Entry 182

Using class number \(4\), in Entry 68 we calculated \(j_{2}(\tau)\) at certain points and found,

$$\begin{align}j_{2}\Big(\tfrac{1+\sqrt{-17}}2\Big) &= -2^{11}\big(4+\sqrt{17}\big)^2 \big({-1}+\sqrt{17}\big)\\ j_{2}\Big(\tfrac{1+\sqrt{-73}}2\Big) &= -2^{9}\cdot3^4\big(111+13\sqrt{73}\big)^3\\ j_{2}\Big(\tfrac{1+\sqrt{-97}}2\Big) &= -2^{11}\cdot3^4\big(59+6\sqrt{97}\big)^4\big({-9}+\sqrt{97}\big)\\ j_{2}\Big(\tfrac{1+\sqrt{-193}}2\Big) &= -2^{11}\cdot3^4\big(208+15\sqrt{193}\big)^4\big(903+65\sqrt{193}\big)\end{align}$$ All the quadratic irrationals above with odd powers are odd fundamental solutions to Pell equations \(x^2-dy^2=-16\). For example, the initial solution to \(x^2-193y^2=-16\) is \((x,y)=(903,\,65)\). And Ramanujan already found

$$\begin{align}G_{17} &=  \sqrt{\frac{-3+\sqrt{17}}8} +\sqrt{\frac{5+\sqrt{17}}8} \\ G_{73} &=  \sqrt{\frac{1+\sqrt{73}}8} +\sqrt{\frac{9+\sqrt{73}}8}\\ G_{97} &=  \sqrt{\frac{5+\sqrt{97}}8} +\sqrt{\frac{13+\sqrt{97}}8}\\ G_{193} &= \sqrt{\frac{22+2\sqrt{193}}8} +\sqrt{\frac{30+2\sqrt{193}}8} \end{align}$$

These were used in Entry 162 to find formulas for \(K(k_d)\). But we can combine these to find new formulas using 

$$K(k_d) = \frac{\pi}2  \frac{(2^{1/4}G_d)^3}{\;\big({-j_2(\tau)}\big)^{1/8}}\, \sqrt{\sum_{n=0}^\infty \frac{(4n)!}{n!^4} \frac1{\big(j_2(\tau)\big)^n}}$$

For example, let \(d = 73\)

$$K(k_{73}) = \frac{\pi}2  \left(\sqrt{\frac{1+\sqrt{73}}8} +\sqrt{\frac{9+\sqrt{73}}8}\right)^3 \frac{2^{3/4}}{\;\big({-j_2(\tau)}\big)^{1/8}}\, \sqrt{\sum_{n=0}^\infty \frac{(4n)!}{n!^4} \frac1{\big(j_2(\tau)\big)^n}}$$

where as given above

$$j_2(\tau) = j_2\Big(\tfrac{1+\sqrt{-73}}2\Big) = {-2^{9}}\cdot3^4\big(111+13\sqrt{73}\big)^3$$

and so on.