Fourier series of |sin x|

  • Thread starter Narcol2000
  • Start date
  • #1
25
0
I'm trying to find the fourier series of |sin x| between -pi and pi.

I've got it down to:

[tex]
a_n = \frac{1}{\pi} \int^{\pi}_{-\pi} |sin x| cos (nx) dx
[/tex]

which i wrote as:

[tex]
a_n = \frac{2}{\pi}\int^{\pi}_0 sin x cos (nx) dx
[/tex]

writing

[tex]
sin x cos (nx) = \frac{1}{2} (sin (n+1)x - sin (n-1)x)
[/tex]

I eventually get

[tex]
a_n = \frac{2(n(-1)^n - 1)}{\pi(n^2 - 1)}
[/tex]

giving

[tex]
f(x) = \frac{2}{\pi} + \sum^{\infty}_{n=1} \frac{2(n(-1)^n - 1)}{\pi(n^2 - 1)} cos(nx)
[/tex]

The answer however gives

[tex]
f(x) = \frac{2}{\pi} + \frac{4}{\pi}\sum^{\infty}_{n=1} \frac{cos (2nx)}{4n^2 - 1}
[/tex]

I don't see how they arrive at this... if anyone can let me know where i've gone wrong or if i'm missing something :S
 
Last edited:

Answers and Replies

  • #2
mathman
Science Advisor
7,984
513
You mad a mistake in calculating an. There is no n in the numerator. It should be (-1)n-1-1. Then all the terms for odd n are 0 and the even terms remain. (I suggest you redo the calculation).
 
  • #3
25
0
yeah i found the mistake,
I get the numerator to be
[tex]
2((-1)^{n-1} - 1)
[/tex]

which does lead to the right answer.

thx for your help.:cool:
 

Related Threads on Fourier series of |sin x|

  • Last Post
Replies
8
Views
70K
  • Last Post
Replies
4
Views
7K
Replies
8
Views
5K
  • Last Post
Replies
7
Views
7K
Replies
1
Views
1K
Replies
2
Views
2K
Replies
2
Views
3K
  • Last Post
Replies
9
Views
2K
  • Last Post
Replies
11
Views
79K
  • Last Post
Replies
2
Views
1K
Top