the ratio of jacobi theta functions and a new conjectured q-continued fraction

Question

Given the squared nome $q=e^{2i\pi\tau}$ with $|q|\lt1$, define

$$\begin{aligned}H(q)=\cfrac{2(1+q^2)}{1-q+\cfrac{(1+q)(1+q^3)}{1-q^3+\cfrac{2q^2(1+q^4)}{1-q^5+\cfrac{q^3(1+q)(1+q^5)}{1-q^7+\cfrac{q^4(1+q^2)(1+q^6)}{1-q^9+\cfrac{q^5(1+q^3)(1+q^7)}{1-q^{11}+\ddots}}}}}}\end{aligned}$$

Question: How do we show that $$H(q)\overset{\color{red}{?}}=\frac{2\,q^{1/2}\,\vartheta_3(0,q^2)}{\vartheta_2(0,q^2)}$$ where $\vartheta_n(0,q)$ are jacobi theta functions

Answer 1

In a related answer, I used a formula by Ramanujan, proved by Adiga et al. (1985): $$\small\frac{(-a;q)_\infty\,(-b;q)_\infty - (a;q)_\infty\,(b;q)_\infty} {(-a;q)_\infty\,(-b;q)_\infty + (a;q)_\infty\,(b;q)_\infty} = \cfrac{a+b}{1-q+\cfrac{(a+bq)(aq+b)}{1-q^3+\cfrac{q\,(a+bq^2)(aq^2+b)} {1-q^5+\cfrac{q^2(a+bq^3)(aq^3+b)}{1-q^7+\cdots}}}}\tag{*}$$ Applying it here can be done by setting $a=q_2^3$, $b=q_2^{-1}$ where $q_n=\exp(2\pi\mathrm{i}\tau/n)$, so $q_n^n=q$. Thus we get $$\begin{align} \frac{H(q)}{2q_2} &= q_2^{-1}\,\cfrac{1+q^2}{1-q+\cfrac{(1+q)(1+q^3)}{1-q^3+\cfrac{2q^2(1+q^4)} {1-q^5+\cfrac{q^3(1+q)(1+q^5)}{1-q^7+\cfrac{q^4(1+q^2)(1+q^6)} {1-q^9+\cdots}}}}} \\ &= \frac{(-q_2^3;q)_\infty\,(-q_2^{-1};q)_\infty - (q_2^3;q)_\infty\,(q_2^{-1};q)_\infty} {(-q_2^3;q)_\infty\,(-q_2^{-1};q)_\infty + (q_2^3;q)_\infty\,(q_2^{-1};q)_\infty} \\ &= \frac{(-q_2^3;q)_\infty^2\,(-q_2^{-1};q)_2 - (q_2^3;q)_\infty^2\,(q_2^{-1};q)_2} {(-q_2^3;q)_\infty^2\,\underbrace{(-q_2^{-1};q)_2}_{(1+q_2)^2/q_2} + (q_2^3;q)_\infty^2\,\underbrace{(q_2^{-1};q)_2}_{-(1-q_2)^2/q_2}} \\ &= \frac{(-q_2;q)_\infty^2 + (q_2;q)_\infty^2} {(-q_2;q)_\infty^2 - (q_2;q)_\infty^2} \\ &\stackrel{[1]}{=} \frac{\vartheta_3(0,q_2) + \vartheta_3(0,-q_2)} {\vartheta_3(0,q_2) - \vartheta_3(0,-q_2)} \\ &\stackrel{[2]}{=} \frac{\vartheta_3(0,q^2)}{\vartheta_2(0,q^2)} \end{align}$$ where [1] follows from the product representation $$\begin{align} \vartheta_3(0,q_2) &= (-q_2;q)_\infty^2\,(q;q)_\infty \end{align}$$ and [2] follows from splitting the series representation of $\vartheta_3$ into odd and even parts, resulting in: $$\begin{align} \vartheta_3(0,q_2) + \vartheta_3(0,-q_2) &= 2\vartheta_3(0,q^2) \\ \vartheta_3(0,q_2) - \vartheta_3(0,-q_2) &= 2\vartheta_2(0,q^2) \end{align}$$ More details and background on the Ramanujan formula can be found at the other answer.

Answer 2

(A partial answer.) Compare the three cfracs of similar form,

$$H(q)=\frac{q^{1/2}\,\vartheta_3(0,q^2)}{\vartheta_2(0,q^2)}=\small\cfrac{(1+q^2)}{1-q+\cfrac{q(1+q^{-1})(1+q^3)}{1-q^3+\cfrac{q^2(1+q^0)(1+q^4)}{1-q^5+\cfrac{q^3(1+q)(1+q^5)}{1-q^7+\cfrac{q^4(1+q^2)(1+q^6)}{1-q^9+\ddots}}}}}\tag1$$

$$U(-q)\; =\; 1\;=\;\small\cfrac{(1+q^1)}{1-q+\cfrac{q(1+q^0)(1+q^2)}{1-q^3+\cfrac{q^2(1+q)(1+q^3)}{1-q^5+\cfrac{q^3(1+q^2)(1+q^4)}{1-q^7+\cfrac{q^4(1+q^3)(1+q^5)}{1-q^{9}+\ddots}}}}}\tag2$$

$$S(q) = \frac{1}{q^{1/2}}\frac{\vartheta_2(0,q^2)}{\vartheta_3(0,q^2)}=\small\cfrac{(1+q^0)}{1-q+\cfrac{q(1+q)(1+q)}{1-q^3+\cfrac{q^2(1+q^2)(1+q^2)} {1-q^5+\cfrac{q^3(1+q^3)(1+q^3)}{1-q^7+\cfrac{q^4(1+q^4)(1+q^4)} {1-q^9+\ddots}}}}}\tag3$$

where the pattern of one cfrac to the next is clear. We have $(1)$ as your proposed equality, $(2)$ is the one in this post (with the small change $q \to -q$ for aesthetics), and $(3)$ was established by Somos in A079006. One can then see the nice fact that,

$$H(q)\,U(-q)\,S(q) = 1$$

In fact, $H(q)$ and $S(q)$ are reciprocals. (This is the second reciprocal pair you have found after this, so I assume you are using a general method?)

P.S. Since $(2)$ belongs to an infinite family involving ratios of form $(1-aq^n)$, then $(1),(3)$ may have an infinite family as well.