Entry 123

In the previous entry, we had the unusual hypergeometric function, call it $\beta$ $$\beta =\,_2F_1\Big(\tfrac12,\tfrac12;1;\lambda\big(1+2i/5\big)\Big) = 2^{1/4}\left(\Big(\tfrac{1+\sqrt5}2\Big)^{12}-\sqrt{\Big(\tfrac{1+\sqrt5}2\Big)^{24}-1}\right)^{1/4}G$$ with Gauss's constant $G$. Ramanujan found the unusual equality $$\sqrt[8]{1+\sqrt{1-\left(\tfrac{-1+\sqrt{5}}{2}\right)^{24}}} = \tfrac{-1+\sqrt{5}}{2}\,\left(\tfrac{1+\sqrt[4]{5}}{\sqrt{2}}\right)$$ and one can notice the $24$th powers in both cases. After some manipulation, we find the similar $$\sqrt[8]{\left(\tfrac{1+\sqrt{5}}2\right)^{12}-\sqrt{\left(\tfrac{1+\sqrt{5}}2\right)^{24}-1}} = \sqrt{\tfrac{1+\sqrt{5}}2}\left(\tfrac{-1+\sqrt[4]{5}}{\sqrt{2}}\right)\quad$$ therefore $$\beta =\,_2F_1\Big(\tfrac12,\tfrac12;1;\lambda\big(1+2i/5\big)\Big) = 2^{1/4}\left(\tfrac{1+\sqrt5}2\right) \left(\tfrac{-1+\sqrt[4]{5}}{\sqrt{2}}\right)^2 G$$