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I need to evaluate the Gaussian hypergeometric function $_2F_1(a;b;c;z)$ for the inputs $a=1,b=1,c\in \left\{\frac{n}{2} : n \in \mathbb{N} \setminus \{0,1\}\right\}$, and $z \in [0,1)\subset \mathbb{R}$. Is there a closed form solution for this? If it makes things easier, I am also willing to make the set of possible inputs of $c$ smaller by only considering $c\in \left\{\frac{n}{2} : n \in \mathbb{N} \setminus \{0,1,\ldots,l\}\right\}$ up to some limit $l$.

If there is no closed-form solution what numeric approach/software would be best suited?

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    $\begingroup$ For even $n$ a formula can be found here: functions.wolfram.com/HypergeometricFunctions/Hypergeometric2F1/… $\endgroup$
    – Johannes Trost
    Commented Jul 4, 2019 at 14:46
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    $\begingroup$ $$\, _2F_1(1,1;m;z)=(m-1) z(z-1)^{-2} \left(\sum _{k=2}^{m-1} \frac{\left(\frac{z-1}{z}\right)^k}{m-k}-\left(\frac{z-1}{z}\right)^m \log (1-z)\right)$$ $\, _2F_1(1,1;m/2;z)$ can also be expressed in terms of elementary functions ($\arcsin\sqrt z$) $\endgroup$
    – Carlo Beenakker
    Commented Jul 4, 2019 at 14:51
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    $\begingroup$ $_2F_1(1,1;\frac{1}{2};z) = (1-z)^{-1} \left(1+\frac{\sqrt{z}\ \arcsin\sqrt{z}}{\sqrt{1-z}}\right)$ and $_1F_2(0,1;c;z)=1$ then make use of dlmf.nist.gov/15.5.E16 . $\endgroup$
    – Johannes Trost
    Commented Jul 4, 2019 at 15:03
  • $\begingroup$ Thanks a lot! When I tried implementing the solution for $m-1 \in \mathbb{N}^+$, suggested by @JohannesTrost and Carlo Beenakker, I stumbled across a problem. The results differ from the implementation in GNU GSL, as well as a naive approximation in which I consider the first 90 summands of the hypergeometric series. For example, for $_2F_1(1,1,4,1/2)$, the formula suggested here returns $5.32$ whereas the other two approaches return $1.59$; for $c=3$, there are no differences; for $c=5$, the suggested formula even returns a negative value, which should be impossible. $\endgroup$
    – Julian Karch
    Commented Jul 5, 2019 at 14:09
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    $\begingroup$ The sum is not including the term $-\left(\frac{z-1}{z}\right)^m \log (1-z)$ obviously. $\endgroup$
    – Johannes Trost
    Commented Jul 5, 2019 at 16:40

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