Entry 27

Here are some Ramanujan-inspired formulas for pi using the complete elliptic integral of the first kind \(K(k_d)\) and the golden ratio \(\phi=\tfrac{1+\sqrt{5}}{2}\)$$\frac{1}{\pi} = \frac{1}{2\,K(k_{5})} \sqrt{\sum_{n=0}^\infty \frac{(2n)!^3}{n!^6} \frac{1}{(2^6\phi^{6})^n}}$$ $$\frac{1}{\pi} = \frac{1}{2\,K(k_{15})} \sqrt{\sum_{n=0}^\infty \frac{(2n)!^3}{n!^6} \frac{1}{(2^{12}\phi^{8})^n}}$$ $$\frac{1}{\pi} = \frac{1}{2\,K(k_{25})} \sqrt{\sum_{n=0}^\infty \frac{(2n)!^3}{n!^6} \frac{1}{(2^6\phi^{24})^n}}$$ The denominators having different powers of \(\phi\) are the exact values of \(\left(\frac{\eta(\tau)}{\eta(2\tau)}\right)^{24}\) for \(\tau=\frac{1+\sqrt{-5}}{2},\,\frac{1+\sqrt{-15}}{2},\,\frac{1+\sqrt{-25}}{2},\) respectively. (Using \(d = -35\) would already need a sextic.) The last implies $$e^{\pi\sqrt{25}} \approx 2^6 \phi^{24}-24.00004\dots$$ And \(K(k_n)\) can be expressed as a product of gamma functions as given in the link above. For example $$K(k_5)=\frac{\phi^{3/4}\sqrt{\Gamma(\frac{1}{20})\,\Gamma(\frac{3}{20})\,\Gamma(\frac{7}{20})\,\Gamma(\frac{9}{20})}}{4\sqrt{10\pi}}=1.576390\dots$$ which can also be numerically evaluated by Wolfram Alpha.