Finding Fourier extension and if it converges

  • Thread starter Thread starter Weilin Meng
  • Start date Start date
  • Tags Tags
    Extension Fourier
Click For Summary

Homework Help Overview

The problem involves extending the function f(x) = sin(x)/x periodically for |x| <= pi and investigating the convergence of its Fourier series at x=0. The original poster expresses uncertainty about the convergence of the series at this point.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to calculate the Fourier coefficients and expresses doubt about the convergence of the series at x=0. Some participants suggest exploring identities related to the sine integral function to simplify the summation. Others question whether it is necessary to compute the Fourier series to determine convergence, referencing existing theorems on convergence.

Discussion Status

The discussion is ongoing, with participants exploring different approaches to understanding the convergence of the Fourier series. Some guidance has been offered regarding theorems that may simplify the analysis, but no consensus has been reached on the convergence issue itself.

Contextual Notes

The problem is situated within a PDE course context, with the original poster noting a limited mathematical background in certain areas, which may affect their approach to the problem.

Weilin Meng
Messages
25
Reaction score
0

Homework Statement


Let f(x) = sin(x)/x for |x| <= pi with the obvious definition at x = 0

Extend it periodically. Will the Fourier series converge at x=0?

Homework Equations


Fourier coefficients:

ao = [tex]1/\pi \int_{-\pi}^{\pi} (sin(x)/x)[/tex]

an = [tex]1/\pi \int_{-\pi}^{\pi} (sin(x)/x) * cos(nx)[/tex]

bn = [tex]1/\pi \int_{-\pi}^{\pi} (sin(x)/x) * sin(nx)[/tex]

The Attempt at a Solution



ao = [tex]1/\pi \int_{-\pi}^{\pi} (sin(x)/x)[/tex] = 2Si(pi)/pi

an = [tex]1/\pi \int_{-\pi}^{\pi} (sin(x)/x) * cos(nx)[/tex] = (-Si((n-1)pi) + Si((n+1)pi))/pi

bn = [tex]1/\pi \int_{-\pi}^{\pi} (sin(x)/x) * sin(nx)[/tex] = 0

So the Fourier series will be:
[tex]f(x)=1/\pi*Si(\pi)+\sum_{1}^{\infty }1/\pi(-Si((n-1)\pi) + Si((n+1)\pi)[/tex]

at x=0 because cos(n(0)) = 1

I figured that:
[tex]\sum_{1}^{\infty }1/\pi(-Si((n-1)\pi) + Si((n+1)\pi)[/tex]
Will converge to zero and we get f(x) = si(pi)/pi...

I don't know what si(pi)/pi is but I don't think that converges to 1 as sin(x)/x does at x=0...did i mess up anything?
 
Physics news on Phys.org
bump, nobody yet? Am I not providing anything?
 
I'm not too familiar with the sine integral function, but seeing how your summand consists of a difference of two evaluations (and the arguments have a fixed difference! Surely that will help simplify whatever you do), maybe there's a "difference of si's" identity you could use? Maybe the difference could be expressed usefully as a differential approximation, or even a full blown Taylor series. Or, maybe there's another way to simplify the infinite sum...
 
Ok, since you are bumping, do you really need to work out the Fourier series to figure out where it converges? Don't you have theorems about where it converges to save you this pain?
 
Dick said:
Ok, since you are bumping, do you really need to work out the Fourier series to figure out where it converges? Don't you have theorems about where it converges to save you this pain?

Haha, I don't know if I mentioned, but this is from a PDE course, and we are asked to compute the Fourier series and see if it converges with the original function...The course doesn't assume a huge background in math except some knowledge in ODE's, linear algebra and multi-calc. Unfortunately I have no more memory of how to do taylor or power series, but I am sure that the question does not ask for that.

Also I did not know about the Si function until I put it in an integrator..I admit that the professor usually goes crazy when coming up with his own problems...

anywho the question asks whether it converges or not...so for all I know I could be right in that it does not converge. Can anybody confirm this?
 

Similar threads

  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
Replies
1
Views
3K
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K