Chapter 8 Integration Using Chebyshev Polynomials In this chapter we show how Chebyshev polynomials and some of their funda- mental properties can be made to play an important part in two key techniques of numerical integration. • Gaussian quadrature estimates an integral by combining values of the integrand at zeros of orthogonal polynomials. We consider the special case of Gauss–Chebyshev quadrature, where particularly simple proce- dures follow for suitably weighted integrands. • One can approximately integrate a function by expanding it in a series and then integrating a partial sum of the series. We show that, for Chebyshev expansions, this process — essentially the Clenshaw–Curtis method — is readily analysed and again provides a natural procedure for appropriately weighted integrands. Although this could be viewed as an ‘applications’ chapter, which in an introductory sense it certainly is, our aim here is primarily to derive further basic properties of Chebyshev polynomials. 8.1 Indefinite integration with Chebyshev series If we wish to approximate the indefinite integral X h(X)= w(x)f(x)dx, −1 where −1 <X≤ 1, it may be possible to do so by approximating f(x)on [−1, 1] by an nth degree polynomial fn(x) and integrating w(x)fn(x) between −1andX, giving the approximation X h(X) hn(X)= w(x)fn(x)dx. (8.1) −1 Suppose, in particular, that the weight w(x) is one of the four functions 1 1 1 w(x)=√ , 1, √ , √ , (8.2) 1 − x2 1 − x 1+x and that we take fn(x) as the partial sum of the expansion of f(x) in Cheb- yshev polynomials of the corresponding one of the four kinds Pk(x)=Tk(x),Uk(x),Vk(x),Wk(x). (8.3) © 2003 by CRC Press LLC Then we can use the fact that (excluding the case where Pk(x)=Tk(x)with k =0) X w(x)Pk (x)dx = Ck(X)Qk(X) − Ck(−1)Qk(−1) −1 where Qk(X)=Uk−1(X),Tk+1(X),Wk(X),Vk(X) (8.4a) and √ √ √ 1 − X2 1 1 − X 1+X Ck(X)=− , , 2 1 , −2 1 , (8.4b) k k +1 k + 2 k + 2 respectively. (Note that Ck(−1) = 0 in the first and fourth cases.) This follows immediately from the fact that if x =cosθ then we have d k cos kθ sin kθ = − , dx sin θ d (k +1)sin(k +1)θ cos(k +1)θ = , dx sin θ d (k + 1 )cos(k + 1 )θ sin(k + 1 )θ = − 2 2 , dx 2 sin θ d (k + 1 )sin(k + 1 )θ cos(k + 1 )θ = 2 2 . dx 2 sin θ In the excluded case, we use d 1 θ = − dx sin θ to give X 1 √ 0 − − − 2 T (x)dx = arccos( 1) arccos X = π arccos X. −1 1 − x Thus, for each of the weight functions (8.2) we are able to integrate the weighted polynomial and obtain the approximation hn(X) explicitly. Suppose that n n fn(x)= akTk(x)[Pk = Tk]or akPk(x)[Pk = Uk,Vk,Wk]. (8.5) k=0 k=0 Then in the first case n X hn(X)= ak w(x)Tk(x)dx = k=0 −1 √ n 2 1 1 − X = 2 a0(π − arccos X) − ak Uk−1(X), (8.6) k=1 k © 2003 by CRC Press LLC while in the second, third and fourth cases n X n X hn(X)= ak w(x)Pk (x)dx = ak [Ck(x)Qk(x)]−1 . (8.7) k=0 −1 k=0 The above procedure is a very reliable one, as the following theorem demonstrates. Theorem 8.1 If f(x) is L2-integrable with respect to one of the weights w(x), as defined by (8.2), and hn(X) is defined by (8.6) or (8.7) as appropriate, if Qk(X) and Ck(X) are defined by (8.4), and if ak are the exact coefficients of the expansion of f(x) in Chebyshev polynomials of the corresponding kind, then hn(X) converges uniformly to h(X) on [−1, 1]. Proof: The idea of the proof is the same in all four cases. We give details of the second case here, and leave the others as exercises (Problems 1 and 2). For Pk = Uk, w =1, X hn(X)= fn(x)dx −1 X n = ak sin(k +1)θ dθ. −1 k=0 Thus the integrand is the partial Fourier sine series expansion of sin θf(cos θ), which converges in L2 and hence in L1 (Theorems 5.2 and 5.5). Now X h − hn∞ =max {f(x) − fn(x)} dx X −1 X ≤ max |f(x) − fn(x)| dx X −1 1 = |f(x) − fn(x)| dx −1 n π θf θ − a k θ θ = sin (cos ) k sin( +1) d 0 k=0 → 0,n→∞. Hence hn converges uniformly to h. •• The coefficients ak in (8.5) have been assumed to be exactly equal to the relevant Chebyshev series coefficients. In practice, we most often approximate these by the corresponding coefficients in a Chebyshev interpolation polyno- mial (see Chapter 6) — effectively evaluating the integral that defines ak by © 2003 by CRC Press LLC the trapezoidal rule (see Section 6.2). In some circumstances, we may need to calculate the Chebyshev coefficients more accurately than this. The method followed above is equivalent to methods well known in the literature. For the first choice (Pk = Tk) the method is that of Clenshaw & Curtis (1960) and for the second choice (Pk = Uk) that of Filippi (1964). The analysis of Section 8.1 is taken mainly from Mason & Elliott (1995, and related papers). 8.2 Gauss–Chebyshev quadrature Suppose that we now wish to calculate a definite integral of f(x)withweight w(x), namely b I = f(x)w(x)dx. (8.8) a Suppose also that I is to be approximated in the form n I Akf(xk) (8.9) k=1 where Ak are certain coefficients and {xk} are certain abscissae in [a, b](all to be determined). The idea of Gauss quadrature is to find that formula (8.9) that gives an exact result for all polynomials of as high a degree as possible. If Jn−1f(x) is the polynomial of degree n − 1 which interpolates f(x)in any n distinct points x1, ..., xn,then n Jn−1f(x)= f(xk)k(x) (8.10) k=1 where k is the Lagrange polynomial (as in (6.5)) n x − xr k(x)= (8.11) − r=1 xk xr r=k The polynomial Jn−1f(x)hastheintegral b In = Jn−1f(x)w(x)dx a n b = f(xk) w(x)k(x)dx k=1 a n = Akf(xk) k=1 © 2003 by CRC Press LLC provided that the coefficients Ak are chosen to be b Ak = w(x)k (x)dx. (8.12) a With any n distinct abscissae, therefore, and with this choice (8.12) of coef- ficients, the formula (8.9) certainly gives an exact result whenever f(x)isa polynomial of degree n − 1 or less. We can improve on this degree, however, by a suitable choice of abscissae. Notice too that, for general abscissae, there is no control over the signs and magnitudes of the coefficients Ak, so that evaluation of the formula (8.9) may involve heavy cancellation between large terms of opposite signs, and consequent large rounding error. When we choose the abscissae to maximise the degree of exactness, however, it can be shown that this problem ceases to arise. Theorem 8.2 If xk (k =1,...,n)arethen zeros of φn(x),and{φk : k = 0, 1, 2,...} is the system of polynomials, φk having the exact degree k,orthog- onal with respect to w(x) on [a, b], then (8.9) with coefficients (8.12) gives an exact result whenever f(x) is a polynomial of degree 2n − 1 or less. Moreover, all the coefficients Ak are positive in this case. Proof: Since φn(x) is a polynomial exactly of degree n,anypolynomialf(x)of degree 2n − 1 can be written (using long division by φn)intheform f(x)=φn(x)Q(x)+Jn−1f(x) where Q(x)andJn−1f(x) are polynomials each of degree at most n − 1. Then b b b f(x)w(x)dx = φn(x)Q(x)w(x)dx + Jn−1f(x)w(x)dx. (8.13) a a a Now φn(x) is orthogonal to all polynomials of degree less than n, so that the first integral on the right-hand side of (8.13) vanishes. Thus b b f(x)w(x)dx = Jn−1f(x)w(x)dx a a n = AkJn−1f(xk) k=1 since the coefficients have been chosen to give an exact result for polynomials of degree less than n. But now f(xk)=φn(xk)Q(xk)+Jn−1f(xk)=Jn−1f(xk), since xk isazeroofφn(x). Hence b n f(x)w(x)dx = Akf(xk), a k=1 © 2003 by CRC Press LLC and so (8.9) gives an exact result for f(x), as required. 2 To show that the coefficients Ak are positive, we need only notice that k(x) is a polynomial of degree 2n − 2, and is therefore integrated exactly, so that n b 2 2 Ak ≡ Aj k(xj ) = k(x) w(x)dx>0 j=1 a for each k. •• Thus we can expect to obtain very accurate integrals with the formula (8.9), and the formula should be numerically stable.