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The simplest example of this is $e^x$ which we could say has period 1 (it is its own derivative). $e^{-x}$ would have period 2.

Using similar constructions, I can get a function that has a derivative of period $n$ by doing $e^{x\cdot (1)^{1/n}}$

Is this the only way to get periodic derivatives?

Note: I am treating sin and cos as special cases of this when $n=4$

Is there a proof to this effect?

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    The space of functions satisfying $f^{(n)}=f$ has dimension $n$ (dimensions of solution spaces is an important consideration in differential equations), and there are $n$ distinct $n$th roots of unity.2012-09-09
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    There aren't any.2012-09-09
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    @AndréNicolas why not though?2012-09-09
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    The answer is presumably being typed by anon. If not, recall how to solve linear homogeneous DE with constant coefficients.2012-09-09
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    Expanding a little on "why not" : If we consider the differential equation $y^{(n)} = y$, we can use the characteristic method to solve this. Namely, make the ansatz $y = e^{\lambda x}$. Then $e^{\lambda x}(\lambda^n -1) = 0$. The zeros are the $n$th roots of unity.2012-09-09
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    The other thing to note is that a solution to a differential equation of order $n$ is determined by initial conditions on $y, y', \ldots, y^{(n-1)}$, and therefore the space of solutions has dimension $n$. Thus once you have solutions of the form $e^{\lambda x}$ for $n$ different $\lambda$ (and these are easily seen to be linearly independent), all solutions are linear combinations of these.2012-09-09

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Assume $f^{(n)}(x)=f(x)$ for all $x$. Let $g(x) = \sum_{k=0}^{n-1}\xi^{-k} f^{(k)}(x)$, where $\xi^n=1$. Then $g'(x)=\sum_{k=0}^{n-1}\xi^{-k} f^{(k+1)}(x) = \xi\sum_{k=1}^{n}\xi^{-k} f^{(k)}(x)=\xi g(x)$. We already know all solutions of $y'=c y$ and conclude that $g(x)=a e^{\xi x}$.

If we additionally assume that $\xi$ is primitive, we find for $0\le m$$ \sum_{k=0}^{n-1}\xi^{-mk} f^{(k)}(x)= a_m e^{\xi^m x}$$ Adding all equations leads to $$n f(x) = \sum_{m=0}^{n-1} a_m e^{\xi^m x}.$$