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Fourierseries of X^2-1

  1. Jul 3, 2008 #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.
     
  2. jcsd
  3. Jul 3, 2008 #2

    nicksauce

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    Well you use your formula for a_n and b_n. Did you try that?
     
  4. Jul 3, 2008 #3
    I read that determining if the function is odd or even might be a good start.
     
  5. Jul 3, 2008 #4

    nicksauce

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    True. So is this function even or odd or neither?
     
  6. Jul 3, 2008 #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.
     
  7. Jul 3, 2008 #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: Jul 3, 2008
  8. Jul 3, 2008 #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.
     
  9. Jul 3, 2008 #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.
     
  10. Jul 3, 2008 #9
    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.
     
  11. Jul 4, 2008 #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?
     
  12. Jul 4, 2008 #11
    partial integration = IBP = integration by parts
     
  13. Jul 4, 2008 #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).
     
  14. Jul 4, 2008 #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?
     

    Attached Files:

  15. Jul 4, 2008 #14
    Thanks that was helpful.
     
  16. Jul 4, 2008 #15

    nicksauce

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    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.
     
  17. Jul 4, 2008 #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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