# 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.

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 :)

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