In mathematics, the Jacobi triple product is the identity:
∏ m = 1 ∞ ( 1 − x 2 m ) ( 1 + x 2 m − 1 y 2 ) ( 1 + x 2 m − 1 y 2 ) = ∑ n = − ∞ ∞ x n 2 y 2 n , {\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},}
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 q {\displaystyle x=q{\sqrt {q}}} and y 2 = − q {\displaystyle y^{2}=-{\sqrt {q}}} . Then we have
ϕ ( q ) = ∏ m = 1 ∞ ( 1 − q m ) = ∑ n = − ∞ ∞ ( − 1 ) n q 3 n 2 − n 2 . {\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 x = q 2 q {\displaystyle x=q^{2}{\sqrt {q}}} , y 2 = − q {\displaystyle y^{2}=-{\sqrt {q}}} and x = q 2 q {\displaystyle x=q^{2}{\sqrt {q}}} , y 2 = − q 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 x = e i π τ {\displaystyle x=e^{i\pi \tau }} and y = e i π z . {\displaystyle y=e^{i\pi z}.}
Then the Jacobi theta function
ϑ ( z ; τ ) = ∑ n = − ∞ ∞ e π i n 2 τ + 2 π i n z {\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
∑ n = − ∞ ∞ y 2 n x n 2 . {\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
ϑ ( z ; τ ) = ∏ m = 1 ∞ ( 1 − e 2 m π i τ ) [ 1 + e ( 2 m − 1 ) π i τ + 2 π i z ] [ 1 + e ( 2 m − 1 ) π i τ − 2 π i z ] . {\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:
∑ n = − ∞ ∞ q n ( n + 1 ) 2 z n = ( q ; q ) ∞ ( − 1 z ; q ) ∞ ( − z q ; q ) ∞ , {\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 ( a ; q ) ∞ {\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 | a b | < 1 {\displaystyle |ab|<1} it can be written as
∑ n = − ∞ ∞ a n ( n + 1 ) 2 b n ( n − 1 ) 2 = ( − a ; a b ) ∞ ( − b ; a b ) ∞ ( a b ; a b ) ∞ . {\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 f x ( y ) = ∏ m = 1 ∞ ( 1 − x 2 m ) ( 1 + x 2 m − 1 y 2 ) ( 1 + x 2 m − 1 y − 2 ) {\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
f x ( x y ) = 1 + x − 1 y − 2 1 + x y 2 f x ( y ) = x − 1 y − 2 f x ( y ) {\displaystyle 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 {\displaystyle f_{x}} is meromorphic for | y | > 0 {\displaystyle |y|>0} , it has a Laurent series
f x ( y ) = ∑ n = − ∞ ∞ c n ( x ) y 2 n {\displaystyle f_{x}(y)=\sum _{n=-\infty }^{\infty }c_{n}(x)y^{2n}}
which satisfies
∑ n = − ∞ ∞ c n ( x ) x 2 n + 1 y 2 n = x f x ( x y ) = y − 2 f x ( y ) = ∑ n = − ∞ ∞ c n + 1 ( x ) y 2 n {\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
c n + 1 ( x ) = c n ( x ) x 2 n + 1 = ⋯ = c 0 ( x ) x ( n + 1 ) 2 {\displaystyle 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 ) ∑ n = − ∞ ∞ x n 2 y 2 n {\displaystyle f_{x}(y)=c_{0}(x)\sum _{n=-\infty }^{\infty }x^{n^{2}}y^{2n}}
Evaluating c0(x) To show that c 0 ( x ) = 1 {\displaystyle c_{0}(x)=1} , use the fact that the infinite expansion
∏ m = 1 ∞ ( 1 + x 2 m − 1 y 2 ) ( 1 + x 2 m − 1 y − 2 ) {\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 y 0 {\displaystyle y^{0}}
= ∑ m = 0 ∞ x 2 m 2 ( 1 − x 2 ) 2 ( 1 − x 4 ) 2 ⋯ ( 1 − x 2 m ) 2 {\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 x 2 {\displaystyle x^{2}} instead of x {\displaystyle x} .
= ∏ m = 1 ∞ ( 1 − x 2 m ) − 1 {\displaystyle =\prod _{m=1}^{\infty }\left(1-x^{2m}\right)^{-1}}
Therefore at y 0 {\displaystyle y^{0}} we have f x ( y ) = 1 {\displaystyle f_{x}(y)=1} , and so c 0 ( x ) = 1 {\displaystyle c_{0}(x)=1} .
Other proofs A different proof is given by G. E. Andrews based on two identities of Euler. For the analytic case, see Apostol.
References
Further reading Peter J. Cameron, Combinatorics: Topics, Techniques, Algorithms, (1994) Cambridge University Press, ISBN 0-521-45761-0 Jacobi, C. G. J. (1829), Fundamenta nova theoriae functionum ellipticarum (in Latin), Königsberg: Borntraeger, ISBN 978-1-108-05200-9, Reprinted by Cambridge University Press 2012 {{citation}}: ISBN / Date incompatibility (help) Carlitz, L (1962), "A note on the Jacobi theta formula", Bulletin of the American Mathematical Society, vol. 68, no. 6, American Mathematical Society, pp. 591–592 Wright, E. M. (1965), "An Enumerative Proof of An Identity of Jacobi", Journal of the London Mathematical Society, London Mathematical Society: 55–57, doi:10.1112/jlms/s1-40.1.55