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Series of Integrals

  1. Jan 28, 2012 #1
    1. The problem statement, all variables and given/known data
    Let [itex]f[/itex] be a continuously differentiable function on the interval [itex][0,2\pi][/itex], where [itex]f(0) = f(2\pi)[/itex] and [itex]f'(0) = f'(2\pi)[/itex]. For [itex]n = 1,2,3,\dotsc[/itex], define
    [itex]
    a_n = \frac{1}{2\pi} \int_0^{2\pi} f(x) \sin(nx) dx.
    [/itex]
    Prove that the series[itex] \sum_{n=1}^\infty |a_n|^2[/itex] converges.



    2. Relevant equations



    3. The attempt at a solution
    So far I've got that
    [itex]\frac{1}{2\pi} \int_0^{2\pi} f(x) \sin(nx) dx = \frac{1}{2\pi n}\int_0^{2\pi} f'(x)\cos(nx) dx[/itex]
    via integration by parts and the conditions on [itex]f[/itex]. I've also got that
    [itex]|a_n| \le \frac{1}{2\pi} \int_0^{2\pi} |f(x)| dx[/itex]
    and
    [itex]|a_n| \le \frac{1}{2\pi n} \int_0^{2\pi} |f'(x)| dx[/itex]

    Both from the boundedness of sine and cosine. I also know that [itex]f[/itex] is a rectifiable curve, although I'm not sure this helps at all, nor what that really means. I'm just stuck as to where to go from here since I've forgotten much of what I'd learned of series. Also I'm wondering whether the bounds on [itex]|a_n|[/itex] are actually helping at all. If you could point me in the right direction I'd be very grateful.

    Thanks.


    Edit: I just realized that the second bound on [itex]|a_n|[/itex] implies that it goes to 0, so the series at least has a chance to converge.
     
  2. jcsd
  3. Jan 28, 2012 #2
    You're on the right track with your integration by parts. You might want to take a look at just the integral [itex]\int_0^{2\pi}|cos(nx)|dx[/itex] for various n. Also, the fact that the terms of the series are [itex]|a_n|^2[/itex] and not just [itex]|a_n|[/itex] is important.

    Knowing that [itex]f'(0) = f'(2\pi)[/itex] seems superfluous to this problem..
     
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