Chebyshev Polynomials Reading Problems Differential Equation and Its Solution The Chebyshev differential equation is written as d2y dy (1 − x2) − x + n2 y = 0 n = 0; 1; 2; 3;::: dx2 dx If we let x = cos t we obtain d2y + n2y = 0 dt2 whose general solution is y = A cos nt + B sin nt or as y = A cos(n cos−1 x) + B sin(n cos−1 x) jxj < 1 or equivalently y = ATn(x) + BUn(x) jxj < 1 where Tn(x) and Un(x) are defined as Chebyshev polynomials of the first and second kind of degree n, respectively. 1 If we let x = cosh t we obtain d2y − n2y = 0 dt2 whose general solution is y = A cosh nt + B sinh nt or as y = A cosh(n cosh−1 x) + B sinh(n cosh−1 x) jxj > 1 or equivalently y = ATn(x) + BUn(x) jxj > 1 The function Tn(x) is a polynomial. For jxj < 1 we have n n p 2 Tn(x) + iUn(x) = (cos t + i sin t) = x + i 1 − x n n p 2 Tn(x) − iUn(x) = (cos t − i sin t) = x − i 1 − x from which we obtain 1 h n ni p 2 p 2 Tn(x) = x + i 1 − x + x − i 1 − x 2 For jxj > 1 we have n nt p 2 Tn(x) + Un(x) = e = x ± x − 1 n −nt p 2 Tn(x) − Un(x) = e = x ∓ x − 1 The sum of the last two relationships give the same result for Tn(x). 2 Chebyshev Polynomials of the First Kind of Degree n The Chebyshev polynomials Tn(x) can be obtained by means of Rodrigue's formula (−2)nn! dn p 2 2 n−1=2 Tn(x) = 1 − x (1 − x ) n = 0; 1; 2; 3;::: (2n)! dxn The first twelve Chebyshev polynomials are listed in Table 1 and then as powers of x in terms of Tn(x) in Table 2. 3 Table 1: Chebyshev Polynomials of the First Kind T0(x) = 1 T1(x) = x 2 T2(x) = 2x − 1 3 T3(x) = 4x − 3x 4 2 T4(x) = 8x − 8x + 1 5 3 T5(x) = 16x − 20x + 5x 6 4 2 T6(x) = 32x − 48x + 18x − 1 7 5 3 T7(x) = 64x − 112x + 56x − 7x 8 6 4 2 T8(x) = 128x − 256x + 160x − 32x + 1 9 7 5 3 T9(x) = 256x − 576x + 432x − 120x + 9x 10 8 6 4 2 T10(x) = 512x − 1280x + 1120x − 400x + 50x − 1 11 9 7 5 3 T11(x) = 1024x − 2816x + 2816x − 1232x + 220x − 11x 4 Table 2: Powers of x as functions of Tn(x) 1 = T0 x = T1 2 1 x = (T0 + T2) 2 3 1 x = (3T1 + T3) 4 4 1 x = (3T0 + 4T2 + T4) 8 5 1 x = (10T1 + 5T3 + T5) 16 6 1 x = (10T0 + 15T2 + 6T4 + T6) 32 7 1 x = (35T1 + 21T3 + 7T5 + T7) 64 8 1 x = (35T0 + 56T2 + 28T4 + 8T6 + T8) 128 9 1 x = (126T1 + 84T3 + 36T5 + 9T7 + T9) 256 10 1 x = (126T0 + 210T2 + 120T4 + 45T6 + 10T8 + T10) 512 11 1 x = (462T1 + 330T3 + 165T5 + 55T7 + 11T9 + T11) 1024 5 Generating Function for Tn(x) The Chebyshev polynomials of the first kind can be developed by means of the generating function 1 1 − tx X = T (x)tn 2 n 1 − 2tx + t n=0 Recurrence Formulas for Tn(x) When the first two Chebyshev polynomials T0(x) and T1(x) are known, all other polyno- mials Tn(x); n ≥ 2 can be obtained by means of the recurrence formula Tn+1(x) = 2xTn(x) − Tn−1(x) The derivative of Tn(x) with respect to x can be obtained from 2 0 (1 − x )Tn(x) = −nxTn(x) + nTn−1(x) Special Values of Tn(x) The following special values and properties of Tn(x) are often useful: n n Tn(−x) = (−1) Tn(x) T2n(0) = (−1) Tn(1) = 1 T2n+1(0) = 0 n Tn(−1) = (−1) 6 Orthogonality Property of Tn(x) We can determine the orthogonality properties for the Chebyshev polynomials of the first kind from our knowledge of the orthogonality of the cosine functions, namely, 8 > 0 (m 6= n) > Z π <> cos(mθ) cos(n θ) dθ = π=2 (m = n 6= 0) 0 > > :> π (m = n = 0) Then substituting Tn(x) = cos(nθ) cos θ = x to obtain the orthogonality properties of the Chebyshev polynomials: 8 0 (m 6= n) > 1 > Z Tm(x) Tn(x) dx < p = π=2 (m = n 6= 0) 2 −1 1 − x > > :> π (m = n = 0) We observe that the Chebyshev polynomials form an orthogonal set on the interval −1 ≤ x ≤ 1 with the weighting function (1 − x2)−1=2 Orthogonal Series of Chebyshev Polynomials An arbitrary function f(x) which is continuous and single-valued, defined over the interval −1 ≤ x ≤ 1, can be expanded as a series of Chebyshev polynomials: f(x) = A0T0(x) + A1T1(x) + A2T2(x) + ::: 1 X = AnTn(x) n=0 7 where the coefficients An are given by 1 Z 1 f(x) dx A0 = p n = 0 2 π −1 1 − x and 1 2 Z f(x) Tn(x) dx An = p n = 1; 2; 3;::: 2 π −1 1 − x The following definite integrals are often useful in the series expansion of f(x): Z 1 dx Z 1 x3 dx p = π p = 0 2 2 −1 1 − x −1 1 − x Z 1 x dx Z 1 x4 dx 3π p = 0 p = 2 2 −1 1 − x −1 1 − x 8 Z 1 x2 dx π Z 1 x5 dx p = p = 0 2 2 −1 1 − x 2 −1 1 − x Chebyshev Polynomials Over a Discrete Set of Points A continuous function over a continuous interval is often replaced by a set of discrete values of the function at discrete points. It can be shown that the Chebyshev polynomials Tn(x) are orthogonal over the following discrete set of N + 1 points xi, equally spaced on θ, π 2π π θi = 0; ; ;::: (N − 1) ; π N N N where xi = arccos θi We have 8 8 > 0 (m 6= n) N−1 > 1 X 1 <> Tm(−1)Tn(−1)+ Tm(xi)Tn(xi)+ Tm(1)Tn(1) = N=2 (m = n 6= 0) 2 2 > i=2 > :> N (m = n = 0) The Tm(x) are also orthogonal over the following N points ti equally spaced, π 3π 5π (2N − 1)π θi = ; ; ;:::; 2N 2N 2N 2N and ti = arccos θi 8 0 (m 6= n) > N > X < Tm(ti)Tn(ti) = N=2 (m = n 6= 0) > i=1 > :> N (m = n = 0) The set of points ti are clearly the midpoints in θ of the first case. The unequal spacing of the points in xi(Nti) compensates for the weight factor W (x) = (1 − x2)−1=2 in the continuous case. 9 Additional Identities of Chebyshev Polynomials The Chebyshev polynomials are both orthogonal polynomials and the trigonometric cos nx functions in disguise, therefore they satisfy a large number of useful relationships. The differentiation and integration properties are very important in analytical and numerical work. We begin with −1 Tn+1(x) = cos[(n + 1) cos x] and −1 Tn−1(x) = cos[(n − 1) cos x] Differentiating both expressions gives −1 1 d[Tn+1(x)] − sin[(n + 1) cos x = p (n + 1) dx − 1 − x2 and −1 1 d[Tn−1(x)] − sin[(n − 1) cos x = p (n − 1) dx − 1 − x2 Subtracting the last two expressions yields 1 d[T (x)] 1 d[T (x)] sin(n + 1)θ − sin(n − 1)θ n+1 − n−1 = (n + 1) dx (n − 1) dx sin θ or 0 0 Tn+1(x) Tn−1(x) 2 cos nθ sin θ − = = 2Tn(x) n ≥ 2 (n + 1) (n − 1) sin θ 10 Therefore 0 T2(x) = 4T1 0 T1(x) = T0 0 T0(x) = 0 We have the formulas for the differentiation of Chebyshev polynomials, therefore these for- mulas can be used to develop integration for the Chebyshev polynomials: Z 1 Tn+1(x) Tn−1(x) Tn(x)dx = − + C n ≥ 2 2 (n + 1) (n − 1) Z 1 T1(x)dx = T2(x) + C 4 Z T0(x)dx = T1(x) + C The Shifted Chebyshev Polynomials For analytical and numerical work it is often convenient to use the half interval 0 ≤ x ≤ 1 instead of the full interval −1 ≤ x ≤ 1. For this purpose the shifted Chebyshev polynomials are defined: ∗ Tn (x) = Tn ∗ (2x − 1) Thus we have for the first few polynomials ∗ T0 = 1 ∗ T1 = 2x − 1 ∗ 2 T2 = 8x − 8x + 1 ∗ 3 2 T3 = 32x − 48x + 18x − 1 ∗ 4 3 2 T4 = 128x − 256x + 160x − 32x + 1 11 ∗ and the following powers of x as functions of Tn (x); ∗ 1 = T0 1 x = (T ∗ + T ∗) 2 0 1 1 x2 = (3T ∗ + 4T ∗ + T ∗) 8 0 1 2 1 x3 = (10T ∗ + 15T ∗ + 6T ∗ + T ∗) 32 0 1 2 3 1 x4 = (35T ∗ + 56T ∗ + 28T ∗ + 8T ∗ + T ∗) 128 0 1 2 3 4 The recurrence relationship for the shifted polynomials is: ∗ ∗ ∗ ∗ Tn+1(x) = (4x − 2)Tn (x) − Tn−1(x) T0 (x) = 1 or 1 1 1 xT ∗(x) = T ∗ (x) + T ∗(x) + T ∗ (x) n 4 n+1 2 n 4 n−1 where ∗ −1 Tn (x) = cos n cos (2x − 1) = Tn(2x − 1) n Expansion of x in a Series of Tn(x) A method of expanding xn in a series of Chebyshev polynomials employes the recurrence relation written as 1 xTn(x) = [Tn+1(x) + Tn−1(x)] n = 1; 2; 3 ::: 2 xT0(x) = T1(x) 12 To illustrate the method, consider x4 2 4 2 x x x = x (xT1) = [T2 + T0] = [T1 + T3 + 2T1] 2 4 1 1 = [3xT1 + xT3] = [3T0 + 3T2 + T4 + T2] 4 8 1 1 3 = T4 + T2 + T0 8 2 8 This result is consistent with the expansion of x4 given in Table 2.