TITLE: Why is it not possible to show that $\frac{d}{dx}e^x=e^x$ by differentiating the Fourier series
QUESTION [6 upvotes]: The Fourier series of $ f (x) = e ^ x $ is given by
$$f(x)= \sinh(1)+2\sinh(1) \sum_{n=1}^{\infty}\frac{(-1)^n}{1+\pi^2n^2}[\cos(n\pi x)-n\pi\sin(n\pi x)]$$
By taking the derivative I get
$$f'(x)=2\sinh(1) \sum_{n=1}^{\infty}\frac{(-1)^n}{1+\pi^2n^2}[-n\pi\sin(n\pi x)-n^2\pi^2\cos(n\pi x)]$$
But I can't get an argument as to why it can't be shown that the derivative of $e ^ x$ is $e ^ x$ when I derive the expanded Fourier series.
Can you help me with an argument?

REPLY [4 votes]: The theorem on differentiating Fourier series (on $[-1,1]$ in this case) says that we may differentiate term-by-term if the original function is (piecewise continuously) differentiable on $(-1,1)$ and continuous at the endpoints $-1$ and $1$. This second hypothesis does not hold for the periodic extension of $e^x$.
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