Jacobi Triple Product

Jacobi Triple Product Identity

Theorem: For $|q| \lt 1$,

\begin{equation} \prod\limits_{n=1}^{\infty} \left((1-q^{2n})(1+q^{2n-1}w)(1+q^{2n-1}w^{-1}) \right) = \sum\limits_{n=-\infty}^{\infty} q^{n^2} w^n \tag{1} \end{equation}

Proof:

The identity in (1) is known as the Jacobi triple product identity and it can be proved in several ways. The proof (presented below) is based on some elementary algebraic manipulations.

First, let us prove the following two identities: \begin{equation} \prod\limits_{n=1}^{\infty} (1+q^{2n-1}w) = \sum\limits_{n=0}^{\infty} {q^{n^2} w^n \over \prod\limits_{k=1}^{n} (1-q^{2k})} \quad \quad\quad\quad\quad (\text{I1}) \end{equation} \begin{equation} \prod\limits_{n=1}^{\infty} (1+q^{2n-1}w)^{-1} = \sum\limits_{n=0}^{\infty} {(-1)^n q^n w^n \over \prod\limits_{k=1}^{n} (1-q^{2k})} \quad \quad\quad\quad (\text{I2}) \end{equation}

We will now prove the first identity. Let us define \( F(q,w)\) as follows: \( F(q,w) \equiv \prod\limits_{n=0}^{\infty} (1+q^{2n-1}w) \). Let us assume that there is some region \( \cal{R}\) where \(F(q,w)\) can be written as a power series in \(w\) as shown below (we will determine the radius of convergence after we evaluate the coefficients of this series): \begin{equation} F(q,w) = A_0(q) + A_1(q) w + A_2(q) w^2 + \cdots = \sum\limits_{n=0}^{\infty} A_n(q) w^n \tag{2}\end{equation} Setting \(~w=0\) we get \(~F(q,0)= 1 \implies A_0(q) =1 \).\(~\)Also, note that $$ (1+q w)F(q,q^2w) = F(q,w) \tag{3}$$ Using the above equation and the series expansion in equation (2), we get $$A_k(q) = {q^{2k -1} \over 1 - q^{2k}} A_{k-1}(q) = {q^{2k -1} q^{2k -3}\over (1 - q^{2k})(1 - q^{2k-2})} A_{k-2}(q) = \cdots , \quad \quad \quad \forall k \ge 1 \tag{4}$$ Solving this recursion relation (with \(~A_0(q)=1\)) we get, $$A_n(q) = {q^{n^2} \over \prod\limits_{k=1}^{n} (1-q^{2k})}, \quad \quad \quad \forall n \ge 1 \tag{5}$$ Substituting the above solution in the series expansion in equation (2) we get, $$F(q,w) \equiv \prod\limits_{n=0}^{\infty} (1+q^{2n-1}w) = \sum\limits_{n=0}^{\infty} {q^{n^2} w^n \over \prod\limits_{k=1}^{n} (1-q^{2k})}$$ We can prove the identity in (I2) in a similar manner. For the sake of completeness, I will present the proof of the identity in (I2) as well. To prove this result, let us define \(~ G(q,w) \equiv \prod\limits_{n=0}^{\infty} (1+q^{2n-1}w)^{-1} \).\(~\) Now, \(G(q,w)\) can be expanded in a power series as shown below: $$ G(q,w) = B_0(q) + B_1(q) w + B_2(q) w^2 + \cdots = \sum\limits_{n=0}^{\infty} B_n(q) w^n \tag{6}$$ Setting \(~w=0\) we get \(~G(q,0)= 1 \implies B_0(q) =1 \). Also, note that $$ (1+q w)G(q,w) = G(q,q^2w) \tag{7}$$ Using the above equation and the series expansion in equation (2), we get $$B_k(q) = -{q \over 1 - q^{2k}} B_{k-1}(q) = {q^2\over (1 - q^{2k})(1 - q^{2k-2})} B_{k-2}(q) = \cdots , \quad \quad \quad \forall k \ge 1 \tag{8}$$ Solving this recursion relation (with \(~B_0(q)=1\)) we get, $$B_n(q) = {(-1)^{n}q^n \over \prod\limits_{k=1}^{n} (1-q^{2k})}, \quad \quad \quad \forall n \ge 1 \tag{9}$$ Substituting the above solution in the series expansion in equation (2) we get, $$G(q,w) \equiv \prod\limits_{n=1}^{\infty} (1+q^{2n-1}w)^{-1} = \sum\limits_{n=0}^{\infty} {(-1)^{n}q^{n} w^n \over \prod\limits_{k=1}^{n} (1-q^{2k})}$$ Now, we will compute \(F(q,w)\) using a different approach that will help us establishing the identity in equation (1). Note that \(A_n(q)\) can be written as follows: $$ A_n(q) = {q^{n^2} \over \prod\limits_{k=1}^{\infty} (1-q^{2k})}\prod\limits_{k=0}^{\infty} (1-q^{2n+2k+2}), \quad \quad \quad \forall n \ge 0$$ We can now write \(F(q,w)\) as follows \begin{equation} F(q,w) = \sum\limits_{n=-\infty}^{\infty} {q^{n^2} \over \prod\limits_{k=1}^{\infty} (1-q^{2k})}\prod\limits_{k=0}^{\infty} (1-q^{2n+2k+2}) w^n \tag{10}\end{equation} Note that the sum over \(n\) runs from \(-\infty\) to \(\infty\). When \(n\) is negative, the \(\prod\limits_{k=0}^{\infty} (1-q^{2n+2k+2})\) vanishes and hence the summand is zero when \(n \) is negative. Now, using the identity in (I1) we get, $$ \prod\limits_{k=0}^{\infty} (1-q^{2n+2k+2}) = \sum\limits_{k=0}^{\infty} {q^{k^2} (-1)^k q^{(2n+1)k} \over \prod \limits_{\ell =1}^{k} (1-q^{2\ell})}$$ Using the above relation in equation (10) we get, \begin{equation} \left(\prod\limits_{k=1}^{\infty} (1-q^{2k})\right)F(q,w) = \sum\limits_{n=-\infty}^{\infty} \sum\limits_{k=0}^{\infty} {{q^{(n+k)^2}}(-1)^k q^{k} \over \prod \limits_{\ell =1}^{k} (1-q^{2\ell})}w^{-k} w^{n+k} \tag{11}\end{equation} Let us define \(m = n + k \). Note that the second summand on the right hand side of the above equation looks similar to the expression on the right hand side of the identity in (I2). Using the identity in (I2) in equation (11) we get, \begin{equation} \left(\prod\limits_{k=1}^{\infty} (1-q^{2k})\right)F(q,w) = \sum\limits_{m=-\infty}^{\infty} {q^{m^2} w^m \over \prod\limits_{\ell=1}^{\infty} (1+q^{2\ell-1}w^{-1})} \tag{12}\end{equation} Multiplying both sides by \(\prod\limits_{\ell=1}^{\infty} (1+q^{2\ell-1}w^{-1})\) we get the identity in (I1). This completes the proof of the Jacobi Triple product identity. \(\Box\)