Casos de producto Triple de Jacobi

Question

Hola quiero una guía Cómo puedo probar estas identidades del producto triple de Jacobi

$$\sum_{n=-\infty}^\infty q^{2n^2+n}=\prod_{n=1}^\infty\frac{(1-q^{2n})^2}{(1-q^n)}$$

$$\sum_{n=-\infty}^\infty q^{n^2}=\prod_{n=1}^\infty\frac{(1-(-q)^n)}{(1+(-q)^n)}$$

$$\sum_{n=-\infty}^\infty q^{n(n+1)}=\prod_{n=1}^\infty\frac{(1-q^{4n})}{(1-q^{4n-2})}$$

Answer 1

Comience con la identidad del triple producto

$$\sum_{n = -\infty}^\infty z^n q^{n^2} = \prod_{n = 1}^\infty (1 - q^{2n})(1 + zq^{2n-1})(1 + z^{-1}q^{2n-1}).$$

1. Sustituir el $q$ $q^2$, luego del ajuste del $z = q$, obtenemos

\begin{align} \sum_{n = -\infty}^\infty q^{2n^2+n} &= \prod_{n = 1}^\infty (1 - q^{4n})(1 + q^{4n-1})(1 + q^{4n-3})\\ & = \prod_{n = 1}^\infty (1 - q^{4n})(1 + q^{2n-1})\\ & = \prod_{n = 1}^\infty (1 - q^{2n})(1 + q^{2n})(1 + q^{2n-1})\\ & = \prod_{n = 1}^\infty (1 - q^{2n})(1 + q^n)\\ & = \prod_{n = 1}^\infty \frac{(1 - q^{2n})^2}{1 - q^n}. \end {Alinee el}

2. Ajuste $z = 1$, obtenemos

\begin{align}\sum_{n = -\infty}^\infty q^{n^2} &= \prod_{n = 1}^\infty (1 - q^{2n})(1 + q^{2n-1})(1 + q^{2n-1})\\ &= \prod_{n = 1}^\infty (1 - (-q)^n)\prod_{n\, \text{odd}}^\infty (1 + q^n)\\ &= \prod_{n = 1}^\infty (1 - (-q)^n)\prod_{n = 1}^\infty \frac{1 + q^n}{1 + q^{2n}}\\ &= \prod_{n = 1}^\infty (1 - (-q)^n)\prod_{n = 1}^\infty \frac{1 - q^{2n}}{(1 - q^n)(1 + q^{2n})}\\ &= \prod_{n = 1}^\infty (1 - (-q)^n)\prod_{n = 1}^\infty \frac{1 - q^{2n}}{(1 - q^{2n})(1 - q^{2n-1})(1 + q^{2n})}\\ &= \prod_{n = 1}^\infty \frac{1 - (-q)^n}{(1 - q^{2n-1})(1 + q^{2n})}\\ &= \prod_{n = 1}^\infty \frac{1 - (-q)^n}{1 + (-q)^n}. \end {Alinee el}

3. Ajuste de resultados de $z = q$

\begin{align}\sum_{n = -\infty}^\infty q^{n^2 + n} &= \prod_{n = 1}^\infty (1 - q^{2n})(1 + q^{2n})(1 + q^{2n-2})\\ & = \prod_{n = 1}^\infty (1 - q^{4n})\prod_{n = 2}^\infty (1 + q^{2n-2})\\ & = \prod_{n = 1}^\infty (1 - q^{4n}) \prod_{n = 1}^\infty (1 + q^{2n})\\ & = \prod_{n = 1}^\infty (1 - q^{4n}) \prod_{n = 1}^\infty \frac{1 - q^{4n}}{1 - q^{2n}}\\ & = \prod_{n = 1}^\infty (1 - q^{4n}) \prod_{n = 1}^\infty \frac{1 - q^{4n}}{(1 - q^{4n})(1 - q^{4n-2})}\\ &= \prod_{n = 1}^\infty \frac{1 - q^{4n}}{1 - q^{4n-2}}. \end {Alinee el}