The solution to an exercise I've done approximates $ f(x) = x(1-x)$ as a Fourier series, but does not mention how I can prove that $f(x)$ is indeed equal to the solution series.
What I've done is :
$$g(n) = \int_0^1 f(x)e^{-2{\pi}inx}dx = \int_0^1 (x-x^2)e^{-2{\pi}inx}dx = \int_0^1 (x-x^2)\frac d{dx}\Big(\frac {e^{-2{\pi}inx}}{-2{\pi}in} \Big)dx $$
Integrating by parts two times, I get that the fourier coefficients are $c_n =\frac {-1}{2\pi^2n^2}$, from there (and we can easily calculate that $c_0 = 1/6$) I am able to conclude that
$$f(x) = \frac{1}{6} - \sum_1^\infty \frac{\cos(2\pi nx)}{\pi^2n^2} $$.
The problem is, while doing this, we have never proved that $f(x)$ indeed does converge to the fourier series that we have calculated. What is the argument I can make to conclude that $f(x) = x(1-x)$ does indeed converge to that series? Thank you
It depends on what kind of convergence do you speak.
The usual criterion is the following
Of course in your case, it is clear that $x(1-x)$ is square integrable on $[a,b]$ so you have the $L^2$-convergence. It is important to note that you don't have a priori the ponctual convergence. Actually in your case, since your function is continuous, you can use the Dirichlet's theorem (https://en.wikipedia.org/wiki/Dirichlet_conditions) to know that your convergence is also ponctual.