# Obtaining One Fourier Series from Another

1. Jul 3, 2013

### FeDeX_LaTeX

1. The problem statement, all variables and given/known data
(a) On (-π,π), find the Fourier series of f(x) = x.
(b) Hence, or otherwise, find the Fourier series of g(x) = x2
(c) Hence, show that $$\sum_{n=1}^{\infty} \frac{1}{n^4} = \frac{\pi^4}{90}$$

2. Relevant equations
$$f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} \left( a_n \cos \frac{n \pi x}{L} + b_n \sin \frac{n \pi x}{L} \right)$$

where

$$a_0 = \frac{1}{L} \int_{-L}^{L} f(x) dx$$

$$a_n = \frac{1}{L} \int_{-L}^{L} f(x) \cos \left( \frac{n \pi x}{L} \right) dx$$

$$b_n = \frac{1}{L} \int_{-L}^{L} f(x) \sin \left( \frac{n \pi x}{L} \right) dx$$

3. The attempt at a solution
I found (a) as

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

but for (b), how do I obtain the Fourier series for x2, without starting from scratch? I can derive that Fourier series fine and I use Parseval's identity for the last part and the result follows -- but I'm just puzzled about the 'hence' part of question (b). Are you supposed to integrate both sides from 0 to x? Doing that doesn't seem to give me the Fourier series for x2...

2. Jul 3, 2013

Yes.

3. Jul 3, 2013

### FeDeX_LaTeX

But when I do that, I get:

$$x^2 = \sum_{n=1}^{\infty} \frac{4}{n^2}(-1)^{n} (\cos(nx) - 1)$$

and it has also not included the first term, a0. What am I doing wrong?

The actual result is supposed to be:

$$x^2 = \frac{\pi^2}{3} + \sum_{n=1}^{\infty} \frac{4}{n^2}(-1)^{n}(\cos(nx))$$

4. Jul 3, 2013

### micromass

$$\sum_{n=1}^{+\infty} \frac{4}{n^2} (-1)^n (\cos(nx) - 1) = - \sum_{n=1}^{+\infty} \frac{4(-1)^n}{n^2} + \sum_{n=1}^{+\infty} \frac{4}{n^2} (-1)^n\cos(nx)$$

So there's your $a_0$.

5. Jul 3, 2013

### FeDeX_LaTeX

Oh, I see... but is that term $a_0$ or $a_0 / 2$?

6. Jul 3, 2013

### micromass

It's $a_0/2$.

7. Jul 3, 2013

### FeDeX_LaTeX

OK, but then with Parseval's identity I don't seem to be able to get the result... what am I doing wrong below:

$$\int_{-\pi}^{\pi} x^4 dx = \pi \sum_{n=1}^{\infty} \frac{32}{n^4} + \pi \sum_{n=1}^{\infty} \frac{16}{n^4} = \pi \sum_{n=1}^{\infty} \frac{48}{n^4}$$

$$\sum_{n=1}^{\infty} \frac{1}{n^4} = \frac{\pi^4}{120}$$

?

8. Jul 3, 2013

### micromass

Please don't tell me you just did

$$\left(\sum x_n\right)^2 = \sum x^2_n$$

Cause it looks like you did.

9. Jul 3, 2013

### FeDeX_LaTeX

Oops... applied Parseval's identity in the wrong sense there.