CHAPTER 11 Numerical Differentiation and Integration

Save this PDF as:
Size: px
Start display at page:

Download "CHAPTER 11 Numerical Differentiation and Integration"

Transcription

1 CHAPTER 11 Numericl Differentition nd Integrtion Differentition nd integrtion re bsic mthemticl opertions with wide rnge of pplictions in mny res of science. It is therefore importnt to hve good methods to compute nd mnipulte derivtives nd integrls. You probbly lernt the bsic rules of differentition nd integrtion in school symbolic methods suitble for pencil-nd-pper clcultions. These re importnt, nd most derivtives cn be computed this wy. Integrtion however, is different, nd most integrls cnnot be determined with symbolic methods like the ones you lernt in school. Another compliction is the fct tht in prcticl pplictions function is only known t few points. For exmple, we my mesure the position of cr every minute vi GPS (Globl Positioning System) unit, nd we wnt to compute its speed. If the position is known s continuous function of time, we cn find the speed by differentiting this function. But when the position is only known t isolted times, this is not possible. The sme pplies to integrls. The solution, both when it comes to integrls tht cnnot be determined by the usul methods, nd functions tht re only known t isolted points, is to use pproximte methods of differentition nd integrtion. In our context, these re going to be numericl methods. We re going to present number of methods for doing numericl integrtion nd differentition, but more importntly, we re going to present generl strtegy for deriving such methods. In this wy you will not only hve number of methods vilble to you, but you will lso be ble to develop new methods, tilored to specil situtions tht you my encounter. 7

2 We use the sme generl strtegy for deriving both numericl integrtion nd numericl differentition methods. The bsic ide is to evlute function t few points, find the polynomil tht interpoltes the function t these points, nd use the derivtive or integrl of the polynomil s n pproximtion to the function. This technique lso llows us to keep trck of the so-clled trunction error, the mthemticl error committed by integrting or differentiting the polynomil insted of the function itself. However, when it comes to roundoff error, we hve to tret differentition nd integrtion differently: Numericl integrtion is very insensitive to round-off errors, while numericl differentition behves in the opposite wy; it is very sensitive to round-off errors A simple method for numericl differentition We strt by studying numericl differentition. We first introduce the simplest method, derive its error, nd its sensitivity to round-off errors. The procedure used here for deriving the method nd nlysing the error is used over gin in lter sections to derive nd nlyse dditionl methods. Let us first mke it cler wht numericl differentition is. Problem 11.1 (Numericl differentition). Let f be given function tht is only known t number of isolted points. The problem of numericl differentition is to compute n pproximtion to the derivtive f of f by suitble combintions of the known vlues of f. A typicl exmple is tht f is given by computer progrm (more specificlly function, procedure or method, depending on you choice of progrmming lnguge), nd you cn cll the progrm with floting-point rgument x nd receive bck floting-point pproximtion of f (x). The chllenge is to compute n pproximtion to f () for some rel number when the only id we hve t our disposl is the progrm to compute vlues of f The bsic ide Since we re going to compute derivtives, we must be cler bout they re defined. Recll tht f () is defined by f f ( + h) f () () = lim. (11.1) h 0 h In the following we will ssume tht this limit exists; i.e., tht f is differentible. From (11.1) we immeditely hve nturl pproximtion to f (); we simply 8

3 pick positive h nd use the pproximtion f () f ( + h) f (). (11.) h Note tht this corresponds to pproximting f by the stright line p 1 tht interpoltes f t nd h, nd then using p 1 () s n pproximtion to f (). Observtion 11.. The derivtive of f t cn be pproximted by f () f ( + h) f (). h In prcticl sitution, the number would be given, nd we would hve to locte the two nerest vlues 1 nd to the left nd right of such tht f ( 1 ) nd f ( ) cn be found. Then we would use the pproximtion f () f ( ) f ( 1 ) 1. In lter sections, we will derive severl formuls like (11.). Which formul to use for specific exmple, nd exctly how to use it, will hve to be decided in ech cse. Exmple Let us test the pproximtion (11.) for the function f (x) = sin x t = 0.5 (using 64-bit floting-point numbers). In this cse we hve f (x) = cos x so f () = This mkes it is esy to check the ccurcy. We try with few vlues of h nd find h ( f ( + h) f () )/ h E1 (f ;,h) where E 1 (f ;,h) = f () ( f (+h) f () )/ h. In other words, the pproximtion seems to improve with decresing h, s expected. More precisely, when h is reduced by fctor of 10, the error is reduced by the sme fctor. 9

4 11.1. The trunction error Whenever we use pproximtions, it is importnt to try nd keep trck of the error, if t ll possible. To nlyse the error in numericl differentition, Tylor polynomils with reminders re useful. To nlyse the error in the pproximtion bove, we do Tylor expnsion of f ( + h). We hve f ( + h) = f () + h f () + h f (ξ h ), where ξ h lies in the intervl (, + h). If we rerrnge this formul, we obtin f f ( + h) f () () = h h f (ξ h ). (11.3) This is often referred to s the trunction error of the pproximtion, nd is resonble error formul, but it would be nice to get rid of ξ h. We first tke bsolute vlues in (11.3), f f ( + h) f () () h = h f (ξ h ). Recll from the Extreme vlue theorem tht if function is continuous, then its mximum lwys exists on ny closed nd bounded intervl. In our setting here, it is nturl to let the closed nd bounded intervl be [, +h]. This leds to the following lemm. Lemm Suppose tht f hs continuous derivtives up to order two ner. If the derivtive f () is pproximted by f ( + h) f (), h then the trunction error is bounded by E(f ;,h) = f f ( + h) f () () h h mx x [,+h] f (x). (11.4) Let us check tht the error formul (11.3) confirms the numericl vlues in exmple We hve f (x) = sin x, so the right-hnd side in (11.4) becomes E(sin;0.5,h) = h sinξ h, 30

5 where ξ h (0.5,0.5 + h). For h = 0.1 we therefore hve tht the error must lie in the intervl [0.05sin0.5, 0.05sin0.6] = [ , ], nd the right end of the intervl is the mximum vlue of the right-hnd side in (11.4). When h is reduced by fctor of 10, the fctor h/ is reduced by the sme fctor. This mens tht ξ h will pproch 0.5 so sinξ h will pproch the lower vlue sin For h = 10 n, the error will therefore tend to 10 n sin0.5/ 10 n 0.397, which is in complete greement with exmple This is true in generl. If f is continuous, then ξ h will pproch when h goes to zero. But even when h > 0, the error in using the pproximtion f (ξ h ) f () is smll. This is the cse since it is usully only necessry to know the mgnitude of the error, i.e., it is sufficient to know the error with one or two correct digits. Observtion The trunction error is pproximtely given by f f ( + h) f () () h h f () The round-off error So fr, we hve just considered the mthemticl error committed when f () is pproximted by ( f ( + h) f () )/ h. But wht bout the round-off error? In fct, when we compute this pproximtion we hve to perform the one criticl opertion f ( + h) f () subtrction of two lmost equl numbers which we know from chpter 5 my led to lrge round-off errors. Let us continue exmple 11.3 nd see wht hppens if we use smller vlues of h. Exmple Recll tht we estimted the derivtive of f (x) = sin x t = 0.5 nd tht the correct vlue with ten digits is f (0.5) If we check vlues of h from 10 7 nd smller we find ( )/ h f ( + h) f () h E(f ;,h)

6 This shows very clerly tht something quite drmtic hppens, nd when we come to h = 10 17, the derivtive is computed s zero. If f () is the floting-point number closest to f (), we know from lemm 5.6 tht the reltive error will be bounded by 5 53 since floting-point numbers re represented in binry (β = ) with 53 bits for the significnd (m = 53). We therefore hve ɛ In prctice, the rel upper bound on ɛ is usully smller, nd in the following we will denote this upper bound by ɛ. This mens tht definite upper bound on ɛ is Nottion The mximum reltive error when rel number is represented by floting-point number is denoted by ɛ. There is hndy wy to express the reltive error in f (). If we denote the computed vlue of f () by f (), we will hve f () = f ()(1 + ɛ) which corresponds to the reltive error being ɛ. Observtion Suppose tht f () is computed with 64-bit floting-point numbers nd tht no underflow or overflow occurs. Then the computed vlue f () stisfies f () = f ()(1 + ɛ) (11.5) where ɛ ɛ, nd ɛ depends on both nd f. The computtion of f ( + h) is of course lso ffected by round-off error, so we hve f () = f ()(1 + ɛ 1 ), f ( + h) = f ( + h)(1 + ɛ ) (11.6) where ɛ i ɛ for i = 1,. Here we should relly write ɛ = ɛ (h), becuse the exct round-off error in f ( + h) will inevitbly depend on h in rther rndom wy. The next step is to see how these errors ffect the computed pproximtion of f (). Recll from exmple 5.11 tht the min source of round-off in subtrction is the replcement of the numbers to be subtrcted by the nerest flotingpoint numbers. We therefore consider the computed pproximtion to be f ( + h) f (). h 3

7 If we insert the expressions (11.6), nd lso mke use of lemm 11.4, we obtin f ( + h) f () f ( + h) f () = h h + f ( + h)ɛ f ()ɛ 1 h = f () + h f (ξ h ) + f ( + h)ɛ f ()ɛ 1 h where ξ h (, + h). This leds to the following importnt observtion. (11.7) Theorem Suppose tht f nd its first two derivtives re continuous ner. When the derivtive of f t is pproximted by f ( + h) f (), h the error in the computed pproximtion is given by f f ( + h) f () () h h M 1 + ɛ h M, (11.8) where M 1 = mx x [,+h] f (), M = mx x [,+h] f (). Proof. To get to (11.8) we hve rerrnged (11.7) nd used the tringle inequlity. We hve lso replced f (ξ h ) by its mximum on the intervl [, + h], s in (11.4). Similrly, we hve replced f (ξ h ) nd f ( + h) by their common mximum on [, +h]. The lst term then follows by pplying the tringle inequlity to the lst term in (11.7) nd replcing ɛ 1 nd ɛ (h) by the upper bound ɛ. The inequlity (11.8) cn be replced by n pproximte equlity by mking the pproximtions M 1 f () nd M f (), just like in observtion 11.8 nd using the mximum of ɛ 1 nd ɛ in (11.7), which we denote ɛ(h). Observtion The inequlity (11.8) is pproxmtely equivlent to f f ( + h) f () () h h f () ɛ(h) + f (). (11.9) h Let us check how well observtion grees with the computtions in exmples 11.3 nd

8 Exmple For lrge vlues of h the first term on the right in (11.9) will dominte the error nd we hve lredy seen tht this grees very well with the computed vlues in exmple The question is how well the numbers in exmple 11.6 cn be modelled when h becomes smller. To estimte the size of ɛ(h), we consider the cse when h = Then the observed error is so we should hve We solve this eqution nd find sin0.5 ɛ( 10 16) = ɛ ( 10 16) = ( ) sin 0.5. = If we try some other vlues of h we find ɛ ( 10 11) = , ɛ ( 10 13) = , ɛ ( 10 15) = We observe tht ll these vlues re considerbly smller thn the upper limit which we mentioned bove. Figure 11.1 shows plots of the error. The numericl pproximtion hs been computed for the vlues n = 0.01i, i = 0,..., 00 nd plotted in log-log plot. The errors re shown s isolted dots, nd the function g (h) = h sin0.5 + ɛ h sin0.5 (11.10) with ɛ = is shown s solid grph. It seems like this choice of ɛ mkes g (h) resonble upper bound on the error Optiml choice of h Figure 11.1 indictes tht there is n optiml vlue of h which minimises the totl error. We cn find this mthemticlly by minimising the upper bound in (11.9), with ɛ(h) replced by the upper bound ɛ. This gives g (h) = h f () ɛ + f (). (11.11) h To find the vlue of h which minimises this expression, we differentite with respect to h nd set the derivtive to zero. We find f g () (h) = ɛ f (). h If we solve the eqution g (h) = 0, we obtin the pproximte optiml vlue. 34

9 Figure Numericl pproximtion of the derivtive of f (x) = sin x t x = 0.5 using the pproximtion in lemm The plot is log 10 -log 10 plot which shows the logrithm to bse 10 of the bsolute vlue of the totl error s function of the logrithm to bse 10 of h, bsed on 00 vlues of h. The point 10 on the horizontl xis therefore corresponds h = 10 10, nd the point 6 on the verticl xis corresponds to n error of The plot lso includes the function given by (11.10). 10 Lemm Let f be function with continuous derivtives up to order. If the derivtive of f t is pproximted s in lemm 11.4, then the vlue of h which minimises the totl error (trunction error + round-off error) is pproximtely ɛ f () h f (). It is esy to see tht the optiml vlue of h is the vlue tht blnces the two terms in (11.11)l, i.e., the trunction error nd the round-off error re equl. In the exmple with f (x) = sin x nd = 0.5 we cn use ɛ = which gives h = ɛ = Summry of the generl strtegy Before we continue, let us sum up the derivton nd nlysis of the numericl differentition method in section 11.1, since we will use this over nd over gin. The first step ws to derive the numericl method. In section 11.1 this ws very simple since the method cme stright out of the definition of the derivtive. Just before observtion 11. we indicted tht the method cn lso be de- 35

10 rived by pproximting f by polynomil p nd using p () s n pproximtion to f (). This is the generl pproch tht we will use below. Once the numericl method is known, we estimte the mthemticl error in the pproximtion, the trunction error. This we do by performing Tylor expnsions with reminders. For numericl differentition methods which provide estimtes of derivtive t point, we replce ll function vlues t points other thn by Tylor polynomils with reminders. There my be chllenge to choose the degree of the Tylor polynomil. The next tsk is to estimte the totl error, including round-off error. We consider the difference between the derivtive to be computed nd the computed pproximtion, nd replce the computed function evlutions by expressions like the ones in observtion This will result in n expression involving the mthemticl pproximtion to the derivtive. This cn be simplified in the sme wy s when the trunction error ws estimted, with the ddition of n expression involving the reltive round-off errors in the function evlutions. These expressions cn then be simplified to something like (11.8) or (11.9). As finl step, the optiml vlue of h cn be found by minimising the totl error. Algorithm To derive nd nlyse numericl differentition method, the following steps re necessry: 1. Derive the method using polynomil interpoltion.. Estimte the trunction error using Tylor series with reminders. 3. Estimte the totl error (trunction error + round-off error) by ssuming ll function evlutions re replced by the nerest floting-point numbers. 4. Estimte the optiml vlue of h A simple, symmetric method The numericl differentition method in section 11.1 is not symmetric bout, so let us try nd derive symmetric method Construction of the method We wnt to find n pproximtion to f () using vlues of f ner. To obtin symmetric method, we ssume tht f ( h), f (), nd f ( + h) re known 36

11 vlues, nd we wnt to find n pproximtion to f () using these vlues. The strtegy is to determine the qudrtic polynomil p tht interpoltes f t h, nd + h, nd then we use p () s n pproximtion to f (). We write p in Newton form, p (x) = f [ h] + f [ h, ](x ( h)) We differentite nd find Setting x = yields + f [ h,, + h](x ( h))(x ). (11.1) p (x) = f [ h, ] + f [ h,, + h](x + h). p () = f [ h, ] + f [ h,, + h]h. To get prcticlly useful formul we must express the divided differences in terms of function vlues. If we expnd the second expression we obtin p f [, + h] f [ h, ] f [, + h] + f [ h, ] () = f [ h, ]+ h = h The two first order differences re (11.13) f [ h, ] = f () f ( h), f [, + h] = h f ( + h) f (), h nd if we insert this in (11.13) we end up with p f ( + h) f ( h) () =. h Lemm Let f be given function, nd let nd h be given numbers. If f ( h), f (), f ( + h) re known vlues, then f () cn be pproximted by p () where p is the qudrtic polynomil tht interpoltes f t h,, nd + h. The pproximtion is given by f () p f ( + h) f ( h) () =. (11.14) h Let us test this pproximtion on the function f (x) = sin x t = 0.5 so we cn compre with the method in section

12 Exmple We test the pproximtion (11.14) with the sme vlues of h s in exmples 11.3 nd Recll tht f (0.5) with ten correct decimls. The results re ( )/ h f ( + h) f ( h) (h) E(f ;,h) If we compre with exmples 11.3 nd 11.6, the errors re generlly smller. In prticulr we note tht when h is reduced by fctor of 10, the error is reduced by fctor of 100, t lest s long s h is not too smll. However, when h becomes smller thn bout 10 6, the error becomes lrger. It therefore seems like the trunction error is smller thn in the first method, but the round-off error mkes it impossible to get ccurte results for smll vlues of h. The optiml vlue of h seems to be h 10 6, which is lrger thn for the first method, but the error is then bout 10 1, which is smller thn the best we could do with the first method Trunction error Let us ttempt to estimte the trunction error for the method in lemm The ide is to do replce f ( h) nd f ( + h) with Tylor expnsions bout. We use the Tylor expnsions f ( + h) = f () + h f () + h f () + h3 6 f (ξ 1 ), f ( h) = f () h f () + h f () h3 6 f (ξ ), where ξ 1 (, + h) nd ξ ( h, ). If we subtrct the second formul from the first nd divide by h, we obtin f ( + h) f ( h) h = f () + h ( f (ξ 1 ) + f (ξ ) ). (11.15) 1 38

13 This leds to the following lemm. Lemm Suppose tht f nd its first three derivtives re continuous ner, nd suppose we pproximte f () by f ( + h) f ( h). (11.16) h The trunction error in this pproximtion is bounded by E (f ;,h) = f () f ( + h) f ( h) h h 6 mx x [ h,+h] f (x). (11.17) Proof. Wht remins is to simplify the lst term in (11.15) to the term on the right in (11.17). This follows from f (ξ 1 ) + f (ξ ) mx f (x) + mx f (x) x [,+h] x [ h,] mx f (x) + mx f (x) x [ h,+h] x [ h,+h] = mx f (x). x [ h,+h] The lst inequlity is true becuse the width of the intervls over which we tke the mximums re incresed, so the mximum vlues my lso increse. The error formul (11.17) confirms the numericl behviour we sw in exmple for smll vlues of h since the error is proportionl to h : When h is reduced by fctor of 10, the error is reduced by fctor Round-off error The round-off error my be estimted just like for the first method. When the pproximtion (11.16) is computed, the vlues f ( h) nd f ( + h) re replced by the nerest floting point numbers f ( h) nd f ( + h) which cn be expressed s f ( + h) = f ( + h)(1 + ɛ 1 ), f ( h) = f ( h)(1 + ɛ ), where both ɛ 1 nd ɛ depend on h nd stisfy ɛ i ɛ for i = 1,. Using these expressions we obtin f ( + h) f ( h) = h f ( + h) f ( h) h + f ( + h)ɛ 1 f ( h)ɛ. h 39

14 We insert (11.15) nd get the reltion f ( + h) f ( h) h = f () + h ( f (ξ 1 ) + f (ξ ) ) + f ( + h)ɛ 1 f ( h)ɛ. 1 h This leds to n estimte of the totl error if we use the sme technique s in the proof of lemm Theorem Let f be given function with continuous derivtives up to order three, nd let nd h be given numbers. Then the error in the pproximtion f f ( + h) f ( h) (), h including round-off error nd trunction error, is bounded by f f ( + h) f ( h) () h h 6 M 1 + ɛ h M (11.18) where M 1 = mx x [ h,+h] f (x), M = mx x [ h,+h] f (x). (11.19) In prctice, the interesting vlues of h will usully be so smll tht there is very little error in mking the pproximtions M 1 = mx x [ h,+h] f (x) f (), M = mx x [ h,+h] f (x) f (). If we mke this simplifiction in (11.18) we obtin slightly simpler error estimte. Observtion The error (11.18) is pproximtely bounded by f f ( + h) f ( h) () h h f () ɛ f () +. (11.0) 6 h A plot of how the error behves in this pproximtion, together with the estimte of the error on the right in (11.0), is shown in figure

15 Figure 11.. Log-log plot of the error in the pproximtion to the derivtive of f (x) = sin x t x = 1/ for vlues of h in the intervl [0,10 17 ], using the method in theorem The function plotted is the right-hnd side of (11.0) with ɛ = , s function of h Optiml choice of h As for the first numericl differentition method, we cn find n optiml vlue of h which minimises the error. The error is minimised when the trunction error nd the round-off error hve the sme mgnitude. We cn find this vlue of h if we differentite the right-hnd side of (11.18) with respect to h nd set the derivtive to 0. This leds to the eqution h 3 M 1 ɛ h M = 0 which hs the solution h = 3 3ɛ M 3 M1 3 3ɛ f () f (). 3 At the end of section we sw tht resonble vlue for ɛ ws ɛ = The optiml vlue of h in exmple 11.15, where f (x) = sin x nd = 1/, then becomes h = For this vlue of h the pproximtion is f (0.5) with error A four-point method for differentition In wy, the two methods for numericl differentition tht we hve considered so fr re the sme. If we use step length of h in the first method, the 41

16 pproximtion becomes f f ( + h) f () (). h The nlysis of the symmetric method shows tht the pproximtion is considerbly better if we ssocite the pproximtion with the midpoint between nd + h, f f ( + h) f () ( + h). h At the point +h the pproximtion is proportionl to h rther thn h, nd this mkes big difference s to how quickly the error goes to zero, s is evident if we compre exmples 11.3 nd In this section we derive nother method for which the trunction error is proportionl to h 4. The computtions below my seem overwhelming, nd hve in fct been done with the help of computer to sve time nd reduce the risk of misclcultions. The method is included here just to illustrte tht the principle for deriving both the method nd the error terms is just the sme s for the simple symmetric method in the previous section. To sve spce we hve only included one highlight, of the pproximtion method nd the totl error Derivtion of the method We wnt better ccurcy thn the symmetric method which ws bsed on interpoltion with qudrtic polynomil. It is therefore nturl to bse the pproximtion on cubic polynomil, which cn interpolte four points. We hve seen the dvntge of symmetry, so we choose the interpoltion points x 0 = h, x 1 = h, x = + h, nd x 3 = + h. The cubic polynomil tht interpoltes f t these points is p 3 (x) = f (x 0 ) + f [x 0, x 1 ](x x 0 ) + f [x 0, x 1, x ](x x 0 )(x x 1 ) nd its derivtive is p 3 (x) = f [x 0, x 1 ] + f [x 0, x 1, x ](x x 0 x 1 ) + f [x 0, x 1, x, x 3 ](x x 0 )(x x 1 )(x x ). + f [x 0, x 1, x, x 3 ] ( (x x 1 )(x x ) + (x x 0 )(x x ) + (x x 0 )(x x 1 ) ). If we evlute this expression t x = nd simplify (this is quite bit of work), we find tht the resulting pproximtion of f () is f () p 3 f ( h) 8f ( h) + 8f ( + h) f ( + h) () =. (11.1) 1h 4

17 11.4. Trunction error To estimte the error, we expnd the four terms in the numertor in (11.1) in Tylor series, f ( h) = f () h f () + h f () 4h3 3 f () + h4 3 f (i v) () 4h5 15 f (v) (ξ 1 ), f ( h) = f () h f () + h f () h3 6 f () + h4 4 f (i v) () h5 10 f (v) (ξ ), f ( + h) = f () + h f () + h f () + h3 6 f () + h4 4 f (i v) () + h5 10 f (v) (ξ 3 ), f ( + h) = f () + h f () + h f () + 4h3 3 f () + h4 3 f (i v) () + 4h5 15 f (v) (ξ 4 ), where ξ 1 ( h, ), ξ ( h, ), ξ 3 (, + h), nd ξ 4 (, + h). If we insert this into the formul for p 3 () we obtin f ( h) 8f ( h) + 8f ( + h) f ( + h) = 1h f () h4 45 f (v) (ξ 1 ) + h4 180 f (v) (ξ ) + h4 180 f (v) (ξ 3 ) h4 45 f (v) (ξ 4 ). If we use the sme trick s for the symmetric method, we cn combine ll lst four terms in nd obtin n upper bound on the trunction error. The result is f f ( h) 8f ( h) + 8f ( + h) f ( + h) () 1h h4 18 M (11.) where M = mx x [ h,+h] f (v) (x) Round-off error The trunction error is derived in the sme wy s before. The quntities we ctully compute re f ( h) = f ( h)(1 + ɛ 1 ), f ( + h) = f ( + h)(1 + ɛ 3 ), f ( h) = f ( h)(1 + ɛ ), f ( + h) = f ( + h)(1 + ɛ 4 ). We estimte the difference between f () nd the computed pproximtion, mke use of the estimte (11.), combine the function vlues tht re multiplied by ɛs, nd pproximte the mximum vlues by function vlues t. We sum up the result. 43

18 Figure Log-log plot of the error in the pproximtion to the derivtive of f (x) = sin x t x = 1/, using the method in observtion 11.19, with h in the intervl [0,10 17 ]. The function plotted is the right-hnd side of (11.3) with ɛ = Observtion Suppose tht f nd its first five derivtives re continuous. If f () is pproximted by f () f ( h) 8f ( h) + 8f ( + h) f ( + h), 1h the totl error is pproximtely bounded by f f ( h) 8f ( h) + 8f ( + h) f ( + h) () 1h h 4 18 f (v) () 3ɛ + f (). (11.3) h A plot of the error in the pproximtion for the sin x exmple is shown in figure Optiml vlue of h From observtion we cn compute the optiml vlue of h by differentiting the right-hnd side with respect to h nd setting it to zero, h 3 9 f (v) () 3ɛ h f () = 0 44

19 which hs the solution h = 5 7ɛ f () f (v) (). 5 For the cse bove with f (x) = sin x nd = 0.5 the solution is h For this vlue of h the ctul error is Numericl pproximtion of the second derivtive We consider one more method for numericl pproximtion of derivtives, this time of the second derivtive. The pproch is the sme: We pproximte f by polynomil nd pproximte the second derivtive of f by the second derivtive of the polynomil. As in the other cses, the error nlysis is bsed on expnsion in Tylor series Derivtion of the method Since we re going to find n pproximtion to the second derivtive, we hve to pproximte f by polynomil of degree t lest two, otherwise the second derivtive is identiclly 0. The simplest is therefore to use qudrtic polynomil, nd for symmetry we wnt it to interpolte f t h,, nd + h. The resulting polynomil p is the one we used in section 11.3 nd it is given in eqution (11.1). The second derivtive of p is constnt, nd the pproximtion of f () is f () p () = f [ h,, + h]. The divided difference is esy to expnd. Lemm The second derivtive of function f t cn be pproximted by f f ( + h) f () + f ( h) () h. (11.4) The trunction error Estimtion of the error goes s in the other cses. The Tylor series of f ( h) nd f ( + h) re f ( h) = f () h f () + h f () h3 6 f () + h4 4 f (i v) (ξ 1 ), f ( + h) = f () + h f () + h f () + h3 6 f () + h4 4 f (i v) (ξ ), 45

20 where ξ 1 ( h, ) nd ξ (, + h). If we insert these Tylor series in (11.4) we obtin f ( + h) f () + f ( h) h = f () + h ( f (i v) (ξ 1 ) + f (i v) (ξ ) ). 4 From this obtin n expression for the trunction error. Lemm Suppose f nd its first three derivtives re continuous ner. If the second derivtive f () is pproximted by f () f ( + h) f () + f ( h) h, the error is bounded by f f ( + h) f () + f ( h) () h h 1 mx x [ h,+h] f (x). (11.5) Round-off error The round-off error cn lso be estimted s before. Insted of computing the exct vlues, we compute f ( h), f (), nd f ( + h), which re linked to the exct vlues by f ( h) = f ( h)(1 + ɛ 1 ), f () = f ()(1 + ɛ ), f ( + h) = f ( + h)(1 + ɛ 3 ), where ɛ i ɛ for i = 1,, 3. The difference between f () nd the computed pproximtion is therefore f f ( + h) f () + f ( h) () h = h ( f (ξ 1 ) + f (ξ ) ) ɛ 1 f ( h) ɛ f () + ɛ 3 f ( + h) 4 h. If we combine terms on the right s before, we end up with the following theorem. Theorem 11.. Suppose f nd its first three derivtives re continuous ner, nd tht f () is pproximted by f () f ( + h) f () + f ( h) h. 46

21 Figure Log-log plot of the error in the pproximtion to the derivtive of f (x) = sin x t x = 1/ for h in the intervl [0,10 8 ], using the method in theorem 11.. The function plotted is the right-hnd side of (11.3) with ɛ = Then the totl error (trunction error + round-off error) in the computed pproximtion is bounded by f f ( + h) f () + f ( h) () h h 1 M 1 + 3ɛ h M. (11.6) where M 1 = mx x [ h,+h] f (i v) (x), M = mx x [ h,+h] f (x). As before, we cn simplify the right-hnd side to h f (i v) () + 3ɛ f () 1 h (11.7) if we cn tolerte slightly pproximte upper bound. Figure 11.4 shows the errors in the pproximtion to the second derivtive given in theorem 11. when f (x) = sin x nd = 0.5 nd for h in the rnge [0,10 8 ]. The solid grph gives the function in (11.7) which describes the upper limit on the error s function of h, with ɛ = For h smller thn 10 8, the pproximtion becomes 0, nd the error constnt. Recll tht for the pproximtions to the first derivtive, this did not hppen until h ws bout This illustrtes the fct tht the higher the derivtive, the more problemtic is the round-off error, nd the more difficult it is to pproximte the derivtive with numericl methods like the ones we study here. 47

22 Figure The re under the grph of function Optiml vlue of h Agin, we find the optiml vlue of h by minimising the right-hnd side of (11.6). To do this we find the derivtive with respect to h nd set it to 0, h 6 M 1 6ɛ h 3 M = 0. As usul it does not mke much difference if we use the pproximtions M 1 f () nd M = f (). Observtion The upper bound on the totl error (11.6) is minimised when h hs the vlue 4 36ɛ f () h = f (i v) (). 4 When f (x) = sin x nd = 0.5 this gives h = if we use the vlue ɛ = Then the pproximtion to f () = sin is with n ctul error of Generl bckground on integrtion Our next tsk is to develop methods for numericl integrtion. Before we turn to this, it is necessry to briefly review the definition of the integrl. 48

23 () (b) (c) (d) Figure The definition of the integrl vi inscribed nd circumsribed step functions. Recll tht if f (x) is function, then the integrl of f from x = to x = b is written f (x)dx. This integrl gives the re under the grph of f, with the re under the positive prt counting s positive re, nd the re under the negtive prt of f counting s negtive re, see figure Before we continue, we need to define term which we will use repetedly in our description of integrtion. Definition Let nd b be two rel numbers with < b. A prtition of [,b] is finite sequence {x i } n i=0 of incresing numbers in [,b] with x 0 = nd x n = b, = x 0 < x 1 < x < x n 1 < x n = b. The prtition is sid to be uniform if there is fixed number h, clled the step length, such tht x i x i 1 = h = (b )/n for i = 1,..., n. 49

24 The trditionl definition of the integrl is bsed on numericl pproximtion to the re. We pick prtition {x i } of [,b], nd in ech subintervl [x i 1, x i ] we determine the mximum nd minimum of f (for convenience we ssume tht these vlues exist), m i = min f (x), M i = mx f (x), x [x i 1,x i ] x [x i 1,x i ] for i = 1,,..., n. We use these vlues to compute the two sums n n I = m i (x i x i 1 ), I = M i (x i x i 1 ). i=1 To define the integrl, we consider lrger prtitions nd consider the limits of I nd I s the distnce between neighbouring x i s goes to zero. If those limits re the sme, we sy tht f is integrble, nd the integrl is given by this limit. More precisely, I = f (x)dx = sup I = inf I, where the sup nd inf re tken over ll prtitions of the intervl [,b]. This process is illustrted in figure 11.6 where we see how the piecewise constnt pproximtions become better when the rectngles become nrrower. The bove definition cn be used s numericl method for computing pproximtions to the integrl. We choose to work with either mxim or minim, select prtition of [,b] s in figure 11.6, nd dd together the res of the rectngles. The problem with this technique is tht it cn be both difficult nd time consuming to determine the mxim or minim, even on computer. However, it cn be shown tht the integrl hs very convenient property: If we choose point t i in ech intervl [x i 1, x i ], then the sum Ĩ = i=1 n f (t i )(x i x i 1 ) i=1 will lso converge to the integrl when the distnce between neighbouring x i s goes to zero. If we choose t i equl to x i 1 or x i, we hve simple numericl method for computing the integrl. An even better choice is the more symmetric t i = (x i + x i 1 )/ which leds to the pproximtion I n f ( (x i + x i 1 )/ ) )(x i x i 1 ). (11.8) i=1 This is the so-clled midpoint method which we will study in the next section. 50

25 x 1 () x 1 x 3 x 5 x 7 x 9 (b) Figure The midpoint rule with one subintervl () nd five subintervls (b). In generl, we cn derive numericl integrtion methods by splitting the intervl [, b] into smll subintervls, pproximte f by polynomil on ech subintervl, integrte this polynomil rther thn f, nd then dd together the contributions from ech subintervl. This is the strtegy we will follow, nd this works s long s f cn be pproximted well by polynomils on ech subintervl The midpoint method for numericl integrtion We hve lredy introduced the midpoint rule (11.8) for numericl integrtion. In our stndrd frmework for numericl methods bsed on polynomil pproximtion, we cn consider this s using constnt pproximtion to the function f on ech subintervl. Note tht in the following we will lwys ssume the prtition to be uniform. Algorithm Let f function which is integrble on the intervl [, b] nd let {x i } n be uniform prtition of [,b]. In the midpoint rule, the integrl of f is pproximted i=0 by f (x)dx I mid (h) = h n f (x i 1/ ), (11.9) i=1 where x i 1/ = (x i + x i 1 )/ = + (i 1/)h. This my seem like strngely formulted lgorithm, but ll there is to it is to compute the sum on the right in (11.9). The method is illustrted in figure 11.7 in the cses where we hve one nd five subintervls. 51

26 Exmple Let us try the midpoint method on n exmple. As usul, it is wise to test on n exmple where we know the nswer, so we cn esily check the qulity of the method. We choose the integrl 1 0 cos x dx = sin where the exct nswer is esy to compute by trditionl, symbolic methods. To test the method, we split the intervl into k subintervls, for k = 1,,..., 10, i.e., we hlve the step length ech time. The result is By error, we here men h I mid (h) Error f (x)dx I mid (h). Note tht ech time the step length is hlved, the error seems to be reduced by fctor of Locl error nlysis As usul, we should try nd keep trck of the error. We first focus on wht hppens on one subintervl. In other words we wnt to study the error f (x)dx f ( 1/ ) (b ), 1/ = ( + b)/. (11.30) Once gin, Tylor polynomils with reminders help us out. We expnd both f (x) nd f ( 1/ ) bout the the left endpoint f (x) = f () + (x )f (x ) () + f (ξ 1 ), f ( 1/ ) = f () + ( 1/ )f () + ( 1/ ) f (ξ ), 5

27 where ξ 1 is number in the intervl (, x) tht depends on x, nd ξ is in the intervl (, 1/ ). If we multiply the second Tylor expnsion by (b ), we obtin (b ) f ( 1/ )(b ) = f ()(b ) + f (b )3 () + f (ξ ). (11.31) 8 Next, we integrte the Tylor expnsion nd obtin f (x)dx = (f () + (x )f (x ) ) () + f (ξ 1 ) dx = f ()(b ) + 1 [ (x ) ] b f () + 1 (x ) f (ξ 1 )dx (b ) = f ()(b ) + f () + 1 (x ) f (ξ 1 )dx. (11.3) We then see tht the error cn be written f (x)dx f ( 1/ )(b ) = 1 b (x ) f (b )3 (ξ 1 )dx f (ξ ) 8 1 (x ) f (ξ 1 )dx (b )3 + f (ξ ). 8 (11.33) For the lst term, we use our stndrd trick, f (ξ ) M = mx x [,b] f (x). (11.34) Note tht since ξ (, 1/ ), we could just hve tken the mximum over the intervl [, 1/ ], but we will see lter tht it is more convenient to mximise over the whole intervl [,b]. The first term in (11.33) needs some mssging. Let us do the work first, nd explin fterwords, 1 b (x ) f (ξ 1 )dx 1 (x ) f (ξ 1 ) dx = 1 M = M 1 3 (x ) f (ξ 1 ) dx (x ) dx [ (x ) 3 ] b = M 6 (b )3. (11.35) 53

28 The first inequlity is vlid becuse when we move the bsolute vlue sign inside the integrl sign, the function tht we integrte becomes nonnegtive everywhere. This mens tht in the res where the integrnd in the originl expression is negtive, everything is now positive, nd hence the second integrl is lrger thn the first. Next there is n equlity which is vlid becuse (x ) is never negtive. The next inequlity follows becuse we replce f (ξ 1 ) with its mximum on the intervl [,b]. The lst step is just the evlution of the integrl of (x ). We hve now simplified both terms on the right in (11.33), so we hve The result is the following lemm. f (x)dx f ( 1/ ) (b ) M 6 (b )3 + M 8 (b )3. Lemm Let f be continuous function whose first two derivtives re continuous on the intervl [.b]. The the error in the midpoint method, with only one intervl, is bounded by f (x)dx f ( 1/ ) (b ) 7M 4 (b )3, where M = mx x [,b] f (x) nd 1/ = ( + b)/. The importnce of this lemm lies in the fctor (b ) 3. This mens tht if we reduce the size of the intervl to hlf its width, the error in the midpoint method will be reduced by fctor of 8. Perhps you feel completely lost in the work tht led up to lemm The wise wy to red something like this is to first focus on the generl ide tht ws used: Consider the error (11.30) nd replce both f (x) nd f ( 1/ ) by its qudrtic Tylor polynomils with reminders. If we do this, number of terms cncel out nd we re left with (11.33). At this point we use some stndrd techniques tht give us the finl inequlity. Once you hve n overview of the derivtion, you should check tht the detils re correct nd mke sure you understnd ech step Globl error nlysis Above, we nlysed the error on one subintervl. Now we wnt to see wht hppens when we dd together the contributions from mny subintervls; it should not surprise us tht this my ffect the error. 54

29 We consider the generl cse where we hve prtition tht divides [,b] into n subintervls, ech of width h. On ech subintervl we use the simple midpoint rule tht we nlysed in the previous section, I = The totl error is then f (x)dx = I I mid = n i=1 n i=1 ( xi xi x i 1 f (x)dx n f (x i 1/ )h. i=1 x i 1 f (x)dx f (x i 1/ )h But the expression inside the prenthesis is just the locl error on the intervl [x i 1, x i ]. We therefore hve n ( xi ) I I mid = f (x)dx f (x i 1/ )h i=1 x i 1 n xi f (x)dx f (x i 1/ )h x i 1 i=1 n 7h 3 i=1 4 M i (11.36) where M i is the mximum of f (x) on the intervl [xi 1, x i ]. To simiplify the expression (11.36), we extend the mximum on [x i 1, x i ] to ll of [,b]. This will usully mke the mximum lrger, so for ll i we hve M i = mx f (x) mx f (x) = M. x [x i 1,x i ] Now we cn simplify (11.36) further, i=1 x [,b] n 7h 3 n 4 M 7h 3 i 4 M = 7h3 nm. (11.37) 4 i=1 Here, we need one finl little observtion. Recll tht h = (b )/n, so hn = b. If we insert this in (11.37), we obtin our min result. ). Theorem Suppose tht f nd its first two derivtives re continuous on the intervl [,b], nd tht the integrl of f on [,b] is pproximted by the midpoint rule with n subintervls of equl width, I = f (x)dx I mid = n f (x i 1/ )h. i=1 55

30 Then the error is bounded by where x i 1/ = + (i 1/)h. I I mid (b ) 7h 4 mx f (x) (11.38) x [,b] This confirms the error behviour tht we sw in exmple 11.6: If h is reduced by fctor of, the error is reduced by fctor of = 4. One notble omission in our discussion of the midpoint method is round-off error, which ws mjor concern in our study of numericl differentition. The good news is tht round-off error is not usully problem in numericl integrtion. The only sitution where round-off my cuse problems is when the vlue of the integrl is 0. In such sitution we my potentilly dd mny numbers tht sum to 0, nd this my led to cncelltion effects. However, this is so rre tht we will not discuss it here. You should be wre of the fct tht the error estimte (11.38) is not the best possible in tht the constnt 7/4 cn be reduced to 1/4, but then the derivtion becomes much more complicted Estimting the step length The error estimte (11.38) lets us ply stndrd gme: If someone demnds tht we compute n integrl with error smller thn ɛ, we cn find step length h tht gurntees tht we meet this demnd. To mke sure tht the error is smller thn ɛ, we enforce the inequlity which we cn esily solve for h, (b ) 7h 4 mx f (x) ɛ x [,b] 4ɛ h 7(b )M, M = mx f (x). x [,b] This is not quite s simple s it my look since we will hve to estimte M, the mximum vlue of the second derivtive. This cn be difficult, but in some cses it is certinly possible, see exercise A detiled lgorithm Algorithm 11.5 describes the midpoint method, but lcks lot of detil. In this section we give more detiled lgorithm. 56

31 Whenever we compute quntity numericlly, we should try nd estimte the error, otherwise we hve no ide of the qulity of our computtion. A stndrd wy to do this for numericl integrtion is to compute the integrl for decresing step lengths, nd stop the computtions when difference between two successive pproximtions is less thn the tolernce. More precisely, we choose n initil step length h 0 nd compute the pproximtions I mid (h 0 ), I mid (h 1 ),..., I mid (h k ),..., where h k = h 0 / k. Suppose I mid (h k ) is our ltest pproximtion. Then we estimte the reltive error by the number I mid (h k ) I mid (h k 1 I mid (h k ) nd stop the computtions if this is smller thn ɛ. To void potentil division by zero, we use the test I mid (h k ) I mid (h k 1 ) ɛ I mid (h k ). As lwys, we should lso limit the number of pproximtions tht re computed. Algorithm Suppose the function f, the intervl [,b], the length n 0 of the intitil prtition, positive tolernce ɛ < 1, nd the mximum number of itertions M re given. The following lgorithm will compute sequence of pproximtions to f (x)dx by the midpoint rule, until the estimted reltive error is smller thn ɛ, or the mximum number of computed pproximtions rech M. The finl pproximtion is stored in I. n := n 0 ; h := (b )/n; I := 0; x := + h/; for k := 1,,..., n I := I + f (x); x := x + h; j := 1; I := h I ; bser r := I ; while j < M nd bser r > ɛ I j := j + 1; I p := I ; 57

32 x 0 x 1 () x 0 x 1 x x 3 x 4 x 5 (b) Figure The trpezoid rule with one subintervl () nd five subintervls (b). n := n; h := (b )/n; I := 0; x := + h/; for k := 1,,..., n I := I + f (x); x := x + h; I := h I ; bser r := I I p ; Note tht we compute the first pproximtion outside the min loop. This is necessry in order to hve meningful estimtes of the reltive error (the first time we rech the top of the while loop we will lwys get pst the condition). We store the previous pproximtion in I p so tht we cn estimte the error. In the coming sections we will describe two other methods for numericl integrtion. These cn be implemented in lgorithms similr to Algorithm In fct, the only difference will be how the ctul pproximtion to the integrl is computed The trpezoid rule The midpoint method is bsed on very simple polynomil pproximtion to the function f to be integrted on ech subintervl; we simply use constnt pproximtion by interpolting the function vlue t the middle point. We re now going to consider nturl lterntive; we pproximte f on ech subintervl with the secnt tht interpoltes f t both ends of the subintervl. The sitution is shown in figure The pproximtion to the integrl is 58

33 the re of the trpezoidl figure under the secnt so we hve f (x)dx f () + f (b) (b ). (11.39) To get good ccurcy, we will hve to split [,b] into subintervls with prtition nd use this pproximtion on ech subintervl, see figure 11.8b. If we hve uniform prtition {x i } n with step length h, we get the pproximtion i=0 f (x)dx = n i=1 xi x i 1 f (x)dx n f (x i 1 ) + f (x i ) h. (11.40) We should lwys im to mke our computtionl methods s efficient s possible, nd in this cse n improvement is possible. Note tht on the intervl [x i 1, x i ] we use the function vlues f (x i 1 ) nd f (x i ), nd on the next intervl we use the vlues f (x i ) nd f (x i+1 ). All function vlues, except the first nd lst, therefore occur twice in the sum on the right in (11.40). This mens tht if we implement this formul directly we do lot of unnecessry work. From the explntion bove the following observtion follows. i=1 Observtion (Trpezoid rule). Suppose we hve function f defined on n intervl [,b] nd prtition {x i } n of [.b]. If we pproximte f by its i=0 secnt on ech subintervl nd pproximte the integrl of f by the integrl of the resulting piecewise liner pproximtion, we obtin the pproximtion ( f () + f (b) f (x)dx h n 1 + i=1 ) f (x i ). (11.41) In the formul (11.41) there re no redundnt function evlutions Locl error nlysis Our next step is to nlyse the error in the trpezoid method. We follow the sme recipe s for the midpoint method nd use Tylor series. Becuse of the similrities with the midpoint method, we will skip some of the detils. We first study the error in the pproximtion (11.39) where we only hve one secnt. In this cse the error is given by f () + f (b) f (x)dx (b ), (11.4) 59

34 nd the first step is to expnd the function vlues f (x) nd f (b) in Tylor series bout, f (x) = f () + (x )f (x ) () + f (ξ 1 ), f (b) = f () + (b )f (b ) () + f (ξ ), where ξ 1 (, x) nd ξ (,b). The integrtion of the Tylor series for f (x) we did in (11.3) so we just quote the result here, (b ) f (x)dx = f ()(b ) + f () + 1 If we insert the Tylor series for f (b) we obtin (x ) f (ξ 1 )dx. f () + f (b) (b ) (b ) = f ()(b ) + f (b )3 () + f (ξ ). 4 If we insert these expressions into the error (11.4), the first two terms cncel ginst ech other, nd we obtin f () + f (b) f (x)dx (b ) 1 (x ) f (b )3 (ξ 1 )dx f (ξ ) 4 These expressions cn be simplified just like in (11.34) nd (11.35), nd this yields f () + f (b) f (x)dx (b ) M 6 (b )3 + M 4 (b )3. Let us sum this up in lemm. Lemm Let f be continuous function whose first two derivtives re continuous on the intervl [.b]. The the error in the trpezoid rule, with only one line segment on [,b], is bounded by where M = mx x [,b] f (x). f () + f (b) f (x)dx (b ) 5M 1 (b )3, This lemm is completely nlogous to lemm 11.7 which describes the locl error in the midpoint method. We prticulrly notice tht even though the trpezoid rule uses two vlues of f, the error estimte is slightly lrger thn the 60

35 estimte for the midpoint method. The most importnt feture is the exponent on (b ), which tells us how quickly the error goes to 0 when the intervl width is reduced, nd from this point of view the two methods re the sme. In other words, we hve gined nothing by pproximting f by liner functions insted of constnt. This does not men tht the trpezoid rule is bd, it rther mens tht the midpoint rule is unusully good Globl error We cn find n expression for the globl error in the trpezoid rule in exctly the sme wy s we did for the midpoint rule, so we skip the proof. We sum everything up in theorem bout the trpezoid rule. Theorem Suppose tht f nd its first two derivtives re continuous on the intervl [,b], nd tht the integrl of f on [,b] is pproximted by the trpezoid rule with n subintervls of equl width h, ( f () + f (b) I = f (x)dx I tr p = h Then the error is bounded by n 1 + i=1 ) f (x i ). I Itr p 5h (b ) 1 mx f (x). (11.43) x [,b] As we mentioned when we commented on the midpoint rule, the error estimtes tht we obtin re not best possible in the sense tht it is possible to derive better error estimtes (using other techniques) with smller constnts. In the cse of the trpezoid rule, the constnt cn be reduced from 5/1 to 1/1. However, the fct remins tht the trpezoid rule is disppointing method compred to the midpoint rule Simpson s rule The finl method for numericl integrtion tht we consider is Simpson s rule. This method is bsed on pproximting f by prbol on ech subintervl, which mkes the derivtion bit more involved. The error nlysis is essentilly the sme s before, but becuse the expressions re more complicted, it pys off to pln the nlysis better. You my therefore find the mteril in this section more chllenging thn the tretment of the other two methods, nd should 61

36 mke sure tht you hve good understnding of the error nlysis for these methods before you strt studying section Deriving Simpson s rule As for the other methods, we derive Simpson s rule in the simplest cse where we use one prbol on ll of [,b]. We find the polynomil p tht interpoltes f t, 1/ = ( + b)/ nd b, nd pproximte the integrl of f by the integrl of p. We could find p vi the Newton form, but in this cse it is esier to use the Lgrnge form. Another simiplifiction is to first contstruct Simpson s rule in the cse where = 1, 1/ = 0, nd b = 1, nd then use this to generlise the method. The Lgrnge form of the polyomil tht interpoltes f t 1, 0, 1, is given by x(x 1) (x + 1)x p (x) = f ( 1) f (0)(x + 1)(x 1) + f (1), nd it is esy to check tht the interpoltion conditions hold. To integrte p, we must integrte ech of the three polynomils in this expression. For the first one we hve 1 1 Similrly, we find 1 x(x 1)dx = (x x)dx = 1 (x + 1)(x 1)dx = 4 3, [ 1 3 x3 1 x] 1 (x + 1)x dx = = 1 3. On the intervl [ 1,1], Simpson s rule therefore gives the pproximtion 1 1 f (x)dx 1 3( f ( 1) + 4f (0) + f (1) ). (11.44) To obtin n pproximtion on the intervl [,b], we use stndrd technique. Suppose tht x nd y re relted by x = (b ) y (11.45) We see tht if y vries in the intervl [ 1,1], then x will vry in the intervl [,b]. We re going to use the reltion (11.45) s substitution in n integrl, so we note tht d x = (b )d y/. We therefore hve 1 ( ) b f (x)dx = f (y + 1) + d y = b 1 f (y)d y, (11.46) 1 1 6

Polynomial Functions. Polynomial functions in one variable can be written in expanded form as ( )

Polynomial Functions. Polynomial functions in one variable can be written in expanded form as ( ) Polynomil Functions Polynomil functions in one vrible cn be written in expnded form s n n 1 n 2 2 f x = x + x + x + + x + x+ n n 1 n 2 2 1 0 Exmples of polynomils in expnded form re nd 3 8 7 4 = 5 4 +

More information

Babylonian Method of Computing the Square Root: Justifications Based on Fuzzy Techniques and on Computational Complexity

Babylonian Method of Computing the Square Root: Justifications Based on Fuzzy Techniques and on Computational Complexity Bbylonin Method of Computing the Squre Root: Justifictions Bsed on Fuzzy Techniques nd on Computtionl Complexity Olg Koshelev Deprtment of Mthemtics Eduction University of Texs t El Pso 500 W. University

More information

Example A rectangular box without lid is to be made from a square cardboard of sides 18 cm by cutting equal squares from each corner and then folding

Example A rectangular box without lid is to be made from a square cardboard of sides 18 cm by cutting equal squares from each corner and then folding 1 Exmple A rectngulr box without lid is to be mde from squre crdbord of sides 18 cm by cutting equl squres from ech corner nd then folding up the sides. 1 Exmple A rectngulr box without lid is to be mde

More information

Algebra Review. How well do you remember your algebra?

Algebra Review. How well do you remember your algebra? Algebr Review How well do you remember your lgebr? 1 The Order of Opertions Wht do we men when we write + 4? If we multiply we get 6 nd dding 4 gives 10. But, if we dd + 4 = 7 first, then multiply by then

More information

Integration by Substitution

Integration by Substitution Integrtion by Substitution Dr. Philippe B. Lvl Kennesw Stte University August, 8 Abstrct This hndout contins mteril on very importnt integrtion method clled integrtion by substitution. Substitution is

More information

4.11 Inner Product Spaces

4.11 Inner Product Spaces 314 CHAPTER 4 Vector Spces 9. A mtrix of the form 0 0 b c 0 d 0 0 e 0 f g 0 h 0 cnnot be invertible. 10. A mtrix of the form bc d e f ghi such tht e bd = 0 cnnot be invertible. 4.11 Inner Product Spces

More information

6.2 Volumes of Revolution: The Disk Method

6.2 Volumes of Revolution: The Disk Method mth ppliction: volumes of revolution, prt ii Volumes of Revolution: The Disk Method One of the simplest pplictions of integrtion (Theorem ) nd the ccumultion process is to determine so-clled volumes of

More information

Factoring Polynomials

Factoring Polynomials Fctoring Polynomils Some definitions (not necessrily ll for secondry school mthemtics): A polynomil is the sum of one or more terms, in which ech term consists of product of constnt nd one or more vribles

More information

PROF. BOYAN KOSTADINOV NEW YORK CITY COLLEGE OF TECHNOLOGY, CUNY

PROF. BOYAN KOSTADINOV NEW YORK CITY COLLEGE OF TECHNOLOGY, CUNY MAT 0630 INTERNET RESOURCES, REVIEW OF CONCEPTS AND COMMON MISTAKES PROF. BOYAN KOSTADINOV NEW YORK CITY COLLEGE OF TECHNOLOGY, CUNY Contents 1. ACT Compss Prctice Tests 1 2. Common Mistkes 2 3. Distributive

More information

Integration. 148 Chapter 7 Integration

Integration. 148 Chapter 7 Integration 48 Chpter 7 Integrtion 7 Integrtion t ech, by supposing tht during ech tenth of second the object is going t constnt speed Since the object initilly hs speed, we gin suppose it mintins this speed, but

More information

5.2. LINE INTEGRALS 265. Let us quickly review the kind of integrals we have studied so far before we introduce a new one.

5.2. LINE INTEGRALS 265. Let us quickly review the kind of integrals we have studied so far before we introduce a new one. 5.2. LINE INTEGRALS 265 5.2 Line Integrls 5.2.1 Introduction Let us quickly review the kind of integrls we hve studied so fr before we introduce new one. 1. Definite integrl. Given continuous rel-vlued

More information

Reasoning to Solve Equations and Inequalities

Reasoning to Solve Equations and Inequalities Lesson4 Resoning to Solve Equtions nd Inequlities In erlier work in this unit, you modeled situtions with severl vriles nd equtions. For exmple, suppose you were given usiness plns for concert showing

More information

Graphs on Logarithmic and Semilogarithmic Paper

Graphs on Logarithmic and Semilogarithmic Paper 0CH_PHClter_TMSETE_ 3//00 :3 PM Pge Grphs on Logrithmic nd Semilogrithmic Pper OBJECTIVES When ou hve completed this chpter, ou should be ble to: Mke grphs on logrithmic nd semilogrithmic pper. Grph empiricl

More information

Example 27.1 Draw a Venn diagram to show the relationship between counting numbers, whole numbers, integers, and rational numbers.

Example 27.1 Draw a Venn diagram to show the relationship between counting numbers, whole numbers, integers, and rational numbers. 2 Rtionl Numbers Integers such s 5 were importnt when solving the eqution x+5 = 0. In similr wy, frctions re importnt for solving equtions like 2x = 1. Wht bout equtions like 2x + 1 = 0? Equtions of this

More information

2005-06 Second Term MAT2060B 1. Supplementary Notes 3 Interchange of Differentiation and Integration

2005-06 Second Term MAT2060B 1. Supplementary Notes 3 Interchange of Differentiation and Integration Source: http://www.mth.cuhk.edu.hk/~mt26/mt26b/notes/notes3.pdf 25-6 Second Term MAT26B 1 Supplementry Notes 3 Interchnge of Differentition nd Integrtion The theme of this course is bout vrious limiting

More information

Operations with Polynomials

Operations with Polynomials 38 Chpter P Prerequisites P.4 Opertions with Polynomils Wht you should lern: Write polynomils in stndrd form nd identify the leding coefficients nd degrees of polynomils Add nd subtrct polynomils Multiply

More information

Math 135 Circles and Completing the Square Examples

Math 135 Circles and Completing the Square Examples Mth 135 Circles nd Completing the Squre Exmples A perfect squre is number such tht = b 2 for some rel number b. Some exmples of perfect squres re 4 = 2 2, 16 = 4 2, 169 = 13 2. We wish to hve method for

More information

Basic Analysis of Autarky and Free Trade Models

Basic Analysis of Autarky and Free Trade Models Bsic Anlysis of Autrky nd Free Trde Models AUTARKY Autrky condition in prticulr commodity mrket refers to sitution in which country does not engge in ny trde in tht commodity with other countries. Consequently

More information

Lecture 3 Gaussian Probability Distribution

Lecture 3 Gaussian Probability Distribution Lecture 3 Gussin Probbility Distribution Introduction l Gussin probbility distribution is perhps the most used distribution in ll of science. u lso clled bell shped curve or norml distribution l Unlike

More information

LINEAR TRANSFORMATIONS AND THEIR REPRESENTING MATRICES

LINEAR TRANSFORMATIONS AND THEIR REPRESENTING MATRICES LINEAR TRANSFORMATIONS AND THEIR REPRESENTING MATRICES DAVID WEBB CONTENTS Liner trnsformtions 2 The representing mtrix of liner trnsformtion 3 3 An ppliction: reflections in the plne 6 4 The lgebr of

More information

All pay auctions with certain and uncertain prizes a comment

All pay auctions with certain and uncertain prizes a comment CENTER FOR RESEARC IN ECONOMICS AND MANAGEMENT CREAM Publiction No. 1-2015 All py uctions with certin nd uncertin prizes comment Christin Riis All py uctions with certin nd uncertin prizes comment Christin

More information

SPECIAL PRODUCTS AND FACTORIZATION

SPECIAL PRODUCTS AND FACTORIZATION MODULE - Specil Products nd Fctoriztion 4 SPECIAL PRODUCTS AND FACTORIZATION In n erlier lesson you hve lernt multipliction of lgebric epressions, prticulrly polynomils. In the study of lgebr, we come

More information

Mathematics. Vectors. hsn.uk.net. Higher. Contents. Vectors 128 HSN23100

Mathematics. Vectors. hsn.uk.net. Higher. Contents. Vectors 128 HSN23100 hsn.uk.net Higher Mthemtics UNIT 3 OUTCOME 1 Vectors Contents Vectors 18 1 Vectors nd Sclrs 18 Components 18 3 Mgnitude 130 4 Equl Vectors 131 5 Addition nd Subtrction of Vectors 13 6 Multipliction by

More information

Section 5-4 Trigonometric Functions

Section 5-4 Trigonometric Functions 5- Trigonometric Functions Section 5- Trigonometric Functions Definition of the Trigonometric Functions Clcultor Evlution of Trigonometric Functions Definition of the Trigonometric Functions Alternte Form

More information

9 CONTINUOUS DISTRIBUTIONS

9 CONTINUOUS DISTRIBUTIONS 9 CONTINUOUS DISTIBUTIONS A rndom vrible whose vlue my fll nywhere in rnge of vlues is continuous rndom vrible nd will be ssocited with some continuous distribution. Continuous distributions re to discrete

More information

Use Geometry Expressions to create a more complex locus of points. Find evidence for equivalence using Geometry Expressions.

Use Geometry Expressions to create a more complex locus of points. Find evidence for equivalence using Geometry Expressions. Lerning Objectives Loci nd Conics Lesson 3: The Ellipse Level: Preclculus Time required: 120 minutes In this lesson, students will generlize their knowledge of the circle to the ellipse. The prmetric nd

More information

g(y(a), y(b)) = o, B a y(a)+b b y(b)=c, Boundary Value Problems Lecture Notes to Accompany

g(y(a), y(b)) = o, B a y(a)+b b y(b)=c, Boundary Value Problems Lecture Notes to Accompany Lecture Notes to Accompny Scientific Computing An Introductory Survey Second Edition by Michel T Heth Boundry Vlue Problems Side conditions prescribing solution or derivtive vlues t specified points required

More information

The Definite Integral

The Definite Integral Chpter 4 The Definite Integrl 4. Determining distnce trveled from velocity Motivting Questions In this section, we strive to understnd the ides generted by the following importnt questions: If we know

More information

Appendix D: Completing the Square and the Quadratic Formula. In Appendix A, two special cases of expanding brackets were considered:

Appendix D: Completing the Square and the Quadratic Formula. In Appendix A, two special cases of expanding brackets were considered: Appendi D: Completing the Squre nd the Qudrtic Formul Fctoring qudrtic epressions such s: + 6 + 8 ws one of the topics introduced in Appendi C. Fctoring qudrtic epressions is useful skill tht cn help you

More information

www.mathsbox.org.uk e.g. f(x) = x domain x 0 (cannot find the square root of negative values)

www.mathsbox.org.uk e.g. f(x) = x domain x 0 (cannot find the square root of negative values) www.mthsbo.org.uk CORE SUMMARY NOTES Functions A function is rule which genertes ectl ONE OUTPUT for EVERY INPUT. To be defined full the function hs RULE tells ou how to clculte the output from the input

More information

AREA OF A SURFACE OF REVOLUTION

AREA OF A SURFACE OF REVOLUTION AREA OF A SURFACE OF REVOLUTION h cut r πr h A surfce of revolution is formed when curve is rotted bout line. Such surfce is the lterl boundr of solid of revolution of the tpe discussed in Sections 7.

More information

Review Problems for the Final of Math 121, Fall 2014

Review Problems for the Final of Math 121, Fall 2014 Review Problems for the Finl of Mth, Fll The following is collection of vrious types of smple problems covering sections.,.5, nd.7 6.6 of the text which constitute only prt of the common Mth Finl. Since

More information

MODULE 3. 0, y = 0 for all y

MODULE 3. 0, y = 0 for all y Topics: Inner products MOULE 3 The inner product of two vectors: The inner product of two vectors x, y V, denoted by x, y is (in generl) complex vlued function which hs the following four properties: i)

More information

Review guide for the final exam in Math 233

Review guide for the final exam in Math 233 Review guide for the finl exm in Mth 33 1 Bsic mteril. This review includes the reminder of the mteril for mth 33. The finl exm will be cumultive exm with mny of the problems coming from the mteril covered

More information

Experiment 6: Friction

Experiment 6: Friction Experiment 6: Friction In previous lbs we studied Newton s lws in n idel setting, tht is, one where friction nd ir resistnce were ignored. However, from our everydy experience with motion, we know tht

More information

Regular Sets and Expressions

Regular Sets and Expressions Regulr Sets nd Expressions Finite utomt re importnt in science, mthemtics, nd engineering. Engineers like them ecuse they re super models for circuits (And, since the dvent of VLSI systems sometimes finite

More information

Lecture 5. Inner Product

Lecture 5. Inner Product Lecture 5 Inner Product Let us strt with the following problem. Given point P R nd line L R, how cn we find the point on the line closest to P? Answer: Drw line segment from P meeting the line in right

More information

Ostrowski Type Inequalities and Applications in Numerical Integration. Edited By: Sever S. Dragomir. and. Themistocles M. Rassias

Ostrowski Type Inequalities and Applications in Numerical Integration. Edited By: Sever S. Dragomir. and. Themistocles M. Rassias Ostrowski Type Inequlities nd Applictions in Numericl Integrtion Edited By: Sever S Drgomir nd Themistocles M Rssis SS Drgomir) School nd Communictions nd Informtics, Victori University of Technology,

More information

Applications to Physics and Engineering

Applications to Physics and Engineering Section 7.5 Applictions to Physics nd Engineering Applictions to Physics nd Engineering Work The term work is used in everydy lnguge to men the totl mount of effort required to perform tsk. In physics

More information

Econ 4721 Money and Banking Problem Set 2 Answer Key

Econ 4721 Money and Banking Problem Set 2 Answer Key Econ 472 Money nd Bnking Problem Set 2 Answer Key Problem (35 points) Consider n overlpping genertions model in which consumers live for two periods. The number of people born in ech genertion grows in

More information

EQUATIONS OF LINES AND PLANES

EQUATIONS OF LINES AND PLANES EQUATIONS OF LINES AND PLANES MATH 195, SECTION 59 (VIPUL NAIK) Corresponding mteril in the ook: Section 12.5. Wht students should definitely get: Prmetric eqution of line given in point-direction nd twopoint

More information

and thus, they are similar. If k = 3 then the Jordan form of both matrices is

and thus, they are similar. If k = 3 then the Jordan form of both matrices is Homework ssignment 11 Section 7. pp. 249-25 Exercise 1. Let N 1 nd N 2 be nilpotent mtrices over the field F. Prove tht N 1 nd N 2 re similr if nd only if they hve the sme miniml polynomil. Solution: If

More information

Physics 43 Homework Set 9 Chapter 40 Key

Physics 43 Homework Set 9 Chapter 40 Key Physics 43 Homework Set 9 Chpter 4 Key. The wve function for n electron tht is confined to x nm is. Find the normliztion constnt. b. Wht is the probbility of finding the electron in. nm-wide region t x

More information

Week 7 - Perfect Competition and Monopoly

Week 7 - Perfect Competition and Monopoly Week 7 - Perfect Competition nd Monopoly Our im here is to compre the industry-wide response to chnges in demnd nd costs by monopolized industry nd by perfectly competitive one. We distinguish between

More information

Section 7-4 Translation of Axes

Section 7-4 Translation of Axes 62 7 ADDITIONAL TOPICS IN ANALYTIC GEOMETRY Section 7-4 Trnsltion of Aes Trnsltion of Aes Stndrd Equtions of Trnslted Conics Grphing Equtions of the Form A 2 C 2 D E F 0 Finding Equtions of Conics In the

More information

Vectors 2. 1. Recap of vectors

Vectors 2. 1. Recap of vectors Vectors 2. Recp of vectors Vectors re directed line segments - they cn be represented in component form or by direction nd mgnitude. We cn use trigonometry nd Pythgors theorem to switch between the forms

More information

Numerical Methods of Approximating Definite Integrals

Numerical Methods of Approximating Definite Integrals 6 C H A P T E R Numericl Methods o Approimting Deinite Integrls 6. APPROXIMATING SUMS: L n, R n, T n, AND M n Introduction Not only cn we dierentite ll the bsic unctions we ve encountered, polynomils,

More information

9.3. The Scalar Product. Introduction. Prerequisites. Learning Outcomes

9.3. The Scalar Product. Introduction. Prerequisites. Learning Outcomes The Sclr Product 9.3 Introduction There re two kinds of multipliction involving vectors. The first is known s the sclr product or dot product. This is so-clled becuse when the sclr product of two vectors

More information

FUNCTIONS AND EQUATIONS. xεs. The simplest way to represent a set is by listing its members. We use the notation

FUNCTIONS AND EQUATIONS. xεs. The simplest way to represent a set is by listing its members. We use the notation FUNCTIONS AND EQUATIONS. SETS AND SUBSETS.. Definition of set. A set is ny collection of objects which re clled its elements. If x is n element of the set S, we sy tht x belongs to S nd write If y does

More information

Or more simply put, when adding or subtracting quantities, their uncertainties add.

Or more simply put, when adding or subtracting quantities, their uncertainties add. Propgtion of Uncertint through Mthemticl Opertions Since the untit of interest in n eperiment is rrel otined mesuring tht untit directl, we must understnd how error propgtes when mthemticl opertions re

More information

4 Approximations. 4.1 Background. D. Levy

4 Approximations. 4.1 Background. D. Levy D. Levy 4 Approximtions 4.1 Bckground In this chpter we re interested in pproximtion problems. Generlly speking, strting from function f(x) we would like to find different function g(x) tht belongs to

More information

Binary Representation of Numbers Autar Kaw

Binary Representation of Numbers Autar Kaw Binry Representtion of Numbers Autr Kw After reding this chpter, you should be ble to: 1. convert bse- rel number to its binry representtion,. convert binry number to n equivlent bse- number. In everydy

More information

1. Find the zeros Find roots. Set function = 0, factor or use quadratic equation if quadratic, graph to find zeros on calculator

1. Find the zeros Find roots. Set function = 0, factor or use quadratic equation if quadratic, graph to find zeros on calculator AP Clculus Finl Review Sheet When you see the words. This is wht you think of doing. Find the zeros Find roots. Set function =, fctor or use qudrtic eqution if qudrtic, grph to find zeros on clcultor.

More information

The Velocity Factor of an Insulated Two-Wire Transmission Line

The Velocity Factor of an Insulated Two-Wire Transmission Line The Velocity Fctor of n Insulted Two-Wire Trnsmission Line Problem Kirk T. McDonld Joseph Henry Lbortories, Princeton University, Princeton, NJ 08544 Mrch 7, 008 Estimte the velocity fctor F = v/c nd the

More information

Module 2. Analysis of Statically Indeterminate Structures by the Matrix Force Method. Version 2 CE IIT, Kharagpur

Module 2. Analysis of Statically Indeterminate Structures by the Matrix Force Method. Version 2 CE IIT, Kharagpur Module Anlysis of Stticlly Indeterminte Structures by the Mtrix Force Method Version CE IIT, Khrgpur esson 9 The Force Method of Anlysis: Bems (Continued) Version CE IIT, Khrgpur Instructionl Objectives

More information

QUADRATURE METHODS. July 19, 2011. Kenneth L. Judd. Hoover Institution

QUADRATURE METHODS. July 19, 2011. Kenneth L. Judd. Hoover Institution QUADRATURE METHODS Kenneth L. Judd Hoover Institution July 19, 2011 1 Integrtion Most integrls cnnot be evluted nlyticlly Integrls frequently rise in economics Expected utility Discounted utility nd profits

More information

Distributions. (corresponding to the cumulative distribution function for the discrete case).

Distributions. (corresponding to the cumulative distribution function for the discrete case). Distributions Recll tht n integrble function f : R [,] such tht R f()d = is clled probbility density function (pdf). The distribution function for the pdf is given by F() = (corresponding to the cumultive

More information

Module Summary Sheets. C3, Methods for Advanced Mathematics (Version B reference to new book) Topic 2: Natural Logarithms and Exponentials

Module Summary Sheets. C3, Methods for Advanced Mathematics (Version B reference to new book) Topic 2: Natural Logarithms and Exponentials MEI Mthemtics in Ection nd Instry Topic : Proof MEI Structured Mthemtics Mole Summry Sheets C, Methods for Anced Mthemtics (Version B reference to new book) Topic : Nturl Logrithms nd Eponentils Topic

More information

Introduction to Integration Part 2: The Definite Integral

Introduction to Integration Part 2: The Definite Integral Mthemtics Lerning Centre Introduction to Integrtion Prt : The Definite Integrl Mr Brnes c 999 Universit of Sdne Contents Introduction. Objectives...... Finding Ares 3 Ares Under Curves 4 3. Wht is the

More information

Treatment Spring Late Summer Fall 0.10 5.56 3.85 0.61 6.97 3.01 1.91 3.01 2.13 2.99 5.33 2.50 1.06 3.53 6.10 Mean = 1.33 Mean = 4.88 Mean = 3.

Treatment Spring Late Summer Fall 0.10 5.56 3.85 0.61 6.97 3.01 1.91 3.01 2.13 2.99 5.33 2.50 1.06 3.53 6.10 Mean = 1.33 Mean = 4.88 Mean = 3. The nlysis of vrince (ANOVA) Although the t-test is one of the most commonly used sttisticl hypothesis tests, it hs limittions. The mjor limittion is tht the t-test cn be used to compre the mens of only

More information

Chapter 2 The Number System (Integers and Rational Numbers)

Chapter 2 The Number System (Integers and Rational Numbers) Chpter 2 The Number System (Integers nd Rtionl Numbers) In this second chpter, students extend nd formlize their understnding of the number system, including negtive rtionl numbers. Students first develop

More information

A.7.1 Trigonometric interpretation of dot product... 324. A.7.2 Geometric interpretation of dot product... 324

A.7.1 Trigonometric interpretation of dot product... 324. A.7.2 Geometric interpretation of dot product... 324 A P P E N D I X A Vectors CONTENTS A.1 Scling vector................................................ 321 A.2 Unit or Direction vectors...................................... 321 A.3 Vector ddition.................................................

More information

MATH 150 HOMEWORK 4 SOLUTIONS

MATH 150 HOMEWORK 4 SOLUTIONS MATH 150 HOMEWORK 4 SOLUTIONS Section 1.8 Show tht the product of two of the numbers 65 1000 8 2001 + 3 177, 79 1212 9 2399 + 2 2001, nd 24 4493 5 8192 + 7 1777 is nonnegtive. Is your proof constructive

More information

addition, there are double entries for the symbols used to signify different parameters. These parameters are explained in this appendix.

addition, there are double entries for the symbols used to signify different parameters. These parameters are explained in this appendix. APPENDIX A: The ellipse August 15, 1997 Becuse of its importnce in both pproximting the erth s shpe nd describing stellite orbits, n informl discussion of the ellipse is presented in this ppendix. The

More information

COMPARISON OF SOME METHODS TO FIT A MULTIPLICATIVE TARIFF STRUCTURE TO OBSERVED RISK DATA BY B. AJNE. Skandza, Stockholm ABSTRACT

COMPARISON OF SOME METHODS TO FIT A MULTIPLICATIVE TARIFF STRUCTURE TO OBSERVED RISK DATA BY B. AJNE. Skandza, Stockholm ABSTRACT COMPARISON OF SOME METHODS TO FIT A MULTIPLICATIVE TARIFF STRUCTURE TO OBSERVED RISK DATA BY B. AJNE Skndz, Stockholm ABSTRACT Three methods for fitting multiplictive models to observed, cross-clssified

More information

Physics 6010, Fall 2010 Symmetries and Conservation Laws: Energy, Momentum and Angular Momentum Relevant Sections in Text: 2.6, 2.

Physics 6010, Fall 2010 Symmetries and Conservation Laws: Energy, Momentum and Angular Momentum Relevant Sections in Text: 2.6, 2. Physics 6010, Fll 2010 Symmetries nd Conservtion Lws: Energy, Momentum nd Angulr Momentum Relevnt Sections in Text: 2.6, 2.7 Symmetries nd Conservtion Lws By conservtion lw we men quntity constructed from

More information

The Riemann Integral. Chapter 1

The Riemann Integral. Chapter 1 Chpter The Riemnn Integrl now of some universities in Englnd where the Lebesgue integrl is tught in the first yer of mthemtics degree insted of the Riemnn integrl, but now of no universities in Englnd

More information

Euler Euler Everywhere Using the Euler-Lagrange Equation to Solve Calculus of Variation Problems

Euler Euler Everywhere Using the Euler-Lagrange Equation to Solve Calculus of Variation Problems Euler Euler Everywhere Using the Euler-Lgrnge Eqution to Solve Clculus of Vrition Problems Jenine Smllwood Principles of Anlysis Professor Flschk My 12, 1998 1 1. Introduction Clculus of vritions is brnch

More information

RIGHT TRIANGLES AND THE PYTHAGOREAN TRIPLETS

RIGHT TRIANGLES AND THE PYTHAGOREAN TRIPLETS RIGHT TRIANGLES AND THE PYTHAGOREAN TRIPLETS Known for over 500 yers is the fct tht the sum of the squres of the legs of right tringle equls the squre of the hypotenuse. Tht is +b c. A simple proof is

More information

Homework 3 Solutions

Homework 3 Solutions CS 341: Foundtions of Computer Science II Prof. Mrvin Nkym Homework 3 Solutions 1. Give NFAs with the specified numer of sttes recognizing ech of the following lnguges. In ll cses, the lphet is Σ = {,1}.

More information

Bayesian Updating with Continuous Priors Class 13, 18.05, Spring 2014 Jeremy Orloff and Jonathan Bloom

Bayesian Updating with Continuous Priors Class 13, 18.05, Spring 2014 Jeremy Orloff and Jonathan Bloom Byesin Updting with Continuous Priors Clss 3, 8.05, Spring 04 Jeremy Orloff nd Jonthn Bloom Lerning Gols. Understnd prmeterized fmily of distriutions s representing continuous rnge of hypotheses for the

More information

6 Energy Methods And The Energy of Waves MATH 22C

6 Energy Methods And The Energy of Waves MATH 22C 6 Energy Methods And The Energy of Wves MATH 22C. Conservtion of Energy We discuss the principle of conservtion of energy for ODE s, derive the energy ssocited with the hrmonic oscilltor, nd then use this

More information

Thinking out of the Box... Problem It s a richer problem than we ever imagined

Thinking out of the Box... Problem It s a richer problem than we ever imagined From the Mthemtics Techer, Vol. 95, No. 8, pges 568-574 Wlter Dodge (not pictured) nd Steve Viktor Thinking out of the Bo... Problem It s richer problem thn we ever imgined The bo problem hs been stndrd

More information

Radius of the Earth - Radii Used in Geodesy James R. Clynch February 2006

Radius of the Earth - Radii Used in Geodesy James R. Clynch February 2006 dius of the Erth - dii Used in Geodesy Jmes. Clynch Februry 006 I. Erth dii Uses There is only one rdius of sphere. The erth is pproximtely sphere nd therefore, for some cses, this pproximtion is dequte.

More information

Object Semantics. 6.170 Lecture 2

Object Semantics. 6.170 Lecture 2 Object Semntics 6.170 Lecture 2 The objectives of this lecture re to: to help you become fmilir with the bsic runtime mechnism common to ll object-oriented lnguges (but with prticulr focus on Jv): vribles,

More information

Helicopter Theme and Variations

Helicopter Theme and Variations Helicopter Theme nd Vritions Or, Some Experimentl Designs Employing Pper Helicopters Some possible explntory vribles re: Who drops the helicopter The length of the rotor bldes The height from which the

More information

15.6. The mean value and the root-mean-square value of a function. Introduction. Prerequisites. Learning Outcomes. Learning Style

15.6. The mean value and the root-mean-square value of a function. Introduction. Prerequisites. Learning Outcomes. Learning Style The men vlue nd the root-men-squre vlue of function 5.6 Introduction Currents nd voltges often vry with time nd engineers my wish to know the verge vlue of such current or voltge over some prticulr time

More information

Warm-up for Differential Calculus

Warm-up for Differential Calculus Summer Assignment Wrm-up for Differentil Clculus Who should complete this pcket? Students who hve completed Functions or Honors Functions nd will be tking Differentil Clculus in the fll of 015. Due Dte:

More information

PROBLEMS 13 - APPLICATIONS OF DERIVATIVES Page 1

PROBLEMS 13 - APPLICATIONS OF DERIVATIVES Page 1 PROBLEMS - APPLICATIONS OF DERIVATIVES Pge ( ) Wter seeps out of conicl filter t the constnt rte of 5 cc / sec. When the height of wter level in the cone is 5 cm, find the rte t which the height decreses.

More information

INTERCHANGING TWO LIMITS. Zoran Kadelburg and Milosav M. Marjanović

INTERCHANGING TWO LIMITS. Zoran Kadelburg and Milosav M. Marjanović THE TEACHING OF MATHEMATICS 2005, Vol. VIII, 1, pp. 15 29 INTERCHANGING TWO LIMITS Zorn Kdelburg nd Milosv M. Mrjnović This pper is dedicted to the memory of our illustrious professor of nlysis Slobodn

More information

AA1H Calculus Notes Math1115, Honours 1 1998. John Hutchinson

AA1H Calculus Notes Math1115, Honours 1 1998. John Hutchinson AA1H Clculus Notes Mth1115, Honours 1 1998 John Hutchinson Author ddress: Deprtment of Mthemtics, School of Mthemticl Sciences, Austrlin Ntionl University E-mil ddress: John.Hutchinson@nu.edu.u Contents

More information

Lectures 8 and 9 1 Rectangular waveguides

Lectures 8 and 9 1 Rectangular waveguides 1 Lectures 8 nd 9 1 Rectngulr wveguides y b x z Consider rectngulr wveguide with 0 < x b. There re two types of wves in hollow wveguide with only one conductor; Trnsverse electric wves

More information

Rotating DC Motors Part II

Rotating DC Motors Part II Rotting Motors rt II II.1 Motor Equivlent Circuit The next step in our consiertion of motors is to evelop n equivlent circuit which cn be use to better unerstn motor opertion. The rmtures in rel motors

More information

COMPONENTS: COMBINED LOADING

COMPONENTS: COMBINED LOADING LECTURE COMPONENTS: COMBINED LOADING Third Edition A. J. Clrk School of Engineering Deprtment of Civil nd Environmentl Engineering 24 Chpter 8.4 by Dr. Ibrhim A. Asskkf SPRING 2003 ENES 220 Mechnics of

More information

1.2 The Integers and Rational Numbers

1.2 The Integers and Rational Numbers .2. THE INTEGERS AND RATIONAL NUMBERS.2 The Integers n Rtionl Numers The elements of the set of integers: consist of three types of numers: Z {..., 5, 4, 3, 2,, 0,, 2, 3, 4, 5,...} I. The (positive) nturl

More information

Small Business Networking

Small Business Networking Why network is n essentil productivity tool for ny smll business Effective technology is essentil for smll businesses looking to increse the productivity of their people nd business. Introducing technology

More information

2 DIODE CLIPPING and CLAMPING CIRCUITS

2 DIODE CLIPPING and CLAMPING CIRCUITS 2 DIODE CLIPPING nd CLAMPING CIRCUITS 2.1 Ojectives Understnding the operting principle of diode clipping circuit Understnding the operting principle of clmping circuit Understnding the wveform chnge of

More information

3 The Utility Maximization Problem

3 The Utility Maximization Problem 3 The Utility Mxiiztion Proble We hve now discussed how to describe preferences in ters of utility functions nd how to forulte siple budget sets. The rtionl choice ssuption, tht consuers pick the best

More information

Small Business Networking

Small Business Networking Why network is n essentil productivity tool for ny smll business Effective technology is essentil for smll businesses looking to increse the productivity of their people nd processes. Introducing technology

More information

Small Business Networking

Small Business Networking Why network is n essentil productivity tool for ny smll business Effective technology is essentil for smll businesses looking to increse the productivity of their people nd business. Introducing technology

More information

19. The Fermat-Euler Prime Number Theorem

19. The Fermat-Euler Prime Number Theorem 19. The Fermt-Euler Prime Number Theorem Every prime number of the form 4n 1 cn be written s sum of two squres in only one wy (side from the order of the summnds). This fmous theorem ws discovered bout

More information

Hillsborough Township Public Schools Mathematics Department Computer Programming 1

Hillsborough Township Public Schools Mathematics Department Computer Programming 1 Essentil Unit 1 Introduction to Progrmming Pcing: 15 dys Common Unit Test Wht re the ethicl implictions for ming in tody s world? There re ethicl responsibilities to consider when writing computer s. Citizenship,

More information

10.6 Applications of Quadratic Equations

10.6 Applications of Quadratic Equations 10.6 Applictions of Qudrtic Equtions In this section we wnt to look t the pplictions tht qudrtic equtions nd functions hve in the rel world. There re severl stndrd types: problems where the formul is given,

More information

Online Multicommodity Routing with Time Windows

Online Multicommodity Routing with Time Windows Konrd-Zuse-Zentrum für Informtionstechnik Berlin Tkustrße 7 D-14195 Berlin-Dhlem Germny TOBIAS HARKS 1 STEFAN HEINZ MARC E. PFETSCH TJARK VREDEVELD 2 Online Multicommodity Routing with Time Windows 1 Institute

More information

Small Business Networking

Small Business Networking Why network is n essentil productivity tool for ny smll business Effective technology is essentil for smll businesses looking to increse the productivity of their people nd processes. Introducing technology

More information

Exponential and Logarithmic Functions

Exponential and Logarithmic Functions Nme Chpter Eponentil nd Logrithmic Functions Section. Eponentil Functions nd Their Grphs Objective: In this lesson ou lerned how to recognize, evlute, nd grph eponentil functions. Importnt Vocbulr Define

More information

Small Business Cloud Services

Small Business Cloud Services Smll Business Cloud Services Summry. We re thick in the midst of historic se-chnge in computing. Like the emergence of personl computers, grphicl user interfces, nd mobile devices, the cloud is lredy profoundly

More information

P.3 Polynomials and Factoring. P.3 an 1. Polynomial STUDY TIP. Example 1 Writing Polynomials in Standard Form. What you should learn

P.3 Polynomials and Factoring. P.3 an 1. Polynomial STUDY TIP. Example 1 Writing Polynomials in Standard Form. What you should learn 33337_0P03.qp 2/27/06 24 9:3 AM Chpter P Pge 24 Prerequisites P.3 Polynomils nd Fctoring Wht you should lern Polynomils An lgeric epression is collection of vriles nd rel numers. The most common type of

More information

Small Business Networking

Small Business Networking Why network is n essentil productivity tool for ny smll business Effective technology is essentil for smll businesses looking to increse the productivity of their people nd processes. Introducing technology

More information