11.9_ Representations as Series-solutions

# 2r1 r2 4 2r1 r2 2 is 1 1 thus on 1 1 1

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Unformatted text preview: he series n=0 ∞ ∞ cn z n x (−1)n t2n dt = = 2, n=0 0 n=0 is ∞ = n=0 (−1)n 2n + 1 x2n+1 . 1 (i) convergent when |z | < , and 2 1 (ii) divergent when |z | > . 2 Consequently, on (−1, 1) ∞ f ( x) = n=0 008 (−1)n 2n+5 x . 2n + 1 On the other hand, since lim 10.0 points Compare the radius of convergence, R1 , of the series n→∞ (n + 1)cn+1 ncn = lim n→∞ cn+1 , cn the Ratio Test ensures also that the series ∞ ∞ cn z n n cn z n−1 n=0 with the radius of convergence, R2 , of the series n=1 is ∞ n cn z n−1 n=1 1 (i) convergent when |z | < , and 2 stiurca (mas7745) – 11.9: Representations as Series – meth – (91825) 1 (ii) divergent when |z | > . 2 Thus ∞ f ( x) = x3 Consequently, 009 n=0 1 . 2 R1 = R2 = x3 ∞ n=2 ∞ 2. f (x) = n=3 ∞ 3. f (x) = n=3 ∞ 4. f (x) = n=3 ∞ 5. f (x) = n=2 (6 − x)2 ∞ . n−1 n x 6n 1 6n−3 xn nn x 6n n−2 n x correct 6n−1 1 6n−1 xn Explanation: By the known result for geometric series, 1 1 = x 6−x 6 1− 6 = 1 6 ∞ x 6 n=0 n ∞ 1 = n=0 6n+1 xn . This series converges on (−6, 6). On the other hand, 1 1 d , = 2 (6 − x) dx 6 − x and so on (−6, 6), d 1 = (6 − x)2 dx ∞ = n=1 n 6n+1 ∞ n=0 xn 6n+1 ∞ x n−1 = n=0 ∞ n=0 n + 1 n+3 x . 6n+2 Consequently, Find a power series representation centered at the origin for the function 1. f (x) = n+1 n x= 6n+2 10.0 points f ( x) = 6 n+1 n x. 6n+2 f ( x) = n=3 n−2 n x. 6n−1...
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## This document was uploaded on 09/17/2013.

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