Jacobi triple product

Remarks

Proof idea: call the product $F(z)$; it is holomorphic on $\C \setminus \set{0}$, so it has a Laurent expansion $\sum_n c_n z^n$. The product satisfies $F(q^2 z) = (q z)^{-1} F(z)$, which forces $c_n = q^{n^2} c_0$. The constant $c_0 = 1$ can be found by specializing $z$, for instance with Gauss's or Euler's identities, or by a limit argument. Example: substituting $q \mapsto q^{3/2}$ and $z \mapsto -q^{1/2}$ in the first form gives Euler's pentagonal number theorem $(q; q)_\infty = \sum_{n \in \Z} (-1)^n q^{n(3n-1)/2}$. With $z = 1$ the first form gives $\sum_{n \in \Z} q^{n^2} = \prod_{m \ge 1} (1 - q^{2m})(1 + q^{2m-1})^2$, a product for a theta function. The condition $\abs{q} < 1$ is needed for the infinite product and the series to converge.

Going deeper

For a student

It is the basic identity behind the product formulas for the Jacobi theta functions and the Dedekind eta function. It is a limiting case of Ramanujan's 1psi1 summation. Together with its relative, the quintuple product identity, it gives many partition identities, and it is a step in proofs of the Rogers–Ramanujan identities. In Lie theory it is the Macdonald identity for the affine root system of type $A_1^{(1)}$.