Correction terms in the asymptotic expansion of hypergeometric function

I am interested in obtaining the asymptotic expansion of $$r(\rho)$$ $$($$which is the inverse of $$\rho$$ below$$)$$,

$$\rho=\frac{2b}{1-q}\left(1-\left(\frac br\right)^{1-q}\right)^{1/2}\left(_2F_1\left(\frac{1}{2},1-\frac{1}{q-1},\frac{3}{2},1-\left(\frac br\right)^{1-q}\right)\right)$$

where $$b$$ is just some positive constant while $$-\infty.

Basically I want to series expand $$\rho$$ for large $$r$$ $$($$i.e. as $$r\to \infty)$$ and then invert the series to obtain $$r(\rho)$$. I have tried some readily available asymptotic expansion of $$_2F_1$$.

Basically I got,

$$\rho\sim r\sqrt{1-(b/r)^{1-q}}$$

as the leading term in the expansion. I am not that sure what to pick as the next two correction terms from the link given the allowed values $$-\infty.Is there a detailed easy way of finding the next two correction terms in the expansion, given the allowed values of $$q$$? Thank you.

Since for all $$\alpha\in\mathbb{R}$$ and $$z\in(0,1)$$, by Euler integral representation,
\begin{align*} f(\alpha,z)&:=(1-z)^{1/2} {_2F_1}\left(\frac{1}{2},1+\alpha;\frac{3}{2};1-z\right)\\ &=\frac{(1-z)^{1/2}}{2}\int_{0}^1x^{-1/2}(1-(1-z)x)^{-1-\alpha}\,dx\\ &=\frac{1}{2}\int_{z}^{ 1}\frac{v^{-1-\alpha}}{\sqrt{1-v}}\,dv=\frac{1}{2}\sum_{k\ge 0}\frac{(1/2)_k}{k!}\int_{z}^1\frac{\,dv}{v^{1-k+\alpha}}\\ &=\sum_{\substack{k\ge 0\\ k\neq \alpha}}\frac{(1/2)_k}{2k!(k-\alpha)}\left(1-z^{k-\alpha}\right)-{\bf 1}_{\alpha\in\mathbb{N}}\frac{(1/2)_{\alpha}}{2\alpha!}\log z. \end{align*} Namely, for all $$z\in(0,1)$$, $$f(\alpha,z)=C_{\alpha}-{\bf 1}_{\alpha\in\mathbb{N}}\frac{(1/2)_{\alpha}}{2\alpha!}\log z-\sum_{\substack{k\ge 0\\ k\neq \alpha}}\frac{(1/2)_k}{2k!(k-\alpha)}z^{k-\alpha}$$ with $$C_{\alpha}=\sum_{\substack{k\ge 0\\ k\neq \alpha}}\frac{(1/2)_k}{2k!(k-\alpha)}$$; it is a convergence series expansion. In particular, if $$\alpha\in(0,1)$$, \begin{align} f(\alpha,z)&=\sum_{k\ge 0}\frac{(1/2)_k}{2k!(k-\alpha)}-\sum_{k\ge 0}\frac{(1/2)_k}{2k!(k-\alpha)}z^{k-\alpha}\\ &=\frac{1}{2\alpha}z^{-\alpha}+\sum_{k\ge 0}\frac{(1/2)_k}{2k!(k-\alpha)}-\frac{1}{4(1-\alpha)}z^{1-\alpha}+O(z^{2-\alpha}). \end{align}
Now by set $$\alpha=1/(1-q)$$ and $$z=(b/r)^{1-q}$$ with $$b>0$$, and if $$q<0$$ then \begin{align} \rho &=\frac{2b}{1-q}\left(\frac{1-q}{2b}r+\frac{1-q}{2}\sum_{k\ge 0}\frac{(1/2)_k}{k!((1-q)k-1)}+\frac{1-q}{4q}\left(\frac{b}{r}\right)^{-q}+O\left(\left(\frac{b}{r}\right)^{1-2q}\right)\right)\\ &=r+b\sum_{k\ge 0}\frac{(1/2)_k}{k!((1-q)k-1)}+\frac{b^{1-q}}{2q}r^q+O(r^{2q-1}). \end{align}
• Hi @Zhou, I am not sure how to implement your answer. How do I incorporate the nature of the values of $q$? Can you show me the first three terms in the expansion when $q<0$? – user583893 Jan 27 at 4:19
• Right, my problem now is the inversion of $\rho$ to obtain $r$. I don't think Lagrange inversion method is convenient here. Can you suggest a way @Zhou? – user583893 Jan 27 at 6:08