# Inversion of a hypergeometric function

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.

• If this doesn't get answers, maybe Math.SE would have an answer? Aug 11, 2019 at 8:48
• What did you try in Mathematica? Aug 11, 2019 at 11:46
• @Ulrich, I just tried the routine Solve[\rho[r_,b_,q_]==x,r] Aug 11, 2019 at 11:51
• @MichaelE2, noted. Aug 11, 2019 at 12:08
• Aug 11, 2019 at 12:40

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][] & /@ 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[]] == 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]}}*)

• I'll check this out. Thanks Aug 11, 2019 at 12:29
• you did not include the terms outside the hypergeometric function. And that makes the inversion difficult still. Aug 11, 2019 at 12:41
• @user583893. Transcendental equation can't be solved analytically.Try numerics. Aug 11, 2019 at 12:52
• @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. Aug 11, 2019 at 13:36
• @user583893 See mathematica.stackexchange.com/questions/13767/… Aug 11, 2019 at 13:48