What function satisfies this Fourier series integral?

Click For Summary

Discussion Overview

The discussion revolves around identifying a function f(x) that satisfies a specific integral involving a Fourier series and sine functions. The context includes mathematical reasoning related to Fourier transforms and series.

Discussion Character

  • Mathematical reasoning, Technical explanation

Main Points Raised

  • One participant asks for clues about a function f(x) that meets the integral condition involving sine functions and an alternating series.
  • Another participant suggests that the function may exhibit alternating behavior between +1 and -1.
  • A third participant elaborates that the function's Fourier series could be represented as a sum of sine functions with alternating signs.
  • A later participant proposes a more complex definition involving a weight function W(x) and discusses the implications of this definition on the integral and its relation to Fourier sine transforms.
  • This participant concludes with a statement about the resulting function being real, though the implications of this are not fully explored.

Areas of Agreement / Disagreement

The discussion shows some initial agreement on the nature of the function being related to alternating behavior, but it evolves into a more complex mathematical exploration without clear consensus on the final form of the function.

Contextual Notes

The discussion includes assumptions about the properties of the function and the weight function W(x), which may not be fully defined or agreed upon. The mathematical steps leading to the conclusions are not resolved, leaving some ambiguity.

gonzo
Messages
277
Reaction score
0
Anyone have any clues what function f(x) satisfies the following integral (for an integer n > 0):

[itex]\int^{\pi}_{0} f(x)sin(nx)dx=(-1)^{n+1}[/itex]
 
Last edited:
Physics news on Phys.org
So an alternating function at +1 and -1?
 
Yes. In other words, the function whose Fourier series would be:

sin(x)-sin(2x)+sin(3x)...
 
Nevermind, figured it out.
 
If you define:

[tex]\int_{0}^{\infty}dxW(x)f(x)sin(nx)=(-1)^{n+1}[/tex]

where x is W(x)=0 iff x>2pi, and W8x)=1 iff x<2pi, the integral above is just a Fourier sine transform with inverse:

[tex]W(x)f(x)=-\frac{2}{\pi}\int_{0}^{\infty}dne^{n\pi i}sin(nx)[/tex]

which is equal to [tex]f(x)W(x)= 2i(\delta (x+i \pi)-\delta (x-i \pi ))[/tex]

which is real...:bigrin:
: :bigrin:
 
Last edited:

Similar threads

  • · Replies 33 ·
2
Replies
33
Views
4K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 23 ·
Replies
23
Views
6K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 8 ·
Replies
8
Views
5K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 139 ·
5
Replies
139
Views
11K