The function ${}_2F_1\Big(\frac{a-b}{2},\frac{a+b-1}{2};c;y\Big)$ can be transformed (as reported by A. Erdélyi) by the following formula

${}_2F_1\Big(\frac{a-b}{2},\frac{a+b-1}{2};c;y\Big)= (1-z)^{1-b}{}_2F_1\Big(a,a-1;b;z\Big) $


$y =4z(1-z)$, and $\quad a,b,c \in\mathbb{R}$

The problem is that solving the equation $y =4z(1-z)$ for the value of $z$ yields two solutions

$z_1 = \frac{1}{2} \left(1-\sqrt{1-y}\right)$


$z_2 = \frac{1}{2} \left(1+\sqrt{1-y}\right)$

Therefore, should the resultant transformation include both values of $z_1$ and $z_2$? or just use one value of them and ignore the second one.

How to include both values in one answer?

  • 3
    $\begingroup$ Can you cite a source? These formulas seems incorrect - for one, the lhs depends on c while the rhs does not. It will be simpler to examine the full region of valididity of the formula once it is correct... $\endgroup$ Commented Feb 19, 2012 at 13:54
  • $\begingroup$ Handbook of Special Functions by Yury A. Brychkov chapter 8, Eq.(40) $\endgroup$
    – Remy
    Commented Feb 19, 2012 at 17:11
  • $\begingroup$ As it stands, the lhs and rhs satisfy different (2nd order) linear ODEs, so they really can't be the same. Seems like you have found a typo in that book. $\endgroup$ Commented Feb 20, 2012 at 1:18

1 Answer 1


Here is a general explanation (which I can make more precise once you fix the formulas): the identity will hold "everywhere" except on the branch cut for the hypergeometric function (i.e. on the real line, from 1 to $+\infty$). On the branch cut, the choice of sign will flip the orientation of the cut, so that you get continuity in a clockwise/counter-clockwise manner.

  • $\begingroup$ @Jacues: Thanks I noticed the same thing about the parameter c and why it is not showing in the transformation. The equation can be found in Handbook of Special Functions by Yury A. Brychkov. amazon.com/Handbook-Special-Functions-Derivatives-Integrals/dp/… The transformation is given by equation (40) in chapter 8 as follows ${}_2F_1\Big(a,1-a;b;z\Big)=(1-z)^{b-1}{}_2F_1\Big(\frac{b-a}{2},\frac{b+a-1}{2};c;4z(1-z)\Big)$ $\endgroup$
    – Remy
    Commented Feb 19, 2012 at 15:13
  • $\begingroup$ My problem is that I am using this transformation to simplify an integrand that contains 2F1 within an integral equation that I encountered. Therefore,I was thinking which solution of the transformation of 2F1 I need to plug back in my original integral equation. $\endgroup$
    – Remy
    Commented Feb 19, 2012 at 15:18

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