Entry 37

There is an elegant continued fraction associated with the Ramanujan G and g functions $$G_n =\frac{2^{-1/4}\,\eta^2(\tau)}{\eta\big(\tfrac{\tau}{2}\big)\,\eta(2\tau)}$$ $$g_n =\frac{2^{-1/4}\,\eta\big(\tfrac{\tau}{2}\big)}{\eta(\tau)}$$ where $\tau=\sqrt{-n}$. Recall that $(G_n g_n)^8(G_n^8-g_n^8) = \tfrac{1}{4}$. So perhaps it is no surprise that they can be expressed by the octic continued fraction. Given the nome $q=e^{\pi i \tau}$, then $$\beta(\tau)=\left(\frac{\sqrt{1+G_n^{-12}}\pm\sqrt{1-G_n^{-12}}}{2}\right)^{1/4} =\cfrac{\sqrt{2}\, q^{1/8}}{1 + \cfrac{q}{1 + q + \cfrac{q^2}{1 + q^2 + \cfrac{q^3}{1 + q^3 + \ddots}}}}$$ where the $\pm$ sign changes beyond a bound $n$. Alternatively, $\beta(\tau)=\left(-g_n^{12}+\sqrt{g_n^{24}+1}\right)^{1/4}$. For example, since $G_{1/4} = \frac{(1+\sqrt{2})^{1/4}}{2^{3/16}}$ this implies the identity $$\left(\sqrt{\frac{1}{4}+\frac{2^{1/4}}{(1+\sqrt{2})^3}}-\sqrt{\frac{1}{4}-\frac{2^{1/4}}{(1+\sqrt{2})^3}}\right)^{1/4}=\sqrt{\frac{1}{1+\sqrt{2}}}$$ and is the value of the continued fraction when $q=e^{\pi\, i\sqrt{-1/4}}$.