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 11. f (2) = -x -L AnswerV Solution(a) The figure shows the case L = 1.(b) The Fourier series is of the formf(e) = + om().+ bm sinCOSwhere the coefficients are computed form Eqs.(8)-(10). Substituting for f(z) in these equations yieldsao = (1/L) /(-x) da = 0andam =(-2) cosdz = 0,m = 1,2,... (these can be shown by direct integration, or using the fact thatg(x) dr = 0 when g(x) is an odd function). Finally,1bm =1(-x) sindr =deCoSCos2L cos maL2L(-1)"sin--since cos ma = (-1)". Substituting these terms in the above Fourier series for f(z) yields the desired answer:

Answered: Image /qna-images/answer/30fbe80a-21fe-45a0-ae15-e009a53a23bd.jpg

Answered: (æ, -n

Math

Advanced Math

(æ, -n

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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Question

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

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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