Rogers-Ramanujan continued fraction in terms of Jacobi theta functions?

The Rogers-Ramanujan cfrac is,

$$r = r(\tau)= \cfrac{q^{1/5}}{1+\cfrac{q}{1+\cfrac{q^2}{1+\ddots}}}$$

If $q = \exp(2\pi i \tau)$, then it is known that,

$$\frac{1}{r}-r =\frac{\eta(\tau/5)}{\eta(5\tau)}+1\tag1$$

$$\frac{1}{r^5}-r^5 =\left(\frac{\eta(\tau)}{\eta(5\tau)}\right)^6+11\tag2$$

with the Dedekind eta function, $\eta(\tau)$.

Q: Is there a similar simple identity known using ratios of the Jacobi theta functions $\vartheta_n(0,q)$?

$\color{brown}{Edit}$: In response to a comment, here are some details. Suppose we don't know $(2)$. One way to find such relations is to use known identities. Given the j-function $j(\tau)$, we have,

$$j(\tau)=-\frac{(r^{20} - 228r^{15} + 494r^{10} + 228r^5 + 1)^3}{r^5(r^{10} + 11r^5 - 1)^5}\tag3$$

$$j(\tau)=\frac{(5x^2+10x+1)^3}{x^5}\tag4$$

where $x = \left(\frac{\eta(\tau)}{\sqrt{5}\,\eta(5\tau)}\right)^6$. Equate $(3),\,(4)$,

$$-\frac{(r^{20} - 228r^{15} + 494r^{10} + 228r^5 + 1)^3}{r^5(r^{10} + 11r^5 - 1)^5} = \frac{(5x^2+10x+1)^3}{x^5}\tag5$$

and using symbolic software like Mathematica to factor $(5)$, we find one factor is given by,

$$\frac{1}{r^5}-r^5 =125x+11\tag6$$

and it only takes minor tweaking to make $(6)$ have the form of $(2)$. Thus, all we need is to express $j(\tau)$ not by an eta quotient $x$, but by a theta quotient $y=\left(\frac{\vartheta_n(0,q)}{\vartheta_n(0,q^5)}\right)^k$ so that,

$$j(\tau) = \frac{f_1(y)}{f_2(y)}\tag7$$

is a ratio of polynomials in $y$.

$\color{brown}{P.S.}$ The motivation for this is that the general quintic is solvable by the Rogers-Ramanujan cfrac via the eta quotient above as described in this post.

Solution 1:

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 $n=0,1,2$.

Solution 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 $$ H(e^{-x})=e^{-x/2}\frac{\vartheta_4(3ix/2,\,e^{-4x})}{\vartheta_4(ix/2,\,e^{-4x})}\textrm{, }\;x>0 $$

Solution 3:

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.

Solution 4:

For the cubic $V(q)$ cfrac is $$ V(q):=\frac{q^{1/3}}{1+}\frac{q+q^2}{1+}\frac{q^2+q^4}{1+}\frac{q^3+q^6}{1+}\ldots\textrm{, }|q|<1 $$ and $$ V(e^{-x})=e^{-x/3}\frac{\vartheta_4(ix/4,e^{-3x})}{\vartheta_4(0,e^{-3x})}\textrm{, }x>0 $$ See more formulas in https://arxiv.org/pdf/1107.2393v2.pdf

Solution 5:

Set $T=\sqrt{1-8V(q)^3}$, $q=e^{-\pi\sqrt{r}}$, $r>0$, then $$ k_r^2=\frac{(1-T)(3+T)^3}{(1+T)(3-T)^3} $$ where $k_r$ is the elliptic singular modulus. (see https://arxiv.org/pdf/1008.1304v2.pdf)