Fourierseries of X^2-1

  • #1
Determine the fourierseries of the function

f(x)=x^2−1

defined on the interval [-pi,pi]. What will the coefficient a/b be in front of the cos3x in the fourierseries?


I am not sure how to start this solving this problem. Pls help.
 

Answers and Replies

  • #2
nicksauce
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Well you use your formula for a_n and b_n. Did you try that?
 
  • #3
I read that determining if the function is odd or even might be a good start.
 
  • #4
nicksauce
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True. So is this function even or odd or neither?
 
  • #5
Its even if f(-x)=f(x) and odd if f(-x)=-x so in my case:

f(-x)=(-x)^2−1=x^2-1
so it must be even.
 
  • #6
Hence I must use the cosine series on the interval -pi<x<pi. as

f(x) ~ a_0/2+ sum (n=1 to inf) a_n*cos((n*pi)/Pi)*x where:
a_0=2/pi*int(0 to pi) f(x)dx and
a_n=2/pi*int(0 to pi) f(x)cos(n*pi)/pi*x dx

But do I just start integrating from 0 to Pi to get a_0? Why not -Pi to Pi as indicated by the range of the problem? Is it because the integral will go to zero then?
 
Last edited:
  • #7
matt grime
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It's an even integrand. The integral from -K to K is twice that of the integral from 0 to K.

There are 2 things here.

1) Can you use the formulae you've been given?
2) Do you understand the formulae you've been given?

If you focus on 2), then 1) becomes easier.
 
  • #8
nicksauce
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You could integrate from 0 to Pi, or from -Pi to Pi and take out the factor of 2. Since f(x) is even, it'll be the same either way.
 
  • #9
It's an even integrand. The integral from -K to K is twice that of the integral from 0 to K.

There are 2 things here.

1) Can you use the formulae you've been given?
2) Do you understand the formulae you've been given?

If you focus on 2), then 1) becomes easier.

Thanks. I did find this out. Since I integrated and what you guys are saying is surely correct.
But once I try to get a_n I end up with nasty partial integrations. I have to write back once I got my final results.
 
  • #10
HallsofIvy
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I have no idea what you mean by "partial integration". Basically you are integrating [itex](x^2- 1)cos(nx)[/itex]. That should be easy using integration by parts. Was that what you meant?
 
  • #11
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I have no idea what you mean by "partial integration". Basically you are integrating [itex](x^2- 1)cos(nx)[/itex]. That should be easy using integration by parts. Was that what you meant?

partial integration = IBP = integration by parts
 
  • #12
HallsofIvy
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So the whole problem is to integrate
[tex]\int_{-\pi}^{\pi} (x^2-1)cos(3x)dx[/tex]

That should be easy using integration by parts (twice).
 
  • #13
Yes dirk_mec1 and HallsofIvy, I was wrong...I was thinking of integration by parts. I have used IBP to get the solution of the integrand you specified and my solution matched with the one I got from a computer software, as attached. I can see that the cos(3x) term appears in my solution but the coefficient of (2/9)Pi in front of the cos(3x) term is incorrect. This gives me reason to question what I am doing wrong. I do understand the procedure of this problem, but I am a little confused. Can you pls point me in the right direction?
 

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  • #14
You could integrate from 0 to Pi, or from -Pi to Pi and take out the factor of 2. Since f(x) is even, it'll be the same either way.

Thanks that was helpful.
 
  • #15
nicksauce
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Yes dirk_mec1 and HallsofIvy, I was wrong...I was thinking of integration by parts. I have used IBP to get the solution of the integrand you specified and my solution matched with the one I got from a computer software, as attached. I can see that the cos(3x) term appears in my solution but the coefficient of (2/9)Pi in front of the cos(3x) term is incorrect. This gives me reason to question what I am doing wrong. I do understand the procedure of this problem, but I am a little confused. Can you pls point me in the right direction?

Well for now I'll assume your integration is correct, without checking it myself. The sin terms dissapear because sin(0) = sin(n*pi) = 0. The x=0 boundary dissapears for the first term, because x=0, so for the whole integration, you have 2/9 * pi * cos(3pi) = -2pi/9. To get the a_3 term, you must multiply by 2/pi, giving you -4/9, not (2/9)Pi as you said.
 
  • #16
Nick...thanks so much. What you said made perfect sense. I was disregarding the terms for n=3 and that's what I did wrong. So, now my answer is correct; a=-4 and b=9. Thus, This problem is solved:)
 

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