Find the complex Fourier series of the $2\pi$ periodic function $f(x)=x^2$ and determine the function which the Fourier series converges to.

Hey all. I've calculated the coefficients and found that $x^2 \sim \frac{\pi^2}{3}+\displaystyle\sum_{n=-\infty}^\infty \frac{2(-1)^n}{n^2}e^{inx} $.

Now I am having issues with the notion of the convergence of Fourier series. According to Dirichlet, (or at least according to theorem we've been taught in our course) if $f$ is piecewise continuously differentiable on $[-\pi,\pi]$ then $\forall x\in[-\pi,\pi]\ \displaystyle\lim_{n\to\infty} S_nf(x)=\frac {f(x^-)+f(x^+)}{2} $. Can I conclude by this theorem, since $f(x)=x^2$ is piecewise continuously differentiable on the periodic extension of $f$, and in particular continuous on $\mathbb{R}$, that its Fourier series converges to $f$? (is this correct?)

Thanks in advance :)


Yes, I think you can deduce that. BTW, observe that

$$x=\pi\implies \pi^2=\frac{\pi^2}3+\sum_{n=-\infty}^\infty\frac{2(-1)^n}{n^2}(-1)^n=\frac{\pi^2}3+4\sum_{n=1}^\infty\frac1{n^2}\implies\sum_{n=1}^\infty\frac1{n^2}=\frac{\pi^2}6\ldots$$

  • 1
    $\begingroup$ Thank you Antonio! this is a very nice and elegant proof of the Basel problem $\endgroup$ – Noy Perel Apr 2 '18 at 22:42

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