\alpha^{n+\beta-k} \Gamma(n+\beta-k) [x^k] ({\mathcal{A}}^\alpha_\beta f)(x) + {\mathcal{O}}\left(\alpha^{n}\Gamma(n+\beta-R)\right) && \forall R\in {\mathbb{N}}_0. \end{aligned}$$

${\mathcal{A}}^\alpha_\beta$ is linear.

In the realm of resurgence such an operator is called *alien derivative* or *alien operator* [@mitschi2016divergent II.6].

The power series $f\in {\mathbb{R}}[[x]]$ associated to the sequence $f_n = n!$ clearly fulfills the requirements of Definition \[def:Fpowerseries\] with $\alpha=1$ and $\beta=1$. Therefore, $f \in {{\mathbb{R}}[[x]]^{1}_{1}}$ and $({\mathcal{A}}^1_1 f)(x) = 1$.

The asymptotic expansion in eq.  is normalized such that shifts in $n$ can be absorbed by shifts in $\beta$. More specifically,

\[prop:betashift\]

-   If $f\in {{\mathbb{R}}[[x]]^{\alpha}_{\beta}}$ and the first $m$ coefficients of $f(x)$ vanish, then $\frac{f(x)}{x^m} \in {{\mathbb{R}}[[x]]^{\alpha}_{\beta+m}}$ and $$\begin{aligned}
     ({\mathcal{A}}^\alpha_\beta f)(x) = \left( {\mathcal{A}}^\alpha_{\beta+m} \frac{f(x)}{x^m}\right)(x). \end{aligned}$$

-   If $f\in {{\mathbb{R}}[[x]]^{\alpha}_{\beta}}$ and $m \in {\mathbb{N}}_0$ with $\beta > m$, then $x^m f(x) \in {{\mathbb{R}}[[x]]^{\alpha}_{\beta-m}}$ and $$\begin{aligned}
     ({\mathcal{A}}^\alpha_\beta f)(x) = \left({\mathcal{A}}^\alpha_{\beta-m} x^m f(x)\right)(x). \end{aligned}$$

Suppose the first $m$ coefficients of $f$ vanish. Set $g(x) = \frac{f(x)}{x^m}$ or $g_n = f_{n+m}$. Eq.  gives, $$\begin{aligned}
 g_n &= \sum _{k=0}^{R-1} \alpha^{n+m+\beta-k} \Gamma(n+m+\beta-k) [x^k] ({\mathcal{A}}^\alpha_\beta f)(x) + {\mathcal{O}}\left(\alpha^{n+m}\Gamma(n+m+\beta-R)\right) && \forall R\in {\mathbb{N}}_0. \end{aligned}$$ Therefore $({\mathcal{A}}^\alpha_{\beta+m} g)(x) = \left({\mathcal{A}}^\alpha_{\beta+m} \frac{f(x)}{x^m}\right)(x) = ({\mathcal{A}}^\alpha_\beta f)(x)$. The second statement follows analogously.

Note that this gives an ascending chain of subspaces of ${\mathbb{R}}[[x]]$: $$\begin{aligned}
 {{\mathbb{R}}[[x]]^{\alpha}_{\beta}} \subset 