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I have been trying to invert the hypergeometric function $$\rho(r)=\frac{2b}{1-q}\sqrt{1-\left(\frac br\right)^{1-q}}\,_2F_1\left(\frac{1}{2},1-\frac{1}{q-1};\frac{3}{2};1-\left(\frac br\right)^{1-q}\right)$$

Unfortunately, it cannot be inverted exactly so I tried feeding up the function with random values of $q$ using the routine below

rho[r_, b_, q_] := (2 b/(1 - q)) (1 - (b/r)^(1 - q))^(1/2) Hypergeometric2F1[1/2, 1 - 1/(q - 1), 3/2, 1 - (b/r)^(1 - q)]
Solve[rho[r_, b_, q_]==x,r]

Luckily, there are two values where this function can be inverted exactly, that is $q=-1$ and $q=1/3$. Is there an effective way or routine to find $q$ values in which the function has an exact inverse? By the way, $b$ is just some positive constant while $-\infty<q<1$.

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  • $\begingroup$ If this doesn't get answers, maybe Math.SE would have an answer? $\endgroup$ – lirtosiast Aug 11 at 8:48
  • $\begingroup$ What did you try in Mathematica? $\endgroup$ – Ulrich Neumann Aug 11 at 11:46
  • $\begingroup$ @Ulrich, I just tried the routine Solve[\rho[r_,b_,q_]==x,r] $\endgroup$ – user583893 Aug 11 at 11:51
  • $\begingroup$ @MichaelE2, noted. $\endgroup$ – user583893 Aug 11 at 12:08
  • $\begingroup$ Related: math.stackexchange.com/questions/3009945/… $\endgroup$ – Michael E2 Aug 11 at 12:40
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Not full answer.

We can find more $q$ values in range:1/3..1:

func = Hypergeometric2F1[1/2, 1 - 1/(q - 1), 3/2, 1 - (b/r)^(1 - q)] //FullSimplify
(*Hypergeometric2F1[1/2, (-2 + q)/(-1 + q), 3/2, 1 - (b/r)^(1 - q)]*)

M = 20;(*You can increase this value*)
sol = Solve[(-2 + q)/(-1 + q) == #/2, q][[1]] & /@ Table[2 x - 1, {x, 3, M}]
(* q values*)

func /. sol // FunctionExpand
(*large expression*)

rho[r_, b_, q_] := (2 b/(1 - q)) (1 - (b/r)^(1 - q))^(1/2)*func
Solve[rho[r, 1, q /. sol[[2]]] == x, r, Reals](*for b=1 ,q=3/5*)

(*{{r -> ConditionalExpression[
Root[-1073741824 - 754974720 x^2 - 212336640 x^4 - 29859840 x^6 - 
   2099520 x^8 - 
   59049 x^10 + (754974720 + 424673280 x^2 + 89579520 x^4 + 
      8398080 x^6 + 295245 x^8) #1^2 + (279183360 + 
      48660480 x^2 - 2877120 x^4 - 590490 x^6) #1^4 + (37219840 - 
      4966920 x^2 + 590490 x^4) #1^6 + (2304855 - 
      295245 x^2) #1^8 + 59049 #1^10 &, 2], x > 0]}}*)
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  • $\begingroup$ I'll check this out. Thanks $\endgroup$ – user583893 Aug 11 at 12:29
  • $\begingroup$ you did not include the terms outside the hypergeometric function. And that makes the inversion difficult still. $\endgroup$ – user583893 Aug 11 at 12:41
  • $\begingroup$ @user583893. Transcendental equation can't be solved analytically.Try numerics. $\endgroup$ – Mariusz Iwaniuk Aug 11 at 12:52
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    $\begingroup$ @user583893 Add some conditions on r and x: Simplify[ Table[Solve[ Simplify[FunctionExpand@rho[r, 1, (qq - 2)/qq], r > 1] == x && r > 1 && x > 0, r], {qq, 1, 2M, 2}], x > 0] for M solutions. $\endgroup$ – Michael E2 Aug 11 at 13:36
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    $\begingroup$ @user583893 See mathematica.stackexchange.com/questions/13767/… $\endgroup$ – Michael E2 Aug 11 at 13:48

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