I have a question about this type of question. Will cos(mpix) will always be 0? and how do we determine it will be 0 and when sin(mpix) part also will be 0? moreover, for the sin(mpix) when we do the integration, why the negative sign is canceled? I did not follow.
I have a question about this type of question. Will cos(mpix) will always be 0? and how do we determine it will be 0 and when sin(mpix) part also will be 0? moreover, for the sin(mpix) when we do the integration, why the negative sign is canceled? I did not follow.
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. Will cos(mpix) will always be 0? and how do we determine it will be 0 and when sin(mpix) part also will be 0? moreover, for the sin(mpix) when we do the
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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