Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
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
2  
@GEdgar: Usually for $q$-functions, we have the condition $|q| < 1$. –  J. M. Apr 28 '11 at 16:20
add comment

1 Answer

up vote 4 down vote accepted

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.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.