[Math] best approximation of $\sqrt{2}$

Question

The approximation
\begin{align}
\sqrt{2} &\approx
\frac{1}{8} \operatorname{csch}\left(\frac{3\pi}{2}\right) \operatorname{sech}^3(\pi) \, \left[2+3 \, \sinh\left(\frac{\pi}{2}\right)-\sinh\left(\frac{3\pi}{2}\right) \right. \\
& \hspace{10mm} \left. +3 \, \sinh\left(\frac{5\pi}{2}\right)+\sinh\left(\frac{9\pi}{2}\right)-2 \,\cosh(\pi)+2 \,\cosh(2\pi)+2 \,\cosh(4\pi)\right]
\end{align}
gives the first $8$ correct digits of $\sqrt{2}$.
Is this the best approximation of square root 2 in terms of hyperbolic functions?
If not,then please find more examples of this type.

Answer 1

By exploiting the elliptic lambda function, we have: $$ \sqrt{2} = 2\cdot \frac{\theta_2^2(0,e^{-\pi})}{\theta_3^2(0,e^{-\pi})} $$ and by exploting the expansions of the Jacobi theta functions we have: $$ \sqrt{2} \approx 2\,\left(\frac{2-2e^{5\pi}+2e^{8\pi}-e^{9\pi}}{2+2 e^{5 \pi }+2 e^{8 \pi }+e^{9 \pi }}\right)^2$$ that looks way better and is right up to $14$ digits.