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Insights Using the Fourier Series To Find Some Interesting Sums - Comments

  1. Dec 22, 2016 #1

    Svein

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  2. jcsd
  3. Dec 25, 2016 #2

    micromass

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  4. Dec 26, 2016 #3

    Svein

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    Yes - but I thought it best to keep it reasonably simple the first time through.
     
  5. Jan 1, 2017 #4

    stevendaryl

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    Very nice. I have done similar tricks for evaluating summations, but I didn't know the trick of using Parseval’s formula. My favorite trick is using representations of the Dirac delta function, and that's how I would evaluate [itex]\sum \frac{1}{n^2}[/itex]. However, it's a lot more convoluted.

    Start with the representation for the delta function:

    [itex]\delta(x) = \frac{1}{2\pi} + \frac{1}{\pi} \sum_n cos(nx)[/itex]

    Now, integrate both sides from [itex]-x[/itex] to [itex]+x[/itex].

    [itex]sign(x) = \frac{x}{\pi} + \frac{2}{\pi} \sum_n \frac{1}{n} sin(nx)[/itex]

    (where [itex]sign(x) = \pm 1[/itex] depending on whether [itex]x>0[/itex] or [itex]x < 0[/itex])

    Integrate again, this time from [itex]0[/itex] to [itex]x[/itex]:

    [itex]|x| = \frac{x^2}{2\pi} + \frac{2}{\pi} \sum_n \frac{1}{n^2} (1 - cos(nx))[/itex]

    Using a trig identity, [itex]1-cos(nx) = 2 sin^2(\frac{n}{2} x)[/itex]. So we have:

    [itex]|x| = \frac{x^2}{2\pi} + \frac{4}{\pi} \sum_n \frac{1}{n^2} sin^2(\frac{n}{2} x)[/itex]

    Now, we set [itex]x = \pi[/itex] to get the identity:

    [itex]\pi = \frac{\pi}{2} + \frac{4}{\pi} \sum_n \frac{1}{n^2} sin^2(\frac{n}{2} \pi) = \frac{4}{\pi} \sum_{odd\ n} \frac{1}{n^2}[/itex]

    So:

    [itex]\frac{\pi}{2} = \frac{4}{\pi} \sum_n \frac{1}{n^2} sin^2(\frac{n}{2} \pi) = \frac{4}{\pi} \sum_{odd\ n} \frac{1}{n^2}[/itex]

    Drat! The sum on the right side is only over odd values of [itex]n[/itex], because [itex]sin^2(\frac{n}{2} \pi) = 0[/itex] when [itex]n[/itex] is even. But all is not lost. We can reason as follows:

    [itex]\sum_n \frac{1}{n^2} = \sum_{odd\ n} \frac{1}{n^2} + \sum_{even\ n} \frac{1}{n^2}[/itex]

    but [itex]\sum_{even\ n} \frac{1}{n^2} = \sum_{n} \frac{1}{(2n)^2} = \frac{1}{4} \sum_{n} \frac{1}{n^2}[/itex]. So we have:

    [itex]\sum_n \frac{1}{n^2} = \sum_{odd\ n} \frac{1}{n^2} + \frac{1}{4}\sum_{n} \frac{1}{n^2}[/itex]

    So [itex]\sum_{odd\ n} \frac{1}{n^2} = \frac{3}{4} \sum_n \frac{1}{n^2}[/itex]. Putting this back into our result, we have:

    [itex]\frac{\pi}{2} = \frac{4}{\pi} \sum_n \frac{1}{n^2} sin^2(\frac{n}{2} \pi) = \frac{4}{\pi} \frac{3}{4} \sum_{n} \frac{1}{n^2}[/itex]

    So [itex]\sum_n \frac{1}{n^2} = \frac{\pi^2}{6}[/itex]

    Using Parseval is a lot simpler.
     
  6. Feb 2, 2017 #5
    Nice Insight @Svein!
     
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