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.
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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 08:20
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@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 05:30
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@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 05:33
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(Like Euler pentagonal theorem, this identity follows from Jacobi triple product.) – Grigory M Dec 01 '13 at 13:14
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@GrigoryM Do you know where I can see it derived from the Jacobi triple product? – User Feb 21 '16 at 06:24
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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!
Igor Pak
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Also, two little typos in §5.5.3: The $\prod\limits_{n=0}^\infty$ on the LHS should be a $\prod\limits_{m=1}^\infty$. – darij grinberg Jul 31 '17 at 13:02
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@IgorPak I just read your survey and checked the bijection. There are two situations I'd like to clarify. First, when $s<f$ we add a column to the right, but then it is possible that $s=f=q$ and it does not return. For example, $8=2+2+2+2$, $s=2$, $f=4$, the lower right square is marked; its involution is $3+3+2$ with $s=f=q=2$, and this diagram's involution is $4+4$, right? Second, you said $r\times (r+1)$ unmarked and $(r+1)\times r$ marked have undefined $\alpha$. But it seems the involution of $3\times 4$ unmarked (f=3<4=s) is just $4\times 3$ marked (f=4>3=s), right? – Haoran Chen Sep 08 '24 at 15:13
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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.
joriki
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Thanks joriki! My gut feeling is that a combinatorial proof might be quite difficult. – Kally Feb 20 '12 at 22:37
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The typical analytic proof is not difficult and is an easy consequence of Jacobi's triple product $$\sum_{n=-\infty} ^{\infty} z^{n} q^{n^{2}}=\prod_{n=1}^{\infty}(1-q^{2n})(1+zq^{2n-1})(1+z^{-1}q^{2n-1})$$ for all $z, q$ with $z\neq 0,|q|<1$. Let's put $z=-1$ to get the sum in question. The corresponding product is equal to $$\prod(1-q^{2n})(1-q^{2n-1})^{2}=\prod(1-q^{n})(1-q^{2n-1})=\prod \frac{(1-q^{2n})(1-q^{2n-1})} {1+q^{n}}=\prod\frac{1-q^{n}} {1+q^{n}}$$ which completes the proof.
The proof for Jacobi's triple product is non-trivial / non-obvious and you may have a look at the original proof by Jacobi in this blog post.
On the other hand Franklin obtained a nice and easy combinatorial proof of the Euler's Pentagonal theorem which is equivalent to Jacobi Triple Product.
Paramanand Singh
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It seems you prove in quite an elementary way an equivalent form there. Also for a periodic function analytic on $\Im(z) \in (a,b)$ the Fourier series converging there is unique, in the same way that a Laurent expansion on $|z| \in (r,R)$ is unique – reuns Aug 20 '17 at 19:41
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I think there is a problem in your proofs : $a_0$ (or $G$) is a function of $q$, not a constant. – reuns Aug 23 '17 at 04:00
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@reuns: $G$ is further expressed in terms of another function $A(q) $ times the product $\prod(1-q^{2n})$ and then we show that $A(q) $ is constant and equal to $1$. It a standard proof taken from standard sources and unless I have typos it can't be wrong. At least I did not find any fault with it. – Paramanand Singh Aug 23 '17 at 04:07
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When looking at the Fourier series of Jacobi triple product I obtain $\prod_{n=1}^\infty \frac{1-q^n}{1+q^n}=a_0(q)\sum_{n=-\infty}^\infty (-1)^n q^{n^2}$ for $|q| < 1$. How do you show $a_0(q)$ is constant ? – reuns Aug 23 '17 at 04:11
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@reuns: see the post http://paramanands.blogspot.com/2011/02/elliptic-functions-theta-functions-contd.html where this is shown to be constant. This part is tricky/non-obvious and I believe is done by Gauss – Paramanand Singh Aug 23 '17 at 05:22
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It is unclear. You didn't define $A(q)$ (it is analytic for $|q| < 1$ ?) and it is not clear how you obtain $A(q) = A(q^4)$. Are you using the modularity of something, or that doubly periodic entire functions are constant ? – reuns Aug 23 '17 at 05:27
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The second part "[..] Jacobi begins with the series expansion" seems more clear – reuns Aug 23 '17 at 05:33
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@reuns: I have defined $A(q) $ by equation $G=A(q) \prod_{n=1}^{\infty}(1-q^{2n})$ and then shown $A(q) =A(q^{4})$ via elementary manipulation of infinite products. This is based on equality $\theta_{3}(\pi/4,q)=\theta_{3}(\pi/2,q^{4})$. – Paramanand Singh Aug 23 '17 at 05:36
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@reuns: Jacobi's proof in second part is a gem and I try to advertise it everywhere on MSE. This is how proofs should be. Simple and clear and requiring least mathematical machinery. Mathematical enjoyment should not be restricted to elite few University dons and instead it should cater to masses like music/art – Paramanand Singh Aug 23 '17 at 05:42