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A function $f$ has derivative cycle of length $n$ if it satisfies $f^n(x) = f(x)$ , $f^n$ being the $n^{th}$ derivative.

There are solutions for $n = 1, n=2 $ and $n=4$ which are $f(x) = e^x$, $f(x) = e^{-x}$ and $f(x)=sin(x)$ respectively. But I can't seem to find one for $n=3$.

Does anyone know a solution, or a proof there isn't one?
One may also ask for which $n$ is there a solution?

Thank you for your help!

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Let $\omega = \sqrt[3]{1} = \frac{-1+i\sqrt{3}}{2}$ be the cube root of unity, then $f(x)=e^{\omega x}$ will have the property that it equals its third derivative. If you require the function to be real add its conjugate. Thus we have
\begin{eqnarray*} f(x) = e^{-x/2} \cos \left( \frac{x\sqrt{3} }{2} \right). \end{eqnarray*} A similar construction will work for any value of $n$.

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