Does Fourier series of x^2 converge?

gauss mouse
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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.
 
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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.
 
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!
 
gauss mouse said:
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.
 
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.
 
Oh ok, I did not know that result. Nice!
 
Yeah it's pretty sweet. It's not too restrictive.
 
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