title: Jacobi triple product
text: # Jacobi triple product
  
  In mathematics, the Jacobi triple product is the mathematical 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 $A_1,$ 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 $y^{2}=-{\sqrt {q}}$. 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 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 we can then write the theta function 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 $c_{0}(x)=1$ 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.
  
  For the analytic case, see Apostol.
  
  