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Partial sum of Fourier Coefficients

  1. Jun 10, 2008 #1
    1. The problem statement, all variables and given/known data

    f (x) = 0 -pi<x<0

    x^2 0<x<pi

    Find the fourier series and use it to show that

    (pi^2)/6=1+1/2^2+1/3^2+...

    2. Relevant equations


    N/A

    3. The attempt at a solution

    I was able to find the fourier series and my answer matched with the back of the book. But I don't understand how I am supposed to use it to prove the expansion of pi^2/6 . I tried plugging in (pi^2)/6 for f(x) but that didn't work out.


    Thanks!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jun 10, 2008 #2

    nicksauce

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    What do you find to be the fourier series for f(x)?
     
  4. Jun 10, 2008 #3

    Dick

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    Just to echo nicksauce, if you found the fourier series as you said, what is it?
     
  5. Jun 10, 2008 #4
    I didn't want to type out the fourier series since I am really new to latex but here it is

    f(x) =pi^2/6 +[tex]\Sigma[/tex] { (2(-1)^n/n^2) cos nx + ((-1)^(n+1) pi/n + 2/(pi * n^3) [(-1)^n-1] sin nx }

    I hope that made sense and that I didn't make any typos.
     
  6. Jun 10, 2008 #5
    P.S: and n goes from 1 to infinity
     
  7. Jun 10, 2008 #6

    Dick

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    Thanks. Since the mean value of the function over the interval is pi^2/6, I was hoping I could just say put x=0. But you've got that alternating sign and factor of 2 in the cosine part. Are you sure of the series? Otherwise, I'll have to work it out. Didn't want to do that.
     
  8. Jun 11, 2008 #7
    I am pretty sure about the series. I verified with the back of the book. We were also supposed to show pi^2/12=1-1/2^2+ etc. I plugged in x=o and was able to prove it.

    For the pi^2/6, I tried setting x to be pi/2 so that the cosine term goes away but that didn't work out.
     
  9. Jun 11, 2008 #8

    Dick

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    It looks to me like you should be putting x=pi. That turns cos(nx) into (-1)^n makes the sin(nx) vanish. I'm trying to check your series. I get something a lot like it. But I can't seem to get all the parts quite right. Guess I'm not very good at this...
     
  10. Jun 11, 2008 #9

    Dick

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    Doh. I've been being stupid. Regarded as a periodic function f(x) is discontinuous at x=pi. The series doesn't converge to f(pi). It converges to (f(pi)+f(-pi))/2. Or pi^2/2. Do you know why? Now try x=pi in your series.
     
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