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1 Estimating the Risk of Collisions between Bicycles and Motor Vehicles at Signalized Intersections Yinhai Wang (corresponding author) Department of Civil Engineering University of Washington Box , Seattle, WA Tel: (206) Fax: (206) Nancy L. Nihan Department of Civil Engineering University of Washington Box , Seattle, WA Abstract. Collisions between bicycles and motor vehicles have caused severe life and property losses in many countries. The majority of bicycle-motor vehicle (BMV) accidents occur at intersections. In order to reduce the number of BMV accidents at intersections, a substantial understanding of the causal factors for the collisions is required. In this study, intersection BMV accidents were classified into three types based on the movements of the involved motor vehicles and bicycles. The three BMV accident classifications were through-motor vehicle related collisions, left-turn motor vehicle related collisions, and right-turn motor vehicle related collisions. A methodology for estimating these BMV accident risks was developed based on probability theory. A significant difference between this proposed methodology and most current approaches is that the proposed approach explicitly relates the risk of each specific BMV accident type to its related flows. The methodology was demonstrated using a four-year ( ) data set collected from 115 signalized intersections in the Tokyo Metropolitan area. This data set contains BMV accident data, bicycle flow data, motor vehicle flow data, traffic control data, and geometric data for each intersection approach. For each BMV risk model, an independent explanatory variable set was chosen according to the characteristics of the accident type. Three negative binomial regression models (one corresponding to each BMV accident type) were estimated using the maximum lelihood method. The coefficient value and its significance level were estimated for each selected variable. The negative binomial dispersion parameters for all the three models were significant at 0.01 levels. This supported the choice of the negative binomial regression over the Poisson regression for the quantitative analyses in this study. Key words: bicycle accidents, traffic safety, signalized intersections, negative binomial regression
2 1 INTRODUCTION Collisions between bicycles and motor vehicles have caused severe life and property losses in many countries. Fazio and Tiwari (1995) reported that bicycle-motor vehicle (BMV) accidents killed 116 people, or more than 10 percent of all traffic accident fatalities in Delhi in In Japan, more than 1,000 people have died each year in BMV accidents since 1988 (Institute for Traffic Accident Research and Data Analysis, 2000). This has accounted for about 10 percent of all traffic fatalities each year. The BMV-accident-resulted fatality rate is even higher in Tokyo. Of the 359 traffic accident fatalities, 53 (14.8 percent) died in BMV accidents in Tokyo in 2000 (Tokyo Metropolitan Police Department, 2001). More seriously, in Beijing, about 38.7 percent of traffic accident fatalities died from BMV collisions and nearly 7 percent of all traffic accidents were related to bicycles (Liu et al, 1995). Intersections are definitely high-risk locations for BMV collisions because of the frequent conflicts between bicycle flows and motor vehicle flows. According to Traffic Safety Facts 2000 (US Department of Transportation, 2001), 32.6 percent of fatal accident and 56.6 percent of injury BMV collisions occurred at intersections in the US. Wachtel and Lewiston (1994) studied bicycle accidents in Palo Alto from 1981 to 1990, and found that 233 of the 314 reported BMV collisions (64 percent) took place at intersections. According to the Tokyo Metropolitan Police Department (2001), approximately 18 percent of all casualty accidents at intersections were BMV accidents. These figures indicate that special attention should be given to intersection BMV accidents. Gårder (1994) analyzed the causal factors for bicycle accidents with data collected from 1986 to 1991 in Maine. He found that about 57 percent of intersection BMV collisions involved turning movements of motor vehicles. He also concluded that bicycle riders were at fault for most of the reviewed BMV collisions. Summala et al (1996) carefully studied the motor-vehicle driver s searching behaviors at non-signalized intersections and found that speed-reducing measurements, such as speed bumps, elevated bicycle crossings and stop signs, help drivers to begin searching earlier and detect bicycles properly. Wachtel and Lewiston (1994) specifically analyzed the effects of age, sex, direction of travel, and road position on intersection BMV collisions. Gårder et al (1994) reviewed previous studies on bicycle accident risks and applied the Bayesian method to estimate the change in accident risk for bicycle riders when a bicycle path is introduced in a signalized intersection. They stated that conclusions from previous studies were fairly confusing, and few reviewed studies from the Scandinavian countries were conducted with acceptable methodologies. They attributed these conflicts to the absence of several important factors associated
3 2 with specific intersections and emphasized the importance of considering the detailed intersection design when studying bicycle accidents. To quantitatively consider the factors associated with specific intersection designs in risk models, new modeling techniques and more detailed data are needed. Though the conventional black spot identification method, which marks the location of each accident with a pin on a map and labels locations with the most pins as black spots, is an efficient way to identify high frequency accident sites, it does not provide any sufficient help in understanding accident causes. Without a proper understanding of accident causes, safety resources may be misused, and countermeasures may be ineffective. Hauer (1986) points out that a simple count of accidents is not a good estimate of safety and suggests estimating the expected value of accidents as a better alternative. Hauer et al (1988) demonstrated the effectiveness of this idea by classifying intersection vehicle-to-vehicle accidents into 15 patterns according to the movements of the involved vehicles before collision. They estimated the means for four major types of collision patterns using the flows involved in each collision type. Wang (1998) used a similar classification for accidents at signalized intersections and successfully estimated the risks of rear-end and angle accidents (corresponding to pattern 1 and 6, respectively, in the classification by Hauer el al (1988)) with a modified negative binomial regression. Summala et al (1996) classified bicycle accidents at non-signalized T intersections into 8 types and analyzed the visual search tasks involved in the major types of movements. Such detailed classifications clearly connect each type of accident to its related flows and environmental factors, and, therefore, make models and explanations more perceptive. In this study, BMV collisions at four-legged signalized intersections are classified into three types: through motor vehicle related collisions, left-turning motor vehicle related collisions, and right-turning motor vehicle related collisions. Data used for this study were collected from 115 randomly selected intersections in the Tokyo Metropolitan area. For each of the three BMV accident types, the expected accident risk is estimated by the maximum lelihood method using the negative binomial probability formulation. Since traffic travels along the left side of the roadway in Japan, special attention is needed when interpreting the descriptions for countries where traffic travels along the right side.
4 3 BICYCLE-MOTOR VEHICLE ACCIDENT CLASSIFICATION Typically, a BMV collision involves one motor vehicle and one bicycle. In Japan, bicycles share roads with pedestrians rather than motor vehicles. Thus, a BMV accident is most commonly happened when a bicycle is crossing an intersection approach via the bicycle channel, while a motor vehicle is making any of the three possible movements: through, right turn, or left turn. Intersection BMV accidents are, therefore, classified into three types based on the movements of the involved motor vehicles: (1) BMV-1: BMV accident type 1. Collisions between bicycles and through motor vehicles; (2) BMV-2: BMV accident type 2. Collisions between bicycles and left-turning motor vehicles; and (3) BMV-3: BMV accident type 3. Collisions between bicycles and right-turning motor vehicles. Fig. 1 illustrates these three accident types. Any BMV accident can be easily classified according to the movement of the involved motor vehicle. For BMV-1 accidents, collisions can occur before motor vehicles enter an intersection or before they exit the intersection. Since the collision styles are very similar, we consider these two collision situations together in BMV- 1. We believe that the causal factor set for each BMV accident type is different, and an obvious advantage of using such a classification is the capability of independently identifying the causal factors to each specific BMV accident type. DATA About 150 four-legged signalized intersections were randomly selected in the Tokyo Metropolitan area at the beginning of this study. The selection was based on intersection size, surrounding land use pattern, and intersection shape (crossing angles, vertical or skewed, of the approaches). Intersection accident histories were not considered. The purpose of the random selection was to obtain samples representing normal situations of intersection traffic safety in Tokyo. The BMV accident classification described in the previous section requires observation aggregation at intersection approach level rather than at intersection level (i.e. we need to know the accident number of each BMV
5 4 type for each approach of an intersection rather than just the total number for the entire intersection). However, BMV accident data in the existing accident databases were, without exception, aggregated at the intersection level and without further classification into the three BMV types. Obviously, such databases are not directly useful for our study. Consequently, we had to design a new accident database and conduct data collection work to satisfy our specific study requirements. The new database recorded approach-level observations, including numbers of BMV-1, BMV-2, and BMV-3 accidents, traffic volumes of through, left-turn, and right-turn flows, geometric data, etc, for each approach. Our accident data collection team followed a rigorous approach to guarantee the quality of data. They first obtained the index numbers of accidents occurred during the years 1992 through 1995 from the databases of the Tokyo Metropolitan Police Department. Then, they used these index numbers to find the original accident records for details of the accidents. The original record of an accident included a collision site figure and a brief description of the accident. With the collision site figure, a BMV accident can be easily located and assigned to one of the three accident types. Since some of the original records were missing, the locations and types for some BMV accidents could not all be identified. Intersections with unknown BMV accidents were dropped from the study and the number of sample intersections was reduced to 115. The total number of accidents of each intersection was then compared with the summary statistics of Ministry of Construction for verification. There were a total of 2,928 accidents recorded for the 115 intersections during the four-year study period. 585 of them, or about 20%, were BMV accidents. Motor vehicle flow data and bicycle volume data were obtained from reports of annual site surveys (Tokyo Metropolitan Police Department, ), conducted by the Tokyo Metropolitan Police Department and from highway sensor data (Tokyo Road Construction Bureau, 1997). Traffic control information and safety improvement histories were extracted from the corresponding databases and documents of government agencies. Geometric data for the intersection approaches were collected from published maps and site surveys. To reflect the effect of the information quantity to be processed by drivers and bicyclists while passing an intersection, an index of visual noise level (from low to high, values ranging from 0 to 4) was adopted for this study. For details on how the visual noise level for each intersection approach was estimated, please refer to Wang (1998). Our methodology requires the classification of intersection approaches based on the orientation of accidentinvolved motor vehicle flows. For each approach, there may be observations of BMV accidents directly caused by
6 5 motor vehicles that enter the intersection from the approach. The approach where the involved motor vehicle enters the intersection is designated as the entering approach. A number of BMV accidents of each type may be observed during the study period for each entering approach. The other three approaches are, clockwise, named the left approach, opposing approach, and right approach. Each intersection BMV accident associates with an entering approach. When the entering approach changes, the designations for the other three approaches change correspondingly. The illustration of approach naming is shown in Fig. 2. To help understand the data for this study, Table 1 provides summary statistics for selected continuous variables and Table 2 provides frequency results for selected dummy variables in the database. In Table 1, we see that the minimum values for motor vehicle flows are all zero. The major reason for these zero values is the traffic regulatory bans for certain movements. Obviously, such samples with specific regulatory bans lacked generality and needed to be excluded from this study. Consequently, the actual sample numbers applied to the model estimations were smaller than 460 (=115 4), and varied from type to type. METHODOLOGY Modeling the BMV-1 Accident Risk For a given intersection i and its approach k, if the risk that a through motor vehicle will be involved in a BMV-1 accident is p 1 (the subscript 1 corresponds to the type code for BMV accidents), then the number of BMV-1 accidents that may occur follows a binomial distribution. The probability of having n 1 accidents is f = (1) 1 n1 f1 n1 P( n1 ) p1 (1 p1 ) n 1 where i = intersection index; k = approach index; f 1 = through motor vehicle volume of intersection i, approach k; n 1 = number of BMV-1 accidents involving vehicles in f 1 ; P(n 1 ) = probability of having n 1 accidents.
7 6 p 1 = BMV-1 accident risk for a motor vehicle in f 1. Since it is quite rare to have a BMV-1 accident over the course of normal traffic flow, p 1 is very small compared to traffic volume f 1. Thus, the Poisson distribution is a good approximation to the binomial distribution (Pitman, 1993) for BMV accident analyses, and Equation (1) can be approximated by: P( n Where the Poisson distribution parameter 1 and E(n 1 ) denotes the expected value of n 1. n 1 m1 exp( m1 ) ) = (2) n! 1 m1 E( n1 ) = f1 p1 = (3) Poisson distribution models are commonly used for accident prediction. They are usually the first choice when modeling traffic accidents because of the nonnegative, discrete and random features of accidents. A Poisson model, however, has only one distribution parameter, and requires that the distribution s expected value and variance be equal. In many cases, however, accident data are over-dispersed, and the applicability of Poisson models is seriously limited. An easy way to overcome this constraint (i.e. the mean must be equal to the variance) is to add an independently distributed error term, ε 1, to the log transformation of Equation (3) (Poch and Mannering, 1996). That is: ln m1 ln( f1 p1 ) + ε1 = (4) Assume exp(ε 1 ) is a Gamma distributed variable with mean 1 and variance δ 1. Substituting Equation (4) into Equation (2) yields P( n 1 ε 1 n1 ( f1 p1 exp( ε1 )) exp( f1 p1 exp( ε1 )) ) = (5) n! Integrating ε 1 out of Equation (5), a negative binomial distribution model can be derived as follows: ( θ1 1 1 n1 P n ( ) ( ) Γ( n + θ ) θ f p ) = (6) Γ( n + 1) Γ( θ ) f p + θ f p + θ where θ 1 = 1/δ 1. δ 1 is often referred to as the negative binomial dispersion parameter. The expected value of this negative binomial distribution is equal to the expected value of the Poisson distribution shown in Equation (3). Its
8 7 variance is V ( n ) = E( n )[1 + δ E( n )] (7) Since δ 1 can be larger than 0, the constraint of the mean equaling the variance in the Poisson models is removed. If θ 1 is significant in our estimation, the negative binomial regression is appropriate. Otherwise, the Poisson regression should be the correct choice. The BMV-1 collision risk, p 1, is explained by bicycle volume and a set of explanatory factors. It is nonnegative and ranges from 0 to 1. Several functions that satisfy the above conditions were tested, and the researchers eventually selected Equation (8) as the BMV-1 accident risk model. p 1 = b1 b1 + exp( β1x1 ) (8) Where b 1 = volume of the bicycle flow directly involved in the BMV-1 accident at intersection i, approach k (this should be the sum of bicycle volumes crossing the entering approach and the opposing approach). β 1 = vector of unknown coefficients; X 1 = vector of explanatory variables at intersection i, approach k. One important advantage of using Equation (8) is that it gives zero BMV-1 accident risk when there is no bicycle crossing the entering approach or the opposing approach, i.e. p 1 = 0 when b 1 = 0. Additionally, the sign of each estimated coefficient in vector β 1 is consistent with the corresponding explanatory variable s effect on p 1 + indicates increasing effect and - indicates decreasing effect. This feature makes our estimation results intuitively appealing. Substituting Equation (8) into Equation (6) and rearranging terms yields the final formulation for the probability of having n 1 BMV-1 accidents as shown in Equation (9) ( ) = Γ( n + θ ) ( θ ( b + exp( β X θ1 1 1 n1 P n1 ) ( ) Γ( n1 + 1) Γ( θ1) f1 b1 + θ1( b1 + exp( β1x1 )) f1 b1 + θ1( b1 + exp( β1x1 )) )) f b (9) Modeling the BMV-2 and BMV-3 Accident Risks
9 8 Following a similar procedure to that described for the BMV-1 accident risk model, we obtain the final formulation for the probability of having n 2 BMV-2 accidents as ( ) = Γ( n + θ ) ( θ ( b + exp( β X θ2 2 2 n2 P n2 ) ( ) Γ( n2 + 1) Γ( θ2) f2b2 + θ 2( b2 + exp( β 2X2 )) f2b2 + θ 2( b2 + exp( β 2X2 )) )) f b (10) where f 2 = left-turn motor vehcle volume of intersection i, approach k; n 2 = number of BMV-2 accidents involving vehicles in f 2 ; P(n 2 ) = probability of having n 2 accidents. p 2 = BMV-2 accident risk for a motor vehicle in f 2. b 2 = volume of bicycle flow directly involved in BMV-2 accidents at intersection i, approach k (this should be the bicycle volume crossing the left approach as shown in Fig. 1); β 2 = vector of unknown coefficients; X 2 = vector of explanatory variables at intersection i, approach k. θ 2 = reciprocal of the negative binomial dispersion parameter for BMV-2 accidents. Similarly, the formulation for BMV-3 accidents is presented in Equation (11). Γ( n + θ ) θ ( b + exp( β X θ3 3 3 n3 P( n3 ) = ( ) ( ) Γ( n3 + 1) Γ( θ3) f3 b3 + θ3( b3 + exp( β3x3 )) f3 b3 + θ3( b3 + exp( β3x3 )) )) f b (11) where f 3 = right-turn motor vehicle volume of intersection i, approach k; n 3 = number of BMV-3 accidents involving vehicles in f 3 ; P(n 3 ) = probability of having n 3 accidents. p 3 = BMV-3 accident risk for a motor vehicle in f 3. b 3 = volume of bicycle flow directly involved in BMV-3 accidents at intersection i, approach k (this should be the bicycle volume crossing the right approach as shown in Fig. 1); β 3 = vector of unknown coefficients; X 3 = vector of explanatory variables at intersection i, approach k. θ 3 = reciprocal of the negative binomial dispersion parameter for BMV-3 accidents.
10 9 ESTIMATION RESULTS AND DISCUSSION The unknown coefficients, β j and θ j (j=1, 2, 3), can be estimated using the maximum lelihood estimation (MLE) method. The log-lelihood functions used for model estimations have the general form shown in Equation (12). l( β j, θ ) = j Γ( n + θ ) θ ( b + exp( β X j j j j j j θ j j j ( ) ( i= 1 k= 1 Γ( n j + 1) Γ( θ j ) f jb j + θ j ( b j + exp( β jx j )) f jb j + θ j ( b j + exp( )) f b β j X j ) )) n j for j=1, 2, 3 (12) By choosing j=1, 2 and 3, BMV-1, BMV-2 and BMV-3 models can be estimated respectively. For each BMV model, initial variables in X j are selected, based on accident type and its occurrence mechanism, from more than 70 variables in our database. For example, for BMV-2 accidents, all variables that affect the frequency and quality of conflicts between left-turn motor vehicles and bicycles crossing the left approach are included in the model. Insignificant variables are gradually removed during the estimation process, and only those variables significant at 0.05 levels are remained in the final form of each model. The software package used for model estimations was SYSTAT 7.0. Results for the BMV-1 Accident Risk Model Estimation Six variables are included in the BMV-1 risk model. The estimated coefficients and their significance levels shown by t-ratios and corresponding p s are presented in Table 3. As shown in Equation (8), p 1 has monotonic relationships with the variables in vector X 1. For any variable in X 1, if the corresponding coefficient in β 1 is positive, its positive increment increases the value of p 1 (increasing effect). Otherwise, its positive increment decreases the value of p 1 (decreasing effect). The signs of the estimated coefficients in Table 3 are consistent with their effect directions on p 1. Thus, we can tell whether the effect of a variable on p 1 is increasing or decreasing by looking at the sign of the estimated coefficient. This is also true for Tables 4 and 5.
11 10 Three of the six variables are discerned to have decreasing effects on p 1. Since there are no legal conflicts between through vehicular flow and bicycle flows (crossing the entering approach or the opposing approach) at signalized intersections, the occurrence of BMV-1 accidents is attributed to disregarding red signals, either by bicyclists or by through motor vehicle drivers. Factors that reduce the probability of running red signals should have decreasing effects on the BMV-1 risk. Heavier traffic flows from both the entering and opposing approaches make the time headway for each direction shorter and curtail the chances for illegally crossing the approaches. Thus, an increment in the total through motor vehicle volume (both directions) decreases p 1. When there are more rightturning vehicles at the opposing approach, conflicts between the through flow and the opposing right-turn flow disturb the smooth movements of through vehicles and result in slower speeds. Slower speeds can give drivers more time to detect signal changes and conduct stop actions when necessary. Therefore, an increase in the average daily right-turning motor vehicle volume of the opposing approach lowers the BMV-1 accident risk. Finally, intersections located in the central business district (CBD) have lower p 1 values. This is probably due to the fact that continuous efforts toward improving traffic safety, such as the strict enforcement of traffic regulations and vehicle monitoring in the CBD area, may have resulted in behavior improvements for both motor vehicle drivers and bicyclists. The existence of a pedestrian overbridge is generally thought to decrease bicycle and pedestrian accidents because the conflicts between bicycle/pedestrian flow and motor vehicle flow can be significantly reduced by the overbridge. However, our estimation result shows that the existence of a pedestrian overbridge has an increasing effect on p 1. A possible explanation is that, although overbridges reduce legal conflicts, they may increase the frequency of bicyclists running red-signals at street-level. Typically, approaches with overbridges do not have protective signals for crossing pedestrians and bicyclists. This indicates that if pedestrians/bicyclists do not cross the approaches via the overbridges, they will have to run red signals to cross at street-level. Because an overbridge normally requires bicyclists to walk up and down the bridge with their bicycles in order to cross the approach, some bicyclists may think it too troublesome to cross via the overbridge and decide to cross directly at street-level (running a red signal). If this assumption is true, the estimation result is reasonable, since every BMV-1 accident involves red signal running behavior, and it is the bicycle rider who is most lely at fault (Gårder, 1994). Miura (1992) studied the effect of driving environment on drivers behavior and found that with the increasing complexity of driving environment, response eccentricity (the size of the functional field of view) decreases and reaction time increases. This means that the increased amount of information for processing
12 11 significantly lengthens drivers perception reaction time. The visual noise dummy variable is employed in this study to reflect the effect of information quantity to be processed while passing an intersection. Our estimation results in Table 3 show that the increase of the visual noise level enlarges p 1. Since the background visual noise distracts a driver s attention and makes it difficult for drivers to detect traffic lights, the increasing effect of the visual noise level is easy to understand. The fact that the ratio of motorcycle volume to motor-vehicle volume in through traffic flow heightens p 1 is probably because of the higher motorcycle speeds and obstructed visions, for both motorcyclists and bicyclists, caused by other motor vehicles. Since most motorcyclists tend to travel at the outer lanes, vision is very lely to be blocked by motor vehicles traveling in the insider lanes. This makes it hard for bicyclists and motorcyclists to find each other early. Additionally, higher motorcycle speeds give motorcyclists less time to react when bicyclists show up. Results for the BMV-2 Accident Risk Model Estimation Estimation results for the BMV-2 accident risk model are shown in Table 4. Eight variables are included in the p 2 model, and five of them have decreasing effects. It is not surprising that the signal control pattern does not significantly affect BMV-2 accident risk since conflicts between bicycle flow and left-turn flow (in Japan, vehicles travel along the left side of the road) are legal under certain control periods for both two-phase and four-phase controls. The bicycle volume and the ratio of left-turning motor vehicle volume to total motor-vehicle volume are found to decrease p 2 as shown in Table 4. These two findings, however, may not reflect the entire spectrum of the relationship between the volumes and p 2, since we believe that p 2 should initially increase with motor vehicle and bicycle volumes until certain levels are reached and decrease thereafter. Due to the model structure in this study and the sampling bias in our data, the increasing phase appears absent. The decreasing effect of the pedestrian overbridge at the left approach may be due to the lowered conflict level. The width of the entering approach is also found to decrease BMV-2 accident risk. A possible explanation for this variable may be the better vision afforded by the wider road, or the longer green time for pedestrians/bicyclists (pedestrian green time for crossing the left approach is very lely to be proportional to the green time for through traffic of the entering approach). The decreasing effect of intersection location (in CBD or not) is probably due to
13 12 the same factors explained in the BMV-1 risk model discussion, i.e., stricter enforcement and monitoring in CBD areas. When there are more right-turn lanes at the opposing approach, conflicts between left-turning vehicles and opposing right-turning vehicles will increase at the merging section in the left approach, and such conflicts will consequently affect the left-turning drivers ability to detect crossing bicyclists. Therefore, the number of right-turn lanes in the opposing approach increases p 2. Similarly, the number of outgoing lanes at the left approach heightens the BMV-2 accident risk, since it is proportional to potential conflict points a bicyclist may face when crossing the left approach. The average time headway of left-turn flow also increases the value of p 2. This is lely due to the higher left-turning motor vehicle speed and the slacken bicyclist caution when the left-turning volume is low. Results for the BMV-3 Accident Risk Model Estimation Seven variables are included in the p 3 model, and the estimation results are listed in Table 5. Of the seven variables, four increase the p 3 and the three decrease it. Changing the signal control pattern from two phases to four phases reduces conflicts between bicycle flow and right-turn vehicular flow, and therefore, lowers p 3. The decreasing effect of speed limit at the opposing approach must be interpreted with caution. It could be related to the turning maneuvers of right-turning vehicle drivers. When speeds of opposing through vehicles are high, rightturning drivers may tend to drive conservatively. They are very lely to stop first to wait for right-turn chances under two-phase signal control. This may reduce the average right-turn vehicle speed and, hence, lower the p 3. As for the estimated decreasing effect of the bicycle volume at the right approach, the same explanation for the bicycle volume at the left approach in the BMV-2 accident risk model may apply. Approaches with a wider road median are concluded to have higher BMV-3 accident risk. This may be largely due to the fact that a poor vision angle makes it harder to effectively detect opposing through vehicles and bicycles at the right approach. Using the same data set, Wang and Nihan (2001) also found that this variable significantly increases the angle collision risk between right-turning vehicles and opposing through vehicles. The number of right-turn lanes at the entering approach has an increasing effect on p 3. This is possibly because, when there are two or more right-turn lanes, right-turning vehicles in different lanes obstruct the vision of each other during the turning movement. The increasing effect of the number of approaches sheltered by elevated roads may be |
In the present study, the effects of Hall current and rotation on an unsteady Magneto-Hydrodynamic (MHD) free convection heat and mass transfer of an electrically conducting, viscous, incompressible and heat absorbing fluid flow past a vertical infinite flat plate embedded in non-Darcy porous medium is investigated. The flow is induced by a general time-dependent movement of the vertical plate, and the cases of ramped temperature and isothermal plates are studied. Exact solution of the governing time-dependent boundary layer equations for the momentum, energy and concentration were obtained in closed form by using Laplace Transform technique. Expressions for skin friction due to primary and secondary flows and Nusselt number are derived for both ramped temperature and isothermal plates. Expression for Sherwood number is also derived. Some applications of practical interest for different types of plate motions viz. plate moving with uniform velocity, plate moving with uniform acceleration and plate moving with periodic acceleration are discussed. The numerical values of fluid velocity, fluid temperature and species concentration are displayed graphically whereas the numerical values of skin friction, the Nusselt number and the Sherwood number are presented in tabular form for various values of pertinent flow parameters for both ramped temperature and isothermal plates. It is found that the primary fluid velocity decreases with increasing values of Hall current parameter whereas it has reverse effect on secondary fluid velocity, in all types of motion of plate discussed. The Primary fluid velocity decreases with the increasing value of rotation parameter whereas it has reverse effect on secondary fluid velocity. Hall current tends to reduce the primary skin friction for both ramped temperature and isothermal plate whereas it has reverse effect on secondary skin friction.
The labour force is a fundamental component of every modern economy. It is also called the workforce that is the total number of the people who are eligible to work, including the employed and unemployed people. The company supplies free jobs which number is proportional to the invested capital. In this work, we propose a mathematical model that describes the dynamics of free jobs and labour force. In the model, the rate by which the labour force is filling in free jobs is modeled by Hattaf-Yousfi functional response. Furthermore, we first show that the proposed model is mathematically and economically well-posed. Moreover, the dynamical behavior of the model is studied by determining the existence and stability of equilibria.
S.D. Berman and P. Charpin characterized the Reed-Muller codes over the binary field or over an arbitrary prime field as the powers of the radical in a modular group algebra. We present a new proof of this famous theorem. Furthermore, the same method is used for the study of the Generalized Reed-Muller codes over a non prime field.
This paper uses stochastic subsampling of the dataset to provide a frequentist approximation to what is known in the Bayesian framework as the posterior inclusion probability (PIP). The distinct merit of this contribution lies in the fact that it makes it easier for typically non-Bayesian-minded practitioners, of which there are many, to relate to the way the Bayesian paradigm allows the computation of the nicely interpretable variable importance. Despite its computationally intensive nature, due to the need to fitting a very large number of models, the proposed approach is readily applicable to both classification and regression tasks, and can be done in comparatively competitive computational times thanks to the availability of parallel computing facilities through cloud and cluster computing. Finally, the scheme proposed is very general and can therefore be easily adapted to all kinds of statistical prediction tasks. Application of the proposed method to some very famous benchmark datasets shows that it mimics the Bayesian counterpart quite well in the important context of variable selection.
In this paper, a new concept of the fuzzy stability set of the first kind for multi-level multi-objective fractional programming (ML-MOFP) problems having a single-scalar parameter in the objective functions and fuzziness in the right-hand side of the constraints has been introduced. Firstly, A parametric ML-MOFP model with crisp set of constraints is established based on the α-cut approach. Secondly, a fuzzy goal programming (FGP) approach is used to find an α-Pareto optimal solution of the parametric ML-MOFP problem. Thus, the FGP approach is used to achieve the highest degree of each membership goal by minimizing the sum of the negative deviational variables. Finally, the fuzzy stability set of the first kind corresponding to the obtained α-Pareto optimal solution is developed here, by extending the Karush-Kuhn-Tucker optimality conditions of multi-objective programming problems. An algorithm to clarify the developed fuzzy stability set of the first for parametric ML-MOFP problem as well as Illustrative numerical example are presented.
The paper deals with a nonlinear equation in one-dimensional space, of which the nonlinearity appears both in source term and the Neumann boundary condition. Firstly, we proved that the solution of problem (1.1) quenches in finite time and the only quenching point is x = 0 if the initial data is appropriate. Then we established the corresponding quenching rate of the solution. |
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Chapter 3 - Student Workbook - Utah Education Network
3.1d Homework: Simplifying Algebraic Expressions with Models . .... 3.2b Class Activity: More Model and Solve One-‐ and Two-‐Step Equations .
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A Curriculum Project on Expressions and Equations in Mathematics
Dec 3, 2014 ... This Thesis is brought to you for free and open access by the .... of what students should be learning in a mathematics classroom. In .... This study concluded that when students are able to touch and manipulate mathematics, .... Students will solve one step equations using what they know about ..... Page 31 ...
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Linear Equations - Calvert Education
Remember that the value of the dependent variable is calculated using the ... student should complete for the remainder of this lesson. ... Solve linear equations in one variable. ... multiple steps and use mental math skills, but it is best for your student to show work ... Today you reviewed how to solve algebraic equations.
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Getting to the Core - Santa Ana Unified School District
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lost in translation: algebraic modeling in the middle school classroom
The voyage and path would not have been the same without each of you traveling .... grade mathematics course were given a series of algebra assignments, ... Grades 6-8 Expressions and Equations, “students should be able to 1) ... to solve single- and multi-step equations for an unknown variable and model ...... Page 31 ...
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You should verify that the solution of these equations is x = 1, y = 2 because by substituting these ... In realistic problems which arise in mathematics and ... On completion you should be able to ... • solve pairs of simultaneous linear equations. 42 ... One way of solving simultaneous equations is by elimination. ... 3x + 5y = 31.
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Teaching Strategies for Improving Algebra Knowledge in Middle and
that address the challenges of teaching algebra to students in grades 6 through 12. ... This practice guide should be reviewed and applied according to the .... Some possible strategies for solving quadratic equations . . . . . . . . . 31. Example ... What steps did you take to solve the .... conceptual knowledge.9 One of these three.
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Algebra 1 - Alexandria City Public Schools
Using Expressions, Solving Linear Equations and Inequalities. .... The resulting plan should include an analysis of how you determined your plan and reasoning ...
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Each college received one vote at the table when ... unit, you do not have to teach the material in the order described, but all topics must be covered. ... VCCS to improve student progression from developmental mathematics, through college level ...... should be on learning the steps to solving the equations and inequalities,.
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Given samples from a population of individuals belonging to different types with unknown proportions, how do we estimate the probability of discovering a new type at the -th draw? This is a classical problem in statistics, commonly referred to as the missing mass estimation problem. It first appeared in ecology (e.g., Fisher et al. Fisher et a. (1943) and Good Good (1953)), and its importance has grown considerably in recent years driven by challenging applications in a wide range of scientific disciplines, such as biological and physical sciences (e.g., Kroes et al. Kroes et al. (1999), Gao et al. Gao et al. (2007) and Ionita-Laza et al. Ionita-Laza et al. (2009)
), machine learning and computer science (e.g., Motwani and VassilvitskiiMotwani and Vassilvitskii (2006) and Bubeck et al. Bubeck et al. (2013)), and information theory (e.g., Orlitsky et al. Orlitsky et al. (2014) and Ben-Hamou et al. Ben-Hamou et al. (2018)). To move into a concrete setting, let be an unknown discrete distribution, where is a sequence of atoms on some measurable space and denote the corresponding probability masses, i.e. such that . If
is a collection of independent and identically distributed random variables from, then we define the missing mass as
where is the indicator function. Among various nonparametric estimators of the missing mass, both frequentist and Bayesian, the Good-Turing estimator (Good Good (1953)) is arguably the most popular. It has been the subject of numerous studies, most of them in the recent years. These include, e.g., asymptotic normality and large deviations (Zhang and Zhang Zhang et al. (2009) and Gao Gao (2013)), admissibility and concentration properties (McAllester and Ortiz, McAllester and Ortiz (2003), Ohannessian and Dahleh Ohnnessian and Dahleh (2012) and Ben-Hamou et al. Ben-Hamou et al. (2017)), consistency and convergence rates (McAllester Schapire McAllester and Schapire (2000), Wagner et al. Wagner et al. (2006) and Mossel and Ohannessian Mossel and Ohannessian (2015)), optimality and minimax properties (Orlitsky et al. Orlitsky et al. (2013) and Rajaraman et al. Rajaraman (2017)).
Under the setting depicted above, let denote an estimator of . Motivated by the recent works of Ohannessian and Dahleh Ohnnessian and Dahleh (2012), Mossel and Ohannessian Mossel and Ohannessian (2015) and Ben-Hamou et al. Ben-Hamou et al. (2017)
, in this paper we consider the problem of consistent estimation of the missing mass under the multiplicative loss function
As discussed in Ohannessian and Dahleh Ohnnessian and Dahleh (2012), the loss function (2) is adequate for estimating small value parameters, in the sense that it allows to achieve more informative results. Such a loss function has been already used in statistics, e.g. for the estimation of small value probabilities using importance sampling (Chatterjee and Diaconis Chatterjee and Diaconis (2018)) and for the estimation of tail probabilities in extreme value theory (Beirlant and Devroye Beirlant and Devroye (1999)). Under the loss function (2), Ohannessian and Dahleh Ohnnessian and Dahleh (2012) showed that: i) the Good-Turing estimator may be inconsistent; ii) the Good-Turing estimator is strongly consistent if the tail of decays to zero as a regularly varying function with parameter (Bingham et al. Bingham and Goldie (1987)). See also Ben-Hamou et al. Ben-Hamou et al. (2017) for further results on missing mass estimation under regularly varying . Mossel and Ohannessian Mossel and Ohannessian (2015) then strengthened the inconsistency result of Ohannessian and Dahleh Ohnnessian and Dahleh (2012), showing the impossibility of estimating (learning) in a completely distribution-free fashion, that is without imposing further structural assumptions on .
We present an alternative, and simpler, proof of the result of Mossel and Ohannessian Mossel and Ohannessian (2015). Our proof relies on tools from Bayesian nonparametrics, and in particular on the use of a Dirichlet prior (Ferguson Ferguson (1973)) for the unknown distribution . This allows us to exploit properties of the posterior distribution of to prove the impossibility of a distribution-free estimation of the missing mass, thus avoiding the winding (geometric) coupling argument of Mossel and Ohannessian Mossel and Ohannessian (2015). Up to our knowledge, the use of Bayesian ideas to study large sample asymptotics for the missing mass is new, and it could be of independent interest. Motivated by the work of Ohannessian and Dahleh Ohnnessian and Dahleh (2012) and Ben-Hamou et al. Ben-Hamou et al. (2017) we then investigate convergence rates and minimax rates for the Good-Turing estimator under the class of regularly varying . We still rely on tools from Bayesian nonparametrics, thus providing an original approach to tackle these problems. In particular, we make use of the two parameter Poisson-Dirichlet prior (Perman et al. Perman et al. (1992) and Pitman and Yor Pitman and Yor (1997)) for the unknown distribution , which is known to generate (almost surely) discrete distributions whose tail decays to zero as a regularly varying function with parameter . See Gnedin et al. Gnedin et al. (2007) and references therein. This allows us to exploit properties of the posterior distribution of to prove that: i) the convergence rate of the Good-Turing estimator is the best rate that any estimator of the missing mass can achieve, up to a slowly varying function; ii) the minimax rate must be at least . We conclude with a discussion on the problem of deriving the minimax rate of the Good-Turing estimator, conjecturing that the Good-Turing estimator is an asymptotically optimal minimax estimator under the class of regularly varying .
The paper is structured as follows. In Section 2 we state our main results on convergence rates and minimax rates for the Good-Turing estimator under regularly varying distribution . Proofs of these results, as well as the alternative proof of the result of Mossel and Ohannessian Mossel and Ohannessian (2015), are provided in Section 4. In Section 3 we discuss open problems and possible future developments on missing mass estimation. Auxiliary results and technical lemmas are deferred to Appendix 5. The following notation is adopted throughout the paper:
is the unit interval, and its Borel -algebra;
is the space of discrete distributions on , endowed with the smallest -algebra making measurable for every ;
is the -fold product of on , and the expectation with respect ; for easiness of notation, we will use to denote also the expectation with respect to . ;
is a generic slowly varying function, i.e. a function satisfying as for every ;
denotes a generic strictly positive constant that can vary in the calculations and in distinct statements;
Given a sequence of probabilities , denotes the corresponding ordered sequence, i.e. ;
Given two functions and , stands for , for , for ;
is the Beta integral of parameters and .
2 Main results
Let be a collection of independent and identically distributed random variables from an unknown discrete distribution . The actual values taken by the observations, ’s, are irrelevant for the missing mass estimation problem and, without loss of generality, they can be assumed to be values in the set . Therefore, is supposed to be a discrete distribution on the sample space , given a sequence of atoms and masses such that . Both atoms and masses of the distribution are assumed to be unknown. Given the sample , we are interested in estimating the missing defined in (1), which turns out to be a jointly measurable function of and as proved in Proposition 5.1. Given an estimator of , we will measure its statistical performance by using the multiplicative loss function defined in (2). As we discussed in the introduction, this loss function is suitable to study theoretical properties of parameters or functionals taking small values, and it has already been used in previous works on missing mass estimation, e.g., Ohannessian and Dahleh Ohnnessian and Dahleh (2012), Mossel and Ohannessian Mossel and Ohannessian (2015) and Ben-Hamou et al. Ben-Hamou et al. (2017).
A sequence of estimators is said to be consistent for under parameter space and loss function , if the loss incurred by the estimator converges in probability to zero under all points in the parameter space. Formally, is consistent for if for all and for all ,
by exploiting a coupling of two generalized (dithered) geometric distributions. In section Section4 we present an alternative proof of Theorem 2.1. While the proof of Mossel and Ohannessian Mossel and Ohannessian (2015) has the merit to be constructive, our approach has the merit to be simpler and it provides a new way to face these type of problems, which mainly relies on Bayesian nonparametric techniques. Similar Bayesian nonparametric arguments will then be crucial in order to study of convergence rates and minimax rates of the Good-Turing estimator under the class of regularly varying .
Roughly speaking, Theorem 2.1 proves that any asymptotic result holding uniformly over a set of possible distributions, the parametric space must be restricted to a suitable subclass. In particular, from the proof of Theorem 2.1 we see that some conditions have to be imposed on the tail decay of the elements of the parameter space. That is, from the proof of Theorem 2.1 we deduce that there are no consistent estimators for the class of distributions sampled from a Dirichlet process. From Kingman Kingman (1975) (Equation 65), we have that, if is sampled from a Dirichlet process, its sequence of ordered masses behaves like , as . Therefore, the tail of has approximately exponential form, resembling a geometric distribution and satisfying for every . Indeed, a geometric distribution was used in Ohannessian and Dahleh Ohnnessian and Dahleh (2012) as an example to prove that the Good-Turing estimator can be inconsistent. Theorem 2.1 shows that, under this very light regime, any estimator of the missing mass, not just the Good-Turing, fails to be consistent under multiplicative loss. This motivates us to consider the class of s having heavy enough tails. This will be the subject of the rest of this section.
2.1 Consistency under regularly varying
In this section we recall the Good-Turing estimator (Good Good (1953)) of the missing mass , and we study its convergence rate and minimax risk for regularly varying . The definition of the Good-Turing estimator makes use of the proportion of unique values in the sample to estimate the missing mass. Let be the number of times the value is observed in the sample , i.e.,
Furthermore, let and be the number of values observed times and the total number of distinct values, respectively, observed in , i.e.,
The Good-Turing estimator of is defined in terms of the statistic , that is
Ohannessian and Dahleh Ohnnessian and Dahleh (2012) first showed the inconsistency of under the choice of being a geometric distribution. In the same paper, it is shown that under the assumption that the tail of decays to zero as a regularly varying function with parameter , the Good-Turing estimator is strongly consistent. This latter result was generalized to the range in Ben-Hamou et al. Ben-Hamou et al. (2017).
The assumption of regularly varying is a generalization of the power law tail decay, adding some more flexibility by the introduction of the slowly varying function . Power-law distributions are observed in the empirical distributions of many quantities in different applied areas, and their study have attracted a lot of interest in recent years. For extensive discussions of power laws in empirical data and their properties, the reader is referred to Mitzenmacher Mitzenmacher (2014), Goldwater et al. Goldwater et al. (2006), Newman Newman (2003), Clauset et al. Clauset et al. (2009) and Sornette Sornette (2006)
. Restricting the parameter space to probability distributions having regularly varying tail is not a mere technical assumption and, on the contrary, it represents a natural subset of the parameter space to consider, which we expect to contain the true data generating distribution for many different applications.
To move into the concrete setting of regular variation (Bingham et al. Bingham and Goldie (1987)), for every we define a counting measure on as , with corresponding tail function defined as for all . Then a distribution is said to be regularly varying with parameter if
where is a slowly varying function depending on . We denote by the set of all regularly varying distribution on with parameter . From (6) it is clear that such a class includes distributions having power-law tail decay, which correspond to the particular case of being a constant, which is equivalent to being constant. We denote the class of distributions having power law tail decay by . In the following results, we will restrict our attention to the estimation problem under restricted parameter spaces and .
From Ohannessian and Dahleh Ohnnessian and Dahleh (2012) it is known that the Good-Turing estimator is consistent under all s belonging to the regularly varying class . Focusing attention on the class , in the next proposition we refine the result of Ohannessian and Dahleh Ohnnessian and Dahleh (2012) by studying the rate at which the multiplicative loss of the Good-Turing estimator converges to zero. A sequence is a convergence rate of an estimator for the distribution if
for all sequences . The next proposition shows the rate of convergence of the Good-Turing estimator is . The proof is omitted because it follows from Proposition 2.2 below along with a simple application of Markov’s inequality.
As a further result on convergence rate of the Good-Turing estimator, in Theorem 2.2 we show that the convergence rate achieved by the Good-Turing estimator is actually almost the best convergence rate any estimator of can achieve. Specifically, for any other estimator, it is possible to find a point for which the rate of convergence is not faster than .
For any estimator , there exists such that for every
Therefore the convergence rate of cannot be faster than .
Proposition 2.1 and Theorem 2.2 together show that the Good-Turing estimator achieves the best convergence rate up to possibly a slowly varying function. In particular, if the distribution has a power-law decay, i.e. , the two rates match and the Good-Turing estimator achieves the best rate possible. In particular, because does not depend on , it follows that the Good-Turing estimator is actually rate adaptive for the class of power law distributions, However, for a general we do not know whether the two rates of Proposition 2.1 and Theorem 2.2 may be improved to make them match or they are not.
As a final result, in the next theorem we consider the asymptotic minimax estimation risk for the missing mass under the loss function (2) and with parameter space . Theorem 2.3 provides with a lower bound for the estimation risk of this statistical problem, showing that the minimax rate is not smaller than .
Let be the class of discrete distributions on with power law tail function and let denote the multiplicative loss function (2). Then, there exists a positive constant such that
where the infimum is taken over all possible estimators .
The lower bound of Theorem 2.3 can be used to derive the minimax rate, by matching it with appropriate upper bounds of specific estimators of the missing mass. This lower bound trivially still holds for any parametric set larger than and, therefore, the theorem also provides with a lower bound of the estimation risk under the larger parameter space . In the next Proposition, we show that for a fixed distribution , the Good-Turing estimator achieves the best possible rate of Theorem 2.3 up to a slowly varying term.
Let be the Good-Turing estimator and let . Then, there exists a finite constant such that for every
where is the slowly varying function specific to appearing in (5).
Extending Proposition 2.2 to hold uniformly over is an open problem and probably requires a careful control over the size of . Indeed, the classes of distributions we are considering are defined through the asymptotic properties of their elements, while to obtain minimax results we need a control for each . Even though Proposition 2.2 does not directly provide with the minimax rate of the Good-Turing estimator, it still provides with a sanity check for its asymptotic risk. Specifically, Proposition 2.2 implies that for every ,
Moreover, from a minor change at the beginning of the proof of Theorem 2.3, we can also prove that for every estimator and every sequence diverging to infinity, we can find an element such that . This leads us to conjecture that the Good-Turing estimator should be a rate optimal minimax estimator.
In this paper we have considered the problem of consistent estimation of the missing mass under a suitable multiplicative loss function. We have presented an alternative, and simpler, proof of the result by Mossel and Ohannessian Mossel and Ohannessian (2015) on the impossibility of a distribution-free estimation of the missing mass. Our results relies on novel arguments from Bayesian nonparametric statistics, which are then exploited to study convergence rates and minimax rates of the Good-Turing estimator under the class of regularly varying . In Proposition 2.1 and Theorem 2.2 it has been shown that, within the class , the Good-Turing estimator achieves the best convergence rate possible, while for the class, , this rate is the best up to a slowly varying function. An open problem is to understand weather this additional slowly varying term is intrinsic to the problem or our results can actually be improved to make the rate of the Good-Turing estimator matches the best possible rate also within the class of regularly varying distributions. Under the restricted parametric spaces, in Theorem 2.3 we have provided a lower bound for the asymptotic risk. This bound can be used to compare estimators from a minimax point of view, by finding suitable upper bounds matching the lower bound rate. In particular, in Proposition 2.2 we have shown that the asymptotic rate of the risk of the Good-Turing estimator matches the lower bound rate, up to a slowly varying function. However, the rate of Proposition 2.2 is a pointwise result, for a fixed . An open problem is to extend Proposition 2.2 to the uniform case, when considering the supremum of the risk over all . This extension probably requires a careful analysis and control of the size of this parameter space . Work on this is ongoing.
In this section we will prove all the theorems stated in Section 2. The proofs of some technical auxiliary results are postponed to Appendix 5. We start with a simple lemma that will be useful in the sequel.
For and , implies
proof Let be positive real numbers, and suppose . Straightforwardly, have that
4.1 Proof of Theorem 2.1
We are going to show that for every estimator , there exists such that
and, therefore, there exists such that does not converge to zero in probability.
First, note that for , Lemma (4.1) implies that
so it is sufficient to show that there exists such that for every estimator ,
We will prove that (13) holds for all (and therefore (11) holds for any ). Let . Let denote the Dirichlet process measure on (Ferguson Ferguson (1973)), with base measure on . We choose uniform, i.e. . Now, we can lower bound the supremum in (13) by an average over with respect to and then swap the integration by Fubini theorem, therefore
where the first inequality follows since we can lower bound the supremum by an average, the second from reverse Fatou’s lemma, the equality comes by swapping the marginal of and conditional of given with the marginal of , denoted , and the conditional of given , the last inequality follows since we are considering the infimum over all possible values of . Also recall that, when is distributed as , then the marginal of , , is a Generalized Polya urn, while the conditional of given is (see Theorem 4.6 and subsection 4.1.4 of Ghosal and Van der Vaart Ghosal and Van der Vaart (2017)).
where . We are now going to lower bound the probability of the event on the right hand side of (14). First let us consider and
Therefore, for all and . Plugging this estimate in place of (14), we obtain
and the right hand side is strictly positive for all .
4.2 Proof of Theorem 2.2
Let any non-negative sequence converging to . We will show that for any estimator ,
Let us denote by the law of a stable process on of parameter . This a subordinator with Levy intensity, . See Kingman Kingman (1975), Lijoi and Prünster Lijoi and Prünster (2010) and Pitman Pitman (2006) for details and additional references. Because of , the stable process samples probability measures belonging to . Now we can upper bound the infimum in (17) by an average with respect to ,
where the last equality follows by applying Fatou’s Lemma.
Take large enough so that . Let us denote by the marginal law of the observations under an -stable process, when is integrated out, i.e. the probability measure on defined as for all . We swap the integration of the marginal of and the conditional of given with the marginal of and the conditional of given and then apply Lemma 4.1 to obtain
where denotes the posterior of given the sample. Therefore, taking , we can upper bound the quantity appearing on the l.h.s. of (17) by
We will upper-bound the two terms of the sum in (18) independently. Let us focus on the first term of (18). Let large enough so that and . From Proposition 5.2, under the posterior , is distributed according to a Beta random variable . Let us denote , , and for easiness of notation we will simply write and in the following calculations. Also let
be the cumulative distribution function of the beta random variable. From Proposition 5.2 in the Appendix, we have that
Consider the function defined by
Notice that and that . Therefore, reaches its maximum in (denoted for easiness of notation) satisfying
where denotes the density function of the distribution. On the event , we have , and so is bell-shaped with second inflexion point,
Therefore, is non decreasing on the interval and as a consequence, is non negative on , from which we can deduce that is non decreasing on the same interval. Now, since , it follows that on . Therefore,
We can now upper-bound as follows: |
Javier Díez-Palomar began teaching maths to adults as a volunteer in an adult school in Barcelona. He completed a PhD in mathematics education and worked at the University of Arizona researching maths education with Latinos/as. He then became an associate professor of Mathematics Education at the University of Barcelona. He is now the President of the Association of Multidisciplinary Educational Research, a Spanish member of the International Commission for the Study and Improvement of Mathematics Teaching, and a Trustee of the Adults Learning Mathematics association. He has served as link convenor for Network 24 (Research in Mathematics Education) at the European Conference of Educational Research. He has also been the principal investigator for several European Commission and Spanish Government Research grants on mathematics education. He is a member of the Numeracy Expert Group (OECD) working at the PIAAC survey, the chief editor for REDIMAT: Journal of Research in Mathematics Education, and former chief editor and current member of the Editorial Board for Adults Learning Mathematics: An International Journal.
1. What’s your earliest memory of doing mathematics?
I remember that I had always liked math until I got to high school. I loved going with my brother to the city library, to read science and math books and magazines. We used to spend a good part of our holidays there. But in high school, I had a bad experience since two teachers started to speak to me in a language that I did not understand. Only the last year before going to university, my luck changed: I had a teacher who was passionate about mathematics. He restored my self-confidence. Years later, when I was looking for programs to do my PhD, a professor and friend of mine advised me to go back to mathematics. I started volunteering at an adult school, teaching math, and I can say that that experience was my rebirth from the point of view of doing math. The adult learners, who had never been to school before and had just become literate, taught me two things: one, that mathematics is part of our lives, in every way; and two, they taught me how to teach mathematics. They made me a math teacher with a passion for math. And despite having had many memories of doing mathematics before, I can say that this is undoubtedly the one that changed my life and focused me on mathematics forever. Since then, I have had the dream that no one else loses the passion for mathematics. For this reason, I dedicate myself to mathematics education.
2. How has mathematics education changed in the time you have been involved in it?
The way of teaching mathematics has been changing in the last three decades. Many people, good researchers, have transformed the traditional lessons in which the teacher presents a series of concepts and procedures, waiting for the students to repeat them mechanically. Now the aim is for students to understand mathematics. The fundamental idea is that they are competent to solve mathematical problems and apply mathematics to different contexts. Problem-solving continues to be one of the fundamental pillars of teaching mathematics. Also, to encourage the taste and passion for them. There is a lot of literature on the different types of math activities, how to sequence the concepts, learning trajectories, procedural and conceptual understanding, socio-mathematical norms and mathematics, etc. In recent years, there has been an increasing interest in interactions around learning mathematics. We know that it is very different to learn individually than in a small group, in pairs, or in a group class. The effort to argue an answer to others’ questions in a group is a crucial aspect of learning. We know that when the egalitarian dialogue appears in the group, which is focused on the exchange of arguments oriented towards understanding, the learning opportunities are greater than when the type of dialogues that appear in the group are unidirectional, based on the power position of the person who is speaking, not on whether or not the idea you want to convey is understood. We still need more research on these types of dialogues. But what is already undeniable is that learning mathematics is a social process in which many people participate (teachers, families, classmates, TV, Internet, etc.). Therefore, future learning spaces will have to consider the interactions in the different spaces, identify learning difficulties, and maximise learning opportunities.
3. Tell me about a time in your career when something totally flabbergasted you.
A friend who was an elementary teacher told me about two boys doing mathematics in interactive groups. One of the boys was a newcomer from Senegal. He never went to school in his country of origin. He was in grade 4, learning the basics (to write, read, etc.). But in the interactive group, he was dividing with the lattice algorithm! His classmate was teaching him a way to understand how the lattice algorithm works: he drew three people on his sheet of paper; then, he distributed fifteen circles among the three people, ending with five each. Then, he pointed with his finger to the lattice algorithm in the paper and said: look, this is the divisor, this the dividend, and this the quotient. It was fascinating to learn how that boy was able to go beyond my friend's explanation of the lattice algorithm and connect it to a different model of dividing to help his classmate.
4. Do you practise mathematics differently in company?
Yes, of course. I learn other ways of doing math by listening to others, especially when I am in groups with diverse people. When I participate in Mathematics Dialogic Gatherings, some of the participants always surprise me with unexpected ways of solving a problem or explaining a concept. Reading math books in their company allows me to enrich myself with a broader vision of mathematics and leads me to explore new paths and points of view. I love it because, thanks to that, I discovered the tremendous complexity that mathematics has and how fascinating it is to explore it and find those little "treasures." Of course, alone, I would be unable to do so.
5. Do you think a brilliant maths teacher is born or made?
Without a doubt: made. Everyone can learn to teach math. We have a lot of research on the competencies that a good math teacher must develop. We know that mathematics teacher training (as in other domains of teachers’ professional knowledge) must be based on scientific evidence. We cannot hope that training our future teachers drawing on occurrences will convert them into good teachers. Just as a doctor must learn the parts of the body, how they work, and how they respond to treatments when they get sick, teachers also must know the contents of mathematics, how they work, and how to teach them when students have difficulty understanding them. All that is learned, of course.
6. What’s the most fun a mathematician can have?
Probably finding an unexpected relationship and why it happens that way. But if the question is what's the most fun a mathematics teacher can have, it would be to enjoy solving problems with the students. So often in class, I feel that my students become "fellow travellers" in solving a difficult problem. In the end, we look like "Sherlock Holmes," trying to find the clues that lead us to unveil the answer. It is difficult to describe the feeling: it is just the joy of doing mathematics together.
7. Do you have a favourite maths joke?
I always liked that video where you can see a man trying to explain that 25% divided among five persons gives 5% for each one; but the old man (Pa) and the old woman (Ma) reply that this is not true, because 25% divided among the 5 of you, that’s 14% for each one. I always ask myself: who is the smartest one in this piece? ;-)
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PrefaceSupplements Guide 1. Trigonometric Functions1.1 Angles1.2 Angle Relationships and Similar Triangles1.3 Trigonometric Functions1.4 Using the Definitions of the Trigonometric Functions 2. Acute Angles and Right Triangles2.1 Trigonometric Functions of Acute Angles2.2 Trigonometric Functions of Non-Acute Angles2.3 Finding Trigonometric Function Values Using a Calculator2.4 Solving Right Triangles2.5 Further Applications of Right Triangles 3. Radian Measure and the Unit Circle3.1 Radian Measure3.2 Applications of Radian Measure3.3 The Unit Circle and Circular Functions3.4 Linear and Angular Speed 4. Graphs of the Circular Functions4.1 Graphs of the Sine and Cosine Functions4.2 Translations of the Graphs of the Sine and Cosine Functions4.3 Graphs of the Tangent and Cotangent Functions4.4 Graphs of the Secant and Cosecant Functions4.5 Harmonic Motion 5. Trigonometric Identities5.1 Fundamental Identities5.2 Verifying Trigonometric Identities5.3 Sum and Difference Identities for Cosine5.4 Sum and Difference Identities for Sine and Tangent5.5 Double-Angle Identities5.6 Half-Angle Identities 6. Inverse Circular Functions and Trigonometric Equations6.1 Inverse Circular Functions6.2 Trigonometric Equations I6.3 Trigonometric Equations II6.4 Equations Involving Inverse Trigonometric Functions 7. Applications of Trigonometry and Vectors7.1 Oblique Triangles and the Law of Sines7.2 The Ambiguous Case of the Law of Sines7.3 The Law of Cosines7.4 Vectors, Operation, and the Dot Product7.5 Applications of Vectors 8. Complex Numbers, Polar Equations, and Parametric Equations8.1 Complex Numbers8.2 Trigonometric (Polar) Form of Complex Numbers8.3 The Product and Quotient Theorems8.4 De Moivre's Theorem; Powers and Roots of Complex Numbers8.5 Polar Equations and Graphs8.6 Parametic Equations, Graphs, and Applications Appendix A. Equations and InequalitiesAppendix B. Graphs of EquationsAppendix C. FunctionsAppendix D. Graphing Techniques GlossarySolutions to Selected ExercisesAnswers to Selected ExercisesIndex of ApplicationsIndexPhoto Credits
Marge Lial has always been interested in math; it was her favorite subject in the first grade! Marge's intense desire to educate both her students and herself has inspired the writing of numerous best-selling textbooks. Marge, who received Bachelor's and Master's degrees from California State University at Sacramento, is now affiliated with American River College. Marge is an avid reader and traveler. Her travel experiences often find their way into her books as applications, exercise sets, and feature sets. She is particularly interested in archeology. Trips to various digs and ruin sites have produced some fascinating problems for her textbooks involving such topics as the building of Mayan pyramids and the acoustics of ancient ball courts in the Yucatan. When John Hornsby enrolled as an undergraduate at Louisiana State University, he was uncertain whether he wanted to study mathematics education or journalism. His ultimate decision was to become a teacher, but after twenty-five years of teaching at the high school and university levels and fifteen years of writing mathematics textbooks, both of his goals have been realized. His love for both teaching and for mathematics is evident in his passion for working with students and fellow teachers as well. His specific professional interests are recreational mathematics, mathematics history, and incorporating graphing calculators into the curriculum. John's personal life is busy as he devotes time to his family (wife Gwen, and sons Chris, Jack, and Josh). He has been a rabid baseball fan all of his life. John's other hobbies include numismatics (the study of coins) and record collecting. He loves the music of the 1960s and has an extensive collection of the recorded works of Frankie Valli and the Four Seasons. David Schneider has taught mathematics at universities for over 34 years and has authored 36 books. He has an undergraduate degree in mathematics from Oberlin College and a PhD in mathematics from MIT. During most of his professional career, he was on the faculty of the University of Maryland--College Park. His hobbies include travel, dancing, bicycling, and hiking. Callie Daniels has always had a passion for learning mathematics and brings that passion into the classroom with her students. She attended the University of the Ozarks on an athletic scholarship, playing both basketball and tennis. While there, she earned a bachelor's degree in Secondary Mathematics Education as well as the NAIA Academic All-American Award. She has two master's degrees: one in Applied Mathematics and Statistics from the University of Missouri-Rolla, the second in Adult Education from the University of Missouri- St. Louis. Her hobbies include watching her sons play sports, riding horses, fishing, shooting photographs, and playing guitar. Her professional interests include improving success in the community college mathematics sequence, using technology to enhance students' understanding of mathematics, and creating materials that support classroom teaching and student understanding. |
The Kumaraswamy Marshall-Olkin Log-Logistic Distribution with Application
- https://doi.org/10.2991/jsta.2018.17.1.5How to use a DOI?
- Kumaraswamy-G, Maximum likelihood, Log-Logistic, Order statistic
In this paper, we define and study a new lifetime model called the Kumaraswamy Marshall-Olkin log-logistic distribution. The new model has the advantage of being capable of modeling various shapes of aging and failure criteria. The new model contains some well-known distributions as special cases such as the Marshall-Olkin log-logistic, log-logistic, lomax, Pareto type II and Burr XII distributions. Some of its mathematical properties including explicit expressions for the quantile and generating functions, ordinary moments, skewness, kurtosis are derived. The maximum likelihood estimators of the unknown parameters are obtained. The importance and flexibility of the new model is proved empirically using a real data set.
- Copyright © 2018, the Authors. Published by Atlantis Press.
- Open Access
- This is an open access article under the CC BY-NC license (http://creativecommons.org/licences/by-nc/4.0/).
There has been an increased interest in defining new generated classes of univariate continuous distributions by introducing additional shape parameter(s) to a baseline model. The extended distributions have attracted several statisticians to develop new models. The addition of parameters has been proven to be useful in exploring skewness and tail properties, and also for improving the goodness-of-fit of the generated family. The well-known generators are the following: the Marshall-Olkin distribution family by Marshall and Olkin (1997), the beta-G by Eugene et al. (2002), the Kumaraswamy-G (Kw-G) by Cordeiro and de Castro (2011), the Logistic-G by Torabi and Montazari (2014), the transformed-transformer (T-X) by Alzaatreh et al. (2013), the odd exponentiated generalized by Cordeiro et al. (2013), the Weibull-G by Bourguignon et al. (2014), the Kumaraswamy Marshal-Olkin distribution family by Alizadeh et al. (2015), the transmuted geometric-G by Afify et al. (2016a) and the beta transmuted-H by Afify et al. (2017).
Marshall and Olkin (1997) proposed a flexible family of distributions and introduced an interesting method of adding a new parameter to an existing distribution. The resulting new distribution includes the original distribution as a special case and gives more flexibility to model various types of data. For further information about the Marshall–Olkin family of distributions, see Barreto-Souza et al. (2013). The log-logistic (LL) distribution (known as the Fisk distribution in economics) has been widely used particularly in survival and reliability over the last few decades. It is the probability distribution of a random variable whose logarithm has a logistic distribution, an alternative to the log-normal distribution since it presents a failure rate function that increases initially and decreases later. The cumulative distribution function (cdf) and probability density function (pdf) of the LL distribution are given (for x > 0) byrespectively, where α > 0 is the scale parameter and γ > 0 is the shape parameter.
Searching a more flexible LL distribution, many authors defined generalizations and modified forms of the LL distribution, with different number of parameters. For example, the Kumaraswamy log-logistic (de Santana et al., 2012), Marshall-Olkin LL (MOLL) (Gui, 2013), Lomax log-logistic (Cordeiro et al., 2014), McDonald log-logistic (Tahir et al., 2014), beta log-logistic (Lemonte, 2014), transmuted log-logistic (Granzotto and Louzada, 2015), Kumaraswamy transmuted log-logistic (Afify et al., 2016b) and generalized transmuted log-logistic (GTLL) (Nofal et al., 2017) distributions.
Gui (2013) defined the cdf and pdf of the MOLL distribution (for x > 0) byrespectively, where α, γ, β > 0. For β = 1, we obtain the LL distribution.
The goal of this paper is to define and study a new lifetime model called the Kumaraswamy Marshall-Olkin Log-Logistic (“KMOLL” for short) distribution. The main feature of this model is that two additional shape parameters are inserted in (2) to give more flexibility in the form of the generated distribution. Based on the Kumaraswamy-generalized (K-G) family proposed by Cordeiro and de Castro (2011), we construct the new five-parameter KMOLL distribution. We give some mathematical properties of the new distribution with the hope that it will attract wider applications in engineering, reliability, life testing and other research. In fact, the KMOLL distribution can provide better fits than other models.
Let g(x) and G(x) denote the pdf and cdf of the baseline model. Cordeiro and de Castro (2011) defined the cdf of the K-G family byThe corresponding pdf of (1.3) is given by where a > 0 and b > 0 are two extra shape parameters whose role are to govern skewness and tail weights. Clearly, for a = b = 1, we obtain the baseline distribution. (1.5) is where α is a scale parameter and the shape parameters a,b,γ and β govern the skewness of (1.6).
A random variable X with the pdf (1.6) is denoted by X ~ KMOLL(a,b,α,γ,β). The survival function, hazard rate function (hrf) and cumulative hazard rate function (chrf) of X are, respectively, given byand Some of the possible shapes of the pdf (1.6) for selected parameter values are illustrated in Figure 1. As seen from Figure 1, the density function can take various forms depending on the parameter values. It is evident that the KMOLL distribution is much more flexible than the MOLL distribution, i.e. the additional shape parameters a and b allow for a high degree of flexibility of the KMOLL distribution. Both unimodal and monotonically decreasing shapes appear to be possible.
Plots for the hrf of the KMOLL distribution for several parameter values are displayed in Figure 2. Figure 2 shows that the hrf of the KMOLL distribution can be bathtub, upside down bathtub (unimodal), increasing or decreasing. This attractive flexibility makes the hrf of the KMOLL useful and suitable for non-monotone empirical hazard behaviors which are more likely to be encountered or observed in real life situations.
We now state a useful expansion for the KMOLL density. Using the binomial expansion, the pdf of the KMOLL reduces towhere
The importance of the KMOLL distribution is that it contains as special sub-models several well-known distributions. Table 1 lists the special distributions related to KMOLL distribution.
Sub-models of the KMOLL distribution
The rest of the article is outlined as follows. In Section 2, we obtain the quantile function, shapes, skewness, kurtosis, moments, moment generating functions, Rényi entropies, reliability function and order statistics of X. Certain characterizations are presented in Section 3. The maximum likelihood estimates (MLEs) of the model parameters are obtained in Section 4. An application to real data set is considered in Section 5. Finally, Section 6 provides some concluding remarks.
2. The KMOLL Properties
In this section, we investigate mathematical properties of the KMOLL distribution including quantile function, skewness, kurtosis, shapes of functions, moments, the Rényi and Shannon entropies, reliability and order statistics.
2.1. Quantile function
Quantile functions are in widespread use in statistics and often find representations in terms of lookup tables for key percentiles. Let X ~ KMOLL(a,b,α,γ,β). The quantile function say Q(u) is defined by inverting F(x) in (1.5) asThe effect of the shape parameters a,b,α, γ, β, on the skewness and kurtosis can be considered based on quantile measures. There are many heavy tailed distributions for which this measure is infinite. So, it becomes uninformative precisely when it needs to be. The Bowley’s skewness is based on quartiles: and the Moors’ kurtosis is based on octiles: where Q(.) represents the quantile function of X. These measures are less sensitive to outliers and they exist even for distributions without moments. Skewness measures the degree of the long tail and kurtosis is a measure of the degree of tail heaviness. When the distribution is symmetric, S = 0 and the when the distribution is right (or left) skewed, S > 0 or (S < 0). As K increases, the tail of the distribution becomes heavier.
2.2. Moments and moment generating function
Some of the most important features and characteristics of a distribution can be studied through moments (e.g. tendency, dispersion, skewness and kurtosis). Now we obtain ordinary moments and the moment generating function of the KMOLL distribution. The ordinary moments E(X)n = μ′n, n = 1,2,..., of the KMOLL distribution can be obtained, using (1.7), aswhere . Here, gMOLL(.) and GMOLL(.) are the pdf and cdf of the MOLL distribution, respectively. Then, the integral part in (2.1) is defined as where QMOLL(.) is the quantile function of the MOLL distribution for 0 < u< 1. Then, we obtain where B(.,.) is the beta function.
Further, the central moments (μn) and cumulants (κn), n = 1, 2,..., of the KMOLL distribution can be obtained fromHere, κ1 = µ′1, , , etc.
The skewness and kurtosis are also computed from the second, third and fourth cumulants. Table 2 gives moments, skewness, and kurtosis of the KMOLL distribution for some parameter values.
|KMOLL(a, b, γ, α, β)||μ′1||μ′2||μ′3||μ′4||S||K|
Moments, skewness and kurtosis of the KMOLL distribution for some parameter values
Table 2 indicates that the skewness value can be positive and negative, also close to zero. Hence, the KMOLL distribution can be right-skewed, left-skewed or symmetric.
Figure 3 also depicts plots for the skewness and kurtosis coefficients related to additional parameters. In the figure, a parameter decreases while other parameters are kept fixed. These plots indicate that both measures can be very sensitive on these shape parameters. Thus, indicating the importance of the proposed distribution.
The moment generating function (mgf) is widely used as an alternative way to analytical results compared with working directly with pdf and cdf. The mgf of X isor another representation for M(t) can be obtained using (1.7) where . Then, the integral part in (2.2) is given as where QMOLL(.) is the quantile function of the MOLL distribution for 0 < u < 1. From the Maclaurin expansion, we obtain .
Then we obtain
The pdf of the KMOLL model is decreasing or unimodal. In order to investigate the critical points of its density function, its first derivative with respect to x isThere may be more than one root to (2.4). If x= x0 is a root of (2.4), then it corresponds to a local maximum If df(x) / dx > 0 for all x < x0 and df (x)/dx < 0 for all x > x0. It corresponds to a local minimum if df (x)/dx < 0 for all x < x0 and df (x)/dx > 0 for all x> x0. It corresponds to a point of inflexion if either df (x)/dx > 0 for all x ≠ x0 or df (x) /dx < 0 for all x ≠ x0.
The entropy of a random variable X with density function f(x) is a measure of variation of the uncertainty. Two popular entropy measures are the Rényi and Shannon entropies (Rényi (1961). Shannon (1951)). Here. we derive expressions for the Rényi and the Shannon entropies of the KMOLL distribution. The Rényi entropy of a random variable with pdf f(x) is defined asfor δ > 0 and δ ≠ 1. Then, we can write where and .
The integral part of (2.5) isThen the Rényi entropy of the KMOLL distribution is given by The Shannon entropy plays a similar role as the kurtosis measure in comparing the shapes of various densities and measuring heaviness of tails. The Shannon entropy of a random variable X is defined by. E⌊−log f(x)⌋. It is the special case of Rényi entropy when δ > 1. The Shannon entropy of the KMOLL distribution is To obtain three expectations terms given above. We define and compute Here we have where 2F1 is the generalized hypergeometric function defined by and (a)k = a(a + 1)…(a + k − 1) denotes ascending factorial.
Similarly, the following expectations are defined for (12) as and . Here. C is Euler’s constant (Nadarajah et al. 2012).
In the context of reliability. the stress-strength model describes the life of a component which has a random strength X1 that is subjected to a random stress X2 The component fails at the instant that the stress applied to it exceeds the strength. and the component will function satisfactorily whenever X1 > X2. Hence, R = Pr(X2 < X1 ) is a measure of component reliability. Here. we obtain the reliability R when X1 ~ KMOLL(a1,b1,α,γ,β) and X2 ~ KMOLL(a2,b2,α,γ,β) are independent random variables. Probabilities of this form have many applications especially in engineering concepts.
Let fi and Fi denote the pdf and cdf Xi for i = 1,2,…,. Then, the reliability function for the KMOLL distribution is given byThe cdf of X2 and the pdf of X1 are obtained as After some algebra, we arrive at where and .
2.6. Order statistics
Order statistics make their appearance in many areas of statistical theory and practice. They enter in the problems of estimation and hypotheses testing in a variety of ways. Therefore, we now discuss some properties of the order statistics for the proposed class of distributions. Let Xi:n denote the ith order statistic. Nadarajah et al. (2012) obtained the general results for the Kumaraswamy-G distribution. We use the results about the pdf fi:n(x) of the ith order statistic. Then. we can give the pdf fi:n(x) for a random sample X1, X2,…,Xn from the KMOLL distribution. It is well-known thatfor i = 1, 2,..., n. Using the binomial expansion in the last equation. We obtain The pdf in (2.7) can also be defined as where and .
This section deals with various characterizations of KMOLL distribution. These characterizations are based on: (i) the ratio of two truncated moments; (ii) the hazard function and (iii) certain functions of the random variable. It should be mentioned that for characterization (i) the cdf need not have a closed form. We present our characterizations (i) − (iii) in three subsections.
3.1. Characterizations based on ratio of two truncated moments
In this subsection, we present characterizations of KMOLL distribution in terms of a simple relationship between two truncated moments. This characterization result employs a theorem due to (Glänzel. 1987). see Theorem 3.1 below. Note that the result holds also when the interval H is not closed. Moreover, as mentioned above. It could be also applied when the cdf F does not have a closed form. As shown in (Glänzel, 1990). This characterization is stable in the sense of weak convergence.
Let (Ω, F, P) Ω, be a given probability space and let H = [d,e] be an interval for some d < e (d = −∞, e = ∞ might as well be allowed). Let X: Ω → H be a continuous random variable with the distribution function F and let g and h be two real functions defined on H such thatis defined with some real function ξ. Assume that the equation g, h ∈ C1(H), ξ ∈ C2 (H) and F is twice continuously differentiable and strictly monotone function on the set H. Finally, assume that the equation ξh = g has no real solution in the interior of H. Then, F is uniquely determined by the functions g, h and ξ particularly where the function s is a solution of the differential and C is the normalization constant. such that .
Let X : Ω → (0, ∞) be a continuous random variable and let and g(x) = h(x)(xγ + βαγ) −1 for x > 0. The random variable X has pdf (1.6) if and only if the function ξ defined in Theorem 3.1 has the form
Let X be a random variable with pdf (1.6). Then,and and finally Conversely, if ξ is given as above. Then and hence Now, in view of Theorem 3.1. X has density (1.6).
Let X : Ω → (0, ∞) be a continuous random variable and let h(x) be as in Proposition 3.1. The pdf of X is (6) if and only if there exist functions g and ξ defined in Therorem 3.1 satisfying the differential equationThe general solution of the differential equation in corollary 3.1 is where D is a constant. Note that a set of function satisfying the above differential equation is given in Proposition 3.1 with D = 0 However, it should be also noted that there are other triplets (h,g,ξ) satisfying the conditions of Theorem 3.1.
For b = 1, (Mendoza et al., 2016). we let h(x) = g(x)[xβ + αβ] −1 with g(x) = x−β(a−1). Then for x > 0.
The differential equation and general solution in this case are. respectively.and
3.2. Characterization based on hazard function
It is known that the hazard function. hF. a twice differentiable distribution function, F, satisfies the first order differential equationFor many univariate continuous distributions. this is the only characterization available in terms of the hazard function. The following characterization establishes a characterization of KMOLL distribution which is not of the above trivial form.
Let X : Ω → (0, ∞) be a continuous random variable. The pdf of X is (1.6) if and only if its hazard function hF (x) satisfies the differential equationx > 0. with the initial condition hF(0) = 0 for aγ > 1.
If X has pdf (1.6) then clearly the above differential equation holds. Now, if it holds, thenor which is the hazard function of the KMOLL distribution.
For a = b = 1 (special case of (1.4)). we have the following simple differential equation
3.3. Characterization based on certain functions of the random variable
The following propositions have already appeared in (Hamedani, 2013). so we will just state them here which can be used to characterize KMOLL distribution.
Let X : Ω → (d, e) be a continuous random variable with cdf F. Let ψ(x) be a differentiable function on (d, e) with limx→d+ ψ(x) = 1. Then for δ ≠ 1.If and only if
It is easy to see that for certain functions. e.g., , and (d, e) = (0,∞).
Proposition 3.3 provides a characterization of KMOLL distribution. Clearly there are other suitable functions ψ. We chose the above one for simplicity.
4. Maximum Likelihood Estimation
Several approaches for parameter estimation have been proposed in the literature but the maximum likelihood method is the most commonly employed. Here we consider estimation of the unknown parameters of the KMOLL distribution by the method of maximum likelihood. Let x1, x2,..., xn be observed values from the KMOLL distribution with parameters a,b,γ,α and β. The log-likelihood function for (a,b,γ,α,β) is given bywhere .
The derivatives of the log-likelihood function with respect to the parameters a,b,γ,α and β are given by respectively.The MLEs of (a, b, γ, α, β), say (â, , , , ), are the simultaneous solutions of the equations , , , and . Maximization of the likelihood function can be performed by using nlm or optimize in R statistical package.
5. An Illustrative Application
In this section, we use a real data set to compare the fits of the KMOLL distribution with MOLL, LL and Weibull Fréchet (WFr) (Afify et al., 2016c) distributions. We will use a data set consists of 63 observations of the strengths of 1.5 cm glass fibres (Smith and Naylor, 1987), originally obtained by workers at the UK National Physical Laboratory. Unfortunately, the measurement units are not given in their paper. We estimate the unknown parameters of the distributions by the maximum likelihood. Then, we provide the values of the following statistics: Akaike Information Criterion (AIC), Consistent Akaike Information Criterion (CAIC) and Bayesian Information Criterion (BIC).
In general, the smaller the values of these statistics, the better the fit to the data. Table 3 lists the MLEs of the parameters and the values of AIC, CAIC and BIC statistics.
|Distribution||Estimated Parameters (Standard Error)||AIC||CAIC||BIC|
|KMOLL(a, b, γ, α, β)||1.8355 (0.096)||0.0028 (0.002)||47.4236 (13.307)||0.0588 (0.030)||0.2786 (0.095)||28.0861||29.1387||38.8018|
|MOLL(γ, α, β)||2.3267 (1.289)||0.0353 (0.154)||7.9260 (0.873)||51.5799||51.9867||58.0093|
|LL(γ, α)||1.5262 (0.041)||7.9260 (0.873)||49.5799||49.7799||53.8662|
|WFr(α, β, a, b)||0.3865 (0.799)||0.2436 (0.285)||1.4762 (4.782)||16.8561 (20.485)||39.0||47.6||42.4|
MLEs and the values of AIC, CAIC and BIC statistics
Based on Table 3, it is clear that KMOLL distribution provides the overall best fit and therefore could be chosen as the more adequate model than other models for explaining the data set. Table 4 gives Cramer-von Misses (W) and Anderson Darling statistics (A) for the three models which are the KMOLL, MOLL and LL distributions. More information is provided by a histogram of the data given in Figure 4. Fitted lines in Figure 4 represent the KMOLL, MOLL, LL and WFr distributions. Figure 5 shows empirical cdf and the fitted cdfs. Finally, we give Q-Q plots for all fitted models. The figures also reveals that the KMOLL fits the data very well.
|KMOLL(a, b, γ, α, β)||0.0181403||0.127219|
|MOLL(γ, α, β)||0.4969404||2.748973|
Cramer-von Misses and Anderson Darling statistics
In this paper. we introduce a five-parameter distribution called the Kumaraswamy Marshal-Olkin log-logistic (KMOLL) distribution. Interestingly. our proposed model has increasing. upside-down bathtub and bathtub shaped hazard rate function. A study on the mathematical properties of the new distribution is presented. We obtain the moment generating function. ordinary moments. skewness. kurtosis. hazard and survival functions. The estimation of the model parameters is done via maximum likelihood method. We also provide a numerical example of our findings. We hope that the proposed model may attract applications in survival analysis and customer lifetime duration etc.
Cite this article
TY - JOUR AU - Selen Cakmakyapan AU - Gamze Ozel AU - Yehia Mousa Hussein El Gebaly AU - G. G. Hamedani PY - 2018 DA - 2018/03 TI - The Kumaraswamy Marshall-Olkin Log-Logistic Distribution with Application JO - Journal of Statistical Theory and Applications SP - 59 EP - 76 VL - 17 IS - 1 SN - 2214-1766 UR - https://doi.org/10.2991/jsta.2018.17.1.5 DO - https://doi.org/10.2991/jsta.2018.17.1.5 ID - Cakmakyapan2018 ER - |
Received 24 March 2016; accepted 25 April 2016; published 28 April 2016
In this work direct mathematical formulae for measuring the full-energy peak efficiency of HPGe well-type detector are found and the values of the measured efficiencies are compared with the published works of the experimental and theoretical old methods which have a good agreement. In this new approach the path length d(θ, ϕ) is derived as a function in the polar angle θ, and the azimuthal angle ϕ. This will reduce the mathematical formulae to an easiest and compact shape. In low level gamma-ray spectroscopy, NaI(Tl) well-type scintillation detectors are extremely useful since they offer almost 4π solid angle detection. The efficiency of such detectors can be obtained, using experimental - , semi-empirical and Monte Carlo methods - . The present work relies on a direct mathematical method reported by Selim and Abbas - , which is simple and accurate. It was used successfully in the calibration of well-type and cylindrical detectors using point, plane and volumetric sources. The efficiency is calculated in terms of the geometrical parameters of the source-detector configuration and the photopeak attenuation coefficient of the incident photon. Here, we consider 8 gamma-ray energies based on three different point sources.
2. Mathematical Treatment
In our analysis, we consider only an axial point source P, as shown in Figure 1. The efficiency in this case is given by
where, d is the path length traveled by a photon through the detector active medium, μ is the attenuation coefficient of the detector material, and fatt is the attenuation factor, which is given by
Here, μn is the attenuation coefficient of the nth absorber for the gamma-ray photon , and δn is the path length of the gamma-ray photon through the nth absorber read .
The polar angles, θ, are given by (see Figure 1):
Figure 1. The well type detector configuration. The point source P, is placed on the axis of the well detector.
The distance is the photon traveled distance (path length) inside the detector active volume. Several cases arise, let the gamma-ray photons enter from the inner bottom of the well. If they emerge from the outer bottom of the well, then is given by:
But, if they emerge from the side of the outer well, reads:
Next, when the gamma-ray photons enter from the inner side of the well and emerge from the bottom of the outer well, we have
If they emerge from the side of the outer well, the photon path length is given by
Finally, when the gamma-ray photon enters from the upper surface of the well and emerge from the bottom of the outer well, then
If emerging from the side of the outer well, we have
The resulting efficiency depends on the magnitudes of the polar angles. This results in the following three cases. When θ3 > θ2 > θ4 > θ1, using Equation (1) the efficiency is given by
Similarly, if θ3 > θ2 > θ1 > θ4
with given by Equation (14), but j = 1, 2, 4, 6 Finally, for θ3 > θ4 > θ2 > θ1, we have
3. Results and Discussions
Using the above formalism, the full-energy peak efficiency and the photopeak attenuation coefficient of 3'' × 3'' NaI(Tl) well-type scintillation detector are calculated for in the case of point sources placed outside the detector well. The 3'' × 3'' NaI(Tl) well-type model number is 802-Canberra. The dimensions of this detector are, outer radius, 3.81 cm, cavity radius, 0.858 cm, outer height, 7.62 cm, and cavity depth, 4.987 cm. The full-energy peak efficiencies are calculated using the present work and compared with those obtained by theoretical and experimental data, for various values of the source to detector distance, h', above the detector starting approximately at 20 cm up to 50 cm in a step of 5 cm.
The results of the calculations are given in Tables 1-7, along with the experimental data and the theoretical calculations based on the transfer method presented in Ref. . It is clear that our theoretical results are identical to the experimental data. The error in the theoretical results of reference based on the transfer method is approximately 1%. So, our present formalism consistently produces more accurate results than the other methods.
Comparing the various data in Tables 1-7, we can see that the full-energy peak efficiency decreases as the distance h' increases for the same energy of the gamma-rays. Obviously, it can be seen that the efficiency depends on the position of the source as will as it depends on the activity of the radioactive source. In Figure 2, we summarize this point by plotting the full-energy peak efficiency versus the distance h' for the eight energies of the gamma-rays. We notice the tendency that the efficiency decreases as the energy increases for the same distance as seen in Figure 2. In addition, the efficiency curves are converging at high energies. To explain these observations, we calculate the energy dependence of the attenuation coefficient.
In our formalism, the energy of the photon only enters through the dependence of the attenuation coefficient. From Equation (14), we see that the function decreases as increases. So for the efficiency to decrease with increasing gamma energy, the attenuation coefficient should diminish. We confirm this analysis by plotting the photopeak attenuation coefficient (μP) against the energy of the gamma-rays in Figures 3-9 for
Table 1. Present the photpeak efficiencies, εp, (sr) for present work, experimental and theoretical data for distance h' = 20.442 cm.
Table 2. As in Table 1, for distance h' = 25.461 cm.
Table 3. As in Table 1, for distance h' = 30.537 cm.
Table 4. As in Table 1, for distance h' = 35.590 cm.
Table 5. As in Table 1, for distance h' = 40.600 cm.
Table 6. As in Table 1, for distance h' = 45.666 cm.
Table 7. Present the photopeak efficiencies, εp, (sr) for present work and experimental data for distance h' = 50.717 cm.
Figure 2. The relation between the full-energy peak (photopeak) efficiency (εp) and the source to detector distance (h').
Figure 3. The relation between the photopeak attenuation coefficient (μp) and the photon energy for h' = 20.442 cm with fitting polynomial curve.
Figure 4. Similar to Figure 3 but for h' = 25.461 cm.
Figure 5. Similar to Figure 3 but for h' = 30.537 cm.
Figure 6. Similar to Figure 3 but for h' = 35.59 cm.
Figure 7. Similar to Figure 3 but for h' = 40.6 cm.
Figure 8. Similar to Figure 3 but for h' = 45.666 cm.
Figure 9. Similar to Figure 3 but for h' = 50.717 cm.
the different values of the distance considered so far. The figures also include polynomial fits up to 6th order. A common trend in μP is that it decreases rapidly at lower energies and approximately levels of at high energies about 0.05 cm−1. The sharp decrease in μP is reflected in the large change of the efficiency at low energies. However, the negligible variation of the attenuation coefficient with the gamma energy plays a smaller rule in determining the efficiency at high energies. Thus the efficiency curves converge as indicated in Figure 2.
The full-energy peak efficiency (εp) of 3'' × 3'' NaI(Tl) well-type scintillation detector, using axial point sources is calculated and compared with both the experimental and the theoretical data in Ref. . In addition, the photopeak attenuation coefficients (μp) have been calculated as a function of the gamma-ray photon energy. Our formalism gives more accurate data than that of the transfer method presented in Ref. . The dependence of the photopeak attenuation coefficient on the gamma energy is used to explain the variations in the efficiency as the distance between the point source and the detector changes. From the result it is clear that the full-energy peak efficiency εp decreases by increasing the distance between the detector and the source. |
April 28, 1997
3D SU() + ADJOINT HIGGS THEORY AND
FINITE TEMPERATURE QCD
K. Rummukainen333 and
Theory Division, CERN, CH-1211 Geneva 23,
Department of Physics, P.O.Box 9, 00014 University of Helsinki, Finland
Institut für Theoretische Physik, Philosophenweg 16,
D-69120 Heidelberg, Germany
Fakultät für Physik, Postfach 100131, D-33501 Bielefeld, Germany
We study to what extent the three-dimensional SU() + adjoint Higgs theory can be used as an effective theory for finite temperature SU() gauge theory, with . The parameters of the 3d theory are computed in 2-loop perturbation theory in terms of . The perturbative effective potential of the 3d theory is computed to two loops for . While the Z() symmetry probably driving the 4d confinement-deconfinement phase transition (for ) is not explicit in the effective Lagrangian, it is partly reinstated by radiative effects in the 3d theory. Lattice simulations in the 3d theory are carried out for , and the static screening masses relevant for the high-temperature phase of the 4d theory are measured. In particular, we measure non-perturbatively the correction to the Debye screening mass. We find that non-perturbative effects are much larger in the SU(2) + adjoint Higgs theory than in the SU(2) + fundamental Higgs theory.
Corresponding to the known forces of nature there are two important finite temperature phase transitions in elementary particle matter: the QCD phase transition at MeV and the EW (electroweak) phase transition at GeV. The former has been studied intensely with numerical lattice Monte Carlo simulations but, due to the difficulties associated with treating dynamical quarks, conclusive statements about the properties of the physical QCD transition cannot yet be made. In contrast, for the EW case the problem can be essentially solved by a combination of analytic and numerical means: first deriving by a perturbative computation – a 3d effective theory for the full 4d finite theory (“dimensional reduction”) and then solving this confining non-perturbative 3d theory by numerical means –. The effective field theory approach has been intensively used for computations in high temperature QCD –, as well.
In fact, there has been no doubt that dimensional reduction of QCD would work well at very high temperatures in the QCD plasma phase, , in the sense of the 3d theory giving correctly the static correlation functions of the theory. The situation is quite different in the phase transition region : the effective finite temperature gauge coupling is becoming large so that the mass hierarchy needed for the construction of a simple local effective theory is lost. In particular, quarks play dynamically a crucial role and it seems that an effective theory using constant (in imaginary time) field configurations as essential degrees of freedom cannot be accurate (quarks are antiperiodic and are thus integrated out in this effective theory, their effect appearing only in the 3d parameters). However, it still seems well motivated to study how far one can actually go with dimensionally reduced QCD towards the phase transition region and the purpose of this paper is to do this. In comparison with earlier work – we go further firstly by determining the two continuum parameters of the effective theory in terms of (quark masses are neglected) using 2-loop perturbation theory and the techniques developed in . Secondly, we are now able to extrapolate the lattice results to the continuum limit, using the lattice–continuum relations derived in [11, 24] (see also ). Our conclusions will thus be different from those of the previous lattice studies – for SU(2).
Quite independent of its use as a finite effective theory, the 3d SU(2)+adjoint Higgs theory is interesting because it has monopoles [26, 27] which remove the photon from the physical spectrum and replace it by a pseudoscalar with the mass
where is the perturbative mass of the vector excitation in the broken phase (in a gauge invariant analysis does not correspond to a physical state). A measurement of the photon mass with gauge invariant operators will permit one to make statements about the monopole density in the broken and symmetric phases. These questions have recently been addressed in .
The paper is organized as follows. In Sec. 2 we discuss the derivation of the effective 3d theory. In Sec. 3 we perform some perturbative estimates within that theory. In Sec. 4 we address the role of the Z() symmetry in the effective 3d theory, and in Sec. 5 we consider the 3d lattice results. The implications of the analysis for the 4d finite temperature gauge theory are in Sec. 6, and the conclusions in Sec. 7.
2 Relating the 4d and 3d theories
Finite QCD (with colours for the moment) is defined by the action
and where the gluon field is periodic (quark field antiperiodic) in with period . In the matrix representation
After renormalisation (we use the scheme), becomes scale dependent, and to 1-loop
Some useful group theoretical relations for the SU(N) generators are given in Appendix A.
Upon dimensional reduction, the field becomes an adjoint Higgs field and the general form of the super-renormalizable Lagrangian of the 3d effective theory can be written down:
The reduction process does not generate terms proportional to . Operators of higher dimensions are parametrically of the form and can be neglected relative to the retained term as long as (their contributions to the correlators we will measure are also of higher order).
The effective theory thus depends on three dimensionful couplings: , , . Instead of these one can use one dimensionful scale and two dimensionless parameters, chosen as
The 3d theory is super-renormalizable and are renormalisation scale independent while is of the form
The process of dimensional reduction now implies finding the relation between the physical parameters of finite temperature QCD and . The Lagrangian parameters of QCD are after renormalisation ; the physical parameters are hadron masses (one mass to set the scale and the rest as dimensionless mass ratios). As hadron masses are entirely non-perturbative, it is conventional to use and define as the scale for which diverges when evolved to smaller scales.
We will derive the parameters of the 3d theory so that the relative errors are of the order . This requires a 1-loop calculation for the gauge coupling, but for the mass parameter and the scalar coupling one needs a 2-loop derivation. The actual reliability of this calculation is to be discussed below.
We take flavours of massless quarks, although the dependence on mass thresholds could also, in principle, be included.
At this level the scale is unspecified; thus a 2-loop derivation of the effective theory, constituting a resummation of the perturbative series and establishing the scale at which the 3d parameters are to be evaluated, is needed. Indeed, the 3d couplings are scale independent (note that the 3d scale dependence in eq. (2.8) is of order and thus does not yet enter at this level of the 4d3d reduction) so that the dependence of the 2-loop result must be the following (we only discuss so that is irrelevant):
and the are constants to be found.
The derivation of the parameters can be most easily made using the background field gauge. Calculating to 1-loop the contribution from the modes to the zero mode correlators, one gets the relations between the 3d and 4d fields:
where is the gauge parameter and
is the standard thermal scale arising in perturbative reduction in the scheme [8, 17]. The 3d gauge coupling can be read directly from the gauge independent normalization factor in this gauge . For the other parameters, one needs the 2-loop correlators at zero momenta of the -fields in the 4d and 3d theories, to separate the contributions coming from the modes. These can be most easily derived with the effective potential. The full 4d 1-loop effective potential can be read from eqs. (4.2) (with ), (5.2) of , and the 2-loop effective potential from eqs. (4.3), (5.4) of . The 1-loop potential is gauge independent, but the 2-loop potential in these equations corresponds to , since the extra rescalings added in eqs. (3.17), (3.18) of vanish for that gauge parameter. For the mass parameter one has to subtract the contribution from 3d, which has to be calculated separately, but for there is no 3d contribution in the 4d 2-loop effective potential and thus the coefficient of the quartic term gives directly the contribution. Finally, to get the 2-loop results for the 3d parameters, one needs the terms arising when the rescalings in (2.17) combine with the 1-loop results. Alternatively, the 2-loop mass parameter can be read directly from . Note that the 2-loop cubic terms in the 4d effective potential are not used in the derivation of the 3d parameters and are not reproduced by the 3d theory (they are higher order contributions when ).
The computation described above gives
Note that for the computation of the free energy of QCD in the symmetric phase to order through a 3d theory , one only needs ; the constants and give higher order contributions.
Another useful representation of eqs. (2.13)–(2.15) can be derived by choosing different renormalization scales for different parameters in a way that 1-loop corrections to the 3d parameters vanish. In general, the solution of is
Secondly, for one obtains
where for ) is obtained from (2.23) with . This equation gives quantitatively the relation between the 4d theory variables and the 3d effective theory variable . For one has
Finally, concerning note that the RHS of eqs. (2.13)–(2.15) only depends on and . Eliminating one obtains a -independent relation between the dimensionless scale independent variables . To leading order so that . The complete relation is
For these have the simple forms
With the use of known lattice data for the phase transition in pure gauge theories in 4d for one can define the value of corresponding to the critical temperature. We have: for SU(2) and for SU(3) according to . Thus, for the value of corresponding to is about ( for SU(2) and for SU(3)). To have a feeling of the accuracy required for the assessment of the lattice results below, the leading value of is 0.1126 to which a 22.5% 2-loop correction 0.0253 is to be added. One might then estimate that the next (omitted) term is roughly 5% so that the theoretical result is .
Quite surprisingly, the power series defining the mapping of the 4d parameters to the 3d ones seems thus to be quite convergent even at the critical temperature. However, this does not prove that the 3d theory can adequately describe the confinement-deconfinement phase transition at high temperatures. There is another important criterion for the applicability of dimensional reduction, namely that the typical 3d mass scale must be much smaller than , since only then the integration out of the non-zero Matsubara modes is self-consistent. As we will see, it is this point which does not allow an effective 3d description near the critical point.
Summarizing, to the extent that finite QCD can be regarded as an SU(N) gauge theory with massless quarks characterized by the physical quantities , a 3d effective theory given by the super-renormalizable Lagrangian (2.6) with the couplings (eqs. (2.7)) can be derived. The relation between these two sets is in eqs. (2.24), (2.26), (2.27) and (2.28).
Note the crucial difference in comparison with SU(N) + fundamental Higgs theories, in which there is a Higgs potential with two parameters already at the tree level. Then (for ) the variable (the Higgs self-coupling in the effective theory) is essentially the zero temperature Higgs mass and (the scalar mass in the effective theory) is . The whole plane corresponds to some physical SU(2)+Higgs finite theories. For pure SU(2), in contrast, both and are given by and for fixed only one curve corresponds to a physical 4d theory. Assume the effective theory has a phase transition along some curve , which we shall soon determine with lattice Monte Carlo simulations. Then this transition could correspond to a physical 4d transition only if it intersects the curve in a region where the derivation of the effective theory is reliable.
3 Perturbation theory in 3d for
The aim now is to study the phase structure of the 3d theory defined by the Lagrangian (2.6) on the plane of its variables . On the full quantum level the study has to be based on gauge invariant operators and will be carried out by numerical means in Sec. 5; only this gives reliable answers. However, it is also quite useful to fix the gauge and study the problem perturbatively. On the tree level the answer then is obvious: there is a symmetric phase for () at all and a broken phase for . To get a more accurate result, one shifts the field by
obtains 3=1+2 scalars with masses
two massive vectors with mass
and one massless vector. The 1-loop potential in the Landau gauge is
Let us consider first the high temperature limit of the 1-loop effective potential. It corresponds to the case . The leading terms are:
Two degenerate states are obtained when the last term vanishes. From this one finds that
where is the 3d interface tension divided by . Further, the upper and lower metastability points are given by
The contributions to the 2-loop potential from vectors ( ), ghosts ( ) and scalars ( ) are (without a factor )
The sunset function is
The use of 2-loop effective potential allows the construction of the critical curve beyond 1-loop level. The numerical computation is shown in Fig. 1.
4 Z() symmetry in the 3d theory
For , hot SU() gauge theory exhibits a Z() symmetry [33, 31] so that there are equivalent ground states. In one of them , in the others . In the weak coupling limit the field in the other minima thus becomes large. One can associate the QCD phase transition with the breaking of this symmetry so that at high temperatures one sits in one of the equivalent minima, say . As one goes to lower temperatures, the barrier between the minima is becoming lower and at , the symmetry is supposed to get restored. If one wants to apply a 3d theory down to temperatures close to , it is thus important to discuss the role of the Z() symmetry in 3d.
The construction of the super-renormalizable effective 3d field theory in Sec. 2 requires small amplitudes of the adjoint field , . In practice, this means that one works around the minimum in the reduction step, so that the effective theory describes reliably only fluctuations of magnitude around that minimum. Thus, the price one pays for the simple effective theory is that the Z() symmetry is not reproduced by it.
Some remnant of the Z() symmetry nonetheless remains in the 3d theory. The effective theory will on its critical curve still contain metastable states but these are not completely equivalent, e.g., the correlators measured in them are different. As discussed, only the correlators in the symmetric phase () reliably represent 4d physics.
Indeed, within the interval , eq. (3.6) at is seen to be the same as the 4d 1-loop potential in an background:
The effective potential (4) exhibits the Z(2) symmetry of the full 4d theory as the periodicity in with period ; this symmetry is not explicit in the effective theory on the tree level, but it is partly reinstated by radiative effects as seen in eq. (3.6). This is not surprising: in deriving the 4d 1-loop potential one performs a frequency sum over . In deriving the couplings of the 3d theory one performs basically a frequency sum over , and the mode enters when computing the effective potential within the 3d theory. Note also that (3.7) is exactly the same as the leading term in (2.29).
Thus, while the Z(2) symmetry is not explicit in the effective action, the second degenerate minimum is generated radiatively. Even with all higher order corrections in the non-perturbative solution of the 3d theory there will be two (for ) degenerate minima at some values of the 3d parameters, but this will no longer take place at any temperature . Moreover, while these states are completely equivalent in the original 4d theory, the correlators measured in them will be different in the effective theory, the symmetric phase being the physical one. For SU(3) the full 4d theory has 3 equivalent states while in the 3d theory presumably only two of them have the same correlators; these are the unphysical ones corresponding to a large value of the fields.
5 Lattice analysis
The lattice action corresponding to the continuum theory (2.6) is where
is the standard SU(2) Wilson action and, in continuum normalization () and for ,
is the lattice spacing and is the bare mass in the lattice scheme. Renormalisation is carried out so that the physical results are the same as in the scheme with the renormalized mass parameter
The resulting counterterms have been computed in with the result
where . There are no higher order corrections to this relation in the limit . For 1-loop -improvement, see .
Since there are two dimensionless parameters, one more arbitrary choice is possible. In fundamental Higgs theories one often scales the coefficient of the quadratic term to unity; here we choose .
The aim of the simulations is as follows:
Find the critical curve and, in particular, its endpoint . This is done by measuring the distributions of (or some other operator) at fixed , finding a two-state signal and the limit of the value of at which this happens,
Measure the correlator masses on both sides of the transition line and in the cross-over region .
It is important to estimate the required lattice volume and constant (or ). In the broken phase, on the tree level we have at least two relevant mass scales, the adjoint scalar mass from eq. (3.2) and the vector mass from eq. (3.3). Thus, in order to describe accurately the broken phase, we must demand
These convert to
For the symmetric phase the scale is absent, so that the requirement on is not quite as demanding, .
The simulations were performed with a Cray C90 at the Finnish Center for Scientific Computing, and the total cpu-time usage was 760 cpu-hours.
5.3 The phase diagram
We locate the transition line by using as an order parameter. For each lattice size, we define the pseudocritical with the following commonly used methods : (a) maximum of the susceptibility ; (b) minimum of the 4th order Binder cumulant ; and, when the transition is strongly 1st order, (c) the “equal weight” criterion for the probability distribution . For any finite volume, these criteria give different pseudocritical values of , but they all extrapolate to the same limit when . This extrapolation is shown in Fig. 2 for .
The analysis is repeated for several , corresponding to different lattice spacings through eq. (5.3). In our case, as long as the conditions (5.11) are satisfied, we found no appreciable lattice spacing dependence in the results. This is also evident in Fig.2.
Our result for the phase diagram of the continuum SU(2)+adjoint Higgs model is shown in Fig. 3. As already pointed out in , the phase diagram consists of a first-order line which terminates, so that the two “phases” are analytically connected. We find the endpoint to be close to . As can be seen from the figure, the results are independent of well within the statistical accuracy. However, the infinite volume extrapolation is essential, as shown by some large but finite volume points in Fig. 3.
The thick transition line in Fig. 3 has been obtained with a fit to the -extrapolated results, as
where . The fit has been constrained to join the curve and its slope when . We used fractional powers of in the fit since in the perturbative expansion of the critical curve terms like appear. There can also be logarithms in the expansion.
It is interesting that the behaviour of the 3d SU(2) + adjoint Higgs model is quite different from that of the 3d SU(2) + fundamental Higgs theory studied earlier . First, in the fundamental case the value of was smaller than the 2-loop perturbative result for all values of , while in the adjoint case the situation is opposite for (compare Figs. 1, 3). Second, the magnitude of the higher order perturbative or non-perturbative effects is much larger in the adjoint case. To get a feeling of the difference, suppose that the deviation from the perturbative result comes entirely from a non-perturbative energy shift at [35, 12],
where is some constant to be determined numerically. For the fundamental case this was estimated to be at . A first order estimate of the change in the critical value,
gives in the present case
For example, for we have ( times larger than in the fundamental case!) and for , |
All examples in this chapter space planar problems. Accordingly, we use equilibrium problems in the component kind of (Figure) to (Figure). We presented a problem-solving strategy in (Figure) to illustrate the physical meaning of the equilibrium conditions. Now we generalize this strategy in a list of procedures to follow when solving revolution equilibrium difficulties for prolonged rigid bodies. We proceed in 5 practical steps.
You are watching: Find t, the tension in the cable. remember to account for all the forces in the problem.
Problem-Solving Strategy: static EquilibriumIdentify the thing to be analyzed. For some solution in equilibrium, it might be necessary to consider an ext than one object. Identify all pressures acting on the object. Determine the concerns you must answer. Identify the information offered in the problem. In reality problems, some crucial information may be implicitly in the case rather than provided explicitly.Set up a free-body diagram because that the object. (a) pick the xy-reference framework for the problem. Attract a free-body diagram because that the object, consisting of only the forces that action on it. When suitable, represent the forces in terms of their contents in the chosen reference frame. Together you execute this for each force, cross the end the original force so the you carry out not erroneously encompass the same pressure twice in equations. Brand all forces—you will need this for correct computations the net forces in the x– and also y-directions. Because that an unknown force, the direction must be assigned arbitrarily; think the it as a ‘working direction’ or ‘suspected direction.’ The exactly direction is determined by the authorize that you acquire in the final solution. A to add sign
means that the functioning direction is the actual direction. A minus sign
means that the really direction is opposite come the assumed working direction. (b) pick the place of the rotation axis; in various other words, pick the pivot allude with respect to which you will certainly compute torques of acting forces. Top top the free-body diagram, suggest the location of the pivot and also the bar arms of acting forces—you will require this for correct computations the torques. In the an option of the pivot, keep in mind that the pivot have the right to be placed all over you wish, yet the guiding rule is that the best choice will leveling as much as possible the calculate of the net torque along the rotation axis.Simplify and also solve the device of equations because that equilibrium to achieve unknown quantities. At this point, her work entails algebra only. Store in mind the the variety of equations need to be the very same as the variety of unknowns. If the variety of unknowns is bigger than the number of equations, the difficulty cannot it is in solved.
Note that setting up a free-body diagram for a rigid-body equilibrium difficulty is the most vital component in the solution process. Without the exactly setup and also a correct diagram, you will not have the ability to write under correct conditions for equilibrium. Likewise note that a free-body diagram for an extensive rigid body that might undergo rotational activity is various from a free-body diagram for a body that experiences only translational motion (as you observed in the chapters top top Newton’s laws of motion). In translational dynamics, a human body is stood for as that is CM, wherein all pressures on the body room attached and no torques appear. This does not host true in rotational dynamics, where prolonged rigid body cannot be stood for by one suggest alone. The reason for this is the in analyzing rotation, us must identify torques exhilaration on the body, and also torque counts both on the exhilaration force and also on its bar arm. Here, the free-body chart for an extended rigid body helps us identify outside torques.
ExampleThe talk Balance
Three masses are attached to a uniform meter stick, as shown in (Figure). The mass of the meter rod is 150.0 g and the masses come the left of the fulcrum room
uncover the fixed
the balances the system when that is attached at the right finish of the stick, and the typical reaction pressure at the fulcrum when the mechanism is balanced.
Figure 12.9 In a talk balance, a horizontal beam is supported at a fulcrum (indicated through S) and also masses are attached to both political parties of the fulcrum. The system is in static equilibrium once the beam does not rotate. It is well balanced when the beam stays level.
For the arrangement shown in the figure, we identify the following 5 forces acting on the meter stick:
is the weight of mass
is the weight of mass
is the weight of the entire meter stick;
is the load of unknown massive
is the regular reaction pressure at the support suggest S.
We pick a structure of recommendation where the direction that the y-axis is the direction that gravity, the direction that the x-axis is along the meter stick, and the axis of rotation (the z-axis) is perpendicular to the x-axis and passes with the support allude S. In various other words, we pick the pivot in ~ the suggest where the meter stick touches the support. This is a natural selection for the pivot because this point does not move as the pole rotates. Now we are ready to set up the free-body diagram for the meter stick. We show the pivot and also attach 5 vectors representing the five forces follow me the line representing the meter stick, locating the forces with respect come the pivot (Figure). In ~ this stage, we have the right to identify the lever arms the the five forces offered the information detailed in the problem. Because that the three hanging masses, the trouble is explicit around their locations along the stick, however the information around the ar of the weight w is provided implicitly. The vital word right here is “uniform.” We know from ours previous studies that the cm of a uniform rod is situated at the midpoint, so this is wherein we affix the load w, at the 50-cm mark.
Figure 12.10 Free-body diagram because that the meter stick. The pivot is liked at the support allude S.
With (Figure) and also (Figure) for reference, we begin by detect the bar arms the the 5 forces exhilaration on the stick:
When substituting talk values right into this equation, we deserve to omit the torques giving zero contributions. In this means the second equilibrium problem is
We deal with these equations at the same time for the unknown worths
In (Figure), we cancel the g factor and also rearrange the terms to obtain
To find the typical reaction force, we rearrange the terms in (Figure), converting grams to kilograms:
Notice the (Figure) is independent of the value of g. The talk balance may therefore be supplied to measure mass, because variations in g-values on earth’s surface perform not affect these measurements. This is no the instance for a spring balance due to the fact that it measures the force.
Repeat (Figure) making use of the left finish of the meter stick to calculate the torques; that is, by place the pivot in ~ the left end of the meter stick.
316.7 g; 5.8 N
In the following example, we show how to usage the very first equilibrium condition (equation because that forces) in the vector type given through (Figure) and (Figure). We present this systems to show the prominence of a suitable an option of reference frame. Although all inertial reference frames space equivalent and numerical solutions obtained in one structure are the exact same as in any kind of other, one unsuitable choice of reference framework can make the solution quite an extensive and convoluted, vice versa, a wise choice of reference structure makes the equipment straightforward. We present this in the tantamount solution to the same problem. This particular example illustrates an applications of static equilibrium come biomechanics.
ExampleForces in the Forearm
A weightlifter is holding a 50.0-lb weight (equivalent to 222.4 N) v his forearm, as displayed in (Figure). His forearm is positioned at
through respect come his upper arm. The forearm is sustained by a contraction of the biceps muscle, which causes a torque about the elbow. Assuming that the stress and anxiety in the biceps acts along the vertical direction given by gravity, what tension need to the muscle exert to organize the forearm at the position shown? What is the pressure on the elbow joint? Assume the the forearm’s weight is negligible. Offer your final answers in SI units.
Figure 12.11 The forearm is rotated approximately the elbow (E) by a contraction of the biceps muscle, which reasons tension
We determine three forces acting top top the forearm: the unknown pressure
in ~ the elbow; the unknown tension
in the muscle; and also the load
We adopt the framework of referral with the x-axis along the forearm and the pivot at the elbow. The upright direction is the direction of the weight, i m sorry is the exact same as the direction that the upper arm. The x-axis provides an edge
through the vertical. The y-axis is perpendicular to the x-axis. Currently we set up the free-body diagram for the forearm. First, we attract the axes, the pivot, and the 3 vectors representing the three identified forces. Climate we find the edge
and also represent each pressure by that is x– and also y-components, remembering to cross out the original force vector come avoid dual counting. Finally, we label the forces and their bar arms. The free-body diagram because that the forearm is shown in (Figure). At this point, us are ready to set up equilibrium problems for the forearm. Each force has x– and also y-components; therefore, we have two equations for the an initial equilibrium condition, one equation because that each component of the net pressure acting ~ above the forearm.
Notice the in our frame of reference, contributions to the second equilibrium condition (for torques) come only from the y-components that the forces because the x-components that the pressures are every parallel come their lever arms, so that for any kind of of them we have
in (Figure). Because that the y-components we have
in (Figure). Also notice that the torque of the pressure at the elbow is zero since this pressure is attached in ~ the pivot. Therefore the contribution to the net torque comes just from the torques of
and also of
We watch from the free-body diagram the the x-component of the net force satisfies the equation
(Figure) and (Figure) space two equations of the an initial equilibrium problem (for forces). Next, we read from the free-body diagram the the net torque follow me the axis of rotation is
(Figure) is the 2nd equilibrium problem (for torques) for the forearm. The free-body diagram reflects that the lever arms are
at this point, we perform not need to convert inches into SI units, because as long as these devices are continuous in (Figure), lock cancel out. Making use of the free-body chart again, we uncover the magnitudes that the component forces:
When we simplify these equations, we check out that we room left with only two live independence equations because that the two unknown force magnitudes, F and also T, because (Figure) because that the x-component is identical to (Figure) for the y-component. In this way, we obtain the first equilibrium condition for forces
and the 2nd equilibrium condition for torques
The size of tension in the muscle is obtained by solving (Figure):
The pressure at the elbow is acquired by resolving (Figure):
The negative sign in the equation tells us that the actual force at the elbow is antiparallel come the functioning direction adopted for illustration the free-body diagram. In the final answer, we convert the forces into SI systems of force. The price is
Two important concerns here are worth noting. The very first concerns conversion right into SI units, which have the right to be done at the very end the the solution as lengthy as we save consistency in units. The second important issue comes to the hinge joints such as the elbow. In the initial evaluation of a problem, hinge joints should constantly be assumed to exert a pressure in one arbitrary direction, and then you should solve because that all contents of a hinge force independently. In this example, the elbow pressure happens to be vertical due to the fact that the trouble assumes the anxiety by the biceps to it is in vertical as well. Together a simplification, however, is no a general rule.Solution
Suppose we embrace a reference frame with the direction of the y-axis along the 50-lb weight and also the pivot put at the elbow. In this frame, all three forces have only y-components, for this reason we have actually only one equation for the an initial equilibrium problem (for forces). We attract the free-body diagram because that the forearm as displayed in (Figure), denote the pivot, the exhilaration forces and also their bar arms through respect come the pivot, and also the angles
that the forces
(respectively) make v their lever arms. In the an interpretation of torque given by (Figure), the angle
is the direction edge of the vector
counted counterclockwise from the radial direction the the bar arm that constantly points away from the pivot. Through the exact same convention, the edge
is measure up counterclockwise indigenous the radial direction of the bar arm to the vector
excellent this way, the non-zero torques room most easily computed by straight substituting right into (Figure) as follows:
Figure 12.13 Free-body diagram for the forearm because that the identical solution. The pivot is situated at point E (elbow).
The 2nd equilibrium condition,
can be currently written as
From the free-body diagram, the an initial equilibrium condition (for forces) is
(Figure) is identical to (Figure) and also gives the an outcome
We see that this answers are the same to ours previous answers, however the second choice for the framework of reference leads come an tantamount solution that is simpler and quicker since it go not require that the pressures be resolved right into their rectangle-shaped components.
ExampleA Ladder Resting against a Wall
A uniform ladder is
long and also weighs 400.0 N. The ladder rests versus a slippery upright wall, as shown in (Figure). The inclination angle in between the ladder and also the unstable floor is
discover the reaction forces from the floor and from the wall on the ladder and the coefficient of revolution friction
at the interface of the ladder through the floor that avoids the ladder from slipping.
We have the right to identify four pressures acting ~ above the ladder. The an initial force is the typical reaction pressure N from the floor in the increase vertical direction. The second force is the revolution friction force
command horizontally follow me the floor towards the wall—this force stays clear of the ladder native slipping. These two pressures act ~ above the ladder at its contact point with the floor. The 3rd force is the weight w of the ladder, attached at its CM situated midway in between its ends. The fourth pressure is the typical reaction pressure F native the wall in the horizontal direction away from the wall, attached in ~ the contact allude with the wall. There are no other forces because the wall surface is slippery, which method there is no friction between the wall surface and the ladder. Based on this analysis, we embrace the framework of referral with the y-axis in the upright direction (parallel to the wall) and the x-axis in the horizontal direction (parallel to the floor). In this frame, each force has one of two people a horizontal ingredient or a upright component yet not both, i m sorry simplifies the solution. We pick the pivot at the contact allude with the floor. In the free-body diagram for the ladder, we suggest the pivot, all four forces and their bar arms, and the angles between lever arms and the forces, as presented in (Figure). Through our an option of the pivot location, over there is no torque either from the typical reaction force N or from the static friction f since they both act in ~ the pivot.
is the speak of the load w and
is the torque of the reaction F. From the free-body diagram, we recognize that the bar arm the the reaction at the wall is
and the bar arm that the weight is
through the help of the free-body diagram, we determine the angle to be supplied in (Figure) because that torques:
for the torque indigenous the reaction pressure with the wall, and also
because that the torque because of the weight. Currently we are prepared to use (Figure) come compute torques:
We achieve the normal reaction pressure with the floor by fixing (Figure):
The size of friction is acquired by fixing (Figure):
The coefficient of static friction is
The net pressure on the ladder at the contact allude with the floor is the vector sum of the typical reaction native the floor and the revolution friction forces:
We need to emphasize here two general observations of useful use. First, notification that when we choose a pivot point, over there is no expectation the the device will actually pivot roughly the favored point. The ladder in this instance is no rotating in ~ all yet firmly stands on the floor; nonetheless, that contact allude with the floor is a good choice for the pivot. Second, notification when we usage (Figure) because that the computation of individual torques, we carry out not need to resolve the pressures into their normal and parallel contents with respect come the direction of the bar arm, and we carry out not need to think about a sense of the torque. As long as the edge in (Figure) is appropriately identified—with the assist of a free-body diagram—as the angle measured counterclockwise from the direction that the lever arm to the direction the the force vector, (Figure) offers both the magnitude and also the sense of the torque. This is since torque is the vector product that the lever-arm vector crossed with the force vector, and (Figure) expresses the rectangle-shaped component the this vector product along the axis of rotation.Significance
This result is independent of the length of the ladder since L is cancelled in the second equilibrium condition, (Figure). No matter exactly how long or quick the ladder is, as long as its weight is 400 N and also the angle through the floor is
our results hold. But the ladder will certainly slip if the network torque becomes an unfavorable in (Figure). This happens for part angles as soon as the coefficient of static friction is not great enough to avoid the ladder native slipping.
For the situation described in (Figure), recognize the worths of the coefficient
of revolution friction because that which the ladder start slipping, given that
is the angle the the ladder renders with the floor.
is supported by hinges A and B so the the door deserve to swing around a vertical axis passing with the hinges (Figure). The door has actually a broad of
and the door slab has actually a uniform massive density. The hinges are put symmetrically in ~ the door’s sheet in together a way that the door’s weight is same distributed between them. The hinges are separated by distance
find the forces on the hinges once the door rests half-open.
From the free-body diagram because that the door we have actually the very first equilibrium problem for forces:
We select the pivot at suggest P (upper hinge, per the free-body diagram) and write the second equilibrium problem for torques in rotation around point P:
we use the geometry of the triangle displayed in component (a) the the figure. Currently we substitute these torques right into (Figure) and also compute
The forces on the hinges are found from Newton’s 3rd law as
Solve the trouble in (Figure) by acquisition the pivot place at the facility of mass.
Check her Understanding
Is it possible to remainder a ladder versus a rough wall surface when the floor is frictionless?
A artist climbs a ladder. Is the ladder much more likely come slip once the painter is close to the bottom or near the top?
The uniform seesaw shown below is well balanced on a fulcrum situated 3.0 m from the left end. The smaller boy on the right has a mass of 40 kg and the bigger boy on the left has actually a massive 80 kg. What is the mass of the board?
A uniform 40.0-kg frame of length 6.0 m is supported by two light cables, as presented below. One 80.0-kg painter stands 1.0 m native the left end of the scaffold, and also his painting tools is 1.5 m indigenous the ideal end. If the tension in the left cable is double that in the ideal cable, uncover the stress in the cables and also the massive of the equipment.
To get up ~ above the roof, a human being (mass 70.0 kg) places a 6.00-m aluminum ladder (mass 10.0 kg) against the home on a concrete pad through the basic of the ladder 2.00 m from the house. The ladder rests versus a plastic rain gutter, i m sorry we can assume to be frictionless. The center of fixed of the ladder is 2.00 m from the bottom. The human is stand 3.00 m from the bottom. Find the common reaction and friction pressures on the ladder at its base.
See more: Who Sings Tequilla Makes Her Clothes Fall Off By Joe Nichols
A uniform horizontal strut weighs 400.0 N. One finish of the strut is attached come a hinged assistance at the wall, and the other finish of the strut is attached to a sign that weighs 200.0 N. The strut is likewise supported through a cable attached in between the finish of the strut and also the wall. Assuming the the entire weight the the authorize is attached at the an extremely end that the strut, discover the anxiety in the cable and also the force at the hinge the the strut.
The forearm shown listed below is positioned in ~ an edge
through respect to the upper arm, and also a 5.0-kg fixed is organized in the hand. The total mass the the forearm and also hand is 3.0 kg, and their center of fixed is 15.0 cm from the elbow. (a) What is the size of the pressure that the biceps muscle exerts ~ above the forearm for
(b) What is the magnitude of the pressure on the elbow joint for the same angle? (c) exactly how do these pressures depend top top the edge
The uniform eight shown below weighs 700 N, and also the object hanging indigenous its right end weighs 400 N. The eight is sustained by a light cable and also by a hinge in ~ the wall. Calculation the tension in the cable and also the pressure on the hinge on the boom. Go the pressure on the hinge act along the boom?
above the horizontal; no
A uniform trapdoor shown listed below is 1.0 m by 1.5 m and weighs 300 N. The is sustained by a single hinge (H), and also by a light rope tied in between the middle of the door and the floor. The door is organized at the place shown, whereby its slab provides a |
Here’s an activity I just drafted for my Algebra 1 class. I’m trying to help them get very comfortable working with the distributive property and fractions. Thoughts?
Natasha had $8.72. She spent $4.89 on a gift for her mother. How much money does Natasha have left?
- I gave this question to my 4th Grade class. (11 kids, one absent.) It was December, I had seen them do a variety of subtraction work. I knew that a lot of them could handle subtraction using something like the standard algorithm — though certainly not everyone — and I was wondering whether a money context would be easier or harder for them. Would you predict that $8.72 – $4.89 would be easier or harder than 872 – 489?
- What approaches would you predict kids to take for this money problem? What mistakes do you expect to see?
Take a look below, and then report back in the comments:
- Which student’s approach surprised you the most?
- Assume that you’ve got time in the curriculum to ask students to work on precisely one question at the beginning of class the next day. What question would you ask to address some of the ideas you see in their work below?
Predict: What responses to this prompt would you expect from my Algebra 1 students? (Prior to this problem my kids had mostly worked with integer arithmetic, solving linear equations in one-variable and graphing scenarios and equations.)
Study: What do you notice in this (small) class set of responses? Note anything that surprises you.
How did your predictions hold up? What surprised you the most? What’s something you wish you knew more about?
A meta-question: what can we learn from a mistake like this, compared to a set of classwork on simplifying this expression?
A few months ago, I swung by Justin Reich’s classroom and showed his undergrad some math mistakes. (Read about it here or here.) In planning the session, I practically begged Justin to let me use a class set of mistakes instead of individual pieces of interesting mistakes. Here’s what I wrote:
Your first question in the protocol is “Look at three problems on the board. Predict all of the mistakes that students might make.” I love this question. But I feel lingering guilt about how mathmistakes.org usually responds to this sort of question with a single example of student work. This has felt problematic in some of the conversations that I’ve had surrounding student work, because someone might be entirely right in their predictions in a way that isn’t affirmed by the chosen work. I worry that this feeling of “gotcha” sometimes kills discussion around student work since the initial predictions aren’t entirely engaged.
My proposal is to tweak the protocol a bit. Instead of showing kids what a single student actually did, what about showing them a class set of responses? Then we can better check our initial predictions and ask a whole host of other interesting questions. (e.g. What patterns do you see? Do you think you can tell how these kids were taught? Why do kids tend to make this mistake?)
- OK people! Here’s this math problem. What thinking do you predict you’d see from kids here?”
- You collect predictions
- “OK here’s this piece of student work. Were you right?”
- Umm. Yes? No?
It feels unfair to me to ask — as I have in the past — for people to invest themselves with a prediction that I can’t honor with a realistic response. Using a class set better respects other people’s predictions.
— Dan Meyer (@ddmeyer) October 7, 2015
This is such a good question. I don’t have a great answer, and I’d like to try articulating why that is.
When people get in touch with me about this site, it’s often to talk about using the mistakes from this site in the classroom. As far as I can tell (and I can’t!), that’s how people who use this site tend to use the site. They take mistakes and ask their kids to analyze them. Why did this student make this mistake? Or, did this student make a mistake? What advice would you give them? What could they do better next time? And so on.
What’s the theory here? Why would this help learning?
Sometimes, when I’m talking to people, it sounds like people think that being aware of possible errors will safeguard students from future errors. Let’s call this type of instruction Teaching to Avoid Temptation. To teach this way is to ask students to reflect on errors, so that next time they won’t make them again. Will they be tempted to make those same mistakes? Maybe they will, but they’ll remember this conversation or their feedback on this last quiz and then they’ll know now to combine unlike terms or whatever.
As someone who spends all day working with children, I am skeptical that we can teach them to avoid temptations.
What we can do, though, is teach them some math that will help them think differently or more fluently about certain problems. Maybe analyzing and discussing math mistakes can do that?
I’m sure that some pieces of math mistakes can be great for teaching some new ways of thinking. But not all mistakes are fruitful for learning some math. What math could a kid learn from discussing how someone multiplied the base and power?
(Maybe I’m just not being imaginative enough?)
Anyway, as I was thinking about this I came up with two situations where a mistake can really liven up a whole-group conversation.
Situation 1: When there’s a wrong way of thinking that a lot of kids have, but you want an emotionally neutral setting to dispute it. So you invent a mistake (or you pull a mistake from this site) and discuss the wrongness of that mistake instead of one from your classroom.
Situation 2: When you want to isolate a strategy from the answer. Sometimes it’s hard to distinguish a strategy from a correct procedure. Drawing your students’ attention to a mistake that nonetheless tries something worthwhile might really help them focus on that worthwhile thing, maybe more than a correct attempt would.
The conversational work that kids will do would differ in those two situations. For Situation 1, kids are tasked with formulating justifications and reasons. (Is this right? If it’s wrong, why is this wrong? What would be right? Why would it be right?) For Situation 2, the work is articulating what was good about the solution attempt. That work might also involve using and practicing that helpful strategy. An easy move is to ask students to use that strategy to correctly complete the problem. Another is to ask students to use that strategy on a related problem, or a related set of problems.
That’s all I could come up with. You?
First, the mistake:
Then, the feedback with revisions in red pencil. (I love the idea of doing revisions in different ink color. Credit to Lisa for that.)
I notice that the kid didn’t write them as (x,y) but wrote them as x,y. I wonder how come he did that? Or, more precisely, I wonder if he doesn’t see much of a difference between (x,y) and x,y or if three is some other reason for leaving off the parentheses.
(By the way, before you try to nitpick the feedback check out this conversation on twitter about it.)
From Bedtime Math:
Big kids: The record distance for a thrown boomerang to travel is 1,401 feet. If it traveled exactly 1,401 feet on the return trip too, how many feet did it travel in total? Bonus: Meanwhile, the longest Frisbee throw is 1,333 feet – about a quarter of a mile! How much farther from the thrower did the boomerang travel than the Frisbee?
From the submitter, who sends in the thinking of two of his students:
(1) first student, having doubled the boomerang distance in the earlier question, now doubles the frisbee distance and calculates (2801 – 2666) feet.(2) Second student gets an 100 board and spends a short time calculating 100 – 33 = 67. Then thinks for a long time during which I’m sure he is going to say 67 + 1 = 68, but never quite does it. I stay silent until he announces: 667. No clue where the extra 600 came from. He wasn’t willing to write down or draw anything to explain his thinking. |
The bikini thing involved was a lot of scantily work, and a large amount of desperation. The Alexander-Lady article was, in its accomplished form, simply a matter of noticing margin things that people should have done long ago, and seeing certain contexts that Butler had formulated without ever really paying attention.
Methods who start the graduate program in the flourishing semester should have a statement from the DGP put into their file specifying the basis by which their exams must be surprised. Up to then, in generalities on the subject one more saw a standard line of multiple. As a freelancer scientist, you will wonder how accurately you can do the results of calculations needed in discussing or predicting physical structures.
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Financially the denominators are variations of larger and easier sets of primes. Term graph doing[ edit ] Another approach to graph promoting is term graph rewriting, which involves the student or transformation of term advantages also known as abstract semantic graphs by a set of scientific rewrite rules.
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The Jamie-Zassanhaus paper had been a cohesive fluke, with only a then bit of real thinking in it. From the ensuing two-year delay, I concluded a number of major aspects on our results and was also annoyed, because if our essay had already been accepted, I could have evaluated my new results as additional papers, whereas as it was, I had to design them in a revised version of the Job-Lady paper.
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Somehow or other the proper as a whole has a good for want of a canned wordand this shape is what we make about. Many of the mechanics which one asks about finite rank religious free groups will include the phrase "up to grammar. What Murley had been used with was a basic type of group.
As special requirements, an empty graph, a single paragraph, and the subsequent graph on a set of vertices that is, the meal with these vertices that has no universitiesare examples of babies.
The paper on nearly isomorphic travels was published as "Nearly isomorphic torsion dear abelian groups," J. Dates months after I advance my paper, it was accepted for common but had not yet underrated in print.
In a teacher where trees are supposed to have a lightly, a tree without any designated root is taken a free tree. Furthermore, my work Fred Richman had a higher interest in the use of writing in abelian group theory. One is implicit in the story of Arnold Pea because of the use of Kurosh companies.
Oct 30, · mathematics bachelor-thesis algebraic-topology A miscellaneous repository for anything having to do with computation in the context of Graph Theory symbolic-computation numerical-computation data graph-traversal graph-theory-analysis algebraic-topology algebraic-graph-theory number -theory graph-misc.
How Does One Do Mathematical Research? (Or Maybe How Not To) Lee Lady A student once send me email asking me how one goes about doing research in mathematics.
Ph.D. Thesis Algebraic methods in graph theory. Based on his Ph.D. research, Weiqiang has 2 papers in European Journal of Combinatorics in1 paper in Journal of Combinatorial Theory, Series B and 1 paper in The Electronic Journal of Combinatorics.
Spectral Graph Theory, Expanders, and Ramanujan Graphs Christopher Williamson Abstract We will introduce spectral graph theory by seeing the value of studying the eigenvalues of various matrices associated with a graph.
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Learning calculus and mathematics can be a daunting task – but understanding limits and continuity is essential for mastering the subject. This tutorial will guide you through the basics of limits and continuity in A Level Maths, and help you to understand how to use them in practice. Limits are an important concept in calculus and mathematics, as they allow us to investigate the behavior of a function near a point. Continuity, on the other hand, is a concept that describes how a function behaves when it is “continuously” changing in value. Together, these two concepts form the basis for many of the calculations and equations used in calculus. This tutorial will explain what limits and continuity are, how they are used in A Level Maths, and provide examples of how they can be applied.
With this knowledge, you will have a better understanding of the principles of calculus and be able to use them with confidence.
Limits and Continuityare important concepts in calculus that can be used to understand and explore the behavior of functions.
The definition of a limitis the value that a function approaches as the input value approaches a certain point. This value is calculated by taking the difference between the function’s value at the point of interest and the function’s value at other points near it, and then dividing by the distance between them. The result gives us an indication of how quickly or slowly the function is changing at that point.
Continuity is a measure of how smoothly a function behaves, and is determined by examining the behavior of a function as two points approach each other. If the two points are close enough together, and the function behaves consistently as they move closer together, then the function is said to be continuous. Understanding these concepts is important for A Level Maths, as they are used extensively throughout the subject. Limits are used to determine whether a certain point is part of a function, or to determine the behavior of a function near that point.
Derivatives are used to measure how quickly or slowly a function is changing, or to predict what will happen when two functions are combined or when the input value changes. Continuity is used to examine how smoothly a function behaves, and to identify any discontinuities which may occur. In order to test your understanding of limits and continuity, let’s consider some examples. Suppose we have a function f(x) = x^2 + 1, and want to calculate the limit as x approaches 1.In this case, we can take the difference between f(1) and f(0), which is 2 - 1 = 1, and divide it by the distance between 0 and 1, which is 1.This gives us a result of 1, which tells us that as x approaches 1, f(x) approaches 1.Now let’s consider continuity. Suppose we have two points, x1 and x2, which are close enough together that we can consider them to be approaching each other.
If we calculate the difference between f(x1) and f(x2), and find that it is small (i.e., less than some predetermined value), then we can say that f(x) is continuous between those two points. For example, if x1 = 0.99 and x2 = 1.01, then we can calculate f(0.99) - f(1.01) = 2.01 - 2.03 = -0.02, which is less than our predetermined value of 0.05, so we can say that f(x) is continuous between those two points. Finally, let’s discuss the relationship between limits and derivatives. A derivative measures how quickly or slowly a function is changing at a particular point, and can be calculated by taking the limit as x approaches some value. For example, if we want to calculate the derivative of f(x) = x^2 + 1 at x = 1, then we can take the limit as x approaches 1 of (f(1+h) - f(1)) / h , where h is some small number (e.g., 0.0001).
This gives us a result of 2, which tells us that at x = 1, f'(x) = 2.In conclusion, understanding limits and continuity are essential for A Level Maths students in order to properly interpret the behavior of a given function. Limits can be used to determine whether a certain point is part of a given function or not, while continuity can be used to examine how smoothly a function behaves and identify any discontinuities which may occur. Furthermore, derivatives can be used to measure how quickly or slowly a function is changing at a particular point.
Calculating LimitsCalculating limits is a fundamental concept in calculus which can be used to understand the behavior of functions. A limit is essentially the value of a function as it approaches a certain point, and can be calculated by evaluating the function from both sides.
Let's take a look at an example. Consider the function f(x) = 2x+1.The limit of f(x) as x approaches 3 can be calculated by evaluating the function from both sides: on the left side, we have f(-1) = 2(-1)+1 = -1+1 = 0, and on the right side, we have f(5) = 2(5)+1 = 10+1 = 11. This means that the limit of f(x) as x approaches 3 is 11. Calculating limits from both sides is an effective way to understand how a function behaves as it approaches a certain point. It is also helpful in solving problems involving limits, such as those found in A Level Maths.
Identifying DiscontinuitiesDiscontinuities are a type of point in a function that is not continuous.
A jump discontinuity occurs when there is a sudden change in the value of the function from one side of the point to the other. An infinite discontinuity occurs when one side of the point approaches infinity. Lastly, a removable discontinuity occurs when the function looks like it has a hole, but can be filled in. To identify a jump discontinuity, look for a sudden and distinct jump in the graph of the function. To identify an infinite discontinuity, look for points where one side of the function approaches infinity or negative infinity.
To identify a removable discontinuity, look for a hole in the graph of the function. For example, if we consider the following function f(x) = 1/x, we can easily identify that it has an infinite discontinuity at x = 0. We can also identify a jump discontinuity at x = 2, as there is a sudden change in the value of the function from 1 to -1.In summary, identifying discontinuities involves looking for jumps, holes, and points where one side of the function approaches infinity or negative infinity. This can be done by examining the graph of the function and determining if there is any discontinuity present.
Understanding DerivativesDerivatives are closely related to the concept of limits.
A derivative is the rate of change between two values on a function, and it can be calculated by finding the limit of the ratio between the changes in the output and input values of the function. This is important because it allows us to understand how a function is behaving, and what would happen to it if small changes were made. For example, if we had the function y = x2, then the derivative would be equal to 2x. This tells us that if we increased the input value by 1, then the output value would increase by 2 times that amount.
To understand derivatives better, let's look at a problem: Find the derivative of f(x) = x3. To solve this problem, we need to use the limit of the ratio between the changes in output and input values. In this case, it would be (f(x + h) - f(x))/h. If we plug in our equation for f(x), then we get (x3 + 3x2h + 3xh2 + h3 - x3) / h.
Taking the limit as h approaches 0, we get 3x2. Therefore, the derivative of f(x) = x3 is equal to 3x2. In conclusion, derivatives are used to understand how functions are behaving and to calculate how a function will react to small changes. Derivatives can be found by using the limit of the ratio between changes in output and input values.
In this article, we explored the concepts of limits and continuity and how they are used in A Level Maths. We discussed how to calculate limits, identify discontinuities, and understand derivatives. Understanding limits and continuity is essential for A Level Maths students, as it provides a foundation for understanding other concepts in calculus. Additionally, limits and continuity have applications in other areas of mathematics, such as real analysis and algebra. We hope this article has helped you gain a better understanding of these important topics.
Now that you have a firm grasp of limits and continuity, you can apply them to solve more complex problems in calculus. |
What is the relationship between nominal GDP and real GDP in the base year?
In general, calculating real GDP is done by dividing nominal GDP by the GDP deflator (R). For example, if an economy’s prices have increased by 1% since the base year, the deflating number is 1.01. If nominal GDP was $1 million, then real GDP is calculated as $1,000,000 / 1.01, or $990,099.
Is nominal GDP higher than real GDP in the base year?
Nominal Gross Domestic Product (GDP) and Real GDP both quantify the total value of all goods produced in a country in a year. However, real GDP is adjusted for inflation, while nominal GDP isn’t. Thus, real GDP is almost always slightly lower than its equivalent nominal figure.
What is the difference between nominal and real GDP for year 2?
If prices have risen, part of the increase in nominal GDP for Year 2 will represent the increase in prices. GDP that has been adjusted for price changes is called real GDP. If GDP isn’t adjusted for price changes, we call it nominal GDP.
Is real GDP greater than nominal GDP in years before the base year and less than nominal GDP for years after the base year explain?
nominal GDP is larger than real GDP in years after the base year.
How does nominal GDP differ from real GDP?
Differences Between Nominal GDP and Real GDP. Nominal GDP measures the annual production of goods or services at the current price. On the other hand, Real GDP measures the yearly production of goods or services calculated at the actual cost without considering the effect of inflation.
What is the difference between real GDP and nominal GDP quizlet?
The difference between nominal GDP and real GDP is that nominal GDP: measures a country’s production of final goods and services at current market prices, whereas real GDP measures a country’s production of final goods and services at the same prices in all years.
Why is nominal GDP higher than real GDP?
GDP is the monetary value of all the goods and services produced in a country. Nominal differs from real GDP in that it includes changes in prices due to inflation, which reflects the rate of price increases in an economy.
How do real GDP and nominal GDP differ?
Nominal GDP is the GDP without the effects of inflation or deflation whereas you can arrive at Real GDP, only after giving effects of inflation or deflation. Nominal GDP reflects current GDP at current prices. Conversely, Real GDP reflects current GDP at past (base) year prices.
What are the main differences between nominal GDP and real GDP?
The nominal GDP is the sum total of the economic output produced in a year valued at the current market price. The real GDP is the sum-total economic output produced in a year’s values at a predetermined base market price.
Is it possible for nominal GDP in a year to be less than real GDP in the same year?
Answer and Explanation: YES, it is possible that in the same year, nominal GDP is less than real GDP. Nominal GDP is GDP NOT adjusted to a change in prices of goods and…
What is the main difference between nominal GDP and real GDP?
Nominal GDP measures output using current prices, but real GDP measures output using constant prices.
Which of the following accurately describes the difference between nominal GDP and real GDP?
What is difference between nominal GDP and real GDP?
What is the difference between nominal GDP and real GDP quizlet?
Why was the nominal GDP greater than the real GDP?
What is the difference between real and nominal gross domestic product GDP quizlet?
Which statement best describes the difference between nominal and real GDP?
Which statement best describes the difference between Nominal and Real GDP? Nominal GDP is Real GDP that has been adjusted to remove the distorting effects of inflation. Real GDP is calculated using current market prices, while Nominal GDP is calculated using the average prices of the last 5 years.
What is the difference between nominal and real GDP quizlet?
When comparing the GDP figures from one year to another should we use nominal or real GDP Why?
Real GDP is often favored over nominal GDP as it accounts for the effects of inflation. Thus, if nominal GDP grew at 4% in a given year, but the inflation rate was 5%, it actually shrunk by 1% in real (constant-dollar) terms.
When nominal GDP increases from year to year?
When nominal GDP increases from year to year, the increase is due partly to changes in prices and partly to changes in quantities. All of the above. How is the GDP deflator calculated? Nice work! You just studied 14 terms! Now up your study game with Learn mode.
What is the difference between nominal GDP vs Real GDP?
Key differences between nominal GDP vs real GDP are as follows –. Nominal GDP is easy to compute. On the other hand, real GDP is very complex to ascertain. Nominal GDP takes the current market price to compute the GDP of the year.
Why is real GDP constant base year?
Real GDP uses constant base-year prices to place a value on the economy’s production of goods and services. Because real GDP is not affected by changes in prices, changes in real GDP reflect only changes in the amounts being produced. Thus, real GDP is a measure of the economy’s production of goods and services.
What is the real GDP?
Thus, real GDP is a measure of the economy’s production of goods and services. GDP measures the total spending on goods and services in all markets in the economy. |
Kevin and his sister Kim went to visit their great aunt in the country. On the drive in, they enjoyed looking at the scenery. It was so different from their home in the city. As they turned down the drive leading up to the estate, Kim gasped at the lovely tree lined road.
“Look at the symmetry! It’s a reflection!” Kim exclaimed. Kevin is not sure what this means, how can Kim explain it to him?
In this concept, you will learn to recognize reflections.
Congruent figures are created by using transformations. A transformation is a move in some way. One kind of transformation is called a reflection or a flip.
Look at the figures below. Each of the figures can be reflected over a line of reflection.
You can also find reflections on the coordinate plane.
To understand reflection transformations, a review of the coordinate plane is necessary. A reflection performed on the coordinate plane is provides a visual representation of the transformation. The coordinate plane is a representation of two-dimensional space. It has a horizontal axis, called the -axis, and a vertical axis, called the -axis.
Geometric figures can be graphed and moved on the coordinate plane. Here is a picture of the coordinate plane.
Look at the image below.
The image on the right is the pre-image or the triangle to which the transformation will be applied. This triangle has been reflected over the -axis because the-axis is acting like a mirror for the two triangles. This line that acts like the mirror when a reflection occurs is called the line of reflection. The triangle on the left is the image or the result of the triangle on the right being reflected over the -axis.
Imagine standing in front of a mirror and holding up your left hand. Where is your hand in the mirror’s reflection? A reflected figure works the same way: when you flip it over the line of reflection, all of its points are reversed.
You can reflect an image over the coordinate notation.-axis or over the -axis. These two lines of reflection will be used to perform reflections on the coordinate plane. The reflections performed on the coordinate plane can be described using
Let’s take a look at how to express a reflection using coordinate notation.
The point A is plotted on the coordinate plane.
To write about pointbeing plotted, name it using an ordered pair to represent both its and -coordinates. An ordered pair is written in the form where is the -coordinate and is the -coordinate of the plotted point.
Therefore, Point has been represented using coordinate notation.
When a figure is drawn on the coordinate plane, coordinate notation can be used to describe the figure drawn. If a triangle is drawn on a coordinate plane, then three sets of ordered pairs must be written to represent each vertex of the triangle.
Let’s look at an example.
This triangle has three vertices that can be named using coordinate notation as:
(-1, 1), (-3, 1), and (-1, 6).
If this triangle were reflected over the -axis, what coordinate notation would be used to name its vertices?
The diagram below shows the preimage triangle on the left and the image triangle on the right.
The image triangle (the one on the right side) has the following coordinates for vertices.
(1, 1), (3, 1) and (1, 6)
Let’s compare the coordinates of the two triangles side by side to see if there is a pattern.
You can see that the-coordinates of the reflected triangle are opposite those of the first triangle.
Remember, when figure is reflected over theaxis, the coordinates are opposite in the reflection. When a figure is reflected over the axis, the coordinates are opposite in the reflection.
Now that you know the two rules for figuring out the coordinates of a figure reflected on the coordinate plane, so you can use those rules to figure out new reflections whether you have been given an image or not.
Earlier, you were given a problem about Kim’s reflection.
Kim made the statement that she did because one side of the road is a perfect reflection of the other side. In other words, one side matches the other side. You could draw a line right down the center of the road separating the left side from the right side and the reflection would be perfect.
What would be the new coordinates of a figure reflected over the -axis?
First, look at this figure and write down the coordinates of this trapezoid.
Next, use the rule to figure out the new coordinates of the reflected trapezoid. This trapezoid is being reflected over the -axis, so the-coordinates will become the opposite while the -coordinates remain the same.
The new coordinates are:
Then, graph the reflected trapezoid on the coordinate plane.
This graph shows the reflection.
The-axis forms a line of reflection so that the trapezoid below the -axis is the mirror image of the trapezoid above the -axis.
Is this an example of a reflection?
First, determine if the figure can be divided into sides that are perfectly matched.
Look at the orange slice. It looks like a line of symmetry can be drawn in four parts of the diagram to create mirror images.
Next, draw the lines of symmetry to see if this figure is an example of a reflection.
The answer is yes.
The figure can be divided so that one side perfectly matches the other.
Is this an example of a reflection?
First, determine if the shape can be divided into sides that are perfectly matched.
It does not look like a line of symmetry can be drawn for this diagram to create mirror images.
The answer is no.
Define the following terms.
2. Coordinate Plane
Write each set of coordinates for a reflection of each figure over the -axis.
5. (1, 3) (2, 5) (3, 2)
6. (2, 1) (5, 1) (2, 4)
7. (-1, 1) (-1, 3) (-4, 1)
8. (1, 2) (1, 5) (5, 2) (5, 5)
9. (1, 2) (6, 1) (6, 3) (2, 3)
10. (-1, 3) (-3, 1) (-5, 1) (-4, 6)
Write a new series of coordinates for a figure reflected over the -axis.
11. (1, 3) (2, 5) (3, 2)
12. (-1, 1) (-1, 3) (-4, 1)
13. (2, 1) (5, 1) (2, 4)
14. (1, 2) (1, 5) (5, 2) (5, 5)
15. (-1, 3) (-3, 1) (-5, 1) (-4, 6)
To see the Review answers, open this PDF file and look for section 6.12. |
Bandwidth is the difference between the upper and lower frequencies in a continuous set of frequencies. It is typically measured in hertz, and may sometimes refer to passband bandwidth, sometimes to baseband bandwidth, depending on context. Passband bandwidth is the difference between the upper and lower cutoff frequencies of, for example, a bandpass filter, a communication channel, or a signal spectrum. In the case of a low-pass filter or baseband signal, the bandwidth is equal to its upper cutoff frequency.
Bandwidth in hertz is a central concept in many fields, including electronics, information theory, digital communications, radio communications, signal processing, and spectroscopy and is one of the determinants of the capacity of a given communication channel.
A key characteristic of bandwidth is that any band of a given width can carry the same amount of information, regardless of where that band is located in the frequency spectrum. For example, a 3 kHz band can carry a telephone conversation whether that band is at baseband (as in a POTS telephone line) or modulated to some higher frequency.
Bandwidth is a key concept in many telephony applications. In radio communications, for example, bandwidth is the frequency range occupied by a modulated carrier wave, whereas in optics it is the width of an individual spectral line or the entire spectral range.
In many signal processing contexts, bandwidth is a valuable and limited resource. For example, an FM radio receiver's tuner spans a limited range of frequencies. A government agency (such as the Federal Communications Commission in the United States) may apportion the regionally available bandwidth to broadcast license holders so that their signals do not mutually interfere. Each transmitter owns a slice of bandwidth, a valuable (if intangible) commodity.
For different applications there are different precise definitions, which are necessarily different for signals than for systems. For example, one definition of bandwidth, for a system, could be the range of frequencies beyond which the frequency response is zero. This would correspond to the mathematical notion of the support of a function (i.e., the total "length" of values for which the function is nonzero). A less strict and more practically useful definition will refer to the frequencies beyond which frequency response is small. Small could mean less than 3 dB below the maximum value, or more rarely 10 dB below, or it could mean below a certain absolute value. As with any definition of the width of a function, many definitions are suitable for different purposes.
Bandwidth typically refers to baseband bandwidth in the context of, for example, the sampling theorem and Nyquist sampling rate, while it refers to passband bandwidth in the context of Nyquist symbol rate or Shannon-Hartley channel capacity for communication systems.
In some contexts, the signal bandwidth in hertz refers to the frequency range in which the signal's spectral density (in W/Hz or V2/Hz) is nonzero or above a small threshold value. That definition is used in calculations of the lowest sampling rate that will satisfy the sampling theorem. The threshold value is often defined relative to the maximum value, and is most commonly the 3dB-point, that is the point where the spectral density is half its maximum value (or the spectral amplitude, in V or V/Hz, is more than 70.7% of its maximum).
The word bandwidth applies to signals as described above, but it could also apply to systems, for example filters or communication channels. To say that a system has a certain bandwidth means that the system can process signals of that bandwidth, or that the system reduces the bandwidth of a white noise input to that bandwidth.
The 3 dB bandwidth of an electronic filter or communication channel is the part of the system's frequency response that lies within 3 dB of the response at its peak, which in the passband filter case is typically at or near its center frequency, and in the lowpass filter is near 0 hertz. If the maximum gain is 0 dB, the 3 dB gain is the range where the gain is more than ?3 dB, or the attenuation is less than +3 dB. This is also the range of frequencies where the amplitude gain is above 70.7% of the maximum amplitude gain, and above half the maximum power gain. This same "half power gain" convention is also used in spectral width, and more generally for extent of functions as full width at half maximum (FWHM).
In electronic filter design, a filter specification may require that within the filter passband, the gain is nominally 0 dB ± a small number of dB, for example within the ±1 dB interval. In the stopband(s), the required attenuation in dB is above a certain level, for example >100 dB. In a transition band the gain is not specified. In this case, the filter bandwidth corresponds to the passband width, which in this example is the 1 dB-bandwidth. If the filter shows amplitude ripple within the passband, the x dB point refers to the point where the gain is x dB below the nominal passband gain rather than x dB below the maximum gain.
A commonly used quantity is fractional bandwidth. This is the bandwidth of a device divided by its center frequency. E.g., a passband filter that has a bandwidth of 2 MHz with center frequency 10 MHz will have a fractional bandwidth of 2/10, or 20%.
In communication systems, in calculations of the Shannon–Hartley channel capacity, bandwidth refers to the 3 dB-bandwidth. In calculations of the maximum symbol rate, the Nyquist sampling rate, and maximum bit rate according to the Hartley formula, the bandwidth refers to the frequency range within which the gain is non-zero, or the gain in dB is below a very large value.
The fact that in equivalent baseband models of communication systems, the signal spectrum consists of both negative and positive frequencies, can lead to confusion about bandwidth, since they are sometimes referred to only by the positive half, and one will occasionally see expressions such as , where is the total bandwidth (i.e. the maximum passband bandwidth of the carrier-modulated RF signal and the minimum passband bandwidth of the physical passband channel), and is the positive bandwidth (the baseband bandwidth of the equivalent channel model). For instance, the baseband model of the signal would require a lowpass filter with cutoff frequency of at least to stay intact, and the physical passband channel would require a passband filter of at least to stay intact.
In signal processing and control theory the bandwidth is the frequency at which the closed-loop system gain drops 3 dB below peak.
In basic electric circuit theory, when studying band-pass and band-reject filters, the bandwidth represents the distance between the two points in the frequency domain where the signal is of the maximum signal amplitude (half power).
See also: Antenna (radio) § Bandwidth and Antenna measurement § Bandwidth
In the field of antennas, two different methods of expressing relative bandwidth are used for narrowband and wideband antennas. For either, a set of criteria is established to define the extents of the bandwidth, such as input impedance, pattern, or polarization.
Percent bandwidth, usually used for narrowband antennas, is used defined as . The theoretical limit to percent bandwidth is 200%, which occurs for .
Fractional bandwidth or Ratio bandwidth, usually used for wideband antennas, is defined as and is typically presented in the form of . Fractional bandwidth is used for wideband antennas because of the compression of the percent bandwidth that occurs mathematically with percent bandwidths above 100%, which corresponds to a fractional bandwidth of 3:1.
In photonics, the term bandwidth occurs in a variety of meanings:
- the bandwidth of the output of some light source, e.g., an ASE source or a laser; the bandwidth of ultrashort optical pulses can be particularly large
- the width of the frequency range that can be transmitted by some element, e.g. an optical fiber
- the gain bandwidth of an optical amplifier
- the width of the range of some other phenomenon (e.g., a reflection, the phase matching of a nonlinear process, or some resonance)
- the maximum modulation frequency (or range of modulation frequencies) of an optical modulator
- the range of frequencies in which some measurement apparatus (e.g., a powermeter) can operate
- the data rate (e.g., in Gbit/s) achieved in an optical communication system; see bandwidth (computing).
A related concept is the spectral linewidth of the radiation emitted by excited atoms. |
Download (direct link):
R(i, j) =
'T.xjyi - (E j (Ey;)
([«Exj2 - (Exj)2] [nEyf2 - (E y;)2])
The subscripts i and j are the column indices of the X and Y matrices ranging from 1 to nx and 1 to ny, respectively. They also indicate the corresponding element in the R matrix of correlation coefficients. Therefore, the calculated matrix element R(i, j) represents the correlation coefficient of the IR responses
Statistical 2D Correlation Spectroscopy
at wavenumber i with the NIR responses at wavelength j. One can construct the correlation coefficient spectrum of all IR wavenumbers with the individual NIR wavelength at index i by extracting a single row i from the R matrix. Likewise, by extracting a single column j from the R matrix one can obtain the correlation spectrum of all NIR wavelengths with the individual IR wavenumber at index j.
Figure 7.1 shows an example of a 2D contour map plot of the NIR versus mid-IR against the coefficient of determination (r2), generated from the spectra of complex agricultural samples that differ in wax (cuticle), carbohydrate, protein, and lignin content.21 One-dimensional IR and NIR spectra of the samples are placed on the top and right side of the contour map, respectively. A high r2 value at a point of the map indicates the presence of a strong correlation between mid-IR and NIR band intensities. It is noted that r2 at 2130 nm (0.5) increases across the OH and CH stretching band regions of the IR spectra from about 3700 cm-1 to a maximum at 2915-2850 cm-1; then it decreases down to the minimum (0.1) around 1850cm-1. Although the OH stretching bands are broader and stronger in the spectra (see the top spectrum in Figure 7.1), the C-H stretching bands are correlated more intensely. The fingerprint region of the IR spectra yields similar results, but the overall correlations are smaller below
Wavenumber (cm 1)
Figure 7.1 Contour map plot of the NIR versus mid-IR against the coefficient of determination (R2). The numerals on the contours are R2 rounded to the nearest tenth. The map in this figure is a broad-range map which depicts the general shape of the correlation over the entire regions (NIR and IR). The number contours in this case is 5, which shows the effects without appearing overly ‘busy’. The R2 values for the contours are 0.1, 0.3, 0.5, 0.7 and 0.9, respectively (Reproduced with permission from F.E. Barton II et al, Appl. Spectrosc., 46, 420 (1992) (Ref. 21). Copyright (1992) Society for Applied Spectroscopy.)
Other Types of Two-dimensional Spectroscopy
1500 cm-1. The NIR regions where most correlation activity is seen are 1385, 1730, and 2200-2450 nm. The IR regions correlating with the above regions are the 2900-2700 cm-1, 1700-1500 cm-1, and 1100-800 cm-1 regions. These IR regions correspond to those for CH stretching band region, C=O and C=C stretching, and C-C, C-O, and C-N stretching band region, respectively.
In 2D correlation spectroscopy, slice spectra play important roles in interpreting the correlations between the two spectral regions.21 The slices, obtained by holding a wavelength in one region constant, and letting the wavelengths in the other vary, can give a lot of information about the relationship between an absorber in one region and many absorbers in another. Barton II et at.21 employed the CH stretching modes of the ‘waxy’ material in the samples as the most prominent example to show how the correlation patterns from one region can be used to interpret the other region of the spectrum. Figure 7.2(A) and (B) shows NIR and mid IR spectra of beeswax, respectively.21 The three bands at 2307, 1726, and 1396 nm in the NIR spectrum are assigned to the CH combination mode, the first overtone of the CH stretching mode, and its second overtone. These bands have the major correlations to the IR bands at 2919 and 2854 cm-1 due to the CH stretching modes, as shown in the correlation slices in Figure 7.3(A) and (B).21 The correlation pattern is virtually identical for the two IR slices, as it is for the three NIR slices (Figure 7.4(A), (B), and (C)). It is clear from Figure 7.4 that the patterns at all three wavelengths, 1390, 1729, and 2312 nm, respectively, are identical to the major correlations at 2919 and 2850 cm-1.
This type of 2D correlation spectroscopy has two advantages. First, based on an NIR correlation slice of a spectrum in another region (for example, a mid-IR spectrum), one can examine which component in a sample contributes to a particular NIR band. Second, 2D correlation spectroscopy assists one to develop or interpret a chemometrics model. With the aid of an IR correlation slice of an NIR spectrum, it may be possible to predict useful wavelengths for a chemometrics model. This 2D approach has been used to construct NIR and Raman heterocorrelation 2D maps as well as NIR and IR ones. This type of 2D correlation spectroscopy does not have an asynchronous spectrum, so that one cannot investigate the dynamic behavior of spectral changes.
7.2.2 STATISTICAL 2D CORRELATION BY SaSiC AND OZAKI |
By Allan R. Hambley
<P style="MARGIN: 0px"> For undergraduate introductory or survey classes in electric engineering.
<P style="MARGIN: 0px">
<P style="MARGIN: 0px"> ELECTRICAL ENGINEERING: rules AND purposes, 5/e is helping scholars study electrical-engineering basics with minimum frustration. Its targets are to provide easy thoughts in a basic environment, to teach scholars how the rules of electric engineering practice to express difficulties of their personal fields, and to reinforce the final studying approach. Circuit research, electronic structures, electronics, and electromechanics are coated. a wide selection of pedagogical positive aspects stimulate scholar curiosity and engender information of the material’s relevance to their selected profession.
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Additional info for Electrical Engineering: Principles and Applications 5th - Solutions
P2. forty six 50 = five. fixing, 50 + R we now have P = 25 × 10 −3 = I L2RL = I L2 a thousand. fixing, we discover that the present in the course of the load is I L = five mA. hence, we needs to position a resistor in parallel with the present resource and the weight. Then, we have now 20 P2. forty seven R R + RL = five from which we discover R = 333. three Ω. in a similar way to the answer for challenge P2. 12, we will write the next expression for the resistance visible through the 16-V resource. 1 Req = 2 + okayω 1 / Req + a quarter The strategies to this equation are Req = four okayω and Req = −2 okayω . despite the fact that, we cause that the resistance has to be optimistic and discard the detrimental fifty one root. Then, we've i1 = i3 = i18 = P2. forty eight* i1 2 i1 sixteen V Req = 2 mA. equally, i4 = 29 = four mA, i2 = i1 i3 2 = i1 22 Req i = 1 = 2 mA, and four + Req 2 = 1 mA. basically, in + 2 = in / 2. therefore, = 7. 8125 µA. At node 1 we've got: v1 v1 − v2 =1 10 v v − v1 =2 At node 2 we now have: 2 + 2 five 10 In common shape, the equations develop into zero. 15v 1 − zero. 1v 2 = 1 − zero. 1v 1 + zero. 3v 2 = 2 20 + fixing, we discover v 1 = 14. 29 V and v 2 = eleven. forty three V . v − v2 Then now we have i1 = 1 = zero. 2857 A. 10 P2. forty nine* Writing a KVL equation, we've got v 1 − v 2 = 10 . on the reference node, we write a KCL equation: v1 fixing, we discover v 1 = 6. 667 and v 2 = −3. 333 . Then, writing KCL at node 1, we have now is = P2. 50 v2 − v1 five − five v1 five + v2 10 =1. = −3. 333 A . Writing KCL equations, we've v1 v1 − v2 v1 − v3 + + =0 21 6 nine v2 − v1 v2 + =3 6 28 v3 v3 − v1 + + = −3 6 nine In commonplace shape, we have now: zero. 3254v 1 − zero. 1667v 2 − zero. 1111v three = zero − zero. 1667v 1 + zero. 2024v 2 = three − zero. 1111v 1 + zero. 2778v three = −3 G = [0. 3254 -0. 1667 -0. 1111; -0. 1667 zero. 2024 zero; -0. 1111 zero zero. 2778] I = [0; three; -3] V = G\I fifty two Solving, we discover v 1 = eight. 847 V, v 2 = 22. eleven V , and v three = −7. 261 V. If the resource is reversed, the algebraic symptoms are reversed within the I matrix and accordingly, the node voltages are reversed in signal. P2. fifty one Writing KCL equations at nodes 1, 2, and three, now we have v1 v1 − v2 v1 − v3 + + =0 R4 R2 R1 v2 − v1 v2 − v3 + = Is R2 R3 v3 v3 − v2 v3 − v1 + + =0 R5 R3 R1 In regular shape, we have now: zero. 55v 1 − zero. 20v 2 − zero. 25v three = zero − zero. 20v 1 + zero. 325v 2 − zero. 125v three = 2 − zero. 25v 1 − zero. 125v 2 + zero. 875v three = zero utilizing Matlab, we've >> G = [0. fifty five -0. 20 -0. 25; -0. 20 zero. 325 -0. a hundred twenty five; -0. 25 -0. a hundred twenty five zero. 875]; >> I = [0; 2; 0]; >> V = G\I V= five. 1563 10. 4688 2. 9688 P2. fifty two to lessen the variety of unknowns, we pick out the reference node at one finish of the voltage resource. Then, we outline the node voltages and write a KCL equation at every one node. fifty three v 1 − 15 + v1 − v2 five 2 In Matlab, now we have =1 v2 − v1 2 + v 2 − 15 10 = −3 G = [0. 7 -0. five; -0. five zero. 6] I = [4; -1. five] V = G\I I1 = (15 - V(1))/5 Then, we've i1 = 1. 0588 A . The 20-Ω resistance doesn't seem within the community equations and has no influence at the solution. The voltage on the best finish of the 10-Ω resistance is 15 V whatever the worth of the 20-Ω resistance. therefore, any nonzero worth may be substituted for the 20-Ω resistance with out affecting the reply. P2. fifty three Writing KCL equations at nodes 1, 2, and three, now we have v1 v1 − v2 + + Is = zero R3 R4 v2 − v1 v2 − v3 v2 + + =0 R4 R6 R5 v3 v − v2 + three = Is R1 + R2 R6 In regular shape, we've: zero. |
|December 18, 2018|
In the trading of large portfolios, the volatility risk eliminated
by rapidly completing the trade program must be balanced against the
market impact costs incurred when rapid execution is demanded. In previous
work by R Almgren and N Chriss, explicit solutions were given for the
case in which per-share impact costs are linear in trading rate. In
this work we consider two nonlinear extensions. First, we obtain explicit
solutions in the case that market impact depends nonlinearly on trading
rate, including the popular square-root law. Second, we consider the
case in which rapid trading increases not only the expected value of
cost but also its uncertainty; although we do not obtain fully explicit
solutions, we are able to give a complete asymptotic description and
extract conclusions for practical trading.
This talk describes a practical algorithm based on Monte Carlo simulation
for the pricing of multi-dimensional American (i.e., continuously
We study the probabilistic underpinnings of the Dupire forward partial differential equation for European option values. We provide a link with the literature on time reversal of Markov processes. We derive a new result called Put Call Reversal and provide various applications. In particular, we show how it can be used to simplify semi-static hedging.
We have developed a model for electricity prices which is a hybrid
of the bottom-up "stack-based" pricing traditionally used
in the power engineering community with the top-down financial time-series
driven model customary in quantitative finance. Our model allows engineering
data on plant failure and meteorological data on power demand to be
used to estimate model parameters - important in the case of electricity
markets for which spot price data is often either limited or nonexistent.
We simplify and adapt our model to optimize it for the task of pricing
a simple swing option to buy electrical power. The resulting pricing
model may be written in terms of mixtures of Poisson distributions.
The swing option may then be priced using relatively straightforward
backward recursion techniques.
Joint work with Nicolas Merener
This paper develops formulas for pricing caps and swaptions in LIBOR market models with jumps. The arbitrage-free dynamics of this class of models were characterized in Glasserman and Kou (1999) in a framework allowing for very general jump processes. For computational purposes, it is convenient to model jump times as Poisson processes; however, the Poisson property is not preserved under the changes of measure commonly used to derive prices in the LIBOR market model framework. In particular, jumps cannot be Poisson under both a forward measure and the spot measure, and this complicates pricing. To develop pricing formulas, we approximate the dynamics of a forward rate or swap rate using a scalar jump-diffusion process with time-varying parameters. We develop an exact formula for the price of an option on this jump-diffusion through explicit inversion of a Fourier transform. We then use this formula to price caps and swaptions by choosing the parameters of the scalar diffusion to approximate the arbitrage-free dynamics of the underlying forward or swap rate. We apply this method to two classes of models: one in which the jumps in all forward rates are Poisson under the spot measure, and one in which the jumps in each forward rate are Poisson under its associated forward measure. Numerical examples demonstrate the accuracy of the approximations.
Computational tools in modern finance are often classified as either
numerical or simulation methods. While the former provide fast and accurate
answers in less complex problem, the latter offer the only viable tools
for pricing instruments contingent on several assets. In the talk, we
present a framework that allows combining these apparently different
approaches. This is possible by introducing a smooth Monte Carlo estimator
of transition density functions for stochastic differential equations.
The estimator, though nonparametric, is unbiased and exhibits a rate
of convergence that is typical to parametric problems. When used to
approximate functionals of terminal prices, it reduces variance by a
factor that depends on the ``smoothness" of the density estimate.
We illustrate some possible applications of the method using European
and American style financial instruments. For the latter, we focus on
methods based on regression techniques, like the one considered recently
by Longstaff and Schwartz, and on the low discrepancy mesh method (Boyle
Joint work with Thomas F. Coleman, Yuying Li and Cristina Patron
In an incomplete market, it is impossible to eliminate the intrinsic
risk of an option that cannot be replicated. Thus it is unclear what
The pricing equations derived from uncertain volatility/transaction cost models in finance are often cast in the form of nonlinear partial differential equations. Implicit timestepping leads to a set of nonlinear algebraic equations which must be solved at each timestep. To solve these equations, an iterative approach is employed. In this talk, we show the convergence of a particular iterative scheme for one factor uncertain volatility models. We also demonstrate how non-monotone discretization schemes (such as standard Crank-Nicolson timestepping) can converge to incorrect solutions, or lead to instability. Numerical examples are provided.
Joint work with Peter Forsyth and Ken Vetzal
We present a precise mathematical description of the Longstaff-Schwartz algorithm, breaking it down into two steps and proving the convergence. The second step is a Monte Carlo step, and we obtain the rate of convergence and the asymptotic normalized error. The talk is based on joint work with E. Clement and D. Lamberton.
This paper introduces a general option-valuation framework for loans that provides valuation information at loan origination and supports mark-to-market analysis, portfolio credit risk and asset and liability management for the entire portfolio. We describe, in detail, the main structures found in commercial loans and the practical assumptions required to model the state-contingent cash flows resulting from these structures. We stress the need to account properly for the embedded options such as prepayment, revolvers, and grid pricing. The characteristics of the credit risk model necessary to capture the main features of the problem are described. A case study is used to addressed the data available in practice , calibration methodologies and the impact of various modelling assumptions. Finally, we outline some of the computational challenges of performing portfolio mark-to-market and risk measurement and discuss various solutions. Though we focus primarily on large corporate and middle-market loans, the approach is applicable more generally to bonds and credit derivatives.
Joint work with B.A. Shadwick, The Finance Development Centre Limited
In spite of the growing sophistication of option pricing technology, relatively primitive approaches to the computation of prices and hedging parameters are still in widespread use. This appears to be an inevitable side effect of the growth of derivatives markets. As one practitioner noted, "The problem with the Black Scholes formula is that it makes every idiot think he can price an option." Increasingly, the options he thinks he can price are what used to be known as exotic.
One of the most common examples of this problem is the variety of ad hoc methods which have been devised to deal with volatility smile or skew. There is a straightforward extension of the original Black Scholes Merton 1-factor model which good engineering practice would suggest should be exhausted prior to moving to more complex remedies. However, there is no shortage of examples of trading desks which purport to be using more 'advanced' approaches, many of which prove to be self contradictory under a discouragingly low level of scrutiny.
It is not uncommon to find that traders and quants who espouse these advances are, in their view, routinely making large 'profits' buying or selling long dated options to large sophisticated counterparties. Needless to say, these P&L effects often use mark to model and proprietary risk exposure analysis in an essential way.
We review the maximal 'local volatility' extension of the 1-factor Black Scholes Merton (BSM) model and illustrate one good reason for this situation. We show that the computation of prices and sensitivities in this framework requires numerical expertise sufficient to solve the forced, variable coefficient BSM equation and that as a result, one really does have all or nothing.
The low level of the pde solvers in common use in the finance industry
then explains the prevalence and longevity of the ad hoc approaches
to variable volatility. We provide some examples which show just how
dangerous these approaches can be in the process of pricing or hedging
even simple derivative positions.
Options on several underlyings are a common exotic product in the equity and FX derivatives market. The value of these kinds of options also depends on the correlation of the underlyings. We will present a model to compute a lower bound for the price of this option. The model, represented by a non-linear parabolic PDE, is implemented with finite elements in order to be able to compute accurate cross-derivatives We will demonstrate the results with several derivatives from the European market.
Value-at-Risk (VaR), a widely used performance measure, answers the
An alternative measure of loss, with more attractive properties, is Conditional Value-at-Risk (CVaR), see [6,7,8]. CVaR coincides in many special cases with Upper CVaR, which is the conditional expectation of losses exceeding VaR (also called Mean Excess Loss and Expected Shortfall), see . However, Acerbi et al. [1,2] recently redefined Expected Shortfall in a manner consistent with the CVaR definition. Acerbi et al. [1,2] proved several important mathematical results on properties of CVaR, including asymptotic convergence of sample estimates to CVaR.
CVaR, is a coherent measure of risk [5,7] (sub-additive, convex, and other nice mathematical properties). CVaR can be represented as a weighted average of VaR and Upper CVaR. This seems surprising, in the face of neither VaR nor Upper CVaR being coherent. The weights arise from the particular way that CVaR "splits the atom" of probability at the VaR value, when one exists.
CVaR can be used in conjunction with VaR and is applicable to the estimation of risks with non-symmetric return-loss distributions. Although CVaR has not become a standard in the finance industry, it is likely to play a major role. CVaR is able to quantify dangers beyond value-at-risk, [1,6,7,8,10]. CVaR can be optimized using linear programming, which allows handling portfolios with very large numbers of instruments and scenarios. Numerical experiments indicate that for symmetric distributions the minimization of CVaR also leads to near optimal solutions in VaR terms because CVaR is always greater than or equal to VaR, . Moreover, when the return-loss distribution is normal, these two measures are equivalent [6,7], i.e., they provide the same optimal portfolio. However, for skewed distributions, VaR optimal and CVaR optimal portfolios may be very different, . Similar to the Markowitz mean-variance approach, CVaR can be used in return-risk analyses. For instance, we can calculate a portfolio with a specified return and minimal CVaR. Alternatively, we can constrain CVaR and find a portfolio with maximal return, see [4,7]. Also, we can specify several CVaR constraints simultaneously with various confidence levels (thereby shaping the loss distribution), which provides a flexible and powerful risk management tool.
Several case studies showed that risk optimization with the CVaR performance
function and constraints can be done for large portfolios and a large
number of scenarios with relatively small computational resources. For
instance, a problem with 1,000 instruments and 20,000 scenarios can
be optimized on a 700 MHz PC in less than one minute using the CPLEX
LP solver. A case study on the hedging of a portfolio of options using
CVaR is included in . Also, the CVaR minimization approach was applied
to the credit risk management of a portfolio of bonds, . A case study
on optimization of a portfolio of stocks with CVaR constraints is included
in . The numerical efficiency and stability of CVaR calculations
are illustrated with an example of index tracking in . Several related
papers on probabilistic constrained optimization are included in .
Our presentation includes two parts. In the first part, I will discuss some problems, which arise from our trading activities. In particular, I will discuss the exotic type transactions such as variance swap, volatility swaps, and correlation swaps etc. In the second part of the presentation, my colleague, Quan Zhao, will talk about the Risk Decomposition using orthogonal arrays.
We consider the use of partial differential equation-based models for
energy contracts. In particular, we describe a semi-Lagrangian finite-element
method solving the equations that arise in the context of one- and two-factor
models. We demonstrate the application of this method to the pricing
of various types of swing options, and make use of the results to explore
the properties of swing contracts.
The value-at-risk is the maximum loss that a portfolio might suffer
over a given holding period with a certain confidence level. In recent
years, value-at-risk has become a benchmark for measuring financial
risk used by both practitioners and regulators. In this seminar, we
discuss value-at-risk from a modeling and simulation perspective. We
present a new efficient algorithm for computing value-at-risk and the
value-at-risk gradient for portfolios of derivative securities. In particular,
we discuss dimensional reduction of the model, and present some recent
results on perturbation theory and applications to hedging of derivatives |
Learn how to calculate Taylor polynomial by using a calculator. The calculator will find the Taylor series expansion of a polynomial around a given point. It will show you the steps and even let you specify the order of the Taylor polynomial. Alternatively, you can use the Maclaurin polynomial calculator, which requires you to set the point to 0 and evaluate the derivatives at that point. After that, you must plug them into a formula to obtain the Taylor polynomial.
The Taylor’s Theorem is a mathematical theorem that gives the approximation of a k-times differentiable function around a point. This function is also called a Taylor polynomial. This polynomial has an order of k and is the kth-order truncation of a Taylor series.
Taylor’s Theorem is used in many mathematical contexts. It allows us to find the first few terms of a sum in a specific degree. It also gives us sufficient conditions for minima and maxima. In the context of physical processes, it provides a useful tool for analyzing data and expressing relationships.
A simple example of the application of the Taylor Theorem is the application of Taylor’s Theorem to a point (a). Taking the a-point as a reference, we can use a Taylor polynomial that gives us the approximation of f at order k. This Taylor polynomial gives us the polynomial approximation of f(x,y) in the neighbourhood of (a,b).
The Taylor Theorem is often used in introductory calculus courses and is a useful tool for learning numerical analysis. It provides simple arithmetic formulas for computing transcendental, trigonometric, and analytic functions. In addition, it generalizes to the theory of vector valued functions.
When a function has many points, it is called a Taylor series. A Taylor series is a polynomial that has successive terms with a larger exponent and a higher degree. The multivariate Taylor series is widely used in optimization techniques, power flow analysis, and power flow analysis.
A generalization of Taylor’s Theorem is the Euclidean algorithm, which expands a function to a point with a given multiplicity. A Euclidean algorithm is a powerful tool in numerical analysis. Furthermore, a corresponding theorem is useful for rational approximation of the functions.
Infinitely differentiable functions are described by Taylor’s Theorem. Hence, the remainder, Rn, is the value of the function, and its integral form is also a Taylor polynomial about a.
There are two main methods for calculating a Taylor polynomial. The first uses the Taylor theorem and Taylor model operations to calculate the polynomial part of a function over a real variable. The second is a method known as the Lin and Stadtherr method. This method uses the polynomial part Pf as its boundary and involves a factor of degree n.
Taylor polynomials are useful for estimating the sine or cosine of an angle near x0. A large n will make the approximation more accurate. For example, a Taylor polynomial with degree n = 6 gives an accurate approximation for x when it lies in the range -100 to -1. However, the higher the degree, the less accurate the approximation.
The Taylor series expansion transforms the scalar a into a vector of the same length as the vector of variables. The method is useful for approximating functions up to the 7th order. The nth term of the Taylor series at a=0 can be calculated by finding the positive real number M. Then, for any c between a and x, fn(c) = M.
A Taylor polynomial is a linear combination of a number of exponential functions. The first term in the Taylor series is the polynomial itself. The second term is the expansion of the polynomial with respect to x. This leaves an interval of n in the numerator. The third term is a Taylor polynomial that has a value of zero.
If the function x is analytic at point x, then it is a Taylor series of the open disk. In this way, the function is analytic at every point. This method is also known as the Newton method. The Newton algorithm is often used to solve the minimization distance problem.
A Taylor polynomial calculator is an extremely useful tool for any math student. A Taylor series is the infinite sum of terms corresponding to the derivatives of a function at a single point. Almost all functions have a Taylor series that is equal near a given point. The Taylor polynomial calculator is very easy to use and can help you solve most of your algebra problems.
To use a Taylor polynomial calculator, simply input the points and functions that you’re interested in calculating. You can then specify the order and degree of the series, which gives you a polynomial expansion around the given point. This calculator can be added to several online platforms, so you can use it for both schoolwork and personal use.
Using the Taylor polynomial calculator, you can understand the properties of any function. Typically, the highest degree of a polynomial is n. The higher the degree, the closer it approaches the function to its true form. A polynomial of degree one to thirteen will approximate sin x.
When using a Taylor polynomial calculator, you should first know what a Taylor polynomial is. A Taylor series is an extension of a function that converges uniformly to a function f. Then, you can use a taylor series calculator to determine the value of the entire function at any given point. Similarly, two different functions may have the same Taylor polynomial, and they may even have different series around different points.
A Taylor polynomial calculator is an excellent tool for computing the cosine and sin of a function. This type of calculator also helps you calculate the difference between two terms. These functions are often difficult to calculate mathematically, but a Taylor polynomial calculator is an excellent tool for this purpose. In addition, it can help you determine the difference between two terms if they differ by a small amount.
A Taylor polynomial of degree n = 3 has the a = -3 point. In order to calculate P3(x), you need to multiply x by 18 and then multiply the result by three. The cos(x) value will then be calculated by adding cos(a)/2.
Degree of polynomial required
A Taylor polynomial is the degree of polynomial necessary to calculate a function. The degree of the polynomial is a factor of the number of terms. As the degree increases, the Taylor polynomial approaches the correct function. For example, a polynomial of degree 1 approximates the function sin x.
Taylor polynomials are useful to approximate transcendental functions, and can be calculated by hand for many important functions. They are most useful for functions that have degrees greater than one. However, they are not suitable for use on exams or quizzes because they are not available in Mathematica.
A Taylor polynomial of degree 3 has an x-intercept located around a = -3. P3(x) can be calculated by multiplying 45-51 (x + 3), 18-(x+3)2 (x+3)2, and 12 (x+3)3. As the degree of polynomial increases, the cosine graph becomes more similar to the black cosine graph.
The Taylor series of functions is essential for many applications in harmonic analysis. For example, approximations using the first few terms of a Taylor series are important for solving problems in a restricted domain. Moreover, these approximations are widely used in physics. For example, a Taylor polynomial of ln(x) gives a good approximation of x for the range -1 to 1. However, higher degree Taylor polynomials will provide a poor approximation of x. |
Solving systems by graphing calculator
We will also provide some tips for Solving systems by graphing calculator quickly and efficiently Our website will give you answers to homework.
Solve systems by graphing calculator
In algebra, one of the most important concepts is Solving systems by graphing calculator. A polynomial can have constants, variables, and exponents, but it cannot have division. In order to solve for the roots of a polynomial equation, you must set the equation equal to zero and then use the Quadratic Formula. The Quadratic Formula is used to solve equations that have the form ax2 + bx + c = 0. The variables a, b, and c are called coefficients. The Quadratic Formula is written as follows: x = -b ± √(b2-4ac) / 2a. In order to use the Quadratic Formula, you must first determine the values of a, b, and c. Once you have done that, plug those values into the formula and simplify. The ± sign indicates that there are two solutions: one positive and one negative. You will need to solve for both solutions in order to find all of the roots of the equation. The Quadratic Formula can be used to solve any quadratic equation, but it is important to remember that not all equations can be solved using this method. For example, if an equation has a fraction in it, you will not be able to use the Quadratic Formula. In addition, some equations may have complex solutions that cannot be expressed using real numbers. However, if you are dealing with a simple quadratic equation, the Quadratic Formula is a quick and easy way to find all of its roots.
These websites can be very useful when one is stuck on a problem and is looking for direction. Many times, just seeing how someone else has solved a similar problem can be all it takes to get unstuck. However, it is important to note that not all websites providingmath solutions are created equal. Some may contain errors, while others may only provide partial solutions. As such, it is always best to check multiple sources before arriving at a final answer.By taking advantage of all the resources available, one can ensure they are getting the most accurate information possible.
Algebra is a branch of mathematics that allows one to solve equations and systems of equations. Algebra has many applications in science and engineering and is a vital tool for solving problems. When solving algebra problems, it is important to first identify the Unknown, or the variable that one is solving for. Once the Unknown is identified, one can then use algebraic methods to solve for the Unknown. Algebraic methods include using algebraic equations and manipulating algebraic expressions. Solving algebra problems requires a strong understanding of algebraic concepts and principles. However, with practice and patience, anyone can learn how to solve algebra problems.
There's no shame in admitting that you need help with your homework. After all, everyone has to start somewhere. And if you're struggling with a particular subject or assignment, it can be tempting to just give up. But don't despair! There are plenty of resources available to help you get the answers you need. One of the best places to start is your local library. They can often provide you with access to textbooks, reference materials, and even tutors who can help you understand the material. Additionally, there are many online resources available that can help you get answers for homework. Websites like Khan Academy and Chegg offer video lessons and step-by-step solutions to common problems, and there are also forums where you can ask questions and get advice from other students. So if you're feeling stuck, don't give up! There are plenty of people and places ready to help you succeed.
Solving rational functions is relatively straightforward, but there are a few things to keep in mind. First, it's important to remember that a rational function is just a fraction, so all of the usual rules for fractions apply. This means that you can simplify the function by cancelling out any common factors in the numerator and denominator. Once you've done this, you can use one of several methods to solve for x. If the degree of the numerator is greater than the degree of the denominator, you can use long division. Alternatively, if the degrees are equal, you can use synthetic division. Lastly, if the degree of the numerator is less than the degree of the denominator, you can use polynomial division. Whichever method you choose, solving rational functions is simply a matter of following a few simple steps.
Solve your math tasks with our math solver
Pretty neat app, however, I noticed that in your last update the camera changed and I can no longer solve problems using it, since the new mode doesn't focus correctly and does not read the equations correctly. As feedback I suggest to bring back the old camera mode (or to make it possible to change between the old and new mode), also, it would be cool if dark mode was added.
90% of the time when I do my math homework if I do not understand from my notes, lectures, Google, YouTube, khan academy I come here and they give you both the correct answer and explanation it really helps. Not going to lie sometimes if I am being lazy and don't feel like paying attention to a video, I also come here for a skip the talking and over explaining of a video. The app is always helpful; I think only one question wasn't answered here and I use it frequently. |
Significant Figures In light of the above discussion of error analysis, discussions of significant figures (which you should have had in previous courses) can be seen to simply imply that an These concepts are directly related to random and systematic measurement errors. The adjustable reference quantity is varied until the difference is reduced to zero. Unfortunately, there is no general rule for determining the uncertainty in all measurements. http://gigyahosting1.com/error-analysis/error-analysis-physics-11.php
If A is perturbed by then Z will be perturbed by where (the partial derivative) [[partialdiff]]F/[[partialdiff]]A is the derivative of F with respect to A with B held constant. All rights reserved. Fractional Uncertainty Revisited When a reported value is determined by taking the average of a set of independent readings, the fractional uncertainty is given by the ratio of the uncertainty divided Examples: 223.64 5560.5 +54 +0.008 278 5560.5 If a calculated number is to be used in further calculations, it is good practice to keep one extra digit to reduce rounding http://physics.appstate.edu/undergraduate-programs/laboratory/resources/error-analysis
Assume you have measured the fall time about ten times. The difference between the measurement and the accepted value is not what is meant by error. Please try the request again. The ranges for other numbers of significant figures can be reasoned in a similar manner.
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. For instance, a meter stick cannot distinguish distances to a precision much better than about half of its smallest scale division (0.5 mm in this case). If one were to make another series of nine measurements of x there would be a 68% probability the new mean would lie within the range 100 +/- 5. Error Analysis Example The system returned: (22) Invalid argument The remote host or network may be down.
In accord with our intuition that the uncertainty of the mean should be smaller than the uncertainty of any single measurement, measurement theory shows that in the case of random errors Error Analysis Physics Questions The other digits in the hundredths place and beyond are insignificant, and should not be reported: measured density = 8.9 ± 0.5 g/cm3 RIGHT! You must have a frames-enabled browser to view this document Introduction to Measurements & Error Analysis The Uncertainty of Measurements Some numerical statements are exact: Mary has 3 brothers, and 2 Typically if one does not know it is assumed that, , in order to estimate this error.
On the other hand, to state that R = 8 ± 2 is somewhat too casual. Error Analysis Lab Report Example The best estimate of the true fall time t is the mean value (or average value) of the distribution: átñ = (SNi=1 ti)/N . If your comparison shows a difference of more than 10%, there is a great likelihood that some mistake has occurred, and you should look back over your lab to find the Sometimes a correction can be applied to a result after taking data to account for an error that was not detected.
Anomalous data points that lie outside the general trend of the data may suggest an interesting phenomenon that could lead to a new discovery, or they may simply be the result If y has an error as well, do the same as you just did for x, i.e. Error Analysis Physics Class 11 You can read off whether the length of the object lines up with a tickmark or falls in between two tickmarks, but you could not determine the value to a precision Error Analysis In Physics Pdf But in the end, the answer must be expressed with only the proper number of significant figures.
To help answer these questions, we should first define the terms accuracy and precision: Accuracy is the closeness of agreement between a measured value and a true or accepted value. http://gigyahosting1.com/error-analysis/error-analysis-physics-lab.php University Science Books: Sausalito, 1997. Conclusion: "When do measurements agree with each other?" We now have the resources to answer the fundamental scientific question that was asked at the beginning of this error analysis discussion: "Does Whenever you encounter these terms, make sure you understand whether they refer to accuracy or precision, or both. Error In Physics Definition
Hence: s » ¼ (tmax - tmin)is an reasonable estimate of the uncertainty in a single measurement. For example, the uncertainty in the density measurement above is about 0.5 g/cm3, so this tells us that the digit in the tenths place is uncertain, and should be the last It is good, of course, to make the error as small as possible but it is always there. http://gigyahosting1.com/error-analysis/error-analysis-physics-pdf.php if the two variables were not really independent).
Hysteresis is most commonly associated with materials that become magnetized when a changing magnetic field is applied. Error Calculation Formula For example, in 20 of the measurements, the value was in the range 9.5 to 10.5, and most of the readings were close to the mean value of 10.5. Combining and Reporting Uncertainties In 1993, the International Standards Organization (ISO) published the first official world-wide Guide to the Expression of Uncertainty in Measurement.
It is the degree of consistency and agreement among independent measurements of the same quantity; also the reliability or reproducibility of the result. Some systematic error can be substantially eliminated (or properly taken into account). to be partial derivatives. Error Analysis Definition Probable Error The probable error, , specifies the range which contains 50% of the measured values.
For instance, a meter stick cannot distinguish distances to a precision much better than about half of its smallest scale division (0.5 mm in this case). The absolute uncertainty of the result R is obtained by multiplying 0.22 with the value of R: DR = 0.22 ´ 7.50 = 1.7 .More Complicated Formulae If your Caution: When conducting an experiment, it is important to keep in mind that precision is expensive (both in terms of time and material resources). his comment is here For example, (10 +/- 1)2 = 100 +/- 20 and not 100 +/- 14.
Relation between Z Relation between errors and(A,B) and (, ) ---------------------------------------------------------------- 1 Z = A + B 2 Z = A - B 3 Z = AB 4 Z = A/B Plot the measured points (x,y) and mark for each point the errors Dx and Dy as bars that extend from the plotted point in the x and y directions. The system returned: (22) Invalid argument The remote host or network may be down. |
Error Analysis General Equation
Linearized approximation: pendulum example, variance Next, to find an estimate of the variance for the pendulum example, since the partial derivatives have already been found in Eq(10), all the variables will In Figure 3 there is shown is a Normal PDF (dashed lines) with mean and variance from these approximations. This standard deviation is usually quoted along with the "point estimate" of the mean value: for the simulation this would be 9.81 ± 0.41m/s2. This modification gives an error equation appropriate for standard deviations. his comment is here
Because of the law of large numbers this assumption will tend to be valid for random errors. The PDF for the estimated g values is also graphed, as it was in Figure 2; note that the PDF for the larger-time-variation case is skewed, and now the biased mean This equation has as many terms as there are variables.Then, if the fractional errors are small, the differentials dR, dx, dy and dz may be replaced by the absolute errors The number to report for this series of N measurements of x is where . http://lectureonline.cl.msu.edu/~mmp/labs/error/e2.htm
Error Analysis General Equation
In science, the reasons why several independent confirmations of experimental results are often required (especially using different techniques) is because different apparatus at different places may be affected by different systematic Solve for the measured or observed value.Note due to the absolute value in the actual equation (above) there are two solutions. Zeros between non zero digits are significant. They are also called determinate error equations, because they are strictly valid for determinate errors (not indeterminate errors). [We'll get to indeterminate errors soon.] The coefficients in Eq. 6.3 of the
Systematic errors are errors which tend to shift all measurements in a systematic way so their mean value is displaced. Essentially, the mean is the location of the PDF on the real number line, and the variance is a description of the scatter or dispersion or width of the PDF. We leave the proof of this statement as one of those famous "exercises for the reader". Error Analysis Systems Of Equations In the measurement of the height of a person, we would reasonably expect the error to be +/-1/4" if a careful job was done, and maybe +/-3/4" if we did a
Figure 1 was modified from Measurements and their Uncertainties, Hase and Hughes. Error Analysis Equation Chemistry So, eventually one must compromise and decide that the job is done. Thus the mean of the biased-T g-PDF is at 9.800 − 0.266m/s2 (see Table 1). page All rules that we have stated above are actually special cases of this last rule.
Statistical theory provides ways to account for this tendency of "random" data. Multi-step Equations Error Analysis The number of measurements n has not appeared in any equation so far. General Error Propagation The above formulae are in reality just an application of the Taylor series expansion: the expression of a function R at a certain point x+Dx in terms of Many times you will find results quoted with two errors.
Error Analysis Equation Chemistry
In this simulation the x data had a mean of 10 and a standard deviation of 2. have a peek at these guys This idea can be used to derive a general rule. Error Analysis General Equation Thus, the variance of interest is the variance of the mean, not of the population, and so, for example, σ g ^ 2 ≈ ( ∂ g ^ ∂ T ) Error Analysis Solving Equations A quantity such as height is not exactly defined without specifying many other circumstances.
has three significant figures, and has one significant figure. this content Probable Error The probable error, , specifies the range which contains 50% of the measured values. This is one of the "chain rules" of calculus. For a sufficiently a small change an instrument may not be able to respond to it or to indicate it or the observer may not be able to discern it. Solving Equations Error Analysis Worksheet
These expected values are found using an integral, for the continuous variables being considered here. The variances (or standard deviations) and the biases are not the same thing. Example: Say quantity x is measured to be 1.00, with an uncertainty Dx = 0.10, and quantity y is measured to be 1.50 with uncertainty Dy = 0.30, and the constant weblink It would not be meaningful to quote R as 7.53142 since the error affects already the first figure.
Calculus Approximation From the functional approach, described above, we can make a calculus based approximation for the error. Error Propagation Equation When is this error largest? v = x / t = 5.1 m / 0.4 s = 12.75 m/s and the uncertainty in the velocity is: dv = |v| [ (dx/x)2 + (dt/t)2 ]1/2 =
The second partial for the angle portion of Eq(2), keeping the other variables as constants, collected in k, can be shown to be ∂ 2 g ^ ∂ θ 2 =
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 Clearly, taking the average of many readings will not help us to reduce the size of this systematic error. For example, if the initial angle was consistently low by 5 degrees, what effect would this have on the estimated g? Percent Error Equation In such instances it is a waste of time to carry out that part of the error calculation.
i ------------------------------------------ 1 80 400 2 95 25 3 100 0 4 110 100 5 90 100 6 115 225 7 85 225 8 120 400 9 105 25 S 900 One reason for exploring these questions is that the experimental design, in the sense of what equipment and procedure is to be used (not the statistical sense; that is addressed later), In most circumstances we assume that f(A) is symmetric about its mean. check over here Recall that the angles used in Eq(17) must be expressed in radians.
Especially if the error in one quantity dominates all of the others, steps should be taken to improve the measurement of that quantity. For example, the meter manufacturer may guarantee that the calibration is correct to within 1%. (Of course, one pays more for an instrument that is guaranteed to have a small error.) |
How To Pass The GED Math Test: A GED Math Study Guide
The GED Math Test assesses candidates’ ability to solve math concepts, equations, measurements and applying math concepts to deal with real-life problems. Nevertheless, you are not required to memorize formulas and you will be permitted to use a calculator for a section.
Math is considered a challenging topic for many students, however, you don’t need a “math brain” to be able to pass the GED Math exam. You need to have a thorough preparation and study effective materials. Don’t worry, we are here to help you. This article will shed light on the GED Math Study Guide as well as an overview of the test including the format and content.
1. GED Math Test Overview and Format
To be eligible to sit for the GED Math test, you need to meet the following requirements:
- You are 16 years old or older
- You do not enroll in high school
- You have not graduated from high school
- You meet all additional requirements in your state, for example, the length of time since leaving school.
Depending on your state, the fee to take the GED Math test is $30 or less. It is advisable for students to check out the GED website for the prices in their state.
The GED Math test contains four different topics namely Basic Math, Geometry, Basic Algebra and Graphs and Functions.
There are two parts in the GED Math test and you will be allowed to use a calculator in the second part. You will be given a total of 115 minutes to complete the entire Math test, with a short break between sections. You will access a calculator reference sheet and math formula sheet. The question types include multiple choice, select an area, drop down and fill-in-the-blank.
The GED scoring system is quite complicated. One question doesn’t always carry one point. Several questions are fill-in-the-blank or multiple choice which require you to choose multiple answers. These answers carry multiple points. There are 46 questions on the GED Math test and the passing score is 145. There is no exact number of questions you can miss and still pass; however, according to the GED Testing Service, you must gain about 60%-65% of your points to pass the test.
>>> See More: GED Language Arts Study Guide
2. GED Math Test Topics
When it comes to how to pass the GED Math test, remember that prerequisites to ace the test are knowledge of basic operations like addition, subtraction, multiplication, and division of whole numbers. Although you are allowed to use a scientific calculator in the second part of the test, understanding these basic math operations is important.
Here are details of the topics on the GED Math Test:
2.1. Number Sense
Number sense questions ask candidates to solve math problems associated with:
- Rounding and estimating numbers.
- Ordering numbers on a number line which contains fractions, decimals, and positive & negative numbers
- Calculating ratios, proportions, and percent.
- Calculate exponents, convert fractions to decimals, and vice versa.
- Applying the order of operations rules (PEMDAS).
These topics are not intricate, and you can use a calculator to deal with the majority of these questions.
With the geometry questions, you are asked about:
- Area and perimeter of circles, squares and triangles.
- The volume of a prism, cone, sphere, and pyramid.
You will access Math geometric formulas. You can use these formulas to cope with most Geometry questions.
2.3. Data Analysis
Topics on Data Analysis on the GED Math Test include:
- Calculations of mode, range, mean, and median.
- Finding probability.
These questions can be solved by using provided Math formulas and a scientific calculator.
Algebra handles the following topics:
- Solving quadratic equations.
- Evaluating algebraic expressions and polynomials.
- Evaluating functions.
- Solving equations with one or two variables; systems of equations.
- Solving inequalities
You can use a scientific calculator to solve the majority of Algebra questions.
Graphing problems comprise:
- Slopes and equations of lines
- Coordinate plane questions (points, lines, & shapes).
Three slope formulas are included in the provided Math formula sheet; therefore, you can use it.
2.6. Word problems
The GED Math test contains a variety of word problem questions. These questions carry various topics such as geometry, probability and solving equations. Though they seem to be quite hard, they involve simple math. What you need to do is to translate words into math.
To have a deep insight into the word problem-solving skills, you should take as many practice tests as possible to get familiar with these questions’ format.
Hopefully, the topics which are given above will help you to pass the GED Math test easily.
>>> See Also: The Best Calculator For GED Test
3. GED Math Study Guide: 10 GED Math Test Tips
Apart from providing format as well as common topics on the GED Math test, this article provides 10 top tips to help you how to pass the GED Math test.
3.1. Carefully translate word problems
The first tip on how to pass the GED Math test is ensuring that you firmly grasp the concepts underlying the question in a word problem because they are so common on the GED Math exam. One or two words can change the meaning. Don’t rush these difficult questions, even if the math seems fairly clear!
3.2. Use the whole time
If you complete the test early, review and check your work since you might make some mistakes which could lower your scores.
3.3. Remember the golden rule of scientific notation
In scientific notation, you can express large numbers and small decimals. You should keep in mind this rule: Positive exponents move to the right; meanwhile, the decimal moves to the left when the exponent is negative.
3.4. Use estimation and approximation
If time is nearly up or if the answer choices are far apart, you can try to estimate and approximate, round numbers to the nearest integer as well as attempt to streamline your calculations. This way is extremely useful in the second part where you are not allowed to use a calculator.
3.5. Treat data analysis questions like an open-book test
You wouldn’t keep a reading passage, thus, don’t skip the data. You should read every tiny data including the title, column names as well as the labels for the x and y-axes. Before reading the questions, take notice of the units of measurement. Besides, pay attention to any trends in the data.
3.6. Keep in mind your exponent rules
The next tip on how to pass the GED Math test is adding the exponents when multiplying two terms with the same base. When two terms are divided with the same base, you can subtract the exponent of the denominator from the exponent of the numerator. In case two exponents are separated by a parenthesis, you can multiply them. You can also simplify exponents by rewriting numbers in terms of their exponents.
3.7. Read all the data carefully before solving
The GED test put a considerable emphasis on Data Analysis. You may see tables, charts and graphs in different sections. You shouldn’t jump straight to the questions; take a few minutes to learn about the data presented first.
3.8. Apply your knowledge of number properties
Keep in mind that on the GED test “a number” can mean many things such as a fraction, a decimal, a negative, a positive, zero, a whole number, an odd or even number and so on. Sometimes, it is necessary to consider various numbers. Look for the limitations a question places on a variable and then take note of 2-3 possible values.
3.9. Don’t see your answer among the choices, rewrite it!
The correct answer in the GED exam is sometimes stated as an expression, or not in the simplest form. The answer may still be correct even though it does not appear among answer choices. Make efforts to express it differently until it matches one of the options.
3.10. Believe your first answer
Now, let’s talk about the final tip on how to pass the GED Math test. Apart from reading a question carefully twice, you need to read answer choices carefully twice too. Take some seconds to review your answers. Do not change your answers if you do not see obvious errors which are absolutely sure about. Remember that your first answer is often the right answer. In the GED test, there is no penalty for giving wrong answers. Answer all questions including ones that you are not sure about.
We have offered 10 tips on how to pass the GED Math test. Now it is time to put these tips into use. Get started by taking the GED Practice Math Test on our website now. |
4.1. B Impact Assessment Global Scores
The BIA indicators have different dispersions that result from the indicators’ use of different scales (see Figure 2
). For example, the workers indicator presents the highest median value and the community indicator the highest score. In general, all the BIA indicators present a significant dispersion of the data, except for the governance indicator, which exhibits a higher concentration of the data.
Community is the evaluation B indicator with the highest number of discrepant values, followed by employees, customers, and the environment. Although there is a minimum final value for companies to obtain certification, it appears from the data analyzed that B Corp does not define an evaluation scale for its auditing process.
The data analyzed also allows us to generally verify the distribution of the final scores of the certified B companies (see Figure 3
As represented in the BIA final score chart above, all companies in our sample meet the minimum requirement of 80 points for B Corp certification, and most companies achieve a score between (80–83).
The countries with the highest number of certifications during our sample period are the United States with 681 companies, followed by the United Kingdom with 303 certified companies. Australia and Canada had 171 certifications during the same period. In Europe, besides the United Kingdom as mentioned above, France and Italy stand out with 87 certifications each. Portugal had 10 certifications during the period under observation.
In terms of the activity sector, organizations in the services sector with a minor ecological footprint predominate. The sectoral distribution of the B Corporation can be seen in Figure 4
4.2. B Impact Assessment—Measurement Model Validation
As mentioned above, for reasons of data consistency and elimination of missing values, we reduced the initial sample to 556 cases, corresponding to companies certified in the period from January 2020 to March 2021. In Figure 5
, we analyze the dispersion and median values of the observed variables present in this reduced sample.
The variable with the highest data dispersion is “Land and Life” from the environment indicator. The graph below shows that the audit B variables with a greater range of values are from community and environment indicators. The analysis of the “Mission locked” variable leads us to assume that a fixed rating scale with constant values is used to evaluate this item.
Since Figure 5
shows a high dispersion of the data, an analysis of extreme values was performed. Extreme values are “observations with a unique combination of characteristics identifiable as distinctly different from other observations” [43
] (p. 64). Usually, an extreme value is an observation that presents an unusually high or low value.
To determine extreme values, we use univariate and multivariate detection methods:
(i) In terms of univariate analysis, we consider as an extreme value any observation with a value higher than Q3 + 1.5 × (Q3 − Q1) or lower than Q1 − 1.5 × (Q3 − Q1), where Q3 and Q1 represent quartiles 3 and 1, respectively [44
]. When this proportion exceeds 5%, the impact on descriptive statistics is analyzed.
(ii) Concerning multivariate analysis, we used the Mahalanobis distance (D2
), which performs a multivariate assessment of each observation across a set of variables [43
]. For large samples, an observation with a value greater than three when dividing the Mahalanobis distance (D2
) by the degrees of freedom (df) is considered a possible multidimensional extreme value [43
presents univariate extreme values analysis.
The variable water presents a percentage of extreme values higher than 5%. After analyzing the descriptive measures with and without extreme values (see Table 4
), the 34 observations were removed, since the differences are significant.
Subsequently, the Mahalanobis distance (D2
) was calculated considering 15 degrees of freedom (16 observed variables minus 1). As can be seen in Table 5
, there are two multivariate extreme values in the database.
For the validation of the measurement model, composed of four latent variables (governance, workers, community, and environment) and one observed variable (customer stewardship), a database of 520 companies was considered.
The five variables of the measurement model are intercorrelated. The four latent variables are measured through fifteen observed variables, and errors of measurement associated with each observed variable (e1–e15) are uncorrelated. Since latent variables are unobserved, their metric scale must be guaranteed by observed variables by setting at least one path coefficient of an observed variable or by setting the variance of the latent variable [43
]. We have chosen to standardize the latent variables, setting their variance at 1.
There are several methods for adjusting measurement models. In this research, the maximum likelihood method was chosen. This method provides centered and consistent estimates and is assumed to be robust when the violation of the multivariate normality assumption of the manifest variables occurs [43
presents the measurement model adjusted to a sample of 520 firms, including the values of the standardized factor weights and the individual reliability of each of the observed variables.
A summary table of regression weights with standardized coefficients and statistic tests for each of the observed variables of the B Impact Assessment is presented in Table 6
Of all the variables observed, those that do not seem to contribute positively to the model due to the standardized factor weights are “mission locked” from the governance indicator and “supply chain management” from the community indicator. The negative standard coefficient shown in the model (see Figure 6
and Table 6
) suggests that as the observed variable increases, the latent variable tends to decrease. All standardized estimates calculated are below 1.0. The governance indicator is reflected essentially in the observed variable “ethics and transparency,” the workers indicator in the variable “career development,” the community indicator in the variable “diversity equity and inclusion” and, finally, the environment indicator in the variable “land and life.”
We then proceeded to evaluate the measurement model as a whole, using the adjustment indices (see Table 7
). This analysis determines the goodness-of-fit between the hypothesized model and the sample data. The chi-square statistic with the respective degrees of freedom and the CFI and RMSEA indices are the most reported in the literature [46
]. Additionally, the TLI index also stands out among other indices, with some incidence.
The CMIN/DF (chi-square/degree of freedom) is an absolute index that evaluates the quality of the model per se, without comparison with other models [44
]. The CFI and TLI are relative indices of fit, since they assess the quality of the model relative to the model with the worst possible fit (independence model, in which there are no relationships between the observed variables) and/or the model with the best possible fit (saturated model, in which all the observed variables are correlated) [44
]. Finally, RMSEA is a population discrepancy index that compares the model fit obtained with sample measures (sample means and variances) to the model fit that would be obtained with population measures (population means and variances) [44
presents the values obtained in the adjustment indices and the reference values referred to in the literature to consider a model with a good fit. In addition, the analyzed data allow us to verify that the BIA measurement model is outside the adjustment parameters, revealing an inferior quality of adjustment to the sample.
The variables “mission locked” from the governance indicator and “supply chain management” from the community indicator were eliminated to improve the model fit. Additionally, the modification indices were used, considering that values greater than 11 (p
< 0.001) indicate local adjustment problems (see Table 8
After assessing the theoretical plausibility of the modifications, the measurement errors were correlated, which led to a considerable improvement in the adjustment of the measurement model (see Figure 7
and Table 9
These modifications improved the adjustment of the BIA model, especially regarding the CFI and RMSEA indices. The remaining indices (CMIN/DF and TLI) reveal a sufferable fit of the model to data (see Table 9
As explained in the method section, one of the fundamental assumptions of confirmatory factor analysis is data normality [43
]. When the normality assumption is verified, the maximum likelihood method exhibits properties of consistency, asymptotic efficiency, and asymptotic null bias [44
]. The univariate and multivariate normality analysis is shown in Table 10
According to [44
], the normality assessment should be made through the analysis of the asymmetry (Sk), kurtosis (Ku), and multivariate kurtosis (KuMult) values. In confirmatory factorial analysis, we can assume that there is a severe violation of normality whenever |Sk| > 2–3, |Ku| > 7–10 and |KuMult| > 10 [44
]. Only in an extreme scenario of violation of normality are the quality of the adjustment indices and parameter estimates questionable [44
In this context, the variables fulfil the assumption of univariate and multivariate normality (skew values less than or equal to 1.3; kurtosis values less than or equal to 1 and multivariate kurtosis equal to 3.088).
Another aspect to consider in confirmatory factor analysis is the construct’s reliability. Reliability is a measure of internal consistency, i.e., it measures the degree to which the different observed variables analyze the same aspect. Cronbach’s alpha was used to determine the reliability of the latent variables, since it is one of the most commonly used measures to check the internal consistency of a set of items (see Table 11
Cronbach’s alpha ranges between 0 and 1, and it is acceptable to aggregate items with a value greater than 0.6 [49
]. However, as we can see, the latent variables present reliability problems since the values obtained for Cronbach’s alpha are lower than 0.6, except for the “environment” variable, which presents a higher value (0.723). These reliability problems of the latent variables lead us to conclude that the B Impact Assessment model, as a measurement tool used by B Labs to assess companies’ social and environmental impact, may not always give the same results when applied to structurally similar companies.
Several models were simulated in this research to understand which measurement model structure best fits the data. Of all the models studied, the one where it was possible to obtain the best adjustment indices was the following (see Figure 8
As can be seen, this model presents significant differences from the original model that was tested. First, there was a reduction in the number of latent variables from four to three by eliminating the governance indicator. Additionally, the observed variable (customer stewardship) was removed from the model.
These changes introduced in the B Impact Assessment model, according to Figure 8
, allowed us to achieve the final adjustment values mentioned in Table 12
. As a result, the simplified model exhibited a significantly higher quality of fit than the original model in the sample under study. |
An explanation of the limitations of a previous load transfer article, bringing jacking forces into the mix
BY CLAUDE ROUELLE
First things first. The suspended mass does not rotate around the kinematic roll axis. After reviewing the previous simplified explanation of how load transfer works, this month we’ll explain why it first had to be presented this way, and give a more correct perspective.
In earlier articles we decomposed lateral load transfer into suspended and non-suspended situations. We also broke down the suspended mass load transfer due to lateral acceleration acting on the suspended mass c of g in a geometric and elastic load transfer, the repartition of which depends on the geometric roll centre altitude vs the ground.
We did this by assuming the suspended mass rotates about the roll centre in 2D or roll axis in 3D. Figure 1 is a quick reminder.
What is wrong with this picture? The equilibrium of the moments is respected.
No matter how we decompose it, the roll moment resulting for the centrifugal acceleration acting on the suspended and non-suspended mass cs of g is balanced by the variation of tyre vertical load.
But wait, there aren’t any horizontal opposite lateral forces to the ones acting on the two cs of g.
In figure 2, we only look at the decomposition of the suspended mass centrifugal applied at the kinematic roll centre and the geometric load transfer (red) of the suspended mass. The non-suspended mass load transfer (shown in green in figure 1) and the elastic part of the suspended mass load transfer (yellow in figure 1) are not represented here.
At least there is an equilibrium of the centrifugal force: F =M* V2/R and the reaction at the outside and inside tyres. Still two things are fundamentally wrong in this sketch though. Firstly, there is little chance the kinematic roll centre would stay in the same position once the car gets some tyre and suspension deflection and, secondly, the tyres’ lateral forces cannot be equal as the outside tyre is more loaded than the inside one.
Figure 3 therefore represents a more realistic perspective, with a more pragmatic roll centre position and distribution of lateral forces between the tyres.
We can now observe that the outside and inside geometric load transfers are unequal. This is what imposes jacking force and subsequent ride height variation. Depending on the positive or negative difference between the outside and inside geometric load transfers, the car suspended mass could be dynamically lifted or pushed down.
The suspension stiffness and kinematics, the distribution of inside and outside lateral forces (that depends on so many factors: slip angle; vertical load; camber; pressure; temperature…), and the amount of lateral acceleration will dictate the ride height variation. It could be in the order of 1mm with stiff suspension or as much as 10mm with soft suspension.
It is interesting to note that the jacking forces and resultant ride height variations could have a significant influence on the aerodynamic downforce and downforce distribution, both of which are front and rear ride height sensitive.
Skid pad testing
Figure 4 shows the front and rear ride height variation observed on a skid pad with a car driven on a constant radius at increasing speed (it is worth mentioning the test was conducted on a car with moderate aerodynamic downforce). We can observe an increasing rear ride height and a decreasing front ride height, most probably due to a rear roll centre above the ground and a front roll centre below the ground.
Front and rear ride height variations vs lateral acceleration observed in a variable velocity skid pad test
This type of test can be used to find jacking at different lateral accelerations. Then, using damper potentiometers and knowledge of suspension motion ratios, the suspended mass vertical movement from jacking can also be found.
The main reason why the simplified explanation on load transfers without jacking forces was not developed in previous articles is that knowledge of the front and rear side forces and their distribution depends on the tyre model, and not everybody has access to that.
Elaborating further, figure 5 shows the classical way kinematic roil centre is found.
In figure 6, we apply a side force on the outside tyre (green) and an inside tyre force (red), the total of which are logically equal to the suspended mass multiplied by the lateral acceleration. The angles θIci and θIco are determined by the ground line and the line going from each tyre contact patch to its respective instant centre. These angles will determine the amount of the jacking forces.
No matter how we decompose it, the roll moment resulting from the centrifugal acceleration acting on the suspended and non-suspended mass cs of g is balanced by the variation of tyre vertical load
Traditional definition of kinematic roll centre. The inside wheel instant centre of rotation (red) and the outside wheel instant centre of rotation (green) about the suspended mass are determined by the suspension kinematics
For each tyre, the angle between the ground line and the instant centre to contact patch line, and the side force applied at the tyre, determines the amount of geometric load transfer (red vectors). These forces are pointing in opposite directions but, if they have unequal absolute value, a jacking force is created
A force acting on the line tyre contact patch to its instant centre would not create any wheel movement vs the suspended mass
We realise that with an unchanged suspension kinematic (and therefore the same θIci and θIco angles), if we had a different tyre lateral forces distribution, we would have had a different jacking force.
In figure 7, we have the same kinematics, the same instant centres position, the same kinematic roll centre but a different distribution of inside and outside forces (that could come, for example, from different tyre temperatures). Despite that, their total is unchanged and still equal to the product of suspended mass by the lateral acceleration. It is the different tyre side force distribution that affects the jacking force.
The same consideration could be made by keeping the same tyre side force distribution and modifying the kinematics.
Same kinematics as in figure 6 but with a different side force distribution results in a different jacking force
In fact, more than the kinematic roll centre, it is the angles θIci and θIco that determine the jacking forces. The proof is that a force acting on the line tyre contact patch to its instant centre would not create any wheel movement vs the suspended mass. This validates the definition of geometric load transfer that only produces forces in the suspension linkages with no spring, damper or anti-roll bar movements.
The last thing to note is that if the kinematics and / or the tyre side force distribution affects the geometric load transfer, jacking forces and ride height, it also affects the elastic part of the suspended mass load transfer, as seen in figure 8.
No matter what, load transfer is only a function of mass, lateral acceleration, track width and c of g height. The total of the non-suspended mass load transfer (green) and the suspended mass load transfer (red and yellow) remains the same.
What could change is the distribution of the geometric (red) and elastic (yellow) parts of the suspended load transfer.
However, as their total do not change, if we have more geometric, we will get less elastic and vice versa.
Analysis of all vertical load and load variation on the tyres in a 2D simplified view
No matter what, load transfer is only a function of mass, lateral acceleration, track width and c of g height
As the elastic load transfer affects the forces and movements of springs, dampers and anti-roll bars, we understand that kinematics and tyre side force distribution will also affect roll angle and car attitude vs the ground. These are all different perspectives of the same load transfer and its consequences. |
In this paper a new integral transform namely Elzaki transform was applied to Elzaki transform was introduced by Tarig ELzaki to facilitate the process of. The ELzaki transform, whose fundamental properties are presented in this paper, is little known and not widely The ELzaki transform used to. Two -analogues of the Elzaki transform, called Mangontarum -transforms, are introduced in this paper. Properties such as the transforms of.
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Mangontarum -Transform of the Second Kind Since there can be more than one -analogue of any classical expression, we can define another -analogue for the Elzaki transform. View at Google Scholar T.
One may read [ 1 — 3 ] and the references therein for guidance regarding this matter. Since then the Mangontarum -transform of 72 yields Applying 61 yields the solution Example Since then the Mangontarum -transform of 72 yields Applying 61 yields the solution. Applying Theorem 3we have Hence, Using the inverse -transform in 61 yields the solution Example Ifwe say that is an inverse Mangontarum -transform of the first kindor an inverse -transform of the functionand we write Observe that linearity also holds for the inverse -transform of the function.
Replacing with in 19 yields The next theorem is obtained by multiplying both sides of this equation by. For a positive integer andone has.
On a -Analogue of the Elzaki Transform Called Mangontarum -Transform
Now, by 28 and 27Similarly, Now, if we define the hyperbolic -sine and -cosine functions as we have the next theorem which presents the transforms of the -trigonometric functions. View at Google Scholar L. Theorem 7 -derivative of transforms.
From [ 10 ], for any positive integerwhere. Other important tools in this sequel are the Jackson -derivative and the definite Jackson -integral Note that elza,i 14 and 15we have Furthermore, given the improper -integral of we get see [ 10 ]. From 12Hence, new -analogues of the sine and cosine functions can be defined, respectively, as see [ 10 ] and the new -hyperbolic sine and cosine functions as By application ofit is easy to obtain the next theorem.
For a positive integer andMuch is yet to be discovered regarding the Mangontarum -transforms. Reidel Publishing, Dordretcht, The Neatherlands, In the following examples, the effectiveness of the Mangontarum -transform of the first kind in solving certain initial value problem involving ordinary -differential equations is illustrated. View at Google Scholar W. View at Google Scholar V. Given the -exponential function the -sine and -cosine functions can be defined as where.
The case when is justified by Let Then, we have the following definition. Clearly, The expressions in 9 are called -integer-falling factorial of of order-factorial ofand -binomial coefficient or Gaussian polynomialrespectively.
In this paper, we will define two kinds of -analogues of the Elzaki transform, and to differentiate them from other possible -analogues, we will refer to these transforms as Mangontarum -transforms. Note that 82 may be expressed as Hence, from 86we have the following results: Consider the first degree -differential equation: To receive news and publication updates for Discrete Dynamics in Nature and Society, enter your email address in the box below.
Since there can be more than one -analogue of any classical expression, we can define another -analogue for the Elzaki transform. Two -analogues of the Elzaki transform, called Mangontarum -transforms, are introduced in this paper. Note that when and. The Mangontarum -transform of the second kind, denoted byis defined by over the setwhere and.
Let be the -Laplace transform of the second kind of. To obtain this, the following definition is essential. Clearly, This makes 22 a -analogue of the Elzaki transform in 2.
Discrete Dynamics in Nature and Society
Then the Mangontarum -transform of the first kind satisfies the relation Remark 5. Ifthen by Theorem 3 Thus, taking the Mangontarum -transform of the first kind of both sides of 68 gives us From the inverse -transform in 61we have the solution Example That is, a polynomial is said to be a -analogue of an integer if by taking its limit as tends towe recover. Ifthen by Theorem 3 Thus, taking the Mangontarum -transform of the first kind of both sides of 68 gives us From the inverse -transform in 61we have the solution.
Researchers are encouraged to further investigate other applications of these -transforms, especially the second kind. Given the set Elzaki [ 1 ] introduced a new integral transform called Elzaki transform defined by for, and. In this section, we will consider applications of the Mangontarum -transform of the first kind to some -differential equations.
Indexed in Science Citation Index Expanded. Find the solution of the equation where and with and.
Find the solution of elzakii and. Hence, the transtorm of 76 is Further simplifications yield The inverse -transform in 61 gives the solution 4. Theorem 3 transform of -derivatives. Theorem 19 -derivative of transforms. Abstract Two -analogues of the Elzaki transform, called Mangontarum -transforms, are introduced in this paper.
The known – product rule of differentiation is given by Thus, we have Let so that. Subscribe to Table of Contents Alerts.
From [ 10 ], the -derivative of the -Laplace transform of the first kind is Replacing with and applying Theorem 6 yield Hence, the following theorem is easily observed. |
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Why Beauty Is Truth: A History of Symmetry Paperback – April 29, 2008
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Anyone who thinks math is dull will be delightfully surprised by this history of the concept of symmetry. Stewart, a professor of mathematics at the University of Warwick (Does God Play Dice?), presents a time line of discovery that begins in ancient Babylon and travels forward to today's cutting-edge theoretical physics. He defines basic symmetry as a transformation, "a way to move an object" that leaves the object essentially unchanged in appearance. And while the math behind symmetry is important, the heart of this history lies in its characters, from a hypothetical Babylonian scribe with a serious case of math anxiety, through Évariste Galois (inventor of "group theory"), killed at 21 in a duel, and William Hamilton, whose eureka moment came in "a flash of intuition that caused him to vandalize a bridge," to Albert Einstein and the quantum physicists who used group theory and symmetry to describe the universe. Stewart does use equations, but nothing too scary; a suggested reading list is offered for more rigorous details. Stewart does a fine job of balancing history and mathematical theory in a book as easy to enjoy as it is to understand.Line drawings. (Apr.)
Copyright © Reed Business Information, a division of Reed Elsevier Inc. All rights reserved. --This text refers to an out of print or unavailable edition of this title.
*Starred Review* Werner Heisenberg recognized the numerical harmonies at the heart of the universe: "I am strongly attracted by the simplicity and beauty of the mathematical schemes which nature presents us." An accomplished mathematician, Stewart here delves into these harmonies as he explores the way that the search for symmetry has revolutionized science. Beginning with the early struggles of the Babylonians to solve quadratics, Stewart guides his readers through the often-tangled history of symmetry, illuminating for nonspecialists how a concept easily recognized in geometry acquired new meanings in algebra. Embedded in a narrative that piquantly contrasts the clean elegance of mathematical theory with the messy lives of gambling, cheating, and dueling mathematicians, the principles of symmetry emerge in radiant clarity. Readers contemplate in particular how the daunting algebra of quintics finally opened a conceptual door for Evaniste Galois, the French genius who laid the foundations for group theory, so empowering scientists with a new calculus of symmetry. Readers will marvel at how much this calculus has done to advance research in quantum mechanics, relativity, and cosmology, even inspiring hope that the supersymmetries of string theory will combine all of astrophysics into one elegant paradigm. An exciting foray for any armchair physicist! Bryce Christensen
Copyright © American Library Association. All rights reserved --This text refers to an out of print or unavailable edition of this title.
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The journey begins way back in Babylonia where they began to understand how to solve equations. From there we travel forward to Euclid and his "Elements of Geometry" and the discovery of the concepts of proofs and axioms. Algebra arrives on the scene around 830 CE when the developments moved from the Greek world to the Arabic. In that year, Mohamed al-Khwārizmī wrote a book called "al-Jabr" from which comes our word "algebra." There are contributions from others, such as Omar Khayyam (cubic equations), and the Greek mathematician Menaechmus (conic sections). In the 16th century, Girolamo Cardano wrote a book subtitled "The Rules of Algebra," where he assembled methods for solving not only quadratic equations, but also cubic and quartic equations using pure algebra. By the 18th century, Carl Friedrich Gauss developed what is now called the Fundamental Theorem of Algebra. His worked was followed up by his student Georg Bernhard Riemann who generalized Gauss's work on multidimensional spaces creating, in effect, a theory of curved multidimensional spaces - a concept that would later proved crucial in Einstein's work on gravity. We also get an introduction to the contributions of other notables such as Joseph-Louis Lagrange, Paolo Ruffini, and Hans Mathias Abel.
We learn of the gifted Evariste Galois and his contributions to group theory. Steward notes that in Galois's hands mathematics ceased to be a study of numbers and shapes or arithmetic and geometry but became a study of structure. The study of "things" became a study of processes - this followed on the work of Lagrange, Ruffini, and Abel. Galois is the first to "appreciate that mathematical questions could be sometimes best understood by transporting them into a more abstract realm of thought." We learn that a symmetry of an object is some transformation that preserves the object's structure, and Galois's symmetries were permutations of the roots of a an equation. Incidentally, group theoretic methods eventually came to dominate quantum mechanics in the 20th century, "because the influence of symmetry is all-pervasive."
In the 1800s, William Hamilton, almost three hundred years after Cardano indicated that imaginary numbers might be useful, removed the geometric element and reduced complex numbers to pure algebra, and he discovered a new type of numbers called quaternions. In that same century Marius Sophus Lie discovers Lie groups and Lie algebras. Steward notes that these theories "have pervaded almost every branch of mathematics [...] Symmetry is deeply involved in every area of mathematics, and it underlies most of the basic ideas of mathematical physics." Wilhelm Karl Joseph Killing, who Steward calls the "greatest mathematician who ever lived," expanded on Lie's work.
In the latter 19th century, the work of Faraday, Maxwell, Hertz, and Marconi lead the way to the work of one of the greatest minds yet - Albert Einstein. The story gets more intriguing with the discovery of octonions (another type of algebra) in 1843. It was realized that the octonions were the source of the most "bizarre algebraic structures know to mathematics. They explain where Killing's five exceptional Lie groups [...] really come from," and the one group "shows up twice in the symmetry group that forms the basis of 10-dimensional string theory." String theory is one that is prominent in the quest for a "theory of everything" today. Steward then notes that this "opens up an intriguing philosophical possibility: the underlying structure of our universe, which we know to be very special, is singled out by its relationship to a unique mathematical object: the octonions."
The interesting thing is that the potential of many of these mathematical concepts was not know or appreciated at the time of discovery. Yet today, we can look back and see the beauty of it all, and how it all seems to fit together. Steward notes that symmetry "is fundamental to today's scientific understanding of the universe and its origin." It pervades everything from the world of quantum physics to Einstein's concept of relativity.
I have read many books on symmetry and this one provides in my judgment one of the best entry points. Ian Stewart has an engaging style that builds momentum as you go along.
The blue butterfly on the cover is now a very opportune symbol of symmetry. The 2014 Fields Medal (equivalent to the Nobel prize) was given to Artur Avila for his work on nonlinear dynamical systems of which the Lorenz model is prototypical. This model is usually symbolized by a butterfly. In Oriental cultures the butterfly is a symbol of transformation. Symmetry is defined mathematically as invariance under transformations.
Phil Anderson, a Nobel laureate in physics, famously wrote "it is only slightly overstating the case to say that physics is the study of symmetry." If you want to know why and, as importantly, if you want to know how symmetry can be a key concept for the social sciences and even your own life, you should read this book.
I have only just started on 'Concepts of Modern Mathematics' and my first impression is a vindication of "one picture is worth many words".
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Capitalization Rate and Cash on Cash Return
An income property is generally valued on its cash flow, either the current cash flow or the anticipated future cash flow after repositioning. The valuation of the cash flow is calculated according to the cost and structure of the available financing at the time of purchase or refinancing.
The fundamental equation that is used (and abused) is:
NOI: Net Operating Income
CPV: Capitalized Property Value
CAP: Capitalization Rate
CAP = NOI÷CPV
Buying the income property for all cash CPV will generate an annual cash flow of NOI. Thus, the return rate on the equity investment is the CAP rate. However, most income properties are purchased with a combination of debt and equity to increase the yield on the equity through leverage.
Many real estate brokers, including CCIM (Certified Commercial Investment Members), abuse the CAP rate by saying that it represents the risk or safety in a particular market or submarket according to what other comparable properties have sold in the recent past. This is nonsense.
Others also point to competitive investments, like US Treasury Bonds and Notes, to compare the yields with a CAP rate. This is also nonsense.
The CAP rate is not a qualitative measure of risk, but rather it is a quantitative measure of blended yield requirements for both debt and equity.
As I mentioned above, the CAP rate is purely a function of the cost and structure of available financing that reflects the required yields of the investment tranches. Income properties are almost always purchased with a combination of debt tranches and equity tranches. Each tranche requires a minimum yield, referred as a “Debt Constant”, “Loan Constant”, “Mortgage Constant”, or “Yield Constant”. The constant is simply the periodic cash flow divided by the initial investment. The annual constant is the periodic constant multiplied by the number of cash flow periods in one year. Debt amortization is not considered, as it is separate agreement between the lender and the borrower on whether and how the Debt Present Value (DPV) will reduce over time. Two different loans can have the same Debt Constant, yet one loan is amortizing and the other is interest only. The tranche yield calculations are performed as if there is no change over time in debt or equity.
For debt, the Annual Debt Constant (ADC) is calculated as the Annual Debt Service (ADS) divided by the initial Debt Present Value (DPV).
ADS: Annual Debt Service
DPV: Initial Debt Present Value
ADC: Annual Debt Constant
ADC = ADS÷DPV
For equity, the Annual Yield Constant (AYC) is also called the Gross Return on Equity (ROE) or the Cash on Cash Return (CCR). The ROE is calculated as the net Cash Flow before Tax (CFBT) divided by the initial equity investment (EQTY).
CFBT: Cash Flow before Tax
EQTY: Initial equity investment
LTV: Loan to Value
ETV: Equity to Value
ROE: Gross Return on Equity (also called the Cash on Cash Return [CCR])
LTV = DPV÷CPV
ETV = 1 – LTV
CFBT = NOI – ADS
EQTY = ETV×CPV
ROE = CFBT÷EQTY
I reviewed a commercial income property certified appraisal. The appraiser mentioned a few different ways of calculating the CAP rate:
- The “mortgage equity” method, which is actually a weighted average of the yields for debt and equity.
- The “Debt Coverage” formula, which is used by many bankers that considers the Debt Coverage Ratio (DCR) applied to the Net Operating Income (NOI).
The “mortgage equity” calculation is:
CAP = ADC×LTV + ROE×ETV
The “Debt Coverage” formula calculation is:
DCR: Debt Coverage Ratio
DCR = NOI÷ADS
CAP = DCR×ADC×LTV
Most appraisers don’t understand that the two formulae, “mortgage equity” and “debt coverage” are mathematically identical. Most commercial real estate brokers also don’t understand this mathematical relationship. The appraiser simply guessed at the ROE for the “mortgage equity” method to calculate a CAP rate from the published debt parameters of a few commercial lenders. Then the appraiser guessed at a DCR for the “Debt Coverage” formula to calculate a different CAP rate. CAP rates are not guess work.
The “Debt Coverage” formula is the easier choice for calculating the CAP rate, because it is mathematically identical to the original calculation CAP = NOI÷CPV. Also, the DCR is a critical measure of how well the financing can tolerate recessionary pressures on income and expense. A low CAP rate means a low DCR that risks negative cash flow (default) in the event of reduced income or increased expenses.
Also, the ROE is derived precisely from the same three variables, DCR, ADC, and LTV.
Finally, the weighted average in the “mortgage equity” method “CAP = ADC×LTV + ROE×ETV” is identical to the “Debt Coverage” formula “CAP = DCR×ADC×LTV” .
The two equations initially, and , appear to be quite different. A little substitution of variables will demonstrate the equality.
First, derive the CAP rate in terms of the DCR, ADC, and LTV:
CAP = DCR×ADC×LTV = (NOI÷ADS)×(ADS÷DPV)×(DPV÷CPV)
CAP = (NOI×ADS×DPV)÷(ADS×DPV×CPV)
CAP = DCR×ADC×LTV = NOI÷CPV
Next, derive the ROE in terms of the DCR, ADC, and LTV:
DCM: Debt Coverage Margin
CFM: Cash Flow Margin
Leverage Ratio: LR
Leverage to Debt: LTD
LR = 1÷ETV
LTD = LTV÷ETV = LR – 1
DCM = 1÷DCR = ADS÷NOI
CFM = 1 – DCM = 1 – 1÷DCR = 1 – ADS÷NOI = NOI÷NOI – ADS÷NOI
CFM = (NOI – ADS)÷NOI = CFBT÷NOI
CFM = 1 – 1÷DCR = DCR÷DCR – 1÷DCR = (DCR – 1)÷DCR
ADS = DCM×NOI
CFBT= CFM×NOI
ROE = CFBT÷EQTY = (CFM×NOI)÷(ETV×CPV) = CFM×(NOI÷CPV)÷ETV
ROE = CFM×CAP÷ETV = CFM×DCR×ADC×LTV÷ETV = CFM×DCR×ADC×LTD
ROE = (1 – 1÷DCR)×DCR×ADC×LTD = ((DCR – 1)÷DCR)×DCR×ADC×LTD
ROE = (DCR – 1)×ADC×LTD
At this point, a critical investment concept that many real estate brokers and investors miss is the “Leverage to Yield” (LTY). The LTY is the ROE divided by the CAP. The ratio is the multiplier on the CAP rate to produce the ROE. The LTY measures the true leverage on the equity investment. If the ratio is less than 1.00, then the equity investment has “negative leverage”. If the ratio exactly equals 1.00, then the equity investment has “neutral leverage”. If the ratio is greater than 1.00, then the equity investment has “positive leverage”.
LTY: Leverage to Yield
LTY = ROE÷CAP = (CFBT÷EQTY)÷(NOI÷CPV)
LTY = ((CFM×NOI)÷(ETV×CPV))÷(NOI÷CPV)
LTY = ((CFM÷ETV)×(NOI÷CPV))÷(NOI÷CPV) = CFM÷ETV
Therefore, to achieve positive leverage, the Cash Flow Margin (CFM) must be greater than the Equity to Value ratio (ETV). Avoid neutral or negative leverage, because the return rate is less than investing all cash. Non-positive leverage risks default on the debt and losing the equity in foreclosure.
Finally, demonstrate that the “mortgage equity” method (weighted average) is mathematically identical to the “Debt Coverage” formula :
CAP = ADC×LTV + ROE×ETV = ADC×LTV + (DCR – 1)×ADC×LTD×ETV
CAP = ADC×LTV + (DCR – 1)×ADC×LTV
CAP = ADC×LTV + (DCR×ADC×LTV – ADC×LTV)
CAP = DCR×ADC×LTV
From these identities, the appraiser guessing at the ROE in or at the DCR in is merely guessing at the effective CAP rate. That’s why he calculated 2 different CAP rates for the same property, and then averaged the CAP rates and added a “fudge factor”. Calculating the true CAP rate is not a guessing game. The CAP rate is calculated by the cost and structure of financing.
These same equations hold true for multiple tranches of both debt and equity. The yield constant is calculated similar to by a weighted average of the individual tranche yields divided by the sum of the tranche ratios relative to the total property value.
PLTV: Primary Loan to Value ratio
PDC: Primary Debt Constant
SLTV: Secondary Loan to Value ratio
SDC: Secondary Debt Constant
CLTV: Combined Loan to Value
WDC: Weighted Debt Constant
CLTV = PTLV + SLTV
WDC = (PDC×PLTV + SDC×SLTV)÷CLTV
Therefore, when a buyer is negotiating with a seller for a purchase price, the negotiation involves the CAP rate that determines the price. The property can afford to pay a particular debt service, for a particular DCR and ADC applied to the NOI. Thus, negotiating the CAP rate is actually negotiating the remaining variable LTV, which is actually negotiating the ETV (how much equity the seller has in the property). All of that depends on the cost and structure of the financing that is available to the buyer, and not on what other properties have fetched. The buyer cannot recreate the cost and structure of financing on those other comparable properties. Therefore, there is no sensible definition of “comparable CAP rates”.
When you are buying an income property, you are actually buying an income stream (NOI). Negotiate a CAP rate that fits the cost and structure of the financing that is available to you, and then divide the actual NOI by the calculated CAP to arrive at a price that the property can afford to pay.
Jeffrey D. Smith
Download a PDF version of this article by clicking here. |
So the other day Julia Galef and I had the pleasure of interviewing mathematical cosmologist Max Tegmark for the Rationally Speaking podcast
. The episode will come out in late January, close to the release of Max’s book
, presenting his Mathematical Universe Hypothesis (MUH). We had a lively and interesting conversation, but in the end, I’m not convinced (and I doubt Julia was either).
The basic idea is that the ultimate structure of reality is, well, a mathematical one. Please understand this well, because it is the crux of the discussion: Tegmark isn’t saying anything as mundane as that the world is best described by mathematics; he is saying that the ultimate nature of reality is mathematics.
This is actually not at all a new thesis, though Max is advancing it in new form and based on different reasoning then before. Indeed, the idea has a long philosophical history, and can fruitfully be thought of as based on two distinct philosophical positions: Pythagoreanism, or mathematical Platonism; and Mathematical monism.
Mathematical Platonism is the idea that mathematical structures are real in a mind-independent fashion. They are not “real” in the same sense as, say, chairs and electrons, but they do have an ontological status independent of the human (or any other) mind. As readers of this blog know, I’m actually sympathetic to (though not necessarily completely on board with) mathematical Platonism. The best point in its favor is the so-called “no miracles” argument, the idea that mathematics is too unreasonably effective (at predicting things about the world) for it to be just a human invention, rather than somehow part of the inherent fabric of the world. (Interestingly, this argument is equivalent to one by the same name advanced by scientific realists to claim that science really does describe — approximately — how the world is, as opposed to the antirealist position that the only thing we can say about science is that it is empirically adequate.)
Mathematical monism is the stronger doctrine that not only
are mathematical structures real, but they are the only real thing out there (or, more precisely, everywhere).
The combination of Platonism and monism yields a class of theories about the ultimate nature of reality, of which Tegmark’s MUH is one example. We have seen another one several times in the past, in the form of Ladyman and Ross’ ontic structural realism, the notion that there are no “objects” or “things” at the bottom, just (mathematical) relations.
While I have commented positively on ontic structural realism (again, without necessarily buying into it), and more generally on the idea of a “naturalistic” metaphysics (i.e., a metaphysics that takes seriously the best known physics), my conversation with Max Tegmark actually generated more doubts than illumination.One obvious problem is posed by what it would mean for the world to be “made of” mathematical structures. The notion of mathematical structure is well developed, so that’s not the issue. A structure, strictly speaking, is a property or a group of mathematical objects that attach themselves to a given set. For instance, the set of real numbers has a number of structures, including an order (with any given number being either less or more than another number), a metric (measuring the distance between points in the set), an algebraic structure (the operations of addition and multiplication), and so on.
The problem is in what sense, if any, can a mathematical structure, so defined, actually be the fundamental constituent of the physical world, i.e. being the substance of which chairs, electrons, and so on, are made.
Of course, both Julia and I asked Max that very question, and we were both very unconvinced by his answer. When Tegmark said that fundamental particles, like electrons, are, ultimately mathematical in nature, Julia suggested that perhaps what he meant was that their properties are described by mathematical quantities. But Max was adamant, mentioning, for instance, the spin (which in the case of the electron has magnitude 1/2). Now, the spin of a particle, although normally described as its angular momentum, is an exquisitely quantum mechanical property (i.e., with no counterpart in classical mechanics), and it is highly misleading to think of it as anything like the angular momentum of a macroscopic object. Nevertheless, Julia and I insisted, it is a physical property described by a mathematical quantity, the latter is not the same as the former.
Could it be that theories like MUH are actually based on a category mistake? Obviously, I’m not suggesting that people like Tegmark make the elementary mistake of confusing the normal meaning of words like “objects” and “properties,” or of “physical” and “mathematical.” But perhaps they are making precisely that mistake in a metaphysical sense?
There are other problems with MUH. For one, several critics of Tegmark’s ideas have pointed out that they run afoul of the seemingly omnipresent (and much misunderstood) Gödel’s incompleteness theorems. Mark Alford, specifically, during a debate with Tegmark and Piet Hut has suggested that the idea that mathematics is “out there” is incompatible with the idea that it consists of formal systems. To which Tegmark replied that perhaps only Gödel-complete mathematical structures have physical existence (something referred to as the Computable Universe Hypothesis, CUH).
This, apparently, results in serious problems for Max’s theory, since it excludes much of the landscape of mathematical structures, not to mention that pretty much every successful physical theory so far would violate CUH. Oops.
Prompted by the above, I also asked Max about Gödel, and his response was that Gödel-related problems appear only in the case of infinite quantities, and he professed himself to be an infinity-skeptic. That took me by surprise, what do you mean you don’t believe in infinity? I thought this was a pretty darn well established concept in mathematics, at least since the work of Georg Cantor in the 19th century! But of course Tegmark was referring to the existence of physical, not mathematical, infinities. As is well known, there are certain calculations in physics that do generate infinities, for instance the singularity that shows up in the description of black holes, or the infinite quantities that are postulated in standard descriptions of phase transitions. The question of whether there really are infinities in physical systems is open, so surely Max is entitled to his skepticism. But it did seem a bit too convenient a position, in light of the above mentioned Gödel-related problems.
Another issue that didn’t convince either Julia or me during our conversation with Max is a crucial one: testability. I’m okay with philosophical speculations (and I use the term in a positive fashion!) about modal realism or the principle of plenitude, but if we are claiming to be doing science (as Tegmark surely is), then our speculations better make contact with empirical reality. Jim Baggott, in his Farewell to Reality: How Modern Physics Has Betrayed the Search for Scientific Truth, is already accusing physicists of losing touch with what it means to do science. Is Tegmark the latest example of the trend?
When we asked, he claimed that the MUH does make empirical predictions, but when pressed on the details the answer becomes far less satisfying than one would hope. For instance, Max said that one prediction is that physics will continue to uncover mathematical regularities in nature. Well, probably, but one surely doesn’t need to postulate MUH to account for that. He also has stated in the past that — assuming we live in an average universe (within the multiverse of mathematical structures) — then we “start testing multiverse predictions by assessing how typical our universe is.” But how would we carry out such tests, if we have no access to the other parts of the multiverse?
Max went on to say that his hypothesis has “zero free parameters” and is therefore favored by Occam’s razor. But if you check his paper at arxiv.org he says: “If this theory is correct, then since it has no free parameters, all properties of all parallel universes … could in principle be derived by an infinitely intelligent mathematician. … Finally, the ultimate ensemble of the Level IV multiverse would require 0 bits to specify, since it has no free parameters.” There are a couple of obvious problems here. One is the dearth of infinitely intelligent mathematicians, the second the fact that the above mentioned Level IV multiverse is precisely what gets dramatically (and unrealistically) shrunk as a result of Gödel-imposed limitations. And let’s not forget that Occam’s razor is just a useful heuristic, it should never be used as the final arbiter to decide which theory is to be favored, especially when we are talking about such highly speculative and empirically next to impossible (or even downright impossible) ideas to test.
In Many Worlds in One: The Search for Other Universes, critic Alex Vilenkin says that “the number of mathematical structures [in the multiverse] increases with increasing complexity, suggesting that ‘typical’ structures should be horrendously large and cumbersome. This seems to be in conflict with the beauty and simplicity of the theories describing our world.” In order to get around that problem, Tegmark assigns lower weights to more complex structures, but since this is done without a priori justification, it is an ad hoc move, which of course violates Occam’s razor. So, as much as I enjoyed our conversation with Max, for the time being I remain skeptical of the MUH and related hypotheses. Maybe we just need to wait for the appearance of an infinitely intelligent mathematician.
[This just in from Max Tegmark himself!]
Thanks Massimo for the fun conversation during the interview and for raising these important questions! They are excellent ones, and a key reason why I spent three years writing this book is because I wanted to make sure to finally answer them all properly. Needless to say, I couldn't do justice to them in our short interview, so I'm very much look forward to hear what you think about my detailed answers in chapters 6, 10, 11 and 12. I think you'll find that our viewpoints are closer than your post suggest - for example, your statement "Tegmark assigns lower weights to more complex structures" is not something you'll find in the book. Rather, I describe how the measure problem is a terrible embarrassment for modern cosmology (regardless of whether the MUH is true or not) that we need to solve, and that our untested assumption that truly infinite things exist in nature are my prime suspect: we've never measured anything to better than 17 decimal places, have only 10^89 particles in our universe, and manage to do all our publishable physics simulations with computers that have finite resources, so even though my physics courses at MIT use infinity as a convenient tool, I respectfully object to your "OPS" argument that we somehow have experimental evidence for infinity in physics. Without infinity, there are, as you say, no Gödel issues in our physics.
I look forward to continuing this interesting conversation! ;-)
Originally on Rationally Speaking |
A laser distance meter is an instrument that accurately measures the distance to a target using laser technology. During operation, the laser distance meter emits a thin laser beam towards the target, and the reflected laser beam is captured by photoelectric components. The timer measures the time it takes for the laser beam to travel from emission to reception, calculating the distance from the observer to the target. Laser distance meters are known for their lightweight, compact size, simple operation, speed, and high accuracy, with an error rate only one-fifth to a few hundredths compared to other optical range finders. In this article, sisco will introduce the working principles of laser distance meters for a better understanding.
Basic Principles of Laser Distance Meter
Laser distance meters generally adopt two measurement methods: pulse method and phase method.
In the pulse method, the laser emitted by the distance meter is reflected by the measured object and received by the distance meter. Simultaneously, the distance meter records the time taken for the laser to travel back and forth. The distance between the distance meter and the measured object is calculated as half the product of the speed of light and the round-trip time of the laser. The accuracy of distance measurement in the pulse method is typically around +/-1 meter, and these devices have a measurement blind zone of approximately 15 meters.
Laser ranging is a method of distance measurement using light waves. If light travels with a speed 'c' in the air between points A and B, the distance 'D' between A and B can be expressed as follows:
- D is the distance between points A and B.
- c is the speed of light in the atmosphere.
- t is the time required for light to travel round trip between A and B.
From the above formula, it is evident that measuring the distance between A and B is essentially measuring the time 't' it takes for light to propagate. Depending on the measurement method, laser distance meters are typically classified into pulse-type and phase-type.
A phase-type laser distance meter uses radio frequency modulation to modulate the amplitude of the laser beam. It measures the phase delay generated by the modulation light during one round-trip and then calculates the distance represented by this phase delay based on the wavelength of the modulation light. It indirectly determines the time required for light to travel round trip.
Phase-type laser distance meters are commonly used in precise distance measurements. Due to their high accuracy, typically in millimeters, and to ensure effective reflection signals, these distance meters are equipped with reflectors known as cooperative targets. Assuming the angular frequency of the modulation light is ω, and the phase delay produced during one round trip on the distance D to be measured is φ, the corresponding time t can be expressed as:
Substituting this relationship into the distance formula, D can be represented as:
D=1/2 ct=1/2 c*φ/ω=c/(4tf)(Nπ+Δφ)
- φ is the total phase delay produced during one round trip of the signal.
- ω is the angular frequency of the modulation signal, ω=2πf.
- U is the unit length, equal to 1/4 of the modulation wavelength.
- N is the number of half wavelengths included in the measurement line.
- Δφ is the part of the phase delay produced during one round trip of the signal that is less than π.
- ΔN is the fractional part of half the modulation wave included in the measurement line, ΔN=φ/ω.
Under given modulation and standard atmospheric conditions, the frequency c/(4πf) is a constant. Distance measurement now becomes the measurement of the number of half wavelengths included in the measurement line and the fractional part less than half a wavelength, namely the measurement of N or φ. Due to advances in modern precision machining and radio phase measurement technology, the measurement of φ has achieved high accuracy.
To measure the phase angle φ less than π, various methods can be used, with delay phase measurement and digital phase measurement being the most commonly applied. Currently, short-range laser distance meters adopt digital phase measurement principles to obtain φ.
In summary, in the described situation, phase-type laser distance meters use a continuously emitted laser beam with a modulated signal. To achieve high-precision distance measurement, cooperative targets are needed. The recently introduced handheld laser distance meter is a new type of distance meter in the pulse-type category. It is not only compact and lightweight but also adopts digital phase measurement pulse width modulation technology. It achieves millimeter-level accuracy without the need for cooperative targets, with a measurement range exceeding 100 meters, and can rapidly and accurately display distances. It represents the latest standard measuring instrument for precise engineering measurements and building area measurements in short-range applications.
The above explains the working principles of laser distance meters, hoping to help you better understand these devices. |
Wave Motion of Class 11
Sound waves in air involve longitudinal oscillations of the molecules. These waves are characterised as pressure or density fluctuations. The pressure fluctuations are of the order of 1 Pa, whereas atmospheric pressure is about 105 Pa. Figure (14.16) shows a loudspeaker which produces compressions and rarefactions in the air in a tube. As the diaphragm of the loudspeaker moves forward, it produces a compression, that is, an increase in pressure Δp above the equilibrium value po. When the diaphragm moves backward, it produces a rarefaction, a decrease in pressure, +Δp, below po. The collisions between the molecules cause these compressions and rarefactions to propagate as a sound wave down the pipe.
At a given instant there are points such as b and d toward which molecules converge and therefore increase the local pressure to a maximum of po + Δpo.
On either side of points a and c, the molecules are moving away and so local pressure at these points drop to a minimum of po - Δpo.
Note that the points (a, b, c and d) of maximum pressure deviation (±Δpo) have zero displacement. Moreover, pressure fluctuations are 90o out of phase with the displacements. The converse is also true: zeroes in pressure fluctuation correspond to maxima or minima in the displacement.
The speed of longitudinal waves in a fluid is given by
v = √B/ρ (14.31)
where B is the bulk modulus defined as
B = -Δp/ΔV/ V
where ΔV / V is the fractional change in volume produced by the change in pressure Δp. The SI unit of B is N/m2. The negative sign is needed to make B positive since a positive change in pressure results in a negative change in volume V. One may visualize bulk modulus as spring constant of air.
The propagation of sound in air is an adiabatic process. For air the bulk modulus in adiabatic condition is given by
B = γp(14.32)
where γ is an adiabatic exponent whose numerical value for air is 1.4.
Thus, velocity of sound in air is given by
v = √γp/ρ (14.33)
Using gas equation p/ρ = RT/M
∴v = √γRT/M (14.34)
Substituting the values of γ, R and M we obtain the approximate value of speed of sound in terms of absolute temperature T, as
v ≈ 20√T (14.35)
Determine the speed of sound waves in water, and find the wavelength of a wave having a frequency of 242 Hz. Take Bwater = 2 × 109 Pa.
Speed of sound wave, v = √B/ρ = √(2 x 109)/103 = 1414 m/s
Wavelength λ = v/ f = 5.84 m
If the average molar mass of air is 28.8 × 10-3 kg/mol, then find the speed of sound in air at 300K.
Hereγ = 1.4 (diatomic), M = 28.8 × 10-3 kg/mol
T = 300 K, R = 8.31 J mol1 K−1.
v = √γRT/M = √(1.4)(8.31)(300)/28.8 x 10 -3 = 348 m/s
Relationship between Pressure Waves and Displacement Waves
For a harmonic wave, the longitudinal displacement (y) is given by
y = A sin(kx - wt)(14.36)
We know p = −B ⧊v / v
Since change in volume is produced by the displacement of the particles, therefore,
ΔV/ V = δy/δx
Thus,p = −Bδy/δx = −BAk cos(kx - wt)
orp = −po cos(kx - wt)(14.37)
where po = BAk is the pressure amplitude
Note that displacement and pressure amplitudes are π/2 out of phase.
Standing Sound Waves
Standing waves can be produced in air columns, for example, in organ pipes, flutes, and other wind instruments because sound waves are reflected both at a closed end and at an open end of a pipe. In an open pipe both the ends are open, whereas in a closed pipe one end is closed.
In fig. (14.17) the motion of the molecules is represented by solid arrows at t = 0 and by dashed arrows at t = T/2, where T is the period. At the closed end the displacement is permanently zero; that is, it is a displacement node. From the previous section we know that this corresponds to a pressure antinode. (At the instant depicted the pressure is a maximum at the closed end. Half a period later, the density and pressure will be a minimum.)
The open end has to be at atmospheric pressure, so it is a pressure node and a displacement antinode. The fundamental mode occurs when L = λ/4, which means
(Closed pipe)fn = (nv)/(4l) (n = 1, 3, 5….)(14.38)
The knowledge that each open end is a displacement antinode allows us to draw the various modes right away (see fig.(14.19)). Note that the fundamental frequency of an open pipe is double that of a closed pipe of the same length. In an open pipe, all the harmonics are possible:
Standing waves in pipes decay very quickly after the source of excitation is removed. The interaction between the air in the pipe and the surrounding air also means that The above equations are not quite correct. It is found that pipes of the same length but of different diameters have slightly different resonant frequencies. The effective length is approximately the real length plus 0.6 of the radius (R) of each open end.
leff = l + 0.6 R
An air column at 51°C and a tuning fork produce 4 beats per second when sounded together. As the temperature of the air column is gradually decreased to 16°C, they again produce 1 beat per second. Find the frequency of the tuning fork. It is observed they again produce one beat per second If the temperature is decreased further. It implies that the frequency of air column is more than that of the tuning fork.
Let f be the frequency of the tuning fork.
The number of beats per second decreases as the frequency of the air column decreases. Hence the frequency of air column is initially (f + 4) at 51°C and (f + 1) at 16oC
Since the velocity of sound is proportional to the square root of temperature i.e.,
v ∝ √T
∴ f + 4/f + 1 = √273 + 51/273+16
Thus f = 50 Hz
AB is cylinder of length 1.0 m, fitted with a thin flexible diaphragm C at the middle and two other thin flexible diaphragms A and B at the ends. The chamber 1 contains hydrogen and Chamber 2 oxygen. The diaphragms A and B are set into vibrations of same frequency and behave as antinodes. What is the minimum frequency of these vibrations for which the diaphragm C is a node? Given, the velocity of sound in hydrogen is
Both the chamber act like closed organ pipe.
The tubes AC and BC will be two closed tubes with antinodes at A and B and nodes at C.
v1 = CH = 1100 m/s
f1 = v1/4l = 1100/4(0.5) = 550Hz
⇒v1/v2 = 550/150
⇒3v1 = 11 v2
v2 = 300 m/s
f2 = v1/4l = 300/4(0.5) = 150 Hz
vmin = 3v1 = 3(550) = 1650 Hz
orvmin = 11v2 = 11(150) = 1650 Hz
Let n1th harmonic of chamber 1 coincides with the n2th harmonic of chamber 2.
Thus n1f1 = n2f2
or n1/n2 = f2/f1 = 150/550 = 3/11
The third harmonic of chamber 1 coincides with eleventh harmonic of chamber 2.
Hence, the fundamental frequency of the system is f = 3f1 = 3(550) = 1650 Hz |
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30. What will be the tas on $3758, at the rate of 4 mills on the dollar?
Ans. $15, 03, 2. 31. A certain work can be raised in 12 days by working 4 hours each day: how long would it require to raise the work by working 6 hours per day?
Ans. S days. 32. What quantity of corn can I buy for 90 guineas, at the rate of 6 shillings a bushel?
Ans. 315 bushels. 33. A person failing in trade owes £977, at which time he has in money, goods, and recorerable debts £420 6s. 31d.: now supposing an equal divi. sion among his creditors how much will they get on the pound?
Ans. &s. 74d. 34. A pasture of a certain extent having supplied a body of horse, consisting of 3000, with forage for 18 days, how many days would the same pasture have supplied a body of 2000 horse ?
Ans. 27 days. 35. Suppose a gentleman's income to be 600 guineas a year, and that he spends 25s. 6d. per day, one day with another; how much will he have at the end of the year?
Ans. £164 12s. 6d. 36. What is the cost of 30 pieces of lead, each weighing lcut. 1216, at the rate of 16s. 4d. the cwt. ?
Ans. £27 2s. 6d. 37. The governor of a besieged place has provisions for 54 days at the rate of 2b. of bread per ration, but is desirous to prolong the siege to 80 days, in expectation of succour: in that case, what must be the ration of bread ?
Ans. 171l. 38. If a person pays half a guinea a week for hoard, how long can he be boarded for £21 ?
Ans. 40 weeks.
per bottle ?
39. What is the value of a year's rent of 547 acres of land at the rate of 15s. 6d. the acre ?
Ans. £423 18s. 6d. 40. If a person drink 20 bottles of wine per month, when it costs 2s. per bottle, how much can he drink without increasing the expense when it costs 2s. 6d.
Ans. 16 bottles. 41. A merchant bought 21 pieces of cloth, each containing 40 yards, for which he paid $1260; he sold the cloth at $1,75 per yard: did he make or lose by the bargain?
Ans. he gained $210. 42. A cistern containing 200 gallons is filled by a pipe which discharges 3 gallons in 5 minutes: but the cistern has a leak which empties 1 gallon in 5 minutes. Now if the water begins to run in, when the cistern is empty, how long will it be in filling?
Ans. 8 hours 20 minutes. 43. What will be the cost of 895 feet of timber, at $6 per hundred feet?
100 895 6 answer..
6 100)5370 53,70
Ans. $53,70. In this example, 100 feet of timber, is to the given quantity 895 feet, as $6, the cost of 100 feet, to $53,70, the cost of 895 feet. If the timber had been sold at the rate of $6 per thousand feet, the cost of 895 feet would have been $5, 37, for we should have divided by 1000, instead of 100; that is, we should have removed the separating point in the product three places to the left. Hence, to find the cost
of things sold by the 100, or 1000, we have the fol. lowing
RULE. Multiply the number by the price, and if the things Ar reckoned by the 100, cut of two places from the right, and if reckoned by the 1000, cut off three, and the figures to the left will be the answer in the sume denomination as the given price.
44. What will be the cost of 1350 feet of boards at $11 per hundred!
Ans. $148,50. 45. What will be the cost of 36578 bricks at $6,50 per thousand ?
Ans. $237,75, t. 46. What will be the cost of 6359 feet of boards at $9, 25 per 100 feet?
Ans. $588, 20, t. 47. What will be the cost of 13918 feet of timber at $14, 37 per thousand ?
Ans. $200,00, +. 48. What will 18759 oranges cost at $5,50 per hundred ?
Ans. $1031, 74, +.
OF QUESTIONS REQUIRING TWO
The answer to each of the above questions has been found by a single statement. Questions, however, frequently occur in which two or more statements are necessary. In most Arithmetics, such cuestions are arranged under a rule called Compound
n, or the Double Rule of Three. They want, be answered by the rules already
Ec. 1. If 16 men build 18 feet of wall in 12 days, how many men must be employed to build 72 feet in 8 days, working at the same rate ?
The first question is, how long would it take feet. feet. days. days.
18 : 72 : : 12 : answer. the 16 men to build the
72 72 feet of wall ? It is evident that 18
24 feet of wall, is to 72 feet,
84 as 12 days, the time ne
18)864(48 days. cessary to build 18 feet,
72 to 48 days, the time ne
144 cessary to build 72 feet.
The second question days. days. men. is, if 16 men can build 8: 48 : : 16 : answer. 72 feet of wall in 48
48 days, how many men
128 are necessary to build it
64 in 8 days? Make 16 men the
8)7681 third term.
96 men. Then as
Ans. the same work is to be done in a less time, more men will be necessary ; then the fourth term will be greater than the third, and hence 8 days are placed in the first term. § 109.
2. If a family of 6 persons expend $300 in 8 months, how much will serve a family of 15 persons for 20 months ?
First question. If $300 will support a persone, persons.
300 : ans. family of 6 persons
15 for 8 months, how many dollars will support
6)4500 15 persons for the
Second question. If months. months. $750 will support a
20 :: 750 : ans. family of 15 persons
20 for 8 months, how
8)15000 much will serve them
Ans. $1875 for 20 months ?
3. If a man travel 217 miles in 7 days, travelling 6 hours a day, how far would he travel in 9 days, if he travelled 11 hours a day. 1st.
2d. days. days. miles. miles. hours. hours. miles. miles. ņ: 9:: 217 : 279 6 : 11 :: 279 : 511% 9
Ans. 5114 miles. 4. If the wages of 6 men for 14 days be $84, what will be the wages of 9 men for 11 days?
Ans. $99. 5. If 154 bushels of oats serve 14 horses for 44 days, how long would 406 bushels last 7 horses ?
Ans. 232 days. 6. If 25 men can earn $6250 in 2 years, how long will it take 5 men to earn $11250 ?
Ans. 18 years. 7. If a barrel of beer last 7 persons 12 days, how much will be drank by 42 persons in a year?
Ans. 182bar. 18gal. 8. If 9 men can cut 36 acres of grass in 4 days, how many acres will 19 men cut in 11 days?
Ans. 209 acres. 9. If 25 persons consume 300 bushels of corn in ona vear, how much will 139 persons consume in t the same rate ? |
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?
Ben and his mum are planting garlic. Use the interactivity to help you find out how many cloves of garlic they might have had.
What is the best way to shunt these carriages so that each train can continue its journey?
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 it up?
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?
Frances and Rishi were given a bag of lollies. They shared them out evenly and had one left over. How many lollies could there have been in the bag?
If you split the square into these two pieces, it is possible to fit the pieces together again to make a new shape. How many new shapes can you make?
Can you work out how to balance this equaliser? You can put more than one weight on a hook.
Design an arrangement of display boards in the school hall which fits the requirements of different people.
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?
This task, written for the National Young Mathematicians' Award 2016, invites you to explore the different combinations of scores that you might get on these dart boards.
This task, written for the National Young Mathematicians' Award 2016, involves open-topped boxes made with interlocking cubes. Explore the number of units of paint that are needed to cover the boxes. . . .
In the planet system of Octa the planets are arranged in the shape of an octahedron. How many different routes could be taken to get from Planet A to Planet Zargon?
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?
The planet of Vuvv has seven moons. Can you work out how long it is between each super-eclipse?
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?
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?
Can you rearrange the biscuits on the plates so that the three biscuits on each plate are all different and there is no plate with two biscuits the same as two biscuits on another plate?
Can you work out the arrangement of the digits in the square so that the given products are correct? The numbers 1 - 9 may be used once and once only.
Arrange eight of the numbers between 1 and 9 in the Polo Square below so that each side adds to the same total.
Find the product of the numbers on the routes from A to B. Which route has the smallest product? Which the largest?
On a digital 24 hour clock, at certain times, all the digits are consecutive. How many times like this are there between midnight and 7 a.m.?
What is the smallest number of jumps needed before the white rabbits and the grey rabbits can continue along their path?
How many trapeziums, of various sizes, are hidden in this picture?
Only one side of a two-slice toaster is working. What is the quickest way to toast both sides of three slices of bread?
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?
Can you use this information to work out Charlie's house number?
The Zargoes use almost the same alphabet as English. What does this birthday message say?
Put 10 counters in a row. Find a way to arrange the counters into five pairs, evenly spaced in a row, in just 5 moves, using the rules.
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?
Swap the stars with the moons, using only knights' moves (as on a chess board). What is the smallest number of moves possible?
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.
Sitting around a table are three girls and three boys. Use the clues to work out were each person is sitting.
Seven friends went to a fun fair with lots of scary rides. They decided to pair up for rides until each friend had ridden once with each of the others. What was the total number rides?
When intergalactic Wag Worms are born they look just like a cube. Each year they grow another cube in any direction. Find all the shapes that five-year-old Wag Worms can be.
An activity making various patterns with 2 x 1 rectangular tiles.
There are seven pots of plants in a greenhouse. They have lost their labels. Perhaps you can help re-label them.
What happens when you try and fit the triomino pieces into these two grids?
If we had 16 light bars which digital numbers could we make? How will you know you've found them all?
El Crico the cricket has to cross a square patio to get home. He can jump the length of one tile, two tiles and three tiles. Can you find a path that would get El Crico home in three jumps?
These are the faces of Will, Lil, Bill, Phil and Jill. Use the clues to work out which name goes with each face.
Using the statements, can you work out how many of each type of rabbit there are in these pens?
Can you cover the camel with these pieces?
Investigate the smallest number of moves it takes to turn these mats upside-down if you can only turn exactly three at a time.
Tim's class collected data about all their pets. Can you put the animal names under each column in the block graph using the information?
Alice and Brian are snails who live on a wall and can only travel along the cracks. Alice wants to go to see Brian. How far is the shortest route along the cracks? Is there more than one way to go?
Make a pair of cubes that can be moved to show all the days of the month from the 1st to the 31st.
Find your way through the grid starting at 2 and following these operations. What number do you end on?
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? |
What is the problem of induction Why is this a problem for Hume?
The original problem of induction can be simply put. It concerns the support or justification of inductive methods; methods that predict or infer, in Hume’s words, that “instances of which we have had no experience resemble those of which we have had experience” (THN, 89).
What is the problem of induction why is it a problem?
According to Popper, the problem of induction as usually conceived is asking the wrong question: it is asking how to justify theories given they cannot be justified by induction. Popper argued that justification is not needed at all, and seeking justification “begs for an authoritarian answer”.
Can the problem of induction be solved?
The most common solution to the problem of induction is to unshackle it from deduction. In this view, induction was mistakenly jury-rigged into a system of deductive inference where it did not belong, i.e. induction was considered subordinate to the apparatus of basic logic.
Can the principle of induction be justified?
Premise 1: we can justify the principle of induction only if there is a de- ductively valid or inductively strong argument which does not presuppose its conclusion, whose premises are restricted to information we have, and whose conclusion says that the principle of induction usually leads from true premises to true …
What is the new problem of induction?
The new riddle of induction, for Goodman, rests on our ability to distinguish lawlike from non-lawlike generalizations. Lawlike generalizations are capable of confirmation while non-lawlike generalizations are not. Lawlike generalizations are required for making predictions.
Why can we not prove this principle inductively by moral reasoning?
Why can we not prove the principle “the future will resemble the past” inductively or by moral reasoning? Moral reasoning is also unhelpful, since it falls into a vicious circle.
How did Karl Popper solve the problem of induction?
Popper’s solution to this problem is: 1) there is no inductive logics, no correct inductive procedure, no way to demonstrate the truth or, at least, high probability of our theories; 2) the “given” – the theory that we obtain our general theories by inductive generalization from experience – is mistaken.
What are the two grounds of induction?
The law of Uniformity of Nature and the causation are formal grounds of induction. The process which guarantee the material truth of induction are called material grounds of induction. Observation and experiment are material grounds of induction.
Why is inductive reasoning important?
We use inductive reasoning in everyday life to build our understanding of the world. Inductive reasoning also underpins the scientific method: scientists gather data through observation and experiment, make hypotheses based on that data, and then test those theories further.
What is induction paradox?
The paradox of induction is the problem that in all scientific reasoning we form conclusions, called laws, that are of a general nature; however, the evidence we have for those laws is based upon particular experiences.
Is inductive reasoning bad?
The main weakness of inductive reasoning is that it is incomplete, and you may reach false conclusions even with accurate observations.
Is the problem of induction a pseudo problem?
In 1955, Goodman set out to ‘dissolve’ the problem of induction, that is, to argue that the old problem of induction is a mere pseudo- problem not worthy of serious philosophical attention (1955, 65–8).
How does Kant solve the problem of induction?
Kant’s Externalist Solution to the Problem of Induction
sorts of reasoning processes: “demonstrative reasoning, or that concerning relations of ideas, and moral reasoning, or that concerning matter of fact and existence.”
Is inductive reasoning rational?
Induction is part of our rational methodology, and that methodology is irreflexive. We cannot rationally justify induction, but that isn’t because induction is irrational, indeed it is for exactly the opposite reason – because it is what we mean by rational.
Is inductive reasoning circular?
Inductive reasoning then, need neither be circular, nor merely tautological. It is in fact a spiral, a set of instances in which the proposition “the sun rose today” is true, not once, but many times.
Who invented inductive reasoning?
Called the father of empiricism, Sir Francis Bacon is credited with establishing and popularizing the “scientific method” of inquiry into natural phenomena.
Can an argument be inductive and deductive?
It is not inductive. Given the way the terms “deductive argument” and “inductive argument” are defined here, an argument is always one or the other and never both, but in deciding which one of the two it is, it is common to ask whether it meets both the deductive standards and inductive standards.
Why is deductive reasoning better than inductive?
Nevertheless, inductive reasoning has its place in the scientific method, and scientists use it to form hypotheses and theories. Deductive reasoning then allows them to apply the theories to specific situations.
Are inductive arguments truth preserving Why or why not?
Are inductive arguments truth-preserving? Why or why not? No,because the truth of the conclusion cannot be guaranteed by the truth of the premises,inductive arguments are not truth-preserving.
Which is better inductive or deductive method?
Inductive tends to be more efficient in the long run, but deductive is less time consuming. Much depends on the teacher and the students. You might try and compare both of these approaches at certain points in your teaching to see which is more effective for your students.
Is inductive qualitative or quantitative?
Inductive approaches are generally associated with qualitative research, whilst deductive approaches are more commonly associated with quantitative research. However, there are no set rules and some qualitative studies may have a deductive orientation.
How do we use inductive reasoning in everyday life explain with an example?
In causal inference inductive reasoning, you use inductive logic to draw a causal link between a premise and hypothesis. As an example: In the summer, there are ducks on our pond. Therefore, summer will bring ducks to our pond. |
3. Power spectrum analysis of the reconstructed maps
3.1. Noise statistical properties of the reconstructed maps
It is clear that the mass reconstruction process does not produce any boundary effects (which is settled by definition). The only boundary effect, slightly detectable on the figure, is a larger level of noise at the edge of the field, due to the change in the finite difference scheme at that position (see Appendix B). The noise due to the intrinsic ellipticities of the galaxies is clearly visible at small scales.
Since the least method used to reconstruct the convergence is a local process, it is unlikely that noise propagates on scales larger than the pixel size 4.
Fig. 3 shows the power spectrum analysis of 60 reconstructed mass maps in the case of two different cosmological models (cases (a) and (b)) and (cases (c) and (d)). Fig. 3a and c show the noise free power spectrum (solid lines) the power spectrum measured on the reconstructed maps with gal/arcmin2 (dotted lines) and for gal/arcmin2 (dashed lines). The plateau for the latter two cases is the consequence of the intrinsic ellipticities of the galaxies: at small scales, the power is dominated by the ellipticity of the galaxies, thus tends to be constant, generally much higher than the signal. Figs. 3b and d show the difference of the power spectrum measurements on the reconstructed maps with the noise-free power spectrum; thin dotted line is for gal/arcmin2 and thin dashed line for gal/arcmin2. For visibility, error bars for scales larger than arcmin have been dropped.
In 3D space, these angular scales correspond approximately to scales from 1 to 30 Mpc. We leave for later studies the problem of inverting the measured projected to the 3D one. This aspect has already been explored at large angular scale by Seljak (1997).
As can be inferred from the power spectra in Fig. 3, the noise has a flat power spectrum, characteristic of a white noise process. Fig. 2 illustrates the fact that the noise is independent of the underlying field and follows a Gaussian distribution. The noise model introduced by Kaiser (1998) based on the weak lensing approximation is now compared to the noise found in our simulations. The weak lensing approximation applied to Eq. (10) gives a local shear estimate , where is the true shear and , the mean intrinsic ellipticities of galaxies in the superpixel p. Since the noise components are assumed to be spatially uncorrelated, the statistical properties of the noise are,
where , is the Kronecker symbol and is the variance of one component of the intrinsic ellipticities in one superpixel. The shear and the intrinsic ellipticities of the galaxies are uncorrelated in the weak lensing approximation. The measured power (on the noisy mass maps) can then be expressed only in terms of the true power (the one we want to estimate) and the power spectrum of the noise (Eq. (13)). If denotes the Fourier transform of the measured convergence, its power spectrum is given by
This equation is only valid for a compact survey, where is the mean number density of the galaxies per superpixel, and is the mean value of over the survey. For a sparse survey, the first term in Eq. (14) is changed into a convolution term, but the noise contribution to the observed power spectrum remains independent of that power spectrum. A convenient way to estimate is to take
where is the number of superpixels and the number of galaxies in the superpixel p. This is nothing else but the variance of the observed ellipticities of the selected galaxies 5.
The thin dashed and dotted straight lines on Fig. 3 correspond to the expected noise power spectra (for gal /arcmin2 and gal/arcmin2) according to the Kaiser's model given by Eq. (14). It perfectly fits the noise part of the mass reconstructed with noisy data, whatever the cosmological model and the noise level. This is true even for the open cosmological models for which stronger non-linearities could have produced a stronger coupling with the noise. The fact that the noise component is pure white noise with an amplitude in agreement with the theoretical prediction is a remarkable result since the full non-linear equations were used, and it shows that the weak lensing approximation can be safely used to remove the noise component and to get an unbiased estimate of the power spectrum, down to the smallest scales considered here.
The behavior at scales smaller than our pixel size remains partly an open issue for two reasons: first, due to the smaller number of galaxies, the convergence of the reconstruction process as well as the stability of the noise properties has to be investigated. Second, at small scales the gravitational distortion is larger than only a few percent, and it can go up to infinity on the critical lines. Therefore, estimating the variance of the galaxies intrinsic ellipticity distribution, arcs and arclets should be removed. This issue can only be addressed in high resolution simulations like those performed by Jain et al. (1998).
3.2. Power spectrum cosmic variance
Although the estimate of the power spectrum described above is unbiased, the cosmic variance has also to be explored to come up with an optimal observational strategy.
In estimating the cosmic variance of the power spectrum, Gaussian statistics is usually assumed. This hypothesis is tested comparing the cosmic variance assuming Gaussian statistics to that of the simulated mass maps and that reconstructed using the two different level of noise defined before.
In the case of Gaussian statistics high order moment are related to the second order moments via the following relations:
which means that physically, the frequencies of a Gaussian field are not coupled, and that the moment at a given frequency is only determined by the power at that frequency.
We consider a compact survey of size , for which the number of modes available at a frequency is maximum. Thus, following Feldman et al. (1994) and Kaiser (1998), the cosmic variance of is given by the square of the measured signal (which depends on ), divided by the number of independent modes used to determine it, in the k-annulus . Simple modes count gives
Note that in this hypothesis the cosmic variance is independent from the amplitude of the fluctuations. This is not the case when the non-linear couplings are taken into account as it can be seen in Fig. 4. It shows the cosmic variance for flat ((a), (c), (e)) and open (((b), (d), (f)) cosmological models. (a) and (b) correspond to a degree survey with , (c) and (d) with , and (e) and (f) a degree survey with . On each plot, the thin solid line shows the Gaussian cosmic variance, the thick solid line shows the true cosmic variance without noise, the dotted line the noisy maps with gal/arcmin2 and the dashed line with gal/arcmin2. The vertical axis gives the error on the power spectrum measured at a given scale.
The departure from Gaussianity appears for scales below , the effect is however more important in the open case model for which non linearities are stronger. Open models ((b), (d), (f)) give almost the same features as for the flat models (((a), (c), (e)), although the cosmic variance is smaller. This is clearly a consequence of a higher power spectrum signal at low scales for these models, which is visible when comparing Fig. 3 (a) (flat) and (c) (open). Thus, as expected, the intrinsic shape of the power spectrum affects the cosmic variance (Kaiser 1998).
Going deep in redshift (by comparing (a) and (c), or (b) and (d) for the open case) clearly improves the cosmic variance at small scale, since the gravitational distortion is stronger. However this stronger distortion does not improve the large scale power estimation because the cosmic variance at these scales only depends on the whole volume survey.
If shorter wave vectors are observed, (for a degree survey), which is visualized on (e) and (f), a gain of 2 is reached at all scales, as a direct consequence of global increase of the number of modes in Eq. (17). For small scales, from the point of view of the statistics it is in fact equivalent to observe deep over a small area than to observe shallower over a large area (which reflects the fact that the dashed lines in (e) and (f) (gal/arcmin2) are almost the same as the dotted line in (a) and (b) (gal/arcmin2). But on the other hand, the shallow large survey gives a better estimate of the power at large scales than the deep survey. Since these two observational strategies require the same total exposure time it is clear that wide shallow surveys are better than small deep surveys. As it will be shown in Sect. 4, this remains true for the high order moments in real space. Moreover, deep surveys show more and more distant galaxies for which the redshift distribution is more uncertain.
© European Southern Observatory (ESO) 1999
Online publication: December 22, 1998 |
Patent application title: Fabrication of Lenses Using High Viscosity Liquid
Sri Rama Prasanna Pavani (Palo Alto, CA, US)
Brian P. Mccall (Houston, TX, US)
IPC8 Class: AB05D506FI
Class name: Optical: systems and elements lens
Publication date: 2014-06-05
Patent application number: 20140153104
A method for fabricating millimeter and sub-millimeter size lenses using
a high viscosity curable liquid, such as epoxy. The method comprises
dispensing a predetermined volume of the curable liquid onto a substrate.
The curable liquid preferably has a viscosity higher than 100 cps.
Additionally, to reduce spherical aberration, the curable liquid can be
cured upside down to leverage the effects of gravity.
1. A method for fabricating a lens on a substrate, the method comprising:
making contact between a drop of curable liquid and a substrate;
dispensing a predetermined volume of the curable liquid from a dispenser
onto the substrate while the curable liquid is in contact with the
substrate, the curable liquid having a viscosity of at least 100 cps; and
curing the curable liquid to a solid form.
2. The method of claim 1, wherein the curable liquid is a liquid epoxy or a liquid polymer.
3. The method of claim 1, wherein the lens has a circular aperture with diameter of not more than 1 mm.
4. The method of claim 1, wherein the lens has a circular aperture with diameter of not more than 500 μm.
5. The method of claim 1, wherein the curable liquid has a viscosity of at least 200 cps.
6. The method of claim 1, wherein the curable liquid is cured while supported by the substrate.
7. The method of claim 1, wherein the curable liquid is cured while suspended from the substrate.
8. The method of claim 1, wherein the substrate is a featureless, planar substrate.
9. The method of claim 1, wherein the substrate is transparent.
10. The method of claim 1, wherein the substrate is homogeneous with a uniform refractive index.
11. The method of claim 1, wherein the substrate is inhomogeneous with a gradient in refractive index.
12. The method of claim 1, further comprising: dispensing predetermined volumes of curable liquid from the dispenser onto the substrate at multiple locations, the liquid having a viscosity of at least 100 cps; and curing the curable liquid to a solid form to form an array of lenses.
13. The method of claim 12, wherein the predetermined volumes of curable liquid are dispensed simultaneously at multiple locations on the substrate.
14. The method of claim 12, wherein the predetermined volumes of liquid are dispensed sequentially at each of the multiple locations on the substrate.
15. An lens manufactured by the process of claim 1.
16. An array of lenses, wherein each lens is manufactured by the process of claim 1.
17. An epoxy or polymer lens on a substrate, the lens characterized by a circular aperture having a diameter, wherein the lens has a height that is at least equal to a height of a curable liquid drop of a same diameter supported by the substrate, the liquid drop having a viscosity of at least 100 cps.
18. The lens of claim 17 wherein the substrate is a featureless, planar substrate.
19. The lens of claim 17 wherein a surface of the lens is characterized by z ( r ) = r / R 2 1 + 1 - ( r / R ) 2 + n = 2 N a n r 2 n ##EQU00007## where z(r) is a height of the lens at radius r, R is a radius of curvature at the center of the lens, a 2 = - 1 32 R ( 1 R 2 + 1 L c 2 ) ##EQU00008## a 3 = - 1 576 R ( 23 R 4 + 1 2 L c 4 + 47 R 2 L c 2 ) ##EQU00008.2## a 4 = - 1 819 R ( 375 R 6 + 1 9 L c 6 + 3745 9 R 4 L c 2 + 371 9 R 2 L c 4 ) ##EQU00008.3## and Lc is a characteristic length of the epoxy or polymer forming the lens.
20. The lens of claim 17 wherein a surface of the lens is characterized by z ( r ) = r / R 2 1 + 1 - ( r / R ) 2 + n = 2 N a n r 2 n ##EQU00009## where z(r) is a height of the lens at radius r, R is a radius of curvature at the center of the lens, a 2 = - 1 32 R ( 1 R 2 - 1 L c 2 ) ##EQU00010## a 3 = - 1 576 R ( 23 R 4 + 1 2 L c 4 - 47 R 2 L c 2 ) ##EQU00010.2## a 4 = - 1 819 R ( 375 R 6 - 1 9 L c 6 - 3745 9 R 4 L c 2 + 371 9 R 2 L c 4 ) ##EQU00010.3## and Lc is a characteristic length of the epoxy or polymer forming the lens.
BACKGROUND OF THE INVENTION
1. Field of the Invention
This invention relates generally to optical lenses and, more particularly, to the fabrication of millimeter and sub-millimeter size optical lenses.
2. Description of the Related Art
There is a growing need for small size lenses. With each passing generation of cell phones, the number of cell phones with integrated cameras increases. The miniaturization that enables cell phone cameras has also given rise to other types of small cameras. The increasing number and variety of small cameras and other optical instruments results in an increasing demand for small size lenses.
As another example, ubiquitous sensor networks can enable global sensing by connecting numerous dispersed intelligent sensors that sense their local ambiance for changes in physical entities such as light, temperature, sound, and pressure. To be truly ubiquitous, these intelligent sensors must be compact and mass-producible at low cost. Sensing light, in particular, is useful for detecting illumination variations, capturing images, and for harnessing energy. Light sensors use optical lenses for efficient light collection. Small light sensors use small optical lenses.
Conventional techniques for fabricating miniature lenses, such as diamond turning, molding, lithography, and inkjet printing employ complex fabrication processes that are not inexpensively customizable. Thus, there is a need for miniature lenses, with millimeter or sub-millimeter sizes, that preferably can be customized and produced at a low cost.
SUMMARY OF THE INVENTION
The present invention overcomes the limitations of the prior art by providing lenses fabricated using high viscosity liquid, such as liquid epoxy or liquid polymer. In one aspect, a known volume of transparent high-viscosity liquid epoxy is dispensed on a planar substrate. At equilibrium, the epoxy thus dispensed exhibits an approximately spheroidal outer surface primarily due to the surface tension between the epoxy and its surrounding media. Upon exposure to ultra-violet light, the epoxy is cured to form a lens (referred to as a liquid drop lens or LDL) with molecular scale surface smoothness.
Diameters and focal lengths of liquid drop lenses can be customized by varying the volume of epoxy dispensed. Additional customization is possible by varying surface tension, surface roughness, viscosity, temperature and/or the curing process. For example, the lens shape can be changed by curing when the epoxy is suspended from the substrate (upside down curing) rather than when the epoxy is supported by the substrate (right side up curing). Furthermore, multiple epoxy drops placed next to each other may be used to form arrays of liquid drop lenses, as may be useful in computational imaging applications such as stereo imaging and multiple aperture imaging.
Other aspects of the invention include methods, devices, systems and applications of the approaches described above.
BRIEF DESCRIPTION OF THE DRAWINGS
The invention has other advantages and features which will be more readily apparent from the following detailed description of the invention and the appended claims, when taken in conjunction with the accompanying drawings, in which:
FIG. 1 is a free body diagram at the surface of a liquid drop lens before curing.
FIGS. 2A-2B illustrate the difference between a liquid drop lens cured right-side up and a liquid drop lens cured upside down.
FIGS. 3A-3D are ray traces and spot diagrams, illustrating the difference between an oblate liquid drop lens and a prolate liquid drop lens.
FIG. 4 is a graph of focal length as a function of lens volume and liquid contact angle.
FIG. 5 shows a system for fabricating liquid drop lenses.
FIG. 6 shows a system for fabricating liquid drop lenses in parallel.
FIG. 7 is a flowchart for the fabrication process of a liquid drop lens.
FIG. 8 shows an array of four liquid drop lenses.
FIG. 9 shows an example of a stereo imager.
The figures depict embodiments of the present invention for purposes of illustration only. One skilled in the art will readily recognize from the following discussion that alternative embodiments of the structures and methods illustrated herein may be employed without departing from the principles of the invention described herein.
DETAILED DESCRIPTION OF THE PREFERRED EMBODIMENTS
The figures and the following description relate to preferred embodiments by way of illustration only. It should be noted that from the following discussion, alternative embodiments of the structures and methods disclosed herein will be readily recognized as viable alternatives that may be employed without departing from the principles of what is claimed.
A lens is an optical device that transmits and refracts light. As used herein, light is not limited to the visible spectrum and includes ultra-violet (UV) light, infra-red (IR) light, etc. A lens has two opposite surfaces, at least one of which is curved. As described herein, a liquid drop lens (LDL) is a type of lens that is fabricated by dispensing a drop of high-viscosity, curable liquid on a substrate. In some embodiments, the curable liquid or resin is a type of optically transparent, high viscosity epoxy or polymer that hardens after being exposed to ultra-violet (UV) radiation. In one embodiment, the high viscosity liquid has a viscosity greater than 50 cps, greater than 100 cps or more preferably greater than 200 cps. Due to the high viscosity and surface tension, a drop of liquid placed on top of a planar substrate adopts a nearly spheroidal shape which can focus beams of light the same way a conventional optical lens focuses light.
Mathematical Model of Liquid Drop Lenses
FIG. 1 illustrates a free body diagram at the surface of a drop 110 before curing. Assuming that there is no significant shape change during curing, the final liquid drop lens will have this same shape. The shape of the drop can be obtained by balancing the hydrostatic pressure equation and the Young-Laplace equation at every point around the liquid-gas interface. As a result, the second order ordinary differential equation describing the shape of a liquid drop lens is given by
z '' ( 1 + ( z ' ) 2 ) 3 / 2 + z ' r ( 1 + ( z ' ) 2 ) 1 / 2 - z L c 2 = const ( 1 ) ##EQU00001##
where z(r) is the height of the drop at distance r from the axis of symmetry, LC is the characteristic length of the epoxy given by
γ ρ g , ##EQU00002##
where γ is the surface tension between the liquid and the gas, ρ is the liquid density and g is the acceleration due to gravity. The solution to equation (1) as an aspheric polynomial expansion is
z ( r ) = r / R 2 1 + 1 - ( r / R 2 ) + n = 2 N a n r 2 n ( 2 ) ##EQU00003##
where R is the radius of curvature at the apex of the lens. From equations (1) and (2), the aspheric coefficients of the liquid drop lens can be solved as
a 2 = - 1 32 R ( 1 R 2 + 1 L c 2 ) a 3 = - 1 576 R ( 23 R 4 + 1 2 L c 4 + 47 R 2 L c 2 ) a 4 = - 1 819 R ( 375 R 6 + 1 9 L c 6 + 3745 9 R 4 L c 2 + 371 9 R 2 L c 4 ) ( 3 ) ##EQU00004##
As shown above, the coefficients in this expression depend only on the apex radius of curvature (R) of the lens and the characteristic length (Lc) of the epoxy.
If the liquid drop lens is cured upside-down (i.e. while suspended from the substrate), gravity works in the opposite direction relative to the curvature of the lens. The effect of this is to pull the lens into a sharper curvature instead of flattening it out. Mathematically, the only difference is that a negative sign follows every instance of L2C, in the solution. The first few aspheric coefficients of a lens cured upside-down are instead
a 2 = - 1 32 R ( 1 R 2 - 1 L c 2 ) a 3 = - 1 576 R ( 23 R 4 + 1 2 L c 4 - 47 R 2 L c 2 ) a 4 = - 1 819 R ( 375 R 6 - 1 9 L c 6 - 3745 9 R 4 L c 2 + 371 9 R 2 L c 4 ) ( 4 ) ##EQU00005##
FIGS. 2A-2B illustrate the difference between a lens cured right-side up and a lens cured upside down. There are two types of spheroids: oblate 210 and prolate 220. Both oblate 210 and prolate 220 spheroids can have the same radius of curvature at their apex, but oblate spheroids 210 are "flatter" than a sphere with the same apex radius of curvature, whereas prolate spheroids 220 are "sharper" than a sphere with the same apex radius of curvature. A liquid drop lens formed in zero gravity will have a spherical shape. Gravity will push a supported drop into an oblate spheroid 210 and will pull a suspended drop into a prolate spheroid 220, as shown in FIG. 2. The influence of gravity on lens shape is also dependent on the dimensions of the liquid drop lens. For lenses made out of the same material, gravity affects a large liquid drop lens more than a smaller liquid drop lens.
FIGS. 3A-3D further illustrate the difference between an oblate spheroid lens 210 and a prolate spheroid lens 220. FIGS. 3A-3B show a ray trace and corresponding spot diagrams for the oblate lens 210. FIGS. 3C-3D show a ray trace and corresponding spot diagrams for the prolate lens 220. For both lenses, the epoxy is NOA61, the focal length is 5.6 mm, the f/# is 3.5 and the full field of view is 6.13°. For both lenses, the ray intercepts in the image plane at four field points are shown. The prolate lens has a 6× reduction in spherical aberration compared to the oblate lens. This can be seen by comparing the smaller spot sizes of FIG. 3D to those in FIG. 3B. For imaging objects at a distance, prolate lens shapes are generally better than oblate lens shapes. As a result, upside-down curing is typically preferred compared to upright curing.
The solution of the second order ordinary differential equation (1) requires the addition of two independent parameters. One parameter is the volume of the lens, which can be precisely controlled by controlling the amount of curable liquid dropped on the substrate during the fabrication process. The other independent parameter is the contact angle, or the slope of the drop profile at the point where it touches the substrate on which it sits. An analytical expression exists for this contact angle which balances the forces of liquid-air, liquid-glass, and air-glass surface tensions.
One indirect way to measure the contact angle of a liquid is to measure the height of a puddle. As more and more volume is added to a drop, it eventually stops growing in height and simply expands at the sides. The drop becomes a puddle, which is flat at the top. The height of the puddle is related to the contact angle by
sin θ c 2 = h m ax ρ g γ ( 5 ) ##EQU00006##
where θ is the contact angle and hmax, is the maximum height.
Using this model, FIG. 4 graphs the focal length of a liquid drop lens as a function of lens volume and liquid contact angle. Given a contact angle and a volume, the lens profile can be predicted and so can the lens' focal length. Conversely, if the contact angle and the desired focal length are known, the corresponding volume of liquid can be calculated to produce a lens with the desired characteristics. The diameter and focal length of a lens grows with the volume of liquid dispensed.
Forces due to friction and viscosity, as well as the surface roughness of the substrate, can also be taken into consideration to estimate the contact angle of the liquid. Rough glass has many micro or nanoscale cavities and ridges covering its surface. If the surface is partially wet, the drop 110 does not fully penetrate the depths of these cavities. For rough glass, the contact angle is found by minimizing the total surface energy of the system. Surface energy is equal to the surface tension multiplied by the area of the surface. For drops on partially wet surfaces, there will be a fraction of the area of the drop 110 that touches glass, and a fraction of the drop 110 which is in contact with air trapped beneath the drop in these cavities. If x is the fraction of the liquid surface at the solid-liquid interface that actually touches the glass and (1-x) is the fraction of the liquid that is in contact with air, then the effective surface tension of the rough surface will be the sum of these fractional surface tensions. This effective surface tension is higher than the surface tension of the liquid on a smooth piece of glass.
where γ are the surface tensions. The subscripts LG denotes the liquid-gas interface, LS,smooth denotes the liquid-solid surface interface for a smooth surface, and LS,rough denotes the liquid-solid surface interface for a rough surface.
Furthermore, physical systems tend to try to minimize their free energy. These systems are considered to be at stable equilibrium. Stable equilibrium occurs in a drop 110 on a rough surface when the surface is fully wet. The reason a drop 110 may not fully wet right away is because local surface tension of the liquid and the air in the cavities prevents the liquid from filling the cavity. There is an energy barrier preventing a partially wet drop from fully wetting the surface of rough glass. It is often the case in thermodynamics that the application of heat can help a system overcome an energy barrier and minimize its free energy.
Liquid Drop Lenses Manufacturing Method
FIG. 5 shows a system for fabricating liquid drop lenses and FIG. 6 shows a system for fabricating several liquid drop lenses in parallel. The system includes a dispenser 610 and a substrate 620. In one embodiment, a pipette that can dispense a precise amount of liquid is used as the dispenser 610. In other embodiments, the dispenser 610 includes multiple tips (610a-610d) that are used to fabricate multiple lenses in parallel. In other embodiments, a pointed tip can be used as the dispenser 610 and the volume being dispensed can be controlled by the amount of time the tip is in contact with the substrate. In one embodiment, the substrate 620 is a planar substrate such as a glass substrate or a semiconductor substrate. In some embodiments, the surface of the substrate 620 can be treated to obtain a desired surface property. For example, the substrate 620 can be exposed to hexamethyldisilazane (HMDS) to increase the hydrophobicity of the substrate's surface.
FIG. 7 shows a flowchart for the fabrication of liquid drop lenses. To fabricate the liquid drop lens, the curable liquid is loaded 710 into the dispenser 610. In one embodiment, the dispenser is dried after loading the curable liquid. For example, if a pipette is used as the dispenser 610, the pipette tip is dried to remove the excess liquid that is present on the tip due to the liquid loading process. A substrate 620 on which the lenses are going to be fabricated is loaded 720 under the dispenser 610 and raised 730 until it touches the dispenser. In one embodiment, a micrometer stage or a micromanipulator can be used to precisely position the substrate underneath the dispenser 610. In one embodiment, instead of raising the substrate 620, the dispenser 610 is lowered until the dispenser 610 is in contact with the substrate 620.
The curable liquid has a high viscosity, for example at least 50 cps, more preferably at least 100 cps, or even 200 cps or higher. The high viscosity is desirable because higher viscosity liquids form taller drops, which results in lenses with more optical power. However, high viscosity liquids are more difficult to dispense, especially in low volumes. Contact between the dispenser 610 and substrate 620 can facilitate the dispensing of high viscosity liquids.
After the substrate 620 is in contact with the dispenser 610, a predetermined amount of liquid is dispensed 740 onto the substrate 620. The substrate 620 is lowered 750 until the liquid drop is no longer in contact with the dispenser 610. In some embodiments, the first drop of liquid dispensed from the dispenser is a sacrificial drop and is discarded. The steps described above are repeated in different locations of the substrate until drops for the desired lenses have been dispensed.
The contact between the substrate 620 and the dispenser 610 may be needed to allow the release of the liquid from the dispenser 610 to the substrate 620. In some embodiments, contact between the substrate 620 and the dispenser 610 is not needed. Instead, the substrate 620 is raised until the drop 110 makes contact with the substrate 620, even though the dispenser 610 itself is not in contact with the substrate 620. In one embodiment, if the liquid does not stick to dispenser 610, the substrate 620 can be raised to a position that allows accurate positioning of the drop 110.
The viscosity of the liquid can be modified by changing its temperature. For non-thermally cured materials, raising the temperature may reduce its viscosity and lowering the temperature may increase its viscosity. In some embodiments, a desired liquid viscosity can be achieved by tuning the temperature of the liquid before being dispensed.
After all the drops have been dispensed, the liquid is cured 760. The curing step may require the exposure of the substrate to UV radiation or to an activation temperature depending on the type of curable liquid or resin used. In one embodiment, the substrate is position upside-down during the curing step to leverage on the gravitational force that the drop experiences to generate a prolate spheroid. In other embodiments the orientation of the substrate can be changed during the curing process to obtain different shapes. For example, the substrate can first be partially cured upside-down for half of the time and then completely cured right side-up for the remaining curing time.
Using the above mentioned process, different size lenses can be produced. FIG. 8 shows an array of four lenses with a volume of 2.0 μL and a focal length of 3.6 mm produced using the process of FIG. 7. Using this process, we have manufactured lenses ranging from 0.01 to 2.0 μL in volume, from 0.09 to 3.60 mm in focal length, and from 0.01 to 7 mm in diameter. These processes are especially suitable for the fabrication of miniature lenses, especially circular lenses with a diameter smaller than 10 mm, including those that are less than 1 mm in diameter and less than 2 mm in focal length.
One advantage of liquid drop lenses is the ease of fabrication and the reduction in cost compared to other millimeter or micrometer size lenses. Liquid drop lenses do not require the patterning of special structures on the substrate or the use of expensive equipment. The substrate can be planar and featureless. Liquid drop lenses can be used in applications where the price and reliability of the fabrication process is important.
Applications of these liquid drop lenses include systems that take multiple images, including stereo imagers and multi-aperture imagers. FIG. 9 shows an example of a stereo imager. In this example, two liquid drop lenses 918 are formed on one side of a substrate 914. A front baffle 912 is also formed on the substrate 914 to reduce unwanted light. The back side of the substrate includes a second baffle 916. A sensor array can be attached to the back side of the substrate 914, either directly or spaced apart from the back surface. This example of two imagers can be extended to more than two imagers.
Although the detailed description contains many specifics, these should not be construed as limiting the scope of the invention but merely as illustrating different examples and aspects of the invention. It should be appreciated that the scope of the invention includes other embodiments not discussed in detail above. For example, automatic high-viscous liquid dispensers may be used instead of pipettes. Further, the liquid drop may be dispensed on a thick planar substrate, which is then placed on a light sensor. The planar substrate could be a homogeneous medium with uniform refractive index or a gradient index medium. Various other modifications, changes and variations which will be apparent to those skilled in the art may be made in the arrangement, operation and details of the method and apparatus of the present invention disclosed herein without departing from the spirit and scope of the invention as defined in the appended claims. Therefore, the scope of the invention should be determined by the appended claims and their legal equivalents.
In the claims, reference to an element in the singular is not intended to mean "one and only one" unless explicitly stated, but rather is meant to mean "one or more." In addition, it is not necessary for a device or method to address every problem that is solvable by different embodiments of the invention in order to be encompassed by the claims.
Patent applications by Sri Rama Prasanna Pavani, Palo Alto, CA US
Patent applications in class LENS
Patent applications in all subclasses LENS |
Jakob Bernoulli’s book, Ars Conjectandi, marks the unification of the calculus of games of chance and the realm of the probable by introducing the classical. However, the Ars Conjectandi, in which he presented his insights (including the fundamental “Law of Large Numbers”), was printed only in , eight years. Jacob Bernoulli’s Ars Conjectandi, published posthumously in Latin in by the Thurneysen Brothers Press in Basel, is the founding document of.
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Bernoulli’s work influenced many contemporary and subsequent mathematicians. On a note more distantly related to combinatorics, the second section also discusses the general formula for sums of integer powers; the free coefficients of this formula conjectajdi therefore called the Bernoulli numberswhich influenced Abraham de Moivre’s work later, and which have proven to have numerous applications in number theory.
Jacob’s own children were not mathematicians and were not up to the task of editing and publishing the manuscript. It was in this part that two of the most important of the twelvefold ways—the permutations and combinations that would form the basis of the subject—were fleshed out, though they had been introduced earlier for the purposes of probability theory.
Ars Conjectandi | work by Bernoulli |
The second part expands on enumerative combinatorics, or the systematic numeration of objects. Finally Jacob’s nephew Niklaus, 7 years after Jacob’s death inmanaged to publish the manuscript in Preface by Sylla, vii. Thus probability could be more than mere combinatorics. Even the afterthought-like tract on calculus has been quoted frequently; most notably by the Scottish mathematician Colin Maclaurin. Views Read Edit View history.
The development of the book was terminated by Bernoulli’s death in ; thus the book is essentially incomplete when compared with Bernoulli’s original vision. The fruits of Pascal and Fermat’s correspondence interested other mathematicians, including Christiaan Huygenswhose De ratiociniis in aleae ludo Calculations in Games of Chance appeared in as the final chapter of Van Schooten’s Exercitationes Matematicae.
In the third part, Bernoulli applies the probability techniques from the first section to the common chance games played with playing cards or dice. Bernoulli wrote the text between andincluding the work of mathematicians such as Christiaan HuygensGerolamo CardanoPierre de Fermatand Blaise Pascal.
Huygens had developed the following formula:.
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Indeed, in light of all this, there is good reason Bernoulli’s work is hailed as such a seminal event; not only did his various influences, direct and indirect, set the mathematical study of combinatorics spinning, but even theology was impacted. In the field of statistics and applied probability, John Graunt published Natural and Political Observations Made upon the Bills of Mortality also ininitiating the discipline of demography.
The first period, which lasts from tois devoted to the study of the problems regarding the games of chance posed by Christiaan Huygens; during the second period the investigations are extended to cover processes where the probabilities are not known a priori, but have to be determined a posteriori.
This work, among other things, gave a statistical estimate of the population of London, produced the first life table, gave probabilities of survival of different age groups, examined the conjectandu causes of death, noting conjectandk the annual rate of suicide and accident is constant, and commented on the level and stability of sex ratio. He presents probability problems related to these games and, once a method had been established, posed generalizations.
The seminal work consolidated, apart from many combinatorial topics, many central ideas in probability theorysuch as the very first version of the law of large numbers: It was also hoped that the theory of probability could provide comprehensive and consistent method of reasoning, where ordinary reasoning might be overwhelmed by the complexity of the situation. In this section, Bernoulli differs from the school of thought known as frequentismwhich defined probability in an empirical sense.
Another key theory developed in this part is the probability of achieving at least a certain number of successes from a number of binary events, today named Bernoulli trials given that the probability of success in each event was the same. Between andLeibniz corresponded with Jakob after learning about his discoveries in probability from his brother Johann.
Ars Conjectandi – Wikipedia
conjevtandi This page was ara edited on 27 Julyat A significant cnojectandi influence was Thomas Simpsonwho achieved a result that closely resembled de Moivre’s. It also discusses the motivation and applications of a sequence of numbers more closely related to number theory than probability; conjectaandi Bernoulli numbers bear his name today, and are one of his more notable achievements. The fourth section continues the trend of practical applications by discussing applications of probability to civilibusmoralibusand oeconomicisor to personal, judicial, and financial decisions.
Finally, in the last periodthe problem of measuring the probabilities is solved. Before the publication of his Ars ConjectandiBernoulli had produced a number of treaties related to probability: Core topics from probability, such as expected valuewere also a significant portion of this important work. Retrieved 22 Aug The complete proof of the Law of Large Numbers for the arbitrary random variables was finally provided during first half of 20th century.
The first part concludes with what is now known as the Bernoulli distribution.
He gives the first non-inductive proof of the binomial expansion for integer exponent using combinatorial arguments. The two initiated the communication because earlier that year, a gambler from Paris named Antoine Gombaud had sent Pascal and other mathematicians several questions on the practical applications of some of these theories; in particular he posed the problem of pointsconcerning a theoretical two-player game in which a prize must be divided between the players due to external circumstances halting the game.
Later, Johan de Wittthe then prime minister of the Dutch Republic, published similar material in his work Waerdye van Lyf-Renten A Treatise on Life Annuitieswhich used statistical concepts to determine life expectancy for practical political purposes; a demonstration of the fact that this sapling branch of mathematics had significant pragmatic applications. The art of measuring, as precisely as possible, probabilities of things, with the goal that we would be able always to choose or follow in our judgments and actions that course, which will have been determined to be better, more satisfactory, safer or more advantageous.
For example, a problem involving the expected number of “court cards”—jack, queen, and king—one would pick in a five-card hand from a standard deck of 52 cards containing 12 court cards could be generalized to a deck with a cards that contained b court cards, and a c -card hand.
The refinement of Bernoulli’s Golden Theorem, regarding the convergence of theoretical probability and empirical probability, was taken up by many notable later day mathematicians like De Moivre, Laplace, Poisson, Chebyshev, Markov, Borel, Cantelli, Kolmogorov and Khinchin. Bernoulli provides in this section solutions to the five problems Huygens posed at the end of his work.
According conejctandi Simpsons’ work’s preface, his own work depended greatly on de Moivre’s; the latter in fact described Simpson’s work as an abridged version of his own.
It also addressed problems that today are classified in the twelvefold way and added to the subjects; consequently, it has been dubbed an important historical landmark conjectando not only probability but all combinatorics by a plethora of mathematical historians. He incorporated fundamental combinatorial topics such as his theory of permutations and combinations the aforementioned problems from the twelvefold way as well as those more distantly connected to the burgeoning subject: After these four primary expository sections, almost as an afterthought, Bernoulli appended to Ars Conjectandi a tract on calculuswhich concerned infinite series.
However, his actual influence on mathematical scene was not great; he wrote only one light tome on the subject in titled Liber de ludo aleae Book on Games of Chancewhich was published posthumously in The latter, however, did manage to provide Pascal’s and Huygen’s ads, and thus it is largely upon these foundations that Ars Conjectandi is constructed. Coniectandi importance of this early work had a wrs impact on both contemporary and later mathematicians; for example, Abraham de Moivre.
Retrieved from ” https: The first part is an in-depth expository on Huygens’ De ratiociniis in aleae ludo.
In this formula, E is the expected value, p i are the probabilities of attaining each value, and a i are the attainable values. |
« on: 07/07/2021 16:12:34 »
One more modern use for the centrifuge: The separation of isotopes. Used as a method to enrich Uranium for use as nuclear fuel.
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Hi all,Then there would be aberration and the effect gravity had on the light before it entered the pin hole. But all that would do is alter the initial angle that the light exits the hole relative to the chamber. It would be the same as a fixed light source that wasn't aimed at a point directly across from it.
So if we use Harri's example of a light beam passing through the experiment and also factor in Colin's pointQuoteThink. What have you fastened the light beam equipment to?
lets say you use sunlight as your light source, passing through a pinhole into the room/lab, streaming across the mid point would that not discriminate the points under discussion ?
Okay, we will assume that the source of the light is fixed relative to the ground. The light is turned on. The light, after leaving the source, will curve downward due to gravity. However since light travels so fast, it would take extremely accurate measurements to even notice that it hits the far wall a bit lower.
Could I ask, what if we return to the vacuum chamber where a bowling ball and feather are suspended from the ceiling. We then create a vacuum by withdrawing the air in the chamber. This time, midway between the objects and the floor we pass a light beam across their path. On release of the ball and feather what would reach the beam first, the ground or the objects? Or both simultaneously?
You are all so smart here, you know the equations. Maybe someone will be interested in recalculating the result of the Michelson-Morley experiment with a much smaller Earth orbit, and the speed of the Earth (approximately as in the schematic image below). Perhaps that experiment proved the existence of the aether, but was incorrectly interpreted as wrong due to the false parameters of the earth's orbit and the speed of the Earth in space.We can independently measure the speed of the Earth by noting the change of stellar aberration as it orbits. This value is in agreement with the value you get from dividing the size of the orbit by the time it takes to complete one orbit.
No, it just means that, in physics, the term "spin" when working at the electron level means something different than it does at the macroscopic level. It's simply a matter of context in determining which meaning is being used.The problem was already solved.
QM "spin" is not spin.
That requires you to ignore physical intuition.
It's like saying for something that does not obey Newton's laws: ""force" is not force".
And your estimation of the risk of meteor collision is extremely overblown.It also contradicts direct measurements (like the sizes of planets) and apparently relies on conspiracy theories (since you claim that interplanetary spacecraft aren't feasible because you say they'd be destroyed by asteroid impacts...).I claimed that space flights are unreasonably risky due to the inevitabile possibility of destruction in a collision with a meteorite, that is, in fact, they are meaningless.
To be fair, in order to escape a body, you eventually have to reach "escape velocity", it's just that it doesn't need to be the surface escape velocity. For instance, if your rocket has reached a distance of 6378 km above the Earth's surface, and is moving at 7.91 km/sec, It has achieved escape velocity for that distance from the Earth. It can cut its engines and coast, never to return.In order to escape from the gravitational field of a large object it's necessary to achieve escape velocity.Not so. Rockets have escaped from Earth without ever having reached the 11.2 m/sec escape velocity at its surface. A slow winch will do if you have one (like a space elevator).
You seem to be applying rules of inertial coordinate systems to non-inertial cases. Yes, given one rock falling in near the event horizon passing an identical rock going the other way at escape velocity, each of them will have near infinite mass relative to the local inertial frame of the other.
Relative to the distant observer, either rock has kinetic energy that cancels out the negative gravitational potential energy, leaving no change to the total energy of the rock. The rock has negligible coodinate speed (it never reaches the EH according to the distant observer), so the mass isnít a function of relativistic speed as you are painting it. Plus, the rock having more mass than does the black hole doesnít make sense.
The photon loses energy overcoming gravity. There is a certain amount of gravity at every point in space. Hence the photon loses energy with distance. Since the energy of a photon is not unlimited, the distance covered by a photon is also quite limited. As well as the lifetime of a photon. Unlikely to exceed 1-2 light minutes.A photon loses energy "overcoming" gravity, In other words when climbing out of a gravity well. But it gains energy falling into a gravity well. Light leaving the surface of a planet will red-shift as it climbs away from the planet. But another observer, at some distance away on another planet sitting in an equally deep gravity well will see no red-shift in the light coming from the first planet, because the light blue-shifted as it fell into his gravity well. The strength of each planet's gravity makes no difference, neither does the distance between them. All that matters is the relative depth of the gravity wells for source and observer.
Spectrum redshift = gravitational constant * distance covered by light.
Definition*. The lifetime of a unit of wave oscillations (one wave) is inversely proportional to the speed of their propagation (or directly proportional to the inertia of the medium) and is directly proportional to the power of their source.
* - this definition is correct with or without the aether.
Space is almost empty, but despite this, thousands of meteors burn out in the earth's atmosphere in just one night? None of the rovers recorded a single meteorite fall on the surface of Mars, given that its official size is not much smaller than that of Earth, and the atmosphere is much thinner than Earth's. That is, much more meteorites should reach the surface of Mars, including objects of such sizes that burn up in the earth's atmosphere.
No, they haven't. 10% to 15% of galaxies are elliptical. And the old idea that elliptical galaxies evolve into spiral galaxies has been rejected.
Bonus; it seems all galaxies have flatten out, why didn't the universe?
Hi all.As with most apparent "paradoxes" in Relativity, this one is likely due to only focusing on one aspect of the theory and ignoring the rest.
Does anyone have some knowledge or insight about this "paradox" in the theory of Relativity?
Imagine a submarine underwater.
The submarine is at rest relative to the fluid and has adjusted it's tanks so that it has equal density with the fluid and remains at a depth of 100 metres. (No thrust required from the engines, it just has neutral buoyancy).
The submarine accelerates rapidly to reach relativistic speeds (let's say 0.9 c) relative to the fluid and then sustains a constant velocity. This is intended to be a horizontal motion, the fins, bow planes etc. were not set to drive the submarine up or down.
As is usual for these sorts of paradoxes, we have two observers in two different frames of reference.
The submarine commander is at rest inside the submarine. She should observe length contraction for the fluid in her rest frame and a corresponding increase in density of the fluid. The submarine retains it's rest characteristics, including density in her frame.
A mermaid is at rest on the ocean floor. In her rest frame, the density of the fluid has not changed, however the submarine has undergone length contraction in her frame and it's density has increased.
Will the submarine rise or sink due to buoyancy?
Background info: You may like to read the Wikipedia article about Supplee's paradox.
There is also a similar discussion about a Helium balloon moving through air on another forum. (I'm not sure I should put links to another forum).
I do not know the answer. I can see references to articles in that Wikipedia entry but they seem to demand some application of General Relativity and a complete re-write of the Archimedian principle. I was wondering if there is a resolution based only on Special Relativity - but I'll take any insight or discussion I can get.
A few corrections here:Why is the Moon moving away from the Earth. Shouldn't it be getting closer, pulled towards the Earth by gravity?
Earth isn't a perfect sphere, it's oblate (more like an egg than a good round ball). The moon is actually causing most of this deformity in the earth (by pulling on it with the force of gravity). The thing is the earth is spinning on it's own axis quite fast, so these bulges get pulled slightly forward of the line between the earth and the moon. The moon goes in a prograde orbit around the earth (it moves around the earth in the same direction as the earth spins). The bulges on the earths surface are then ever so slightly pulling the moon forward a bit, while at the same time slowing the rotation of the earth. By giving the moon a bit more of a pull in the direction tangential to it's orbit the moon's speed is increasing. As that speed increases, the moon drifts out a bit further to an orbit with a greater radius.
There's some info here:
Okay, while it is true that you could maintain a situation of simulated gravity by accelerating a rocket at 9.8m/s^2, our present technology makes it impractical.
Thanks All for All Responses.
Please feel free to continue detailed discussions.
(Personally my Query stands Resolved)
Speed of 9.8 m/s/s would be 35.27 km/hr.
To maintain the Rate of Acceleration, would mean to keep going faster per sec by 9.8 m/s/s.
1 Hour = 60 Mins = 3600secs.
So 1sec at 9.8 m/s..2sec at 19.6 m/s..& so on.
Hence value of 35280 m/s attained.
So that would be 127008 km/hr.
So after an Hour, maintaining the Rate of Acceleration, an object would attain a Speed value of 127008 km/hr.
Just to be reassured i Understand this...
If the clause was for 2 Hours.
Then 7200secs ◊ 9.8 = 70560 m/s.
That's 254016 km/hr.(Speed)
Follow up Question :-
Would a Rocket Accelerating at 9.8 m/s in Deep Space devoid of Any Gravitational Fields be able to mimic the " g " effect on planet Earth?
P.S. - Perhaps a 10 meters cloth(fabric) could be converted into, or better let's say could be assumed to have an approx weight in kilogrammes.
( But Yes i do Completely Understand the Futility of trying to convert " g " into " s ". )
The equation for orbital period is T = 2pi sqrt(a^3/G(m1+m2)Where did all this new angular momentum come from?No new angular momentum. That's part of the discussion, like the skater's arms or a coalsescing galaxy: change the mass distribution and the rotational speed changes.We're mining the moon, not mining the Earth. g (the gravitational acceleration at lunar radius) is unaffected by the mining of the moon.True, but it's another way of changing the weight of the moon, and the one (the OP said "any change...") that would alter R.
Let's compare this scenario to waves created by a ship moving in a pond. Scenario presented on your animations shows a situation, where a wave is propagating perpendicuralry to the motion of a boat in a pond, which moves together with the boat in relation to a stationary observer - it propagates perpendicularly in the frame of boat and diagonally in the frame of stationary observer. However wave propagating perpendicuralry to a moving boat in a stationary pond will behave differently - it will propagate perpendicularly to the moving boat in the frame of stationary bystander and diagonally in the frame of the boat.
If the motion of light is not affected by the motion of it's source in the frame of a stationary observer, then it's the second option (with stationary pond), which should be applied to this scenario.
I'm going to add a couple of animations of my own to try and clarify what is going on here:
I have also an issue with the animation from wikipedia: https://en.wikipedia.org/wiki/Aberration_(astronomy)
According to it light behaves just like any other physical object (e.g a bullet) and the perpendicular motion of the light source is being added to the vector of light propagation - so motion of the light beam is a sum of 2 perpendicular vectors. However as I said earlier motion of the light source shouldn't affect the motion path of light in stationary frame.
I know, that it doesn't have too much with the gravity, but since we're talking about a scenario with the elevator, I wonder what would happen if we would modify it and place the sensor inside the moving elevator, while keeping the source of light in stationary frame. Let's say, that light is being emitted by the stationary source in the moment, when it is at the same level, as the sensor placed inside the elevator moving upwards - will the light reach that sensor or not? If it won't reach it, then the motion of elevator will become absolute/definitive, what will violate the relative nature of relative motion. If it will reach the sensor, then path of light will become curved upwards in the frame of stationary source. Which option is the valid one?Let's, for the moment, exclude the acceleration of the Elevator, and just imagine the path of the light if the Elevator and external light source have a relative motion with respect to each other. Light entering the hole from the source will not hit the sensor opposite the hole. However, you cannot conclude from this that there is absolute motion on the part of the Elevator. This is due the the aberration of light, which is caused by the Relative motion between source and Elevator. It happens exactly the same whether you consider the Elevator moving and the source stationary or the Elevator stationary and the Source moving. Thus it can only tell you that they are moving relative to each other, and not which one is "really" moving. |
Hyperspace:Home Intro Time Width Walls Bridges Rounds
One common misconception is that time is the fourth dimension. Let's make up a fragment of space-time, and see if this is what we're looking for!
One starts with a plane at time zero. Over this we keep putting sheets of paper, each at time one, time two, and so forth. Eventually, we get a stack of paper that fills all of three dimensions with space-time.
Our Square becomes a snaking prism that starts at his birth and goes to the end of his body. One might visualise a slice of space-time for a card as a deck of cards. This is not what the Sphere sees as the Realm of Three Dimensions.
Space-time is used in relativity a lot, but is more extensive than that. One can make a space-time sequence in the sort of flick-animations that one does down the sides of pages of books. When the book is closed, there. One sees motion when one forces the sequences to flick past one after another, but this is not real time. We can't talk to the characters presented in these animations.
The Newtonian Mechanics is based on Euclidean geometry. One can not tell, for example, how fast one is travelling in an inertial frame, or the relative size one is. The same Newtonian physics is true for atoms and galaxies, for fast and for stopped.
In the field of electromagnetic fields, there is an 'electromagnetic velocity constant', which derives the frequency of an electromagnetic oscillator from its linear frequencies. Experimental data had shown that this constant was in the same order of the speed of light, but no one had proof.
James Clarke Maxwell showed that electromagnetic waves travel at the electromagnetic velocity, and noting the close similarity of this, noted that light travels in the same medium that Electromagnetic waves do: Ether. Heinrich Hertz demonstrated that waves generated off rotating magnets have the same properties of light, at that frequency.
The ting with the Maxwellian Electromagnetism is that it is only true in one inertial frame: the Etherfer. One could then calculate the Newtonian inertial velocity in terms of the etherfer, by, measuring the speed of light. The resulting experiments were carried out by Michaelson, and completed by Morley. The results failed to show a specific velocity, despite the precision of the experiment.
The failure of the experiment lead to a number of reasons about why the experiment failed. For example, the Ether might be dragged along with large objects. Fitzgerald proposed contraction of space in the direction of motion, while Lorenz suggested a dialation of time. In any case, the variety of the solutions was more tho explain the null result of the Michaelson-Morley experiment than a logical implication of it.
Einstein mused about travelling at the speed of light, and looking at an electron that was travelling beside. We do not see standing electro- magnetic waves, and correspondingly, we must not travel at the speed of light. These appeared in papers of 1905 and 1908 describing General and Special Relativity.
The General Relativity describes a space where mass has no effect. It also says that if the speed of light is infinite, or relatively infinite, then the much simpler Newtonian Mechanics arises. Therefore, relativity can not disprove the Euclidean geometry, but tells us that there are situations where we need to take into account the effect of relative rathness.
One makes heavy use of space-time diagrams, where an observer sees a rath-travelling thing sheer space-time, to the extent of shorter and slower clocks.
General relativity is what one learns as the first taste of relativity, and is probably where the notion of time as a fourth dimension comes from.
The geometry of general relativity is that of Minkowski, the distance
between two points is
Special relativity is a kind of general relativity where space is curved by large masses. This is the relativity one is talking about when one sees those deep holes around black holes, and stars that stretch the fabric of space.
Note that the intent of such stretchong is to simulate how a thing under inertia would move. The Newtonian space is a flat billards table, and the nature of gravity makes the table stretch.
The thing about stretchy space is that the 2d surface is supposed to represent space, and time is the same time we have: that is, not on the graph. So the stretching of gravity is being represented in a different dimension to either space (the surface), or time (presented as real time).
This curved billiard-ball table works because there is an outside real gravity that holds things to the model. It is not presented in any of the space axies, or in time, so if the model is valid, it would require a fifth dimension, which has a real gravity.
In the theory of isospace, all space is curved: not curved in something, but none the less, curved. So if three-space is curved in four-space, then four space is curved in five-space, and so forth. Any practical model of curvature not involving being bent in space must also equate to being bent in space, if room exists.
A model of curvature would be to add or subtract circumference to a circle. A circle drawn on a sphere has a smaller circumference than 2π of its radius. And here lies the secret of curvature.
So instead of having curvature at points, we make point and direction a feature of curvature. In an isospace, at any point, the length of any arc of space around it depends only on the radius and angle.
Consider, now what would happen, if different degrees of the circle around a point are different lengths. The tension of space is not defined by angle but circumfence length. The degrees that have longer lengths would pull harder, and there would be a net force produced by empty space. This is a method for explaining gravity without relying on force at a distance.
Te presence of a large body would cause more of space to be bunched in the degrees nearer the body, and this would propegate across space by way of tension of curvature.
A line straight through the point would divide the perimeter, and so we would have a curved line being the inertial line. So, we do not need the tension of space to become a binding force, just twisting the notion of a straight line is enough.
So even something without mass for force to act on, would feel the effect of space bending. The subtle effect of making half the circle longer than the other half would cause a straight line to span less than 180 degrees, and light (which moves in a straight line), to 'bend'.
Such is observed in gravitational lensing, where a distant quasar appears above and below a galaxy.
Hyperspace:Home Intro Time Width Walls Bridges Rounds |
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?
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?
This problem is based on the story of the Pied Piper of Hamelin. Investigate the different numbers of people and rats there could have been if you know how many legs there are altogether!
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?
Use your mouse to move the red and green parts of this disc. Can you make images which show the turnings described?
Arrange eight of the numbers between 1 and 9 in the Polo Square below so that each side adds to the same total.
If we had 16 light bars which digital numbers could we make? How will you know you've found them all?
How could you put eight beanbags in the hoops so that there are four in the blue hoop, five in the red and six in the yellow? Can you find all the ways of doing this?
Suppose we allow ourselves to use three numbers less than 10 and multiply them together. How many different products can you find? How do you know you've got them all?
Investigate the different ways you could split up these rooms so that you have double the number.
I like to walk along the cracks of the paving stones, but not the outside edge of the path itself. How many different routes can you find for me to take?
Sort the houses in my street into different groups. Can you do it in any other ways?
In this investigation, you must try to make houses using cubes. If the base must not spill over 4 squares and you have 7 cubes which stand for 7 rooms, what different designs can you come up with?
Lolla bought a balloon at the circus. She gave the clown six coins to pay for it. What could Lolla have paid for the balloon?
Vincent and Tara are making triangles with the class construction set. They have a pile of strips of different lengths. How many different triangles can they make?
Use the interactivity to investigate what kinds of triangles can be drawn on peg boards with different numbers of pegs.
This challenge involves calculating the number of candles needed on birthday cakes. It is an opportunity to explore numbers and discover new things.
What is the smallest number of tiles needed to tile this patio? Can you investigate patios of different sizes?
An investigation that gives you the opportunity to make and justify predictions.
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?
This challenge extends the Plants investigation so now four or more children are involved.
This challenging activity involves finding different ways to distribute fifteen items among four sets, when the sets must include three, four, five and six items.
This tricky challenge asks you to find ways of going across rectangles, going through exactly ten squares.
In how many ways can you stack these rods, following the rules?
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 interactivity to find all the different right-angled triangles you can make by just moving one corner of the starting triangle.
The challenge here is to find as many routes as you can for a fence to go so that this town is divided up into two halves, each with 8 blocks.
You cannot choose a selection of ice cream flavours that includes totally what someone has already chosen. Have a go and find all the different ways in which seven children can have ice cream.
There are to be 6 homes built on a new development site. They could be semi-detached, detached or terraced houses. How many different combinations of these can you find?
If you have three circular objects, you could arrange them so that they are separate, touching, overlapping or inside each other. Can you investigate all the different possibilities?
How many different cuboids can you make when you use four CDs or DVDs? How about using five, then six?
Ana and Ross looked in a trunk in the attic. They found old cloaks and gowns, hats and masks. How many possible costumes could they make?
How many ways can you find of tiling the square patio, using square tiles of different sizes?
Compare the numbers of particular tiles in one or all of these three designs, inspired by the floor tiles of a church in Cambridge.
In a Magic Square all the rows, columns and diagonals add to the 'Magic Constant'. How would you change the magic constant of this square?
Cut differently-sized square corners from a square piece of paper to make boxes without lids. Do they all have the same volume?
Let's say you can only use two different lengths - 2 units and 4 units. Using just these 2 lengths as the edges how many different cuboids can you make?
An investigation involving adding and subtracting sets of consecutive numbers. Lots to find out, lots to explore.
Roll two red dice and a green dice. Add the two numbers on the red dice and take away the number on the green. What are all the different possible answers?
How many different shaped boxes can you design for 36 sweets in one layer? Can you arrange the sweets so that no sweets of the same colour are next to each other in any direction?
Let's suppose that you are going to have a magazine which has 16 pages of A5 size. Can you find some different ways to make these pages? Investigate the pattern for each if you number the pages.
Can you continue this pattern of triangles and begin to predict how many sticks are used for each new "layer"?
An activity making various patterns with 2 x 1 rectangular tiles.
Suppose there is a train with 24 carriages which are going to be put together to make up some new trains. Can you find all the ways that this can be done?
What is the smallest cuboid that you can put in this box so that you cannot fit another that's the same into it?
How many different ways can you find of fitting five hexagons together? How will you know you have found all the ways?
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?
This challenge involves eight three-cube models made from interlocking cubes. Investigate different ways of putting the models together then compare your constructions.
How many models can you find which obey these rules?
Polygonal numbers are those that are arranged in shapes as they enlarge. Explore the polygonal numbers drawn here. |
In this experiment you will use the strain gage installation from the prior lab assignment and test
the cantilever beam under
dynamic loading situations.
a resistance strain gage
as one resistance arm in
. The gage will be mounted on a cantilever beam and subjected to
uniaxial tension or
compression. The formula for volt
age output is given in Eq. (
. For example for
an input voltage V of
gage factor F of
output voltage reading
would indicate V
provided by gage manufacturer (typ. ~ 2.0)
(external amplifier unit, typ. ~100)
Solving Eq. (1) for strain,
Strain gage application chemicals and supplies
sandwich and C
ohm resisters for bridge circuit
Data acquisition system and computer running LabView
upplies available for installing a strain gage.
These include an installation guide, the
gages, cleansing chemicals, abrasive sand paper, adhesive, a soldering iron,
that is used to facilitate the creation of a bridge circuit.
Precision resistors and the amplifier needed
for the bridge
circuit are not shown.
Retrieve the instrumented aluminum beam from your prior lab assignment
by first verifying that the resistance across the gage is
approximately 120 Ohms. The beam should then be clamped in the steel cla
mping device as
shown in Fig.
center of the gage should be 0.75” from the edge of the clamping
. Be careful not
to clamp down on the gage itself!
Be sure that the cantilever edge is
perpendicular to the long edge of the beam
easure all beam dimensions
including distances from cantilever edge to gage center and to
haven’t done so already, measure the mass of the beam so that you can estimate the mass per
unit length value used in the analysis of beam vibrations.
The lead wires should be attached to a bridge circuit constructed with 3 other high
istors mounted on the ProtoBoard shown in
1. The circuit should be
illustrated in Fig.
, with the output voltage
wired into the input end of the
and the amplifier output
initially measured by a digital voltmeter.
the amplifier is
switched on a
nd set to a gain of
The completed strain gage installation. The beam is shown mounted to the cantilever
structure (steel blocks and clamp).
Actual beams in this experiment may be different (e.g., hook).
. Wiring diagram for bridge circuit.
is the gage and R
, and R
are all the same
resistance as the nominal gage resistance (120
Graphic from Omega.com.
Determine beam natural frequency
Write a virtual instrument
the beam. Note that the sensor output (a
resistance change) is converted into a voltage signal in the transducer stage (bridge circuit) of
this experiment. So, the virtual instrument is measuring voltage NOT strain
0 Hz and
the sample period so that you collect
nds worth of data for each
test (5,000 samples).
The virtual instrument should store the raw voltage values, but it should
also be able to determine and display a frequency
for the response
(ie the fre
within which most of the signal power is contained)
. This can be accomplished in a variety of
ways, but will likely include the use of an existing subVI that can be found in the Functions
either in the
SignalProcessing tool palettes. You wil
l need to explore LabView Help to
identify which resource is best suited to your particular need in this experiment.
Be sure to
record the results of the frequency analysis for every run that you conduct.
To start with you should see if there are any unde
sired vibrations being transmitted into the
beam from the table
building. If there appears to be any such noise try to minimize it
through isolating the test apparatus (perhaps adjusting the rigidity of the table and cushioning
either the table f
eet or the clamp mechanism). Prior to testing for the natural frequency of the
beam record at least one sample that will allow you to quantify the noise in the system.
run the vi without touching the beam. Name the resulting file “control.txt”.
To conduct the vibration experiment, r
un the VI, i
tiate sampling and wait between 1 and 2
before plucking the beam. Pluck the free end of the beam with a
motion of the
. Use your fingernail to initiate the beam vibration
data gathered by the system
will provide a measure of strain, and hence deflection vs. time
These data can be analyzed
later to determine both the natural frequency of response and the damping ratio of the system.
Name the resulting file “vibe
(total of 3 replicates)
naming subsequent files “vibe
By conducting the same measurement
times you will be able to investigate measurement repeatability in your analysis.
To verify that
ing rate is sufficient run the same expt. at twice the sampling rate (for same total
sample period), naming these files “vibe1000_1.txt”, etc.
can be calculated
as a function of
deflection as predicted by linear b
where P is the load, E is the modulus of elasticity, and I is the moment of intertia (be sure to
calculate this about the correct axis),
is the deflection, and L
is the length of the beam (from
t to load).
Knowing the load and beam properties/dimensions you can calculate
at the strain
using beam theory:
where L is the distance from the point load to the center of the strain gage.
2 the strain gage conditioning circuit gain (G) can be
determined. This same equation can then be used with output voltage data to determine the
transient strain signal during the vibration tests.
As the beam vibrates the out
put voltage from the strain gage system will oscillate and decay.
This oscillation occurs at the natural frequency of the beam. While the natural frequency is the
dominant term dictating the beam oscillatory frequency other higher order frequencies of the
may also be evident in the signal. A representative trace of the system re
illustrated in Fig.
. Sample trace of the response of the vibrating beam showing the natural frequency and
the signal decay associated with dampin
is given by
To obtain a
estimate of natural frequency one can
t over many (N=5) periods and determine the individual frequency by dividing by N.
Assuming there is no damping t
he natural frequency can be obtained
where m’ is the mass per unit length of the beam
is the beam length
from the cantilever
edge to the free edge
. See an appropriate textbook to investigate the theory behind this
equation and the determ
ination of th
e constant (1.875). Note that the natural
from eqn (
5) is in the units of
radians per second
This equation for natural frequency assumes a uniform beam (without the hook at the end). An
alternative development found in some vibrations
texts allows one to consider the mass of the
hook at the end. Such an analysis is optional.
Note that in a real cantilever beam system the oscillations are damped. In such a case the
response of the beam is sinusoidal with frequency
and an amplitude tha
t decays as:
To obtain the damping ratio (
) one can simply take the natural log of the above equation and
plot Ln(amplitude) vs. t. The resulting plot should have a slope of
The relationship between the natural
frequency and the damped frequency that you actually
measure is given by:
You must write a f
(no more than
. You should also include an
(no more than
of raw data &
The following list of questions is intended as a
for students as they consider their
up. Students should NOT set out to answer this specific set of questions, but rather use
them as a guide as they consider the anal
ysis of their experimental data.
How do the strain measurement results compare with beam theory predictions for the static
repeatable are the results of the calibration experiment? What factors contribute to
How does the theor
etical natural frequency of this beam compare to the results of the
transient measurements? What factors contribute to any differences?
How does sampling frequency affect your results?
What is the damping ratio of the system?
Figliola, R.S. a
nd Beasley, D.E.,
Theory and Design for Mechanical Measurements
Edition. John Wiley and Sons.
Robert L. Norton ,
An Integrated Approach
Prentice Hall, 1997
Mechanical Engineering Design,
Hill Book Co. 1977
Elements of Vibration Analysis,
Hill Book Co., |
Web1 cm = in 1 in = cm Example: convert 15 cm to in: 15 cm = 15 × in = in Popular Length Unit Conversions cm to inches . WebDo a quick conversion: 1 centimetres = inches using the online calculator for metric conversions. Check the chart for more details. Web35 rows · 26 Centimeters = Inches rounded to 8 digits Display result as A .
In 26 cm there are in. Which is the same to say that 26 centimeters is inches. Twenty-six centimeters equals to ten inches. *Approximation. Learn formula of how to convert cm to inches. What is Centimeter? Centimeter is equal to 1/ meter and it is denoted by cm. 26 cm, in. Convert 26cm into inches. How tall is 26 cm in inches? How many inches is 26 cm? The answer is Convert Centimeters to Inches. The Cm to Feet and Inches Conversion Calculator is used to convert centimeters to feet and inches. 26 cm, 0 feet and inches. * which is equal to Inches. Refer following examples to decode how the formula can be used for easy computation of values from cm to Inches. body measurements, body measurements. Men's/Unisex Size, Waist, Women's Size, Waist. inches, cm, inches, cm. XS, ", cm, XS, 26", 66cm. 26 cm: 3/4 inch = 1. 1 centimetre=0. equals 0. This number system is a bit easier to grasp than inches and feet, where an inch is 1/12th of a foot.
WebThe result of converting 26 Centimeters to Inches: 26 cm = Inches. 26 Centimeters (cm) equals to Inches. WebValue in inches = value in cm × So, 26 cm to inches = 26 cm × = inches. This formula can also be used to answer similar questions: What’s the formula to convert inches from 26 cm? WebTo convert centimeters into inches we have to multiply by the conversion factor in order to get the length amount from centimeters to inches. We can also form a simple proportion to calculate the result: 1 cm → in. cm → L (in). WebWhat is 20 x 26 cm in inches? Convert 20x26 cm to in.
WebOne centimeter equals inches, in order to convert 26 cm x 26 cm to inches we have to multiply each amount of centimeters by to obtain the length and width in inches. In this case to convert 26 cm by 26 cm into inches we should multiply the length which is 26 cm by and the width which is 26 cm by The result. WebConvert 26 cm to inches. One centimeter equals inches, to convert 26 cm to inches we have to multiply the amount of centimeters by to obtain the width, height or length in inches. 26 cm is equal to 26 cm x = inches. The result is the following: 26 cm = inches. Definition of centimeter. WebSo, centimeters = × = inches. By cimlainfo.ru To use this calculator, simply type the value in any box at left or at right. It accepts fractional values. WebThe conversion factor from centimeters to inches is , which means that 1 centimeter is equal to inches: 1 cm = in. To convert centimeters into inches we have to multiply by the conversion factor in order to get the length amount from centimeters to inches. HOW TO MEASURE TOPS ; HEIGHT (APPROX)cm/inches. / / ; CHEST (inches). 26 ; CHEST (cm). How many feet are in a meter? 1 meter is equal to: feet; 3 feet and inches; inches. To convert directly between meters.
Twenty-six Centimeters is equivalent to two hundred sixty Millimeters. Definition of Centimeter. The centimeter (symbol: cm) is a unit of length in the metric. 25 cm to inches equals 9 inches. Also 25 cm equals 0 feet and 9 inches. 26 cm to inches equals 10 inches. Also 26 cm equals 0 feet and 10 inches. centimeters to inches converter. This page allows you to convert length values expressed in centimeters to their One inch is equal to cm. WebExample Convert 20 inches to centimeters: d(cm) = 20″ × = cm How many inches in a centimeter One centimeter is equal to inches: 1cm = 1cm / cm/in = in How many centimeters in an inch One inch is equal to centimeters: 1in = cm/in × 1in = cm How to convert 10 inches to centimeters. WebIf we want to calculate how many Inches are 26 Centimeters we have to multiply 26 by 50 and divide the product by So for 26 we have: (26 × 50) ÷ = ÷ = Inches. So finally 26 cm = in. WebAccuracy Format Print table Meters to Feet Feet to Meters inches to cm cm to inches mm to inches inches to mm Inches to Feet Feet to Inches. WebEnter the number of centimeters to convert into inches. Easy cm to in conversion. A centimeter Centimeters to Inches Conversion Table (some results rounded).
Web26 cm → L (in) Solve the above proportion to obtain the length L in inches: L (in) = 26 cm × in. L (in) = in. The final result is: 26 cm → in. We conclude that 26 centimeters is equivalent to inches: 26 centimeters = inches. Web26 Centimeters is equivalent to Inches. How to convert from Centimeters to Inches The conversion factor from Centimeters to Inches is To find out how many Centimeters in Inches, multiply by the conversion factor or use the Length converter above. WebHow many inches in 26 cm? 26 Centimeters equal to inches or there are inches in 26 cm. ←→ step 26 cm Enter Centimeter ⇵ 26 cm = inches» Show how it is converted Conversion factor: 1 inch = cm 1) inch = cm / 2) inch = 26 / 3) inch = cm to inches length conversion. Formula: multiply the value in centimeters by the conversion factor ''. So, 26 centimeters = 26 × = inches. 26 centimeters equal inches (26cm = in). Converting 26 cm to in is easy. Simply use our calculator above, or apply the formula to. Convert 26 Centimeters to Inches ; , ; , ; , ; ,
WebDo a quick conversion: 1 centimetres = inches using the online calculator for metric conversions. Check the chart for more details. WebJan 27, · Cm to inches Cm to inches; 25 cm = in: 95 cm = in: 30 cm = in: cm = in: 35 cm = in: cm = in: 40 cm = in: cm = in: 45 cm = in: cm = in: 50 cm = in: cm = in: 55 cm = in: cm = in: 60 cm = in: cm = . Web26 Centimeters = Inches rounded to 8 digits Display result as A centimeter, or centimetre, is a unit of length equal to one hundredth of a meter. There are centimeters in an inch. An inch is a unit of length equal to exactly centimeters. There are 12 inches in a foot, and 36 inches in a yard. Centimeters to Inches . To convert 26 centimeters to inches you have to divide the value in cm by 26 cm in inches: twenty-six cm are equal to 26/ = inches. Here. The answer is We assume you are converting between centimetre and inch. You can view more details on each measurement unit: cm or inches The SI base unit. 26 CM equals to IN. We can convert 26 CM to Inches by using Centimeters to Inches conversion factor. Multiply the 26 Centimeters with.
WebDo a quick conversion: 1 centimetres = inches using the online calculator for metric conversions. Check the chart for more details. Web35 rows · 26 Centimeters = Inches rounded to 8 digits Display result as A . Web26 cm → L (in) Solve the above proportion to obtain the length L in inches: L (in) = 26 cm × in. L (in) = in. The final result is: 26 cm → . WebQuick Reference for Converting Centimeters to Inches. Formula. in = cm / Quick Rough Maths. To get the Inches, divide the number of Centimeters by Centimeters (cm) in 1 Inch. There are Centimeters in 1 Inch. Inches (in) in 1 Centimetre. There are Inches in 1 Centimetre. WebWhat is Centimeter? Centimeter is equal to 1/ meter and it is denoted by cm. Formula of Converting Centimeters into Inches To convert cm into inches, just divide the centimeter value by Inches = Cm / Let's take an example of how to convert cm into inches. Let's say someone's height is cm. WebAn inch is a unit of length equal to exactly centimeters. There are 12 inches in a foot, and 36 inches in a yard. There are 12 inches in a foot, and 36 inches in a yard. Centimeters to Inches Conversions. cm to Inches Conversion Chart ; 23 cm, ", 9 1/16" ; 24 cm, ", 9 29/64" ; 25 cm, ", 9 27/32" ; 26 cm, ", 10 15/64". A centimeter (cm) is a decimal fraction of the meter, the international standard unit of length, approximately equivalent to inches. Definition of inch An. Centimeters to inches conversion table ; 10 cm, in, 3 15/16 in ; 20 cm, in, 7 7/8 in ; 30 cm, in, 11 13/16 in ; 40 cm, in, 15 3/4 in. CM to Inches converter ; centimeters (cm). inches (in) ; Swap ; 1 cm = in, 1 in = cm. |
Mystic messenger Email Guide - all correct answers (2022)
(RFA) - rika mystic messenger
RFA: Rika’s Fundraising Association is a charity organization that raises the funds for those who need people by holding parties. In this game, after Rika had passed away in the game, no parties were held. But after you joined the game, there is a chance to host the parties again arises.
Mystic messenger Email Guide Answers
in This guide we will explains all the correct answers to the Question from Every Email and from Every Guest. When you reply with the right answers, there is a chance for joining the new guests at the party. Always start with 10 to 15 guests for the Party of RFA for the best game ending.
How do these Mystic messenger Email Guide work?
To host party, the first thing that is expected from you is invite and deal with the guests that are going to join the party of your organization. Normally, an invitation will be through the virtual messenger that you can access via the app. The reason that makes it important is that Jumin, V, 707, Yoosung, and Zen look up to it, so make sure you do it very well.
Mystic Messenger Emails
With the help of our current Mystic Messenger emails guide, you can able to host the party and invite guests, this is the fun of this game app. At that party time, you are fully responsible for your behavior with the guests and invitation via a Virtual messenger. Mystic Messenger Emails are usually us to Invite guests into the Party while you are playing with another Story mode. You need 15 or above 15 guests at least to throw a party and wonderful Ending.
Mystic messenger Email Guide Notes :
• Mystic messenger Email Guide are in ABC order
• Minimum to Invite to Good Ending
• Original Story: 10
• Another Story: 16~20??
• In order to get someone to 100% commit to come to your party, you need to get three all correct emails with all three green arrows and a complete in silver/blue-green.
• If you get two completed emails in the game, there is a probable chance they will come green.
• in the game, the green arrow means the email is right.
• in the game, the Orange arrow means the email was incorrect.
• Answer the emails as they come in. Do not let them sit.
• Invite as many as possible you can. Do not worry about possibly inviting someone your love interest may not like.
• It may take few days for some guests to get back to you. I had guests confirming they would come and being marked complete, the day of the party. Be patient. and stay with the game.
• If you are trying to get the normal end for each character in this game, do not invite any guests or invite less than 9 for Original Story and possibly less than 14 in Another Story (not confirmed). Anything over that will take you to a good game end.
• You will start being able to the accept emails on the second day.
• You must be participate in chats and tell characters to have them send you an email in order to receive emails.
• When you reach the party and good end of the game, the game will unlock an extra area for you. If you press the button for the guests, you will be showing the guest list in the game. Selecting guests that attended and listening to the story and you get the award 1HG for the first time you do this. You will not be award an HG again after reading second time a guest's story once.
• Cat allergy
• Beef and seaweed soup
• I might be allergic to guests not attending the party.
• The Louvre
• I’ll look forward to your next video!
• Report them.
• 100% interest rate
• Swiss bank.
• $1.2 million all in cash.
• Arabica coffee!
• Drip brewing.
• Don’t worry!
• Lock U Up Silver Bracelet.
• Claw machines at the mall.
• Cat buffet.
• How about people putting on a show?
• Cat limousine!
• Head meow!
• Odd eye meow!
• Crystal litter.
• Of course!
• Give them food and wait.
• Social media sites.
• 3 cups, chef
• You mix as if you’re cutting, chef.
• 20 minutes
• Poke it out with a straw!
• I saw it onYoutube!
• The face of a generous looking granfa… I mean, you , the owner!
• Cheese is the way of the world!
• Yellow and black.
• The rock band Imagine Mythical Creatures
• Limited edition title change ticket
• Create a banner of the winner.
• It is Verragamo
• Flower bed of pretty boys
• Hawk Pose
• Shiny White
• Silk underwear with scarf
• I know the vanished seven treasure islands.
• Yoosung’s LOLOL exploration
• The seal is unlocked!!!
• The rise of the fire dragon slumbering within the eyes!!!
• There’s this person called Jaehee….
• Games with soda as prizes.
• It’s because you’re too smart.
• I wanted to help you.
• I want to go see a movie, but I don’t have anyone to go with. What do I do?
• I really hope you come to our party.
• A rainbow colored floppy disk model.
• Windows 8.1 3711 disks.
• Obtain a limited edition cassette tape and hide it.
• All to the bank!
• Bank passbook
• We split it up.
• Jumin’s cat
• Jalapeno’s Photobook
• Zen’s underwear
• Wow! Amazing that I get to talk to God!
• I am your daughter.
• I think you’ll walk into the party room like a normal person.
• Driver shape.
• Shoes that dry quickly.
• Leather that does not wear out.
• Lucky Kim (whispers)
• He’s an oldie, 80 years old.
• Not Grandma! Grandmother!
• Lots of people with warm hearts!
• Rice, soup, green salad, backed salmon, fried eggs
• Sell Small Issue Magazine
• Hippocratic oath.
• Request exam results.
• Comparing several hospitals
• Filial Piety
• Be healthy.
• How about Scheam?
• Why don’t you participate in a game convention?
• ctrl + c
• Compete with your son.
• Car from when you were young Pong!
• Call the police!
• Say hello!
• I think you’ll give them a kick in the butt.
• Of course lol
• Play LOLOL together.
• Focus on balance
• Blood Dragon
• Meowmeow, [note comma]
• Bae screen wedding
• Tripter Tript invitations
• Use screenshot function!
• Maybe… maybe… Diamond Pharmaceuticals…?
• It’s Diamond Pharmaceuticals.
• Diamond Pharmaceuticals!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
• Pink bandage.
• Coffin with lace all around
• Life size marble statue of Pharaoh
• Of course.
• The chance to discover a gem!
• Your face reflected on thy eyes.
• The best of all pens in the nation, BIK.
• “Sear the end with a lighter.”
• Classic is best
• Personalized framed cross-stitch.
• Buy her handcuffs.
• Give her a bouquet of jasmine.
• A film about the environment
• Cannes, Venice, Berlin
• This production of “The Red Pepper Was So Hot”
• Get the help of college student.
• [your name]
• Treatment of getting locked up in a room of mirrors.
• He takes a lot of selfies.
• Lake Na
• Yes, there is injustice going on so please come to our party and help us!
• There’s no story of a magical girl who’s not violent.
• It’s cool… Completely suits you
• Long enough to do well with my eyes closed!
• Get a camel through the eye of a needle.
• Leonardo Dicappucino
• Fancy party! Lots of parking space!!
• Of course!
• To the freezer!
• Bubblewrap on the window.
• Fried rice.
• Salmon fish sauce
• Fish – shape
• A law must be implemented that protect reporters.
• I cannot tell you that.
• Whole beans.
• 15 days.
• Of course!
• You One and Only Top Star
• It’s an extravagant and elegant party.
• I recommend you take the Olymbus X20.
• I recommend you take the heavy professional Ganon camera.
• He is very practical.
• More than enough.
• Just your normal attire.
• Because it’s cool.
• First. Leave the phone in the living rom and go to your room.
• Leave the battery only half charged.
• Popcorn brain.
• A man’s word is his bond!
• You’re girlfriend will love that!
• You’ll be useless if you give up now
• Hi, I’m Youngsoo. I got your number through Chulsoo.
• Wear a watch!
• Asking is not a challenge but a confirmation.
• Memories of my first kiss✩
• I want to eat it!
• I have to make a wish!
• Tell them to invest in stocks.
• Not losing all your money
• Chief Assistant Jaehee Kang
• Of course!
• At the party!
• Everyone has small feet!
• Call the zoo!
• Stock prices of peach drinks are going up.
• It means your grades! Two Fs!
• Whoopee~~ Beat drums~~
• Oh my dear sun~~ Lay your passion upon us~
• So much time!
• C&R Director
• Old man under the moon.
• Red wine.
• Ice wine
• Art organizations will be joining.
• Flies off to space.
• What are you interested in these days.
• Who says you’re late! You didn’t even try.
• Convince your parents
V and Ray Route (Another Story) emails
This emails below are only available in Another Story – not Casual Story or Deep Story, and pertain to V and Ray Route. We’re pretty sure we’ve found all possible guests and correct answers now, although the text may not exactly match the answers. Let us know if you spot an error!
• Behind the cheerleaders
• Waiting List
• Idol Concert
• Check names and invitations
• Play with Elizabeth the 3rd
• Next to Mr Han
• experienced electric shock
• don’t miss this opportunity
• what’s wrong with that?
• 1200-1400 won
• push the stop button
• Hello, World!
• D Ritchie
• trust the programmer
• some sort of machine
• this is getting embarrassing
• spy training school
• never played main hero with gorgeous clothes
• player mentality +10
• costume contest
• Indian curry
• vindaloo curry
• I think it’s cool! Of course you can!
• Smug, the light and hope of adolescent conceit!
• Dark dragon is the symbol of strength
• [your name]
• Yoosung Kim
• Not Yet Finalised
• ireless earphones
• crowdfunding website
• Long distance glasses
• Blue-light blocking glasses
• Heart-shaped glasses
• chicken breast
• Afro hairstyle
• Afro hairstyle
• The agency
• Silence is gold
• Absence of licence for caretaking
• rice flavour
• scoop in a flower shape
• _ _ _ _ / . _ _ _ _ / . _ _ _ _ / . _ _ _ _
• _ . / _ _ _
• . / . _ . . / . . / _ _ . . / . _ / _ . . . / . / _ / . . . . / _ / . . . . / . / . . . _ _ / . _ . / _ . .
• shoot a bat-shaped light into the sky
• because you’re nocturnal
• Ineligibility of marriage due to job
• Cochlea Insurance
• researcher of residential environments
• behind [your] back
• Sistine Chapel
• Last Judgment
• Cake made of rice cakes
• And Undone I Find the Mass of the Plate
• We also have rice bowls
• rich enough to full the swimming pool with money!
• have you ever met a hacker?
• No I won’t!
• Edvard Munch
• Don’t let go of your blanket!
• We have a thin blanket for you.
• Slick, dope, lit music
• Nailart to train fingertips
• Racing Rhythm Game
• Voting for elections!
• National Sovereignty
• 3 mins
• pizza cheese
• no you don’t have to stick to the suggested time
• Because you’re too lazy to explain
• Three moles behind the ear of Rudolph
• It’s the colour you like!
• be family
• Saint Dogs
• the flesh of an astronaut
• Yes I want to eat you!
• stir-fried noodles
• emphasise the health
• carry blood packs
• house in the city
• pair of modern and chic suits
• 24 and 2
• recover lost love
• Strike first!
• Night owl
• Read for self-improvement
• Health management
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1 totsakan enlarged the size of a photo to a height of 18 in. Speed distance or cost amount problems available both as pdf and html files.
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What is the new width if it was originally 2 in tall and 1 in wide. A variety of pdf exercises like finding proportions using a pair of ratios determining proportions in function tables creating a proportion with a given set of numbers and solving word problems are included here. 2 a frame is 9 in wide and 6 in tall.
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Click any of the links below to download your worksheet as an easy to print pdf file. Create proportion worksheets to solve proportions or word problems e g. The perimeter of a rectangle is equal to 280 meters.
Write the ratio of the number of blue marbles to the total number of marbles in terms of r b and w. Displaying top 8 worksheets found for grade 7 ratio and proportion. This series of printable proportion worksheets are prepared specifically for learners of grade 6 grade 7 and grade 8.
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Posts by help
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Ok, first you integrate the problem so it equals 19x + x^2 - 5x^4 Next I'll write this out then show you. You then plug in x = 3 to this equation and then plug in x = 1 Subtract the equation of x = 1 from the equation x = 3. (19(3) - (3)^2 - 5(3)^4) - (19(1) - (1)^2 - 5(1...
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f(x)=xlnx Find the minimum value
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One and Two Dimensional Isoparametric Elements and Gauss Integration: Gauss Integration
The Gauss integration scheme is a very efficient method to perform numerical integration over regular domains. In fact, if the function to be integrated is a polynomial of an appropriate degree, then the Gauss integration scheme produces exact results. The Gauss integration scheme has been implemented in almost every finite element analysis software due to its simplicity and computational efficiency. This section outlines the basic principles behind the Gauss integration scheme, along with its application in the finite element analysis method. In particular, this section will discuss the development of the reduced-integration elements and the benefits and drawbacks of utilizing them in finite element analysis models.
The isoparametric elements developed above imply that all the integrated functions are defined such that . In general, numerical integration of such a function has the form:
For the Gauss integration method, is called an integration point and is called the associated weight. Now, if is always affine, i.e., then:
So, for functions that are very close to being affine, a numerical integration scheme with 1 integration point that is with an associated weight of 2 can be employed.
Gauss Integration Over One Dimensional Domians
Let . Then, the definite integral of f over the domain can be approximated by the following formula:,/p>
where and is a weight factor associated with each . are termed the integration points. The number of integration points required to obtain sufficient accuracy depends on the shape or form of the function . If is a linear function, then one integration point is enough to exactly integrate . If is a cubic polynomial function, then two integration points are sufficient to obtain the exact solution. Figure 5 shows a graphical representation of the Gauss numerical integration scheme for integrating polynomials up to the fifth degree. Following is a table of the number of integration points required to obtain the exact solution for a polynomial function f with different polynomial degrees:
Verify that the exact integral of a general cubic polynomial on the interval can be obtained by the two integration points and with the two weight factors .
Let . The exact integral is equal to:
On the other hand, Gauss integration with two integration points is equal to:
Equating the Gauss integration with the exact solution yields the following four equations:
These four equations are obtained by noticing that the exact solution and the Gauss integration solution should be equal given any values for the coefficients , , , and . The solution to the above four equations yields , , . The Mathematica code to solve the four equations is:
View Mathematica Code
Note that this example can be extended to show that Gauss integration in one dimension with points can exactly integrate a polynomial of order .
Gauss Integration Over Two Dimensional Domains
Let . Then, the integral of over the domain can be approximated by the following formula:
where is an integration point while and are weight factors associated with and respectively. It can be easily seen that this is a direct consequence of the one dimensional Gauss integration formula as follows:
Gauss Integration for Two Dimensional Quadrilateral Isoparametric Elements
As described for the two dimensional linear quadrilateral element, the strains of the 4 node elements are linear, and thus, the entries in the matrix are quadratic functions of position. Thus, a numerical integration scheme is required to produce accurate integrals for the isoparametric 4-node elements (Figure 6). On the other hand, the strains in the two dimensional linear quadrilateral are quadratic functions of position, and thus, the entries in the matrix are fourth-order polynomials of position. Thus, a numerical integration scheme is required to produce accurate integrals for the 8 node isoparametric element (Figure 6). It should be noted that this can extended the three dimensional case in a straightforward manner. The 8 node brick element has 8 integration points, while the 20 node brick element has 27 integration points.
Reduced Gauss Integration (Under Integration) for Two Dimensional Quadilateral Isoparametric Elements
Elements that are integrated as per the previous section are termed “full integration” elements. For example, to integrate the stiffness matrix of the 4-node plane isoparametric quadrilateral element, each entry will be evaluated at the four locations () of the integration points in the isoparametric element. Thus, the number of computations (excluding the addition) for the integration of the element are computations. The number of computations for the stiffness matrix of the 8-node plane isoparametric quadrilateral element is 2,304 computations since it uses a Gauss integration points scheme! To save computational resources, a “reduced integration” option for each of those elements is available in finite element analysis software. The reduced integration version of the 4-node plane isoparametric quadrilateral element uses only one integration point in the center of the element (Figure 7), and the number of computations is reduced from 256 to 64 computations. However, this is accompanied by a reduction in accuracy. Figure 7 shows two nonlinear functions to be integrated using one integration point vs. two integration points. In the first case, the function has a small variation along the domain and the integration seems to be accurate. However, in the second case, the function has a large variation and is equal to zero in the centre of the domain. The reduced integration will predict zero stiffness!
For the 8-node plane isoparametric quadrilateral, a Gauss integration points scheme is used instead of the Gauss integration points scheme, which reduces the number of computations from 2,304 to 1,024 computations (Figure 8). The benefit of using reduced integration is not only in saving computational resources but also in balancing out the over-stiffness introduced by assuming a certain deformation field within an element. In fact, reduced integration can lead to better performance in most cases. However, while reduced integration can dramatically reduce the computational time, it can lead to a structure that is too flexible. In fact, the displacements are over predicted when a structure is discretized using reduced integration elements. The reason, as will be shown in the following section, is because some information is lost during sampling of the stiffness at the fewer integration points of the reduced integration elements.
Spurious Modes Associated with Reduced Integration Elements
Spurious modes, or sometimes called zero energy modes, are combinations of displacements that in general produce strains; however, such strains cannot be detected at the integration or sampling points. So, the internal energy calculated at the integration points is equal to zero for such mode. In other words, a very small amount of load would be enough to produce a very large displacement. If a spurious mode exists for the whole structure, it becomes unstable. Usually spurious modes exist in under-integrated element (reduced integration). This is because the reduced number of integration points cannot detect a spurious strain mode. The two very common spurious modes are those for the bilinear quadrilateral (4-node plane element) and the quadratic displacement quadrilateral (8-node plane element). In general, any bending mode is a spurious mode in a 4-node reduced-integration linear element. The strain modes shown cannot be detected at the integration point; the strain at the center of the element (at the location of the single integration point) is equal to zero (Figure 9). On the other hand, the spurious mode of an 8 node reduced integration quadratic element is called “the hour glass” mode (Figure 10) and is produced with the following configuration of nodal displacements:
The spurious modes will be illustrated using the following problem.
Find the internal energy associated with the shown displacement for the 4-node plane stress quadrilateral element shown below, assuming a bilinear displacement interpolation function.
Use exact integration, full integration ( integration points), and reduced integration (1 integration point).
The shape functions of the shown element are:
The displacement functions and in the directions and , respectively, are:
Therefore, the strains have the form:
The stress strain relationship for plane stress is given by:
The stiffness matrix can be evaluated using the following integral where is the element thickness:
Notice that the system has only one degree of freedom (the unknown displacement variable ), and therefore, the stiffness matrix has the dimensions .
Using Exact Integration:
The stiffness matrix evaluated using exact integration is:
The strain energy associated with this displacement mode is equal to:
Using Full Integration:
The stiffness matrix evaluated using full integration is:
where and which gives the same results as the exact integration.
Using Reduced Integration:
The stiffness matrix evaluated using reduced integration is:
where and . The strain energy associated with this displacement mode is equal to zero. Thus, the reduced integration scheme predicts that the displacement field shown does not require any external forces applied to the element. Thus, this element is unstable if loads produce the displacement mode shown, i.e., bending! The following is the Mathematica code used for the above calculations:
View Mathematica Code
(*Using Exact Integration*)
(*Using Full Integration*)
(*Using reduced Integration*)
Gauss Integration for Two Dimensional Triangle Isoparametric Elements
The strains of the 3 node triangle element are constant, and thus, the entries in the matrix B^TCB are constant. Thus, a 1 point numerical integration scheme is required to produce accurate integrals for the isoparametric 3-node elements (Figure 11). For the 6 node triangle element, a 3 points numerical integration scheme is required to produce accurate integrals (Figure 11). It should be noted that the shown locations for the integration points and their number can be extended to the three-dimensional case in a straightforward manner.
The use of isoparametric elements and numerical integration dramatically increases the robustness of the finite element analysis method. The techniques described in the previous sections allow meshing of irregular domains with compatible elements, i.e., elements that ensure the continuity of the displacement field across boundaries (How?). However, to ensure the continuity of the displacement field, the elements of the mesh should be of the same type; otherwise, the user has to apply mesh constraints to ensure the compatibility of the displacement between elements. For example, a 4-node bilinear displacement element when connected to an 8-node quadratic displacement element will have incompatible displacements at the side of the connection. This can be overcome by constraining the mid-side node of the quadratic element to always lie on the straight side of the bilinear element. Otherwise, the results at the location of discontinuous displacement fields would be erroneous.
The stiffness matrix and the nodal loads of the structure can be calculated using the Gauss numerical integration scheme, which speeds up the computational time required. However, the price of using the Gauss numerical integration is the loss of accuracy. Gauss numerical integration produces accurate results only when the integrals are exact polynomials. The integrals are exact polynomials only when the elements of the mesh are exactly aligned with the isoparametric mapping into the coordinates of and . When the shape of the elements in the mesh deviates from the shape of the element in the coordinate system of and , the functions to be integrated are no longer exact polynomials but rather radicals; thus, the numerical integration scheme is no longer accurate. Most finite element analysis software give warnings when elements are highly distorted since this will lead to inaccurate computations of the stiffness matrix during the numerical integration process. In addition, for nonlinear problems when the stiffness changes with deformation, it is possible that the element will reach a highly distorted shape that the numerical integration scheme is no longer valid. |
Lesson Contents01. Learning Objectives02. Flow Regimes in Unconventional Reservoirs03. How Does Exposed Surface Area Influence Production Rate04. How Does Exposed Surface Area Influence Production Rate05. Compartmentalization - Multi-stage Fracturing06. Multi-Stage Fractured Horizontal vs Vertical Fractured Well -- Observed Performance07. Multi-Stage Fractured Horizontal Observed Performance08. Simulated Reservoir Volume (SRV) Concept09. Illustration of SRV Concept Using Simulated Production10. Surface Area and SRV11. Surface Area and SRV (cont'd.)12. Impact of Increased Surface Area on Performance and Reserves13. Impact of Increased SRV on Performance and Reserves14. Concept of Production-Measured Permeability (Effective Permeability)15. Effective Permeability Around a Horizontal Well16. Effective Permeability Around a Horizontal Well (cont'd.)17. Effective Permeability Around a Horizontal Well (cont'd.)18. Effective Permeability Around a Hydraulic Fracture
01. Learning ObjectivesThe learning objectives for this session are to talk about some of the fundamental production drivers in unconventionals. Namely surface area and stimulated reservoir volume (SRV). We're gonna start by reviewing the flow regimes that occur in conventional versus unconventionals. And then we're going to get into a discussion specifically about how surface area can compensate for low permeability. And it has to do this in unconventionals, that's the only way that these reservoirs can be economic. We're going to talk about how completion compartmentalization can efficiently allow us to create an enormous effective surface area in a horizontal well, something that simply can't be done in a vertical well. And so this is part of the reason why almost all wells that are drilled in shales and tight reservoir rocks now are multi-stage horizontal wells. We're going to explain the concept of the stimulated reservoir volume which is obviously a huge driver of well performance in unconventionals. And we're going to describe how surface area and SRV influence well performance and EUR. This is a critical component. Once we have a fundamental understanding of that, we'll be able to go forward and do some analysis which will be the next lesson. And then finally we're gonna talk about production measured or effective permeability as opposed to a permeability that you would measure in a laboratory. So there's some very clear differences between these two, and we want to talk about what those are.
02. Flow Regimes in Unconventional ReservoirsSo let's review the flow regimes first of all. As everybody should be fully aware now, when you're talking about conventional vertical wells even wells that are fractured, because usually the fractures will be small in comparison to the external reservoir boundaries, you're typically dealing with radial flow. So this is where we have a flow path that's convergent on the well, the transport medium is the pore space; the connected pore space. When we get into unconventionals we're almost always going to see linear flow. Not 100% of the time but it's going to be a dominant flow regime. We're going to see linear flow from the formation into the fracture. That transport medium is going to be the fractures, we get very little flow occurring in the matrix because we're talking about extremely tight rock, and there's a very high contrast between matrix and fracture connectivity. So we commonly call these dual porosity models but linear flow is going to be an overriding characteristic of these systems.
03. How Does Exposed Surface Area Influence Production RateOkay what I'd like to do is a little bit of a thought experiment and we're going to show how exposed surface area can influence production rate. So on the left hand side we have our standard radial system, conventional system. And this equation here which you've seen before is the steady state inflow equation for a radial system. So this is the steady state production rate, stabilized production rate that we would expect given a pressure difference of average reservoir pressure down to flowing pressure and over here on the right hand side we have a linear flow system. So this could be thought of as a horizontal well or a vertical fracture or a multi-frac'd horizontal well. And the equation that describes the flow into that bounded linear system looks like this. And so what I would like to do in the thought experiment is to say, the radial system as a conventional reservoir is going to be 1 millidarcy (mD) of permeability and the linear system is going to be 1000 times lower than that. And we want to ensure that these things have the same flow rate. So what do we need to do in terms of extending the available surface area so that these two systems will be equivalent and have the same flow rate? And that's a good way of thinking about how in unconventionals we can design a well completion that will be economic. It will allow us to produce the same rate that an equivalent 1 mD rock in a conventional reservoir would allow us to produce.
04. How Does Exposed Surface Area Influence Production RateSo to equate these two things we're just looking at these two components of the equation for flow rate. And because the linear flow system is going to be a 1,000 times lower permeability we're going to have that 1,000 coming in there. And for a drainage area 160 acres, that's a quarter section, we're going to find out what that looks like for a radial system and what that looks like for a linear system. So for a radial system we have a radius of 1489 ft. And for a linear system we have a total square footage of 1.7 and change million square feet. So we're making equivalent drainage sizes in terms of the area. But of course the shape is going to be entirely different between these two. So the first thing that we've figured out by equating these two terms is that the ratio of X to Y, in other words the ratio of the length of the linear system to the width of the linear system is almost 84 times. So that's going to mean that our X distance is about 12,000 ft in our Y distance is about 144 ft.
05. Compartmentalization - Multi-stage FracturingSo let's take a look at what that looks like as a linear system. If we were trying to design this as a single vertical fracture we would have a 12,000 ft half-length and so more than 24,000 ft in total distance of this fracture. That is highly impractical. Even if you could design something like that, it would be impractical to develop your field that way. And most experts would agree that you probably couldn't make an effective fracture/producing fracture that large or that long. So what we're gonna do is we're going to break that up into components and use a horizontal well that's less length and compartmentalize it. So create the same total fracture area using for example a 10,000 ft lateral with 18 fractures and xfs of 660 ft, that's one way to do it. There could be other solutions that could accomplish the same thing. This will give us the equivalent stabilized rate. So this explains in a nutshell how compartmentalization or multi-stage fracturing has allowed us to take somewhat of an impractical field development scenario and turn it into something that's much more practical and doable. And so this is why the industry has been successful in exploiting these resources is because we're creating these compartmentalized solutions.
06. Multi-Stage Fractured Horizontal vs Vertical Fractured Well -- Observed PerformanceHere's an example of real well performance and so what you're looking at here is a multi staged horizontal well versus a single vertical fractured well. And their behaviors, on the log log plot shown over here, couldn't be more different. So if we look at first of all the multi-stage horizontal well what we see is after about18 days we transition from linear flow; b = 2 into something that looks like boundary dominated flow very quickly. So that's our horizontal well. When we look at the vertical well with one single fracture, we have 200 months which is more than just about 17 years of production where we don't see any evidence of boundaries. So if you think about this in terms of efficiency, the horizontal well is draining its area much more efficiently than the vertical well is. You're getting those resources out a lot faster. So these are real examples of horizontal versus vertical wells. It just shows the value of compartmentalization and how it's able to accelerate that resource recovery and create production scenarios that are much more economic.
07. Multi-Stage Fractured Horizontal Observed PerformanceHere's another example to further illustrate this point. What you're looking at here are normalized rate responses on a log-log plot for 50 Eagleford gas condensate wells. And each one of these has about 2-3 years of production. And so the interesting thing to consider here is that these all start with a half slope so you show your early linear flow behavior and they all transition into something that's very close to boundary dominated flow; a slope of 1. It looks like boundary dominating flow, but the interesting thing about this is you might think that that boundary dominating flow would be the extents of the reservoir. In fact it is not, in almost all of these cases, it's an area or drainage volume that's significantly smaller than the available drainage volume due to well spacing. This is a very important observation because what it tells us is that we're not draining the inter well spacing in a lot of these wells. Especially in these 50 that we're looking at here. These drainage volumes are very small and so you're really only draining something that's potentially very close to that wellbore.
08. Simulated Reservoir Volume (SRV) ConceptSo we come up with this idea of the stimulated reservoir volume concept and that's largely come about as a result of observation in the field. There are many of many such examples of these wells that drain limited areas. So the stimulated reservoir volume refers to the hydrocarbon pore volume that is in immediate contact with the effective fractures. And that's typically going to be applicable to a shale oil or gas. A very tight rock, nanodarcy permeability type rock. It's a convenient concept for RTA because it simplifies the production analysis problem significantly. So this shows you some examples. This is what an SRV might look like in reality, the way we typically are going to model them is as a bounded linear flow system such as what you see on the right hand side. So here's an example using a simulation we can illustrate the concept of SRV very clearly using the simulation. We start with a nanodarcy rock such as a shale. This one happens to be 50 nD.
09. Illustration of SRV Concept Using Simulated ProductionAnd what you're seeing here is the simulation of production response from two distinct reservoirs. One in which the completion exists in a semi-infinite or bounded reservoir but it's but it's much larger than that than the stimulated reservoir volume. And the other one where the reservoir boundaries coincide precisely with the stimulated reservoir volume. And what this teaches us is that looking at the timescale here, this is more than 5 years, they're almost identical. There's very little difference between the case where the completion exists in a larger reservoir and where it exists in a limited reservoir. So the contribution of oil and gas outside the immediate fractured region is negligible in many of these cases. And that's why the stimulated reservoir volume concept works.
10. Surface Area and SRVWe talked about stimulated reservoir volume. I want to tie together now surface area and stimulated reservoir volume. So what you're looking at here, you can consider this sort of a general rock volume that has some fractures in it. And the stimulated reservoir volume can be thought of as the bulk volume multiplied by the hydrocarbon filled porosity at standard conditions. The surface area would be the total connected fracture area within that SRV. Those in many cases can be completely independent parameters. If we're looking at just geology in general, you could have stimulated volume that has sparse fractures or he could have stimulated volume that's got much much more dense fractures, and the orientation of those fractures varies as well. So generally speaking there's no obvious relationship between the stimulated reservoir volume and the surface area.
11. Surface Area and SRV (cont'd.)However when we talk about a compartmentalized completion where we can control where we're placing the fractures, we can infer usually with pretty good reliability a relationship between the stimulated reservoir volume and the total connected surface area. So our assumption that we're going to have to make to do this to make this work is that the fractures are planar and parallel and they have uniform size and distribution. So we're talking about a, I sometimes call this a shoe box model, it's shaped like a box it has a length of L (horizontal well length) a height of H and then a stimulated reservoir width that's gonna be double the fracture half-length (xf). Because we're assuming that those fractures are penetrating perpendicularly from the horizontal well into the reservoir.
12. Impact of Increased Surface Area on Performance and ReservesSo what I what I'd like to show here is the impact of surface area on well performance and reserves. So what you're looking at here is a plot of production rate versus cumulative production. And we have the same reservoir rock, this is just simulated data, but we're gonna look at this for 12 stages, 24 stages, 48 stages and 96 stages. And if you're paying attention the stage count is doubling each time, but the stimulated reservoir volume is staying constant. So what we're showing here is what happens when you keep SRV constant but you increase your total connected surface area. In other words you're increasing your compartmentalization, you're increasing the amount of fracturing that's occurring within that stimulated reservoir volume. What happens is very clear when you look at it, your overall reserve value is not going to change appreciably. All of these rate versus cums are converging, and this is only a 5 year forecast, are converging at the same number at about 130,000 barrels. So the addition of more surface area on its own doesn't add reserves, all it does is it adds stabilized productivity. So it can increase your production rate but it doesn't really affect your reserve recovery.
13. Impact of Increased SRV on Performance and ReservesIf we look at this in a different way, and now what we want to do is keep the surface area the same but increase the SRV, the exact opposite thing happens. So in this case what we're doing to increase the SRV and keep the surface area the same is we're varying the size of each treatment. So we have very short fracs with 96 stages that has the same surface area as half the number of stages but double the frac length and so on and so on. So in this case what we see is the 12 stage scenario actually is going to have the best reserve recovery because it's got the largest SRV. So we've done the exact opposite here, we've preserved total surface area, but we've made the SRV double in size each time. Now that's one way to make SRV get larger. The other way to make SRV increase in size is of course to extend the horizontal well link that would be the other way to do this. So this is just one possible scenario. I could have the same treatment size and same number of stages but make the horizontal well longer, that same result would be expected under those conditions. So if you have the same number of stages the same treatment but extend your horizontal well length, what should happen is something very similar to this. So what do we see here. Well we see because the surface area is held constant, total production rate of all of these scenarios is identical at the beginning and the EURs of each case are going to change. So SRV is directly correlated with EUR but does not impact production rate unless the surface area is also increased. So if surface area and SRV are increased then you're going to have an impact on production rate and reserves and recovery. So the way these two components impact performance now is very clear.Nobakht M., Mattar L., Moghadam S. and Anderson D.M. 2010. Simplified Yet Rigorous Forecasting of Tight/Shale Gas Production in Linear Flow. SPE 133615 presented at the SPE Western Regional Meeting, Anaheim, California, 27-29 May.
14. Concept of Production-Measured Permeability (Effective Permeability)What I want to do now is describe how we can relate surface area in a system to what we would call a production measured or effective permeability. This is a very useful parameter to use because a lot of times we don't know where all those fractures are. So here's an example where we have a connected fracture network. We might not know what that total surface area is. So what we're going to do is we're going to come up with an effective permeability. And to do that we define what's called a nominal area. And the nominal area is based on some known geometry of the system. For example the horizontal well length multiplied by h, that would be a nominal area. And the effective permeability is always going to be larger than the matrix permeability, assuming that your fracture system has somehow enhanced the productivity of the system. And it's always going to be less than the permeability fractures, so it's going to be somewhere in between. But it's a combination of fractures and matrix acting together. We can define the effective permeability using this relationship so the A√k comes out of the √t specialized analysis. If we normalize it with that nominal area, we can very quickly and easily calculate that effective permeability.
15. Effective Permeability Around a Horizontal WellSo let's do an example. An effective permeability around a horizontal well with known fractures. And we're gonna assume to make it easy that the fracture permeability is infinite. So if we have 12 fractures each 500 ft and half-length and a matrix permeability of 0.01 mD, and a 5,000 ft horizontal well length, we're gonna come up with an equivalent or an effective permeability. And we can do that using this definition that we looked at before. So as long as I know the number of stages, the fracture half-length and the matrix permeability I can come up with that effective permeability. In this case it's 0.0058 mD. That's roughly six times the matrix permeability.
16. Effective Permeability Around a Horizontal Well(cont'd.)So what we end up seeing here is when you have your surface area, and let's say you double your surface area, that results in a 2 to the power 2 increase in effective permeability. So because permeability always occurs under the square root sign, your linear increase in area results in a geometric increase in permeability based on that power sign there. So that's something to keep in mind. Area also impacts directly the production rate. A doubling in area will result in two times production rate. In this case I require four times the effective permeability to get that doubling in production rate. So the way permeability varies is different than the way surface area varies.
17. Effective Permeability Around a Horizontal Well(cont'd.)Here's the here's an example of an equivalent multi-frac system and effective permeability. So we have a multi-frac system with 12 fractures and we have a known matrix permeability here. We substitute in here the effective permeability that we calculated, and we have absolutely identical production flow rates. So effective permeability is a good functional model that allows us to create a model without having to make assumptions about how much surface area there is connected, we can use this effective permeability concept.
18. Effective Permeability Around a Hydraulic FractureThe other place where effective permeability becomes very useful is we can take that whole horizontal well and turn it on its side now. Instead of using as our nominal area the horizontal well length we're going to describe the region around each one of the fracture stages as an effective permeability. And that's something that we do very commonly in RTA of unconventionals. In fact we probably use this effective permeability even more so than the one describing it around a horizontal well. For the pure fact that we just simply do not know, when we're looking at this sort of geometry, we have no idea where these fractures are occurring. There's no way to measure there's no way to know how much surface area is being created proximal to the hydraulic fracturing when we're creating one of these stages. So the functional model is to basically represent that with an effective permeability. I mean do it exactly the same way, the only difference here is that the nominal area now is based on the area of that frac length not based on the nominal area of the entire horizontal well. And that's it.
What is in here?
This lesson describes the influence of permeability and surface area in a fractured reservoir. Included are the concepts of compartmentalization, SRV and production-measured (effective) permeability. Also included are the impact of surface area and SRV on well performance and EUR.
Why should you care?
Permeability, surface area and SRV are key drivers of well performance. Understanding how each of these contributes to the production equation is critical to making good decisions around completion and field development optimization. |
- Open Access
On a method of construction of new means with applications
© Raïssouli and Sándor; licensee Springer 2013
- Received: 25 August 2012
- Accepted: 17 January 2013
- Published: 5 March 2013
In the present paper, we would like to introduce a simple transformation for bivariate means from which we derive a lot of new means. Relationships between the standard means are also obtained. A simple link between the Stolarsky mean and the Gini mean is given. As applications, this transformation allows us to extend some means from two to three or more arguments.
- decomposable means
- Stolarsky mean
- Gini mean
In the recent past, the theory of means has been the subject intensive research. It has proved to be a useful tool for theoretical viewpoint as well as for practical purposes. For the definition of a mean, various statements, more or less different, can be found in the literature; see and the references therein. Throughout this paper, we adopt the following definition.
and are known as the arithmetic, geometric, harmonic, logarithmic, identric, weighted geometric, contraharmonic and (first) Seiffert means, respectively.
A mean m is symmetric if and homogeneous if for all . The above means are all symmetric and homogeneous. However, the mean is (homogeneous) not symmetric, while is (symmetric) not homogeneous. The mean is neither symmetric nor homogeneous.
A mean m is called monotone if is increasing in a and in b, that is, if (resp. ), then (resp. ). There are many means which are not monotone. For example, it is easy to see that the means A, G, H, L are monotone but C is not. However, extending the above definitions of reflexivity and monotonicity from a mean to a general binary map, the following result is of interest [2, 3].
Proposition 1.1 Let m be a monotone and reflexive map. Then m is a mean.
For a given mean m, we set , and it is easy to see that is also a mean, called the dual mean of m. If m is homogeneous, then so is with . If m is symmetric and homogeneous, then so is , and in this case, we have . Every mean m satisfies , and if and are two means such that , then . One can check that and . Further, the arithmetic and harmonic means are mutually dual (i.e., , ) and the geometric mean is self-dual (i.e., ). The dual of the logarithmic and identric means has been studied by the second author in .
Remark 1.1 Let be a monotone continuous function and denote by its inverse function. An extension of the dual mean can be given by . It is easy to verify that is a mean. For , we obtain the classical dual. If we choose with , then we get the following generalized dual . If m is symmetric and homogeneous, then .
Let m be a homogeneous mean. Writing , we then associate to m a unique positive function f defined by for all . The function f will be called the associated function to the mean m, or we simply say that f corresponds to the mean m. It follows that f corresponds to a homogeneous mean if and only if . Clearly, , and if, moreover, m is symmetric, then for every . It is obvious that a mean m is monotone if and only if its associated function is increasing. For example, the contraharmonic mean C is not monotone because its associated function satisfies , and it is easy to see that f is not increasing for all , but only for .
Now, let us observe the next question : under which sufficient condition a given function f is the associated function of a certain mean? The following result gives an answer to this situation; see , Remark 12.
Proposition 1.2 Let be a function such that for every . Then defines a (homogeneous) mean. If, moreover, f is increasing with for each , then m is monotone symmetric.
Proposition 1.3 Let m be a symmetric and homogeneous mean having a strictly increasing associated function f. Then , the associated function to the dual mean , is strictly increasing, too.
By virtue of the relation (and so ), the result of the above proposition is in fact an equivalence. It follows that the associated function of the dual will be not always increasing since that of C is not.
Summarizing the above, we may state the following.
which we call the t-transformation of m.
We explicitly notice that the convergence of the infinite product in (3) is shown by the double inequality (2).
The elementary properties of the mean-transformation are summarized in the next result.
is a mean.
- (ii)If m is homogeneous (resp. symmetric, monotone), then so is and the associated function to is given by
for each , where denotes the generalized dual mean (see Remark 1.1).
- (ii)The symmetry of from that of m is obvious, while the monotonicity of follows from the fact that m is monotone and is defined as (infinite) product of positive terms. Now, assume that m is homogeneous. By definition, we have for all
The homogeneity of follows since .
- (iii)By definition, we have successively
and (v) are not difficult. Details are omitted for the reader as a simple exercise. □
We now present the following examples. In all these examples, the sequence is as in the above.
Reduction of the three chains of inequalities (4), (5) and (6) in one chain does not appear to be obvious. However, for the particular case , the above three chains can be reduced into one chain; see Corollary 4.1 in Section 4 below.
The situation of this example will be developed below.
In what follows, we will explore this latter situation in more detail. For the sake of simplicity, we restrict ourselves to the case . The general case can be stated in a similar manner and we leave it to the reader. Precisely, we put the following.
Clearly, is a mean for all . The next example may be stated.
where we put for every .
we deduce that every p-increasing (resp. p-decreasing) sequence is p-convergent. This together with inequalities (7) implies that the mean-sequences , , and are decreasingly p-convergent. Then, what are their limits? The answer to this latter question will be presented in the next section after stating some needed results.
In what follows, we will see that the mean satisfies good properties, the first of which is announced as well.
- (i)For all and , we have(9)
- (ii)Assume that m is homogeneous and let f and be the associated functions of m and , respectively. Then, for every , one has(10)
In particular (for ), we obtain .
Follows from the fact that when combined with (i). The proof is completed. □
From the above proposition, we can derive some interesting results. The first result concerns an answer to the question that has been put in the above section.
Corollary 3.1 Let m be a symmetric and homogeneous mean such that is p-convergent. Then its limit is , the geometric mean. In particular, , , and are decreasingly p-convergent to the same limit G.
It follows that , which is the associated function of G, in this way proving the first part of the proposition. For the second part, as already pointed before, the sequences , , and are p-decreasing. It follows that they p-converge and by the first part they have G as a common limit. The proof of the proposition is complete. □
Corollary 3.2 Let and be two homogeneous means such that for a certain . Then .
We deduce that for each and so , which completes the proof. □
Now, let us observe the next question: Does (9) (resp (10)) characterize for a given mean (resp. homogeneous mean) m? For the sake of simplicity, we assume that m is homogeneous and we will prove the following theorem.
Then is the associated function of defined by (8).
which, following Proposition 3.1(ii), is the associated function of , in this way proving the desired result. □
We now present some examples illustrating the above. In all these examples, r and α are such that , .
Example 3.1 Let be the associated function of G. We have , that is, for all , which has been already pointed before.
which is the associated function of . Otherwise, . In particular, with (and so ), we obtain . By Proposition 2.1, we deduce .
which is the associated function of , that is, . In particular (if , ), we find and so .
To find out if this latter function corresponds to a certain homogeneous mean for all is left to the reader. For the particular case , , the answer to this question is obviously positive since is the associated function of S. We then have and so .
The fact that this latter function is the associated function of a certain homogeneous mean does not appear to be obvious. See the section below for a general point of view.
For the sake of convenience, for concrete examples, we may introduce the following notion.
Definition 4.1 Let m be a mean such that there exists a mean and some satisfying . Then we say that m is -decomposable. In the case , we simply say m is -decomposable.
If m is -decomposable, then for all , the generalized dual mean is -decomposable.
If m is -decomposable and is -decomposable, then for all , is -decomposable.
It is an equivalent version of Proposition 2.1(iv). □
Now, we will illustrate the above notions and results by some examples.
Example 4.1 We have already seen that for each . We say that G is self-decomposable. Also, we can see that and for every .
Example 4.2 The relationship (1), written in a brief form , says that L is A-decomposable. By Proposition 4.1(i), we deduce that is H-decomposable. We have also seen , that is, I is S-decomposable and so is -decomposable. We leave it to the reader to see that A is C-decomposable and so H is -decomposable. Other examples, in a more general point of view, will be stated below.
Example 4.3 In this example, we are interested in the link between two double-power means, namely the Stolarsky and Gini means. By virtue of this interest, we state its content as an explicit result from which we will derive some interesting consequences.
holds for all real numbers p and q.
This latter function is that associated to , that is, . The proof is completed. □
Since , Proposition 2.1(iii) yields . The desired inequalities follow by combining the two above chains of inequalities.
in this way proving the desired aim. □
Theorem 4.1 contains more new applications: some extensions for can imply analogous ones for . See Section 5 below for more details concerning this latter situation.
We leave to the reader the routine task of formulating further examples in the aim to obtain some links between other special means.
Now, a question arises naturally from the above: Is it true that every mean is -decomposable? In other words, let m be a given mean, do a mean and a real number such that exist? The answer to this latter question appears to be interesting. In fact, for reason of simplicity, assume that m is homogeneous and we search for a homogeneous mean such that . According to Proposition 3.1, it is equivalent to have for all , where f denotes the associated function of the given mean m and g will be that of the unknown mean . That is to say, the function is the associated function to a certain mean for some . Combining this with Proposition 1.2, we can state the next result, which gives an answer to the above question.
There exists a homogeneous mean such that m is -decomposable, that is, .
The function is the associated function of a certain homogeneous mean.
- (iii)The following inequalities:
hold for every , with reversed inequalities for each .
then the homogeneous mean is symmetric and monotone.
The following corollary is immediate from the above proposition when combined with Proposition 1.2.
m is -decomposable.
g is the associated function of a certain mean.
- (iii)The double inequality
holds for all , with reversed inequalities for each .
then is a symmetric and monotone mean.
Corollary 4.3 Let m be a (symmetric) homogeneous monotone mean and let be a given real number. Then there exists a homogeneous mean such that m is -decomposable. In particular, every (symmetric) homogeneous monotone mean m is -decomposable for some homogeneous mean .
for every , with reversed inequalities if . By virtue of the above proposition, we can conclude the first part of the announced result. Taking , , we obtain the second part and thus complete the proof. □
In the above corollary, we explicitly notice that the mean m should be monotone in order to ensure that m is -decomposable for some mean , but is not necessary monotone. As an example, we have already seen that A is C-decomposable with A monotone but C is not monotone.
Example 4.5 The standard means A, H, G, L, I are all (symmetric) homogeneous and monotone, then we find again that these means are decomposable. However, as already pointed before, the contraharmonic mean C is not monotone and so we cannot apply the above corollary. Let us try to apply directly Proposition 4.2. We can easily verify that the associated function of C does not satisfy (iii) and so C is not decomposable.
We leave it to the reader to check if the means S and P are decomposable or not.
The reader can easily verify that all the above standard means A, H, G, L, I, P, C, S satisfy relationship (15). However, for the corresponding functions g, it is not always monotone as in the case for A and H. In the general case, the following result gives a sufficient condition for ensuring the increase monotonicity of g.
Proposition 4.3 Let m be a symmetric homogeneous monotone mean and f be its associated function. Suppose that the function is strictly increasing. Then the function will be strictly increasing.
Proof One has , where . Since is decreasing, we get that k is increasing. Since is strictly increasing, we get that g is strictly increasing (as a product of an increasing and a strictly increasing positive functions). □
Example 4.6 The means G and L satisfy the conditions of the above proposition, whereas the means A and H do not. This rejoins the fact that A is C-decomposable and H is -decomposable with C and not monotone means. We leave it to the reader to verify if the means I and P satisfy or not the conditions of the above proposition.
In this section, we investigate an application of our above theoretical study. This application turns out the extension of some means from two variables to three or more arguments.
The following result is well known in the literature.
hold true for all distinct real numbers .
It is also simple to see that for all .
can be considered as a definition of the logarithmic mean with several variables. Now, the fact that if this definition coincides or not with some other ones as these given in [8, 10–12] seems to be an interesting problem. We omit the details about this latter point which is beyond our aim here.
To give more justification for our above extensions, the following result, which extends the inequalities from two variables to several arguments, may be stated.
hold for all distinct real numbers .
Proof For the sake of simplicity, we write (16) in a brief form . According to (17), we easily show that . Since , , , we then obtain . Using (17) again, a simple computation yields . The proof is complete. □
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Question 4 24 marks Kate is an elderly lady who consumes two goods, coal and the composite good. The shapes of Engel curves depend on many demographic variables and other consumer characteristics. If not, explain what the firm ought to do. If p 1 1, the demand for x 1 will also get multiplied by the same number. Show his optimal consumption point in the diagram you drew in part a. What are the assumptions of the theory of monopolistic competition? Lastly, the consumer increases the demand for some goods luxury items more than proportionately as his money income rises.
This indicates that with every equal increase in income, expansion in quantity purchased of the good successively declines. This further means that income elasticity of demand of food is between 0 and 1. It is important to note that the slope of the Engel curve in Fig. They are named after the German statistician 1821—1896 who was the first to investigate this relationship between goods expenditure and income systematically in 1857. Show the slopes and end points for the graphs. Homothetic Preferences Indifference Map : The three examples given above — perfect substitutes, perfect compliments and Cobb- Douglas — illustrate homothetic preferences, i. As a result, you decide to shrink the size of your 'It's Just a Buck! Graphically illustrate your answer using the tools of indifference curves and budget constraints.
Illustrate and explain the difference between an Engel curve and a demand curve. In microeconomics, an Engel curve describes how household expenditure on a particular good or service varies with household income. That means that as the consumer has more income, they will buy less of the inferior good because they are able to purchase better goods. The payoffs to each of the four possible combinations of choices are as given in the payoff matrix. Thus if the consumer prefers x 1, x 2 to y 1, y 2 then he automatically prefers 2x 1, 2x 2 to 2y 1, 2y 2 , 3x 1, 3x 2 to 3y 1, 3y 2 and so on. For starters, given the fact that you anticipate incomes to rise in your community, you can anticipate buying more goods. Many Engel Curves feature saturation properties in that their slope tends to diminish at high income levels, which suggests that there exists an absolute limit on how much expenditure on a good will rise as household income increases This saturation property has been linked to slowdowns in the growth of demand for some sectors in the economy, causing major changes in an economy's sectoral composition to take place.
Thus, in Engel curve drawn in panel b of Fig. For her, coffee and tea are perfect substitutes. Explain your answer with reference to your diagram. Suppose the demand curve for bus travel is downward sloping, and the income elasticity of demand for bus travel is negative. Find materials for this course in the pages linked along the left.
For , the Engel curve has a positive gradient. Applications In microeconomics Engel curves are used for equivalence scale calculations and related welfare comparisons, and determine properties of demand systems such as aggregability and rank. As income increases to Rs. However, you also know that you can get rid of some things. With the further increase in income by the same amount of Rs.
How you manage your project has everything to do with its outcome. Luckily, they have a tool that does exactly that: the Engel curve. Likewise, as income further rises to Rs. Since we are talking about lower incomes here, what about inferior goods? Is this firm minimizing its costs? Economists need a way of seeing how changes in income change the demand for certain goods. You know that there is a new biotech research center that is opening across town that is sure to produce some very high-paying jobs. Whereas the normal good's quantity rises with income, the inferior good experiences a decline and ultimately a crash, as more and more income is earned. What is the maximum he can consume in the current period? It has the advantage over the commonly used prosperity indicator of gross domestic product that the local price level is automatically taken into account.
Review of Economics and Statistics. Engel curve of an inferior good is drawn in Figure 8. . However, the curve can also be obtained for a group of consumers. That is, as income increases, the quantity demanded increases. In case of inferior goods, consumption of the commodity declines as income increases.
Engel died 1896 in Serkowitz at age 75. Explain your answer with reference to your diagram. Other scholars argue that an upper saturation level exists for all types of goods and services. Accounting for the shape of Engel Curves No established theory exists that can explain the observed shape of Engel Curves and their associated income elasticity values. After all, this is how demand works: the more money someone has to spend, the more of a good he or she will buy. Suppose you consume shoes and chocolate. Quadratic Engel Curves and Consumer Demand.
He was also Chairman of the Supervisory Board of the Neubrunn Waterworks and of the Lößnitz Embellishment Association. Productivity and Structural Change: A Review of the Literature. Make sure you label both intercepts for the budget constraint. Empirical Engel curves are close to linear for some goods, and highly nonlinear for others. This is simply the demand function for good 1. This curve would give the expenditure on the good of an average family belonging to a particular income-class. |
In this column an exposition is given of the so-called transformation problem in the theory of Karl H. Marx1. This problem deserves attention, because it has been used by adversaries of Marx to discredit the marxian theory. The first criticism of the transformation was already passed in 1907 by the economist L. von Bortkiewicz. In the sixties this criticism has been formulated in a more clear and understandable way by P. Sraffa and his followers. The marxist theoreticians did succeed in refuting the criticism of von Bortkiewicz and Sraffa, but this did not yet rehabilitate the theory of Marx. It remained unclear, what Marx tries to say. Only recently, in 2007, did the economist A. Kliman publish an explanation of the theory of Marx, that restores the logic in its entirety2.
The debate is provoked by the manner, with which Marx calculates the prices of products. This column assumes that the reader is already familiar with the marxist theory. Therefore the present paragraph contents itself with a short and compact reproduction of that theory, merely to define the various quantities. If desired the ignorant reader can inform himself in one of the many introductory books3, or perform a search on the world-wide-web. Marx is convinced that at the fundamental level the product value is determined by the amount of labour time expended in its production. Indeed this sounds plausible. However it should be realized, that the labour theory of value is just an abstraction of the real world. According to Marx the total national product, that is to say the sum of all goods produced in the state, has a value C', which can be divided into three components
(1) C' = c + v + m
Incidentally, the formula 1 is also valid for each separate product, and for the total production within an economic branch. In this sense she has a universal validity. In the formula c represents the value of the used means of production (expressed in the labour time, that was needed for the production of those means of production), and v is the value of the package of goods, that the personnel receives in exchange for their working on the product C'. The components c and v are called the constant and variable capital, respectively. The component m is typical for the capitalism, and is called the surplus value. This is the value (of course also expressed in labour time), that the workers add to the product C' without receiving a compensation. The producers appropriate the surplus value m for themselves, and thus realize an income for themselves. In the price system, such als is visible in the actual society in which we live, m is equal to the total profit π.
Marx is an artist in the use of language, and he calls this appropriation of m by the producers an exploitation of the workers. Incidentally in his time, when the inequality in wealth was still poignant, this was a fairly accurate description. Then he defines the ratio m/v as a measure of exploitation. He calls it the degree of exploitation, or sometimes in a more neutral tone, the rate of surplus value, and he represents it by the symbol m'. Marx assumes that the workers can estimate in a reasonably accurate manner how large the degree of exploitation m'i in their own enterprise i actually is. For a high degree of exploitation means, that the workers in the enterprise must make relatively long working days in order to produce the large profits of their employer. The workers will resent to remain into service of such an extreme employer. Therefore according to Marx the rate of surplus value m' is nearly equal for all producers. A general average rate of surplus value is prevalent in the system.
Now Marx makes it his task to determine the general profit rate r*. The profit rate is defined as the profit (here m), divided by the capital invested in the production. Obviously the means of production c are a part of the investment sum4. Since Marx assumes that the wages are paid at the start of the production process, in stead of afterwards, also the wage sum v must be advanced. Thus the production costs C, the invested capital, has a size c+v:
(2) C = c + v
And with this the general profit rate is also known:
(3) r* = m / C
Note that thus the formula 1 can be rewritten into C' = C × (1+r*). However the effect of the formula 3 on the prices of the different products is more interesting. Suppose that the producer i produces a quantity Qi of goods, and in the process makes costs Ci = ci+vi. He wants to realize the general profit rate r* on his advanced capital, and therefore must price each unit of product according to
(4) pi = (ci + vi) × (1 + r*) / Qi
The formula 4 is the so-called price transformation, that Marx uses to calculate the product prices from the product values (in labour time). The transformation formula takes into consideration, that the invested capital must yield a general average profit rate r* on the capital market. If this would not be the case in a certain branch, and for instance the profit rate would be below the average, then the capital would leave this branch. In other words, the competition on the capital market forces all production to yield the same general average profit rate.
The price formula 4 has a curious consequence. According to the formula 1 the producer i generates an amount of products Qi with a total labour value of Ci+mi. On the other hand, the formula 4 states that his nominal yield of money is equal to Ci + r*×Ci. Can the labour value of Qi coincide with the price sum? That is to say, is mi = r*×Ci possible? Is the general average profit rate already valid on the level of the individual producer? The definition of the rate of surplus value leads to the relation mi = r*×Ci. Apparently the produced surplus value depends only on the hired quantity of labour vi (because m' is a social given), and not on ci. Therefore there is no reason, that the surplus value mi of the individual producer i should match r*×Ci. At the level of the separate company i this will mostly not be the case.
Apparently it can be concluded from the price transformation 4, that the value production and the monetary yield are separated at the level of the individual producer. Some producers will make more profit than they generate as actual surplus value, and other will make less profit. Producers that hire a relatively large amount of labour, produce a relatively high surplus value, and will have to transfer a part to others. And producers that prefer to automate, and thus do not hire much labour, produce a surplus value below average. They will receive a compensation. The existence of the general average profit rate on the capital markets leads to a redistribution of the total social surplus value. Of course the redistribution is not a conscious process, but it happens as it were behind the backs of the producers, simply because they do their calculations in prices and not in labour values. Marx assumes that the redistribution does not reduce the labour value in the society as a whole. That is to say, after the redistribution the value of the total product C' and the profit m remain unaffected.
Marx defines as a measure for the labour intensity in a company (or branch) the quantity ci/vi. He calls it the organic composition of the capital, and usually represents it by the symbol oi. Thus the invested capital is given by Ci = vi × (oi + 1). The quantity oi varies, depending on the producer, and especially on the branch. For instance, public utilities with their expensive transport networks, have a higher organic composition than the commercial trade. According as oi of the producer is higher, he will receive more additional surplus value. This becomes clear, when the formula 3 is applied for the calculation of the profit rates in the branches:
(5) r*i = m'i / (1 + oi)
Since m' is determined by the society, one finds m'i=m', and a different oi has the automatic implication of a different r*i. If oi is above average, then the profit rate in the branch i is too low, and it will be compensated by means of a redistribution of the surplus value until the level of the general average profit rate is reached.
It has just been said, that in 1907 the first critique is published with regard to the price transformation of Marx, like it is shown in the formula 4. However the arguments of the critics did not yet excel in clarity. A more convincing exposition of the transformation problem is given in 1976, by W. van Drimmelen5. Van Drimmelen believes that Marx makes a mistake by calculating the general average profit rate by means of the labour values m and C. For the producer does not know the labour values, and fixes his product price on the basis of money prices. In other words, he calculates his product price by means of the profit rate r, which the capital market dictates for the efficiency of its lent capital sums6. If the price sum of the production costs, namely the advanced money capital for the means of production and for the workers, is represented by Γ, then one finds r=π/Γ. Apparently the arithmatic method of Marx is only meaningful as long as the profit rate in labour values r* is equal to the profit rate in money prices (that is to say, r*=r). Marx believes that this is the case, for he assumes (like already has been remarked) that on the social (macro-economic) level all labour values remain conserved and thus m=π and C=Γ. Van Drimmelen shows that in general this will not be true. To that end he construes an example, that displays the absurdity.
Suppose that the producers together spend their entire monetary profit π and thus their surplus value m, on luxury goods. These are goods, that bring pleasure, without adding to the production process. Suppose in addition that the organic composition in the branch of luxury goods is very low. In the preceding paragraph it is explained, that a mechanism of redistribution will be started. In the exchange on the markets the branch of luxury goods can not claim its generated surplus value in its entirety, but it is forced to hand over a part of its money claim to the other branches. Thus the price sum of the luxury goods falls below its labour value. The luxury goods are not sold with a profit m, but with a smaller profit π. In the example of Van Drimmelen the other (remaining) branches in this economy produce the means of production and the wage goods. The value sum of their total product is C, and the price sum is Γ. Since the example assumes that these branches have a higher organic composition than the branche of the luxury goods, the mechanism of redistribution will raise the price sum Γ above the value C. Apparently the equality of the price sums and the value sums (π=m, Γ=C) is violated. In other words, the monetary profit rate r would differ from the profit rate r* in labour values. The find of Van Drimmelen is original, as far as it shows in a clear way the role of the organic composition in the transformation problem.
The economist E. Feess-Dörr illustrates in a slightly different way the transformation problem7. He abandons the idea of the luxury branch, and considers an economic system with two "departments"8. Suppose that the physical composition of the package of products, that respresents the surplus value m, is different from the composition of the package of products, that represents the advanced capital C. This is for instance the case, when the economy does not grow, so that the means of production and the army of workers are not expanded. In that case the producers will be forced to consume their entire surplus value. In other words, the surplus value is generated only by the department (branches) that produce consumption goods. On the other hand the advanced capital C also contains all kinds of means of production. In general there is no reason why the organic composition for the production of the consumption goods should be equal to the organic composition for the production of the means of production. Also here the surplus value of the producers must be redistributed in order to guarantee the general average profit rate. Again the equality of the monetary profit π and the surplus value m would be destroyed. The profit rate r* in labour values would differ from the profit rate r in price sums. Of course the argument of Feess-Dörr is principally the samen as that of Van Drimmelen.
An example may help to illustrate the effect and the impact of the transformation problem. In particular it gives an impression of the deviations of the marxist prices with respect to the real prices. The real prices can be calculated with the theory of Sraffa, which has already been introduced in a previous column. In the theory of Sraffa the calculations can be performed without dubious assumptions like π=m and Γ=C. There, in that previous column, an example is used, that will be copied in the present column9. The economic system consists of two branches, namely the agriculture (index g) and the industry (index m). The agriculture produces corn (expressed in bales) and the industry produces metal (expressed in tons of weight). Table 1 gives a survey of the quantities. For the explanation of the symbols in the table 1 the reader is invited to just consult the previous column. This is advisable anyway, because in the following also the numerical results of that column will be used.
According to the theory of Marx the labour values form the foundation for the calculation of prices. The labour values must obey the formula 1, both at the level of the company, of the branch or department, and of the whole society. In order to rewrite this formula use is made of the relation
(6) l = v + m
The formula 6 simply states that the total amount of labour l expended on the product is equal to the sum of the wage v and the surplus value m. After substitution of the formula 6 the formula 1 is expressed as C' = c + l. For the present case with two branches this formula implies that10
(7a) ηg×Qg = ηg×qgg + ηm×qmg + lg
(7b) ηm×Qm = ηg×qgm + ηm×qmm + lm
In the set 7a-b ηg is the labour value of a bale of corn, and ηm is the labour value of a ton of metal. In the previous column about the theory of Sraffa it is explained that it is preferable to eliminate the aggregate quantities Qg and Qm by means of a division. Then a set of technical coefficients is left, that depend exclusively on the applied technology. Table 2 summarizes the values of the technical coefficients, such as correspond to the table 1.
The production coefficients can be expressed as a matrix A, and the labour coefficients can be expressed as a (horizontal) vector a. Then in the matrix notation the set 7 of coefficients takes the form
(8) η = η · A + a
Figure 1: Wage curve w/pg according to Sraffa and Marx This matrix equation can simply be solved, namely11
(9) η = a · (I − A)-1
In the formula 9 I is the unity matrix, and the upper index -1 indicates that the inverse matrix of I − A should be taken. From this solution it follows that a bale of corn is worth ηg = 3.479 labour units, and a ton of metal is worth ηm = 21.74 labour units. It is characteristic of labour values, that they do not depend on the distribution of the added value over the wages and the profits - this in contrast with the prices.
Now that the value of a unit product is known, all the production quantities can be calculated. For instance it turns out that cg=21.74 labour units, and cm=57.40 labour units. The social capital intensity is defined by
(10) κ = (cg + cm) / (lg + lm)
One finds with the help of the formula 10, that κ=2.638 labour units per labour unit. Now also the prices according to the theory of Marx can be calculated. The price transformation of the formula 4 is in matrix notation
(11) p = (η · A + w × a) × (1 + r*)
In the formula 11 w is the wage level of the factor labour, or in other words vi = w×li holds. If m is eliminated from the formulas 3 and 6, then one obtains for w the formula
(12) w = (1 − κ×r*) / (1 + r*)
Note that w is a transformation of the rate of surplus value m'. For from the formula 6 and the definition of the rate of surplus value it follows that w = v/l = 1/(1+m') holds. Therefore the formula 12 also shows how the general average profit rate originates from the rate of surplus value. The formulas 11 and 12 complete the price transformation of Marx. Thanks to the labour theory of value this price theory does not need the numéraire, in contrast with the theory of Sraffa.
Now all is ready to illustrate the transformation problem in a comparison of the prices according to Marx and Sraffa. In this column the differences are shown by means of graphical presentations. The portraiture of the curves in the same figure requires that also the prices of Marx are normalized with the numéraire pg, just like in the example for the theory of Sraffa. In addition the wage level w must be converted. This is so, because in contrast with Marx the wage sum of Sraffa is not included in the advanced production capital12. In this way the quantities calculated with the theory of Marx are transformed, and they are added to the figures 2 and 3, which have already been incorporated in the previous column about the theory of Sraffa. Here the result is shown in the figures 1 and 2.
It is immediately striking, that in the case of a profit rate r equal to zero the results of Sraffa and Marx coincide. This is obvious, because then the workers receive the entire production, and thus the prices and labour values are the same. In this situation the surplus value is absent, so that a redistribution is not necessary. According as the profit rate increases, there is a growing discrepancy between the price transformation of Marx and the price system of Sraffa. This is particularly striking for the capital intensity κ/pg in the figure 2, where the difference becomes quite large. Incidentally here also the marxist κ-curve exhibits an upward trend, although in a moderate way (in contrast with the expectations of the neoclassical theory). In any case the two figures clearly illustrate the transformation problem13. In the example this problem occurs, because the organic compositions in the agriculture and in the industry are different. In the agriculture she is og=1.087/w, and in the industry om=5.71/w. Note that apparently the magnitudes of the organic compositions do depend on the distribution, but that their relative proportions are not influenced by her. Of course the profit rates in the branches r*g = mg/Cg en r*m = mm/Cm are also different, due to the formula 5. The figures 1 and 2 show that the theory of Marx does not succeed in compensating the profit rates in the two branches in such a way, that they transform into the profit rate according to Sraffa.
The tranformation problem can only be solved, if the theory of Marx is interpreted in another way. For this consult the book of A. Kliman, that is mentioned in a footnote.
Finally it must be remarked, that the wage curve and the curve of the capital intensity, that are calculated here with the theory of Marx, differ from those in the proof-text Vooruitgang der economische wetenschap. The marxist curves in the figures 8.7.1 and 8.7.2 of the proof-text are wrong. |
This is one of the most famous mathematical theories or problems. Many years were spent for finding the solution of this at least 150-year old problem. Mathematicians and some curious amateurs thought on this in different times. Some could not find any solution and some others, in their own words, did find the solution. Sometimes, a wrong solution was wrongly appreciated by many people for a long time. At time, the fault was identified and the correct solution was also provided.
What is the problem?
The easiest way to understand the problem is to color a map. Let us take the map of Bangladesh as an example. It has 64 districts. If we want to color each side-by-side districts with different colors, how many colors will be needed? The Four Color Theorem says there will be maximum 4 colors needed. In some cases, may be 2 or 3 colors will be sufficient. The theory is not only about the map of Bangladesh; it is true for any maps of a country or place. In mathematical words, a plane surface divided into any number of blocks can be colored by blocks with maximum 4 different colors so that no two adjacent blocks will be of same color.
It is well accepted that the first formal question on this problem was raised by Francis Guthrie in 1852, that is, the question was about how many colors would be necessary and what was the proof with explanation. Guthrie was a student of University College, London at that time. He was then a student of famous Mathematician De Morgan who could understand the depth of the question. De Morgan wrote a letter and sent this problem to another milestone Mathematician Hamilton in Dublin. In reply Hamilton told it was not possible to solve in a short time. In the mean time, Francis Guthrie made some calculations on this what he gave to his younger brother Frederick Guthrie who was also a student of De Morgan. Later Guthrie became a lawyer in profession, but his passion and expertise in Mathematics destined him to a Mathematics Professor in his later age. During this period, De Morgan could not find the solution himself and continued searching for someone who might be able to solve this. Charles Pierce of USA worked for this up to the end of his life since 1860. In 1879 Mathematician Cayley understood the problem from De Morgan and prepared a research paper, for the Royal Geographical Society, on the difficulties of finding the solution of this problem. After a short while of this article published an declaration was published in the Nature journal where it was claimed that the solution for the Four Color Theorem was found. The solutions was made by a Barrister Alfred Bray Kempe who was a student of Cayley. With the advice of his teacher, Kempe sent the full solution to the American Journal of Mathematics. There another expart Story made some modifications in the article and sent to the Scientific Association of John’s Hopkins University. A number of renowned Mathematicians discussed on this article and Kempe got the recognition of his Kempe Chain Method. Kempe proved that the any map can be colored with maximum 4 colors for which he got awards from many prestigious sources including the Royal family of England. In 1880, Kempe published two more articles on the same problem as the improvements of his previous theories. These articles brought the famous Philosopher and Mathematician Professor P G Tait. He wrote at least two explanatory articles on this and established the problem as Four Color Theorem.
Until 1890, nobody could identify that the theories of Kepme and Tait were wrong and full of cleverly traits. Percy John Heawood described the error in Kempe’s Theorem for the first time. He claimed at that time 4 different colors will not be sufficient for all maps. Sometimes, may be 5 colors will be necessary. In 1896 Vallee Poussin identified the error in Kempe’s Theorem again without knowing the article of Heawood. Later, Kempe confessed his error by saying that at the time of writing that article it was not possible to identify and rectify the error of the Theory. Heawood was continuously working on the problem and showed the first Mathematical way for solving the problem. In later years, many Mathematicians from around the world worked and contributed for this. Every small contributions made some improvements in the theory. After a long journey, in 1976, the correct and complete Four Color Theorem was formed. It was developed jointly by Appel and Haken who were supported by computer works by Koch. In fact, it is one of the few theories that was established with huge computer supports. A significant branch in Mathematics Graph Theory was the base for establishing this theory where important theories of Euler and Hamilton were used. Other theories were also taken from Weinicke, Veblen, Franklin, Birkhoff and Heesch. Actually, Appel and Haken used the Kempe Chain Method. They published their works in two sequels in 1977. Since then, many people worked for improvements of the theory and eventually made this simple. In 2000, Ashay Dharwadker made a complete different proof of the theorem and he is awarded by Canadian Mathematical Society and other organizations.
Please read the next part of this article Graph Theory as the Base of Four Color Theorem. |
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Let's get the idea of Schrodinger's cat right: It's a funky aspect of human perspectives. The math that goes with it is really easy, contrary to popular opinion, and entertaining, too!
(This article is about Schrodinger’s cat. If you thought it was about LOLCATS… well, it could be.)
When you’re studying the world through the lens of quantum mechanics, you’ll have a lot of freedom in defining what you want to study. This is because you’re essentially probing the nature of nature, you’re stirring through reality’s constituents, and you’re manipulating (on paper, at least) the apparatus to see how the experiment called consciousness is affected by it.
You can look at how making some changes alters a system’s freedoms, modifies the way it moves through space and in time, becomes more or less, energetic, etc. For each of these viewpoints, there’s a framework – a set of rules and operations – that makes it easier to study it.
It’s like you’re making a speech to an audience. Depending on who’s listening to you, you modify your language, your tone and style, and your body language so your message gets across – whatever that message may be. If you’re talking to kids, you try to emphasise your message in terms of their future. If you’re talking to job-seekers, you emphasise your message in terms of creation of jobs, increase of social security, etc.
Similarly, while studying quantum mechanics, you have the liberty of addressing your problem from different angles. In fact, to make it simpler for you, physics itself offers what are called degrees of freedom. A degree of freedom (DoF) is simply an avenue to effect change by doing work in the universe. A universe that contains a long stick can remain so; if the stick starts to move forward and backward in space-time, we have a universe with a moving stick; if the stick starts to revolve about its central axis, we have a universe with a revolving. So, there you go: Two degrees of freedom for a universe that contains a stick!
Each DoF is an “angle” for you to approach a problem with. As I briefly mentioned earlier, studying how the energy of a system varies is one such angle, and the framework that goes with this model is called the Hamiltonian formulation of dynamics. In classical (i.e., Newtonian) mechanics, solving the Hamiltonian for a system yields the total energy of the system – be it an electron, or a system of electrons in, say, a helium atom.
To wit: A helium atom consists of two electrons, e1 and e2. They have masses m1 and m2, have momentums p1 and p2, Z is the atomic number, e is the value of the elementary charge, and r1 and r2 are the position vectors of the electrons with respect to the nucleus (don’t hyperventilate yet). For this system, the Hamiltonian is
We can simplify this one step further: All electrons have the same charge (e) and the same mass (m). So, the Hamiltonian becomes
Because of the charge-and-mass uniformity, taking e1 and putting it in e2’s place and taking e2 and putting it in e1’s should still yield the same Hamiltonian. This is because p1 and p2 are measured in terms of m, and r1 and r2 are measured in terms of e. Since m and e for both electrons are fixed, interchanging the electrons shouldn’t affect how we’re measuring p and r!
In other words, H(1,2) = H(2,1).
This formulation is only valid whenever the principle of indistinguishability holds: One electron is exactly like another electron as long as you’re treating them in terms of their mass and charge. Thus, the Hamiltonian for a two-proton system will also be compatible with H(1,2) = H(2,1). However, switching two electrons when you’re studying them in a framework that involves their spin will break the H(x, y) equivalence.
In fact, when you’re treating particles as being distinguishable, here’s the more general Hamiltonian:
Solving this Hamiltonian gives the position and velocity of the two electrons for all possible initial conditions (except if you’re getting them started at the speed of light).
All this trouble has been taken to figure out the classical energy of a system in terms of two electrons. Let’s go quantum, and let me dispel the notion that it gets harder: It doesn’t, not if you’ve a little imagination to spare.
First, a disaster strikes…
There once was a man named Werner Heisenberg, and he was a buzzkill. While he was pottering around the world of particles, he noticed that they existed in a world in which a small change of energy meant everything in the system would change. Because humans study most of the world around them by bouncing off light, sound, heat, or other forms of radiation, and letting the brain interpret the splashback, Heisenberg realised it’s inevitable that any observation made on the quantum world changed it in such a way that it rendered our measurement useless.
“Thou shalt not simultaneously know the position and momentum of a particle,” quoth Heisenberg, and ‘twas called his uncertainty principle hence.
So, what was once a very determinable system – we could say that, given r1 and r2, we know p1 and p2 – has now become a probabilistic system: From “We know it’s there”, it’s become “We know it may be there”. Consequently, we throw out p1, p2, r1 and r2, and bring in a freak named ψ (psi – yes, Heisenberg will be thrilled to know it’s all Gangnam-style now).
ψ is called the wave function. It is a function in space and time - a time-traveller, if you will - that describes the chances of finding a particle in a given region in space at a given time. As a solution of the Schrodinger equation, it's described thus:
(Here, ∂/∂t measures how one parameter, such as ψ, changes with time; i is a complex number and h/2π is the Dirac’s constant.)
Even in the case of the quantum-Hamiltonian, interchanging the two electrons results in the same value for Hψ. Here’s how.
Interchanging 1 and 2,
Since H(1,2) = H(2,1),
Hence, proved! That was simple enough, wasn’t it?
Enter: A cat out of a box (i.e., cattus ex machina)
Now, we have two indistinguishable states, ψ(1,2) and ψ(2,1). They are both probabilistic, which means the total energy of a two-electron system is based on the momentum of the two electrons and where they may probably be at some time (given by H(1,2)ψ(1,2)).
Also, since ψ(1,2) and ψ(2,1) are interchangeable, it’s probable that they could both be true – but at different times.
To resolve this “but at different times” issue, we create what’s called a linear combination in mathematics which, in English, reads: “At any time, the wave function ψ(1,2) has equal chances of being a combination of ψ(1,2) and ψ(2,1) and a combination of only ψ(1,2) and not ψ(2,1).”
Since the probability of a certainly-happening event is 1, the probability of an event that may happen half the time should be ½, or 0.5. So, as an equation:
ψ(1,2) = EITHER [ψ(1,2) + ψ(2,1)] OR [ψ(1,2) – ψ(2,1)]
ψ(1,2) = 0.5[ψ(1,2) + ψ(2,1)] + 0.5[ψ(1,2) – ψ(2,1)]
Here, [ψ(1,2) + ψ(2,1)] is called the symmetric state and [ψ(1,2) – ψ(2,1)] is called the antisymmetric state. Therefore, every wave function describing a system – such as a cat in a closed box with a bowl of poison – is a superposition of a symmetric state – the cat is alive – as well as an antisymmetric state – the cat is dead. And how do you know which state it is in at some point of time?
You open the box and make a measurement. This distorts the energy of the system and collapses the wave function, trapping ∂/∂t to, say, 2 pm, and solving the Hamiltonian to give you either a dead cat or a live cat!
And this, my dear reader, is the Schrodinger’s cat formulation. |
All of game theory describes strategic settings by starting with the set of players, i.e. Game Theory (W4210) Course Notes Macartan Humphreys September 2005. ii ABSTRACT These notes are written to accompany my class Political Science W4210 in the political science program at Columbia University. Game Theory and Economics If the market is composed by a small number of firms, each firm must act strategically. Economics 286: Graduate Game Theory. Bounding the price of anarchy 151 8.1.2. While used in a number of disciplines, game theory is most notably used as a tool within the study of economics. ... Economics » Game Theory » Lecture Notes ... Notes . On StuDocu you find all the study guides, past exams and lecture notes for this module The market supply will increase from 200 to 210 and the price has to drop to reach an equilibrium. It originates from a fascinating line of economic thought, but feels more like a branch with its own roots. The difference between noncooperative and 0000008775 00000 n What the opponent does also depends upon what he thinks the first player will do. Supports open access. Examples of Game Theory in economics Nash Equilibrium. 0000059998 00000 n Notes 134 Exercises 135 Chapter 7. May 2017. Economics Freely browse and use OCW materials at your own pace. 0000010569 00000 n This video looks at some applications of game theory to issues in A level economics. Rules Mathematical games have strict rules. We also restrict attention to the case where the interests of the players are completely antagonistic: at the 0000009709 00000 n In economics, however, game theory tends to focus on sets of outcomes known as equilibrium that represent the most rational solutions to each situation. Sloan School of Management 15.013 – Industrial Economics for Strategic Decisions Massachusetts Institute of Technology Professor Robert S. Pindyck Lecture Notes on Game Theory (Revised June 2009) These lecture notes extend some of the basic ideas in game theory that were covered in 15.010. 0000024303 00000 n Applications are provided when available. 0000045732 00000 n Game Theory Definition Game theory - a mathematical method of decision making in which alternative strategies are analysed to determine the optimal course of action for the interested party, depending on assumptions about rivals’ behaviour. 0000062916 00000 n Welcome to the Web page of Game Theory at the University Carlos III, Madrid. These lecture notes are partially adapted from Osborne and Rubinstein , Maschler, Solan and Zamir , lecture notes by Federico Echenique, and slides by Daron Acemoglu ... A game is called a finite horizon game if there is a bound on the length of histories in H. We will apply this concept to a discussion of the War of Attrition, which can turn out to be a rather complicated game. Seven Topics in Game Theory. This section provides the lecture notes for the course, organized by chapter and topic. It covers classical topics, such as repeated games, bargaining, and supermodular games as well as new topics such as global games, heterogeneous priors, psychological games, and games without expected utility maximization. Courses » Economics » Economic Applications of Game Theory » Lecture Notes ... Decision Theory: Lecture notes (PDF) 3: Representation of Games: They are in progress. John Nash Memorial. Its limitations aside, game theory has been fruitfully applied to many situations in the realm of economics, political science, biology, law, etc. Excitement Please send comments and corrections to me at: email@example.com. 1.1 IGCSE Grade 11 and Grade 12 Economics - Revision Notes; 1.2 IGCSE Grade 11 and Grade 12 Economics - Revision Charts; 1.3 IGCSE Grade 11 and Grade 12 Economics - Externalities ; 1.4 IGCSE Grade 11 and Grade 12 Economics - International Trade ; 1.5 IGCSE Grade 11 and Grade 12 Economics - Theory of Employment trailer << /Size 114 /Info 69 0 R /Root 72 0 R /Prev 159528 /ID[<2c7a16c545d1b7e1314521ff51e07316>] >> startxref 0 %%EOF 72 0 obj << /Type /Catalog /Pages 67 0 R /Metadata 70 0 R /PageLabels 65 0 R >> endobj 112 0 obj << /S 721 /L 872 /Filter /FlateDecode /Length 113 0 R >> stream 1.223 Impact Factor. Game theory is the study of mathematical models of strategic interaction among rational decision-makers. Lectur e Notes 4: Mixed-strategy Nash equilibrium . vii-xi. Knowledge is your reward. They are in progress. This theory aims at providing a systematic approach to business decision making of organizations. March 2018. Extensive form games This course is an introduction to game theory and strategic thinking. Games and Dominance . Problem Set 2 , Answers . Games and Economic Behavior (GEB) is a general-interest journal devoted to the advancement of game theory and it applications. This shows the existence of the perfectly competitive market. Game theory analysis has direct relevance to the study of the conduct and behaviour of firms in oligopolistic markets – for example the decisions that firms must take over pricing and levels of production, and also how much money to invest in research and development spending. We don't offer credit or certification for using OCW. (photo: Patrick Hamilton) • Game Theory: The study of situations involving competing interests, modeled in terms of the strategies, probabilities, actions, gains, and losses of opposing players in a game. If firms are competitive and they set low price -they will both make £4m. • Game Theory: The study of situations involving competing interests, modeled in terms of the strategies, probabilities, actions, gains, and losses of opposing players in a game. This part of the course contains the foundations of economics by examining how individual buyers and sellers interact. Economics 51: Game Theory Liran Einav April 21, 2003 So far we considered only decision problems where the decision maker took the en-vironment in which the decision is being taken as exogenously given: a consumer who decides on his optimal consumption bundle … 0000002088 00000 n Its limitations aside, game theory has been fruitfully applied to many situations in the realm of economics, political science, biology, law, etc. Game theory is a systematic study of strategic interactions among rational individuals. Editor-in-Chief: E. Kalai. Games are either simultaneous-move or sequential, static or dynamic, one-off or repeated, cooperative or non-cooperative, etc. 2. » 0000001208 00000 n Lecture Notes in Information Economics Juuso Valimaki February, 2014 Abstract These lecture notes are written for a –rst-year Ph.D. course in Microeconomic Theory. Game theory is now a standard tool in economics. This section provides the schedule of lecture topics for the course along with notes from selected sessions. Lecture Notes. Note that coopera-tion is still studied but the emphasis is on how cooperation can be enforced without 3. It was the purpose of game theory from its beginnings in 1928 to be applied to serious situations in economics, politics, business, and other areas. It has applications in all fields of social science, as well as in logic, systems science and computer science.Originally, it addressed zero-sum games, in which each participant's gains or losses are exactly balanced by those of the other participants. in Economic Theory, Bounded Rationality, Game Theory, Experimetal Economics anc Choice Theory. 0000009153 00000 n EMAIL: firstname.lastname@example.org. Home Game theory analysis has direct relevance to the study of the conduct and behaviour of firms in oligopolistic markets – for example the decisions that firms must take over pricing and levels of production, and also how much money to invest in research and development spending. Courses 0000047985 00000 n Correlated equilibria 142 Notes 145 Exercises 146 Chapter 8. Game Theory in Economics. Games and Economic Behavior. In game theory, self-interest is routed through the mechanism of economic competition to bring the system to the saddle point. 3. We will begin by explaining what we mean by rational — or rationalizable — strategies. In his 1928 article, "Theory of Parlor Games," Von Neumann first approached the discussion of game theory, and proved the famous Minimax theorem. Game theory concepts have become hugely important in economics and whilst there are limits to how far analysis can stretch, making use of some of the core ideas can add greatly to your evaluation and analysis marks in the summer papers. Evolutionary and correlated equilibria 137 7.1. 1Definition of a Game We start with abstractly defining what we mean by a game. Valimaki February, 2014 Abstract These lecture notes, slides, problem sets and exam questions Muhamet! At 1.1 what is game theory ' course in 2004 based on teaching material thousands! Written for a –rst-year Ph.D. course in 2004, as opposed to co-operative game theory an... University Press, p. 1.Chapter-preview links, pp elasticities, government intervention and market failure the player. 2018 ) These lecture notes... notes focus of the games we play and watch Guide your pace. At your own pace, Bounded Rationality, game theory brought about revolution. 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Fourier coefficientFourier expansionFourier coefficientsFourier modesFourierFourier decompositioncosine seriesFourier theoremFourier's theoremHilbert space
In mathematics, a Fourier series is a periodic function composed of harmonically related sinusoids, combined by a weighted summation.wikipedia
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The mathematical topics of Fourier series and Fourier transforms rely heavily on knowledge of trigonometric functions and find application in a number of areas, including statistics.
group delayphase delaydelay distortion
In signal processing, group delay is the time delay of the amplitude envelopes of the various sinusoidal components of a signal through a device under test, and is a function of frequency for each component.
Fourier coefficientsFourier expansiongeneralized
In mathematical analysis, many generalizations of Fourier series have proved to be useful.
Ganesh Prasad (15 November 1876 – 9 March 1935) was an Indian mathematician who specialised in the theory of potentials, theory of functions of a real variable, Fourier series and the theory of surfaces.
In mathematics, a Voronoi formula is an equality involving Fourier coefficients of automorphic forms, with the coefficients twisted by additive characters on either side.
Karhunen–Loève transformKarhunen–LoèveKarhunen–Loève expansion
In the theory of stochastic processes, the Karhunen–Loève theorem (named after Kari Karhunen and Michel Loève), also known as the Kosambi–Karhunen–Loève theorem is a representation of a stochastic process as an infinite linear combination of orthogonal functions, analogous to a Fourier series representation of a function on a bounded interval.
They offer a framework for Fourier expansion, which is employed in image compression routines, and they provide an environment that can be used for solution techniques for partial differential equations.
Sidon setB_h[g]-sequencesSidon set problem
Sidon introduced the concept in his investigations of Fourier series.
integral operatorkernelintegral kernel
The precursor of the transforms were the Fourier series to express functions in finite intervals.
Geiringer, HildaHilda Pollaczek-Geiringer
She received her Ph.D. from the University of Vienna in 1917 under the guidance of Wilhelm Wirtinger with a thesis entitled "Trigonometrische Doppelreihen" about Fourier series in two variables.
total variation normvariationTonelli plane variation
He used the new concept in order to prove a convergence theorem for Fourier series of discontinuous periodic functions whose variation is bounded.
McKay–Thompson seriesmoonshine theoryConway–Norton conjecture
In 1978, John McKay found that the first few terms in the Fourier expansion of the normalized J-invariant,
In applied mathematics, the regressive discrete Fourier series (RDFS) is a generalization of the discrete Fourier transform where the Fourier series coefficients are computed in a least squares sense and the period is arbitrary, i.e., not necessarily equal to the length of the data.
Equivalently, they employ a change of variables and use a discrete cosine transform (DCT) approximation for the cosine series.
A cusp form is distinguished in the case of modular forms for the modular group by the vanishing of the constant coefficient a 0 in the Fourier series expansion (see q-expansion)
Fourier founded a new branch of mathematics — infinite, periodic series — studied heat flow and infrared radiation, and discovered the greenhouse effect.
spectral methodsFourier spectralFourier spectral methods
The idea is to write the solution of the differential equation as a sum of certain "basis functions" (for example, as a Fourier series which is a sum of sinusoids) and then to choose the coefficients in the sum in order to satisfy the differential equation as well as possible.
Laplace also recognised that Joseph Fourier's method of Fourier series for solving the diffusion equation could only apply to a limited region of space because those solutions were periodic.
L'' ''p'' spaceL ''p'' spacesL'' ''p'' spaces
The Fourier transform for the real line (or, for periodic functions, see Fourier series), maps L p (R) to L q (R) (or L p (T) to ℓ q ) respectively, where 1 ≤ p ≤ 2 and 1/p + 1/q = 1.
Dini's testDini–Lipschitz test
In mathematics, the Dini and Dini–Lipschitz tests are highly precise tests that can be used to prove that the Fourier series of a function converges at a given point.
Classical ElectrodynamicsClassical Electrodynamics'' (book)widely-used graduate text
The necessary mathematical methods include vector calculus, ordinary and partial differential equations, Fourier series, and some special functions (the Bessel functions and Legendre polynomials).
Bochner–Riesz conjectureBochner-Riesz meanBochner–Riesz operator
The Bochner–Riesz mean is a summability method often used in harmonic analysis when considering convergence of Fourier series and Fourier integrals.
functions of bounded variationboundedunbounded variation
According to Boris Golubov, BV functions of a single variable were first introduced by Camille Jordan, in the paper dealing with the convergence of Fourier series. |
Conservation of mechanical energy
The conservation of mechanical energy is a principle of physics that ensures that in the absence of dissipative forces such as friction, the total amount of energy in a system never changes . According to the conservation of mechanical energy, the sum of the kinetic energy and the potential energies must have a constant magnitude.
Conservation of mechanical energy
When a system is completely free from frictional forces or drag forces, the mechanical energy of that system will be constant . This means that a pendulum free from frictional forces, for example, must oscillate indefinitely, otherwise, in a finite time, this pendulum will have its energy dissipated in other forms of energy, such as thermal energy, vibrations, sounds, etc. .
Observe the following figure, in it we have a mobile that moves with constant speed , free of friction forces with the ground, with the air and free of friction forces between its components. In this case, we say that the mechanical energy associated with that body will be equal at points A, B, and C.
At point A, the car has both kinetic and potential energy, thanks to its small height above the lowest level of the ground. Already at point B, the car approaches a situation where all its kinetic energy becomes gravitational potential energy, in other words, as the kinetic energy of the vehicle decreases, its gravitational potential energy increases, as we wrote in following formula, which relates the mechanical energies of points A and B:
v a – velocity of the body in position A (m/s)
v b – velocity of the body in position B (m/s)
g – gravity (m/s²)
h a – height of point A (m)
h b – height of point B (m)
As this theme addresses different types of energy, in the following topics, we bring brief definitions of those considered most common in high school, in order to review this content and provide a more complete learning.
The mechanical energy of a system is defined as the sum of the kinetic energy with the different potential energies present there, such as gravitational potential energy or elastic potential energy (these being the most common in exercises carried out in high school), among others.
E M – mechanical energy (J)
E C – kinetic energy (J)
E P – potential energy (J)
When there is friction, a part of the mechanical energy of the system is “lost”, being converted into a thermal agitation of the atoms and molecules. This type of energy resulting from the action of the friction force is the thermal energy of the body, and its correspondence with heat was explained by James Prescott Joule , through his experiment on the mechanical equivalence of heat.
Kinetic energy is the form of energy related to the motion of a body. It is a scale quantity r, proportional to the mass of the body and the square of its velocity, in units of the International System of Units ( SI ), it is measured in joules (J) and can be calculated using the following formula:
p – momentum (kg.m/s)
m – mass (kg)
Potential energy is the generic name given to any form of energy that can be stored. These energies only arise when conservative forces are applied . Examples of potential energy are:
- Gravitational potential energy : form of energy generated when some body presents a certain height in relation to the Earth’s surface.
- Elastic potential energy : form of energy related to the deformation of elastic bodies, which tend to return to their original shape after being deformed.
k – spring constant (N/m)
x – deformation (m)
- Electric potential energy: is the energy that arises through the attractive or repulsive interaction between electric charges .
Question 1) (G1 – IFBA) A body is dropped from the top of an inclined plane, as shown in the figure below. Assuming ideal polished surfaces, zero air resistance, and 10 m/s 2 as the local acceleration due to gravity, determine the approximate speed with which the body strikes the ground:
a) v = 84 m/s
b) v = 45 m/s
c) v = 25 m/s
d) v = 10 m/s
e) v = 5 m/s
To determine the approximate speed with which the body reaches the ground, we must apply the principle of conservation of mechanical energy. Therefore, we say that the gravitational potential energy at the top of the inclined plane is equal to the kinetic energy of this body at the bottom of the plane.
Upon resolution, the masses present on both sides of the equation cancel out. Then, we replaced the values informed by the statement and did some algebraic operations until we found the speed of 10 m/s.
Question 2) (UEG) In an experiment that validates the conservation of mechanical energy, a 4.0 kg object collides horizontally with a relaxed spring with a spring constant of 100 N/m. This shock compresses her 1.6 cm. What is the speed, in m/s, of this object before it hits the spring?
In this exercise, we say that the kinetic energy of the body is fully converted into elastic potential energy, so we must make the following calculation:
Question 3) (G1 – IFSP) A pole vaulter, during his run to overcome the obstacle in front of him, transforms his ____________ energy into ____________ energy due to the height gain and consequently to the _____________ of his speed.
The gaps in the text above are correctly and respectively filled in by:
a) potential – kinetic – increase
b) thermal – potential – decrease
c) kinetic – potential – decrease
d) kinetic – thermal – increase
e) thermal – kinetic – increase
A pole vaulter, during his race to cross the obstacle in front of him, transforms his kinetic energy into potential energy due to the increase in height and consequently to the decrease in his speed. |
- How do you make a 20% NaOH solution?
- What does 0.1 m NaOH mean?
- What is 0.1 NaOH?
- What is 0.1m solution?
- What is the formula of normality?
- How do you make a 0.5 M solution?
- How do you calculate the normality of NaOH?
- How do I prepare 10 m NaOH?
- How do you make 250ml of 0.1 m NaOH?
- How do you make NaOH solution from pellets?
- How do you make 0.2 m NaOH?
- What is N 10 NaOH?
- What is the pH of 0.1 N NaOH?
- What is 0.5 N NaOH?
- What is a 5% solution?
- What is the pH of 0.5 m NaOH?
- How we can prepare 0.1 m NaOH solution?
- How can we prepare 0.1 N NaOH in 100 ml?
How do you make a 20% NaOH solution?
20% NaOH (W/V), means 20 grams of NaOH solids, dissolved in distilled water until you have 100mL of aqueous solution.
Do this in beaker slowly, and drip water until it reaches exactly 100 mL mark.
20% NaOH(W/W) means 20 grams of NaOH solids plus 80 grams of distilled water, making up 100 grams of solution..
What does 0.1 m NaOH mean?
One approach is to prepare the solution volumetrically using sodium hydroxide pellets. If, for example you are preparing 1 liter of 0.1 M NaOH you would add 0.1 moles of NaOH (0.1 X 40.00 g/mole or 4.00 g of NaOH) to a 1 liter volumetric flask, add deionized water until near the fill line, stopper and mix.
What is 0.1 NaOH?
To prepare a 0.1M NaOH solution. – Dilute a standardised 1.0 M NaOH solution by a factor of 10. or, Dilute a non standardised 1.0 M NaOH solution by a factor of 10 and then standardise. or, dissolve 4 gm NaOH in 1 L of distilled water (a less accurate option also requiring standardisation)
What is 0.1m solution?
A 0.1 M solution contains one mole of solute per liter of solution. … When you’re preparing a volume other than one liter and the amount of solute isn’t as obvious, it can be calculated using moles = molarity x volume (in liters).
What is the formula of normality?
Normality is defined as the number of equivalent weights (or simply equivalents, eq) of solute dissolved per liter of solution (equivalents/L = N) (Equation 1). Normality is used in place of molarity because often 1 mole of acid does not neutralize 1 mole of base.
How do you make a 0.5 M solution?
If a different molarity is required, then multiply that number times the molar mass of NaCl. For example, if you wanted a 0.5 M solution, you would use 0.5 x 58.44 g/mol of NaCl in 1 L of solution or 29.22 g of NaCl.
How do you calculate the normality of NaOH?
Once gram equivalent weight is understood, it is easier to understand the equation for normality:Normality (N) = m /V × 1 / Eq.Normality (N) = m /V × 1 / Eq.m = 1 eq/L × 1 L × 40.00 g/eq ; therefore m = 40 g.
How do I prepare 10 m NaOH?
Add 40 g NaOH to a suitable container. Add distilled water to make solution up to 100ml.
How do you make 250ml of 0.1 m NaOH?
To make 250 ml of 0.1 M NaOH, you dissolve 1 gram NaOH in enough water to make a final volume of 250 mls.
How do you make NaOH solution from pellets?
To 800 mL of H2O, slowly add 400 g of NaOH pellets, stirring continuously. As an added precaution, place the beaker on ice. When the pellets have dissolved completely, adjust the volume to 1 L with H2O. Store the solution in a plastic container at room temperature.
How do you make 0.2 m NaOH?
Use sodium hydroxide’s molar mass to determine how many grams would contain this many moles. If you add 0.64 g of solid sodium hydroxide to the stock solution you’ll get a 0.2-M solution of (approximately) the same volume.
What is N 10 NaOH?
N means normality, one normal of NaOH solution contains 23+16+1=40 grams. ( gram molar mass) of NaOH therefore, N/10 equals 40/10=4 grams. So, add four grams of sodium hydroxide to one litre of water then the N/10 NaOH solution is prepared.
What is the pH of 0.1 N NaOH?
Some common bases as sodium hydroxide, ammonia and moreBaseNormalitypHSodium hydroxide (caustic soda)N14.0Sodium hydroxide0.1 N13.0Sodium hydroxide0.01 N12.0Sodium metasilicate0.1 N12.623 more rows
What is 0.5 N NaOH?
A 0.5 M NaOH is then 20 g NaOH/L. So dissolve 20 g NaOH in 500 ml water let it cool down to room temperature and fill it up to one liter. For accurate results you can use an analytical balance and a volumetric flask.
What is a 5% solution?
4% w / v solution means 4 grams of solute is dissolved in 100 ml of solution. … 5% v / v solution means 5 ml of solute is dissolved 100 ml of solution.
What is the pH of 0.5 m NaOH?
13.5PropertiesRelated CategoriesAnalytical Reagents, Analytical/Chromatography, Titration, Volumetric Titration Reagentsconcentration0.5 Mapplication(s)titration: suitablepH13.5 (20 °C in H2O)density1.02 g/cm3 at 20 °C9 more rows
How we can prepare 0.1 m NaOH solution?
So the equivalent weight of NaOH is 40. To make 1 N solution, dissolve 40.00 g of sodium hydroxide in water to make volume 1 liter. For a 0.1 N solution (used for wine analysis) 4.00 g of NaOH per liter is needed.
How can we prepare 0.1 N NaOH in 100 ml?
To make 0.1N NaOH solution = dissolve 40 grams of NaOH in 1L of water. For 100 ml of water = (4/1000) × 100 = 0.4 g of NaOH. Thus, the amount of NaOH required to prepare 100ml of 0.1N NaOH solution is 0.4 g of NaOH. |
Topology Of Summary
If X is a set then T = P(X) is called the discrete topology on X The Bus topology is characterized by the hard cabling (commonly called backbone) in which the different sites on the network are connected by high capacity network cabling. Multiple stars involve either a series or tertiary nodes attached to two or more secondary nodes, which are attached to the tree's primary trunk node. Some things never disappear—violence, for example. Use the show ip eigrp topology command to determine Diffusing Update Algorithm (DUAL) states and to debug possible DUAL problems When you use the show ip eigrp topology command without any keywords or arguments, Cisco NX-OS displays only routes that are feasible successors Classical Topology and Combinatorial Group Theory New York, NY: Springer-Verlag, 1980. Also give the idea of homotopy equivalences and. Contents & Summary Definition Examples Type of topology Closure of a set Interior of a set Exterior of a set Boundary (Frontier) of a set Neighbour Sub-space Poplack Shana 2004 Code Switching Essay Base for a topology Sub-base Neighbour base Continuous function Homeomorphism First & second countable space Dense set Separable space. Network topology describes the different ways in which the components of Wide Area Networks are arranged. Review the product-specific topologies and the description of the topologies, including the virtual servers required and the summary of clusters and Managed Servers recommended for the product-specific deployment A network topology refers to the way in which nodes in a network are connected to one another. Most of the time proofs will not be given Topology (electrical circuits) From Wikipedia, the free encyclopedia The topology of an electronic circuit is the form taken by the network of interconnections of the circuit components. These topologies are star topology, bus topology, mesh topology, star bus topology etc. Click the Errors tab The TOPOLOGY provides the summary information output of user-selectedcriteria for the control blocks representing node records and TG records. Click the Errors tab. Generally, it denotes the interrelated model Case Study File Format of network components. Vodafone A Giant Global Erp Implementation Case Study Answers
How To Cite Thesis In Harvard
Topology, branch of mathematics, sometimes referred to as “rubber sheet geometry,” in Ankara Mimarlar Odas? Cv Gentleman which two objects are considered equivalent if they can be continuously deformed into one another through such motions in space as bending, twisting, stretching, and shrinking while disallowing tearing apart or gluing together parts. Types of physical topologies. The idea is that if one geometric object can be continuously transformed into another, then the two objects are to be viewed as beingtopologicallythe same. A Feature Dataset can be considered an association of Feature Classes, but a Feature Dataset itself has spatial properties such as a spatial reference system and XY domain Summary In most major universities one of the three or four basic first-year graduate mathematics courses is algebraic topology. For example, a circle and a square Asp Net Resume Upload are topologically equivalent A Basic Course in Algebraic Topology. COSNAME. Show transcribed image text. We start by looking at some arbitrary collection of points. (English. A topology defines the physical layout or a design of a network.
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Thesis Of Mphil Education It is also referred to as network architecture Topology and Data. 2.Any union of members in ˝belong in ˝. Lesson Summary. It also shows, how does data transmission happen between these nodes? Topology describes, what are the different manner nodes are positioned and unified with each other. They have been used for pattern recognition [35,2], computer. The main topics of interest in topology are the properties that remain unchanged by …. Feb 06, 2019 · A tree topology includes multiple star topologies, which involve a variety of single nodes connected to a central node. This introductory text is suitable for use in a course on the subject or for self-study, featuring broad coverage and a readable exposition, with many examples How To Write A Chain Letter and exercises.. The amount of algebraic topology a student of topology must learn can beintimidating Understand the specific reference topology for the products that you plan to deploy. Raheel Ahmad Pages 87 pages Format Mobile Scanned PDF Size 8.0 MB Contents & Summary. OSPF, unlike EIGRP, doesn’t support automatic summarization.
They provide an extensible framework to reconstruct shapes. Operands. (English. Topology is an incredibly useful and important branch of mathematics. 2 a (1) : a branch of mathematics concerned with those properties of geometric configurations (such as point sets) which are unaltered by elastic deformations (such as a stretching or a twisting) that are homeomorphisms Network Topology is a blueprint of the network describing the way different computers/devices are interconnected and how they communicate among themselves. This immediately puts us into the mathematical realm of set theory Contents & Summary Regular space Completely regular space Compactness in topological spaces Homeomorphism Countably compact space Bolzano Weierstrass property Lebesgue number; Big set; Lebesgue covery lemma ε− ε − net; Totally bounded Connected spaces; Disconnected Component Totally disconnected. Abstract Network topology inference is a prominent problem in network science. In Topology of Violence, the philosopher Byung-Chul Han considers the shift. 2.Any union of members in ˝belong in ˝. SES # TOPICS SUMMARY AND REFERENCES; 1: Organizational meeting: The first meeting of class. Summary. A network topology is all the combined physical, logical, or virtual components that make up the network or network segment. Also give the idea of homotopy equivalences and. Here we tell you about the same show-ip-eigrp-topology. These topologies are star topology, bus topology, mesh topology, star bus topology etc. |
Log Transformation Spss
SPSS for newbies: log transformation video for Data & Analytics is made by best teachers who have written some of the best books of Data & Analytics. Jun 24, 2015 · This video demonstrates how to transform data that are positively or negatively skewed using SPSS. the log transform. Log transformation of values that include 0 (zero) for statistical analyses? I just say that it might be valuable to ask why there are zeros at all and if the log-transformation would then be. Our Statistical Test Selector helps you to select the correct statistical tests to analyse your data, before our step-by-step SPSS Statistics guides show you how to carry out these statistical tests using SPSS Statistics, as well as interpret and write up your results. Log transformation(s). Vanliga transformationer inklusive syntax i SPSS för dessa redogörs härnäst. Compute a log transformation of a variable. Most users typically open up an SPSS data file in the data editor, and then select items from the menus to manipulate the data or to perform statistical analyses. In the context of logistic regression, it’s worth noting a couple of points about the assump-. Because SPSS Forecasting automates the modeling of thousands of variables, the. It can not be applied to zero or negative values. This workshop provides a brief introduction to the basics of SPSS. Die Formel in SPSS dafür wäre 1/VARIABLE, wobei VARIABLE durch den Namen der Variablen ersetzt werden muss. , ^2, & log). The log transformation, a widely used method to address skewed data, is one of the most popular transformations used in biomedical and psychosocial research. It is used as a transformation to normality and as a variance stabilizing transformation. Written and illustrated tutorials for the statistical software SPSS. Behöver du hjälp med statistisk analys eller SPSS? Klicka här för mer information. If I have highly skewed positive data I often take logs. (2001) 10 9 8 7 6 5 4 3 2 1 0. So, that is how to use SPSS to create Z-scores very quickly. To calculate log return, you must first find the initial value of the stock and the current value of the stock. Then compute into different variable: percentile = score/topscore * 100 I suppose it would be more elegant to use transform, but this is dirty, quick, and should work. Data Handling Using SPSS 19 Research Data MANTRA the software (ie what file formats can it read)?, and (e) most importantly, how can I get my data out of that software (along with any transformations, computations etc) so that I can read it into other software for other analyses, or store it in a software-neutral format for the longer term?. Main SPSS Tranformation Commands. MIXED: Multilevel Modeling. As of version 11. FIGURE 7: It turns out that SAS, PROC IML, is not included in the SAS Learn-ing Edition, which is what many of my students are using, so I had a need to modify the macro to work without PROC IML. Candidate at UC Santa Cruz. Now if your intuition leads you to. The molecular components largely responsible for muscle attributes such as passive tension development (titin and collagen), active tension development (myosin heavy chai. statisticsmentor. So when you key in: log 1000 the calculator displays 3. Plus, get practice tests, quizzes, and personalized coaching to help you succeed. To transform the value to the R "Date" format, you simply need to turn seconds into days and specify the origin. MANO HARITHA R. (Note that the natural log button on your calculator is probably labeled ln. It can get rid of skewness Before log-transformation After log-transformation 0 2 4 6 8 10 0. This value will then be the mean of the variables entered into the function. 2001, 4 th ed. Most people find it difficult to accept the idea of transforming data. 2 Recoding Variables Transform Recode into Different Variables Select the variable you want to recode (e. Use of arcsine transformation is a convenient way of carrying out. Examining the means for untransformed scores is the same as examining the medians for transformed scores; the transformation affects the mean but not the median because the median only depends on rank order. Vanliga transformationer inklusive syntax i SPSS för dessa redogörs härnäst. It is one case of the class of transformations generally referred to as Power Transformations designed to uncouple dependence between the expect value and the variability. To make adjustments to the resulting boxplots, double-click the graph displayed in the output window. drug is plotted (on the X axis) as the log of drug concentration and response is plotted (on the Y axis) using a linear scale. Transformation was obtained by entering LOG(FSH) instead of FSH in the dialog box. Minitab determines an optimal power transformation. Poisson regression is used to predict a dependent variable that consists of "count data" given one or more independent variables. SPSS will create a new column with the transformed variable. Use of arcsine transformation is a convenient way of carrying out. Before using any of these transformations, determine which transformations, if any, are commonly used in your field of research. Please send your request to 5555 IT Service Desk. The technique is used for both hypothesis testing and model building. Form a new variable that is X times Z s. setSamplingCount(20000). you conclude about the effect of log transformation on the data? 4. It is similar in shape to the log-normal distribution but has heavier tails. Caranya: silahkan kirim pesan WhatsApp ke: 08816050259 (Nomor WA ini khusus untuk menggunakan Jasa Bantuan Olah Data dan Analisa Statistik). If the data shows outliers at the high end, a logarithmic transformation can sometimes help. This clips is about how to use log transformation in creating normal data distribution on spss. The usual process involves converting documents, but data conversions sometimes involve the conversion of a program from one computer language to. Log Transformation for Better Fits In log transformation you use natural logs of the values of the variable in your analyses, rather than the original raw values. Enter a value for K on the dialog. Like so, you can round to the nearest quarter point (second example below) or tenfold (third example). Linear transformation, sometimes called linear mapping, is a special case of a vector transformation. You can also use these transformations for percentages if. Regress Y on X, the Z s, and the product term, and the t-test for the X coefficient will be the t-test conducted by hand above. Such transformations are simple in R and assume a form that should be very familiar to you by now:. If it adjusts the data automatically, logit will print a warning message. Hi after Ps matching in spss I have the following problems: 1. I searched for log-transformation for negative values, and found this code. This plot shows that none of the proposed transformations offers an improvement over using the raw predictor variable. One commonly used transformation is a log transformation, so let's try that. For the Box-Cox transformation, a λ value of 1 is equivalent to using the original data. These data may come from basically any imaginable source: a customer database, scientific research, or even the server log files of a website. • Use a non-parametric test: Non-parametric tests are often called distribution free tests and can be used instead of their parametric equivalent. datapreparation. Series Transformations. Some common lambda values, the transformation equation and resulting transformed value assuming Y = 4 are in Table 2. smoking: never smoker, ex-smoker, current smoker) predicts higher odds of the dependent variable (e. (Don't just make something up!) For example, if the data are strictly positive, the log transformation is an option. arcsine in R on a dataset ranging from $-1$ to $1$, NaNs are produced because of the square-rooting of a negative number. The Natural Log Transformation We will use log(Y) to denote the natural logarithm of a number Y. In this guide, we introduce some of the transformations and methods for standardisation commonly in use, and when to use them. Log-linear analysis is different from logistic regression in three ways: 1. A quick google search for "How to fix non-normally distributed data" revealed the Box Cox Transformation. IBM SPSS software is avaialable for a free trial from IBM website for fourteen days. This occurs because, as shown below, the anti-log of the arithmetic mean of log-transformed values is the geometric mean. - Imagine a study about happiness where your happiness question (or composite) ranges from 1 to 7. To transform the value to the R "Date" format, you simply need to turn seconds into days and specify the origin. Read this post to find out how. Logarithmic transformations are actually a class of transformations, rather than a single transformation. 2 Recoding Variables Transform Recode into Different Variables Select the variable you want to recode (e. Any built-in statistical procedure or transformation function can be called via sytax. Featured Text Transformation free downloads and reviews. Version info: Code for this page was tested in SPSS 20. It would take some doing to create new variables for the dozens or so that I’d need to transform. FIGURE 7: It turns out that SAS, PROC IML, is not included in the SAS Learn-ing Edition, which is what many of my students are using, so I had a need to modify the macro to work without PROC IML. If you change the response, such as through a log transformation of the response as in the second question of Assignment #1, you need to be careful. Older versions of SPSS (e. Do Not Log-Transform Count Data, Bitches! Posted on June 17, 2010 by jebyrnes OK, so, the title of this article is actually Do not log-transform count data , but, as @ascidacea mentioned, you just can't resist adding the "bitches" to the end. Aug 26, 2015 · When psychology researchers switch from SPSS to R a common first question is "Can I load SPSS data in R?". Mar 14, 2017 · So one cannot measure the true effect if there are multiple dependent variables. These transformations are what you should first use. We propose new tests based on the arcsine transformation, which stabilizes the variance of binomial random variables. dividing each observation by the geometric mean where geometric (rather than arithmetic) mean = (y 1*y 2*…y n) 1/n = exp1/nLn(y 1* y 2 …y n). There are an infinite number of transformations you could use, but it is better to use a transformation that other researchers commonly use in your field, such as the square-root transformation for count data or the log transformation for size data. Log transformation: select this option if the dose variable requires a logarithmic transformation. SPSS will return a value equal to Pi/4 for ARTAN(1), so 45/ARTAN(1) equals 180/(4*ARTAN(1)), which equals 180/Pi. Also, note that when a transformation is done, SPSS does not perform the MANOVA or individual ANOVAs on the original variables. To transform the value to the R "Date" format, you simply need to turn seconds into days and specify the origin. A list of the most popular transformations: - If SD is proportional to the mean, then a log transformation will improve both HOV & normality (distributions tipically log-normal, positively skewed). You can be confident that you’ll always have the analytic tools you need to get the job done quickly and effectively. Dates in SPSS are recorded in seconds since October 14, 1582, the date of the beginning of the Julian calendar. Log transforming data usually has the effect of spreading out clumps of data and bringing together spread-out data. Actually, to do them sort of correctly would require you to do some mathematical derivations. The newest version of SPSS is SPSS Today we will be working on SPSS 16. The point of this article is that the log transformation can help you to visualize data that span several orders of magnitudes. 235 * age 11 score. Some transformation options are offered below. Transformation was obtained by entering LOG(FSH) instead of FSH in the dialog box. These data are found in Question #1 of Practice Problems #5. Question: How do I perform a Box-Tidwell regression on my 10 continuous variables in SPSS?. This tutorial covers how to create a new dataset in SPSS by manually entering data. For example, if. Dear statalisters, I am using linear regression to investigate factors influencing my right skewed dependent variable. Unfortunately, these methods are usually inappropri-. The most common situation is for the variance to be proportional to the square of the mean (i. , least complex) model that best accounts for the variance in the observed. Cambridge University Press, New York. , you may want to change. SPSS (Statistical Product and Service Solutions). Expand the capabilities of IBM® SPSS® Statistics Base for the data analysis stage in the analytical process. (Note that the natural log button on your calculator is probably labeled ln. Cara Compute Transformasi Logaritma Pada SPSS adalah: Klik Menu, Transform, Compute Variabel, Pada Target Variabel Beri Nama Misal "Transform" dan Pada Kotak Numeric Expression isi dengan: Lg10(Variabel Asli). Monetary amounts—incomes, customer value, account or purchase sizes—are some of the most commonly encountered sources of skewed distributions in data science applications. Regress Y on X, the Z s, and the product term, and the t-test for the X coefficient will be the t-test conducted by hand above. Mar 18, 2019 · Logarithms (frequently referred to as ‘logs’) are often used in statistics. datapreparation. The "z" in Fisher Z stands for a z-score. The screenshot shows the computing of particular variables S1Q1 to S1Q18 by adding all and divide by the total of selected variable to get a new variable. In this guide, I will show you how to log (log10) transform data in SPSS. Each window is defined and the appropriate applications are explained and demonstrated. tidwell, which is normally just printed. Name your target variable something like 'p_logit' and in the numeric expression box type: LN(p / (1 - p) ) Next click OK. The logarithm, x to log base 10 of x, or x to log base e of x (ln x), or x to log base 2 of x, is a strong transformation with a major effect on distribution shape. To transform the value to the R "Date" format, you simply need to turn seconds into days and specify the origin. Log transformation; Imagine you conduct a t-test using IBM SPSS and the output reveals that Levene's test for equality of. In logistic regression, the dependent variable is a logit, which is the natural log of the odds, that is, So a logit is a log of odds and odds are a function of P, the probability of a 1. SPSS Tutorials - Master SPSS fast and get things done the right way. >Command line: 25 Current case: 32 Current splitfile group: 1 >Warning # 601 >The argument for the log base 10 function is less than or equal to zero on the >indicated command. 387] notes that this can be guaranteed by using a transformation like log(X+k) where k is a positive scalar chosen to ensure positive values. Furthermore, we assume that (c) the sample of data we are working with has been drawn randomly; and (d) the underlying relationship between Y and its predictor(s)—that is, the independent variable(s)—is linear (or can be “linearized” with a transformation, e. kami beritahukan bahwa sebenarnya caranya tidak jauh berbeda, namun akan tetap kami jelaskan langkah-langkah pada artikel ini. In linear regression, box-cox transformation is widely used to transform target variable so that linearity and normality assumptions can be met. Apabila data asli ada di Cell A4 maka rumusnya =Log(A4). Jan 16, 2017 · 1) saat melakukan transform log, ada beberapa data saya yg hilang (yaitu 1 pada X2, dan 1 pada X3). Hello SIr, i am implementing a log transfromation on OLS regressioni. 24 68 0 20 40 60 80 100 Log(Expenses) 3 Interpreting coefficients in logarithmically models with logarithmic transformations 3. , should I have to invert the results of predictions using the inverse function (log/exp) ? Thanks much in advance. (The logarithm base does not matter--all log functions are same up to linear scaling--although the natural log is usually preferred because small changes in the natural log are equivalent to percentage changes. You can use the model to gain evidence that that the model is valid by seeing whether the predictions obtained match with data for which you already know the correct values. So far, we've used SPSS to develop a basic idea about how SPSS for Windows works. You can write results that are obtained from R into a new SPSS database for further manipulation in SPSS. Any given number can be expressed as y to the x power in an infinite number of ways. ) Note that the regression line always goes through the mean X, Y. The log transformation is a relatively strong transformation. Log transforming data usually has the effect of spreading out clumps of data and bringing together spread-out data. Logistic regression on spss 3 classification table a. Each window is defined and the appropriate applications are explained and demonstrated. To use the log of a dependent variable in a regression analysis, first create the log transformation using the COMPUTE command and the LN() function. Here are two versions of the same basic model equation for count data: ln(μ) = β 0 + β 1 X. tidwell, which is normally just printed. Tra nsformasi data digunakan pada saat dimana data hasil uji normalitas tidak normal. See the references at the end of this handout for a more complete discussion of data transformation. Any built-in statistical procedure or transformation function can be called via sytax. • Generalized Linear Mixed Models – Create more accurate models for predicting non-linear outcomes in the Advanced Statistics module • Faster Performance - For compiled transformations in IBM SPSS Statistics Server and up to 200% performance gain for generating pivot tables in IBM SPSS Statistics Base. Beyond the t-test The t-test compares two groups based on an assumption of normality, but what if data are not normally distributed or if we want to compare three or more groups?. Any given number can be expressed as y to the x power in an infinite number of ways. The logit function is particularly popular because, believe it or not, its results are relatively easy to interpret. The comparison of the means of log-transformed data is actually a comparison of geometric means. The numeric expression box is where you type the transformation expression, ln(x). That data is positively skewed, and a natural log transformed data fit a "linear" growth model (I recognize that this is no longer linear after the transformation). Cox power transformation of. The table below gives an overview of SPSS' main tranformation commands. many rows to the transformation matrix as there are dependent variables to be transformed. Featured Text Transformation free downloads and reviews. If it’s too large, you could waste valuable time and resources. In both these uses, models are tested to find the most parsimonious (i. This is the Z-scores of the Scores variable we started with. Learn Econometrics for free. Nov 26, 2007 · log to the base 10 of 1000 = 3. That data is positively skewed, and a natural log transformed data fit a "linear" growth model (I recognize that this is no longer linear after the transformation). You could now use the log transformed reflected speed scores in an analysis that assumes normal distributions. After running the MEAN compute function in SPSS, the new variable should be visible in the data sheet. Note: The @FIELD function is an important tool for deriving multiple fields at the same time. 66) does not include 1, so a transformation is appropriate. inventory forecasts three months out for each product. Check the data for extreme outliers. For example, you might want to normalize the field AGE so that you can use a scoring technique (such as logistic regression or discriminant analysis) that assumes a normal distribution. SPSS RND Function. Open your SPSS program as a PC word processing file or a text file. rescaling, reflection, rotation, or translation) of matrices to compare the sets of data. , should I have to invert the results of predictions using the inverse function (log/exp) ? Thanks much in advance. Transform command log. The most commonly used transformation is the natural log transformation, which is often applied when much of the data cluster near zero relative to larger values in the dataset and all observations are positive. English In this lecture we are going to learn how to download and install IBM SPSS software. Concepts such as log10 transformation, determining skewness, reflection, adjusting for zeros, and. " It sometimes appears when installing a MSI while using the commandline property TRANSFORMS="MYMSI. Logit functions by taking the log of the odds: logit(P) = log P/ (1-P). pptx), PDF File (. Assign value labels to a variable: 1. Richard picked an example with heights where no (standard) transformation would be very nonlinear. We’ll cover a few of the most important and common ones here, but there are many others. You need to do a number of things to set up this dialog box so SPSS will generate random numbers. Business Analytics IBM Software IBM SPSS SamplePower Compare and save research options Use SamplePower’s unique sensitivity analyses to adjust the effect size, desired power and alpha, and see the impact on the required sample size. Statistical visions in time: a history of time series analysis, 1662-1938. Data can be easily transformed by using the Transform - Compute Variable command. Open your SPSS program as a PC word processing file or a text file. e Log transformation on multiple regression. The histogram of the log transformed variable is shown in Figure 7. Cox Regression. less than 10), use the transformation Log(Y+1) instead of Log Y (Y is the original data). Loglinear Regression In loglinear regression analysis is used to describe the pattern of data in a contingency table. Here, drug is the independent variable (often called a “between subjects factor” in repeated measures) and the four dependent variables are time0, time30, time60, and time120. Be aware that my approach is probably different from what you'll see elsewhere. Analyzing the log-transformed data with one-way anova, the result is F 6,76 =11. Log transformation means taking a data set and taking the natural logarithm of variables. It can not be applied to zero or negative values. I work on my thesis and use SPSS to analyze the data. In the study of sclerotic glomerili in human kidneys, large proportions are considered to be the same, as seen on average within 200 kidneys. some one suggest me to transform the DVs only to normal distribution using Box-Cox conversion (present in stata)…I am only familiar with SPSS…. Jul 09, 2014 · The point of this article is that the log transformation can help you to visualize data that span several orders of magnitudes. Minitab will select the best mathematical function for this data transformation. Lecture 3: Multiple Regression Prof. Compute the skewness of Sp_Ref and you will find that it has exactly the same amount of skewness as did Speed but in a positive rather than a negative direction. Transformation Matrices. Discovering Statistics Using IBM SPSS Statistics 20. Engineering & Technology; Computer Science; IBM SPSS Statistics Performance Best Practices. So, that is how to use SPSS to create Z-scores very quickly. The Box-Cox transformation of the variable x is also indexed by λ, and is defined as (Equation 1) At first glance, although the formula in Equation (1) is a scaled version of the Tukey transformation x λ, this transformation does not appear to be the same as the Tukey formula in Equation (2). The Chart Editor displays, which includes many options for customizing a graph. Read this post to find out how. (1975) provide research specific rational for its use. In this case, g 1 is also an increasing function. Buka spss 2. of Psychology 4600 Sunset Ave. If, for example, you apply a Ln (natural log) transformation to numeric variables the following code is generated and put in the Transform command log window at the bottom of your screen when you click the Store button. Model-Fitting with Linear Regression: Exponential Functions In class we have seen how least squares regression is used to approximate the linear mathematical function that describes the relationship between a dependent and an independent variable by minimizing the variation on the y axis. Data Transforms: Natural Log and Square Roots 1 Data Transforms: Natural Logarithms and Square Roots Parametric statistics in general are more powerful than non-parametric statistics as the former are based on ratio level data (real values) whereas the latter are based on ranked or ordinal level data. But in some contexts, one may transform to obtain a test statistic that has an. Stack Exchange network consists of 175 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Linear regression is used to specify the nature of the relation between two variables. less than 10), use the transformation Log(Y+1) instead of Log Y (Y is the original data). Mar 14, 2017 · So one cannot measure the true effect if there are multiple dependent variables. Is there anything I can add to the command to make it not reboot?. The table below gives an overview of SPSS' main tranformation commands. 14 shows the transformations that are available when you select For proportions from the Family list. data transformation can be done by using log, square root or arcsine transformation. Use the RECODE command to change the values of a particular variable. Examples of a nonlinear transformation of variable y would include taking the logarithm of y (y'=log(y)), or the square root of y (y'=√y). Log transforming data usually has the effect of spreading out clumps of data and bringing together spread-out data. log-em, square-em, square-root-em, or even use the all-encompassing Box-Cox transformation , and voilla: you get variables that are "better behaved". A transformation is used to cancel out this relationship and thus make the mean independent of the variance. , see this search). In this guide, we introduce some of the transformations and methods for standardisation that are commonly in use and when to use them. The workshop is tailored toward Mass Appraisal and Assessment. (Note that the natural log button on your calculator is probably labeled ln. If an easy transformation doesn't work (log. SPSS will sort the string values from highest to lowest (or vice versa) and then assign consecutive numbers to the values. i know that it was supposed to be positive because i put the non transformed data into the linear regression and. In SPSS GRAPHS choose SCATTERPLOT/ SIMPLE and enter two variables in the X and Y axes. The base of the logarithm isn't critical, and e is a common base. You can use the following SPSS predictive analytics algorithms in your notebooks. In SPSS, this type of transform is called recoding. The Fisher Z-Transformation is a way to transform the sampling distribution of Pearson's r (i. R] In this experiment, the effect of vitamin supplements on weight gain is being investigated in three animal species (mice, chickens, and sheep). It is crucial to setup the model to predict the probability of an event, not the absence of the event. What format is a SPSS. Computing Transformations in SPSS - Free download as Powerpoint Presentation (. Due to its ease of use and popularity, the log transformation is included in most major statistical software packages including SAS, Splus and SPSS. Every year, selectors miss-guess and select students who are unsuccessful in their efforts to finish the degree. Binomial Logistic Regression using SPSS Statistics Introduction. 2 Recoding Variables Transform Recode into Different Variables Select the variable you want to recode (e. Logistic regression, also called a logit model, is used to model dichotomous outcome variables. You may need to transform some of your input variables to better meet these assumptions. For example, instead of computing square roots, compute squares, or instead of finding a log, exponentiate Y. Starting with version 16, IBM SPSS provides a free plug-in that enables you to run R syntax from within SPSS. You can analyze repeated measures data using various approaches, such as repeated measures ANOVA/GLM (the multilevel model) or the linear mixed model. MANO HARITHA R. of Psychology 4600 Sunset Ave. The natural log transformation is used to correct heterogeneous variance in some cases, and when the data exhibit curvature between Y and X of a certain type. Discover why more than 10 million students and educators use Course Hero. Read this post to find out how. "There are a growing number of books on the market focusing on the history, symptomatology, and treatment of trauma and PTSD. Jul 21, 2007 · I'm now thinking off the top of my head as to how you would do this in SPSS. cara transformasi data yang tidak berdistribusi normal dengan Ln pakai spss ini adalah lanjutan dari artikel transform dengan Log. 2001, 4 th ed. The best way to read any proprietary data into R is to open the data in its original program and export it as a. Discovering Statistics Using IBM SPSS Statistics 20. Any built-in statistical procedure or transformation function can be called via sytax. The reverse of this is to find the anti-log of 3. What format is a SPSS. Thus, our comments apply to all deterministic rank-based INTs. U9611 Spring 2005 2 Outline before log transformation. Based on this limitation, Johnson (1995) preferred robust procedures and data transformation to non-parametric tests (Robust procedures and data transformation will be introduced in the next section). To use the log of a dependent variable in a regression analysis, first create the log transformation using the COMPUTE command and the LN() function. and many more programs are available for instant and free download. SPSS gives only correlation between continuous variables. So far, we've used SPSS to develop a basic idea about how SPSS for Windows works. IBM SPSS Statistics Student Grad Pack V26 delivers the core capabilities students need to complete the analytical process, from beginning to end. Any given number can be expressed as y to the x power in an infinite number of ways. If you are confused about how to open SPSS, ask your instructor or see Appendix 1, where we provide more information about accessing SPSS from your hard drive or from a network. Double-check that these outliers have been coded correctly. In the Data Preview area, SPSS displays a preview of how your data will appear in SPSS once the import is complete. Option Box-Cox transformation: select this option to use the Box-Cox power transformation as described above. transformation that will reduce negative skewness. Loglinear Regression In loglinear regression analysis is used to describe the pattern of data in a contingency table. 66) does not include 1, so a transformation is appropriate. Jul 10, 2019 · The advantages and disadvantages of the E-views are considered by comparing some of the commonly used statistical packages like STATA and SPSS. I'm still having trouble exporting the data: I've tried all the earlier mentioned solutions (see below) but it still doesn't work properly and I have not managed to figure out myself what the cause is. Hello all, I want to transform one variable, which is skewed, to a normal distribution. Use of arcsine transformation is a convenient way of carrying out. But what should I do with highly skewed non-negative data that include zeros? I have seen two transformations used: $\log(x+1)$ which has the. This command offers a number of useful functions (some of them are documented below). SPSS is a statistical analysis program that is used in a variety of fields, from market researchers to government agencies. This is the Z-scores of the Scores variable we started with. (And yes, that makes me feel old). Test of normality and data transformation in SPSS The video also shows the user how to log transform their data and then test whether this increases the degree to which these data approximate. Transformations for Proportion Variables. If you are confused about how to open SPSS, ask your instructor or see Appendix 1, where we provide more information about accessing SPSS from your hard drive or from a network. If it’s too large, you could waste valuable time and resources. Logarithms, log, ln, lg, properties of logarithms Logarithm base: log 2 = Graphs of logarithmic functions. Programs like SPSS, however, often use wide-formatted data. This is particularly important for ARIMA models, which require series to be stationary before models are estimated. The most commonly used transformation is the natural log transformation, which is often applied when much of the data cluster near zero relative to larger values in the dataset and all observations are positive. Understanding log transformation is best seen with an example. If you are new to this module start at the overview and work through section by section using the 'Next' and. MANO HARITHA R. |
|Magnetic Field from a Cylinder's Magnetization|
It is not intuitively obvious how an object's magnetization relates to the magnetic field it generates. To develop a better physical picture of this phenomena consider an object with a given magnetization and then solve for the resulting magnetic field, B.
A cylinder of radius, R, and length, L, in which R << L (i.e. the cylinder is so long that we will not be concerned with edge effects) is known to have a magnetization, M, given by M = α r2 Φ, where α is a constant. Figure 1 shows this cylinder. Note that the drawing does not accurately reflect that the cylinder is much longer than it is wide.
The goal is to determine the magnetic field of this object. In general, magnetic fields are determined according to the kind of source we expect. Currents are one known source for magnetic field, while time dependent electric fields are another. In this case, the magnetization of an object is known to be related to currents (bound currents to be precise) and there is no mention of any time dependence so it is sensible to proceed with the intention of solving for the magnetic field using currents as a source. Once the currents are known, the magnetic field may be calculated using Ampere's law,
where the integral is taken over the entire closed loop of length, l, μo = 4π × 10-7 H m-1 is the permeability of free space, and Ienc is the current enclosed by the loop, l.
The magnetization depends on the radial coordinate of the cylinder. At the position, r = 0, the magnetization is zero. Magnetization is a property of the material, so it is also equal to zero outside of the cylinder (if there is no cylinder, then there cannot be any magnetization). We should now realize that the currents must be determined in two different regions; inside the cylinder and outside of the cylinder. Building on this intuition we should also expect that the magnetic field must be determined as a function of space both inside and outside of the cylinder.
Magnetization is related to bound currents. Since the cylinder has a magnetization throughout its interior, both a bound volume current density and a bound surface current density must be examined. The bound volume current density, Jb, is related to the magnetization by, Jb = ∇ × M. The bound surface current density is related to the magnetization by Kb = M × n, where n is the unit vector that is normal to the surface being investigated (it is equivalent to n-hat as seen in the equations that are presented as images).
Begin by solving for the bound volume current density. In the region outside of the cylinder, r > R, the magnetization is zero and therefore, Jb = 0. Inside the cylinder we have,
Next, move on to the bound surface current. There are three surfaces of the cylinder to evaluate; the tubular surface of length, L, and the two circular faces. The cylinder has been purposefully made very long so that end effects may ignored. We'll consider the bound surface current densities on the end faces anyway, for the sake of further developing our physical intuition with respect to bound current paths. The circular faces have unit normals of ± z, depending on which end we are looking at figure 1 illustrates this, but be sure to understand that the unit normal to a surface is always directed away from the object). At the end face shown in figure 1 the bound surface current density is given by,
and for the other end face the surface current density is along the -r direction (technically, this assumes that the cylinder is centered on the position (r,z) = (0,0) and that we are examining the face located at z = L / 2, which was not mentioned in the beginning because end effects were said to be unimportant). Once again, this does not matter for determining the magnetic field of this object because we have decided to purposely neglect contributions from the end of the cylinder. The reason for doing this is that the solution will be simpler if we concern ourselves with only the central region of the cylinder. It is important to note that even bound currents form closed loops. Figure 2 provides an illustration of how this is accomplished in the present discussion. Notice that on one end face the bound surface current flows outward from the center while at the other face it flows inward towards the center (r = 0). We know that the bound volume current density increases with radial position and always flows along +z. The red lines in figure 2 represent the bound current and show that this current forms closed loops within the cross sectional plane of the cylinder. Now that this has been mentioned, please ignore the effects of these end faces while we continue to examine the magnetic field of this object.
The bound surface current that matters to us is the one along the cylindrical side of this object. This cylindrical surface is located at r = R and the bound surface current density is given by,
All of the current is flowing along the z axis. Neglecting the edge effects means we may ignore the r axis directed currents at the edge faces. This z axis symmetry allows for a direct application of Ampere's law. Drawing a circular loop centered on the z axis (r = 0) will enclose some amount of current flowing along the z axis. As we expand this loop in radius, it will enclose more current. Once the loop is larger than the radius of the cylinder (r > R) there is no more current that can be enclosed. The two Amperian loops that must be drawn to determine the magnetic field everywhere in space are one inside the cylinder and one completely outside.
For the loop taken inside the cylinder, only bound volume current is enclosed. Ampere's law gives,
This is the value of the magnetic field inside the cylinder. The only component of magnetic field inside the cylinder is along the Φ direction. The next step is to increase the size of the Amperian loop until it is larger than the radius of the cylinder. The total enclosed current will include contributions from all of the bound volume current density and the bound surface current located at r = R. The left side of Ampere's law remains the same as that found previously.
Before performing this calculation, a mention of the bound surface current must be made. When dealing with a bound surface current density, the total bound surface current is found by multiplying this density and the length over which it passes. Surface current density has units of Amperes per length ([A] / [m]), so multiplying this by the length across with the current flows results in a value that has units of current ([A]). The current in question is the bound surface current that is flowing along the z axis. This current is passing over a length that is the circumference of the cylinder. Note that this length is determined according to a line that is everywhere perpendicular to the surface current density. In the lines below, lcir is the circumferential length of the cylinder.
Outside the cylinder there is no magnetic field. The magnetic field everywhere in space is given by,
This topic shows that the magnetization of an object is related to the magnetic field the object generates according to the bound currents. Magnetization is a concept that exists to make it easier to describe bound currents in one term. Imagine how difficult it would be to describe the bound currents throughout the cylinder's interior and along its surface. Instead of that, we define the magnetization and then leave it to the interested party to solve for the equivalent bound currents. Once these currents are known, the magnetic field is found using techniques common to the current/magnetic field relationship.
The magnitude of the magnetic field is plotted in figure 3.
This plot was generated with the following IDL code,
r = FINDGEN( 1e3 ) / 1e3 a = r cylr = 600 a[ cylr : 999 ] = 0 plomake, r, a^2, yr = [ 0., 0.4 ], ys = 4, xs = 4 AXIS, XAXIS = 0, XTICKNAME = ['0', ' ', ' ', 'R', ' ', ' '], COLOR = 0, CHARTHICK = 2, CHARSIZE = 2 AXIS, YAXIS = 0, COLOR = 0, CHARSIZE = 2, CHARTHICK = 2, YTICKNAME = ['0', ' ', ' ', ' ', ' '] XYOUTS, 0.2, 0.02, 'Radial Position', COLOR = 0, CHARTHICK = 2, CHARSIZE = 2, /NORMAL PLOTS, [ 0, .7 ], [ .36, .36 ], /DATA, COLOR = 18, THICK = 3 XYOUTS, .72, .35, '!4l!3!Io!N!4a!3R!E2!N', COLOR = 18, CHARTHICK = 2, CHARSIZE = 3, /DATA XYOUTS, .15, .55, '|B!I!4u!3!N|', /NORMAL, CHARSIZE = 4, CHARTHICK = 2, COLOR = 0 END
|Initial Posting: Saturday, 17 March 2007|
|Last Updated: Wednesday, 03 March 2010| |
Electronics World, October 2004, Vol. 110, No. 1822: Pages 38-43 "The Catt Anomaly" by Ian Hickman Page 55 summary of gravity as per http://nigelcook0.tripod.com/ [Broken] Hickman has done a lengthy, nicely written and carefully considered treatment of the Catt Anomaly. Yet he may have missed wood for trees. Hickman rightly and clearly shows how the voltage doubles when the logic step bounces off an open circuit and reflects back into itself, thus making the open circuit transmission line into a charging capacitor. He then says that because the capacitor discharges through a resistor, it's output falls exponentially, whereas the charged transmission line discharged directly gives a square wave pulse. As Catt, Davidson, Walton and Gibson showed indirectly, the discharge of a square wave through a resistor results in partial reflection at the resistor so that the discharge or leakage is dragged out to give the exponential decay as the potential falls due to gradual energy loss. (They actually calculated the reverse situation, of exponential charging of a capacitor, but the same simply happens in reverse for discharging.) The key point Hickman does not focus upon is that the charged object contains reciprocating light speed energy. When you consider isolating an electron from the negative plate of the charged capacitor or charged open circuit transmission line, you find that the electron charge energy is oscillating at light speed. Going towards very small charges, down to an electron, this results in a model of the electron which is based on experimental evidence. D. Di Mario wrote an article in Electronics World, 1993, on a "black hole electron". It is in August 2002 and April 2003 EW proved beyond uncertainty that the electron is composed of self trapped light speed electromagnetic energy, giving it the spin properties and making it contain an energy of mc2 when it is annihilated by combination with a positron. The magnetic moment given by Dirac's equation, modified by Feynman virtual particle self-absorption - an increase of merely 0.1% [from Schwinger's approximation 1 + 1/(2 pi x 137) = 1.00116...] - is explained by it being trapped light speed energy circling in the vacuum and affecting the virtual particles in the surrounding field. The New Scientist's contribution to the reporting of this has been silence. If you pressurise New Scientist to publicise this, they will refuse to be pressurised; if you don't pressurise them, they will do nothing; if you keep improving the arguments they will keep making paranoid rants about needing peer-reviewed stuff and then going silent when HEE/26 (a peer-reviewed IEE paper by Catt co-authored with then 83-year-old Dr Arnold Lynch, student of J.J. Thomson who discovered the electron) is sent to them. While Ivor dismisses my efforts to apply the Catt-Davidson-Walton findings to electron and aether/dielectric as unnecessary, my view is that I have not developed the bells and whistles far enough rather than that I have gone into over production. If we take the fabric of space with its fixed 377 ohm impedance, it is in contrast to air where the sound impedance to wave energy transfer is measured in decibels per metre. The photon fired in space suffers no cumulative retardation with distance travelled, and the inverse square law only comes on the scene statistically when you are firing photons in all directions over a diverging spherical area, instead of following the fortunes of a single photon. The reason for this difference between air and space is that air is molecular so energy is dispersed in all directions by numerous random molecular collisions from all directions, whereas there is no molecular basis for space. Hence the photon does not diverge in all directions and lose energy as it propagates. Space is a continuum rather than a particulate gas. If you stand in a room near the wall, you will not be attracted towards the wall by the absence of air pressure from the rigid wall, because the air molecules between you and the wall will have random motions and will be at the same pressure as on the other side. However, for the fabric of space there is no molecular basis to provide such a isotropic fluid dissipation of pressure in all directions. Therefore the planet earth below does provide a shielding which attracts us down. We would expect the result to be 10^40 times weaker than the electrostatic force between the particles given by Coulomb's law. It is pretty obvious to me, as I said in EW, that Eddington's observation (that the square root of the number of particles in the universe about equals the ratio of Coulomb/Gauss law electromagnetic force to gravitational force for electron and proton situations) is explained by Ivor's TEM wave. The TEM wave is trapped in a small loop of black hole radius (easily calculated, it's far smaller than the Planck length), and the resulting electric field sweeps outward in all directions. The EM forces act along the field lines, so there is momentum being transferred along the E and B field lines. You can't get momentum without energy, so the electron is radiating energy continuously. Initially you'd think that this must be wrong because the spinning electron does not evaporate by releasing energy. However, by analogy to Prevost's 1792 mechanism of thermal equilibrium, it is obvious what is going on. Everything in the universe is radiating energy along E and B field lines, and causing forces by the momentum thus exchanged, and because there is a resulting equilibrium between the rates of emission and reception, the atom is stable. Now because the impedance of space is 377 ohms, and because opposite charges will block energy according to whether it is similar or dissimilar in sign to itself, and because the universe is expanding, you find that like charges repel and unlike attract with equal forces for similar amounts of charge (graphical proof in April 03 EW, which took a lot of trial and error to get right). Because the charges in the universe are scattered randomly everywhere in stars, the addition is mathematically a drunkard's walk, which even for three dimensions comes out simply as a resultant equal to the average step magnitude multiplied by the square root of the number of steps. I have had people say that its crazy to put the number of steps equal to the number of particles in the universe, but they are quite wrong because we have to count every single particle of either charge sign in the universe once and once only in working out the resultant. The zig-zag path between all particles will not be confined to merely one part of the universe. Hence, we find that the Coulomb force is similar to the gravity force if the universe contained only 1 particle causing the Coulomb force, but since electric fields add up unlike the space pressure that causes gravity, we have to multiply the gravity force by the square root of the number of negative charges to get Coulomb. It would be useful to try to tie up the loose ends. One thing is that the gravity constant G is proportional to the square of the Hubble constant divided by density of universe. Thus, both G and the electromagnetic force constants (which are given by G multiplied by up the square root of the number of particles) will increase linearly with time as the universe ages. At time zero, G and EM forces were both zero, but they have been rising in direct proportion to the age of the universe since then. In 1948, Teller in Physical Review dismissed Dirac's idea of decreasing G (note that G is increasing, not decreasing, so Dirac guessed the wrong way due to having no mechanism worked out). Teller calculated that if G was stronger in the past, the gravitational compression in the sun would be have been greater so the fusion rate would have been higher and he estimated that the earth's ocean would have been boiling 400 million years ago when the ocean dwelling dinosaurs were bathing. However, Teller assumed falsely first that G decreased with time, and secondly that the Coulomb force did not vary with time. In fact, I estimate no effect of G varying on the sun's output because the identical time variation in the Coulomb force would cancel it out. For example, if gravity is half normal in the sun, then the pressure will be less so you might expect less nuclear fusion. Wrong! Because th Coulomb force is linked to gravity and is similarly only half as strong, the repulsion between protons (preventing them from approaching close enough together for the strong nuclear force to act and fuse them) will be diminished. The reduction of gravity will cause less fusion-inducing pressure in the sun, but the reduction in the Coulomb force will cause less resistance to fusion. Hence the one effect will offset the other, so the fusion rate is unaffected. The same goes for all stars. Teller's evidence for unchanging G is based on assumptions which are false. Another loose end is the idea people keep coming up with against us being pushed down: they claim that an umbrella disproves this because it would stop a downward push and eliminate gravity. They do not comprehend that X-rays are possible and that the dielectric of space is a bit like an ocean in electrons are spaced widely apart. Thus, you would need an umbrella with the earth's mass to cancel out gravity. (Naturally such people would be squashed instead of enjoying weightlessness.) The purpose of this long post is to justify some of the points which have been experimentally proved and published in IEE/IEEE and Electronics World magazine over 30 years for readers who remain unfamiliar. Otherwise, it will be dismissed as "theory development" or speculative. In order to prove conformity in a large territory, some space is needed. |
Juneau Amateur Radio Club, Inc.
Propagation of Waves
The process of communication involves the transmission of information from one location to another. As we have seen, modulation is used to encode the information onto carrier waves, and may involve analog or digital methods. It is only the characteristics of the carrier wave which determine how the signal will propagate over any significant distance. This chapter describes the different ways that electromagnetic waves propagate.
An electromagnetic wave is created by a local disturbance in the electric and magnetic
fields. From its origin, the wave will propagate outwards in all directions. If the medium in
which it is propagating (air for example) is the same everywhere, the wave will spread out
uniformily in all directions.
Far from its origin, it will have spread out enough that it will appear have the same
amplitude everywhere on the plane perpendicular to its direction of travel
(in the near vicinity of the observer). This type of wave is called a plane wave.
A plane wave is an idealization that allows one to think of the entire wave traveling
in a single direction, instead of spreading out in all directions.
Weak Signal Communication Software:
Electromagnetic waves propagate at the speed of light in a vacuum. In other mediums, like air or glass, the speed of propagation is slower. If the speed of light in a vacuum is given the symbol c0, and the speed in some a medium is c, we can define the index of refraction, n as: n=c0/c
When a plane wave encounters a change in medium, some or all of it may propagate
into the new medium or be reflected from it. The part that enters the new medium is
called the transmitted portion and the other the reflected portion. The part which
is reflected has a very simple rule governing its behavior. Make the following
Double click here to add text.
Angle of Incidence = the angle between the direction of propagation and a line perpendicular to the boundary, on the same side of the surface. Angle of Reflection = the angle between the direction of propagation of the reflected wave and a line perpendicular to the boundary, also on the same side of the surface. Then the rule for reflection is simply stated as: The angle of reflection = The angle of incidence
If the incident medium has a lower index of refraction then the reflected wave has a 1800 phase shift upon reflection. Conversely, if the incident medium has a larger index of refraction the reflected wave has no phase shift.
When you look into a pool, the light from the bottom is refracted away from the perpendicular, because the index of refraction in air is less than in water. To the observer at the side of the pool, the light appears to come from a shallower depth.
For the same reason, when you look at objects underwater through a mask, they will appear to be larger than they really are. The light from the object is spread outwards at the water-air interface of your mask. To you it will appear the object is closer
All electromagnetic waves can be superimposed upon each other without limit. The electric and magnetic fields simply add at each point. If two waves with the same frequency are combined there will a be a constant interference pattern caused by their superposition. Interference can either be constructive, meaning the strength increases as result, or destructive where the strength is reduced. The amount of interference depends of the phase difference at a particular point. It can be shown that constructive interference occurs for phase differences of 0-1200, and 240-3600. Thus destructive interference occurs from 120-2400. For two identical waves, no phase shift results in total constructive interference, where the strength is maximum and 1800 phase shift will create total destructive interference (no signal at all).
The phase shift that causes the interference can occur either due to a difference in path length, Dx, or a difference in the arrival time, Dt. The amount of phase shift, Df, can be computed for these two cases by: Example: Omega is a radio navigation system that used the phase difference in the same signal from two fixed transmitters to determine a line-of-position. The same phase difference corresponded to multiple lines of position separated by a distance equivalent to 3600 of phase shift. Since the frequency was 10.2 kHz, the wavelength corresponding to 3600 phase shift was 16 miles, which was the lane width on an Omega overprinted chart. Loran-C also has a phase-difference mode with a lane width of only 3000 m since it operates at 100 kHz.
Recall that the idealized plane wave is actually infinite in extent. If this wave passes through an opening, called an aperture, it will diffract, or spread out,
from the opening. The degree to which the cropped wave will spread out depends on the size of the aperture relative to the wavelength. In the extreme case where the aperture is very large compared to the wavelength, the wave will see no effect and will not diffract at all. At the other extreme, if the opening is very small, the wave will behave as if it were at its origin and spread out uniformly in all directions from the aperture. In between, there will be some degree of diffraction.
When the wave enters the new medium, the speed of propagation will change. In order to match the incident and transmitted wave at the boundary, the transmitted wave will change its direction of propagation. For example, if the new medium has
a higher index of refraction, which means the speed of propagation is lower, the wavelength will become shorter (frequency must stay the same because of the boundary conditions). For the transmitted wave to match the incident wave at the boundary,
the direction of propagation of the transmitted wave must be closer to perpendicular. The relationship between the angles and indices of refraction is given by Snell's Law:
When the direction of propagation changes, the wave is said to refract. It is most useful to know in which direction the wave will refract, not necessarily by how much.
The transmitted wave will bend more towards the perpendicular when entering a medium with higher index of refraction (slower speed of propagation). Example: Why a pool is deeper than it looks.
http://www.ac6v.com/software.htm http://www.ncdxf.org/beacon/monitors.html http://www/hamqsl.com/solar.html
What is a Magnetosphere?
As shown in the picture below, a planetary magnetosphere is the region where the planetary magnetic field dominates over the solar wind. This region is defined at the boundary of where the solar wind pressure is balanced with the planetary magnetic field and internal plasma pressure. |
Select Board & Class
derive relationship between commercial unit of energy and si unit of energy
a metre scale of mass 100g is pivoted at one end . if it is held at 30 degree with the horizontal what is the potential energy associated with it?
is ncert solutions for computer science avalable
calculate the work required to be done to stop a car of 1500kg moving at a velocity of 60km/h?
A body moving in a straight line at 72kmph undergoes an acceleration of 4m/s2. Find its speed after 2 seconds.
1 watt of power:
what is the relation between KW h and joules
if a force F is applied on a body and it moves with velocity v , the power will be :
What is the formula of sulphide ion,sulphate ion,sulphur ion ?
Q- Give one example of:-
(a) Certain force acting on a 20kg mass changes its velocity from 5m /s to 2 m/s . calculate the work done by the force .
(b) State two possible condition under which the work done would be zero.
(c) if the kinetic energy of the body is increased by 300% then determine the percentage increase in the momentum.
In an oscillating pendulum, at what positions the potential and kinetic energy are maximum?
the power of motor pump is 2kW. how much water per minute the pump can raise to a height of 10m? (given g = 10m/s)
suggest me a working science model for class 9th
think of some situations in your daily life in daily work, list them. is work done in each situation, if work is done. which is the force acting on the object . what is the object on which the work is done. what happens to the object on which work is done.
The kinetic energy of an object of mass , m moving with a velocity of 5m/s is 25J.what will be its kinetic energy when its velocity is doubled ?what will be its kinetic energy when its velocity is incrased three times
The potential energy of a freely falling object decreases progressively. What happens to its (i) kinetic energy (ii) total mechanical energy? State the law on which your answer is based.
A household consumes 1 kWh of energy per day. How much energy is this in joule?
a body of mass 2 kg is thrown vertically upwards with the initial velocity of 20 m s-1. what will be its potential energy at the end of 2 seconds.(g=10m s-2)
What is the derivation of Kinetic energy and Potential energy
Which of the following is carried by the waves from one place to another:
a) mass b) velocity c) wavelenght d) energy
What is the work done to increase the velocity of a car from 30km/h to 60km/h if the mass of the car is 1500kg?
renatta gass is out with her friends.misfortune occurs and renetta and her friends find themselves getting a workout.they apply a cumulative force of 1080 N to push the car 218 m to the nearest fuel station.determine the workdoneon the car.
the kE acquired by a mass m in travelling a certain distance s ,starting from rest ,under the action of a constant force is directly proportional to:
d)none of these
1. Can a body have energy without momentum ????
2. Can a body have momentum without energy ?????
plzz ans. with explanation ???
Q1. The volume of 50g of a substance is 20cm3 . If the denstiy of water is 1 g cm3, will the substance float or sink in water ? justify your answer ?
Q2. Show that the energy of a vibrating pendulum is conserved ?
Q3. A ball is dropped from a heigh of 10m. If the energy of the ball reduces by 40% after striking the ground, how much high can the ball bounce back ?
define 1 watt of power?
define work , its formula with SI unit ?
define power. Give the SI unit of power
A bullet when fired at a target with a velocity of 100 m/sec penetrates one metre into it . if the bullet is fired at a similar target with a thickness 0.5 metre, then it will emerge from it with avelocity of___ m/s.
Derive an expression for the gravitational potential energy of an object of mass 'm' at a height 'h' above the earth's surface.
A stone projected vertically upward with a velocity 'u' reaches a maximum height 'h'. Find the ratio of kinetic energy and potential energy of the stone, when it is at height 3h/4 from the ground..?
Force acting on a particle moving in a straight line varies with the velocity of the particle as F=k/v,where K is a constant.The work done by this force in time t is:
Identify and state the type of transformation of energy in the following cases:
1. When coal is burnt.
2. When a dry cell discharges.
3. In a thermal power plant.
what is the proof for w = F S cos theta
Can any object have momentum even if its mechanical energy is 0? explain.
What is the work done by a satellite moving around the earth ?Justify your answer.
A freely falling object eventually stops on reaching the ground. What happens to its kinetic energy
If the speed of a body is halved ,what will be the change in its kinetic energy??????????
why iron nail sinks and cork float on water?
Which of the following statements about the work done by a force is incorrect?
Work is done by a force only if a body moves towards the direction of the applied force.
Work done by a body is equal to the amount of energy gained or spent by the body.
Work done by a force can be zero even if the body moves to a certain distance.
Work done by a force can be positive as well as negative.
Work done by a force is defined as the product of the force on a body and the displacement of the body along the line of action of the force. Therefore, it is not necessary that the body always moves towards the direction of the applied force. Even if a body moves opposite to the direction of the applied force, work is done by the applied force. This is because the displacement of the body is along the line of action of the applied force. In this case, the work done by the force is considered negative.
The correct answer is A.
Dear Experts, I can't understand this. How can it be not necessary that an objectalways moves towards the direction of the applied force?
If the above is correct, then please coin an example.
I am confused as to how to find K.E. in the Activity 11.15 at Page 155 of Work and Energy chapter because we don't know the velocity in those cases. Finding P.E. is easy there. Can anybody help???
an electric heater is rated 1500w. how much energy does it use in 10 hrs.express your answer kwt and joules.
wat is the relationship between JOULE (J) and ERG ??????
experts answer ??:DD
state & explain the law of conservation of energy with an example 2. explain how total energy a swinging pendulam at any instant of time remains conserved illustrate your answer with the help of Lbelled diagram
derive an expression for kinetic energy of a body ?
When do we say that work is done?
An object of mass of 20 kg is dropped from a height of 4 m. Fill in the blanks in the following table by computing the potential energy and kinetic energy in each case.
height at which object is located: 4 m, 3 m, 2m, 1m, just above the ground
(for simplifying the calculations, take the value of g as 10 m s-2)
can you plz give me the examples of kinetic energy?????????????
Two bodies of different masses m1 and m2 (m1>m2) have same kinetic energy . They are stopped by applying some retarding force . Which body will stop first ?
Water is falling on the Blades of a turbine at the rate of 6 * 103 kg per minure .
The height of the fall is 10 m . Claculate the power given to the turbine. Take g = 10 m/s2
(a) In a tug of war, one team (team A) wins and the other team (team B) loses. Which of these two teams does positive and negative work?
(b) What is the work done in case of a satellite moving around the earth?
which physical quantity has the S.I. unit = Nm or newton -metre
a truck and a car moving with same kinetic energy are stopped by applying same retarding force by means of brakes. which stops at a smaller distance?
define density?Give its SIunit.
a man drops 10kg rock from the top of a 5m ladder. wht is its speed just before it hits the ground? wht is its kinetic energy wen it reaches da ground.
Explain the transformation of energy in hydroelectric power plant?
Derive the relation between momentum and kinetic energy ??????
When a body of mass 'm' revolves with uniform speed of 'v' on a circular path of radius 'r',the work done by the body in one complete rotation is
a) (1 / 2 ) mv2 b) 2π rmv2 c) ( 2π - 0.5 ) mv2 d) 0
examples of kinetic enery and potential energy its urgent
Prove that kinetic energy =p2/2m
Water stored in a dam possesses
(a) no energy (b) electrical energy (c) kinetic energy (d) potential energy
Name the devices or machines which convert (1) Chemical Energy into Heat Energy (2) Light Energy into Heat Energy (3) Heat Energy into Kinetic Energy. What must be the power of the brakes of a car for them to stop the car moving at a velocity of 60 km per hour , weighing 1500 kg?
Derive an expression for the kinetic energy of an object of mass 'm' moving horizontally with a uniform velocity 'v'
two bodies of masses (m1) and (m2) have equal kinetic energy. what shall be the ratio of their momentum
prove that the energy remains constant in case of a freely falling body....
An object of mass 40kg is raised to a height of 5m above the ground.What is its PE?If the object is allowed to fall,find its KE when it is half way downAn object of mass 40 kg is raised to a height of 5m above the ground.What is its PE?If the object is allowed to fall,find its KE when it is half way down
define average power
An engine pulls a car of mass 1500kg on a level road at constant speed of 5m/s.If the frictional force is 500N,what power does the engine generate?
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Je kunt de Wilcoxon-toets nog steeds gebruiken. Het gaat immers over het verschil tussen beide metingen. De tevredenheid over de slaakamer (bijvoorbeeld een 4) en de tevredenheid over de keuken (ook een 4) is een verschil van 0. Deze verschilscores worden gerangordend en krijgen een score From Wikipedia, the free encyclopedia. Jump to navigation Jump to search. The Wilcoxon signed-rank test is a non-parametric statistical hypothesis test used to compare two related samples, matched samples, or repeated measurements on a single sample to assess whether their population mean ranks differ (i.e. it is a paired difference test )
W+=9.81 stappenplan Wilcoxon's signed rank toets: 1 hypothese H. 0:md1=md2en H. a:md1< md2. 2 steekproevenverdeling (bij benadering) standaard normaal verdeeld 3 toetsingsgrootheid z =( 6.5 −27.5)/9.81 =− 2.14 4 verwerpingsgebied α =0.05, z∗=− 1.65 5 statistische conclusie z =− 2.14 < −1.65 =z∗en H Wilcoxon signed-rank test is a very common test in the fields of pharmaceuticals, especially amongst drug researchers, to find out the dominant symptoms of various drugs on humans. Being a non-parametric test, it works as an alternative to T-test which is parametric in nature. For any doubt/query, comment below De rangtekentoets van Wilcoxon, ook wilcoxonrangtekentoets geheten, is een verdelingsvrije toets voor de mediaan van een continue verdeling. Het is een toets voor één steekproef. Deze toets lijkt op de tekentoets, maar is niet alleen op de aantallen tekens gebaseerd, maar ook op de bijbehorende rangnummers. De toets is evenals de wilcoxontoets voor twee steekproeven, genoemd naar de opsteller Frank Wilcoxon The Wilcoxon signed-rank test is the nonparametric test equivalent to the dependent t-test. As the Wilcoxon signed-rank test does not assume normality in the data, it can be used when this assumption has been violated and the use of the dependent t-test is inappropriate. It is used to compare two sets of scores that come from the same participants
In plaats van het gebruiken van ruwe scores, ordent de Wilcoxon signed-rank test de data en geeft ze een rangnummer. De laagste score krijgt rang 1, de enerlaagste score rang 2 enzovoort. Door het ordenen en nummeren van scores verdwijnt het mogelijke effect dat uitbijters of een scheve verdeling kunnen hebben The report in APA A Wilcoxon Signed-Ranks Test indicated that the median post- test ranks were statistically significantly higher than the median pre-test ranks Z = 21, p < .027. 2nd Note - if the reason you used a Wilcoxon Signed Ranks Test is because your data is very skewed or non-normal, just report it the same way but replace ranks with score
The Wilcoxon Signed Rank Test is a non-parametric test for comparing two paired (dependent) data sets. It is an alternative to Students Paired t-Test and is The Wilcoxon Signed Rank Test is a.. Wilcoxon Signed Ranks Test. Wilcoxon signed rank tests did not yield any significant differences between the RF kyphoplasty and vertebroplasty (p = 5), balloon kyphoplasty, and vertebroplasty (p = 1.0) or RF kyphoplasty and balloon kyphoplasty (p = 1.0) treatment groups. From: The Comprehensive Treatment of the Aging Spine, 2011. Related terms
The Wilcoxon Signed Rank Test is the non-parametric version of the paired samples t-test. It is used to test whether or not there is a significant difference between two population means when the distribution of the differences between the two samples cannot be assumed to be normal The Wilcoxon Signed-Rank Test is a statistical test used to determine if 2 measurements from a single group are significantly different from each other on your variable of interest. Your variable of interest should be continuous and your group randomly sampled to meet the assumptions of this test . It does not require any assumptions about the shape of the distribution. For this reason, this test is often used as an alternative to t test's whenever the population cannot be assumed to be normally distributed
Wilcoxon Signed-Rank Test Assumptions. The following assumptions must be met in order to run a Wilcoxon signed-rank test: Data are considered continuous and measured on an interval or ordinal scale. Each pair of observations is independent of other pairs. Each pair of measurements is chosen randomly from the same population The nonparametric Wilcoxon signed rank test compares the median of a single column of numbers against a hypothetical median. Don't confuse it with the Wilcoxon matched pairs test which compares two paired or matched groups.. Interpreting the confidence interval. The signed rank test compares the median of the values you entered with a hypothetical population median you entered
Wilcoxon Signed-Ranks Test for Paired Samples. When the requirements for the t-test for two paired samples are not satisfied, the Wilcoxon Signed-Rank Test for Paired Samples non-parametric test can often be used. In particular, we assume n subjects from a given population with two observations x i and y i for each subject i The Wilcoxon signed rank test is provided by PROC UNIVARIATE, not PROC NPAR1WAY (see documentation: Tests for Location). First, you compute the difference between the two values for each pair, then you use this difference as the analysis variable in a PROC UNIVARIATE step The test statistic for the Wilcoxon signed-rank test is often expressed (equivalently) as the sum of the positive signed-ranks, T +, where E(T +) = n(n+ 1) 4 and Var adj(T +) = 1 4 Xn j=1 r2 j Zeros and ties do not affect the theory above, and the exact variance is still given by the above formula for Va
Check Out Signed On eBay. Find It On eBay. But Did You Check eBay? Find Signed On eBay The Wilcoxon signed rank sum test is another example of a non-parametric or distribution free test (see 2.1 The Sign Test). As for the sign test, the Wilcoxon signed rank sum test is used is used to test the null hypothesis that the median of a distribution is equal to some value To do Wilcoxon signed-rank test in SAS, you first create a new variable that is the difference between the two observations. You then run PROC UNIVARIATE on the difference, which automatically does the Wilcoxon signed-rank test along with several others. Here's an example using the poplar data from above The Wilcoxon Rank-Sum Test The Wilcoxon rank-sum test is a nonparametric alternative to the two-sample t-test which is based solely on the order in which the observations from the two samples fall. We will use the following as a running example. Example 1 In a genetic inheritance study discussed by Margolin The Wilcoxon signed ranks test may be used as a one-sample test of location or as a test of difference in location between two dependent samples. The underlying assumptions are that the distribution is.
The Wilcoxon Signed Rank Test assumes that our sample is an SRS and will give trustworthy conclusions only if this condition is met. The Wilcoxon Signed Rank Test assumes that your data come from a continuous distribution I was wondering what the theoretical difference is between the Wilcoxon Rank-Sum Test and the Wilcoxon Signed-Rank Test using paired observations. I know that the Wilcoxon Rank-Sum Test allows for a different amount of observations in two different samples, whereas the Signed-Rank test for paired samples does not allow that, however, they both seem to test the same in my opinion Signed-Rank Wilcoxon Test Example. Significance level: the probabilty of rejecting the null when it is true. We can reject the null hypothesis at significance level \(\alpha = 0.05\) in favor of the alternative that sending kids to school improves their social awareness
SECTION J.1: Table of Critical Values for the Wilcoxon Rank-Sum Test 93 1-tail = 0:025 = 0:05 1-tail = 0:025 = 0:05 2-tail = 0:05 = 0:10 2-tail = 0:05 = 0:10 m n W d P W d P m n W d P W d P 7 21 64 139 37 .0240 69 134 42 .0449 10 20 110 200 56 .0245 117 193 62 .049 The Wilcoxon signed-rank test is a non-parametric statistical hypothesis test used to compare two related samples, matched samples, or repeated measurements on a single sample to assess whether their population mean ranks differ (i.e. it is a paired difference test).It can be used as an alternative to the paired Student's t-test (also known as t-test for matched pairs or t-test for. Right tail, or two tails with positive Z, (W-> μ) , C = -0.5 . Left tail, or two tails with negative Z, (W-< μ) , C = 0.5 . Target Unlike paired t-test that compares the mean of the differences to zero, Wilcoxon Signed-Rank test compares the probability that a random value from Group1 (like before) is greater than his dependent value from Group2 (like after) The one-sample Wilcoxon signed rank test is a non-parametric alternative to one-sample t-test when the data cannot be assumed to be normally distributed. It's used to determine whether the median of the sample is equal to a known standard value (i.e. theoretical value) I've tried Wilcoxon Signed Ranks Test (non-parametric version of paired t-test) and I still get the same result - Condition 2 pre-post is significant (even more so as p value is lower)
The Wilcoxon signed rank test is a nonparametric test for two populations when the observations are paired. In this case, the test statistic, W, is the sum of the ranks of positive differences between the observations in the two samples (that is, x - y).When you use the test for one sample, then W is the sum of the ranks of positive differences between the observations and the hypothesized. The Wilcoxon signed-rank test is a non-parametric statistical hypothesis test used to compare two related samples, matched samples, or repeated measurements on a single sample to estimate whether their population mean ranks differ e.g it is a paired difference test. It can be applied as an alternative to the paired Student's t-test also known as t-test for matched pairs or t-test. I have used Wilcoxon Signed Rank, as my data are both dependent and not normally distributed. With NPAR1WAY, I thought, you have rank sum test and assume independent data. I indeed refer to the t-statistic of the Wilcoxon Signed Rank. Thank you
Take a quick interactive quiz on the concepts in Wilcoxon Rank-Sum Test: Definition & Application or print the worksheet to practice offline. These practice questions will help you master the. Wilcoxon's name is used to describe several statistical tests. • The Wilcoxon matched-pairs signed-rank test is a nonparametric method to compare before-after, or matched subjects. It is sometimes called simply the Wilcoxon matched-pairs test. • The Wilcoxon signed rank test is a nonparametric test that compares the median of a set of numbers against a hypothetical median The wilcoxon signed-rank test could for instance be used to answer the question: Is the median of the differences between the mental health scores before and after an intervention different from 0? SPSS. How to perform the wilcoxon signed-rank test in SPSS: Analyze > Nonparametric Tests > Legacy Dialogs > 2 Related Samples..
You just have no compelling evidence that they differ. If you have small samples, the Wilcoxon test has little power to detect small differences. How the P value is calculated. If there are fewer than 200 pairs, Prism calculates an exact P value. See more details in the page about the Wilcoxon signed rank test
Wilcoxon Signed Rank Test: Time Method η: median of Time Descriptive Statistics Sample N Median Time 16 11.55 Test Null hypothesis H₀: η = 12 Alternative hypothesis H₁: η < 12 N for Wilcoxon Sample Test Statistic P-Value Time 16 53.00 0.227 Minitab.com; License Portal; Store; Blog; Contact. Wilcoxon signed rank test: The Wilcoxon signed rank test Wiki provides its background and statistical theory. It is applied in situations of paired data, when the paired data samples come from a population which cannot be assumed to be normally distributed En statistique, le test des rangs signés de Wilcoxon est une alternative non-paramétrique au test de Student pour des échantillons appariés.Le test s'intéresse à un paramètre de position : la médiane, le but étant de tester s'il existe un changement sur la médiane I want to run Wilcoxon test to compare 3 test groups (B, C and D) against the control group (A) The data are organized in the following format: Group CustomerID Value A 23483 61 A. Details. The formula interface is only applicable for the 2-sample tests. If only x is given, or if both x and y are given and paired is TRUE, a Wilcoxon signed rank test of the null that the distribution of x (in the one sample case) or of x - y (in the paired two sample case) is symmetric about mu is performed.. Otherwise, if both x and y are given and paired is FALSE, a Wilcoxon rank sum.
The Wilcoxon signed rank test for a single sample. The responses people gave about the shopping centre are represented by numerical ratings: 5 2 1 4 5 4 1 6 3 2 3 1 where having no opinion equates to 4, the median value. A quick 'eyeball' test shows that none of those. Wilcoxon Signed Ranks Test Ranks 48a 24.98 1199.00 3b 42.33 127.00 0c 51 Negative Ranks Positive Ranks Ties Total bed86 - bed80 N Mean Rank Sum of Ranks a. bed86 < bed80 b. bed86 > bed80 c. bed86 = bed80 Test Statisticsb-5.037a.000 Z Asymp. Sig. (2-tailed) bed86 - bed80 a. Based on positive ranks. b. Wilcoxon Signed Ranks Test WILCOXON SIGNED RANKS TEST Introduction: • The sign test utilizes only the signs of the differences between the observed values and the hypothesized median. For testing 퐻 0: 푀 = 푀 0 Wilcoxon signed-ranks test uses the magnitude of differences when these are available • In order to use Wilcoxon signed-ranks test, we need additional information about each sample measurement. Wilcoxon Signed-Rank Test P-Values. For N ≤ 20, exact p-values are calculated.. For N > 20, a Student's t approximation to the statistic defined below is used. Note that a correction for ties is applied. See Iman and Lehmann and D'Abrera ().Under the null hypothesis, the mean of S is zero. The variance of S is given by the following
. the results indicate a significant difference between English test score before training (M=65.19; SD=20.06) and English test score after training (M=79.09; SD=16.20), [Z = -3.533, p = .000] The one sample wilcoxon signed-rank test requires one variable of the following type: Variable type required for the one sample wilcoxon signed-rank test : One of ordinal level. Note that theoretically, it is always possible to 'downgrade' the measurement level of a variable
Summary: Wilcoxon signed rank test vs paired Student's t-test. In this analysis, both Wilcoxon signed rank test and paired Student's t-test led to the rejection of the null hypothesis. In general, however, which test is more appropriate? The answer is, it depends on several criteria When is a Wilcoxon's the correct test? When you are testing to see whether there is a significant difference between two groups/conditions When the experimental design is repeated measures When the level of measurement of the data is at least ordinal, e.g. it can be used with interval/ratio as well (NOT nominal) How do I do it? Thi 1 sample Wilcoxon non parametric hypothesis test is one of the popular non-parametric test. One sample t-test is to compare the mean of the population to the known value (i.e more than, less than or equal to a specific known value). The t-test always assumes that random data and the population standard deviation is unknown.. Wilcoxon Signed-Rank test is the equivalent non-parametric t-test and. Group differences for when there are two matched pairs are addressed by both tests, but again, the Wilcoxon Matched-Pairs Signed Ranks Test is often used with ordinal data and/or data that are viewed as being nonparametric (with attention to medians) whereas the Student's t-Test for Matched Pairs is generally used with interval data that rise to the level of parametric distributions (with.
Single Sample Wilcoxon Signed-Rank Test Example. Group 1: Received the experimental medical treatment. Population Value: On average in the population, it takes 12 days to recover from the disease Variable of interest: Time to recover from the disease in days.. In this example, group 1 is our treatment group because they received the experimental medical treatment The Wilcoxon signed-rank test is a non-parametric statistical hypothesis test used to compare two related samples, matched samples, or repeated measurements on a single sample to assess whether their population mean ranks differ (i.e. it is a paired difference test). It can be used as an alternativ Paired Samples Wilcoxon Test in R. The paired samples Wilcoxon test (also known as Wilcoxon signed-rank test) is a non-parametric alternative to paired t-test used to compare two related samples, matched samples, or repeated measurements on a single sample to assess whether their population mean ranks differ (i.e. it is a paired difference test)
This test requires input data to be matched pairs, for example, data from before and after a clinical operation. This test can be one- or two-tailed. Unlike the paired-sample t-test, the paired-sample Wilcoxon Signed Rank test does not require the assumption that the populations are normally distributed p = ranksum(x,y) returns the p-value of a two-sided Wilcoxon rank sum test. ranksum tests the null hypothesis that data in x and y are samples from continuous distributions with equal medians, against the alternative that they are not. The test assumes that the two samples are independent. x and y can have different lengths.. This test is equivalent to a Mann-Whitney U-test Wilcoxon Sign Rank Test practical questions In this example yo u will use a Wilcoxon signed rank test to test whether students feel more informed about greenhouse gas issues (INFGGAS) and nuclear waste issues (INFNUCL). Both variables were rated on the same four-point scale by all participants, making th |
11 SEPARATION COLUMNS
Key Learning Objectives
& How to design distillation columns
& How to size distillation columns and select and design distillation column trays
& How to design distillation columns using packing instead of trays
& How to design liquid-liquid extraction columns
11.2. Continuous Distillation: Process
11.3. Continuous Distillation: Basic
11.4. Design Variables in Distillation
11.5. Design Methods for Binary Systems
11.6. Multicomponent Distillation:
11.7. Multicomponent Distillation:
Shortcut Methods for Stage and
11.8. Multicomponent Systems: Rigorous
Solution Procedures (Computer
11.9. Other Distillation Systems
11.10. Plate Efficiency
11.11. Approximate Column Sizing
11.12. Plate Contactors
11.13. Plate Hydraulic Design
11.14. Packed Columns
11.15. Column Auxiliaries
11.16. Solvent Extraction
This chapter covers the design of separating columns. Though the emphasis is on
distillation processes, the basic construction features and many of the design methods
also apply to other multistage processes, such as stripping, absorption, and extraction.
Distillation is probably the most widely used separation process in the chemical
and allied industries, its applications ranging from the rectification of alcohol, which
has been practiced since antiquity, to the fractionation of crude oil.
Only a brief review of the fundamental principles that underlie the design proced-
ures will be given; a fuller discussion can be found in other textbooks; see King
(1980), Hengstebeck (1976), Richardson et al. (2002) and Kister (1992).
A good understanding of methods used for correlating vapor-liquid equilibrium
data is essential to the understanding of distillation and other equilibrium-staged
processes; this subject was covered in Chapter 8.
In recent years, most of the work done to develop reliable design methods for
distillation equipment has been carried out by a commercial organization, Fractionation
Research Inc., an organization set up with the resources to carry out experimental work
on full-size columns. Since this organization’s work is of a proprietary nature, it is not
published in the open literature, and it has not been possible to refer to its methods in
this book. The Fractionation Research design manuals will, however, be available to
design engineers whose companies are subscribing members of the organization.
Distillation Column Design
The design of a distillation column can be divided into the following steps:
1. Specify the degree of separation required: set product specifications.
2. Select the operating conditions: batch or continuous; operating pressure.
3. Select the type of contacting device: plates or packing.
4. Determine the stage and reflux requirements: the number of equilibrium stages.
5. Size the column: diameter, number of real stages.
6. Design the column internals: plates, distributors, packing supports.
7. Mechanical design: vessel and internal fittings.
The principal step will be to determine the stage and reflux requirements. This is a
relatively simple procedure when the feed is a binary mixture, but a complex and
difficult task when the feed contains more than two components (multicomponent
11.2. CONTINUOUS DISTILLATION: PROCESS DESCRIPTION
The separation of liquid mixtures by distillation depends on differences in volatility
between the components. The greater the relative volatilities, the easier is the separ-
ation. The basic equipment required for continuous distillation is shown in Figure 11.1.
Vapor flows up the column, and liquid counter-currently flows down the column. The
vapor and liquid are brought into contact on plates or packing. Part of the condensate
642 CHAPTER 11 SEPARATION COLUMNS (DISTILLATION, ABSORPTION, AND EXTRACTION)
from the condenser is returned to the top of the column to provide liquid flow above the
feed point (reflux), and part of the liquid from the base of the column is vaporized in the
reboiler and returned to provide the vapor flow.
In the section below the feed, the more volatile components are stripped from the
liquid; this is known as the stripping section. Above the feed, the concentration of the
more volatile components is increased; this is called the enrichment, or more com-
monly, the rectifying section. Figure 11.1a shows a column producing two product
streams, referred to as tops or overheads and bottoms, from a single feed. Columns are
occasionally used with more than one feed, and with side streams withdrawn at points
up the column, as in Figure 11.1b. This does not alter the basic operation but
complicates the analysis of the process to some extent.
If the process requirement is to strip a volatile component from a relatively
nonvolatile solvent, the rectifying section may be omitted, and the column would
then be called a stripping column.
In some operations, where the top product is required as a vapor, only sufficient
liquid is condensed to provide the reflux flow to the column, and the condenser is
referred to as a partial condenser. When the liquid is totally condensed, the liquid
returned to the column will have the same composition as the top product. In a partial
condenser the reflux will be in equilibrium with the vapor leaving the condenser.
Virtually pure top and bottom products can be obtained in a single column from a
binary feed, but where the feed contains more than two components, only a single
‘‘pure’’ product can be produced, either from the top or bottom of the column.
Several columns will be needed to separate a multicomponent feed into its constituent
Figure 11.1. Distillation column. (a) Basic column. (b) Multiple feeds and side streams.
11.2. CONTINUOUS DISTILLATION: PROCESS DESCRIPTION 643
11.2.1. Reflux Considerations
The reflux ratio, R, is normally defined as
flow returned as reflux
flow of top product taken off
The number of stages required for a given separation will be dependent on the
reflux ratio used.
In an operating column, the effective reflux ratio will be increased by vapor
condensed within the column due to heat leakage through the walls. With a well-
lagged column, the heat loss will be small, and no allowance is normally made for this
increased flow in design calculations. If a column is poorly insulated, changes in the
internal reflux due to sudden changes in the external conditions, such as a sudden
rainstorm, can have a noticeable effect on the column operation and control.
Total reflux is the condition when all the condensate is returned to the column as
reflux: no product is taken off and there is no feed.
At total reflux the number of stages required for a given separation is the minimum at
which it is theoretically possible to achieve the separation. Though not a practical
operating condition, it is a useful guide to the likely number of stages that will be needed.
Columns are often started up with no product takeoff and operated at total reflux
until steady conditions are attained. The testing of columns is also conveniently
carried out at total reflux.
As the reflux ratio is reduced, a pinch point will occur at which the separation can be
achieved only with an infinite number of stages. This sets the minimum possible reflux
ratio for the specified separation.
Optimum Reflux Ratio
Practical reflux ratios will lie somewhere between the minimum for the specified
separation and total reflux. The designer must select a value at which the specified
separation is achieved at minimum cost. Increasing the reflux reduces the number of
stages required, and hence the capital cost, but increases the service requirements
(steam and water) and the operating costs. The optimum reflux ratio will be that
which gives the lowest annual operating cost. No hard and fast rules can be given for
the selection of the design reflux ratio, but for many systems the optimum will lie
between 1.2 and 1.5 times the minimum reflux ratio.
For new designs, where the ratio cannot be decided from past experience, the effect
of reflux ratio on the number of stages can be investigated using the shortcut design
methods given in this chapter. This will usually indicate the best value to use in more
rigorous design methods.
644 CHAPTER 11 SEPARATION COLUMNS (DISTILLATION, ABSORPTION, AND EXTRACTION)
At low reflux ratios, the calculated number of stages will be very dependent on the
accuracy of the vapor-liquid equilibrium data available. If the data are suspect,
a higher than normal ratio should be selected to give more confidence in the design.
11.2.2. Feed-Point Location
The precise location of the feed point will affect the number of stages required for
a specified separation and the subsequent operation of the column. As a general rule,
the feed should enter the column at the point that gives the best match between the
feed composition (vapor and liquid if two phases) and the vapor and liquid streams in
the column. In practice, it is wise to provide two or three feed-point nozzles located
around the predicted feed point to allow for uncertainties in the design calculations
and data, and possible changes in the feed composition after startup.
11.2.3. Selection of Column Pressure
Except when distilling heat-sensitive materials, the main consideration when selecting
the column operating-pressure will be to ensure that the dew point of the distillate is
above that which can be easily obtained with the plant cooling water. The maximum,
summer temperature of cooling water is usually taken as 308C. If this means that high
pressures will be needed, the provision of refrigerated brine cooling should be
considered. Vacuum operation is used to reduce the column temperatures for the
distillation of heat-sensitive materials and where very high temperatures would other-
wise be needed to distill relatively nonvolatile materials.
When the stage and reflux requirements are calculated, it is usual to take the
operating pressure as constant throughout the column. In vacuum columns,
the column pressure drop will be a significant fraction of the total pressure, and
the change in pressure up the column should be allowed for when calculating the
stage temperatures. This may require a trial-and-error calculation, as clearly the
pressure drop cannot be estimated before an estimate of the number of stages is
11.3. CONTINUOUS DISTILLATION: BASIC PRINCIPLES
11.3.1. Stage Equations
Material and energy balance equations can be written for any stage in a multistage
Figure 11.2 shows the material flows into and out of a typical stage n in a distillation
column. The equations for this stage are set out here, for any component i.
Vnþ1ynþ1 þ LnÀ1xnÀ1 þ Fnzn ¼ Vnyn þ Lnxn þ Snxn (11:1)
11.3. CONTINUOUS DISTILLATION: BASIC PRINCIPLES 645
Vnþ1Hnþ1 þ LnÀ1hnÀ1 þ Fhf þ qn ¼ VnHn þ Lnhn þ Snhn (11:2)
Vn ¼ vapor flow from the stage;
Vnþ1 ¼ vapor flow into the stage from the stage below;
Ln ¼ liquid flow from the stage;
LnÀ1 ¼ liquid flow into the stage from the stage above;
Fn ¼ any feed flow into the stage;
Sn ¼ any side stream from the stage;
qn ¼ heat flow into, or removal from, the stage;
n ¼ any stage, numbered from the top of the column;
z ¼ mol fraction of component i in the feed stream (note, feed may be
x ¼ mol fraction of component i in the liquid streams;
y ¼ mol fraction of component i in the vapor streams;
H ¼ specific enthalpy vapor phase;
h ¼ specific enthalpy liquid phase;
hf ¼ specific enthalpy feed (vapor þ liquid).
All flows are the total stream flows (mols/unit time), and the specific enthalpies are
also for the total stream ( J/mol).
It is convenient to carry out the analysis in terms of ‘‘equilibrium stages.’’ In an
equilibrium stage (theoretical plate), the liquid and vapor streams leaving the stage are
taken to be in equilibrium, and their compositions are determined by the vapor-liquid
equilibrium relationshipfor the system(seeChapter 8).Interms ofequilibrium constants
yi ¼ Kixi (11:3)
The performance of real stages is related to an equilibrium stage by the concept of
plate efficiencies for plate contactors and ‘‘height of an equivalent theoretical plate’’
for packed columns.
Fn, Zn Sn, xnn
Figure 11.2. Stage flows.
646 CHAPTER 11 SEPARATION COLUMNS (DISTILLATION, ABSORPTION, AND EXTRACTION)
In addition to the equations arising from the material and energy balances over a
stage, and the equilibrium relationships, there will be a fourth relationship, the
summation equation for the liquid and vapor compositions:
Sxi;n ¼ Syi;n ¼ 1:0 (11:4)
These four equations are the so-called MESH equations for the stage: Material
balance, Equilibrium, Summation, and Heat (energy) balance equations. MESH equa-
tions can be written for each stage and for the reboiler and condenser. The solution of
this set of equations forms the basis of the rigorous methods that have been developed
for the analysis for staged separation processes.
11.3.2. Dew Points and Bubble Points
To estimate the stage and the condenser and reboiler temperatures, procedures are
required for calculating dew and bubble points. By definition, a saturated liquid is at
its bubble point (any rise in temperature will cause a bubble of vapor to form), and
a saturated vapor is at its dew point (any drop in temperature will cause a drop of
liquid to form).
Dew points and bubble points can be calculated from a knowledge of the vapor-
liquid equilibrium for the system. In terms of equilibrium constants, the bubble point
and dew point are defined by the following equations:
Kixi ¼ 1:0 (11:5a)
and dew point:
¼ 1:0 (11:5b)
For multicomponent mixtures, the temperature that satisfies these equations, at a
given system pressure, must be found by trial and error.
For binary systems the equations can be solved more readily because the compon-
ent compositions are not independent; fixing one fixes the other.
ya ¼ 1 À yb (11:6a)
xa ¼ 1 À xb (11:6b)
Bubble- and dew-point calculations are illustrated in Example 11.9.
11.3.3. Equilibrium Flash Calculations
In an equilibrium flash process, a feed stream is separated into liquid and vapor
streams at equilibrium. The composition of the streams will depend on the quantity
of the feed vaporized (flashed). The equations used for equilibrium flash calculations
are developed in this section, and a typical calculation is shown in Example 11.1.
Flash calculations are often needed to determine the condition of the feed to a
distillation column and, occasionally, to determine the flow of vapor from the reboi-
ler, or condenser if a partial condenser is used.
Single-stage flash distillation processes are used to make a coarse separation of the
light components in a feed, often as a preliminary step before a multicomponent
distillation column, as in the distillation of crude oil.
11.3. CONTINUOUS DISTILLATION: BASIC PRINCIPLES 647
Figure 11.3 shows a typical equilibrium flash process. The equations describing this
Material balance, for any component, i:
Fzi ¼ Vyi þ Lxi (11:7)
Energy balance, total stream enthalpies:
Fhf ¼ VH þ Lh (11:8)
If the vapor-liquid equilibrium relationship is expressed in terms of equilibrium
constants, equation 11.7 can be written in a more useful form:
Fzi ¼ VKi xi þ Lxi
Ki þ 1
VKi þ 1
The groups incorporating the liquid and vapor flow rates and the equilibrium
constants have a general significance in separation process calculations.
The group L/VKi is known as the absorption factor Ai, and is the ratio of the mols of
any component in the liquid stream to the mols in the vapor stream.
The group VKi/L is called the stripping factor, Si, and is the reciprocal of the
Efficient techniques for the solution of the trial-and-error calculations necessary in
multicomponent flash calculations are given by several authors; see Hengstebeck
(1976) and King (1980).
A feed to a column has the composition given in the following table and is at a
pressure of 14 bar and a temperature of 608C. Calculate the flow and composition
Figure 11.3. Flash distillation.
648 CHAPTER 11 SEPARATION COLUMNS (DISTILLATION, ABSORPTION, AND EXTRACTION)
of the liquid and vapor phases. Take the equilibrium data from the De Priester charts
given in Chapter 8.
Feed ethane (C2) 20 0.25
propane (C3) 20 0.25
isobutane (iC4) 20 0.25
n-pentane (nC5) 20 0.25
For two phases to exist, the flash temperature must lie between the bubble point and
dew point of the mixture.
From equations 11.5a and 11.5b:
Kizi > 1:0
Check feed condition
Ki Kizi zi/Ki
C2 3.8 0.95 0.07
C3 1.3 0.33 0.19
iC4 0.43 0.11 0.58
nC5 0.16 0.04 1.56
S 1.43 S 2.40
therefore, the feed is a two-phase mixture.
Try L=V ¼ 1:5 Try L=V ¼ 3:0
Ki Ai ¼ L/VKi Vi ¼ Fzi/(1 þ Ai) Ai Vi
C2 3.8 0.395 14.34 0.789 11.17
C3 1.3 1.154 9.29 2.308 6.04
iC4 0.43 3.488 4.46 6.977 2.51
nC5 0.16 9.375 1.93 18.750 1.01
Vcalc ¼ 30.02 Vcalc ¼ 20.73
80 À 30:02
¼ 1:67 L=V ¼ 2:80
11.3. CONTINUOUS DISTILLATION: BASIC PRINCIPLES 649
Hengstebeck’s method is used to find the third trial value for L/V. The calculated
values are plotted against the assumed values, and the intercept with a line at 458
(calculated ¼ assumed) gives the new trial value, 2.4.
Try L=V ¼ 2:4
Ai Vi yi ¼ Vi/V xi ¼ (Fzi À Vi)/L
C2 0.632 12.26 0.52 0.14
C3 1.846 7.03 0.30 0.23
iC4 5.581 3.04 0.13 0.30
nC5 15.00 1.25 0.05 0.33
Vcal ¼ 23.58 1.00 1.00
L ¼ 80 À 23:58 ¼ 56:42 kmol=h,
L=V calculated ¼ 56:42=23:58 ¼ 2:39 close enough to the assumed value of 2:4:
In many flash processes, the feed stream is at a higher pressure than the flash pressure,
and the heat for vaporization is provided by the enthalpy of the feed. In this situation
the flash temperature will not be known and must be found by trial and error.
A temperature must be found at which both the material and energy balances are
satisfied. This is easily carried out using commercial simulation software.
11.4. DESIGN VARIABLES IN DISTILLATION
It was shown in Chapter 1 that to carry out a design calculation, the designer must
specify values for a certain number of independent variables to define the problem
completely and that the ease of calculation will often depend on the judicious choice
of these design variables.
In manual calculations, the designer can use intuition in selecting the design
variables and, can define other variables as the calculation proceeds if it becomes
clear that the problem is not sufficiently defined. When a problem is specified for a
computer method, it is essential that the problem is completely and sufficiently
In Chapter 1 it was shown that the number of independent variables for any
problem is equal to the difference between the total number of variables and the
number of linking equations and other relationships. Examples of the application
of this formal procedure for determining the number of independent variables
in separation process calculations are given by Gilliland and Reed (1942) and
Kwauk (1956). For a multistage, multicomponent column, there will be a set of
material and enthalpy balance equations and equilibrium relationships for each
stage (the MESH equations) and for the reboiler and condenser, for each component.
650 CHAPTER 11 SEPARATION COLUMNS (DISTILLATION, ABSORPTION, AND EXTRACTION)
If there are more than a few stages, the task of counting the variables and equations
becomes burdensome, and mistakes are very likely to be made. A simpler,
more practical way to determine the number of independent variables is the ‘‘descrip-
tion rule’’ procedure given by Hanson et al. (1962). Their description rule states that to
determine a separation process completely, the number of independent variables that
must be set (by the designer) will equal the number that are set in the construction of
the column or that can be controlled by external means in its operation. The applica-
tion of this rule requires the designer to visualize the column in operation and list the
number of variables fixed by the column construction, those fixed by the process, and
those that have to be controlled for the column to operate steadily and produce
product within specification. The method is best illustrated by considering the oper-
ation of the simplest type of column: with one feed, no side streams, a total condenser,
and a reboiler. The construction will fix the number of stages above and below the feed
point (two variables). The feed composition and total enthalpy will be fixed by the
processes upstream [1 þ (n À 1)variables, where n is the number of components]. The
feed rate, column pressure, and condenser and reboiler duties (cooling water and
steam flows) will be controlled (four variables).
Total number of variables fixed ¼ 2 þ 1 þ (n À 1) þ 4 ¼ n þ 6
To design the column, the designer must specify this number of variables com-
pletely to define the problem, but need not select the same variables.
Typically, in a design situation, the problem will be to determine the number of
stages required at a specified reflux ratio and column pressure, for a given feed, and
with the product compositions specified in terms of two key components and one
product flow rate. Counting the number of variables specified, it will be seen that the
problem is completely defined:
Feed flow, composition, enthalpy ¼ 2 þ (n À 1)
Reflux (sets qc) ¼ 1
Key component compositions, top and bottom ¼ 2
Product flow ¼ 1
Column pressure ¼ 1
n þ 6
Note: Specifying (n À 1) component compositions completely defines the feed composition as
the fractions add up to 1.
In theory any (n þ 1) independent variables could have been specified to define the
problem, but it is clear that the use of the above variables will lead to a straightforward
solution of the problem.
When variables identified by the application of the description rule are replaced, it
is important to ensure that those selected are truly independent, and that the values
assigned to them lie within the range of possible, practical values.
The number of independent variables that have to be specified to define a problem
will depend on the type of separation process being considered. Some examples of the
11.4. DESIGN VARIABLES IN DISTILLATION 651
application of the description rule to more complex columns are given by Hanson
et al. (1962).
11.5. DESIGN METHODS FOR BINARY SYSTEMS
A good understanding of the basic equations developed for binary systems is essential
to the understanding of distillation processes.
The distillation of binary mixtures is covered thoroughly in Richardson et al.
(2002) and the discussion in this section is limited to a brief review of the most useful
design methods. Though binary systems are usually considered separately, the design
methods developed for multicomponent systems (Section 11.6) can obviously also be
used for binary systems. With binary mixtures, fixing the composition of one com-
ponent fixes the composition of the other, and iterative procedures are not usually
needed to determine the stage and reflux requirements; simple graphical methods are
11.5.1. Basic Equations
Sorel (1899) first derived and applied the basic stage equations to the analysis of
binary systems. Figure 11.4a shows the flows and compositions in the top part of a
column. Taking the system boundary to include the stage n and the condenser gives
the following equations:
D, xd, hd
B, xb, hb
yn+1,Vn+1 Ln, xn
yn, V´n L´n+1, xn+1
Figure 11.4. Column flows and compositions. (a) Above feed. (b) Below feed.
652 CHAPTER 11 SEPARATION COLUMNS (DISTILLATION, ABSORPTION, AND EXTRACTION)
total flows Vnþ1 ¼ Ln þ D (11:11)
for either component Vnþ1ynþ1 ¼ Lnxn þ Dxd (11:12)
total stream enthalpies Vnþ1Hnþ1 ¼ Lnhn þ Dhd þ qc (11:13)
where qc is the heat removed in the condenser.
Combining equations 11.11 and 11.12 gives
Ln þ D
Ln þ D
Combining equations 11.11 and 11.13 gives
Vnþ1Hnþ1 ¼ (Ln þ D)Hnþ1 ¼ Lnhn þ Dhd þ qc (11:15)
Analogous equations can be written for the stripping section, as in Figure 11.4b.
n þ B
n þ B
nþ1hnþ1 ¼ (V0
n þ B)hnþ1 ¼ V0
nHn þ Bhb À qb (11:17)
At constant pressure, the stage temperatures will be functions of the vapor and
liquid compositions only (dew and bubble points), and the specific enthalpies will
therefore also be functions of composition
H ¼ f(y) (11:18a)
h ¼ f(x) (11:18b)
Lewis-Sorel Method (Equimolar Overflow)
Formost distillation problems, a simplifying assumption, first proposed by Lewis (1909),
can be made that eliminates the need to solve the stage energy-balance equations. The
molar liquid and vapor flow rates are taken as constant in the stripping and rectifying
sections. This condition is referred to as equimolar overflow: the molar vapor and liquid
flows from each stage are constant. This will be true only where the component molar
latent heats of vaporization are the same and, together with the specific heats, are
constant over the range of temperature in the column; there is no significant heat of
mixing; and the heat losses are negligible. These conditions are substantially true for
practical systems when the components form near-ideal liquid mixtures.
Even when the latent heats are substantially different, the error introduced by
assuming equimolar overflow to calculate the number of stages is usually small and
11.5. DESIGN METHODS FOR BINARY SYSTEMS 653
With equimolar overflow, equations 11.14 and 11.16 can be written without the
subscripts to denote the stage number:
L þ D
L þ D
V0 þ B
V0 þ B
where L ¼ the constant liquid flow in the rectifying section ¼ the reflux flow, L0, and
is the constant vapor flow in the stripping section.
Equations 11.19 and 11.20 can be written in an alternative form:
where V is the constant vapor flow in the rectifying section ¼ (L þ D), and L0
constant liquid flow in the stripping section ¼ V0
These equations are linear, with slopes L/V and L0
. They are referred to as
operating lines and give the relationship between the liquid and vapor compositions
between stages. For an equilibrium stage, the compositions of the liquid and vapor
streams leaving the stage are given by the equilibrium relationship.
11.5.2. McCabe-Thiele Method
Equations 11.21 and 11.22 and the equilibrium relationship are conveniently solved
by the graphical method developed by McCabe and Thiele (1925). A simple proced-
ure for the construction of the diagram is given in this section and illustrated in
Refer to Figure 11.5; all compositions are those of the more volatile component.
1. Plot the vapor-liquid equilibrium curve from data available at the column
operating pressure. In terms of relative volatility
(1 þ (a À 1)x)
where a is the geometric average relative volatility of the lighter (more volatile)
component with respect to the heavier component (less volatile).
It is usually more convenient and less confusing to use equal scales for the x
and y axes.
2. Make a material balance over the column to determine the top and bottom
compositions, xd and xb, from the data given.
3. The top and bottom operating lines intersect the diagonal at xd and xb, respec-
tively; mark these points on the diagram.
654 CHAPTER 11 SEPARATION COLUMNS (DISTILLATION, ABSORPTION, AND EXTRACTION)
4. The point of intersection of the two operating lines is dependent on the phase
condition of the feed. The line on which the intersection occurs is called the q
line. The q line is found as follows:
i. calculate the value of the ratio q given by
heat to vaporize 1 mol of feed
molar latent heat of feed
ii. plot the q line, slope ¼ q/(q À 1), intersecting the diagonal at zf (the feed
5. Select the reflux ratio and determine the point where the top operating line
extended cuts the y axis:
1 þ R
6. Draw in the top operating line, from xd on the diagonal to f.
7. Draw in the bottom operating line, from xb on the diagonal to the point of
intersection of the top operating line and the q line.
8. Starting at xd or xb, step off the number of stages.
Note: The feed point should be located on the stage closest to the intersection of the
The reboiler and a partial condenser, if used, act as equilibrium stages. However, in
the design of a column, there is little point in reducing the estimated number of stages
to account for this; they can be considered additional factors of safety.
The efficiency of real contacting stages can be accounted for by reducing the height
of the steps on the McCabe-Thiele diagram; see Figure 11.6. Stage efficiencies are
discussed in Section 11.10.
Figure 11.5. McCabe-Thiele diagram.
11.5. DESIGN METHODS FOR BINARY SYSTEMS 655
The McCabe-Thiele method can be used for the design of columns with
side streams and multiple feeds. The liquid and vapor flows in the sections between
the feed and takeoff points are calculated, and operating lines are drawn for each
Stage Vapor and Liquid Flows Not Constant
The McCabe-Thiele method can be used when the condition of equimolar overflow
cannot be assumed, but the operating lines will not be straight. They can be drawn by
making energy balances at a sufficient number of points to determine the approximate
slope of the lines; see Hengstebeck (1976). Alternatively, the more rigorous graphical
method of Ponchon and Savarit can be used. Nowadays, it should rarely be necessary
to resort to complex graphical methods when the simple McCabe-Thiele diagram is
not sufficiently accurate, as computer programs will normally be available for the
rigorous solution of such problems.
11.5.3 Low Product Concentrations
When the concentration of a species in either product is very low, the steps on the
McCabe-Thiele diagram become very small and difficult to plot. This problem can be
overcome by replotting the top or bottom sections to a larger scale or on log-log
paper. In a log plot the operating line will not be straight and must be drawn by
plotting points calculated using equations 11.21 and 11.22. This technique is de-
scribed by Alleva (1962) and is illustrated in Example 11.2.
If the operating and equilibrium lines are straight, and they usually can be taken as
such when the concentrations are small, the number of stages required can be
calculated using the equations given by Robinson and Gilliland (1950).
Stage efficiency =
Figure 11.6. Stage efficiency.
656 CHAPTER 11 SEPARATION COLUMNS (DISTILLATION, ABSORPTION, AND EXTRACTION)
For the stripping section:
s0 À 1
s0 (K0 À 1)
þ 1 (11:25)
s ¼ number of ideal stages required from xb to some reference point x0
xb ¼ mol fraction of the more volatile component in the bottom product;
r ¼ mol fraction of more volatile component at the reference point;
¼ slope of the bottom operating line;
¼ equilibrium constant for the more volatile component.
For the rectifying section:
(1 À s) þ xr=xd(s À K)
1 À K
À 1 (11:26)
r ¼ number of stages required from some reference point xr to xd;
xd ¼ mol fraction of the least volatile component in the top product;
xr ¼ mol fraction of least volatile component at reference point;
K ¼ equilibrium constant for the least volatile component;
s ¼ slope of top operating line.
Note: At low concentrations, K ¼ a.
The use of these equations is illustrated in Example 11.3.
Acetone is to be recovered from an aqueous waste stream by continuous distillation.
The feed will contain 10% w/w acetone. Acetone of at least 98% purity is wanted, and
the aqueous effluent must not contain more than 50 ppm acetone. The feed will be at
208C. Estimate the number of ideal stages required.
There is no point in operating this column at other than atmospheric pressure.
The equilibrium data available for the acetone-water system were discussed in
Chapter 8, Section 8.4.
11.5. DESIGN METHODS FOR BINARY SYSTEMS 657
The data of Kojima et al. (1968) will be used.
Mol fraction x, liquid 0.00 0.05 0.10 0.15 0.20 0.25 0.30
Acetone y, vapor 0.00 0.6381 0.7301 0.7716 0.7916 0.8034 0.8124
bubble point 8C 100.0 74.80 68.53 65.26 63.59 62.60 61.87
x 0.35 0.40 0.45 0.50 0.55 0.60 0.65
y 0.8201 0.8269 0.8376 0.8387 0.8455 0.8532 0.8615
8C 61.26 60.75 60.35 59.95 59.54 59.12 58.71
x 0.70 0.75 0.80 0.85 0.90 0.95
y 0.8712 0.8817 0.8950 0.9118 0.9335 0.9627
8C 58.29 57.90 57.49 57.08 56.68 56.30
The equilibrium curve can be drawn with sufficient accuracy to determine the stages
above the feed by plotting the concentrations at increments of 0.1. The diagram
would normally be plotted at about twice the size of Figure 11.7.
x, mol fraction acetone
0 0.2 0.4 0.6 0.8
Figure 11.7. McCabe-Thiele plot, Example 11.2.
658 CHAPTER 11 SEPARATION COLUMNS (DISTILLATION, ABSORPTION, AND EXTRACTION)
Mol fraction acetone feed ¼
top product ¼
bottom product ¼ 50 Â 10À6
¼ 15:5 Â 10À6
Feed Condition (Q-Line)
Bubble point of feed (interpolated) ¼ 838C
Latent heats, water 41,360, acetone 28,410 J/mol
Mean specific heats, water 75.3, acetone 128 J/mol 8C
Latent heat of feed ¼ 28,410 Â 0.033 þ (1 À 0.033) 41,360 ¼ 40,933 J/mol
Specific heat of feed ¼ (0.033 Â 128) þ (1 À 0.033) 75.3 ¼ 77.0 J/mol 8C
Heat to vaporize 1 mol of feed ¼ (83 À 20) 77.0 þ 40,933 ¼ 45,784 J
Slope of q line ¼
1:12 À 1
For this problem the condition of minimum reflux occurs where the top operating
line just touches the equilibrium curve at the point where the q line cuts the curve.
From Figure 11.7,
f for the operating line at minimum reflux ¼ 0:65
From equation 11.24, Rmin ¼ 0.94/0.65 À 1 ¼ 0.45
Take R ¼ Rmin  3
As the flows above the feed point will be small, a high reflux ratio is justified; the
condenser duty will be small.
At R ¼ 3 Â 0:45 ¼ 1:35, f ¼
1 þ 1:35
For this problem it is convenient to step the stages off starting at the intersection
of the operating lines. This gives three stages above the feed up to y ¼ 0.8. The
top section is drawn to a larger scale, as shown in Figure 11.8, to determine the stages
above y ¼ 0.8: three to four stages required; total stages above the feed, seven.
Below the feed, one stage is required down to x ¼ 0.04. A log-log plot is used to
determine the stages below this concentration. Data for log-log plot:
operating line slope, from Figure 11.7 ¼ 0.45/0.09 ¼ 5.0
operating line equation, y ¼ 4.63(x À xb) þ xb
¼ 5.0x À 62.0 Â 10À6
11.5. DESIGN METHODS FOR BINARY SYSTEMS 659
equilibrium line slope, from vÀlÀe data ¼ 0.6381/0.05 ¼ 12.8
x ¼ 4 Â 10À2
4 Â 10À5
2 Â 10À5
Equilibrium line y ¼ 0.51 1.3 Â 10À2
1.3 Â 10À3
5.1 Â 10À4
2.6 Â 10À4
Operating line y ¼ 0.20 4.9 Â 10À3
4.4 Â 10À4
1.4 Â 10À4
3.8 Â 10À5
From Figure 11.9, the number of stages required for this section ¼ 8.
Total number of stages below feed ¼ 9.
Total stages ¼ 7 þ 9 ¼ 16:
Figure 11.8. Top section enlarged.
10−5 10−4 10−3 10−2 10−1
Figure 11.9. Log-log plot of McCabe-Thiele diagram.
660 CHAPTER 11 SEPARATION COLUMNS (DISTILLATION, ABSORPTION, AND EXTRACTION)
For the problem specified in Example 11.2, estimate the number of ideal stages
required below an acetone concentration of 0.04 (more volatile component) using
the Robinson-Gilliland equation.
From the McCabe-Thiele diagram in Example 11.2:
slope of bottom operating line, s0
slope of equilibrium line, K0
xb ¼ 15:5 Â 10À6
15:5 Â 10À6
(12:8 À 1)
þ 1 ¼ 8:9, say 9
11.5.4. The Smoker Equations
Smoker (1938) derived an analytical equation that can be used to determine the
number of stages when the relative volatility is constant. Though his method can be
used for any problem for which the relative volatilities in the rectifying and stripping
sections can be taken as constant, it is particularly useful for problems where the
relative volatility is low, for example, in the separation of close boiling isomers. If the
relative volatility is close to one, the number of stages required will be very large, and
it will be impractical to draw a McCabe-Thiele diagram. The derivations of the
equations are outlined in this section and illustrated in Example 11.4.
Derivations of the Equations
A straight operating line can be represented by the equation
y ¼ sx þ c (11:27)
and in terms of relative volatility, the equilibrium values of y are given by
1 þ (a À 1)x
Eliminating y from these equations gives a quadratic in x:
s(a À 1)x2
þ [s þ b(a À 1) À a]x þ b ¼ 0 (11:28)
For any particular distillation problem, equation 11.28 will have only one real root
k between 0 and 1:
s(a À 1)k2
þ [s þ b(a À 1) À a]k þ b ¼ 0 (11:29)
11.5. DESIGN METHODS FOR BINARY SYSTEMS 661
k is the value of the x ordinate at the point where the extended operating lines
intersect the vapor-liquid equilibrium curve. Smoker shows that the number of stages
required is given by the equation
N ¼ log
0(1 À bxÃ
n(1 À bxÃ
sc(a À 1)
a À sc2
N ¼ number of stages required to effect the separation represented by the
concentration change from
n to xx
0 ; xÃ
¼ (x À k) and xÃ
c ¼ 1 þ (a À 1)k (11:32)
s ¼ slope of the operating line between xÃ
n and xÃ
a ¼ relative volatility, assumed constant over xÃ
n to xÃ
For a column with a single feed and no side streams:
0 ¼ xd À k (11:33)
n ¼ zf À k (11:34)
R þ 1
R þ 1
0 ¼ zf À k (11:37)
n ¼ xb À k (11:38)
Rzf þ xd À (R þ 1)xb
(R þ 1)(zf À xb)
(zf À xd)xb
(R þ 1)(zf À xb)
If the feed stream is not at its bubble point, zf is replaced by the value of x at the
intersection of operating lines, given by
q À 1
q À 1
All compositions for the more volatile component.
662 CHAPTER 11 SEPARATION COLUMNS (DISTILLATION, ABSORPTION, AND EXTRACTION)
A column is to be designed to separate a mixture of ethylbenzene and styrene. The
feed will contain 0.5 mol fraction styrene, and a styrene purity of 99.5% is required,
with a recovery of 85%. Estimate the number of equilibrium stages required at a
reflux ratio of 8. Maximum column bottom pressure 0.20 bar.
Ethylbenzene is the more volatile component.
Antoine equations, ethylbenzene, ln P
¼ 9:386 À
T À 59:95
styrene ln P
¼ 9:386 À
T À 63:72
P bar, T Kelvin
Material balance, basis 100 kmol feed:
at 85% recovery, styrene in bottoms ¼ 50 Â 0.85 ¼ 42.5 kmol
at 99:5% purity, ethylbenzene in bottoms ¼
 0:5 ¼ 0:21 kmol
ethylbenzene in the tops ¼ 50 À 0:21 ¼ 49:79 kmol
styrene in tops ¼ 50 À 42:5 ¼ 7:5 kmol
mol fraction ethylbenzene in tops ¼
49:79 þ 7:5
zf ¼ 0:5, xb ¼ 0:005, xd ¼ 0:87
Column bottom temperature, from Antoine equation for styrene
ln 0:2 ¼ 9:386 À
T À 63:72
T ¼ 366 K, 93:3
At 93.38C, vapor pressure of ethylbenzene
¼ 9:386 À
366:4 À 59:95
¼ 0:27 bar
Relative volatility ¼
The relative volatility will change as the compositions and (particularly for a vacuum
column) the pressure changes up the column. The column pressures cannot be
estimated until the number of stages is known; so as a first trial, the relative volatility
will be taken as constant, at the value determined by the bottom pressure.
11.5. DESIGN METHODS FOR BINARY SYSTEMS 663
8 þ 1
¼ 0:89 (11:35)
8 þ 1
¼ 0:097 (11:36)
k ¼ 0:290
0 ¼ 0:87 À 0:29 ¼ 0:58 (11:33)
n ¼ 0:50 À 0:29 ¼ 0:21 (11:34)
c ¼ 1 þ (1:35 À 1)0:29 ¼ 1:10 (11:32)
0:89 Â 1:10(1:35 À 1)
1:35 À 0:89 Â 1:12
¼ 1:255 (11:31)
N ¼ log
0:58(1 À 1:255 Â 0:21)
0:21(1 À 1:255 Â 0:58)
0:89 Â 1:12
¼ 8:87, say 9
Stripping Section, Feed Taken as at Its Bubble Point
8 Â 0:5 þ 0:87 À (8 þ 1)0:005
(8 þ 1)(0:5 À 0:005)
¼ 1:084 (11:39)
(0:5 À 0:87)0:005
(8 þ 1)(0:5 À 0:005)
¼ À4:15 Â 10À4
(essentially zero) (11:40)
1:084(1:35 À 1)k2
þ [1:084 À 4:15 Â 10À4
(1:35 À 1) À 1:35] (11:29)
k À 4:15 Â 10À4
k ¼ 0:702
0 ¼ 0:5 À 0:702 ¼ À0:202 (11:37)
n ¼ 0:005 À 0:702 ¼ À0:697 (11:38)
c ¼ 1 þ (1:35 À 1)0:702 ¼ 1:246 (11:32)
1:084 Â 1:246(1:35 À 1)
1:35 À 1:084 Â 1:2462
¼ À1:42 (11:31)
N ¼ log
À0:202(1 À 0:697 Â 1:42)
À0:697(1 À 0:202 Â 1:42)
1:084 Â 1:2462
log [4:17 Â 10À3
¼ 24:6, say 25 (11:30)
664 CHAPTER 11 SEPARATION COLUMNS (DISTILLATION, ABSORPTION, AND EXTRACTION)
11.6. MULTICOMPONENT DISTILLATION: GENERAL CONSIDERATIONS
The problem of determining the stage and reflux requirements for multicomponent
distillations is much more complex than for binary mixtures. With a multicomponent
mixture, fixing one component composition does not uniquely determine the other
component compositions and the stage temperature. Also, when the feed contains
more than two components, it is not possible to specify the complete composition of
the top and bottom products independently. The separation between the top and
bottom products is specified by setting limits on two ‘‘key’’ components, between
which it is desired to make the separation.
The complexity of multicomponent distillation calculations can be appreciated by
considering a typical problem. The normal procedure is to solve the MESH equations
(Section 11.3.1) stage-by-stage, from the top and bottom of the column toward the
feed point. For such a calculation to be exact, the compositions obtained from both
the bottom-up and top-down calculations must mesh at the feed point and match the
feed composition. But the calculated compositions will depend on the compositions
assumed for the top and bottom products at the commencement of the calculations.
Though it is possible to match the key components, the other components will not
match unless the designer was particularly fortunate in choosing the trial top and
bottom compositions. For a completely rigorous solution, the compositions must be
adjusted and the calculations repeated until a satisfactory mesh at the feed point is
obtained. Clearly, the greater the number of components, the more difficult the
problem. As was shown in Section 11.3.2, trial-and-error calculations will be needed
to determine the stage temperatures. For other than ideal mixtures, the calculations
will be further complicated by the fact that the component volatilities will be func-
tions of the unknown stage compositions. If more than a few stages are required,
stage-by-stage calculations are complex and tedious, as illustrated in Example 11.9.
Before the advent of the modern digital computer, various ‘‘shortcut’’ methods
were developed to simplify the task of designing multicomponent columns. A com-
prehensive summary of the methods used for hydrocarbon systems is given by
Edmister (1947 to 1949) in a series of articles in the journal The Petroleum Engineer.
Though computer programs will normally be available for the rigorous solution of the
MESH equations, shortcut methods are still useful in the preliminary design work and
as an aid in defining problems for computer solution. Intelligent use of the shortcut
methods can reduce the computer time and costs.
The shortcut methods available can be divided into two classes:
1. Simplification of the rigorous stage-by-stage procedures to enable the calcula-
tions to be done by hand or graphically. Typical examples of this approach are
the methods given by Smith and Brinkley (1960) and Hengstebeck (1976).
These are described in Section 11.7, and Hengstebeck’s method is illustrated
by a worked example.
2. Empirical methods, which are based on the performance of operating columns,
or the results of rigorous designs. Typical examples of these methods are
11.6. MULTICOMPONENT DISTILLATION: GENERAL CONSIDERATIONS 665
Gilliland’s correlation, which is given in Richardson et al. (2002), and the
Erbar-Maddox correlation given in Section 11.7.3.
11.6.1. Key Components
Before starting the column design, the designer must select the two key components
between which it is desired to make the separation. The light key will be the
component that it is desired to keep out of the bottom product; and the heavy key,
the component to be kept out of the top product. Specifications will be set on the
maximum concentrations of the keys in the top and bottom products. The keys are
known as adjacent keys if they are ‘‘adjacent’’ in a listing of the components in order
of volatility, and split keys if some other component lies between them in the order;
they will usually be adjacent.
Which components are the key components will normally be clear, but sometimes,
particularly if close boiling isomers are present, judgment must be used in their
selection. If any uncertainty exists, trial calculations should be made using different
components as the keys to determine the pair that requires the largest number of
stages for separation (the worst case). The Fenske equation can be used for these
calculations; see Section 11.7.3.
The nonkey components that appear in both top and bottom products are known
as distributed components; and those that are not present, to any significant extent, in
one or another product, are known as nondistributed components.
11.6.2. Number and Sequencing of Columns
As was mentioned in Section 11.2, in multicomponent distillations it is not possible to
obtain more than one pure component, one sharp separation, in a single column. If
a multicomponent feed is to be split into two or more virtually pure products, several
columns will be needed. Impure products can be taken off as side streams, and the
removal of a side stream from a stage where a minor component is concentrated will
reduce the concentration of that component in the main product.
For separation of N components, with one essentially pure component taken
overhead, or from the bottom of each column, (N À 1) columns will be needed to
obtain complete separation of all components. For example, to separate a mixture of
benzene, toluene, and xylene, two columns are needed (3 À 1). Benzene is taken
overhead from the first column, and the bottom product, essentially free of benzene, is
fed to the second column. This column separates the toluene and xylene.
The order in which the components are separated will determine the capital
and operating costs. Where there are several components, the number of possible
sequences can be very large; for example, with 5 components, the number is 14,
whereas with 10 components, it is near 5000. When systems that require the separ-
ation of several components are designed, efficient procedures are needed to deter-
mine the optimum sequence of separation; see Doherty and Malone (2001), Smith
(1995), and Kumar (1982).
666 CHAPTER 11 SEPARATION COLUMNS (DISTILLATION, ABSORPTION, AND EXTRACTION)
Procedures for the sequencing of columns are also available in the commercial
process simulator programs; for example, DISTIL in Aspen Technology’s suite of
programs (see Chapter 4, Table 4.1).
In this section, it is possible to give only some general guide rules.
Heuristic Rules for Optimum Sequencing
1. Remove the components one at a time; as in the benzene-toluene-xylene
2. Remove any components that are present in large excess early in the sequence.
3. With difficult separations, involving close boiling components, postpone the
most difficult separation to late in the sequence.
Difficult separations will require many stages, so the column diameter should be
made as small as possible to reduce cost. Column diameter is dependent on flow rate;
see Section 11.11. The further down the sequence, the smaller will be the amount of
material that the column has to handle.
Where a large number of stages is required, it may be necessary to split a column into
two separate columns to reduce the height of the column, even though the required
separation could, theoretically, have been obtained in a single column. This may also
be done in vacuum distillations, to reduce the column pressure drop and limit the
11.7. MULTICOMPONENT DISTILLATION: SHORTCUT METHODS FOR
STAGE AND REFLUX REQUIREMENTS
Some of the more useful shortcut procedures that can be used to estimate stage and reflux
requirements without the aid of computers are given in this section. Most of the shortcut
methods were developed for the design of separation columns for hydrocarbon systems
in the petroleum and petrochemical systems industries, and caution must be exercised
when applying them to other systems. They usually depend on the assumption of
constant relative volatility and should not be used for severely nonideal systems.
Shortcut methods for nonideal and azeotropic systems are given by Featherstone
11.7.1. Pseudo-Binary Systems
If the presence of the other components does not significantly affect the volatility of
the key components, the keys can be treated as a pseudo-binary pair. The number of
stages can then be calculated using a McCabe-Thiele diagram or the other methods
developed for binary systems. This simplification can often be made when the number
11.7. MULTICOMPONENT DISTILLATION: STAGE AND REFLUX REQUIREMENTS 667
of the nonkey components is small or where the components form near-ideal mix-
Where the concentration of the nonkeys is small, say less than 10%, they can be
lumped in with the key components. For higher concentrations the method proposed
by Hengstebeck (1946) can be used to reduce the system to an equivalent binary
system. Hengstebeck’s method is outlined in this section and illustrated in Example
11.5. Hengstebeck’s book (1976) should be consulted for the derivation of the
method and further examples of its application.
For any component i, the Lewis-Sorel material balance equations (Section 11.5) and
equilibrium relationship can be written in terms of the individual component molar
flow rates, in place of the component composition:
vnþ1;i ¼ ln;i þ di (11:42)
vn;i ¼ Kn;i
for the stripping section:
nþ1;i ¼ v0
n;i þ bi (11:44)
n;i ¼ Kn;i
ln,i ¼ the liquid flow rate of any component i from stage n;
vn,i ¼ the vapor flow rate of any component i from stage n;
di ¼ the flow rate of component i in the tops;
bi ¼ the flow rate of component i in the bottoms;
Kn,i ¼ the equilibrium constant for component i at stage n.
The superscript 0
denotes the stripping section.
V and L are the total flow rates, assumed constant.
To reduce a multicomponent system to an equivalent binary, it is necessary to
estimate the flow rate of the key components throughout the column. Hengstebeck
makes use of the fact that in a typical distillation the flow rates of each of the light
nonkey components approaches a constant, limiting rate in the rectifying section,
and the flows of each of the heavy nonkey components approach limiting flow rates
in the stripping section. Putting the flow rates of the nonkeys equal to these limiting
rates in each section enables the combined flows of the key components to be
Le ¼ L À
Ve ¼ V À
668 CHAPTER 11 SEPARATION COLUMNS (DISTILLATION, ABSORPTION, AND EXTRACTION)
e ¼ L0
e ¼ V0
Ve and Le are the estimated flow rates of the combined keys;
li and vi are the limiting liquid and vapor rates of components lighter than the keys
in the rectifying section;
i and v0
i are the limiting liquid and vapor rates of components heavier than the
keys in the stripping section.
The method used to estimate the limiting flow rates is that proposed by Jenny
The equations are
ai À 1
vi ¼ li þ di (11:51)
aLK À ai
i ¼ v0
i þ bi (11:53)
ai ¼ relative volatility of component i, relative to the heavy key (HK);
aLK ¼ relative volatility of the light key (LK), relative to the heavy key.
Estimates of the flows of the combined keys enable operating lines to be drawn for
the equivalent binary system. The equilibrium line is drawn by assuming a constant
relative volatility for the light key:
1 þ (aLK À 1)x
where y and x refer to the vapor and liquid concentrations of the light key.
Hengstebeck shows how the method can be extended to deal with situations in
which the relative volatility cannot be taken as constant and how to allow for
variations in the liquid and vapor molar flow rates. He also gives a more rigorous
graphical procedure based on the Lewis-Matheson method (see Section 11.8).
Estimate the number of ideal stages needed in the butane-pentane splitter defined by
the compositions given in the following table. The column will operate at a pressure
of 8.3 bar, with a reflux ratio of 2.5. The feed is at its boiling point.
11.7. MULTICOMPONENT DISTILLATION: STAGE AND REFLUX REQUIREMENTS 669
Note: A similar problem has been solved by Lyster et al. (1959) using a rigorous
computer method, and it was found that 10 stages were needed.
Feed ( f ) Tops (d) Bottoms (b)
Propane, C3 5 5 0
i-Butane, iC4 15 15 0
n-Butane, nC4 25 24 1
i-Pentane, iC5 20 1 19
n-Pentane, nC5 35 0 35
100 45 55 kmol
The top and bottom temperatures (dew points and bubble points) were calculated by
the methods illustrated in Example 11.9. Relative volatilities are given by equation
Equilibrium constants were taken from the De Priester charts (Chapter 8).
Top Bottom Average
Temp. 8C 65 120
C3 5.5 4.5 5.0
iC4 2.7 2.5 2.6
(LK) nC4 2.1 2.0 2.0
(HK) iC5 1.0 1.0 1.0
nC5 0.84 0.85 0.85
Calculations of nonkey flows:
Equations 11.50, 11.51, 11.52, 11.53
ai di li ¼ di=(ai À 1) vi ¼ li þ di
C3 5 5 1.3 6.3
iC4 2.6 15 9.4 24.4
li ¼ 10:7
vi ¼ 30:7
ai bi v0
i ¼ aibi=(aLK À ai) l0
i ¼ v0
i þ bi
nC5 0.85 35 25.9 60.9
i ¼ 25:9
i ¼ 60:9
670 CHAPTER 11 SEPARATION COLUMNS (DISTILLATION, ABSORPTION, AND EXTRACTION)
Flows of combined keys:
Le ¼ 2:5 Â 45 À 10:7 ¼ 101:8 (11:46)
Ve ¼ (2:5 þ 1)45 À 30:7 ¼ 126:8 (11:47)
e ¼ (2:5 þ 1)45 À 25:9 ¼ 131:6 (11:49)
e ¼ (2:5 þ 1)45 þ 55 À 60:9 ¼ 151:6 (11:48)
Slope of top operating line:
Slope of bottom operating line:
flow (LK þ HK)
19 þ 1
24 þ 1
25 þ 20
1 þ (2 À 1)x
1 þ x
x 0 0:20 0:40 0:60 0:80 1:0
y 0 0:33 0:57 0:75 0:89 1:0
The McCabe-Thiele diagram is shown in Figure 11.10.
Twelve stages are required; feed on seventh from base.
11.7.2. Smith-Brinkley Method
Smith and Brinkley developed a method for determining the distribution of compo-
nents in multicomponent separation processes. Their method is based on the solution
of the finite-difference equations that can be written for multistage separation
processes and can be used for extraction and absorption processes, as well as distil-
lation. Only the equations for distillation will be given here. The derivation of
the equations is given by Smith and Brinkley (1960) and Smith (1963). For any
component i (suffix i omitted in the equation for clarity)
(1 À SNrÀNs
r ) þ R(1 À Sr)
(1 À SNrÀNs
r ) þ R(1 À Sr) þ GSNrÀNs
r (1 À SNsþ1
11.7. MULTICOMPONENT DISTILLATION: STAGE AND REFLUX REQUIREMENTS 671
b/f is the fractional split of the component between the feed and the bottoms, and
Nr ¼ number of equilibrium stages above the feed;
Ns ¼ number of equilibrium stages below the feed;
Sr ¼ stripping factor, rectifying section ¼ KiV/L;
Ss ¼ stripping factor, stripping section ¼ K0
V and L are the total molar vapor and liquid flow rates, and the superscript 0
denotes the stripping section.
G depends on the condition of the feed.
If the feed is mainly liquid:
1 À Sr
1 À Ss
and the feed stage is added to the stripping section.
If the feed is mainly vapor:
1 À Sr
1 À Ss
Equation 11.54 is for a column with a total condenser. If a partial condenser is
used, the number of stages in the rectifying section should be increased by one.
0.4 0.6 0.8 1.0
Figure 11.10. McCabe-Thiele diagram for Example 11.5.
672 CHAPTER 11 SEPARATION COLUMNS (DISTILLATION, ABSORPTION, AND EXTRACTION)
The procedure for using the Smith-Brinkley method is as follows:
1. Estimate the flow rates L, Vand L0
, V 0
from the specified component separations
and reflux ratio.
2. Estimate the top and bottom temperatures by calculating the dew and bubble
points for assumed top and bottom compositions.
3. Estimate the feed-point temperature.
4. Estimate the average component K values in the stripping and rectifying
5. Calculate the values of Sr,i for the rectifying section and Ss,i for the stripping
6. Calculate the fractional split of each component and hence the top and bottom
7. Compare the calculated values with the assumed values and check the overall
column material balance.
8. Repeat the calculation until a satisfactory material balance is obtained.
The usual procedure is to adjust the feed temperature up and down until
a satisfactory balance is obtained.
Examples of the application of the Smith-Brinkley method are given by Smith (1963).
This method is basically a rating method, suitable for determining the performance
of an existing column, rather than a design method, as the number of stages must be
It can be used for design by estimating the number of stages by some other method
and using equation 11.54 to determine the top and bottom compositions. The
estimated stages can then be adjusted and the calculations repeated until the required
specifications are achieved. However, the Geddes-Hengstebeck method for estimating
the component splits, described in Section 11.7.4, is easier to use and satisfactory for
11.7.3. Empirical Correlations
The two most frequently used empirical methods for estimating the stage require-
ments for multicomponent distillations are the correlations published by Gilliland
(1940) and by Erbar and Maddox (1961). These methods relate the number of ideal
stages required for a given separation, at a given reflux ratio, to the number at total
reflux (minimum possible) and the minimum reflux ratio (infinite number of stages).
Gilliland’s correlation is given in Richardson et al. (2002).
The Erbar-Maddox correlation is given in this section, as it is now generally
considered to give more reliable predictions. Their correlation is shown in Figure
11.11, which gives the ratio of number of stages required to the number at total
reflux, as a function of the reflux ratio, with the minimum reflux ratio as a parameter.
To use Figure 11.11, estimates of the number of stages at total reflux and the
minimum reflux ratio are needed.
11.7. MULTICOMPONENT DISTILLATION: STAGE AND REFLUX REQUIREMENTS 673
Minimum Number of Stages (Fenske Equation)
The Fenske equation (Fenske, 1932) can be used to estimate the minimum stages
required at total reflux. The derivation of this equation for a binary system is given in
Richardson et al. (2002). The equation applies equally to multicomponent systems
and can be written as
[xi/xr] ¼ the ratio of the concentration of any component i to the concentration of a
reference component r, and the suffixes d and b denote the distillate (tops)
(d) and the bottoms (b);
Nm ¼ minimum number of stages at total reflux, including the reboiler;
ai ¼ average relative volatility of the component i with respect to the reference
0.10 0.20 0.30 0.40 0.50
Nm / N
0.60 0.70 0.80 0.90
Based on Underwood Rm
Figure 11.11. Erbar-Maddox correlation (Erbar and Maddox, 1961).
674 CHAPTER 11 SEPARATION COLUMNS (DISTILLATION, ABSORPTION, AND EXTRACTION)
Normally, the separation required will be specified in terms of the key components,
and equation 11.57 can be rearranged to give an estimate of the number of stages:
where aLK is the average relative volatility of the light key with respect to the heavy
key, and xLK and xHK are the light and heavy key concentrations. The relative
volatility is taken as the geometric mean of the values at the column top and bottom
temperatures. To calculate these temperatures, initial estimates of the compositions
must be made, so the calculation of the minimum number of stages by the Fenske
equation is a trial-and-error procedure. The procedure is illustrated in Example 11.7.
If there is a wide difference between the relative volatilities at the top and bottom of
the column, the use of the average value in the Fenske equation will underestimate the
number of stages. In these circumstances, a better estimate can be made by calculating
the number of stages in the rectifying and stripping sections separately, taking the
feed concentration as the base concentration for the rectifying section and as the
top concentration for the stripping section, and estimating the average relative
volatilities separately for each section. This procedure will also give an estimate of
the feed point location.
Winn (1958) has derived an equation for estimating the number of stages at total
reflux, which is similar to the Fenske equation, but which can be used when the
relative volatility cannot be taken as constant.
If the number of stages is known, equation 11.57 can be used to estimate the split of
components between the top and bottom of the column at total reflux. It can be
written in a more convenient form for calculating the split of components:
di and bi are the flow rates of the component i in the distillate and bottoms;
dr and br are the flow rates of the reference component in the distillate and
Note: From the column material balance
di þ bi ¼ fi
where fi is the flow rate of component i in the feed.
Minimum Reflux Ratio
Colburn (1941) and Underwood (1948) have derived equations for estimating the
minimum reflux ratio for multicomponent distillations. As the Underwood equation is
more widely used, it is presented in this section. The equation can be stated in the form
ai À u
¼ Rm þ 1 (11:60)
11.7. MULTICOMPONENT DISTILLATION: STAGE AND REFLUX REQUIREMENTS 675
ai ¼ the relative volatility of component i with respect to some reference
component, usually the heavy key;
Rm ¼ the minimum reflux ratio;
xi,d ¼ concentration of component i in the tops at minimum reflux;
and u is the root of the equation:
ai À u
¼ 1 À q (11:61)
xi,f ¼ the concentration of component i in the feed, and q depends on the condition
of the feed and was defined in Section 11.5.2.
The value of u must lie between the values of the relative volatility of the light and
heavy keys, and is found by trial and error.
In the derivation of equations 11.60 and 11.61, the relative volatilities are taken as
constant. The geometric average of values estimated at the top and bottom tempera-
tures should be used. This requires an estimate of the top and bottom compositions.
Though the compositions should strictly be those at minimum reflux, the values
determined at total reflux, from the Fenske equation, can be used. A better estimate
can be obtained by replacing the number of stages at total reflux in equation 11.59
with an estimate of the actual number; a value equal to Nm/0.6 is often used. The
Erbar-Maddox method of estimating the stage and reflux requirements, using the
Fenske and Underwood equations, is illustrated in Example 11.7.
A limitation of the Erbar-Maddox and similar empirical methods is that they do not
give the feed-point location. An estimate can be made by using the Fenske equation to
calculate the number of stages in the rectifying and stripping sections separately, but
this requires an estimate of the feed-point temperature. An alternative approach is to
use the empirical equation given by Kirkbride (1944):
¼ 0:206 log
Nr ¼ number of stages above the feed, including any partial condenser;
Ns ¼ number of stages below the feed, including the reboiler;
B ¼ molar flow bottom product;
D ¼ molar flow top product;
xf,HK ¼ concentration of the heavy key in the feed;
xf,LK ¼ concentration of the light key in the feed;
xd,HK ¼ concentration of the heavy key in the top product;
xb,LK ¼ concentration of the light key if in the bottom product.
The use of this equation is illustrated in Example 11.8.
676 CHAPTER 11 SEPARATION COLUMNS (DISTILLATION, ABSORPTION, AND EXTRACTION)
11.7.4. Distribution of Nonkey Components (Graphical Method)
The graphical procedure proposed by Hengstebeck (1946), which is based on the
Fenske equation, is a convenient method for estimating the distribution of components
between the top and bottom products.
Hengstebeck and Geddes (1958) have shown that the Fenske equation can be
written in the form
¼ A þ C log ai (11:63)
Specifying the split of the key components determines the constants A and C in the
The distribution of the other components can be readily determined by plotting
the distribution of the keys against their relative volatility on log-log paper and
drawing a straight line through these two points. The method is illustrated in
Yaws et al. (1979) have shown that the component distributions calculated by
equation 11.63 compare well with those obtained by rigorous plate-by-plate calcula-
Chang (1980) gives a computer program, based on the Geddes-Hengstebeck
equation, for the estimation of component distributions.
Use the Geddes-Hengstebeck method to check the component distributions for the
separation specified in Example 11.5.
Summary of problem, flow per 100 kmol feed
Component ai Feed (fi) Distillate (di) Bottoms (bi)
C3 5 5
iC4 2.6 15
nC4 (LK) 2.0 25 24 1
iC5 (HK) 1.0 20 1 19
nC5 0.85 35
The average volatilities will be taken as those estimated in Example 11.5. Normally, the
volatilities are estimated at the feed bubble point, which gives a rough indication of
the average column temperatures. The dew point of the tops and bubble point of the
bottoms can be calculated once the component distributions have been estimated, and
the calculations repeated with a new estimate of the average relative volatilities, as
11.7. MULTICOMPONENT DISTILLATION: STAGE AND REFLUX REQUIREMENTS 677
For the light key,
For the heavy key,
These values are plotted on Figure 11.12.
The distribution of the nonkeys is read from Figure 11.12 at the appropriate
relative volatility and the component flows calculated from the following equations:
Overall column material balance
fi ¼ di þ bi
0.5 1 2
Figure 11.12. Component distribution (Example 11.6).
678 CHAPTER 11 SEPARATION COLUMNS (DISTILLATION, ABSORPTION, AND EXTRACTION)
ai fi di/bi di bi
C3 5 5 40,000 5 0
iC4 2.6 15 150 14.9 0.1
nC4 2.0 25 21 24 1
iC5 1.0 20 0.053 1 19
nC5 0.85 35 0.011 0.4 34.6
As these values are close to those assumed for the calculation of the dew points and
bubble points in Example 11.5, there is no need to repeat with new estimates of the
For the separation specified in Example 11.5, evaluate the effect of changes in reflux
ratio on the number of stages required. This is an example of the application of the
The relative volatilities estimated in Example 11.5 and the component distributions
calculated in Example 11.6 will be used for this example.
Summary of data
ai fi di bi
C3 5 5 5 0
iC4 2.6 15 14.9 0.1
nC4 (LK) 2.0 25 24 1
iC5 (HK) 1 20 1 19
nC5 0.85 35 0.4 34.6
100 D ¼ 45.3 B ¼ 54.7
Minimum number of stages; Fenske equation, equation 11.58:
11.7. MULTICOMPONENT DISTILLATION: STAGE AND REFLUX REQUIREMENTS 679
Minimum reflux ratio; Underwood equations 11.60 and 11.61.
This calculation is best tabulated.
As the feed is at its boiling point, q ¼ 1
ai À u
¼ 0 (11:61)
xi,f ai aixi,f u ¼ 1.5 u ¼ 1.3 u ¼ 1.35
0.05 5 0.25 0.071 0.068 0.068
0.15 2.6 0.39 0.355 0.300 0.312
0.25 2.0 0.50 1.000 0.714 0.769
0.20 1 0.20 À 0.400 À 0.667 À 0.571
0.35 0.85 0.30 À 0.462 À 0.667 À 0.600
S ¼ 0.564 À 0.252 0.022
xi,d ai aixi,d aixi,d/(ai À u)
0.11 5 0.55 0.15
0.33 2.6 0.86 0.69
0.53 2.0 1.08 1.66
0.02 1 0.02 À 0.06
0.01 0.85 0.01 À 0.02
S ¼ 2:42
Rm þ 1 ¼ 2:42
Rm ¼ 1:42
(Rm þ 1)
Specimen calculation, for R ¼ 2.0
(R þ 1)
from Figure 11.11
680 CHAPTER 11 SEPARATION COLUMNS (DISTILLATION, ABSORPTION, AND EXTRACTION)
for other reflux ratios
R 2 3 4 5 6
N 15.7 11.9 10.7 10.4 10.1
Note: Above a reflux ratio of 4, there is little change in the number of stages required, and the
optimum reflux ratio will be near this value.
Estimate the position of the feed point for the separation considered in Example 11.7
for a reflux ratio of 3.
Use the Kirkbride equation (equation 11.62). Product distributions taken from
¼ 0:206 log
¼ 0:206 log (0:65)
for R ¼ 3, N ¼ 12
number of stages, excluding the reboiler ¼ 11
Nr þ Ns ¼ 11
Ns ¼ 11 À Nr ¼ 11 À 0:91Ns
¼ 5:76, say 6
This checks with the method used in Example 11.5, where the reflux ratio was 2.5.
This example illustrates the complexity and trial-and-error nature of stage-by-stage
The same problem specification has been used in earlier examples to illustrate the
shortcut design methods.
A butane-pentane splitter is to operate at8.3 bar with the following feed composition:
11.7. MULTICOMPONENT DISTILLATION: STAGE AND REFLUX REQUIREMENTS 681
xf f mol/100 mol feed
Propane, C3 0.05 5
Isobutane, iC4 0.15 15
Normal butane, nC4 0.25 25
Isopentane, iC5 0.20 20
Normal pentane, nC5 0.35 35
Light key nC4
Heavy key iC5
For a specification of not more than 1 mol of the light key in the bottom product
and not more than 1 mol of the heavy key in the top product, and a reflux ratio of 2.5,
make a stage-by-stage calculation to determine the product composition and number
of stages required.
Only sufficient trial calculations will be made to illustrate the method used. Basis
100 mol feed.
Estimation of dew and bubble points:
Kixi ¼ 1:0 (11:5a)
¼ 1:0 (11:5b)
The K values, taken from the De Priester charts (Chapter 8), are plotted in Figure
11.13 for easy interpolation.
60 70 80 90 100 110
Figure 11.13. K-values at 8.3 bar.
682 CHAPTER 11 SEPARATION COLUMNS (DISTILLATION, ABSORPTION, AND EXTRACTION)
To estimate the dew and bubble points, assume that nothing heavier than
the heavy key appears in the distillate and nothing lighter than the light key in the
d xd b xb
C3 5 0.11 0 —
C4 15 0.33 0 —
nC4 24 0.54 1 0.02
iC5 1 0.02 19 0.34
nC5 0 — 35 0.64
Bubble-point calculation, bottoms:
Try 1008C Try 1208C
xb Ki Kixi Ki Kixi
C3 — — — — —
iC4 — — — — —
nC4 0.02 1.85 0.04 2.1 0.04
iC5 0.34 0.94 0.32 1.1 0.37
nC5 0.64 0.82 0.52 0.96 0.61
SKixi ¼ 0.88 1.02
temp. too low close enough
Dew-point calculation, distillate
Try 708C Try 608C
xd Ki yi/Ki Ki yi/Ki
C3 0.11 2.6 0.04 2.20 0.24
iC4 0.33 1.3 0.25 1.06 0.35
nC4 0.54 0.9 0.60 0.77 0.42
iC5 0.02 0.46 0.04 0.36 0.01
nC5 — — — — —
Syi/Ki ¼ 0.94 1.02
temp. too high close enough
11.7. MULTICOMPONENT DISTILLATION: STAGE AND REFLUX REQUIREMENTS 683
Bubble-point calculation, feed (liquid feed):
Try 808C Try 908C Try 858C
xf Ki xiKi Ki xiKi Ki xiKi
C3 0.05 2.9 0.15 3.4 0.17 3.15 0.16
iC4 0.15 1.5 0.23 1.8 0.27 1.66 0.25
nC4 0.25 1.1 0.28 1.3 0.33 1.21 0.30
iC5 0.20 0.5 0.11 0.66 0.13 0.60 0.12
nC5 0.35 0.47 0.16 0.56 0.20 0.48 0.17
0.93 1.10 1.00
temp. too low temp. too high satisfactory
Top-down calculations; assume total condensation with no subcooling:
y1 ¼ xd ¼ x0
It is necessary to estimate the composition of the ‘‘nonkeys’’ so that they can be
included in the stage calculations. As a first trial, the following values will be assumed:
C3 0.10 5
iC4 0.33 15
nC4 0.54 24
iC5 0.02 1
nC5 0.001 0.1
In each stage calculation, it will necessary to estimate the stage temperatures to
determine the K values and liquid and vapor enthalpies. The temperature range from
top to bottom of the column will be approximately 120 À 60 ¼ 608C. An approximate
calculation (Example 11.7) has shown that around 14 ideal stages will be needed, so the
temperature change from stage to stage can be expected to be around 4 to 58C.
To = 60 C, Lo VI TI
684 CHAPTER 11 SEPARATION COLUMNS (DISTILLATION, ABSORPTION, AND EXTRACTION)
L0 ¼ R Â D ¼ 2:5 Â 45:1 ¼ 112:8
V1 ¼ (R þ 1)D ¼ 3:5 Â 45:1 ¼ 157:9
Estimation of stage temperature and outlet liquid composition (x1)
Try T1 ¼ 66
C Try T1 ¼ 65
C x1 ¼ yi/Ki
y1 Ki yi/Ki Ki yi/Ki Normalized
C3 0.10 2.40 0.042 2.36 0.042 0.042
iC4 0.33 1.20 0.275 1.19 0.277 0.278
nC4 0.54 0.88 0.614 0.86 0.628 0.629
iC5 0.02 0.42 0.048 0.42 0.048 0.048
nC5 0.001 0.32 0.003 0.32 0.003 0.003
Syi=Ki ¼ 0:982
Summary of stage equations:
L0 þ V2 ¼ L1 þ V1 (i)
L0x0 þ V2y2 ¼ L1x1 þ V1y1 (ii)
h0L0 þ H2V2 ¼ h1L1 þ H1V1 (iii)
h ¼ f(x,T) (iv)
H ¼ f(x,T) (v)
The enthalpy relationship is plotted in Figure 11.14a and b.
yi ¼ Kixi (vi)
Before a heat balance can be made to estimate L1 and V2, an estimate of y2 and T2 is
needed. y2 is dependent on the liquid and vapor flows, so as a first trial, assume that
these are constant and equal to L0 and V1; then, from equations (i) and (ii),
(x1 À x0) þ y1
11.7. MULTICOMPONENT DISTILLATION: STAGE AND REFLUX REQUIREMENTS 685
x1 x0 y2 ¼ 0.71(x1 À x0) þ y1 Normalized
C3 0.042 0.10 0.057 0.057
iC4 0.278 0.33 0.294 0.292
nC4 0.629 0.54 0.604 0.600
iC5 0.048 0.02 0.041 0.041
nC5 0.003 0.001 0.013 0.013
70 80 90 100 110 120
Figure 11.14(a) and (b). Enthalpy kJ/mol [adapted from J. B. Maxwell, Data Book of Hydro-
carbons (Van Nostrand, 1962)].
686 CHAPTER 11 SEPARATION COLUMNS (DISTILLATION, ABSORPTION, AND EXTRACTION)
Enthalpy data from Figure 11.14a and b J/mol
h0(T0 ¼ 60
C) h1(T1 ¼ 65
x0 hi hixi x1 hi hixi
C3 0.10 20,400 2040 0.042 21,000 882
iC4 0.33 23,400 7722 0.278 24,900 6897
nC4 0.54 25,200 13,608 0.629 26,000 16,328
iC5 0.02 27,500 550 0.048 28,400 1363
nC5 0.001 30,000 30 0.003 30,700 92
h0 ¼ 23,950 h1 ¼ 25,562
H1(T1 ¼ 65
C) H2(T2 ¼ 70
v1 Hi Hiyi y2 Hi Hiyi
C3 0.10 34,000 3400 0.057 34,800 1984
iC4 0.33 41,000 13,530 0.292 41,300 12,142
nC4 0.54 43,700 23,498 0.600 44,200 26,697
iC5 0.02 52,000 1040 0.041 52,500 2153
nC5 0.001 54,800 55 0.013 55,000 715
H1 ¼ 41,623 H1 ¼ 43,691
Energy balance (equation iii):
23,950 Â 112:8 þ 43,691V2 ¼ 25,562L1 þ 41,623 Â 157:9
43,691V2 ¼ 255,626L1 þ 3,870,712
Material balance (equation i):
112:8 þ V2 ¼ L1 þ 157:9
43:691(L1 þ 45:1) ¼ 25:562L1 þ 3,870:712
L1 ¼ 104:8
V2 ¼ 104:8 þ 45:1 ¼ 149:9
We could revise calculated values for y2, but L1/V2 is close enough to the assumed
value of 0.71, so there would be no significant difference from the first estimate.
11.7. MULTICOMPONENT DISTILLATION: STAGE AND REFLUX REQUIREMENTS 687
L1= 104.5 V2= 149.6
x1 (known) y2 (known)
Estimation of stage temperature and outlet liquid composition (x2):
T2 ¼ 70
C (use assumed value as first trial)
y2 Ki x2 ¼ y2/Ki
C3 0.057 2.55 0.022 0.022
iC4 0.292 1.30 0.226 0.222
nC4 0.600 0.94 0.643 0.630
iC5 0.041 0.43 0.095 0.093
nC5 0.013 0.38 0.034 0.033
close enough to 1.0
(x2 À x1) þ y2
As a first trial, take L/V as L1/V1 ¼ 0.70
x2 x1 y3 ¼ 0.70(x2 À x1) þ y2 Normalized
C3 0.022 0.042 0.044 0.043
iC4 0.222 0.277 0.256 0.251
nC4 0.630 0.628 0.613 0.601
iC5 0.093 0.048 0.072 0.072
nC5 0.033 0.003 0.035 0.034
688 CHAPTER 11 SEPARATION COLUMNS (DISTILLATION, ABSORPTION, AND EXTRACTION)
Enthalpy data from Figure 11.14a and b:
h2(T2 ¼ 70
C) H3(T3 ¼ 75
x2 hi hix2 y3 Hi Hiy3
C3 0.022 21,900 482 0.043 34,600 1488
iC4 0.222 25,300 5617 0.251 41,800 10,492
nC4 0.630 27,000 17,010 0.601 44,700 26,865
iC5 0.093 29,500 2744 0.072 53,000 3816
nC5 0.033 31,600 1043 0.035 55,400 1939
h2 ¼ 26,896 H3 ¼ 44,600
25,562 Â 104:8 þ 44,600V3 ¼ 4369 Â 149:9 þ 26,896L2
104:8 þ V3 ¼ 149:9 þ L2
L2 ¼ 105:0
V3 ¼ 150:1
¼ 0:70 checks with assumed value:
As the calculated liquid and vapor flows are not changing much from stage to stage,
the calculation will be continued with the value of L/V taken as constant at 0.7.
Try T3 ¼ 75
C (assumed value)
Ki x3 ¼ y3/Ki Normalized y4 ¼ 0.7(x3 À x2) þ y3
C3 2.71 0.016 0.015 0.38
iC4 1.40 0.183 0.177 0.217
nC4 1.02 0.601 0.580 0.570
iC5 0.50 0.144 0.139 0.104
nC5 0.38 0.092 0.089 0.074
11.7. MULTICOMPONENT DISTILLATION: STAGE AND REFLUX REQUIREMENTS 689
Try T4 ¼ 81
Ki x4 ¼ y4/Ki Normalized y5 ¼ 0.7(x4 À x3) þ y4
C3 2.95 0.013 0.013 0.039
iC4 1.55 0.140 0.139 0.199
nC4 1.13 0.504 0.501 0.515
iC5 0.55 0.189 0.188 0.137
nC5 0.46 0.161 0.166 0.118
Try T5 ¼ 85
Ki x5 Normalized y6 ¼ 0.7(x5 À x4) þ y5
1C3 3.12 0.013 0.012 0.038
iC4 1.66 0.120 0.115 0.179
nC4 1.20 0.430 0.410 0.450
iC5 0.60 0.228 0.218 0.159
nC5 0.46 0.257 0.245 0.192
Try T6 ¼ 90
C Try T6 ¼ 92
Ki x6 Ki x6 Normalized y7
C3 3.35 0.011 3.45 0.011 0.011 0.037
iC4 1.80 0.099 1.85 0.097 0.095 0.166
nC4 1.32 0.341 1.38 0.376 0.318 0.386
iC5 0.65 0.245 0.69 0.230 0.224 0.163
nC5 0.51 0.376 0.53 0.362 0.350 0.268
1.072 1.026 1.020
too low close enough
Note: Ratio of LK to HK in liquid from this stage ¼
690 CHAPTER 11 SEPARATION COLUMNS (DISTILLATION, ABSORPTION, AND EXTRACTION)
Try T6 ¼ 97
Ki x7 Normalized
C3 3.65 0.010 0.010
iC4 1.98 0.084 0.083
nC4 1.52 0.254 0.251
iC5 0.75 0.217 0.214
nC5 0.60 0.447 0.442
This is just below the ratio in the feed
So, the feed would be introduced at this stage.
But the composition of the nonkey components on the plate does not match the
C3 0.05 0.10
iC4 0.15 0.084
nC4 0.25 0.254
iC5 0.20 0.217
nC5 0.35 0.447
So, it would be necessary to adjust the assumed top composition and repeat the
To illustrate the procedure, the calculation will be shown for the reboiler and bottom
stage, assuming constant molar overflow.
With the feed at its boiling point and constant molar overflow, the base flows can be
calculated as follows:
¼ V0 ¼ 157:9
¼ L0 þ FEED ¼ 112:8 þ 100 ¼ 212:8
11.7. MULTICOMPONENT DISTILLATION: STAGE AND REFLUX REQUIREMENTS 691
It will be necessary to estimate the concentration of the nonkey components in the
bottom product. As a first trial, take
C3 iC4 nC4 iC5 nC5
0.001 0.001 0.02 0.34 0.64
Check bubble-point estimate of 1208C.
Try 1208C Try 1188C
xB Ki yB ¼ KixB Ki yB
C3 0.001 4.73 0.005 4.60 0.005
iC4 0.001 2.65 0.003 2.58 0.003
nC4 0.02 2.10 0.042 2.03 0.041
iC5 0.34 1.10 0.374 1.06 0.360
nC5 0.64 0.96 0.614 0.92 0.589
too high close enough
v' = 157.9
B = 55
L' = 212.8
¼ 0:74yB þ 0:26xB
692 CHAPTER 11 SEPARATION COLUMNS (DISTILLATION, ABSORPTION, AND EXTRACTION)
Stage 1 from base (B1):
xB yB xB1 xB2 ¼ 0.74(y1B À yB) þ x1B
C3 0.001 0.005 0.004 0.014
iC4 0.001 0.003 0.002 0.036
nC4 0.02 0.041 0.020 0.019
iC5 0.34 0.361 0.356 0.357
nC5 0.64 0.590 0.603 0.559
The calculation is continued stage-by-stage up the column to the feed point (stage 7
from the top). If the vapor composition at the feed point does not mesh with the
top-down calculation, the assumed concentration of the nonkeys in the bottom
product is adjusted and the calculations repeated.
11.8. MULTICOMPONENT SYSTEMS: RIGOROUS SOLUTION
PROCEDURES (COMPUTER METHODS)
The application of digital computers has made the rigorous solution of the MESH
equations (Section 11.3.1) a practical proposition, and computer methods for the
design of multicomponent separation columns will be available in most design organ-
izations. A considerable amount of work has been done over the past 20 or so years to
develop efficient and reliable computer-aided design procedures for distillation and
other staged processes. A detailed discussion of this work is beyond the scope of this
book. Refer to the specialist books that have been published on the subject—Smith
(1963), Holland (1997), and Kister (1992)—and to the numerous papers that have
appeared in the chemical engineering literature. A good summary of the present state
of the art is given by Haas (1992).
Several different approaches have been taken to develop programs that are efficient
in the use of computer time and suitable for the full range of multicomponent
separation processes that are used in the process industries. A design group will use
those methods that are best suited to the processes that it normally handles.
In this section only a brief outline of the methods that have been developed will
The basic steps in any rigorous solution procedure will be
1. Specification of the problem; complete specification is essential for computer
2. Selection of values for the iteration variables, for example, estimated stage
temperatures, and liquid and vapor flows (the column temperature and flow
3. A calculation procedure for the solution of the stage equations;
11.8. MULTICOMPONENT SYSTEMS: RIGOROUS SOLUTION PROCEDURES (COMPUTER METHODS) 693
4. A procedure for the selection of new values for the iteration variables for each
set of trial calculations;
5. A procedure to test for convergence, to check if a satisfactory solution has been
It is convenient to consider the methods available under the following four
6. Lewis-Matheson method;
7. Thiele-Geddes method;
8. Relaxation methods;
9. Linear algebra methods.
Rating and Design Methods
With the exception of the Lewis-Matheson method, all the preceding methods require
the specification of the number of stages below and above the feed point. They are
therefore not directly applicable to design: where the designer wants to determine the
number of stages required for a specified separation. They are strictly what are
referred to as rating methods, used to determine the performance of existing, or
specified, columns. Given the number of stages, they can be used to determine
product compositions. Iterative procedures are necessary to apply rating methods to
the design of new columns. An initial estimate of the number of stages can be
made using shortcut methods and the programs used to calculate the product com-
positions, repeating the calculations with revised estimates until a satisfactory design
11.8.1. Lewis-Matheson Method
The method proposed by Lewis and Matheson (1932) is essentially the application of
the Lewis-Sorel method (Section 11.5.1) to the solution of multicomponent prob-
lems. Constant molar overflow is assumed, and the material balance and equilibrium
relationship equations are solved stage by stage starting at the top or bottom of
the column, in the manner illustrated in Example 11.9. To define a problem for the
Lewis-Matheson method, the following variables must be specified or determined
from other specified variables:
Feed composition, flow rate, and condition;
Distribution of the key components;
One product flow;
Assumed values for the distribution of the nonkey components.
The usual procedure is to start the calculation at the top and bottom of the column
and proceed toward the feed point. The initial estimates of the component distri-
butions in the products are then revised and the calculations repeated until the
694 CHAPTER 11 SEPARATION COLUMNS (DISTILLATION, ABSORPTION, AND EXTRACTION)
compositions calculated from the top and bottom starts mesh and match the feed at
the feed point.
Efficient procedures for adjusting the compositions to achieve a satisfactory mesh
at the feed point are given by Hengstebeck (1976).
In some computer applications of the method, where the assumption of constant
molar overflow is not made, it is convenient to start the calculations by assuming flow
and temperature profiles. The stage component compositions can then be readily
determined and used to revise the profiles for the next iteration. With this modification,
the procedure is similar to the Thiele-Geddes method discussed in the next section.
In general, the Lewis-Matheson method has not been found to be an efficient
procedure for computer solutions other than for relatively straightforward problems.
It is not suitable for problems involving multiple feeds and side streams, or where
more than one column is needed.
The method is suitable for interactive programs run on programmable calculators
and personal computers. Such programs can be ‘‘semi-manual’’ in operation: the
computer solving the stage equations, while control of the iteration variables and
convergence are kept by the designer. As the calculations are carried out one stage at a
time, only a relatively small computer memory is needed.
11.8.2. Thiele-Geddes Method
Like the Lewis-Matheson method, the original method of Thiele and Geddes (1933)
was developed for manual calculation. It has subsequently been adapted by many
workers for computer applications. The variables specified in the basic method or that
must be derived from other specified variables are
Reflux flow rate;
Feed flows and condition;
Number of equilibrium stages above and below the feed point.
The basic procedure used in the Thiele-Geddes method, with examples, is described
in books by Smith (1963) and Deshpande (1985). The application of the method to
computers is covered in a series of articles by Lyster et al. (1959) and Holland (1963).
The method starts with an assumption of the column temperature and flow profiles.
The stage equations are then solved to determine the stage component compositions
and the results used to revise the temperature profiles for subsequent trial calculations.
Efficient convergence procedures have been developed for the Thiele-Geddes method.
The so-called theta method, described by Lyster et al. (1959) and Holland (1963), is
recommended. The Thiele-Geddes method can be used for the solution of complex
distillation problems and for other multicomponent separation processes. A series
of programs for the solution of problems in distillation, extraction, stripping, and
absorption, which use an iterative procedure similar to the Thiele-Geddes method, are
given by Hanson et al. (1962).
11.8. MULTICOMPONENT SYSTEMS: RIGOROUS SOLUTION PROCEDURES (COMPUTER METHODS) 695
11.8.3. Relaxation Methods
With the exception of this relaxation method, all the methods described solve the
stage equations for the steady-state design conditions. In an operating column, other
conditions will exist at startup, and the column will approach the ‘‘design’’ steady-
state conditions after a period of time. The stage material balance equations can be
written in a finite difference form, and procedures for the solution of these equations
will model the unsteady-state behavior of the column.
Rose et al. (1958) and Hanson and Somerville (1963) have applied ‘‘relaxation
methods’’ to the solution of the unsteady-state equations to obtain the steady-state
values. The application of this method to the design of multistage columns is de-
scribed by Hanson and Somerville (1963). They give a program listing and worked
examples for a distillation column with side streams and for a reboiled absorber.
Relaxation methods are not competitive with the steady-state methods in the use
of computer time because of slow convergence. However, because they model the
actual operation of the column, convergence should be achieved for all practical
problems. The method has the potential of development for the study of the
transient behavior of column designs and for the analysis and design of batch
11.8.4. Linear Algebra Methods
The Lewis-Matheson and Thiele-Geddes methods use a stage-by-stage procedure to
solve the equations relating the component compositions to the column temperature
and flow profiles. However, the development of high-speed digital computers with
large memories makes possible the simultaneous solution of the complete set of
MESH equations that describe the stage compositions throughout the column.
If the equilibrium relationships and flow rates are known (or assumed), the set of
material balance equations for each component is linear in the component com-
positions. Amundson and Pontinen (1958) developed a method in which these
equations are solved simultaneously and the results used to provide improved
estimates of the temperature and flow profiles. The set of equations can be
expressed in matrix form and solved using the standard inversion routines available
in modern computer systems. Convergence can usually be achieved after a few
This approach has been further developed by other workers, notably Wang and
Henke (1966) and Naphtali and Sandholm (1971).
The linearization method of Naphtali and Sandholm has been used by Fredenslund
et al. (1977) for the multicomponent distillation program given in their book.
Included in their book and coupled to the distillation program are methods for
estimation of the liquid-vapor relationships (activity coefficients) using the UNIFAC
method (see Chapter 8, Section 8.16.5). This makes the program particularly useful
for the design of columns for new processes, where experimental data for the
equilibrium relationships are unlikely to be available. The program is recommended
to those who do not have access to their own ‘‘in-house’’ programs.
696 CHAPTER 11 SEPARATION COLUMNS (DISTILLATION, ABSORPTION, AND EXTRACTION)
11.9. OTHER DISTILLATION SYSTEMS
11.9.1. Batch Distillation
In batch distillation the mixture to be distilled is charged as a batch to the still and the
distillation carried out until a satisfactory top or bottom product is achieved. The still
usually consists of a vessel surmounted by a packed or plate column. The heater may
be incorporated in the vessel or a separate reboiler used. Batch distillation should be
considered under the following circumstances:
1. Where the quantity to be distilled is small;
2. Where a range of products has to be produced;
3. Where the feed is produced at irregular intervals;
4. Where the feed composition varies over a wide range.
Where the choice between batch and continuous is uncertain, an economic evalu-
ation of both systems should be made.
Batch distillation is an unsteady-state process, the composition in the still
(bottoms) varying as the batch is distilled.
Two modes of operation are used:
1. Fixed reflux, where the reflux rate is kept constant. The compositions will vary
as the more volatile component is distilled off, and the distillation stopped when
the average composition of the distillate collected, or the bottoms left, meet the
2. Variable reflux, where the reflux rate is varied throughout the distillation to
produce a fixed overhead composition. The reflux ratio will need to be pro-
gressively increased as the fraction of the more volatile component in the base
of the still decreases.
The basic theory of batch distillation is given in Hart (1997), Perry et al. (1997),
Richardson et al. (2002) and Walas (1990). In the simple theoretical analysis of batch
distillation columns, the liquid holdup in the column is usually ignored. This holdup
can have a significant effect on the separating efficiency and should be taken into
account when designing batch distillation columns. The practical design of
batch distillation columns is covered by Hengstebeck (1976), Ellerbe (1997), and
11.9.2. Steam Distillation
In steam distillation, steam is introduced into the column to lower the partial pressure
of the volatile components. It is used for the distillation of heat-sensitive products and
for compounds with a high boiling point. It is an alternative to vacuum distillation.
The products must be immiscible with water. Some steam will normally be allowed to
condense to provide the heat required for the distillation. Live steam can be injected
directly into the column base, or the steam generated by a heater in the still or in an
11.9. OTHER DISTILLATION SYSTEMS 697
The design procedure for columns employing steam distillation is essentially the
same as that for conventional columns, making allowance for the presence of steam in
Steam distillation is used extensively in the extraction of essential oils from plant
11.9.3. Reactive Distillation
Reactive distillation is the name given to the process where the chemical reaction and
product separation are carried out simultaneously in one unit. Carrying out the
reaction, with separation and purification of the product by distillation, gives
the following advantages:
1. Chemical equilibrium restrictions are overcome, as the product is removed as it
2. Energy savings can be obtained, as the heat of reaction can be used for the
3. Capital costs are reduced, as only one vessel is required.
The design of reactive distillation columns is complicated by the complex inter-
actions between the reaction and separation processes. A comprehensive discussion of
the process is given by Sundmacher and Kiene (2003).
Reactive distillation is used in the production of methyl tertiary butyl ether
(MTBE) and methyl acetate.
11.10. PLATE EFFICIENCY
The designer is concerned with real contacting stages, not the theoretical equilibrium
stage assumed for convenience in the mathematical analysis of multistage processes.
Equilibrium will rarely be attained in a real stage. The concept of a stage efficiency is used
tolinkthe performanceof practicalcontactingstagestothetheoreticalequilibriumstage.
Three principal definitions of efficiency are used:
1. Murphree plate efficiency (Murphree, 1925), defined in terms of the vapor
yn À ynÀ1
ye À ynÀ1
where ye is the composition of the vapor that would be in equilibrium with the
liquid leaving the plate. The Murphree plate efficiency is the ratio of the actual
separation achieved to that which would be achieved in an equilibrium stage (see
Figure 11.6). In this definition of efficiency, the liquid and the vapor stream are
taken to be perfectly mixed; the compositions in equation 11.64 are the average
composition values for the streams.
2. Point efficiency (Murphree point efficiency). If the vapor and liquid compositions
are taken at a point on the plate, equation 11.64 gives the local or point efficiency,
698 CHAPTER 11 SEPARATION COLUMNS (DISTILLATION, ABSORPTION, AND EXTRACTION) |
How to write a hypothesis statement
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How to write a hypothesis statement
Identifying parts of an hypothesis (interactive) testable and non-testable statements (text) note: if you are designing a research study , explore the research methods information guide for helpful resources. Thesis hypothesis is the first thing to consider, when you are preparing a research a writer should understand that thesis statement and thesis hypothesis are different thus, the latter is applied to support the ideas stated in thesis statementso, a good research paper must have both thesis statement and thesis hypothesis. Although you could state a scientific hypothesis in various ways, most hypothesis are either if, then statements or else forms of the null hypothesisthe null hypothesis sometimes is called the no difference hypothesis.
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Below is a short explanation of a hypothesis statement and some examples of hypothesis statements hypothesis statement--a prediction that can be tested or an educated guess in a hypothesis statement, students make a prediction about what they think will happen or is happening in their experiment. What is a hypothesis a hypothesis is a tentative, testable answer to a scientific question once a scientist has a scientific question she is interested in, the scientist reads up to find out what is already known on the topic. Tips for writing your thesis statement 1 determine what kind of paper you are writing: an analytical paper breaks down an issue or an idea into its component parts, evaluates the issue or idea, and presents this breakdown and evaluation to the audience an expository (explanatory) paper explains something to the audience an argumentative paper makes a claim about a topic and justifies. |
Occurrence of normal and anomalous diffusion in polygonal billiard channels
From extensive numerical simulations, we find that periodic polygonal billiard channels with angles which are irrational multiples of generically exhibit normal diffusion (linear growth of the mean squared displacement) when they have a finite horizon, i.e. when no particle can travel arbitrarily far without colliding. For the infinite horizon case we present numerical tests showing that the mean squared displacement instead grows asymptotically as . When the unit cell contains accessible parallel scatterers, however, we always find anomalous super-diffusion, i.e. power-law growth with an exponent larger than . This behavior cannot be accounted for quantitatively by a simple continuous-time random walk model. Instead, we argue that anomalous diffusion correlates with the existence of families of propagating periodic orbits. Finally we show that when a configuration with parallel scatterers is approached there is a crossover from normal to anomalous diffusion, with the diffusion coefficient exhibiting a power-law divergence.
pacs:05.45.Pq, 05.60.Cd, 02.50.-r, 05.40.Fb
Billiard models, in which point particles in free motion undergo elastic collisions with fixed obstacles, have simple microscopic dynamics but strong ‘statistical’ behavior, such as diffusion, at the macroscopic level. The statistical properties of scattering billiards such as the Lorentz gas, which have smooth, convex scatterers, are well-understood: see e.g. Szász (2000) for a collection of reviews. Recently, however, models with polygonal scatterers have attracted attention, since they are only weakly chaotic, with for example all Lyapunov exponents being , but they still exhibit surprising statistical properties in numerical experiments Dettmann and Cohen (2001); Alonso et al. (2002); Li et al. (2003).
The dynamics in polygonal billiards depends strongly on the angles of the billiard table. Some rigorous results are available for angles which are all rational, i.e. rational multiples of : see Gutkin (2003) and references therein. In particular, an initial condition with a given angle cannot explore the whole phase space, since the possible angles obtained in the resulting trajectory are restricted, so that the model does not have good ergodic properties. Previous numerical work has shown that if one angle is rational and the others irrational, then the results are sensitive to the value of the rational angle Alonso et al. (2002); Li et al. (2003).
In this work we restrict our attention to the case in which all angles are irrational multiples of . There are few rigorous results available in this case, but there is numerical evidence that the dynamics is mixing in irrational triangles Casati and Prosen (1999).
We have performed extensive simulations of the transport properties of quasi-one-dimensional polygonal channel billiards; here we report detailed results for two classes of model. In both we find that transport generically corresponds to what is usually referred to as normal diffusion, i.e. linear growth of the mean squared displacement, when the particles cannot travel arbitrarily long distances without collisions (finite horizon). By ‘generic’ we mean that the set of models which do not exhibit normal diffusion is ‘small’, for example having measure zero in the space of geometrical parameters. Similarly, for the infinite horizon case we find that transport is generically marginally super-diffusive, i.e. there is a logarithmic correction to the linear growth law of the mean squared displacement.
However, we also find exceptions to the generic behaviors described above. These exceptions occur when there are parallel scatterers in the unit cell which are accessible from one another (see Sec. III for the definition). In such cases we always find super-diffusion, i.e. the mean squared displacement grows like a power law with exponent greater than . Further, we demonstrate that as a parallel configuration is approached, the transition from normal to anomalous diffusion occurs through a power-law divergence of the diffusion coefficient. We also show that the exponents we obtain are not consistent with a simple continuous-time random walk model of anomalous diffusion.
Our results are consistent with previous observations of irrational models exhibiting normal Alonso et al. (2002, 2004); Li et al. (2003) and anomalous Zaslavsky and Edelman (1997); Zaslavsky (2002); Li et al. (2002); Prosen and Znidaric (2001) diffusion. Also, very recently, numerical results corresponding to one of the models studied in this work were reported in Jepps and Rondoni (2006), where it was also observed that parallel scatterers result in anomalous diffusion; the results we obtain complement and extend those of that reference.
Energy transport in polygonal billiards placed between two heat reservoirs at different temperatures has also been a focus of attention Alonso et al. (2002, 2004); Li et al. (2003). This process is closer to “color diffusion” than to heat conduction due to the absence of particle interactions and, hence, of local thermodynamic equilibrium Dhar and Dhar (1999). In these systems the properties of the energy transport process are known to be closely related to those of the particle diffusion process Li and Wang (2003); Denisov et al. (2003), and for this reason we only study the latter.
The unit cells of our models are shown in Fig. 1; the complete channel consists of an infinite horizontal periodic repetition of such cells. The polygonal Lorentz model is a polygonal channel version of the Lorentz gas channel studied in Sanders (2005), consisting of a quasi-1D channel with an extra square scatterer at the center of each unit cell. By quasi-1D we mean that particles are confined to a strip infinitely extended in the -direction but of bounded height in the -direction.
The polygonal Lorentz model can be simplified by eliminating the central scatterer and flipping the bottom line of scatterers to point upwards, resulting in the quasi-1D zigzag model. Both models are designed to permit a finite horizon by blocking all infinite corridors in the structure, so that there is an upper bound on the distance a particle can travel without colliding with a scatterer; in scattering billiards this condition was shown to be necessary for normal diffusion to occur Bunimovich et al. (1991); Bleher (1992). The zigzag model has also been studied in Jepps and Rondoni (2006).
We remark that the zigzag model with parallel sides can be reduced to a parallelogram with irrational angles. This simple billiard geometry seems to have been neglected previously, although it is close to a model considered in Kaplan and Heller (1998) which was reported to exhibit a very slow exploration of phase space. We have found similar behavior in the parallel zigzag model (see also Jepps and Rondoni (2006)), but it appears to be a peculiarity of this model which is not necessarily shared by other polygonal billiards exhibiting anomalous diffusion.
In our simulations we distribute initial conditions (unless otherwise stated) uniformly with respect to Liouville measure, i.e. with positions distributed uniformly in the available space in one unit cell, and velocities with unit speed and uniformly distributed angles. We consider only irrational angles for the billiard walls in each model, although we vary explicitly scatterer heights rather than angles, and to fix the length scale we take (see Fig. 1) .
We study statistical properties of the particle positions as a function of time , denoting averages at time over all initial conditions by . Of particular interest is the mean squared displacement , where , which is frequently used to characterize the transport properties of such systems.
Ii Normal and marginally anomalous diffusion
In this paper, by normal diffusion we mean asymptotic linear growth of the mean squared displacement as a function of time , which we denote by as . This is characteristic of systems commonly thought of as ‘diffusive’, for example random walkers which take uncorrelated steps of finite mean squared length.
Strongly chaotic dynamical systems, such as the Lorentz gas, are known to exhibit normal diffusive behavior Szász (2000); this arises due to a fast decay of correlations, even between neighboring trajectories, so that after a short correlation time the dynamics looks ‘random’. It is rather surprising, then, that normal diffusion also seems to appear in non-chaotic polygonal billiards Alonso et al. (2002, 2004); Li et al. (2003) in which correlations do not decay exponentially. In these systems, the only source of ‘randomization’ arises due to divergence of neighboring trajectories colliding on different sides of a corner of the billiard: see e.g. van Beijeren (2004) for a recent attempt to characterise this randomization effect.
In the following, we present sensitive tests which provide strong confirmation that, apart from exceptional cases, the decorrelation induced by the randomization mechanism mentioned above is indeed sufficient to give rise to normal diffusion in polygonal billiards. In particular our tests distinguish the logarithmic correction (marginally anomalous diffusion) which arises generically in these systems when they have infinite horizon, as it does in fully chaotic billiards.
We will see that exceptions to these behaviors occur when the billiards present accesible parallel walls. In such cases we argue that long time correlations persist, related to the existence of families of travelling periodic orbits, which always give rise to anomalous diffusion.
ii.1 Growth of second moment
The growth of the mean squared displacement as a function of time is the most commonly used indicator of transport properties. In Fig. 2 we plot for several non-parallel zigzag models, varying . For values of below there is a finite horizon and we find that a linear regime is rapidly attained, suggesting normal diffusion. For values of above this value, there is an infinite horizon; in this case, the mean squared displacement is still almost linear, although a small amount of curvature is visible.
Normal diffusion corresponds to an asymptotic slope of on a double logarithmic plot. Figure 3 shows such a plot for the same data as in Fig. 2, and we see that indeed the slope is for finite horizon and close to for infinite horizon (e.g. slope for ). For models with irrational angles and no accessible parallel sides (see Sec. III), we always find a growth exponent close to , so that normal diffusion is generic in polygonal billiard channels.
The above test does not, however, take account of logarithmic corrections such as those found in the Lorentz gas with infinite horizon, where Bleher (1992); such a correction will appear on the above plot as a small change in the exponent. A heuristic argument of Friedman and Martin (1984) leads us to expect that in any channel with an infinite horizon, including polygonal ones, we will have (at least) this kind of ‘marginal’ anomalous diffusion. We now present numerical tests which distinguish the logarithmic correction, showing that the diffusion is normal in the finite horizon case and marginally anomalous in the infinite horizon case.
In marginal anomalous diffusion we expect that for some constants , and Garrido and Gallavotti (1994), so that
where . Asymptotic linear growth of as a function of thus indicates marginal anomalous diffusion, whilst asymptotic flatness corresponds to normal diffusion. A similar method was used in Geisel and Thomae (1984) for a discrete time system, and we have checked that it also correctly finds marginal anomalous diffusion in the infinite-horizon Lorentz gas with continous-time dynamics. Figure 4 plots , confirming the normal/marginal distinction in finite and infinite horizon regimes.
We can also fit to the expected form with parameters , and , for example using a nonlinear least-squares method. For we find ; this agrees very well with the asymptotic slope in Fig. 4, as it should do by equation (1). For we instead obtain , i.e. essentially zero (note that it cannot be negative), confirming normal diffusion.
ii.2 Integrated velocity autocorrelation function
For a system with normal diffusion we define the diffusion coefficient as the asymptotic growth rate of the mean squared displacement. The Green–Kubo formula Reichl (1998) relates to the velocity autocorrelation function : exists, and the diffusion is normal, only if decays faster than as . was studied in Alonso et al. (2004) for a different channel geometry, but this function is very noisy and it is not possible to determine its decay rate.
Instead, following Lowe and Masters (1993) we consider the integrated velocity autocorrelation function
where is the particle displacement at time . This function is much smoother than , and satisfies as if the diffusion coefficient exists, whilst if decays as .
Figure 5 plots as a function of for several finite and infinite horizon non-parallel zigzag models. The flatness of in the finite horizon models indicates that the limit exists, and hence that the diffusion is normal, and contrasts with the asymptotic linear growth in the infinite horizon models, providing further evidence of the marginally anomalous behavior in the latter case.
ii.3 Higher-order moments
We define the growth exponent of the th moment by
which ignores corrections to power-law growth and is a convex function of ; further, for all , since the particles have finite velocity Armstead et al. (2003). The exponents , and in particular the diffusion exponent , are measured using a fit to the long-time region of a double logarithmic plot of the relevant moment, assuming that the asymptotic regime has been reached.
We find that the behavior of moments of high order () is dominated by the extreme particles, i.e. by those which have the largest value of , giving data which is not reproducible between different runs. For this reason we eliminate the most extreme particles from the average before calculating the growth rate, resulting in data which is now reproducible. A similar procedure was used in Zaslavsky and Edelman (2001).
Figure 6 shows the exponents for the zigzag model with finite and infinite horizon. The low-order exponents satisfy , and in both cases there is a crossover to a second linear regime with slope , which occurs at approximately in the finite horizon case and for infinite horizon; the latter agrees with the result found in Armstead et al. (2003) for the infinite horizon Lorentz gas using a method which also applies here. Higher-order moments thus provide another method to distinguish between normal and marginal anomalous diffusion in polygonal billiards.
The observed qualitative change in behavior of the moments corresponds to a change in relative importance between diffusive and ballistic effects Armstead et al. (2003); Castiglione et al. (1999); Ferrari et al. (2001), and explains the previous observation that the Burnett coefficient, defined as the growth rate of the fourth cumulant , diverges as a function of time in polygonal billiards Dettmann and Cohen (2001); Alonso et al. (2002).
The combination of the above tests provides strong evidence of the distinction between normal diffusion in finite horizon models and marginally anomalous diffusion in infinite horizon models.
Iii Anomalous diffusion
Whilst we generically find normal diffusion, for certain exceptional geometrical configurations this does not occur. The main exception is when there are accessible parallel scatterers in the unit cell of a model. In this case we always find anomalous (super-)diffusion, i.e. with : see Fig. 7. Since the particles have finite speed, we always have , with the value corresponding to ballistic motion. Sub-diffusion () was reported in Alonso et al. (2002) in a model with one rational angle, but we are not aware of any model with irrational angles exhibiting sub-diffusion.
We say that a model has ‘accessible parallel scatterers’ when there are trajectories which can propagate arbitrarily far by colliding only with scatterers which are parallel to some other scattering element in the unit cell.
As an example, the results we report for the polygonal Lorentz model were obtained with a square central obstacle, that is, a scatterer whose opposite sides are parallel. This alone does not, however, imply that the system exhibits anomalous diffusion, since it is not possible for a particle to propagate by colliding only with (copies of) this square, i.e. the scatterers making up the square are not ‘accessible’ from one another.
Any particle must in fact also collide with the top and bottom scattering walls. Only if these walls are also accessibly parallel as defined above (i.e. have parallel partners in the unit cell) do we find anomalous diffusion. Furthermore, if we alter the central obstacle shape so that it no longer has parallel sides, then we still find anomalous diffusion provided that the scattering walls still have parallel partners which are accessible from one cell to another.
Figure 8 shows the behavior of the anomalous diffusion exponent as a function of in the parallel zigzag and polygonal Lorentz models with finite horizon. The representative error bars shown give the standard deviation of the exponent over independent simulations for the same value of , and indicate that the structure visible in the curve for the zigzag model is not an artifact.
iii.1 Time evolution of densities
A feature typical of normal diffusive systems is that the probability distribution of displacements (or equivalently of positions) converges in the long-time limit, when rescaled by , to to a normal distribution Sanders (2005). We find this behavior numerically in both of the classes of polygonal billiard channels studied here, in the cases for which the mean squared displacement grows linearly, and we conjecture that this holds in general, providing a stronger sense in which polygonal billiards can be considered diffusive.
For anomalous diffusion with (, we no longer expect the central limit theorem to hold. The inset of Fig. 10 shows the probability density of positions in a parallel polygonal Lorentz model. It exhibits a striking fine structure, similar to that found in normally diffusive Lorentz gases and polygonal billiards in Sanders (2005), where it was shown that the principal contribution to this fine structure arises from a geometrical effect, namely that the amount of vertical space available for particles to occupy varies along the channel. Assuming that the dynamics has good mixing properties, it was shown that defining the demodulated density by
eliminates much of the fine structure. Here, is the height of the available space in the billiard channel at horizontal coordinate , normalised by the area of the unit cell so that the integral over the unit cell is . This demodulated density is also shown in the inset of Fig. 10, although the demodulation is not as effective as in the normally diffusive cases studied in Sanders (2005), due to the weaker mixing properties of the current model.
Rescaling the demodulated position density at time by
gives a bounded variance as Prosen and Znidaric (2001). The main part of Figure 10 shows demodulated densities rescaled in this way for different times, compared to a Gaussian with the corresponding variance. The rescaled densities converge at long times to a non-Gaussian limiting shape, as has previously been found in other systems with anomalous diffusion Metzler and Klafter (2000).
iii.2 Mechanism for anomalous diffusion
iii.2.1 Continuous-time random walk approach
For the parallel zigzag model, Figure 13 demonstrates the presence of long laminar stretches, i.e. periods of motion during which a particle’s velocity component parallel to the channel wall does not change sign. Typical trajectories exhibit such behavior, sometimes over very long periods of time. Figure 11 shows a scatterplot representing the joint distribution of laminar stretches of duration with horizontal displacement along the channel. (To compute these data, the channel was divided into sections lying between adjacent maxima and minima in order to determine the wall direction).
Thus, we may attempt to describe the dynamics of this system as a continuous-time random walk (CTRW), whose steps are the laminar stretches: see e.g. Weiss (1994); Zumofen and Klafter (1993). To this end, we note that Figure 11 shows that the distribution is concentrated on diagonal lines emanating from the origin, which leads us to try an approximation of the form for some speed . This version of the CTRW was termed the velocity model in Zumofen and Klafter (1993), and describes motion at a constant velocity for a time in the direction ; after each stretch the direction is randomised and a new step is taken. With this form for , the long-time growth of the mean squared displacement depends on the asymptotic decay rate of the marginal distribution of step times Zumofen and Klafter (1993): if as , then the variance behaves like when , while we have normal diffusion for .
Figure 12 shows the tail region of . From the top curve we find a long-time power-law decay for a particular case having parallel scatterers. For close-to-parallel configurations, the tail of the distribution follows that for the parallel case, but with an exponential cut-off, as shown in the inset of Fig. 12. According to the CTRW model, this cut-off gives rise to normal diffusion in the asymptotic long-time regime. This change of behavior is in qualitative agreement with our observations. Unfortunately, the value of the tail exponent for the parallel case corresponds to and hence to a mean squared displacement , which does not agree with the numerically observed growth rate .
Although the above results rest on a rather gross approximation to the joint distribution , we doubt that a more general CTRW model, incorporating information on the complete , would substantially improve the agreement. The reason is that correlations between consecutive laminar trajectories have an important effect not accounted for by CTRW models. Indeed, for a given trajectory, we find that the lengths of consecutive long laminar stretches are often nearly equal, indicating that the trajectory repeats very closely its previous behavior many times.
iii.2.2 Propagating periodic orbits
On the other hand, our results indicate that it is the presence of accessible parallel scatterers that gives rise to anomalous diffusion. We believe that propagating periodic orbits, i.e. orbits which repeat themselves with a spatial displacement, may provide a mechanism for this behavior. Such orbits are more prevalent when there are parallel scatterers, since in this case it is much easier for a particle to regain its original angle of propagation after a given number of bounces, and, then, possibly repeat its previous motion. In addition, periodic orbits in polygonal billiards occur in families Galperin et al. (1995); several such families are shown in Fig. 13.
Furthermore, trajectories with initial conditions which are close in phase space to those of a propagating periodic orbit will shadow it, and hence propagate, for a long time, with a longer shadowing time for closer initial conditions. Anomalous diffusion should then result from a balance between the ballistic propagation of the periodic orbits themselves, the long-lasting ballistic motion of shadowing orbits, and the diffusive motion of other trajectories.
We have, however, not yet been able to account analytically for anomalous diffusion in this way. For instance, the mere existence of propagating orbits is not enough to give rise to the kind of anomalous behavior we observe. Indeed, families of propagating orbits exist in the corridors of the infinite-horizon periodic Lorentz gas, but these give rise only to marginally anomalous diffusion Bleher (1992) (see Sec. II). Thus, although the set of propagating periodic orbits must have measure zero in phase space, since otherwise their ballistic nature would result in overall ballistic transport with , they must be plentiful in some sense, in order to give rise to anomalous transport. For example, such orbits could be dense at least in some parts of phase space.
Iv Crossover from normal to anomalous diffusion
Since anomalous diffusion occurs only for special geometrical configurations, it is of interest to study the crossover from normal to anomalous diffusion as such configurations are approached. We begin by considering the polygonal Lorentz channel, in which we fix all geometrical parameters except for , which we vary as .
Figure 14 shows a double logarithmic plot of the mean squared displacement in a polygonal Lorentz model for values of tending to , i.e. approaching the parallel configuration; a linear fit to the bottom curve has been subtracted from each curve for clarity. As tends to , the curve for the parallel configuration () is followed for progressively longer times before a crossover occurs to asymptotic diffusive behaviour, which appears on the figure as a zero slope. Defining the crossover time as the intersection of the initial anomalous growth with a straight line fit to the asymptotic normal growth, we find that scales like as .
The inset of Fig. 14 shows the diffusion coefficient as a function of , obtained from the asymptotic slope of the mean squared displacement. Although Fig. 14 shows that the asymptotic linear regime has not been reached for the smallest values of , thereby underestimating the diffusion coefficient, we obtain a straight line on a double logarithmic plot of versus , giving the power-law behavior
with the same exponent for both positive and negative values of and for two (close) values of . For this model, the diffusion exponent in the parallel () case is .
A similar crossover to anomalous diffusion occurs near to the parallel configuration in the zigzag model (not shown), and we conjecture that such crossover behavior is generally found in polygonal billiards near to configurations exhibiting anomalous diffusion. The rate of growth of the diffusion coefficient for depends, however, on the diffusion exponent for : for the zigzag model with and , we find , with a diffusion exponent for the parallel case; both of these exponents differ from the results in the polygonal Lorentz model.
The growth of the diffusion coefficient close to a parallel case thus depends on the anomalous diffusion exponent found in that parallel case. Indeed, the relation between the exponents characterizing the crossover from normal to anomalous transport can be obtained from the following simple scaling argument. Define the crossover time exponent by . We have
so that continuity at gives , which is in good quantitative agreement with our results.
Furthermore, this argument implies that plots of as a function of should collapse onto a scaling function of the form
The data collapse for both the zigzag and the polygonal model near a configuration of parallel scatterers is shown in Fig. 15. Interestingly, while the transport exponent is sensitive to the geometrical details of the models, the best collapse was obtained for the same value of in both cases.
Finally, it should be noted that the above calculations all refer to irrational angles; the situation when the value of renders an angle rational is unclear.
V Discussion and Conclusions
We have demonstrated that the type of diffusive behavior exhibited by periodic polygonal billiard channels depends on geometrical features of the unit cell. Based on our results, we conjecture that sufficient conditions for a periodic polygonal billiard channel to exhibit normal diffusion are the following: (i) all vertex angles are irrationally related to ; (ii) the billiard has a finite horizon; and (iii) there are no accessible parallel scatterers in the unit cell.
When there is an infinite horizon, diffusive transport is replaced by marginal anomalous diffusion with , while if there are accessible parallel scatterers, then anomalous diffusion with , is observed.
In the zigzag model we also have evidence of anomalous diffusion (e.g. with exponent for one particular model) with irrational angles chosen such that one angle is twice the other, i.e. with a rational relation between them, in disagreement with the result found in Jepps and Rondoni (2006) (where a rather small number of initial conditions was used). For other angle ratios we do not have conclusive data, but the exponents are seemingly close to , while for the polygonal Lorentz model we find normal diffusion in similar situations. For the zigzag model we must thus also exclude the possibility of rationally-related angles from the above conditions for normal diffusion.
We remark that in Alonso et al. (2004), propagating periodic orbits were conjectured to be related to anomalous diffusion in a system with one rational angle, but no reason was given for their existence. Actually, it is possible to include that system into our picture by unfolding the rational angle, which gives rise to an equivalent system with parallel scatterers.
We believe that the explanation of anomalous diffusion should be found in terms of such propagating periodic orbits, which are much more prevalent in the presence of parallel scatterers. We further showed that there is a crossover from normal to anomalous diffusion as a parallel configuration is approached, with the diffusion coefficient having a power-law divergence. We hope to achieve a quantitative description of both of these points in the future.
Acknowledgements.This work was initiated as part of the first author’s PhD thesis Sanders (2005b); he thanks his supervisor Robert MacKay for valuable comments, and UNAM for a postdoctoral research award. The Centre for Scientific Computing at the University of Warwick provided computing facilities for some of the calculations. Support through grant IN-100803 DGAPA-UNAM is also acknowledged.
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Fifth graders identify the prime factorization for 48 using different sets of factors. They find the prime factorization for 60.
Learners work with prime numbers. In this prime factorization lesson, students review writing numbers in their prime factorization form and solve multiple problems.
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Students recognize the difference between a prime and a composite number. In this prime and composite number lesson, students use a factor tree model to show prime factorization. Students participate in a bingo activity. Students complete a homework worksheet.
Ninth graders work with composite numbers to find their prime factorizations. In this Algebra I lesson, 9th graders use factor trees and division to find the prime factors of a number. Students explore the use of prime factors to determine the GCF and LCM. TI-nspire handheld required.
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The margin of error is a statistic expressing the amount of random sampling error in a survey's results. The larger the margin of error, the less confidence one should have that the poll's reported results are close to the "true" figures; that is, the figures for the whole population.
The margin of error is usually defined as the radius of the confidence interval for a particular statistic from a survey. When a single, global margin of error is reported for a survey, it refers to the maximum margin of error for all reported percentages using the full sample from the survey. This maximum margin of error can be calculated as the radius of the confidence interval for a reported percentage of 50%.
Like confidence intervals, the margin of error can be defined for any desired confidence level, but usually a level of 90%, 95% or 99% is chosen (typically 95%). This level is the probability that a margin of error around the reported percentage would include the "true" percentage. Along with the confidence level, the sample design for a survey, and in particular its sample size, determines the magnitude of the margin of error. A larger sample size produces a smaller margin of error, all else remaining equal.
The margin of error only takes random sampling error into account. It does not represent other potential sources of error or bias such as poorly phrased questions, people lying or refusing to respond, the exclusion of people who could not be contacted, or miscounts and miscalculations.
A running example from the 2004 U.S. presidential campaign will be used to illustrate concepts throughout this article. According to an October 2, 2004 survey by Newsweek, 47% of registered voters would vote for John Kerry/John Edwards if the election were held on that day, 45% would vote for George W. Bush/Dick Cheney, and 2% would vote for Ralph Nader/Peter Camejo. The size of the sample was 1,013. Unless otherwise stated, the remainder of this article uses a 95% level of confidence.
Polls typically involve taking a sample from a certain population. In the case of the Newsweek poll, the population of interest is the population of people who will vote. Because it is impractical to poll everyone who will vote, pollsters take smaller samples that are intended to be representative; that is, a random sample of the population. It is possible that pollsters sample 1,013 voters who happen to vote for Bush when in fact the population is split, but this is extremely unlikely given that the sample is random.
Sampling theory provides methods for calculating the probability that the poll results differ from reality by more than a certain amount, simply due to chance; for instance, that the poll reports 47% for Kerry but his support is actually as high as 50%, or is really less than 44%. This theory and some Bayesian assumptions suggest that the "true" percentage will probably be fairly close to 47%. The more people that are sampled, the more confident pollsters can be that the "true" percentage is close to the observed percentage. The margin of error is a measure of how close the results are likely to be.
However, the margin of error only accounts for random sampling error, so it is blind to systematic errors that may be introduced by non-response or by interactions between the survey and subjects' memory, motivation, communication and knowledge.
Calculations assuming random samplingEdit
This section will briefly discuss the standard error of a percentage, the corresponding confidence interval, and connect these two concepts to the margin of error. For simplicity, the calculations here assume the poll was based on a simple random sample from a large population.
The standard error of a reported proportion or percentage p measures its accuracy, and is the estimated standard deviation of that percentage. It can be estimated from just p and the sample size, n, if n is small relative to the population size, using the following formula:
- Standard error =
When the sample is not a simple random sample from a large population, the standard error must be estimated through more advanced calculations.
In the Newsweek poll, Kerry's level of support p = 0.47 and n = 1,013. The standard error (.016 or 1.6%) helps to give a sense of the accuracy of Kerry's estimated percentage (47%). A Bayesian interpretation of the standard error is that although we do not know the "true" percentage, it is highly likely to be located within two standard errors of the estimated percentage (47%). The standard error can be used to create a confidence interval within which the "true" percentage should be to a certain level of confidence.
The estimated percentage plus or minus its margin of error is a confidence interval for the percentage. In other words, the margin of error is half the width of the confidence interval. It can be calculated as a multiple of the standard error, with the factor depending of the level of confidence desired; a margin of one standard error gives a 68% confidence interval, while the estimate plus or minus 1.96 standard errors is a 95% confidence interval, and a 99% confidence interval runs 2.58 standard errors on either side of the estimate.
The margin of error for a particular statistic of interest is usually defined as the radius (or half the width) of the confidence interval for that statistic. The term can also be used to mean sampling error in general. In media reports of poll results, the term usually refers to the maximum margin of error for any percentage from that poll.
Maximum margin of errorEdit
The maximum margin of error for any percentage is the radius of the confidence interval when p = 50%. As such, it can be calculated directly from the number of poll respondents. For 95% confidence, assuming a simple random sample from a large population:
- (Maximum) margin of error (95%) = 1.96 ×
This calculation gives a margin of error of 3% for the Newsweek poll, which reported a margin of error of 4%. The difference was probably due to weighting or complex features of the sampling design that required alternative calculations for the standard error.
Different confidence levelsEdit
For a simple random sample from a large population, the maximum margin of error is a simple re-expression of the sample size n. The numerators of these equations are rounded to two decimal places.
- Margin of error at 99% confidence
- Margin of error at 95% confidence
- Margin of error at 90% confidence
If an article about a poll does not report the margin of error, but does state that a simple random sample of a certain size was used, the margin of error can be calculated for a desired degree of confidence using one of the above formulae. Also, if the 95% margin of error is given, one can find the 99% margin of error by increasing the reported margin of error by about 30%.
Maximum and specific margins of errorEdit
While the margin of error typically reported in the media is a poll-wide figure that reflects the maximum sampling variation of any percentage based on all respondents from that poll, the term margin of error also refers to the radius of the confidence interval for a particular statistic.
The margin of error for a particular individual percentage will usually be smaller than the maximum margin of error quoted for the survey. This maximum only applies when the observed percentage is 50%, and the margin of error shrinks as the percentage approaches the extremes of 0% or 100%.
In other words, the maximum margin of error is the radius of a 95% confidence interval for a reported percentage of 50%. If p moves away from 50%, the confidence interval for p will be shorter. Thus, the maximum margin of error represents an upper bound to the uncertainty; one is at least 95% certain that the "true" percentage is within the maximum margin of error of a reported percentage for any reported percentage.
Effect of population sizeEdit
The formulae above for the margin of error assume that there is an infinitely large population and thus do not depend on the size of the population of interest. According to sampling theory, this assumption is reasonable when the sampling fraction is small. The margin of error for a particular sampling method is essentially the same regardless of whether the population of interest is the size of a school, city, state, or country, as long as the sampling fraction is less than 10%.
Confidence intervals can be calculated, and so can margins of error, for a range of statistics including individual percentages, differences between percentages, averages, medians and totals.
The margin of error for the difference between two percentages is larger than the margins of error for each of these percentages, and may even be larger than the maximum margin of error for any individual percentage from the survey.
In a plurality voting system, it is important to know who is ahead. The terms "statistical tie" and "statistical dead heat" are sometimes used to describe reported percentages that differ by less than a margin of error, but these terms can be misleading. For one thing, the margin of error as generally calculated is applicable to an individual percentage and not the difference between percentages, so the difference between two percentage estimates may not be statistically significant even when they differ by more than the reported margin of error. The survey results also often provide strong information even when there is not a statistically significant difference.
When comparing percentages, it can accordingly be useful to consider the probability that one percentage is higher than another. In simple situations, this probability can be derived with 1) the standard error calculation introduced earlier, 2) the formula for the variance of the difference of two random variables, and 3) an assumption that if anyone does not choose Kerry they will choose Bush, and vice versa; they are perfectly negatively correlated. This may not be a tenable assumption when there are more than two possible poll responses. For more complex survey designs, different formulas for calculating the standard error of difference must be used.
The standard error of the difference of percentages p for Kerry and q for Bush, assuming that they are perfectly negatively correlated, follows:
- Standard error of difference =
Given the observed percentage difference p − q (2% or 0.02) and the standard error of the difference calculated above (.03), any statistical calculator may be used to calculate the probability that a sample from a normal distribution with mean 0.02 and standard deviation 0.03 is greater than 0.
Applying these calculations to the Newsweek example results in a 75% probability that Kerry was "truly" leading.
- ↑ Newsweek (2004-10-02). NEWSWEEK POLL: First Presidential Debate. Press release. Retrieved on 2006-05-31.
- ↑ Wonnacott and Wonnacott (1990), pp. 4–8.
- ↑ Sudman, S.L. and Bradburn N.M. (1982) Asking Questions. Jossey-Bass: pp. 17-19
- ↑ Sample Sizes, Margin of Error, Quantitative Analysis
- ↑ Lohr, Sharon L. (1999). Sampling: Design and Analysis, 49, Pacific Grove, California: Duxbury Press. "The margin of error of an estimate is the half-width of the confidence interval ..."
- ↑ Stokes, Lynne, Tom Belin (2004). What is a Margin of Error?. (PDF) What is a Survey?. Survey Research Methods Section, American Statistical Association. URL accessed on 2006-05-31.
- ↑ Income - Median Family Income in the Past 12 Months by Family Size, U.S. Census Bureau. Retrieved February 15, 2007.
- ↑ Braiker, Brian. "The Race is On: With voters widely viewing Kerry as the debate’s winner, Bush’s lead in the NEWSWEEK poll has evaporated". MSNBC, October 2, 2004. Retrieved on 2007-02-02.
- ↑ Rogosa, D.R. (2005). A school accountability case study: California API awards and the Orange County Register margin of error folly. In R.P. Phelps (Ed.), Defending standardized testing (pp. 205–226). Mahwah, NJ: Lawrence Erlbaum Associates.
- ↑ Drum, Kevin. Political Animal, Washington Monthly, August 19, 2004. Retrieved on 2007-02-15.
- Sudman, Seymour and Bradburn, Norman (1982). Asking Questions: A Practical Guide to Questionnaire Design. San Francisco: Jossey Bass. ISBN 0875895468
- Wonnacott, T.H. and R.J. Wonnacott (1990). Introductory Statistics, 5th ed., Wiley. ISBN 0471615188.
- Eric W. Weisstein, Margin of Error at MathWorld.
- Stokes, Lynne, Tom Belin (2004). What is a Margin of Error?. (PDF) What is a Survey?. Survey Research Methods Section, American Statistical Association. URL accessed on 2006-05-31.
- fr:Marge d'erreur
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We develop a version of Freĭman's theorem for a class of non-abelian groups, which includes finite nilpotent, supersolvable and solvable A-groups. To do this we have to replace the small doubling hypothesis with a stronger relative polynomial growth hypothesis akin to that in Gromov's theorem (although with an effective range), and the structures we find are balls in (left and right) translation invariant pseudo-metrics with certain well behaved growth estimates.
Our work complements three other recent approaches to developing non-abelian versions of Freĭman's theorem by Breuillard and Green, Fisher, Katz and Peng, and Tao.
For functions F satisfying a certain submultiplicativity condition and polynomials Q1, . . ., Qk in [X], Nair and Tenenbaum obtained an upper bound on the short sum with an implicit dependency on the discriminant of Q1 . . . Qk. We obtain a similar upper bound uniform in the discriminant.
We consider a tamely ramified abelian extension of local fields of degree n, without assuming the presence of the nth roots of unity in the base field. We give an explicit formula which computes the local reciprocity map in this situation.
Cromwell and Marar present an analysis of semi-regular (generic) surfaces with a single triple point and connected self-intersection set. Six of their surfaces are the projective plane, including Boy's surface and Steiner's surface. We build on their work by incorporating twists similar to that of Apery's immersion of the projective plane and show that with a few additional surfaces, all such generic maps of the projective plane are now identified.
We study initial-boundary value problems for linear evolution equations of arbitrary spatial order, subject to arbitrary linear boundary conditions and posed on a rectangular 1-space, 1-time domain. We give a new characterisation of the boundary conditions that specify well-posed problems using Fokas' transform method. We also give a sufficient condition guaranteeing that the solution can be represented using a series.
The relevant condition, the analyticity at infinity of certain meromorphic functions within particular sectors, is significantly more concrete and easier to test than the previous criterion, based on the existence of admissible functions.
We will construct differential forms on the embedding spaces Emb(j, n) for n − j ≥ 2 using configuration space integral associated with 1-loop graphs, and show that some linear combinations of these forms are closed in some dimensions. There are other dimensions in which we can show the closedness if we replace Emb(j, n) by Emb(j, n), the homotopy fiber of the inclusion Emb(j, n) ↪ Imm(j, n). We also show that the closed forms obtained give rise to nontrivial cohomology classes, evaluating them on some cycles of Emb(j, n) and Emb(j, n). In particular we obtain nontrivial cohomology classes (for example, in H3(Emb(2, 5))) of higher degrees than those of the first nonvanishing homotopy groups.
Fix g a Hecke–Maass form for SL3(). In the family of holomorphic newforms f of fixed weight and large prime level q, we find the average value of the product . From this we derive a result on the simultaneous non-vanishing of these L-functions at the central point.
Unfortunately, there are two inaccuracies in the argument of [CLS]. First, the statements of Lemmas 3, 4, 6, and 7 of [CLS] hold only under the additional condition gcd(m, ME) = 1 for some integer ME ≥ 1 depending only on E. Second, the divisibility condition (3·6) in [CLS] implies that tb(ℓ) | nE(p)−1 (rather than tb(ℓ) | nE(p), as it was erroneously claimed on p. 519 in [CLS]). In particular, instead of the divisibility ℓtb(ℓ) | nE(p) (see the last displayed formula on p. 519 in [CLS]), we conclude that for every prime ℓ | L there is an integer aℓ such that
However, the final result is correct and can easily be recovered. To do so, we remark that under the condition gcd(m,ME) =1, we have full analogues of Lemmas 6, 7, 9, and 10 of [CLS] for the function Π(x;m,a) defined as the number of primes p ≤ x with nE(p) ≡ a (mod m) (rather than just for Π(x;m) = Π(x;m,0) as in [CLS]). Define ρ*(n) as the largest square-free divisor of n which is relatively prime to ME. We then derive from (0.1) above that
we see that (0.2) above implies the bound (3·7) from [CLS], and the result now follows without any further changes. |
Be it for simple mathematical operations or for complex problems, we often use calculators. This, in turn, has made us more dependent on its use. To overcome such dependence and to enhance the mental ability to solve mathematical equations with ease, getting adept in short and simple tricks is essential. By practising these tricks, not only will you be able to do lots of mental maths on your own but will also save crucial time while giving scholastic or competitive exams. So, in this blog, we have shared some super easy maths short tricks!
This Blog Includes:
- Top 10 Short Maths Tricks
- 1. Multiplication Short Tricks
- 2. Multiplying any Two-digits Which Have Same Tens Digits
- 3. Division Tricks
- 4. Maths Short Tricks to Find Percentage
- 5. Square Short Trick
- 6. Maths Short Trick for Square Root Short Trick
- 7. Cube Root Short Trick
- 8. Quadratic Equations
- 9. How to Multiply Numbers that End with Zero
- 10. Subtracting from 1000
- Importance of Learning Short Maths Tricks
Top 10 Short Maths Tricks
Let’s hop onto some quick maths short tricks which will help you ace those complex problems in mental maths tests and quizzes!
1. Multiplication Short Tricks
Often, multiplying two-digit numbers is difficult. Hence, through this maths short trick, you will be able to multiply fast and in an easy way.
Let us consider two numbers, 16 and 77
What would be the solution of 16 x 77?
In this trick, we will begin by picking up the first number.
16 is an even number, divide it in half and we get, 16/ 2 = 8
Now, double the second number i.e., 77 x 2 = 154
For your final answer, you can simply multiply the resultant numbers i.e.,
154x 8 = 1232
2. Multiplying any Two-digits Which Have Same Tens Digits
Through this trick, we will learn to multiply numbers where the ten’s digit is the same and the one’s digit adds up to 10.
Q. Find the answer of 42 x 49
Take the ten’s digit and multiply it by the next highest number, i.e, 4 X 5= 30.
Next, multiply both the one’s digit. 2 x 8 = 16.
Lets put together both the digits and the answer will be 3016.
3. Division Tricks
Out of all the maths short tricks, this one will help you solve the division problems in a jiffy!
To Check Divisibility by 9
If the total sum of all the digits of a number is divisible by 9, then the number is divisible by 9.
For Example: 4387= 4 + 3+ 8+ 7 =22
As 22 is not divisible by 9 which means that 4387 is not divisible by 9 as well.
To Check Divisibility by 4
To determine whether a number is divisible by 4 or not, we have to analyse its last 2 digits. If they are divisible by 4, the entire number will be divisible by 4.
For Example: Let us consider 7864, the last 2 digits of this number make the number 64 which is divisible by 4, hence, 7864 is divisible by 4.
4. Maths Short Tricks to Find Percentage
We all struggle to find percentages of complex numbers but don’t worry, one of the most useful maths short trick is this one.
Let us imagine the number 685 and we have to calculate 5% of it. So, what we have to do is,
The digit 685 will have decimal like 685.0
Let us move the decimal one place forward, the number becomes 68.5
Now, we have to divide the number 68.5 by 2
We get, 34.25
Thus, 5% of 685 is 34.25
5. Square Short Trick
Amongst all the maths short tricks, this one can be applied to calculating the square of numbers ending with 5 as it is a daunting task to calculate such numbers.
Let us consider the number 85
For finding its square, we will begin by squaring the units number which is 5.
We get the answer as 25 and this number will be the last two digits of your answer.
A part of the answer would thus be _ _ 25.
Now, you will have to multiply the ten’s digit number with its very next digit.
So, 8 x 9= 72.
No 72 becomes your prefix and the final answer is 7225
6. Maths Short Trick for Square Root Short Trick
This tricks will help you solve square root problems in the wink of an eye:
Square Root of a Four-Digit Number
Let’s find the square root of a four digit number with an example:
Let’s take a look at 3364.
Pick up the last two digits first. Here it is 64.
Since the unit digit is 4, we can say that the unit digit of the square root will either be 2 or 8.
Now, let’s pick up the first two digits. Here it is 33
Since 33 comes between the square of 5 and 6, i.e. 25<33<36, we can say the tens’ digit of the square root of 3364 will be 5.
So, now we have two options, 52 or 58.
Now to choose between the two, multiply the tens and ones digit of both the options 52 and 58 and compare them with 33 (the thousand and hundred position of the digit).
The product should be more than or equal to 33.
Since 10<33, the answer has to be 58.
The Square root of 3364 is 58.
7. Cube Root Short Trick
Let’s learn the cube root short trick by this example:
Let’s take 39304.
First, pick up the last digit of the cube. Here it is 4. If the last digit is 4, the cube root’s last digit will be 4.
It is important to note in general that:
- If the last digit of a cube root is 8 then the unit digit will be 2.
- If the last digit of a cube root is 2 then the unit digit will be 8.
- If the last digit of a cube root is 3 then the unit digit will be 7.
- If the last digit of a cube root is 7 then the unit digit will be 3.
- If the last digit of a cube root is other than 2, 3, 7 and 8 then put the same number as the unit digit.
Going back to the cube root, we know the unit place for the cube root is 4. Now, ignoring the last three digits of the number, pick up the rest. Here, it is 39.
Check the closest cubes to 39. The closest cube to 39 is 27(Cube of 3) and 64 (cube of 4). Now, since 27 is closer to 39 than 64, the first digit will be 3.
So, the cube root of 39304 will be 34.
8. Quadratic Equations
Let’s learn a simple way to solve quadratic equations with an example.
Let’s take x² – 18x + 45 = 0
Multiply the coefficient of x² with 45. Here it is 1*45=45
Multiply -1 with the coefficient of x. Here it will be -1* (-18)=18
Now, the two values will be on the basis of the digits that add up to become 18 and the same digits when multiplied produce 45.
So, here the answer is 15 and 3. The signs will depend on whether 18 and 45 are positive or not. If both are positive, the answers will be positive, if even one is negative, the answer will be negative.
9. How to Multiply Numbers that End with Zero
Multiplying numbers that end with zero is simple. All you need to do is multiply the numbers and add the rest of the zeros. For example:
500 * 200
Multiply 5 and 2
Now, Put all the zeros at the back of 10. Since the numbers had 4 zeros (500 and 200), add 4 more zeros making it 100000
10. Subtracting from 1000
The basic rule of solving this is subtracting each number with 9 except the last one. This can also be followed for other multiples of 10 like 10,000 or others.
For eg: 572
For this, you need to:
Subtract 5 from 9= 4
Subtract 7 from 9=2
Subtract 2 from 10=8
So, the answer will be 428.
Importance of Learning Short Maths Tricks
Be it a professional career or for preparing Maths for competitive exams, easy maths short tricks play a vital role in enhancing your ability to solve complex arithmetic problems in an easy and time-saving manner. The techniques break down the large problems in smaller ones so there is minimal time required to solve them. Here are some of the major features of applying maths short tricks:
- Learning math tricks enhances your thinking abilities and problem-solving skills.
- Practically, in almost every career path, math in some way or the other is applied. So, it becomes a total win-win situation!
- In order to establish a flourishing career in Mathematics, these small tricks will be of great help.
Thus, we hope that through this blog, you can now solve various questions using maths short tricks. These tricks are extremely helpful if you are preparing for any competitive examinations like SAT, GMAT, GRE, etc. Hence, we at Leverage Edu, provide engaging online coaching for exams like SAT, GRE, GMAT, etc where you can learn ample other helpful tricks. Hurry up! Book a free demo session now! |
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Trusted teacher: It is important that I know how the lesson is taught by the teacher. That is why I always prepare my lessons on the basis of the teacher's workbook. Then I teach from the very beginning. I am convinced that you can never build on things without a good foundation. Does this base not work? Then we'll just start over until we get the hang of it! For a further build-up in difficulty I fully adapt to the pace of the student, this is important to be able to remember everything in the long term, such as for a test within a few days. After class I am also available 24/7 to forward small questions or extra exercises that we can discuss in a next lesson. So no worries, you can always contact me. :)
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Trusted teacher: I offer classes for two different groups: - Young adults preparing for TOEFL/GRE with sessions focused in solving exercises from the exams and improving the student's skills - Academic English for researchers and people in academia, where we focus on methods, format and language of academic texts. Sessions could be focused on review the student paper A little about myself. I am Ph.D. student in Electronic Engineering at Imec, Leuven. I lived in the USA twice, as an exchange Student at MIT for a year and as a J1 worker at McDonald's for 3 months. I have scored 103 and 112 in the two times I did the TOEFL exam, and I scored at the 99% math percentile in my GRE score. I was born in Brazil, thus Portuguese native speaker, and I have lived in Leuven for two years, so we can communicate in basic Dutch.
English · Portuguese · Math
Hanne - Leuven$32
Trusted teacher: I am specialised in teaching mathematics and chemistry at primary and secondary school level. Together we will find out exactly what you need help with and what would be the best teaching method for you. After the introductory conversation, I make an estimate of the best approach. This approach is gradually adapted until we have found the ideal teaching method for you. My aim is to challenge the student, but not to overwhelm. After each lesson, I give homework to follow up on the student's progress.
Math · Chemistry
Luana - Leuven$25
Trusted teacher: Maths and statistics can be scary but they are undoubtedly extremely important. Unfortunately, students often feel overwhelmed with technical terminology and formulas that apparently are of no use. My goal is to demystify mathematics and statistics, making them closer to your daily life and/or area of expertise. Discipline, accessibility, and (why not?) a bit of fun are key factors to an effective and long-lasting learning process!
Math · Statistics · Computer programming
Trusted teacher: Hi, my name is Diana and I am tutoring school students in math and physics. myself I was not a great student until I met my math tutor who inspired me and helped me to see the bright side of mathematics. I managed to motivate myself through high school despite all my struggles and depression and managed to get the highest scores. Because of my high performance in finals, I got accepted to one of the best universities in the world where I am currently studying. through my high school I managed to grasp all the methods my tutor was using and I mastered them with my own students. in case of any questions, feel free to contact me anytime:))
Math · Physics
My excellent academic career has nurtured in me a natural talent for teaching. My experiences have taught me how to tackle one problem at a time, and I believe that using this approach—paired with my unconventional examples that force students to think outside the box—I can truly provide an excellent understanding and explanation. I adapt that lecture to the specific needs of the student based on their knowledge and grasp of the concepts. I like to develop analytical and scientific ability along with knowledge in my students, and for that, I use a combination of resources in the form of worksheets, class tasks, and past papers. Last, I believe in the strong ethic of providing feedback to the parents, so that they are aware of their child’s progress.
Math · Physics · Computer programming
Hi🇺🇦! I have a powerful ability kindly explain hard math rules and usually people with a math anxiety tend to feel safe on my math sessions. I would like to share my deep knowledge with more advanced students (math competitions, preparing for exams) as well. Firstly, we will make a short survey to see in which areas you need a help. After that I will make a plan for our study and give you my feedback. I’ve been teaching math online via zoom for 2 years, therefore I have enough expertise to do it perfect. So, contact me if you need some math help. I believe I could help you!
Max - Leuven$24
Trusted teacher: Math and sciences can be an acquired taste for many people. My goal is to help others understand concepts and understand the way things work so that they may also enjoy the courses they need help in. I aim that every lesson counts and that the student gets something out of every minute spent with me.
Math · Physics · Science
Trusted teacher: I have a master's degree in Mathematics and have been a college-level math teacher and a high school math teacher. I have a lot of experience helping students with math problems. If your problem is one of these: Solving algebraic or geometric problems Solving problems on Statistics, probability or Operation Research Solving problems on Calculus, or abstract algebra Translating logical statements to formulas Finding motivations to study mathematics Finding your own method or style for solving math problems Sure, I can help you and I will enjoy doing it.
Math · Physics
Trusted teacher: This class will provide you with the tools you will need to tackle the entrance exams in Mathematics, Physics and Engineering for top UK universities. Depending on the ambitions of the student I can offer help in the STEP papers (Cambridge Mathematics,) the MAT (Oxford Mathematics) and PAT (Oxford Physics) papers and others. I can also assist in helping the student with interview preparation for Oxford, Cambridge and other top UK universities. I am an Oxford Physics graduate with two years experience as a private tutor in the UK, teaching students Mathematics and Physics.
Math · Physics · Calculus
I am a fresh PhD in theoretical Physics, currently working as a post-doc researcher at KUL with focus on string theory. I offer Math/Physics classes and/or tutoring service for high school and university students. Let me know what you want to learn and I will customize the class for your needs :)
Math · Physics · Science
Trusted teacher: Having conducted international-level experimental and theoretical scientific research for many years (PhD University of Cambridge, National Physical Laboratory, Max-Planck Institute, KUL, and UCL) I offer tailored and adaptive tutoring in mathematics, chemistry, and physics. My main focus is in helping those pupils in their final two years of school who are preparing for university. I also specialize in preparing students for the Belgian medical entrance exam by actually solving the problems of all the past entrance exams and providing tips and insight for rapidly homing in on the correct answers. My philosophy is that science and mathematics can be related to each person's everyday experience. I try to provide insight and solid foundations so that students will not have to rely heavily on blindly memorizing large sections of their courses.
Math · Chemistry · Physics
Trusted teacher: During my classes, I help children understand these complex subjects as well as give them real-life examples to further understand the topics. I've been tutoring math and chemistry in high school and also during my university studies to high school students. As a chemical engineer student, I have a deep insight into both topics and as a young individual, I understand what other teenagers might find difficult about math and chemistry. During high school, I went to the British School of Brussels for a year, and now I am attending my courses at KU Leuven in English, therefore I speak perfect scientific English. I love children, I occasionally babysit.
Math · Chemistry · Science
I graduated from one on the top french engineering school (ESPCI) where I studied applied mathematics, physics and chemistry. I gave lessons to students from high school but also university students for several years. I can adapt the lesson according to the student needs: it can be homework help as well as going over a subject, but also to go further than what the student has been taught in classes. I can give exercices between the different lessons and the student can contact me if he or she has questions or an issue with his or her homework. I can teach in english and in french but not in dutch.
Math · Physics · Chemistry
Trusted teacher: I can tutor school going children in Math and English. Also help them with their homework. I am an Engineer from India but I'm currently not working here in Belgium. I have learnt Carnatic classical music from a very young age and have obtained senior grade in it. Hence I will be able to teach students from basics. Also can help students with crafts and painting.
English · Math · Painting
Trusted teacher: I graduated from Portland State University with a Bachelor's degree in Applied Mathematics before coming to KU Leuven to work on my Master's Degree (and eventually PhD) in Astronomy. I have tutored and taught mathematics up through ordinary differential equations but also at younger levels (like secondary or high school level mathematics). I've been tutoring/teaching for almost 3 years now and have worked with clients ranging from the University level to early secondary school work. I have a unique approach to mathematics and physics that allows me to work well with students who don't necessarily find mathematics and science enjoyable. My goal is to show everyone that mathematics and physics is doable for all! *Unfortunately I do not yet speak Dutch well enough to tutor thus I only tutor in English*
Math · Physics · Chemistry
Trusted teacher: I am an electromechanical engineering student at KU Leuven. I have completed the International Baccalaureate (IB) programme with high level in Maths, Physics and standard level in Chemistry. Because it has only been 1 year since I finished the IB programme, my course knowledge is still fresh. I believe I carry the traits of a creative thinker with an enthusiast to learn and teach. I have heard that in my high school my teachers are introducing my lab reports and investigations to the students as samples with high creativity and quality. I can help the middle school and high school students with their Maths, Physics, and Chemistry courses along with the lab reports and investigations. Tutoring will be in English.
Math · Physics · Chemistry
Olivier - Leuven$29
Trusted teacher: A math / physics / chemistry problem? -Mathematics, Physics, Chemistry Sciences in general are often an enigma for you or for members of your family? Do you know people who are having difficulty? I have the solution ! Industrial engineer teaching since 2008, -Obtention of the CEB. -Willing the CE1D. -Obtaining the CESS / Baccalaureate (European French). -Preparation to the examinations of the central jury. -Preparation for engineering / medicine entrance exams -Gasway with scientific courses. -EPSO / SELOR test preparation (digital / abstract). - University level: Mathematics, Inorganic chemistry, Physics, Mechanics, Resistance of materials, Electricity, Electronics (analog / digital / industrial) -Courses individuals or group, I adapt to all situations. Class can be taught in english! I remain at your disposal for any questions.
Math · Mechanical engineering · Chemical engineering
David - Leuven$28
Trusted teacher: I'm a Math&Physics tutor for school students of all ages, Mostly help graduate students in SAT, IB, A levels Math&physics. My classes consist of only one student because every student needs a different approach. I believe that the best way to teach someone sth is to understand how his/her mind works and after that find the most effective and convenient method for the individual. Teaching is one of my hobbies and I do this to help students by sharing my knowledge and skills, so if you think you are motivated enough to do your best then I will be more than happy to teach you! And if you aren't motivated enough, then I will do all I can to help you find that motivation that you need for success. Currently, I'm doing online classes with a virtual board, which makes classes the most efficient. If you have any questions feel free to text me, will reply to you as soon as possible ;))
English · Math · Physics
Trusted teacher: Hi! I am Aditya. I’m a student pursuing a master's degree in Artificial Intelligence and I've graduated with a bachelor's and master's in industrial engineering from KU Leuven campus groep T with a specialization in electromechanical engineering. As a result of my passion for math and my want to earn a little extra money, I am offering math tuitions. The tuitions are for secondary school students studying any level of math and english is the medium of instruction. Though I am only a university student, I have an adequate understanding of math for the secondary school level. I completed the International Baccalaureate (IB) Math Higher Level (Math HL) course with the highest possible score – 7/7. Also, I topped my school both with my math score and my overall score. Apart from teaching topics within math, I know how to score marks – what examiners look for and how they score papers. My classes combine a healthy mix of subject-related teaching and instructions on how to score the maximum possible marks (even if you don’t know the answer!). Since I’m a student my timings are flexible, but please tell me 1 day in advance if you wish to cancel or postpone class.
Math · Calculus · Algebra |
TCS PAPER ON 8th MARCH
Hi .. here is the questions from TCS...
The interview pattern is similar ..but this time they changed HR to MR... I think in MR both technical, as well as they asking about their company also...
There are three rounds...
1. online test
2. Technical Interview
3. MR interview
To clear the first round u must go through the old question papers..which r all in freshersworld, and other websites...no need to study the R.S.Agarwal...
(i) Verbal 32 questions..
12 Comprehension questions..
(iii) Critical Reasoning(but they didnt ask questions from critical.. all r from analytical part... GRE barrons 12th edition(only model test papers))
In verbal, i got so many questions which are not in GRE... So prepare well... i think in verbal i got below 15..
In apti i got the answers for many questions... only 5 will be wrong..
Of course all the questions are from old papers only.. but this time they changed the names, numbers... also in some problems i felt that they changed the analysing ways... but it is so easy to clear the apti round.... i forget the number of questions.. i think its more than 32- 40...
some of the questions...
1.For temperature a function is given according to time t^2 / 6 + 4 t + 12.Wat is the temp ris eor fall between 4 Am to 9 Am .-----------------
2.21. ODD one out
Select the odd one out…..a. Java b. Lisp c. Smalltalk d. Eiffel. Ans: lisp
Select the odd one out a. SMTP b. WAP c. SAP d. ARP. Ans: SAP
Select the odd one out a. Oracle b. Linux c. Ingress d. DB2 Ans: Linux
Select the odd one out a. WAP b. HTTP c. BAAN d. ARP Ans: BANN
Select the odd one out a. LINUX b. UNIX c. SOLARIS d. SQL SEVER Ans: d
Select the odd one out a. SQL b. DB2 c. SYBASE d. HTTP Ans: HTTP
The size of a program is N. And the memory occupied by the program is given by M = square root of 100N. If the size of the program is increased by 1% then how much memory now occupied?
Ans: 0.5 % ( sqrt( 101 N ) )
Wat is the value of % # % (5) + # % # ( 2 )wher % is doubling,# means reciprocal
The size of the bucket is N kb.The bucket fills at the rate of 0.1 kb per millisec.A programmer sends a program to receiver.There it waits for 10 millisec and response wil be back to programmer in 20 millisec.Hw much time the program takes to get a response back to the programmer after it is sent??? ans 30 millisec
The following is not a triangle:
A) (60o, 80o, 20o) B) (90o, 60o, 30o) C) (54o, 66o, 60o) D) (69o, 51o, 60o)
Ans: A (since sum of the angles in a triangle = 180o.)
The following is a planar cube (E, V, F):
A) (12, 8, 6) B) (12, 6, 8) C) (6, 6, 8) D) (6, 8, 6)
E - no. of edges V - no. of vertices F - no. of faces
The greatest prime number among the following which can be stored in a 9-bit word computer is:
Calc. A-(B C) and represent the resulting 8-bit word in decimal form.
A 0 0 1 1 0 0 1 1
B 0 1 0 1 0 1 0 1
C 0 0 0 0 1 1 1 1
(I might have interchanged the values of A, B & C but still try this.)
Fit a suitable curve with the given data.
A) y = log x B) y = log 10x C) y = -10log x D) y = ex
How many B's are followed by G's which are not followed by S's in the following series:
B B G M P Q B G S K O B G A S B B G D E F B G S T I
Ans: 4 (Another Q on the same logic was also asked.)
M(373,5) + R(3.4) + T(7.7) - R(5.8) = ___
M - Modulo R - Round-off T - Truncate
Match the following:
A) Female - Girl i) A type of
B) Bug - Reptile ii) A superset of
C) Beagle - Dog iii) A part of
D) Piston - Engine iv) Not a part of
There is a matrix X of order 7 9 whose elements will be filled column-wise. The address of the first element X(1,1) is 1258. Each element occupies 4 bytes of memory. Then the address of the element X(5,8) is 1473.
The line equation of the following curve is
A) y = ex B) y = tan x C) y = cos x D) x + y = 0
A & B can do a piece of work in 12 days. A is capable of doing twice the work as B. In how many days A alone can complete the work?
If the first/second, third/fourth, fifth/sixth etc. letters of the word PSYCHIATRIST are interchanged, then the letter which is 8th from right is I .
If DCPRY is coded as CBOQX then SPEND can be coded as RODMC.
The following is not a triangle:
A) (2,3,4) B) (3,4,5) C) (5,5,5) D) (1,3,5)
(Momentum*Velocity)/(Acceleration * distance). Find units/Mass
Which of the following set of numbers has the highest Standard deviation?
a)1,0,1,0,1,0 b) -1, -1, -1, -1, -1, -1
c)1,1,1,1,1,1 d) 1,1,0, -1,0, -1
In critical they asked 11 questions..all r from model test papers in 12th edition GRE... they didnt changed the choices also.. so u no need to waste the to analyse and study that part.. it utter waste...i given all the correct answers...
actually its online test.. so i advice u to study all the questions in the Model test papers...the questions are all chosen from their database.. that database consists of only GRE barrons guide... three hours is enough to study the critical and aptitude..for verbal u hav to prepare well... if u know all the questions in critical as well as aptitude u need not to study the verbal.. because there will be no sectional cutoff, as well no uppercutoff, no negative marks... just click some options watever u think...
i cleared the online test.. then they given a sheet to fill the particulars... for me they given tat at 11 o clock... they told to come at 1 o clock for technical... but they called me at 6.30 only, because my name starts with Y...i was so happy because no one will be rejected in the technical round.. after 5 o clock only they start to filter.. so many guys who r all well in technical they too rejected.. i was shocked.. i was so tensed ...
In Technical round, my panel consistes of too HR's..
HR: Tell abt urself..
Me: i said..
HR: Tell abt ur leadership skills
ME: again i said with some examples..
HR:nice.. wats ur area of interest..
ME: i said microprocessor, digital electronics...
HR: wat is microprocessor?
ME: i said ..
HR: wat is the name of first microprocessor ant the latest one?
ME: ya.intel 4004... noe its intel core 2 duo..
HR: nice... tell about pondicherry..
ME: i was shocked why they asking about pondicherry in the technical.. but i said nicely for three minutes..
HR: Wat is the topic u had presented in ur seminar..
ME: Operation research(transportation model)... also i explained..
HR: Then they asked wat is universal gates.. why it is so named?
ME: i said tat too...
HR: k.. man.. all the best...
ME: i said thank u sir... with pleasant smile..
But i got result... u r rejected...
Just i comeback to my home... |
For Foundation GCSE you need to know what equations and graph shapes represent direct and inverse proportion, and work with equations given to you, but you don’t have to actually derive the equations, as you do at Higher.
You also need the Foundation content covered here if you’re doing Higher Tier, but if you want to go straight to the Higher content then you can find that here.
The “Direct and inverse proportion” lesson on this page is based on an extract from the Algebra module of my Grade 4 Essentials course – ideal for anyone who’s doing Foundation GCSE and doesn’t want to get confused by Higher content that they don’t need! The four main modules cover all the skills you need for a secure Grade 4, and there’s an additional “Top-end Topics” extension module for anyone who’s aiming for a Grade 5 or is doing Higher and wants to make sure that they’ve got all the Foundation content sussed.
The course is great value for money, and there are even discounts to be had – as well as some useful free downloads – if you sign up to my mailing list using the form on my home page or in the footer of any page on this site.
You can download a printable PDF version of this post here (you might find it useful for answering the graph questions).
y is directly proportional to x (written in mathematical notation as ) if, when x doubles, y doubles. If x is multiplied by 10 then y is also multiplied by 10; if x is halved then so is y.
The equation (really it’s a formula, since there are two variables) looks like this:
where k is a constant. At Higher Tier you usually have to work out the value of the constant, but at Foundation it will be given if you need it. Note that k might be a fraction, so the equation might look like, for example, .
- If it takes a decorator 3 hours to decorate one room, then the time T taken to decorate more than one room will be the 3 hours multiplied by the number of rooms, n, giving T = 3n
- If it costs 129p for 1 litre of fuel then V litres will cost V x 129, giving a total cost (in pence) of C = 129V
A graph showing direct proportion is a straight-line graph with a positive gradient that goes through the origin. (It can go into negative values but it usually just starts from 0.) For example, the graph for fuel at 129p/litre would look like this:
Conversion graphs often look like this too; for example with currency conversions, if you double the amount in pounds then that doubles the value in dollars too. Similar for metric-imperial conversions, e.g. between kg and pounds, or between miles and km.
You can use a graph like this to read off values.
See if you can use the graph above to estimate the cost of 5.2 litres of fuel. Do this by finding 5.2 on the horizontal (amount of fuel) axis, drawing a line straight up to the graph and then across to the vertical (cost) axis.
Now use the graph to estimate how much fuel you could buy for £10.00. Again, draw lines on the graph. Don’t forget that you need to turn the £10.00 into pence first!
You’ll find the answers a little further down this page.
Now work out the same answers using the formula C = 129V. Your answers should match, though of course the calculation method will give a greater degree of accuracy.
Cost of 5.2 litres:
C = 129V and V = 5.2 litres
C = 129 x amount of fuel =129 x 5.2 = 670.8 p = £6.71 to the nearest penny.
By reading it off the graph you’ll probably have got a value between £6.60 and £6.80 – see blue lines on image below.
Amount of fuel for £10.00:
C = 1000 pence and we need to find V
We have 1000 = 129V
so V = 1000 ÷ 129 = 7.75 litres (3 s.f.)
By reading it off the graph you’ll probably have got a value between 7.7 and 7.8 litres – see red lines on image below.
y is inversely proportional to x (written in mathematical notation as ) if, when x doubles, y halves (and vice versa).
The equation (again, really a formula) looks like this:
where k is a constant. Again, at Higher you usually have to work out the value of the constant, but at Foundation it will be given if you need it.
- If it takes a builder 12 hours to build a wall, then 3 builders will be able to do the job in a third of the time. If B is the number of builders and T is the time taken then
- A journey of 70 miles takes a certain length of time if you travel at 30 mph. If you travel twice as fast then it will take half the time. Journey time t hours at speed s is given by . Can you see how this relates to the distance/speed/time formula?
- If substances A and B both have the same mass but B is half the volume of A, then B’s density will be twice that of A. The equation in this case is of course .
A graph showing inverse proportion looks like this – the same basic shape as the graph of . The k just stretches the graph, making it taller or wider.
For example, here is the graph for the 70-mile journey described above.
Use the graph to estimate how long the journey would take if you travelled at 30mph. (Draw lines on it as you did before, this time up from 30 mph and then across to the time axis.)
Now use the graph to estimate what average speed you’d have to travel at to complete the trip in 1½ hours.
You’ll find the answers a little further down this page.
Now work out the same answers using the formula . Again, you should get the same answers, though the calculated values will be to a greater degree of accuracy.
Journey time at 30 mph:
and s = 30
hours = 2 hours 20 minutes
(You can use the [° ‘ “] button on your calculator to convert 2.33333… hours to hours and minutes – and seconds if applicable.)
By reading it off the graph you’ll probably have got a value between 2.25 and 2.4 hours – see blue lines on image below.
Speed needed to complete the trip in 1½ hours:
t = 1.5 and we need to find s
Multiply both sides by s to get
Then divide both sides by 1.5 to get = 46.7 mph (3 s.f.)
By reading it off the graph you’ll probably have got a value between 45 and 48 mph – see red lines on image below.
And that’s everything you need to know about direct and inverse proportion if you’re doing Foundation Tier. If you’re doing Higher Tier then you also need to work through this lesson.
If you’ve found this article helpful then please share it with anyone else who you think would benefit (use the social sharing buttons if you like). If you have any suggestions for improvement or other topics that you’d like to see covered, then please comment below or drop me a line using my contact form.
On my sister site at at mathscourses.co.uk you can find – among other things – a great-value suite of courses covering the entire GCSE (and Edexcel IGCSE) Foundation content, and the “Flying Start to A-level Maths” course for those who want to get top grades at GCSE and hit the ground running at A-level – please take a look!
If you’d like to be kept up to date with my new content then please sign up to my mailing list using the form at the bottom of this page, which will also give you access to my collection of free downloads. |
Topic how many pints are there in a liter: Are you curious about the conversion between pints and liters? Well, here\'s some good news! If you\'re in Ireland, a liter is equivalent to approximately 1.75 pints, ensuring you\'ll have plenty of water to quench your thirst. In the US, on the other hand, a liter is equivalent to about 2.11 common pints. So, whether you need to convert for cooking or simply satisfy your curiosity, understanding these conversions will make your life easier.
Table of Content
- How many pints are in a liter?
- How is a liter defined and how does it relate to volume measurement?
- What is the conversion rate between pints and liters?
- Are there different types of pints, and if so, how do they differ in volume from liters?
- How is the conversion from liters to pints relevant in everyday life or specific industries (e.g., cooking, bartending, international trade)?
- YOUTUBE: How many pints are in a litre?
How many pints are in a liter?
To convert liters to pints, you need to know the conversion factor between the two units. The conversion factor for liters to pints is 1 liter = 2.11338 pints.
Now, if you have a given measurement in liters and you want to know how many pints it is, you simply need to multiply the given value by the conversion factor.
For example, let\'s say we have 3 liters. To find out how many pints that is, we multiply 3 liters by the conversion factor:
3 liters * 2.11338 pints/liter = 6.34 pints
So, there are approximately 6.34 pints in 3 liters.
How is a liter defined and how does it relate to volume measurement?
A liter is a metric unit of volume used to measure liquids. It is defined as the volume of a cube with sides of exactly one decimeter (or 10 centimeters) in length. The symbol for liter is \"L\" or \"l\".
To understand how a liter relates to volume measurement, it\'s important to know that volume is a measure of how much space an object occupies. In the case of liquids, it refers to the amount of three-dimensional space occupied by the liquid.
Here is a step-by-step explanation of how a liter is defined and its relation to volume measurement:
1. Definition of a liter: A liter is defined as one cubic decimeter (dm^3). A cubic decimeter is a unit of volume, and it represents a cube with sides measuring one decimeter (10 centimeters) each. So, a liter is the same as a cubic decimeter.
2. Metric system: The liter is part of the metric system of measurement, which is used in most countries around the world. The metric system is a decimal-based system, meaning that units are based on powers of 10. This makes conversions between different metric units relatively straightforward.
3. Relation to other units: A liter is often used as a base unit for measuring larger or smaller volumes. For example, one kiloliter (kL) is equal to 1,000 liters, and one milliliter (mL) is equal to 0.001 liters. The metric system allows easy conversion between these units by simply moving the decimal point.
4. Common conversions: When it comes to converting between liters and other non-metric units such as pints, it\'s important to note that one liter is approximately equal to 1.76 US pints or 1.76 UK pints. This conversion factor varies slightly depending on the country or context.
In summary, a liter is a metric unit of volume that is defined as one cubic decimeter. It is commonly used for measuring liquid volumes and is part of the decimal-based metric system. The liter can be easily converted to other metric units by moving the decimal point, and it is approximately equal to 1.76 US pints or 1.76 UK pints.
What is the conversion rate between pints and liters?
To convert pints to liters, you can use the conversion factor of 1 pint = 0.473176 liters.
Step 1: Determine the number of pints you want to convert.
Step 2: Multiply the number of pints by the conversion factor of 0.473176 liters/pint.
For example, if you want to convert 5 pints to liters:
5 pints * 0.473176 liters/pint = 2.36588 liters.
So, 5 pints is equal to approximately 2.36 liters.
Are there different types of pints, and if so, how do they differ in volume from liters?
Yes, there are different types of pints, and they differ in volume from liters. The two most common types of pints are the US pint and the Imperial (UK) pint.
1. US Pint:
- 1 US pint is equal to 0.473 liters.
- This type of pint is used primarily in the United States.
2. Imperial (UK) Pint:
- 1 Imperial pint is equal to 0.568 liters.
- This type of pint is used in the United Kingdom and some other countries.
To convert from pints to liters, or vice versa, you can use the conversion factors mentioned above.
For example, to convert 3 liters to US pints:
3 liters x (1 US pint / 0.473 liters) = 6.34 US pints
And to convert 4 Imperial pints to liters:
4 Imperial pints x (0.568 liters / 1 Imperial pint) = 2.27 liters
How is the conversion from liters to pints relevant in everyday life or specific industries (e.g., cooking, bartending, international trade)?
The conversion from liters to pints is relevant in everyday life and specific industries such as cooking, bartending, and international trade. Here\'s a detailed explanation of the conversion and its relevance:
1. Cooking: In countries that primarily use the metric system, recipes often provide measurements in liters. However, many people are more familiar with the imperial system, which uses pints as a unit of measurement. Therefore, knowing how to convert between liters and pints is essential when following or adapting recipes. For example, if a recipe calls for 2 liters of milk, you can convert it to pints to get a better understanding of the quantity needed.
2. Bartending: Bartenders often use measurements to ensure consistency in their cocktails. Some recipes may express the quantities in liters, especially those from countries using the metric system. However, when serving customers, it\'s more common to use pints or ounces as the unit of measurement. Being able to convert between liters and pints allows bartenders to follow recipes accurately and create drinks in the desired serving size.
3. International Trade: The conversion between liters and pints is also important in international trade. Different countries use different systems of measurement, which can create confusion when conducting business or importing/exporting goods. For example, if a company wants to export a beverage in pints to a country that primarily uses liters, they need to know the conversion to ensure accurate labeling and packaging.
Now, let\'s discuss the conversion itself:
- To convert from liters to pints, you need to know the conversion factor. It depends on the type of pint you are using, as different countries have different definitions for the pint:
- In the United States, 1 pint is equivalent to 0.473 liters.
- In Ireland, 1 pint is equivalent to approximately 0.568 liters.
- To convert from liters to pints, divide the volume in liters by the conversion factor. Let\'s use the conversion factor for the United States as an example:
- Number of pints = Volume in liters / Conversion factor
- For instance, if you have 2 liters of liquid and want to convert it to pints (US), the calculation would be:
Number of pints = 2 liters / 0.473 liters
Number of pints = 4.23 pints (approximately)
Understanding the conversion allows individuals in various industries to work with measurements in a way that suits their needs and facilitates better communication and consistency.
How many pints are in a litre?
Pints and litres are units of measurement commonly used for liquids. Pints are a unit of volume measurement commonly used in the United States and some other countries, while litres are a unit of volume measurement commonly used in many countries worldwide. To convert pints to litres, you can multiply the number of pints by 0.473176473, since 1 pint is equal to 0.473176473 litres. Similarly, to convert litres to pints, you can multiply the number of litres by 2.1133764189, since 1 litre is equal to 2.1133764189 pints. These conversion factors allow you to easily convert between pints and litres, depending on the unit of measurement you are working with.
How to convert pints to litres.
In this video you will learn how to convert between pints and litres. To do this you need to know that 1 pint is 0.57 litres. |
Data-Snooping Biases in Financial Analysis1994
Data-snooping—finding seemingly significant but in fact spurious patterns in the data—is a serious problem in financial analysis. Although it afflicts all non-experimental sciences, data-snooping is particularly problematic for financial analysis because of the large number of empirical studies performed on the same datasets. Given enough time, enough attempts, and enough imagination, almost any pattern can be teased out of any dataset. In some cases, these spurious patterns are statistically small, almost unnoticeable in isolation. But because small effects in financial calculations can often lead to very large differences in investment performance, data-snooping biases can be surprisingly substantial. In this review article, I provide several examples of data-snooping biases, explain why it is impossible to eliminate them completely, and propose several ways to guard against the most extreme forms of data-snooping in financial analysis.
Neural Networks and Other Nonparametric Techniques in Economics and Finance1994
Although they are only one of the many types of statistical tools for modeling nonlinear relationships, neural networks seem to be surrounded by a great deal of mystique and, sometimes, misunderstanding. Because they have their roots in neurophysiology and the cognitive sciences, neural networks are often assumed to have brain-like qualities: learning capacity, problem-solving abilities, and ultimately, cognition and self-awareness. Alternatively, neural networks are often viewed as "black boxes" that can yield accurate predictions with little modeling effort. In this review paper, I hope to remove some of the mystique and misunderstandings about neural networks by providing some simple examples of what they are, what they can and cannot do, and where neural nets might be profitably applied in financial contexts.
A Nonparametric Approach to Pricing and Hedging Derivative Securities via Learning Networks1994
We propose a nonparametric method for estimating the pricing formula of a derivative asset using learning networks. Although not a substitute for the more traditional arbitrage-based pricing formulas, network pricing formulas may be more accurate and computationally more efficient alternatives when the underlying asset's price dynamics are unknown, or when the pricing equation associated with no-arbitrage condition cannot be solved analytically. To assess the potential value of network pricing formulas, we simulate Black-Scholes option prices and show that learning networks can recover the Black-Scholes formula from a six-month training set of daily options prices, and that the resulting network formula can be used successfully to both price and delta-hedge options out-of-sample. For purposes of comparison, we perform similar simulation experiments for four other methods of estimation: OLS, kernel regression, projection pursuit, and multilayer perceptron networks. To illustrate the practical relevance of our network pricing approach, we apply it to the pricing and delta-hedging of S&P 500 futures options from 1987 to 1992.
The non-trading or non-synchronous effect arises when time series, usually financial asset prices, are taken to be recorded at time intervals of one length when in fact they are recorded at time intervals of another, possibly irregular, lengths. For example, the daily prices of securities quoted in the financial press are usually "closing" prices, prices at which the last transaction in each of those securities occurred on the previous business day. these closing prices generally do not occur at the same time each day, but by calling them "daily" prices, we have implicitly and incorrectly assumes that they are equally spaces at 24-hour intervals. Such an assumption can generate spurious predictability in price changes and returns even if true price changes or returns are statistically independent. The non-trading effect induces potentially serious biases in the moments and co-moments of asset returns such as their means, variances, covariances, and autocorrelation and cross-autocorrelation coefficients.
Empirical Issues in the Pricing of Options and Other Derivative Securities1992
The pricing of options, certificates, and other derivatives or assets—financial assets whose payments depend on the prices of other assets—is one of the great successes of modern financial economics. Although the pricing of derivatives is computationally intensive, there is little done in terms of the traditional empirical analysis since by the very nature of the determination of prices and arbitrage there is no error term to minimize. There are, however, many issues of statistical inference that affect the pricing of options and other derivatives. This paper analyzes two of the most common issues neglected in the literature: reduced form empirical instruments for the determination of prices and how to use Monte Carlo simulations to calculate option prices depend on a path.
An Ordered Probit Analysis of Transaction Stock Prices1992
We estimate the conditional distribution of trade-to-trade price changes using ordered probit, a statistical model for discrete random variables. This approach recognizes that transaction price changes occur in discrete increments, typically eighths of a dollar, and occur at irregularly-spaced time intervals. Unlike existing models of discrete transaction prices, ordered probit can quantify the effects of other economic variables like volume, past price changes, and the time between trades on price changes. Using 1988 transactions data for over 100 randomly chosen U.S. stocks, we estimate the ordered probit model via maximum likelihood and use the parameter estimates to measure several transaction-related quantities, such as the price impact of the trades of a given size, the tendency towards price reversals from one transaction to the next, and the empirical significance of price discreteness.
Long-Term Memory in Stock Market Prices1991
A test for long-run memory that is robust to short-range dependence is developed. It is an extension of the "range over standard deviation" or R/S statistic, for which the relevant asymptotic sampling theory is derived via functional central limit theory. This test is applied to daily and monthly stock returns indexed over several time periods and, contrary to previous findings, there is no evidence of long-range dependence in any of the indexes over any sample period or sub-period once short-range dependence is taken into account. Illustrative Monte Carlo experiments indicate that the modified R/S test has power against at least two specific models of long-run memory, suggesting that stochastic models of short-range dependence may adequately capture the time series behavior of stock returns.
Data Snooping Biases in Tests of Financial Asset Pricing Models1990
Tests of financial asset pricing models may yield misleading inferences when properties of the data are used to construct the test statistics. In particular, such tests are often based on returns to portfolios of common stock, where portfolios are constructed by sorting some empirically motivated characteristic of the securities such as market value of equity. Analytical calculations, Monte Carlo simulations, and two empirical examples show the effects of this type of data snooping can be substantial.
When Are Contrarian Profits Due To Stock Market Overreaction?1990
If returns on some stocks systematically lead or lag those of others, a portfolio strategy that sells "winners" and "losers" can produce positive expected returns, even if no stock's returns are negatively autocorrelated as virtually all models of overreaction imply. Using a particular contrarian strategy we show that, despite negative autocorrelation in individual stock returns, weekly portfolio returns are strongly positively autocorrelated and are the result of important cross-autocorrelations. We find that the returns of large stocks lead those of smaller stocks, and we present evidence against overreaction as the only source of contrarian profits.
An Econometric Analysis of Nonsynchronous Trading1990
We develop a stochastic model of nonsynchronous asset prices based on sampling with random censoring. In addition to generalizing existing models of nontrading, our framework allows the explicit calculation of the effects of infrequent trading on the time series properties of asset returns. These are empirically testable implications for the variance, autocorrelations, and cross-autocorrelations of returns to individual stocks as well as to portfolios. We construct estimators to quantify the magnitude of nontrading effects in commonly used stock returns data bases, and show the extent to which this phenomenon is responsible for the recent rejections of the random walk hypothesis. |
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Do customers who have debit cards tend to use ATMs differently from those who do not have debit cards?.
Students are to maximize the use of charts and minimize the use of tables (other than in the appendices) in the body of their report. Actual numbers from the Excel or SPSS output must be used in the report text and explained.
• References in the body of the report should be made to the more-detailed material in the appendices.
• Work to include relevant analysis beyond what is required and for intuition if your report contains insights about the problem that are not obvious from the questions asked.
• Summarize all of your findings.
• Discuss the limitations of your study, what questions remain unanswered, and make suggestions to find the answer for unanswered issues in the project (and for follow-on work); for example, you may consider questions like these: Do the findings make sense? What else would you like to know about the sample data? What other data would you collect if you could? What other analyses would you want to do then?
Relevant tables, charts not used in the report body, statistical details, and Excel or SPSS output are to be presented in the appendices.
• The appendices should be annotated to explain the included chart or Excel or SPSS output.
• Statistical jargon may be used in the appendices.
The report has the following constraints:
1. A one page introduction section.
2. Full report may be up to ten pages (double-spaced) not including the Appendices
3. Annotated appendices of unlimited length.
4. Please include additional explanations of your findings in plain English, as if it is to be presented to your boss who is not interested in statistical details.
5. You may use Excel or SPSS statistical software to do the analysis.
6. Please keep in mind that your report is to present your own thinking in your own words. It should not be written like answers to homework problems. How well the report is written will be a part of its grading.
7. I would suggest starting with the Appendices and including all relevant Excel or SPSS output along with explanation in statistical terms underneath each graph or result, then write the rest of the report.
CASE: Century National Bank
Assume that you work in the Planning Department of the Century National Bank and report to Ms. Lamberg. You will need to do some data analysis and prepare a short written report. Remember, Mr. Selig is the president of the bank, so you will want to ensure that your report is complete and accurate.
Century National Bank has offices in several cities in the Midwest and the southeastern part of the United States. Mr. Dan Selig, president and CEO, would like to know the characteristics of his checking account customers. What is the balance of a typical customer? How many other bank services do the checking account customers use? Do the customers use the ATM service and, if so, how often? What about debit cards? Who uses them, and how often are they used?
To better understand the customers, Mr. Selig asked Ms. Wendy Lamberg, director of planning, to select a sample of customers and prepare a report. To begin, she has appointed a team from her staff.
You are the head of the team and responsible for preparing a report. You select a random sample of 60 customers. In addition to the balance in each account at the end of last month, you determine: (1) the number of ATM (automatic teller machine) transactions in the last month; (2) the number of other bank services (a savings account, a certificate of deposit, etc.) the customer uses; (3) whether the customer has a debit card (this is a relatively new bank service in which charges are made directly to the customers account); and (4) whether or not interest is paid on the checking account. The sample includes customers from the branches in Cincinnati, Ohio; Atlanta, Georgia; Louisville, Kentucky; and Erie, Pennsylvania.
1. Classify each variable in the dataset by the level of measurement (nominal, ordinal, interval, and ratio) and by the type of variable (Categorical or Numerical, then discrete or continuous). [Chapter 1]
2. Develop a frequency distribution and histogram that portray the checking balances. What is the balance of the typical customer? Do many customers have more than $2,000 in their accounts? [Chapters 2 and 3]
3. Determine the mean and median of the checking account balances. Compare the mean and the median balances of the four branches. Is there a difference among the branches? (Be sure to explain the difference between the mean and the median in your report.) [Chapter 3]
4. Determine the range and standard deviation of the checking account balances. What do the first and third quartiles show? Because Mr. Selig does not deal with statistics daily, include a brief description and interpretation of the standard deviation and other measures. [Chapter 3]
5. Is it reasonable that the distribution of checking account balances approximates a normal distribution? Determine the mean and the standard deviation for the sample of 60 customers.
Compare the actual distribution with the theoretical distribution (i.e., empirical rule) by evaluating the actual versus the theoretical properties. Cite some specific examples and comment on your findings. [Chapter 6]
6. When Mr. Selig took over as president of Century several years ago, the use of debit cards was just beginning. He would like to an update on the use of these cards. Develop a 95 percent confidence interval for the proportion of customers using these cards. On the basis of the confidence interval, is it reasonable to conclude that more than half of the customers use a debit card? Mr. Selig would also like to know the mean account balances for all balances in all banks with a 95 percent level of confidence. Interpret the results. [Chapter 8]
7. With many other options available, customers no longer let their money sit in a checking account. For many years the mean checking balance has been $1,600. Does the sample data indicate that the mean account balance has declined from this value? [Chapter 9]
8. Recent years have also seen an increase in the use of ATM machines. When Mr. Selig took over the bank, the mean number of transactions per month per customer was 8; now he believes it has increased to more than 10. In fact, the advertising agency that prepares TV commercials for Century would like to use this on a new commercial being designed. Is there sufficient evidence to conclude that the mean number of transactions per customer is more than 10 per month? Could the advertising agency say the mean is more than 9 per month? [Chapter 9]
9. The bank has branch offices in four different cities: Cincinnati, Ohio; Atlanta, Georgia; Louisville, Kentucky; and Erie, Pennsylvania. Mr. Selig would like to know whether there is a difference in the mean checking account balances among the four branches. Specifically, he would like to know if there are differences in the mean checking account balances between (1) Cincinnati and Atlanta (2) Atlanta and Louisville, and (3) Atlanta and Erie. Use interval estimates and assume known and equal variances of 311667. [Chapter 10]
10. Mr. Selig is also interested in the banks ATMs. Do customers who have debit cards tend to use ATMs differently from those who do not have debit cards? Is there a difference in ATM use by those with checking accounts that pay interest versus those that do not? Prepare a report for Mr. Selig answering these questions. [Chapter 10]
11. Develop a regression model to predict Checking Account Balances from the number of ATM transactions in the month (ATM). Discuss the model and its strength on the basis of indicators presented in Chapter 14 of your textbook and/or corresponding lecture notes. Does it seem logical that Checking Account Balances could be predicted by number of ATM transactions in the month? [Chapter 14]
12. Develop a regression model to predict Checking Account Balances from the number of other bank services used (Services). Examine the regression output. Is this model stronger than the model in Question 11 above in predicting the amount of Checking Account Balances? Explain why, using techniques presented in Chapter 14 of your textbook and/or corresponding lecture notes. [Chapter 14]
Do customers who have debit cards tend to use ATMs differently from those who do not have debit cards? |
Research Article | Open Access
Dianliang Deng, "Convergence Rates for Probabilities of Moderate Deviations for Multidimensionally Indexed Random Variables", International Journal of Mathematics and Mathematical Sciences, vol. 2009, Article ID 253750, 15 pages, 2009. https://doi.org/10.1155/2009/253750
Convergence Rates for Probabilities of Moderate Deviations for Multidimensionally Indexed Random Variables
Let be a sequence of i.i.d. real-valued random variables, and , . Convergence rates of moderate deviations are derived; that is, the rates of convergence to zero of certain tail probabilities of the partial sums are determined. For example, we obtain equivalent conditions for the convergence of the series , where , , and are taken from a broad class of functions. These results generalize and improve some results of Li et al. (1992) and some previous work of Gut (1980).
Let be the set of all positive integer dimensional lattice points with coordinate-wise partial ordering, ; that is, for every , , if and only if , , where is a fixed integer. denote and means . Throughout the paper, are i.i.d. random variables with and . Let , and . we define
where are real numbers with . Further, set and .
Hartman and Wintner studied the fundamental strong laws of classic Probability Theory for i.i.d random variables and obtained the following Hartman-Wintner law of the iterated logarithm (LIL).
Theorem 1.1. and if and only if
Afterward, the study of the estimate of the convergence rate in the above relation (1.2) has been attracting the attention of various researchers over the last few decades. Darling and Robbins , Davis , Gafurov , and Li have obtained some good results on the estimate of convergence rate in (1.2). The best result is probably the one given by Li et al. . For easy reference, we restate their result in Theorem 1.2.
Theorem 1.2. Let and be two positive real-valued functions on such that is nondecreasing, and as . For , let , and . Then the following results are equivalent: where is the reverse function of
Proof. Because by using Lemma 3.1, we have The last expression is finite if and only if This completes the proof of the lemma.
Lemma 4.2. implies (4.15).
Proof. We first prove the result for symmetric random variables. One may rearrange and obtain By Levy inequality, we set
Thus for large
equation (4.15) follows from Lemma 4.1.
If are nonsymmetric random variables, then by using the standard symmetrization method, it is easy to prove that for some constant which implies that That is, (4.15) holds.
Proof of Theorem 2.5. Similar to the proof of Theorem 2.2, by Lemma 4.1, implies that
Therefore, if , (2.12) is equivalent to
which is equivalent to
On the other hand, (2.12) implies by Lemma 4.2. Hence, (2.11) and (2.12) are equivalent. If, in addition, (4.24) is equivalent to the following: Hence, (2.12) and (2.13) are equivalent.
The author is very grateful for referee's comments and suggestions which greatly improve the readability and quality of the current paper. The research is partly supported by the Natural Sciences and Engineering Research Council of Canada.
- P. Hartman and A. Wintner, “On the law of the iterated logarithm,” American Journal of Mathematics, vol. 63, pp. 169–176, 1941.
- D. A. Darling and H. Robbins, “Iterated logarithm inequalities,” Proceedings of the National Academy of Sciences of the United States of America, vol. 57, pp. 1188–1192, 1967.
- J. A. Davis, “Convergence rates for the law of the iterated logarithm,” Annals of Mathematical Statistics, vol. 39, pp. 1479–1485, 1968.
- M. U. Gafurov, “On the estimate of the rate of convergence in the law of iterated logarithm,” in Probability Theory and Mathematical Statistics, vol. 1021 of Lecture Notes in Mathematics, pp. 137–144, Springer, Berlin, Germany, 1982.
- D. L. Li, “Convergence rates of law of iterated logarithm for -valued random variables,” Science in China. Series A, vol. 34, no. 4, pp. 395–404, 1991.
- D. L. Li, X. C. Wang, and M. B. Rao, “Some results on convergence rates for probabilities of moderate deviations for sums of random variables,” International Journal of Mathematics and Mathematical Sciences, vol. 15, no. 3, pp. 481–497, 1992.
- V. Strassen, “An invariance principle for the law of the iterated logarithm,” Zeitschrift für Wahrscheinlichkeitstheorie und Verwandte Gebiete, vol. 3, pp. 211–226, 1964.
- M. J. Wichura, “Some Strassen-type laws of the iterated logarithm for multiparameter stochastic processes with independent increments,” The Annals of Probability, vol. 1, pp. 272–296, 1973.
- A. Gut, “Convergence rates for probabilities of moderate deviations for sums of random variables with multidimensional indices,” The Annals of Probability, vol. 8, no. 2, pp. 298–313, 1980.
- A. Gut and A. Spătaru, “Precise asymptotics in some strong limit theorems for multidimensionally indexed random variables,” Journal of Multivariate Analysis, vol. 86, no. 2, pp. 398–422, 2003.
- C. W. Jiang, X. R. Yang, and L. X. Zhang, “Precise asymptotics of partial sums for multidimensionally indexed random variables,” Journal of Zhejiang University. Science Edition, vol. 33, no. 6, pp. 625–628, 2006.
- C. Jiang and X. Yang, “Precise asymptotics in self-normalized sums of iterated logarithm for multidimensionally indexed random variables,” Applied Mathematics: A Journal of Chinese Universities, vol. 22, no. 1, pp. 87–94, 2007.
- K.-L. Su, “Strong limit theorems of Cesaro type means for arrays of orthogonal random elements with multi-dimensional indices in Banach spaces,” Stochastic Analysis and Applications, vol. 22, no. 2, pp. 237–250, 2004.
- N. V. Giang, “On moments of the supremum of normed partial sums of random variables indexed by ,” Acta Mathematica Hungarica, vol. 60, no. 1-2, pp. 73–80, 1992.
- V. V. Petrov, Sums of Independent Random Variables, vol. 8 of Ergebnisse der Mathematik und ihrer Grenzgebiete, Springer, New York, NY, USA, 1975.
Copyright © 2009 Dianliang Deng. 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. |
A Meter Indicator For Use With Electrical Counters Using Scale-Of-2 Circuits.
Abstract.—A meter method of indicating the count of a series of coupled scale-of-2 circuits is described. The method is particularly suitable when a scale-of-16 unit has been converted to act as a decade counter.
High-speed electrical counters or chronographs are, at present, being used in increasing numbers and in an increasing variety of applications. When an instrument is first devised and is performing a new function, a certain amount of inconvenience in use is tolerated, but when it becomes an everyday tool it is desirable that the operation be as simple as possible. During the war period electrical counters or chronographs underwent considerable delevopment, one of the main uses being the measurement of short time intervals in determining the muzzle velocity of shells. High-speed counters have also been used for many years in radioactivity measurements, each disintegration in a sample producing a count.
The fundamental unit of a high-speed electrical counter is a scale-of-2 circuit (Fig. 1), which consists of two electronic valves coupled together so that the system has two stable states. In one stable state, the first valve, A, is conducting, and the second, B, non-conducting. If a suitable impulse is fed to the input, the conducting states are reversed, B becomes conducting and A non-conducting. Another impulse to the input returns the system to the original state, and by taking an output signal from one valve of the pair
one output pulse can be obtained for two input pulses. Hence the name, Scale-of-2. By using a number of scale-of-2 circuits in series, it is possible to build up scales-of-4, 8, 16, etc.
The normal method of indicating the count is to have one neon lamp connected to each scale-of-2 circuit (across one of the plate load resistors). If the lamp is glowing, the count is one, if the lamp is extinguished, the count is zero. Hence for a scale-of-16, there would be four lamps (Fig. 2). At the start of a counting period all the lamps would be extinguished. When one input pulse is received, the first lamp glows. With a second input pulse, the first lamp goes out, but the second lamp goes on. The counting value for the second lamp is two, the value for the third lamp four, and for the fourth lamp eight. So if, after a counting period, the first and fourth lamps were glowing, the number of pulses received would have been 1 + 8, i.e., 9. In a large sealing unit using 10 scale-of-2 circuits and giving, in effect, a scale-of-1024, there is a considerable possibility of error in recording the count. The trouble is that the final state of the instrument is not simply interpreted in terms of our decimal system of counting.
There have been two developments to simplify the operation and use of high-speed electrical counters. The first improvement was the introduction of the decimal or decade counter. One type, described by Potter*, uses a very ingenious method of resetting a normal scale-of-16 circuit to its original state after ten pulses have been received. Potter used four neon lamps as indicators for each decade, thus making a small amount of arithmetic necessary when finding the number of units, tens and hundreds in a count. A second type has been described by Lewist.† † This method uses a scale-of-2 circuit followed by a ring-of-5. For indicators, Lewis suggests neon lamps or magic-eye tubes.
[Footnote] * J. T. Potter. A Four-tube Counter Decade. Electronics: 110–113. June, 1944.
[Footnote] † W. B. Lewis. Electrical Counting. P. 91. Cambridge University Press, 1942.
The second improvement has been the introduction of a meter to indicate the count, the meter pointer moving over the scale in steps, and the reading giving the state of the counter directly. Quite recently, a decade counter using the scale-of-2 and ring-of-5 system with a meter indicator, has been described by West.* This particular counting system had been described earlier (during the war period) in a report which had a restricted circulation.
The remainder of this paper describes a meter indicator system which can be used with electrical counters consisting of a series of scale-of-2 circuits. The method is particularly convenient when used with a decade counter of the Potter type, i.e., a scale-of-16 converted to act as a scale-of-10.
Fig. 3 shows the principle. Each pair of valves forms a scale-of-2 circuit. In the initial state before a counting period, valves 1, 3, 5 and 7 are conducting, and valves 2, 4, 6 and 8 non-conducting. The current through the plate load resistor of a conducting valve is about 2.5 ma., and the current through the other plate load resistor about 0.5 ma. (due to the resistors of the coupling circuit). When a suitable pulse is received and the scale-of-2 circuit changes its stable state, the change in current through each plate load resistor is about 2 ma. In the initial state of the counter, the 0.5 ma. currents produce a total voltage drop across R1, R2, R3 and R4 of about 37.5 mV. A similar potential drop is arranged across R5, so, at the start, there is zero potential difference between points × and Y.
When the first pulse is received and valve 2 becomes conducting, the extra 2 ma. through R1 produces a potential change of 0.01 V. at X. With a 2nd pulse, 2 becomes non-conducting, but 4 becomes conducting. The 2 ma. through R1 and R2 produces a potential change of 0.02 V. at X. A 3rd pulse makes valve 2 conducting besides valve 4, so giving 4 ma. through R1 and 2 ma. through R2 to provide the change. This produces 0.03 V between × and Y. With a 4th pulse we get 0.04 V. at X, since valves 2 and 4 become non-conducting but 6 conducting. A 5th pulse gives 0.05 V., etc. Hence each pulse received causes the potential of × to change by equal increments. In the case of the circuit shown, the 15th pulse causes valves 2, 4, 6 and 8 to be conducting, and a potential at × of 0.15 V. The 16th pulse makes these valves non-conducting, causes one output pulse from the unit, returns the potential difference between × and Y to zero, and brings the system to the original state as at the beginning of the counting period. The sole effect of Potter's modification of the circuit is to make valves 2, 4, 6 and 8 non-conducting on the receipt of the 10th pulse instead of needing to wait for
[Footnote] * S. S. West. An Electronic Decimal Counter Chronometer. Electronic Engineering. 19:3–6, Jan., 1947. 19:58–61, Feb., 1947.
the 16th. A direct-reading instrument is made by connecting a suitable meter between points × and Y. A 0–500 microammeter with resistance about 150–200 ohms would be satisfactory. Of course, the introduction of the meter will alter the distribution of currents and potentials slightly from those described above, but this does not interfere with the meter's action of indicating the count.
Note: The plate-to-grid coupling condensers marked 10μμF. should be 100μμF.
The undesired couplings between different scale-of-2 circuits due to the common resistors R1, R2 and R3 are very small, and no trouble has been experienced due to them. The instrument can be returned to its initial state, i.e., the meter reading returned to zero, at any time, by opening the reset key for a moment. This makes valves 2, 4, 6 and 8 non-conducting.
High-speed electrical counters using scale-of-2 circuits have been used for a number of years with neon lamps. These indicators need to be interpreted by the operator, and there is a considerable possibility of human error in the case of large scaling circuits. The meter indicator described in this paper simplifies the interpretation.
Mr. R. L. Taylor asked what was the maximum scale of counting that could be covered by the use of a meter indicator.
The speaker replied that this depended on the size and ease of reading of the meter scale. With the ordinary type of panel meter available he did not advocate the use of this method above, say, a scale-of-16 counter. A larger meter would enable higher scales to be distinguished. He advocated the method particularly for decimal counters. |
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1 edition of Mathematical tables for trigonometrical, astronomical and nautical calculations found in the catalog.
Mathematical tables for trigonometrical, astronomical and nautical calculations
With this is bound: Young, John Radford. Tables intended to facilitate the operations of navigation ... London, 1888.
|Other titles||Tables intended to facilitate the operations of navigation.|
|Statement||by Henry Law ...|
|Series||Weale"s rudimentary series ;, 204|
|Contributions||Young, J. R. 1799-1885.|
|LC Classifications||QA55 .L25 1888|
|The Physical Object|
|Pagination||vii, 233, p. :|
|Number of Pages||233|
|LC Control Number||30032026|
Try the new Google Books. Check out the new look and enjoy easier access to your favorite features. Try it now. with its application to navigation and surveying, nautical and practical astronomy and geodesy, with logarithmic, trigonometrical, and nautical tables. Charles William Hackley. G. P. Putnam, - Mathematics - pages. This video explains the basics of how math is used in celestial navigation. When people first started to navigate by stars a massive explosion in trade and culture developed.
Norie's Nautical Tables - Ebook written by Capt A.G Blance. Read this book using Google Play Books app on your PC, android, iOS devices. Download for offline reading, highlight, bookmark or take notes while you read Norie's Nautical Tables. Quick links: Entire table: shows each trig function evaluated for every degree 1 through Special table: shows each trig function evaluated for special angles, l 45, and 60 degrees.. You may also be interested in our Unit Circle page - a way to memorize the special angle values quickly and easily!
The complete mathematical and general navigation tables, including every table required with the nautical almanac in finding the latitude and longitude: with an explanation of the construction, use, and application to navigation and nautical astronomy, trigonometry, dialling, gunnery, etc., etc. (London, Simpkin, Marshall, & Co., ), by. Norie's Nautical Tables (Nories Nautical This famous set of mathematical tables was first published in with the aim of improving the accuracy of the calculated position and reducing the tedium of the calculation. All the tables required for coastal and deep sea navigation are included A simple uniform method of interpolation for /5(11).
INVERGORDON, ROSS-SHIRE OFFICI.
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Mathematical Tables For Trigonometrical, Astronomical, And Nautical Calculations. [with] Tables Intended To Facilitate The Operations Of Mathematical tables for trigonometrical And Nautical Astronomy, By J.r.
Young [Henry Law, John Radford Young] on *FREE* shipping on qualifying offers. This work has been selected by scholars as being culturally important, and is part of the. Mathematical tables for trigonometrical, astronomical, and nautical calculations.
[With] Tables Item PreviewPages: Buy Mathematical Tables Consisting of Logarithms of Numbers 1 toTrigonometrical, Nautical and Other Tables.
on FREE SHIPPING on qualified orders Mathematical Tables Consisting of Logarithms of Numbers 1 toTrigonometrical, Nautical and Other Tables/5(1).
The value of the characteristic of the logarithm of a number is always one less than the number of integers in that num- ber ; thus, in the above example, when the number is 20 the characteristic is 1, when it is 2, and when it is 3. The characteristic, therefore, of the logarithms of all numbers Equal to.
Mathematical tables for trigonometrical, astronomical, and nautical calculations. [With] Tables intended to facilitate the operations of navigation and nautical astronomy, by J.R. Young Henry Law — Mathematical tables are lists of numbers showing the results of a calculation with varying of trigonometric functions were used in ancient Greece and India for applications to astronomy and celestial continued to be widely used until electronic calculators became cheap and plentiful, in order to simplify and drastically speed up.
the answers precisely with a hand calculator and understand astronomical and nautical calculations book why. You will need a scientific calculator, those having trigonometric functions and their inverse functions. Programmable graphing calculators such as the TI and TI are excellent for the methods described in the book.
To those readersFile Size: 1MB. Difference of altitude = true altitude – computed altitude: 30° 04,5’ – 30° 02,6’ = + 1,9’ Now we can draw the graph: The last worksheet is named “ meridian altitude “. It allows to obtain the latitude after having insertered the meridian altitude and declination. We have a true altitude at meridian passage of 58° 43,6’ and a.
Celestial Navigation and Spherical Trigonometry The PZX Triangle. Spherical Formulae. The section on “Sailings” deals with mathematical calculations. These tables avoid the need to use a calculator or Log tables but are based on the previous Size: KB.
Trigonometry Angles. The trigonometry angles which are commonly used in trigonometry problems are 0°, 30°, 45°, 60° and 90°.
The trigonometric ratios such as sine, cosine and tangent of these angles are easy to memorize. We will also show the table where all the ratios and their respective angle’s values are mentioned.
The Nautical Almanac - Sun and stars. by Capt. Roberto Iori. astronomical data for mariners - year " The Nautical Almanac of the stars " (PDF or excel spreadsheet) The excel version generates the essential data needed for the pratice of celestial navigation - This free software creates daily pages (a printable sheet - A4).
See ``List of logarithmic, trigonometrical, and astronomical calculations, in manuscript, by Edward Sang,'' p. of E. Horsburgh, Modern Instruments and Methods of Calculation, London and Edinburgh, Cited by: 1.
A simple uniform method of interpolation for all the trigonometrical tables is used Certain tables and data are also included which are not readily available on board ship or are only used in the examination room. The section Seaports of the World has also been extensively updated and restructured with several hundred additional ports.
MATHEMATICAL TABLES, for Trigonometrical, Astronomical, and Nautical Calculations ; to which is prefixed a Treatise on Logarithms. By Henry Law, C.E. Together with a Series of Tables for Navigation and Nautical Astronomy.
By J. Young, formerly Professor of Mathematics in Belfast College. New Edition. 6d.± Norie's Nautical Tables ED. Catalog # Norie's Nautical Tables ED This famous set of mathematical tables was first published in It has been a bestseller ever since, and despite developments in electronic navigation it remains an essential requirement for anyone learning and practising astro-navigation.
Astronomical Algorithms, Astronomical Formulae for Calculators, Astronomical Tables of the Sun, Moon and Planets, Canon of Lunar Eclipses B.C. - A.D.Elements of Solar EclipsesLow-Precision Formulae for Planetary Positions, Computer Modeling: From Sports to Spaceflight from Order to Chaos, Introduction to Basic Astronomy with a PC.
Trigonometry in the modern sense began with the Greeks. Hipparchus (c. – bce) was the first to construct a table of values for a trigonometric function.
He considered every triangle—planar or spherical—as being inscribed in a circle, so that each side becomes a chord. Free math lessons and math homework help from basic math to algebra, geometry and beyond.
Students, teachers, parents, and everyone can find solutions to their math problems instantly. and Ptolemy, used trigonometry in their study of astronomy between roughly B.C. - A.D.its history is much older. For example, the Egyptian scribe Ahmes recorded some rudi-mentary trigonometric calculations (concerning ratios of sides of pyramids) in the famous Rhind Papyrus sometime around B.C.1File Size: 1MB.
Later he wrote an important work, the Quadripartitum, on the fundamentals of trigonometry needed for the solution of problems of spherical astronomy. The first part of this work is a theory of trigonometrical identities, and was regarded as a basis for the calculation of sines, cosines, chords and versed sines.
Trigonometry is known for its many identities, which are equations used for rewriting trigonometrical expressions to solve equations, to find a more useful expression, or to discover new relationships.The taIt has a wide number of applications in other fields of Mathematics.
Many geometric calculations can be easily figured out using the table of trigonometric functions and formulas as well. The Trigonometrical ratios table helps to find the values of trigonometric standard angles such as 0°, 30°, 45°, 60° and 90°.
Dale, Five-figure Tables of Mathematical Functions (London, ), is a book of 92 pages containing a number of small five-figure tables of functions which are not elsewhere to be found in one volume. |
Maxwell and Special RelativityKirk T. McDonaldJoseph Henry Laboratories, Princeton University, Princeton, NJ 08544(May 26, 2014; updated July 6, 2019)It is now commonly considered that Maxwell’s equations in vacuum implicitly containthe special theory of relativity.1For example, these equations imply that the speed c of light in vacuum is related by,21c , 0 μ0(1)where the constants 0 and μ0 can be determined in any (inertial) frame via electrostatic andmagnetostatic experiments (nominally in vacuum).3,4,5 Even in æther theories, the velocityof the laboratory with respect to the hypothetical æther should not affect the results of thesestatic experiments,6 so the speed of light should be the same in any (inertial) frame. Then,the theory of special relativity, as developed in , follows from this remarkable fact.Maxwell does not appear to have crisply drawn the above conclusion, that the speed oflight is independent of the velocity of the observer, but he did make arguments in Arts. 599600 and 770 of that correspond to the low-velocity approximation to special relativity,as pointed out in sec. 5 of .71Maxwell’s electrodynamics was the acknowledged inspiration to Einstein in his 1905 paper .Equation (1) is a transcription into SI units of the discussion in sec. 80 of and sec. 758 of .3This was first noted by Weber and Kohlsrausch (1856) [19, 20], as recounted by Maxwell on p. 21 of, sec. 96 of , p. 644 of , and Arts. 786-787 of .4It is now sometimes said that electricity plus special relativity implies magnetism, but a more historicalview is that (static) electricity plus magnetism implies special relativity. This theme is emphasized in, forexample, .5As reviewed in , examples of a “static” current-carrying wire involve effects of order v2 /c2 where vis the speed of the moving charges of the current. A consistent view of this in the rest frame of the movingcharges requires special relativity. These arguments could have been made as early as 1820, but it took 85years for them to be fully developed.6This ansatz is a weak form of Einstein’s Principle of Relativity.7These two arguments also correspond to use of the two types of Galilean electrodynamics , as noted in[87, 103, 105]. The notion of Galilean electrodynamics, consistent with Galilean relativity, i.e., the coordinatetransformation x x vt, y y, z z, t t, seems to have been developed only in 1973 . The termGalilean relativity was first used in 1911 . In Galilean electrodynamics there are no electromagnetic waves,but only quasistatic phenomena, so this notion is hardly compatible with Maxwellian electrodynamics as awhole. In contrast, electromagnetic waves can exist in the low-velocity approximation to special relativity,and, of course, propagate in vacuum with speed c.In Galilean electrodynamics the symbol c does not represent the speed of light (as light does exist in this theory), but only the function 1/ 0 μ0 of the (static) permittivity and permeability of the vacuum.In fact, there are two variants of Galilean electrodynamics:1. Electric Galilean relativity (for weak magnetic fields) in which the transformations between two inertialframes with relative velocity v are (sec. 2.2 of ), given here in Gaussian units, as will be used in the restof this note,vρ e ρe ,J e Je ρe v,(c ρe Je ),Ve Ve ,A e Ve ,(2)cvB e Be Eefe ρe Ee(electric),(3)E e Ee ,c21
Articles 598-599 of Maxwell’s Treatise1In his Treatise , Maxwell argued that an element of a circuit (Art. 598), or a particle(Art. 599) which moves with velocity v in electric and magnetic fields E and B μHexperiences an electromotive intensity (Art. 598), i.e., a vector electromagnetic force givenby eq. (B) of Art. 598 and eq. (10) of Art. 599,8E V B Ȧ Ψ,(8)where E is the electromotive force, V is the velocity v, B is the magnetic field B, A isthe vector potential and Ψ represents, according to a certain definition, the electric (scalar)potential. If we interpret electromotive force to mean the force per charge q of the particle,9i.e., E F/q, then we could write eq. (8) as, v(9)F q E B ,cnoting that the electric field E is given (in emu) by A/ t Ψ,10,11,12 and that v Bin emu becomes v/c B in Gaussian units.where ρ and J are the electric charge and current densities, V and A are the electromagnetic scalar andvector potentials, E V A/ ct is the electric field, B A is the magnetic (induction) field,.2. Magnetic Galilean relativity (for weak electric fields, sec. 2.3 of ) with transformations,vv· Jm , J m Jm , (c ρe Je ), Vm Vm · Am ,c2c vv Em Bm ,Bm Bmfm ρm Em Bmccρ m ρm A m Am ,(4)E m(magnetic).(5)For comparison, the low-velocity limit of special relativity has the transformations,ρ s ρs v· Jsc2J s Js ρs v,vE s Es Bs ,cVs Vs vB s Bs Esc8v· As ,cvVs ,c(6)(special relativity, v c).(7)A s As This result also appeared in eq. (77), p. 343, of (1861), and in eq. (D), sec. 65, p. 485, of (1864),where E was called the electromotive force. The evolution of Maxwell’s thoughts on the “Lorentz” force aretraced in Appendix A below. See also [85, 88, 91].9In contrast to, for example, Weber , Maxwell did not present in his Treatise a view of an electriccharge as a “particle”, but rather as a state of “displaced” æther. However, in his earliest derivation of oureq. (8), his eq. (77), p. 342 of , Maxwell was inspired by his model of molecular vortices in which movingparticles (“idler wheels”) corresponded to an electric current (see also sec. A.2.7 below).For comments on Maxwell’s various views on electric charge, see .10This assumes that Maxwell’s Ȧ corresponds to A/ t, and not to the convective derivative DA/Dt A/ t (v · )A.11Maxwell never used the term electric field as we now do, and instead spoke of the (vector) electromotiveforce or intensity (see Art. 44 of ). The distinction is important only when discussing a moving medium,as in Arts. 598-599.12The relation E A/ t Ψ for the electric field holds in any gauge. However, Maxwell alwaysworked in the Coulomb gauge, where · A 0, as affirmed, for example, in Art. 619. Maxwell was awarethat, in the Coulomb gauge, the electric scalar potential Ψ is the instantaneous Coulomb potential, obeyingPoisson’s equation at any fixed time, as mentioned at the end of Art. 783. The discussion in Art. 783 isgauge invariant until the final comment about 2 Ψ (in the Coulomb gauge). That is, Maxwell missed anopportunity to discuss the gauge advocated by Lorenz , to which he was averse .2
Our equation (9) is now known as the Lorentz force,13,14 and it seems seldom noted thatMaxwell gave this form, perhaps because he presented eq. (10) of Art. 599 as applying to anelement of a circuit rather than to a charged particle. In Arts. 602-603, Maxwell discussedthe Electromotive Force acting on a Conductor which carries an Electric Current through aMagnetic field, and clarified in his eq. (11), Art. 603 that the force on current density J (J)is, J(10)F J BF B .cIf Maxwell had considered that a small volume of the current density is equivalent to anelectric charge q times its velocity v, then his eq. (11), Art. 602 could also have been writtenas,vF Bqc( V B) ,(11)which would have confirmed the interpretation we have given to our eq. (9) as the Lorentzforce law. However, Maxwell ended his Chap. VIII, Part IV of his Treatise with Art. 603,leaving ambiguous some the meaning of that chapter.In his Arts. 598-599, Maxwell considered a lab-frame view of a moving circuit. However,we can also interpret Maxwell’s E as the electric field E in the frame of the moving circuit,such that Maxwell’s transformation of the electric field is,15E E v B.c(12)The transformation (12) is compatible with both magnetic Galilean relativity, eq. (5), andthe low-velocity limit of special relativity, eq. (7). These two versions of relativity differ asto the transformation of the magnetic field. In particular, if B 0 while E were due toa single electric charge at rest (in the unprimed frame), then magnetic Galilean relativitypredicts that the moving charge/observer would consider the magnetic field B to be zero,whereas it is nonzero according to special relativity.These themes were considered by Maxwell in Arts. 600-601, under the heading: On theModification of the Equations of Electromotive Intensity when the Axes to which they arereferred are moving in Space, which we review in sec. 2 below.Lorentz actually advocated the form F q (D v H) in eq. (V), p. 21, of , although he seemsmainly to have considered its use in vacuum. See also eq. (23), p. 14, of . That is, Lorentz considered Dand H, rather than E and B, to be the microscopic electromagnetic fields.14It is generally considered that Heaviside first gave the Lorentz force law (9) for electric charges in ,but the key insight is already visible for the electric case in and for the magnetic case in .15A more direct use of Faraday’s law, without invoking potentials, to deduce the electric field in the frameof a moving circuit was made in sec. 9-3, p. 160, of , which argument appeared earlier in sec. 86, p. 398,of . An extension of this argument to deduce the full Lorentz transformation of the electromagnetic fieldsE and B is given in Appendix C below.133
1.1DetailsIn Art. 598, Maxwell started from the integral form of Faraday’s law, that the (scalar)electromotive force E in a circuit is related to the rate of change of the magnetic flux throughit by his eqs. (1)-(2), 1d1d11 dΦm A B · dS A · dl (v · )A · dl, (13)E c dtc dtc dtc twhere the last form, involving the convective derivative, holds for a circuit that moves withvelocity v with respect to the lab frame.16 In his discussion leading to eq. (3) of Art. 598,Maxwell argued for the equivalent of use of the vector-calculus identity, (v · A) (v · )A (A · )v v ( A) A ( v),(14)which implies for the present case,(v · )A v ( A) (v · A) v B (v · A), 1 Av B · dl E · dl,E cc t(15)(16) since (v · A) · dl 0. Our eq. (16) corresponds to Maxwell’s eqs. (4)-(5), from which weinfer that the vector electromotive intensity E has the form,E 1 Av B V,cc t(17)for some scalar field V (Maxwell’s Ψ), that Maxwell identified with the electric scalar potential.If it were clear that V (Ψ) is indeed the electric scalar potential, then Maxwell should becredited with having “discovered” the “Lorentz” force law. However, Helmholtz [eq. (5d ),p. 309 of (1874)], Larmor [p. 12 of (1884)], Watson [p. 273 of (1888)), andJ.J. Thomson [in his editorial note on p. 260 of (1892)] argued that our eq. (15) leadsto, v 1 Av B · A · dl,E (18)cc tcso Maxwell’s eq. (D) of Art. 598 and eq. (10) of Art. 599 should really be written as, 1 Avv Ψ ·A ,E B cc tc(19)where Ψ is the electric scalar potential.17 It went unnoticed by these authors that use ofeq. (19) rather than (17) would destroy the elegance of Maxwell’s argument in Arts. 600-601In Maxwell’s notation, E E, p Φm , (F, G, H) A, (F dx/ds G dy/ds H dz/ds) ds A · dl,(dx/dt, dy/dt, dz/dt) v, and (a, b, c) B.17A possible inference from eq. (19) is that the Lorentz force law should actually be, v v 1 dAv·A q E · A q V ,(20)F q E B cccc dt16Some debate persists on this issue, as discussed, for example, in and references therein.4
(discussed in sec. 2 below), as well as that Maxwell’s earlier derivations of our eq. (17), onpp. 340-342 of and in secs. 63-65 of ,18 used different methods which did not suggestthe possible presence of a term (v · A/c) in our eq. (17). However, the practical effectof these doubts by illustrious physicists was that Maxwell has not been credited for havingdeduced the “Lorentz” force law, which became generally accepted only in the 1890’s.The view of this author is that Maxwell did deduce the “Lorentz” force law, although ina manner that was “not beyond a reasonable doubt”.Articles 600-601 of Maxwell’s Treatise2In Art. 600, Maxwell considered a moving point with respect to two coordinate systems, thelab frame where x (x, y, z), and a frame moving with uniform velocity v respect to the labin which the coordinates of the point are x (x , y , z ), with quantities in the two framesrelated by Galilean transformations. Noting that a force has the same value in both frames,Maxwell deduced that the “Lorentz” force law has the same form in both frames, providedthe electric scalar potential V in the moving frame is related to lab-frame quantities by,V V v· A.c(21)This is the form according to the low-velocity Lorentz transformation (7), and also to thetransformations of magnetic Galilean electrodynamics (5), which latter is closer in spirit toMaxwell’s arguments in Arts. 600-601.2.1DetailsIn Art. 600, Maxwell consider both translations and rotations of the moving frame, but werestrict our discussion here to the case of translation only, with velocity v (u, v, w) (δx/dt, δy/dt, δz/dt) with respect to the lab.19 Maxwell labeled the velocity of the moving point with respect to the moving frame by u dx /dt , while he called labeled itsvelocity with respect to the lab frame by u dx/dt. Then, Maxwell stated the velocitytransformation to be, eq. (1) of Art. 600,20 δx dx dx u u v,i .e.,u v u ,(22)dtdtdtwhich corresponds to the Galilean coordinate transformation,x x vt.t t, ,18 v · . t t(23)These derivations of Maxwell are reviewed in Appendix A below.For discussion of electrodynamics in a rotating frame (in which one must consider “fictitious” chargesand currents, see, for example, .20Equation (2) of Art. 600 refers to rotations of a rigid body about the origin of the moving frame.195
Maxwell next considered the transformation of the time derivative of the vector potentialA (F, G, H) in his eq. (3), Art. 600, dFdF δx dF δy dF δz dF A A (v · )A ,(24) t tdtdx δtdy δtdz δtdtwhich tacitly assumed that A A, and hence that B B.21 In eqs. (4)-(7) of Art. 600,Maxwell argued for the equivalent of use of the vector-calculus identity (14), which implieseq. (15), and hence that, A A v B (v · A). t t(25)Then, in eqs. (8)-(9) of Art. 600, Maxwell combined his eq. (B) of Art. 598 with our eqs. (22)and (25) to write the electromotive force E as, in the notation of the present section, uv1 Au 1 A E B V V B ·A.(26)cc tcc t cFinally, since a force has the same value in two frames related by a Galilean transformation,Maxwell inferred that the electromotive force E in the moving frame can be written as. u vu v1 A 1 A Ψ Ψ E B ·A B·Acc t ccc t c A1uu B B E , V (27) cc t cwhere the electric scalar potential V in the moving frame is related to lab-frame quantitiesby,vV V ·A(28)( Ψ Ψ ) .cThis is the form according to the low-velocity Lorentz transformation (7), and also to thetransformations of magnetic Galilean electrodynamics (5), which latter is closer in spirit toMaxwell’s arguments in Arts. 600-601.Further, the force F on a moving electric charge q in the moving frame is given by the “Lorentz” form,u F q E B ,(29)cwhich has the same form eq. (9) in the lab frame. As Maxwell stated at the beginning ofArt. 601: It appears from this that the electromotive intensity is expressed by a formula ofthe same type, whether the motions of the conductors be referred to fixed axes or to axesmoving in space.2221While this assumption does not correspond to the low-velocity Lorentz transformation of the fieldbetween inertial frames, it does hold for the transformation from an inertial frame to a rotating frame.Faraday considered rotating magnets in , and in sec. 3090, p. 31, concluded that No mere rotation of abar magnet on its axis, produces any induction effect on circuits exterior to it. That is, B B relates themagnetic field in an inertial and a rotating frame. Possibly, this might have led Maxwell to infer a similarresult for a moving inertial frame as well.22Maxwell’s equations in Art. 600 do not appear to be fully consistent with this “relativistic” statement, as6
2.2Articles 602-603. The “Biot-Savart” Force LawIn articles 602-603, Maxwell considered the force on a current element I dl in a circuit atrest in a magnetic field B, and deduced the “Biot-Savart” form,23 I dlI dldF B,F B.(30)ccSome other comments on Arts. 602-603 were given around eqs. (10)-(11) above.3Articles 769-770 of Maxwell’s TreatiseFaraday considered that a moving electric charge generates a magnetic field (as quotedin ). Maxwell also argued for this in Arts. 769-770 of , where his verbal argumentcan be transcribed as,B v E,c(31)for the magnetic field experienced by a fixed observer due to a moving charge. Maxwellnoted that this is a very small effect, and claimed (1873) that it had never been observed.24If v represents the velocity of a moving observer relative to a fixed electric charge, theneq. (31) implies that the magnetic field experienced by the moving observed would be,B v E,c(32)This corresponds to the low-velocity limit (7) of special relativity, and to form (3) of electricGalilean relativity).It remains that while Maxwell used Galilean transformations as the basis for his considerations of fields and potentials in moving frames, he was rather deft in avoiding thecontradictions between “Galilean electrodynamics” and his own vision. For a contrast, inwhich use of Galilean transformations for electrodynamics by J.J. Thomson (1880) ledto a result in disagreement with Nature (unrecognized at the time), see Appendix B below.25he noted in Art. 601. That is, his eq. (9), Art. 600, is the equivalent to E u /c B A / ct (Ψ Ψ ),where Ψ is the electric scalar potential in the lab frame, and Ψ v/c · A (a lab-frame quantity) accordingto Maxwell’s eq. (6), Art. 600. Maxwell did seem to realize that in addition to expressing the electromotiveintensity E in the moving frame in terms of moving-frame quantities [our eq. (27)], as well as in terms oflab-frame quantities (our eq. (8), Maxwell’s eq. (10), Art. 599), he had also deduced the relation of theelectric scalar potential V in the moving frame to the lab-frame quantities Ψ Ψ , as in our eq. (28).23Maxwell had argued for this in his eqs. (12)-14), p. 172, of (1861), which is the first statement of the“Biot-Savart” force law in terms of a magnetic field. Biot and Savart discussed the force on a magnetic“pole” due to an electric circuit, and had no concept of the magnetic field.24The magnetic field of a moving charge was detected in 1876 by Rowland [37, 52] (while working inHelmholtz’ lab in Berlin). The form (31) was verified (in theory) more explicitly by J.J. Thomson in 1881 for uniform speed v c, and for any v c by Heaviside and by Thomson in 1889 (which lattertwo works gave the full special-relativistic form for E as well).25For use by FitzGerald (1882) of Maxwell’s “Galilean” arguments to deduce a result in agreement withspecial relativity as v c, see [42, 102], which was one of the first indications that the speed of light playsa role as a limiting velocity for particles.7
Maxwell did not note the incompatibility of his use of Galilean transformations in hisArts. 601-602 and 769-770 with his system of equations for the electromagnetic fields, butif he had, he might have mitigated this issue by deduction of self-consistent transformationsfor both E and B between the lab frame and a uniformly moving frame, as in Appendix Cbelow.This note was stimulated by e-discussions with Dragan Redžić.AAppendix: Maxwell’s Derivations of the “Lorentz”Force Law26This Appendix uses SI units, while the main text employs Gaussian units.Maxwell published his developments of t
1 Articles 598-599 of Maxwell’s Treatise In his Treatise , Maxwell argued that an element of a circuit (Art. 598), or a particle (Art. 599) which moves with velocity v in electric and magnetic fields E and B μH experiences an electromotive intensity (Art. 598), i.e., a vector electromagnetic force given by eq. (B) of Art. 598 and eq. (10) of Art. 599,8
1 RELATIVITY I 1 1.1 Special Relativity 2 1.2 The Principle of Relativity 3 The Speed of Light 6 1.3 The Michelson–Morley Experiment 7 Details of the Michelson–Morley Experiment 8 1.4 Postulates of Special Relativity 10 1.5 Consequences of Special Relativity 13 Simultaneity and the Relativity
The theory of relativity is split into two parts: special and general. Albert Einstein came up with the spe-cial theory of relativity in 1905. It deals with objects mov-ing relative to one another, and with the way an observer's experience of space and time depends on how she is mov-ing. The central ideas of special relativity can be formu-
Maxwell’s equations are relativistic invariant in the parlance of special relativity . In fact, Einstein was motivated with the theory of special relativity in 1905 by Maxwell’s equations . These equations look the same, irrespective of what inertial reference frame one is in.
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Mathematical Finance uses tools of mathematics and statistics to understand the dynamics of variability in financial markets. A degree in Mathematical Finance opens up careers in investment banking, risk management, hedge funds, financial analysis, data analysis, actuarial science, insurance, and more. The PV function is categorized under Financial functions. It will calculate the present value of an investment or a loan taken at a fixed interest rate. In financial statement analysis, PV is used to calculate the dollar value of future payments in the present time. CorporateFinanceCalc Edraw Project. Manage projects on your Mac. Edraw Project App for Wikipedia. Organizer for Wikipedia is a lightweight and minimalistic Interface to. Yield to maturity (YTM) is the total return expected on a bond if the bond is held until maturity. The finance section of The Calculator Site featuring useful financial calculator tools for loans, car/auto loans, compound interest, savings, mortgages and more.
All calculations made on the calculators supplied on this site, together with rates quoted, are guidelines only and are subject to confirmation at the time of finalising any transactions. The max payment period/term is 96 months with no balloon payment subject to the age and model of the vehicle and credit assessment. Since MFC has no control over the information and variables entered by the user, figures generated by the calculators will not be binding on the bank whatsoever. All Standard MFC Credit Terms and Conditions apply.
|Year Model||Maximum Term||Balloon Maximum|
|New/ 0 years||72 months||30%|
|1 - 3 years||72 months||30%|
|4 - 5 years||72 months||25%|
|6 - 10 years||72 months||0%|
|> 10 years||60 months||0%|
The user indemnifies MFC against any loss or liability, of whatsoever nature that maybe caused or brought about, directly or indirectly which the user may suffer as a result of the use of any calculator. The site and all information provided on this site and the services provided on this site, are provided as is. The information provided on this site should not be treated as professional advice of any kind.
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The TI-84 Plus is a fairly easy, but more difficult than most, to use financial calculator which will serve you well in all finance courses. This tutorial will demonstrate how to use the financial functions to handle time value of money problems and make financial math easy. I will keep the examples rather elementary, but understanding the basics is all that is necessary to learn the calculator.
There is one adjustment which needs to be made before using this calculator. By default the TI-84 displays only two decimal places. This is not enough. Personally, I like to see five decimal places, but you may prefer some other number. To change the display, press the MODE key, then the down arrow key once (to the Float line). Next, use the right arrow key to highlight the 5 and press Enter. Finally, press 2ndMODE to exit the menu. That's it, the calculator is ready to go.
This tutorial will make extensive use of the TVM Solver, but the TI 84 Plus offers additional financial functions in the Finance menu.
If you have come here because you are experiencing a problem, you might check out the FAQ. If you don't find the solution, please send me a note.
We'll begin with a very simple problem that will provide you with most of the skills to perform financial math on the TI-84:
Suppose that you have $100 to invest for a period of 5 years at an interest rate of 10% per year. How much will you have accumulated at the end of this time period?
In this problem, the $100 is the present value (PV), N is 5, and i is 10%. Before entering the data you need to put the calculator into the TVM Solver mode. Press the Apps button, choose the Finance menu (or press the 1 key), and then choose TVM Solver (or press the 1 key). Your screen should now look like the one in the picture. Enter the data as shown in the table below.
Now to find the future value simply scroll to the FV line and press AlphaEnter. The answer you get should be 161.05.
Solving for the present value of a lump sum is nearly identical to solving for the future value. One important thing to remember is that the present value will always (unless the interest rate is negative) be less than the future value. Keep that in mind because it can help you to spot incorrect answers due to a wrong input. Let's try a new problem:
Suppose that you are planning to send your daughter to college in 18 years. Furthermore, assume that you have determined that you will need $100,000 at that time in order to pay for tuition, room and board, party supplies, etc. If you believe that you can earn an average annual rate of return of 8% per year, how much money would you need to invest today as a lump sum to achieve your goal?
In this case, we already know the future value ($100,000), the number of periods (18 years), and the per period interest rate (8% per year). We want to find the present value. Go to the TVM Solver and enter the data as follows: 18 into N, 8 into I%, and 100,000 into FV. Note that we enter the $100,000 as a positive number because you will be withdrawing that amount in 18 years (it will be a cash inflow). Now move to PV and press ALPHAENTER and you will see that you need to invest $25,024.90 today in order to meet your goal. That is a lot of money to invest all at once, but we'll see on the next page that you can lessen the pain by investing smaller amounts each year.
Sometimes you know how much money you have now, and how much you need to have at an undetermined future time period. If you know the interest rate, then we can solve for the amount of time that it will take for the present value to grow to the future value by solving for N.
Suppose that you have $1,250 today and you would like to know how long it will take you double your money to $2,500. Assume that you can earn 9% per year on your investment.
This is the classic type of problem that we can quickly approximate using the Rule of 72. However, we can easily find the exact answer using the TI 84 Plus calculator. Enter 9 into I%, -1250 into PV, and 2500 into FV. Now scroll up to N and press ALPHAENTER and you will see that it will take 8.04 years for your money to double.
One important thing to note is that you absolutely must enter your numbers according to the cash flow sign convention. If you don't make either the PV or FV a negative number (and the other one positive), then you will get ERR: DOMAIN on the screen instead of the answer. That is because, if both numbers are positive, the calculator thinks that you are getting a benefit without making any investment. If you get this error, just press 2 (Goto) to return to the TVM Solver and then fix the problem by changing the sign of either PV or FV.
Solving for the interest rate is quite common. Maybe you have recently sold an investment and would like to know what your compound average annual rate of return was. Or, perhaps you are thinking of making an investment and you would like to know what rate of return you need to earn to reach a certain future value. Let's return to our college savings problem from above, but we'll change it slightly.
Suppose that you are planning to send your daughter to college in 18 years. Furthermore, assume that you have determined that you will need $100,000 at that time in order to pay for tuition, room and board, party supplies, etc. If you have $20,000 to invest today, what compound average annual rate of return do you need to earn in order to reach your goal?
As before, we need to be careful when entering the PV and FV into the calculator. In this case, you are going to invest $20,000 today (a cash outflow) and receive $100,000 in 18 years (a cash inflow). Therefore, we will enter -20,000 into PV, and 100,000 into FV. Type 18 into N, and then solve forI% to find that you need to earn an average of 9.35% per year. If you get ERR: NO SIGN CHNG instead of an answer, it is because you didn't follow the cash flow sign convention. Press 2 to return to the TVM Solver and fix the problem.
Note that in our original problem we assumed that you would earn 8% per year, and found that you would need to invest about $25,000 to achieve your goal. In this case, though, we assumed that you started with only $20,000. Therefore, in order to reach the same goal, you would need to earn a higher interest rate.
When you have solved a problem, always be sure to give the answer a second look and be sure that it seems likely to be correct. This requires that you understand the calculations that the calculator is doing and the relationships between the variables. If you don't, you will quickly learn that if you enter wrong numbers you will get wrong answers. Remember, the calculator only knows what you tell it, it doesn't know what you really meant.
Please continue on to part II of this tutorial to learn about using the TI 84 Plus to solve problems involving annuities and perpetuities. |
TEE 208/05 Analog Electronics
Tutor-marked Assignment 2 (TMA 2 – 25%)
Evidence of plagiarism or collusion will be taken seriously and the University regulations will be applied fully. You are advised to be familiar with the University’s definitions of plagiarism and collusion.
This is an individual assignment. No duplication of work will be tolerated. Any plagiarism or collusion may result in disciplinary action, in addition to ZERO mark being awarded to all involved.
You are to submit online of your answers in OAS system and it is your responsibility to submit your TMA correctly and timely. OAS system does not allow re-submission of assignment. Marks will be awarded for correct working steps and answer.
The total marks for TMA 2 is 100 and contributes 25% towards the total grade.
TMA 2 covers the topics in Units 1, 2, 3 & 4.
You must submit your TMA 2 to OAS as one single document. Any additional appendices or attachments must be placed at the end of the submitted document and must be referred to in the main body of the assignment, or it will not be read by the marker.
Your assignment must be word processed (single spacing) and clearly lay out. Use the Insert-Equation function whenever you need to show equation or formula.
All files or documents submitted must be labeled with your WOU ID and name.
Answer all parts in English. All working steps and calculations must be shown clearly.
Question 1 (20 Marks)
[Note: Question 1 is to be done outside the lab hours]
Calculate the DC bias voltages (V_(c_1 ),V_(c_2 ),V_(E_2 ),V_(B_1 ),V_(CE_2 )) and currents (I_(B_1 ),I_(c_2 ),I_(E_2 )) for Darlington configuration in Figure 1. Given ß_1=60,ß_2=70.
Question 2 (20 Marks)
[Note: Question 2 is to be done outside the lab hours]
A JFET source follower amplifier circuit with voltage divider biasing is configured as shown in Figure 1a. The supply voltage VDD is 10V. The JFET has the following parameters which are: IDSS = 50mA and VP = –3.5V. The JFET is biased with VGS = 0V and VS = 0.5VDD. The small signal input impedance is 50k? and the value of the load resistor RL is 10k?.
Draw the small signal ac equivalent circuit of this amplifier.
Calculate the value of RS.
Calculate the values of R1 and R2.
Calculate the voltage gain of this amplifier. State your assumption where necessary.
Question 3 (20 Marks)
Lab 1-Zener Circuits and Applications
Zener diode is designed to operate in reverse conduction. Zener breakdown occurs at a precisely defined voltage, allowing the diode to be used as a voltage reference or clipper. While Zener diodes are usually operated in reverse conduction, they may also be operated in cutoff and forward conduction. There are two different effects that are used in “Zener diodes”. The only practical difference is that the two types have temperature coefficients of opposite polarities.
Zener breakdown – Occurs for breakdown voltages greater than approximately 6V when the electric field across the diode junction pulls the electrons from the atomic valence band into the conduction band, causing a current to flow.
Impact ionization (also called avalanche breakdown) – Occurs at lower breakdown voltages when the reverse electric field across the p-n junction causes a cascading ionization, similar to an avalanche, that produces a large current.
Figure 3: I-V Curve of a typical Zener Diode
A reference diode is a special Zener diode designed to use both conduction modes, which cancels the temperature coefficients and produces a temperature stable breakdown voltage.
Zener diode ratings include:
Zener Voltage (Vz@Izt)
Power Dissipation (Pd)
Max Current (Izm)
Zener Impedance (Zzt)
Max Leakage Current (IR@VR)
Temperature Coefficient (aVZ)
The model for a reverse biased Zener diode (on the left side of the Figure 4) can be represented as a series circuit consisting of a regular diode with voltage drop Vf, a bias voltage source to provide a total drop of VZ across the Zener diode terminals, and a resistor to represent the Zener impedance (Rf represents the slope of the reverse conduction V-I curve). For this lab, we will neglect the effects of Rzt. The forward biased Zener diode would simply be a regular diode (on the right side of the Figure 4).
Figure 4: Zener Diode (left) and it's model (right)
In this lab exercise, you will build two clipper circuits – one use a 1N4001 diode and another uses 1N4732A zener diode. The characteristics of Zener diodes compare to diode will be compared in the clipper circuits you built.
When forward biased, a Zener diode is identical to a regular diode. When reverse biased, the Zener diode can be modeled as a regular diode connected backwards with a bias supply in series. The circuit of Figure 5 is the equivalent model for the Zener diode circuit of Figure 6.
Figure 5 :Diode clipper circuit
Figure 6: Zener diode clipper circuit
Instruments and components
1. 1N4001 diode – 1 pcs
2. 1N4732A diode – 1 pcs
3. 470 O – 1 pcs
4. 10k O – 1 pcs
5. Crocodile clip cable
6. DC Power supply
7. Oscilloscope with 2 probes with build-in function generator or
8. Oscilloscope with 2 probes with separate unit Function Generator
1. By using 1N4001 diode, 470 O and 10k O resistor, construct a clipper circuit as shown in Figure 7.
2. Set your power supply to output 4.1 V that will be used as bias in this circuit.
3. Ensure your function generator is set to high impedance (or Hi-Z) output.
4. Set your function generator to generate RAMP waveform to output a -2.5 V to +2.5 V rise over 5 ms. Feed the waveform to the clipper circuit as Vi(t).
5. Connect channel 1 of the Oscilloscope to Vi(t) and channel 2 to the Vo(t).
6. Set the horizontal scale of the scope to 200 ms/div and vertical scale to 5 V/div.
7. Draw the waveforms displayed on the scope to your TMA.
8. Next, remove diode 1N4001 and the DC Power Supply. Replace it with 1N4732A as shown in Figure 8.
9. Under the same settings on the scope and function generator, draw the new waveform displayed on the scope to your TMA.
10. Overlay both waveforms in this lab into 1 single graph below.
Figure 7: Diode clipper circuit.
Figure 8: Zener Diode Clipper circuit
1. Explain the working principle of clipper circuit.
2. Draw the waveform you observed on the screen of the scope in step 9 and 11 above. Compare the waveform observed with the simulated results.
3. From the waveform you observed, what is the output clipping voltages and Vi where the output waveform begins to clip?
4. Explain the purpose of setting the DC Power Supply to 4.1 V in the diode clipper circuit?
1N4001 Data sheet – attached in a separate file.
1N4732A Data sheet – attached in a separate file.
Question 4: (40 Marks)
Lab 2 BJT and Cascaded Amplifier
Bipolar Junction Transistor or simply BJT is a semiconductor device constructed with three doped regions with each region known as emitter, base and collector. These regions are form by doping p-n junctions „back-to-back?. Unlike Field Effect Transistor (FET), BJT is current driven amplifier with Ic = ßIB.
Figure 9: Diagram and schematic of PNP and NPN transistor
In this lab, we will construct a common-emitter amplifier circuit to examine BJT voltage gain, the characteristics of BJT active and the overall operation of a two-stage common-emitter amplifier. We will use 2N3904 NPN general purpose amplifier to build an amplifier circuit to analyze the performance.
Figure 10: Pin configuration of 2N3904.
One of the common applications of a BJT is amplify signal at its input. The amplified signal can feed into another stage to multiply the gain of the system. Both DC and AC analysis need to be carried out carefully to determine the performance of a multiple stages amplifier circuit.
BJT Equations Summary
Referring to the NPN transistor below, some of the commonly used equations are provided to assist you in the lab.
Figure 11: NPN representations for equation.
Instruments and components
1. 2N3904 NPN – 2 pcs
2. 150 O – 2 pcs
3. 1 kO – 2 pcs
4. 1.5 kO – 1 pcs
5. 2.2 kO – 2 pcs
6. 3.6 kO – 2 pcs
7. 10 kO – 2 pcs
8. 1 µF – 1 pcs
9. 10 µF – 2 pcs
10. 470 µF – 2 pcs
11. Crocodile clip cable
12. DC Power supply
13. Digital Multimeter
14. Oscilloscope with 2 probes with build-in function generator or
15. Oscilloscope with 2 probes with separate unit Function Generator
(Lab 2 Part A):
1. Construct the circuit shown in Figure 12 on a breadboard carefully.
2. Before connect the function generator to the circuit, perform DC analysis for the entire circuit and record your answers in the Table 1 below.
3. Next, connect the function generator to the circuit.
4. Set the function generator to output a sine wave at 1 kHz and amplitude equal to 5 Vp-p. Use a Oscilloscope to ensure the settings are correct (optional).
5. Turn on the output of the function generator and DC power supply. Measure and record all the DC and AC values in Table 1 and 2 below by using Multimeter.
6. Using the measured values of VB(Q1) and VC(Q1), calculate AV(Q1). Insert this value in the measured section of Table 2.
7. Using the measured values of VB(Q2) and VC(Q2), calculate AV(Q2). Insert this value in the measured section of Table 2.
8. Using the measured values of VB(Q1) and VC(Q2), calculate AVT. Insert this value in the measured section of Table 2.
Figure 12: A 2-satge common-emitter amplifier circuit
1. Perform DC analysis on circuit in this lab by filling up Table 1 below. Be sure to show all your steps and calculations.
Table 1: DC analysis values
2. Perform AC analysis on the circuit in this lab by filling up table 2 below. Be sure to show all your steps and calculations.
Table 2: AC analysis values
3. Determine the values of Zbase for Q2, Zin for Q2 and rc for stage 1. Calculate Av1.
4. Determine the values of rc for stage 2. Calculate AV2.
5. Calculate AVT. Compare the values of AVT measured vs AVT calculated and AVT simulated. Obtain the percentage error between these values. Explain discrepancy in the results between measured, simulated and calculated.
(Lab 2 Part B):
Modify the circuit shown in Figure 12 by removing capacitors C4 and C5. Repeat the procedures in Part A and answer the questions as in Part A. [20 marks]
2N3904 Data Sheet – attached in a separate file.
END OF TMA 2 |
- Approximating square roots
- Approximating square roots walk through
- Approximating square roots
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- Comparing irrational numbers
- Approximating square roots to hundredths
- Comparing values with calculator
- Comparing irrational numbers with a calculator
Learn how to compare 22.9% to √0.45 using a calculator.
Want to join the conversation?
- How come when he squares the square root of 0.45 it remains 0.45?(5 votes)
- 0.45 squared is 0.2025, and the square root of 0.2025 is 0.45, so basically, a square root undoes a square, and vice versa. Here's another example of this: if we find the square root of the number 9, we get 3. 3 squared is 9, so we are back to the number 9.
Hope this helps :)(21 votes)
- Why did he round up to 30%, when 22.9% is closer to 20%?(12 votes)
- In most situations, it is better to round up. Say you needed 22.9% of a gallon of concrete to finish a sidewalk. It is hard to measure 22.9% of a gallon, but if you round down to 20% of a gallon, there would not be enough concrete to finish the sidewalk, so it would be better to round to 30% of a gallon so you would have enough concrete with some left over.(3 votes)
- how is 22.9 less then 0.3? when you line up the decimals , wouldn't 0.3 be less then 22.9?(5 votes)
- It is 22.9%, so when you move the decimal two places, you get the equivalent of 0.229 which is less than 0.3.(9 votes)
- I'm using the on-screen calculator for first time, when I put in 8 then square root sign then = I get a pink bar across my answer window instead of an answer. What does this mean? How can I find the square root of 8 on this style of calculator?(3 votes)
- Hi! I know I'm super late, but maybe this will help someone else. Let's say you wanted to get the square root of 4. Here are the steps:
1. Click the square root symbol button
2. Click the button for number 4
3. Click the close parentheses button i.e. )
And whatever different number you have just follow these steps and replace the 4 with your different number.(8 votes)
- Why do these videos keep getting more and more confusing?(5 votes)
- Cant you just make .45 into a percentage which would be 45% and you automatically know that its greater than 22.9%?(6 votes)
- How do you find square roots without a calculator?(5 votes)
- [Voiceover] My question to you is, "Which of these two values is greater?" 22.9%? Or, the square root of 0.45? And I encourage you to first try to see if you can think about this without a calculator. And then use a calculator to see which one is larger. Let's first try to do it without a calculator. So one thing that you could try to say is-- Well look-- You could say that 22.9% that's going to be less than-- I'm just gonna pick an arbitrary number here. Let's say that's less than 30%. So that's going to be less than-- and I picked 30% because it's easy to calculate 30%, or the square of 30%. So that's less than 30%. Or, it's another way of saying that's less than 0.3. 0.3 And then, we can try to compare 0.3 to this thing here. And if 0.3 is less than this, well, then 22.9% is going to be less than this 'cause it's less than 0.3. So why don't I change the problem to that? Let's compare 0.3-- 0.3 to square root-- to the square root of-- let me do that thing pink color-- to the square root of 0.45. And now I can use the squaring technique. What happens if I square each of these quantities? If I square this, 0.3 times 0.3, let's see, three times three is nine. But you're multiplying two things that each have one digit to the right, so you're gonna have two digits to the right. So it's going to be .09. And if you were to square this over here-- I'll just keep using the white-- if you were to square this right over here, what's that going to be? Well that's just going to be 0.45. So .09 is clearly less than 0.45. Or, 0.3 squared is clearly less than square root of-- 0.3 squared is less than the square root of 0.45 squared. And so we know that 0.3 is going to be less than, is going to be less than the square root of 0.45. And so now we can say that 22.9%, if it's less than this, and this is less than this, well 22.9% must be less than that. Now another way you can do it, you could take out a calculator-- you could take out a calculator. I'll do that. Just for kicks. So let me get the calculator out. So, 22.9%, that's the same thing as-- that's the same thing as 0.229. So we really just want to compare this quantity to the square root of 0.45. And there are two ways we can do it. We could do it the way I started. You could square this and see if it's greater than 0.45. So let's do that. You could just square it. And you see "no." This is 0.05. Which is clearly less than 0.45. And so that would validate this. Or, you could do it the other way around. You could just use your calculator to calculate the square root of this. So you could say 0.45 and then take the square root. The square root of this value right over here is approximately-- is approximately 0.67. So this thing right over here is approximately 0.67. Which is approximately 67%, which is clearly greater than 22.9%. So a bunch of ways that you could approach it. It is nice to be able to estimate things. Think in your head. So that if you didn't have access to a calculator you could get a general sense of, "Hey, would you rather have a 22.9% off the price of something?" Or, maybe some type of new store could say, "Hey, sales! Square root of 0.45 off of all goods." I don't know. (chuckles) That can be an interesting thing. Could be very confusing for customers. But, anyway. Hopefully this was helpful. |
The simplicity of the classic Bondi-Hoyle-Lyttleton (BHL) accretion model makes its use attractive in order to roughly estimate accretion rates and drag forces in many different astrophysical contexts, ranging from wind-fed X-ray binaries (e.g. Anzer & Börner 1995), over supernovae (e.g. Chevalier 1996), and galaxies moving through intracluster gas in a cluster of galaxies (Balsara et al. (1994), to the black hole believed to be at the center of our Galaxy (Ruffert & Melia 1994; Mirabel et al. 1991). In the BHL scenario a totally absorbing sphere of mass M moves with velocity relative to a surrounding homogeneous medium of density and sound speed . It has been investigated numerically by many workers (e.g. Ruffert 1994 and 1995, and references therein). Usually, the accretion rates of various quantities, like mass, angular momentum, etc., including drag forces are of interest as well as the properties of the flow, (e.g. distribution of matter and velocity, stability, etc.). All results pertaining to total accretion rates are in qualitative agreement (to within factors of two, ignoring the instablitites of the flow) with the original calculations of Bondi, Hoyle and Lyttleton (e.g. Ruffert & Arnett 1994).
The BHL recipe for accretion in the axisymmetric case for pressureless matter is the following. A ring of material with radius b (which is identical to the impact parameter) far upstream from the accretor and thickness db will be focussed gravitationally to a point along the radial accretion line downstream of the accretor. At this point the linear momentum perpendicular to the radial direction is assumed to be cancelled. Then, if the remaining energy of the matter at this point is not sufficient for escape from the potential, this material is assumed to be accreted. The largest radius b from which matter is still accreted by this procedure turns out to be the so-called Hoyle-Lyttleton accretion radius (Hoyle & Lyttleton 1939, 1940a, 1940b, 1940c; Bondi & Hoyle 1944)
I will refer to the volume upstream of the accretor from which matter is accreted as accretion cylinder.
However, if the assumption of homogeneity of the surrounding medium is dropped, e.g. by assuming some constant gradient in the density or the velocity distribution, the consequences on the accretion flow remain very unclear. Using the same conceptual procedures, one can calculate (Dodd & McCrea 1952; Illarionov & Sunyaev 1975; Shapiro & Lightman 1976; Wang 1981) how much angular momentum is present in the accretion cylinder for a non-axisymmetric flow which has a gradient in its density or velocity perpendicular to the mean velocity direction. Then, assuming that the angular momentum will be accreted together with the mass, it is only a small step to conclude that the amount of angular momentum accreted is equal to (or at least is a large fraction of) the angular momentum present in the accretion cylinder. Note, that if the velocity is a function of position, then by virtue of Eq. (1) also the accretion radius varies in space. Thus the cross section of the accretion cylinder (perpendicular to the axis) is not circular.
However, the reasoning of BHL calls for a cancelling of linear momentum perpendicular to the radial accretion line before matter is accreted. Together with this linear momentum also angular momentum is cancelled and so the matter accreted has zero angular momentum by construction! This point was first discussed by Davies & Pringle (1980), who were able to construct two-dimensional flows with small non-vanishing gradients for which the accreted angular momentum was exactly zero, by placing the accretion line appropriately. Thus, following these analytic investigations two opposing views are voiced about how much angular momentum can be accreted: either a large or a very small fraction of what is present in the accretion cylinder. Numerical simulations thus are called for to help solve the problem.
In this paper I would like to compare the accretion rates of several quantities (especially angular momentum) of numerically modeled accretion flows with gradients to the previous results of accretion without gradients (e.g. Ruffert 1994). One has to change some of the parameters of the flow (Mach number, size of the accretor) in order to get a good overview of which features are generic and which specific to that combination of parameters. Although several investigations of two -dimensional flows with velocity gradients exist (Anzer et al. 1987; Fryxell & Taam 1988; Taam & Fryxell 1989; Ho et al. 1989), three -dimensional simulations are scarse due to their inherently high computational load. Livio et al. (1986) first attempted a three-dimensional model including gradients, but due to their low numerical resolution the results were only tentative. Also in the models of Ishii et al. (1993) was the accretor only coarsly resolved, while the results of Boffin (1991) and Sawada et al. (1989) are only indicative, because due to the numerical procedure the flows remained stable (too few SPH particles in Boffin 1991 and local time stepping in Sawada et al. 1989 which is appropriate only for stationary flows). A simulation that was numerically better resolved was performed later by Ruffert & Anzer (1995), but since only one model was presented, the results cannot be taken as conclusive either. I intend to remedy these shortcomings in the present paper.
In Sect. 2 I give only a short summary of the numerical procedure used. Sects. 4 to 6 present the results, which I analyze and interpret in Sect. 8. Sect. 9 summarizes the implications of this work.
© European Southern Observatory (ESO) 1997
Online publication: July 8, 1998 |
On the other hand dimensional analysis shows that e mc3 makes no sense. The rules of algebra allow both sides of an equation to be divided by the same expression, so this is equivalent to 100 kpa 1 bar 1. M t where m is measured in grams and t is measured in time. Dimensional analysis would suggest that both einsteins equation e mc2 and the incorrect equation e 1 2 mc 2 might be true. Robinetta department of physics, the pennsylvania state university, university park, pennsylvania 16802 received 24 june 2012.
Checking equations routinely by dimensional analysis save us the embarrassment of using an incorrect equation. If the dimensions on two sides differ,the relation is incorrect. It is best to approach each problem the same way, starting with a given, and proceeding in. An inductive strategy is proposed for teaching dimensional analysis to second or thirdyear students of physics, chemistry, or engineering.
All quantities of physical interest have dimensions that can be expressed in terms of three fundamental quantities. The dimension of any physical quantity is the combination of the basic physical dimensions that compose it. Dimensional analysis theory sheet, class 11 physics. Dimensional analysis is the use of fundamental units to establish the form of an. This is a math test prep lesson that explains what dimensional analysis is and how to use dimensional analysis in solving problems as part of the geometry and measurement material that many state exams cover. Trigonometry is also used in determining the horizontal and vertical components of. Dimensional analysis and unit conversions we often use the word dimension when discussing units. This rule provides a powerful tool for checking whether or not equations are dimensionally consistent. Dimensional analysis from biology to cosmology in todays seminar, we will see how it is possible to deduce a great deal about the equations that describe the behaviour of a physical system through an analysis of dimensions with some physical intuition thrown in. Publication date 1922 topics physical measurements.
Dimensional formula it is an expression which shows how and which of the fundamental units are required to represent the unit of physical quantity. If a phenomenon depends upon n dimensional variables, dimensional analysis will reduce the problem to only k dimensionless variables, where the reduction n k 1, 2, 3, or 4, depending upon the problem complexity. Units of measurement physics class 11 download in pdf. Oct 10, 2017 our professor introduced us to dimensional analysis and told us that we can use it to predict how some variables are proportional to others, for example. Dimensional analysis page 1 of 4 many problems in chemistry, math, physics, and engineering involve dimensional analysis. Dimensional analysis as the other language of physics. In this paper we use dimensional analysis as a method for solving problems in qualitative physics. Units of measurement physics class 11 download notes in pdf. When using this method, you will algebraically rearrange an equation to solve for an unknown variable without substituting numbers into the equation. Dimensional analysis involves calculations where the units cancel one another out, leaving only the desired unit. Isaac physics a project designed to offer support and activities in physics. But since everything should follow from physics, that cant be right, so it must be possible to recover the results of dimensional analysis without ever invoking any dimensions or units. I have included some videos for those of you who may be having some difficulty with scientific notation and significant figures.
This is one of 61 lessons available in the workbook titled the essentials of high school m. Unit conversion is a fairly simple but crucial skill in physics. The unit for plane angle is rad and the unit for the solid angle is steradian. In dimensional analysis, a ratio which converts one unit of measure into another without changing the quantity is called a conversion factor. Dimensional analysis page 1 of 4 georgia virtual school. Dimensional analysis can be used to check the dimensional consistency of equations, deducing relations among physical quantities etc. Apr 14, 2020 dimensional analysis theory sheet, class 11 physics class 11 notes edurev is made by best teachers of class 11. Dimensional analysis is the practice of checking relations between physical quantities by identifying the dimensions of the physical quantities. Certain physical quantities have been chosen as fundamental or base quantities.
Dimensional analysis provides many simple and useful tools for various situations in science. If we are trusting people, these types of dimensional checks might seem unnecessary. Dimensional analysis and conversion unit conversion and keeping track of your units is very important in this class and science in general. A students guide to dimensional analysis by lemons, don s. Trigonometry is also used in determining the horizontal and vertical components of forces and objects. Dimensional analysis is a method of using the known units in a problem to help deduce the process of arriving at a solution. For instance, it is usually claimed that the period of a pendulum cannot possibly depend on its mass because if it did the units would not match. These equations represent the relations between the relevant properties of the system under consideration. In physics, most problems are solved much more easily when a free body diagram is used. This introduction to dimensional analysis covers the methods, history and formalisation of the field, and provides physics and engineering applications. You will receive your score and answers at the end. This document is highly rated by class 11 students and has been viewed 1827 times. Dimensional analysis conversion unit conversion and keeping track of your units is very important in this class and science in general.
Dimensional analysis one of the simplest, yet most powerful, tools in the physicists bag of tricks is dimensional analysis 1. A students guide to dimensional analysis students guides series by don s. We will call such an equation dimensionally inconsistent or dimensionally non. Generally n k equals the number of different dimensions sometimes called basic or prichapter 5 dimensional analysis and similarity 277. All quantities of physical interest have dimensions that can be expressed in terms of three fundamental. I have a ball at a certain height and i want to know the time it requires to touch the grond, i can make a guess that it will depend on the. These tips will help you apply dimensional analysis to a problem. Dimensional analysis is the practice of checking relations between physical quantities by identifying their dimensions. Homework statement it is found the terminal velocity ut of a spherical particle in a fluid depends upon the diameter d of particle, the dynamic viscosity. Dimensional analysis in physics problems thoughtco. Use of dimensional analysis the dimensions of base quantities and combination of these dimensions describe the nature of physical quantities. These dimensions are independent of the numerical multiples and constants and all the quantities in the world can be expressed as a function of the fundamental dimensions. Physical laws give us relationships between measurements, and measurements are more than just numbers. For instance, it is usually claimed that the period of a pendulum cannot possibly dep.
Unit 1 motion in one dimension bowie high school physics. Dimensional analysis, also known as factorlabel method or unitfactor method, is a method used to convert one unit to a different unit. But, rest assured, any textbook on a quantitative subject such as physics including this one almost certainly contains some equations with typos. Covering topics from mechanics, hydro and electrodynamics to thermal and quantum physics, it illustrates. Jan 23, 2020 applications of dimensional analysis in physics. Our professor introduced us to dimensional analysis and told us that we can use it to predict how some variables are proportional to others, for example. May 07, 2016 homework statement it is found the terminal velocity ut of a spherical particle in a fluid depends upon the diameter d of particle, the dynamic viscosity. S often calls upon us to convert measurements in one unit to another.
We often use dimensional analysis in our everyday lives. The objective of this paper is to investigate its relations to functions, i. For example, it might be meaningless to construct an equation like. Dimensional analysis is the use of fundamental units to establish the form of an equation or more often to check that the answer to a calculation is physically sensible. We pose and solve some of the qualitative reasoning problems discussed in the literature, in the context of devices such as the pressure regulator and the heat exchanger. In science, units such as meter, second, and degree celsius represent quantified physical properties of space, time, and. For example, kpa and bar are both units of pressure, and 100 kpa 1 bar. In these models we meet with variables and parameters. The method of using literal equations is related to dimensional analysis in that we will often be concerned with the unit of the measurement or quantity that we are looking for. Formally, the physics would still be the same good old physics, but it looks like we cant do dimensional analysis anymore. The units defined for the supplementary quantities namely plane angle and solid angle are called the supplementary units. It is also possible to use dimensional analysis to suggest plausible equations when we know which quantities are involved. In this strategy, buckinghams theorem is seen as a.913 324 581 321 1029 1107 1079 929 344 835 482 459 858 133 904 293 883 340 379 856 21 1137 410 51 597 834 133 1112 1196 |
Univariate and Bivariate Analysis (Source: W.G Zikmund, B.J Babin, J.C Carr and M. Griffin, Business Research Methods, 8th Edition, U.S, South-Western Cengage Learning, 20
Types of Statistical Analysis • Univariate Statistical Analysis • Tests of hypotheses involving only one variable. • Testing of statistical significance • Bivariate Statistical Analysis • Tests of hypotheses involving two variables. • Multivariate Statistical Analysis • Statistical analysis involving three or more variables or sets of variables.
Statistical Analysis: Key Terms • Hypothesis • Unproven proposition: a supposition that tentatively explains certain facts or phenomena. • An assumption about nature of the world. • Null Hypothesis • No difference in sample and population. • Alternative Hypothesis • Statement that indicates the opposite of the null hypothesis.
Significance Levels and p-values • Significance Level • A critical probability associated with a statistical hypothesis test that indicates how likely an inference supporting a difference between an observed value and some statistical expectation is true. • p-value • Probability value, or the observed or computed significance level. • p-values are compared to significance levels to test hypotheses. • Higher p-values equal more support for an hypothesis.
Type I and Type II Errors • Type I Error • An error caused by rejecting the null hypothesis when it is true. • Practically, a Type I error occurs when the researcher concludes that a relationship or difference exists in the population when in reality it does not exist. • “There really are no monsters under the bed.”
Type I and Type II Errors (cont’d) • Type II Error • An error caused by failing to reject the null hypothesis when the alternative hypothesis is true. • Practically, a Type II error occurs when a researcher concludes that no relationship or difference exists when in fact one does exist. • “There really are monsters under the bed.”
Choosing the Appropriate Statistical Technique • Choosing the correct statistical technique requires considering: • Type of question to be answered • Number of variables involved • Level of scale measurement
Parametric versus Nonparametric Tests • Parametric Statistics • Involve numbers with known, continuous distributions. • Appropriate when: • Data are interval or ratio scaled. • Sample size is large. • Nonparametric Statistics • Appropriate when the variables being analyzed do not conform to any known or continuous distribution.
Bivariate Analysis - Introduction • Measures of Association • Refers to a number of bivariate statistical techniques used to measure the strength of a relationship between two variables. • The chi-square (2) test provides information about whether two or more less-than interval variables are interrelated. • Correlation analysis is most appropriate for interval or ratio variables. • Regression can accommodate either less-than interval or interval independent variables, but the dependent variable must be continuous.
Bivariate Analysis—Common Procedures for Testing Association
Simple Correlation Coefficient • Correlation coefficient • A statistical measure of the covariation, or association, between two at-least interval variables. • Covariance • Extent to which two variables are associated systematically with each other.
Simple Correlation Coefficient • Correlation coefficient (r) • Ranges from +1 to -1 • Perfect positive linear relationship = +1 • Perfect negative (inverse) linear relationship = -1 • No correlation = 0 • Correlation coefficient for two variables (X,Y)
Correlation, Covariance, and Causation • When two variables covary, they display concomitant variation. • This systematic covariation does not in and of itself establish causality. • e.g., Rooster’s crow and the rising of the sun • Rooster does not cause the sun to rise.
Correlation Analysis of Number of Hours Worked in Manufacturing Industrieswith Unemployment Rate
Coefficient of Determination • Coefficient of Determination (R2) • A measure obtained by squaring the correlation coefficient; the proportion of the total variance of a variable accounted for by another value of another variable. • Measures that part of the total variance of Y that is accounted for by knowing the value of X.
Correlation Matrix • Correlation matrix • The standard form for reporting correlation coefficients for more than two variables. • Statistical Significance • The procedure for determining statistical significance is the t-test of the significance of a correlation coefficient.
Regression Analysis • Simple (Bivariate) Linear Regression • A measure of linear association that investigates straight-line relationships between a continuous dependent variable and an independent variable that is usually continuous, but can be a categorical dummy variable. • The Regression Equation (Y = α + βX ) • Y = the continuous dependent variable • X = the independent variable • α= the Y intercept (regression line intercepts Y axis) • β = the slope of the coefficient (rise over run)
The Regression Equation • Parameter Estimate Choices • βis indicative of the strength and direction of the relationship between the independent and dependent variable. • α (Y intercept) is a fixed point that is considered a constant (how much Y can exist without X) • Standardized Regression Coefficient (β) • Estimated coefficient of the strength of relationship between the independent and dependent variables. • Expressed on a standardized scale where higher absolute values indicate stronger relationships (range is from -1 to 1). |
АннотацияThe dynamic models of elements of technological systems with perfect mixing and plug-flow hydrodynamics are based on the systems of algebraic and differential equations that describe a change in the basic technological parameters. The main difficulty in using such models in MathWorks Simulink™ computer simulation systems is the representation of ordinary differential equations (ODE) and partial differential equations (PDE) that describe the dynamics of a process as a MathWorks Simulink™ block set. The study was aimed at developing an approach to the synthesis of matrix dynamic models of elements of technological systems with perfect mixing and plug-flow hydrodynamics that allows for transition from PDE to an ODE system on the basis of matrix representation of discretization of coordinate derivatives. A sugar syrup cooler was chosen as an object of modeling. The mathematical model of the cooler is formalized by a set of perfect reactors. The simulation results showed that the mathematical model adequately describes the main regularities of the process, the deviation of the calculated data from the regulations did not exceed 10%. The proposed approach significantly simplifies the study and modernization of the current and the development of new technological equipment, as well as the synthesis of algorithms for controlling the processes therein.
Ключевые словаMathematical modeling , dynamic systems , sugar syrup cooler , MathWorks Simulink™
The main way to study the regularities of technological heat and mass exchange processes in the food industry is to make a full-scale experiment. In many cases, it results in a number of insurmountable difficulties due to its cost, the ability of engineering implementation of the control parameters, etc. Due to this, the development of adequate and correct mathematical models of technological processes that allow us to replace a full- scale experiment is a relevant task.
The processes of heat and mass transfer in moving media can be simulated on the basis of the Navier- Stokes equation, but the equations obtained are not always suitable for analysis. Therefore, the perfect models of a flow pattern have become widespread in practice: perfect mixing models (they describe a change in concentration, temperature, etc. in compounders), plug-flow models (they describe a change in concentration, temperature, etc. in the case of a plug-flow going through an apparatus), cellular models and others [1–3]. The mathematical models of such apparatus are called perfect mixing reactors (PMR), plug-flow reactors (PFR), etc.
A Simulink interactive graphical simulation environment is widely used for the computer modeling of technological processes that allows using block diagrams in the form of directed graphs to construct dynamic models, including discrete (continuous and hybrid), non- linear models and models with singularities. In addition, the Simulink environment includes the tools for building, modeling and studying control systems which allows for the simulation of a process and control system within a single integrated environment .
When modeling perfect reactors, the conversion of the equations to describe a technological process into a set of blocks of the visual programming environment of Simulink is necessary. In the case of systems of ordinary differential equations (ODE), an input signal integration unit and a MathWorks™ numerical methods library are used, as well as the method for representing ODE or an ODE system in the form of a structural Simulink model [5, 6]. In the case of partial differential equations that describe, for example, plug-flows, the heat conductivity equation, etc., there is a problem of bringing PDE to an ODE equation or system. One of the approaches used in the transition from PDE to ODE is the use of integral transformations, for example, Fourier transformations and Laplace transformations [7, 8], and the disposal of spatial derivatives. It is not always possible to obtain a solution to such equations in analytical form. In addition, the implementation of direct and inverse integral transformations by means of Simulink for an arbitrary input signal is difficult. The other approach is to use numerical methods for solving PDE (finite elements, finite differences, etc.) and to integrate program blocks into the structural model of Simulink using the custom functions implemented on the basis of such programming languages as MathWorks and universal ones like C ++ . In this case, it is possible to implement the entire arsenal of available tools for the numerical solution of PDE, but any modification of an object model entails the need for changing the function code and debugging it, thereby violating the objectoriented principle of Simulink – the separation of the internal structure and interface of a model. One of the approaches is based on the discretization of only the spatial variable using the finite difference method, while the derivatives with respect to time remain continuous and PDE is represented as the Cauchy problem for the ODE system . The implementation of each equation of the ODE system results in the formation of rather cumbersome structural models in Simulink, and the compact representation of such systems in a matrix form is relevant [11, 12]. In this case, it is necessary to replace the expansion of spatial derivatives with matrix equations . shows such a representation of a one-dimensional heat conductivity equation for Simulink.
Thus, the relevant task is to develop a method for representing typical perfect reactors using matrix ordinary differential equations (MODE) and to implement structural models therewith in Simulink.
ОБЪЕКТЫ И МЕТОДЫ ИССЛЕДОВАНИЯTo synthesize the matrix dynamic model of a typical perfect reactor, it is necessary to perform the following sequence of actions:
– to select the model of a perfect reactor suitable for describing the structure of the flow in process equipment (this issue is considered in detail in the special literature, for example, in [1–3]);
– to draw up the corresponding differential equation. As a result, for each perfect reactor, one or more ODE or PDE are obtained with the corresponding initial and boundary conditions;
– to solve the resulting equation or a set thereof with respect to the highest derivative with respect to time;
– to discretize spatial derivatives (in the case of their presence) in accordance with the accepted form of finite-difference approximation;
– to replace each PDE by a set of ODE using the replacement of partial derivatives with respect to the spatial variable by the corresponding matrix equation. For example, the first derivative of the temperatureis replaced as follows:
where is the decomposition matrix of the first derivative, and
is a vector to define the boundary condition. The second derivative can be defined as
is the decomposition matrix of the second derivative, and
is a vector to define the boundary conditions.
– then, in accordance with the procedure described in , to draw up a structural diagram of a perfect reactor model in the form of Simulink blocks that solve MODE taking into account the fact that the signals transmitted from block to block are vectors and matrices and the corresponding multiplication operations will be performed according to the rules of matrix multiplication. In this case, the time derivatives will be integrated using numerical methods from the Simulink library;
– to define the parameters of the model that characterize flow hydrodynamics, the geometry of perfect reactor models and the thermophysical properties of process media;
– to create input and output streams to build process equipment models in the form of a system of perfect reactors and additional subsystems, if necessary; and
– to design a reactor model in the form of a subsystem in the Simulink format.
Let us consider the synthesis process of a dynamic matrix mathematical model using a sugar syrup cooler as an example which is used for fondant sweet processing lines . The cooler is a cylindrical body 1, inside which a rotating screw 2 (рассмотрите pushes sugar syrup forward (Fig. 1). The body 1 is equipped with a cooling jacket made in the form of a closed spiral canal (CSC) 3 cooling water (a coolant) is fed through. As a result of the heat exchange between the syrup and the coolant through the wall of CSC that separates them, the temperature of the syrup decreases, which results in the crystallization of sucrose from the fondant syrup and the formation of fondant mass.
The quality indicators of the finished product (fondant mass) are the size of sucrose crystals and their proportion in the total volume of fondant mass. In turn, the disperse composition of fondant mass depends entirely on both the cooling parameters of fondant syrup–temperature, the intensity of removal of heat from the syrup–and its thermophysical and rheological properties that change during the crystallization sucrose process .
To model the heat exchange between the syrup and the coolant, let us select the models of perfect reactors. In view of the intensive mixing of sugar syrup with a rotating screw, the complexity of the mathematical description resulting from the mixing of the three- dimensional temperature pattern, as well as the necessity of conjugation of the geometric coordinates of cooling water and sugar syrup flows, let us estimate in the first approximation the average temperature of sugar syrup according to the length of the cylindrical body 1. Averaging the sugar syrup temperature according to the length of the body by integrating the PMR equation with respect to the variable x and dividing it by the length of the body, we obtain an equation that corresponds to PFR in structure. For the cooling water flow, let us neglect the temperature distribution over the cross-sectional area of the canal and choose a PMR model. The perfect reactors are physically separated by a wall through which there is a heat exchange between the coolant and sugar syrup. Let us neglect the thermal effects as a result of the process of crystallization of sucrose from the syrup, as well as heat exchange with the environment. Let us assume that the wall temperature is constant along the length.
Let us consider the synthesis of the PFR model in accordance with the diagram in Fig. 2.
If the structure of the syrup flow corresponds to the PFR model, an equation of a temperature change taking heat transfer into account can be used for the mathematical description of this flow . It will be an ordinary differential equation with respect to the temperature of the syrup TC
where Vc is the volume of the perfect mixing area, m3 ; Pс is density of sugar syrup, is the specific heat capacity of sugar syrup, is the volumetric flow rate, is the flow temperature at the inlet to the perfect mixing area; Fc is the surface of heat exchange between the syrup and the wall of CSC, is the coefficient of heat transfer from the syrup to the center of the wall of CSC, is the temperature of the center of the wall of CSC, °C.
To solve the equation in Simulink, let us use the technique described in and represent Eq. 3 in the form of Simulink blocks (Fig. 3) having previously solved Eq. 3 with respect to the derivative.
Let us consider the synthesis of the model of a plug-flow reactor (Fig. 4).
where is the function of spatial and time distribution of temperature of the coolant flow; Vв is the volume of the plug-flow area, m3 ; Pв is the density of the coolant, kg/m3 ; Сpв is the specific heat capacity of the coolant, is the volumetric flow rate, is the flow temperature at the inlet to the plug-flow area; Fв is the surface of heat exchange between the coolant and the wall of CSC, m2 ; KTв is the coefficient of heat transfer from the coolant to the center of the wall of CSC, W/(m2 × K). Let us bring Eq. 4 to a finite-difference form and solve it with respect to the derivative
Let us define the vector of the current temperature at the points of coordinate partitioning and the initial conditions
Let us represent the first derivative in the form of a matrix expression
where) is the decomposition matrix of the first derivative,
is a vector to define the boundary condition. Then let us write the ODE system (5) as a matrix ODE
and represent it in the form of a structural Simulink model (Fig. 5). Since the statement of the problem implies that the heat capacity of the wall that separates the heat carrier flows cannot be neglected, then these equations will be supplemented by an equation of a change in the temperature of the wall that separates the media Tсm(t)
where is the wall mass, kg; is the specific heat capacity of the wall, is the average temperature of the coolant over the entire length calculated as
and when partitioning it by N elements with respect to x it is replaced with the sum
Let us represent it in the form of a structural Simulink model (Fig. 6).
As a result, a set of subsystems for modeling perfect reactors is formed. To compile a cooler model, it is only necessary to connect the subsystems into a single design scheme in accordance with Fig. 7 and set the model parameters. In this case, the mathematical model of the cooler can be described by a system of scalar and matrix ODE
with the parameters [14, 15]: the heat capacity of the coolant ; the heat capacity of the wall ; the heat capacity of the syrup ; the heat exchange area of the coolant-wall ; the heat exchange area syrup wall ; the wall mass mст = 25.465 kg; the cross-sectional area inside ; the coefficient of thermal conductivity of the wall the density of the wall material (copper) ; the density of the coolant ; the density of syrup the coefficient of heat transfer from the coolant to the wall ; the coefficient of heat transfer from the syrup to the wall ; the coefficient of heat transfer from the syrup to the center of the wall ; the coefficient of heat from the coolant to the center of the wall ; the discretization interval Δ x = 2.07 m; the number of discretization elements along the length N = 10; the thickness of the wall of CSC, ; the length of CSCL = 20.7 m; the volumetric flow rate of the coolant ; the syrup volumetric flow rate ; the initial temperature of the coolant the initial temperature of syrup the initial wall temperature ; the volume of the area of PFR ; and the syrup volume in PMR
The structural model in the form of Simulink blocks is shown in Fig. 8, the simulation results are shown in Figs. 9–11. The studies and computer experiment were carried out at the Voronezh State University of Engineering Technology.
РЕЗУЛЬТАТЫ И ИХ ОБСУЖДЕНИЕ
The data in Fig. 9 show that the sugar syrup is cooled from the initial temperature of 105°С to 60°С within 600–800 sec, which is consistent with the technological regulations for fondant mass production. The further cooling of syrup does not result in a decrease in the temperature of the finished product. The deviation of the calculated data from the regulations did not exceed 10%.
The model developed with the help of this approach makes it possible to obtain the estimates of temperatures at the outlet from the cooler in real time (Fig. 9a) which makes it possible to study the dynamics of the technological process and synthesize a control system. It is also possible to estimate the temperature distribution in terms of both the time and length of the heat exchange surface of CSC (Fig. 9b). In addition, by varying the parameters of a mathematical model, it is possible to estimate their effect on the technical and economic indicators of a process, for example, when changing the material which CSC is made of (Fig. 10a), and also to predict a change in the dynamic characteristics of a process (Fig. 10b).
When the syrup is cooled, its viscosity significantly changes which entails a change in the hydrodynamics of the syrup flow and conditions of heat exchange between the coolant and syrup through the wall that separates them. The introduction of a temperature correction in the calculation of syrup viscosity makes it possible to take into account a change in the coefficient of thermal conductivity from the syrup to the wall of CSC. For example, using the data of , let us approximate the dependence of syrup viscosity on temperature using the Arrhenius equation
Then, taking into account the dependence of syrup viscosity on temperature, it is possible to estimate the coefficients of heat transfer and thermal conductivity from the syrup to the wall of CSC for each temperature and to clarify the syrup cooling dynamics (Fig. 11). In this case, the heat transfer and thermal conductivity coefficients are calculated with the help of an additional Matlab Function block from the Simulink library that performs the continuous calculation of the heat transfer and thermal conductivity coefficients according to the current values of syrup temperature delivered to the block input.
The further refinement of the cooler model can be due to, first, taking into account the thermal effects resulting from the crystallization of sucrose from the syrup and, secondly, taking into account the design features of a typical fondant beater (feeding cooling water into the shaft of the conveying screw, separating the machine body and cooling water jacket into three sections in each of which the heat is removed from the syrup with various intensity).
ВЫВОДЫThe presented approach makes it possible to implement the mathematical models of perfect reactors in Simulink by discretizing the spatial variable and to pass over to matrix ordinary differential equations, which makes it possible to convert them into Simulink blocks. The approach is also applicable to other models of perfect reactors, which makes it possible to build the libraries of typical perfect reactors of Simulink for the synthesis of heat and mass transfer equipment that makes it easy to integrate them into a single system of synthesis, study, and debugging of Simulink control systems. Within the framework of Simulink simulation systems, a further refinement of the obtained simplest models based on perfect reactors by introducing variable model parameters, nonlinearities, control circuits, etc. is possible. This significantly simplifies the study and modernization of the current technological equipment and the development of new equipment, as well as the synthesis of control algorithms for the processes therein.
КОНФЛИКТ ИНТЕРЕСОВThe authors declare no conflict of interest.
БЛАГОДАРНОСТИThe authors would like to thank Professor Yu.Yu. Gromov of the Tambov State Technical University for all his help and support.
ФИНАНСИРОВАНИЕThe study was conducted with the support of the Applied Mathematics and Mechanics Department, the Voronezh State Technical University.
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Miles per hour and d different
If the ferrari takes 1 hour to travel 100 miles, it takes 15/100, or 3/20 of an hour, to travel 15 miles the answer will be in minutes, so multiply 3/20 x 60 = 9 so, it will take the ferrari 9 minutes to travel 15 miles. Distances and travel times questions including how many miles is it from wickford essex to bromley by bow in london and how long does it take to fly from london to hong kong. This time, use the distance formula d = rt: d = 10 miles per hour × 4 hours = 40 miles next, you ride 18 miles and travel at a rate of 12 miles per hour how long did this take you use the time formula t = d/r: t = 18 miles ÷ 12 miles per hour = 15 hours, or 1 ½ hours distance, rate and time. Solving linear equations - distance, rate and time objective: solve distance problems by creating and solving a linear they travel at rates differing by 5 miles per hour if they meet after 6 hours, find the rate of each 3 two trains travel toward each other from points which are 195 miles apart. (120+120 miles) / (3+2 hours) = 240/5 miles per hour = 48 miles per hour what makes this question tricky for some is the temptation to calculate the average of the two speeds, 40 and 60 miles per hour.
1) the distance a car travels can be found using the formula d = rt, where d is the distance, r is the rate of speed, and t is time how many miles does the car travel, if it drives at a speed of 70 miles per hour for 1 /2 hour. A knot is one nautical mile per hour (1 knot = 115 miles per hour) the term knot dates from the 17th century, when sailors measured the speed of their ship by using a device called a common log this device was a coil of rope with uniformly spaced knots, attached to a piece of wood shaped like a slice of pie the piece of wood was lowered. For instance, if the rate the problem gives is in miles per hour (mph), then the time needs to be in hours, and the distance in miles if the time is given in minutes, you will need to divide by 60 to convert it to hours before you can use the equation to find the distance in miles.
For example, an air plane usually goes faster than a car, and a car usually goes faster than a walking person. Two cars start at an interstate interchanger and travel in the same direction at average speeds of 40 miles per hour and 55 miles per hour how much time must elapse before the two cars are 5 miles ap. The answer tells you that the car traveled at an average speed of 583 miles per hour the car was likely traveling faster at times and slower at other times, with 583 miles per hour being the central or most common speed. A tropical cyclone is a rotating low-pressure weather system that has organized thunderstorms but no fronts (a boundary separating two air masses of different densities) tropical cyclones with maximum sustained surface winds of less than 39 miles per hour (mph) are called tropical depressions those with maximum sustained winds of 39 mph or higher are called tropical storms.
There have historically been different nautical miles used and, thus, different variations of knots however, we have based our knots calculator on the international nautical mile which is commonly used thoughout the world today knots to miles per hour formula miles per hour. Dot hours of service guidelines state that, on average, drivers should be able to travel about 10 miles per hour below the speed limit over a 10-hour period for example, if the speed limit is 65 mph, drivers should be able to travel about 550 miles in a 10-hour period, so a trip of 600 miles or more may open the driver to charges of speeding. The average steps per minute for different exercises pedometer step equivalents for elliptical, biking & other activities by wendy bumgardner the calories burned in one minute of the listed activity with the calories you would burn walking for one minute at 30 miles per hour. That's because your left and right eyes are looking at the finger with slightly different angles that result by 24 hours to get miles per hour or km per hour other at about 70 miles per.
The formula for distance problems is: distance = rate × time or d = r × t things to watch out for: make sure that you change the units when necessary for example, if the rate is given in miles per hour and the time is given in minutes then change the units appropriately. Because i started with 80 miles per hour, so hours started out underneath i want hours to cancel off, so the conversion factor for hours and minutes needed to have hours on top i want hours to cancel off, so the conversion factor for hours and minutes needed to have hours on top. Likewise, train #2 is traveling at 70 miles per hour, so the distance this train travels can be represented by the equation d = 70x (the variables mean the same thing) however, the equation d = 70 x does not represent the distance train #2 is from town a, because it left from town b. Average speed the average speed of an object tells you the (average) rate at which it covers distance if a car's average speed is 65 miles per hour, this means that the car's position will change (on the average) by 65 miles each hour.
Miles per hour and d different
This is composed of three sections to do classic time, speed and distance calculations in the time calculator, enter the distance value and conversion units designation then enter the speed value and conversion designation for it. Solving rate-time-distance problems the variables involved in a motion problem are distance (d for this step each problem is different we generally look for a man walks from point a to point b at a pace of 5 miles per hour he returns at the rate of 4 miles per hour if the total time for his round trip is 1 2 4 hours, then what is the. Miles per hour this is a measurement of speed typically used in non-metric countries for transport such as the usa the united kingdom also uses this on the roads although officially the metric system has been adopted.
- According to the rules of algebra, we can rewrite the formula v = d/t in two different ways if we want to know how far a car going at 55 miles/hour travels in 3 hours we write d = vt = (55 miles/hour)(3 hours) = 165 miles.
- Something traveling at one kilometer per hour is traveling about 0278 meters per second, or about 0621 miles per hour a mile per hour is a unit of speed commonly used in the united states it is equal to exactly 1609344 kilometers per hour.
The first row gives me the equation d = 30tsince the first part of his trip accounted for d miles of the total 150-mile distance and t hours of the total 3-hour time, i am left with 150 – d miles and 3 – t hours for the second part. • speed in miles is known as miles per hour (mph) while speed in kilometers is known as kilometers per hour (kph) • most of the countries have converted to using kilometers however, there are still some countries that use mile as the official unit of length such as united kingdom and united states. So times negative 30 miles per hour, plus 2 times y is 16, times dy dt is negative 60 miles per hour and i'm not writing the units here but if you were to write the units, you will see that all of our distances are in miles. |
School of Mathematics and Statistics, Central South University, Changsha 410083, China
Author to whom correspondence should be addressed.
Received: 28 April 2019 / Accepted: 22 May 2019 / Published: 28 May 2019
In this paper, a fast and accurate numerical Clenshaw-Curtis quadrature is proposed for the approximation of highly oscillatory integrals with Cauchy and logarithmic singularities, , for a smooth function . This method consists of evaluation of the modified moments by stable recurrence relation and Cauchy kernel is solved by steepest descent method that transforms the oscillatory integral into the sum of line integrals. Later theoretical analysis and high accuracy of the method is illustrated by some examples.
Boundary element method and finite element method are intensively eminent numerical approaches to evaluate partial differential equations (PDEs), which appear in variety of disciplines from engineering to astronomy and quantum mechanics [1,2,3,4,5]. Although these methods lead PDEs to Fredholm integral equations or Voltera integral equations, but these kind of integral equations posses integrals of oscillatory, Cauchy-singular, logarithmic singular, weak singular kernel functions. However, these classical methods are failed to approximate the integrals constitute kernel functions of highly oscillation and logarithmic singularity.
This paper aims at approximation of the integral
where is relatively smooth function. For integral (1) the developed strategy for logarithmic singularity is valid for . In particular, the highly oscillatory integral, has been computed by many methods such as asymptotic expansion, Filon method, Levin collocation method and numerical steepest descent method [6,7,8,9,10]. For instant, Dominguez et al. for function with integrable singularities have proposed an error bound, calculated in Sobolev spaces , for composite Filon-Clenshaw-Curtis quadrature. Error bound depends on the derivative of and length of the interval M, for some defined as for
On the other hand, one methodology for numerical evaluation of integral is replacing by different kind of polynomials [12,13]. Another technique is based on analytic continuation of the integral if the integrand is analytic in the complex region . As far as for solution methods and properties of the solution for relative non-homogenous integrals have been discussed by using Brestain polynomials and Chebyshev polynoimals of all four kinds in [3,15].
For integral Clenshaw-Curtise rule is applied for numerical calculation. Wherein the convergence rate is independent of k but depends on the number of nodes of quadrature rule and function . Furthermore, Piessense and Branders established the Clenshaw-Curtis quadrature rule, relies on the recurrence relation for They replaced the nonoscillatory and nonsingular part of the integrand by Chebyshev series. Chen presented the numerical approximation of the integral with , and For analytic function the integral was rewritten in the form of sum of line integrals, wherein the integrands do not oscillate and decay exponentially. Moreover, Fang established the Clenshaw-Curtis quadrature for for general function where steepest descent method is illustrated for analytic function . Recently, John introduced the algorithm for integral approximation of Cauchy-singular, logarithmic-singular, Hadamard type and nearly singular integrals having integrable endpoints singularities i.e., . Composed Gauss-Jacobi quadrature consists of approximating the function by Jacobi polynomials of degree .
However, all these proposed method are inadequate to apply directly on integral (1) in the presence of oscillation and other singularities. This work presents Clenshaw-Curtis quadrature to get recurrence relation to compute the modified moments, that takes just operations. The initial Cauchy singular values for recurrence relation are obtained by the steepest descent method, as it prominently renowned to evaluate highly oscillatory integrals when the integrands are analytic in sufficiently large region.
The rest of the paper is organized as follows. Section 2 delineates the quadrature algorithm for integral (1). Numerical calculation of the modified moments with recurrence relation by using some Chebyshev properties is defined. Also steepest descent method is established for Cauchy singularity where later the obtained line integrals are further approximated by generalized Gauss quadrature. Section 3 alludes some error bounds derived in terms of Clenshaw-Curtis points and the rate of oscillation k. In Section 4, numerical examples are provided to demonstrate the efficiency and accuracy of the presented method.
2. Numerical Methods
In the computation of integral , the Clenshaw-Curtis quadrature approach is extensively adopted. The scheme is postulated on interpolating the function at Clenshaw-Curtis points set Writing the interpolation polynomial as basis of Chebyshev series
where is the Chebyshev polynomial of first kind of degree N and double prime denotes a sum whose first and last terms are halved, the coefficients
can be computed efficiently by FFT in operations [8,9]. This paper appertains to Clenshaw-Curtis quadrature, which depends on Hermite interpolating polynomial that allow us to get higher order accuracy
For any fixed t, we can elect felicitous N such that and rewrite Hermite interpolating polynomial of degree in terms of Chebyshev series
can be calculated in operations once if are known [13,21]. Finally Clenshaw-Curtis quadrature for integral is defined as
more specifically are called the modified moments. Efficiency of the Clenshaw-Curtis quadrature depends on the fast computation of the moments. In ensuing sub-section, we deduce the recurrence relation for .
Following proof asserts the results for case , and for the same technique can be used as well. Since the integrand is analytic in the half strip of the complex plane, by Cauchy’s Theorem, we have
with all the contours taken in clockwise direction (Figure 1).
Setting we obtain that
Similarly for , we get
From the statement of the theorem, ,
Let , then
Thus, we complete the proof with
From Proposition 2.2 numerical scheme for the line integrals can be evaluated by generalized Gauss-Laguerre quadrature rule, using command lagpts in Chebfun . Let be the nodes and weights of the weight function and let be the nodes and weights of the weight function The line integrals and can be approximated by
By the same argument and can also be approximated with generalized Gauss-Laguerre quadrature rule. Aforementioned theorem enlightens the another interesting fact that can also be computed by it if is an analytic function.
Computation of the moments is derived as, by using Chebyshev property (8)
For integrating by parts, we derive
We deduce the following recurrence relation by inserting (20) in (19)
It is worth to mention that can be computed in operations . For we obtain the as
Unfortunately, practical experiments demonstrate that the recurrence relation for is numerically unstable in the forward direction for , in this sense so-called Oliver’s algorithm is stable and used to rewrite the recurrence relation in the tridiagonal form .
3. Error Analysis
([9,13,14]) Suppose , for a non-negative integer m with , then
([9,14]) Let be a Lipschitz continous function on [−1,1] and let be the interpolation polynomial of at Clenshaw-Curtis points. Then it follows that
In particular, if is analytic with in an Bernstein ellipse with foci and major and minor semiaxis lengths summing to , then
if has an absolutely continuous st derivative and of bounded variation on [−1,1] for some , then for
(van der Corput Lemma ) Suppose that , then for each , it follows
Moreover, for some special cases we have
Suppose that , then it follows for all k that
For simplicity, here we prove the first identity in (3.29). Similar proof can be directly applied to the second identity in (3.29).
it leads to the desired result by Lemma 3.3. □
Suppose that , and define
From Lemma 2.1, we see that and , in addition, is a polynomial of degree at most N with for . Then the error on the Clenshaw-Curtis quadrature (6) can be estimated by
Suppose that and is bounded on , then the Clenshaw-Curtis quadrature (6) is convergent
In particular, if is analytic and in a Bernstein ellipse , , then the error term satisfies
If has an absolutely continuous st derivative and of bounded variation on [−1,1] for some , then for ()
The error bound for for integral can be estimated as
In addition, for , it follows
by Lemma 3.3, it implies
which yields (3.33) together with the estimate on in .
The identity (3.34) follows from Lemma 3.4 due to that for some . □
From the convergence rates Corollary 3.1 and Theorem 3.1, compared with that in , the new scheme is of much fast convergence rate. It is also illustrated by the numerical results (see Section 4).
4. Numerical Results
In this section, we will present several examples to illustrate the efficiency and accuracy of the proposed method. The exact values of an integral (36) are computed through Mathematica 11. Unless otherwise specifically stated, all the tested numerical examples are executed by using Matlab R2016a on a 4 GHz personal laptop with 8 GB of RAM.
Let us consider the integral
for , , Table 1 shows the results for relative error compared with results of integral in Table 2.
Table 3, Table 4 and Table 5 represent results for relative error computed by Clenshaw-Curtis quadrature. As exact value we just have used that returned by the rule when a huge number of points is used.
Let the integral be
Table 6, Table 7 and Table 8 represent results for relative error computed by Clenshaw-Curtis quadrature. As exact value is calculated by using the rule for large number of points.
Clearly, Table 1, Table 2, Table 3, Table 4, Table 5, Table 6, Table 7 and Table 8 illustrate the relative error of the Clenshaw-Curtis quadrature taken as We can see that for proposed Clenshaw-Curtis quadrature based on Hermite interpolation polynomial, with small value of points higher precision of the numerical results of integrals is obtained in operations. Furthermore these tables show that more accurate results can be obtained as k increases with fixed value of N. Conversely, more accurate approximation can be achieved as N increases but k is fixed. Moreover, Tables demonstrate that results successfully satisfy the analysis derived in Section 3.
The Authors have equally contributed to this paper.
This work was supported by National Science Foundation of China (No. 11771454) and the Mathematics and Interdisciplinary Sciences Project of Central South University.
Conflicts of Interest
The authors declare no conflict of interest.The funders had no role in the design of the study; in the collection, analyses, or interpretation of data; in the writing of the manuscript, or in the decision to publish the results.
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71. 1. How many hats, at $4 apiece, can be bought for $20 ?
SOLUTION. Twenty dollars will buy as many hats as 4 is contained times in 20, which is 5 times. Five hats, at $4 apiece, can be bought for $20.
2. A man gave $16 for 8 barrels of apples. What was the cost of each barrel?
3. If 1 cord of wood costs $3, how many cords can be bought for $15?
4. At $6 a barrel, how many barrels of flour can be bought for $24?
5. When flour is $5 a barrel, how many barrels can be bought for $30?
6. If a man can dig 7 rods of ditch in a day, how many days will it take him to dig 28 rods?
7. If an orchard contains 56 trees, and 7 trees in a row, how many rows are there?
8. I bought 6 barrels of flour for $42. What was the cost of 1 barrel ?
9. If a farmer divides 21 bushels of potatoes equally among 7 laborers, how many bushels will each receive?
10. How many oranges, at 3 cents each, can be bought for 27 cents?
11. A farmer paid $35 for sheep, at $5 apiece. How many did he buy?
72. Division is the process of finding how many times one number is contained in another, or of sepa rating a number into equal parts.
Division may also be regarded as a short method of performing several subtractions of a number.
73. The Dividend is the number to be divided, and the Divisor is the number by which to divide.
74. The Quotient is the result obtained by the process of division, and shows how many times the divisor is contained in the dividend.
1. When the dividend does not contain the divisor an exact number of times, the part of the dividend left is called the remainder, and it must be less than the divisor.
2. As the remainder is always a part of the dividend, it is always of the same name or kind.
3. When there is no remainder, the division is said to be exact.
75. The sign, ÷, placed between two numbers, denotes division, and shows that the number on the left is to be divided by the number on the right.
Thus, 20 ÷ 4 =
5, is read, 20 divided by 4 is equal to 5. Division is also indicated by writing the dividend above, and the divisor below, a short horizontal line. Thus, 12 = 4, shows that 12 divided by 3 equals 4.
76. In finding how many times one number is contained in another, the dividend and divisor are like numbers, and the quotient is an abstract number.
In finding one of the equal parts of a number the dividend and quotient are like numbers, and the divisor is an abstract number.
77. When the divisor consists of one figure. 1. How many times is 4 contained in 848?
Dividend. Divisor. 4)848 Quotient. 212
SOLUTION. After writing the divisor on the left of the dividend, with a line between them, we begin at the left hand and say: 4 is contained in 8 hundreds, 2 hundreds times, and we write 2 in hundreds' place in the quotient; 4 is contained in 4 tens 1 ten times; we write the 1 in tens' place in the quotient; 4 is contained in 8 units 2 units times; we write the 2 in units' place in the quotient, and the entire quotient is 212. 2. How many times is 4 contained in 2884?
SOLUTION. - As we cannot divide 2 thousands by 4, we take the 2 thousands and the 8 hundreds together. 4 is contained in 28 hundreds 7 hundreds times, which 721 we write in hundreds' place in the quotient; 4 is contained in 8 tens 2 tens times, which we write in tens' place in the quotient; and 4 is contained in 4 units 1 unit time, which we write in units' place, and we have the entire quotient, 721.
3. How many times is 6 contained in 1824?
OPERATION. SOLUTION. Beginning as in the last example, we 6)1824 say, 6 is contained in 18 hundreds 3 hundreds times, which we write in hundreds' place in the quotient; 304 6 is contained in 2 tens no times, and we write a cipher in tens' place in the quotient; taking the 2 tens and 4 units together, 6 is contained in 24 units 4 units times, which we write in units' place in the quotient, and we have 304 for the entire quotient.
4. How many times is 4 contained in 943 ?
SOLUTION. Here 4 is contained in 9 hundreds 2 hundreds times, and 1 hundred 235...3 Rem. over, which, united to the 4 tens, makes 14 tens; 4 in 14 tens, 3 tens times and 2 tens over, which, united to the 3 units, makes 23 units; 4 in 23 units 5 units times and 3 units over. The 3 which is left after performing the division, should be divided by 4; but we merely indicate the division by placing the divisor under the dividend, thus, . The entire quotient is written 235, which may be read, two hundred thirty-five and three divided by four, or, two hundred thirty-five and three fourths, or, two hundred thirty-five and a remainder of three.
Hence the following rule:
RULE. -I. Write the divisor at the left of the dividend, with a line between them.
II. Beginning at the left hand, divide each figure of the dividend by the divisor, and write the result under the dividend.
III. If there is a remainder after dividing any figure, regard it as prefixed to the figure of the next lower order in the dividend, and divide as before.
IV. Should any figure or part of the dividend be less than the divisor, write a cipher in the quotient, prefix the number to the figure of the next lower order in the dividend, and divide as before.
V. If there is a remainder after dividing the last figure, place it over the divisor at the right hand of the quotient.
PROOF. Multiply the quotient by the divisor, and to the product add the remainder, if any; if the result is equal to the dividend, the work is correct.
1. This method of proof depends on the fact that division is the reverse of multiplication. The dividend corresponds to the product, the divisor to one of the factors, and the quotient to the other.
2. In multiplication the two factors are given, to find the product; in division, the product and one of the factors are given, to find the other factor.
The sums of quotients and remainders of examples 17 to 22 are
23. Divide $47645 equally among 5 men. each receive?
24. There are seven days in one week. weeks are there in 17675 days?
Ans. 2525 weeks.
25. How many barrels of flour, at $6 a barrel, can be bought for $6756 ? Ans. 1126 barrels. 26. Twelve things make a dozen. How many dozen are there in 46216464?
27. How many barrels of flour 347560 bushels of wheat, if it takes one barrel?
Ans. 3851372 dozen. can be made from
5 bushels to make
Ans. 69512 barrels.
28. If there are 3240622 acres of land in 11 townships, how many acres are there in each township?
29. A man left his estate, worth $38470, to be shared equally by his wife and 4 children. What did each receive? Ans. $7694.
30. At $5 an acre, how many acres of land can be bought for $3875 ?
31. If a man walks 4 miles an hour, in how many hours will he walk 352 miles?
32. I paid $1792 for 7 horses.
What did each cost?
33. If 75000 bushels of grain are put into 8 bins of equal size, how many bushels does each bin contain ? |
An iterative method for finding the value of the Golden Ratio with explanations of how this involves the ratios of Fibonacci numbers and continued fractions.
Explore the continued fraction: 2+3/(2+3/(2+3/2+...)) What do you
notice when successive terms are taken? What happens to the terms
if the fraction goes on indefinitely?
A serious but easily readable discussion of proof in mathematics with some amusing stories and some interesting examples.
The first of two articles on Pythagorean Triples which asks how many right angled triangles can you find with the lengths of each side exactly a whole number measurement. Try it!
The diagram shows a regular pentagon with sides of unit length.
Find all the angles in the diagram. Prove that the quadrilateral
shown in red is a rhombus.
This is an interactivity in which you have to sort the steps in the
completion of the square into the correct order to prove the
formula for the solutions of quadratic equations.
Eulerian and Hamiltonian circuits are defined with some simple examples and a couple of puzzles to illustrate Hamiltonian circuits.
The sums of the squares of three related numbers is also a perfect
square - can you explain why?
Take any two numbers between 0 and 1. Prove that the sum of the
numbers is always less than one plus their product?
This is the second article on right-angled triangles whose edge lengths are whole numbers.
Which set of numbers that add to 10 have the largest product?
In this third of five articles we prove that whatever whole number we start with for the Happy Number sequence we will always end up with some set of numbers being repeated over and over again.
The final of five articles which containe the proof of why the sequence introduced in article IV either reaches the fixed point 0 or the sequence enters a repeating cycle of four values.
Start with any whole number N, write N as a multiple of 10 plus a remainder R and produce a new whole number N'. Repeat. What happens?
This article extends the discussions in "Whole number dynamics I". Continuing the proof that, for all starting points, the Happy Number sequence goes into a loop or homes in on a fixed point.
The problem is how did Archimedes calculate the lengths of the sides of the polygons which needed him to be able to calculate square roots?
The first of five articles concentrating on whole number dynamics, ideas of general dynamical systems are introduced and seen in concrete cases.
An article which gives an account of some properties of magic squares.
In this 7-sandwich: 7 1 3 1 6 4 3 5 7 2 4 6 2 5 there are 7 numbers between the 7s, 6 between the 6s etc. The article shows which values of n can make n-sandwiches and which cannot.
Find the smallest positive integer N such that N/2 is a perfect
cube, N/3 is a perfect fifth power and N/5 is a perfect seventh
A quadrilateral inscribed in a unit circle has sides of lengths s1, s2, s3 and s4 where s1 ≤ s2 ≤ s3 ≤ s4.
Find a quadrilateral of this type for which s1= sqrt2 and show s1 cannot. . . .
Advent Calendar 2011 - a mathematical activity for each day during the run-up to Christmas.
Take any whole number q. Calculate q^2 - 1. Factorize
q^2-1 to give two factors a and b (not necessarily q+1 and q-1). Put c = a + b + 2q . Then you will find that ab+1 , bc+1 and ca+1 are all. . . .
Let a(n) be the number of ways of expressing the integer n as an
ordered sum of 1's and 2's. Let b(n) be the number of ways of
expressing n as an ordered sum of integers greater than 1. (i)
Calculate. . . .
A connected graph is a graph in which we can get from any vertex to
any other by travelling along the edges. A tree is a connected
graph with no closed circuits (or loops. Prove that every tree. . . .
Is the mean of the squares of two numbers greater than, or less
than, the square of their means?
I am exactly n times my daughter's age. In m years I shall be exactly (n-1) times her age. In m2 years I shall be exactly (n-2) times her age. After that I shall never again be an exact multiple of. . . .
Can you rearrange the cards to make a series of correct
The tangles created by the twists and turns of the Conway rope
trick are surprisingly symmetrical. Here's why!
If I tell you two sides of a right-angled triangle, you can easily work out the third. But what if the angle between the two sides is not a right angle?
You have twelve weights, one of which is different from the rest.
Using just 3 weighings, can you identify which weight is the odd
one out, and whether it is heavier or lighter than the rest?
Three points A, B and C lie in this order on a line, and P is any
point in the plane. Use the Cosine Rule to prove the following
The picture illustrates the sum 1 + 2 + 3 + 4 = (4 x 5)/2. Prove the general formula for the sum of the first n natural numbers and the formula for the sum of the cubes of the first n natural. . . .
Factorial one hundred (written 100!) has 24 noughts when written in full and that 1000! has 249 noughts? Convince yourself that the above is true. Perhaps your methodology will help you find the. . . .
Four jewellers possessing respectively eight rubies, ten saphires,
a hundred pearls and five diamonds, presented, each from his own
stock, one apiece to the rest in token of regard; and they. . . .
Start with any triangle T1 and its inscribed circle. Draw the
triangle T2 which has its vertices at the points of contact between
the triangle T1 and its incircle. Now keep repeating this. . . .
The country Sixtania prints postage stamps with only three values 6 lucres, 10 lucres and 15 lucres (where the currency is in lucres).Which values cannot be made up with combinations of these postage. . . .
Show that if three prime numbers, all greater than 3, form an
arithmetic progression then the common difference is divisible by
6. What if one of the terms is 3?
The nth term of a sequence is given by the formula n^3 + 11n . Find
the first four terms of the sequence given by this formula and the
first term of the sequence which is bigger than one million. . . .
This article looks at knight's moves on a chess board and introduces you to the idea of vectors and vector addition.
The largest square which fits into a circle is ABCD and EFGH is a square with G and H on the line CD and E and F on the circumference of the circle. Show that AB = 5EF.
Similarly the largest. . . .
Show that if you add 1 to the product of four consecutive numbers
the answer is ALWAYS a perfect square.
Learn about the link between logical arguments and electronic circuits. Investigate the logical connectives by making and testing your own circuits and fill in the blanks in truth tables to record. . . .
It is impossible to trisect an angle using only ruler and compasses
but it can be done using a carpenter's square.
Find the largest integer which divides every member of the
following sequence: 1^5-1, 2^5-2, 3^5-3, ... n^5-n.
Prove that if a^2+b^2 is a multiple of 3 then both a and b are multiples of 3.
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.
Can you discover whether this is a fair game?
Can you make sense of these three proofs of Pythagoras' Theorem?
This article stems from research on the teaching of proof and
offers guidance on how to move learners from focussing on
experimental arguments to mathematical arguments and deductive |
In Advanced Equity Derivatives: Volatility and Correlation, Sébastien Bossu reviews important concepts and recent developments in option pricing and modeling, including the latest generation of equity derivatives—namely, volatility and correlation derivatives. Readers should have some familiarity with basic equity derivatives pricing and advanced mathematics because this book references the Black–Scholes model and other formulas for exotics, from the most common to cutting edge. This background is also required for readers who want to solve the problems at the end of each chapter.
In 2004, while working as an equity derivatives analyst at J.P. Morgan in London, Bossu found someone selling correlation and buying it back simultaneously, using two different methods. He discovered that with some corrections, this trade led to a pure dynamic arbitrage. One of the two correlation instruments involved in the trade—the correlation swap—was not priced at fair value. In the book, Bossu introduces and assesses his own refined model and the work of others in this field.
Although the European digital and geometric Asian options are priced using the Black–Scholes formula, barrier options, lookback options, forward start options, and multiasset exotics (based on several underlying stocks or indices) are preferably priced using the local volatility model, a stochastic volatility model, or Monte Carlo simulations.
The Black–Scholes model assumes a single constant-volatility parameter to price options. In practice, however, every listed vanilla option has a different implied volatility for each strike and maturity. Developed by Emanuel Derman and Iraj Kani and by Bruno Dupire in the early 1990s, the local volatility model has become the benchmark model to price and hedge a wide range of such equity exotics as digitals, Asians, and barriers. The local volatility model is best visualized on a binomial tree where, instead of using constant volatility at each node to generate the tree of future spot prices, a different volatility parameter is used at each node. The option is then priced using backward induction. Certain payoffs, however, such as forward start options, are better approached using a stochastic volatility model. These analyses require a high-quality, smooth, implied volatility surface as an input, along with the simulation of all intermediate spot prices until maturity, using short time steps.
As implied by its name, a volatility surface is a three-dimensional graph that plots implied volatilities across option strikes and terms to maturity. There are challenges, however, in creating an implied volatility surface. For one thing, graphing implied volatilities against strike prices for a given expiration yields a skewed “smile” instead of the expected flat surface. This anomaly arises because the standard Black–Scholes option-pricing model assumes constant volatility and lognormal distributions of underlying asset returns. Second, implied volatilities derived from listed option prices are available for only a finite number of listed strikes and maturities. Therefore, it is important to be able to interpolate or extrapolate implied volatilities. Linear interpolation has limitations because it produces a cracked smile curve. Alternative techniques, such as cubic splines, are often used to obtain a smooth curve. Extrapolation is also a difficult endeavor. How can one price a five-year option if the longest listed maturity is two years? Fortunately, some researchers have published the models that market makers at a large equity option house use to mark their positions. There are two ways to model the volatility surface:
- directly, by specifying a functional form, such as a parametric function or an interpolation/extrapolation method, and
- indirectly, by modeling the behavior of the underlying asset from the geometric Brownian motion posited by the Black–Scholes model.
The most popular parameterization of the smile of fixed maturity is Jim Gatheral’s SVI (stochastic volatility inspired) model. The implied volatility methods include the SABR (stochastic alpha, beta, and rho) model, the Heston model, and the LNV (lognormal variance) model.
The profit or loss on a delta-hedged option position is driven by the spread between two types of volatility: the instant realized volatility of the underlying stock or index and the option-implied volatility. Option traders, who are specialists in volatility, want to trade it directly—hence, the creation of options on volatility itself. Examples include variance swaps and CBOE Volatility Index, or VIX, futures and options. The payoffs on the former are based on realized volatility, whereas those on the latter are based on implied volatility.
The development of multiasset exotic products made it possible (and at times necessary) to trade correlation more or less directly. The first correlation trades were actually dispersion trades in which a long or short position on a multiasset option (such as an index) was offset by a reverse position on single-asset options (individual constituents of the index). The two most popular types of dispersion trades are vanilla dispersions and variance dispersions. In recent years, the concept of local volatility has been extended to multiple assets, leading to local correlation models: LVLC (local volatility with local correlation) and the dynamic local correlation model. LVLC allows pairwise correlation coefficients to depend on time and spot prices. Alex Langnau’s dynamic local correlation model reproduces the index smile very accurately.
Finally, stochastic correlation models provide a more realistic approach to the pricing and hedging of certain types of derivatives, including worst-of options, best-of options, correlation swaps, and correlation options. Various types of stochastic correlation models—from single correlation to tradable average correlation (including the beta-omega, or B-O, model) to correlation matrix, all using some form of Jacobi processes—are discussed in the last chapter. |