0
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Is there a way of inverting this function to obtain $r(\rho)$?

rho[r_, b0_, q_] := 
  r (1 + (Sqrt[π]Gamma[1/(q - 1)])/((1 - q) Gamma[1/2 ((q + 1)/(q - 1))]) b0 /r + (1 + q)/(2 q) (b0/r)^(1 - q))

Note that $q<0$ and $b0$ is some positive constant.

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The typical way to do this, is to use Solve or Reduce get r to one side of the equality. It seems like Mathematica cannot solve the equation, unfortunately:

Reduce[
 {
  rho == r (1 + (Sqrt[\[Pi]] Gamma[1/(q - 1)])/((1 - q) Gamma[1/2 ((q + 1)/(q - 1))]) b0/r + (1 + q)/(2 q) (b0/r)^(1 - q)),
  q < 0,
  b0 > 0
 },
 r
]

During evaluation of In[2]:= Reduce::nsmet: This system cannot be solved with the methods available to Reduce.

Out[2]= Reduce[{rho == r (1 + ((1 + q) (b0/r)^(1 - q))/(2 q) + ( b0 Sqrt[[Pi]] Gamma[1/(-1 + q)])/((1 - q) r Gamma[(1 + q)/(2 (-1 + q))])), q < 0, b0 > 0}, r]

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  • $\begingroup$ What if I specify the value for $q$ and $b0$? I tried it for $q=-2$ and $b0=1$ but I don't understand what Mathematica spit out. $\endgroup$ – user583893 Feb 6 at 14:56
  • 1
    $\begingroup$ It spits out Root objects, which are symbolic representations of the exact roots of polynomials. In this case, it gives roots of 3rd-degree polynomials, so you can use the option Cubics -> True in Reduce to expand them. $\endgroup$ – Sjoerd Smit Feb 6 at 15:38
  • $\begingroup$ Both Solve and Reduce should give answers in terms of Root for integer values of q. Typically, there are no solutions otherwise. $\endgroup$ – bbgodfrey Feb 6 at 18:38

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