Are these two $q$-continued fractions equivalent?

Question

In this MSE post, Nicco Mnisi defined a particular $q$-continued fraction of order $12$. More generally, define the cfrac found in Ramanujan's Notebooks, Vol III, Chap. 16, page 24, where $|ab|<1$ and $|q|<1$,

$$\begin{aligned}U(q) &= \prod_{n=0}^\infty \frac{\big(1-a^2q^3(q^4)^n\big)\big(1-b^2q^3(q^4)^n\big)}{\big(1-a^2q(q^4)^n\big)\big(1-b^2q(q^4)^n\big)}\\ &= \dfrac{1} {1-ab+\dfrac{(a-bq)(b-aq)} {(1-ab)(1+q^2)+\dfrac{(a-bq^3)(b-aq^3)} {(1-ab)(1+q^4)+\dfrac{(a-bq^5)(b-aq^5)} {(1-ab)(1+q^6)+\ddots }}}} \end{aligned}$$

I extrapolated that the general form of Nicco's cfrac, without a factor $q^{k_1}(1-q^{k_2})$, apparently is,

$$V(q) = \dfrac{1} {1+ab-\dfrac{(a+bq)(b+aq)} {1+(ab)^3+\dfrac{(a-bq^2)(b-aq^2)q} {1+(ab)^5-\dfrac{(a+bq^3)(b+aq^3)q^2} {1+(ab)^7+\dfrac{(a-bq^4)(b-aq^4)q^3} {(1+(ab)^9-\ddots }}}}} $$

Question: If $ab=q$, and $|q|<1$, is it true that $U(q) = V(q)$?

I tested these numerically for various $a,b,q$ and it seems to be true with $V(q)$ converging faster. However, a rigorous proof is needed.

P.S. With the appropriate factor $q^{k_1}(1-q^{k_2})$ affixed, these cfracs are algebraic numbers. For example, in this post,

\begin{align} \frac{1}{N(e^{-2\pi})}+N(e^{-2\pi}) &= \frac{4}{1-\sqrt{3\big(3+\sqrt3-3^{3/4}\sqrt{2+\sqrt3}\big)}}\\ &= \frac{\;8}{\;\sqrt{2}( 3+ \sqrt{2}) - \sqrt[4]{3}(3+\sqrt{3})}\\[4pt] &= 536.4953904\dots\end{align}

Answer 1

In this answer we shall follow a path Ramanujan must have likely taken at some point, though we have no solid evidence that he actually did. We prove the claim in two different ways part I and Part II. Part I is a rather long route while part II is very short and direct.

We make use of the usual notation of a q-pochhammer symbol

$\displaystyle (a;q)_\infty=(a)_\infty=\prod_{n = 1}^{\infty}\left(1-aq^{n-1}\right)$

the well known ramanujan theta function

$f(a,b)=\displaystyle \sum_{n=-\infty}^{\infty} a^{\frac{n(n+1)}{2}} b^{\frac{n(n-1)}{2}}=(ab;ab)_{\infty} (-a;ab)_{\infty} (-b;ab)_{\infty}\tag1$

and the q-binomial theorem

$\displaystyle \sum_{n=0}^{\infty} \frac{(c;q)_n}{(q;q)_n}z^n=\frac{(cz;q)_\infty}{(z;q)_\infty}\tag2$

Part I

In chapter 16 of Ramanujan's second notebook we have entry 11

$\displaystyle \cfrac{(a-b)}{1-q+}\cfrac{(a-bq)(aq-b)}{1-q^3+}\cfrac{q(a-bq^2)(aq^2-b)}{1-q^5+}\cdots=\frac{(-a)_\infty(b)_\infty-(a)_\infty(-b)_\infty}{(-a)_\infty(b)_\infty+(a)_\infty(-b)_\infty}$

which can be rewritten as

$\displaystyle \cfrac{2}{1-}\cfrac{(a-b)}{1-q+}\cfrac{(a-bq)(aq-b)}{1-q^3+}\cfrac{q(a-bq^2)(aq^2-b)}{1-q^5+}\cdots=1+\frac{(-a)_\infty(b)_\infty}{(a)_\infty(-b)_\infty}$

After replacing $a$ by $-a$, we let $ab=q$ and then apply entry 30(ii) $f(a,b)+f(-a,-b)=2f(a^3b,ab^3)$ from chapter 16 of Ramanujan's second notebook, after which we are led to

$\displaystyle \cfrac{1}{1+}\cfrac{(a+b)}{1-(ab)+}\cfrac{(ab)(1+a^2)(1+b^2)}{1-(ab)^3+}\cfrac{(ab)^2(1+a^2(ab))(1+b^2(ab))}{1-(ab)^5+}\cdots=\frac{f(a^3b,ab^3)}{f(a,b)}$

We now introduce the following ramanujan theta function

$bf\Big(a^3b^5,\displaystyle \frac{a}{b}\Big)=\displaystyle \sum_{n=-\infty}^{\infty} (a^n b^{n+1})^{2n+1}\tag3$

which has a nice jacobi triple product

$bf\Big(a^3b^5,\displaystyle \frac{a}{b}\Big)=(a+b)(a^4b^4;a^4b^4)_{\infty}(-a^3b^5;a^4b^4)_\infty(-a^5b^3;a^4b^4)_\infty\tag4$

and satisfies the ff identity

$f(a,b)-f(a^3b,ab^3)=bf\Big(a^3b^5,\displaystyle \frac{a}{b}\Big)\tag5$

To see how, we add entry 30(ii) $f(a,b)+f(-a,-b)=2f(a^3b,ab^3)$ and entry 30(iii) $f(a,b)-f(-a,-b)=2af\Big(\displaystyle \frac{b}{a}, a^5b^3\Big)$ from [1] together and then recall entry 18(i) $f(a,b)=f(b,a)$

Note that if we let $a=\displaystyle \frac{q^{\frac{1}{4}}}{w}$ and $b=\displaystyle q^{\frac{1}{4}}w$, the function $bf\Big(a^3b^5,\displaystyle \frac{a}{b}\Big)$ becomes jacobi theta function $\vartheta_{10}(q,w)$. Thus it generalizes the form of jacobi theta function $\vartheta_{10}(q,w)$ in much the same way ramanujan theta function $f(a,b)$ generalizes $\vartheta_{00}(q,w)$.

To find out how the triple product identity $(4)$ comes about, we use the very general ramanujan theta function $(1)$

$bf\Big(a^3b^5,\displaystyle \frac{a}{b}\Big)=b(a^4b^4;a^4b^4)_{\infty}(-a^3b^5;a^4b^4)_\infty\Big(-\displaystyle \frac{a}{b};a^4b^4\Big)_\infty$

Next up we employ the q-binomial theorem $(2)$

Let $c=q$ and in place of q insert $q^4$ and then we have

$\displaystyle \sum_{n=0}^{\infty} \Big(-\frac{a}{b}\Big)^n=\frac{\Big(\displaystyle -\frac{a}{b}q^4;q^4\Big)_\infty}{\Big(\displaystyle -\frac{a}{b};q^4\Big)_\infty}$

Since $q=ab$

$\displaystyle \sum_{n=0}^{\infty} \Big(\displaystyle -\frac{a}{b}\Big)^n=\frac{(-a^5b^3;a^4b^4)_\infty}{\Big(\displaystyle -\frac{a}{b};a^4b^4\Big)_\infty}$

$\displaystyle \frac{b}{a+b}=\frac{(-a^5b^3;a^4b^4)_\infty}{\Big(\displaystyle -\frac{a}{b};a^4b^4\Big)_\infty}$

therefore

$\displaystyle (a+b)(-a^5b^3;a^4b^4)_\infty=b\Big(\displaystyle -\frac{a}{b};a^4b^4\Big)_\infty$

After employing $(5)$ we are immediately led to the theta quotient

$\displaystyle \cfrac{(a+b)}{1-(ab)+}\cfrac{(ab)(1+a^2)(1+b^2)}{1-(ab)^3+}\cfrac{(ab)^2(1+a^2(ab))(1+b^2(ab))}{1-(ab)^5+}\cdots=\frac{bf\Big(a^3b^5,\displaystyle \frac{a}{b}\Big)}{f(a^3b,ab^3)}=\frac{\displaystyle \sum_{n=-\infty}^{\infty} (a^n b^{n+1})^{2n+1} }{\displaystyle \sum_{n=-\infty}^{\infty} a^{n(2n+1)} b^{n(2n-1)} }$

which then reduces to

$\displaystyle \cfrac{1}{1-(ab)+}\cfrac{(ab)(1+a^2)(1+b^2)}{1-(ab)^3+}\cfrac{(ab)^2(1+a^2(ab))(1+b^2(ab))}{1-(ab)^5+}\cdots=\frac{(-a^5b^3;a^4b^4)_{\infty}(-a^3b^5;a^4b^4)_\infty}{(-a^3b;a^4b^4)_{\infty}(-ab^3;a^4b^4)_\infty}\tag6$

as desired.

Part II

If we change the sign of $b$ to $-b$ in entry 11 and then let $q=ab$, the general continued fraction becomes

$\displaystyle \cfrac{(a+b)}{1-(ab)+}\cfrac{(ab)(1+a^2)(1+b^2)}{1-(ab)^3+}\cfrac{(ab)^2(1+a^2(ab))(1+b^2(ab))}{1-(ab)^5+}\cdots=\frac{(-a;ab)_\infty(-b;ab)_\infty-(a;ab)_\infty(b;ab)_\infty}{(-a;ab)_\infty(-b;ab)_\infty+(a;ab)_\infty(b;ab)_\infty}$

Recalling ramanujan theta function $(1)$ we immediately observe that the right hand side is actually

$\displaystyle \cfrac{(a+b)}{1-(ab)+}\cfrac{(ab)(1+a^2)(1+b^2)}{1-(ab)^3+}\cfrac{(ab)^2(1+a^2(ab))(1+b^2(ab))}{1-(ab)^5+}\cdots=\frac{f(a,b)-f(-a,-b)}{f(a,b)+f(-a,-b)}$

And now dividing entry 30(iii) $f(a,b)-f(-a,-b)=2af\Big(\displaystyle \frac{b}{a}, a^5b^3\Big)$ by entry 30(ii) $f(a,b)+f(-a,-b)=2f(a^3b,ab^3)$ we are directly led to the theta quotient

$\displaystyle \cfrac{(a+b)}{1-(ab)+}\cfrac{(ab)(1+a^2)(1+b^2)}{1-(ab)^3+}\cfrac{(ab)^2(1+a^2(ab))(1+b^2(ab))}{1-(ab)^5+}\cdots=\frac{bf\Big(a^3b^5,\displaystyle \frac{a}{b}\Big)}{f(a^3b,ab^3)}$

which then reduces to

$\displaystyle \cfrac{1}{1-(ab)+}\cfrac{(ab)(1+a^2)(1+b^2)}{1-(ab)^3+}\cfrac{(ab)^2(1+a^2(ab))(1+b^2(ab))}{1-(ab)^5+}\cdots=\frac{(-a^5b^3;a^4b^4)_{\infty}(-a^3b^5;a^4b^4)_\infty}{(-a^3b;a^4b^4)_{\infty}(-ab^3;a^4b^4)_\infty}$

after employing the jacobi triple products $(4)$ and $(1)$

Remark

Note that if we let ${a\to ia}$ and ${b\to -ib}$ our continued fraction $(6)$ becomes

$\displaystyle \cfrac{1}{1-(ab)+}\cfrac{(ab)(1-a^2)(1-b^2)}{1-(ab)^3+}\cfrac{(ab)^2(1-a^2(ab))(1-b^2(ab))}{1-(ab)^5+}\cdots=\frac{(a^5b^3;a^4b^4)_{\infty}(a^3b^5;a^4b^4)_\infty}{(a^3b;a^4b^4)_{\infty}(ab^3;a^4b^4)_\infty}$

which is the q-analog of the ff continued fraction for the ratio of gamma functions

$\displaystyle \cfrac{4(a+b)}{(a+b)+}\cfrac{(2a)(2b)}{3(a+b)+}\cfrac{(3a+b)(a+3b)}{5(a+b)+}\cfrac{(4a+2b)(2a+4b)}{7(a+b)+}\cdots= \frac{\displaystyle\Gamma\Big(\frac{a+3b}{4(a+b)}\Big)\Gamma\Big(\frac{3a+b}{4(a+b)}\Big)}{\displaystyle\Gamma\Big(\frac{3a+5b}{4(a+b)}\Big)\Gamma\Big(\frac{5a+3b}{4(a+b)}\Big)}$

a special case of a continued fraction due to Nörlund, see here, [3], [4] and [5]

And moreover if we replace either $a$ by $-a$ or $b$ by $-b$ in $(6)$, then we have the form suggested by the OP

$V(q)=\displaystyle \cfrac{1}{1+(ab)-}\cfrac{(ab)(1+a^2)(1+b^2)}{1+(ab)^3+}\cfrac{(ab)^2(1-a^2(ab))(1-b^2(ab))}{1+(ab)^5-}\cdots=\frac{(a^5b^3;a^4b^4)_{\infty}(a^3b^5;a^4b^4)_\infty}{(a^3b;a^4b^4)_{\infty}(ab^3;a^4b^4)_\infty}$

Thus to answer the OP's question, if $q=ab$ given $|q|<1$ then U(q)=V(q) i.e. entry 11 and entry 12 in [1] are equal only when $q=ab$

[1] Chapter 16 of Ramanujan's second notebook: Theta-functions and q-series C. Adiga, B.C. Berndt, S. Bhargava and G.N. Watson

[2] A continued fraction of Ramanujan C. Adiga and N. Anitha

[3] Fractions continues et différences réciproques, Acta Math. 34(1911), 1-108. N. E. Nörlund

[4] Ramanujan's notebooks part II B.C. Berndt

[5] Die Lehre von den Kettenbrüchen, Band 2, dritte Auf., B. G. Teubner, Stuttgart, 1957. O. Perron