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Homework Help: Prove only using Fourier Series

  1. May 16, 2005 #1
    Prove only using Fourier Series!!!

    By considering the Fourier Series of [itex]f(x)=x^2[/itex] prove that

    [tex] \sum_{n=1}^{\infty} \frac{1}{n^4} = \frac{\pi^4}{90}[/tex]
     
  2. jcsd
  3. May 16, 2005 #2
    Ok, so I considered the Fourier Series of [itex]f(x)=x^2[/itex] and found that

    [tex]\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{6}[/tex]

    Here is how I did it

    [tex]a_0 = \frac{1}{2\pi}\int_{-\pi}^{\pi}x^2dx = \frac{2\pi^2}{3} = \frac{2\pi^{2m}}{2m+1}[/tex] when [tex]m = 1[/tex]

    [tex]a_0 = \frac{1}{\pi}\int_{-\pi}^{\pi}x^2\cos(nx)dx = \frac{4}{n^2}(-1)^n [/tex]

    Since the function is even, [itex]b_n = 0[/itex].

    Hence the Fourier Series can be written explicitly as

    [tex]x^2 = \frac{\pi^2n}{2n+1} + \sum_{n=1}^{\infty}a_n \cos(nx)[/tex]

    And since [itex]n=1[/itex] we have

    [tex]x^2 = \frac{\pi^2}{3} + 4\sum_{n=1}^{\infty}\frac{(-1)^n\cos(nx)}{n^2}[/tex]

    Since [itex]x \equiv \pi[/itex] we know [itex]\cos(nx)=(-1)^n[/itex], so

    [tex]\pi^2 = \frac{\pi^2}{3} + 4\sum_{n=1}^{\infty}\frac{1}{n^2}[/tex]

    and rearranging

    [tex]\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{6} = \zeta(2)[/tex]
     
    Last edited: May 16, 2005
  4. May 16, 2005 #3
    Then I tried to do the same for [itex]f(x)=x^4[/itex], for the reason that now [itex]n=2[/itex] and I will have an expression for [itex]\sum\frac{1}{n^4}[/itex]. I came close, but not close enough.

    I know that

    [tex]\sum_{n=1}^{\infty} = \frac{\pi^4}{90} = \zeta(4)[/tex]

    but I am only allowed to use results from Fourier Analysis. Is there anyone who can help? Or perhaps has an idea?
     
  5. May 16, 2005 #4
    I am not sure I am 100% correct but what I got is slightly different from what we need.
    We have
    [tex]\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{6}[/tex]

    If we consider the fourier series of [tex]f(x)=x^4[/tex]

    [tex]x^4 = \frac{\pi}{5} + \sum_{n=1}^{\infty}(-1)^n \left( \frac{8 \pi^2}{n^2} - \frac{48}{n^4 \pi} \right) \cos(nx)[/tex]

    when [tex]x = \pi[/tex],

    [tex]\pi^4 = \frac{\pi^4}{5} + \sum_{n=1}^{\infty} \left( \frac{8 \pi^2}{n^2} - \frac{48}{n^4 \pi} \right)[/tex]

    rearraging,

    [tex]\sum_{n=1}^{\infty}\frac{1}{n^4} = \frac{8\pi^4}{15} \times \frac{\pi}{48}=\frac{\pi^5}{90}[/tex]
     
  6. May 16, 2005 #5

    HallsofIvy

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    [tex]a_0 = \frac{1}{2\pi}\int_{-\pi}^{\pi}x^2dx = \frac{2\pi^2}{3} = \frac{2\pi^{2m}}{2m+1}[/tex]

    If this is a0 WHAT is "m"?

    Same question: if
    [tex]x^2 = \frac{\pi^2n}{2n+1} + \sum_{n=1}^{\infty}a_n \cos(nx)[/tex]

    What does the "n" in the term before the sum mean???
     
  7. May 16, 2005 #6
    I think I really made a mistake. The fourier series of [tex]x^4[/tex] should be:

    [tex]x^4 = \frac{\pi^4}{5} + \sum_{n=1}^{\infty} \left( (-1)^n \frac{8(n^2\pi^2-6)}{n^4} \cos(nx) \right)[/tex]

    then put [tex]x = \pi[/tex], the result follows
     
  8. May 16, 2005 #7
    Sorry HallsovIvy, the n in the term before the sum should be "m". The m's are not summed, but the n's are - as you probably can guess.
     
  9. May 16, 2005 #8
    Kelvin, thats is the same Fourier Series as I got. However I can't see the connection between the two. I've played around with substituting [itex]\pi[/itex] for [itex]x[/itex], but I cannot get the required result.

    Is there any chance you (or someone else) could post the last remaining steps in this proof for me.
     
  10. May 17, 2005 #9
    so putting [tex]x=\pi[/tex],

    [tex]\pi^4 = \frac{\pi^4}{5} + \sum_{n=1}^{\infty} \left( (-1)^n \frac{8(n^2\pi^2-6)}{n^4} (-1)^n \right)[/tex]

    [tex]\frac{4\pi^4}{5} = \sum_{n=1}^{\infty} \left( \frac{8(n^2\pi^2-6)}{n^4} \right)[/tex]

    [tex]\frac{4\pi^4}{5} = 8\pi^2 \sum_{n=1}^{\infty} \frac{1}{n^2} - 48 \sum_{n=1}^{\infty}\frac{1}{n^4}[/tex]

    using the result that

    [tex]\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{6}[/tex]

    [tex]\sum_{n=1}^{\infty}\frac{1}{n^4} = \frac{8\pi^4}{15} \times \frac{1}{48}=\frac{\pi^4}{90}[/tex]
     
  11. May 17, 2005 #10
    OMG! I forgot that I could use the result from the first part !!!! Thanks Kelvin.

    I wonder if anyone here knows how to do it the "easy" way? You know, using the general hypergeometric function. I'd be impressed! :0
     
  12. May 17, 2005 #11

    OlderDan

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    Maybe I am missing something, but I am wondering how showing the result based on the Fourier series of [itex]f(x)=x^4[/itex] is satisfying the problem you originally posed

    Did you state the problem incorrectly? Is there some simple connection between the series for [itex]f(x)=x^4[/itex] and [itex]f(x)=x^2[/itex] that I am missing? OR does the original problems still need to be solved?
     
  13. May 18, 2005 #12
    OlderDan,

    From what I can gather, the question asked to to find a Fourier series for a particular function, then notice that it looked almost like what we wanted. Then it required a little but of imagination to guess that the actual answer we want will require us to find the Fourier series of a slightly different function. Lo and behold, [itex]f(x) = x^4[/itex] does the trick because you get the [itex]\sum\frac{1}{n^4}[/itex] instead of the [itex]\sum\frac{1}{n^2}[/itex] you get from the Fourier series of [itex]f(x) = x^2[/itex].

    I thought was that this was going to be hard to solve, since I haven't been formally introduced to the Riemann Zeta Function (which is [itex]\zeta(4) = \frac{\pi^4}{90}[/itex]). But in hindsight, the Fourier analysis involved is not far above second year uni.
     
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