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Question Chebyshev’s equation may be written " # 1 1 dy d (1 − x2 ) 2 + n2 (1 − x2 )− 2 y = 0 dx dx where n is a positive integer. The solutions which satisfy the boundary conditions Tn (−1) = (−1)n and Tn (1) = 1 are polynomials of degree n. Find the first three Chebyshev polynomials. Prove that if m 6= n then Z 1 1 (1 − x2 )− 2 Tm (x)Tn (x)dx = 0 −1 Answer To find the Chebyshev polynomials Tn (x) one substitutes a general polynomial into the differential equation and uses the boundary conditions. We illustrate this with T2 (x). The general form of T2 (x) is a quadratic so that T2 (x) = a0 + a1 x + a2 x2 . Using T2 (1) = 1 gives a0 +a1 +a2 = 1. Using T2 (−1) = 1 gives a0 −a1 +a2 = 1. Subtracting the equations gives a1 = 0 and hence a0 = 1 − a2 . Writing a2 as a we see that T2 (x) = ax2 + (1 − a). We now substitute this into Chebyshev’s equation (with n = 2) and obtain {2a(1 − 2x2 ) + 4ax2 + 4(1 − a)}(1 − x2 )−1/2 = 0, ⇒ 4 − 2a = 0. So that a = 2 and T2 (x) = 2x2 − 1. Now to show orthogonality. since Tn and Tm are solutions of Chebeyshev’s equation we have: " # d dTn (1 − x2 ) + n2 (1 − x2 )−1/2 Tn = 0 dt dx " # d 2 dTm (1 − x ) + m2 (1 − x2 )−1/2 Tm = 0 dt dx (1) (2) Multiplying equation (1) by Tm and equation (2) by Tn , subtracting and integratingZ by parts between −1 and 1 gives: 1 (n2 − m2 ) ( (1 − x2 )−1/2 Tn Tm dx "−1 # " d dTm d dTn Tn = (1 − x2 ) − Tm (1 − x2 ) dx dx dx dx −1 " Ã !#1 dTm dTn = (1 − x2 ) Tn − Tm dx dx −1 Z 1 1 #) dx − Z 1 −1 ( ) dTm dTn dTn dTm − (1 − x2 ) dx (1 − x ) dx dx dx dx 2 =0 Since the first term vanishes at ±1, and the terms and the integral cancel. 2