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11. f(z) = -z -L

11. f(z) = -z -L

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 really confused about how to solve this type of question. How do we know the coefficient of a0, am and bm is 1/L? and what is the answer for a0 from the -L to L?

11. f (x) : = -x -L< x < L; f(x+2L) = f (x) Answer V Solution (a) The figure shows the case L = 1. (b) The Fourier series is of the form ao MTL f(x) 2 +E l + bm sin am cos L L т-1 where the coefficients are computed form Eqs.(8)-(10). Substituting for f(x) in these equations yields (1/L) / (-x) dæ = 0 1 Am = L and (-а) cos dx = 0, L -- -L m = 1, 2, ... (these can be shown by direct integration, or using the fact that g(x) dæ = 0 when g(x) is an odd function). Finally, -a 1 тт 1 (7) bm sin dx = COS CoS dx т -- тт L -L 2L cos mT 2L(-1)" sin тт m²72 т since cos mT = (-1)". Substituting these terms in the above Fourier series for f(x) yields the desired answer:
Transcribed Image Text:11. f (x) : = -x -L< x < L; f(x+2L) = f (x) Answer V Solution (a) The figure shows the case L = 1. (b) The Fourier series is of the form ao MTL f(x) 2 +E l + bm sin am cos L L т-1 where the coefficients are computed form Eqs.(8)-(10). Substituting for f(x) in these equations yields (1/L) / (-x) dæ = 0 1 Am = L and (-а) cos dx = 0, L -- -L m = 1, 2, ... (these can be shown by direct integration, or using the fact that g(x) dæ = 0 when g(x) is an odd function). Finally, -a 1 тт 1 (7) bm sin dx = COS CoS dx т -- тт L -L 2L cos mT 2L(-1)" sin тт m²72 т since cos mT = (-1)". Substituting these terms in the above Fourier series for f(x) yields the desired answer:
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