Entry 101

Level 8. Recall the Level 4 continued fraction $$\begin{align}\text{K}_4(q) &= \sqrt2\,q^{1/8} \prod_{n=1}^\infty\frac{(1-q^{4n-1})(1-q^{4n-3})}{(1-q^{4n-2})(1-q^{4n-2})} = \frac{\sqrt2\,\eta(\tau)\,\eta^2(4\tau)}{\eta^3(2\tau)}\\ &= \cfrac{\sqrt2\,q^{1/8}} {1+\cfrac{q} {1+\cfrac{q+q^2} {1+\cfrac{q^3} {1+\cfrac{q^2+q^4} {1+\ddots}}}}} = \cfrac{\sqrt2\,q^{1/8}} {1+\cfrac{q} {1+q+\cfrac{q^2} {1+q^2+\cfrac{q^3} {1+q^3+\cfrac{q^4} {1+q^4+\ddots}}}}}\end{align}$$ Compare its similarity to the Level 8 version using the ratio of $(a,b)$ from Entry 100 $$\begin{align}\text{K}_8(q) &\,=\,  q^{1/2}\,\prod_{n=1}^\infty\frac{(1-q^{8n-1})(1-q^{8n-7})}{(1-q^{8n-3})(1-q^{8n-5})}\\ &= \cfrac{q^{1/2}}{1+\cfrac{q+q^2}{1+\cfrac{q^4}{1+\cfrac{q^3+q^6}{1+\cfrac{q^8}{1+\ddots}}}}} = \cfrac{q^{1/2}}{1+q+\cfrac{q^2}{1+q^3+\cfrac{q^4}{1+q^5+\cfrac{q^6}{1+q^7+\cfrac{q^8}{1+q^9+\ddots}}}}}\end{align}$$

In fact, they have the quadratic relation $$\quad\frac1{\text{K}_8(q)}-\text{K}_8(q) =\left(\frac{\sqrt2}{\text{K}_4(q^2)}\right)^2 = \left(\frac{\eta^3(4\tau)}{\eta(2\tau)\eta^2(8\tau)}\right)^2$$ while $\text{K}_4(q)$ which is an eta quotient can also be expressed another way $$\frac1{\big(\text{K}_4(q)\big)^2}-\big(\text{K}_4(q)\big)^2 =\frac12\left(\frac{\eta(\tau/2)}{\eta(2\tau)}\right)^4\quad$$