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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

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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) $r=r(\rho)$.

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  • $\begingroup$ Can't I use the Lagrange reversion theorem in this case? $\endgroup$ – user583893 Feb 26 at 10:00
  • $\begingroup$ Sure, I just suggested a software to ease your life but there are also analytic ways to find the inverse of a polynomial, as you say :) $\endgroup$ – Cyclops Feb 26 at 10:04
  • $\begingroup$ Wait, I'm confused with what you said in your answer. Does the last part say I can find the inverse by solving the equation $\rho (r)=0$? Why did you equate the function to zero? $\endgroup$ – user583893 Feb 26 at 10:14
  • $\begingroup$ You are right, there was a mistake in my answer. I edited it, so sorry for the mislead! $\endgroup$ – Cyclops Feb 26 at 10:34

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