TITLE: Motivation for/history of Jacobi's triple product identity
QUESTION [28 upvotes]: I'm taking a short number theory course this summer. The first topic we covered was Jacobi's triple product identity. I still have no sense of why this is important, how it arises, how it might have been discovered, etc. The proof we studied is a bit on the clever side for my taste, giving me no sense of how said proof might have been discovered. Can anyone help?

REPLY [5 votes]: I would like to offer very elementary proof for Jacobi Triple Product Identity.
It is very direct way proof and elementary proof.
Let's define 
 $$ F(z)= \prod_{n = 1}^{\infty}(1 + zq^{2n - 1})(1 + z^{-1}q^{2n - 1})$$
$$ F(zq^2)= \prod_{n = 1}^{\infty}(1 + zq^2q^{2n - 1})(1 + z^{-1}q^{-2}q^{2n - 1})$$
$$ F(zq^2)= \prod_{n = 1}^{\infty}(1 + zq^{2n + 1})(1 + z^{-1}q^{2n - 3})$$
$$ F(zq^2)= (1 + z^{-1}q^{ -1})\prod_{n = 1}^{\infty}(1 + zq^{2n + 1})(1 + z^{-1}q^{2n - 1})$$
Multiple both side by $zq$
$$ zqF(zq^2)= zq(1 + z^{-1}q^{ -1})\prod_{n = 1}^{\infty}(1 + zq^{2n + 1})(1 + z^{-1}q^{2n - 1})$$
$$ zqF(zq^2)= (1 + zq)\prod_{n = 1}^{\infty}(1 + zq^{2n + 1})(1 + z^{-1}q^{2n - 1})$$
$$ zqF(zq^2)= \prod_{n = 1}^{\infty} (1 + zq)(1 + zq^{2n + 1})(1 + z^{-1}q^{2n - 1})$$
$$ zqF(zq^2)= \prod_{n = 1}^{\infty} (1 + zq^{2n - 1})(1 + z^{-1}q^{2n - 1})$$
$$ zqF(zq^2)= F(z)$$
Let's define $G(z)$
$$G(z)= \sum_{n = -\infty}^{\infty}z^{n}q^{n^{2}} $$
$$G(zq^2)= \sum_{n = -\infty}^{\infty}z^{n}q^{2n}q^{n^{2}} $$
$$G(zq^2)= \sum_{n = -\infty}^{\infty}z^{n}q^{n^{2}+2n} $$
Multiple both side by $zq$
$$zqG(zq^2)= zq\sum_{n = -\infty}^{\infty} z^{n}q^{n^{2}+2n} $$
$$zqG(zq^2)= \sum_{n = -\infty}^{\infty} z^{n+1}q^{n^{2}+2n+1} $$
$$zqG(zq^2)= \sum_{n = -\infty}^{\infty} z^{n+1}q^{(n+1)^2} = \sum_{n = -\infty}^{\infty}z^{n}q^{n^{2}}$$
$$G(z)=zqG(zq^2)$$
Because $F(z)$ and $G(z)$ satisfy same function relation, we can write
$$G(z)=A(q)F(z)$$
where A(q)  is only depends on $q$
We got the relation 
 $$ \sum_{n = -\infty}^{\infty}z^{n}q^{n^{2}} =A(q) \prod_{n = 1}^{\infty}(1 + zq^{2n - 1})(1 + z^{-1}q^{2n - 1})  \tag{1}$$
We only need to find $A(q)$ .
$z-->-z$ in Equation $(1)$
$$ \sum_{n = -\infty}^{\infty}(-1)^n z^{n}q^{n^{2}} =A(q) \prod_{n = 1}^{\infty}(1 - zq^{2n - 1})(1- z^{-1}q^{2n - 1})$$
$$ \sum_{n = -\infty}^{\infty}(-1)^n z^{n}q^{n^{2}} \sum_{n = -\infty}^{\infty}z^{n}q^{n^{2}}  =A^2(q) \prod_{n = 1}^{\infty}(1 - zq^{2n - 1})(1- z^{-1}q^{2n - 1})(1 + zq^{2n - 1})(1+ z^{-1}q^{2n - 1})$$
$$ \sum_{n = -\infty}^{\infty}(-1)^n z^{n}q^{n^{2}} \sum_{n = -\infty}^{\infty}z^{n}q^{n^{2}}  =A^2(q) \prod_{n = 1}^{\infty}(1 - z^2q^{2(2n - 1)})(1- z^{-2}q^{2(2n - 1)})$$
$z-->-z^2$
$q-->q^2$
in Equation $(1)$
$$ \sum_{n = -\infty}^{\infty}(-1)^nz^{2n}q^{2n^{2}} =A(q^2) \prod_{n = 1}^{\infty}(1 - z^2q^{2(2n - 1)})(1 - z^{-2}q^{2(2n - 1)})$$
$$ \sum_{n = -\infty}^{\infty}(-1)^n z^{n}q^{n^{2}} \sum_{n = -\infty}^{\infty}z^{n}q^{n^{2}}  =\frac{A^2(q)}{A(q^2)} \sum_{n = -\infty}^{\infty}(-1)^nz^{2n}q^{2n^{2}}$$
if we focus $z^0$  term and we multiply in left side.
We get that
$$\sum_{n = -\infty}^{\infty}(-1)^{n}q^{2n^{2}} =\frac{A^2(q)}{A(q^2)}$$
$z-->-1$
$q-->q^2$
in Equation $(1)$
$$ \sum_{n = -\infty}^{\infty}(-1)^{n}q^{2n^{2}} =A(q^2) \prod_{n = 1}^{\infty}(1 -q^{2(2n - 1)})(1 -q^{2(2n - 1)})$$
$$\frac{A^2(q)}{A(q^2)}= A(q^2) \prod_{n = 1}^{\infty}(1 -q^{2(2n - 1)})(1 -q^{2(2n - 1)})$$
$$\frac{A^2(q)}{A^2(q^2)}=  \prod_{n = 1}^{\infty}(1 -q^{2(2n - 1)})^2$$
$$\frac{A(q)}{A(q^2)}=  \prod_{n = 1}^{\infty}(1 -q^{2(2n - 1)})$$
$$\frac{A(q)}{A(q^2)}=  \prod_{n = 1}^{\infty} \frac{ (1 -q^{2(2n)}) (1 -q^{2(2n - 1)})}{(1 -q^{2(2n)})}$$
$$\frac{A(q)}{A(q^2)}=  \prod_{n = 1}^{\infty} \frac{ (1 -q^{2n}) }{(1 -q^{4n})}$$
$$A(q)=C\prod_{n = 1}^{\infty}(1 -q^{2n})$$  where $C$ is constant
 $$ \sum_{n = -\infty}^{\infty}z^{n}q^{n^{2}} =A(q) \prod_{n = 1}^{\infty}(1 + zq^{2n - 1})(1 + z^{-1}q^{2n - 1})$$
$$ \sum_{n = -\infty}^{\infty}z^{n}q^{n^{2}} =C \prod_{n = 1}^{\infty}(1 -q^{2n}) \prod_{n = 1}^{\infty}(1 + zq^{2n - 1})(1 + z^{-1}q^{2n - 1})$$
if $q=0$  then C can be found as $C=1$
$$ \sum_{n = -\infty}^{\infty}z^{n}q^{n^{2}} = \prod_{n = 1}^{\infty}(1 - q^{2n})(1 + zq^{2n - 1})(1 + z^{-1}q^{2n - 1})$$