Let $f(x)=x\sin(x)$, $x\in[-\pi,\pi$
Find the Fourier series of $g(x)=f'(x)=\sin(x)+x\cos(x)$, $x\in(-\pi,\pi)$ and show that it converges pointwise to $g$.
Try
I found the FS of the function $f(x)=x\sin(x)$, $x\in[-\pi,\pi)$ (in the just previous problem) to $1-\frac{1}{4}\cos(x)+\sum_{n=2}^{\infty} \left (\frac{2(-1)^{n+1}}{n^2-1} \right) \cos(nx)$ and shown that it converges uniformly to $f$. (or written as $$1-\frac{1}{2}\cos(x)-2\sum_{k=2}^{\infty}\frac{(-1)^k}{(k^2-1)}\cos(kx)$$
Is there a trick to finding the Fourier of the derivative of my function given that I have the Fourier for my function? How do I show pointwise convergence to $g$?
*I know about convolutions as I suspect I need that in this problem. I think pointwise convergence is true if it is continuous and periodic.
Let $f(x) = x\sin(x)$ for $ x\in [-\pi,\pi)$, and similarly, denote by $f$ its periodisation. Recall that, since $f$ is differentiable over $(-\pi,\pi)$ and is $2\pi$ periodic function, we can write it as a Fourier series $$ f(x) = \sum_{n=-\infty}^{\infty} c_n(f) e^{-inx} $$ with Fourier coefficients defined as $$ c_n(f) := \frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)e^{inx} dx, \quad n \in \mathbb{Z}. $$ Then, approaching similarly, we can define the Fourier coefficients of the derivative of $f$ as $$ c_n(f') = \frac{1}{2\pi}\int_{-\pi}^{\pi} f'(x) e^{inx} dx \quad n \in \mathbb{Z}. $$ with a corresponding Fourier series.
Relating Fourier Coefficients
Applying integration by parts, and assuming that $f$ is finite at the boundary, we find that $$ \begin{align} c_n(f') &= \frac{1}{2\pi}\int_{-\pi}^{\pi} f'(x) e^{inx} dx \\ &= \frac{1}{2\pi}\left[f(x)e^{inx}\Big\vert_{-\pi}^{\pi} - in\int_{-\pi}^{\pi}f(x)e^{inx} dx\right] \\ \text{$n$ an integer so $e^{in\pi} = (-1)^n$, giving}\\ &= \frac{1}{2\pi}\left[\left(f(\pi^-)e^{in\pi} - f(-\pi^+)e^{-in\pi}\right) - in\int_{-\pi}^{\pi}f(x)e^{inx} dx\right] \\ &= \frac{1}{2\pi}\left[(-1)^n\left(f(\pi^-) - f(-\pi^+)\right) - in\int_{-\pi}^{\pi}f(x)e^{inx} dx\right]\\ \text{if $f$ periodic, then}\\ &=\frac{-in}{2\pi}\int_{-\pi}^{\pi}f(x)e^{inx} dx\\ &= -inc_n(f). \end{align} $$ This gives the nice relation between the Fourier coefficients between a function and its derivative, that being $$ c_n(f') = -in c_n(f), \quad $$ Note: This holds with the current definition of the Fourier coefficient, and may different up to a constant using other definitions.
Using the above you can proceed to find a solution to your question. We can start by first computing the Fourier coefficients of the periodisation of your function $f(x) = x\sin(x)$ when $x\in (-\pi,\pi)$. Indeed, we find that $$ c_n(f) = \begin{cases} 1 & \text{if } n = 0\\ -\frac{1}{4} & \text{if } n = \pm 1\\ -\frac{(-1)^{n}}{(n^2 - 1)} & \text{otherwise}. \end{cases} $$ Using the above work, we therefore find that $$ c_n(g) = \begin{cases} 0 & \text{if } n = 0\\ \frac{i}{4} & \text{if } n = 1\\ -\frac{i}{4} & \text{if } n = -1\\ \frac{in(-1)^{n}}{(n^2 - 1)} & \text{otherwise}. \end{cases} $$ Hence the Fourier series for $g(x)$ would be
$$ \begin{align} g(x) &= \sum_{n=-\infty}^{\infty}c_n(g) e^{-inx} \\ &= \frac{i}{4}(e^{-ix} - e^{ix}) + \sum_{n=2}^{\infty}\frac{in(-1)^{n}}{(n^2-1)}e^{-inx} + \sum_{n=2}^{\infty}\frac{i(-n)(-1)^{-n}}{((-n)^2-1)}e^{inx} \\ &= \frac{i}{4}(e^{-ix} - e^{ix}) + \sum_{n=2}^{\infty}\frac{in(-1)^{n}}{(n^2-1)}(e^{-inx} - e^{inx}) \\ \text{using standard identities}\\ &= \frac{\sin(x)}{2} + 2\sum_{n=2}^{\infty}\frac{n(-1)^{n}}{(n^2-1)}\sin(nx), \end{align} $$ as required.
Note: You should keep in mind that different Fourier coefficient definitions also correspond to different Fourier series definitions.