Finding a general form of the inverse of a function

Question

I have obtained the asymptotic expansion of some function at infinity and the leading orders are of the following form $$\rho (r)=\frac{r^q}{2 q}+\frac{3 (q-1) r^{2 q-1}}{8 q (2 q-1)}+\frac{3 r^{2 q-1}}{8 q (2 q-1)}-\frac{\sqrt{\pi } \Gamma \left(\frac{q}{q-1}\right)}{\Gamma \left(\frac{1}{2}+\frac{1}{q-1}\right)}+r $$ and I want to obtain its inverse, i.e. $r(\rho)$. What technique can I use to explicitly write the inverse for a general $-1\leq q <1$? Thanks

Answer 1

First you have to make sure that there exists an inverse for $-1\leq q<1$, which you will have to find out (hint: a strictly monotonous polynomial in a given range is invertible in this range). Supposing that an inverse exists and baring in mind that this function is rather complicated, I would suggest that you use some computer software to find the inverse, like Mathematica.

You can do it by solving the equation $\rho=\rho(r)$ and obtain the solution(s) $r=r(\rho)$.