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How come the trick used to prove the Basel problem doesn't work for other values?

  1. Aug 13, 2012 #1
    The solution using the Fourier series of x and Parseval's Identity to prove that

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

    doesn't seem to work for exponents higher than 2, why?

    For example, I found the Fourier coefficients of [itex] - \frac{x^2} {4} [/itex] to be
    [itex] \frac{(-1)^{n+1}} {n^2} [/itex].

    So if I use the same reasoning as in the Basel problem:
    [tex] \sum_{n=1}^{∞} \frac{1} {n^4} = \frac{1} {\pi} \int_{-\pi}^{\pi} \frac{x^4} {16} dx = \frac{\pi^4} {40}[/tex]

    which is not even close to the correct value.


    How come when Euler does it, it works like magic, but when I try to do the same thing, it fails? Am I doing something wrong? Is Parseval's Identity valid only for the function f(x) = x?
     
    Last edited: Aug 13, 2012
  2. jcsd
  3. Aug 13, 2012 #2


    What "trick" or "reasoning" are you talking about? There are at least 3 more or less simple functions the Fourier

    series of which can give the value of the series of inverse squares, and the same can be done with any series of inverse even powers.

    DonAntonio
     
  4. Aug 13, 2012 #3
  5. Aug 13, 2012 #4

    Mute

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    Unlike for the Basal problem, the fourier coefficient [itex]a_0[/itex] for the problem you are doing is not zero.

    When you account for the [itex]a_0^2[/itex] term, you should get the correct result.

    (By the way, Euler solved the problem using different methods than Parseval's theorem).
     
    Last edited: Aug 13, 2012
  6. Aug 13, 2012 #5

    micromass

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    Also, I think that your Fourier coefficients are not quite correct. They should be

    [tex]\frac{1}{2\pi}\int_{-\pi}^\pi f(x)e^{-inx}dx[/tex]

    I don't get your answer of [itex](-1)^{n+1}/n^2[/itex]. Rather, I get an extra factor of 1/2.
    Also, as noted, there are problems with n=0, which you must calculate separately.
     
  7. Aug 13, 2012 #6
    Oh I guess that must be my mistake.

    And yet it is not obvious to me how the Bernoulli numbers come up. Anyways, I'll work on it more again.

    But I have a related question:

    is there any sort of analogue of Parseval's identity for the inner product of two functions?
    What I mean is, Parseval's identity relates the squared norm (the integral of f(x) squared on some domain) with the sum of the squares of the Fourier coefficients, similar to the way one finds the squared norm of a vector. Could there be a similar relationship between the integral of f(x)*g(x) on the domain with the sum of the products of the Fourier coefficients of f and g? (analogous to the dot product of two vectors).

    I'm asking this because zeta(3) could be considered the "dot product" of two sequences, one with fourier coefficients 1/n and one with 1/n^2 (or with (-1)^n somewhere) and so the same method could have been used.
     
  8. Aug 13, 2012 #7
    Parseval's theorem proves that the Fourier transform (discrete or continuous) is unitary. So yes.

    Proving that an operator preserves the norm of all vectors is equivalent to proving that the operator preserves all two vector inner products. Furthermore, knowing the quadratic form of any linear operator for all vectors gives you the sesquilinear form of the operator for all pairs of vectors, and thus, completely specifies the operator. Look up the "polarization identity".
     
  9. Aug 13, 2012 #8
    But then could this be used to find the value for the sum of the reciprocals of the cubes?

    Because you could consider it the inner product of x and x^2 (multiplied by a scalar factor)?

    Since the Fourier coefficients of x involve 1/n and the Fourier coefficients of x^2 involve 1/n^2, then the inner product of x and x^2 should involve 1/n^3?
     
  10. Aug 13, 2012 #9
    In the interest of full disclosure, I haven't been paying full attention to this thread. I just knew the answer to this last question off hand. I'm not going to swear on anything given any assumptions or workflows that have been presumed already, but I'll say this:

    There is an 'inner-product Parseval's identity'. The correct form of it can be seen in the third equation of http://en.wikipedia.org/wiki/Parseval's_theorem. The special case of real functions is shown in the fifth equation.

    If you have found functions whose inner-product of fourier coefficients gives you the series you are looking for, by all means go ahead an do the integral to find the value of the sum. Just don't forget the negative coefficients and don't forget the zeroth coefficient. If your using equation 5 from the link I posted, then the negative coefficients are taken care of. Just remember that those equations are talking about complex exponential fourier series, not sines and cosines series.
     
  11. Aug 13, 2012 #10
    As a hint, if you try to apply the Parseval's identity to x and x^2, and do so correctly, you will find 0=0.

    If a_n are the x coefficients and b_n are the x^2 squared coefficients:

    a_0=0
    a_n for n≠0 are imaginary

    b_n are all real.

    a_n*b_n is imaginary or zero (either way the real part is zero, and the real part of any sum of such coefficients is zero).

    x^3 is an odd function, and any integral symmetric about the origin will be zero.

    I suspect you forgot about the factor of i in the coefficients of x.
     
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