g^{-1} \in {{\mathbb{R}}[[x]]^{\alpha}_{\beta}}$ and $$\begin{aligned}
 \label{eqn:asymp_chainrule_normal} ({\mathcal{A}}^\alpha_\beta (f \circ g)) (x) = f'(g(x)) ({\mathcal{A}}^\alpha_\beta g)(x) + \left(\frac{x}{g(x)} \right)^{\beta} e^{\frac{g(x) - x}{\alpha x g(x)}} ({\mathcal{A}}^\alpha_\beta f ) (g(x)), \\ \label{eqn:inverse_asymp} ({\mathcal{A}}^\alpha_\beta g^{-1})(x) = -{g^{-1}}'(x) \left(\frac{x}{g^{-1}(x)}\right)^\beta e^{\frac{g^{-1}(x) - x}{\alpha x g^{-1}(x)}} ({\mathcal{A}}^\alpha_\beta g ) (g^{-1}(x)). \end{aligned}$$

This theorem will be proven by ensuring that if $f,g \in {{\mathbb{R}}[[x]]^{\alpha}_{\beta}}$, then $f \circ g^{-1} \in {{\mathbb{R}}[[x]]^{\alpha}_{\beta}}$ and by constructing the asymptotic expansion of $f \circ g^{-1}$. For this, it turns out to be convenient to work in the ring ${{\mathbb{R}}[[x]]^{\alpha}_{\beta+2}}$ that contains ${{\mathbb{R}}[[x]]^{\alpha}_{\beta}}$ as a subring.

\[lmm:chainrule\_lagrange\] If $f,g$ as above and $A(x):= \frac{\frac{x}{g(x)}-1}{x}$ as well as $B(x):= f(x)g'(x)\left(\frac{g(x)}{x}\right)^{\beta}$, then $A, B \in{{\mathbb{R}}[[x]]^{\alpha}_{\beta+2}}$ and $$\begin{aligned}
 \label{eqn:diffgroupsumrep1} [x^n] f(g^{-1}(x)) &= \sum_{m=0}^n {n+\beta+1 \choose m} [x^{n-m}] B(x) A(x)^m && \forall n\in {\mathbb{N}}_0. \end{aligned}$$

The statement $A \in {{\mathbb{R}}[[x]]^{\alpha}_{\beta+2}}$ follows from $g_0 = 0$, $g_1=1$, Theorem \[thm:chain\_analytic\] and Proposition \[prop:betashift\]. The fact that $B \in {{\mathbb{R}}[[x]]^{\alpha}_{\beta+2}}$ follows additionally from the result $f(x) \in {{\mathbb{R}}[[x]]^{\alpha}_{{\beta}}} \Rightarrow f'(x) \in {{\mathbb{R}}[[x]]^{\alpha}_{{\beta+2}}}$ from Proposition \[prop:ordinary\_derivative\]. Eq.  can be derived using the Lagrange inversion theorem [@flajolet2009analytic A.6]: $$\begin{aligned}
 [x^n] f(g^{-1}(x)) &= \frac{1}{n}[x^{n-1}] f'(x) \left(\frac{x}{g(x)}\right)^{n}= [x^{n}] f(x) \left(\frac{x}{g(x)}\right)^{n-1} \left( \frac{x}{g(x)} - x \frac{\partial}{\partial x}\frac{x}{g(x)} \right) \\ &= [x^{n}] f(x)g'(x)\left(\frac{g(x)}{x}\right)^{\beta} \left(\frac{x}{g(x)}\right)^{n+\beta+1} = [x^{n}] B(x) 