Jacobi triple product

In mathematics, the Jacobi triple product is the identity:

${\displaystyle \prod _{m=1}^{\infty }\left(1-x^{2m}\right)\left(1+x^{2m-1}y^{2}\right)\left(1+{\frac {x^{2m-1}}{y^{2}}}\right)=\sum _{n=-\infty }^{\infty }x^{n^{2}}y^{2n},}$

"Triple product identity" redirects here. For the ternary operation on vector spaces, see Triple product.

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

Jacobi's proof relies on Euler's pentagonal number theorem, which is itself a specific case of the Jacobi triple product identity.

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

${\displaystyle \phi (q)=\prod _{m=1}^{\infty }\left(1-q^{m}\right)=\sum _{n=-\infty }^{\infty }(-1)^{n}q^{\frac {3n^{2}-n}{2}}.}$

The Rogers–Ramanujan identities follow with ${\displaystyle x=q^{2}{\sqrt {q}}}$, ${\displaystyle y^{2}=-{\sqrt {q}}}$ and ${\displaystyle x=q^{2}{\sqrt {q}}}$, ${\displaystyle y^{2}=-q{\sqrt {q}}}$.

The Jacobi Triple Product also allows the Jacobi theta function to be written as an infinite product as follows:

Let ${\displaystyle x=e^{i\pi \tau }}$ and ${\displaystyle y=e^{i\pi z}.}$

Then the Jacobi theta function

${\displaystyle \vartheta (z;\tau )=\sum _{n=-\infty }^{\infty }e^{\pi {\rm {i}}n^{2}\tau +2\pi {\rm {i}}nz}}$

can be written in the form

${\displaystyle \sum _{n=-\infty }^{\infty }y^{2n}x^{n^{2}}.}$

Using the Jacobi triple product identity, the theta function can be written as the product

${\displaystyle \vartheta (z;\tau )=\prod _{m=1}^{\infty }\left(1-e^{2m\pi {\rm {i}}\tau }\right)\left[1+e^{(2m-1)\pi {\rm {i}}\tau +2\pi {\rm {i}}z}\right]\left[1+e^{(2m-1)\pi {\rm {i}}\tau -2\pi {\rm {i}}z}\right].}$

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:

${\displaystyle \sum _{n=-\infty }^{\infty }q^{\frac {n(n+1)}{2}}z^{n}=(q;q)_{\infty }\;\left(-{\tfrac {1}{z}};q\right)_{\infty }\;(-zq;q)_{\infty },}$

where ${\displaystyle (a;q)_{\infty }}$ is the infinite q-Pochhammer symbol.

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

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

Proof

Let ${\displaystyle f_{x}(y)=\prod _{m=1}^{\infty }\left(1-x^{2m}\right)\left(1+x^{2m-1}y^{2}\right)\left(1+x^{2m-1}y^{-2}\right)}$

Substituting xy for y and multiplying the new terms out gives

${\displaystyle f_{x}(xy)={\frac {1+x^{-1}y^{-2}}{1+xy^{2}}}f_{x}(y)=x^{-1}y^{-2}f_{x}(y)}$

Since ${\displaystyle f_{x}}$ is meromorphic for ${\displaystyle |y|>0}$, it has a Laurent series

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

which satisfies

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

so that

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

and hence

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

Evaluating c₀(x)

To show that ${\displaystyle c_{0}(x)=1}$, use the fact that the infinite expansion

${\displaystyle \prod _{m=1}^{\infty }\left(1+x^{2m-1}y^{2}\right)\left(1+x^{2m-1}y^{-2}\right)}$

has the following infinite polynomial coefficient at ${\displaystyle y^{0}}$

${\displaystyle =\sum _{m=0}^{\infty }{\frac {x^{2m^{2}}}{(1-x^{2})^{2}(1-x^{4})^{2}\cdots (1-x^{2m})^{2}}}}$

which is the Durfee square generating function with ${\displaystyle x^{2}}$ instead of ${\displaystyle x}$.

${\displaystyle =\prod _{m=1}^{\infty }\left(1-x^{2m}\right)^{-1}}$

Therefore at ${\displaystyle y^{0}}$we have ${\displaystyle f_{x}(y)=1}$, and so ${\displaystyle c_{0}(x)=1}$.

Other proofs

A different proof is given by G. E. Andrews based on two identities of Euler.^[1]

For the analytic case, see Apostol.^[2]