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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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I'm sorry, could you define your notation here? Maybe everyone else knows it, and if so then it's fine, but I'm not sure about it. Sorry if this is stupid. – BBischof Apr 28 '11 at 14:16
@BBischof, added it now! – quanta Apr 28 '11 at 14:18
What is q ? If a real number (for convergence of the series) then what does $\equiv 1 \pmod 2$ mean? – GEdgar Apr 28 '11 at 14:25
$q$ is a variable, the congruence notation means we're working in $\mathbb Z[q]/2\mathbb Z[q]$. – quanta Apr 28 '11 at 14:26
@GEdgar: Usually for $q$-functions, we have the condition $|q| < 1$. – J. M. Apr 28 '11 at 16:20
up vote 6 down vote accepted

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


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:, wikipedia lists another proof but I have not read it yet.

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Doesn't your substitution give $(q^2;q^2)_{\infty}(q;q^2)_{\infty}(q;q^2)_{\infty}$ rather than the $\rho$ term I have (I don't see how these are equal)? – quanta Apr 28 '11 at 16:09
Thanks a lot! I've studied pentagonal number theorem so I will try to understand that proof. – quanta Apr 28 '11 at 16:20

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