Non combinatorial proof of Jacobi triple product? The jacobi triple product identity in the form given below - 
$$\prod_{n\geq 1}(1-q^n) (1-xq^{n-1}) (1-\frac{1}{x}q^n) = \sum_{k\geq 0} (-x)^k q^{k \choose 2}$$
can be proved as follows,
Let the LHS be denoted by $F(x)$, then note $\frac{F(x)}{F(qx)} = -x$. From this we can determine $F(x)$ up to a constant, i.e. we can say,
$$F(x) = a_0(q) \sum_{k\geq 0} (-x)^k q^{k \choose 2}$$
I am stuck at how to determine $a_0(q)$. Can somebody give me an easy way to do that?
 A: The Jacobi Triple Product can be written as, using $(a)_{\infty} = (a;q)_{\infty}$

$$\sum_{k\geq0}(-x)^{k}q^{\binom{k}{2}} = (q)_{\infty}(-x)_{\infty}(-q/x)_{\infty}$$

To determine $a_{0}(q)$, one needs the term $[x^{0}]$ of $F(x)$ :
\begin{align}a_{0}(q) &= [x^{0}]\;F(x)\\
&= (q)_{\infty}\left([x^{0}]\;(-x)_{\infty}(-q/x)_{\infty}\right)
\end{align}
Generating functions
Let's first recall some generating functions 

  
*
  
*If $p(n)$ denotes the number of possible partitions of $n$, then
  $$\sum_{n\geq 0}p(n)q^{n} = \frac{1}{(q)_{\infty}}$$
  
*If $Q_{\geq 1}(k,n)$ denotes the number of possible partitions of $n$ in $k$ distinct parts $\geq 1$, then
  $$\sum_{k,n\geq 1}Q_{\geq 1}(k,n)x^{k}q^{n} = (-xq)_{\infty}$$
  
*Similarly if $Q_{\geq 0}(k,n)$ denotes the number of possible partitions of $n$ in $k$ distinct parts $\geq 0$
  $$\sum_{k,n\geq 0}Q_{\geq 0}(k,n)x^{k}q^{n} = (-x)_{\infty}$$
  

Proofs of these formulas are pretty much straightforward if you develop each RHS.
Back to the proof of $a_{0}(q)$
Given what we just saw, one can write
$$[x^{0}]\;(-x)_{\infty}(-q/x)_{\infty} = [x^{0}]\;\left[\left(\sum_{k,n\geq 0}Q_{\geq 0}(k,n)x^{k}q^{n}\right)\,\left(\sum_{k',n'\geq 1}Q_{\geq 1}(k',n')x^{-k'}q^{n'}\right)\right]$$
The RHS can be interpreted as the generating function of pairs of partitions $(\lambda, \mu)$ such that :


*

*$\lambda$ is a partition of $n$ in $k$ distinct parts $\geq 0$

*$\mu$ is a partition of $n'$ in $k'$ distinct parts $\geq 1$


According to the Cauchy product of 2 power series, the only way to get $[x^{0}]$ is to extract every term with $k = k'$, which means that $[x^{0}]\;(-x)_{\infty}(-q/x)_{\infty}$ is the generating function of pairs $(\lambda,\mu)$ of partitions of $(n,n')$, both in $k$ distinct parts.
Just like Ferrer diagrams, one can encode any partition of an integer using something called the Frobenius representation

The Frobenius representation of a partition of an integer $n$ is a pair $(\lambda,\mu)$ of partitions in distinct parts (with the same number of parts)
  $$\left(\begin{array}{c}\lambda\\\mu\end{array}\right) = \left(\begin{array}{cccc}\lambda_{1} & \lambda_{2} & \cdots & \lambda_{k}\\\mu_{1} & \mu_{2} & \cdots & \mu_{k}\end{array}\right)$$
  with
  $$n = \sum_{i} \lambda_{i} + \sum_{i} \mu_{i} + k$$

That means that $[x^{0}]\;(-x)_{\infty}(-q/x)_{\infty}$ is the generating function of integer partitions
$$[x^{0}]\;(-x)_{\infty}(-q/x)_{\infty} = \frac{1}{(q)_{\infty}}$$
Conclusion
$$a_{0}(q) = (q)_{\infty}\left([x^{0}]\;(-x)_{\infty}(-q/x)_{\infty}\right) = 1$$
