# Sum using fourier series

• May 28th 2011, 06:51 PM
Ulysses
Sum using fourier series
Hi. Well, I have to demonstrate that $\sum_{n=1}^{\infty}\frac{1}{(2n-1)^2}=\frac{\pi^2}{8}$
Using a previous result of a Fourier series for f(t)=1 if 0<t<1, f(t)=0 if -1<t<0.
I've found that:
$f(t)\sim{}1/2+\sum_{n=1}^{\infty}\frac{2}{(2n-1)\pi}\sin([2k-1]\pi t)$

I think this result, the series expansion for the function is right. But when I've tried to demonstrate what the problem asks me I get to an absurd, so I'm not sure where the error is.

This is what I did:

$S(1/2)=1=\frac{1}{2}+\sum_{n=1}^{\infty}\frac{2}{(2n-1)\pi}\sin([2k-1]\frac{\pi}{2})$
Working from here:

$\frac{1}{2}=\frac{2}{\pi}\sum_{n=1}^{\infty}\frac{ (-1)^{n-1}}{(2n-1)}$
$\frac{\pi}{4}=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{(2n-1)}$
$\frac{\pi^2}{16}=\sum_{n=1}^{\infty}\frac{1}{(2n-1)^2}$

So, I'm doing something wrong. Perhaps you can see more clearly where am I committing the mistake.
• May 28th 2011, 07:25 PM
TheEmptySet
Quote:

Originally Posted by Ulysses
Hi. Well, I have to demonstrate that $\sum_{n=1}^{\infty}\frac{1}{(2n-1)^2}=\frac{\pi^2}{8}$
Using a previous result of a Fourier series for f(t)=1 if 0<t<1, f(t)=0 if -1<t<0.
I've found that:
$f(t)\sim{}1/2+\sum_{n=1}^{\infty}\frac{2}{(2n-1)\pi}\sin([2k-1]\pi t)$

I think this result, the series expansion for the function is right. But when I've tried to demonstrate what the problem asks me I get to an absurd, so I'm not sure where the error is.

This is what I did:

$S(1/2)=1=\frac{1}{2}+\sum_{n=1}^{\infty}\frac{2}{(2n-1)\pi}\sin([2k-1]\frac{\pi}{2})$
Working from here:

$\frac{1}{2}=\frac{2}{\pi}\sum_{n=1}^{\infty}\frac{ (-1)^{n-1}}{(2n-1)}$
$\frac{\pi}{4}=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{(2n-1)}$
$\frac{\pi^2}{16}=\sum_{n=1}^{\infty}\frac{1}{(2n-1)^2}$

So, I'm doing something wrong. Perhaps you can see more clearly where am I committing the mistake.

You can't just square an infinite sum you would be missing all of the cross terms.

You are using the wrong function. You need to use
$f(t)=\begin{cases}0, \text{ if } -1 < t < 0 \\ t, \text{ if } 0 \le t < 1 \end{cases}$
Attachment 21612
You will need to re index the series to get it to match
• May 31st 2011, 01:18 PM
Ulysses
It goes as follows. Exercise 4 asks me to find the Fourier representation for:
http://www.physicsforums.com/attachm...1&d=1306644128
(sorry I link this from outside this forum, but I couldn't attach the image not today, nor even a few days ago)
I've already done this part and verified with Mathematica my results by plotting. I'll post it downwards.

Then problem five says: Using the previous exercise verify that:

$1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{ 1}{5^2}+....=\frac{\pi^2}{6}$
$1+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+\frac{ 1}{9^2}+....=\frac{\pi^2}{8}$
$1-\frac{1}{2^2}+\frac{1}{3^2}-\frac{1}{4^2}+\frac{1}{5^2}-....=\frac{\pi^2}{12}$

So, as I said, I've proved the first using c). My teacher told me that I can use the function c) and its Fourier representation for proving the three asked summations.

Its fourier representation is: $f(t)\sim{} \frac{\pi^2}{6}+\sum_{n=1}^{\infty}\frac{2}{n^2}(-1)^n\cos(nt)+\left ( \frac{2(-1)^n-2}{n^3\pi}-\frac{\pi (-1)^n}{n}\right ) \sin(nt)$

So this is what I did:

$S(\pi)=\frac{\pi^2}{2}=\frac{\pi^2}{6}+\sum_{n=1}^ {\infty}\frac{2}{n^2}(-1)^n\cos(n\pi)\rightarrow{}\frac{\pi^2}{3}=\sum_{n =1}^{\infty} \frac{2}{n^2}\rightarrow{\sum_{n=1}^{\infty}\frac{ 1}{n^2}=\frac{\pi^2}{6}}$

Everything right till there. Now I should prove this: $1-\frac{1}{2^2}+\frac{1}{3^2}-\frac{1}{4^2}+\frac{1}{5^2}-....=\frac{\pi^2}{12}$, and then using some addition I could prove c). But I couldn't do it. Thats how my teacher did it (I don't remember exactly how, which values he used, but in general terms he used something like this), there are probably another ways. I've also tried using a).

The Fourier representations for a) and b) are:

a) $f(t)\sim{} \frac{1}{2}+\sum_{n=1}^{\infty}\frac{-1}{n\pi}\sin(2n\pi t)$
b) $f(t)\sim{} \frac{1}{2}+\sum_{k=0}^{\infty}\frac{2}{(2k+1)\pi} \sin([2k+1] \pi t)$

b) is the one that I've tried to use for proving b at the beginning. So following my teachers indications I've tried to prove the third summation. This is what I've tried, using c) again:
$f(\frac{\pi}{2})=\frac{\pi^2}{4}= \frac{\pi^2}{6}+ \sum_{n=1}^{\infty} \left ( \frac{2(-1)^n-2}{n^3\pi}-\frac{\pi (-1)^n}{n}\right ) (-1)^{n+1}$

The thing is I can't handle the expression, I think I should use some algebra to work this, I've also tried to show some identity between the expression I got there and the one I'm trying to get, but didn't get too far. So I thought that maybe you could make me some suggestions.

When using a) I get some bad results, I don't know why. This is quiet apart, but I'd also like to know whats going wrong. I've tried to make with a)

$f(\frac{1}{4})= \frac{1}{4} = \frac{1}{2} - \frac{1}{\pi} \sum_{n=1}^{\infty} \frac{1}{n} \sin(n\frac{\pi}{2} ) \rightarrow{} \frac{\pi}{4} = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}$
But mathematica gives $\log [2]$, there is something that I'm doing wrong with this.

I've also tried with Parsevals relation.

$||f||^2=\displaystyle\int_{0}^{1}t^2dt=\frac{1}{3}$
$2||f||^2=\frac{2}{3}=\frac{1}{2}+ \sum_{n=1}^{\infty} \frac{1}{n^2\pi^2} \rightarrow{} \frac{\pi^2}{3}= \sum_{n=1}^{\infty}\frac{1}{n^2}$
Which again is wrong, as showed at first $\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$

And I don't know what mistakes I'm committing here.

Thats all. Bye there, and thanks.
• May 31st 2011, 02:06 PM
Ulysses
I have just proved that $\sum_{n=1}^{\infty}\frac{1}{(2n-1)^2}=\frac{\pi^2}{8}$ using the equation b, its fourier representation and the Parseval relation.

Now I'm trying to solve the third summation.
• May 31st 2011, 11:14 PM
chisigma
Quote:

Originally Posted by Ulysses
It goes as follows. Exercise 4 asks me to find the Fourier representation for:
http://www.physicsforums.com/attachm...1&d=1306644128
(sorry I link this from outside this forum, but I couldn't attach the image not today, nor even a few days ago)
I've already done this part and verified with Mathematica my results by plotting. I'll post it downwards.

Then problem five says: Using the previous exercise verify that:

$1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{ 1}{5^2}+....=\frac{\pi^2}{6}$
$1+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+\frac{ 1}{9^2}+....=\frac{\pi^2}{8}$
$1-\frac{1}{2^2}+\frac{1}{3^2}-\frac{1}{4^2}+\frac{1}{5^2}-....=\frac{\pi^2}{12}$

...

After You have found that is...

$S= \sum_{n=1}^{\infty}\frac{1}{n^{2}} = \frac{\pi^{2}}{6}$ (1)

... the second and third relation can be found in elementar way as follows...

$1+\frac{1}{3^{2}}+\frac{1}{5^{2}}+... = 1+\frac{1}{2^{2}}+ \frac{1}{3^{2}}+... -\frac{1}{2^{2}}\ ( 1+\frac{1}{2^{2}}+ \frac{1}{3^{2}}+...) \implies$

$\implies 1+\frac{1}{3^{2}}+\frac{1}{5^{2}}+... = \frac{3}{4}\ S = \frac{\pi^{2}}{8}$ (2)

$1-\frac{1}{2^{2}}+\frac{1}{3^{2}}-... = 1+\frac{1}{2^{2}}+ \frac{1}{3^{2}}+... -\frac{1}{2}\ ( 1+\frac{1}{2^{2}}+ \frac{1}{3^{2}}+...) \implies$

$\implies 1-\frac{1}{2^{2}} +\frac{1}{3^{2}}-... = \frac{1}{2}\ S = \frac{\pi^{2}}{12}$ (3)

Kind regards

$\chi$ $\sigma$
• Jun 1st 2011, 04:46 AM
Ulysses
Thank you. That was easy, I didn't realize it was so easy.

One last question. Why do I get this result when trying this? what am I doing wrong?

From equation a)
$f(\frac{1}{4})= \frac{1}{4} = \frac{1}{2} - \frac{1}{\pi} \sum_{n=1}^{\infty} \frac{1}{n} \sin(n\frac{\pi}{2} ) \rightarrow{} \frac{\pi}{4} = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}$
This is wrong, as I said before, mathematica gives log(2)

Using Perseval relation:
$||f||^2=\displaystyle\int_{0}^{1}t^2dt=\frac{1}{3}$
$2||f||^2=\frac{2}{3}=\frac{1}{2}+ \sum_{n=1}^{\infty} \frac{1}{n^2\pi^2} \rightarrow{} \frac{\pi^2}{3}= \sum_{n=1}^{\infty}\frac{1}{n^2}$

Edit:One more thing:
$\frac{1}{2}S=\sum_{n=1}^{\infty}\frac{1}{2n^2}$
Well, the terms on this summation doesn't look like the ones in the problem, how can I fix that? I know they are equal, but I want them to look equal using some identity. The biggest problem is that the terms in the exercise are alternating, while the sum obtained are not.