Fourier Series

**Deadstar** · February 22nd 2010, 04:03 PM

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

**Opalg** · February 23rd 2010, 12:52 AM

Originally Posted by Deadstar

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.

**Deadstar** · February 23rd 2010, 01:58 AM

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

**Deadstar** · February 23rd 2010, 03:09 AM

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

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

**Opalg** · February 23rd 2010, 03:14 AM

Originally Posted by Deadstar

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.

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