Jacobi Triple Product

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

:$\prod_^\infty \left( 1 - x^\right) \left( 1 + x^ y^2\right) \left( 1 +\frac\right) = \sum_^\infty x^ y^,$
for complex numbers ''x'' and ''y'', with , ''x'', < 1 and ''y'' ≠ 0.

It was introduced by in his work ''Fundamenta Nova Theoriae Functionum Ellipticarum''.

The Jacobi triple product identity is the Macdonald identity for the affine root system of type ''A''₁, 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 $y^2=-\sqrt$. Then we have
:$\phi(q) = \prod_^\infty \left(1-q^m \right) = \sum_^\infty (-1)^n q^.$

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

Let $x=e^$ and $y=e^.$

Then the Jacobi theta function

:$\vartheta(z; \tau) = \sum_^\infty e^$

can be written in the form

:$\sum_^\infty y^x^.$

Using the Jacobi Triple Product Identity we can then write the theta function as the product

:\vartheta(z; \tau) = \prod_^\infty
\left( 1 - e^\right)
\left 1 + e^\right\left 1 + e^\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:

:$\sum_^\infty q^z^n = (q;q)_\infty \; \left(-\tfrac;q\right)_\infty \; (-zq;q)_\infty,$
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

:$\sum_^\infty a^ \; b^ = (-a; ab)_\infty \;(-b; ab)_\infty \;(ab;ab)_\infty.$

Proof

Let $f_x(y) = \prod_^\infty \left( 1 - x^ \right)\left( 1 + x^ y^2\right)\left( 1 +x^y^\right)$

Substituting for and multiplying the new terms out gives

:$f_x(xy) = \fracf_x(y) = x^y^f_x(y)$

Since $f_x$ is meromorphic for $, y, > 0$, it has a Laurent series

:$f_x(y)=\sum_^\infty c_n(x)y^$

which satisfies

:$\sum_^\infty c_n(x)x^ y^=x f_x(x y)=y^f_x(y)=\sum_^\infty c_(x)y^$

so that

:$c_(x) = c_n(x)x^ = \dots = c_0(x) x^$

and hence

:$f_x(y)=c_0(x) \sum_^\infty x^ y^$

Evaluating

Showing that $c_0(x) =1$ is technical. One way is to set $y= 1$ and show both the numerator and the denominator of
:$\frac1 =\frac$

are weight 1/2 modular under $z\mapsto -\frac$, 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.Chapter 14, theorem 14.6 of

References

* Peter J. Cameron
''Combinatorics: Topics, Techniques, Algorithms''
(1994) Cambridge University Press,
*
*
*{{Citation , last=W first=E. M., title= An Enumerative Proof of An Identity of Jacobi, journal=Journal of the London Mathematical Society, pages=55–57, publisher=London Mathematical Society, year=1965, doi=10.1112/jlms/s1-40.1.55

Elliptic functions
Theta functions
Mathematical identities
Theorems in number theory
Infinite products