Extension of Jacobi triple Product Identity for $\sum\limits_{n = - \infty }^ \infty z^n q^{n^2} h^{n^3}$

Question

I asked a question about Extension of Jacobi triple Product Identity for $\sum\limits_{n = - \infty }^ \infty z^n q^{n^2} h^{n^3}$. I have new idea about zeros for $\sum\limits_{n = - \infty }^ \infty z^n q^{n^2} h^{n^3}$ because my previous method does not work for zeros.

If you want ,you can see the previous question here.

The Jacobi triple product identity is: $$\prod\limits_{n=1}^{ \infty }(1-q^{2n})(1+zq^{2n-1})(1+z^{-1}q^{2n-1})=\sum\limits_{n = - \infty }^ \infty z^n q^{n^2} $$

I would like to extend the idea for $\sum\limits_{n = - \infty }^ \infty z^n q^{n^2} h^{n^3}$

$|h|=1$ and $|q|<1$. ( $h=e^{ix}$ where x is a real number)

My Conjecture for the expansion is:

$$G(z,q,h)\prod\limits_{n=1}^{ \infty }(1+\frac{zq^{2n-1}h^{3n^2-3n+1})}{P_n(q,h)})(1+\frac{z^{-1}q^{2n-1}h^{-3n^2+3n-1}}{R_n(q,h)})=\sum\limits_{n = - \infty }^ \infty z^n q^{n^2} h^{n^3} \tag 1 $$

$z\to z^{-1}$ and $h\to h^{-1}$

Because of symmetry :

$$G(z,q,h)= G(z^{-1},q,h^{-1})$$ $$R_n(q,h)=P_n(q, h^{-1})$$

$$G(z,q,h)\prod\limits_{n=1}^{ \infty }(1+\frac{zq^{2n-1}h^{3n^2-3n+1})}{P_n(q,h)})(1+\frac{z^{-1}q^{2n-1}h^{-3n^2+3n-1}}{P_n(q,h^{-1})})=\sum\limits_{n = - \infty }^ \infty z^n q^{n^2} h^{n^3}\tag 2 $$

$z=ZQ^{2}h^{3}$

$q=Qh^{3}$

$$G(ZQ^{2}h^{3},Qh^{3},h)\prod\limits_{n=1}^{ \infty }(1+\frac{ZQ^{2n+1}h^{3n^2+3n+1}}{P_n(Qh^{3},h)})(1+\frac{Z^{-1}Q^{2n-3}h^{-3n^2+9n-7}}{P_n(Qh^{3},h^{-1})})=\sum\limits_{n = - \infty }^ \infty Z^n Q^{2n+n^2}h^{3n+3n^2+n^3} \tag 3 $$

$$G(ZQ^{2}h^{3},Qh^{3},h) ZQh (1+\frac{Z^{-1}Q^{-1}h^{-1}}{P_1(Qh^{3},h^{-1})}) \prod\limits_{n=1}^{ \infty }(1+\frac{ZQ^{2n+1}h^{3n^2+3n+1}}{P_n(Qh^{3},h)})\prod\limits_{n=2}^{ \infty }(1+\frac{Z^{-1}Q^{2n-1}h^{-3n^2+3n-1}}{P_{n+1}(Qh^{3},h^{-1})})=\sum\limits_{n = - \infty }^ \infty Z^{1+n} Q^{1+2n+n^2}h^{1+3n+3n^2+n^3} \tag 4 $$

$$\frac{G(ZQ^{2}h^{3},Qh^{3},h)}{P_1(Qh^{3},h^{-1})} ( 1+ZQhP_1(Qh^{3},h^{-1})) \prod\limits_{n=2}^{ \infty }(1+\frac{ZQ^{2n-1}h^{3n^2-3n+1}}{P_{n-1}(Qh^{3},h)})\prod\limits_{n=1}^{ \infty }(1+\frac{Z^{-1}Q^{2n-1}h^{-3n^2+3n-1}}{P_{n+1}(Qh^{3},h^{-1})})=\sum\limits_{n = - \infty }^ \infty Z^{n} Q^{n^2}h^{n^3} \tag 5 $$

Zeros must be the same with Equation (2) and Equation (5) Thus, $$P_1(qh^{3},h^{-1})=\frac{1}{P_1(q,h)}$$

$$P_{n+1}(q,h)=P_n(qh^3,h)$$

and

$$\frac{G(zq^{2}h^{3},qh^{3},h)}{P_1(qh^{3},h^{-1})}=G(z,q,h)$$

$$G(zq^{2}h^{3},qh^{3},h)=\frac{G(z,q,h)}{P_{1}(q,h)}$$

$$G(z,q,1)=\prod\limits_{n=1}^{ \infty }(1-q^{2n})$$ is known Thus $P_{1}(q,1)=1$

$$G(z,q,h)\prod\limits_{n=1}^{ \infty }(1+\frac{zq^{2n-1}h^{3n^2-3n+1})}{P_1(qh^{3(n-1)},h)})(1+\frac{z^{-1}q^{2n-1}h^{-3n^2+3n-1}}{P_1(qh^{-3(n-1)},h^{-1})})=\sum\limits_{n = - \infty }^ \infty z^n q^{n^2} h^{n^3} \tag 6 $$

$$1+\frac{zqh}{P_1(q,h)}=0$$ $$z=-q^{-1}h^{-1}P_1(q,h)$$ is a zero for $\sum\limits_{n = - \infty }^ \infty z^n q^{n^2} h^{n^3}$

$$\sum\limits_{n = - \infty }^ \infty (-1)^n q^{n^2-n} h^{n^3-n}P_1(q,h)^n=0 \tag 7 $$

Let's assume that $$P_1(q,h)=1+H_1(h)q+H_2(h)q^2+H_3(h)q^3+H_4(h)q^4+....$$

$$......+q^6h^{-6}P_1(q,h)^{-2}-q^2P_1(q,h)^{-1}+1-P_1(q,h)+q^2h^6P_1(q,h)^2-q^6h^{24}P_1(q,h)^3+....$$

First four terms of $P_1(q,h)$ : $$P_1(q,h)=1+q^2(h^6-1)+q^4(h^6-1)(2h^6+1)+....$$

$$1/P_1(q,h)=1-q^2(h^6-1)-q^4(h^6-1)(h^6+2)+....$$

I have not found other terms yet. My questions :

1. Can you please help me find more terms of $P_1(q,h)$ and closed form of $P_1(q,h)$ or $1/P_1(q,h)$? (Maybe a closed form is exist for them)
2. And I also have been working on $G(z,q,h)$. Is it possible that it does not depend on $z$ terms? Can someone help me prove or disprove that $G(z,q,h)$ does not depend on $z$?

Thanks a lot for helps

EDIT: My Proof Attempt for Question 2:

$$G(zq^{2}h^{3},qh^{3},h)=\frac{G(z,q,h)}{P_{1}(q,h)}$$

$$G(z,q,h)=G(zq^{2}h^{3},qh^{3},h)P_{1}(q,h)$$

$z \to zq^{2}h^{3}$

$q \to qh^{3}$

$$G(zq^{2}h^{3},qh^{3},h)=G(zq^{4}h^{12},qh^{6},h)P_{1}(qh^{3},h)$$

$$\frac{G(z,q,h)}{P_{1}(q,h)}=G(zq^{4}h^{12},qh^{6},h)P_{1}(qh^{3},h)$$

$$G(z,q,h)=G(zq^{4}h^{12},qh^{6},h)P_{1}(qh^{3},h)P_{1}(q,h)$$

$z \to zq^{2}h^{3}$

$q \to qh^{3}$

$$G(zq^{2}h^{3},qh^{3},h)=G(zq^{6}h^{27},qh^{9},h)P_{1}(qh^{6},h)P_{1}(qh^{3},h)$$ $$\frac{G(z,q,h)}{P_{1}(q,h)}=G(zq^{6}h^{27},qh^{9},h)P_{1}(qh^{6},h)P_{1}(qh^{3},h)$$

$$G(z,q,h)=G(zq^{6}h^{27},qh^{9},h)P_{1}(qh^{6},h)P_{1}(qh^{3},h)P_{1}(q,h)$$

If we continue it ,

$$G(z,q,h)= \lim\limits_{ n\to \infty }G(zq^{2n}h^{3n^2},qh^{3n},h) \prod\limits_{n=0}^{ \infty } P_{1}(qh^{3n},h) $$

Because of $|h|=1$ and $|q|<1$. ( $h=e^{ix}$ where x is a real number)

$$\lim\limits_{ n\to \infty } zq^{2n}h^{3n^2} = 0$$

$$\lim\limits_{ n\to \infty }G(zq^{2n}h^{3n^2},qh^{3n},h)=Q(q,h)$$

$Q(q,h)$ cannot have a $z$ term.

$$G(z,q,h)=Q(q,h) \prod\limits_{n=0}^{ \infty } P_{1}(qh^{3n},h) $$

$$G(z,q,1)=Q(q,1) \prod\limits_{n=0}^{ \infty } P_{1}(q,1) $$

$$Q(q,1)=G(z,q,1)=\prod\limits_{n=1}^{ \infty }(1-q^{2n})$$

Is it valid proof for Question 2? Thanks a lot for comments

EDIT: 24/03/2020 Because $G(z,q,h)$ does not have z terms, we can write: $$G(q,h)=G(qh^{3},h)P_{1}(q,h)$$

$$P_{1}(q,h)=\frac{G(q,h)}{G(qh^{3},h)}$$

I have found one more term for $P_1(q,h)$ : $$P_1(q,h)=1+q^2(h^6-1)+q^4(h^6-1)(2h^6+1)+(h^{-6}-h^{24}+(h^6-1)(5h^{12}+2h^6+2))q^6+....$$

We can write $$P_{1}(q,h)=1+q^2(h^6-1)+q^4(h^6-1)(2h^6+1)+(h^{-6}-h^{24}+(h^6-1)(5h^{12}+2h^6+2))q^6+....=\frac{1-q^2-q^4+(h^6-2+h^{-6})q^6+...}{1-h^6q^2-h^{12}q^4+(h^6-2+h^{-6})h^{18}q^6+...}=\frac{G(q,h)}{G(qh^{3},h)}$$

Thus , few terms of $G(q,h)$

$$G(q,h)=1-q^2-q^4+(h^6-2+h^{-6})q^6+...$$

I wonder more terms and what the closed form of $G(q,h)$ is.

Could you please help me to find more terms of $G(q,h)$?

Maybe a closed form can be seen for $G(q,h)$?

(Note that: $$G(q,1)=\prod\limits_{n=1}^{ \infty }(1-q^{2n})$$ )