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In this post, the OP seeks a closed-form for, $$A=\,_2F_1\big(\tfrac12,\tfrac12;1;\tfrac19\big)=1.02966\dots$$ Using the transformation, $$\,_2F_1\big(\tfrac12,\tfrac12;1;z\big) = \tfrac2{1+\sqrt{1-z}}\,_2F_1\Big(\tfrac12,\tfrac12;1;\big(\tfrac{1-\sqrt{1-z}}{1+\sqrt{1-z}}\big)^2\Big)$$ or more generally (DLMF 15.8.21), $$\,_2F_1\big(a,b;a-b+1;z\big) = (1+\sqrt{z})^{-2a}\,_2F_1\Big(a,a-b+\tfrac12;2a-2b+1;\tfrac{4\sqrt{z}}{(1+\sqrt{z})^2}\Big)$$ we can transform $A$ to, $$A =6\times\frac{\,_2F_1\big(\tfrac12,\tfrac12;1;\color{blue}{(1-\sqrt2)^8}\big)}{(1+\sqrt2)^2}$$ However, it turns out that the following do have a closed-form, $$\begin{aligned} B&=\,_2F_1\big(\tfrac12,\tfrac12;1;\color{blue}{(1-\sqrt2)^4}\big)=1.00748\dots\\[2.0mm] &=\frac{\big(1+\sqrt2\big)\Gamma^2\big(\tfrac14\big)}{2^{5/2}\,\pi^{3/2}}\end{aligned}$$ $$\begin{aligned} C&=\,_2F_1\big(\tfrac12,\tfrac12;1;\color{blue}{(1-\sqrt2)^2}\big)=1.04760\dots\\[2.0mm] &=\frac{\sqrt{1+\sqrt2}\,\Gamma\big(\tfrac18\big)\Gamma\big(\tfrac38\big)}{2^{9/4}\,\pi^{3/2}}\end{aligned}$$

Q: So, is there a known transformation between $_2F_1\big(\tfrac12,\tfrac12;1;z\big)$ and $_2F_1\big(\tfrac12,\tfrac12;1;z^2\big)$?

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  • $\begingroup$ $$\phantom{}_2 F_1\left(\frac{1}{2},\frac{1}{2};1;z\right)=\frac{2}{\pi} K(z)$$ with Mathematica notation. It follows that such a hypergeometric function can be computed through the AGM mean. $\endgroup$ – Jack D'Aurizio Dec 17 '16 at 16:22
  • $\begingroup$ For instance, $$A = \frac{3}{\operatorname{AGM}(2,4)}.$$ $\endgroup$ – Jack D'Aurizio Dec 17 '16 at 16:34
  • $\begingroup$ What formula did you use to make the first transformation? $\endgroup$ – TreFox Dec 28 '16 at 19:56
  • $\begingroup$ @TreFox: I've edited the post to include the transformation. $\endgroup$ – Tito Piezas III Dec 29 '16 at 5:33
  • $\begingroup$ I think you already answered this question yourself. Suppose there was such a transformation, then the ratio $\,_2F_1\big(\tfrac12,\tfrac12;1;\color{blue}{(1-\sqrt2)^4}\big)/\,_2F_1\big(\tfrac12,\tfrac12;1;\color{blue}{(1-\sqrt2)^2}\big)$ would be algebraic, which in turn would mean that $\Gamma\big(\tfrac18\big)\Gamma\big(\tfrac38\big)/\Gamma^2\big(\tfrac14\big)$ is algebraic, which is most probably not. $\endgroup$ – Nemo Dec 29 '16 at 8:37

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