Rogers-Ramanujan continued fraction in terms of Jacobi theta functions?
Motivated by ccorns's comment, I decided to re-visit this question. After some effort, I found what I was looking for. The Rogers-Ramanujan cfrac is,
$$r = r(\tau)= \cfrac{q^{1/5}}{1+\cfrac{q}{1+\cfrac{q^2}{1+\ddots}}}$$
where $q = \exp(2\pi i \tau)$. It turns out we have the nice relations,
$$\begin{aligned}\frac{1}{r^5}-r^5 &=\,u^3+11\\ &= \frac{v(v-5)^2}{(v-1)^2}+11 \end{aligned}$$
where the three roots of the cubic in $u$ are eta quotients,
$$u_n = \left(\frac{\eta(\tau')}{\eta(5\tau')}\right)^2$$
with $\tau'=\tau+n$, while the three roots of the cubic in $v$ are theta quotients,
$$v_n = \left(\frac{\vartheta_{n+2}(0,p)}{\vartheta_{n+2}(0,p^5)}\right)^2$$
for the nome $p = e^{\pi i \tau}$ and for $n=0,1,2$.
Given the Jacobi theta function $$ \vartheta_4(u,q)=1+2\sum^{\infty}_{n=1}(-1)^nq^{n^2}\cos(2nu) $$ I. Define the Rogers-Ramanujan as $$ R(q)=\frac{q^{1/5}}{1+}\frac{q}{1+}\frac{q^2}{1+}\frac{q^3}{1+}\ldots\textrm{, }\;|q|<1 $$ then $$ R(e^{-x})=e^{-x/5}\frac{\vartheta_4(3ix/4,\,e^{-5x/2})}{\vartheta_4(ix/4,\,e^{-5x/2})}\textrm{, }\;x>0 $$ II. Define the Ramanujan-Gordon-Gollniz as $$ H(q)=\frac{q^{1/2}}{1+q+}\frac{q^2}{1+q^3+}\frac{q^4}{1+q^5+}\frac{q^6}{1+q^7+}\ldots\textrm{, }\;|q|<1 $$ then $$ H(e^{-x})=e^{-x/2}\frac{\vartheta_4(3ix/2,\,e^{-4x})}{\vartheta_4(ix/2,\,e^{-4x})}\textrm{, }\;x>0 $$
I'm sorry about the octic cfrac. Yes it seems to be non trivial. In Berndt's book (see: B.C. Berndt. "Ramanujan Notebooks III". Springer Verlag, New York (1991), In chapter 19 Entry 1) one can find the expansion $$ M(q)=\tfrac{1}{2}\,q^{1/8}\frac{\sum^{\infty}_{n=-\infty}q^{n^2/2+n/2}}{\sum^{\infty}_{n=-\infty}q^{n^2}}=\frac{q^{1/8}}{1+}\frac{q}{1+q+}\frac{q^2}{1+q^2+}\frac{q^3}{1+q^3+}\ldots $$ Hence if $|q|<1$ and $$ \vartheta_3(z,q):=\sum^{\infty}_{n=-\infty}q^{n^2}e^{2niz} $$ For $x>0$ we have $$ M(e^{-x})=\tfrac{1}{2}\,e^{-x/8}\frac{\vartheta_3(ix/4,e^{-x/2})}{\vartheta_3(0,e^{-x})} $$ Equivalently, let $q = e^{2\pi i z}$, then $$M(q) = \tfrac{1}{2}\,q^{1/8} \frac{\vartheta_3\big(i\ln(q^{1/4}),q^{1/2}\big)}{\vartheta_3(0,q)}=\frac{\eta(z)\,\eta^2(4z)}{\eta^3(2z)}$$ where $\eta(z)$ is the Dedekind eta function.