s(t) = 3. - of 80 ft/s. As the ball moves upward, gravity slows it. Eventually the ball begins to fall back to the ground. The height h of the ball after t seconds in the air is 16t2 + 801. given by the quadratic function h(t) a. How high does the ball go? b. For how many seconds is the ball in the air before it hits the ground? - Apes unit 8 quizlet Omscs 6200 midtermact of making people leave a place because of danger 7. blizzard - 5.a very bad snowstorm 8. tsunami - 3.a big wave that can destroy towns near the sea 9. aftershock - 10.a small earthquake after a larger one 10. avalanche - 7.a disaster in which snow and ice move quickly down a mountain.Natalie found a tennis ball outside a tennis court. She picked up the ball and threw it over the 12-foot fence into the court. The height of the ball, h, and time t seconds is given by the equation . h(t) = -16t 2 + 18t + 5. Will she make it over the fence? Sig sauer p238 hd california - The height h in feet of a ball after t seconds is given by h80t 16t2 ... - Then after an hour was spent in tying up his finger Uncle Podger wondered where the hammer had disappeared to. This opportunity has given me wisdom but only when I was patient enough to hear a child. - hook they throw is given by the function h(t) = —16t2 + 32t + 5. 2) If a toy rocket is launched vertically upward from ground level with an initial velocity of 128 feet per second, the its height h after t seconds is given by the equation h(t) ——16t2 + 128t (if air resistance is neglected). How long will it take the rocket to retum to the ... Wpf telerik themesJohn deere ct322 fuel system - Then after an hour was spent in tying up his finger Uncle Podger wondered where the hammer had disappeared to. This opportunity has given me wisdom but only when I was patient enough to hear a child.What is the maximum height of the ball? maximum or minimum height when #h' =0#. #h' = -32 t + 64#. #-16t(t-4)=0#. #rArrt=0" or " t=4#. Since the parabola is symmetrical the axis of symmetry will pass through the maximum.Surah yusuf pdf - velocity of 48 ft/s. It height h in feet after t seconds of books y. sold in a bookstore x days after an award- winning author appeared at an autograph-signing reception. What was copies of the book were sold? - too A 00 O = C=-SS 12 Z ISS 16t2 + 48t + 4. is given by the function h (t) =-- i.) What height will the ball be when 2 seconds has ...Pennsylvania gamefowl farms - Determine the speed of the second ball if the two balls are to meet at a height h/2 above the ground. The height of the building is 40 m. How much later must a second stone be dropped from the rest at the same initial height so that the two stones hit the ground at the same time?Mfm prayer points pdf - The height of the ball h in feet can be desœibed by the following equation where t is the time in seconds aner the ball was 16t2 + 47t + 79. released; h(t) = - Round off your answers to the include units in your answer. a.) How long will It take until the ball reaches its highest point? 2 secoüS -32 b.) What will be the maximum height of the ...Aliner explorer - feet per second, then its height h after t seconds is given by the equation h(t) — —16t2 + 128t (if air resistance is ignored). How long will it take for the rocket to return to the grounë? O Yosemite Falls in California is made of three smaller falls. The upper fall drops 1450 feet. The height h in feet of a water droplet fallingHyper tough ht200 manual
« ΠροηγούμενηΣυνέχεια » 1. What is the amount of $628.37 for 1 yr. 5 mo. 5 da. at 71 per cent? SOLUTION.- a = $628.37 = Prin. I of a = b = 52.364 = Int. 16 mo. 20 da. at 6 per cent. of .01 of a = c = 1.571 = " 15 da. at 6 per cent. btes d = $53.935 = " 17 mo. 5 da. at 6 per cent. of d = e = 13.484 = “ 17 mo. 5 da. at 1} per cent. a + d + e= $695.789 = Amt. 17 mo. 5 da. at 77 per cent. (b.) What is the interest of 2. $847.98 for 1 yr. 10 mo. 15 da. at 5 per cent? .3. $3865 for 2 yr. 9 mo. 10 da. at 7 per cent? 4. $594.32 for 9 mo. 19 da, at 41 per cent? 5. $219.74 for 3 mo. 4 da. at 3 per cent? 6. $392.58 from June 9, 1846, to June 3, 1847, at 8 per cent? 7. $936.20 from Oct. 1, 1855, to Nov. 17, 1855, at 1} per cent? (c.) What is the amount of — 8. $540 for 8 mo. 19 da. at 7 per cent? 9. $3600 for 15 mo. 25 da. at 9 per cent? 10. $8293 for 29 mo. 29 da. at 54 per cent? 11. $95.63 from Jan. 1, 1855, to Oct. 31, 1856, at 84 per cent? 12. $289.46 from March 27, 1852, to June 4, 1857, at 52 per cent? (d.) Although the preceding method may always be employed, it will often, if not usually, be better to Compute the interest at the given rate for any time for which it may be easily computed, and then take such multiples and parts of this as will give the interest for the required time. 13. What is the amount of $976.25 for 1 yr. 3 mo. 18 da. at 7 per cent ? SOLUTION. a = $976.25 = Prin. I of b = c = 17.084 = “ 3 mo. į of c= d = 3.416 = " 18 da. a+b+c+ d = $1056.087 = Amt. 1 yr. 8 mo. 31 da, (e.) What is the interest of - 28. Jan. 1, 1857, I bought 1000 bbl. of flour at $8.75 per bbl., borrowing the money at 6 per cent to pay for it. Feb. 1, I sold 1 the flour at $9.25 per bbl., and immediately put the money on interest at 7 per cent. Feb. 15, I sold the remainder at $9.62 per bbl., and put the money received for it on interest at 74 per cent. July 1, I collected the money due me and paid the amount of that which I had borrowed. What was my gain ? 113. Compound Interest. (a.) When interest is to be paid at regular intervals, or, if unpaid, is to be added to the principal to form a new principal on which interest is to be computed, it is called COMPOUND INTEREST. Note. — The difference between Simple and Compound Interest is that, by the latter, interest is reckoned on interest, while by the former it is not. (b.) To compute compound interest, we find the amount of the principal at simple interest to the time when the first interest is due; then the amount of this amount to the time when the second interest is due; and so on to the time of settlement. The last amount found is the required amount, and the difference between it and the given principal is the compound interest. 1. To what sum will $432 amount in 1 yr. 8 mo. 15 da. at 4 per cent, payable semi-annually, and what is its compound interest for that time? 2. To what sum will $950 amount in 2 yr. 6 mo. at 6 per cent, payable semi-annually? 3. What is the compound interest of $175.50 for 4 yr. 8 mo. 12 da. at 4 per cent, payable annually ? 4. What is the compound interest of $450 for 3 yr. 9 mo at 9 per cent, payable quarterly ? 5. To what sum will $875.50 amount in 5 yr. 8 mo. 10 da. at 3 per cent, payable annually! 6. To what sum will $1500 amount in 2 yr. 2 mo, at 10 per cent, payable semi-annually? (c.) The following table shows what part of any principal is equal to its amount at 3, 4, 5, 6, 7, and 8 per cent annual compound interest, for any number of years not exceeding 25. (d.) The amount of any sum of money, est annual compound interest, for any time and rate mentioned in the table, may be found by multiplying the principal by the appropriate number selected from the table. 7. What is the amount of $562.48 for 8 yr. 4 mo. at 7 per cent? SOLUTION. — By the table, it appears that the amount of a sum of money for 8 years, at 7 per cent compound interest, is 1.718186 times the principal. Multiplying $562.48 by 1.718186, we have $966.44526128, or, omitting the denominations below mills, a = $966.445 = Amount for 8 .years. 2} per cent. of a = b = 22.55 = Interest for 4 mo. a + b = c = $988.995 = Amount for 8 yr. 4 mo.. What is the amount at compound annual interest of — PRACTICAL APPLICATIONS OF INTE REST AND PERCENTAGE. 114. Promissory Notes. (a.) A PROMISSORY Note, a NOTE OF Hand, or, as it is more commonly called, a Note, is a written promise to pay a specified sum of money. (b.) The person who signs a note is called the PROMISOR, or MAKER of the note, and the one to whom it is made payable is called the PROMISEE. The one who has legal possession of it is called the HOLDER of the note. The FACE OF A NOTE is the sum for which it is given. ILLUSTRATIONS. - In Form 1, Elbridge Clapp is the Promisor or Maker of the note, and Hiram A. Pratt is the Promisee. Mr. Pratt will be the Holder of the note until it is paid, or till he transfers it to some other person. The FACE OF THE NOTE is $500. (c.) The words “VALUE RECEIVED" should always be placed in a note, as an acknowledgment that the maker gave it in conside ration of something of equivalent value.
This worked for me! Number sets symbols in LaTeX What could an aquatic civilization use to write on/with? Also, correctly setting the PDF viewer options (Options -> Configure Texmaker -> Commands -> PDF Viewer) Built-in Viewer and Embed, I was able to preview the pdf in Texmaker (View -> Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count). http://accessdtv.com/file-not/tftp-file-not-found-error.html I then uninstalled and reinstalled Texmaker allowing it to use the default directory on the C drive. Why cast an A-lister for Groot? Then you can reopen the texmaker and it will work. Given that ice is less dense than water, why doesn't it sit completely atop water (rather than slightly submerged)? http://tex.stackexchange.com/questions/63999/texmaker-windows-version-first-use-error-log-file-not-found I tried opening my tex file in another latex compiling software (Texshop) and it is still not producing a pdf. –esmitex Sep 4 '14 at 8:26 @esmitex Do you My .tex file runs fine. Not the answer you're looking for? click on the folder icon and find this file: C:/Program Files/MiKTeX 2.9/miktex/bin/x64/latex.exe Texmaker -> Options/Configure Texmaker/Commands/PdfLaTeX... The error message is log file not found. Pdflatex Log File Not Found Mac Would you like to answer one of these unanswered questions instead? And a new document must be saved with a .tex extension (without spaces or special characters in the name) before being compiled (if the message "log file not found" is displayed, this website Also, correctly setting the PDF viewer options (Options -> Configure Texmaker -> Commands -> PDF Viewer) Built-in Viewer and Embed, I was able to preview the pdf in Texmaker (View -> I installed the Windows version of Texmaker changing the C drive to D and leaving the rest of the install path the same. Texmaker Pdflatex Or any other file? I was able to solve this by creating an empty file in the same directory as my .tex file. Why don't miners get boiled to death at 4 km deep? Raise equation number position from new line How to say each other on this sentence How do we play with irregular attendance? I don't know if this works for you, but you could try... Texmaker View Pdf File Not Found Please add a minimal working example (MWE) that illustrates your problem. –Marco Daniel Aug 11 '12 at 11:39 1 @Rachelle What exactly is the workflow here? Log File Not Found Texstudio Player claims their wizard character knows everything (from books). Chances are that there is some error(s) in your tex file and TeXStudio cannot cope. check over here Problem occurred when I tried to run my bib file. make it a simple filename and let us know what happens. For me this path is: C:\Program Files\MiKTeX 2.9\miktex\bin\x64 share\|improve this answer edited May 13 '13 at 21:31 Peter Jansson 4,52872842 answered May 13 '13 at 21:13 Brian 11112 2 Welcome Texmaker Could Not Start The Command Pdflatex If I am told a hard percentage and don't get it, should I look elsewhere? It is true that you don't need the extensions. Why does removing Iceweasel nuke GNOME? http://accessdtv.com/file-not/texmaker-error-file-not-found.html In that case I recommend MikTeX if you are running Windows (miktex.org/), or Live TeX if you run an UNIX system (www.tug.org/texlive/) -> Or because you didn't add it's path to It is a semicolon-separated list of folders from where it will look-up any executable file that you try to use (such as pdflatex). Then it can be compiled by Texmaker without errors. When I typed pdflatex sample2e on the command line window, it says pdflatex is not recognized as an internal or external command, operable program or batch file. –user32229 Jun 14 '13 Configure Texmaker To work around this, you should provide the full commands in the settings, escaping them properly: "C:\Users\YOUR_USERNAME\AppData\Local\Programs\MiKTeX 2.9\miktex\bin\latex.exe" -interaction=nonstopmode %.tex "C:\Users\YOUR_USERNAME\AppData\Local\Programs\MiKTeX 2.9\miktex\bin\pdflatex.exe" -synctex=1 -interaction=nonstopmode %.tex These paths might differ for you, As happened to me this morning :S The fix is to simply as MikTex to update. Feb 26 '12 at 23:10 add a comment\| 1 Answer 1 active oldest votes up vote 22 down vote accepted The "View PDF" button in TeXmaker only opens an already existing DDoS: Why not block originating IP addresses? http://accessdtv.com/file-not/texmaker-error-log-file-not-found.html Make sure that the document you are trying to compile has a .tex prefix, it will not work otherwise. I've just "mv"ed a 49GB directory to a bad file path, is it possible to restore the original state of the files? I assume you are creating a .tex file, then trying to run something, then getting the warning? –Joseph Wright♦ Aug 11 '12 at 12:35 Yes, exactly when i run It gives information about what the document is and how to process it, like for example the name and version number of the compiler to use. Linked 10 Latex Error “log file not found” 223 How do I update my TeX distribution? 162 What are the advantages of TeX Live over MiKTeX? 14 Texmaker-log file not found Encode the alphabet cipher Does the reciprocal of a probability represent anything? Not the answer you're looking for? What could an aquatic civilization use to write on/with? It's a mix. –texenthusiast Aug 5 '13 at 15:00 \| show 1 more comment up vote 4 down vote Texmaker -> Options/Configure Texmaker/Commands/LaTeX... Feb 26 '12 at 22:55 Also, welcome to TeX.sx! What do you call someone without a nationality? Linked 25 Texmaker (Windows version) - first use - Error: Log file not found! 4 Error when compiling xcolor.ins in TeXmaker to generate xcolor.sty file Related 25Texmaker (Windows version) - first See also How do I update my distribution. –Alan Munn Jul 19 '12 at 23:59 add a comment\| 6 Answers 6 active oldest votes up vote 25 down vote accepted You share\|improve this answer answered Jun 7 at 20:57 Luc 1136 add a comment\| up vote 0 down vote I had the same issues and I was able to get it to
- 13.12., morning: Graduate meeting - 13.12., afternoon: LMS regional meeting for the South and South West UK - 14.-15.12.: Workshop Algebraic Structures and Quantum Physics The event location is School of Mathematics CF24 4AG Cardiff (UK) To get to the department from Cardiff central station on foot, see this map. Alternatively, take a taxi. Organizers: Gandalf Lechner, Mathew Pugh, Simon Wood Contact: LechnerG, PughMJ, WoodSi at cardiff.ac.uk Registration: If you are interested to attend this meeting, please write an email with your name, affiliation and travel dates to one of the organisers (see above). We have some funds to support travel/accommodation of LMS members. 1) Graduate Meeting In the Graduate Meeting, there will be two lectures introducing the topics of the talks of the regional meeting to a general mathematical audience. Furthermore, PhD students have the opportunity to give talks about their PhD projects. We are still looking for people interested to speak (depending on the number of talks, duration will be between 20 and 30 minutes). There are funds available to support your travel/accommodation. For funding, preference will be given to those students that give talks, but likely we will also have some funds to support participants. If you are interested to give such a talk, or simply to participate, please write an email to one of the organisers (see above). 2) LMS Regional Meeting - Shahn Majid (Queen Mary) Braided algebra and dual bases of quantum groups - Ingo Runkel (Hamburg) Categorification and field theory The Regional Meeting will be followed by a dinner at 7.30pm (venue to be confirmed). To reserve a place at the dinner, please email the organisers. The cost of the dinner will be approximately £40, including drinks. 3) Workshop Algebraic Structures and Quantum Physics - Chris Fewster (York) An analogue of the Coleman-Mandula theorem for QFT in curved spacetimes - Veronique Fisher (Bath) Pseudo-differential operators on Lie groups - Christian Korff (Glasgow) The Verlinde Algebra and quantum Bäcklund transformations - Pieter Naaijkens (Aachen) Stability of superselection sectors of topologically ordered models - Ulrich Pennig (Cardiff) Yang-Baxter representations of the infinite symmetric group - Ko Sanders (Dublin) Modular nuclearity and entanglement entropy in algebraic QFT - Anne Taormina (Durham) - Michael Tuite (Galway) Zhu reduction for vertex operator algebras on Riemann surfaces Abstracts for the LMS MEETING Ingo Runkel: Categorification and field theory The exchange of ideas between mathematics and physics has been an important source of progress. Group theory and symmetries in physical models provide an important example of this. A more recent mathematical tool, category theory, has also started to make its way into the description of physical theories. In this talk I would like to speak about topological quantum field theory, and how it links two ideas: On the mathematical side, there is categorification, a fancy way of saying that when one sees a non-negative integer one should try to think of it as the dimension of some vector space. On the physical side, there is the idea to study simplified models in fewer dimensions first and then try to pass to higher dimensions. Shahn Majid: Braided algebra and dual bases of quantum groups It is well-known that quantum groups can be used to generate braided categories and hence invariants of knots and braids. In this talk I will focus on the opposite, i.e. doing braided algebra and braided-Hopf algebra at the level of braid diagrams as a way to construct new quantum groups and gain insight into their deeper structure. It is known that every braided category has at its heart one of these braided-Hopf algebras. Moreover, every rigid braided-Hopf algebra in the category of representations of a quantum group leads to a new braided category and a new quantum group (called the `double-bosonization’ of the braided-Hopf algebra) of which this is the category of representations. I will illustrate how this inductively constructs q-deformed quantum groups u_q(g) of semisimple Lie algebras at certain special roots of unity in a way that automatically provides the dual basis to their PBW basis. This also provides new formulae for their quantum double R-matrix needed for applications of in quantum gravity and quantum computing, as well as a way to construct new and exotic quantum groups. Abstracts for the WORKSHOP Chris Fewster: An analogue of the Coleman-Mandula theorem for QFT in curved spacetimes The Coleman--Mandula (CM) theorem states that the Poincar\'e and internal symmetries of a Minkowski spacetime quantum field theory cannot combine nontrivially in an extended symmetry group. In this talk I will describe some of the background to the CM theorem and then establish an analogous result for a general class of quantum field theories in curved spacetimes. Unlike the CM theorem, our result is valid in dimensions $n\ge 2$ and for free or interacting theories. It makes use of a general analysis of symmetries induced by the action of a group $G$ on the category of spacetimes. Such symmetries are shown to be canonically associated with a cohomology class in the second degree nonabelian cohomology of $G$ with coefficients in the global gauge group of the theory. The main result proves that the cohomology class is trivial if $G$ is the universal cover $S$ of the restricted Lorentz group. Among other consequences, it follows that the extended symmetry group is a direct product of the global gauge group and $S$, all fields transform in multiplets of $S$, fields of different spin do not mix under the extended group, and the occurrence of noninteger spin is controlled by the centre of the global gauge group. Veronique Fischer: Pseudo-differential operators on Lie groups This talk will start with a personal and partial overview of harmonic analysis and its relations with quantum mechanics, especially via micro-local and semi-classical analysis (pseudo-differential theory). We will then discuss recent developments in the pseudo-differential theory on Lie groups. Christian Korff: The Verlinde Algebra and quantum Bäcklund transformations Starting from the canonical quantisation of the Ablowitz-Ladik chain we discuss the quantum analogue of a Baecklund transformation which defines a discrete time evolution for the resulting quantum system of q-bosons, where q is the quantisation parameter. We show that the composition of multiple Baecklund transformation defines a 2D TQFT which at $q=0$ describes the su(n) WZW Fusion Ring. At q=1 one obtains free bosons but the underlying combinatorics of the TQFT fusion coefficients is still interesting: they are related to cylindric reverse plane partitions and the generalised symmetric group. Pieter Naaijkens: Stability of superselection sectors of topologically ordered models Topologically ordered quantum spin models, such as Kitaev’s toric code, have many interesting properties. One of them is that they have anyonic excitations, i.e. excitations which have braided statistics. These excitations and their properties can be studied in the spirit of the Doplicher-Haag-Roberts programme in algebraic quantum field theory, to recover the full modular tensor category describing them. This analysis is done with respect to a reference representation, typically coming from a translation invariant ground state of the system, which plays the role of the vacuum. An important question is if this structure is stable. That is, if we perturb the dynamics, and obtain a new ground state, do we get the same modular tensor category? In my talk I will discuss recent results in this direction. Joint work with Matthew Cha and Bruno Nachtergaele. Ulrich Pennig: Yang-Baxter representations of the infinite symmetric group The Yang-Baxter equation was introduced as a consistency equation in statistical mechanics, but has since then appeared in many other research areas, for example integrable quantum field theory, knot theory, the study of Hopf algebras and quantum information theory. Its solutions are called R-matrices. In this talk I will discuss a joint project with G. Lechner and S. Wood, in which we found a complete classification of all unitary involutive R-matrices up to a natural equivalence relation. Each such R-matrix defines a representation and an extremal character of the infinite symmetric group and we identify the ones that come from R-matrices among all such characters using techniques involving operator algebras. Ko Sanders: Modular nuclearity and entanglement entropy in algebraic QFT Entanglement is an important experimental resource in quantum physics, and entanglement entropy is a measure for the amount of entanglement (which may be infinite). In the context of algebraic quantum field theory (AQFT) I will describe how the entanglement entropy can be estimated in terms of nuclearity properties of suitable modular operators. Such nuclearity properties have been established in several toy model cases and there are reasons to believe that they hold quite generally. This talk is based on joint works with G. Lechner, with S. Hollands and with O. Islam. Michael Tuite: Zhu reduction for vertex operator algebras on Riemann surfaces A Vertex Operator Algebra (VOA) is an algebraic formulation of a chiral conformal field theory. One of the most important tools in VOA theory is Zhu reduction which allows one to expand genus one correlation functions in terms of elliptic functions. In this talk I will discuss recent progress in extending Zhu reduction to Riemann surfaces of genus two and beyond. In particular, Zhu reduction is explicitly expressed in terms of certain Green's functions and holomorphic differentials. I also discuss some applications such as an Sp(4,Z) invariant differential equation describing the genus two partition functions for the (2,5) minimal model generalizing the Rogers-Ramanujan functions to genus two.
This action might not be possible to undo. Are you sure you want to continue? 1007/s10670-011-9300-4 ORIGINAL ARTICLE Evidential Holism and Indispensability Arguments Received: 17 February 2010 / Accepted: 26 May 2011 / Published online: 22 July 2011 Ó Springer Science+Business Media B.V. 2011 Abstract The indispensability argument is a method for showing that abstract mathematical objects exist (call this mathematical Platonism). Various versions of this argument have been proposed (§1). Lately, commentators seem to have agreed that a holistic indispensability argument (§2) will not work, and that an explanatory indispensability argument is the best candidate. In this paper I argue that the dominant reasons for rejecting the holistic indispensability argument are mistaken. This is largely due to an overestimation of the consequences that follow from evidential holism. Nevertheless, the holistic indispensability argument should be rejected, but for a different reason (§3)—in order that an indispensability argument relying on holism can work, it must invoke an unmotivated version of evidential holism. Such an argument will be unsound. Correcting the argument with a proper construal of evidential holism means that it can no longer deliver mathematical Platonism as a conclusion: such an argument for Platonism will be invalid. I then show how the reasons for rejecting the holistic indispensability argument importantly constrain what kind of account of explanation will be permissible in explanatory versions (§4). 1 Mathematical Platonism is a realist ontological position: it asserts that mathematical statements are true, that mathematical statements are committed to the existence of mathematical objects (such as numbers, sets, functions etc.), and thus that these mathematical objects exist (Maddy 1990). In addition to realism, Platonism insists that these mathematical objects are abstract; the Platonist maintains that there exist J. Morrison (&) Department of Philosophy, University of Birmingham, Birmingham B15 2TT, UK e-mail: firstname.lastname@example.org pp. or those who think that ontology can’t be done. one might be horriﬁed to discover that one’s ontology has been furnished with such bizarre entities as abstracta. models and functions. p. A major contemporary source of motivation for Platonism is the indispensability argument (Colyvan 2001. giving an alternative semantics or a reason for understanding mathematical claims non-literally will allow one to accept the indispensability of mathematical claims while trying to avoid their ontological commitments. and more speciﬁcally. Showing that science could be done without mathematics is one option available to those scientiﬁc realists who wish to avoid Platonism about mathematical objects. p. but deny that those claims commit to mathematical abstracta. this is because most people agree that it ‘‘appears to make no sense to ask where numbers or sets are located. p. the speciﬁc heat capacity of H2O is 4. Michael Resnik (Resnik 1997. The argument proceeds from the fact that we make indispensable use of mathematics and mathematical claims in our scientiﬁc theorising to the conclusion that we ought to be ontologically committed to the existence of such things as numbers. without argument. Morrison non-spatiotemporal. pp. 6). Alternatively one could agree that science depends on mathematical claims. But for scientiﬁc realists. It is inaccurate to say that there is such a thing as the indispensability argument since there are many different forms. causally inefﬁcacious mathematical objects (Hale 1994. the argument does not attempt to convince either those who think that matters of ontology are the proprietary domain of metaphysicians. 93–95) critically discusses her position. 299. such as: 1 If most mathematical realists are Platonists. Maddy maintains that sets can be perceivable (Maddy 1990. since it suggests that scientists are committed to the existence of abstracta in the same way as they are to atoms and genes. One exception is Penelope Maddy. Since indispensability arguments explicitly involve the claim that scientiﬁc theories can be a source of ontological commitments. that mathematical claims are indispensable to scientiﬁc theorising. 123 . 545). 58–63). and that the ontological commitments of sentences like: 1. then they are abstract. This is unpalatable to contemporary metaphysical tastes on at least two counts: one might repudiate the idea that scientists can furnish one’s ontology at all. Linnebo 2006 p.200 J/kg/K can be understood directly and literally from their paraphrases into canonical notation. This paper will assume. who attributes physical/causal properties to some mathematical objects. 299). or when they came into existence’’ (Hale 1994. sets.1 So the argument is controversial. In addition. The characteristic that is common to all of them is the dialectical procedure from the premise that science indispensably depends on mathematics to the conclusion that mathematical objects exist. I will not question the initial premise that scientiﬁc theorising indispensably depends on mathematical claims. the indispensability argument does stand as a challenge.264 J. it suggests that normal scientiﬁc practices commit us to the existence of abstract objects. there is general consensus that if such mathematical objects exist at all. who maintain that science can discover what exists. Regardless of how such a distinction is drawn. p. for example. It says that we should accept the ontological commitments of our best scientiﬁc theories. Let’s call such an argument a Semantic Indispensability Argument (SIA) (see Liggins 2008 and Wagner 1996). ignoring the many differences between tools and results. the present point is that various types of indispensability argument attempt to either remove it or redraw it. and that those mathematical claims are committed to the existence of abstracta. and that as such SIA does not commit her to mathematical Platonism. ontologically-committing results. Examples of the former are Holistic Indispensability Arguments—these approaches have the consequence that the distinction between 2 Clearly. she simply takes actual scientiﬁc practice to licence a restricted notion of scientiﬁc realism: no scientist would think that we should be wholesale scientiﬁc realists. 280). there exists an x such that x = speciﬁc heat capacity of H2O = the number of Joules energy required to raise one kilogramme of water by one degree Kelvin = 4. various arguments can be given for putting particular cases of apparent ontological commitment into either the ‘tool’ or ‘result’ categories. For example. If this is right. Our scientiﬁc realist feels no need to provide any alternative or non-literal semantics for the instrumental parts of a theory. that our best scientiﬁc theories indispensably appeal to mathematical claims. the ontologically committing propositions are just those that can act as explanations. Such a distinction is a consequence of prior philosophical theory.Evidential Holism and Indispensability Arguments 265 2. the invocation of a tool.2 So. Our restricted scientiﬁc realist maintains that numbers are tools. a distinction between those claims which are tools and those which are results. but it doesn’t licence drawing this line arbitrarily. 123 . or as attempts to demonstrate that her restricted version of realism is actually committed to numbers after all. then we can understand other varieties of indispensability argument either as attempts to force the restricted realist to be a wholesale realist. hence we should accept the existence of abstracta. our scientiﬁc realist points out that the overwhelming majority of scientiﬁc theories involve explicit idealizations and simpliﬁcations: real-world complex systems are reduced to partial models in which only one or two variables contribute any causal inﬂuence. Rather. Not all parts of a scientist’s theory should be treated equally. etc. of all the propositions indispensable to science. perhaps.200 It may be that these two assumptions. or that identify causes. along with the commitment to scientiﬁc realism. or that are falsiﬁable. Being as respectably-well-informed about actual scientiﬁc practices as she is. But our scientiﬁc realist who has no taste for Platonic entities may not feel the pull of SIA. while talk of numbers is merely instrumental. it’s not enough just to stipulate that some claims are ontologically committing while others are only instrumental—appealing to scientiﬁc practice might allow us to invoke such a distinction (see Maddy 1990. are sufﬁcient to secure mathematical Platonism. and which parts are merely instrumental devices for getting those conclusions. (Thanks to an anonymous referee for encouraging this point). She notes that while scientists might indispensably depend upon idealizations. SIA will only convert wholesale scientiﬁc realists into Platonists. she might try to argue that scientiﬁc discussion of atoms and genes is talk about substantial. the scientiﬁc realist looks to make some sort of distinction between the parts of the theory which are ontologicallycommitting conclusions. neither they nor she are thereby committed to thinking that the world is anything like the way it is represented in those models. The reasoning is as follows: if the relata of falsiﬁcation are observations and theories. Falsiﬁcation Thesis (FT): Observations only falsify whole theories. Epistemic naturalism tells us that only scientiﬁcally acceptable methods of belief-formation are sufﬁcient to form justiﬁed beliefs. The argument then proceeds as follows: epistemic naturalism tells us that scientiﬁc inquiry is the only source of knowledge or justiﬁed belief. Quine gave an indispensability argument for mathematical Platonism that invokes evidential holism as a premise. So holism’s role in this indispensability argument is to secure this premise: holism (FT) is taken to imply epistemic naturalism. I return to explanatory indispensability arguments in §4. then no propositions are known a priori. 2 Many commentators maintain that W. any mathematical or logical parts). since her ontologically privileged category of scientiﬁc results is really no different from the instrumentally useful mathematical tools— both are importantly instrumental for giving explanations of physical phenomena. because it is not entirely clear quite what Quine’s evidential holism amounts to. Recently. The indispensability thesis tells us that scientiﬁc inquiry cannot proceed without existentially quantifying over mathematical objects. Variations on such an argument are used to provide support for epistemic naturalism. the falsiﬁcation thesis has long since been invoked in Quinean-style arguments against the possibility of a priori knowledge. Conﬁrmation Thesis (CT): Observations conﬁrm entire theories. Explanatory Indispensability Arguments work in the other way: a distinction between tools and results might seem to be permissible. Morrison tools and results is not available to the scientiﬁc realist: the tools are actually a lot more like results.266 J. For example. and if any propositions can in principle be empirically defeated.V. In §2 and §3 I explore holistic indispensability arguments. it remains in principle possible to blame the empirical failures of a theory on any of its a priori constituents (i. It follows that the existence of 3 Strictly speaking there are only two distinct theses here. the argument maintains that a restricted scientiﬁc realist will have to be a mathematical Platonist. If a necessary condition for a proposition to be known a priori is that it is empirically indefeasible.O. This is. I have identiﬁed three distinct evidentially holistic theses that are commonly treated interchangeably (Morrison 2010)3: Prediction Thesis (PT): Only whole theories imply observations. These distinctions are useful as they enable us to identify candidate roles for holism in indispensability arguments. not individual parts thereof. and perhaps it is also acceptable that mathematical talk is more tool-like than result-like. As such.e. not individual parts thereof. in part. but the role which holism is supposed to play is not immediately clear. since I show that PT and FT are contapositives. Nevertheless. then any particular observation underdetermines which of a theory’s constitutive claims is false. 123 . scientists have empirical conﬁrmation of the mathematical claims on which they (indispensably) depend. The idea is that when a scientist carries out an experiment. it is sufﬁcient to note that this holistic/naturalistic indispensability argument need not convert our restricted scientiﬁc realist into a wholesale scientiﬁc realist. The ﬁrst premise of the argument is the indispensability thesis. it is epistemic naturalism. She can agree that claims which quantify over mathematical abstracta are a necessary condition for scientiﬁcally justiﬁed beliefs. This issue has been much discussed. Morrison 2010. which we saw above: scientiﬁc theorising indispensably depends upon claims that existentially quantify over mathematical abstracta. Penelope Maddy claims that the argument involves the premise that ‘‘as holists. the mathematical along with the physical hypotheses’’ (Maddy 2005b. But for our purposes here. (Resnik 1995. Quine’s holism—his interpretation of Duhem’s thesis—asserts that theories are conﬁrmed only as totalities. Likewise. the question as to whether holism entails mathematical Platonism is dependent on whether holism entails the repudiation of apriority. we take a scientiﬁc theory to be conﬁrmed as a whole. If mathematics is an inextricable part of 123 . A more widely recognised role for evidential holism in indispensability arguments invokes the conﬁrmation thesis (CT). Others argue that the argument sketched above turns on a pivotal and fallacious inference from the underdetermination of falsiﬁcation to the possibility of revising a mathematical or logical belief (Klee 1992. We should note that the pivotal premise of this holistic/naturalistic-indispensability argument is not evidential holism. and she can agree that there are no other sources of epistemic justiﬁcation. Resnik and Orlandi 2003). p. and so it receives conﬁrmation only as a whole. Elliott Sober tells us that [t]his indispensability argument for mathematical realism gives voice to an attitude towards conﬁrmation elaborated by Quine. Some maintain that respectable analyses of a priori knowledge should not make empirical indefeasibility a necessary condition (Rey 1998. So this indispensability argument will only work if it is true that evidential holism entails that there is no a priori knowledge. which is purported to be a consequence of holism (in the form of FT). Jenkins 2008). all the while maintaining that none of this compels her think that mathematical claims are anything more than (importantly) useful tools. 444). p. the results conﬁrm not only the experimental hypotheses at stake but also all of the mathematical claims (along with all other additional theoretical claims). 166) Similarly. The next part of the argument invokes evidential holism. As such. instead. We can see this move being made when Michael Resnik claims that indispensability arguments involve the following premise: Evidential holism: The evidence for a scientiﬁc theory bears directly upon its theoretical apparatus as a whole and not upon its individual hypotheses.Evidential Holism and Indispensability Arguments 267 mathematical objects becomes a necessary condition for the possibility of any beliefs being justiﬁed. As such. A theory makes contact with experience only as a whole. By invoking holism. (Sober 1993. The last step of the argument is to show that this empirical support means that a scientist cannot fail to be committed to the existence of mathematical objects. 35) The argument starts by claiming that scientists often must use mathematical claims. But my 4 I do not mean to imply that Hilary Putnam should be understood as having argued for the existence of abstract mathematical objects. 166. and that it misdescribes actual mathematical and scientiﬁc practices and the important differences between them (Maddy 1992). Elliott Sober thinks that holism so expressed is an incorrect theory of conﬁrmation. rather than restricted. about her realism. then. p. holism generates a problem: how are we to account for the empirical successes of our mathematical claims? And a constraint on a reply is that it seems that we should explain the empirical successes of mathematics in the same way that we explain the empirical successes of any other scientiﬁc hypothesis. Commentators often refer to indispensability arguments as ‘Quine-Putnam indispensability arguments’. So … the existence of mathematical objects is as well grounded as that of the other entities posited by science. where he discusses the ‘‘intellectual dishonesty of denying the existence of what one daily presupposes.’’ (Putnam 1971. p. and yet b) believing ourselves to be committed only to the objects invoked by the physical claim and not to the entities of the mathematical. then we should give a similar explanation of the empirical successes of mathematics. Penelope Maddy thinks that holism so expressed is empirically false as it makes a mistaken claim about the nature of evidence in mathematics. (1995.268 J. we can infer that we can have empirical support for these mathematical claims. any scientiﬁc realist should properly be wholesale. one physical and one mathematical. p. he too suggests that it mischaracterises actual scientiﬁc practice. 3 It should come as no surprise that the premise expressing evidential holism has come under ﬁre. Resnik goes on to describe how this inference might work: [I]f mathematics is an indispensable component of science. and that any overlap is illusory. 123 . by holism. namely that the entities they discuss are real.4 In effect. This is consonant with Putnam’s account of indispensability arguments. then the empirical success of the theory conﬁrms the entire theory—mathematics and all. As such. I disagree that CT gives a mistaken account of actual scientiﬁc and mathematical practices. If we explain the success of physical hypotheses by thinking that the entities they discuss are real. Morrison a physical theory. to the same extent. my emphasis) Resnik’s description of the argument suggests that there would be something remiss in a) supposing that an experiment could conﬁrm two different claims. David Liggins (2008) makes it perfectly clear that the elision is incorrect. and he raises several problems in conﬁrmation theory which he claims it cannot answer (Sober 1993). 347). whatever evidence we have for science is also evidence for the mathematical objects and mathematical principles it presupposes. The Bayesian formula (plus some argument about the correct measure of conﬁrmation) is just one example of the kind of supplementary presupposition which might enable an observation to be said to support one particular theoretical claim more than another.Evidential Holism and Indispensability Arguments 269 assessment of the holistic indispensability argument has this much in common with both Sober and Maddy: evidential holism. But since scientists do not (and indeed. and that CT is a consequence of the mistaken assumption that PT and FT will yield holistic conclusions about inductive relations such as conﬁrmation and disconﬁrmation. She suggests that if CT is correct we should expect mathematicians to be looking to developments in science to tell them which of their theories are conﬁrmed: ‘‘[i]f this is correct. and that if they want to distribute the conﬁrmation of an entire theory amongst its individual parts then they cannot do so differentially—every part of a theory is conﬁrmed to the same extent by the evidence. not individual parts thereof. should not) think about evidence this way. In this section I will argue that the holistic argument just sketched does fail. set theorists should be eagerly awaiting the outcome of debate over 123 . Sober seems to think that by deﬁnition such a position would no longer be ‘holistic’ (Sober 2000. that there are no good reasons for endorsing CT. Maddy’s concerns about CT and mathematical practice are as follows. CT must have something wrong. The conﬁrmation thesis is a claim about the relata of conﬁrmation: observations taken by themselves only support entire theories. taken individually. With: No individual theoretical claims can be differentially supported. He maintains that evidential holists are committed to CT. Following the line taken in (Morrison 2010). will be able to identify particular propositions as being better supported than others. is false. 268 footnote 28). p. and that there are plenty of good reasons for rejecting it. Sober’s objections depend on his giving a particular construal of how he thinks that the holist must ‘distribute’ conﬁrmation accruing to an entire theory among its constitutive parts. Morrison (2010) argues that Quine’s evidential holism is properly understood as a combination of only the prediction and falsiﬁcation theses (PT and FT). as it appears in that argument. But this is entirely consistent with the possibility that observations. observations don’t single out any particular propositions/hypotheses for support. I think that CT is consistent with differential assignments of evidential values among the parts of a theory. The real problem with the holistic indispensability argument given above is that it depends on CT. That is to say that even if one held CT. taken in conjunction with various auxiliary assumptions about the experimental set-up and about the relative likelihoods of various hypotheses. and that its failure is a direct consequence of its invocation of CT. but that is all it denies. I maintain that evidential holism proper describes the holistic nature of the deductive relations that must hold between observations and hypotheses. one could still endorse differential distribution of support. We should not conﬂate: Conﬁrmation Thesis (CT): Observations conﬁrm entire theories. But CT is a holistic position: it denies that observations taken by themselves admit differential support for the constituent parts of a theory. 281). Quine’s evidential holism tells us that ‘‘our statements about the external world face the tribunal of sense experience not individually but only as a corporate body’’ (Quine 1951. but only because this was the sentence he was wondering about and is prepared to reject. p. And if CT is consistent with a notion of differential support. without differential support.’’ (1990a. But this is not the case. set theorists do not regularly keep an eye on developments in fundamental physics. If practicing scientists do not think about evidence holistically. If CT is consistent with a notion of differential support. when in fact they do not. 123 . I do not dispute her assessment of the practice of scientists. A proper diagnosis would require an examination of the motivations behind CT to see if there are any particular reasons for thinking that holism about conﬁrmation should be understood in the stronger form which rules out any notion of differential support. then we should have no reason to expect scientists to think that all parts of their theories receive equal conﬁrmation. I do not dispute her assessment of the practice of mathematicians. if CT can be compatible with a notion of differential support. p.5 I think that the expectation is misplaced—CT does not entail that each individual part of a theory is supported by observation when the theory-as-a-whole is supported by observation. Maddy argues that evidential holism is descriptively false: it mischaracterises the actual nature of the evidential relation. preparing to tailor the practice of set theory to the nature of the resulting applications of continuum mathematics. 14). p. then ‘‘[i]f we remain true to our naturalistic principles. Maddy’s argument about CT and scientiﬁc practice is as follows. if they operate as though not all parts of a theory are equally conﬁrmed. The mere fact that a theory that indispensably relies on mathematical claims is supported by observation is no reason to think that the mathematical claims are thereby supported.270 J. as I’ve suggested above. 280). However. then we should have no reason to expect mathematicians to look to scientiﬁc results to ﬁnd support for their theories. in so far as scientists have withheld their assent to the existence of some of the posits of well-conﬁrmed theories until claims about those particular objects have received ‘direct veriﬁcation’. 41). 289). describing scientists as acting as though all parts of a theory are equally conﬁrmed. it was not necessary that it be consonant with the descriptions that scientists might use to describe their own activities. what is surprising is the scarcity of any arguments motivating CT in either form.’’ (Maddy 1992. I think that the expectation is misplaced—CT does not entail that each individual part of a theory is supported by observation when the theory-as-a-whole is supported by observation. as I’ve suggested above. p.’’ (ibid. although it is hard to see why that should be the case. Such an examination is undertaken in Morrison (2010). we must allow a distinction to be drawn between parts of a theory that are true and parts that are merely useful. Morrison quantum gravity. But ‘‘the actual practice of science presents a very different picture’’ (Maddy 1992. CT must get something wrong. Since they do not. p. then the holistic indispensability argument will not work. I can only speculate that the many commentators on holistic indispensability arguments have assumed a stronger version of CT that rules out the possibility of differential support. 5 Quine clearly does not place too much emphasis in what scientists think that they are up to when espousing his holism: ‘‘the scientist thinks of his experiment as a test speciﬁcally of his new hypothesis. So while he was concerned to give an accurate explanation of scientiﬁc practice. So we can see that in order to establish any sort of deductive link between holism about prediction and holism about conﬁrmation we will need to appeal to something which makes the same connection between prediction and conﬁrmation as this HD principle. which seems to be well motivated. for example. and that inductive support for these laws can be garnered from particular instances of these laws. which is to say that the testable observation is deductively derived from an entire theory. for that very reason it shows up in paradoxes to do with conﬁrmation. indeed. we need to have a principled reason for thinking that whatever predicts an observation gets conﬁrmed by it. if the holist additionally endorses this principle of hypothetico-deductivism then they should also accept that the observation conﬁrms the entire theory (CT). This rather remarkable discrepancy can be explained. and at the same time it is regarded as highly controversial in discussions of evidence in the philosophy of science. By endorsing PT. As such. where it is rarely given any explicit defence or motivation. What motivates endorsing the HD principle as an additional. I have suggested that the prediction thesis is broadly equivalent to the claim that scientiﬁc observations are theory-laden. p. This supposition makes it seem as though the credibility associated with the prediction thesis. the principle seems an intuitive and eminently plausible theory of how conﬁrmation relations might work. The inferential link between CT and PT is at once prima facie plausible and fallacious—indeed. at ﬁrst glance to anyone who is not familiar with the debates that go on in the philosophy of science. The holist who accepts the prediction thesis needs some reason to think that they should be committed to this hypothetico-deductive principle of conﬁrmation before endorsing the conﬁrmation thesis. which is itself in need of independent motivation. will transfer to the conﬁrmation thesis. I suspect that adherents of the conﬁrmation thesis suppose that there is an intimate link between the prediction thesis and the conﬁrmation thesis. the prediction thesis (PT) seems to be straightforwardly motivated. we think that seeing white swans lends inductive support for the claim that all swans are white. In order to make an inference from the prediction thesis to the conﬁrmation thesis. A minimum condition for thinking that the prediction thesis implies the conﬁrmation thesis will involve accepting a principle of the hypothetico-deductive (HD) theory of conﬁrmation. It is important to note that any such adherent of CT is thereby relying upon some extra-holistic machinery. Indeed. As a result. A crude account of this HD principle maintains that for a to conﬁrm b it sufﬁces for a to be deductively derivable from b. the general idea that ‘instances support their generalisations’ seems innocent enough. 35). Sober says it is something like an ‘‘unexceptional observation’’ (Sober 1993. This HD principle does precisely that: it equates conﬁrming instances with deductive consequences. the holist is committed to the idea that an entire theory predicts a testable observation. extra-holistic component? Arguments for this HD principle stem from the view that scientiﬁc laws are universal generalisations. and thus that whatever intuitive plausibility attaches to the theory-ladenness of observation is thereby motivation for the prediction thesis (Morrison 2010). In contrast. it is no surprise that 123 . So. Prima facie.Evidential Holism and Indispensability Arguments 271 The conﬁrmation thesis (CT) is openly endorsed in many mainstream epistemological discussions. no contemporary account of hypotheticodeductivism would maintain the HD principle stated above. the Quinean doctrines which could plausibly be employed to produce CT tend to be more controversial than CT itself. His account of evidence says that only entire theories yield observational predictions. and if those predictions are incorrect then the observations. so such arguments won’t be straightforwardly motivating to most. 123 . simply because it is ¨ve position that it results in these unintuitive consequences. Should he also be committed to the existence of mathematical abstracta? His theory of evidence makes it look as though the mathematical claims are evidentially on a par with the scientiﬁc claims. Moreover. that we should reject the holistic indispensability argument that depends on CT. and is willing to accept those ontological commitments of his best as-yet-unfalsiﬁed theory. In general. both being as-yetunfalsiﬁed. the formalisation that any observed deductive consequence of b thereby conﬁrms b quickly leads to problems. Carl Hempel. whether it is acceptable for him to (a) suppose that two different claims. for example. If so. the evidential holist who accepts PT and FT while repudiating CT—we might think that his theory of evidence is somewhat more like a holistic falsiﬁcationism. Suppose that mathematical claims are necessary constitutive parts of the theory. but it sounds strange to think that an observation of a swan that is a swan conﬁrms or lends inductive support to the claim that all swans are white. and yet (b) believe himself to be committed only to the objects invoked by the physical claim and not to the entities of the mathematical? The relevant consideration here seems to be whether he can insist on treating mathematical statements as mere instrumental tools in his theory rather than substantive ontologically-committing parts of his theory. while PT and FT are well motivated and defensible statements of evidential holism. only falsify the entire theory. we could ask. Jerry Fodor and Ernie Lepore suggested that something like CT is a consequence of Quine’s dissolution of the analytic/synthetic distinction or his semantic holism. they cannot can be employed to generate indispensability arguments which should compel a restricted scientiﬁc realist to be a wholesale scientiﬁc realist.272 J. All candidate such a naı examples of sophisticated hypothetico-deductive positions. rather than any particular concerns about mathematical or scientiﬁc practice. This issue alone is suggestive that the hypothetico-deductive principle as it is stated above requires modiﬁcation. as we did above.6 It is for this reason. Suppose further that this holistic falsiﬁcationist is a scientiﬁc realist. For example. How could this restricted scientiﬁc realism be defended? 6 It might be that various Quinean doctrines are supposed to yield CT. Once we see that CT is not a consequence of ‘unexceptional observations’ about the nature of prediction. Furthermore. Morrison someone might think that the intuitive appeal of PT is sufﬁcient to motivate CT. from its founder. onwards have tried to capture the intuition that ‘instances support their generalisations’ while avoiding the formalisation that equates conﬁrming instances with deductive consequences precisely because of these problems. one physical and one mathematical. it becomes difﬁcult to discern any other motivations to accept it. Samir Okasha (2000) explains why this gets the direction of ﬁt the wrong way round. Indeed. Consider. these are not the only problems for hypothetico-deductive accounts of evidence. but this seems unlikely (Fodor and Lepore 1992). taken alone. might be evidentially on a par. all swans are white entails that all swans are swans. However. His holistic falsiﬁcationism can provide a license for treating strictly mathematical claims differently from the strictly scientiﬁc. p. since it adds nothing to what S would logically imply anyway. and that as 123 . as failed predictions fail to determine which of our theory’s claims is at fault. With: No individual theoretical claims can be falsiﬁed. which he thinks are purely instrumental.Evidential Holism and Indispensability Arguments 273 One strategy is to explore what would happen if a whole theory (scientiﬁc and mathematical claims combined) made a false prediction. and sundry irrelevant sentences in S will be exempted as well. we shouldn’t conﬂate: Falsiﬁcation thesis: Observations only falsify theories. In this narrow sense. We should separate FT from the claim that no individual claims can be falsiﬁed. and recognize that FT is consistent with understanding failed predictions as allowing for a notion of differential evidential consequences for theoretical claim. he can consistently maintain that those mathematical parts of his theory. So evidential holism. are in no danger of being falsiﬁed. in the form of PT and FT but not CT. Again. That is. this demonstrates that FT does not entail that every claim in a theory must be epistemically on a par. does not generate indispensability arguments that compel a restricted scientiﬁc realist to be a wholesale scientiﬁc realist—quite to the contrary. Doing so enables the holistic falsiﬁcationist to treat some claims differently from others: for example. and that in principle any one of the constituent claims of the theory could be responsible for generating the false prediction. it provides a principled method for resisting going wholesale.’’ (Quine 1990b. Quine has long since advertised this consequence of (falsiﬁcationist) evidential holism: ‘‘We exempt some members of [theory] S from this threat [of falsiﬁcation] on determining that the fateful implication still holds without their help. Any purely logical truth is thus exempted. It follows that the falsiﬁcation thesis (FT) does not entail that every claim in a theory must be epistemically on a par. PT tells us that the whole theory has been falsiﬁed. not individual parts thereof. Does it follow that all of the parts of the theory are epistemically equal? We get the result that the falsifying-observation alone underdetermines which part of the theory is to blame. 4 Holistic indispensability arguments are supposed to suggest that a distinction between the tools of a theory and a theory’s results is not defensible. But this is consistent with the possibility that failed predictions in addition with some other auxiliary might allow us to single out particular claims for refutation. In this case. But it doesn’t follow that they are epistemically on a par more generally. the parts of the theory are epistemically equal: they are observationally on a par. 11). Individual theoretical claims cannot be falsiﬁed by failed predictions alone. and in its stead they have turned to examination of explanatory indispensability arguments. However. Many commentators have agreed that evidential holism is false. Many commentators have been convinced that Maddy’s concerns about mathematical practice are sufﬁciently problematic to repudiate holistic indispensability arguments. but for all that I agree that CT is indefensible. Bangu 2008. Rather. So. The idea is that even if the scientiﬁc realist wishes to maintain that claims involving mathematical objects are mere tools. for example. the reason she willingly accepts ontological commitments to genes and atoms is principally that they are indispensably instrumental in explanations of physical phenomena. but if true. Such a symmetry thesis is not without intuitive appeal: it seems to make sense that where an observation of litmus paper turning blue is taken to conﬁrm that the liquid is alkali. In these. the fact (if it is one) that the liquid is alkali explains why the litmus paper turned blue (see Bird 2010a for discussion). largely because of the sorts of criticisms suggested by Maddy and Sober against CT. Most contemporary discussions of indispensability arguments no longer turn on issues to do with holism. the observation of litmus paper turning blue might be taken to conﬁrm that the liquid will be dangerous if ingested in large quantities. then she should grant them the same ontological status as she does to other theoretical posits. Leng 2005. could explain some physical phenomena is generally taken to be a reason for thinking that observing the physical phenomena adduces some degree of empirical conﬁrmation on the hypothesis. the live concern is whether there are any genuine mathematical explanations of physical phenomena. I maintain that evidential holism is true. Current interests have turned to explanatory indispensability arguments (Baker 2005. Alan Baker has suggested two such physical explanada: the primenumbered periodic life cycles of North American cicadas. While this is the right result. I have disagreed with their speciﬁc criticisms of CT. I have given reasons above for thinking that it has come about for the wrong reasons. After all. and that as such the holistic indispensability arguments are unsound. That is. There are also good reasons for thinking that the biconditional does not hold without exception—most counterexamples put pressure on the conditional that if e conﬁrms h it follows that h. Morrison such scientiﬁc realists should be as committed to mathematical abstracta as they are to genes and atoms. if true. if it can be shown that those tools are indispensable to explaining physical phenomena. I do not intend to contribute to that issue. there is no valid indispensability argument from this form of evidential holism to the conclusion that scientiﬁc realists should be as committed to mathematical abstracta as they are to genes and atoms. Colyvan 2002. The converse conditional. I think the preceding discussion about evidential holism might constrain what can be said about explanation in these new indispensability arguments. would explain e. would explain e. if true. in the form of PT and FT. if true.274 J. and the relative efﬁciency 123 . however. this claim does not seem to explain why the liquid turned the litmus blue. Hempel thought that explanation is a relation that is symmetrical with conﬁrmation. e conﬁrms h if and only if h. It is this consequence that is signiﬁcant. seems much more defensible: that a hypothesis. Melia 2002). In contrast. Much of these discussions concern whether there are any genuine mathematical explanations of physical phenomena. 2009. or if they have good reasons for not doing so. or that mathematicians should look for empirical support for their theories. IBE tells us that. 2009). (Bird 2010b. I have not argued that mathematicians do not take considerations of the applied parts of their theories into account. scientiﬁc realists will not want to reject this prima facie plausible link between explanation and conﬁrmation. If we take Maddy’s objections about mathematical practice seriously enough to repudiate holistic indispensability arguments. which in turn appeals to a direct link between explanation and conﬁrmation. then we should expect mathematicians to be looking to developments in science to tell them which of their theories are conﬁrmed. If Baker is correct that the mathematical propositions are the best explanations of the physical phenomena. explanatory indispensability arguments show that mathematical claims are conﬁrmed. several competing hypotheses each provide possible explanations of the evidence. those who are adherents of explanatory indispensability arguments because of Maddy’s criticisms of holistic indispensability arguments should reject the prima facie plausible link between explanation and conﬁrmation. Thus the fact that Einstein’s general theory of relativity could explain the anomalous precession of the perihelion of Mercury was a reason to think that Einstein’s theory is correct. the argument must get something wrong. then the fact that explanatory indispensability arguments offend against those same concerns is equally problematic. As such. the argument here proceeds by parallel steps with the case against holistic indispensability arguments: if we do take Maddy’s concerns about mathematical practice seriously enough to repudiate holistic indispensability arguments. then that evidence provides some degree of conﬁrmation to that hypothesis. and if the relationship between explanation and conﬁrmation just described is plausible. For each physical phenomenon he argues that the correct explanans is a mathematical claim. If this is the case. In some cases. the evidence most strongly conﬁrms that hypothesis which best explains the evidence. Alexander Bird expresses this point succinctly: The basic idea behind IBE is that if a putative hypothesis would explain some evidence. Since they do not. subject to various constraints. 11) Since explanatory indispensability arguments are aimed at scientiﬁc realists. 123 . we should take care not to offend against them with explanatory indispensability arguments. and since scientiﬁc realists appeal to IBE.Evidential Holism and Indispensability Arguments 275 of using hexagonal structures in building hives (Baker 2005. Maddy’s concerns about mathematical practice should follow: mathematicians should augment their normal standards and methods of evidence with the additional consideration that if scientists successfully use some mathematical propositions in explaining physical phenomena then those propositions are better conﬁrmed than their rivals. Maddy’s objection from mathematical practice is seen to reapply: if scientiﬁc evidence can conﬁrm mathematical statements. For the most part. Thus. since it goes to the heart of their primary motivation for scientiﬁc realism: inference to the best explanation (IBE). Rather. p. then we have a reason for thinking that in each case the physical phenomenon conﬁrms the mathematical claim.7 7 Maddy’s mathematical practice objection has little force if mathematicians either do look to developments in science to see which of their theories is conﬁrmed. the issue is not so much of a debate about the metaphysics of mathematics. Considered like this.276 J. then the objection gives a principled reason for doing so: mathematicians don’t take scientiﬁc evidence as conﬁrmation of mathematical claims.8 The tension that is raised by the mathematical practice objection is as follows: naturalistic philosophers of science. entails the conﬁrmation of mathematical claims. who think that the epistemology of science should follow from the actual evidential standards employed by scientists. Bangu 2008. the mathematical practice objection stands as a disagreement between competing methodologies: should the question ‘‘do mathematical objects exist?’’ be answered by scientiﬁc or mathematical ways and means. so neither should scientiﬁc realists. should not consider a mathematical theory conﬁrmed as a result of its ability to explain physical phenomena. why should a scientiﬁc realist be committed to mathematical abstracta? What’s stopping a scientiﬁc realist from accepting IBE. but denying that the mathematical parts are conﬁrmed? If it seems ad hoc or arbitrary to think that best explanations provide conﬁrmation in all cases except those involving mathematical claims. Supposing that IBE is one of the methods of science. Mathematical naturalists. The epistemological premise in the indispensability arguments considered here has changed from conﬁrmational holism to IBE. 358) 9 123 .9 Rather. but the objection from practice remains: since actual mathematicians don’t look to science to discover which of their theories is conﬁrmed. who think that the epistemology of mathematics should follow from the actual evidential standards employed in maths. given that they disagree? The holistic indispensability argument can be understood as an attempt to deny that there are entirely distinct evidential methodologies available for mathematics and science—but I’ve shown how evidential holism (properly construed) fails to support such a conclusion. Penelope Maddy represents the debate this way in her 2005a (see p. (Thanks to an anonymous referee for encouraging this point). nor is it a disagreement between whether one should inquire into the metaphysical questions prior to or only after settling the methodological questions. then we have uncovered a methodological difference between mathematics and science: scientiﬁc theorising involves inferring the conﬁrmation of propositions from their status as best explanations of physical phenomena. where mathematics involves no such inference. Baker 2009). The principled reason for denying a link between explanation and conﬁrmation in the case of mathematical explanations of physical phenomena is that mathematicians don’t consider scientiﬁc applicability to be a source of support for their theories. should endorse inferential methods such as IBE. Consistent application of IBE. endorsing mathematical claims as best explanations. The explanatory indispensability 8 Mathematicians might still employ IBE within mathematics for inferring the conﬁrmation of mathematical theories or propositions from their status as best explanations of mathematical phenomena. and commitment to the existence of mathematical objects. Such applications of IBE wouldn’t be sufﬁcient to compel scientiﬁc realists to accept the existence of mathematical abstracta since in these inferences the explananda aren’t sufﬁciently connected to the general motivations of scientiﬁc realists (see also Leng 2005. following the explanatory indispensability argument. Morrison The wider point of this attack on the explanatory indispensability argument is as follows. Leng. Many thanks to my colleagues and the audience at the University of Birmingham. (2002). Kirk Surgener. (2010a). 275–289. Philosophy of Science. S. Further thanks for discussion and comments go to Darragh Byrne. Journal of Philosophy. (2005a).). 437–459). 89. Philosophy of logic. Mathematical realism: The indispensability argument. Just how controversial is evidential holism? Synthese. Ø. Duncan Pritchard. Baker. The indispensability of mathematics. Maddy. (2001). (2005b). (1971). J. 166–174. 129. V. (1992). C. Melia. (1992). Liggins. (1990a). 111. Chris Hookway. Resnik. Is platonism epistemologically bankrupt? Philosophical Review. Three forms of naturalism. (2002). 436–450. 123 . 69–74. (1992). Oxford: Oxford University Press. New York: Harper. V. W. The Philosophical Review. P.). 545–574. May 24th 2010. Putnam. O. Colyvan. (1951). The epistemology of science – a bird’s-eye view. A. Quine. (1990). Quine. D. Philosophy Compass. Holism about meaning and about evidence: in defence of W. Paul Faulkner. I’m also grateful to the editors of this journal for having been so responsive with keeping communications clear. Perspectives on Quine (pp. Fodor. B. Scientiﬁc vs. but which conﬂicts with the evidential norms found in mathematics. 60. M. Synthese. W. Erkenntnis.. heuristics and the development of mathematics (pp. Bangu. Pritchard (Eds. D. London: King’s College Publications. Cambridge. Mind. 60. P. (2008). clear and incisive comments. 173(3). (2000). (1990b). In this sense. (2010b). Mathematics and aesthetic considerations in science. M. In C. Two Dogmas of empiricism.). (2005). Bird. Erkenntnis. London: Routledge. M. 167–189). P. V. Oxford: Oxford University Press. 52. 351–376. explanatory indispensability arguments make the same mistake as holistic ones: they are insufﬁciently attentive to the different standards of evidence at work in mathematics and in science. Bernecker & D. Linnebo. 175. Quine. In S. 3(3). Colyvan. Three indeterminacies. J. which appeals to a link between explanation and evidence. Quine. British Journal for the Philosophy of Science. (1994). Acknowledgments Thanks to Jacob Busch for his feedback on earlier drafts of this work and ongoing encouragement. Hale. 13–20. & Lepore. Okasha. V. P. 39–61. 59. S. Maddy. Klee. J. (2008). Inference to the best explanation and mathematical realism. Oxford handbook of philosophy of mathematics and logic (pp. Putnam. R. References Baker.Evidential Holism and Indispensability Arguments 277 argument invokes a different evidential norm that’s found in science: IBE. David Liggins. 611–633. E. 335–352. A. Gillies (Eds. A priori knowledge: debates and developments. Philosophical Studies. Joe Melia. 487–491. In S. Oxford: Oxford University Press. Oxford: Blackwell Publishers. Maddy. Pursuit of truth. Mathematical reasoning. Shapiro (Ed. 1–16). Realism in mathematics. In Barrett & R. Are there genuine mathematical explanations of physical phenomena? Mind. Gibson (Eds. 11(3). 75–79. 5–16. Arash Pessian. Morrison. 103(2). David Walker. Sean Cordell. (2006). (2005). Response to Colyvan. I’m indebted to the anonymous reviewers for this journal for having provided such useful. W. and the ‘Quine-Putnam’ indispensability argument. Nick Wiltsher and Rich Woodward. Quine. Cellucci & D. The Routledge companion to epistemology. The Bulletin of Symbolic Logic. Gerry Hough. A. Jenkins. 68(1). 3(2). Oxford: Blackwell Publishers.). 20–43. 299–325. In defense of the Quine-Duhem thesis: a reply to Greenwood. Inductive knowledge. H. Mary Leng. Mathematical existence. Synthese. Philosophia Mathematica. Bird. (2008). Epistemological challenges to mathematical Platonism. Mass: MIT Press. Mind. (2010). 113–127. 223–238. Mathematical explanation. 160. Bob Plant. (2009). A. (1995). M. O. Maddy. 111. Mathematical explanation in science. 114. timely and relevant. Holism: A shopper’s guide. Indispensability and practice. where this work was presented to a research seminar. In Morton & S. Benacerraf and his critics (pp. D. Philosophical Forum. 92. 34 (3 & 4). G. Oxford: Blackwell. Prospects for platonism. (1996). 301–316. Proceedings of the Aristotelian Society. S. Sober. Wagner. A naturalistic a priori. 102(1). Mathematics and indispensability. Rey. E. M. 73–99). E. Morrison Resnik. 35–57. & Orlandi. Quine: I—Elliott Sober. Sober. Holistic realism: a response to Katz on holism and intuition. N. (1998). 123 . (2000).). Stich (Eds. Philosophical Review. (2003). Philosophical Studies. (1993). Quine’s Two Dogmas. 74 (Supplementary Volume).278 J. 237–280. 25–43. This action might not be possible to undo. Are you sure you want to continue? We've moved you to where you read on your other device. Get the full title to continue listening from where you left off, or restart the preview.
1 AP Physics 1 Midterm Exam Review 1. The graph above shows the velocity v as a function of time t for an object moving in a straight line. Which of the following graphs shows the corresponding displacement x as a function of time t for the same time interval? 2. A target T lies flat on the ground 3 m from the side of a building that is 10 m tall, as shown above. A student rolls a ball off the horizontal roof of the building in the direction of the target. Air resistance is negligible. The horizontal speed with which the ball must leave the roof if it is to strike the target is most nearly (A) 3/10 m/s (B) 2 m/s (C) 3 m/s (D) 3 m/s 2 3. The graph above shows velocity v versus time t for an object in linear motion. Which of the following is a possible graph of position x versus time t for this object? (A) (B) (C) (D) 2 4. Starting from rest at time t = 0, a car moves in a straight line with an acceleration given by the accompanying graph. What is the speed of the car at t = 3 s? (A) 1.0 m/s (B) 2.0 m/s (C) 6.0 m/s (D) 10.5 m/s Questions 5-6 A car starts from rest and accelerates as shown in the graph below. 5. At what time would the car be moving with the greatest velocity? (A) 2 seconds (B) 4 seconds (C) 6 seconds (D) 8 seconds 6. At what time would the car be farthest from its original starting position? (A) 2 seconds (B) 4 seconds (C) 6 seconds (D) 8 seconds 7. Which of the following sets of graphs might be the corresponding graphs of Position, Velocity, and Acceleration vs. Time for a moving particle? (A) (B) (C) (D) 3 8. An arrow is aimed horizontally, directly at the center of a target 20 m away. The arrow hits m below the center of the target. Neglecting air resistance, what was the initial speed of the arrow? (A) 20 m/s (B) 40 m/s (C) 100 m/s (D) 200 m/s 9. A 40 kg box is dragged across a frictionless horizontal surface by a rope. The tension in the rope is 65 N and it is at an angle of 42 o to the horizontal. What is the box's acceleration? A. 1.6 m/s 2 B. 1.1 m/s 2 C. 0.6 m/s 2 D. 1.2 m/s 2 E. 0.9 m/s What is the acceleration of a block on a ramp inclined 35 degrees to the horizontal if µ k = 0.4? A. 3.5 m/s 2. B. 0.9 m/s 2. C. 2.4 m/s 2. D. 9.8 m/s 2. E. not enough information given. 11. A block of mass 5 kilograms lies on an inclined plane, as shown above. The horizontal and vertical supports for the plane have lengths of 4 meters and 3 meters, respectively. The coefficient of kinetic friction between the plane and the block is 0.3. The magnitude of the force F necessary to pull the block up the plane with constant speed is most nearly (A) 30 N (B) 41 N (C) 49 N (D) 50 N (E) 58 N 12. When an object of weight W is suspended from the center of a massless string as shown above, the tension at any point in the string is (A) 2Wcos (B) ½Wcos (C) Wcos (D) W/(2cos ) (E) W/(cos ) 4 13. A block of mass 3m can move without friction on a horizontal table. This block is attached to another block of mass m by a cord that passes over a frictionless pulley, as shown above. If the masses of the cord and the pulley are negligible, what is the magnitude of the acceleration of the descending block? (A) Zero (B) g/4 (C) g/3 (D) 2g/3 (E) g 14. A rope of negligible mass supports a block that weighs 30 N, as shown above. The breaking strength of the rope is 50 N. The largest acceleration that can be given to the block by pulling up on it with the rope without breaking the rope is most nearly (A) 6 m/s 2 (B) 6.7 m/s 2 (C) 10 m/s 2 (D) 15 m/s 2 (E) 16.7 m/s 15. A simple Atwood's machine (above) remains motionless when equal masses M are placed on each end of the chord. When a small mass m is added to one side, the masses have an acceleration a. What is M? You may neglect friction and the mass of the cord and pulley. (see sketch above) (A) m(g a) 2a (B) 2m(g a) a (C) 2m(g+a) a (D) m(g+a) 2a (E) m(a g) 2a 5 16. A block of mass M is initially at rest on a frictionless floor. The block, attached to a massless spring with spring constant k, is initially at its equilibrium position. An arrow with mass m and velocity v is shot into the block. The arrow sticks in the block. What is the maximum compression of the spring? ( A) v m k ( B) v m M k ( m M ) v ( C) mk ( D) mv ( m M ) k 17. Two 5 kg masses are attached to opposite ends of a long massless cord which passes tautly over a massless frictionless pulley. The upper mass is initially held at rest on a table 50 cm from the pulley. The coefficient of kinetic friction between this mass and the table is 0.2. When the system is released, its resulting acceleration is closest to which of the following? (see sketch above) (A) 9.8 m/s 2 (B) 7.8 m/s 2 (C) 4.9 m/s 2 (D) 3.9 m/s 2 (E) 1.9 m/s A spring is compressed between two objects with unequal masses, m and M, and held together. The objects are initially at rest on a horizontal frictionless surface. When released, which of the following is true? (A) The total final kinetic energy is zero. (B) The two objects have equal kinetic energy. (C) The speed of one object is equal to the speed of the other. (D) The total final momentum of the two objects is zero. 19. Two football players with mass 75 kg and 100 kg run directly toward each other with speeds of 6 m/s and 8 m/s respectively. If they grab each other as they collide, the combined speed of the two players just after the collision would be: (A) 2 m/s (B) 3.4 m/s (C) 4.6 m/s (D) 7.1 m/s 20. A rubber ball is held motionless a height h o above a hard floor and released. Assuming that the collision with the floor is elastic, which one of the following graphs best shows the relationship between the total energy E of the ball and its height h above the surface. 6 (A) (B) (C) (D) 21. A mass m has speed v. It then collides with a stationary object of mass 2m. If both objects stick together in a perfectly inelastic collision, what is the final speed of the newly formed object? (A) v / 3 (B) v / 2 (C) 2v / 3 (D) 3v / A tennis ball of mass m rebounds from a racquet with the same speed v as it had initially as shown. The magnitude of the momentum change of the ball is (A) 0 (B) 2mv (C) 2mv sin (D) 2mv cos 23. From the top of a high cliff, a ball is thrown horizontally with initial speed v o. Which of the following graphs best represents the ball's kinetic energy K as a function of time t? (A) (B) (C) (D) 24. The following graphs, all drawn to the same scale, represent the net force F as a function of displacement x for an object that moves along a straight line. Which graph represents the force that will cause the greatest change in the kinetic energy of the object from x = 0 to x = x 1? (A) (B) (C) (D) 25. A pendulum bob of mass m on a cord of length L is pulled sideways until the cord makes an angle θ with the vertical as shown in the figure to the right. The change in potential energy of the bob during the displacement is: (A) mgl (1 cos θ) (B) mgl (1 sin θ) (C) mgl sin θ (D) mgl cos θ 7 26. A child holds a ball in their hand, taking this position to be zero. They toss the ball upwards and catch it at the same level. Select the correct row of signs: Ascent Peak of flight Descent s v a s v a s v a a b c d e At time zero, two runners run side-by-side for an instant as one passes the other. What is true of the time marked t? a. The runners are again side-by-side b. Runner A is ahead c. Runner B is ahead d. The runners have the same acceleration e. None of the above 28. Two objects begin at the same position and the same time zero. Subsequent times are labeled for each. When do the two have the same speed? a. Before one second b. It will not happen until after six seconds c. Just after two seconds d. Between three and four seconds e. They never have the same speed and never will 8 29. In the graph above, during how many times spans is there positive acceleration and slowing down? a. Zero b. One c. Two d. Three e. Four 30. In Cartesian coordinates, what is 1 2 A + B 2C? a. < -5, 4 > b. < 4, 4 > c. < -8, 4 > d. < -7, 5 > 31. A projectile is fired from left to right as shown above. During this first half of the flight diagrammed, the speed of the projectile is: a. remaining constant b. increasing c. decreasing 9 32. On a smooth, level track, three strongmen can accelerate a train car at an acceleration a. What acceleration can six total strongmen produce when two similar train cars are added to the first? a. 2a b. a c. 3 2 a d. 2 3 a e. 1 2 a 33. How many of the graphs above demonstrate a net force of zero? a. four b. two c. three d. one 34. On the frictionless surface above, what is the force of the 4kg block on the 1kg block? a. 26N b. 2N c. 8N d. 18N 35. A 2.0kg block slides down a 30º incline at a constant velocity. What is the coefficient of kinetic friction between the block and incline? a b c d 10 36. The coefficient of static friction between the wall and block is μ. What force, F, must be applied to keep the block from sliding down? a. mg/μ b. mg c. μ/mg d. μmg 37. Each of the spherical masses is 1kg. Rank the tensions in the cords. a. B = D > A = C = E b. A = B = C = D = E c. B > A = C = D = E d. B = C = D > A = E 38. What is the net gravitational force on the center mass? a. 4Gm2 3D 2 28Gm2 28Gm2 b. c. d. 12Gm2 3D2 3D 2 D 2 11 39. A rider spins in an amusement park ride so that they do not slide down. Which set of physical forces acts on the passenger? a. a force of gravity, a force of static friction, and a centrifugal force b. a force of gravity, a normal force from the wall, and a centripetal force c. a force of gravity, a force of static friction, and a normal force from the wall d. a force of gravity, a normal force from the wall, and a centrifugal force 40. A satellite of mass m orbits a planet of mass M at an orbital radius R and speed v. Which of the following is true for the satellite? I. the acceleration of the satellite is GM R 2 II. the acceleration of the satellite is v2 R III. the period of the satellite s orbit is 2πR a. I and III b. II and III c. I and II d. I and II and III v 41. A bungee jumper has a mass of 50kg and falls straight 40m downward from a bridge. If they are then traveling at -10m/s, the work done by gravity has been and the work done by the bungee cord has been. a. -20,000 N m, 20,000 N m b. -20,000 N m, 2,500 N m c. 2,500 N m, 0 N m d. 20,000 N m, -17,500 N m 12 42. A cart with a mass of 4.0kg has a velocity of 2m/s when at a position of 3m. What is the velocity of the cart when at a position of 8m? a. 3 m/s b. 7 m/s c. 2 m/s d. - 2 m/s 43. A toy car with a mass of 500g has a velocity of 6m/s when at a position of 2m. What is the velocity of the toy car when at a position of 4m? a. 20 m/s b. 4 m/s c. 1 m/s d. 5 m/s 44. An 1800kg car is capable of accelerating from 0m/s to 10m/s in 1.5 seconds. If the power output of the car is constant, how much time will it take the car to accelerate from 0m/s to 40m/s? a. 3s b. 6s c. 12s d. 24s 45. A person with a mass m stands on the left end a wooden board with a mass of 10m, which is on frictionless ice. The person walks to the right until they reach the right end of the board. Relative to the stationary post shown on the left, the person, the board, and the center of mass of the person/board system. a. moves right, moves right, remains stationary b. moves right, moves left, moves left c. moves right, moves left, moves right d. moves right, moves left, remains stationary 13 46. What are the coordinates of the center of mass of the system above? a. 4.25m, -1.00m b. -1.5m, 1.00m c. 3.11m, 0.00m d m, 2.13m 47. Taking the bottom-left corner as the origin, what are the coordinates of the center of mass of the system above? a. 18L, 12L b. 6L 13, 8L 13 c. 9L, 24L d. 9L, 12L The left diagram shows the momentum vector of a firecracker as it flies through the air. It then explodes into two pieces. Which of the remaining vector diagrams represents a possibility for the momenta of the two fragments immediately after the explosion? 14 49. A block of mass 100m sits at rest on a frictionless surface. A bullet with mass m strikes the block at 100v and leaves the block at 10v. What is the velocity of the block after the bullet has passed through? a v b v c v d v 50. Is the collision diagrammed above perfectly elastic? a. yes b. no c. cannot be determined 15 1. A model rocket is launched vertically with an engine that is ignited at time t = 0, as shown above. The engine provides an upward acceleration of 30 m/s 2 for 2.0 s. Upon reaching its maximum height, the rocket deploys a parachute, and then descends vertically to the ground. a. Determine the speed of the rocket after the 2 s firing of the engine. b. What maximum height will the rocket reach? c. At what time after t = 0 will the maximum height be reached? 2. A world-class runner can complete a 100 m dash in about 10 s. Past studies have shown that runners in such a race accelerate uniformly for a time t and then run at constant speed for the remainder of the race. A world-class runner is visiting your physics class. You are to develop a procedure that will allow you to determine the uniform acceleration a and an approximate value of t for the runner in a 100 m dash. By necessity your experiment will be done on a straight track and include your whole class of eleven students. a. By checking the line next to each appropriate item in the list below, select the equipment, other than the runner and the track, that your class will need to do the experiment. Stopwatches Tape measures Rulers Masking tape Metersticks Starter's pistol String Chalk b. Outline the procedure that you would use to determine a and t, including a labeled diagram of the experimental setup. Use symbols to identify carefully what measurements you would make and include in your procedure how you would use each piece of the equipment you checked in part (a). c. Outline the process of data analysis, including how you will identify the portion of the race that has uniform acceleration, and how you would calculate the uniform acceleration. 16 3. A mass-less string passes over a frictionless pulley hanging from the ceiling. (a classic two mass Atwood s machine) The string connects two objects of mass 15.0 kg and 20.0 kg. When the masses are released (a) find their acceleration and (b) the tension in the string between them. 4. A track consists of a frictionless arc XY, which is a quarter-circle of radius R, and a rough horizontal section YZ. Block A of mass M is released from rest at point X, slides down the curved section of the track, and collides instantaneously and inelastically with identical block B at point Y. The two blocks move together to the right, sliding past point P, which is a distance L from point Y. The coefficient of kinetic friction between the blocks and the horizontal part of the track is Express your answers in terms of M, L,, R, and g. a. Determine the speed of block A just before it hits block B. b. Determine the speed of the combined blocks immediately after the collision. c. Assuming that no energy is transferred to the track or to the air surrounding the blocks. Determine the amount of energy transferred in the collision d. Determine the additional thermal energy that is generated as the blocks move from Y to P 17 5. Two objects of masses m1 = 0.65 kg and m2 = 1.00 kg are placed on a horizontal frictionless surface and a compressed spring of spring constant K = 290 N/m is placed between them. Neglect the mass of the spring. The spring is not attached to either object and is compressed a distance of 11 cm. If the objects are released from rest, find the final velocity of each mass. 6. From near ground-level, a cannon fires a cannonball at a velocity of (90m/s, 30º). The cannonball strikes a castle wall 500m from the cannon. a. At what height does it strike the wall? Show work for partial credit. b. Sketch six graphs versus time: x-position, x-velocity, x-acceleration, y-position, y-velocity, y-acceleration. 7. The system above has a block of mass 2m hanging on the right and tied to a cord. The cord runs over two pulleys and is tied on the other end to an L-brace with a mass of m which sits on the floor. Placed on the L- brace is a bowling ball of mass 3m. a. What is the tension in the cord? b. What is the normal force of the floor on the L-brace? c. Draw a free-body-diagram for the L-brace, labeling all forces. d. If the bowling ball then rolls off of the L-brace, with what acceleration will it rise? 18 8. The orbital radii and orbital periods for several planets are provided below. Complete the chart by calculating R 3 and T 2. Planet R (m) T (s) R 3 T 2 Venus 1.08 x x 10 7 Earth 1.50 x x 10 7 Mars 2.28 x x 10 7 b. Sketch a graph of R 3 versus T 2. Include axes labels and units. c. Draw a linear regression line. Write the slope of the line, including units, next to the graph. d. Derive an equation for the mass of the sun in terms of T, R, and any fundamental constants. 19 9. A block with a mass of m begins at rest along the left side of a loop with radius R. The block slides down the loop and then across a rough surface with a coefficient of kinetic friction μ. a. What is the speed of the block at the base of the circle? b. How far, L, along the rough portion of the flat track will the block slide before stopping? c. How much thermal energy will be generated by friction as the block slows down? d. Check what effects each of the following would have on the distance, L, the block slides. Increasing block mass Increasing R Decreasing μ Decreasing g Increases Decreases Has no effect 20 10. A motion sensor and a force sensor record the motion of a cart along a track, as shown above. The cart is given a push so that it moves toward the force sensor and then collides with it. The two sensors record the values shown in the following graphs. a. Determine the cart's average acceleration between t = 0.33 s and t = 0.37 s. b. Determine the magnitude of the change in the cart's momentum during the collision. c. Determine the mass of the cart. d. Determine the energy lost in the collision between the force sensor and the cart. 1. If the kinetic energy of an object is 16 joules when its speed is 4.0 meters per second, then the mass of the objects is (1) 0.5 kg (3) 8.0 kg (2) 2.0 kg (4) 19.6 kg Base your answers to questions 9 1. Which one of the following statements concerning kinetic energy is true? A) Kinetic energy can be measured in watts. B) Kinetic energy is always equal to the potential energy. C) Kinetic energy is always Physics 2A, Sec B00: Mechanics -- Winter 2011 Instructor: B. Grinstein Final Exam INSTRUCTIONS: Use a pencil #2 to fill your scantron. Write your code number and bubble it in under "EXAM NUMBER;" an entry Physics 23 Exam 2 Spring 2010 Dr. Alward Page 1 1. A 250-N force is directed horizontally as shown to push a 29-kg box up an inclined plane at a constant speed. Determine the magnitude of the normal force, 1. A car enters a horizontal, curved roadbed of radius 50 m. The coefficient of static friction between the tires and the roadbed is 0.20. What is the maximum speed with which the car can safely negotiate Chapter 5 Using Newton s Laws: Friction, Circular Motion, Drag Forces Units of Chapter 5 Applications of Newton s Laws Involving Friction Uniform Circular Motion Kinematics Dynamics of Uniform Circular Name: Class: _ Date: _ AP Physics C Fall Final Web Review Multiple Choice Identify the choice that best completes the statement or answers the question. 1. On a position versus time graph, the slope of Practice Midterm 1 1) When a parachutist jumps from an airplane, he eventually reaches a constant speed, called the terminal velocity. This means that A) the acceleration is equal to g. B) the force of 1. What is the average speed of an object that travels 6.00 meters north in 2.00 seconds and then travels 3.00 meters east in 1.00 second? 9.00 m/s 3.00 m/s 0.333 m/s 4.24 m/s 2. What is the distance traveled 1. Two boys with masses of 40 kg and 60 kg are holding onto either end of a 10 m long massless pole which is initially at rest and floating in still water. They pull themselves along the pole toward each Physics: Principles and Applications, 6e Giancoli Chapter 4 Dynamics: Newton's Laws of Motion Conceptual Questions 1) Which of Newton's laws best explains why motorists should buckle-up? A) the first law Chapter 10 Energy and Work 10.1 Quantitative 1) A child does 350 J of work while pulling a box from the ground up to his tree house with a rope. The tree house is 4.8 m above the ground. What is the mass * By request, but I m not vouching for these since I didn t write them Exam 2 is at 7 pm tomorrow Conflict is at 5:15 pm in 151 Loomis There are extra office hours today & tomorrow Lots of practice exams MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics Physics 8.01T Fall Term 2004 Practice Exam Three Solutions Problem 1a) (5 points) Collisions and Center of Mass Reference Frame In the lab frame, AP Physics C Oscillations/SHM Review Packet 1. A 0.5 kg mass on a spring has a displacement as a function of time given by the equation x(t) = 0.8Cos(πt). Find the following: a. The time for one complete Week 3 homework IMPORTANT NOTE ABOUT WEBASSIGN: In the WebAssign versions of these problems, various details have been changed, so that the answers will come out differently. The method to find the solution Chapter 7: Momentum and Impulse 1. When a baseball bat hits the ball, the impulse delivered to the ball is increased by A. follow through on the swing. B. rapidly stopping the bat after impact. C. letting 1. A body of mass m slides a distance d along a horizontal surface. How much work is done by gravity? A) mgd B) zero C) mgd D) One cannot tell from the given information. E) None of these is correct. 2. Exam Three Momentum Concept Questions Isolated Systems 4. A car accelerates from rest. In doing so the absolute value of the car's momentum changes by a certain amount and that of the Earth changes by: Work, Energy & Momentum Homework Packet Worksheet 1: This is a lot of work! 1. A student holds her 1.5-kg psychology textbook out of a second floor classroom window until her arm is tired; then she releases 1. An archer pulls his bow string back 0.4 m by exerting a force that increases uniformly from zero to 230 N. The equivalent spring constant of the bow is: A. 115 N/m B. 575 N/m C. 1150 N/m D. 287.5 N/m Midterm Solutions I) A bullet of mass m moving at horizontal velocity v strikes and sticks to the rim of a wheel a solid disc) of mass M, radius R, anchored at its center but free to rotate i) Which of L06-1 Name Date Partners LAB 6 - GRAVITATIONAL AND PASSIVE FORCES OBJECTIVES And thus Nature will be very conformable to herself and very simple, performing all the great Motions of the heavenly Bodies Worksheet #1 Free Body or Force diagrams Drawing Free-Body Diagrams Free-body diagrams are diagrams used to show the relative magnitude and direction of all forces acting upon an object in a given situation. Tennessee State University Dept. of Physics & Mathematics PHYS 2010 CF SU 2009 Name 30% Time is 2 hours. Cheating will give you an F-grade. Other instructions will be given in the Hall. MULTIPLE CHOICE. Sample Questions for the AP Physics 1 Exam Sample Questions for the AP Physics 1 Exam Multiple-choice Questions Note: To simplify calculations, you may use g 5 10 m/s 2 in all problems. Directions: Each MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 1) Vector A has length 4 units and directed to the north. Vector B has length 9 units and is directed 55 Name Date Partners LAB 6: GRAVITATIONAL AND PASSIVE FORCES And thus Nature will be very conformable to herself and very simple, performing all the great Motions of the heavenly Bodies by the attraction AP Physics B Practice Workbook Book 1 Mechanics, Fluid Mechanics and Thermodynamics. The following( is applicable to this entire document copies for student distribution for exam preparation explicitly Phys 131 Fall 2015 Fundamental Mechanics: Supplementary Exercises 1 Motion diagrams: horizontal motion A car moves to the right. For an initial period it slows down and after that it speeds up. Which of Chapter 4 Forces and Newton s Laws of Motion continued 4.9 Static and Kinetic Frictional Forces When an object is in contact with a surface forces can act on the objects. The component of this force acting Name: ate: 1. How much work is required to lift a 2-kilogram mass to a height of 10 meters?. 5 joules. 20 joules. 100 joules. 200 joules 5. ar and car of equal mass travel up a hill. ar moves up the hill Exam Name SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question. 1) A person on a sled coasts down a hill and then goes over a slight rise with speed 2.7 m/s. Chapter 9 9. Figure 9-36 shows a three particle system. What are (a) the x coordinate and (b) the y coordinate of the center of mass of the three particle system. (c) What happens to the center of mass Ph\sics 2210 Fall 2012 - Novcmbcr 21 David Ailion Unid: Discussion T A: Bryant Justin Will Yuan 1 Place answers in box provided for each question. Specify units for each answer. Circle correct answer(s) MASSACHUSETTS INSTITUTE OF TECHNOLOGY Physics Department 8.01L: Physics I November 7, 2015 Prof. Alan Guth Problem Set #8 Solutions Due by 11:00 am on Friday, November 6 in the bins at the intersection 1. Which of the following statements about a spring-block oscillator in simple harmonic motion about its equilibrium point is false? (A) The displacement is directly related to the acceleration. (B) The A cannon shoots a clown directly upward with a speed of 20 m/s. What height will the clown reach? How much time will the clown spend in the air? Projectile Motion 1:Horizontally Launched Projectiles Two BHS Freshman Physics Review Chapter 2 Linear Motion Physics is the oldest science (astronomy) and the foundation for every other science. Galileo (1564-1642): 1 st true scientist and 1 st person to use Force Force as a Vector Real Forces versus Convenience The System Mass Newton s Second Law Outline Force as a Vector Forces are vectors (magnitude and direction) Drawn so the vector s tail originates at Page 1 of 12 CTEnergy-1. A mass m is at the end of light (massless) rod of length R, the other end of which has a frictionless pivot so the rod can swing in a vertical plane. The rod is initially horizontal Lecture 6 Weight Tension Normal Force Static Friction Cutnell+Johnson: 4.8-4.12, second half of section 4.7 In this lecture, I m going to discuss four different kinds of forces: weight, tension, the normal VELOCITY, ACCELERATION, FORCE velocity Velocity v is a vector, with units of meters per second ( m s ). Velocity indicates the rate of change of the object s position ( r ); i.e., velocity tells you how chapter 5 Newton s Law of Motion Static system 1. Hanging two identical masses Context in the textbook: Section 5.3, combination of forces, Example 4. Vertical motion without friction 2. Elevator: Decelerating PS/PHYSICS The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION PHYSICAL SETTING PHYSICS Tuesday, June 22, 2010 9:15 a.m. to 12:15 p.m., only The answers to all questions in this examination Physics 1401 - Exam 2 Chapter 5N-New 2. The second hand on a watch has a length of 4.50 mm and makes one revolution in 60.00 s. What is the speed of the end of the second hand as it moves in uniform circular Problem Set 1 1.1 A bicyclist starts from rest and after traveling along a straight path a distance of 20 m reaches a speed of 30 km/h. Determine her constant acceleration. How long does it take her to CHAPTER 1 SECTION Matter in Motion 4 Gravity: A Force of Attraction BEFORE YOU READ After you read this section, you should be able to answer these questions: What is gravity? How are weight and mass different? Physics Assignment KEY Dynamics Chapters 4 & 5 ote: for all dynamics problem-solving questions, draw appropriate free body diagrams and use the aforementioned problem-solving method.. Define the following Review Chapters 2, 3, 4, 5 4) The gain in speed each second for a freely-falling object is about A) 0. B) 5 m/s. C) 10 m/s. D) 20 m/s. E) depends on the initial speed 9) Whirl a rock at the end of a string 8. Potential Energy and Conservation of Energy Potential Energy: When an object has potential to have work done on it, it is said to have potential energy, e.g. a ball in your hand has more potential energy 5. Forces and Motion-I 1 Force is an interaction that causes the acceleration of a body. A vector quantity. Newton's First Law: Consider a body on which no net force acts. If the body is at rest, it will CHAPTER 6 WORK AND ENERGY CONCEPTUAL QUESTIONS. REASONING AND SOLUTION The work done by F in moving the box through a displacement s is W = ( F cos 0 ) s= Fs. The work done by F is W = ( F cos θ). s From CTA-1. An Atwood's machine is a pulley with two masses connected by a string as shown. The mass of object A, m A, is twice the mass of object B, m B. The tension T in the string on the left, above mass Practice Test SHM with Answers MPC 1) If we double the frequency of a system undergoing simple harmonic motion, which of the following statements about that system are true? (There could be more than one Physics: Principles and Applications, 6e Giancoli Chapter 2 Describing Motion: Kinematics in One Dimension Conceptual Questions 1) Suppose that an object travels from one point in space to another. Make FRICTION, WORK, AND THE INCLINED PLANE Objective: To measure the coefficient of static and inetic friction between a bloc and an inclined plane and to examine the relationship between the plane s angle Chapter 3: Falling Objects and Projectile Motion 1. Neglecting friction, if a Cadillac and Volkswagen start rolling down a hill together, the heavier Cadillac will get to the bottom A. before the Volkswagen. Chapter 4: Newton s Laws: Explaining Motion 1. All except one of the following require the application of a net force. Which one is the exception? A. to change an object from a state of rest to a state Lecture 07: Work and Kinetic Energy Physics 2210 Fall Semester 2014 Announcements Schedule next few weeks: 9/08 Unit 3 9/10 Unit 4 9/15 Unit 5 (guest lecturer) 9/17 Unit 6 (guest lecturer) 9/22 Unit 7, PHYSICS HOMEWORK #1 DISPLACEMENT & VELOCITY KINEMATICS d v average t v ins d t verysmall / error d t d t v a ave t 1. You walk exactly 50 steps North, turn around, and then walk exactly 400 steps South. Supplemental Questions The fastest of all fishes is the sailfish. If a sailfish accelerates at a rate of 14 (km/hr)/sec [fwd] for 4.7 s from its initial velocity of 42 km/h [fwd], what is its final velocity? 4277(a) Semester 2, 2011 Page 1 of 9 THE UNIVERSITY OF SYDNEY EDUH 1017 - SPORTS MECHANICS NOVEMBER 2011 Time allowed: TWO Hours Total marks: 90 MARKS INSTRUCTIONS All questions are to be answered. Use STE Physics Intro Name 1. Mass, Force and Gravity Before attempting to understand force, we need to look at mass and acceleration. a) What does mass measure? The quantity of matter(atoms) b) What is the Chapter 9 9.2 Figure 9-37 shows a three particle system with masses m 1 3.0 kg, m 2 4.0 kg, and m 3 8.0 kg. The scales are set by x s 2.0 m and y s 2.0 m. What are (a) the x coordinate and (b) the y coordinate circular motion & gravitation physics 111N uniform circular motion an object moving around a circle at a constant rate must have an acceleration always perpendicular to the velocity (else the speed would Forces When an object is pushed or pulled, we say that a force is exerted on it. Forces can Cause an object to start moving Change the speed of a moving object Cause a moving object to stop moving Change PHY101 Solved MCQ'S For Exam www.vuattach.ning.com Name: Date: 1. A right circular cylinder with a radius of 2.3 cm and a height of 1.4 cm has a total surface area of: A) 1.7 10 3 m 2 B) 3.2 10 3 m 2 C) PhysicsFactsheet September 2000 Number 05 Work Energy & Power 1. Work If a force acts on a body and causes it to move, then the force is doing work. W = Fs W = work done (J) F = force applied (N) s = distance AP Physics Circular Motion Practice Test B,B,B,A,D,D,C,B,D,B,E,E,E, 14. 6.6m/s, 0.4 N, 1.5 m, 6.3m/s, 15. 12.9 m/s, 22.9 m/s Answer the multiple choice questions (2 Points Each) on this sheet with capital MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department o Physics Physics 8.1 Fall 1 Problem Set 5 Work and Kinetic Energy Solutions Problem 1: Work Done by Forces a) Two people push in opposite directions on Candidates should be able to : Derive the equations of motion for constant acceleration in a straight line from a velocity-time graph. Select and use the equations of motion for constant acceleration in Pendulum Force and Centripetal Acceleration 1 Objectives 1. To calibrate and use a force probe and motion detector. 2. To understand centripetal acceleration. 3. To solve force problems involving centripetal Chapter 7 Momentum and Impulse Collisions! How can we describe the change in velocities of colliding football players, or balls colliding with bats?! How does a strong force applied for a very short time Physics Midterm Review Packet January 2010 This Packet is a Study Guide, not a replacement for studying from your notes, tests, quizzes, and textbook. Midterm Date: Thursday, January 28 th 8:15-10:15 Room: Revised form 081695R Force Concept Inventory Originally published in The Physics Teacher, March 1992 by David Hestenes, Malcolm Wells, and Gregg Swackhamer Revised August 1995 by Ibrahim Halloun, Richard
Take a rectangle of paper and fold it in half, and half again, to make four smaller rectangles. How many different ways can you fold 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? 10 space travellers are waiting to board their spaceships. There are two rows of seats in the waiting room. Using the rules, where are they all sitting? Can you find all the possible ways? An irregular tetrahedron is composed of four different triangles. Can such a tetrahedron be constructed where the side lengths are 4, 5, 6, 7, 8 and 9 units of length? Use the interactivity to play two of the bells in a pattern. How do you know when it is your turn to ring, and how do you know which bell to ring? Starting with four different triangles, imagine you have an unlimited number of each type. How many different tetrahedra can you make? Convince us you have found them all. What is the greatest number of counters you can place on the grid below without four of them lying at the corners of a square? What is the best way to shunt these carriages so that each train can continue its journey? Can you work out how many cubes were used to make this open box? What size of open box could you make if you had 112 cubes? Can you shunt the trucks so that the Cattle truck and the Sheep truck change places and the Engine is back on the main line? Using different numbers of sticks, how many different triangles are you able to make? Can you make any rules about the numbers of sticks that make the most triangles? In how many ways can you fit two of these yellow triangles together? Can you predict the number of ways two blue triangles can be fitted together? Design an arrangement of display boards in the school hall which fits the requirements of different people. Take 5 cubes of one colour and 2 of another colour. How many different ways can you join them if the 5 must touch the table and the 2 must not touch the table? The ancient Egyptians were said to make right-angled triangles using a rope with twelve equal sections divided by knots. What other triangles could you make if you had a rope like this? How many triangles can you make on the 3 by 3 pegboard? How many different ways can you find of fitting five hexagons together? How will you know you have found all the ways? An activity making various patterns with 2 x 1 rectangular tiles. Swap the stars with the moons, using only knights' moves (as on a chess board). What is the smallest number of moves possible? You have 4 red and 5 blue counters. How many ways can they be placed on a 3 by 3 grid so that all the rows columns and diagonals have an even number of red counters? These practical challenges are all about making a 'tray' and covering it with paper. A 2 by 3 rectangle contains 8 squares and a 3 by 4 rectangle contains 20 squares. What size rectangle(s) contain(s) exactly 100 squares? Can you find them all? How many models can you find which obey these rules? 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? How many different triangles can you make on a circular pegboard that has nine pegs? Building up a simple Celtic knot. Try the interactivity or download the cards or have a go on squared paper. A dog is looking for a good place to bury his bone. Can you work out where he started and ended in each case? What possible routes could he have taken? 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! Zumf makes spectacles for the residents of the planet Zargon, who have either 3 eyes or 4 eyes. How many lenses will Zumf need to make all the different orders for 9 families? In a square in which the houses are evenly spaced, numbers 3 and 10 are opposite each other. What is the smallest and what is the largest possible number of houses in the square? Arrange eight of the numbers between 1 and 9 in the Polo Square below so that each side adds to the same total. Only one side of a two-slice toaster is working. What is the quickest way to toast both sides of three slices of bread? 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? How can you put five cereal packets together to make different shapes if you must put them face-to-face? When newspaper pages get separated at home we have to try to sort them out and get things in the correct order. How many ways can we arrange these pages so that the numbering may be different? How many different symmetrical shapes can you make by shading triangles or squares? Find out what a "fault-free" rectangle is and try to make some of A game for 2 people. Take turns placing a counter on the star. You win when you have completed a line of 3 in your colour. Just four procedures were used to produce a design. How was it done? Can you be systematic and elegant so that someone can follow This practical challenge invites you to investigate the different squares you can make on a square geoboard or pegboard. Place eight queens on an chessboard (an 8 by 8 grid) so that none can capture any of the others. The planet of Vuvv has seven moons. Can you work out how long it is between each super-eclipse? Place the numbers 1 to 10 in the circles so that each number is the difference between the two numbers just below it. Make your own double-sided magic square. But can you complete both sides once you've made the pieces? An extra constraint means this Sudoku requires you to think in diagonals as well as horizontal and vertical lines and boxes of Can you put the numbers 1 to 8 into the circles so that the four calculations are correct? How many shapes can you build from three red and two green cubes? Can you use what you've found out to predict the number for four red and two green? Can you find all the different triangles on these peg boards, and find their angles? You have two egg timers. One takes 4 minutes exactly to empty and the other takes 7 minutes. What times in whole minutes can you measure and how? 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?
All Slader step-by-step solutions are FREE. Can you find your fundamental truth using Slader as a completely free Algebra 2 (Volume 2) solutions manual? Now is the time to redefine your true self using Slader’s free Algebra 2 (Volume 2) answers. Shed the societal and cultural narratives holding you back and let free step-by-step Algebra 2. Evaluate: Homework and Practice Find the rational zeros of each polynomial function. Then write each function in Online Hints Extra factored form.This Homework Practice Workbook gives you additional problems for the concept exercises in each lesson. The exercises are designed to aid your study of mathematics by reinforcing important mathematical skills needed to succeed in the everyday world. The materials are organized by chapter and lesson, with one Practice worksheet for every lesson in Glencoe Algebra 1. To the Teacher These.Algebra 2 Honors. The purpose of this course is to deepen your understanding of advanced algebra and statistics while preparing you for the SAT and higher level mathematics. You should be very comfortable with the concepts you learned in Algebra 1 since we will build upon them. All students will need to bring their computer, note-taking guide with paper, and calculator to class every time we. Homework and Practice 2-3 Equations and Their Solutions LESSON 1. 26 w 18 for w 8 3. 42 g 15 for g 26 5. 32 z 8 for z 4 7. d 3 4 for d 15 9. 36 m 81 for m 45 11. v 31 13 for v 45 13. 4r 107 for r 27 15. n 0 70 for n 70 2. a 3 19 for a 58 4. k 6 108 for k 18 6. u 23 41 for u 17 8. 13y 65 for y 5 10. 7 j 84 for j 12 12. 69 c 121 for c 52 14. 9 q. A few items on the Problem Set and Homework Assignments may vary slightly due to the fact that our students are using recently updated materials. Learning Gets Easy with Personal Math Trainer Evaluate Homework and Practice Answers Though every subject comes with a challenge, but math is a subject which can actually give shivers to everyone. I must quote here that most of the students avoid doing math homework! Interpreting Vertex Form and Standard Form Graphing Quadratic Functions solutions, Houghton Mifflin Harcourt Algebra 1, 2015. Download Mathleaks app to gain access to the solutions. Start at the beginning of the module. Teach for approximately 20 - 30 minutes. This includes the four-step teaching process. At the conclusion of your lesson, summarize and evaluate. You will evaluate yourself and your peers will be given the opportunity to provide feedback. The instructor will provide feedback. Full Presentation. PW page 1-10. Understanding Piecewise-Defined Functions Piecewise-Defined Functions solutions, Houghton Mifflin Harcourt Algebra 1, 2015. Download Mathleaks app to gain access to the solutions. LOGARITHMS AND THEIR PROPERTIES Definition of a logarithm: If and is a constant, then if and only if. In the equation is referred to as the logarithm, is the base, and is the argument. The notation is read “the logarithm (or log) base of .” The definition of a logarithm indicates that a logarithm is an exponent. EngageNY math 6th grade 6 Eureka, worksheets, Equivalent Ratios, The Structure of Ratio Tables, Additive and Multiplicative, From Ratios Tables to Double Number Line Diagrams, Common Core Math, by grades, by domains, examples and step by step solutions. Proportions and MODULE 5 Percent Get immediate feedback and help as you work through practice sets. Personal Math Trainer Interactively explore key concepts to see how math works. Animated Math Go digital with your write-in student edition, accessible on any device. my.hrw.com Scan with your smart phone to jump directly to the online edition. Helping Your Students With Homework A Guide for Teachers. Helping Your StudentsWith Homework A Guide for Teachers By Nancy Paulu Edited by Linda B. DarbyIllustrated by Margaret Scott Office of Educational Research and Improvement U.S. Department of Education. Foreword Homework practices vary widely. Some teachers make brilliant assignments that combine learning and pleasure. Others use. I expect my students will have more difficulty solving equations for x, then evaluating the function when given x. With about ten minutes left in class, I will assign the remainder of the Independent Practice as homework if students have not completed the handout. Then, I will distribute the Exit Slip for students to complete before they leave. Evaluate each expression. 8. 19 48 9. 63b, for b 15 10. w 178, for w 226 11. a b, for a 253 12. h k 84, for h 46 13. r(s), for r 109 and b 11 and k 73 and s 33 Mixed Review Multiply or divide. 18. Use the table at the right. If the pattern continues, how many laps in all will 8 swimmers swim on the fourth day? 14. 18 1,854 15. 16. 490 17. 54. The module will use a variety of strategies for teaching and learning, involving a mixture of group-work, class discussion and analysis, and independent work between sessions. Students will complete regular exercises as homework and will also do a variety of exercises in class. Students will complete regular exercises as homework and will also do a variety of exercises in class. Students may also undertake an academic placement, through which they will learn how to apply the knowledge and skills gained in studying for this module in a professional context outside the University.
Can you see why 2 by 2 could be 5? Can you predict what 2 by 10 will be? 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. Can you work out what is wrong with the cogs on a UK 2 pound coin? A game for 1 or 2 people. Use the interactive version, or play with friends. Try to round up as many counters as possible. What happens when you turn these cogs? Investigate the differences between turning two cogs of different sizes and two cogs which are the same. What is the greatest number of squares you can make by overlapping three squares? Investigate how the four L-shapes fit together to make an enlarged L-shape. You could explore this idea with other shapes too. How can the same pieces of the tangram make this bowl before and after it was chipped? Use the interactivity to try and work out what is going on! A tetromino is made up of four squares joined edge to edge. Can this tetromino, together with 15 copies of itself, be used to cover an eight by eight chessboard? Imagine a wheel with different markings painted on it at regular intervals. Can you predict the colour of the 18th mark? The 100th mark? 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. You have 4 red and 5 blue counters. How many ways can they be placed on a 3 by 3 grid so that all the rows columns and diagonals have an even number of red counters? Mr McGregor has a magic potting shed. Overnight, the number of plants in it doubles. He'd like to put the same number of plants in each of three gardens, planting one garden each day. Can he do it? Seeing Squares game for an adult and child. Can you come up with a way of always winning this game? A game for 1 person. Can you work out how the dice must be rolled from the start position to the finish? Play on line. 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 remove them. The winner is the last one to remove a counter. How you can make sure you win? Three beads are threaded on a circular wire and are coloured either red or blue. Can you find all four different combinations? Use the interactivities to fill in these Carroll diagrams. How do you know where to place the numbers? An interactive game for 1 person. You are given a rectangle with 50 squares on it. Roll the dice to get a percentage between 2 and 100. How many squares is this? Keep going until you get 100. . . . An interactive activity for one to experiment with a tricky tessellation A game for 2 people that everybody knows. You can play with a friend or online. If you play correctly you never lose! Can you use small coloured cubes to make a 3 by 3 by 3 cube so that each face of the bigger cube contains one of each colour? A and B are two interlocking cogwheels having p teeth and q teeth respectively. One tooth on B is painted red. Find the values of p and q for which the red tooth on B contacts every gap on the. . . . 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? Use the Cuisenaire rods environment to investigate ratio. Can you find pairs of rods in the ratio 3:2? How about 9:6? Use the interactivity to investigate what kinds of triangles can be drawn on peg boards with different numbers of pegs. What are the coordinates of the coloured dots that mark out the tangram? Try changing the position of the origin. What happens to the coordinates now? Try entering different sets of numbers in the number pyramids. How does the total at the top change? 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? Place the numbers 1 to 10 in the circles so that each number is the difference between the two numbers just below it. Train game for an adult and child. Who will be the first to make the train? 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. Can you explain the strategy for winning this game with any target? Choose 13 spots on the grid. Can you work out the scoring system? What is the maximum possible score? Have a go at this well-known challenge. Can you swap the frogs and toads in as few slides and jumps as possible? Choose a symbol to put into the number sentence. This 100 square jigsaw is written in code. It starts with 1 and ends with 100. Can you build it up? Here is a chance to play a version of the classic Countdown Game. These interactive dominoes can be dragged around the screen. If you have only four weights, where could you place them in order to balance this equaliser? Try to stop your opponent from being able to split the piles of counters into unequal numbers. Can you find a strategy? A train building game for 2 players. Find out what a "fault-free" rectangle is and try to make some of your own. Can you find all the different ways of lining up these Cuisenaire rods? Use the interactivity or play this dice game yourself. How could you make it fair? Ahmed has some wooden planks to use for three sides of a rabbit run against the shed. What quadrilaterals would he be able to make with the planks of different lengths? Is it possible to place 2 counters on the 3 by 3 grid so that there is an even number of counters in every row and every column? How about if you have 3 counters or 4 counters or....? 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? 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?
Triangle numbers can be represented by a triangular array of squares. What do you notice about the sum of identical triangle numbers? Is it possible to rearrange the numbers 1,2......12 around a clock face in such a way that every two numbers in adjacent positions differ by any of 3, 4 or 5 hours? An irregular tetrahedron is composed of four different triangles. Can such a tetrahedron be constructed where the side lengths are 4, 5, 6, 7, 8 and 9 units of length? We start with one yellow cube and build around it to make a 3x3x3 cube with red cubes. Then we build around that red cube with blue cubes and so on. How many cubes of each colour have we used? 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. 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. 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? 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? How many winning lines can you make in a three-dimensional version of noughts and crosses? How could Penny, Tom and Matthew work out how many chocolates there are in different sized boxes? Watch these videos to see how Phoebe, Alice and Luke chose to draw 7 squares. How would they draw 100? How can the same pieces of the tangram make this bowl before and after it was chipped? Use the interactivity to try and work out what is going on! An extension of noughts and crosses in which the grid is enlarged and the length of the winning line can to altered to 3, 4 or 5. 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? Which hexagons tessellate? ABC is an equilateral triangle and P is a point in the interior of the triangle. We know that AP = 3cm and BP = 4cm. Prove that CP must be less than 10 cm. A huge wheel is rolling past your window. What do you see? A standard die has the numbers 1, 2 and 3 are opposite 6, 5 and 4 respectively so that opposite faces add to 7? If you make standard dice by writing 1, 2, 3, 4, 5, 6 on blank cubes you will find. . . . Show that among the interior angles of a convex polygon there cannot be more than three acute angles. Exchange the positions of the two sets of counters in the least possible number of moves 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? A game for 2 players. Can be played online. One player has 1 red counter, the other has 4 blue. The red counter needs to reach the other side, and the blue needs to trap the red. Here is a solitaire type environment for you to experiment with. Which targets can you reach? Charlie and Alison have been drawing patterns on coordinate grids. Can you picture where the patterns lead? Can you see why 2 by 2 could be 5? Can you predict what 2 by 10 will be? 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? 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? Can you describe this route to infinity? Where will the arrows take you next? 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? 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. A and B are two interlocking cogwheels having p teeth and q teeth respectively. One tooth on B is painted red. Find the values of p and q for which the red tooth on B contacts every gap on the. . . . Imagine a large cube made from small red cubes being dropped into a pot of yellow paint. How many of the small cubes will have yellow paint on their faces? Investigate how the four L-shapes fit together to make an enlarged L-shape. You could explore this idea with other shapes too. A game for 2 players. Given a board of dots in a grid pattern, players take turns drawing a line by connecting 2 adjacent dots. Your goal is to complete more squares than your opponent. Slide the pieces to move Khun Phaen past all the guards into the position on the right from which he can escape to freedom. Square numbers can be represented as the sum of consecutive odd numbers. What is the sum of 1 + 3 + ..... + 149 + 151 + 153? Can you cross each of the seven bridges that join the north and south of the river to the two islands, once and once only, without retracing your steps? A 2 by 3 rectangle contains 8 squares and a 3 by 4 rectangle contains 20 squares. What size rectangle(s) contain(s) exactly 100 squares? Can you find them all? Use the interactivity to play two of the bells in a pattern. How do you know when it is your turn to ring, and how do you know which bell to ring? 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? This article is based on some of the ideas that emerged during the production of a book which takes visualising as its focus. We began to identify problems which helped us to take a structured view. . . . Three circles have a maximum of six intersections with each other. What is the maximum number of intersections that a hundred circles could have? A game for two players on a large squared space. Which of these dice are right-handed and which are left-handed? How many moves does it take to swap over some red and blue frogs? Do you have a method? It is possible to dissect any square into smaller squares. What is the minimum number of squares a 13 by 13 square can be dissected into? Can you fit the tangram pieces into the outlines of the watering can and man in a boat? A Hamiltonian circuit is a continuous path in a graph that passes through each of the vertices exactly once and returns to the start. How many Hamiltonian circuits can you find in these graphs? Can you fit the tangram pieces into the outlines of Mai Ling and Chi Wing? Can you fit the tangram pieces into the outline of Little Ming and Little Fung dancing?
Follow the clues to find the mystery number. Sweets are given out to party-goers in a particular way. Investigate the total number of sweets received by people sitting in different positions. Can you put the numbers from 1 to 15 on the circles so that no consecutive numbers lie anywhere along a continuous straight line? Investigate the smallest number of moves it takes to turn these mats upside-down if you can only turn exactly three at a time. Have a go at this game which has been inspired by the Big Internet Math-Off 2019. Can you gain more columns of lily pads than your opponent? Place the numbers 1 to 8 in the circles so that no consecutive numbers are joined by a line. There is a long tradition of creating mazes throughout history and across the world. This article gives details of mazes you can visit and those that you can tackle on paper. Is it possible to place 2 counters on the 3 by 3 grid so that there is an even number of counters in every row and every column? How about if you have 3 counters or 4 counters or....? An investigation involving adding and subtracting sets of consecutive numbers. Lots to find out, lots to explore. Number problems at primary level that require careful consideration. Can you find the chosen number from the grid using the clues? A game for 2 people. Take turns placing a counter on the star. You win when you have completed a line of 3 in your colour. 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. Use the numbers and symbols to make this number sentence correct. How many different ways can you find? 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? 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? What two-digit numbers can you make with these two dice? What can't you make? What do the numbers shaded in blue on this hundred square have in common? What do you notice about the pink numbers? How about the shaded numbers in the other squares? What do the digits in the number fifteen add up to? How many other numbers have digits with the same total but no zeros? How long does it take to brush your teeth? Can you find the matching length of time? My local DIY shop calculates the price of its windows according to the area of glass and the length of frame used. Can you work out how they arrived at these prices? Ben and his mum are planting garlic. Can you find out how many cloves of garlic they might have had? 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. What happens when you try and fit the triomino pieces into these two grids? In this calculation, the box represents a missing digit. What could the digit be? What would the solution be in each case? What is the best way to shunt these carriages so that each train can continue its journey? The planet of Vuvv has seven moons. Can you work out how long it is between each super-eclipse? Can you substitute numbers for the letters in these sums? Can you put the 25 coloured tiles into the 5 x 5 square so that no column, no row and no diagonal line have tiles of the same colour in them? Choose four different digits from 1-9 and put one in each box so that the resulting four two-digit numbers add to a total of 100. 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? Have a go at balancing this equation. Can you find different ways of doing it? Can you complete this calculation by filling in the missing numbers? In how many different ways can you 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 jugs. What happens when you round these three-digit numbers to the nearest 100? Can you work out some different ways to balance this equation? Can you fill in this table square? The numbers 2 -12 were used to generate it with just one number used twice. 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? Using the statements, can you work out how many of each type of rabbit there are in these pens? This multiplication uses each of the digits 0 - 9 once and once only. Using the information given, can you replace the stars in the calculation with figures? Zumf makes spectacles for the residents of the planet Zargon, who have either 3 eyes or 4 eyes. How many lenses will Zumf need to make all the different orders for 9 families? Use your logical-thinking skills to deduce how much Dan's crisps and ice-cream cost altogether. What happens when you add three numbers together? Will your answer be odd or even? How do you know? A challenging activity focusing on finding all possible ways of stacking rods. 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. This challenge extends the Plants investigation so now four or more children are involved. Have a go at this well-known challenge. Can you swap the frogs and toads in as few slides and jumps as possible? Can you order pictures of the development of a frog from frogspawn and of a bean seed growing into a plant? Can you use the information to find out which cards I have used? In this problem it is not the squares that jump, you do the jumping! The idea is to go round the track in as few jumps as possible.
'Information Technology and Society' Second Life class transcript, on Harvard's virtual island [11:06] Aphilo Aarde: Did all of you receive the notecard I sent to the "Soc and Info Tech - Aphilo on Berkman" group list here [11:06] Ju Roussel is Online [11:06] Aphilo Aarde: Who did not receive it - this notecard has information relevant to this course. [11:06] Aphilo Aarde: Great [11:07] JonathanE Cortes: brb have to re boot [11:07] Aphilo Aarde: :) [11:07] JonathanE Cortes is Offline [11:07] Aphilo Aarde: Here's the course wiki again: http://socinfotech.pbworks.com/ [11:07] JonathanE Cortes is Online [11:08] sandhya2 Patel accepted your inventory offer. [11:08] XiuJuan Ying accepted your inventory offer. [11:08] JonathanE Cortes accepted your inventory offer. [11:08] Kevin Menczel is Offline [11:08] XiuJuan Ying: ty [11:08] JonathanE Cortes: thx [11:08] Melchizedek Blauvelt accepted your inventory offer. [11:08] Aphilo Aarde: yw [11:08] Ju Roussel accepted your inventory offer. [11:08] Aphilo Aarde: I just passed out the notecard again [11:09] Daisyblue Hefferman is Offline [11:09] Aphilo Aarde: We'll begin shortly - but I'd just like to bring to your attention http://webnographers.org again, as well as the course wiki - where I'm posting notes - http://socinfotech.pbworks.com [11:10] Marian Dragovar is Online [11:11] Aphilo Aarde: Are there any questions from the last times? [11:12] Marian Dragovar is Offline [11:12] Ju Roussel: [We are listening...] [11:13] Aphilo Aarde: I'll address nanotechnology a little today [11:13] Aphilo Aarde: So in this revolution [11:13] Aphilo Aarde: of information technology KNOWLEDGE IS VALUE [11:14] Aphilo Aarde: And Where is knowledge? [11:14] Aphilo Aarde: It's sometimes in universisites and in research labs [11:14] Aphilo Aarde: but it's particularly in 'scientists' [11:14] Aphilo Aarde: Knowledge is generated by human minds [11:14] Aphilo Aarde: and bodies [11:15] Aphilo Aarde: human minds are the critical source of value [11:15] Aphilo Aarde: in the Information TEchnology reovlution [11:15] Aphilo Aarde: This is the founding concept of this course [11:15] Aphilo Aarde: So, to recap a little further [11:16] Aphilo Aarde: the founding discovery of the IT revolution is the transistor in 1947 at Bel Labs in NJ [11:16] Aphilo Aarde: And this led to a Nobel prize for 3 people - [11:16] Jules Mandelbrot is Online [11:16] Aphilo Aarde: William Shockley was the leader of this team [11:17] Aphilo Aarde: ANd he saw extraordinary possiblities for the transistor [11:17] Aphilo Aarde: which was created by a whole network of scientists at Bell [11:17] Aphilo Aarde: But Bell Labs couldn't take advantage of it [11:17] Aphilo Aarde: because they were a telecommunications monolpoly [11:18] Aphilo Aarde: and they couldn't go into another business, due to antitrust regulation in the U.S. [11:18] Aidan Aquacade is Online [11:18] Aphilo Aarde: Other businesses would have had to pick it up, but businesses were not interested. [11:19] Aphilo Aarde: So Shockley treid to create his own lab, after asking RCA and Raytheon corporations - they said no, - that vacuum tubes were enough. [11:19] Aphilo Aarde: (They were using vacuum tubes for early stereo equipment, for example. [11:19] Aphilo Aarde: Other possibilities? [11:19] Aphilo Aarde: Shockley didn't see many ... [11:20] Aphilo Aarde: So he came to Palo Alto, California, because his mother was there. [11:20] Aphilo Aarde: Shockley was depressed - He had a Nobel prize, but nothing to do! [11:20] Aphilo Aarde: He LATER became a Stanford Professor. [11:21] Aphilo Aarde: Beckman Labs in the Silicon VAlley area said that the TRANSISTOR sounds interesting. [11:21] Aphilo Aarde: So Beckman Labs helped Shockley start a conductor business [11:22] Aphilo Aarde: They could do that because a whole network of Palo Alto electronic companies was forming. [11:22] Aphilo Aarde: And because Stanford University was supporting entreprenial companies - [11:22] Aphilo Aarde: Stanford had an entrepreneurial attitude. [11:22] Aphilo Aarde: But all of this was ACCIDENTAL [11:23] Aphilo Aarde: Actors are necessary for this story [11:23] Aphilo Aarde: (My method here is to tell the story, analyze the story, and look at the factors, including the actors). [11:24] JenzZa Misfit is Online [11:24] Aphilo Aarde: A man named TERMAN, who was a graduate student in the 1920s at MIT, wanted a Ph.D. in electrical engineering [11:24] Aphilo Aarde: worked there after getting his Ph.D. got tuberculosis, and then went back to California. [11:25] Aphilo Aarde: He decided to take a position at Stnaford, and later became Dean. [11:25] Aphilo Aarde: He wasn't a great researcher or engineer, by many accounts, but he could identify great MINDS. [11:25] Aphilo Aarde: In his researcher, he was experimenting on radar technology leading to electronics. [11:26] Aphilo Aarde: There was a lot of research on radar at the time, including by Hewlett and Packard. [11:26] Aphilo Aarde: Terman was asked why they didn't make something practical. [11:27] JenzZa Misfit is Offline [11:27] Aphilo Aarde: But Hewlett and Packard - young researchers at the time - didn't have any money [11:27] Aphilo Aarde: So Terman took $700 from his own pocket and gave it to Hewlett and Packard, who started a company. [11:27] JenzZa Misfit is Online [11:28] Aphilo Aarde: World War II started and their radar-related and other devices became very valuable. [11:28] sensu20 whispers: watch that third sip [11:28] Aphilo Aarde: They sold millions of these to the Defense Department. [11:28] Aphilo Aarde: After WWII, Hewlett Packard was a very established company. [11:28] Aphilo Aarde: And Terman becaome a provost at Stanford. [11:29] Aphilo Aarde: This example could be amplified - many companies in Silicon VAlley had similar beginnings. [11:29] Kazuhiro Aridian is Online [11:29] JenzZa Misfit is Offline [11:29] Aphilo Aarde: Terman convinced Stanford to use assets they had at the time - LAND - to start companies. [11:30] Ju Roussel: Do you think the same process is happening now? [11:30] Aphilo Aarde: And in 1951, they started the Stanford Industrial Park, emerging out of Stanford University [11:30] Aphilo Aarde: Good question - JU. [11:31] Aphilo Aarde: Well, Stanford doesn't have nearly as much land to lease to engineers emerging, in part, out of Stanford, so we have a very different landscape ... and the culture of milieu [11:31] Aphilo Aarde: which we'll look at in a while is less 'emergent' but I do think there's a lot of innovation still happening. [11:32] Melchizedek Blauvelt: Right now the opposite is happening it seems, people seem to be extremely risk-averse (free content) and on the lookout for instant gratification (Twitter) [11:32] Aphilo Aarde: I'd like to mention World University & School - http://worlduniversity.wikia.com/wiki/World_University - which I'm developing, as an example [11:32] Aphilo Aarde: This class is the first at it - although WUaS does link the 1900 courses of MIT Open Course Ware, and the [11:32] Ju Roussel: Yes, did you notice any more interest after I mentioned WU at Metanomics Community Forum Thursday? ;) [11:33] Aphilo Aarde: free open courses at Berkeley and Yale, for example ... all of these might be examples of contemporary kinds of bases for innovation. [11:33] Aphilo Aarde: Thank you, Ju! [11:34] Michele Mrigesh is Offline [11:34] Aphilo Aarde: I think World University & School will grow gradually ... through word of mouth, such as what you did the other day at teh Metanomics Community Forum [11:34] Aphilo Aarde: Interesting Mel [11:35] Aphilo Aarde: I would say that innovation around Open Source and Free Ware carries on - alongside business - and that Twitter, even, is yet another example. [11:35] Spider Mycron is Offline [11:35] Ju Roussel: Sorry for being disruptive - the so called 'triple helix' relationships and success of Silicon Valley does not give peace to European politicians (and researchers, as well!) [11:36] Aphilo Aarde: of open source and free ware, as is Second Life ... but it's how people make new things with these technologies that's so fascinating - and I'll make the case that so much has been done by accident and hacking, as you'll see). [11:36] Aphilo Aarde: Yes, Ju - Japan tried to recreate the Silicon Valley culture in the 70s and 80s - but it's not very reproducible. [11:37] Ju Roussel: You can't 'make' them. [11:37] Aphilo Aarde: So, back to the Stanford Industrial Park in the 1950s. [11:38] Aphilo Aarde: Terman convinced Stanford to use assets they had at the time - LAND - to start companies. And in 1951, they started the Stanford Industrial Park, emerging out of Stanford University [11:39] Aphilo Aarde: And there were a lot of Stanford engineers and computer scientists in these newly forming companies [11:39] Aphilo Aarde: in the openness of California culture in the 1950s [11:39] Jules Mandelbrot is Offline [11:39] Aphilo Aarde: And the Stanford Industrial Park also had access to Stanford faculty and students. [11:40] Aphilo Aarde: Any new enterprise had to be approved by Stnaford - they alone could offer long term leases to companies. [11:40] Aphilo Aarde: So Terman was able to pick and establish the first companies to be tenants - [11:41] Aphilo Aarde: and the first was Hewlett Packard. [11:42] Aphilo Aarde: and the other was Varian [11:42] Aphilo Aarde: which also made radar parts [11:42] Aphilo Aarde: During the 1950s, under the initiative of Stanford, a cluster of innovative companies [11:42] Aphilo Aarde: around Stanford [11:43] Aphilo Aarde: emerged - and, remember, it's this clustering which has made industrial revolutions so far-reaching. [11:43] Aphilo Aarde: So into this context comes Shockley. [11:44] Profdan Netizen is Offline [11:44] Aphilo Aarde: And the cluster of emerging companies on the West Coast, however, was not up to the level of east coast US companies. [11:44] Profdan Netizen is Online [11:44] Aphilo Aarde: (Like RCA and Raytheon, and a variety of companies relating to MIT, for one). [11:45] Bon McLeod is Offline [11:45] Aphilo Aarde: Shockley decided to go into Microelectronics, and brought with him the best MINDS from Bell Labs - [11:45] Aphilo Aarde: The argument for this course is that MINDS are the raw material of the Information TEchnology revolution, again [11:46] Aphilo Aarde: Eight young engineers came west from the East Coast [11:46] Aphilo Aarde: 6 came from Bell Labs [11:46] Aphilo Aarde: and 2 from other labs. [11:46] Daisyblue Hefferman is Online [11:46] Aphilo Aarde: They all wanted to work with Shockley [11:47] Aphilo Aarde: 1 was Bob Noyce, who later became known as the Mayor of Silicon Valley, and who worked on integrated circuits at Intel [11:47] Aphilo Aarde: Schockley's company failed [11:47] Aphilo Aarde: He was a genius, but a horrible person. [11:47] Aphilo Aarde: He later became a prominent professor. [11:47] Aphilo Aarde: He tried to demonstrate the biological inferiority of women and African Americans, for example. [11:48] Aphilo Aarde: But he was a genius - he was stubborn, actually [11:48] Aphilo Aarde: He wanted to work on micro-circuits, but not in SILICON [11:49] Aphilo Aarde: - we're still in the 50s, before silicon was used for chips - [11:49] Aphilo Aarde: Shockley wanted to work in GALLISUM ARSENIDE [11:49] Aphilo Aarde: *GALLIUM ARSENIDE [11:50] Aphilo Aarde: Because of Shockley's wish to work with this material, the 8 engineers left Shockley Semiconductor - leaving the company an empty shell. [11:50] Aphilo Aarde: And Shockley became a professor at Stanford. [11:50] Aphilo Aarde: But due to the transistor, Shockley and others invented MICROELECTRONICS [11:51] Aphilo Aarde: and migrated KNOWLEDGE to Silicon Vally. [11:51] Aphilo Aarde: These 8 young engineers, and others, started Fairchild Semiconductors, now in the mid 1950s [11:51] Aphilo Aarde: And each of these engineers alter split to create their own companies [11:52] Aphilo Aarde: Fairchild led to INTEL and AMD, etc. - still the key processor manufacturers [11:53] Aphilo Aarde: and 140 COMPANIES WERE SPUN OUT of Fairchild - ASTOUNDING, in a way [11:53] Aphilo Aarde: In 1959, integrated circuits were invented. [11:53] Aphilo Aarde: And the Defense Department made up 50% of th market. [11:53] Aphilo Aarde: Research programs for Microelectronics developed. [11:54] Aphilo Aarde: In 1957, something else happened [11:54] Aphilo Aarde: to put all these developments in context [11:54] Aphilo Aarde: Can anyone say what happened? [11:55] Aphilo Aarde: Sputnik - the Russian satellite system - was launched. [11:55] Ju Roussel: :) [11:55] Aphilo Aarde: It was the 1st human made satellite circling the earth [11:55] XiuJuan Ying: were the americans pleasd [11:56] Aphilo Aarde: It also suggested that the backward Soviet system was overtaking the Americans into SPACE [11:56] Aphilo Aarde: :) [11:56] Aphilo Aarde: So, the Defense Department responded with money for TECHNOLOGY [11:56] Aphilo Aarde: - that's how pleased they were [11:57] Aphilo Aarde: And in 1959 - a race to the moon emerged [11:57] Aphilo Aarde: "Nothing to be done there" [11:57] Aphilo Aarde: "No Night Life" [11:57] Aphilo Aarde: "Cannot Moonlight" [11:57] Aphilo Aarde: Humans do interest things - don't you think :) [11:57] XiuJuan Ying: no atmosphere [11:58] Aphilo Aarde: After the extraordinary events of 1959, the military THEN became involved with technology [11:58] Ju Roussel: Sputnik vs. First Man on the Moon. SL'ers were discussing where else in the Solar system there can be life. [11:59] Aphilo Aarde: So, to recap - we have Microelectronics, Computing, and Telecommunications emerging on the one hand [11:59] Aphilo Aarde: and Genetic Engineering, on the other hand [12:00] Aphilo Aarde: Yes... SETI projects - Search for Extraterrestrial Life are legitimate science projects - if life emerged on earth ... genetic engineering - then it's logical that it may have emerged elsewhere [12:01] Aphilo Aarde: but of course, life on earth emerged in very specific conditions 3.5. billion years ago, which may not be easily replicable [12:01] Aphilo Aarde: GENETIC ENGINEERING - as another key aspect of the INFORMATION TECHNOLOGY revolution [12:02] Aphilo Aarde: In 1953, it started - in hospitals, and with Crick and Watson decoding life~ [12:02] Aphilo Aarde: And it became TECHNOLOGY when UCSF and Stanford experimented with splicing [12:03] Aphilo Aarde: (Remember the definition of technology for this course involves replicability, which gene splicing builds on). [12:04] Aphilo Aarde: So Stanford, Maryland, DC, Virginia and Harvard were all major centers for gene splicing. [12:04] Mab MacMoragh is Online [12:04] Aphilo Aarde: Back to SILICON VALLEY [12:04] Ju Roussel: Interestingly, Watson was always @ Cold Spring Harbor? [12:05] Aphilo Aarde: Another key aspect to the information TEchnology reovlution were CULTURAL developments. [12:05] Aphilo Aarde: not sure, JU - cold spring? [12:05] Ju Roussel: CSHL - the 'Meca' of geneticists [12:05] Ju Roussel: NY State [12:06] Aphilo Aarde: The cultural development that was so significant on a wide scale - was the ability to innovate by THINKING NEW APPLICATIONS [12:06] Aphilo Aarde: that weren't there before. [12:06] Aphilo Aarde: This emerged directly from the cultures of 1960s [12:06] Aphilo Aarde: Thinking Differently [12:06] Aphilo Aarde: involves rebelling against the establishment [12:07] Aphilo Aarde: Counterculture didn't want revolution; they wanted THEIR revolution. [12:07] Aphilo Aarde: Culture inthis context was extremely important, especially in the San Francisco Bay Area [12:08] Aphilo Aarde: where alternative forms of COMPUTING emerged, linked to the personal computer. [12:08] Aphilo Aarde: This computer, which was originally called the microcomputer, came directly from the Homebrew Computer Club - in Sausalito (Marin county), Menlo Park (in the south SF Bay), and San Francisco itself. [12:09] Aphilo Aarde: They were about finding codes that could work on computers. [12:09] Aphilo Aarde: They created a world around new languages - computer languages. [12:10] Aphilo Aarde: And countercultural scientists such as Stewart Brand, who edited and wrote the Whole Earth Catalog [12:10] Aphilo Aarde: became public figures for counterculture [12:10] Aphilo Aarde: And offered a 'bridge' betwen society and hippies [12:11] Aphilo Aarde: So culture was very important - and there was a lot of thinking about what to do with it. [12:11] Profdan Netizen is Offline [12:11] Aphilo Aarde: This thinking had to to with [12:11] Aphilo Aarde: 1 decentralization [12:11] Aphilo Aarde: 2 making things small -/ micro [12:12] Aphilo Aarde: 3 opening up source code to the public. [12:12] JonathanE Cortes: Iv got an old coppy of that [12:12] Zinnia Zauber is Online [12:12] Aphilo Aarde: These ways of thinking were ENTIRELY different from the thinking of IBM and ATT, etc. [12:12] Aphilo Aarde: of what Jonathan? [12:13] JonathanE Cortes: whole earth [12:13] Aphilo Aarde: So to bring things together, [12:13] Ju Roussel: Oh ues IBM could not figure it out for next 2 decades [12:13] Aphilo Aarde: The Whole Earth Catalog is great - and worth perusing again and again t0 this day, for the way of thinking. [12:14] Aphilo Aarde: So to bring things together, the most extraordinary aspects of the Information TEchnology Revolution were due to [12:14] Aphilo Aarde: 1 an entrepreneurial attitude, partly relating to Stanford [12:15] Aphilo Aarde: 2 that technology migrated from major research centers (like Bell Labs) and that then military money 'watered' all of these innovations. [12:16] Aphilo Aarde: 3 the cultural revolution from counterculture - leading to thinking differently, esepcially around technology [12:16] Aphilo Aarde: ANALYTIC AND SYSTEMATIC [12:16] Aidan Aquacade is Offline [12:16] Aphilo Aarde: What created Silicon Valley as the site of this information revolution? [12:16] Zinnia Zauber is Offline [12:17] Aphilo Aarde: As background: [12:17] Aphilo Aarde: concerning regional economies ... [12:17] Aphilo Aarde: 1 All regional economies are based on [12:17] Aphilo Aarde: A. raw materials [12:17] Aphilo Aarde: B. capital (money) [12:17] Aphilo Aarde: C. labor [12:17] Gentle Heron is Offline [12:18] Aphilo Aarde: What was specifc to Silicon Valley? [12:18] Aphilo Aarde: There were specific forms of each of these. [12:18] Christinapsu5152 Palianta is Online [12:18] Ju Roussel: specialization/concentration of specific resources? [12:18] Aphilo Aarde: A the raw material was knowledge and information generating capacity [12:19] Zinnia Zauber is Online [12:19] Aphilo Aarde: so, not exactly universities, and this knowledge came from Bell Labs, and then diffused. [12:20] Aphilo Aarde: B Capital first came significantly from the military - but these were PULBIC monies, not private [12:21] Aphilo Aarde: (yes - Ju - around knowledge) [12:21] Aphilo Aarde: C labor - was hihgly skilled, technical and scientific labor [12:21] Aphilo Aarde: with origins at Stanford and Berkeley - and with market origins, as well. [12:22] Aphilo Aarde: It was part of these university's policies to produce more Ph.D.s in these years [12:23] Aphilo Aarde: to catch up with Harvard and MIT. [12:23] Aphilo Aarde: Berkeley and Stanford produced less than half the Ph.D.s that Harvard and MIT produced in the 1950s and 60s. [12:24] Aphilo Aarde: And then Stanford decided to fund engineering, and Berkeley got money from the State (of California and the Federal Government). [12:25] Aphilo Aarde: In the 1960sHarvard and MIT produced 4 times as many Ph.D.s - and had a deliberate government policty to support this. [12:25] Aphilo Aarde: So the government started to support financially more Ph.D.s in California, as an Inveestment in the Technology revolution. [12:26] Aphilo Aarde: Regarding B - CAPITAL again [12:26] Aphilo Aarde: There was a problem here [12:26] Aphilo Aarde: These technology businesses were RISKY [12:26] Aphilo Aarde: They have to try something NEW [12:27] Aphilo Aarde: They also have to be stubborn, and keep trying [12:27] Aphilo Aarde: A startup on average fails 7 times. [12:27] Aphilo Aarde: Venture capitalists liked to give money to failures - you'd have to conclude [12:28] Aphilo Aarde: In the 1960s, when the technology revolution was about to explode, there was no capital available. [12:28] Michele Mrigesh is Online [12:28] Aphilo Aarde: So, two kinds of special capital developed. [12:28] Aphilo Aarde: 1 Capital which was designed to be lost [12:28] Aphilo Aarde: 2 Capital which was defined as speculative [12:29] Aphilo Aarde: Money started to work as capital. [12:29] Aphilo Aarde: And the one who gives the money did not ever expect a return. [12:29] Aphilo Aarde: And who can do that? [12:29] Aphilo Aarde: The government. [12:29] Aphilo Aarde: But why? [12:30] Aphilo Aarde: (International pride and competition?) [12:30] Aphilo Aarde: Actually, because they expect a return down the line - [12:30] Ju Roussel: Cold War [12:30] Aphilo Aarde: that is, because the capital investment comes back ... [12:31] Aphilo Aarde: (the cold war - sputnik above - did spur the military to invest heavily in technology , but we're talking abou capital - so different institutions, as well). [12:32] Aphilo Aarde: It's the part of givernment that doesn't care about cost, only performance [12:32] Aphilo Aarde: > MILITARY INVESTMENT [12:32] Aphilo Aarde: (rarely does capital not care about cost, or risk, another way of thinking about cost). [12:33] Marian Dragovar is Online [12:33] Aphilo Aarde: When you have to survivie, you don't count money - and this was the concern during the cold war, for some, in the US Defense Department). [12:33] Aphilo Aarde: What made [12:34] Aphilo Aarde: Silicon Valley original was the unlimited funds from the military. [12:34] Aphilo Aarde: Companies could fail. [12:34] Aphilo Aarde: ... in this context. [12:34] Aphilo Aarde: If, finally, you might get a CHIP, [12:34] Aphilo Aarde: you get military superiority. [12:35] Aphilo Aarde: And it worked out exactly that way. [12:35] Aphilo Aarde: 20 years later, the U.S. outperformed the Soviety Union completely - in 1984. [12:36] Aphilo Aarde: So, you have a purely military strategy in the 1970s, with all the money the military poured into informatio technology - and it paid off. [12:36] Aphilo Aarde: 2nd Source of capital [12:36] Aphilo Aarde: - the capital you're ready to lose, because when you win, you win really, really big. [12:37] Aphilo Aarde: Overall, in the 1990s, venture capitalists had increased their bets by a factor of seven, counting potentially losses. [12:37] Aphilo Aarde: The reward came a little bit later. [12:37] Aphilo Aarde: When industry developed, people from industry became rich. [12:38] Aphilo Aarde: Venture capital money came originally from inside industry. [12:38] Aphilo Aarde: ... by starting to work together with other established firms. [12:39] Aphilo Aarde: And all these special RAW MATERIALS, CAPTIAL, AND LABOR came together in one area - the SF Bay Area. [12:39] Aphilo Aarde: And something else happens, as well. [12:40] Aphilo Aarde: UC Berkeley Professor AnnaLee Saxenian in her book "Regional Advantage" [12:40] Aphilo Aarde: compares Silicon Valley with the Boston rt 125 area [12:40] Tutti Jupiter gave you villa cristine. [12:41] Aphilo Aarde: She shows how in the 1960s adn 70s, the Boston area was far ahead in [12:41] Aphilo Aarde: Microelectronics, computers, telecommunications, the internet [12:41] Aphilo Aarde: leading to new networks of technology and genetic engineering. [12:41] Aphilo Aarde: Around these technologies, there are 2 key factors [12:42] Aphilo Aarde: 1 industry structure of specific companies which are suppliers [12:42] Marian Dragovar is Offline [12:42] Aphilo Aarde: Here networks of companies are developing knowledge of what they do best. [12:43] Aphilo Aarde: Universities do the same. [12:43] Aphilo Aarde: Complex organizations organized around networks constitute themselves again and again through history. [12:43] Aphilo Aarde: 2 social networks also formed [12:44] Aphilo Aarde: THe world of silicon valley was made up of individuals, building their own companies, doing their own thing [12:44] Aphilo Aarde: People were required to sign confidentiality statements [12:44] Aphilo Aarde: But in Silicon Valley, these were limited to 6 months only ! [12:45] Aphilo Aarde: Otherwise, you wouldn't find people to work for you! [12:45] Aphilo Aarde: The edge shifted so quickly. [12:45] Aphilo Aarde: Social Networks - where peopl emet after work to talk about work sprang up. [12:45] Aphilo Aarde: The key thing was a process of excitement about work [12:46] Aphilo Aarde: among engineers, students [12:46] Aphilo Aarde: that then network and lead to people creating their own companies [12:46] Aphilo Aarde: This created a milieu of permeability - where nothing was stable [12:47] Aphilo Aarde: And this is the missing link for understanding CULTURE in Silicon Valley when there was so much creativity - and which Saxenian documented. [12:47] Aphilo Aarde: This leads to synergy where 2+2 = 5, not 4 [12:47] Aphilo Aarde: Why? [12:48] Aphilo Aarde: Putting things together whose added value is more that things as separate entities, due to INTERACTION [12:48] Michele Mrigesh is Offline [12:48] Aphilo Aarde: and the added value du to elementes in the process. [12:49] Aphilo Aarde: The FACT that Silicon Valley culture [12:49] Aphilo Aarde: people and companies could talk to one another generated a MILIEU OF INNOVATION [12:50] Aphilo Aarde: equaling a cluster set of research centers, companies, venture capital companies, labor markets, and [12:50] Aphilo Aarde: professional organizations which - through INTERACTIONS - create synergy [12:51] Aphilo Aarde: Despite the traffic, congestion of Silicon Valley, people keep coming to SV [12:51] Aphilo Aarde: People get hooked and never get out. [12:51] Aphilo Aarde: This is the machine creating this revolution. [12:51] Aphilo Aarde: The Japanese tried to reproduce this pattern unsuccessfully ... because it wasn't organic. [12:51] Aphilo Aarde: LAST STAGE [12:52] Aphilo Aarde: GLOBAL NETWORKING [12:52] Hydra Shaftoe is Online [12:52] Aphilo Aarde: connecting to othe rplaces around the wolrd with the same SYNERGY generating capacity. [12:52] Aphilo Aarde: This didn't happen the same way as in Silicon Valley. [12:53] Aphilo Aarde: Electronic companies originally started to produce chips in low cost comopanies, particularly in the southeast ASIA [12:53] Aphilo Aarde: The ADDED VALUE is really in the abiity to ADD KNOWLEDGE for profits. [12:53] Aphilo Aarde: CRITICAL: is the ability to connect around the world, such as Taiwan does. [12:54] Aphilo Aarde: So, how NETWORKING works in SOFTWARE [12:54] Ju Roussel: or, knowledge without commercialization of ideas does not go far.. [12:54] Hydra Shaftoe is Offline [12:55] Aphilo Aarde: yes, Ju, yet OPEN SOURCE and FREE WARE and NONMARKET INFORMATION PRODUCTION (Benkler) seems to be a new part of economic processes, fascinatingly. [12:55] Aphilo Aarde: So, how NETWORKS works in SOFTWARE [12:56] Aphilo Aarde: e.g. in Banglaore - These networks are generally organized by IMMIGRANT ENTREPRENEURS [12:57] Aphilo Aarde: Someone from India, China or Taiwan comes with an Electrical Engineering degree, stays, learns the trade,and creates a company [12:57] Aphilo Aarde: and then reestablishes contact with the country they came from, through their company [12:58] Aphilo Aarde: Through personal connections, Silicon Valley has established worldwide networks, but not to all places in the world. [12:58] Aphilo Aarde: So there's a different kind of production system worldwide. [12:58] Aphilo Aarde: SUMMARY [12:59] Mab MacMoragh is Offline [12:59] Aphilo Aarde: How the information technology revolution lead to a new PARADIGM OF Technology and Socioeconomic organization. [12:59] Aphilo Aarde: PARADIGM [12:59] Aphilo Aarde: A paradigm here is a cluster of interrlated innovations, technoloies, managerial [13:00] Aphilo Aarde: innovations. This cluster is able to generate new products and processes leading to synergies and improving productivity. [13:01] Aphilo Aarde: PARADIGM the system needs each other for progress in every field to come together. [13:01] Aphilo Aarde: So, to close - here are the five aspects of this new information technology paradigm. [13:01] Kate Miranda is Offline [13:01] Aphilo Aarde: 1 It's about INFORMATION GENERATION AND PROCESSING [13:02] Profdan Netizen is Online [13:02] Aphilo Aarde: 2 It's PERVASIVE, AND IT INVADES AND INFLUENCES EVERY DOMAIN OF SOCIOECONOMIC ACTIVITY [13:02] Aphilo Aarde: 3 CHARACTERIZED BY NETWORKING - mentalities, companies, people, etc. [13:03] Breeze Underwood is Offline [13:03] Aphilo Aarde: 4 there's a quality of FLEXIBILITY - the system is such that it can reorganize and reprogram components without disintegration [13:04] Breeze Underwood is Online [13:04] Aphilo Aarde: 5 There's a TECHNOLOGICAL CONVERGENCE in integrating this OPEN SYSTEM, not closing, and which is bounded only by technological devleopments. [13:04] Aphilo Aarde: NEXT week [13:05] Aphilo Aarde: 1 Context of braoder social and political circumstances of IT REVOLUTION [13:05] Aphilo Aarde: 2 DEVELOPMENT OF THE INTERNET [13:05] Aphilo Aarde: So, that's what I wanted to relate to you this week. [13:05] Ju Roussel: Great story, amazingly put together by you. Applause! [13:05] Aphilo Aarde: Thanks :) [13:06] Melchizedek Blauvelt: yay! [13:06] sandhya2 Patel: thank you [13:06] JonathanE Cortes: thx very good [13:06] Aphilo Aarde: The information technology revolution is fascinating - and I'd also like to acknowledge Professor Manuel Castells' research and contribution to this. [13:06] XiuJuan Ying: thanx Aphilo, interesting as usual :-) [13:07] Cat Abeyante is Offline [13:07] Aphilo Aarde: I'd also like to invite you to begin to talk about World University and School - to explore how the open teaching and learning possibilities there might begin [13:07] Dusty Artaud is Online [13:07] Aphilo Aarde: to generate and ongoing culture of innovation ... [13:07] Aphilo Aarde: There are already a lot of resources there. [13:08] Mab MacMoragh is Online [13:08] Cat Abeyante is Online [13:08] Aphilo Aarde: And thanks for your observations ... type chat allows for multiple concurrent conversations ... [13:08] Ju Roussel: Yes, very interesting, let me know if there is any way to help you from here out-/in- world [13:08] Ju Roussel: so I went to the spreadsheets today [13:08] Aphilo Aarde: Thanks ... will do ... I'd like to invite you all to join this Google Group [13:09] Ju Roussel: and shamelessly added myself [13:09] Ju Roussel: [posted via g group] [13:09] Aphilo Aarde: http://groups.google.com/group/World-University-and-School [13:09] Aphilo Aarde: Great! [13:09] Aphilo Aarde: Ju [13:09] Ju Roussel: The most amazing thing is - I followed your blog from before Christmas [13:10] Aphilo Aarde: As part of this Google Group, there are spreadsheets with beginning lists of moderators for World Univ a& Sch ubjects [13:10] Ju Roussel: That is, before I realized you were in-world [13:10] Aphilo Aarde: 'Subjects,' 'Languages,' and "Nation States' [13:11] Aphilo Aarde: Great Ju, as well - here's my blog, too - http://scott-macleod.blogspot.com/ [13:11] Aphilo Aarde: For the Google Groups World Univ spreadsheets, they're an opportunity to engage what you know, or what you'd like to learn [13:11] Aphilo Aarde: Adding your name to them is a little like becoming a Wikipedia moderator [13:12] Aphilo Aarde: And Wikipedia has something like 14 million articles in 272 languages [13:12] Ju Roussel: I have a team of 900+ business admin./management studies authors potentially contributing. [13:12] Aphilo Aarde: all shaped by us, so I welcome you to teach something to your web camera at World University & School, about whatever you like :) [13:13] Aphilo Aarde: or learn something - great free software there - [13:13] Aphilo Aarde: http://worlduniversity.wikia.com/wiki/Educational_Software [13:13] Aphilo Aarde: Interesting, Ju. [13:13] Tarek String is Online [13:14] Aphilo Aarde: World Univ & Sch is about fee and open teaching and learning - with an academic focus on great universities' open, free content. [13:14] Ju Roussel: :) I coordinate a semi-open source :) project. Learned so much from them so far! [13:14] Aphilo Aarde: Please add links to WUaS by clicking 'edit this page' [13:14] Aphilo Aarde: I'm interested in generating community [13:14] Aphilo Aarde: So, thank you all very much for coming. [13:14] Kazuhiro Aridian is Offline [13:15] Aphilo Aarde: And I'll be here next week [13:15] sandhya2 Patel: thank you for your presentation [13:15] sandhya2 Patel: : )) [13:15] Aphilo Aarde: the web is remarkable for learning opportunities [13:15] Aphilo Aarde: yw, Sandhya! [13:15] Tarek String is Offline [13:15] Ju Roussel: Have a great weekend, everyone. [13:15] XiuJuan Ying: thany you for a great talk [13:16] Aphilo Aarde: yw - see you next week - and I have office hours now, if you have any thoughts [13:16] Aphilo Aarde: obsrvations, questions. [13:16] XiuJuan Ying: i have now joined yr google grou [13:16] Kymsara Rayna is Offline [13:16] Aphilo Aarde: Great, Xiu! Thanks [13:16] Aphilo Aarde: http://socinfotech.pbworks.com/ is where this transcript can be found Welcome to Information Technology and Society In this class we'll focus on how the information technology revolution developed, especially vis-a-vis long time Berkeley Professor Manuel Castells' research on the Network Society, as well as webnographers.org - a wiki bibliography on virtual ethnography. I invite your questions, and I'll post a version of the text from each class to socinfotech.pbworks.com over the weeks. There's already a lot of information on this wiki, which will develop with time. Please join the Google Group for World University and School - like Wikipedia with MIT Open Course Ware - For more information: (http://scott-macleod.blogspot.com/2010/02/mantis-shrimp-information-technology.html - February 16, 2010)
Exotic smoothness, noncommutative geometry and particle physics. We investigate how exotic differential structures may reveal themselves in particle physics. The analysis is based on the A. Connes’ construction of the standard model. It is shown that, if one of the copies of the spacetime manifold is equipped with an exotic differential structure, compact object of geometric origin may exist even if the spacetime is topologically trivial. Possible implications are discussed. An gauge model is constructed. This model may not be realistic but it shows what kind of physical phenomena might be expected due to the existence of exotic differential structures on the spacetime manifold. There is no interesting topology on , the Euclidian four-dimensional space (or to be more precise it is topologically equivalent to a single point space). The counter-intuitive results that may be given infinitely many exotic differential structures raised question of their physical consequences . An exotic differential structure on a manifold is, by definition, a differential structure that is not diffeomorphic to the one considered as a standard one, . This means that the sets od differentiable functions are different. For example, there are functions on that are not differentiable on some exotic which is homeomorphic but not diffeomorphic to . Here we would like to investigate the role that exotic differential structures on the spacetime manifold may play in particle physics. Our starting point will be the A. Connes’ noncommutative geometry based construction of the standard model . A. Connes managed to reformulate the standard notions of differential geometry in a pure algebraic way that allows to get rid of the differentiability and continuity requirements. The notion of spacetime manifold can be equivalently described by the (commutative) algebra of smooth functions on and can be generalized to (a priori) an arbitrary noncommutative algebra. Fiber bundles became projective modules in this language. A properly generalized connection can describe gauge fields on these objects. This allows to incorporate the Higgs field into the gauge field so that the correct (that is leading to spontaneously broken gauge symmetry) form of the scalar potential is obtained. The reader is referred to for details. We shall consider the algebra : where denotes matrices over the ring or . The hat denotes that the functions are smooth with respect to some nonstandard differential structure on . The free Dirac operator has the form: here, as before, the hat denotes the ”exoticness” of the appropriate differential structure. The parameters describe the fermionic mass sector. Let be a (self-adjoint) one-form in here denotes the universal differential algebra of : We will use the following notation for an with belonging to the appropriate matrix algebra in (1). The physical bosonic fields are defined via the representation in terms of (bounded) operators in the appropriate Hilbert space : Standard calculations lead to Note, that the and are given in terms of the exotic differential structure. They will be the source of the part of the gauge group. The additional term is the price we have to pay for the ”exactness” of the gauge symmetry: noncommutative geometry prefers broken gauge symmetries. It is still an open question if noncommutative geometry provides us with new unbroken symmetries, see Ref. 8-11 for details. There is one subtle step in the reduction of the gauge symmetry from to . Namely, one should require that the part of the associated connection is equal to , the part of the connection and the ”exotic” factor is equal to . A more elegant but equivalent treatment can be found Ref. 9. But these are defined with respect to different differential structures! This can be done only locally as the exotic differential structure defines different set of smooth function than the standard one (and vice versa). We will return to this problem later. This defines the algebraic structure of the standard model. To obtain the Lagrangian, we have to calculate the curvature , . This can be easily done. The bosonic part of the action is given by the formula where is the Diximier trace defined by Here denotes the th eigenvalue of the (compact) operator . The Diximier trace gives the logarithmic divergencies, and gives zero for operators in the ordinary trace class. We will use the heat kernel method . For a second order positive pseudodifferential operator , where denotes the space square integrable functions on the vector bundle , the operator is well defined for . Then the Mellin transformation provides us with the formula: Now, we have to restrict ourselves to the case in (2) so that the free Dirac operator takes the form where is defined with respect to an exotic differential structure. This allows us to calculate the Diximier trace and the notion of a point retains its ordinary spacetime sense. This is not very restrictive as the gauge symmetry is unbroken. Calculation of the Diximier trace in the general case is more involved (if possible) and we would loose the convenient spacetime interpretation. The formula leads to the following asymptotic formula: where are the spectral coefficients , is the metric tensor, dots denote the finite terms in the limit and the hat distinguishes between the standard and exotic structures. For the Dirac Laplace’ans we have and is equal to the curvature . This gives the the following value of the Yang-Mills (bosonic) action (roughly speaking this is the ”logarithmic divergence” term): where the the trace is taken over the Clifford algebra and the matrix structure. As before, the hat is used to distinguish the ”exotic” part of the curvature from the ”non-exotic” one. Note, that due to continuity, the two integrals do not feel the different differential structures, so formally, the action looks the same as in the ordinary case. Now, standard algebraic calculations (after elimination of spurious degrees of freedom by hand or by going to the quotient space ) lead to the following Lagrangian (in the Minkowski space): The stress tensor is defined with respect to the exotic differential structure. We will not need the concrete values of the traces in (18) so will not quote them (they are analogous to those in ). Fermion fields are added in the usual way : where we have included the term into . The quark fields are defined with respect to the exotic differential structure. To proceed, let us review some results concerning exotic differential structures on . An exotic consists of a set of points which can be globally continuously identified with the set four coordinates . These coordinates may be smooth locally but they cannot be globally continued as smooth functions and no diffeomorphic image of an exotic can be given such global coordinates in a smooth way. There are uncountable many of different . C. H. Brans has proved the following theorem : Theorem 1. There exist smooth manifolds which are homeomorphic but not diffeomorphic to and for which the global coordinates are smooth for , but not globally. Smooth metrics exist for which the boundary of this region is timelike, so that the exoticness is He has also conjectured that such localized exoticness can act as an source for some externally regular field, just as matter or a wormhole can. Of course, there are also whose exoticness cannot be localized. They might have important cosmological consequences. We also have Theorem 2. If is a smooth connected 4-manifolds and is a closed submanifold for which , then any smooth, time-orientable Lorentz metric defined over can be continued to all of . Now we are prepared to analyse the Lagrangian given by (18). Despite the fact that it looks like an ordinary one we should remember that the strongly interacting fields are defined with respect to an exotic differential structure. This means that, in general, these fields not be smooth with respect to the standard differential structure, although they are smooth solutions with respect to the exotic one. certainly are continuous. In general, only those ”exotic” fields that vanish a compact set (not necessary containing the exotic region) can be expected to be differentiable with respect to the standard differential structure is because manifolds are locally Euclidean and constant functions are differentiable ) and consistent with the derivation of the Lagrangian (18). Theorem 2. suggests that it might possible to continue a Lorentz structure to all of spacetime so that (18) make sense (e.g. for a non-compact manifold , submanifolds for which fulfil the required conditions ). This means that strongly interacting fields probably must vanish outside a compact set to be consistent with the standard (?) differential structure that governs electroweak sector. One can say that the exotic geometry confines strongly interacting particles to inside bag-like structures. We do not claim (although it might be so) that we have found a solution to the confinement problem, but these results are really astonishing. estimation of the size of such an object is not possible without at the moment not available information on the global structure of exotic manifolds. A priori, they may be as small as baryon or as big as What important is is the fact that such object are not black-hole-like ones. It is possible to ”get inside such an object and go back”. There is no topological obstruction that can prevent us entering the exotic region: everything is smooth but some fields must have compact supports. One may investigate its structure as one does in the case of baryons via electroweak interactions. Of course, the above analysis is classical: we do not know how to quantize models that noncommutative geometry provides us with. Let us conclude by saying exotic differential structures over spacetime may play important role in particle physics. They may provide us with ”confining forces” of pure geometrical origin: one do not have to introduce additional scalar fields to obtain bag-like models. We have discussed only exotic but there are also other exotic 4-manifolds. The proposed model is probably far from being a realistic one but it is the only constructed. We have conncted the geometrical exoticness with strong interactions. We can give only one reason for doing so. A. Conne’s construction provides us with spontaneously broken gauge symmetries. Exact gauge symmetries are ”out of the way” so we have made the sector One can also ask the question if one writes a Lagrangian, must all of its terms be defined with respect to the same differential structure on spacetime? The answer is not so obvious. Obviously, the topic deserves further investigation. One of the most important questions is how do exotic differential structures influence quantum theory? This is under investigation. Aknowledgement: I greatly enjoyed the hospitality extended to me during a stay at the Physics Department at the University of Wisconsin-Madison, where the final version of the paper was discussed and written down. The stay at Madison was possible due to the means provided by the II Joint M. Skłodowska-Curie USA-Poland Fund. This work was supported in part by the grant KBN-PB 2253/2/91. M. Freedman, J. Diff. Geom. 17, 357 (1982). S. K. Donaldson, J. Diff. Geom. 18, 279 (1983). R. E. Gompf, J. Diff. Geom. 18, 317 (1983). S. DeMichelis and M. Freedman, J. Diff. Geom. 35, 219 (1992). R. E. Gompf, J. Diff. Geom. 37, 199 (1993). C. H. Brans and D. Randall, Gen. Rel. Grav. 25, 205 (1993). C. H. Brans, IAS-preprint IASSNS-HEP-94/22. A. Connes, Publ. Math. IHES 62 (1983) 44;Non-Commutative Geometry (Academic Press, 1993). J. G. Várilly and J. M. Garcia-Bondía, J. Geom. Phys. 12, 223 (1993). A. Connes, in The interface of mathematics and physics (Claredon, Oxford, 1990) eds . D. Quillen, G. Segal and S. Tsou. A. Connes and J. Lott, Nucl. Phys. B Proc. Suppl. 18B, 29 (1990). A. H. Chamseddine, G. Felder and J. Fröhlich, Phys. Lett. B296, 109 (1992). A. H. Chamseddine, G. Felder and J. Fröhlich, Nucl. Phys. B395, 672 (1993). J. Sładkowski, to be published in Int. J. Theor. Phys., Bielefeld University preprint, BI-TP93/26 (1993). J. Sładkowski, Bielefeld University preprint, BI-TP93/64 (1993), Proceedings of the XVII Silesian School of Theoretical Physics, Szczyrk (1993): Acta Phys. Pol B 25 (1994) 1255. P. B. Gilkey, The Index Theorem and the Heat Equation, (Princeton Univ. Press, Princeton 1984). P. B. Gilkey, Invariance Theory the Heat Equation, and the Atiyah-Singer Index Theorem, (Publish or Perish, Wilmington 1984). P. B. Gilkey, Inv. Math. 26, 231 (1974). N. Hurt, Geometric Quantization in Action, (Reidel, Dordrecht, 1983). R. Mańka and J. Sładkowski, Phys. Lett. B224, 97 (1989). D. Kastler and T. Schücker, Theor. Math. Phys. 92, 223 (1992); (English translation p. 1075 (1993). J. M. Garcia-Bondía, preprint; hep-th/9404075.
Prev - Geometry Module 1, Topic A, Lesson 4 Next - Geometry Module 1, Topic B, Overview Geometry Module 1, Topic A, Lesson 5 Student Outcome Students become familiar with vocabulary regarding two points of Like 137. 2015-16 Lesson 1: 1An Experience in Relationships as Measuring Rate 7• 1 The order of the ratios is important. In this case, it is stated that the ratio is shirts to pants, which means the first number in the ratio represents shirts and. Prev - Geometry Module 1, Topic B, Lesson 8 Next - Geometry Module 1, Topic B, Lesson 10 Geometry Module 1, Topic B, Lesson 9 Student Outcome Students write unknown angle proofs, which does not require any new 223. Learn for free about math, art, computer programming, economics, physics, chemistry, biology, medicine, finance, history, and more. Khan Academy is a nonprofit with the mission of. 2015-16 1•1 Homework Helper G1-M1-Lesson 4 By the end of first grade, students should know all their addition and subtraction facts within 10. The homework for Lesson 4 provides an opportunity for students to create flashcards. The goal of Eureka Math is to produce students who are not merely literate, but fluent, in mathematics. This teacher edition is a companion to Eureka Math online and EngageNY. Sequence of Geometry Modules Module 1. 2016/01/05 · Eureka Math Grade 1 Module 3 Duane Habecker 13 videos 11,691 views Last updated on Jan 5, 2016 Play all Share Loading. Save Sign in to YouTube Sign in Eureka Math Grade 1 Module 3 Lesson 1 by Duane Habecker 7:27. May 18, 2017 - This board is a compilation of tools and resources to assist curriculum users with Eureka Math implementation. See more ideas about Eureka math, Math and 8th grade math. 2014/08/13 · Common Core Eureka Math for Grade 8, Module 5 Created by teachers, for teachers, the research-based curriculum in this series presents a comprehensive, coherent sequence of thematic units for teaching the skills outlined in the. Geometry Module 5, Topic B, Lesson 8 Jan 26, 2016 - Student Outcomes Congruent chords have congruent arcs, and the converse is true. Arcs between parallel chords are congruent. Print Materials Eureka Math is pleased to offer a full complement of printed materials for the 2018-2019 school year. Request a Quote Student Editions Our student editions, K—12, are bound workbooks containing module. 7/28/2014 Page 8 of 29 HS Geometry Semester 1 Module 1: Congruence, Proof, and Constructions 45 days Topic C: Transformations/Rigid Motions 10 instructional Days In Topic C, students are reintroduced to rigid. T GEOMETRY Lesson 6: Solve for Unknown Angles—Angles and Lines at a Point This work is licensed under a 51 This work is derived from Eureka Math and licensed by Great Minds. ©2015 Great Minds. eureka- This. Eureka Math: Module 2 Lesson 21-24 12/9/2014 Page 2 of 58 12/9/2014 Page 3 of 58 MP.1 Mathematically proficient students start by explaining to themselves Make sense of problems and persevere in solving. For more EngageNY/Eureka Math resources, visit EMBARC.online. Lessons 1–8 Eureka Math Homework Helper 2015–2016 2015-16 Lesson 1: Interpreting Division of a Fraction by a Whole Number Visual Models. 2015/05/25 · Common Core Eureka Math for Grade 2, Module 8 Created by teachers, for teachers, the research-based curriculum in this series presents a comprehensive, coherent sequence of thematic units for teaching the skills outlined in the. Lesson 13: Properties of Similarity Transformations This file derived from GEO S.85 This work is derived from Eureka Math and licensed by Great Minds. ©2015 Great Minds. eureka- -M2 TE 1.3.0 08.2015 This work is. 1 8 2 16 3 24 4 32 9 12 Complete the table. If Gabby wants to make a regular octagon with a side length of 20 inches using wire, how much wire does she need? Justify your reasoning with an explanation of whether perimeter. Great Minds is a non-profit organization founded in 2007 by teachers and scholars who want to ensure that all students receive a content-rich education. Eureka Math Module 1 Selection File type icon File name Description Size Revision Time User ċ d. Grade 8 Module 1 Midmodule Assessment.docx View May 10, 2014, 4:21 PM Caddo Common ċ e. Grade 8 Module 1 End. 2014/08/13 · Eureka Math, A Story of Ratios: Grade 8, Module 7: Introduction to Irrational Numbers Using Geometry by Great Minds Paperback $17.95 Only 4 left in stock - order soon. Ships from and sold by Walrus Book Co. HIGLEY UNIFIED SCHOOL DISTRICT INSTRUCTIONAL ALIGNMENT 7/27/2014 Page 1 of 67 8th Grade Algebra I Semester 1 Module 1: Relationships Between Quantities and Reasoning with Equations and Their Graphs 40. Eureka_Math_Grade_3_Module_1_Parent_Tip_Sheets.pdf. Sign in. 2015/03/31 · These images are taken from Eureka Math Grade 8, Module 2 in the free Module PDFs on eureka Then in Grade 10, Module 1, precise language and notation is used to define each of the rigid motions. A. Hisense Tv 40インチH3シリーズ Clive Cussler Fargo Novels In Order Dkny Juicy Perfume Ryzen 5 2600 Vs Intel I5 Cbs All Access Number of Subscribers Asia Cup Schedule 2018ダウンロード Whisper Light Fit＆Flare Dressフレンチコネクション Ek Tha Tiger Full Movie Hd Salman Khan Macos Disk Creator
The misinterpretation of psi as a physical wave in ordinary space is possible only because the most common applications of quantum mechanics are to one-particle states, for which configuration space and ordinary space are isomorphic. Is the curl of Figure 2 positive or negative, and in what direction? That is, if we know a vector field then we can evaluate the curl at any point - and the result will be a vector representing the x- y- and z-directions. In Figure 1, we have a vector function V and we want to know if the field is rotating at the point D that is, we want to know if the curl is zero. The slope and y-intercept can be obtained directly from an equation in this form. In Figure 2, we can see that the water wheel would be rotating in the clockwise direction. That is, if we know a vector field then we can evaluate the curl at any point - and the result will be a vector representing the x- y- and z-directions. That is, every ordered pair that is a solution of the equation has a graph that lies in a line, and every point in the line is associated with an ordered pair that is a solution of the equation. The graph of a first-degree equation in two variables is a straight line. To understand this, we will again use the analogy of flowing water to represent a vector function or vector field. The green vector in Figure 4 will try to rotate the water wheel in the clockwise direction, but the black vector will try to rotate the water wheel in the counter-clockwise direction - therefore the green vector and the black vector cancel out and produce no rotation. Substituting into Equation 1 yields Note that we get the same result if we subsitute -4 and 2 for x2 and y2 and 3 and 5 for x1 and y1 Lines with various slopes are shown in Figure 7. We could plot some points, and draw the graph, but this is a lot of work. The overlapping waves from the two slits cancel each other out in some locations, and reinforce each other in other locations, causing a complex pattern to emerge. Example of a Vector Field Surrounding a Point. Because we are observing the curl that rotates the water wheel in the x-y plane, the direction of the curl is taken to be the z-axis perpendicular to plane of the water wheel. Using the intercepts to graph an equation is called the intercept method of graphing. Particles as waves[ edit ] Main articles: We then shade this half-plane. The more we go down in this situation, for every step we move to the right, the more downward sloping will be, the more of a negative slope we'll have. In general, if two lines have slopes and m2: Often, 0, 0 is a convenient test point. However, is undefined, so that a vertical line does not have a slope. The Algebra Rule works like this: Notice, x is 0. If so, we shade the half-plane containing the test point; otherwise, we shade the other half-plane. In words, Equation says: So let's say our line looks something like that. For more information, see the partial derivative page. It is related to the distribution of energy: So this is our y-intercept. Now, they tell us what the slope of this line is. So far, H is only an abstract Hermitian operator. Cannot find rigorous solution, just the obvious. The correspondence principle does not completely fix the form of the quantum Hamiltonian due to the uncertainty principle and therefore the precise form of the quantum Hamiltonian must be fixed empirically. The solutions you gave 2, 4 and 4, 2 are in integers. This is the earliest known reference to the Many-worlds interpretation of quantum mechanics. So the y-intercept, this point right over here, this is where the line intersects with the y-axis. In Figure 2, we can see that the water wheel would be rotating in the clockwise direction. And in what direction is it? Historical background and development[ edit ]. It is related to the distribution of energy:Straight lines. Suppose you have "y = 3x + 2".Since this has just "x", as opposed to "x 2" or "\|x\|", this graphs as just a plain straight line (because it is a linear equation).The first thing you need to do is draw what is called a "T-chart". How to Solve a Cubic Equation. The first time you encounter a cubic equation (which take the form ax3 + bx2 + cx + d = 0), it may seem more or less unsolvable. However, the method for solving cubics has actually existed for centuries!. This is called the slope-intercept form because "m" is the slope and "b" gives the y-intercept. (For a review of how this equation is used for graphing, look at slope and graphing.). I. This is called the slope-intercept form because "m" is the slope and "b" gives the y-intercept. (For a review of how this equation is used for graphing, look at slope and graphing.). I like slope-intercept form the best. X-Intercept and Y-Intercept from Graphs Worksheet. The following Worksheet can be done as an Online Worksheet. Questions 1 to 10 are on Y-Intercept, and Questions 11 to 20 are on X-Intercept. The equation of a line is typically written as y=mx+b where m is the slope and b is the y-intercept. If you know the slope (m) any y-intercept (b) of a line, this page will show you how to find the equation of the line.Download
A rectilinear motion is a motion of an object in a path not necessarily in a straight line. Some objects may travel with a uniform velocity without any acceleration called a uniform rectilinear motion whereas in some cases the speed of the object may vary along the path. Here is a list of rectilinear motion examples we are going to discuss below:- Fruit Falling Down the Tree As the fruit gets ripened, it gets detached and falls on the ground; and even when the heavy wind blows you find many fruits fallen on the ground around the tree. Due to the gravitational force of attraction, a fruit falls linearly towards the ground as the fruit detached from the node of a branch of a tree. You must have seen a group of soldiers marching on the ground or you must have marched during the occasion. The speed of soldier marches in a row remains constant throughout the marching session, hence it is an example of a uniform rectilinear motion. A ball thrown from a height moves in a path towards the bowling pins depending on the force applied to accelerate the center of mass of the bowling ball. As the ball is thrown towards the pin, the bowling ball will follow a path to collide with the bowling pins. While we are running, we either maintain our speed constant or vary accordingly. If a person running on a stadium in a rectilinear motion maintains a constant speed at every interval of time then we say that a person is in a uniform rectilinear motion. Arrow Hitting the Object An arrow released from an archer follows a straight path until it hits the target. It follows the rectilinear motion with a speed of an arrow gradually decreasing as it travels through the medium. Car Travelling on the Road A car traveling on a road accelerates at a constant speed, or increases or decreases its velocity is also an example of rectilinear motion. As we plot a graph of displacement v/s time, the graph may be an inclined line or a straight line. Read more on 10+ Motion Energy Examples: Detailed Facts Pushing a Load A man pushes a load of 45kg in weight and displaces a load at a distance of one meter per second. A force is applied on the object to drag the object every one meter in one second, such that the displacement of the object remains the same throughout, hence it is said to be in a uniform rectilinear motion. Boulder Sliding Down A boulder sliding down travels almost in a straight inclined path towards the horizontal land. The velocity of a boulder increases gradually and then decreases as it touches the horizontal surface. If we plot a graph of velocity versus time then we will trace a parabolic curve on a graph. A train moving on a track is also an example of rectilinear motion. The speed of a train depends upon the combustion of coal to boost the engine. The velocity of a train may increase or decrease along the path. A swimmer swimming in a swimming pool travels in a rectilinear motion. A person can swim in the water because of the buoyant force exerted on the body of a person. While swimming, a person follows a path in the water. Kicking the Ball Upon kicking a ball, the potential energy associated with a ball is converted into kinetic energy and sets into motion. The ball moves in a rectilinear motion upon kicking the ball. The electric lifts move in a vertically upward or downward direction converting the electrical energy into mechanical energy. The motion of the lift is also a rectilinear motion example. A boy riding a bicycle maintains the momentum upon giving the velocity to the bicycle. The momentum of a bicycle is equal to the total mass of the bicycle and a rider and the velocity of the bicycle. The motion of a cycle is in a rectilinear motion. Pulling the Trolley You must have used a shopping trolley at the malls. The motion of a trolley is equal to the rectilinear motion as a trolley covers a specific path on the push or pull force incident on the handle of a trolley. A person walking in a path covers an equal distance at every interval of time. Hence, the motion of a person is said to be in a uniform rectilinear motion. As the condensed molecules gain high potential energy and become unstable, this potential energy is turned into kinetic energy and the water molecules accelerate towards the ground. This is due to the gravitation potential energy of the Earth that brings the raindrops down on the surface. The motion of the raindrop falling on the ground follows the rectilinear motion. Flow of Water The water flows from the upper level to the lower level land in the form of mechanical energy. The flow of water carries the sediments along with it. This venturing of water in a kinetic motion is also in a rectilinear motion. The velocity of the water varies depending upon the volume of the water. The rockets flying in space follow the third law of Newton. To release the rocket in space, an equal force has to be generated to oppose the gravitational pull of the Earth. A rocket engine is mostly made of Hydrogen fuel that supplies enough energy to produce trust on a ground opposing gravity. The rockets move vertically upward in a rectilinear motion and leave into space. A girl sliding down moves down in an inclined path converting her gravitational potential energy into kinetic energy. Pulling the Object with the help of Pulley When you pull the object with the help of a pulley, for example, a man drawing water from a well, the motion of an object is rectilinear. Frequently Asked Questions What is the distance covered by an object moving in a rectilinear motion with a velocity of 20km/hr in 35 minutes? We know that v=d/t d=20km/h x 35 min= 20/60 x 35=11.67kms Hence, in 35 minute the object will cover 11.67 km. On what factors does the rectilinear motion of an object depend? The object sets into motion due to the external impedance of force. The rectilinear motion of an object depends upon the force applied, the momentum gained by the object, and the gravitational force exerted on the object.
The angular velocity is the rate of change in the angular displacement of a particle. The angular velocity has two words: angular and velocity. Velocity means displacement upon a time, and angular velocity means angular displacement upon a time. If any particle is making a circular motion, then suppose the particle equals p and center equal to q. Now p comes to q during t time. Now it will create an angle in the center which is theta. This angle theta is also known as angular displacement. However, angular displacement travels per unit time, or angle sustains per unit time. However, we know this ratio is angular velocity. Angular velocity’s symbol is omega. Next, omega equals the angle subtended per unit time. Let particles’ direction of motion be anti-clockwise. Angular velocity and angular displacement are the vector quantities. So the direction of angular displacements is the same as angular velocity. Angular velocity formula Angular velocity formula is omega equals 2 pi N over 60 rad per second. We know the velocity formula in linear motion is the general formula; velocity is nothing but the distance by time. If your body covers two beats in one second, then we can say that the velocity in the body in linear motion is two minutes per second. Suppose we apply this technique in the angular motion also, so in this case, the angular displacement unit per second time. So at first, if your body is rotating, let N be the body’s rpm. The rpm means revolutions per minute. If we want per second, then we need to divide by 60. Because one minute will contain 60 seconds and revolutions per minute converts into per second by dividing by 60, for one revolution, the body rotates one complete revolution; we know that it will cover 360 degrees which is equal to 2 pi radians. If you take 1 minute, it will be N revolution. N revolution means 2 pi into N radians. Next, you take one second that is 2 pi N divided by 60 radians per second. This we got the angular velocity formula, omega equals 2 pi N by 60. ω= 2πN/60 rad/s Velocity in linear motion v= d/t= 2/1 m/s v=2 m/s, t=1 In angular motion If N=rpm=revolutions per minute N/60=rps=revolutions per second 1 revolution=360 degree=2π radians Next, 1 minute= (N) revolutions =2π into N radians Next, 1 second=2πN/60 rad/s ω= 2πN/60 rad/s Angular velocity unit The unit of angular velocity is radians per second. Angular velocity symbol Omega (ω) is the symbol of angular velocity. Angular velocity to linear velocity Let, v is linear velocity and ⍵ is angular velocity. At first, a body which is moving on the circular path of radius r, then the position changes from p to q point. Next, that’s why theta is made in the centre. However, theta is angular displacement. We know, angular displacement theta equals l by r. Next, we can write l equals to r into theta. Now, we need to differentiate both sides with respect to time t. So we can write dl/dt equal to r into dθ/dt where r is constant. Next, dl/dt will be v and dθ/dt will be ⍵. However, angular displacement is now angular velocity and rate of change of linear displacement by time is linear velocity. Now, this is an scaler form and in vector form we can write v equals to ⍵ ⨯ r. Now, v is the linear velocity, ⍵ is the angular velocity and r is the radius vector or position vector. Then, v= ⍵ ⨯ r. Angular velocity dimensional formula The dimensional formula of angular velocity is omega equals angular displacement by time. Now the SI unit of angular velocity is radian per second. Now theta is the dimensionless quantity. If a circle of radius r, initially a particle is p, after some time interval the position of the particle is p. Formula for angular displacement is equal to arc upon radius. The arc length is considered as s. So theta equals to s by r. Nows measures the intern of the meter, and r measures the intern of the meter too. So both meters will be canceled, and the result is 1. Now the dimension of angular displacement is M power zero, L power zero, and T power zero. It is the dimensionless quantity. So the place of angular displacement is written as 1 upon t, and t is nothing but time, so the SI unit of time is second. Now omega equals 1 upon second is expressed in terms of dimension T. Now T will be the numerator which will be T to the power minus 1. Next, the dimensional formula’s format of angular velocity is M to the power absent L to the power absent and T to the power minus 1. Angular Velocity = Angular displacement × [Time]-1…(1) Now, the dimensional formula of Angular displacement = [M^0 L^0 T^0]….(2) And, the dimensions of time = [M^0 L^0 T^1]……(3) Now, on substituting equation (2) and (3) in equation (1) we get, Next, angular Velocity = Angular displacement × [Time]-1 Or, v = [M^0 L^0 T^0] × [M^0 L^0 T^1]-1 = [M^0 L^0 T^-1] However, the angular velocity is dimensionally represented as [M^0 L^0 T^-1]. Angular velocity of the earth We know the formula of angular velocity, which is omega equals 2 pi divided by T. We are assuming that it is rotating in a circular path. However, T is nothing but how much time it takes to rotate that is T. However, we know earth takes 24 hours to complete one rotation. But we need to transform the hour to the minute. Therefore we need to convert 24 hours which equals 24, into 3600. Now omega equals to 2 pi upon 86400. Angular velocity of the earth’s equation dt = 24Hours Now, The SI unit of angular velocity is rad/sec so we need to convert our time to seconds Next, 1 hour = 3600 sec Now, ω=2π/24×3600 rad/sec Angular velocity and speed Let’s say we have some object that’s moving in a circular path. So the object is moving in a circular path that looks something like a center clockwise circular path. Now, the object is making five revolutions every second. However, radians are just one way to measure angles. You could do with degrees per second. Now, If we do it with radians, we know that each revolution is 2 pi radians. Next, If we go all the way around a circle, we have gone 2 pi radians. There is 2 pi per revolution, so we can do a little bit of dimensional analysis. Next, we get 5 times 2 pi which gets us 10 pi radians per second. And it works out the dimensional analysis, and it also makes sense. However, If we are doing five revolutions a second, each of these revolutions is 2 pi radians. So we are doing 10 pi radians/second. So either 5 revs/second or 10 pi radians/second, they are both essentially measuring the same thing. However, this measure of how fast you are orbiting around a central point is called angular velocity. Next, it is angular velocity because if you think about it, it is telling us how fast our angle is changing or the speed of the angle changing. However, angular velocity tends to be treated as angular speed. Next, It is a vector quantity, and it is a little unintuitive that the vector’s actually popping out of the page. However, it is actually a pseudo vector. So it’s a vector quantity, and the direction of the vector is dependent on which way it’s spinning. For example, when it is spinning in a counterclockwise direction, there is a vector; the real angular vector does pop out of the page. If it is going clockwise, the angular velocity vector will pop into the page. When we are just thinking about a two-dimensional plane, we can really think of an angular velocity as the pseudo-scalar. But we can include that as a scalar quantity, as long as we specify which way it is rotating. So 10 pi radian per second, we could call it angular velocity. And this tends to be denoted by an omega. However, We could say angular velocity is equal to a change in angle over a change in time. So after the calculation, we could tell that omega is equal to speed which we are using v for, divided by the radius. Angular velocity equation Q. Calculate the angular velocity of a particle moving along the straight line given by θ = 3t3 + 6t + 2 when t = 5s. - So, we know θ = 3t3 + 6t + 2 and t= 5 second However, ⍵= dθ/dt= 9t^2+6. Next, ⍵= 9(5^2)+6 = 231units/second. Some frequently asked questions What is the angular velocity in simple terms? Angular velocity is the rate of change in the angular displacement of a particle. However, angular velocity has two words: angular and velocity. Velocity means displacement upon time, and angular velocity means angular displacement upon a time. What is called angular speed? Angular velocity tends to be treated as angular speed. It is a vector quantity, and it is a little unintuitive that the vector’s actually popping out of the page. However, it is actually a pseudovector. Is Omega angular velocity? Yes, omega is the symbol of angular velocity. What is the symbol of angular acceleration? Alpha (α) is the symbol of angular acceleration. What is omega equal to? ω = 2πf, omega is equal to 2 pi f. What is the formula of angular velocity? Angular velocity formula is omega equals to 2 pi N over 60 rad per second. However, we know the velocity formula in linear motion is the general formula; velocity is nothing but the distance by time. What is omega in physics class 11? Average angular speed = ΔΘ/Δt. However, angular speed, ω = dΘ / dt. V = w r, in which v – linear speed of particle transferring in a circle of radius r.
Intro Quant Reasoning Intro Quant Reasoning MATH 1030 Popular in Course Popular in Mathematics (M) This 3 page Class Notes was uploaded by Miss Noel Mertz on Monday October 26, 2015. The Class Notes belongs to MATH 1030 at University of Utah taught by Staff in Fall. Since its upload, it has received 8 views. For similar materials see /class/229949/math-1030-university-of-utah in Mathematics (M) at University of Utah. Reviews for Intro Quant Reasoning Report this Material What is Karma? Karma is the currency of StudySoup. Date Created: 10/26/15 MATH 103002SPRING 2005 7 NOTES ON PERCENTAGE INCREASE Consider the following FALSE statements 1 lfl earn 20 less than you do then you earn 20 more than 1 do 2 1f the actual price of a car is 66 more than the labeled price due to tax then the labeled price is 66 less than the actual price We have discussed in class why the the rst statement is false The second statement is false for the same reason One way to see this is to substitute actual numbers into the statement Suppose that 1 make 20 less than you do and you make 10hri Then my wage is given by the formula 10h r 7 10h r 10h r 7 2h7 8h7quot because my wage is found by taking your wage 10hr and subtracting 20 of that which is 2hri On the other hand 20 more than 8hr is 8h7quot 8h7quot 8h7quot 7 1160h7quot 9160h39ri This difference is accounted for by the fact that a percentage increase or decrease depends on the starting value 20 of the large number is more than 20 of the smaller number This can be explained by the following fact Percentage change is a relative change ie it depends on the starting value Quiz 3 problem 3 asked you to calculate the pretax cost of a car given the price after taxes Funda mentally this problem just statement 2 above Several students calculated the pretax price of the car as if statement 2 were correct but for the same reason as before statement 2 is not true So the calculation 19187i997010661918719917921158 is incorrect We can check this assuming the pretax cost was 17921158 the actual cost would be that amount plus tax which is 17921158 0106617 921158 19 104140 19 187199 A good mental exercise would be to try and explain why the number we arrived at 19104i40 using the wrong method is smaller than the given number 19187i99i Hint consider what the starting values are and try to put the problem into the context of the boxed fact above By the way the correct solution to problem 3 can be found with the rest of the quiz solutions on our webpage MATH 103002SPRING 2005 7 INDUCTIVE VS DEDUCTIVE ARGUMENTS As the title suggests these notes are meant to be a supplement to the lecture on inductive and deductive reasoning An argument is composed of a series of premises or supposed facts and a conclusion which should be derived from the premises The textbook gives formal de nitions of the two types of arguments An inductive argument makes a case for a general conclusion from more speci c examples A deductive argument makes a case for a speci c conclusion from more general premises These are nice de nitions but in practice it is dif cult to determine if a speci c argument is inductive or deductive Consider the following situation Linda is 31 years old single outspoken and very bright She majored in philosophy As a student she was deeply concerned with issues of discrimination and social justice and also participated in antinuclear rallies Rank the following possibilities from 1 most likely to 5 least likely 7 Linda is a teacher 7 Linda works in a bookstore and takes yoga classes 7 Linda is a bank teller 7 Linda sells insurance 7 Linda is a bank teller and is active in the feminist movement A response to this question would be a conclusion The reasons for such a conclusion would constitute the premises For example I could argue as follows Premise Teachers are bright Premise Teachers are college graduates Premise A college degree isn7t needed to work in a bookstore Premise Linda graduated from college Conclusion Linda is more likely to be a teacher than to work at a bookstore In other words Linda is a teacher77 ranks higher than Linda works in a bookstore and takes yoga classes77 The following is another possible argument This is patterned what a group in our class did Premise There are ve possible scenarios Premise I wrote each on a sheet and randomly picked that Linda works in a bookstore Conclusion Linda is more likely to work in a bookstore than anything else These are both examples of inductive arguments This can be ascertained using the de nitions but perhaps an easier way to know that these are inductive arguments is to ask How can I analyze the argument An inductive argument can only be analyzed in terms of its strength It requires personal judgment as to whether or not the conclusion follows from the premises ln examples above do you think the arguments are sound 2 MATH 1030702SPRING 2005 7 INDUCTIVE VS DEDUCTIVE ARGUMENTS On the other hand a deductive argument can be analyzed in terms of its validity whether or not the conclusion is direct result of the premises and its soundness in addition to being valid the premises are true Contrast the previous two inductive arguments with the following deductive arguments Premise All people who are bank tellers and active feminists are bank tellersi Premise Linda is a person Conclusion It is more likely that Linda is a bank teller than that she is a bank teller AND active in the feminist movement Premise All women who sell insurance are deeply concerned with issues of social justice Premise Linda is deeply concerned with issues of social justice Conclusion Linda sells insurance With each of these examples we can analyze their validity and soundnessi Given the premise all bank tellers who are active in the feminist movement are bank tellers77 the conclusion is correct Therefore the argument is valid Moreover the premise is true so we conclude that the argument is sound The second argument is invalid as can be veri ed by drawing a Venn Diagrami Thus even though the conclusion may still be correct the argument is not sound Exercise 1 Analyze the argument Premise All women who are deeply concerned with issues of social justice sell insurance Premise Linda is deeply concerned with issues of social justice Conclusion Linda sells insurance Exercise 2 Come up with three inductive arguments and three deductive arguments related to what Linda does For each of the inductive arguments analyze the strength of the argument For each of the deductive arguments draw the accompanying Venn diagram and determine the validity and soundness possible of the argument Mathematicians tend to be skeptical of inductive argumentsi Even if the argument is very strong they are never satis ed until a deductive proof is given Consider the following example Conclusion If n is a natural number then the sum of the rst it numbers is nltn1gt 2 ile123n71ni A possible inductive argument could be 22 l Premise For n 2 we have 1 2 3 7 Premise Forn3wehavel236 4 4 l Premise For n 4we havel234 10 Premise I plugged it into my calculator for lots of other numbers and it always worked If you7re interested in seeing a deductive proof I would be more than happy to show you one
Appendix 2 - Mathematical requirements Mathematics, the bane of many students lives! Apart from the four rules of number (addition, subtraction, multiplication and division) there are other mathematical techniques and concepts that are essential at this level. Paper 1 (HL and SL) does not allow calculators and so any calculations use simple numerical relationships. Ensure that you are familiar with all of the concepts below. Calculations and basic arithmetic functions - the four rules of number The four rules of number are addition, subtraction, multiplication and division. In paper 1 no calculators are allowed, so the basic maths requirement is simple manipulation of these rules. For instance, you will be expected to be able to divide and multiply by 2, but not by 1.3 (unless the number you are dividing is clearly a multiple of 1.3, eg 4.6) Manipulation of equations is something that some students get confused with. Subtraction and addition Whatever operation is done to the left hand side MUST be done to the right hand side. In other words, if something (a number or a symbol) is subtracted from the left hand side it must also be subtracted from the right hand side. Add to the left and we must also add to the right. For the equation: A + B = C + D If we subtract 'A' from the left it must also be subtracted from the right for the equation to remain correct A + B - A = C + D - A Which cancels down to: B = C + D - A We have taken a term (in this case 'A') from the left to the right CHANGING THE SIGN as we did so. This is an allowed mathematical manipulation. Multiplication and division Take the equation: Whatever you do to one side of the relationship (i.e. quantities to the left of the equals sign or the right) you MUST do to the other side for the relationship to remain valid. Divide both sides by B gives which simplifies to: It seems as if the B has gone from the top left to the bottom right, and indeed any transformation like this is allowed PROVIDING that there are no additions or subtractions on either side. i.e. In this equation 'B' cannot be taken to the bottom right and cancelled at the same time on the left, as the term produced on the left by dividing both sides by B: does not equal A. The result would be: The equation to calculate the relative molecular mass of an ideal gas may be stated as: Where m = mass in grams, Mr is the relative molecular mass, P is th epressure, T is the absolute temperature, R is the universal gas constant and V is the volume in litres. To rearrange the equation for Mr it must be taken from the bottom right to the top left (by multiplying both sides by Mr): And then the P and V terms must be taken from the top left to the bottom right (by dividing both sides by P and then by V) Decimals, fractions and percentages Decimals, fractions and percentages are three ways or representing a part of a whole. The circle is shown divided up into parts. One fourth part may be represented as the decimal 1.5, or the fraction 1/4 or the percentage of the whole 25%. To convert from fraction to decimal the numberat the bottom of the fraction ( the denominator) must be divided into the number at the top (the numerator). Hence, 2/5 is worked as follows: five into 2 doesn't go, add in the next digit (the one after the decimal point). Five into 20 goes 4, therefore the decimal = 0.4 A ratio shows how one number is related to another. The ratio between two numbers is represented by a colon ':' value 1 : value 2 The ratio of 35chlorine atoms to 37chlorine atoms is 3:1 This means that for every four chlorine atoms there are 3 35chlorine atoms for every 1 37chlorine atom (on average). A relationship may be given as a ratio, for example, the relationship between the mass of a hydrogen atom and the mass of a helium atom is in a ratio of 1:4. This tells us that one helium atom has a mass four times that of a hydrogen atom. Approximation is a way of expressing numbers that are almost exact. For example the number 4.99 is approximately equal to 5. As chemistry is an experimental science it is important to realise that measurements are often not exact and that approximations need to be applied. The value of absolute zero is approxiamtely equal to -273 ºC The value of atmospheric pressure in kPa is approximately equal to 100. Avogadro's number is approximately 6.02 x 1023 The mathematical constant 'Pi' is approximately equal to 3.142 Most of the accepted values used for the constants of chemistry are available to a higher degree of accuracy than give by the exam boards. They have been approximated for easy usage. For example the value of atmospheric pressure at STP is quoted as 100.0 kPa, when clearly the atmospheric pressure varies from day to day. In calculations to find the empirical formula of a compound from percentage composition data, the numbers that are obtained are almost never EXACT ratios of number of atoms. However, if they are approximately a simple ratio and we know that atomic ratios have to be integral (whole numbers) then this approximation is good enough. Proportionality is when the value of one thing is affected in some way by the value of a second quantity. It is expressed using the proportionality symbol . A man's happiness may be proportional to the number of beers drunk.... of course, so could his unhappiness! Proportionality is very important in understanding how factors are interlinked. Proportionality may be: Directly proportional means that there is a straighforward mathematical relationship between two quantities. The mass of a sample is directly proportional to the number of moles of particles present. Direct proportionality is easily expressed as a mathematical relationship: This means that whatever happens to the mass also happens to the moles. Double the mass and you double the moles etc Indirect proportionality means that one factor affects the other but not in a direct manner. There may be a logarithmic relationship, a squared relationship or a more complex one. Inverse proportionality means that an increase in one factor produces a corresponding decrease in another factor. For example an increase in the pressure of a gas causes a decrease in the gas volume. We say that the gas pressure is inversely proportional to the volume. Rate is inversely proportional to the time taken for any given process The reciprocal of any number is 1 divided by that number. The reciprocal of 2 is 1/1. The reciprocal may be useful when analysing data. For example the Arrhenius equation may be used to find out the activation energy from a series of kinetics experiments: k = Ae-Ea/RT Taking natural logs: lnk = lnA - Ea/RT If a graph is plotted of k values against the reciprocal temperature values (1/T), a straight line is produced with slope equal to -Ea/R. This allows us to find the activation energy of the reaction. The speed of a car and the time taken for a journey have a reciprocal relationship Speed is proportional to the reciprocal of the time taken (also called inverse proportionality) speed 1/time taken Pay attention to the number of significant figures required in the answer. Significant figures means the number of non-zero digits in a number, counted from the first to the last. The IB usually requires answers to 3 significant figures. Write the number 1.04037 to three, four and five significant figures. 1.04037 = 1.04 to three sig figs 1.04037 = 1.040 to four sig figs 1.04037 = 1.0404 to five sig figs (notice the final digit has been rounded up) Write the number 203507 to three, four and five significant figures. 203507 = 204000 to three sig figs 203507 = 203500 to four sig figs (notice the final non-zero digit has been rounded up) 203507 = 203510 to five sig figs (notice the final non-zero digit has been rounded up) This means the number of digits after the decimal point. Is should not be confused with significant figures. Like significant figures, decimal places can be rounded up, or unchanged by consideration of the next decimal place. The same rules apply as for significant figure rounding. Calculators present a display with many decimal places. This does not mean that they are extremely accurate, rather that they process the data input precisely. If the original data input had an accuracy of 2 decimal places then the calculator readout can only have this degree of accuracy, no more than that. You should not write down long strings of decimal places in your exam solutions. If you have measured a solution volume to an accuracy of ±0.01 cm3 as 20cm3 and then you wish to divide this by 3, the calculator will give you a readout of 6.6666666667 (or something like that depending on the calculator). However, the original accuracy was only 0.01 and so the calculated value should only be written down as 6.67. A decimal number can be approximated by rounding up or down by considering the digit AFTER the last digit required. If this digit is 5 or greater then the previous digit must be rounded up by one. If not then it is left unchanged 0.4354 = 0.435 if only three decimal places are required 0.4356 = 0.436 if only three decimal places are required Knowledge of logarithms is useful in some areas of the syllabus. A logarithm is the number to which ten has to be raised to obtain a given value. What does this mean exactly? A number may be squared and written as follows 42 = 16 The small superscript 2 refers to the fact that the digit 4 is squared. We say that the number 4 has been raised to the power of 1. This superscript is also called the index to which 4 is raised. When we raise the number 10 to different powers we obtain larger numbers. If the power is non-integral (i.e. not a whole number) then we can obtain ANY other number. If we just express a number by the number to which 10 has been raised then this is called the logarithm of that number. (abbreviated to log10 or just log) As 102 = 100 then log 100 = 2 As 104 = 10000 then log 10000 = 4 What advantage does this have? The advantage of using log forms of numbers is twofold: Example: Calculate the value of the folowing sum using logarithms: 100 x 1000 log 100 + log 1000 2 + 3 = 5 Therefore log 5 = 100000 These are logarithms in which the base is not 10, rather the natural number e which is equal to 1.303 i.e. a natural logarithm is the number to which 1.303 has to be raised to achieve the desired value. In certain calculations these natural logarithms emerge naturally (hence the name). They are abbreviated as 'ln' (loge). The natural log of 2, ln2 = 0.693 This means that 1.3030.693 = 2 Natural logs appear in some areas of the syllabus, such as the Arrhenius equation for the rate constant relationship with temperature (HL only): k = AeEa/RT Taking natural logs throughout: lnk = lnA Ea/RT (notice how terms that were previously multiplied together are now summed in the log form) This equation now takes the form of a straight line, y = mx + c Where 'y' is the vertical axis, 'x' in the horizontal axis, 'm' is the slope or gradient and 'c' is the intercept with the 'y' axis. In this case by rearranging the equation to give lnk = Ea/RT + lnA If a graph is plotted of lnk against 1/T, the gradient is the activation energy Ea and the intercept is 'ln A' (the natural log of the Arrhenius constant). The gradient has a negative slope i.e. it slopes down to the right. Following on from the section on logarithms is should be noted that any value can be produced (to three significant figures) by multiplying a number between 1 and 10 by 10, raised to the power of an integer (whole number) This is a way of expressing very large or very small values consistently. 123000 may be written as 1.23 x 100000 In standard form as 1.23 x 105 Notice that dividing by a large number gives a negative index. This is because: 0.0045 may be written as 4.5 x 1/1000 (i.e. 4.5 divided by 1000) In standard form 4.5 x 10-3 Standard form requires that a number be written as a value between 1 and 10, multiplied by 10 raised to the power of an integer. The number of decimal places used in standard form usually depends on the precision required. It is often 2 decimal places, for example Avogadro's number 6.02 x 1023 Units and dimensions Experiments require that values be calculated and measured. Each measurement has two components, the first being the magnitude (the actual number recorded) and the second being the units (the dimensions). A measurement of distance from the earth to the sun is usually given in km as 400,000,000; This same distance measured in another astronomical unit, the light year, is 4.2 x 10-5 light years. Clearly the two numbers are very different and without the units, or dimensions, they are fiarly meaningless. Dimensions in the answer to a problem may be found by considering the individual parts of any equation and combining the dimensions of each individual component. When a solution is made a mass of solute is measured out and dissolved in a volume of water. The mass is measured in grams and the solvent volume in cm3 or dm3. The concentration of the solution is then: concentration = mass/volume units of concentration = units of mass / units of volume units of concentration = g/dm3 If the solution is required in moles/dm3 then the mass must first be converted to moles by dividing by the relative mass of the solute. When writing the dimensions or units on the same line (i.e. not as a fraction such as g/dm3) the index of any factor at the bottom of the fraction (the denominator) must be changed to the opposite sign, i.e. g/dm3 becomes g dm-3. What are the units of the rate constant in the rate law expression: Rate = k[A]2[B] Substituting the dimensions into the equation gives: mol dm-3 s-1 = k (mol dm-3)2 x (mol dm-3) Now cancel out the similar parts And bring the parts at the bottom to the top, changing the sign of the index (powers) k = dm6 mol-2 s-1 You must be familiar with the various different ways to represent information such as graphs and other data forms (pie charts, histograms, bar charts etc.) In chemistry this ALWAYS refers to line graphs. Bar graphs (histograms) are almost never used in chemistry. Bar graphs cannot be used to analyse information further, they only present data for visual comparison. You must be familiar with the terms intercept, gradient and area as applied to line graphs. Any graph line can be related to an equation and the simplest is a straight line. All straight lines have the form y = mx + c, where 'y' is the value on the y axis, 'x' is the value on the x axis, 'm' is the gradient and 'c' is the intercept with the y axis (i.e. the y value at the point where the line crosses the y axis). The gradient (slope) is obtained by measuring the change in y values over a change in x values between two points on the line. Common mistakes in graphing include axes that are not coherent, i.e. the number of squares that represent a value is different over two parts of the axis line. Students sometimes do this to 'squeeze' the graph into a given size. Be familiar with the display on your calculator. There are several different ways that calculators accept scientific notation input and display it. Your calculator may display standard form as 1.24 E-10, meaning 1.24 x 10-4 or it could do so without the capital E. On the keypad there may be an "exp" button or it may simply say "E". Make sure you understand your own calculator! Show each step of the calculation. If an examiner has difficulty reading your answer he can give credit for the method used. Each step that you have taken should be written down to help you own thought processes. It is far easier to revisit a solution that you have set out correctly for later checking. Important: If you make an error early in the calculation you may still get credit if you have written evidence that you have used the correct method. This is called 'carrying forward' an error. You will not have the answer correct numerically but the method you have used will demonstrate that you know how to do it.
3 edition of Simplified rate-law integration for reactants which are first order in each of two reactants found in the catalog. Simplified rate-law integration for reactants which are first order in each of two reactants by National Aeronautics and Space Administration, Ames Research Center, Research Institute for Advanced Computer Science in [Moffett Field, Calif.] Written in English \|Other titles\|\|Simplified rate law integration for reactants which are first order in each of two reactants.\| \|Statement\|\|E. Levin, J.G. Eberhart.\| \|Series\|\|RIACS technical report -- 88.23., NASA-CR -- 185408., RIACS technical report -- TR 88-23., NASA contractor report -- NASA CR-185408.\| \|Contributions\|\|Eberhart, James G., Research Institute for Advanced Computer Science (U.S.)\| \|The Physical Object\| Mmixture = mol/L x mL/ mL = mol/L As shown for Reaction Mixture 1 in the table. Calculate the rest of the reactant concentrations to complete the table. To obtain the values of m, n, and p, the orders of the reactants, use the ratios of the measured relative rates in such a way that the concentrations of some reactants cancel out and the required values are obtainedFile Size: 98KB. In a zero order reaction, the rate=k since anything to the power of 0 is 1. Therefore the rate of reaction does not change over time and the [A] (for example) changes linearly. In a first order reaction, the rate and concentration are proportional. This means that if the concentration is doubled, the rate will double. simple rate law, which takes the form ν = k [A]a[B]b[C]c () i.e. the rate is proportional to the concentrations of the reactants each raised to some power. The constant of proportionality, k, is called the rate constant. The power a particular concentration is raised to is the order of the reaction with respect to that reactant. Note File Size: KB. A reaction is of first order in reactant A and of second order in reactant B How is the rate of this reaction affected when (i) the concentration of B alone is increased to three times (ii) the concentrations of A as well as B are doubled - Chemistry /5(36). So to find a rate equation, given the order of the reactants, we must know that the order with respect to each reactant is the exponent in the rate equation. That being said, we are told that the reaction is first-order with respect to [X], and the reaction is second-order with respect to [Y]. If the values of k are found to be constant for all the sets, the reactions is supposed to obey that particular rate law and follows the order suggested by that integrated rate law. In case, the values of k are not constant, the data is used in the rate equation of other order. The rate equation for first and second order reactions are as follows. On being aware and expanding your consciousness Phoenix Park murders glossary of later Latin to 600 A.D. Florence [and] some Tuscan cities A Prayer for Owen Meany The design of a celestial navigation microcomputer with thoughts on an integrated information distribution system The Time Is Right:todays Tools for Tomorrows Leaders The heartland of divinity Canon law collections in early ninth-century Salzburg. Experiments on the scattering of electrons by ionized mercury vapor Teaching the child to read SIMPLIFIED RATE-LAW INTEGRATION FOR REACTANTS WHICH ARE FIRST ORDER IN EACH OF TWO REACTANTS E. Levin J. Eberhart* Research Institute for Advanced Computer Science NASA Ames Research Center RIACS Technical Report September The purpose of this paper is to present a simple procedure for integrating the rate. Get this from a library. Simplified rate-law integration for reactants which are first order in each of two reactants. [E Levin; J G Eberhart; Research Institute for Advanced Computer Science (U.S.)]. rate law (first, second, zero order) how does temperature affect rxn rates collision theory (reactants moving faster) and integrated rate law (for every 10 degrees rxn rate doubles because molecules move faster and collide more often). The rate law or rate equation for a chemical reaction is an equation that links the initial or forward reaction rate with the concentrations or pressures of the reactants and constant parameters (normally rate coefficients and partial reaction orders). For many reactions the initial rate is given by a power law such as = where [A] and [B] express the concentration of the species A and B. Determining the Rate Law When there Are Multiple Reactants • Changing each reactant will effect the overall rate of the reaction • By changing the initial concentration of one reactant at a time, the effect of each reactant’s concentration on the rate can be determined • In examining results, we compare differences in rate for reactions that only differ in the concentration of one. Compare tirals 2 and 3 to see that tripling nitrate concentration triples the rate. The reaction is first order for nitrate. The final rate law is. The sum of the orders for each reactant gives you the overall order of the reaction. In this case, the reaction is first order with regard to each. depend on overall reaction order of the rate law; equal to units of rate divided by units of concentration first-order integrated rate law. ln[A]t-ln[A]0 = -kt, when units of [A] are the same reaction whose rate depends on the reactant concentration raised to the second power or on the concentrations or two different reactants, each. Book, lecture notes, and Powerpoints Pg. Terms in this set () an integrated rate law of first order, relates its initial concentration [A]₀ to_ for which the rate depends either on a reactant concentration raised to the second power or on the concentrations of two reactants each raised to the first. Determine order for the second element or compound 4. Write out the rate order. Solve for k, by plugging in values for all the other variables. Units for k should be determined as so: mol/(L*s) = (mol^x / L^x) x is the order, for that element. On the right side of. First, let's take experiments 1 and 2, because two of the reactants are the same concentration, and only [B] is different. [B] is multiplied by 4 from experiment 1 to 2. The rate of reaction is also multiplied by 4 from experiment 1 to 2. Therefore the reaction is first order for reactant B. Let's figure out the order for reactant A next. second order reaction is one whose overall reaction order is 2. Its rate either depends on one second order reactant or two separate first order reactants. if the reactants for the reaction are A and B, the rate will be either: rate = k(A)^2 or rate= k(B)^2 or rate= k(A) (B). Dr. Shields discusses how to determine the order of each reactant in the rate law using information given regarding the change in concentration of. For the N 2 O 5 decomposition with a rate law of k[N 2 O 5], this exponent is 1 (and thus is not explicitly shown); this reaction is therefore a first order reaction. It can also be said that the reaction is "first order in N 2 O 5". For more complicated rate laws, the overall reaction order and the orders with respect to each component are used. Order of Reactants and Rate Constant example from ’s AP Chemistry class. Want more video tutorials. Our full lesson includes in-depth. Reaction Rate for Multiple Reactants Consider the reaction and the initial concentration and initial rate data below. a) Determine the rate law for this reaction b) What is the overall order of the rate law. c) Calculate the rate constant. If you use an integrated rate law, you need to first determine the order of the reaction. Pick the integrated rate law for whatever order you have. Determine the order for each of the reactants, NO and H2, from the data given and show your reasoning. (ii) Write the overall rate law for the reaction. Kinetics Questions – 2/22/ reactants. The integrated rate law expresses reactant concentrations as a function of time. The differential rate law is generally just called the rate law. The rate constant k is a constant that allows one to equate the rate of a reaction to the concentration of reactants. The order is theFile Size: KB. To determine the reaction order for each species, it is important to compare how the rate of the reaction changed as the concentration of one of the reactants was changed, and the other reactants were kept constant. Which two experiments can be used to determine the order for oxalic acid. Justify your answer. Which two experiments can be used. The rate law for a certain reaction is second order with respect to one of the reactants, R. Suppose you study this reaction, observing the absorbance of light at the analytical wavelength for R, and record the data with respect to elapsed time. Also suppose that the concentrations of all the other reactants are in large excess, and that R is the only colored species involved. The rate law for a certain reaction is second order with respect. to one of the reactants,R. Suppose you study this reaction, observing the absorbance of light at the analytical wave length for R, and record the data with respect to elapsed time. Simplified rate-law integration for reactions that are first-order in each of two reactants, Levin, E.; Eberhart, J. G. p. Icosahedral matrix representations as a fuuction of Eulerian angles, Martinez-Torres, Emilio, et al. pp. They are proportional to different powers because they may have different species involved in the rate determining step. A stoichiometric equation only tells you the ratio in moles by which species react. It tells you nothing about the reaction me. As result the rate law simplifies to a pseudo m-th order rate law: rate = k∙[A]^m∙[B]^n ≈ k∙[A]^m∙[B]₀^n = k'∙[A]^m. with pseudo m-th order rate constant. k' = k∙[B]₀^n. That allows you treat the results as single reactant rate law. To find m just compare your the .
15 3.3. Parabola and parallel lines This example is similar in spirit to the last, except this time the 10 parabola is held fixed and a line of slope 2 is moved parallel to itself. The objective is to see how the number of intersec- tions changes. 5 At the right is the graph of the parabola y = x(6–x) and the line 0 y = 2x. -202468 Finding the intersections We begin by finding the points of intersection of the parab- ola and the line. We have two geometric objects here, and we When we ask the students how to emphasize that by putting labels on the two y’s. go about finding the intersec- tions of the parabola and the yP = x(6–x) line, we are a bit dismayed by their answers. Most of them pro- y = 2x. L duce a bunch of half remem- So at any x, yP is the height of the parabola at that x, and yL is bered fragments from too long the height of the line. Now at any point where the parabola ago. That prompts us to intro- and the line intersect, the graphs will have the same height. duce the labels yP and yL. That So we want the values of x for which helps them to recall that an equation is actually a set of in- y = y P L structions––in this case instruc- and that gives us the x-equation tions to find the height of the parabola or the line at any x. x(6–x) = 2x And what we want are the values and what we want are the solutions. of x at which these two heights We expand the LHS to get 6x–x2 = 2x How do we solve the equation? –x2 + 4x = 0 x(6–x) = 2x ? Factor: x(–x + 4) = 0 The class instructs us to cancel the x’s: x(4–x) = 0 6–x = 2 and we get the solutions x=0 and x=4. 4–x = 0 and we get the solution x=4. Note that these make sense from the picture, as we expect That’s fine but isn’t there another intersections at the origin and at some point between x=3 and intersection––x=0? What hap- x=6. pened to it? We “lost” it when we cancelled the x’s. parabola and parallel lines 9/3/2007 1 Moving the line ––geometric analysis. Now move the line parallel to itself. What that gives us is the family of all lines of slope 2. We ask the question: How does the number of intersections between the line and 15 the parabola change as the line moves? We begin with a conjecture based on the information we get 10 from the picture: 5 Geometric conjecture: Exactly one of the lines is tangent to the parabola and has one intersection with it. Those lines that are below this tangent intersect the parabola twice, and 0 those that are above the tangent don't intersect it at all. -202468 Moving the line ––algebraic analysis. Now we describe the family of lines algebraically, and verify the correctness of the geometric conjecture. Since the lines all have slope 2, they are described by the Parameters general equation Again we have a parameter, and this time it serves to keep yL = 2x + b. track of the different lines. It’s The parameter b we have used here to keep track of the dif- always nice when the parame- ferent family members is the y-intercept of the line. ter has a physical or geometric interpretation, and here it is Let’s hit the algebra. For each line (for each b), we want to simply the y-intercept of the find the points of intersection with the parabola yP = x(6–x). line. Our interest is in discover- As before we equate the y-expressions ing how the results of our analysis (the number of inter- yP = yL sections) change as the value of x(6–x) = 2x+b. the parameter changes. If we expand the LHS, we have a quadratic equation in x: 6x–x2 = 2x+b. Collect like terms and put everything on the left: –x2 + 4x – b = 0. Actually, we generally like the coefficient of x2 to be posi- tive: x2 – 4x + b = 0. We want the solutions of this equation––the values of x for which it holds. Let’s just first notice that the b is still there and that’s important because we expect that the solutions will depend on b. Indeed what we want to study is the question of exactly how the solutions depend on b. parabola and parallel lines 9/3/2007 2 Because of the b we cannot factor this expression anymore, so we find the solutions with the quadratic formula: There’s that b still hanging in. That’s good––different 4 ± 16 − 4b b’s give different lines, and x = . 2 different intersection points. The precise question we are interested in is this: for each particular b, how many solutions do we have? The key lies in the discriminant, the expression under the root sign. If this is positive, the ± sign will give us two solu- tions, if it’s zero, we’ll just get one, and if it’s negative, we won’t get any. Thus: if 16–4b > 0 we have two solutions if 16–4b = 0 we have one solution if 16–4b < 0 we have no solutions. 15 Solving for b: b < 4 two solutions b = 4 one solution 10 b > 4 no solutions. And this fits exactly with our geometric conjecture. It tells 5 us that low lines have two intersections and high lines have none, and the transitional case turns out to have y-intercept 4. 0 That seems to match the diagram. -202468 -5 A tangent of slope 2 Use your results to find the value of x at which the tangent to the parabola has slope 2. The value of x at which the parabola has slope 2 will be found at the point of contact of the tangent line. Now the 15 quadratic formula has already given us the x-coordinates of the intersection points of the line (for any b) and the curve. So we just need to use that for the case b=4, which is the 10 transitional case. We get: 4 ± 16 − 4b 4 ± 0 5 x = = = 2 . 2 2 When we set b=4 we get a zero discriminant. Of course–– 0 that’s what gave us the tangent in the first place. The value -202468 of x that we get is 2, and this checks out with the diagram. -5 How do we find the intersection point of the tangent? Well that's exactly what the x-formula at the top of the page does––it gives us the x-coordinates of all intersec- tion points. The tangent is the case b=4. parabola and parallel lines 9/3/2007 3 1. On the same set of axes, draw the graph of the parabola y = x2 and the line y = 4x. Observe that the line intersects the parabola exactly twice, once at x=0 and once at x=4. What you are to do is imagine the family of all lines of slope 4 and ask yourself the question––how many intersections does each line have with the parabola? (a) Study your picture and make a conjecture which answers the question. (b) Describe the family of lines algebraically, and verify the correctness of your conjecture. Illustrate your solution with a picture. (c) Use the result of (b) to find the value of x at which the tangent to the curve has slope 4. 2. Consider the parabola y = (x+1)2 + 1 and the family of lines of slope –2. (a) Introduce a parameter to describe this family of lines, and find, in terms of the parameter, the number of intersections between the line and the parabola. Illustrate your solution with a picture. [Hint: the form of the equation of the parabola allows you to find its vertex at a glance.] (b) Use the result of (a) to find the value of x at which the tangent to the curve has slope –2. 3. At the right is the graph of the curve 6 y = 1/x y = 1/x and the line y=-x /4 4 y = –x/4. 2 Observe that the line fails to intersect the curve. What you are to do is imagine the family of all lines of slope –1/4 and ask 0 yourself the question––how many intersections does each line -6 -4 -2 0 2 4 6 have with the curve? -2 (a) Study the picture and make a conjecture which answers -4 the question. -6 (b) Describe the family of lines algebraically, and verify the correctness of your conjecture. Illustrate your solution with a picture. (c) Use the result of (c) to find two values of x at which the tangent to the curve has slope –1/4. parabola and parallel lines 9/3/2007 4 4.(a) Find in terms of b the number of intersections of the curve x y = 1 + x with the line y = x+b. (b) Use the above to find all values of x at which the curve has slope 1. (c) Perhaps you've done this problem "blind" so far. If so, now is the time to try to illustrate your results on a sketch of the graph. Now you may find the prospect of drawing the graph y = x/(1+x) a bit daunting. But #3 above contains a picture of y=1/x and you might note that this graph is closely related––all you need are translations and sign changes. Indeed: x x +1−1 1 = = 1− 1+ x 1+ x 1+ x Can you take it from here? 5. Find in terms of b the number of intersections of the circle x 2 + y 2 = 2 . with the line y = b–x. Draw a picture to illustrate your solution and show that it is geometrically expected. 6. Sketch a number of members of the family of parabolas y = k − (x −1)2 , indexed by k. In terms of k, how many intersections are there with the parabola y = x 2 ? In particular, for what k are the curves tan- gent? Use your diagram to illustrate your solution. 7. How do different members of the family y = x 2 + k of parabolas intersect the circle x 2 + y 2 = 1? Before you do the algebraic analysis, The last 4 problems are all use your intuition to try to guess the form of the answer. Illustrate your challenging, and any one solution with a picture. of them can make a great project. 8. In terms of the parameter a, when does the pair of equations x2 – y2 = 0 #8 is an interesting prob- (x–a)2 +y2 = 1 lem because it appears in A Source Book for College have 0, 1, 2, 3, 4, or 5 solutions? Mathematics Teaching (Mathematical Association [It’s important to get the right picture. The first equation is simpler of America 1990, Alan than it looks––write it as x2 = y2 and take square roots. And the sec- Schoenfield, Ed.) as an ond equation is a moving circle.] example of a problem with which most students, armed 9. Find the radius of the largest circle that lies above the x-axis and with the traditional calcu- 2 underneath the parabola y = 4 − x . [Answer: r = 3/2] lus skills, will not really manage to make much 10. Find the value of k for which the parabolas y = x 2 + k and headway. But that's not us. x = y 2 are tangent. [Answer: k = 3/(44/3) ] parabola and parallel lines 9/3/2007 5
The circuit of an operational amplifier (non- inverting) the input impedance of which is very high and out impedance of which is low, and the voltage gain of which in case of a closed loop is unity (i.e. one), is known as voltage follower. In simple words, a closed loop non- inverting operational amplifier, the voltage gains of which is unity without phase reversal, is called voltage follower. As output voltages (Vout) of such an amplifier follow input voltages (Vin), that’s the reason it is called a voltage follower. This circuit reveals a very simple operation of an operational amplifier and it resembles an emitter follower (common collector amplifier of a transistor) circuit to a great extent. The only exception being that it works in a better manner. Remember that voltage follower circuit is not normally used for amplification due to its being unity gain. Rather, just like an emitter follower such a circuit is mostly used for impedance matching and isolation. That’s why voltage follower is often known as buffer or isolation amplifier as well. In figure 8.56 (a) a voltage follower has been depicted. Input voltages (Vin) are provided on amplifiers’ positive input or non- inverting terminal whereas all output voltages (Vout) have been fed back directly on negative input terminal (or inverting input) (that’s output voltage feedback directly with the negative input). As output has been feedback on negative input, therefore it is called negative feedback. As feedback resistance is zero therefore, this negative feedback is usually maximum, due to which this type of circuit is very close to an ideal circuit. This negative feedback sounds quite weird in the beginning, because output helps in ascertaining the value of differential input (which tends to amplify in order to produce this output). As a virtual short exists between amplifier’s inputs, therefore in such a situation, output voltages are equal to input voltages i.e. Vout = Vin The situation of input voltage and output voltage being mutually equivalent means that value of closed loop voltage gain is unity or one i.e. The mathematical method of ascertaining closed loop voltage gain is that if the value of R2 is considered zero, and value of R1 as being infinity, voltage follower gain can be determined by using the voltage gain equation of non- inverting amplifier. ACL = R2/R1 +1 = 0/ + = 0+1= 1 Thus, voltage follower is a perfect follower circuit which generates output voltages which are exactly equal to input voltages. Besides, maximum negative feedback produces such closed loop input impedance, the value of which is quite high as compared to opened loop input impedance. Thus, closed loop impedance also ensues via this maximum negative feedback, the value of which is quite low as compared to opened loop output impedance. Thus, we almost get a complete technique for the purpose of converting a high impedance source to a low impedance source. Figure 8.56 – (a). Voltage follower has unity gain and maximum bandwidth (b) voltage follower allows high-impedance to drive low-impedance load with no loss of voltage. In figure (b) the concept has been elucidated wherein an input AC source consists of high input impedance R high while load has low impedance RLOW. As negative feedback of a voltage follower is maximum (that’s 100% of output feedback on inverting input), therefore closed loop input impedance (Zin (CL) is high to an unbelievable extent whereas closed loop output impedance Zout (CL) is incredibly low. As a consequence, all input source voltages appear parallel to load resistor. Thus, input and output voltages of a voltage follower are considered identical for practical purposes. In figure 8.57, a practical application of a voltage follower has been shown for enlightenment. In figure (A) a signal source (output resistance of which is 50KΩ), has been connected with an amplifier (the input resistance of which is 15KΩ). If the value of an applied signal is 1V, only 230mV will appear on amplifier’s input while remaining 770mV drop parallel to the output resistance of signal source. Figure 8.57 page 411 – Practical applications of a voltage follower However, if the same signal source is connected with some specific amplifier voltage follower (input resistance of which is 100MΩ), as has been exemplified in the figure (B), the whole input signal will be received on amplifier’s input. As output resistance of a voltage follower is very low, therefore any load fixed on its output, will receive full-fledged signal (that’s high input impedance of a voltage follower circuit enables it to receive even the weakest of signal from output without any signal loss. Further, output impedance of a voltage follower is so low, that this low impedance provides a full signal on loads) Thus, a voltage follower, basically converts a high impedance voltage source to a low impedance voltage source and due to its this very characteristics, it is used as an ideal buffer amplifier for the purpose of interfacing (that’s for keeping both isolated) between high impedance sources and low impedance loads. Previous Topic: Summing Amplifier with Equation Example For electronics and programming-related projects visit my YouTube channel.
A survey on fuzzy fractional differential and optimal control nonlocal evolution equations We survey some representative results on fuzzy fractional differential equations, controllability, approximate controllability, optimal control, and optimal feedback control for several different kinds of fractional evolution equations. Optimality and relaxation of multiple control problems, described by nonlinear fractional differential equations with nonlocal control conditions in Banach spaces, are considered. keywords:fuzzy differential equations, Caputo and Riemann–Liouville fractional derivatives, numerical solutions, fractional evolution equations, controllability, approximate controllability, fractional optimal control, optimal feedback control, nonlocal control conditions. Msc: 26A33, 26E50, 34A07, 34A08, 49J15, 49J45, 93B05, 93C25. R. P. Agarwal, D. Baleanu, J. J. Nieto, D. F. M. Torres and Y. Zhou [This is a preprint of a paper whose final and definite form is with Journal of Computational and Applied Mathematics, ISSN: 0377-0427. Submitted 17-July-2017; Revised 18-Sept-2017; Accepted for publication 20-Sept-2017.] Memory and hereditary properties of different materials and processes in electrical circuits, biology, biomechanics, electrochemistry, control, porous media and electromagnetic processes, are widely recognized to be well predicted by using fractional differential operators Baleanu et al. (2012a); Diethelm (2010); Miller and Ross (1993); Podlubny (1999); Tarasov (2010). During the last decades, the subject of fractional calculus, and its potential applications, have gained an increase of importance, mainly because it has become a powerful tool with accurate and successful results in modeling several complex phenomena in numerous seemingly diverse and widespread fields of science and engineering Agarwal et al. (2010); El-Sayed (1996); Nieto (2010); Shen et al. (2014). Fractional calculus is not only a productive and emerging field, it also represents a new philosophy how to construct and apply a certain type of nonlocal operators to real world problems. The ones possessing both nonlocal effects as well as uncertainty behaviors represent an interesting phenomena. Researchers started to combine, in an intelligent way, the notions of fractional with fuzzy, therefore a hybrid operator called fuzzy fractional operator emerges. In this manuscript, we begin by presenting a review on fractional differential equations under uncertainty. In one of the earliest works, Agarwal et al. Agarwal et al. (2010) took the initiative and introduced fuzzy fractional calculus to handle fractional-order systems with uncertain initial values or uncertain relationships between parameters. Arshad and Lupulescu Arshad and Lupulescu (2011) utilized the results reported in Agarwal et al. (2010) and they proved the existence and uniqueness of fractional differential equations with uncertainty. Afterward, Allahviranloo et al. Allahviranloo et al. (2012) employed the Riemann–Liouville generalized H-differentiability in order to solve the fuzzy fractional differential equations (FFDEs) and presented some new results under this notion. Salahshour et al. Salahshour et al. (2012) apply the technique of fuzzy Laplace transforms and solved some types of FFDEs based on the Riemann–Liouville fuzzy derivative. Based on the delta-Hukuhara derivative for fuzzy valued functions, Fard et al. established stability criteria for hybrid fuzzy systems on time scales in the Lyapunov sense Fard et al. (2016). In Fard et al. (in press), Fard et al. solve a class of fuzzy fractional optimal control problems, where the coefficients of the system can be time-dependent. M ore precisely, they establish a weak version of the Pontryagin maximum principle for fuzzy fractional optimal control problems depending on generalized Hukuhara fractional Caputo derivatives Fard et al. (in press). Generally, the majority of the FFDEs as same as FDEs do not have exact solutions. As a result, approximate and numerical procedures are important to be developed Debbouche and Torres (2013). On the other hand, because many of the parameters in mathematical models often do not appear explicitly, modeling of natural phenomena using fuzzy fractional models plays an important role in various disciplines. Hence, it motivates the researchers to investigate effective numerical methods with error analysis to approximate the FFDEs. As a result, researchers started to develop numerical techniques for FFDEs. Mazandarani and Vahidian Kamyad Mazandarani and Kamyad (2013) introduced a fuzzy approximate solution using the Euler method to solve FFDEs. Ahmadian et al. Ahmadian et al. (2013a) adopted the operational Jacobi operational matrix based on the fuzzy Caputo fractional derivative using shifted Jacobi polynomials. The clear advantage of the usage of this method is that the matrix operators have the main role to find the approximate fuzzy solution of FFDEs instead of considering the methods required the complicated fractional derivatives and their calculations. Ghaemi et al. Ghaemi et al. (2013) adapted a spectral method for the numerical solution of fuzzy fractional kinetic equations. The proposed method is characterized by its simplicity, efficiency, and high accuracy. Using the proposed method, they could reach a suitable approximation of the amount of the concentration value of xylose after a determined time that is important to analyze the kinetic data in the chemical process. Ahmadian et al. Ahmadian et al. (2013b) exploited a cluster of orthogonal functions, named shifted Legendre functions, to solve FFDEs under Caputo type. The benefit of the shifted Legendre operational matrices method, over other existing orthogonal polynomials, is its simplicity of execution as well as some other advantages. The achieved solutions present satisfactory results, obtained with only a small number of Legendre polynomials. Fuzzy theory provides a suitable way to objectively account for parameter uncertainty in models. Fuzzy logic approaches appear promising in preclinical applications and might be useful in drug discovery and design. In this regards, Ahmadian et al. Ahmadian et al. (2013c) developed a tau method based on the Jacobi operational matrix to numerically solve the pharmacokinetics-pharmacodynamic equation, arising from drug assimilation into the bloodstream. The comparison of the results shows that the present method is a powerful mathematical tool for finding the numerical solutions of a generalized linear fuzzy fractional pharmacokinetics-pharmacodynamic equation. Balooch Shahriyar et al. Balooch Shahriyar et al. (2013) investigated an analytical method (eigenvalue-eigenvector) for solving a system of FFDEs under fuzzy Caputo’s derivative. To this end, they exploited generalized H-differentiability and derived the solutions based on this concept. Ahmadian et al. Ahmadian et al. (2014a) were confined with the application of Legendre operational matrix for solving FFDEs arising in the drug delivery model into the bloodstream. The main motivation of this research is to recommend a suitable way to approximate fuzzy fractional pharmacokinetics-pharmacodynamic models using a shifted Legendre tau approach. This strategy demands a formula for fuzzy fractional-order Caputo derivatives of shifted Legendre polynomials. Mazandarani and Najariyan Mazandarani and Najariyan (2014) introduced two definitions of differentiability of type-2 fuzzy number valued functions of fractional order. The definitions are in the sense of Riemann–Liouville and Caputo. The methods, under type-2 fuzzy sets theory, will lead to an increase in the computational cost, although it is closer to the originality of the model. Salahshour et al. Salahshour et al. (2015a) developed the notion of Caputo’s H-differentiability, based on the generalized Hukuhara difference, to solve the FFDE. To this end, they revisited Caputo’s derivatives, and proposed novel fuzzy Laplace transforms and their inverses, with an analytical method to tackle the deficiencies in the state-of-the-art methods. Experimental results using some real-world problems (nuclear decay equation and Basset problem, illustrated the effectiveness and applicability of the proposed method). Simultaneously, the authors in Ahmadian et al. (2014b) investigated an effective numerical method with error analysis to approximate the fuzzy time-fractional Bloch equations (FTFBE) (3.4) on the time interval , with a view to be employed in the image processing domain in near time. Employing Laplace transforms, the authors in Salahshour et al. (2016) proposed a novel efficient technique for the solution of FFDEs that can efficiently make the original problem easier to achieve the numerical solution. The suggested algorithm for the FFDEs use the fuzzy fractional derivative of Caputo type in the range of and is potentially useful in solving fractional viscoelastic problems under uncertainty. Chehlabi and Allahviranloo Chehlabi and Allahviranloo (2016) studied fuzzy linear fractional differential equations of order under Riemann–Liouville H-differentiability. Also, it is corrected some previous results and obtained new solutions by using fractional hyperbolic functions and their properties. There has been a significant development in nonlocal problems for fractional differential equations or inclusions: see, for instance, Benchohra et al. (2003); Debbouche and Baleanu (2011); Debbouche et al. (2012); Debbouche and Torres (2013); N’Guérékata (2009); Wang et al. (2011a); Zhou and Jiao (2010a). Indeed, nonlinear fractional differential equations have, in recent years, been object of an increasing interest because of their wide applicability in nonlinear oscillations of earthquakes, many physical phenomena such as seepage flow in porous media, and in fluid dynamic traffic model Kilbas et al. (2006); Lakshmikantham (2008); Lakshmikantham and Vatsala (2008). On the other hand, there could be no manufacturing, no vehicles, no computers, and no regulated environment, without control systems. Control systems are most often based on the principle of feedback, whereby the signal to be controlled is compared to a desired reference signal and the discrepancy used to compute corrective control actions Franklin et al. (2014). The idea of controllability is an essential characteristic to a control framework exhibiting numerous control issues such as adjustment of unsteady frameworks by input control. Recently, control issues have been addressed by many physicists, engineers and mathematicians, and significant contribution on theoretical and application aspects of the topic can be found in the related literature Debbouche and Torres (2015). As is well-known, the problems of exact and approximate controllability are to be distinguished Debbouche and Torres (2014). In general, in infinite dimensional spaces, the concept of exact controllability is usually too strong. Therefore, the class of evolution equations consisting of fractional diffusion equations must be treated by the weaker concept of controllability, namely approximate controllability Yang and Machado (2017). Recently, many works pay attention to study approximate controllability of different types of fractional evolution systems Debbouche and Torres (2013, 2014); Kerboua et al. (2014). Over the last years, one of the fields of science that has been well established is the fractional calculus of variations: see Malinowska and Torres (2012); Almeida et al. (2015); Malinowska et al. (2015) and references therein. Moreover, a generalization of this area, namely the fractional optimal control, is a topic of research by many authors Agrawal (2004); Frederico and Torres (2008). The fractional optimal control of a distributed system is an optimal control problem for which the system dynamics is defined with partial fractional differential equations Özdemir et al. (2009); Yang et al. (2017). The calculus of variations, with constraints being sets of solutions of control systems, allow us to justify, while performing numerical calculations, the passage from a nonconvex optimal control problem to the convexified optimal control problem. We then approximate the latter problem by a sequence of smooth and convex optimal control problems, for which the optimality conditions are known and methods of their numerical resolution are well developed. The delay evolution systems is an important class of distributed parameter systems, and optimal control of infinite dimensional systems is a remarkable subject in control theory Debbouche and Nieto (2014); Sidi Ammi and Torres (2007, 2012). In the last years, fractional evolution systems in infinite dimensional spaces attracted many authors. When the fractional differential equations describe the performance index and system dynamics, an optimal control problem reduces to a fractional optimal control problem Almeida and Torres (2015); Pooseh et al. (2014). The fractional optimal control of a distributed system is a fractional optimal control for which system dynamics are defined with partial fractional differential equations. There has been very little work in the area of fractional optimal control problem in infinite dimensional spaces, especially optimal controls of fractional finite time delay evolution system. See Sections 6 and 7. Sobolev type semilinear equations serve as an abstract formulation of partial differential equations, which arise in various applications such as in the flow of fluid through fissured rocks, thermodynamics, and shear in second order fluids. Further, the fractional differential equations of Sobolev type appear in the theory of control of dynamical systems, when the controlled system and/or the controller is described by a fractional differential equation of Sobolev type. Furthermore, the mathematical modeling and simulations of systems and processes are based on the description of their properties in terms of fractional differential equations of Sobolev type. These new models are more adequate than previously used integer order models, so fractional order differential equations of Sobolev type have been investigated by many researchers: see, for example, Fec̆kan, Wang and Zhou Fečkan et al. (2013) and Li, Liang and Xu Li et al. (2012). In Debbouche and Nieto (2014); Debbouche and Torres (2014), the notion of nonlocal control condition is introduced and a new kind of Sobolev type condition presented. Kamocki Kamocki (2014) studied the existence of optimal solutions to fractional optimal control problems. Liu et al. Liu et al. (2014) established the relaxation for nonconvex optimal control problems described by fractional differential equations. In Section 8, a kind of Sobolev type condition and a nonlocal control condition for nonlinear fractional multiple control systems is considered. The Sobolev condition is given in terms of two linear operators and requires formulating two other characteristic solution operators and their properties, such as boundedness and compactness. Further, we consider an optimal control problem of multi-integral functionals, with integrands that are not convex in the controls. An interrelation between the solutions of the original problem and the relaxation one is given. Under certain assumptions, it is shown that the relaxed problem has a solution with interesting convergence properties Ekeland and Temam (1976); Mainardi (1996); Debbouche et al. (2017). 2 Basic definitions and notations Here we review some essential facts from fractional calculus Podlubny (1999); Kilbas et al. (2006), basic definitions of a fuzzy number and fuzzy concepts Kaleva (1987); Bede and Gal (2005); Bede et al. (2007), semigroup theory Pazy (1983); Zaidman (1979), and multi-valued analysis Aubin and Cellina (1984); Hu and Papageorgiou (1997). The fractional integral of order of a function is given by where is the classical gamma function. If , then we can write , where and, as usual, denotes convolution. Moreover, with the delta Dirac function. The Riemann–Liouville fractional derivative of order , , , is given by where function has absolutely continuous derivatives up to order . The Caputo fractional derivative of order , , , is given by where function has absolutely continuous derivatives up to order . Let , . The following properties hold: If , then The Caputo derivative of a constant function is equal to zero; The Riemann–Liouville derivative of a constant function is given by We denote the set of all real numbers by and the set of all fuzzy numbers on is indicated by . A fuzzy number is a mapping with the following (i) is upper semi-continuous, (ii) is fuzzy convex, i.e., for all (iii) is normal, i.e., for which , (iv) is the support of the , and its closure cl(supp ) is compact. Definition 2.4 (See Kaleva (1987)). We define a metric on () by a distance, namely the Hausdorff distance as follows: It is shown that is a complete metric space. The concept of Hukuhara-difference, which is recalled in the next definition, was initially generalized by Markov Markov (1979) to introduce the notion of generalized Hukuhara-differentiability for the interval-valued functions. Afterwards, Kaleva Kaleva (1987) employed this notion to define the fuzzy Hukuhara-differentiability for the fuzzy-valued functions. Definition 2.5 (See Kaleva (1987)). Let . If there exists such that , then is called the Hukuhara-difference of and , and it is denoted by . We denote the Caputo fractional derivatives by the capital letter with upper-left index , and the Caputo fractional derivatives of order is defined as where ; ; and is the classical differential operator of order . Let and . We denote as the space of all continuous fuzzy functions defined on . Also let . Then we say that if and only if Anastassiou (2011a). In the rest of the paper, the above notations will be used frequently. The fuzzy Caputo derivatives of order for a fuzzy-valued function are given as follows. Definition 2.7 (See Ahmadian et al. (2017a)). Let be a fuzzy set-value function. Then is said to be Caputo’s fuzzy differentiable at when We now proceed with some basic definitions and results from multivalued analysis. For more details on multivalued analysis we refer to the books Aubin and Cellina (1984); Hu and Papageorgiou (1997). We use the following symbols: is the set of all nonempty closed subsets of ; is the set of all nonempty, closed and bounded subsets of . On , we have a metric, known as the Hausdorff metric, defined by where is the distance from a point to a set . We say that a multivalued map is -continuous if it is continuous in the Hausdorff metric . Let be a multifunction. For , we define a.e. on . We say that a multivalued map is measurable if for every closed set . If , then the measurability of means that , where is the -algebra of subsets in generated by the sets , , , and is the -algebra of the Borel sets in . Suppose that and are two Hausdorff topological spaces and . We say that is lower semicontinuous in the sense of Vietoris (l.s.c., for short) at a point , if for any open set , , there is a neighborhood of such that for all . Similarly, is said to be upper semicontinuous in the sense of Vietoris (u.s.c., for short) at a point , if for any open set , , there is a neighborhood of such that for all . For more properties of l.s.c and u.s.c, we refer to the book Hu and Papageorgiou (1997). Besides the standard norm on (here, is a separable reflexive Banach space), , we also consider the so called weak norm: The space furnished with this norm will be denoted by . 3 Fuzzy fractional differential equations with initial condition and the matrix Also, let and be fuzzy vectors. Note that the coefficients of are expressed as follows: where and are fractional time constants. The predictor-corrector method is investigated in Ahmadian et al. (2014b). Particularly, the fractional Adams–Bashforth as a predictor and the fractional Adams–Moulton as a corrector are exploited. Moreover, a new variant of the fuzzy fractional Adams–Bashforth–Moulton (FFABM) method is introduced Ahmadian et al. (2014b). Finally, Ahmadian et al. (2014b) demonstrates the capability of the developed numerical methods for fuzzy fractional-order problemscin terms of accuracy and stability analysis. Ahmadian et al. Ahmadian et al. (2015a) have dealt with the application of FFDEs to model and analyze a kinetic model of diluted acid hydrolysis under uncertainty as follows. When water is added to the Hemicellulose xylane, Xylose is formed through the hydrolysis reaction. Furfural is the main degradation product obtained through the degradation of a molecule of Xylose by the releasing of three water molecules. Scheme 3.5 demonstrates the depolymerization of Xylan. The simplest kinetic model for the hemicellulose hydrolysis was firstly proposed by Saeman Saeman (1945). He discovered that a straightforward two-step reaction model sufficiently explained the generation of sugars during wood hydrolysis. Saeman’s model assumed pseudo homogeneous irreversible first-order consecutive reactions: In real problems, we firstly choose the initial conditions as starting points. Indeed, initial conditions for such models are determined by analyzer systems, which are not adequate for high accuracy results. So, instead of using deterministic values, it is better to employ uncertain conditions. In order to consider the original problem in a new sense, the authors used the fuzzy initial value instead of the crisp initial value. In this direction, they reconstructed the original problem based on fuzzy fractional calculus: where is a continuous fuzzy-valued function and indicates the fuzzy Caputo fractional derivative of order . Also, the Xylose concentration is defined at time as a fuzzy number, i.e., , . Additionally, the notations mean addition and multiplication, respectively in . This approach can also be a feasible alternative to ordinary differential equation models under uncertainty. As a case study, they have formulated a mathematical model to analyze the Xylose concentration in the acid hydrolysis as a promising source of Xylose. The proposed model has been described by a FFDE of order with the aim of incorporating uncertainty into the initial values and function of the model. This model provides a more realistic view taking into account the variations of Xylose concentration. Moreover, it constitutes a new approach for modeling a chemical reaction. This approach is used to obtain reliable data of acid concentration immediately during the process (see, Figs. 1, 2 and 3). Therefore, Ahmadian et al. Ahmadian et al. (2015b) proposed a new iterative method for solving FFDEs of Caputo type. The basic idea is to convert FFDEs to a type of fuzzy Volterra integral equation. Then the obtained Volterra integral equation is exploited with some suitable quadrature rule to get a fractional predictor-corrector method. Salahshour et al. Salahshour et al. (2015b) investigated the solution of a class of fuzzy sequential fractional differential equation of order . For this purpose, some basic results were developed in terms of the fuzzy concept. The main scope of this study is to present the solution of fuzzy problems based on the contraction principle in the new fuzzy complete metric space. Afterwards, Ahmadian et al.Ahmadian et al. (2015c) conducted a research that has been devoted to solve linear FFDEs of Caputo sense. The basic idea is to develop a fractional linear multistep method for solving linear FFDEs under fuzzy fractional generalized differentiability. The authors proposed fractional linear multistep methods and investigated the consistence, convergence and stability properties of the method. A new fractional derivative, with some simplifications in the formula and computations, was proposed under interval uncertainty in Salahshour et al. (2015c). The authors have developed the highlights of the conformable fractional derivative, which are more influential for the solution of fractional interval differential equations (FIDEs) under generalized Hukuhara differentiability. The reader interested on conformable fractional differentiation is refereed to Benkhettou et al. (2016); Lazo and Torres (2017) and references therein. In practice, a simple fractional derivative satisfying the main rules for uncertain fractional differential equations, such as the product rule and the chain rule, simplifies considerably the cumbersome mathematical expressions of the mathematical modeling in engineering sciences. At the same time, several authors developed theoretical aspects of this concept for the solution of FDEs under uncertainty in their researches Lupulescu (2015); Hoa (2015a); Malinowski (2015); Hoa (2015b), which were a continuation of the concepts proposed in the monographs and papers of fuzzy setting theory Kaleva (1987); Buckley and Feuring (2000); Bede and Gal (2005); Bede et al. (2007); Khastan et al. (2011); Anastassiou (2010); Ahmadian et al. (2016) and fractional calculus Podlubny (1999); Agrawal et al. (2004); Das (2008); Miller and Ross (1993); Baleanu et al. (2012b); Mainardi (2012); Anastassiou (2011b). Very recently, Salahshour et al. Salahshour et al. (2017) made a meaningful contribution to study a theory of FDEs under interval uncertainty. The new interval fractional derivative has several impressive properties that each of them lonely can improve significantly the modeling of real life systems using FDEs such as non-singularity kernel. It was devoted to research on this new and applicable notion in terms of interval uncertainty. The authors proposed adequate conditions for the uniqueness of solution of fractional interval differential equations (FIDEs) under Caputo–Fabrizio fractional derivative. Some interesting properties of this derivative were studied to make the easiest way for dealing with mathematical models based on Caputo–Fabrizio fractional interval equations. Ahmadian et al. Ahmadian et al. (2017a) presented numerical simulations and introduced fuzzy mathematical models that can be represented in terms of FDEs under certainty. Consider the following linear FFDE: in which is a continuous fuzzy-valued function, indicates the fuzzy Caputo’s fractional derivative of order and . Compared with the extensive amount of work put into developing FDE schemes in the literature, we found out that only a little effort has been put into developing numerical methods for FFDE. Even so, most of the solutions are based on a rigorous framework, that is, they are often tailored to deal with specific applications and are generally intended for small-scale fuzzy fractional systems. In this research, the authors deployed a spectral tau method based on Chebyshev functions to reduce the FFDE to a fuzzy algebraic linear equation system to address fuzzy fractional systems. The main advantage of this technique, using shifted Chebyshev polynomials in the interval , is that only a small number of the shifted Chebyshev polynomials is required as well as the good accuracy that will be acquired in one time program running. Thus, it greatly simplifies the problem and reduces the computational costs. The solution is expressed as a truncated Chebyshev series and so it can be easily evaluated for arbitrary values of time using any computer program without any computational effort. The algorithm of the technique is as follows: we generate fuzzy linear equations by applying where , is a shifted Chebyshev polynomial and is a fuzzy-like residual operator for (3.7), which is defined in the matrix operator form Using the definition of fuzzy-like inner product, we have: and . Then, in order to acquire the approximation using the shifted Chebyshev tau approximation, we should find the unknown vector . Therefore, (3.10) can be stated as follows: Then, using the the matrix form and their defined elements, (3.11) can be written in the following matrix form: Finally, system (3.12) can be solved based on the following lower-upper representation by any direct or numerical method: The analysis of the behaviors of physical phenomena is important in order to discover significant features of the character and structure of the mathematical models. In a very recent time, Ahmadian et al. Ahmadian et al. (2017b) defined a new fuzzy approximate solution and fuzzy approximate functions formed on the generalized fractional Legendre polynomials (GFLPs) introduced in Chen et al. (2014), and then fuzzy Caputo fractional-order derivatives of GFLPs in terms of GFLPs themselves are stated and proved. They derived an effective spectral tau method under uncertainty by applying these functions to solve two important fractional dynamical models via the fuzzy Caputo-type fractional derivative. They proposed a new model based on fractional calculus to deal with the Kelvin-Voigt (KV) equation and non-Newtonian fluid behavior model with fuzzy parameters (for Caputo fractional Voigt models see Chidouh et al. (2017)). Numerical simulations are carried out and the analysis of the results highlights the significant features of the new technique in comparison with the previous findings. The homogeneous strain relaxation equation with memory can be defined as a differential equation of fractional order under uncertainty as follows: where is a continuous fuzzy-valued function, represents the Caputo-type fractional derivative under fuzzy notion with order and . Since the (3.13) is a general model of viscosity behavior for non-Newtonian fluid under uncertainty, it is more convenient to define and as a fuzzy number and fuzzy set-valued function, respectively. Therefore, can be a fuzzy number that alter the conditions of the model. In addition, they developed the fuzzy fractional KV model under the Caputo gH-differentiability that offers fuzzy models for mathematical systems of natural phenomena as follows: in which presents a continuous fuzzy function, specifies the Caputo-type derivative for and . Also, and can be defined as a fuzzy parameter and a fuzzy set-valued function, respectively, based on the conditions of the model. To test the proposed technique, practically, they solved the following two problems: where is the fuzzy stress-strain function. In addition, the generalized viscous coefficient and the modulus of elasticity are assumed to be one (. Also, at first, they considered that . The next model is as follows: in which it is assumed that , and . Using a fixed point strategy is one of the fruitful methods to establish the controllability of nonlinear dynamical control system. In the last few years, some interesting and important controllability results, concerning semilinear differential systems involving Caputo fractional derivatives, were proved. Consider the following Sobolev-type fractional evolution system: where is the Caputo fractional derivative of order , . Here is a separable Banach space with the norm ; are two closed linear operators and the pair generates an exponentially bounded propagation family of to . The state takes values in and the control function is given in , the Banach space of admissible control functions, where and is a Banach space with the norm . Operator is bounded and linear from into , and will be specified later. 4.1 Characteristic solution operators We recall the concept of exponentially bounded propagation family. Definition 4.1 (See Liang and Xiao (2001)). A strongly continuous operator family of to a Banach space such that is exponentially bounded, which means that there exist and such that for any and , is called an exponentially bounded propagation family for the abstract degenerate Cauchy problem when . In this case, we say that problem (4.18) has an exponentially bounded propagation family . Moreover, if (4.19) holds, then we also say that the pair generates an exponentially bounded propagation family . where is the Wright function with . We can introduce the following definition of mild solution for system (4.17). For each and , by a mild solution of system (4.17) we mean a function satisfying 4.2 Controllability results In this subsection, we study the controllability of system (4.17) by utilizing the theory of propagation family. We pose the following assumptions: the pair generates an exponentially bounded propagation family of to ; is a norm continuous family for and for ; the control function takes from , the Banach space of admissible control functions, either for with or for , where is Banach space; is a bounded linear operator and the linear operator defined by has a bounded right inverse operator . It is easy to see that and is well defined due to the following fact: for any , where Next, we assume the hypothesis satisfies the following two conditions: for each the function is strongly measurable and, for each , the function is continuous; for each , there is a measurable function such that and, for some , there exists sufficiently large such that there exists a positive constant such that for any bounded set and a.e., where is the measure of noncompactness. Based on our assumptions, it is suitable to define the following control formula for an arbitrary function : Theorem 4.1 (See Wang et al. (2014)). Assume – are satisfied. Furthermore, assume that where . Then system (4.17) is controllable on . 5 Approximate controllability Let be a Hilbert space with a scalar product and the corresponding norm . We consider the following Sobolev-type fractional evolution system: where is the Caputo fractional derivative of order , and are two linear operators with domains contained in and ranges still contained in , the pre-fixed points satisfy and are real numbers. In order to guarantee that generate a semigroup , we consider that the operators and satisfy the following conditions: and are linear, is closed; and is bijective; Now we note that implies that is closed; it follows from , , and the closed graph theorem that is bounded, which generates a uniformly continuous semigroup of bounded linear operators from to itself. Denote by the resolvent set of . If we assume that the resolvent is compact, then is a compact semigroup (see Pazy (1983)). The state takes values in and the control function is given in , the Banach space of admissible control functions, where for with and is a Hilbert space. Moreover, is a bounded linear operator and will be specified later. Define the following two operators: where is the Wright function. Now we introduce Green function: for , where Hence, we have that for and for . Now, we introduce a suitable definition of mild solution. For each , by a mild solution of system (5.25) we mean a function satisfying