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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 have a question about this type of question. Why does the graph draw like in the pictures? which equation is it based on? and how do we determine its period or cycle and others? 

x. X, -T <x < 0, 13. f (x) f (x + 27) = f (x) 10, 0 < x < T; Answer Solution (a) -3x -2
Transcribed Image Text:x. X, -T <x < 0, 13. f (x) f (x + 27) = f (x) 10, 0 < x < T; Answer Solution (a) -3x -2
11. f (2) = -x -L <a < L; f(2 + 2L) = f (x) (> Answer V Solution (a) The figure shows the case L = 1. (b) The Fourier series is of the form f(e) = + om (). + bm sin COS where the coefficients are computed form Eqs.(8)-(10). Substituting for f(z) in these equations yields ao = (1/L) / (-x) da = 0 and am = (-2) cos dz = 0, m = 1,2,... (these can be shown by direct integration, or using the fact that g(x) dr = 0 when g(x) is an odd function). Finally, 1 bm = 1 (-x) sin dr = de CoS Cos 2L cos ma L 2L(-1)" sin -- since cos ma = (-1)". Substituting these terms in the above Fourier series for f(z) yields the desired answer:
Transcribed Image Text:11. f (2) = -x -L <a < L; f(2 + 2L) = f (x) (> Answer V Solution (a) The figure shows the case L = 1. (b) The Fourier series is of the form f(e) = + om (). + bm sin COS where the coefficients are computed form Eqs.(8)-(10). Substituting for f(z) in these equations yields ao = (1/L) / (-x) da = 0 and am = (-2) cos dz = 0, m = 1,2,... (these can be shown by direct integration, or using the fact that g(x) dr = 0 when g(x) is an odd function). Finally, 1 bm = 1 (-x) sin dr = de CoS Cos 2L cos ma L 2L(-1)" sin -- since cos ma = (-1)". Substituting these terms in the above Fourier series for f(z) yields the desired answer:
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