f(x i )l i (x)dx = w i f(x i ) := I n [f] where w i = i=0

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1 Polynomil Interpolting Qudrture Atkinson Chpter 5, Stoer & Bulirsch Chpter 3, Dhlquist & Bjork Chpter 5 Sections mrked with re not on the exm 1 Qudrture Will first consider 1D f C[, b], then singulrities, then multivrite The method is to use polynomil to pproximte f, then to integrte the polynomil exctly We know how to do minimx pproximtions, L pproximtions, nd interpoltion Minimx pproximtions re hrd to compute; L pproximtions don t led to esy error nlysis; we stick to interpolting qudrtures For n interpolting qudrture, I[f] := f(x)dx p(x)dx = n f(x i )l i (x)dx = n w i f(x i ) := I n [f] where w i = l i (x)dx The bove shows how to construct n interpolting qudrture, but it doesn t tell us how good the pproximtion is An obvious pproch is to use the interpoltion error: f(x)dx p(x)dx = f(x) p(x)dx = Ψ(x) (n + 1)! f (n+1) (ξ(x))dx We lredy know tht the interpoltion error doesn t necessrily go to s n This pproch is therefore not widely used nd piecewise polynomil interpoltion is used insted Tht being sid, there re ctegories of method where the function is pproximted globlly by single polynomil tht is then integrted exctly: Gussin qudrture Clenshw-Curtis nd Fejér qudrture Both of these methods converge becuse specil nodes x i re chosen crefully so tht Runge phenomenon, ie non-convergence of the interpolnt, is voided Interestingly, in both of these cses the convergence nlysis is not bsed on direct nlysis of the interpoltion error This is generl feture of the error nlysis of interpolting qudrtures: you cn nlyze the error using the interpoltion error formul, but there re often better methods Interpoltory qudrture Simple exmple: piecewise liner Suppose you hve f i = f(x i ) where = x < x 1 < < x n = b, nd you wnt f(x)dx First mke piecewise-liner interpolting polynomil Define x i+1 x i = h i for nottionl convenience The interpoltion error on the i th intervl is p(x) = 1 h i [f i+1 (x x i ) + f i (x i+1 x)] for x i x x i+1 f(x) p(x) = (x x i)(x x i+1 ) f (ξ i ) The integrl of f is pproximted by the integrl of p The integrl over single subintervl is xi+1 p(x)dx = 1 [f i+1 h i + f i h i ] = f i+1 + f i x i h i This is the re of trpezoid (drw picture), so it s clled the trpezoid rule Adding up ll the subintervls yields I[f] I[p] = n 1 f i+1 + f i h i h i 1

2 This is sometimes clled the composite trpezoid rule, or just the trpezoid rule The error on single intervl is xi+1 (x x i )(x x i+1) f (ξ i (x))dx x i Now since (x x i )(x x i+1 ) is non-positive on the intervl we cn use the integrl men vlue theorem: xi+1 x i (x x i )(x x xi+1 i+1) f (ξ i (x))dx = f (ξ i ) x i (x x i )(x x i+1) dx = f (ξ i ) h3 i 1 If f is bounded, the error cn be bounded s follows xi+1 x i (x x i )(x x i+1) f (ξ i (x))dx mx f (x x i )(x x i+1 ) dx Now notice tht (x x i )(x x i+1 ) is strictly negtive over the integrl, so we cn just integrte it nd then chnge sign fter the fct The nswer is (esy, just some clc & lgebr) xi+1 x i (x x i )(x x i+1) f (ξ i (x))dx mx f h 3 i 1 The totl error is the sum of the errors on ech subintervl, so, ssuming f is bounded, the totl error is bounded by n 1 error mx f h 3 i 1 If ll the intervls hve the sme size h i = h then error mx f nh3 1 = (b )h 1 mx f To construct the qudrture rule you just use interpoltion To get n error estimte you use the interpoltion error estimte, but then you do little extr clever work when bounding the integrl of the error The method is second order since the error is bounded by h s h (it s still second order for unequl h i, s long s the lrgest one goes to zero) 3 Simplest exmple: piecewise constnt Suppose you hve = x < x 1 < < x n = b, but you insted hve f i = f(x i+1/ ) where x i+1/ = (x i+1 + x i )/ is the midpoint of ech subintervl We cn pproximte f using piecewise-constnt polynomil on ech intervl The interpoltion error on ech subintervl is p(x) = f i for x i x < x i+1 f(x) p(x) = (x x i+1/ )f (ξ i (x)) for x i x < x i+1 The integrtion error on ech subintervl is just the integrl of the interpoltion error The obvious thing to do is to tke bsolute vlues, ssume f is bounded, nd then do the integrl, which gives order h error on ech subintervl Adding the errors up leds to n overll order h error This is techniclly correct: the error is bounded bove by constnt times h s h But better bound is possible Consider tht the piecewise constnt pproximtion is lso Tylor pproximtion, whose error formul cn be expnded to so f(x) = f i + (x x i+1/ )f (x i+1/ ) + (x x i+1/) f (ξ i (x)) f(x) f i = f(x) p(x) = (x x i+1/ )f (x i+1/ ) + (x x i+1/) f (ξ i (x)) for x i x < x i+1

3 If we integrte this error, the first term drops out (integrtes to ), leving xi+1 x i f(x) p(x)dx = xi+1 x i (x x i+1/) f (ξ i (x))dx As long s f is bounded on the subintervl, this cn be bounded by xi+1 f(x) p(x)dx mx f h 3 i x i 4 Supposing tht h = h i nd dding up the error bounds cross ll n subintervls we hve tht the error for the composite midpoint rule is bounded by (b )h mx f 4 This is better bound thn the trpezoid rule! Of course tht doesn t men tht the midpoint rule error will lwys be smller thn the trpezoid rule error, just tht we derived better bound The midpoint method is second order precisely becuse it uses the midpoint of the intervl; ny other piecewise-constnt pproximtion will be just first order This whole nlysis shows tht using the interpoltion is just one wy to nlyze the qudrture error, nd tht it s often not the best wy 4 Simpson s rule It uses piecewise-qudrtic pproximtion nd equispced points The min point of tody s lecture is to introduce new wy to compute the error without recourse to the stndrd interpoltion error formul WLOG we ll consider single subintervl where x = h, x 1 =, x = h, nd f i = f(x i ) The qudrtic interpolting polynomil is p(x) = f + (x + h)f[ h, ] + x(x + h)f[ h,, h] The exct integrl, fter some simplifiction, is p(x)dx = h 3 (f + 4f 1 + f ) Just like trpezoid nd midpoint, you cn form composite rule out of this Atkinson derives n error bound directly from the interpoltion error formul, but it requires some d hoc tricks 1 5 There is nother pproch to estimting (bounding) the error tht pplies to ny qudrture bsed on polynomil interpoltion, including Hermite interpoltion (S&B 3) It is formulted not in terms of interpoltion, but just in terms of being ble to integrte polynomils exctly up to certin degree So we need some preliminries Note tht if you hve n + 1 distinct interpoltion points, nd if f(x) is polynomil of degree n then the interpolting polynomil will recover f(x) exctly In which cse the interpoltion-bsed qudrture will be exct But there s more Wht if f(x) is polynomil of degree n+1: f(x) = n+1 x n+1 +? The interpoltion error is f(x) p(x) = n+1 (x x ) (x x n ) Assume tht the points x i re distributed symmetriclly round the midpoint of the intervl (x n + x )/ Then If n is even then the interpoltion error is n odd-degree polynomil tht is ntisymmetric bout the midpoint of the intervl nd must therefore integrte to If n is odd then the interpoltion error doesn t necessrily integrte to 1 d hoc just mens tht it s not systemtic pproch tht cn be pplied in generl 3

4 So: An interpolting qudrture with n + 1 nodes tht re symmetriclly plced round the center of the intervl will integrte polynomils up to degree n exctly when n is odd, nd up to degree n + 1 exctly when n is even 6 Let Q[f] denote the qudrture rule Define the error s (note sign convention) R[f] = Q[f] I[f] (S&B 33) Suppose R[p] = for every polynomil of degree n f C n+1 [, b] R[f] = f (n+1) (t)k(t)dt where the Peno kernel is K(t) = 1 n! R[(x t)n +] Then for ll functions Before proving the theorem, we ll see wht it mens nd how it s useful Apply to Simpson s rule First note tht I n [f] for Simpson s rule is Q[f] = h (f( h) + 4f() + f(h)) 3 Simpson s rule integrtes polynomils of degree 3 exctly (becuse of the quirk noted bove), so the kernel is [ ] K(t) = 1 3! R[(x t)3 +] = 1 6 Brek it down: Q[(x t) 3 +] ( h t) 3 + = for every t [ h, h] h h ( t) 3 + = for t > nd = t 3 for t < (h t) 3 + = (h t) 3 for every t [ h, h] h t (x t)3 dx = (h t) 4 /4 This implies Plot: Compct support We cn now bound the error by K(t) = (x t) 3 +dx { 1 7 (h t)3 (h + 3t) for t K( t) for t R[f] f (4) h h = h 18 (( h t)3 ++4( t)) 3 ++(h t) 3 +) 1 6 K(t) dt = f (4) h 5 9 h t (x t) 3 dx Alterntively, since K(t) hs single sign on the intervl you cn use the integrl men vlue theorem to write h R[f] = f (4) (ξ) K(t)dt = f (4) (ξ) h5 h 9 for some ξ in the intervl If you form composite Simpson s rule nd dd the error bounds you find tht the method is 4th order The originl nottion, following S&B, ws to let I n[f] denote the qudrture rule However, Erin Ellefsen pointed out tht we ve lredy used I n[f] = n w if i where n + 1 is the number of points In the discussion tht follows, n is the degree of polynomil tht is integrted exctly by the qudrture, which is not necessrily the sme s the n from our previous definition 4

5 Notice tht midpoint ws n = nd second order; trpezoid ws n = 1 nd second order; Simpson s is n = nd fourth order This behvior turns out to be generic: formul bsed on interpoltion tht uses equispced nodes nd interpolting polynomils of degree n gives even-order globl ccurcy, either n + if n is even or n + 1 if n is odd Formuls bsed on equispced interpoltion re clled Newton-Cotes formuls We lredy don t expect them to converge s n becuse of Runge phenomenon, but they cn be useful in piecewise interpoltion context 7 Proof of S&B 33 Note tht the interpoltion-bsed qudrture rule is liner It is just the composition of liner functions First tke the function nd construct the pproximting polynomil; this process is liner since the polynomil is constructed by solving liner system where the RHS is just the function dt Next you exctly integrte the interpolting polynomil, nd integrtion is liner Becuse the qudrture is liner, the error is liner too Let Q[f] denote the qudrture The error is (note sign convention) R[f] = Q[f] I[f] is liner opertor becuse it s sum of liner opertors Now consider the Tylor expnsion with Peno reminder f(x) = f() + f ()(x ) + + f (n) () (x ) n + 1 n! n! x f (n+1) (t)(x t) n dt This is something you should hve seen elsewhere You cn look it up; I will not prove the Peno form of the Tylor reminder But I will note tht 1 n! x f (n+1) (t)(x t) n dt = 1 n! f (n+1) (t)(x t) n +dt Apply R to the Tylor expnsion of f(x) nd use linerity nd the fct tht the reminder is for the polynomil prt to get [ ] R[f] = + 1 b n! R f (n+1) (t)(x t) n +dt Now we need to be ble to commute R with the integrl, nd then we ll hve the desired result S&B considers Hermite interpoltion, which is why their discussion is more complicted For our purpose we just need to show tht [ ] b Q f (n+1) (t)(x t) n +dt = f (n+1) (t)q[(x t) n +]dt nd [ ] b I f (n+1) (t)(x t) n +dt = f (n+1) (t)i[(x t) n +]dt All tht Q[ ] does is evlute t prticulr vlues of x, multiply by constnts, nd dd This clerly commutes with the integrtion For the second prt we just need to switch the order of integrtion f (n+1) (t)(x t) n +dtdx = f (n+1) (t)(x t) n +dxdt which is clerly vlid becuse of continuity of the integrnd over the D rectngle (Fubini) This proves tht R[f] = 1 n! f (n+1) (t)r[(x t) n +]dt 5

6 8 We just sw tht you cn derive error formuls without mking ny reference to interpoltion; insted you just require the qudrture formul to integrte polynomils exctly So let s derive new kind of qudrture bsed on integrting polynomils exctly The integrl of polynomil of degree n is + 1 x + + n x n dx = (b ) + 1 b b n+1 n n n + 1 Suppose we hve n + 1 points nd n + 1 weights nd pply our qudrture to the sme polynomil I n [ + 1 x + + n x n ] = w ( + 1 x + + n x n ) + + w n ( + 1 x n + + n x n n) We wnt these expressions to be zero for ny polynomil, so we need w + w w n = b w x + w 1 x w n x n = b w x + w 1 x w n x n = b3 3 w x n + w 1 x n w n x n n = bn+1 n+1 n + 1 The coefficient mtrix is the trnspose of Vndermonde mtrix (which is sometimes clled Vndermonde mtrix): V T w = b, b i = so the solution is exists & is unique provided tht the nodes re distinct Overll, this pproch gives method for computing weights for generl nodes nd polynomil orders Vndermonde mtrices cn be ill-conditioned, so in generl don t use this for lrge systems (with some cvets; some node loctions give OK conditioning) Suppose x = = h, x 1 =, nd x = b = h; then the solution for the weights is just the lredy-found Simpson s rule In fct, this is not new kind of qudrture t ll; the weights derived in this mnner re exctly the sme s the weights derived for interpolting qudrture Recll tht you cn derive the weights from the Lgrnge form of the interpolting polynomil: x i dx 3 p(x) = i p(x)dx = i w i = f(x i )l i (x) f(x i ) l i (x)dx l i (x)dx Also recll tht the Lgrnge polynomils re l i (x) = (V 1 ) 1i + (V 1 ) i x + + (V 1 ) (n+1)i x n where V is the Vndermonde mtrix So in the interpolting qudrture derivtion of the weights we hve w i = (V 1 ) 1i (b ) + (V 1 b ) i + + (V 1 b n+1 n+1 ) (n+1)i n + 1 This is exctly the sme s weights obtined vi the solution of the trnsposed Vndermonde system derived using the condition tht the qudrture exctly integrtes polynomils (without reference to interpoltion) w = V T b So we bsiclly now hve wys to compute weights: integrte the Lgrnge polynomils, or solve the trnsposed Vndermonde system 6

7 Asymptotic Error Formule & Extrpoltion 1 Recll tht the error for the (simple) trpezoid rule is I[f] I 1 [f] = h3 1 f (η) for some η in the intervl The error in the composite rule (with equl spcing, for simplicity) is thus Notice the following curious fct E n [f] 1 lim n h = lim n 1 E n [f] = h3 1 n n f (η i ) f (η i )h = 1 1 f (x)dx = f () f (b) 1 (Riemnn sum limits to n integrl) This gives us n symptotic error formul E n [f] h 1 [f () f (b)] s n IF f () = f (b) = or if f () = f (b) the convergence is fster thn h One wy to use this is to define corrected (composite) trpezoid rule Corrected Trp Rule = Trp Rule + h 1 [f () f (b)] Nturlly you cn only do this if you know f () nd f (b) This rule should be more ccurte thn the stndrd Trp Rule for most functions s long s n is lrge Actully, s we ll see lter, the Corrected Trp Rule is fourth order (globlly), nd you cn keep correcting it The symptotic error formul obtined bove for the composite Trp rule hs the form I[f] I n [f] ch for constnt c independent of n Tody we will derive formuls of the form I[f] I n [f] c h + c 4 h c n h n It tkes some effort though, so we ll strt with bckground (S&B 33) Consider tht 1 f(t)dt = 1 ( ) d (t + c) f(t)dt = [(t + c)f(t)] 1 dt We wnt to relte this to the Trpezoid Rule, which hs the form 1 If we pick c = 1/ in the first formul we get 1 f(t)dt 1 [f() + f(1)] 1 f(t)dt = 1 1 [f() + f(1)] (t 1 )f (t)dt (t + c)f (t)dt 7

8 This gives us n error formul for the simple Trp Rule, though not very useful one Let s keep going 1 (t 1 )f (t)dt = 1 [( t t + c ) f (t) ] ( t t + c ) f (t)dt If we choose c ppropritely we ll get the symptotic error formul for the Trpezoid Rule; viz c = 1/6 implies 1 f(t)dt = 1 1 [f() + f(1)] 1 [f (1) f ()] ( t t + 1 ) f (t)dt 6 Here we recognize the symptotic error formul for the Trp Rule with h = 1, nd s bonus we get n exct reminder term Let s keep going 1 ( t t + 1 ) f (t)dt = 1 [(t t + 1 ) ] 1 t + c f (t) (t 3 3 t + 1 t + c ) f (3) (t)dt How should we choose c? It turns out to be very convenient to choose it so tht the polynomil is t both endpoints, ie c =, so 1 ( t t + 1 ) f (t)dt = 1 1 (t t + 1 ) t f (3) (t)dt 1 (t 3 3 t + 1 ) t f (3) (t)dt = 1 [ (t 4 t 3 + t + c ) 1 f (t)] (3) ( t 4 t 3 + t + c ) f (4) (t)dt Note tht for ny c > the polynomil in the reminder term is positive on [, 1], so we cn use the integrl men vlue theorem to set 1 ( t 4 t 3 + t + c ) 1 f (4) (t)dt = f (4) ( (ξ) t 4 t 3 + t + c ) dt = ( c)f (4) (ξ) We should choose c = which gives the best bound Coincidentlly this lso sets the boundry terms to If we put everything bck together, we get n even better symptotic estimte for the Trpezoid rule error; it would be order h 5 if we hd used [, h] insted of [, 1] 3 Now let s generlize the previous pproch We hve f(t)dt = B 1 (t)f(t) 1 1 B 1 (t)f (t)dt = 1 B (t)f (t) 1 1 B 1 (t)f (t)dt 1 B k 1 (t)f (k 1) (t)dt = 1 k B k(t)f (k 1) (t) 1 1 k B (t)f (t)dt 1 B k (t)f (k) (t)dt The B k (t) re polynomils tht stisfy B k+1 (t) = (k + 1)B k(t) nd B 1 = t 1/ There re bunch of unknown constnts of integrtion which re completely rbitrry s fr s the bove sequence of integrtionsby-prts is concerned We will mke some very specil choices tht led to convenient formuls Specificlly, we wnt B k+1 () = B k+1 (1) = for k > 1 This will gurntee tht the boundry terms t odd orders will be, leving 1 f(t)dt = 1 [f(1) + f()] + even-order boundry terms + the lst integrl 8

9 It lso specifies the B k (t) uniquely If you hve B k (t) where k is odd, then first you integrte to get B k+1 (t) with free constnt, then you integrte gin to get B k+ with two free constnts You then set those constnts by the conditions bove, which gives you exct expressions for B k+1 (t) nd B k+ (t) These polynomils re the Bernoulli polynomils, nd the numbers B k = B k () re the Bernoulli numbers Collecting ll the nottion together we find 1 f(t)dt = 1 m [f()+f(1)]+ B l (l)! [f (l 1) () f (l 1) 1 (1)]+ (m + )! l=1 1 (B m+ (t) B m+ )f (m+) (t)dt Not bene This is version of the Euler-McLurin formul If you re tking the prelim, you would do well to memorize the formul (549) from Atkinson tht corresponds to the Euler-McLurin formul for the composite rule nd for n integrl over [, b] rther thn [, 1] It just so hppens (s result of the specific choice of constnts of integrtion; proof in S&B 33) tht the kernel B m+ (t) B m+ is sign-definite on [, 1] So, by the integrl men vlue theorem 1 f(t)dt = 1 [f() + f(1)] + m l=1 B l (l)! [f (l 1) () f (l 1) (1)] + f (m+) (ξ) (m + )! 1 (B m+ (t) B m+ )dt When you rescle the intervl from [, 1] to [, h], you get n symptotic formul for the simple (not composite) trpezoid rule error h f(t)dt h [f() + f(h)] = c h + c 4 h c m h m + c m+ (h)h m+ The c m+ (h) is relted to the term f (m+) (ξ) nd the h m+ comes from the integrl of the kernel, which we didn t prove In the lst term c m+ (h) is bounded s h 4 Comments: Briefly note tht if you form composite rule nd dd ll the symptotic error estimtes, the terms correspoding to interior nodes will cncel except in the lst term, leving n symptotic error formul of the form (see comment bove; 549 in Atkinson) f(t)dt = I n [f] + C h + C 4 h C m h m + C m+ (h)h m+1 (The coefficients c p in the formul for the simple rule re different from the coefficients C p in the composite rule, but both re independent of h) If f C [, b] nd ll its derivtives go to zero t the endpoints, then the trpezoid rule will converge fster thn ny power of h If f is periodic nd C then the trpezoid rule will converge fster thn ny power of h 5 The composite trpezoid rule hs n symptotic error formul of the form error = C h + C 4 h C m h m + C m+ (h)h m+1 s long s the integrnd hs sufficient number of continuous derivtives Lots of other qudrture rules hve symptotic error formuls tht look like this, but not ll For exmple, pplying the trpezoid rule to infinitely-smooth periodic functions implies tht ll the constnts c p re zero Gussin Qudrture does not hve this kind of symptotic error formul But suppose we hve such formul We cn use it to extrpolte the qudrture The simplest extrpoltion (Aitken s) requires E n [f] c n p 9

10 (eg Trp rule hs p =, Simpson s rule hs higher-order symptotic error formul, nd the corrected trp rule hs p = 4) You might not know whether there is n symptotic error formul, but you cn check empiriclly If there is n symptotic error formul then R 4n := I n I n = I I n (I I n ) I 4n I n I I n (I I 4n ) c/np c/(n) p c/(n) p c/(4n) p = p So we cn estimte p log (R 4n ) If we compute log (R 4n ) for severl different n nd find the sme (or similr) result, then we cn ssume tht there s n symptotic error formul; otherwise not Now, ssuming we hve n symptotic error behvior, consider Setting left equl to right nd solving for I yields I I n I I n p I I n I I 4n I A 4n = I 4n (I 4n I n ) I 4n I n + I n Notice tht you re correcting the most-ccurte estimte I 4n This is Aitken s extrpoltion; it s similr to Aitken s extrpoltion for fixed points The bove derivtion doesn t show wht the error of the extrpolted qudrture method is Let s now ssume tht the originl formul hd n symptotic error formul of the form E n [f] c n p + c 1 n q with q > p (this is still consistent with the originl formul, BTW) Plug in I n = I E n to the expression I A 4n nd simplify You ll find tht the new qudrture hs symptotic error formul I A 4n c 14 q ( p q ) ( p 1) n q s n As expected, we ve corrected the leding-order term in the symptotic error expnsion, so the next one now becomes the leding-order term 6 Richrdson Extrpoltion We re deling with Trp Rule, so we hve the following I[f] = I n [f] + C h + C 4 h C m+1 (h)h m+1 I[f] = I n [f] + C h h C C m+1(h/) hm+1 m+1 where h is the spcing for the rule with n + 1 points (I n [f]) Notice tht we cn eliminte the leding-order error term by multiplying the bottom line by 4, then subtrcting from the top: We cn write this s 3I[f] = I n [f] 4I n [f] C 4h error I[f] = I (1) n [f] + d 4h 4 + d 6 h error, where I (1) n [f] := I n[f] 4I n [f] 3 This is clled Richrdson extrpoltion It s interesting to note tht this is ctully just Simpson s rule We cn see immeditely tht it s fourth order We cn do the sme thing gin: I[f] = I (1) n [f] + d 4h 4 + d 6 h error h 4 I[f] = I (1) 4n [f] + d d different error 6 h 6 1

11 If we multiply the bottom line by 4 nd subtrct from the top line we hve I[f] = I () 4n [f] + e 6h 6 + e 8 h 8 + where I () I(1) n 4n [f] = [f] 4 I (1) 4n [f] 1 4 Actully this is just Boole s rule, nd we cn see immeditely tht it s sixth-order Clerly the process cn continue, nd in generl it does not produce more Newton-Cotes rules We cn write the following formul I n (k) [f] = 4k I n (k 1) [f] I (k 1) n/ [f] 4 k 1 where n is even Assuming the function is smooth enough, the error hs symtotic order h k+ 7 Romberg integrtion I () 1 I () I (1) I () 4 I (1) 4 I () 4 I () 8 I (1) 8 I () 8 I (3) 8 I () 16 I (1) 16 I () 16 I (3) 16 I (4) 16 The number of points n+1 increses downwrds, the order of Richrdson extrpoltion increses rightwrds Romberg integrtion uses the digonl of the bove tble J k [f] = I (k) k Since this is just Richrdson extrpoltion, the error is known nd hs symptotic order h k+ Previously we considered k to be fixed while we decrese h, but here h is decresing s k is incresing h = (b )/( k ) To put things on n even footing, note tht k = log (b ) log (h), so the error is symptoticlly of the form h log ((b )/h)+ s h The error is thus going to fster thn ny power of h 8 Summry The Euler-McLurin (nd similr) symptotic error formul leds to Improved-rte qudrtures like corrected-trpezoid Improved-rte qudrtures like Aitken extrpoltion nd/or Richrdson extrpoltion Qudrtures tht converge fster-thn-lgebriclly for functions tht re infinitely smooth: (i) Romberg, nd (ii) Trpezoid for periodic functions Gussin Qudrture & Chebyshev Methods 1 We ve looked t choosing the weights w i so tht we chieve the highest possible order of integrtion We lso sw tht it s sometimes possible to choose the node loctions to chieve order higher thn n, eg midpoint rule hs n = but integrtes polynomils up to degree 1 exctly; similrly for Simpson s rule n = but it integrtes polynomils of degree 3 exctly Recll tht the Newton-Cotes (equispced) rules lwys hve even globl order of ccurcy: midpoint is second order, trpezoid is second order, Simpson s is fourth order, the next one is fourth order, etc The generl ide now is to choose both nodes x i nd weights w i to chieve s high-order ccurcy s possible We now hve (n + 1) degree of freedom, so we should be ble to integrte polynomils of degree n + 1 exctly You cn enforce this by requiring i w i(x i ) k = xk dx for ll k n + 1, but this is 11

12 nonliner system! Generlly we don t ttck the nonliner system directly; there re better wys tht we will discuss But first, let s briefly expnd our rnge of integrls to something like this f(x)w(x)dx where w(x) is weight function with the usul properties (non-negtive nd xk w(x)dx < for ll k Z + ) One reson for looking t this is to llow singulrities; eg if you wnt to compute 1 1 e x 1 x dx then techniclly you cn t use ny of the rules we ve studied so fr becuse we lwys ssume tht f is continuous Insted you cn just let f(x) = e x nd w(x) = (1 x ) 1/ We ll return to this ide lter Bewre: We didn t derive Peno kernel error formul for weighted integrls In the mentime, Proof: (D&B Theorem 513) Let n w if(x i ) be qudrture tht integrtes polynomils of degree n exctly, nd define ψ(x) = (x x ) (x x n ) Then the interpoltory qudrture rule integrtes ll polynomils of degree n k if nd only if for ll polynomils p of degree k w(x)p(x)ψ(x)dx = First prove tht if the rule integrtes polynomils with degree n k exctly, then the integrl bove is zero Note tht ψ(x)p(x) is polynomil of degree n+1+k, so it will be integrted exctly Then plug in the qudrture formul lst equlity is becuse ψ(x i ) = ψ(x)p(x)w(x)dx = Now prove tht the integrl implies exct qudrture Let n w i ψ(x i )p(x i ) = p(x) = q(x)ψ(x) + r(x) where q(x) hs degree k nd r(x) hs degree n p(x)w(x)dx = q(x)ψ(x)w(x)dx + r(x)w(x)dx By ssumption, the first integrl on the RHS is Also by ssumption, the second integrl cn be computed exctly using the qudrture p(x)w(x)dx = r(x)w(x)dx = i w i r(x i ) Notice tht p(x i ) = r(x i ), so p(x)w(x)dx = i w i r(x i ) = i w i p(x i ) 1

13 QED The ide is to pick the nodes x i so tht the polynomil ψ(x) is orthogonl to ll polynomils of degree k Eg let w(x) = 1, [, b] = [ 1, 1] Suppose we hve 3 nodes (n = ) nd wnt to be ble to integrte quintics (k = ) exctly We need to choose the nodes so tht ψ(x) is cubic tht is orthogonl to ll qudrtics So ψ(x) the third-order Legendre polynomil, which we lredy know hs exctly 3 simple roots in [ 1, 1] Once we know the roots/nodes, we cn compute the weights using the stndrd lgorithms Note tht ψ(x) hs degree n + 1, so by choosing ψ(x) proportionl to the n + 1-st orthogonl polynomil, it will be orthogonl to ll polynomils of degree n, nd will hve exctly n + 1 simple roots in the intervl [, b] This shows tht you cn choose the nodes first nd then find the weights so tht the qudrture rule integrtes polynomils of order n + 1 exctly, just s expected This is clled Gussin qudrture Sometimes for w(x) = 1 it is clled Guss-Legendre; for other weight functions it is sometimes clled Guss-Christoffel qudrture For smll to moderte n you cn just look the nodes nd weights up For lrger n there re purpose-built lgorithms tht will ccurtely nd rpidly compute them for you Now tht we ve constructed Gussin qudrture, let s nlyze the error The most bsic convergence result (without rte of convergence) is bsed on the fct tht the weights re positive We lredy know tht the weights re the integrls of the Lgrnge polynomils w i = w(x)l i (x)dx Notice tht l i is polynomil of degree n, so l i is polynomil of degree n, nd will be integrted exctly So < l j (x) w(x)dx = w i l j (x i ) = w j i Now tht we know Gussin qudrture weights re positive, we need to know why tht mtters The qudrtures derived so fr re liner opertors Suppose you hve method I n [f] tht gives you n exct nswer for ll polynomils up to degree n, ie I n [p] = I[p] for ny polynomil p of degree n Well, we know tht there is polynomil of degree n tht optimlly pproximtes f in the L norm Denote this by p Then I[f] I n [f] = I[f] I[p ] + I n [p ] I n [f] ie we cn relte the error in integrting f to the error in pproximting f by p Now one prt is esy I[f] I[p ] = I[f p ] (b )ρ n (f) The Weierstrss pproximtion theorem sys tht this will go to s n for ny continuous f Wht bout the other piece? All our intepoltion-bsed qudrtures re liner opertors If they re bounded then We now derive bound on the opertor I n [f] I n [p ] = I n [f p ] I n ρ n (f) I n = mx f =1 I n[f] = mx i w i f(x i ) mx f(x i ) i i w i i w i If we cn now show tht there is n f with f = 1 tht chieves this upper bound then we ll hve shown tht this is ctully the norm of the opertor Let integrble f be ny function st f(x i ) =signw i, then I n [f] = i w i 13

14 Now the fct tht the weights hve to integrte the polynomil 1 exctly imply tht i w i = b for ny n The weights clerly depend on n nd on the loction of the nodes Consider wht hppens s you increse n If you choose the nodes in such wy tht the weights remin positive, then I n = (b ) for ny n, nd by the rguments bove the qudrture will converge to the true integrl Otherwise it s possible tht the weights cn grow cusing I n to grow nd possibly cusing the qudrture to diverge This is nother explntion of why the equispced Newton-Cotes formuls re not relible: the weights re not positive This leds to very nice conclusion for Gussin qudrture: f C([, b]) it will converge s n for ny 3 The bove nlysis tells us tht GQ converges for lots of functions, but it doesn t tell us how fst it converges The Peno kernel gives one wy of deriving n error formul for Gussin qudrture (Guss- Legendre, nwys; we didn t derive the Peno form for weighted integrls) The following is nother wy The ide is to show tht Hermite qudrture t the Guss-Christoffel nodes is the sme s Gussin qudrture, nd then use the Hermite error formul Consider globl Hermite interpolting polynomil The qudrture rule is where w i = p(x) = n f(x i )h i (x) + w(x)f(x)dx i w(x)h i (x)dx, w i = n f (x i ) h i (x) f(x i )w i + i f(x i ) w i w(x) h i (x)dx We need to show tht this is the sme s Gussin qudrture when the nodes re the roots of the n + 1-st degree orthogonl polynomil We do this by showing tht w i = nd w i = w(x)(l i (x)) dx, which re the sme s the Gussin qudrture weights First w i = h i (x) = (x x i )(l i (x)) = [(x x ) (x x n )][(x x ) (x x i 1 )(x x i+1 ) (x x i )] (x i x ) (x x i 1 ) (x x i+1 ) (x x i ) ] h i is ψ(x) times polynomil of degree n 1 By definition w i = w(x) h i (x)dx = w(x)ψ(x)poly of lower degree dx = The lst equlity is only true becuse we ve chosen the nodes to be the roots of the n + 1-st degree orthogonl polynomil: the w i in Hermite qudrture rule re not zero for generl node loctions, only for these specil ones Now show tht the weights re w i = w(x)(l i (x)) dx, which is the sme s the GQ weights w i = w(x)h i (x)dx = h i (x) = [1 l i(x i )(x x i )](l i (x)) w(x)(l i (x)) dx l i(x i ) The lst integrl is just w i, so it s zero by the rgument bove w(x)(x x i )(l i (x)) dx 14

15 We ve shown tht Hermite qudrture t the GQ nodes is the sme s Gussin qudrture, so we cn use the Hermite error formul to derive n error formul for GQ The Hermite interpoltion error is (ssuming sufficient smoothness) f(x) p(x) = (x x ) (x x n ) f (n+) (ξ(x)) (n + )! The qudrture error cn be nlyzed by integrting the interpoltion error f(x)dx (x x ) (x x n ) (x w i f(x i ) = f (n+) (ξ(x))dx = f (n+) x ) (x x n ) (ξ) dx (n + )! (n + )! i where the second equlity is men vlue theorem In prctice this isn t terribly useful, so we quote more-useful but hrder-to-prove theorem from ATAP Chpter 19 (Theorem 194): If f nd its derivtives through p 1 re bsolutely continuous on [, b] nd f (p) is of bounded vrition (with totl vrition V ) then the n + 1 point GQ pplied to f stisfies I[f] I n [f] 3 V 15 πp(n p 1) p+1 for n > p + 1 (Integrl is ssumed to be over [ 1, 1]) This implies tht GQ converges quickly s n even when f doesn t hve n infinite number of derivtives (ie when we cn t use the interpoltion error formul) You cn lso prove tht convergence is fster thn lgebric for certin nlytic functions f (ATAP Theorem 193) 4 Sometimes you wnt to pre-specify few node loctions, nd then let the rest be free Eg Guss-Lobtto includes both endpoints nd Guss-Rdu includes the left endpoint Guss-Kronrod nests the nodes so tht when you increse n you get to re-use old function vlues (cf Guss-Legendre, where the ll the nodes & weights chnge when you increse n) Let x,, x p be the fixed nodes nd z,, z q be the free nodes, with weights w i nd v i, respectively The qudrture rule will hve the form f(x)dx p w i f(x i ) + q v i f(z i ) (You cn put weight function in if you wnt) We hve p (q + 1) degrees of freedom (nodes & weights), so we could hope to integrte ll polynomils of degree n = p + q + exctly Now let φ (x),, φ n (x) be bsis functions for the spce of polynomils with degree n (eg monomils) The nonliner system for weights nd nodes is p w i φ k (x i ) + q v i φ k (z i ) = φ k (x)dx, k =,, n where the RHS is known As with GQ this is not the best wy to solve the problem There is firly strighforwrd lgorithm in D&B to solve the generl problem s posed here Insted of going over it (not prelim mteril) we will discuss the specil cse of Guss-Lobtto qudrture with w(x) = 1, ie we will set x = 1 nd x 1 = 1 nd try to choose nodes x 1,, x n 1 nd weights w,, w n to chieve the highest polynomil exctness possible The method is bsed on the theorem we proved erlier: (D&B Theorem 513) Let n w if(x i ) be qudrture tht integrtes polynomils of degree n exctly, nd define ψ(x) = (x x ) (x x n ) Then the interpoltory qudrture rule integrtes ll polynomils of degree n k if nd only if for ll polynomils p of degree k w(x)p(x)ψ(x)dx = 15

16 We developed GQ by mking ψ(x) proportionl to n orthogonl polynomil, which mkes it stisfy the condition of the theorem Now we wnt to use the weight function w(x) = 1, nd we hve pre-specified the nodes x nd x n So we wnt to choose the nodes x 1,, x n 1 so tht (x x )(x x 1 )p(x)(x x 1 ) (x x n 1 )dx = for ll polynomils p of degree k for k s lrge s possible This is the subtle prt: Notice tht ψ(x) = (x x 1 ) (x x n 1 ) is polynomil of degree n 1 nd tht w(x) = (x x )(x x n ) = (x )(x b) is the weight function for the (1, 1) Jcobi polynomils If we chose ψ(x) = (x x 1 ) (x x n 1 ) proportionl to the n 1-degree (1, 1) Jcobi polynomil then the integrl p(x)ψ(x)dx = w(x)p(x) ψ(x)dx = for ll polynomils p of degree n The theorem then gurntees tht the Guss-Lobtto qudrture will be exct for polynomils up to degree n 1 The sme kind of rgument cn be pplied to Guss-Rdu qudrture, but with (1, ) or (, 1) Jcobi polynomil (depending on which endpoint you use) TL;DR: Guss-Legendre-Lobtto qudrture nodes re roots of (1, 1) Jcobi polynomil; it integrtes polynomils of degree n 1 exctly 5 Recll tht if we interpolte t the roots of Chebyshev polynomil, then the node polynomil is proportionl to tht Chebyshev polynomil, nd we hve the following bound on the interpoltion error f(x) p(x) f (n+1) n (n + 1)! In fct, we stted in the section on interpoltion (without proof nd without detils) tht the interpoltion error goes to zero for lrge clss of functions when interpolting t the Chebyshev nodes If we use n interpoltory qudrture where the nodes re the roots of Chebyshev polynomil, then the qudrture error is (s usul) just the integrl of the interpoltion error, which is bounded by (n+1) f(x)dx I n [f] (b ) f n (n + 1)! Tht s relly smll! Fejér s first rule is exctly s bove: qudrture nodes re the roots of Chebyshev polynomil Fejér s second rule uses the criticl points of Chebyshev polynomil (ie the roots of Chebyshev polynomil of the second kind) s nodes The Clenshw-Curtis qudrture uses the sme nodes s Fejér s second rule plus the endpoints, ie the extrem of the Chebyshev polynomil on [ 1, 1] In every cse there re nlyticl expressions for the weights, ll of which re positive There re lso efficient implementtions bsed on the FFT Clenshw-Curtis nd Fejér s second rule hve the benefit tht you cn re-use points if you move from n + 1 to n + 1 points The fct tht the weights re positive implies tht the qudrture converges for ny continuous integrnd This is nother exmple of the gp between qudrture error nd interpoltion error: interpoltion t the Chebyshev points is not gurnteed to converge for ny continuous f, but the qudrture is gurnteed to converge The rte of convergence depends on how mny continuous derivtives the function hs The following theorem shows surprising property of Clenshw-Curtis qudrture: 16

17 (ATAP Theorem 195) If f nd its derivtives through p 1 re bsolutely continuous on [, b] nd f (p) is of bounded vrition (with totl vrition V ) then the n + 1 point Clenshw-Curtis qudrture pplied to f stisfies I[f] I n [f] 3 V 15 πp(n p 1) p+1 for n > N where N is threshold tht depends on p but not f (Integrl is ssumed to be over [ 1, 1]) As with the GQ result, this cn t be bsed on interpoltion error becuse it only ssumes f hs finite number of derivtives, while the interpoltion error ssumes tht f hs n + 1 derivtives More surprisingly: Clenshw-Curtis hs the sme bound on the rte of convergence s GQ despite the fct tht it only integrtes polynomils up to degree n exctly Miscellneous 1 Singulr integrls: either f(x) (or derivtive) hs singulrity on the boundry of [, b], or the intervl is infinite If the integrnd is piecewise-smooth, just split the integrl into subintervls where the integrnd is smooth on ech subintervl Chnge vribles to n integrl without singulrity Put the singulrity in the weight function Chnge vribles to problem with finite intervl Exmple: Chnge from n integrnd tht is only C[, 1] to one tht is C [, 1] 1 1 xdx = t dt Exmple: Chnge from singulr integrnd to smooth one 1 e x x dx = Exmple: Put the singulrity in the weight function 1 1 e t dt e x x dx, w(x) = x 1/ Let s find Guss-Christoffel qudrture with this weight function We need to compute the orthogonl polynomils; just use Grm-Schmidt 1, x φ (x) = 1, φ 1 (x) = x 1 1 = x 1 w 3, φ (x) = x φ 1, x φ 1 φ 1 (x) 1, x w 1 1 = x 6 ( x 1 ) 1 w φ 3 (x) = = x φ (x) 5 7 The roots of φ 3 re vilble in closed form becuse it s cubic; ( x 1 3 ) 1 7 x , x , x We cn find the weights by integrting the Lgrnge polynomils w = 1 w(x)l (x)dx 93588, w 1 = 1 w(x)l 1 (x)dx 7153, w = 17 1 w(x)l (x)dx 34649

18 Notice tht the sum of the weights equls 1 w(x)dx = The nswer is bout 9 nd the error is bout Tht s pretty good for 3-point qudrture, but then gin we re essentilly pproximting e x by qudrtic on [, 1], which is pretty ccurte Exmple: Chnge n infinite intervl to finite one (Could use GQ on this one too) e x π/ 1 + x dx = e 1 sec (θ) dθ π/ True vlue 13439, Trp rule with n = 4, 8, 16 points: 157, 13398, Trp rule with 3 points hs error This is n exmple showing tht the Trp Rule is extremely ccurte for functions tht re C with ll derivtives equl to zero t the endpoints This section ws originlly written for pproximtions on [ π, π], to mtch the mteril on pproximtion theory It hs been re-written for pproximtion on [, π) so tht the DFT/FFT connection to interpoltion is more cler (Our Trig interpoltion theory ws on [, π)) We hve seen tht the trpezoid rule will converge fster-thn-lgebriclly for infinitely-smooth periodic functions Recll tht in Fourier L pproximtion theory the Fourier coefficients re given by j = 1 π b j = 1 π π π f(x) cos(jx)dx, j 1 f(x) sin(jx)dx, j 1 More generlly, the coefficients of the complex Fourier series re c j = 1 π π f(x)e ijx dx It s certinly convenient if you cn evlute these integrls nlyticlly, but often you cn t Enter the Trpezoid Rule If you use the trpezoid rule to compute Fourier coefficients the results won t be perfect, but the errors will converge (s h ) extremely fst for smooth functions f Applied to the complex Fourier integrl bove, c j 1 n 1 f(x k )e ijx k, x k = k π n n k= Note tht there s no hlving of the endpoints becuse the function is periodic Note wht hppens if you ttempt to compute c j+n c j+n 1 n 1 f(x k )e i(j+n)x k, (j + n)x k = π jk n n + πk k= e i(j+n)x k = e ijx k e πki = e ijx k This implies tht the computed coefficients c j nd c j+n re the sme This is clled lising We cn therefore only compute n distinct Fourier coefficients using n equispced points, nd we compute Fourier coefficients for vlues of j =,, n 1 If we wnt to compute ll the Fourier coefficients from j =,, n 1 we cn write them s c = Ff 18

19 where the elements of F re F j,k = e πijk/n /n This mtrix is the sme s the DFT mtrix we sw in the section on equispced trigonometric interpoltion (trig interpoltion section 3 from the notes) It is unitry up to constnt scling fctor, nd there is fst lgorithm (costing O(n log(n)) rther thn O(n ) opertions) clled the FFT to pply it to vectors Suppose tht f(x) hs the following exct Fourier representtion f(x) = Applying the Trpezoid Rule to this we see tht ccording to the Trp Rule c j 1 n n m= k=1 m= c m e imx c m e i(m j)x k = 1 n m= c m n k=1 e i(m j)x k Look t the inner sum: n e i(m j)x k = k=1 n e i(m j)(k π n ) = k=1 n k=1 ( e (m j) π n i) k This is geometric sum with the vlue n whenever m j is divisible by n, nd zero otherwise Returning to the overll sum ccording to the Trp Rule c j c j+qn So the computed coefficient is the sum of coefficients, c j n, c j, c j+n, q= The error between the Fourier coefficient computed using the Trp Rule nd the true Fourier coefficient is the sum of ll the true coefficients in the lising set 3 Adptive qudrture Integrte(f(x),,b,tol) Compute I n [f] Estimte the error If the error is smll enough, return I n [f] Else return Integrte(f(x),,(+b)/,tol ) + Integrte(f(x),(+b)/,b,tol ) Drw Picture The bsolute vlue of the totl error is the sum of the bsolute errors on ech subintervl The prcticl pproch is to estimte the error by computing two different qudrtures on ech intervl, then subtrcting them For exmple, error I n [f] I n [f] The min questions re (i) wht s the underlying qudrture I n, nd how do you estimte the locl error, nd (ii) wht is tol? There re lots of wys to do this We will not go into detil on ny methods Bsiclly you just need be wre tht dptive methods exist 4 Multivrite The simplest wy to pproch multivrite integrl is to reduce it to sequence of univrite integrls Eg d c f(x, y)dydx = F (x)dx where F (x) = d c f(x, y)dy 19

20 The difficulty here is tht you re not ble to evlute F (x) exctly To develop n error bound for the whole integrl you couldn t use our stndrd estimtes becuse our stndrd estimtes ssume tht you re evluting the integrnd exctly Our qudrtures re ll of the form I n [f] = w i f(x i ) i If you cn t evlute the integrnd exctly, then you hve insted w i (f(x i ) + ɛ i ) = w i f(x i ) + i i i w i ɛ i Now the usul error estimtes pply to the first term, nd we cn bound the second term by i w i ɛ i i w i ɛ i The errors ɛ i re bounded using the stndrd estimtes for the inner integrl, so overll you get bounded error An lterntive is to develop qudrture bsed on multivrite interpoltion Our only multivrite interpoltion strtegy (in this clss) is to use tensor product grid In D the n n interpoltion problem reduces to sequence of n one-dimensionl interpoltion problems Then, once you hve the interpolting polynomil you cn compute its integrl which gives you the qudrture The min difficulty is tht tensor product grid requires rectngulr domin, so if you re integrting over circle or tringle or something you cn t exctly fill it with rectngles (Drw picture) To del with this, suppose tht Ω is your domin of integrtion, nd tht you fill it s much s possible with rectngles so tht the qudrture domin is Ω h Ω Under pproprite ssumptions on Ω (eg compctness), we hve tht Ω h Ω s you refine the grid Now consider the integrl f(x)dx = f(x)dx + f(x)dx Ω Ω h Ω\Ω h Assuming f is continuous on the compct set Ω then the second integrl will go to s h (ie s the grid is refined) Conversely, if we use convergent qudrture on ech of the rectngles in Ω h, then the qudrture will converge to the first integrl s the grid is refined There re mny other multidimensionl qudrture rules, eg you cn use tringles rther thn rectngles s your bsic grid You cn lso use Monte-Crlo estimtes for high-dimensionl problems

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