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.

Gauss came up with some bizarre identities, namely $$ \sum_{n\in\mathbb{Z}}(-1)^nq^{n^2}=\prod_{k\geq 1}\frac{1-q^k}{1+q^k}. $$

How can this be interpreted combinatorially? It strikes me as being similar to many partition identities. Thanks.

share|improve this question
    
The first one (at least the $q^{n^2}$ part) just says that the divisors of a non-square can be divided into pairs $\{a,b\}$ such that $ab=1$. The second identity says something similar. –  André Nicolas Feb 11 '12 at 8:20
    
@André: Could you explain that a bit please? How do these pairs of divisors occur? And did you really mean $ab=1$? Modulo something? –  joriki Feb 20 '12 at 5:30
    
@Lando: If you don't already know them, you might be interested in the Jacobi triple product and the pentagonal number theorem. –  joriki Feb 20 '12 at 5:33
    
(Like Euler pentagonal theorem, this identity follows from Jacobi triple product.) –  Grigory M Dec 1 '13 at 13:14
add comment

2 Answers

Here's a combinatorial interpretation, but I have no idea how to turn it into a combinatorial proof.

$\prod(1+q^k)$ is the generating function that counts the number of partitions into distinct parts. $\prod(1-q^k)$ is the generating function that counts the excess of partitions into an even number of distinct parts over the partitions into an odd number of distinct parts. Thus

$$\sum_{n\in\mathbb{Z}}(-1)^nq^{n^2}\prod_{k\geq 1}\left(1+q^k\right)=\prod_{k\geq 1}\left(1-q^k\right)$$

considers partitions into a square and distinct parts and states that the excess of such partitions with an odd square over such partitions with an even square (where each non-zero square occurs in two colours, positive and negative) is equal to the excess of partitions into an even number of distinct parts over the partitions into an odd number of distinct parts.

share|improve this answer
    
Thanks joriki! My gut feeling is that a combinatorial proof might be quite difficult. –  Kally Feb 20 '12 at 22:37
add comment

There is a classical proof by Andrews which you can find in my survey here (section 5.5). There is also a bijective proof of a more general identity I gave in this paper (section 2.2). Enjoy!

share|improve this answer
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.