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, $F(q)\,G(q)=1$.)

  • $\begingroup$ Your additional info makes it much, much clearer. Don't forget those impt details. (I have also made some editing changes.) $\endgroup$ Commented Sep 10, 2015 at 16:21
  • $\begingroup$ Also, be careful with terminology. The nome is just $q_1 = \exp(\pi i \tau)$, while nowadays it is more common to use its square, $q_2 = \exp(\color{red}2\pi i \tau)$. And note that using two dollar signs "$$" at each end centers an expression. If centering is not needed, then use only one dollar sign "$" at each end, ok? $\endgroup$ Commented Sep 10, 2015 at 16:26
  • 2
    $\begingroup$ More generally, you are asking, define, $$C_{m,n}(q)=\cfrac{(1-q^n)}{1-q^m+\cfrac{q^m(1-q^{m-n})(1-q^{m+n})}{1-q^{3m}+ \cfrac {q^{2m}(1-q^{2m-n})(1-q^{2m+n})}{1-q^{5m}+\cfrac{q^{3m}(1-q^{3m-n})(1-q^{3m+n})} {1-q^{7m}+\ddots}}}}$$ If $\,2m=n_1+n_2,\,$ then, $$C_{m,n_1}(q)\;C_{m,n_2}(q) = 1$$ Your post was just the case $m=3$, and $n_1,n_2 = 2,4$. $\endgroup$ Commented Sep 13, 2015 at 7:19

2 Answers 2


(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,

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


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

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 $(1)$.

  • $\begingroup$ @ Tito PiezasIII:very nice analysis,I would have given you the bounty ,had you shown that the q-cfrac is indeed equal to the infinite product. $\endgroup$
    – Nicco
    Commented Sep 15, 2015 at 4:45
  • $\begingroup$ @Nicco: Thanks. And typo has been fixed. Yes, sometimes it is understandly hard to prove a q-frac is equal to a certain q-series. $\endgroup$ Commented Sep 15, 2015 at 6:57
  • $\begingroup$ @Nicco: I've been familiar with the q-series/q-cfrac $C_{m,n}(q)$ for some time now. (See this.) However, I didn't realize they contain aesthetic reciprocal cfrac pairs until your posts. $\endgroup$ Commented Sep 15, 2015 at 7:11
  • $\begingroup$ @ Tito PiezasIII:if we note that $m=\frac{n_{1}+n_{2}}{2}$ is the arithmetic mean,and applying the AM-GM inequality $\frac{n_{1}+n_{2}}{2}\ge \sqrt{n_{1}\,n_{2}}$ ,then we are naturally led to the inequality $$m\ge\sqrt{n_{1}\,n_{2}}$$ $\endgroup$
    – Nicco
    Commented Sep 18, 2015 at 5:29
  • $\begingroup$ The inequality implies that $m$ is bounded below $\endgroup$
    – Nicco
    Commented Sep 18, 2015 at 9:34

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.


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