g)(x) + f(x) (\alpha^{-1} - \beta x + x^2 \frac{\partial}{\partial x} ) ({\mathcal{A}}^\alpha_{\beta} g)(x) \Bigg). \end{gathered}$$ Using ${\mathcal{A}}^\alpha_{\beta+2}= x^2 {\mathcal{A}}^\alpha_{\beta}$ (Proposition \[prop:betashift\]) as well as $[x^k] f(x) \left( x \frac{\partial}{\partial x} g(x) \right)= k [x^k] f(x) g(x) - [x^k] \left( x \frac{\partial}{\partial x} f(x) \right) g(x)$ to reexpress the derivative of $({\mathcal{A}}^\alpha_{\beta} g)(x)$ gives, $$\begin{aligned}
 &= [x^k]x^2 e^{\frac{\frac{x}{g(x)}-1}{\alpha x} } \left(\frac{g(x)}{x}\right)^{\beta-k+1} \left( g'(x)({\mathcal{A}}^\alpha_{\beta} f)(x) - f'(x) ({\mathcal{A}}^\alpha_{\beta} g)(x) \right). \end{aligned}$$ The $x^2$ prefactor shows that $f \circ g^{-1}$ is actually in the subspace ${{\mathbb{R}}[[x]]^{\alpha}_{\beta}} \subset {{\mathbb{R}}[[x]]^{\alpha}_{\beta+2}}$. In accord with Proposition \[prop:betashift\]: $$\begin{aligned}
 \left[x^k\right] \left( {\mathcal{A}}^\alpha_{\beta} f \circ g^{-1} \right)(x)&= [x^k]e^{\frac{\frac{x}{g(x)}-1}{\alpha x} } \left(\frac{g(x)}{x}\right)^{\beta-k-1} \left( g'(x)({\mathcal{A}}^\alpha_{\beta} f)(x) - f'(x) ({\mathcal{A}}^\alpha_{\beta} g)(x) \right). \end{aligned}$$ As $f \circ g^{-1} \in {{\mathbb{R}}[[x]]^{\alpha}_{\beta}}$, the subset $\operatorname{Diff}_{\operatorname{id}}({\mathbb{R}},0)^\alpha_\beta$ is a subgroup of $\operatorname{Diff}_{\operatorname{id}}({\mathbb{R}},0)$. Another application of the Lagrange inversion theorem, $[x^n] p(q^{-1}(x)) =[x^{n}] p(x) q'(x) \left(\frac{x}{q(x)}\right)^{n+1}$, transforms the expression for $\left[x^k\right] \left( {\mathcal{A}}^\alpha_{\beta} f \circ g^{-1} \right)(x)$ into an explicit generating function: $$\begin{aligned}
 \label{eqn:conjugate_asymp_equation} ({\mathcal{A}}^\alpha_{\beta} f \circ g^{-1} )(x) &= e^{\frac{\frac{g^{-1}(x)}{x}-1}{\alpha g^{-1}(x)} } \left(\frac{x}{g^{-1}(x)}\right)^{\beta} \left( ({\mathcal{A}}^\alpha_{\beta} f)(g^{-1}(x)) -\frac{f'(g^{-1}(x))}{g'(g^{-1}(x))} ({\mathcal{A}}^\alpha_{\beta} g)(g^{-1}(x)) \right). \end{aligned}$$ The special case $f(x)=x$ with application of $g'(g^{-1}(x)) = \frac{1}{{g^{-1}}'(x)}$ results in eq. . Solving eq.  for $({\mathcal{A}}^\alpha_{\beta} g)(g^{-1}(x))$ and substituting the result into eq.  