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Math Help - Fourier Series

  1. #1
    Super Member Deadstar's Avatar
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    Fourier Series

    Ugg this is driving me mad...

    Consider the function f : [-\pi, \pi] \rightarrow \mathbb{C} defined by f(\theta) = |\theta|. Use Parseval’s Identity to prove

    \sum_{k=0}^{\infty} \frac{1}{(2k+1)^4} = \frac{\pi^4}{96} DONE

    and also

    \sum_{k=1}^{\infty} \frac{1}{n^4} = \frac{\pi^4}{90}

    Ideas...

    I thought since it doesn't say you HAVE to use the |\theta| function (which I used for the first part) than I could do the second part using f(\theta) = \theta^4 but that didn't work as Parsevals complicated things way too much.

    So I thought maybe find a sum for all the even terms? Like, find \sum_{k=0}^{\infty} \frac{1}{(2k + 2)^4}
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  2. #2
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    Quote Originally Posted by Deadstar View Post
    Ugg this is driving me mad...

    Consider the function f : [-\pi, \pi] \rightarrow \mathbb{C} defined by f(\theta) = |\theta|. Use Parsevalís Identity to prove

    \sum_{k=0}^{\infty} \frac{1}{(2k+1)^4} = \frac{\pi^4}{96} DONE

    and also

    \sum_{k=1}^{\infty} \frac{1}{n^4} = \frac{\pi^4}{90}

    Ideas...

    I thought since it doesn't say you HAVE to use the |\theta| function (which I used for the first part) than I could do the second part using f(\theta) = \theta^4 but that didn't work as Parsevals complicated things way too much.

    So I thought maybe find a sum for all the even terms? Like, find \sum_{k=0}^{\infty} \frac{1}{(2k + 2)^4}
    Write s = \pi^4/96 for convenience. Then s=\textstyle\sum\frac1{n^4}, where the sum is taken over all odd values of (the positive integer) n. Multiply both sides by 2^4 to see that 2^4s = \textstyle\sum\frac1{n^4} if the sum is taken over all odd multiples of 2. Then 2^8s = \textstyle\sum\frac1{n^4} if the sum is taken over all odd multples of 4, and so on. Since every positive integer is (uniquely) equal to some power of 2 times an odd integer, it follows that (1+2^4+2^8+2^{12}+\ldots)s = \textstyle\sum\frac1{n^4}, where the sum is now taken over all positive integers. Sum the geometric series to get the result.
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  3. #3
    Super Member Deadstar's Avatar
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    Holy god thanks for that.

    I've used 22 A4 pages trying to solve it, at one point I though I had even found an expression for \sum_n \frac{1}{n^3}!
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  4. #4
    Super Member Deadstar's Avatar
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    My heevans an even easier way is to see that...

    \sum_{n=1}^{\infty} \frac{1}{n^4} = \sum_{n=1}^{\infty} \frac{1}{(2n)^4} + \sum_{n=0}^{\infty} \frac{1}{(2n+1)^4}.

    So,

    \sum_{n=1}^{\infty} \frac{1}{n^4} = \frac{1}{16}\sum_{n=1}^{\infty} \frac{1}{n^4} + \sum_{n=0}^{\infty} \frac{1}{(2n+1)^4}.

    Hence \frac{15}{16} \sum_{n=1}^{\infty} \frac{1}{n^4} = \frac{\pi^4}{96}.

    => \sum_{n=1}^{\infty} \frac{1}{n^4} = \frac{\pi^4}{90}
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  5. #5
    MHF Contributor
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    Quote Originally Posted by Deadstar View Post
    My heevans an even easier way is to see that...

    \sum_{n=1}^{\infty} \frac{1}{n^4} = \sum_{n=1}^{\infty} \frac{1}{(2n)^4} + \sum_{n=0}^{\infty} \frac{1}{(2n+1)^4}.

    So,

    \sum_{n=1}^{\infty} \frac{1}{n^4} = \frac{1}{16}\sum_{n=1}^{\infty} \frac{1}{n^4} + \sum_{n=0}^{\infty} \frac{1}{(2n+1)^4}.

    Hence \frac{15}{16} \sum_{n=1}^{\infty} \frac{1}{n^4} = \frac{\pi^4}{96}.

    => \sum_{n=1}^{\infty} \frac{1}{n^4} = \frac{\pi^4}{90}
    I can't imagine how I overlooked that. It's a technique I've known for years, but I never thought of using it.
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