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In mathematics, the Jacobi triple product is the 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

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

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

Let $x=e^{i\pi \tau }$ 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, the theta function can be written 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 $c 0 (x)$

Showing that $c_{0}(x)=1$ (the polynomial of x of $y^{0}$ is 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.^[1]

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

References

^ Andrews, George E. (1965-02-01). "A simple proof of Jacobi's triple product identity". Proceedings of the American Mathematical Society. 16 (2): 333. doi:10.1090/S0002-9939-1965-0171725-X. ISSN 0002-9939.
^ Chapter 14, theorem 14.6 of Apostol, Tom M. (1976), Introduction to analytic number theory, Undergraduate Texts in Mathematics, New York-Heidelberg: Springer-Verlag, ISBN 978-0-387-90163-3, MR 0434929, Zbl 0335.10001