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Chem Differential Eq HW Solutions Fall 2011 17

# Chem Differential Eq HW Solutions Fall 2011 17 - 1 2 1 2...

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Section 2.5 Mean Square Approximation and Parseval’s Identity 17 13. For the given function, we have b n = 0 and a n = 1 n 2 . By Parseval’s identity, we have 1 2 π π - π f 2 ( x ) dx = 1 2 n =1 1 n 4 π - π f 2 ( x ) dx = π n =1 1 n 4 = πζ (4) = π 5 90 , where we have used the table preceding Exercise 7 to compute ζ (4). 17. For the given function, we have a 0 = 1 , a n = 1 3 n , b n = 1 n for n 1 . By Parseval’s identity, we have 1 2 π π - π f 2 ( x ) dx = a 2 0 + 1 2 n =1 ( a 2 n + b 2 n ) = 1 + 1 2 n =1 ( 1 (3 n )
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Unformatted text preview: + 1 2 + 1 2 ∞ ² n =1 1 9 n + 1 2 ∞ ² n =1 1 n 2 = 1 2 + 1 2 ∞ ² n =0 1 9 n + 1 2 ∞ ² n =1 1 n 2 . Using a geometric series, we ±nd ∞ ² n =0 1 9 n = 1 1-1 9 = 9 8 . By Exercise 7(a), ∞ ² n =1 1 n 2 = π 2 6 . So ± π-π f 2 ( x ) dx = 2 π ( 1 2 + 1 2 9 8 + 1 2 π 2 6 ) = 17 π 8 + π 3 6 ....
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