# Extension of the Jacobi triple product identity

The Jacobi triple product and the mathematical identity of it 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

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

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

$q=Qh^{3}$

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

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

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

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

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

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

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

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

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

If we continue the rule, we can get $$G(ZQ^{2}h^{3},Qh^{3},h)=G(ZQ^{2}h^{3}Q^2h^{6}h^{3},Qh^{3}h^{3},h)=G(ZQ^{4}h^{12},Qh^{6},h)$$

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

I 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 $z=x^3$,$q=x^3$,$h=x$

$$G(x^{12},x^6,x)=G(x^3,x^3,x)=G(1,1,x) \tag 2$$

I thought If I can find a few term of $G(z,q,h)$ by hand and maybe it can be seen what the pattern of $G(z,q,h)$ has. I 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 $G(z,q,h)$ are:

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

If we order it little bit .

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

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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Try expanding $G$ as a series in positive powers of $q$, with coefficients that are Laurent polynomials in $z$ and $h$. –  S. Carnahan Apr 12 '13 at 11:00
if you typed all this up yourself i can see you do love math. –  John Jiang Apr 13 '13 at 5:23
@John Jiang : I love mathematics very much. I wrote my works above. I wanted to share it. I thought maybe someone can give a hand to solve it or can give a link about the subject. –  Mathlover Apr 15 '13 at 6:23
2) For a conceptual setting, consider the work of Victor Kac (and later Howard Garland and Jim Lepowsky) which explained in a nice way the formal derivaiton by Ian G. Macdonald of identities involving Dedekind's $\eta$-function: these arise naturally in the representation theory of affine Lie algebras. In particular, Jacobi's triple product follows from the study of the affine Lie algebra built on a three dimensional simple Lie algebra.