ourier Sine and Cosine Series. In each of Problems 29 througn 36: (a) Find the required Fourier series for the given function. (b) Sketch the graph of the function to which the series converges over three periods. ´1, 0 Answer V Solution (a) Using Equation (29) with L = 2, we have 2 sin(nt/2) ao = dx = 1 and dx = 2 an = Cos 2/nn for n = 1,5, 9, ... and an = -2/nn for n = 3,7,11,.... Hence we may write a2n = 0 and - 1)T, which when substituted into the series gives the desired answer. We obtain that Thus, an = 0 for n even, an = a2n-1 = 2(-1)*+1/(2n f(=) =- (-1)" Cos 2n – 1 1 2 (2n – 1)Tx 2 (b)
ourier Sine and Cosine Series. In each of Problems 29 througn 36: (a) Find the required Fourier series for the given function. (b) Sketch the graph of the function to which the series converges over three periods. ´1, 0 Answer V Solution (a) Using Equation (29) with L = 2, we have 2 sin(nt/2) ao = dx = 1 and dx = 2 an = Cos 2/nn for n = 1,5, 9, ... and an = -2/nn for n = 3,7,11,.... Hence we may write a2n = 0 and - 1)T, which when substituted into the series gives the desired answer. We obtain that Thus, an = 0 for n even, an = a2n-1 = 2(-1)*+1/(2n f(=) =- (-1)" Cos 2n – 1 1 2 (2n – 1)Tx 2 (b)
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 need help with this one. After computing the an, how do we get rid of the sin on the final answer? I did not get it. Please show me the work and give me as many details as possible. And how to draw the graph? should we just draw the even extension?
Transcribed Image Text:Fourier Sine an d Cosine Sei
les. In each of Problems 29 thrOugn 36:
(a) Find the required Fourier series for the given function.
(b) Sketch the graph of the function to which the series converges over three periods.
(1, 0 < а <1,
0, 1 <х <2;
29. f (x) =
cosine series, period 4
Answer
Solution
(a) Using Equation (29) with L = 2, we have
1
2 sin(nт/2)
dx
1 and
dx
2
ao
An =
COS
Thus, an
= 0 and
1,5, 9, ... and an
a2n-1 = 2(-1)"+/(2n – 1)T, which when substituted into the series gives the desired answer. We obtain that
O for n even, an = 2/nn for n =
n+1
— 2/пп for n %3
3, 7, 11, .... Hence we may write a2n
(-1)"
COS
(2n — 1) тӕ
1
|
f(x) =
2
-
2n – 1
T
n=1
(b)
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