# Numerical Integration

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1 Chpter 1 Numericl Integrtion Numericl differentition methods compute pproximtions to the derivtive of function from known vlues of the function. Numericl integrtion uses the sme informtion to compute numericl pproximtions to the integrl of the function. An importnt use of both types of methods is estimtion of derivtives nd integrls for functions tht re only known t isolted points, s is the cse with for exmple mesurement dt. An importnt difference between differentition nd integrtion is tht for most functions it is not possible to determine the integrl vi symbolic methods, but we cn still compute numericl pproximtions to virtully ny definite integrl. Numericl integrtion methods re therefore more useful thn numericl differentition methods, nd re essentil in mny prcticl situtions. We use the sme generl strtegy for deriving numericl integrtion methods s we did for numericl differentition methods: We find the polynomil tht interpoltes the function t some suitble points, nd use the integrl of the polynomil s n pproximtion to the function. This mens tht the trunction error cn be nlysed in bsiclly the sme wy s for numericl differentition. However, when it comes to round-off error, integrtion behves differently from differentition: Numericl integrtion is very insensitive to round-off errors, so we will ignore round-off in our nlysis. The mthemticl definition of the integrl is bsiclly vi numericl integrtion method, nd we therefore strt by reviewing this definition. We then derive the simplest numericl integrtion method, nd see how its error cn be nlysed. We then derive two other methods tht re more ccurte, but for these we just indicte how the error nlysis cn be done. We emphsise tht the generl procedure for deriving both numericl dif- 79

2 Figure 1.1. The re under the grph of function. ferentition nd integrtion methods with error nlyses is the sme with the exception tht round-off errors re not of must interest for the integrtion methods. 1.1 Generl bckground on integrtion Recll tht if f (x) is function, then the integrl of f from x = to x = b is written f (x)d x. The 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 1.1. Before we continue, we need to define term which we will use repetedly in our description of integrtion. Definition 1.1 (Prtition). Let nd b be two rel numbers with < b. A prtition of [,b] is finite sequence {x i } n of incresing numbers in [,b] i= with x = nd x n = b, = x < 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. 8

3 () (b) (c) (d) Figure 1.. The definition of the integrl vi inscribed nd circumsribed step functions. The trditionl definition of the integrl is bsed on numericl pproximtion to the re. We pick prtition {x i } n of [,b], nd in ech subintervl i= [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 cn then compute two obvious pproximtions to the integrl by pproximting f by two different functions which re both ssumed to be constnt on ech intervl [x i 1, x i ]: The first hs the constnt vlue m i nd the other the vlue M i. We then sum up the res under ech of the two step functions n end up with the two pproximtions I = m i (x i x i 1 ), I = M i (x i x i 1 ), (1.1) to the totl re. In generl, the first of these is too smll, the other too lrge. To define the integrl, we consider lrger prtitions (smller step lengths) nd consider the limits of I nd I s the distnce between neighbouring x i s goes 81

4 to zero. If those limits re the sme, we sy tht f is integrble, nd the integrl is given by this limit. Definition 1. (Integrl). Let f be function defined on the intervl [, b], nd let {x i } n i= be prtition of [,b]. Let m i nd M i denote the minimum nd mximum vlues of f over the intervl [x i 1, x i ], respectively, ssuming they exist. Consider the two numbers I nd I defined in (1.1). If sup I nd inf I both exist nd re equl, where the sup nd inf re tken over ll possible prtitions of [,b], the function f is sid to be integrble, nd the integrl of f over [,b] is defined by I = f (x)d x = sup I = inf I. This process is illustrted in figure 1. 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 1., 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 property tht is very useful when it comes to numericl computtion. Theorem 1.3. Suppose tht f is integrble on the intervl [,b], let {x i } n i= be prtition of [,b], nd let t i be number in [x i 1, x i ] for i = 1,..., n. Then the sum Ĩ = f (t i )(x i x i 1 ) (1.) will converge to the integrl when the distnce between ll neighbouring x i s tends to zero. Theorem 1.3 llows us to construct prcticl, numericl methods for computing the integrl. We pick prtition of [,b], choose t i equl to x i 1 or x i, nd compute the sum (1.). It turns out tht n even better choice is the more symmetric t i = (x i + x i 1 )/ which leds to the pproximtion I f ( (x i + x i 1 )/ ) (x i x i 1 ). (1.3) 8

5 This is the so-clled midpoint rule which we will study in the next section. 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 for deriving more dvnced numericl integrtion methods, nd this works s long s f cn be pproximted well by polynomils on ech subintervl. Exercises 1 In this exercise we re going to study the definition of the integrl for the function f (x) = e x on the intervl [,1]. ) Determine lower nd upper sums for uniform prtition consisting of 1 subintervls. b) Determine the bsolute nd reltive errors of the sums in () compred to the exct vlue e 1 = of the integrl. c) Write progrm for clculting the lower nd upper sums in this exmple. How mny subintervls re needed to chieve n bsolute error less thn 3 1 3? 1. The midpoint rule for numericl integrtion We hve lredy introduced the midpoint rule (1.3) 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 1.4. Let f be 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= by f (x)d x I mid (h) = h f (x i 1/ ), (1.4) where x i 1/ = (x i 1 + x i )/ = + (i 1/)h. This my seem like strngely formulted lgorithm, but ll there is to do is to compute the sum on the right in (1.4). The method is illustrted in figure 1.3 in the cses where we hve 1 nd 5 subintervls. 83

6 () (b) Figure 1.3. The midpoint rule with one subintervl () nd five subintervls (b) A detiled lgorithm Algorithm 1.4 describes the midpoint rule, but lcks lot of detil. In this section we give more detiled lgorithm. Whenever we compute quntity numericlly, we should try nd estimte the error, otherwise we hve no ide of the qulity of our computtion. We did this when we discussed lgorithms for finding roots of equtions in chpter 1, nd we cn do exctly the sme here: We compute the integrl for decresing step lengths, nd stop the computtions when the difference between two successive pproximtions is less thn the tolernce. More precisely, we choose n initil step length h nd compute the pproximtions I mid (h ), I mid (h 1 ),..., I mid (h k ),..., where h k = h / 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, so we count the number of times we divide the subintervls, nd stop when we rech predefined limit which we cll M. 84

7 Algorithm 1.5. Suppose the function f, the intervl [,b], the length n 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)d x 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 ; h := (b )/n; I := ; 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 ; n := n; h := (b )/n; I := ; 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 two times we rech the while loop (the first time we rech the while loop we will lwys get pst the condition). We store the previous pproximtion in I p nd use this to estimte the error in the next itertion. In the coming sections we will describe two other methods for numericl integrtion. These cn be implemented in lgorithms similr to Algorithm 1.5. In fct, the only difference will be how the ctul pproximtion to the integrl is computed. Exmple 1.6. Let us try the midpoint rule 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 85

8 of the method. We choose the integrl cos x d x = 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,,..., 1, i.e., we hlve the step length ech time. The result is By error, we here men h I mid (h) Error f (x)d x I mid (h). Note tht ech time the step length is hlved, the error seems to be reduced by fctor of The error Algorithm 1.5 determines numericl pproximtion to the integrl, nd even estimtes the error. However, we must remember tht the error tht is computed is not lwys relible, so we should try nd understnd the error better. We do this in two steps. First we nlyse the error in the sitution where we use very simple prtition with only one subintervl, the so-clled locl error. Then we use this result to obtin n estimte of the error in the generl cse this is often referred to s the globl error. Locl error nlysis Suppose we use the midpoint rule with just one subintervl. We wnt to study the error f (x)d x f ( 1/ ) (b ), 1/ = ( + b)/. (1.5) 86

9 Once gin, Tylor polynomil with reminder helps us out. We expnd f (x) bout the midpoint 1/ nd obtin, f (x) = f ( 1/ ) + (x 1/ )f ( 1/ ) + (x 1/) f (ξ), where ξ is number in the intervl ( 1/, x) tht depends on x. Next, we integrte the Tylor expnsion nd obtin f (x)d x = (f ( 1/ ) + (x 1/ )f ( 1/ ) + (x 1/) = f ( 1/ )(b ) + f ( 1/ )[ (x 1/ ) ] b + 1 ) f (ξ) d x (x 1/ ) f (ξ)d x = f ( 1/ )(b ) + 1 (x 1/ ) f (ξ)d x, (1.6) since the middle term is zero. This leds to n expression for the error, f (x)d x f ( 1/ )(b ) = 1 (x 1/ ) f (ξ)d x. (1.7) Let us simplify the right-hnd side of this expression nd explin fterwords. We hve 1 b (x 1/ ) f (ξ)d x 1 (x 1/ ) f (ξ) d x = 1 (x 1/ ) f (ξ) d x M (x 1/ ) d x = M 1 [ (x 1/ ) 3] b 3 = M ( (b 1/ ) 3 ( 1/ ) 3) 6 = M 4 (b )3, (1.8) where M = mx x [,b] f (x). 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 1/ ) is never negtive. The next inequlity follows becuse we replce f (ξ) with its mximum on the 87

10 intervl [,b]. The next step is just the evlution of the integrl of (x 1/ ), nd the lst equlity follows since (b 1/ ) 3 = ( 1/ ) 3 = (b ) 3 /8. This proves the following lemm. Lemm 1.7. Let f be continuous function whose first two derivtives re continuous on the intervl [.b]. The error in the midpoint rule, with only one intervl, is bounded by f (x)d x f ( 1/ ) (b ) M 4 (b )3, where M = mx x [,b] f (x) nd 1/ = ( + b)/. Before we continue, let us sum up the procedure tht led up to lemm 1.7 without focusing on the detils: Strt with the error (1.5) nd replce f (x) by its liner Tylor polynomil with reminder. When we integrte the Tylor polynomil, the liner term becomes zero, nd we re left with (1.7). At this point we use some stndrd techniques tht give us the finl inequlity. The importnce of lemm 1.7 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 rule will be reduced by fctor of 8. 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. 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 just nlysed, I = The totl error is then f (x)d x = I I mid = ( xi xi x i 1 f (x)d x f (x i 1/ )h. x i 1 f (x)d x f (x i 1/ )h We note tht the expression inside the prenthesis is just the locl error on the 88 ).

11 intervl [x i 1, x i ]. We therefore hve ( xi I I mid = f (x)d x f (x i 1/ )h x i 1 xi f (x)d x f (x i 1/ )h x i 1 h 3 4 M i (1.9) where M i is the mximum of f (x) on the intervl [xi 1, x i ]. The first of these inequlities is just the tringle inequlity, while the second inequlity follows from lemm 1.7. To simplify the expression (1.9), we extend the mximum on [x i 1, x i ] to ll of [,b]. This cnnot mke the mximum smller, 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 (1.9) further, h 3 4 M i h 3 x [,b] ) 4 M = h3 nm. (1.1) 4 Here, we need one finl little observtion. Recll tht h = (b )/n, so hn = b. If we insert this in (1.1), we obtin our min error estimte. Theorem 1.8. 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 = Then the error is bounded by where x i 1/ = + (i 1/)h. f (x)d x I mid = f (x i 1/ )h. I I mid (b ) h 4 mx f (x), (1.11) x [,b] This confirms the error behviour tht we sw in exmple 1.6: If h is reduced by fctor of, the error is reduced by fctor of = 4. 89

12 One notble omission in our discussion of the error in the midpoint rule 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. In such sitution we my potentilly dd mny numbers tht sum to, nd this my led to cncelltion effects. However, this is so rre tht we will not discuss it here Estimting the step length The error estimte (1.11) 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 ) h 4 mx f (x) ɛ x [,b] 4ɛ h (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, over the whole intervl of integrtion [,b]. This cn be difficult, but in some cses it is certinly possible, see exercise 3. Exercises 1 Clculte n pproximtion to the integrl π/ sin x 1 + x d x = with the midpoint rule. Split the intervl into 6 subintervls. In this exercise you re going to progrm lgorithm 1.5. If you cnnot progrm, use the midpoint lgorithm with 1 subintervls, check the error, nd skip (b). ) Write progrm tht implements the midpoint rule s in lgorithm 1.5 nd test it on the integrl e x d x = e 1. 9

13 x 1 () x 1 4 (b) Figure 1.4. The trpezoidl rule with one subintervl () nd five subintervls (b). b) Determine vlue of h tht gurntees tht the bsolute error is smller thn 1 1. Run your progrm nd check wht the ctul error is for this vlue of h. (You my hve to djust lgorithm 1.5 slightly nd print the bsolute error.) 3 Repet the previous exercise, but compute the integrl 6 ln x d x = ln(11664). 4 Redo the locl error nlysis for the midpoint rule, but replce both f (x) nd f ( 1/ ) by liner Tylor polynomils with reminders bout the left end point. Wht hppens to the error estimte? 1.3 The trpezoidl rule The midpoint rule is bsed on very simple polynomil pproximtion to the function f to be integrted on ech subintervl; we simply use constnt pproximtion tht interpoltes 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 1.4. The pproximtion to the integrl is the re of the trpezoidl polygon under the secnt so we hve f () + f (b) f (x)d x (b ). (1.1) To get good ccurcy, we will hve to split [,b] into subintervls with prtition nd use the trpezoidl pproximtion on ech subintervl, s in figure 1.4b. If we hve uniform prtition {x i } n with step length h, we get the pproximtion i= f (x)d x = xi x i 1 f (x)d x 91 f (x i 1 ) + f (x i ) h. (1.13)

14 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 (1.13). This mens tht if we implement this formul directly we do lot of unnecessry work. From this the following observtion follows. Observtion 1.9 (Trpezoidl 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= 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)d x I trp (h) = h n 1 + ) f (x i ). (1.14) Once we hve the formul (1.14), we cn esily derive n lgorithm similr to lgorithm 1.5. In fct the two lgorithms re identicl except for the prt tht clcultes the pproximtions to the integrl, so we will not discuss this further. Exmple 1.1. We test the trpezoidl rule on the sme exmple s the midpoint rule, cos x d x = sin As in exmple 1.6 we split the intervl into k subintervls, for k = 1,,..., 1. The resulting pproximtions re h I trp (h) Error

15 where the error is defined by f (x)d x I trp (h). We note tht ech time the step length is hlved, the error is reduced by fctor of 4, just s for the midpoint rule. But we lso note tht even though we now use two function vlues in ech subintervl to estimte the integrl, the error is ctully twice s big s it ws for the midpoint rule The error Our next step is to nlyse the error in the trpezoidl rule. We follow the sme strtegy s for the midpoint rule nd use Tylor polynomils. Becuse of the similrities with the midpoint rule, we skip some of the detils. The locl error We first study the error in the pproximtion (1.1) where we only hve one secnt. In this cse the error is given by f () + f (b) f (x)d x (b ), (1.15) nd the first step is to expnd the function vlues f (x), f (), nd f (b) in Tylor series bout the midpoint 1/, f (x) = f ( 1/ ) + (x 1/ )f ( 1/ ) + (x 1/) f (ξ 1 ), f () = f ( 1/ ) + ( 1/ )f ( 1/ ) + ( 1/) f (ξ ), f (b) = f ( 1/ ) + (b 1/ )f ( 1/ ) + (b 1/) f (ξ 3 ), where ξ 1 ( 1/, x), ξ (, 1/ ), nd ξ 3 ( 1/,b). The integrtion of the Tylor series for f (x) we did in (1.6) so we just quote the result here, f (x)d x = f ( 1/ )(b ) + 1 (x 1/ ) f (ξ 1 )d x. (1.16) We note tht 1/ = (b )/ nd b 1/ = (b )/, so the sum of the Tylor series for f () nd f (b) is (b ) f () + f (b) = f ( 1/ ) + f (b ) (ξ ) + f (ξ 3 ). (1.17)

16 If we insert (1.16) nd (1.17) in the expression for the error (1.15), the first two terms cncel, nd we obtin f () + f (b) f (x)d x (b ) = 1 b (x 1/ ) f (b )3 (ξ 1 )d x f (b )3 (ξ ) f (ξ 3 ) b (x 1/ ) f (ξ 1 )d x (b )3 + f (b )3 (ξ ) + f (ξ 3 ) The lst reltion is just n ppliction of the tringle inequlity. The first term we estimted in (1.8), nd in the lst two we use the stndrd trick nd tke mximum vlues of f (x) over ll of [,b]. Then we end up with f () + f (b) f (x)d x (b ) M 4 (b )3 + M 16 (b )3 + M (b )3 16 = M 6 (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 error in the trpezoidl rule, with only one secnt bsed t nd b, is bounded by where M = mx x [,b] f (x). f () + f (b) f (x)d x (b ) M 6 (b )3, This lemm is completely nlogous to lemm 1.7 which describes the locl error in the midpoint rule. We prticulrly notice tht even though the trpezoidl rule uses two vlues of f, the error estimte is slightly lrger thn the estimte for the midpoint rule. The most importnt feture is the exponent on (b ), which tells us how quickly the error goes to 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 function insted of constnt. This does not men tht the trpezoidl rule is bd, it rther mens tht the midpoint rule is surprisingly good. 94

17 Globl error We cn find n expression for the globl error in the trpezoidl rule in exctly the sme wy s we did for the midpoint rule, so we skip the proof. Theorem 1.1. 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 trpezoidl rule with n subintervls of equl width h, ( f () + f (b) I = f (x)d x I trp = h Then the error is bounded by n 1 + ) f (x i ). I Itrp h (b ) 6 mx f (x). (1.18) x [,b] The error estimte for the trpezoidl rule is not best possible in the sense tht it is possible to derive better error estimte (using other techniques) with the smller constnt 1/1 insted of 1/6. However, the fct remins tht the trpezoidl rule is bit disppointing compred to the midpoint rule, just s we sw in exmple 1.1. Exercises 1 Clculte n pproximtion to the integrl π/ sin x 1 + x d x = with the trpezoidl rule. Split the intervl into 6 subintervls. In this exercise you re going to progrm n lgorithm like lgorithm 1.5 for the trpezoidl rule. If you cnnot progrm, use the trpezoidl rule mnully with 1 subintervls, check the error, nd skip the second prt of (b). ) Write progrm tht implements the midpoint rule s in lgorithm 1.5 nd test it on the integrl e x d x = e 1. b) Determine vlue of h tht gurntees tht the bsolute error is smller thn 1 1. Run your progrm nd check wht the ctul error is for this vlue of h. (You my hve to djust lgorithm 1.5 slightly nd print the bsolute error.) 95

18 3 Fill in the detils in the derivtion of lemm 1.11 from (1.16) nd (1.17). 4 In this exercise we re going to do n lterntive error nlysis for the trpezoidl rule. Use the sme procedure s in section 1.3.1, but expnd both the function vlues f (x) nd f (b) in Tylor series bout. Compre the resulting error estimte with lemm When h is hlved in the trpezoidl rule, some of the function vlues used with step length h/ re the sme s those used for step length h. Derive formul for the trpezoidl rule with step length h/ tht mkes it esy to void recomputing the function vlues tht were computed on the previous level. 1.4 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, we omit it here Derivtion of Simpson s rule As for the other methods, we derive Simpson s rule in the simplest cse where we pproximte f by one prbol on the whole intervl [,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 simplifiction is to first construct Simpson s rule in the cse where = 1, 1/ =, nd b = 1, nd then generlise fterwrds. Simpson s rule on [ 1,1] The Lgrnge form of the polynomil tht interpoltes f t 1,, nd 1 is given by x(x 1) (x + 1)x p (x) = f ( 1) f ()(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 x(x 1)d x = 1 1 1(x x)d x = 1 [ 1 3 x3 1 x] 1 = Similrly, we find 1 (x + 1)(x 1)d x = 4 3, (x + 1)x d x = 1 3.

19 On the intervl [ 1, 1], Simpson s rule therefore corresponds to the pproximtion f (x)d x 1 ) f ( 1) + 4f () + f (1). (1.19) 3( 1 Simpson s rule on [,b] To obtin n pproximtion of the integrl on the intervl [,b], we use stndrd technique. Suppose tht x nd y re relted by x = (b ) y (1.) so tht when y vries in the intervl [ 1,1], then x will vry in the intervl [,b]. We re going to use the reltion (1.) s substitution in n integrl, so we note tht d x = (b )d y/. We therefore hve where f (x)d x = b 1 ( ) b f (y + 1) + d y = b f (y)d y, (1.1) 1 ( ) f b (y) = f (y + 1) +. To determine n pproximtion to the integrl of f on the intervl [ 1,1], we use Simpson s rule (1.19). The result is 1 f (y)d y 1 ) f ( 1) + 4f () + f 1( (1) = f () + 4f (1/ ) + f (b) 3( ), 3 since the reltion in (1.) mps 1 to, the midpoint to 1/ = ( +b)/, nd the right endpoint b to 1. If we insert this in (1.1), we obtin Simpson s rule for the generl intervl [,b], see figure 1.5. Observtion Let f be n integrble function on the intervl [,b]. If f is interpolted by qudrtic polynomil p t the points, 1/ = ( + b)/ nd b, then the integrl of f cn be pproximted by the integrl of p, f (x)d x p (x)d x = b ( f () + 4f (1/ ) + f (b) ). (1.) 6 We my just s well derive this formul by doing the interpoltion directly on the intervl [,b], but then the lgebr becomes quite messy. 97

20 x 1 () x 1 4 (b) Figure 1.5. Simpson s rule with one subintervl () nd three subintervls (b). x 1 4 Figure 1.6. Simpson s rule with three subintervls Composite Simpson s rule In prctice, we will usully divide the intervl [, b] into smller subintervls nd use Simpson s rule on ech subintervl, see figure 1.5b. Note though tht Simpson s rule is not quite like the other numericl integrtion techniques we hve studied when it comes to splitting the intervl into smller pieces: The intervl over which f is to be integrted is split into subintervls, nd Simpson s rule is pplied on neighbouring pirs of intervls, see figure 1.6. In other words, ech prbol is defined over two subintervls which mens tht the totl number of subintervls must be even, nd the number of given vlues of f must be odd. If the prtition is {x i } n i= with x i = + i h, Simpson s rule on the intervl 98

21 [x i, x i ] is xi x i f (x)d x h 3 ( f (xi ) + 4f (x i 1 ) + f (x i ) ). The pproximtion of the totl integrl is therefore f (x)d x h 3 ( (f (xi ) + 4f (x i 1 ) + f (x i ) ). In this sum we observe tht the right endpoint of one subintervl becomes the left endpoint of the neighbouring subintervl to the right. Therefore, if this is implemented directly, the function vlues t the points with n even subscript will be evluted twice, except for the extreme endpoints nd b which only occur once in the sum. We cn therefore rewrite the sum in wy tht voids these redundnt evlutions. Observtion Suppose f is function defined on the intervl [, b], nd let {x i } n be uniform prtition of [,b] with step length h. The composite i= Simpson s rule pproximtes the integrl of f by f (x)d x I Simp (h) = h ( n 1 f () + f (b) + f (x i ) ) f (x i 1 ). With the midpoint rule, we computed sequence of pproximtions to the integrl by successively hlving the width of the subintervls. The sme is often done with Simpson s rule, but then cre should be tken to void unnecessry function evlutions since ll the function vlues computed t one step will lso be used t the next step. Exmple Let us test Simpson s rule on the sme exmple s the midpoint rule nd the trpezoidl rule, cos x d x = sin As in exmple 1.6, we split the intervl into k subintervls, for k = 1,,..., 1. 99

22 The result is where the error is defined by h I Simp (h) Error f (x)d x I Simp (h). When we compre this tble with exmples 1.6 nd 1.1, we note tht the error is now much smller. We lso note tht ech time the step length is hlved, the error is reduced by fctor of 16. In other words, by introducing one more function evlution in ech subintervl, we hve obtined method with much better ccurcy. This will be quite evident when we nlyse the error below The error An expression for the error in Simpson s rule cn be derived by using the sme technique s for the previous methods: We replce f (x), f () nd f (b) by cubic Tylor polynomils with reminders bout the point 1/, nd then collect nd simplify terms. However, these computtions become quite long nd tedious, nd s for the trpezoidl rule, the constnt in the error term is not the best possible. We therefore just stte the best possible error estimte here without proof. Lemm 1.16 (Locl error). If f is continuous nd hs continuous derivtives up to order 4 on the intervl [,b], the error in Simpson s rule is bounded by (b )5 E(f ) mx f (i v) (x). 88 x [,b] We note tht the error in Simpson s rule depends on (b ) 5, while the error in the midpoint rule nd trpezoidl rule depend on (b ) 3. This mens tht 3

23 the error in Simpson s rule goes to zero much more quickly thn for the other two methods when the width of the intervl [,b] is reduced. The globl error The pproch we used to deduce the globl error for the midpoint rule, see theorem 1.8, cn lso be used to derive the globl error in Simpson s rule. The following theorem sums this up. Theorem 1.17 (Globl error). Suppose tht f nd its first 4 derivtives re continuous on the intervl [, b], nd tht the integrl of f on [, b] is pproximted by Simpson s rule with n subintervls of equl width h. Then the error is bounded by E(f ) h 4 (b ) 88 mx f (i v) (x). (1.3) x [,b] The estimte (1.3) explins the behviour we noticed in exmple 1.15: Becuse of the fctor h 4, the error is reduced by fctor 4 = 16 when h is hlved, nd for this reson, Simpson s rule is very populr method for numericl integrtion. Exercises 1 Clculte n pproximtion to the integrl π/ sin x 1 + x d x = with Simpson s rule. Split the intervl into 6 subintervls. ) How mny function evlutions do you need to clculte the integrl d x 1 + x with the trpezoidl rule to mke sure tht the error is smller thn 1 1. b) How mny function evlutions re necessry to chieve the sme ccurcy with the midpoint rule? c) How mny function evlutions re necessry to chieve the sme ccurcy with Simpson s rule? 3 In this exercise you re going to progrm n lgorithm like lgorithm 1.5 for Simpson s rule. If you cnnot progrm, use Simpson s rule mnully with 1 subintervls, check the error, nd skip the second prt of (b). 31

24 ) Write progrm tht implements Simpson s rule s in lgorithm 1.5 nd test it on the integrl e x d x = e 1. b) Determine vlue of h tht gurntees tht the bsolute error is smller thn 1 1. Run your progrm nd check wht the ctul error is for this vlue of h. (You my hve to djust lgorithm 1.5 slightly nd print the bsolute error.) 4 ) Verify tht Simpson s rule is exct when f (x) = x i for i =, 1,, 3. b) Use () to show tht Simpson s rule is exct for ny cubic polynomil. c) Could you rech the sme conclusion s in (b) by just considering the error estimte (1.3)? 5 We wnt to design numericl integrtion method f (x)d x w 1 f () + w f ( 1/ ) + w 3 f (b). Determine the unknown coefficients w 1, w, nd w 3 by demnding tht the integrtion method should be exct for the three polynomils f (x) = x i for i =, 1,. Do you recognise the method? 1.5 Summry In this chpter we hve derived three methods for numericl integrtion. All these methods nd their error nlyses my seem rther overwhelming, but they ll follow common thred: Procedure The following is generl procedure for deriving numericl methods for integrtion of function f over the intervl [,b]: 1. Interpolte the function f by polynomil p t suitble points.. Approximte the integrl of f by the integrl of p. This mkes it possible to express the pproximtion to the integrl in terms of function vlues of f. 3. Derive n estimte for the locl error by expnding the function vlues in Tylor series with reminders bout the midpoint 1/ = ( + b)/. 4. Derive n estimte for the globl error by using the technique leding up to theorem

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