Does Fourier series of x^2 converge?

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SUMMARY

The Fourier series of the function f(x) = x^2, defined on the interval [-π, π], converges uniformly to x^2. The series is expressed as \(\frac{\pi^2}{3} + 4\sum_{n=1}^\infty \frac{(-1)^n}{n^2}\cos nx\). The discussion confirms that the absolute convergence of the Fourier series, as established by the Weierstrass M-test and referenced in "Fourier Analysis" by Stein and Shakarchi, guarantees uniform convergence to the original function. This conclusion is critical for understanding the behavior of Fourier series for continuous functions.

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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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