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How could I prove that $(-q;q^2)_\infty (q;q)_\infty = 1 + 2 \sum_{i=1}^\infty (-1)^i q^{2 i^2}?$ If that is too difficult is there a way to show $(-q;q^2)_\infty (q;q)_\infty \equiv 1 \pmod 2?$

edit The identity $(-q;q^2)_\infty (q;q)_\infty = (q^2;q^2)_\infty (q^2;q^4)_\infty$ is easily found and lets you substitute $\rho = q^2$ to reduce the problem to $(\rho;\rho)_\infty (\rho;\rho^2)_\infty = \sum_{i = -\infty}^{\infty} (-1)^{i} \rho^{i^2}.$


Definition: $(a;q)_n = \prod_{k=0}^n \left(1 - aq^k\right)$

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    @GEdgar: Usually for $q$-functions, we have the condition |q| < 1.2011-04-28

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Have you tried to find it as a specialization of the Jacobi triple product?

$(q;q)_{\infty}(-xq;q)_{\infty}(-1/x;q)_{\infty}=\sum_{k=-\infty}^{\infty}x^kq^{k(k+1)/2}$

The specialization $q \to q^2$ and $x \to -1/q$ gives your simplified version.

(Note that $(q^2;q^2)_{\infty}(q;q^2)_{\infty}= (1-q^2)(1-q^4)\dots(1-q)(1-q^3)\dots=(1-q)(1-q^2)\dots=(q;q)_{\infty}$. )

One common proof uses the pentagonal number theorem which has a nice combinatorial proof: http://en.wikipedia.org/wiki/Pentagonal_number_theorem, wikipedia lists another proof http://en.wikipedia.org/wiki/Jacobi_triple_product but I have not read it yet.

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    Thanks a lot! I've studied pentagonal number theorem so I will try to understand that proof.2011-04-28