Unformatted text preview: of f ( x ) =  x  , for1 < 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 n2)! 2 2 n +1 (( n1)!) 2 n ´ 4 n + 1 n + 1 µ ....
View
Full Document
 Fall '11
 StuartChalk
 Calculus, Riemann integral, Henstock–Kurzweil integral, Legendre Series Expansions

Click to edit the document details