Extension of the Jacobi triple product identity

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} $

My idea is below for extension:

Let's assume we define $G(z,q,h)$ as

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

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

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

$q=Qh^{3}$

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

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

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

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

$$G(ZQ^{2}h^{3},Qh^{3},h)\sum\limits_{n = - \infty }^ \infty Z^n Q^{n^2} h^{n^3}=G(Z,Q,h)ZQh\sum\limits_{n = - \infty }^ \infty Z^n Q^{2n+n^2}h^{3n+3n^2+n^3} $$

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

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

$$G(ZQ^{2}h^{3},Qh^{3},h)\sum\limits_{n = - \infty }^ \infty Z^n Q^{n^2} h^{n^3}=G(Z,Q,h)\sum\limits_{n = - \infty }^ \infty Z^n Q^{n^2} h^{n^3} $$

$$G(ZQ^{2}h^{3},Qh^{3},h)=G(Z,Q,h) \tag 1$$

If $h=1$ then $G(z,q,1)=\prod\limits_{n=1}^{ \infty }(1-q^{2n})$ can be gotten from Jacobi_triple_product.

I really wonder how I can find the function $G(z,q,h)$. Please help me which Technics can be applied to find it. Also If you know there is other works about this subject, please share links and references.

Thanks a lot for responses.

Note: If $z=x^3$,$q=x^3$,$h=x$

$$G(x^3,x^3,x)\prod\limits_{n=1}^{ \infty }(1+x^3x^{6n-3}x^{3n^2-3n+1})(1+x^{-3}x^{6n-3}x^{-3n^2+3n-1})=\sum\limits_{n = - \infty }^ \infty x^{3n} x^{3n^2} x^{n^3} $$

$$xG(x^3,x^3,x)\prod\limits_{n=1}^{ \infty }(1+x^{3n^2+3n+1})(1+x^{-3n^2+9n-7})=\sum\limits_{n = - \infty }^ \infty x^{1+3n+3n^2+n^3} $$

$$xG(x^3,x^3,x)\frac{1}{(1+x)}\prod\limits_{n=1}^{ \infty }(1+x)(1+x^{3n^2+3n+1})(1+x^{-1})(1+x^{-3n^2+3n-1})=\sum\limits_{n = - \infty }^ \infty x^{(n+1)^3} $$

$$xG(x^3,x^3,x)\frac{(1+x^{-1})}{(1+x)}\prod\limits_{n=1}^{ \infty }(1+x^{3n^2-3n+1})(1+x^{-3n^2+3n-1})=\sum\limits_{n = - \infty }^ \infty x^{n^3} $$

$$G(x^3,x^3,x)\prod\limits_{n=1}^{ \infty }(1+x^{3n^2-3n+1})(1+x^{-3n^2+3n-1})=\sum\limits_{n = - \infty }^ \infty x^{n^3} $$

The relation is below for $G()$ from Equation 1 If $z=x^3$,$q=x^3$,$h=x$

$$G(x^{12},x^6,x)=G(x^3,x^3,x) \tag 2$$

$$G(x^{12},x^6,x)=G(x^3,x^3,x)=G(1,1,x) \tag 3$$

EDIT: I thought If I can find a few term of $G(z,q,h)$ by hand and maybe seen what the pattern of $G(z,q,h)$. I really wonder if we can find $G(z,q,h)$ in product terms as Jacobi did in his product formula.

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

$$G(z,q,h)(1+zqh)(1+z^{-1}q^{1}h^{-1})(1+zq^3h^7)(1+z^{-1}q^{3}h^{-7})....=\sum\limits_{n = - \infty }^ \infty z^n q^{n^2} h^{n^3} $$

$$G(z,q,h)(1+q^2+q(zh+z^{-1}h^{-1}))(1+q^6+q^3(zh^7+z^{-1}h^{-7}))...=\sum\limits_{n = - \infty }^ \infty z^n q^{n^2} h^{n^3}$$

$$G(z,q,h)(1+q^2+q^6+q^8+q(zh+z^{-1}h^{-1})+q^7(zh+z^{-1}h^{-1})+q^3(zh^7+z^{-1}h^{-7})+q^5(zh^7+z^{-1}h^{-7})+q^4(zh+z^{-1}h^{-1})(zh^7+z^{-1}h^{-7}))...=\sum\limits_{n = - \infty }^ \infty z^n q^{n^2} h^{n^3} $$

$$G(z,q,h)( 1+q^2+q^6+q^8+q(zh+z^{-1}h^{-1})+q^7(zh+z^{-1}h^{-1})+q^3(zh^7+z^{-1}h^{-7})+q^5(zh^7+z^{-1}h^{-7})+q^4(z^2h^8+z^{-2}h^{-8})+q^4(h^{6}+h^{-6}))...=\sum\limits_{n = - \infty }^ \infty z^n q^{n^2} h^{n^3} $$

$$G(z,q,h)( 1+q^2+q^6+q^8+q(zh+z^{-1}h^{-1})+q^7(zh+z^{-1}h^{-1})+q^3(zh^7+z^{-1}h^{-7})+q^5(zh^7+z^{-1}h^{-7})+q^4(z^2h^8+z^{-2}h^{-8})+q^4(h^{6}+h^{-6}))...= 1+q (zh+z^{-1}h^{-1})+q^4 (z^2h^8+z^{-2}h^{-8})+.... $$

EDIT: (Updated on 15th April)

We can see first 3 term of $G(z,q,h)$ easily.To find 4th term: $$G(z,q,h)=1-q^2+q^3\left( (zh+z^{-1}h^{-1}) - (zh^7+z^{-1}h^{-7}) \right) +a_4 q^4 +.... $$

$$G(z,q,h)(1+q^2+q^6+q^8+q(zh+z^{-1}h^{-1})+q^7(zh+z^{-1}h^{-1})+q^3(zh^7+z^{-1}h^{-7})+q^5(zh^7+z^{-1}h^{-7})+q^4(z^2h^8+z^{-2}h^{-8})+q^4(h^{6}+h^{-6}))...= 1+q (zh+z^{-1}h^{-1})+q^4 (z^2h^8+z^{-2}h^{-8})+.... $$

$$(1-q^2+q^3\left( (zh+z^{-1}h^{-1}) - (zh^7+z^{-1}h^{-7}) \right) +a_4 q^4 +.... ) (1+q^2+q^6+q^8+q(zh+z^{-1}h^{-1})+q^7(zh+z^{-1}h^{-1})+q^3(zh^7+z^{-1}h^{-7})+q^5(zh^7+z^{-1}h^{-7})+q^4(z^2h^8+z^{-2}h^{-8})+q^4(h^{6}+h^{-6}))...= 1+q (zh+z^{-1}h^{-1})+q^4 (z^2h^8+z^{-2}h^{-8})+.... $$

$a_4=-1+(z^2h^8+z^{-2}h^{-8})-(z^2h^2+z^{-2}h^{-2})$

Thus first 4 terms of $G(z,q,h)$ is:

$$G(z,q,h)=1-q^2+q^3\left( (zh+z^{-1}h^{-1}) - (zh^7+z^{-1}h^{-7}) \right) + q^4 \left(-1+(z^2h^8+z^{-2}h^{-8})-(z^2h^2+z^{-2}h^{-2}) \right) +.... $$

If we order it little bit more .

$$G(z,q,h)=1-q^2-q^4 +q^3\left( zh(1-h^6) + z^{-1}h^{-1}(1-h^{-6}) \right) + q^4 \left(z^2h^2(h^6-1)+z^{-2}h^{-2}(h^{-6}-1) \right) +.... $$

I will update If I find more terms of $G(z,q,h)$

EDIT: (Updated on 17th April)

I have found 5th term . $a_5= (zh+z^{-1}h^{-1}) - (z h^{19}+z^{-1}h^{-19}) +(z^3 h^3+z^{-3}h^{-3}) - (z^3 h^9+z^{-3}h^{-9})$

$$G(z,q,h)=1-q^2+q^3\left( (zh+z^{-1}h^{-1}) - (zh^7+z^{-1}h^{-7}) \right) + q^4 \left(-1+(z^2h^8+z^{-2}h^{-8})-(z^2h^2+z^{-2}h^{-2}) \right) + q^5\left( (zh+z^{-1}h^{-1}) - (z h^{19}+z^{-1}h^{-19}) +(z^3 h^3+z^{-3}h^{-3}) - (z^3 h^9+z^{-3}h^{-9}) \right)+ .... $$

$$G(z,q,h)=1-q^2-q^4 +q^3\left( zh(1-h^6) + z^{-1}h^{-1}(1-h^{-6}) \right) + q^4 \left(z^2h^2(h^6-1)+z^{-2}h^{-2}(h^{-6}-1) \right) + q^5 \left( zh(1-h^{18}) + z^{-1}h^{-1}(1-h^{-18})+z^3h^3(1-h^6)+z^{-3}h^{-3}(1-h^{-6}) \right)+.... $$

I have not seen a general pattern of the terms yet but I believe there is very beautiful pattern in it. If you help me to find more terms , I will be very appreciated. Maybe the pattern of terms of $G(z,q,h)$ can be seen more .Thanks.

EDIT: 27/01/2020

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

If we apply $z-->z^{-1}$ and $h-->h^{-1}$

$$B_n(q,h)=A_n(q,h^{-1})$$ $$C(q,h)=C(q,h^{-1})$$

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

If we apply

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

$q=Qh^{3}$

$$\frac{\sum\limits_{n = - \infty }^ \infty z^n q^{n^2} h^{n^3}}{\prod\limits_{n=1}^{ \infty }(1+zq^{2n-1}h^{3n^2-3n+1})(1+z^{-1}q^{2n-1}h^{-3n^2+3n-1})}= C(qh^3,h)+\sum\limits_{n = 2 }^ \infty \frac{A_{n-1}(qh^3,h)}{(1+zq^{2n-1}h^{3n^2-3n+1})}+\frac{A_1(qh^3,h^{-1})}{(1+z^{-1}q^{-1}h^{-1})}+\sum\limits_{n = 1 }^ \infty \frac{A_{n+1}(qh^3,h^{-1})}{(1+z^{-1}q^{2n-1}h^{-3n^2+3n-1})}$$

$$A_{n+1}(q,h)=A_{n}(qh^3,h)$$ for $n>0$ $$C(q,h)=C(qh^3,h)-\frac{A_1(q,h)}{(1+zqh)}+\frac{A_1(qh^3,h^{-1})}{(1+z^{-1}q^{-1}h^{-1})}$$

$$C(q,h)=C(qh^3,h)-\frac{A_1(q,h)}{(1+zqh)}+\frac{A_1(qh^3,h^{-1})zqh}{(1+zqh)}$$ $$A_1(q,h)=C(qh^3,h)-C(q,h)$$ $$A_n(q,h)=C(qh^{3n},h)-C(qh^{3(n-1)},h)$$

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

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

$q=qh^{-3/2}$

$$C(qh^{3/2},h)-C(qh^{-3/2},h)=\frac{\sum\limits_{n = - \infty }^ \infty (-1)^n q^{n^2-n} h^{-3/2(n^2-n)} h^{n^3-n}}{\prod\limits_{n=2}^{ \infty }(1-q^{2n-2}h^{-3(n-1)}h^{3n^2-3n})\prod\limits_{n=1}^{ \infty }(1-q^{2n}h^{-3n^2})}$$

$$C(qh^{3/2},h)-C(qh^{-3/2},h)=\frac{\sum\limits_{n = - \infty }^ \infty (-1)^n q^{n^2-n} h^{-3/2(n^2-n)} h^{n^3-n}}{\prod\limits_{n=1}^{ \infty }(1-q^{2n}h^{3n^2})\prod\limits_{n=1}^{ \infty }(1-q^{2n}h^{-3n^2})}$$

$$C(qh^{3/2},h)-C(qh^{-3/2},h)=\frac{\sum\limits_{n = - \infty }^ \infty (-1)^{n+1} q^{n(n+1)} h^{\frac{n(n+1)(2n+1)}{2}}}{\prod\limits_{n=1}^{ \infty }(1-q^{2n}h^{3n^2})\prod\limits_{n=1}^{ \infty }(1-q^{2n}h^{-3n^2})}$$

$q-->q^{1/2}$, $h-->h^{1/3}$

$$[C(\sqrt{q/h},\sqrt[3] {h})-C(\sqrt{qh},\sqrt[3] {h})]\prod\limits_{n=1}^{ \infty }(1-q^{n}h^{n^2})\prod\limits_{n=1}^{ \infty }(1-q^{n}h^{-n^2})=\sum\limits_{n = - \infty }^ \infty (-1)^{n} q^{\frac{n(n+1)}{2}} h^{\frac{n(n+1)(2n+1)}{6}}$$

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

I do not know how to find $C(q,h)$ but few terms of it : $$C(q,h)=C(q,h^{-1})=1-q^2-q^4+....$$

Maybe someone can help me to find more terms and its closed form.

Thanks a lot for helps

EDIT: 30/01/2020

I have new idea about zeros of $\sum\limits_{n = - \infty }^ \infty z^n q^{n^2} h^{n^3}$

The direction of this question will not give zeros for$\sum\limits_{n = - \infty }^ \infty z^n q^{n^2} h^{n^3}$ as @mr_e_man showed in his answer. Thus I will go in different way in another question.

I focused on zeros in a new question. My new question can be found here:

Solution 1:

$$\sum_{n\in\mathbb Z}z^nq^{n^2}h^{n^3}$$

By your conjecture, this function contains $(1+zqh)$ as a factor. Let's generalize a bit to $(1+zqh^c)$, for some constant $c$. But if that were true, then it would have to vanish when

$$z=-q^{-1}h^{-c},$$

that is,

$$0=\sum_{n\in\mathbb Z}(-1)^nq^{n^2-n}h^{n^3-nc}$$

$$=\cdots+q^6h^{-8+2c}-q^2h^{-1+c}+q^0h^0-q^0h^{1-c}+q^2h^{8-2c}-q^6h^{27-3c}+q^{12}h^{64-4c}-\cdots$$

$$=\big(1-h^{1-c}\big)q^0-\big(h^{-1+c}-h^{8-2c}\big)q^2+\big(h^{-8+2c}-h^{27-3c}\big)q^6-\cdots.$$

Since this is supposed to vanish for all $|q|<1$, all coefficients of powers of $q$ must vanish:

$$h^{1-c}=1,\quad h^{8-2c}=h^{-1+c},\quad h^{27-3c}=h^{-8+2c},\quad\cdots$$

$$h^{1-c}=1,\quad h^{9-3c}=1,\quad h^{35-5c}=1,\quad\cdots$$

and since $h=e^{i\theta}$ is also a variable, while the right sides of the equations are constant, the exponents must vanish:

$$c=1,\quad c=3,\quad c=7,\quad\cdots$$

which is clearly impossible. (More explicitly, if $h^p=e^{i\theta p}=1$, then $\theta p=n2\pi$ for some integer $n$; the right side takes discrete values, while the left side takes a continuum of values unless $p=0$.)

Therefore, $(1+zqh^c)$ is not a factor of the function.

By similar reasoning, for any constants $a,b,c$ with $a\neq0$ and either choice of sign, $(1\pm z^aq^bh^c)$ is not a factor. (There may be factors with $a=0$, but a product of these doesn't depend on $z$, while the original function does depend on $z$.)