1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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]


    [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


    User Avatar
    Science Advisor

    [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


    User Avatar
    Science Advisor
    Homework Helper

    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

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook