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Given the square of the nome $q=e^{2i\pi\tau}$ and ramanujan theta function $f(a,b)=\sum_{k=-\infty}^{\infty}a^{k(k+1)/2}b^{k(k-1)/2}$ with $|q|\lt1$, define,

$$\begin{aligned}M(q)=\cfrac{1-q^3}{1-q^2+\cfrac{q^2(1-q^{-1})(1-q^5)}{1-q^6+\cfrac{q^4(1-q)(1-q^7)}{1-q^{10}+\cfrac{q^6(1-q^3)(1-q^{9})}{1-q^{14}+\ddots}}}}\overset{\color{blue}{?}}=\prod_{n=1}^\infty\frac{(1-q^{8n-3})(1-q^{8n-5})}{(1-q^{8n-1})(1-q^{8n-7})}\\[1.5mm]&\end{aligned}$$ $$=\frac{f(-q^3,-q^5)}{f(-q,-q^7)}$$ and $$\begin{aligned}N(q)=\cfrac{1-q^3}{1+q^2-\cfrac{q^2(1+q^{-1})(1+q^5)}{1+q^6+\cfrac{q^4(1-q)(1-q^{7})}{1+q^{10}-\cfrac{q^6(1+q^3)(1+q^{9})}{1+q^{14}+\ddots}}}}\overset{\color{blue}{?}}=\prod_{n=1}^\infty\frac{(1-q^{8n-3})(1-q^{8n-5})}{(1-q^{8n-1})(1-q^{8n-7})}\\[1.5mm]&\end{aligned}$$ $$=\frac{f(-q^3,-q^5)}{f(-q,-q^7)}$$

Q: How do we prove that the two q-continued fractions are equal to the q-products such that the continued fractions are equivalent $M(q)=N(q)$?

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  • $\begingroup$ I have given an answer, but I suppose that you can do that yourself. There are several cfrac questions of yours which I won't answer because there is nothing new to tell. I'd appreciate it if you answer those remaining questions. That would demonstrate progress and unclutter the Unanswered queue. $\endgroup$
    – ccorn
    Nov 8, 2015 at 17:06

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For $n\in\mathbb{N}$, let $q_n=\exp\frac{2\pi\mathrm{i}\tau}{n}$, so $q_n^n=q$.

Use formula $(***)$ from that post, but with $q$ replaced with $q^2$, so it reads $$\small\cfrac{1}{1-q^2+\cfrac{(a+bq^2)(aq^2+b)} {1-q^6+\cfrac{q^2(a+bq^4)(aq^4+b)} {1-q^{10}+\cfrac{q^4(a+bq^6)(aq^6+b)}{1-q^{14}+\cdots}}}} = \frac{(-a^2q^6;q^8)_\infty\,(-b^2q^6;q^8)_\infty} {(-a^2q^2;q^8)_\infty\,(-b^2q^2;q^8)_\infty} \qquad(ab=q^2)\tag{1}$$

Now set $a=-\mathrm{i}q_2^5$, $b=\mathrm{i}q_2^{-1}$. This implies $ab=q^2$, $a/b=-q^3$, $-a^2=q^5$, $-b^2=q^{-1}$, hence $$\underbrace{\cfrac{1}{1-q^2+\cfrac{q^2(1-q^{-1})(1-q^5)} {1-q^6+\cfrac{q^4(1-q)(1-q^7)} {1-q^{10}+\cfrac{q^6(1-q^3)(1-q^9)}{1-q^{14}+\cdots}}}}}_{\frac{M(q)}{1-q^3}} = \frac{(q^{11};q^8)_\infty\,(q^5;q^8)_\infty} {(q^7;q^8)_\infty\,(q;q^8)_\infty}$$ which implies your statement for $M(q)$.

In $(1)$, replace $q$ with $\mathrm{i}q$. The formula requires $ab=-q^2$ then, and odd powers of $q^2$ in $(1)$ change sign. Now set $a=-q_2^5$, $b=q_2^{-1}$. This implies $ab=-q^2$, $a/b=-q^3$, $-a^2=-q^5$, $-b^2=-q^{-1}$, hence $$\underbrace{\cfrac{1}{1-q^2-\cfrac{q^2(1+q^{-1})(1+q^5)} {1+q^6+\cfrac{q^4(1-q)(1-q^7)} {1+q^{10}-\cfrac{q^6(1+q^3)(1+q^9)}{1+q^{14}+\cdots}}}}}_{\frac{N(q)}{1-q^3}} = \frac{(q^{11};q^8)_\infty\,(q^5;q^8)_\infty} {(q^7;q^8)_\infty\,(q;q^8)_\infty}$$ which implies your statement for $N(q)$.

From a bird's view, this works because the right-hand side in (1) is preserved if the signs of $q^2$, $a^2$ and $b^2$ are changed simultaneously. The corresponding left-hand sides must then be equivalent too.

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