Talk:Derivative of Periodic Function

What needs to be established here?
Our starting point is $D_x f \left({x}\right) = D_x f \left({x + L}\right)$, which is nothing more than a statement of the result we want. The rest is just manipulation into a different form.

And we have not established that it is actually true. In light of the equivalent situation for Primitives, can we indeed be so certain?

And I have just thought of a function the period of whose derivative is half its period. So I don't even believe the result itself is even true. --prime mover (talk) 01:18, 31 October 2016 (EDT)


 * As long as the set is open, and the function is differentiable at every point in the domain, then the theorem is true. --kc_kennylau (talk) 06:42, 31 October 2016 (EDT)


 * Not trivial, though, and so definitely needs to be proved. --prime mover (talk) 09:07, 31 October 2016 (EDT)

In the reverse direction, just prove that $f'(z+K) = f'(z)$ for all $z$ means $f(z+K) = f(z)$ for all $z$ via $\int_z^{z+K} f'(x) \mathrm dx = f(z+K) - f(z)$. Of course, this requires existence of $f'(x)$ for every $x$ because $z$ is arbitrary. &mdash; Lord_Farin (talk) 14:47, 31 October 2016 (EDT)
 * I think this approach only works for real valued functions since all periods are commensurable with the fundamental period (i.e. L = nK for some n and K is the fundamental period of the derivative and then the periodicity of f will force the integral of the derivative over an interval of length L to be zero which then forces the integral over length K to be zero which makes the original function periodic over K and so n=1). But for complex functions, I think you can have two periods and it isnt clear to me whether differentiation can produce a 'new' period that is linearly independent of the first.  Though in this case, that might create a pole or a singularity in the parallelogram defined by the two periods. But this involves way more complex analysis than I can remember (maybe stuff like an analytic doubly periodic function is constant combined with the derivative of an analytic function is also analytic)  --Wanfactory (talk) 16:49, 16 December 2016 (EST)