$q$-analogs confusion in some steps

67 Views Asked by At

I am understanding the proof of the general version of the Jacobi triple product, that is $$\prod_{k=1}^{\infty}(1+xq^{2k-2})=\sum_{k=0}^{\infty}\frac{q^{k(k-1)}}{(q^2)_k}x^k$$

In the proof to this identity, there is this one step where it goes;

$$\sum_{l=0}^{\infty}(-1)^l \frac{q^{l^2+l+2kl}}{(q^2)_l}-\sum_{k=-\infty}^{\infty}q^{k^2}x^k\sum_{l=0}^{\infty}(-1)^l \frac{q^{l^2+l+2kl}}{(q^2)_l}=\sum_{l=0}^{\infty}\sum_{k=-\infty}^{\infty}\frac{(-q)^lq^{(k+l)^2}}{(q^2)_l}x^k$$

Can anyone help me with the understanding of this part, if I just understand this part, then I am nearly done with understanding the whole proof.

1

There are 1 best solutions below

1
On BEST ANSWER

I get $$\sum_{l=0}^\infty\sum_{k=-\infty}^\infty\frac{(-q)^lq^{(k+l)^2}}{(q^2)_l}x^k =\sum_{k=-\infty}^\infty \sum_{l=0}^\infty(-1)^l \frac{q^{k^2+2kl+l^2+l}}{(q^2)_l}x^k =\sum_{k=-\infty}^\infty q^{k^2}x^k \sum_{l=0}^\infty(-1)^l \frac{q^{l^2+l+2kl}}{(q^2)_l}$$ which isn't what you have on the left, but maybe should be?