Entry 108

Let \(q = e^{2\pi i \tau}\) and the Dedekind eta function \(\eta(\tau)\). Define the following $$\text{K}_{10}(\tau) = \text{K}_{10}(q) = q^{3/5}\prod_{n=1}^\infty\frac{(1-q^{10n-1})(1-q^{10n-9})}{(1-q^{10n-3})(1-q^{10n-7})}$$ While no continued fraction is yet known for this, it is connected to the mod 5 version, $$R(\tau) =  \text{K}_{5}(q) = q^{1/5}\prod_{n=1}^\infty\frac{(1-q^{5n-1})(1-q^{5n-4})}{(1-q^{5n-2})(1-q^{5n-3})}$$

or the Rogers-Ramanujan continued fraction \(R(\tau)\) by the simple relation

$$\text{K}_{10}(q) = R(q)\,R(q^2)\\ \text{K}_{10}(\tau) = R(\tau)\,R(2\tau)$$

It is known that$$\frac1{R(\tau)}-R(\tau) = \frac{\eta(\tau/5)}{\eta(5\tau)}+1$$ Thus a quadratic root $$R(\tau) = \frac{- \frac{\eta(\tau/5)}{\eta(5\tau)}-1+\sqrt{\left( \frac{\eta(\tau/5)}{\eta(5\tau)}+1\right)^2+4}}2$$ which allows for easily computation of \(\text{K}_{10}(\tau)\). For example, given the golden ratio \(\phi = \frac{1+\sqrt5}2\) and \(\sqrt{-1}=i\), $$\begin{align}R\big(\tfrac12 i\big) &= 0.511428\dots = \tfrac12(\sqrt{5\phi\,}-\phi^2)(+\sqrt[4]5\sqrt{\phi}+\phi^2)\\ R\big(i\big) &= 0.284079\dots = (\sqrt[4]5\sqrt{\phi}-\phi)\\ R\big(2i\big) &=  0.081002\dots =\tfrac12(\sqrt{5\phi\,}-\phi^2)(-\sqrt[4]5\sqrt{\phi}+\phi^2) \end{align}$$which implies the exact values$$\begin{align}\text{K}_{10}\big(\tfrac12 i\big) &= R\big(\tfrac12 i\big)\,R\big(i\big) = 0.145286\dots\\ \text{K}_{10}\big(i\big) &= R\big(i\big)\,R\big(2i\big) \,= 0.203011\dots \end{align}$$