Proving the Sum of a Fourier Series Using Parseval's Relation

[Telemachus]

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.

 

[Char. Limit]

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.

Right there. You can't square a sum that easily, because you would need to use an infinite binomial expansion. It's a sum, after all.

 

[MisterX]

Maybe this is too much of a hint:
The problem seems similar to the Basel problem, which can be solved using Fourier series.

 

[vela]

That's not the series you want to use. The sine will keep flipping sign, so you'll get an alternating series. Also, you want (2n-1)² in the denominator. You only have the first power.

EDIT: Oh, I see what you tried. That doesn't work because

\sum_{n=1}^\infty a_n^2 \ne \left(\sum_{n=1}^\infty a_n\right)^2

 

[Telemachus]

Thanks. There are two more function that I think I can use, from the previous exercise. Actually it asks me to use the Fourier series found in problem 4), which have three different functions. The one I've tried to use was one, the other is the one which you helped me to solve before.

I've used c) in the previous problem for solving \pi^2/6, I think I've used more elementary elements there (I don't have my notebook right now).
I think I should use a). I'll do it tomorrow.

 

[Telemachus]

Well, the problem 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.

For the first, you can see the function https://www.physicsforums.com/showthread.php?t=502460"

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 what's 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.

 

[vela]

I suppose you might be able to use c) to do all of the sums, but it seems kind of complicated for at least one of them.

Your instinct to use b) is good because that series has only the odd terms. The problem is to get (2n+1)² on the bottom. The trick is to integrate the series (and f(t) as well).

 

[Telemachus]

I get it using b and parseval relation, it was quiet easy. That for proving the second summation, now I have to prove the one that gives \frac{\pi^2}{12}.

Thanks vela :)