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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 $k\in \{1,\ldots,n-1\}$?

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    $\begingroup$ Isn’t this equivalent to solving for particular cases of the differential equation $y^{(n)}=y$, so that, generically, the functions you are looking for could be determined from the general solution $$f=\sum_{j=0}^{n-1}c_k\exp(\zeta^jx)\,,$$ where $\zeta=\exp(2\pi i/n)$ for coefficients $c_k$ that make $f$ real-valued? $\endgroup$
    – Jack L.
    Commented Nov 25, 2020 at 8:15
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    $\begingroup$ Oh - that's right, thanks @JackL.! Will remove the question - or do you want to post this as an answer with a few additional remarks? $\endgroup$
    – Dominic van der Zypen
    Commented Nov 25, 2020 at 8:17
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    $\begingroup$ The only possibility are $n=4$. The functions $f:\mathbb{R}\rightarrow \mathbb{C}$ satisfying $f^{(n)}(t)=f(t)$ form a vector space with basis $(e^{\zeta t})$, where $\zeta $ runs through the $n$-th roots of 1. The function $e^{\zeta t})$ is periodic of period $2\pi i/\zeta $. Since you want the period to be real, you must have $\zeta =\pm i$, giving $n=4$, $f(t)=a\cos t+b\sin t$. $\endgroup$
    – abx
    Commented Nov 25, 2020 at 8:59
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    $\begingroup$ @abx: it appears you are/would be right if the OP is asking for periodic functions; I had taken “periodical” to be derivative-wise—that is, the derivative is periodic with period $n$—rather than $f$ itself being a periodic function. My comment above and answer below had been in this regard; perhaps the OP could clarify this as a re-edit to the question, and I will remove my answer accordingly if it’s as you have understood it. $\endgroup$
    – Jack L.
    Commented Nov 25, 2020 at 9:16

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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 $\mathbb{Z}/n\mathbb{Z}$.

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