In mathematics, the Jacobi triple product is the identity:
infty \left( | |
\prod | |
m=1 |
1-x2m\right) \left(1+x2m-1y2\right) \left(1+
x2m-1 | |
y2 |
\right) =
infty | |
\sum | |
n=-infty |
n2 | |
x |
y2n,
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.
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\sqrtq
y2=-\sqrt{q}
\phi(q)=
infty | |
\prod | |
m=1 |
\left(1-qm\right)
infty | |
= \sum | |
n=-infty |
(-1)n
| ||||
q |
.
The Rogers–Ramanujan identities follow with
x=q2\sqrtq
y2=-\sqrt{q}
x=q2\sqrtq
y2=-q\sqrt{q}
The Jacobi Triple Product also allows the Jacobi theta function to be written as an infinite product as follows:
Let
x=ei\pi
y=ei\pi.
Then the Jacobi theta function
\vartheta(z;\tau)=
infty | |
\sum | |
n=-infty |
e\pi
can be written in the form
infty | |
\sum | |
n=-infty |
y2n
n2 | |
x |
.
Using the Jacobi triple product identity, the theta function can be written as the product
\vartheta(z;\tau)=
infty \left( | |
\prod | |
m=1 |
1-e2m
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:
infty | |
\sum | |
n=-infty |
| ||||
q |
zn= (q;q)infty \left(-\tfrac{1}{z};q\right)infty (-zq;q)infty,
(a;q)infty
It enjoys a particularly elegant form when expressed in terms of the Ramanujan theta function. For
|ab|<1
infty | |
\sum | |
n=-infty |
| ||||
a |
| ||||
b |
=(-a;ab)infty (-b;ab)infty (ab;ab)infty.
Let
fx(y)=
infty | |
\prod | |
m=1 |
\left(1-x2m\right)\left(1+x2m-1y2\right)\left(1+x2m-1y-2\right)
Substituting for and multiplying the new terms out gives
fx(xy)=
1+x-1y-2 | |
1+xy2 |
fx(y)=x-1y-2fx(y)
Since
fx
|y|>0
fx(y)=\sum
infty | |
n=-infty |
2n | |
c | |
n(x)y |
which satisfies
infty | |
\sum | |
n=-infty |
2n+1 | |
c | |
n(x)x |
y2n=xfx(xy)=y-2fx(y)=\sum
infty | |
n=-infty |
cn+1(x)y2n
so that
cn+1(x)=
2n+1 | |
c | |
n(x)x |
=...=c0(x)
(n+1)2 | |
x |
and hence
fx(y)=c0(x)
infty | |
\sum | |
n=-infty |
n2 | |
x |
y2n
Showing that
c0(x)=1
y0
y=1
1{c | |
0(e |
2i\pi)}=
| ||||||||||||||
|
are weight 1/2 modular under
z\mapsto-
1 | |
4z |
c0(x)=c0(0)=1
A different proof is given by G. E. Andrews based on two identities of Euler.[1]
For the analytic case, see Apostol.[2]