This is what my textbook says, and I see the examples that support this fact. However, I don't understand why it's true. The derivative is a linear operator. The math should work out because the Fourier Series is just a sum representation of a signal.
Why can't you take the derivative of a function's fourier series to get its derivative?
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1Just to give some intuition, recall that the Fourier series will converge for a very wide class of functions - in particular the continuous periodic functions. So if we can have a Fourier series converge to *any* continuous periodic function, even a nowhere differentiable one, even though all the terms of the infinite series are differentiable, then we shouldn't expect the series of the derivatives, when it exists, necessarily to converge to the derivative of the series, again when that exists. – 2017-02-26
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A Fourier series is not merely a sum. It is an infinite sum, which it turns out is not a sum at all, but a limit of a sequence of partial sums. In particular you can't apply a linear operator to an infinite sum unless the linear operator is continuous.
You would expect the derivative to be continuous, right? Well, it's not. The easiest example I have is this:
The sequence of functions $\frac{\sin{nx}}n$ converges to $0$. The sequence of derivatives is $\cos{nx}$. For the derivative to be continuous this would have to converge to $0$, but clearly it doesn't converge to anything.