Problem for Fourier series

  • Thread starter fufufuha
  • Start date
  • #1
6
0
http://www4.okfoto.co.kr/S_storage4/314500/A05120618494148_t.jpg [Broken] http://www4.okfoto.co.kr/S_storage4/314500/A05120618494175_t.jpg [Broken]
It's solution is http://www4.okfoto.co.kr/_Inc/Process/proSendbinaryHow.asp?w=400&h=400&path=\\100.100.100.11\link_album\pic_root\storage4\314500\A05120618494172.jpg [Broken]
What I know is that insert 2n-1 instead of n
plz... explain to me
I don't know this reason
 

Attachments

  • redfake_1133843847.jpg
    redfake_1133843847.jpg
    1,006 bytes · Views: 297
  • redfake_1133844265.jpg
    redfake_1133844265.jpg
    768 bytes · Views: 317
  • redfake_1133844222.jpg
    redfake_1133844222.jpg
    2.6 KB · Views: 324
Last edited by a moderator:

Answers and Replies

  • #2
Integral
Staff Emeritus
Science Advisor
Gold Member
7,201
56
Please make an effort to provide a more complete statement of your problem.
 
  • #3
6
0
First I'm sorry that I can't write english well.
I have a problem's solution.
But I don't understand solution
For example Another problem's solution is n instead of 2n-1
But this problem's solution uses 2n-1 instead of n
I understood that 2n-1 is odd
but why do 2n-1 instead of n ?
 
  • #4
HallsofIvy
Science Advisor
Homework Helper
41,833
964
When x= 1 or -1, 1-|x|= 0. Suppose you had just [itex]\frac{n\pi}{2}x[/itex]. Then for even values of n, x= 1, you would have [itex]cos(m\pi)[/itex] (m= n/2) which is either -1 or 1. By restricting the numerator to be odd, you are making sure that you always have, for x= 1, [itex]cos(\frac{\pi}{2})[/itex], [itex]cos(\frac{3\pi}{2})[/itex], etc.

By the way, how does one do Fourier Series in precalculus?
 
Last edited by a moderator:

Related Threads on Problem for Fourier series

  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
23
Views
4K
Replies
71
Views
10K
  • Last Post
Replies
23
Views
2K
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
4
Views
6K
  • Last Post
Replies
3
Views
1K
Top