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Math Help - Fourier series

  1. #1
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    Fourier series

    Hey guys.
    I wrote the problem in word, it's much easier for me
    Thanks in advance.
    Attached Thumbnails Attached Thumbnails Fourier series-1.jpg  
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  2. #2
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    Quote Originally Posted by asi123 View Post
    The problem is the other one, (x^3), how can I get \sum_{n=1}^\infty\frac1{n^3} from that?
    You'll be famous if you succeed. \sum_{n=1}^\infty\frac1{n^3} is Apéry's constant, for which there is no known closed form. Even less is known about \sum_{n=1}^\infty\frac1{n^5}.
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    Quote Originally Posted by Opalg View Post
    You'll be famous if you succeed. \sum_{n=1}^\infty\frac1{n^3} is Apéry's constant, for which there is no known closed form. Even less is known about \sum_{n=1}^\infty\frac1{n^5}.
    Well, the question is find the sum of the series \sum_{n=1}^\infty\frac1{n^5}, are you sure that can not be done?
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    Quote Originally Posted by asi123 View Post
    Well, the question is find the sum of the series \sum_{n=1}^\infty\frac1{n^5}, are you sure that can not be done?
    He is sure. And so am I.

    Some research for you to do:

    Riemann zeta function.

    \zeta(2n)=\frac{(-1)^{k+1}(2\pi)^{2n}B_{2n}}{2(2n)!}.

    Very little is known about \zeta(2n + 1) apart from what has already been mentioned for the case n = 1.
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    Quote Originally Posted by mr fantastic View Post
    He is sure. And so am I.

    Some research for you to do:

    Riemann zeta function.

    \zeta(2n)=\frac{(-1)^{k+1}(2\pi)^{2n}B_{2n}}{2(2n)!}.

    Very little is known about \zeta(2n + 1) apart from what has already been mentioned for the case n = 1.
    Just in case the original poster want to see \zeta(2n) derivation, it is here.
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  6. #6
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    Quote Originally Posted by asi123 View Post
    Hey guys.
    I wrote the problem in word, it's much easier for me
    Thanks in advance.
    I think you might need to combine the Fourier series using the convolution theorem for Fourier series, though if it will give anything usefull I don't know.

    .
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