TITLE: a conjecture of certain q-continued fractions
QUESTION [9 upvotes]: Given the squared nome $q=e^{2i\pi\tau}$ with $|q|\lt1$, define,
$$\begin{aligned}F(q)=\cfrac{1-q^2}{1-q^3+\cfrac{q^3(1-q)(1-q^5)}{1-q^9+\cfrac{q^6(1-q^4)(1-q^8)}{1-q^{15}+\cfrac{q^9(1-q^7)(1-q^{11})}{1-q^{21}+\ddots}}}}\overset{\color{red}{?}}=\prod_{n=1}^\infty\frac{(1-q^{12n-2})(1-q^{12n-10})}{(1-q^{12n-4})(1-q^{12n-8})}\\[1.5mm]&\end{aligned}$$
and
$$\begin{aligned}G(q)=\cfrac{1-q^4}{1-q^3+\cfrac{q^3(1-\frac{1}{q})(1-q^7)}{1-q^9+\cfrac{q^6(1-q^2)(1-q^{10})}{1-q^{15}+\cfrac{q^9(1-q^5)(1-q^{13})}{1-q^{21}+\ddots}}}}\overset{\color{red}{?}}=\prod_{n=1}^\infty\frac{(1-q^{12n-4})(1-q^{12n-8})}{(1-q^{12n-2})(1-q^{12n-10})}\\[1.5mm]&\end{aligned}$$
Q: How do we prove rigorously that the two q-continued fractions are equal to the q-series?
(if true, then the two q-continued fractions are reciprocals such that their product is unity, $F(q)\,G(q)=1$.)

REPLY [3 votes]: I write $q_n = \exp\frac{2\pi\mathrm{i}\tau}{n}$, thus $q_n^n=q$.
In a related answer
and another one,
I used a formula by Ramanujan, proved by Adiga et al. (1985):
$$\small\frac{(-a;q)_\infty\,(-b;q)_\infty - (a;q)_\infty\,(b;q)_\infty}
{(-a;q)_\infty\,(-b;q)_\infty + (a;q)_\infty\,(b;q)_\infty}
= \cfrac{a+b}{1-q+\cfrac{(a+bq)(aq+b)}{1-q^3+\cfrac{q\,(a+bq^2)(aq^2+b)}
{1-q^5+\cfrac{q^2(a+bq^3)(aq^3+b)}{1-q^7+\cdots}}}}\tag{1}$$
In both cases, we had been able to simplify the result to a quotient of
Jacobi thetanulls, which implies that it is an eta quotient.
We will encounter that pattern again, so let us work it out in general.
Let me recall some two-variable Jacobi theta functions.
For $z\in\mathbb{C}$, let $w=\exp(\mathrm{i}z)$, so that expressions with
$w$ can be considered functions of $z$. Similarly, expressions with some $q_n$
can be considered functions of $\tau$.
I will need the following Theta functions:
$$\begin{align}
 \vartheta_3(z\mid\tau) = \vartheta_3(z,q_2)
 &= \sum_{k\in\mathbb{Z}} q_2^{k^2}\,w^{2k}
\\  &= (-w^2q_2;q)_\infty\,(-w^{-2}q_2;q)_\infty\,(q;q)_\infty
\\ \vartheta_4(z\mid\tau) = \vartheta_4(z,q_2)
 &= \sum_{k\in\mathbb{Z}} (-1)^k\,q_2^{k^2}\,w^{2k}
\\  &= (w^2q_2;q)_\infty\,(w^{-2}q_2;q)_\infty\,(q;q)_\infty
\\  \vartheta_2(z\mid\tau) = \vartheta_2(z,q_2)
 &= \sum_{k\in\mathbb{Z}} q_8^{(2k+1)^2}\,w^{2k+1}
 = q_8 w \sum_{k\in\mathbb{Z}} q^{k\,(k+1)/2}\,w^{2k}
\\  &= q_8 w\,(-w^2q;q)_\infty\,(-w^{-2};q)_\infty\,(q;q)_\infty
\end{align}$$
where the $q$-Pochhammer representations are due to the
triple product identity.
There is another Theta function, known as $\vartheta_1$,
but we will not need it.
If you split the series for $\vartheta_3$ and $\vartheta_4$ in parts
with even resp. odd summation index $k$, you get
$$\begin{align}
 \vartheta_3(z\mid\tau)
 &= \vartheta_3(2z\mid4\tau) + \vartheta_2(2z\mid4\tau)
\\  \vartheta_4(z\mid\tau)
 &= \vartheta_3(2z\mid4\tau) - \vartheta_2(2z\mid4\tau)
\end{align}$$
and therefore
$$\frac{\vartheta_3(z\mid\tau) - \vartheta_4(z\mid\tau)}
{\vartheta_3(z\mid\tau) + \vartheta_4(z\mid\tau)}
= \frac{\vartheta_2(2z\mid4\tau)}{\vartheta_3(2z\mid4\tau)}$$
Plugging in the product representations, we get
$$\frac{(-w^2q_2;q)_\infty\,(-w^{-2}q_2;q)_\infty
- (w^2q_2;q)_\infty\,(w^{-2}q_2;q)_\infty}
{(-w^2q_2;q)_\infty\,(-w^{-2}q_2;q)_\infty
+ (w^2q_2;q)_\infty\,(w^{-2}q_2;q)_\infty}
\\= w^2q_2\,\frac{(-w^4q^4;q^4)_\infty\,(-w^{-4};q^4)_\infty}
{(-w^4q^2;q^4)_\infty\,(-w^{-4}q^2;q^4)_\infty}$$
Suppose we are given $a,b\in\mathbb{C}$ with $ab=q$.
Then we can set $w^2=a/q_2=q_2/b$ or as well $w^2=b/q_2=q_2/a$,
so the above identity takes the form
$$\begin{align}
    \frac{(-a;q)_\infty\,(-b;q)_\infty
    - (a;q)_\infty\,(b;q)_\infty}
    {(-a;q)_\infty\,(-b;q)_\infty
    + (a;q)_\infty\,(b;q)_\infty}
    &= a\,\frac{(-a^2q^3;q^4)_\infty\,(-b^2q^{-1};q^4)_\infty}
    {(-a^2q;q^4)_\infty\,(-b^2q;q^4)_\infty}
\\  &= b\,\frac{(-a^2q^{-1};q^4)_\infty\,(-b^2q^3;q^4)_\infty}
    {(-a^2q;q^4)_\infty\,(-b^2q;q^4)_\infty}
    \tag{2}
\end{align}$$
Applying the above to your question:

For $F(q)$: In $(1)$ and $(2)$, replace $q$ with $q^3$.
So $(2)$ requires $ab=q^3$ now.
Now set $a=-\mathrm{i}q_2^5$, $b=\mathrm{i}q_2$.
Thus $ab = q^3$ and $a/b = -q^2$. This yields
$$\begin{align}
\mathrm{i}q_2\,F(q) &= \mathrm{i}q_2\,\cfrac{1-q^2}
{1-q^3+\cfrac{q^3(1-q)(1-q^5)}
{1-q^9+\cfrac{q^6(1-q^4)(1-q^8)}
{1-q^{15}+\cfrac{q^9(1-q^7)(1-q^{11})}{1-q^{21}+\cdots}}}}
\\&\stackrel{(1)}{=}
\frac{(\mathrm{i}q_2^5;q^3)_\infty\,(-\mathrm{i}q_2;q^3)_\infty
- (-\mathrm{i}q_2^5;q^3)_\infty\,(\mathrm{i}q_2;q^3)_\infty}
{(\mathrm{i}q_2^5;q^3)_\infty\,(-\mathrm{i}q_2;q^3)_\infty
+ (-\mathrm{i}q_2^5;q^3)_\infty\,(\mathrm{i}q_2;q^3)_\infty}
\\&\stackrel{(2)}{=}
\mathrm{i}q_2\,\frac{(q^{2};q^{12})_\infty\,(q^{10};q^{12})_\infty}
{(q^8;q^{12})_\infty\,(q^4;q^{12})_\infty}
\end{align}$$
which confirms the claim for $F(q)$.
For $G(q)$: Again, use $(1)$ and $(2)$ with $q$ replaced by $q^3$.
Now set $a=-\mathrm{i}q_2^7$, $b=\mathrm{i}q_2^{-1}$.
Thus $ab = q^3$ and $a/b = -q^4$. This yields
$$\begin{align}
\mathrm{i}q_2^{-1}\,G(q) &= \mathrm{i}q_2^{-1}\,\cfrac{1-q^4}
{1-q^3+\cfrac{q^3(1-q^{-1})(1-q^7)}
{1-q^9+\cfrac{q^6(1-q^2)(1-q^{10})}
{1-q^{15}+\cfrac{q^9(1-q^5)(1-q^{13})}{1-q^{21}+\cdots}}}}
\\&\stackrel{(1)}{=}
\frac{(\mathrm{i}q_2^7;q^3)_\infty\,(-\mathrm{i}q_2^{-1};q^3)_\infty
- (-\mathrm{i}q_2^7;q^3)_\infty\,(\mathrm{i}q_2^{-1};q^3)_\infty}
{(\mathrm{i}q_2^7;q^3)_\infty\,(-\mathrm{i}q_2^{-1};q^3)_\infty
+ (-\mathrm{i}q_2^7;q^3)_\infty\,(\mathrm{i}q_2^{-1};q^3)_\infty}
\\&\stackrel{(2)}{=}
\mathrm{i}q_2^{-1}\,\frac{(q^4;q^{12})_\infty\,(q^8;q^{12})_\infty}
{(q^{10};q^{12})_\infty\,(q^2;q^{12})_\infty}
\end{align}$$
which confirms the claim for $G(q)$.

Have fun applying the above formulae to more such continued fractions.