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Fourier series and summations

  1. May 1, 2010 #1
    1. The problem statement, all variables and given/known data
    I'm trying to find the sum of the infinite series:

    1/n^4

    using the Fourier series of x^2 on [-pi, pi] which I have as

    PI^2/3 + 4*sum(n=1..infinity) (-1)^n/n^2*cos(nx)

    2. Relevant equations



    3. The attempt at a solution

    So far all my attempts have been focused on finding a the series at various points in [-pi, pi] and seeing if I can add all the series together to cancel out the terms I don't need or leave the series I want with another series that I know the sum for. However, so far all these attempts have been unsuccessful and I think that since most of these equations seem fairly unwieldy that I'm just missing a small detail that would make this problem easily solvable. Anyone have any suggestion as to where to start on this problem so I can see if I can take it from there?
     
    Last edited: May 1, 2010
  2. jcsd
  3. May 1, 2010 #2

    gabbagabbahey

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    I don't see how you can possibly use that series to find the sum you are looking for...are you sure that's what you are supposed to do?
     
  4. May 1, 2010 #3
    That's the trouble I'm having, I can't really figure out how to go about it. And all the attempts I've made so far leave me with series and sums that are interesting but of no real use in solving the problem.
     
  5. May 1, 2010 #4

    gabbagabbahey

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    Are you sure it isn't a typo and you are really supposed to find [tex]\sum_{n=1}^{\infty}\frac{1}{n^2}[/itex] instead?
     
  6. May 1, 2010 #5
    I honestly don't know, we proved that identity in class, and it's a relatively trivial problem once you have the Fourier series so I don't think he intended for us to find it again.
     
  7. May 1, 2010 #6

    gabbagabbahey

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    What is the exact wording on your original problem statement?
     
  8. May 1, 2010 #7
    Find the Fourier series for f(x) = x^2 on [-pi,pi] and use it to find sum(n=1..infinity) 1/n^4
     
  9. May 1, 2010 #8

    gabbagabbahey

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    Hmmm...have you also found the Fourier series for [itex]x^4[/itex] on the same interval?
     
  10. May 1, 2010 #9
    That was actually my first instinct when my first attempts failed, I thought perhaps the original function was a typo. The integral to find the Fourier coefficients of x^4 on this interval (which I obtained from Wolfram as I didn't want to do all the integration by parts on paper) is:

    [ 4x*(n^2*x^2 - 6)*cos(nx) ] / (n^4)

    which looks even worse as x is now a function outside the cosines. I omitted the sines in the integral as they're irrelevant on this interval.
     
  11. May 1, 2010 #10

    gabbagabbahey

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    Well, the question only makes sense if you've already shown that

    [tex]\begin{aligned}x^4 &= \frac{\pi^4}{5}+8\sum_{n=1}^{\infty}\frac{(-1)^n(n^2\pi^2-6)}{n^4}\cos(nx) \\ &= \frac{\pi^4}{5}+8\pi^2\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}\cos(nx)-48\sum_{n=1}^{\infty}\frac{(-1)^n}{n^4}\cos(nx)\end{aligned}[/tex]

    either on a previous question in this assignment, or in your text/notes.

    Why do I say this? Simple; as you can see above, the series for [itex]x^4[/itex] involves the sum you are trying to calculate, and also the sum in the series for [itex]x^2[/itex]
     
  12. May 1, 2010 #11
    I see, and then taking these series at x = pi will give me the sum after some simple rearrangement and reduction.

    Thanks a lot. I can't believe I didn't even mess around with that series when I found it originally, I just kind of dismissed it even though it was simple enough to use. Much appreciated, I worked out the series on paper myself and I'll fill in all the other details.
     
    Last edited: May 1, 2010
  13. May 3, 2010 #12
    Let's take a step back and look at the theory of Fourier series again. When you expand a function like x^2 in a Fourier series, this is analogous as expanding a vector in terms of basis vectors in linear algebra. It is more convenient to work with exp(i n x) as the basis functions, but let's instead work with the set of functions:

    e0(x) = 1/sqrt(2)

    en(x) = cos(n x) for n >=1

    Define an inner product as follws:

    (f,g) = 1/pi Integral from -pi to pi of f(x)g(x)dx

    We don't need to take the complex conjugate of h if we're dealing with real functions. It is easy to check that the en are orthonormal.

    You can now expand any function f(x) as:

    f(x) = Sum over n of (f,en) en(x)

    If I take the series you obtained:

    PI^2/3 + 4*sum(n=1..infinity) (-1)^n/n^2*cos(nx)

    I can make the above form explicit:

    x^2 = sqrt(2) Pi^2/3 eo(x) + sum(n=1..infinity) 4 (-1)^n/n^2 en(x)


    You can now ask: What is the norm of the function x^2 considered as a vector of the Hilbert space? Of course, you can simply compute the inner product of x^2 with itself, to obtain:

    (x^2,x^2) = 1/pi Integral from -pi to pi of x^4 dx =

    2/5 pi^4

    But, of course, if we have that f = sum over n of cn en(x)

    then the squared norm will also be the sum of the squares of the cn. This is analogous to Pythagoras' theorem, you can see this by writing

    (c0 e0 + c1 e1 + c2 e2..., c0 e0 + c1 e1 + c2 e2+ ..)

    using linearity of the inner product and using the fact that (en,em) = 0 if n and m are different while it is equal to 1 if they are equal, you obtain the result.

    So, the squared norm of x^2 is also given as:

    2 Pi^4/9 + 16*desired summation.

    Equating this to 2/5 pi^4 gives:

    Desired summation = pi^4/90
     
    Last edited: May 3, 2010
  14. May 3, 2010 #13
    B.t.w. the expression for the squared norm expressed as te summation over the squared coefficients is called Parceval's theorem.
     
  15. May 3, 2010 #14
    Thanks a lot Iblis, your proof made a lot of sense. We never really discussed much of the theory of Fourier series, it was mostly a couple of lectures intended to demonstrate an application of the study of Hilbert spaces and orthogonal fuctions.
     
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