Chem Differential Eq HW Solutions Fall 2011 16

Chem Differential Eq HW Solutions Fall 2011 16 - 16 Chapter...

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16 Chapter 2 Fourier Series Solutions to Exercises 2.5 1. We have f ( x )= ± 1i f 0 <x< 1 , - f - 1 0; The Fourier series representation is f ( x 4 π ² k =0 1 2 k +1 sin(2 k +1) πx. The mean square error (from (5)) is E N = 1 2 ³ 1 - 1 f 2 ( x ) dx - a 2 0 - 1 2 N ² n =1 ( a 2 n + b 2 n ) . In this case, a n = 0 for all n , b 2 k =0 , b 2 k +1 = 4 π (2 k +1) , and 1 2 ³ 1 - 1 f 2 ( x ) dx = 1 2 ³ 1 - 1 dx =1 . So E 1 - 1 2 ( b 2 1 )=1 - 8 π 2 0 . 189 . Since b 2 = 0, it follows that E 2 = E 1 . Finally, E 1 - 1 2 ( b 2 1 + b 2 3 - 8 π 2 - 8 9 π 2 0 . 099 . 5. E N = 1 2 ³ 1 - 1 f 2 ( x ) dx - a 2 0 - 1 2 N ² n =1 ( a 2 n + b 2 n ) - 1 2 N ² n =1 b 2 n - 8 π 2 ² 1 n odd N 1 n 2 . With the help of a calculator, we ±nd that E 39 = . 01013 and E 41 = . 0096. So take N = 41. 9. f ( x π 2 x - x 3 for - π<x<π and, for n 1, b n = 12 n 3 ( - 1) n +1 .By Parseval’s identity
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This note was uploaded on 12/22/2011 for the course MAP 3305 taught by Professor Stuartchalk during the Fall '11 term at UNF.

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