# Jacobi triple product

In mathematics, the Jacobi triple product is the mathematical identity:

$\prod _{m=1}^{\infty }\left(1-x^{2m}\right)\left(1+x^{2m-1}y^{2}\right)\left(1+{\frac {x^{2m-1}}{y^{2}}}\right)=\sum _{n=-\infty }^{\infty }x^{n^{2}}y^{2n},$ for complex numbers x and y, with |x| < 1 and y ≠ 0.

It was introduced by Jacobi (1829) in his work Fundamenta Nova Theoriae Functionum Ellipticarum.

The Jacobi triple product identity is the Macdonald identity for the affine root system of type A1, and is the Weyl denominator formula for the corresponding affine Kac–Moody algebra.

## Properties

The basis of Jacobi's proof relies on Euler's pentagonal number theorem, which is itself a specific case of the Jacobi Triple Product Identity.

Let $x=q{\sqrt {q}}$ and $y^{2}=-{\sqrt {q}}$ . Then we have

$\phi (q)=\prod _{m=1}^{\infty }\left(1-q^{m}\right)=\sum _{n=-\infty }^{\infty }(-1)^{n}q^{\frac {3n^{2}-n}{2}}.$ The Jacobi Triple Product also allows the Jacobi theta function to be written as an infinite product as follows:

Let $x=e^{i\pi \tau }$ and $y=e^{i\pi z}.$ Then the Jacobi theta function

$\vartheta (z;\tau )=\sum _{n=-\infty }^{\infty }e^{\pi {\rm {i}}n^{2}\tau +2\pi {\rm {i}}nz}$ can be written in the form

$\sum _{n=-\infty }^{\infty }y^{2n}x^{n^{2}}.$ Using the Jacobi Triple Product Identity we can then write the theta function as the product

$\vartheta (z;\tau )=\prod _{m=1}^{\infty }\left(1-e^{2m\pi {\rm {i}}\tau }\right)\left[1+e^{(2m-1)\pi {\rm {i}}\tau +2\pi {\rm {i}}z}\right]\left[1+e^{(2m-1)\pi {\rm {i}}\tau -2\pi {\rm {i}}z}\right].$ There are many different notations used to express the Jacobi triple product. It takes on a concise form when expressed in terms of q-Pochhammer symbols:

$\sum _{n=-\infty }^{\infty }q^{\frac {n(n+1)}{2}}z^{n}=(q;q)_{\infty }\;\left(-{\tfrac {1}{z}};q\right)_{\infty }\;(-zq;q)_{\infty },$ where $(a;q)_{\infty }$ is the infinite q-Pochhammer symbol.

It enjoys a particularly elegant form when expressed in terms of the Ramanujan theta function. For $|ab|<1$ it can be written as

$\sum _{n=-\infty }^{\infty }a^{\frac {n(n+1)}{2}}\;b^{\frac {n(n-1)}{2}}=(-a;ab)_{\infty }\;(-b;ab)_{\infty }\;(ab;ab)_{\infty }.$ ## Proof

Let $f_{x}(y)=\prod _{m=1}^{\infty }\left(1-x^{2m}\right)\left(1+x^{2m-1}y^{2}\right)\left(1+x^{2m-1}y^{-2}\right)$ Substituting xy for y and multiplying the new terms out gives

$f_{x}(xy)={\frac {1+x^{-1}y^{-2}}{1+xy^{2}}}f_{x}(y)=x^{-1}y^{-2}f_{x}(y)$ Since $f_{x}$ is meromorphic for $|y|>0$ , it has a Laurent series

$f_{x}(y)=\sum _{n=-\infty }^{\infty }c_{n}(x)y^{2n}$ which satisfies

$\sum _{n=-\infty }^{\infty }c_{n}(x)x^{2n+1}y^{2n}=xf_{x}(xy)=y^{-2}f_{x}(y)=\sum _{n=-\infty }^{\infty }c_{n+1}(x)y^{2n}$ so that

$c_{n+1}(x)=c_{n}(x)x^{2n+1}=\dots =c_{0}(x)x^{(n+1)^{2}}$ and hence

$f_{x}(y)=c_{0}(x)\sum _{n=-\infty }^{\infty }x^{n^{2}}y^{2n}$ ### Evaluating c0(x)

Showing that $c_{0}(x)=1$ is technical. One way is to set $y=1$ and show both the numerator and the denominator of

${\frac {1}{c_{0}(e^{2i\pi z})}}={\frac {\sum \limits _{n=-\infty }^{\infty }e^{2i\pi n^{2}z}}{\prod \limits _{m=1}^{\infty }(1-e^{2i\pi mz})(1+e^{2i\pi (2m-1)z})^{2}}}$ are weight 1/2 modular under $z\mapsto -{\frac {1}{4z}}$ , since they are also 1-periodic and bounded on the upper half plane the quotient has to be constant so that $c_{0}(x)=c_{0}(0)=1$ .

### Other proofs

A different proof is given by G. E. Andrews based on two identities of Euler.

For the analytic case, see Apostol.