a conjecture of certain q-continued fractions

Solution 1:

(A half-answer.) Let $|q|<1$ and,

$$p = q^m,\quad \alpha = \pm q^{-n},\quad \beta = \pm q^n,\quad \alpha\beta = 1$$

We propose that,

$$\small C_{m,n}(q)=\prod_{k=1}^\infty\frac{(1-\alpha\, p^{4k})(1-\beta\, p^{4k-4})}{(1-\alpha\, p^{4k-2})(1-\beta\, p^{4k-2})} \overset{\color{red}{?}}=\cfrac{(1-\beta\,p^0)}{ 1-p+\cfrac{p(1-\alpha\, p)(1-\beta\, p)}{1-p^3+\cfrac{p^2(1-\alpha\, p^2)(1-\beta\, p^2)}{1-p^{5}+\cfrac{p^3(1-\alpha\, p^3)(1-\beta\, p^3)}{1-p^{7}+\ddots}}}}\tag1 $$

It takes only some algebraic manipulation to show that given,

$$\alpha_1 = \pm q^{\large -n_1},\quad \beta_1 = \pm q^{\large n_1},\quad \alpha\beta = 1$$

then the two products are reciprocals,

$$\frac{(1-\alpha_1\, p^{4k})(1-\beta_1\, p^{4k-4})}{(1-\alpha_1\, p^{4k-2})(1-\beta_1\, p^{4k-2})} \times \frac{(1-\alpha_2\, p^{4k})(1-\beta_2\, p^{4k-4})}{(1-\alpha_2\, p^{4k-2})(1-\beta_2\, p^{4k-2})}=1$$

if,

$$m =\frac{n_1+n_2}{2}\tag2$$

For example, let $m,\,n_1,\,n_2 = 3,\,2,\, 4$, and using $(1)$ and the positive case of $\pm$, we get,

$$C_{3,2}(q) = \prod_{n=1}^\infty\frac{(1-q^{12n-2})(1-q^{12n-10})}{(1-q^{12n-4})(1-q^{12n-8})} = \small\cfrac{1-q^2}{1-q^3+\cfrac{q^3(1-q)(1-q^5)}{1-q^9+\cfrac{q^6(1-q^4)(1-q^8)}{1-q^{15}+\cfrac{q^9(1-q^7)(1-q^{11})}{1-q^{21}+\ddots}}}}$$

$$C_{3,4}(q) = \prod_{n=1}^\infty\frac{(1-q^{12n-4})(1-q^{12n-8})}{(1-q^{12n-2})(1-q^{12n-10})} = \small\cfrac{1-q^4}{1-q^3+\cfrac{q^3(1-q^{-1})(1-q^7)}{1-q^9+\cfrac{q^6(1-q^2)(1-q^{10})}{1-q^{15}+\cfrac{q^9(1-q^5)(1-q^{13})}{1+q^{21}+\ddots}}}}$$

the same as in the post and,

$$C_{3,2}(q)\,C_{3,4}(q) = 1$$

The reciprocality is true for general $m,n_1,n_2$ that obey $(2)$. However, what remains is to show that the cfrac is indeed equal to the infinite product $(1)$.

Solution 2:

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:

1. 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 $F(q)$.

2. 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.