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Chem Differential Eq HW Solutions Fall 2011 93

Chem Differential Eq HW Solutions Fall 2011 93 - of f x = |...

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Section 5.6 Legendre Polynomials and Legendre Series Expansions 93 For n = 0, we use integration by parts. The integral is a convergent improper integral (the integrand has a problem at 1) 1 - 1 ln(1 - x ) P 0 ( x ) dx = 1 - 1 ln(1 - x ) dx = - (1 - x ) ln(1 - x ) - x 1 - 1 = - 2 + 2 ln2 . To evaluate the integral at x = 1, we used lim x 1 (1 - x ) ln(1 - x ) = 0. 29. Call the function in Exercise 28 g ( x ). Then g ( x ) = 1 2 ( | x | + x ) = 1 2 ( | x | + P 1 ( x )) . Let B k denote the Legendre coefficient of g and A k denote the Legendre coefficient
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Unformatted text preview: of f ( x ) = | x | , for-1 < x < 1. Then, because P 1 ( x ) is its own Legendre series, we have B k = ³ 1 2 A k if k ± = 1 1 2 ( A k + 1) if k = 1 Using Exercise 27 to compute A k , we ±nd B = 1 2 A = 1 4 , B 1 = 1 2 + 1 2 A 1 = 1 2 + 0 = 1 2 ,B 2 n +1 = 0 , n = 1 , 2 , ..., and B 2 n = 1 2 A 2 n = (-1) n +1 (2 n-2)! 2 2 n +1 (( n-1)!) 2 n ´ 4 n + 1 n + 1 µ ....
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