Does Fourier series of x^2 converge?

[gauss mouse]

I'm trying to show that the Fourier series of $f(x)=x^2$ converges and I can't. Does anybody know if it actually does converge? (I'm assuming that $f(x)=x^2$ for $x\in [-\pi,\pi]$).
The Fourier Series itself is $\displaystyle\frac{\pi^2}{3}+4\sum_{n=1}^\infty \frac{(-1)^n}{n^2}\cos nx$
I tried using Dirichlet's test but it wasn't working for me, though that may be because I'm doing something wrong.

 

[micromass]

Isn't it easy to show absolute convergence?? Since $\sum \frac{1}{n^2}$ converges

Do you also want to show that it converges to $x^2$?? There are many theorems out there that give you that, so it depends on what you have seen.

 

[gauss mouse]

I was just looking to see if it converged. So yes, you're absolutely right. I was going well off-beam with my attempt.

And about converging to $x^2$; I guess it does so uniformly since the function is continuous on the circle and the Fourier series converges absolutely.

Thank you!

 

[micromass]

I was just looking to see if it converged. So yes, you're absolutely right. I was going well off-beam with my attempt.

And about converging to $x^2$; I guess it does so uniformly since the function is continuous on the circle and the Fourier series converges absolutely.

Thank you!

It's not because a function is continuous and because the Fourier series converges absolutely, that you can have uniform convergence (I think).
Here, I think you can infer uniform convergence from the Weierstrass M-test.

 

[gauss mouse]

No I think it is. I quote Corollary 2.3 from "Fourier Analysis" by Stein and Shakarchi -

"Suppose that $f$ is a continuous function on the circle and that the Fourier series of $f$ is absolutely convergent, $\sum_{n=-\infty}^\infty |\hat{f}(n)|<\infty.$ Then, the Fourier series converges uniformly to $f$, that is
$\displaystyle \lim_{N\to\infty}S_N(f)(\theta)=f(\theta)$ uniformly in $\theta.$

 

[micromass]

Oh ok, I did not know that result. Nice!

 

[gauss mouse]

Yeah it's pretty sweet. It's not too restrictive.