"Bad" Fourier Series derivation

Question

Let $f(\theta)$ $2\pi$-periodic such that $f(\theta)=e^{\theta}$ for $-\pi<0<\pi$, and $$e^{\theta}=\sum_{n=-\infty}^{\infty}c_{n}e^{in\theta}\,\,\, \mathrm{for}\,\, |\theta|<\pi $$ it's Fourier series. If we formally differentiate this equation, we obtain

$$e^{\theta}=\sum_{n=-\infty}^{\infty}inc_{n}e^{in\theta}. $$

But this implies $c_{n}=inc_{n}$ or $(1-in)c_{n}=0$, so, $c_{n}=0\,\forall n\in\mathbb{Z}$. This is obviously wrong.

Where is the mistake?

The only thing that I could contest is the derivative of $f$. I know a theorem saying that a sufficient condition is $f$ continuous and piecewise smooth, and $f$ is not continous. But it don't implies that I can't derivate term by term.

Answer 1

Note that $c_n=\frac{\sinh(\pi)}{\pi}\frac{(-1)^n}{1-in}$ and the Fourier series for $e^\theta$ on $-\pi<\theta<\pi$ is

$$e^\theta =\frac{\sinh(\pi)}{\pi}\sum_{-\infty}^\infty \frac{(-1)^n}{1-in}\,e^{in\theta}$$

Clearly, the series $\sum_{n=-\infty}^\infty inc_ne^{in\theta}=\frac{\sinh(\pi)}{\pi}\sum_{-\infty}^\infty \frac{(-1)^n\,in}{1-in}\,e^{in\theta}$, which is obtained by formal differentiation under the summation, is divergent. Hence, differentiation under the summation is illegitimate and does not lead to a Fourier series representation of $e^\theta$ for $-\pi<\theta<\pi$.

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EDIT:

We may derive the derivative of the Fourier series representation of $e^{\theta}$ for $\theta\in (-\pi,\pi)$. Proceeding we have

$$\begin{align} \sum_{n=-\infty}^\infty \frac{(-1)^n}{1-in}e^{in\theta}&=1+\sum_{n\ne0}\frac{(-1)^n}{1-in}e^{in\theta}\\\\ &=1+\sum_{n\ne0}\frac{(-1)^n(in-1+1)}{(1-in)(in)}e^{in\theta}\\\\ &=1+\sum_{n\ne0}\frac{(-1)^n}{in}e^{in\theta}+\sum_{n\ne0}\frac{(-1)^n}{(1-in)(in)}e^{in\theta} \end{align}$$

We may differentiate the second series under the summation to obtain

$$\begin{align} \frac{d}{d\theta}\sum_{n=-\infty}^\infty \frac{(-1)^n}{1-in}e^{in\theta} &=\underbrace{\frac{d}{d\theta}\sum_{n\ne0}\frac{(-1)^n}{in}e^{in\theta}}_{=1}+\sum_{n\ne0}\frac{(-1)^n}{(1-in)}e^{in\theta}\\\\ &=\sum_{n=-\infty}^\infty \frac{(-1)^n}{1-in}e^{in\theta} \end{align}$$

So, we have shown that the derivative of the Fourier series $\sum_{n=-\infty}^\infty \frac{(-1)^n}{1-in}e^{in\theta}$ for $\theta\in(-\pi,\pi)$ is in fact itself. This is no surprise since $\frac{d}{d\theta}e^\theta=e^\theta$.

Answer 2

You have to think in terms of distributional derivatives, because in this sense you can switch series and derivative. Think $f$ as defined on the 1-torus. Then the distributional derivative of $f$ it is not $f$ itself, but $f-(e^\pi-e^{-\pi})\delta_{[\pi]}$, where $\delta_{[\pi]}$ is the delta Dirac in the point $[\pi]$ of the 1-torus. So, in distributional sense: $$\sum_{n=-\infty}^\infty inc_ne_n = f'=f-(e^\pi-e^{-\pi})\delta_{[\pi]}=\sum_{n=-\infty}^\infty c_ne_n - (e^\pi-e^{-\pi})\delta_{[\pi]}$$ i.e.: $$\sum_{n=-\infty}^\infty (in-1)c_ne_n=-(e^\pi-e^{-\pi})\delta_{[\pi]}=-(e^\pi-e^{-\pi})\sum_{n=-\infty}^\infty e^{in\pi}e_n \\ = -(e^\pi-e^{-\pi})\sum_{n=-\infty}^\infty \frac{(-1)^n}{2\pi}e_n.$$ Then: $$\forall n\in \mathbb{Z}, (in-1)c_n=-\frac{(-1)^{n}}{2\pi}(e^\pi-e^{-\pi})$$ i.e.: $$\forall n\in\mathbb{Z}, c_n=\frac{(-1)^{n}(e^\pi-e^{-\pi})}{2\pi(1-in)}.$$