Dymore User's Manual Chebyshev polynomials Olivier A. Bauchau August 27, 2019 Contents 1 Definition 1 1.1 Zeros and extrema....................................2 1.2 Orthogonality relationships................................3 1.3 Derivatives of Chebyshev polynomials..........................5 1.4 Integral of Chebyshev polynomials............................5 1.5 Products of Chebyshev polynomials...........................5 2 Chebyshev approximation of functions of a single variable6 2.1 Expansion of a function in Chebyshev polynomials...................6 2.2 Evaluation of Chebyshev expansions: Clenshaw's recurrence.............7 2.3 Derivatives and integrals of Chebyshev expansions...................7 2.4 Products of Chebyshev expansions...........................8 2.5 Examples.........................................9 2.6 Clenshaw-Curtis quadrature............................... 10 3 Chebyshev approximation of functions of two variables 12 3.1 Expansion of a function in Chebyshev polynomials................... 12 3.2 Evaluation of Chebyshev expansions: Clenshaw's recurrence............. 13 3.3 Derivatives of Chebyshev expansions.......................... 14 4 Chebychev polynomials 15 4.1 Examples......................................... 16 1 Definition Chebyshev polynomials [1,2] form a series of orthogonal polynomials, which play an important role in the theory of approximation. The lowest polynomials are 2 3 4 2 T0(x) = 1;T1(x) = x; T2(x) = 2x − 1;T3(x) = 4x − 3x; T4(x) = 8x − 8x + 1;::: (1) 1 and are depicted in fig.1. The polynomials can be generated from the following recurrence rela- tionship Tn+1 = 2xTn − Tn−1; n ≥ 1: (2) 1 0.8 0.6 0.4 0.2 0 −0.2 −0.4 CHEBYSHEV POLYNOMIALS −0.6 −0.8 −1 −1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1 XX Figure 1: The seven lowest order Chebyshev polynomials It is possible to give an explicit expression of Chebyshev polynomials as Tn(x) = cos(n arccos x): (3) This equation can be verified by using elementary trigonometric identities. For instance, it is clear, 2 2 that T2 = cos [2 arccos x] = 2 cos (arccos x) − 1 = 2x − 1, as expected from eq. (1). 1.1 Zeros and extrema It is now easy to verify that Tn(x) possesses n zeros within the interval x 2 [−1; +1]: Tn(x) = cos(n arccos x) = 0 implies n arccos x = (2k − 1)π=2. Hence, the zeros of Chebyshev polynomial Tn(x) are π(2k − 1) x¯ = cos ; k = 1; 2; 3; : : : ; n: (4) k 2n p p 2 For instance,p since T3 = x(4x − 3), its zeros arep 3=2, 0, and − 3=2, which can be written as cos π=6 = 3=2, cos 3π=6 = 0, and cos 5π=6 = − 3=2. The value of Chebyshev polynomial Ti(x) 2 at the zeros of Tn(x) is easily found from eq. (3) as i(2k − 1)π T (¯x ) = cos ; i < n: (5) i k 2n It is also easy to find the extrema of a Chebyshev polynomialp by imposing the vanishing of its 2 derivative, dTn=dx = 0. This leads to [n sin(n arccos x)] = 1 − x = 0, or sin [n arccos x] = 0. The extrema of Chebyshev polynomial Tn(x) are kπ x^ = cos ; k = 0; 1; 2; 3; : : : ; n: (6) k n p 2 For instance,p dT4=dx = x(2x − 1) = 0 leads to extrema cos π=4 = 2=2, cos π=2 = 0, and cos 3π=4 = − 2=2. The additional extrema, cos 0 = 1 and cos π = −1, occur at the ends of the interval. The value of Chebyshev polynomial Ti(x) at the extrema of Tn(x) is easily found from eq. (3) as ikπ T (^x ) = cos ; i < n: (7) i k n 1.2 Orthogonality relationships Chebyshev polynomials are orthogonal within the interval x 2 [−1; +1] with a weight of (1−x2)−1=2, i.e. 8 0 i 6= j Z +1 T (x)T (x) < pi j dx = π=2 i = j 6= 0 : (8) 2 −1 1 − x : π i = j = 0 In addition to the orthogonality property defined by eq. (8), Chebyshev polynomials also enjoy the following discrete orthogonality relationship 8 n 0 i 6= j X < Ti(¯xk)Tj(¯xk) = n=2 i = j 6= 0 : (9) k=1 : n i = j = 0 wherex ¯k, k = 1; 2; 3; : : : ; n are the zeros of Tn as given by eq. (4), and i; j < n. To prove this orthogonality relationship, trigonometric identities are used n n X X i(2k − 1)π j(2k − 1)π T (¯x )T (¯x ) = cos cos i k j k 2n 2n k=1 k=1 n−1 1 X (i + j)(2k + 1)π (i − j)(2k + 1)π = cos + cos ; 2 2n 2n k=0 2 n (i+j)π h (i+j)π (i+j)π i n (i−j)π h (i−j)π (i−j)π i3 1 sin 2 n cos 2n + (n − 1) 2n sin 2 n cos 2n + (n − 1) 2n = + 2 4 (i+j)π (i−j)π 5 sin 2n sin 2n " # 1 sin (i+j)π cos (i+j)π sin (i−j)π cos (i−j)π = 2 2 + 2 2 = 0: 2 (i+j)π (i−j)π sin 2n sin 2n 3 The trigonometric identity, eq. (12) was used to eliminate the summation; the last equality results from the fact that cos(i+j)π=2 = cos(i−j)π=2 = 0. If i = j 6= 0, or i = j = 0, similar developments yield the discrete orthogonality given by eq. (9). Chebyshev polynomials also enjoy an additional discrete orthogonality relationship 8 n 0 i 6= j X 00 < Ti(^xk)Tj(^xk) = n=2 i = j 6= 0 : (10) k=0 : n i = j = 0 wherex ^k, k = 0; 1; 2; ··· ; n are the extrema of Tn as given by eq. (6), and i; j < n. The double prime after the summation sign indicates that the first and last terms of the summation must be halved. To prove this orthogonality relationship, trigonometric identities are used n n X X ikπ jkπ T (^x )T (^x ) = cos cos ; i k j k n n k=0 k=0 n 1 X (i + j)π (i − j)π = cos k + cos k 2 n n k=0 " # 1 sin n+1 (i+j)π cos n (i+j)π sin n+1 (i−j)π cos n (i−j)π = 2 n 2 n + 2 n 2 n 2 (i+j)π (i−j)π sin 2n sin 2n " # 1 sin 2n+1 (i+j)π + sin (i+j)π sin 2n+1 (i−j)π + sin (i−j)π = 2 n 2n + 2 n 2n 4 (i+j)π (i−j)π sin 2n sin 2n 2 h (i+j)π i h (i−j)π i3 1 sin (i + j)π + 2n sin (i − j)π + 2n = 2 + + 4 4 (i+j)π (i−j)π 5 sin 2n sin 2n 1 1 = [2 + cos(i + j)π + cos(i − j)π] = [1 + cos iπ cos jπ] : 4 2 The first term in the last bracket is the term of the sum corresponding to k = 0, whereas the second term in the last bracket is that corresponding to k = n. Bringing these two terms to the left hand side is identical to replacing the summation sign, P, by P00. Here again, the trigonometric identity, eq. (12) was used to eliminate the summation. If i = j 6= 0, or i = j = 0, similar developments yield the discrete orthogonality given by eq. (10). The following trigonometric identities were used in the derivation of the above discrete orthog- onality relationships (n + 1)b nb n sin sin(a + ) X sin(a) + sin(a + b) + ::: + sin(a + nb) = sin(a + kb) = 2 2 ; (11) b k=0 sin 2 (n + 1)b nb n sin cos(a + ) X cos(a) + cos(a + b) + ::: + cos(a + nb) = cos(a + kb) = 2 2 : (12) b k=0 sin 2 4 1.3 Derivatives of Chebyshev polynomials The following expression for the derivatives of Chebyshev polynomials 0 2n [Tn−1 + Tn−3 + ::: + T1] n even; Tn = (13) 2n [Tn−1 + Tn−3 + ::: + T2] + nT0 n odd; where the notation (·)0 indicates a derivative with respect to x, can be proved by mathematical 0 0 0 induction. Indeed, they are verified for the lowest polynomials, T1 = T0, T2 = 2 × 2 T1, T3 = 0 2 × 3 T2 + 3T0, T4 = 2 × 4 (T3 + T1), etc. It then remains to prove that if it is correct for n it is still correct for n + 1. Taking a derivative of the basic recurrence for Chebyshev polynomials, eq. (2), 0 0 0 leads to Tn+1 = 2xTn + 2Tn − Tn−1. Introducing eq. (13) into this recurrence, it is then easy to show that eq. (13) is true for n + 1. 1.4 Integral of Chebyshev polynomials The following recurrence relationship is easy to prove T 0 T 0 2T (x) = n+1 − n−1 ; (14) n n + 1 n − 1 with the help of eq. (13). It then follows that Z +1 2 2 n+1 − n−1 n even; 2 Tn(x) dx = (15) −1 0 n odd These two equations are easily combined to yield Z +1 2 T2n(x) dx = − 2 : (16) −1 4n − 1 1.5 Products of Chebyshev polynomials The product of two Chebyshev polynomials satisfies the following relationship 2Tn(x)Tm(x) = Tn+m(x) + Tn−m(x); n ≥ m: (17) This relationship is an identity for m = 0; indeed, since T0 = 1, it then follows that 2TnT0 = Tn +Tn. Multiplying both sides of this equation by 2x yields 2Tn(2xT0) = 2(2xTn).