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Starting with Fourier cosine series for $f(x)=x^2$ in the interval $(0,l),$ use Parseval identity to compute $\sum_{n=1}^{\infty}\frac{1}{n^4}$.

Attempt: I have that the coefficient $A_n$ for cosine series is $$A_n=(1/(n^3 \pi^3))2 l^2 (2 n \pi \cos( n \pi) + (-2 + n^2 \pi^2) \sin(n \pi))$$

By other hand I have that $\int_0^lx^2dx<\infty$ so the convergence is in $L^2$ sense, so I have the parseval identity, i.e.

$\frac{l^5}{5}=\int_0^l|x^2|^2dx=\sum_{n=1}^\infty A_n^2\Vert \cos\frac{n\pi x}{l}\Vert^2=\sum_{n=1}^\infty A_n^2 (l (n \sin(2 \pi n) + n \sinh(2 \pi n)))/(4 \pi n^2)$

Now I have to take out the $\frac{1}{n^4}$ term from the series, which is definetely not an easy task.

Am I correct so far?

Is there a easy way to proceed? If not, Is there an alternative (easier) solution to this exercise?


Edit

Noticing the cosine multiples of $\pi$ I get

$\frac{\pi^5}{20}=\sum_{n=1}^{\infty}\frac{\sinh(2\pi n)}{n^5}$

but this doesn't looks like $\sum_{n=1}^{\infty}\frac{1}{n^4}$ at all... Unless $\frac{\sinh(2\pi n)}{n}=1$

What am I doing wrong?

Please help me.

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marked as duplicate by Jack D'Aurizio real-analysis May 12 '18 at 9:54

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  • $\begingroup$ Why not take $l=1$? $\endgroup$ – Lord Shark the Unknown May 12 '18 at 1:56
  • $\begingroup$ @LordSharktheUnknown Wouldn't it alter the solution? I mean the exercise says: $(0,l)$ not $(0,1)$ $\endgroup$ – Al t. May 12 '18 at 1:57
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    $\begingroup$ You also have things like $Cos[n\pi]$. For a start, this is peculiar notation: the standard notation is $\cos(n\pi)$. But you should know what the cosines of the multiples of $\pi$ are. $\endgroup$ – Lord Shark the Unknown May 12 '18 at 2:02
  • $\begingroup$ @LordSharktheUnknown Sorry I already edited it. What are the cosines of the multiples of π ? $\endgroup$ – Al t. May 12 '18 at 2:40
  • $\begingroup$ @LordSharktheUnknown I already remembered.. you mean this $\forall n \in \mathbb Z: \cos n \pi = \left({-1}\right)^n $ ? $\endgroup$ – Al t. May 12 '18 at 3:02