# Math Help - Fourier Series

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

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

3. 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 $\sum_n \frac{1}{n^3}$!

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

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

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.