TL;DR
Define an analytic function $f \colon \mathbb{R} \to \mathbb{R}$ by the following function series (uniformly convergent over any bounded subset $S \subset \mathbb{R}$): $$f := \frac{6}{\pi^2} \sum_{n = 1}^\infty{\frac{1}{n^2} f_n}\text{ ,}$$ where each $f_n$ is given by $$f_n(x) = \mathfrak{R}\left(\exp\left(e^{i\frac{2\pi}{n}} x\right)\right) = \exp\left(x \cos\left(\frac{2 \pi}{n}\right)\right)\cos\left(x \sin\left(\frac{2\pi}{n}\right)\right)\text{ .}$$ ($\mathfrak{R} \colon \mathbb{C} \to \mathbb{R}$ is the real part function.)
Does this function have a name? Is it expressible in more common functions, without the series?
Is it the case (as I believe) that for all $n \in \mathbb{Z}^+$ we have $f^{(n)} \neq f$? ($f^{(n)}$ is the $n$th derivative of $f$.)
Further, is there some sense in which we may say that $f^{(n)} \overset{n \to \infty}{\longrightarrow} f$?
Background and Motivation
Let $\mathbb{R}[[X]]$ denote the ring of formal power series in $X$ with real coefficients, considered as a vector space over $\mathbb{R}$.
Let $\mathrm{D} \in \mathcal{L}(\mathbb{R}[[X]])$ be the ‘derivative’ operator, such that $$\mathop{\mathrm{D}} \sum_{n=0}^\infty{a_n X^n} = \sum_{n=1}^\infty{n a_n X^{n - 1}}\text{ .}$$ Using the canonical ordered basis of $\beta = \{1, X, X^2, X^3, \dots\}$, we can informally write this as $$[\mathrm{D}]_\beta^\beta = \begin{bmatrix}0 &1 &0 &0 &\cdots\\ 0 &0 &2 &0 &\cdots\\ 0 &0 &0 &3 &\cdots\\ \vdots &\vdots &\vdots &\vdots &\ddots\end{bmatrix}\text{ .}$$
I was looking for elements $f \in \mathbb{R}[[X]]$ which generated $\mathrm{D}$-cyclic subspaces of some dimension $n$. I will denote the $T$-cyclic subspace generated by $f$ as $Z(f,T)$.
If we find an element $f_n \in \mathbb{R}[[X]]$ such that $\mathop{\mathrm{D}}^n f_n = f_n$, then we know that the subspace $\mathop{\mathrm{span}}\{f_n, \mathop{\mathrm{D}} f_n, \mathop{\mathrm{D}}^2 f_n, \dots \mathop{\mathrm{D}}^{n - 1} f_n\} < \mathbb{R}[[X]]$ is a $\mathrm{D}$-invariant subspace of dimension at most $n$.
Taking $n = 1$, we can construct a function $f_1$ such that $\mathop{\mathrm{D}} f_1 = f_1$. Unsurprisingly, we get $$f_1(X) = \lambda + \lambda X + \frac{\lambda}{2} X^2 + \frac{\lambda}{6} X^3 + \cdots = \lambda \sum_{n = 0}^\infty{\frac{X^n}{n!}} = \lambda\exp(X)\text{ .}$$ And for $\lambda \neq 0$, we have $\dim Z(\lambda\exp, \mathrm{D}) = 1$. So far so good. On this space, with $\beta = \{\exp\}$, we have $[\mathrm{D}]_\beta^\beta = \begin{bmatrix}1\end{bmatrix} = \mathbf{I}_1$.
Taking $n = 2$, we can similarly construct a function $f_2$ such that $\mathop{\mathrm{D}}^2 f_2 = f_2$: $$f_2(X) = \alpha + \beta X + \frac{\alpha}{2} X^2 + \frac{\beta}{6} X^3 + \dots\text{ ,}$$ and defining $\lambda_1 := (\alpha + \beta)/2$ and $\lambda_2 := (\alpha - \beta)/2$, we have $\lambda_1 + \lambda_2 = \alpha$ and $\lambda_1 - \lambda_2 = \beta$, so, $$\begin{aligned}f_2(X) &= (\lambda_1 + \lambda_2) + (\lambda_1 - \lambda_2) X + \frac{\lambda_1 + \lambda_2}{2} X^2 + \frac{\lambda_1 - \lambda_2}{6} X^3 + \cdots\\ &= \lambda_1 \sum_{n = 0}^\infty{\frac{X^n}{n!}} + \lambda_2 \sum_{n = 0}^\infty{\frac{(-X)^n}{n!}} = \lambda_1 \exp(X) + \lambda_2 \exp(-X)\text{.}\end{aligned}$$ And indeed, for nonzero $\lambda_1, \lambda_2$, we have $\dim Z(f_2, \mathrm{D}) = 2$. On this space, with basis $\beta = \{\exp, x \mapsto \exp(-x)\}$, we have
$$[\mathrm{D}]_\beta^\beta = \begin{bmatrix}1 &0\\ 0 &-1\end{bmatrix}\text{ ,}$$
and so ${[\mathrm{D}]_\beta^\beta}^2 = \mathbf{I}_2$.
Noticing the pattern, we can hypothesize that $f_4(X) = \lambda_1 \exp(X) + \lambda_2 \cos(X) + \lambda_3 \exp(-X) + \lambda_4 \sin(X)$, and indeed it is so. For interest's sake, with basis $\beta = \{\exp, \cos, x \mapsto \exp(-x), \sin\}$, we have, on this space
$$[\mathrm{D}]_\beta^\beta = \begin{bmatrix}1 &0 &0 &0\\ 0 &0 &0 &1\\ 0 &0 &-1 &0\\ 0 &-1 &0 &0\end{bmatrix}\text{ ,}$$
and so, as we've come to expect, ${[\mathrm{D}]_\beta^\beta}^4 = \mathbf{I}_4$.
It turns out that the functions we need to generate these $\mathrm{D}$-cyclic spaces are so-called ‘cyclodifferential’ functions. Each is of the form
$$f_m(x) = \sum_{n = 0}^{m-1}\mathfrak{R}\left(\exp\left(z_{n,m} x\right)\right)\text{ ,}$$
where $z_{n,m} \in \mathbb{C}$ is the $n$th $m$th root of unity: $z_{n,m} = e^{\frac{n}{m}2 \pi i}$.
It seems clear that the sum of two cyclodifferential functions of orders $m$ and $n$ is another cyclodifferential function of order less than or equal to $mn/\gcd(m,n)$.
I was wondering what an ‘infinitieth-order’ cyclodifferential function might be. My first attempt to construct one was the limit of the sum $$f = \lim_{m \to \infty} \sum_{n=1}^m{\frac{1}{m}f_n}\text{ ,}$$ where $f_n(x) = \mathfrak{R}\left(\exp\left(e^{i\frac{2\pi}{n}} x\right)\right)$, but this quite obviously limits to $\exp$.
My second attempt is the function series $$f := \frac{6}{\pi^2} \sum_{n = 1}^\infty{\frac{1}{n^2} f_n}\text{ .}$$
The coefficient of $6/\pi^2$ is so that $f(0) = 1$.
My Attempt and the Question
By the Weierstraß $M$-test, this function series converges uniformly over any bounded subset $S \subset \mathbb{R}$, as for $M_n = \frac{\sup_{x \in S}(e^{\lvert x\rvert})}{n^2}$, we have $\lVert f_n \rVert_\infty \le M_n$ and $$\sum_{n = 1}^\infty{M_n} = \frac{\pi^2}{6}\sup_{x \in S}(e^{\lvert x\rvert})\text{ .}$$
As this series converges uniformly on any bounded subset of $\mathbb{R}$, it converges locally-uniformly on $\mathbb{R}$, and thus converges pointwise and so is well-defined.
For uniformly-convergent sequences, one may exchange the order of summation and differentiation, so $${\mathop{\mathrm{D}}}^m f = {\mathop{\mathrm{D}}}^m \frac{6}{\pi^2} \sum_{n = 1}^\infty{\frac{1}{n^2} f_n} = \frac{6}{\pi^2} \sum_{n = 1}^\infty{\frac{1}{n^2} {\mathop{\mathrm{D}}}^m f_n}\text{ ,}$$ notably, for factorial-order derivatives, as $n \mid m \implies \mathop{\mathrm{D}}^m f_n = f_n$, and for all positive integers $n < m$ we have $n \mid m!$, so $${\mathop{\mathrm{D}}}^{m!} f = \frac{6}{\pi^2} \sum_{n = 1}^\infty{\frac{1}{n^2} {\mathop{\mathrm{D}}}^{m!} f_n} = \frac{6}{\pi^2} \left(\sum_{n = 1}^m{\frac{1}{n^2} {f_n}} + \sum_{n = m + 1}^\infty{\frac{1}{n^2} {\mathop{\mathrm{D}}}^{m!} f_n}\right) \text{ .}$$
From this, it seems to me that $\lim_{n \to \infty} {\mathop{\mathrm{D}}}^{n!} f = f$. Does this hold pointwise? Locally-uniformly? Does this mean that we can say that $f^{(n)} \overset{n \to \infty}{\longrightarrow} f$ in some sense?
Further, to me it seems that intuitively for any $n$ we have $\mathop{\mathrm{D}}^n f$ differing from $f$ in the $f_{\nu(n)}$ term, where $\nu(n)$ is the least nondivisor of $n$ – the function $\nu \colon \mathbb{Z}^+ \to \mathbb{Z}^+$ being defined by $\nu(n) = \min\big\{\,m \in \mathbb{Z}^+ \;\big|\; m \not \mid n\,\big\}$ – and that should not be everywhere-cancellable by later terms, due to the reciprocal square coefficient. Thus, I hypothesize that for any positive integer $n$, ${\mathop{\mathrm{D}}}^n f \neq f$, that is to say that there exists some $x \in \mathbb{R}$ such that $({\mathop{\mathrm{D}}}^n f)(x) \neq f(x)$. Does this hold?
I did not make any progress on an expression of this function in terms of more common functions, without a series.
