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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.

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  • $\begingroup$ This a very interesting result, it's hard to check but I will see when I have some time $\endgroup$ – Elaqqad Apr 3 '15 at 12:47

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