Advnced Computtionl Fluid Dynmics AA215A Lecture 3 Polynomil Interpoltion: Numericl Differentition nd Integrtion Antony Jmeson Winter Qurter, 2016, Stnford, CA Lst revised on Jnury 7, 2016
Contents 3 Polynomil Interpoltion: Numericl Differentition nd Integrtion 2 3.1 Interpoltion Polynomils................................. 2 3.2 Error in the Interpolting Polynomil.......................... 3 3.3 Error of Derivtive with Lgrnge Interpoltion..................... 3 3.4 Error of the k th Derivtive of the Interpoltion Polynomil.............. 4 3.5 Interpoltion with Tringle of Polynomils....................... 5 3.6 Newton s Form of the Interpoltion Polynomil..................... 6 3.7 Hermite Interpoltion................................... 8 3.8 Integrtion Formuls using Polynomil Interpoltion.................. 10 3.9 Integrtion with weight function............................ 10 3.10 Newton Cotes formuls.................................. 10 3.11 Guss Qudrture..................................... 11 3.12 Formuls for Guss integrtion with constnt weight function............. 13 3.13 Error bound for Guss integrtion............................ 13 3.14 Discrete orthogonlity of orthogonl polynomils.................... 14 3.15 Equivlence of interpoltion nd lest squres pproximtion using Guss integrtion 14 3.16 Guss Lobto Integrtion................................. 15 3.17 Portrits of Lgrnge, Newton nd Guss........................ 16 1
Lecture 3 Polynomil Interpoltion: Numericl Differentition nd Integrtion 3.1 Interpoltion Polynomils So fr we hve not required P n (x) to equl f(x) t ny points in the intervl. This permits some smoothing! A nturl pproch is to mke P n (x) = 0 + 1 x +... + n x n = f(x) t n + 1 points x 0, x 1,..., x n. Then we hve n + 1 equtions for the n + 1 coefficients. Such polynomil is clled n interpoltion polynomil. A solution exists becuse the determinnt of the equtions is the Vndemonde determinnt 1 x 0... x n 0 1 x 1... x n 1 D =..... =. i x j ) 0 (3.1) i>j(x 1 x n... x n n This cn be seen by noting tht D vnishes if ny x i equls ny x j, but D is polynomil of just the sme order s the product. The interpoltion polynomil is unique since if there were 2 such polynomils P n (x), Q n (x), then P n (x i ) Q n (x i ) = 0 for i = 0, 1,..., n (3.2) But then P n (x) Q n (x) is n n th degree polynomil with n + 1 roots, so it must be zero. In prctice it is esiest to construct P n (x) indirectly s sum of specilly chosen polynomils rther thn solve for the j. To do this note tht φ n,j (x) = (x x 0)(x x 1 )...(x x j 1 )(x x j+1 )...(x x n ) (x j x 0 )(x j x 1 )...(x j x j 1 )(x j x j+1 )...(x j x n ) = 1 when x = x j (3.3) Then we cn get P n (x) = f(x j )φ n,j (x) (3.4) which equls f(x) t x = x 0, x 1,..., x n, nd is the only such polynomil s hs just been shown. The φ n,j (x) re clled Lgrnge interpoltion coefficients. We cn write φ n,j (x) = w n (x) (x x j )w n(x j ) (3.5) 2
LECTURE 3. NUMERICAL DIFFERENTIATION AND INTEGRATION 3 w n (x) = (x x 0 )(x x 1 )...(x x n ) (3.6) 3.2 Error in the Interpolting Polynomil We shll show tht if f(x) hs n (n + 1) th derivtive nd P n (x) is n interpolting polynomil then the reminder is f(x) P n (x) = (x x 0)(x x 1 )...(x x n ) f (n+1) (ξ) (3.7) (n + 1)! ξ is in the intervl defined by x 0, x 1,..., x n. Let S n (x) be defined by f(x) P n (x) = w n (x)s n (x) (3.8) Define w n (x) = (x x 0 )(x x 1 )...(x x n ) (3.9) F (z) = f(z) P n (z) w n (z)s n (x) (3.10) This is continuous in z nd vnishes t n + 2 points x 0, x 1,..., x n, x. Thus by Rolle s theorem F (z) vnishes t n + 1 points, F (z) vnishes t n points,..., nd finlly F (n+1) (z) vnishes t one point ξ. But d n+1 dz n+1 P n(z) = 0 (3.11) Thus nd setting z = ξ d n+1 dz n+1 w n(z)s n (x) = (n + 1)!S n (x) (3.12) F (n+1) (z) = f n+1 (z) (n + 1)!S n (x) (3.13) S n (x) = f(x) P n (x) = 1 (n + 1)! f (n+1) (ξ) (3.14) w n(x) (n + 1)! f (n+1) (ξ) (3.15) Since the error depends on w n (x) we nturlly sk how the x i should be distributed to minimize w n (x). 3.3 Error of Derivtive with Lgrnge Interpoltion Suppose tht Then f (x) = P n(x) + f(x) = P n (x) + w n (n + 1)! f (n+1) (ξ) (3.16) w n (n + 1)! f (n+1) (ξ) + w n (n + 1)! f (n+2) (ξ) dξ dx (3.17)
LECTURE 3. NUMERICAL DIFFERENTIATION AND INTEGRATION 4 n w n = (x x k ) (3.18) Then t n interpoltion point, f (x j ) = P n(x j ) + k j n (x j x k ) f (n+1) (ξ) (3.19) (n + 1)! k j with equl intervls h since With one intervl, f (x 0 ) = P (x 0 ) + hn n + 1 f (n+1) (ξ) (3.20) (x 1 x 0 )(x 2 x 0 )...(x n x 0 ) = n!h n (3.21) f (x 0 ) = f(x 1) f(x 0 ) h + h 2 f (ξ) (3.22) 3.4 Error of the k th Derivtive of the Interpoltion Polynomil Let f n+1 (x) be continuous nd let Then R n (x) = f(x) P n (x) (3.23) n k R n (k) (x) = (x ξ j ) f (n+1) (ξ) (n + 1 k)! (3.24) the distinct points ξ j re independent of x nd lie in the intervls x j < ξ j < x j+k for j = 0, 1,..., n k nd ξ(x) is some point in the intervl contining x nd the ξ j. Proof: R n = f P n hs n+1 continuous derivtives nd vnishes t n+1 points x = x j, for j = 0, 1,..., n. Apply Rolle s theorem k n times. The zeros re then distributed s illustrted.
LECTURE 3. NUMERICAL DIFFERENTIATION AND INTEGRATION 5 This leds to the following tble for the distribution of the n k + 1 zeros ξ j of R (k) n. Define R R (1) R (2) R (k) x 0 x 1 (x 0, x 1 ) x 2 (x 1, x 2 ) (x 0, x 2 ). x k (x k 1, x k ) (x k 2, x k ) (x 0, x k ). x n (x n 1, x n ) (x n 2, x n ) (x n k, x n ) For ny x distinct from the ξ j choose α such tht n k F (z) = R n (k) (z) α (z ξ j ) (3.25) F (x) = 0 (3.26) Then F (z) hs n k + 2 zeros nd by Rolle s theorem F (n k+1) (z) hs one zero, η, sy, in the intervl contining x nd the ξ j. Thus 0 = F (n k+1) (η) (3.27) = Rn n+1 (η) α(n k + 1)! (3.28) = f n+1 (η) α(n k + 1)! (3.29) or α = It follows tht for mesh width bounded by h f n+1 (η) (n k + 1)! (3.30) M is proportionl to sup f n+1 (x) in the intervl. R (k) n Mh n k+1 (3.31) 3.5 Interpoltion with Tringle of Polynomils Set w 0 (x) = (x x 0 ) w 1 (x) = (x x 0 )(x x 1 ) w k (x) = (x x 0 )(x x 1 )...(x x k ) Now to interpolte f(x) with vlues f i t x i we set p(x) = 0 + 1 (x x 0 ) + 2 (x x 0 )(x x 1 )... (3.32)
LECTURE 3. NUMERICAL DIFFERENTIATION AND INTEGRATION 6 Then p 0 (x 0 ) = f 0 = 0 p 1 (x 1 ) = f 1 = 0 + 1 (x 1 x 0 ) so we cn solve for the i in succession p 2 (x 2 ) = f 2 = 0 + 1 (x 2 x 0 ) + 2 (x 2 x 0 )(x 2 x 1 ). k = f k 0 1 (x k x 0 )... k 1 (x k x 0 )...(x k x k 2 ) (x k x 0 )(x k x 1 )...(x k x k 1 ) = f k p k 1 (x k ) w k 1 (x k ) (3.33) 3.6 Newton s Form of the Interpoltion Polynomil The Lgrngin form of the interpoltion polynomil hs the disdvntge tht the coefficients hve to be recomputed if new interpoltion point is dded. To void this let P n (x) = k w k 1 (x) (3.34) k=0 w k (x) = (x x 0 )(x x 1 )...(x x k ) nd w 1 (x) = 1 (3.35) We cn determine the k so tht Suppose this hs been done for P k 1, nd we now dd x k. Then P n (x j ) = f(x j ) for j = 0, 1,..., n (3.36) P k (x) = P k 1 (x) + k w k 1 (x) (3.37) P k (x j ) = P k 1 (x j ) = f(x j ) for j = 0, 1,..., k 1 (3.38) P k (x k ) = P k 1 (x k ) + k w k 1 (x k ) = f(x k ) (3.39) k = f(x k) P k 1 (x k ) w k 1 (x k ). On the other hnd the coefficient of the highest order term in the Lgrnge expression is so by comprison n = f(x j ) w n(x j ) f(x j ) w n(x j ) Thus n is liner combintion of f(x j ) for j = 0, 1,..., n. Multiply by (3.40) (3.41) x n x 0 = x n x j + x j x 0 (3.42)
LECTURE 3. NUMERICAL DIFFERENTIATION AND INTEGRATION 7 Then n (x n x 0 ) = = f(x j ) w n(x j ) (x j x n ) + f(x j ) k j (x j x k ) + j=1 j=1 f(x j ) w n(x j ) (x j x 0 ) f(x j ) k j (x j x k ) (3.43) These re just the expressions for n 1 in the intervls x 0 to x n 1 nd x 1 to x n. Thus if we define f[x 0,..., x n ] = n = f(x j ) k j (x j x k ) (3.44) we find tht nd strting from we hve f[x 0,..., x n ](x 0 x n ) = f[x 1,..., x n ] f[x 0,..., x n 1 ] (3.45) giving Newton s form of the interpoltion polynomil f[x 0 ] = f(x 0 ) (3.46) f[x 0, x 1 ] = f[x 1] f[x 0 ] x 1 x 0 (3.47) f[x 0, x 1, x 2 ] = f[x 1, x 2 ] f[x 0, x 1 ] x 2 x 0 (3.48) P n (x) = f[x 0 ] + (x x 0 )f[x 0, x 1 ] +... + (x x 0 )...(x x n 1 )f[x 0,..., x n ] (3.49) The squre brcketed expressions re Newton s divided differences, nd we now hve Newton s form for the interpoltion polynomil: P n (x) = f[x 0 ] + (x x 0 )f[x 0, x 1 ] +... + (x x 0 )...(x x n 1 )f[x 0,..., x n ] (3.50) ech new term is independent of the previous terms. To estimte the mgnitude of the higher divided differences we cn use the result lredy obtined for the reminder. Theorem: Let x 0, x 1,..., x k 1 be distinct points nd let f be continuous in intervl contining ll these points. Then for some point ξ in this intervl Proof: From Newton s formul f[x 0,..., x k 1, x] = f (k) (ξ) k! (3.51) f(x) P k 1 (x) = (x x 0 )...(x x k 1 )f[x 0,..., x k 1, x] = (x x 0 )...(x x k 1 ) f (k) (ξ) k! (3.52) by the reminder theorem. But x is distinct from x 0, x 1,..., x k 1 so the theorem follows on dividing out the fctors.
LECTURE 3. NUMERICAL DIFFERENTIATION AND INTEGRATION 8 3.7 Hermite Interpoltion We cn generlize interpoltion by mtching derivtives s well s vlues t the interpoltion points. Such polynomil is clled the osculting polynomil nd the procedure is clled Hermite interpoltion. Let H 2n+1 (x) be polynomil of degree 2n + 1 such tht H 2n+1 (x) = f(x j ) for j = 0, 1,..., n (3.53) H 2n+1(x) = f (x j ) for j = 0, 1,..., n (3.54) The 2n + 2 coefficients re needed to stisfy 2n + 2 conditions. H 2n+1 (x) cn be found indirectly s in Lgrnge interpoltion. Let ψ n,j (x) nd γ n,j (x) be polynomils of degree 2n + 1 such tht ψ n,j (x i ) = δ ij, ψ n,j(x i ) = 0, Then the Hermite interpoltion polynomil is for i = 0, 1,..., n γ n,j (x i ) = 0, γ n,j(x i ) = δ ij, for i = 0, 1,..., n H 2n+1 (x) = ( f(xj )ψ n,j (x) + f (x j )γ n,j (x) ) (3.55) It cn be directly verified by differentition tht the required polynomils re nd φ n,j (x) re the Lgrnge polynomils stisfying The error in Hermite interpoltion is ψ n,j (x) = (1 2φ n,j(x)(x x j ))φ 2 n,j(x) (3.56) γ n,j (x) = (x x j )φ 2 n,j(x) (3.57) φ n,j (x i ) = δ ij for i = 0, 1,..., n (3.58) f(x) H 2n+1 (x) = ω 2 n(x) f 2n+2 (ξ) (2n + 2)! (3.59) ω n (x) = (x x 0 )(x x 1 )...(x x n ) (3.60) This cn be proved in the sme wy s the error estimte for Lgrnge interpoltion now F (ξ) = f(ξ) H 2n+1 (ξ) ω 2 n(ξ)s n (x) (3.61) nd on the first ppliction of Rolle s theorem F (ξ) hs 2n + 2 distinct zeros. The Hermite polynomil cn be obtined vi pssge to the limit from the Lgrnge polynomil P 2n+1 (x) which interpoltes f(x) t the 2n + 2 points x 0, x 0 + h, x 1, x 1 + h,..., x n, x n + h (3.62)
LECTURE 3. NUMERICAL DIFFERENTIATION AND INTEGRATION 9 s h 0 to produce n + 1 pirs of coincident points. Define N(h) = (x x k + h) (3.63) nd D(h) = k=1,n, k=1,n, Then the contribution of the points x j nd x j + h to P 2n+1 (x) is Expnding f(x j + h) in Tylor series k j k j Q(x) = f(x j ) N(0) N( h) x (x j + h) D(0) D( h) x j (x j + h) + f(x j + h) N(0) D(h) (x j x k + h) (3.64) N( h) x x j D(0) h Q(x) = f(x j ) N(0) N( h) x (x j + h) D(0) D( h) x j (x j + h) + ( f(x j ) + hf (x j ) + O(h 2 ) ) N(0) N( h) x x j D(h) D(0) h = f(x j ) N(0) [ N( h) D(0) D( h) x x ( j 1 N( h) h D( h) 1 )] D(h) + f (x j )(x x j ) N(0) N( h) D(0) D(h) + O(h) Also Thus in the limit s h 0 1 D( h) 1 D(h) = 2h D (0) D 2 (0) + O(h2 ) (3.65) Q(x) = f(x j ) N 2 ( ) (0) D 2 1 2(x x j ) D (0) (0) D(0) + f (x j )(x x j ) N 2 (0) D 2 (0) nd D (0) equls d dx N(0) evluted t x j, so tht N(0) D(0) = φ n,j(x) (3.66) D (0) D(0) = φ n,j(x j ) (3.67)
LECTURE 3. NUMERICAL DIFFERENTIATION AND INTEGRATION 10 3.8 Integrtion Formuls using Polynomil Interpoltion Let P n 1 (x) interpolte f(x) t x 0, x 1,..., x n. Then P n (x) = f(x j )φ n,j (x) (3.68) so tht φ n,j (x) = k j (x x k) k j (x j x k ) (3.69) φ n,j (x k ) = δ jk (3.70) Now pproximte f(x)dx by P n(x)dx typiclly = x 0, b = x n. Then E(f) denotes the error, nd f(x)dx = A j = A j f(x j ) + E(f) (3.71) φ n,j (x)dx (3.72) The result is exct if f(x) is polynomil of degree n, since then P n (x) = f(x). This is clled precision of degree n. 3.9 Integrtion with weight function We my include weight function w(x) 0 in the integrl. Then now f(x)w(x)dx = A j = A j f(x j ) + E(f) (3.73) j=n φ n,j (x)w(x)dx (3.74) Depending on w(x), nlyticl expressions re not necessrily vilble for evluting the coefficients A j. 3.10 Newton Cotes formuls These re derived by pproximting f(x) by n interpoltion formul using n + 1 eqully spced points in [, b] including the end points. The first 3 such formuls re: Trpezoidl rule f(x)dx = h (f() + f(b)) + E(f), h = b (3.75) 2
LECTURE 3. NUMERICAL DIFFERENTIATION AND INTEGRATION 11 Simpson s rule f(x)dx = h 3 f() + 4h 3 f( + h) + h b f(b) + E(f), h = 3 2 (3.76) nd Simpson s 3 8 formul f(x)dx = 3h 8 f() + 9h 8 f( + h) + 9h 8 f( + 2h) + 3h 8 3.11 Guss Qudrture f(b) + E(f), h = b 3 (3.77) When the interpoltion points x re fixed the integrtion formul hs n + 1 degrees of freedom corresponding to the coefficients A j for j = 0,..., n. Accordingly we cn find vlues of A j which yield exct vlues of the integrl of ll polynomils of degree n, P n (x) = 0 + 1 x +... + n x n, (3.78) since these lso hve n + 1 degrees of freedom corresponding to the coefficients j for j = 0,..., n. If we re lso free to choose the integrtion points x j, then we hve 2n + 2 degrees of freedom corresponding to x j nd A j, for j = 0,..., n. Now it is possible to find vlues of x j nd A j which enble exct integrtion of polynomils of degree 2n + 1. One could try to find the required vlues by solving 2n + 2 nonliner equtions A j f(x j ) = f(x)dx (3.79) f(x) = 1, x,..., x 2n+1 in turn. However the required vlues cn be found indirectly s follows. Let φ i (x) be orthogonl polynomils for the weight function w(x), so tht φ j (x)φ k (x)w(x) = 0 for j k (3.80) Choose x j s the zeros of φ n+1 (x). Then integrtion using polynomil interpoltion is exct for polynomils up to degree 2n + 1. Proof: If f(x) is polynomil of degree 2n + 1 it cn be uniquely expressed s q(x) nd R(x) re polynomils of degree n. Then f(x)w(x)dx = f(x) = q(x)φ n+1 (x) + R(x) (3.81) q(x)φ n+1 (x)w(x)dx + R(x)w(x)dx = since φ n+1 (x) is orthogonl to ll polynomils of degree < n + 1. R(x)w(x)dx (3.82)
LECTURE 3. NUMERICAL DIFFERENTIATION AND INTEGRATION 12 Also the pproximte integrl is I = = = A j f(x j ) A j q(x j )φ n+1 (x j ) + A j R(x j ) A j R(x j ) (3.83) since φ n+1 (x j ) is zero for j = 0, 1,..., n. But then I = R(x)w(x)dx (3.84) exctly, since R(x) is polynomil of degree n. The degree of precision 2n + 1 is the mximum ttinble. Consider the polynomil of φ 2 n+1 (x) of degree 2n + 2. Then I = 0 (3.85) since φ n+1 (x j ) = 0 for j = 0, 1,..., n. But φ 2 n+1(x)w(x)dx > 0 (3.86) becuse φ 2 n+1 (x) > 0 except t the zeros x 0, x 1,..., x n. When pplied to n rbitrry smooth function f(x) which cn be expnded s f() + (x )f () +... + (x )2n+2 f (2n+2) (ξ) (3.87) (2n + 2)! the error is proportionl to bound on f (2n+2) (x) becuse the preceding terms in the Tylor series re integrted exctly.
LECTURE 3. NUMERICAL DIFFERENTIATION AND INTEGRATION 13 3.12 Formuls for Guss integrtion with constnt weight function Number of Points x j A j Precision Order 1 0 2 1 2 2 ± 1 3 1 3 4 3 4 5 ± ± 8 0 ± 15 5 r 5 9 q 3 2 6 5 r 7 q 3+2 6 5 18 30 7 36 128 0 225 ± 5 2 ± 5 + 2 10 7 10 7 9 5 6 18+ 30 36 7 8 322+13 70 900 322 13 70 900 9 10 Tble 3.1: Formuls for Guss integrtion with constnt weight function 3.13 Error bound for Guss integrtion Let H 2n+1 (x) be the Hermite interpoltion polynomil to f(x) t the integrtion points. The Guss integrtion formul is exct for H 2n+1 (x). Hence Thus H 2n+1 (x)w(x)dx = f(x)w(x)dx A j f j = i=0 A j Q(x j ) = According to the error formul for Hermite interpoltion A j f j (3.88) f(x) H 2n+1 (x) = P 2 n(x) f (2n+2)(ξ) (f(x) H 2n+1 (x))w(x)dx (3.89) (2n + 2)! (3.90) P n (x) = (x x 0 )(x x 1 )...(x x n ) (3.91) is the orthogonl polynomil normlized so tht its leding coefficient is unity. Then the error is 1 b E = w(x) (2n + 2)! P n(x)f 2 (2n+2) (ξ)dx (3.92) Since w(x) ˆP 2 n(x) 0 the error lies in the rnge between min f (2n+2) (ξ)a nd mx f (2n+2) (ξ)a (3.93)
LECTURE 3. NUMERICAL DIFFERENTIATION AND INTEGRATION 14 nd hence for some vlue of η in [, b]. A = 1 (2n + 2)! w(x) P 2 n(x)dx (3.94) E = Af (2n+2) (η) (3.95) 3.14 Discrete orthogonlity of orthogonl polynomils Let φ j (x) be orthogonl polynomils stisfying (φ j, φ k ) = φ j (x)φ k (x)w(x)dx = 0 for j k (3.96) Let x i be the zeros of φ n+1 (x). Then if j n, k n, φ j (x)φ k (x) is polynomil of degree 2n. Accordingly (φ j, φ k ) is evluted exctly by Guss integrtion. This implies tht the φ j stisfy the discrete orthogonlity condition A i φ j (x i )φ k (x i ) = 0, given j k, j n, k n (3.97) i=0 A i re the coefficients for Guss integrtion t the zeros of φ n+1. 3.15 Equivlence of interpoltion nd lest squres pproximtion using Guss integrtion Let f (x) be the lest squres fit to f(x) using orthogonl polynomils φ j (x) for the intervl [, b] with weight function w(x) 0. Then f (x) = c jφ j (x) (3.98) c j = (f, φ j) (φ j, φ j ) = f(x)φ j(x)w(x)dx φ2 j (x)w(x)dx (3.99) Let x i be the zeros of φ n+1 (x), nd let P n (x) be the interpoltion polynomil to f(x) stisfying P n (x i ) = f(x i ) for i = 0, 1,..., n (3.100) P n (x) cn be expnded s P n (x) = Then the ĉ k re determined by the conditions ĉ k φ k (x) (3.101) k=0 ĉ k φ k (x i ) = f(x i ) for i = 0, 1,..., n (3.102) k=0
LECTURE 3. NUMERICAL DIFFERENTIATION AND INTEGRATION 15 Also the polynomils φ j (x) stisfy the discrete orthogonlity condition A i φ j (x i )φ k (x i ) = 0 given j k, j n, k n (3.103) i=0 A i re the coefficients for Guss integrtion. Now multiplying (3.102) by A i φ j (x i ) nd summing over i, it follows tht ĉ j = n i=0 A if(x i )φ j (x i ) n i=0 A iφ 2 j (x i) (3.104) This is exctly the formul (3.99) for evluting the lest squres coefficient c j by Guss integrtion. Alterntive Proof: The coefficients ĉ j re ctully the lest squres coefficients which minimize in the discrete semi-norm in which J = f f(x) c j φ j 2 (3.105) c j φ j (x) is evluted by Guss integrtion t the zeros x i of φ n+1 (x) s J = i=0 A i f(x i ) 2 c j φ j (x i ) w(x)dx (3.106) 2 w(x 0 ) (3.107) But s consequence of the interpoltion condition (3.99) the coefficients ĉ j yield the vlue J = 0, thus minimizing J. 3.16 Guss Lobto Integrtion To integrte f(x) over [-1,1] choose integrtion points t 1, 1, nd the zeros of Q n 1, which re the zeros of (1 + x)(1 x)φ n 1 (x) (3.108) Then let f(x) = q(x)(1 x 2 )φ n 1 (x) + r(x) (3.109) if f(x) is polynomil of degree 2n 1, q(x) nd r(x) re polynomil of degree n 2. Now let the polynomils φ j be orthogonl for the weight function w(x) = 1 x 2 (3.110) Then I = 1 1 f(x)dx = 1 1 1 1 q(x)φ n 1 (x)w(x)dx + r(x)dx = r(x)dx (3.111) 1 1
LECTURE 3. NUMERICAL DIFFERENTIATION AND INTEGRATION 16 since φ n 1 is orthogonl to ll polynomils of lower degree. Also the pproximte integrl is Then Î = Î = A j = A j f(x j ) (3.112) 1 1 A j q(x j )(1 x 2 j)φ n 1 (x j ) + since (1 x 2 j )φ n 1(x j ) = 0 for j = 0, 1,..., n. But then φ n,j (x)dx (3.113) A j r(x j ) = A j r(x j ) (3.114) Î = I (3.115) since r(x) is polynomil of degree < n. Here φ j re the Jcobi polynomils P (1,1) j, P (α,β) j is orthogonl for the weight (1 x) α (1 + x) β (3.116) 3.17 Portrits of Lgrnge, Newton nd Guss Figure 3.1: Joseph-Louis Lgrnge (1736-1813)
LECTURE 3. NUMERICAL DIFFERENTIATION AND INTEGRATION 17 Figure 3.2: Sir Isc Newton (1642-1726) Figure 3.3: Crl Friedrich Guss (1777-1855)