For any infinitely differentiable function $f: \mathbb{R}\to \mathbb{R}$ and positive integer $k\in\mathbb{N}$, let $f^{(k)}$ denote the $k$-th derivative of $f$.


For which $n\in\mathbb{N}$, $n>1$, is there a periodical function $f:\mathbb{R}\to \mathbb{R}$ with the property that $f^{(n)} = f$, but $f^{(k)} \neq f$ for $k\in \{1,\ldots,n-1\}$?



Periodical functions with $f^{(n)} = f$, but $f^{(k)} \neq f$ for $k\in \{1,\ldots,n-1\}$
You are asking for solutions $f$ to the simple differentiable equation $$f^{(n)}=f$$ that make $f$ real-valued for real-values of $f$ and not of lower degree $k$th-derivative-wise. As is well-known, the solution is given by $$f(x)=\sum\_{j=0}^{n-1}c\_j\exp(\zeta^jx)\,,$$ for some constants $c\_j$ and $x\in\mathbb{R}$, where $\zeta=\exp(2\pi i/n)$. Because you desire $f$ to be real-valued, this becomes equivalent to $f(x)=\overline{f(x)}$, which becomes equivalent to $c\_{n-j}=\overline{c\_j}$ for all $j$; and because you require $f$ not to be of lower degree $k$th-derivative-wise, that will mean the indices of the non-vanishing coefficients should not form a proper subgroup of $\mathbb{Z}/n\mathbb{Z}$.


