fourier summation

**schinb64** · October 16th, 2007, 03:46 PM

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.

**topsquark** · October 16th, 2007, 03:57 PM

Originally Posted by schinb64

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

**schinb64** · October 16th, 2007, 04:46 PM

Originally Posted by topsquark

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.

**ThePerfectHacker** · October 16th, 2007, 07:17 PM

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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