Entry 107

Given $q = e^{2\pi i \tau}$ and the two $q$-continued fractions $$\begin{align}\quad\text{K}_4(q) &= \sqrt2\,q^{1/8} \prod_{n=1}^\infty\frac{(1-q^{4n-1})(1-q^{4n-3})}{(1-q^{4n-2})(1-q^{4n-2})}\\ & = \cfrac{\sqrt2\,q^{1/8}} {1+\cfrac{q} {1+q+\cfrac{q^2} {1+q^2+\cfrac{q^3} {1+q^3+\ddots }}}}\quad\end{align}$$ $$\qquad \begin{align}\text{K}_8(q)\, &=\,  q^{1/2}\,\prod_{n=1}^\infty\frac{(1-q^{8n-1})(1-q^{8n-7})}{(1-q^{8n-3})(1-q^{8n-5})}\\ &=  \cfrac{q^{1/2}}{1+q+\cfrac{q^2}{1+q^3+\cfrac{q^4}{1+q^5+\cfrac{q^6}{1+q^7+\ddots}}}}\end{align}$$

then we can evaluate them using the modular lambda function $\lambda(\tau)$ as $$\text{K}_4(\tau) = \big(\lambda(2\tau)\big)^{1/8}\quad$$ $$\quad\frac1{\text{K}_8(\tau)}-\text{K}_8(\tau)  = \frac{2}{\big(\lambda(4\tau)\big)^{1/4}}$$ Mathematica can calculate $\lambda(\tau)$ as ModularLambda[tau] and there is a list of exact values in Mathworld. We have previously given examples for $\text{K}_4(\tau) = \big(\lambda(2\tau)\big)^{1/8}$. For $\text{K}_8(\tau)$, then for $d = 4m$ with class number $h(-d)=2$ and $m=10,58$, $$\begin{align}\frac1{\text{K}_8\big(\tfrac14\sqrt{-10}\big)}-\text{K}_8\big(\tfrac14\sqrt{-10}\big) &= 2(1+\sqrt2)\sqrt{3+\sqrt{10}} \;=\; 2\,U_2\sqrt{U_{10}}\\ \frac1{\text{K}_8\big(\tfrac14\sqrt{-58}\big)}-\text{K}_8\big(\tfrac14\sqrt{-58}\big) &= 2(1+\sqrt2)^3\sqrt{99+13\sqrt{58}} =2\,U_2^3\sqrt{U_{58}}\end{align}$$ where $U_n$ are fundamental units.