Motivation for/history of Jacobi's triple product identity

The full appreciation of Jacobi's triple product identity can not be done without some understanding of the elliptic functions. However, it is possible to develop some parts of the theory of elliptic functions without any complex analysis.

Anyways back to the triple product identity, it says that

$$ \sum_{n = -\infty}^{\infty}z^{n}q^{n^{2}} = \prod_{n = 1}^{\infty}(1 - q^{2n})(1 + zq^{2n - 1})(1 + z^{-1}q^{2n - 1})$$

This is a highly non obvious identity which represents equality of a series with a product and to a beginner it might be very difficult to establish. One of the reasons I feel it is important is that it can be used to prove many different identities which have very nice and surprising applications. I will give a famous example here.

If we replace $q$ by $q^{3/2}$ and $z$ by $-q^{1/2}$ then we get

$$ \prod_{n = 1}^{\infty}(1 - q^{3n})(1 - q^{3n - 1})(1 - q^{3n - 2}) = \prod_{n = 1}^{\infty}(1 - q^{n}) = \sum_{n = -\infty}^{\infty}(-1)^{n}q^{(3n^{2} + n)/2}$$

This is Euler's famous Pentagonal theorem which he obtained by multiplying the product by hand to obtain first few terms of the series and then he guessed the pattern of exponents in the series as $(3n^{2} + n)/2$. But it took him some years to prove the identity.

This can be used to evaluate partitions of a positive integer. If $n$ is a positive integer and $p(n)$ denotes the number of partitions of $n$ (i.e. number of ways in which $n$ can be expressed as sum of positive integers without taking into account order of summands) then it can be easily established that

$$1 + \sum_{n = 1}^{\infty}p(n)q^{n} = \frac{1}{{\displaystyle \prod_{n = 1}^{\infty}(1 - q^{n})}} = \frac{1}{{\displaystyle \sum_{n = -\infty}^{\infty}(-1)^{n}q^{(3n^{2} + n)/2}}}$$

so that

$$ \left(1 + \sum_{n = 1}^{\infty}p(n)q^{n}\right)\sum_{n = -\infty}^{\infty}(-1)^{n}q^{(3n^{2} + n)/2} = 1$$

Equating coefficients of $q^{n}$ we get the recursive formula for $p(n)$ as

$$ p(n) = p(n - 1) + p(n - 2) - p(n - 5) - p(n - 7) + p(n - 12) + p(n - 15) - \cdots$$

which is the simplest possible way to calculate the number of partitions of a positive integer. This is one of the applications of Jacobi's triple product identity which can be understood without any reference to complex analysis or elliptic function theory.

I will try to give a rather elementary motivation which I believe to be close to how this identity was actually discovered.

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Let us consider the sine function $f(z)=\sin z$.

• As a function of complex argument, it has one period: $f(z+2\pi)=f(z)$. Further, it is holomorphic in the whole complex plane with only simple zeroes given by $\pi\mathbb{Z}$.

• It has a (finite) exponential series representation $$f(z)=\frac{1}{2i}\left(e^{iz}-e^{-iz}\right).\tag{1}$$

• It has an infinite product representation $$f(z)=z\prod_{n=1}^{\infty}\left(1-\frac{z^2}{\pi^2 n^2}\right).\tag{2}$$

Note that the representation (2) almost follows from the analytic properties of $f(z)$ by Liouville theorem-type arguments ("almost" is due to the need to control the infinite point).

Now a natural generalization of (2) comes out when we try to construct doubly-periodic functions. Liouville theorem forbids to such functions being holomorphic, therefore one can go in two directions:

• Allow poles and keep double periodicity; this leads to Weierstrass elliptic function.

• Keep the functions holomorphic at the price of relaxing periodicity; this will lead to Jacobi theta functions.

Let us choose the 2nd direction. The simplest generalization of (2), obtained by further periodization, is $$\vartheta_A(z)=\sin z\prod_{n=1}^{\infty}\left(\cos2\pi n\tau-\cos2z\right).\tag{3}$$ Naively, the function $\vartheta_A(z)$ is doubly-periodic and has simple zeros at $\pi\mathbb{Z}+\pi\tau\mathbb{Z}$. However, there is a problem - the product in (3) is very ill-defined. This can be cured by multiplying it by another ill-defined product independent of $z$ to get a well-defined quantity $$\vartheta_{B}(z)=\sin z\prod_{n=1}^{\infty}\left(1-2q^{2n}\cos2z+q^{4n}\right),\tag{4}$$ with $q=e^{i\pi\tau},\tau\in \mathbb{H}$. The price to pay for well-definedness is the loss of double-periodicity: though we still have $\vartheta_{B}(z+\pi)=-\vartheta_{B}(z)$, the second period is no longer a true one: $$\vartheta_{B}(z+\pi\tau)=-q^{-1}e^{-2iz}\vartheta_{B}(z).\tag{5}$$

So, once again, (4) defines a holomorphic almost doubly periodic function having simple zeros at $\pi\mathbb{Z}+\pi\tau\mathbb{Z}$. This is an analog of (2). Now the question is: what is the corresponding analog of (1)?

The answer is obtained relatively easily. Consider the function $$\vartheta_{C}(z)=\sum_{n\in\mathbb{Z}}(-1)^n q^{(n+1/2)^2}e^{i(2n+1)z}.\tag{6}$$ It is straightforward to verify that this function has the same periodicity properties as $\vartheta_{B}(z)$: $$\vartheta_{C}(z+\pi)=-\vartheta_{C}(z),\qquad \vartheta_{C}(z+\pi \tau)=-q^{-1}e^{-2iz}\vartheta_{C}(z).\tag{7}$$ Both functions are odd and in particular vanish at $z=0$. Further, $\vartheta_{C}(z)$ has exactly one zero (just as $\vartheta_{B}(z)$) inside the fundamental parallelogram. The last property is obtained by integrating $\vartheta_{C}'(z)/\vartheta_{C}(z)$ over its boundary and using quasiperiodicity (for holomorphic functions, the result should be equal to the number of zeros inside the fundamental parallelogram counted with their multiplicities).

Now, since we have two functions with the same zeros and the same quasiperiodicity properties, they can only be proportional because of Liouville theorem: $$\vartheta_{B}(z)=c(q)\cdot \vartheta_{C}(z).\tag{9}$$ But, if we recall (4) and (6), this is nothing but the Jacobi triple product identity - it remains only to fix the $z$-independent coefficient $c(q)$. In other words, this identity arises by comparing two natural ways (analogous to (1) and (2)) of writing a doubly-(quasi)periodic function. Actually, $\vartheta_{B,C}(z)$ are proportional to the Jacobi theta function $\vartheta_1(z,q)$.

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 $C=1$

$$ \sum_{n = -\infty}^{\infty}z^{n}q^{n^{2}} = \prod_{n = 1}^{\infty}(1 - q^{2n})(1 + zq^{2n - 1})(1 + z^{-1}q^{2n - 1})$$