Entry 17

Here is another family (now based on \(d=-10\)) of pi approximations with a form consistent to the one (based on \(d=-58\)) in Entry 16. Define the fundamental units $$\begin{aligned}
U_{2} &= 1+\sqrt{2}\\
U_{5} &= \frac{1+\sqrt{5}}{2}\\
U_{10} &= 3+\sqrt{10}\\
U_{30} &= 11+2\sqrt{30}\end{aligned}$$ Then with increasing precision, accurate to \(5,10,14,18\) digits $$\begin{aligned}
\pi &\approx \frac{1}{\sqrt{10}} \ln \Big[ 2^6 (U_{5})^{12} \Big]\\
\pi &\approx \frac{1}{2\sqrt{10}} \ln \left[ 2^9 \left(U_2\, U_{5} \sqrt{U_{10}} \,\right)^6 \right]\\
\pi &\approx \frac{1}{3\sqrt{10}} \ln \left[ 2^6 (U_{5})^{12}\, (U_{30})^2 \left( \sqrt{\frac{3+\sqrt{6}}{4} } + \sqrt{\frac{-1+\sqrt{6}}{4}}\right)^{24}\right]\\
\pi &\approx \frac{1}{4\sqrt{10}} \ln \Big[ 2^9 \left(U_2\, U_{5} \sqrt{2\,U_{10}} \,\right)^3 \big(\sqrt{v+1} +\sqrt{v}\big)^{12}\Big]\end{aligned}$$ where \(v = 2^{-1/2}(U_2)^2\,(U_{5})^3\).

If I remember correctly, I think Ramanujan found the first two members. The expression inside the log function is again the exact value of \(\left(\frac{\eta(\tau)}{\eta(2\tau)}\right)^{24}\) where  \(\eta(\tau)\) is the Dedekind eta function, and  \(\tau = \frac{\sqrt{-10}}{2},  \tau = \frac{2\sqrt{-10}}{2},  \tau = \frac{3\sqrt{-10}}{2},  \tau = \frac{4\sqrt{-10}}{2}\), respectively.