Chem Differential Eq HW Solutions Fall 2011 8

Chem Differential Eq HW Solutions Fall 2011 8 - 2 n sin n 2...

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8 Chapter 2 Fourier Series 21. (a) Interpreting the integral as an area (see Exercise 16), we have a 0 = 1 2 π · 1 2 · π 2 = 1 8 . To compute a n , we frst determine the equation oF the Function For π 2 <x<π . ±rom ±igure 16, we see that f ( x )= 2 π ( π - x )i F π 2 <x<π . Hence, For n 1, a n = 1 π ± π π/ 2 2 π u ² ³´ µ ( π - x ) v ± ² ³´ µ cos nx dx = 2 π 2 ( π - x ) sin nx n π π/ 2 + 2 π 2 ± π π/ 2 sin nx n dx = 2 π 2 · - π 2 n sin 2 ¸ - 2 π 2 n 2 cos nx π π/ 2 = - 2 π 2 · π 2 n sin 2 + ( - 1) n n 2 - 1 n 2 cos 2 ¸ . Also, b n = 1 π ± π π/ 2 2 π u ² ³´ µ ( π - x ) v ± ² ³´ µ sin nx dx = - 2 π 2 ( π - x ) cos nx n π π/ 2 - 2 π 2 ± π π/ 2 cos nx n dx = 2 π 2 · π 2 n cos 2 + 1 n 2 sin 2 ¸ . Thus the ±ourier series representation oF f is f ( x )= 1 8 + 2 π 2 ¹ n =1 º - · π 2 n sin 2 + ( - 1) n n 2 - 1 n 2 cos 2 ¸ cos nx + · π 2 n cos 2 + 1 n 2 sin 2 ¸ sin nx » . (b) Let g ( x )= f ( - x ). By perForming a change oF variables x ↔- x in the ±ourier series oF f , we obtain (see also Exercise 24 For related details) Thus the ±ourier series representation oF f is g ( x )= 1 8 + 2 π 2 ¹ n =1 º - · π
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Unformatted text preview: 2 n sin n 2 + (-1) n n 2-1 n 2 cos n 2 cos nx- 2 n cos n 2 + 1 n 2 sin n 2 sin nx . 25. or (a) and (b), see plots. (c) We have s n ( x ) = n k =1 sin kx k . So s n (0) = 0 and s n (2 ) = 0 For all n . Also, lim x + f ( x ) = 2 , so the dierence between s n ( x ) and f ( x ) is equal to / 24 at x = 0. But even we look near x = 0, where the ourier series converges to f ( x ), the dierence | s n ( x )-f ( x ) | remains larger than a positive number, that is about . 28 and does not get smaller no matter how large n . In the fgure, we plot | f ( x )-s 150 ( x ) | ....
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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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