Find the Fourier Series for f(x)=x^2 evaluate f(0) and show that the summation

[Yamahonda450]

Find the Fourier Series for f(x)=x^2 evaluate f(0) and show that the summation
$$\sum^{\infty}_{n=1}\frac{1}{(2n-1)^2}=\frac{\pi^2}{8}$$

The first part of this problem asked that I find $$a_{n}$$ and $$b_{n}$$

Since x^2 is an even function b_n=0 and for a_n I got $$\frac{4(-1)^2}{n^2}$$

for a_0 I got $$\frac{\pi^2}{3}$$

I was able to do a related problem that asked to evaluate f(pi) and show that...
$$\sum^{\infty}_{n=1}\frac{1}{(n)^2}=\frac{\pi^2}{6}$$

I was able to show this one easily but the other one...$$\sum^{\infty}_{n=1}\frac{1}{(2n-1)^2}=\frac{\pi^2}{8}$$

Is a lot more challenging. I attached my work for the related problem if someone could please show me were to start with this one I'd greatly appreciate it.

 

[micromass]

Hi Yamahonda450!

What happens if you do:

$$\sum_{n=1}^{+\infty}{\frac{1}{n^2}}-\sum_{n=1}^{+\infty}{\frac{(-1)^n}{n^2}}$$

 

[Yamahonda450]

$$\sum^{\infty}_{n=1} \frac{1}{(n)^2}$$-$$\sum^{\infty}_{n=1} \frac{(-1)^n}{(n)^2}$$

Is this what you are saying? For some reason it's showing this on 3 lines. It should be the summation minus the second summation.

 

[Yamahonda450]

Ok, now your post is displaying properly. What does the +infinity mean? Is that similar to limits, as in from the right?

 

[micromass]

Yes, that's what I mean to say. Try to work that thing out.

 

[micromass]

Ok, now your post is displaying properly. What does the +infinity mean? Is that similar to limits, as in from the right?

No, sorry for the confusion. That's just me writing things in a weird way. You can just read it as

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

 

[Yamahonda450]

I know the first summation is equal to pi^2/6 so I just replaced the summation with it.

$$\frac{\pi^2}{6}$$-$$\sum^{\infty}_{n=1} \frac{(-1)^n}{(n)^2}$$

Are you saying these two summations are equivalent to...

$$\sum^{\infty}_{n=1} \frac{1}{(2n-1)^2}$$

 

[micromass]

I know the first summation is equal to pi^2/6 so I just replaced the summation with it.

$$\frac{\pi^2}{6}$$-$$\sum^{\infty}_{n=1} \frac{(-1)^n}{(n)^2}$$

You know the second summation too, you worked it out in your attachments.

Are you saying these two summations are equivalent to...

$$\sum^{\infty}_{n=1} \frac{1}{(2n-1)^2}$$

They're almost equal, yes.

 

[Yamahonda450]

I think I have it. If you look at the attached where I say, If it's true that...

Are those two summations equivalent?

 

[micromass]

Looks good. Congrats!

 

[Yamahonda450]

Thanks for the help. I'm still not sure if those summations are equivalent. The one in the attached that I pointed out but it works.

 

[micromass]

Well, just add them:

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

If n is odd, then (-1)ⁿ=-1, and thus

$$\frac{1-(-1)^n}{n^2}=\frac{2}{n^2}$$

if it's even, then we get

$$\frac{1-(-1)^n}{n^2}=0$$

Thus we get

$$\sum_\text{n is odd}{\frac{2n}{n^2}}$$

But odd terms have the form 2k-1, hence we obtain

$$\sum_{k=1}^{\infty}{\frac{1}{(2k-1)^2}}$$

 

[Yamahonda450]

You said add them but then you subtracted instead. Did you mean to write...

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

With the subtracted summation

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

Which would look like...

$$\sum_{n=1}^{\infty}{\frac{1}{n^2}}+\sum_{n=1}^{\infty}{\frac{(-1)^n}{n^2}}=\sum_{n=1}^{\infty}{\frac{1-(-1)^n}{n^2}}$$

Using the equality from (1), the above then becomes...

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

 

[micromass]

Sorry, I made some typo's, they're corrected now...

 

[Yamahonda450]

That makes more sense, but I'm not yet seeing how that relates to my question. I wanted to make sure the following is true...

$$\sum_{n=1}^{\infty}{\frac{(-1)^n}{n^2}}=\sum_{n=1}^{\infty}\left({\frac{1}{2n^2}-\frac{1}{(2n-1)^2}}\right)$$

The summation on the right side is what allows me to reach the solution to this problem as seen in my last attachment. I'm just not sure if the above is true.

 

[micromass]

Ah yes, I see, sorry.

Hmm, it is certainly true, but showing that it is true is not evident.
Basically, what you do is applying associativity and commutativity, that is, you claim that

$$-1+\frac{1}{4}-\frac{1}{9}+\frac{1}{16}+...=(\frac{1}{4}-1)+(\frac{1}{16}-\frac{1}{9})+...$$

Doing this is not always allowed, for example

$$(-1+1)+(-1+1)+(-1+1)+...\neq -1+(1-1)+(1-1)+(1-1)+...$$

But now it is because the series $\sum_{n=1}^\infty{\frac{1}{n^2}}$ converges. Thus the series $\sum_{n=1}^\infty{\frac{(-1)^n}{n^2}}$ converges absolutely. And you can do with absolute convergent series what you like.
For a formal proof on why you can do such things with absolute convergent series, I'll have to direct you to a real analysis textbook...

 

[Yamahonda450]

I don't need to show the proof, I just wanted to make sure. Thank you for your help