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 $-\infty<q<1$.

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 $-\infty<q<1$. Is there a detailed easy way of finding the next two correction terms in the expansion, given the allowed values of $q$? Thank you.


1 Answer 1


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}

  • $\begingroup$ 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 $q<0$? $\endgroup$
    – user583893
    Jan 27, 2019 at 4:19
  • $\begingroup$ @user583893, see the update. $\endgroup$
    – Zhou
    Jan 27, 2019 at 4:58
  • $\begingroup$ 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? $\endgroup$
    – user583893
    Jan 27, 2019 at 6:08

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