Triple Product Identity, Gauss

\displaystyle \sum_{n=-\infty}^{\infty} q^{n^{2}} x^{n}=\prod_{n=1}^{\infty}\left(1+x q^{2 n-1}\right)\left(1+x^{-1} q^{2 n-1}\right)\left(1-q^{2 n}\right)

Here is a proof of the Jacobi’s Triple Product Identity due to Gauss:

Consider

\displaystyle F(n)=1+\frac{a^{n}-1}{a-1} t+\frac{\left(a^{n}-1\right)\left(a^{n}-a\right)}{(a-1)\left(a^{2}-1\right)} t^{2}+\cdots+\frac{\left(a^{n}-1\right)\left(a^{n}-a\right) \cdots\left(a^{n}-a^{n-1}\right)}{(a-1)\left(a^{2}-1\right) \cdots\left(a^{n}-1\right)} t^{n}

Notice that

\displaystyle (1+a^nt)F(n+1)=1+a^{n} t+\frac{a^{n}-1}{a-1} t+\frac{a^{n}\left(a^{n}-1\right)}{a-1} t^{2}+\frac{\left(a^{n}-1\right)\left(a^{n}-a\right)}{(a-1)\left(a^{2}-1\right)} t^{2}+ \cdots+ \quad\quad \quad \frac{a^{n}\left(a^{n}-1\right)\left(a^{n}-a\right) \cdots\left(a^{n}-a^{n-1}\right)}{(a-1)\left(a^{2}-1\right) \cdots\left(a^{n}-1\right)} t^{n+1}

\displaystyle =1+\frac{a^{n+1}-1}{a-1}t+\frac{\left(a^{n+1}-1\right)\left(a^{n+1}-a\right)}{(a-1)\left(a^{2}-1\right)} t^{2}+\cdots+\frac{\left(a^{n+1}-1\right)\left(a^{n+1}-a\right) \cdots\left(a^{n+1}-a^{n}\right)}{(a-1)\left(a^{2}-1\right) \cdots\left(a^{n+1}-1\right)} t^{n+1}

\displaystyle = F(n+1)

Thus by repeated application of the above identity, we get \displaystyle F(n) =(1+t)(1+a t)\left(1+a^{2} t\right) \cdots\left(1+a^{n-1} t\right)

And so we proved the identity

\displaystyle 1+\frac{a^{n}-1}{a-1} t+\frac{\left(a^{n}-1\right)\left(a^{n}-a\right)}{(a-1)\left(a^{2}-1\right)}t^{2}+\cdots+\frac{\left(a^{n}-1\right)\left(a^{n}-a\right) \cdots\left(a^{n}-a^{m-1}\right)}{(a-1)\left(a^{2}-1\right) \cdots\left(a^{m}-1\right)} t^{m}+ \cdots+\frac{\left(a^{n}-1\right)\left(a^{n}-a\right) \cdots\left(a^{n}-a^{n-1}\right)}{(a-1)\left(a^{2}-1\right) \cdots\left(a^{n}-1\right)} t^{n} =(1+t)(1+a t)\left(1+a^{2} t\right) \cdots\left(1+a^{n-1} t\right)

Dividing this identity by

\displaystyle \frac{\left(a^{n}-1\right)\left(a^{n}-a\right) \cdots\left(a^{n}-a^{m-1}\right)}{(a-1)\left(a^{2}-1\right) \cdots\left(a^{m}-1\right)} t^{m} where m =n/2 (Assume n is even)

and substituting \displaystyle a = q^2, t =q^{1-n}x, we get

1+\frac{1-q^{n}}{1-q^{n+2}}q\left(x+x^{-1}\right)+\frac{\left(1-q^{n}\right)\left(1-q^{n-2}\right)}{\left(1-q^{n+2}\right)\left(1-q^{n+4}\right)} q^{4}\left(x^{2}+x^{-2}\right)+ \cdots+\frac{\left(1-q^{n}\right)\left(1-q^{n-2}\right) \cdots\left(1-q^{2}\right)}{\left(1-q^{n+2}\right)\left(1-q^{n+4}\right) \cdots\left(1-q^{2 n}\right)} q^{(n / 2)^{2}}\left(x^{n / 2}+x^{-n / 2}\right)

=(1+qx)\left(1+q x^{-1}\right)\left(1+q^{3} x\right)\left(1+q^{3} x^{-1}\right) \cdots\left(1+q^{n-1} x\right)\left(1+q^{n-1} x^{-1}\right) \times \frac{\left(1-q^{2}\right)\left(1-q^{4}\right) \cdots\left(1-q^{n}\right)}{\left(1-q^{n+2}\right)\left(1-q^{n+4}\right) \cdots\left(1-q^{2 n}\right)}

Taking n \to \infty, we get the triple product identity

\displaystyle \sum_{n=-\infty}^{\infty} q^{n^{2}} x^{n}=\prod_{n=1}^{\infty}\left(1+x q^{2 n-1}\right)\left(1+x^{-1} q^{2 n-1}\right)\left(1-q^{2 n}\right)

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