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Math Help - fourier summation

  1. #1
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    fourier summation

    T^4= (16Pi^4)/5 + 16∑{n=1, ∞}[((2Pi^2)/n^2 ) - (3/n^4)] Cos[nt] + 16 Pi ∑{n=1,∞} ((3/n^3)-(Pi^2/n))Sin[nt]

    From the fourier series above deduce the summations

    ∑{n=1,∞} (1/n^4) = (Pi^4)/90 ; ∑{n=1,∞} ((-1)^n+1)/n^4= (7Pi^4)/720

    ∑{n=odd} 1/n^4 = (Pi^4)/96

    I do not understand what I am being asked to do with the problem. Do I need to find a value for t to figure this out? Any help would be appricated. Thank you for your time.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by schinb64 View Post
    T^4= (16Pi^4)/5 + 16∑{n=1, ∞}[((2Pi^2)/n^2 ) - (3/n^4)] Cos[nt] + 16 Pi ∑{n=1,∞} ((3/n^3)-(Pi^2/n))Sin[nt]

    From the fourier series above deduce the summations

    ∑{n=1,∞} (1/n^4) = (Pi^4)/90 ; ∑{n=1,∞} ((-1)^n+1)/n^4= (7Pi^4)/720

    ∑{n=odd} 1/n^4 = (Pi^4)/96

    I do not understand what I am being asked to do with the problem. Do I need to find a value for t to figure this out? Any help would be appricated. Thank you for your time.
    Let me attempt to translate this:
    Given the Fourier series:
    T^4 = \frac{16 \pi ^4}{5} + 16 \sum_{n = 1}^{\infty} \left ( \frac{2 \pi ^2}{n^2} - \frac{3}{n^4} \right ) cos(nt) + 16 \pi \sum_{n = 1}^{\infty} \left ( \frac{3}{n^3} - \frac{\pi ^2}{n} \right ) sin(nt)

    And you need to find expressions for
    \sum_{n = 1}^{\infty} \frac{1}{n^4} = \frac{\pi ^4}{90}

    \sum_{n = 1}^{\infty} \frac{(-1)^n + 1}{n^4} = \frac{7 \pi^4}{720}

    \sum_{n, odd} \frac{1}{n^4} = \frac{\pi ^4}{96}

    -Dan
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  3. #3
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    Quote Originally Posted by topsquark View Post
    Let me attempt to translate this:
    Given the Fourier series:
    T^4 = \frac{16 \pi ^4}{5} + 16 \sum_{n = 1}^{\infty} \left ( \frac{2 \pi ^2}{n^2} - \frac{3}{n^4} \right ) cos(nt) + 16 \pi \sum_{n = 1}^{\infty} \left ( \frac{3}{n^3} - \frac{\pi ^2}{n} \right ) sin(nt)

    And you need to find expressions for
    \sum_{n = 1}^{\infty} \frac{1}{n^4} = \frac{\pi ^4}{90}

    \sum_{n = 1}^{\infty} \frac{(-1)^n + 1}{n^4} = \frac{7 \pi^4}{720}

    \sum_{n, odd} \frac{1}{n^4} = \frac{\pi ^4}{96}

    -Dan
    that is correct.
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  4. #4
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    Given the Fourier series:
    t^4 = \frac{16 \pi ^4}{5} + 16 \sum_{n = 1}^{\infty} \left ( \frac{2 \pi ^2}{n^2} - \frac{3}{n^4} \right ) cos(nt) + 16 \pi \sum_{n = 1}^{\infty} \left ( \frac{3}{n^3} - \frac{\pi ^2}{n} \right ) sin(nt)
    Let t=0 we get:
    0 = \frac{16\pi^4}{5} + 16\sum_{n=1}^{\infty} \left( \frac{2\pi^2}{n^2} - \frac{3}{n^4} \right)
    Thus,
    \sum_{n=1}^{\infty} \left( \frac{3}{n^4} - \frac{2\pi^2}{n^2} \right) = \frac{\pi^4}{5}
    Now use the fact that,
    \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6} \implies \sum_{n=1}^{\infty} \frac{2\pi^2}{n^2} = \frac{2\pi^4}{6}.
    (It seems we need this fact in this problem).

    Now you can solve for \sum_{n=1}^{\infty} \frac{1}{n^4}
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