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31. f (x) = 1, 0 Answer Solution (a) L = T. For an even extension of the function, the sine coefficients are zero. The cosine coefficients are given by 2 ao = 2 f(x) dæ dx = 2, = - and for n > 0, 2 an = L 2 f(x) cos dx = - L cos nx dx = 0.

31. f (x) = 1, 0 Answer Solution (a) L = T. For an even extension of the function, the sine coefficients are zero. The cosine coefficients are given by 2 ao = 2 f(x) dæ dx = 2, = - and for n > 0, 2 an = L 2 f(x) cos dx = - L cos nx dx = 0.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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I did not understand the following questions. Why one question the coefficient for cos(npix/L) is 1/L, but the other is 2/L. I did not understand, please explain. 

31. f (x) = 1, 0 <x < T; cosine series, period 27 Answer Solution (a) L = T. For an even extension of the function, the sine coefficients are zero. The cosine coefficients are given by T 2 f(x) dæ 2 dx = 2, and for n > 0, 2 f(x) cos 2 dx An COS пх dx — 0. L
Transcribed Image Text:31. f (x) = 1, 0 <x < T; cosine series, period 27 Answer Solution (a) L = T. For an even extension of the function, the sine coefficients are zero. The cosine coefficients are given by T 2 f(x) dæ 2 dx = 2, and for n > 0, 2 f(x) cos 2 dx An COS пх dx — 0. L
(b) Find the Fourier series for the given function. 11. f (x) = -x -L<x < L; ƒ (x + 2L) = f (x) Answer Solution (a) The figure shows the case L = 1. (b) The Fourier series is of the form 8. ( ao ттс f(x): +Σ ат COS + bm sin 2 m=1 where the coefficients are computed form Eqs.(8)-(10). Substituting for f(x) in these equations yields 1 (1/L) / I (-x) cos ao = (-x) dx = 0 and am dx = 0, -- m = 1, 2, ... (these can be shown by direct integration, or using the fact that | g(x) da = 0 when g(x) is an odd function). Finally, -a 1 L (-г) sin (" bm dx = CoS COS dx тт L L 2L cos mT L 2L(-1)" sin тт L тт
Transcribed Image Text:(b) Find the Fourier series for the given function. 11. f (x) = -x -L<x < L; ƒ (x + 2L) = f (x) Answer Solution (a) The figure shows the case L = 1. (b) The Fourier series is of the form 8. ( ao ттс f(x): +Σ ат COS + bm sin 2 m=1 where the coefficients are computed form Eqs.(8)-(10). Substituting for f(x) in these equations yields 1 (1/L) / I (-x) cos ao = (-x) dx = 0 and am dx = 0, -- m = 1, 2, ... (these can be shown by direct integration, or using the fact that | g(x) da = 0 when g(x) is an odd function). Finally, -a 1 L (-г) sin (" bm dx = CoS COS dx тт L L 2L cos mT L 2L(-1)" sin тт L тт
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