Differentiation of a Fourier Series: $f(x)=x$ for $-\pi<x<\pi$

Question

Consider the function $f(x)=x$ where $-\pi < x < \pi$ and extend it periodically to a function defined on $\mathbb{R}$.

The Fourier series of $f$ is $$ 2\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \sin(nx).$$

We are asked to differentiate [the partial sums of] both sides to find an expression for $$ \sum_{n=1}^{\infty} {(-1)^{n+1}} \cos(nx).$$

This however feels very counterintuitive, as this series doesn't converge (especially when putting $x=(2k+1)\pi$ we obtain a result that is clearly wrong.)

How should we approach this?

Answer 1

The periodic extension of $f(x)$ has discontinuities, thus, its Fourier Series converges $\textbf{pointwise}$. When a function series converges pointwise to $S(x)$, there's no guarantee that the series of the derivatives converges to $S'(x)$. You can guarantee the convergence of the series of derivatives if your original series converges $\textbf{uniformly}$.

For Fourier Series, the differentiation theorem holds true when the periodic extension of $f(x)$ is continuous in $(-\infty ; +\infty)$, and $f'(x)$ and $f''(x)$ are piecewise continuous.

So, for the excercise you were given, it's true what you've found about the divergence of the last series.