Jacobi Triple Product Identity

(Jacobi’s triple product identity): For ${|q|<1}$ and ${x \neq 0}$ , we have

$\displaystyle \sum_{n=-\infty}^{\infty} q^{n^{2}} x^{n}=\prod_{n=1}^{\infty}\left(1+x q^{2 n-1}\right)\left(1+x^{-1} q^{2 n-1}\right)\left(1-q^{2 n}\right)$

Let

$F(x) =\sum_{n=-\infty}^{\infty} q^{n^{2}} x^{n}.$

Switching the terms ${a_n}$ and ${a_{-n},}$ we find that

$F(x) = F(\frac{1}{x})$

Shifting the indices we find

$\displaystyle F(x) =\sum_{n=-\infty}^{\infty} q^{(n+1)^{2}} x^{n+1}=q x F\left(q^{2} x\right)$

Now these functional equations with ${x = -\frac{1}{q}}$ , we get ${F(-q) =-F(-q) = F(-\frac{1}{q}).}$
Therefore

$\displaystyle F(-q) =F(-\frac{1}{q})=0.$

In fact apply the functional relations again to get,

$\displaystyle F(-q^{2n-1})=0, \quad \mbox{for all}~~ n \in \mathbb{Z}$

Therefore

$\displaystyle \displaystyle G(x) =\dfrac{F(x)}{\displaystyle \prod_{n=1}^{\infty}\left(1+x q^{2 n-1}\right)\left(1+x^{-1} q^{2 n-1}\right)}$

has no poles.

We have

$\displaystyle G(q^2x)= G(x)$

Writing we

$\displaystyle G(x) =\sum_{k=0}^{\infty} A_k(q) x^n$ ,

and using the functional equation we get

$\displaystyle (1-q^2k)A_{k}(q)=0$

Therefore we get ${A_k(q)=0}$ for ${k \neq 0}$ and

$\displaystyle G(x) = A_0(q).$

Thus we have

$\displaystyle F(x) = \sum_{n=-\infty}^{\infty} q^{n^{2}} x^{n}= A_0(q)\prod_{n=1}^{\infty}\left(1+x q^{2 n-1}\right)\left(1+x^{-1} q^{2 n-1}\right)$

Plug in ${x=-1}$ , we get

$\displaystyle \sum_{n=-\infty}^{\infty} q^{4 n^{2}}(-1)^{n}=A_{0}\left(q^{4}\right) \prod_{n=1}^{\infty}\left(1-q^{8 n-4}\right)^{2}$

For ${x=i}$ , we get

$\displaystyle \sum_{n=-\infty}^{\infty} q^{4 n^{2}}(-1)^{n}=A_{0}(q) \prod_{n=1}^{\infty}\left(1+q^{4 n-2}\right)$

Thus we see,

$\displaystyle \dfrac{A_{0}(q)}{\displaystyle \prod_{n=1}^{\infty}\left(1-q^{2 n}\right)}=\dfrac{A_{0}\left(q^{4}\right)}{\displaystyle \prod_{n=1}^{\infty}\left(1-q^{8 n}\right)}$

Iterate this map ${q \rightarrow q^4}$ and using that ${A_0(q)=1}$ , we see that

$\displaystyle \dfrac{A_{0}(q)}{\displaystyle \prod_{n=1}^{\infty}\left(1-q^{2 n}\right)}=\dfrac{A_{0}\left(q^{4^k}\right)}{\displaystyle \prod_{n=1}^{\infty}\left(1-q^{2.4^k n}\right)}\rightarrow_{k \rightarrow \infty} 1$

Therefore

$\displaystyle A_0(q) =\displaystyle \prod_{n=1}^{\infty}\left(1-q^{2 n}\right)$

We therefore get

$\displaystyle \sum_{n=-\infty}^{\infty} q^{n^{2}} x^{n}=\prod_{n=1}^{\infty}\left(1+x q^{2 n-1}\right)\left(1+x^{-1} q^{2 n-1}\right)\left(1-q^{2 n}\right)$

Another proof using Euler’s Formulae:

$\displaystyle \begin{array}{l} \displaystyle \prod_{n=1}^{\infty}\left(1+q^{2 n-1} x\right)=\displaystyle \sum_{n=0}^{\infty} \dfrac{\displaystyle q^{n^{2}} x^{n}}{\displaystyle \prod_{k=1}^{n}\left(1-q^{2 k}\right)} \\ \displaystyle \prod_{n=1}^{\infty}\left(1+q^{2 n-1} x\right)^{-1}=\displaystyle \sum_{n=0}^{\infty} \dfrac{\displaystyle (-1)^{n} q^{n} x^{n}}{\displaystyle \prod_{k=1}^{n}\left(1-q^{2 k}\right)} \end{array}$

These formula are obtained by using the following functional relations and expanding in to power series with respect to ${x.}$

$\displaystyle \displaystyle A(q, x) =\prod_{n=1}^{\infty}\left(1+q^{2 n-1} x\right) = (1+qx) A(q, q^2x)$

$\displaystyle \displaystyle B(q,x) = \prod_{n=1}^{\infty}\left(1+q^{2 n-1} x\right)^{-1}= (1+qx)^{-1} B(q, q^2x)$

We now use these identities to get the triple product identity:

$\displaystyle \displaystyle A_{n}(q)= \frac{q^{n^{2}} }{\prod_{k=1}^{n}\left(1-q^{2 k}\right)}=\frac{q^{n^{2}}}{\prod_{k=1}^{\infty}\left(1-q^{2 k}\right)} \prod_{k=0}^{\infty}\left(1-q^{2 n+2 k+2}\right)$

By the first identity we get

$\displaystyle \prod_{k=0}^{\infty}\left(1-q^{2 n+2 k+2}\right)=\sum_{k=0}^{\infty} \frac{q^{k^{2}}(-1)^{k} q^{(2 n+1) k}}{\prod_{l=1}^{k}\left(1-q^{2 \ell}\right)}$

Therefore we get

$\displaystyle \displaystyle \left(\prod_{k=1}^{\infty}\left(1-q^{2 k}\right)\right) A(q, x)=\sum_{n=-\infty}^{\infty} \sum_{k=0}^{\infty} \frac{q^{(n+k)^{2}}(-1)^{k} q^{k}}{\prod_{\ell=1}^{k}\left(1-q^{2 \ell}\right)} x^{-k} x^{n+k}$

$\displaystyle \displaystyle \left(\prod_{k=1}^{\infty}\left(1-q^{2 k}\right)\right) A(q, x)= \frac{\displaystyle \sum_{n=-\infty}^{\infty} q^{n^{2}} x^{n}}{B(q,x)}$

$\displaystyle \displaystyle \left(\prod_{k=1}^{\infty}\left(1-q^{2 k}\right)\right) A(q, x)B(q,x) =\displaystyle \sum_{n=-\infty}^{\infty} q^{n^{2}} x^{n}$

We arrive at

$\displaystyle \sum_{n=-\infty}^{\infty} q^{n^{2}} x^{n}=\prod_{n=1}^{\infty}\left(1+x q^{2 n-1}\right)\left(1+x^{-1} q^{2 n-1}\right)\left(1-q^{2 n}\right)$

Application: We

Euler’s Pentagonal theorem:

Substitute $q \rightarrow q^{3/2}, x= -q^{1/2}$ into the triple product identity, we get

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

Of course, we can prove the identity directly, we provide Euler’s direct proof.

Euler’s Proof: You can directly prove it using the functional equation for

$\displaystyle f(q, x)=1-\sum_{n=1}^{\infty}(1-x q)\left(1-x q^{2}\right) \cdots\left(1-x q^{n-1}\right) x^{n+1} q^{n}$
given by

$\displaystyle f(q, x)=1-x^{2} q-x^{3} q^{2} f(q, qx)$

and the fact that

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

The functional equation shows that

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

and we are done.

There are of course many proofs of the above identities including purely combinatorial ones using decompositions of partitions of a number.

Ramanujan’s Theta function:

$\displaystyle f(a, b) \equiv \sum_{n=-\infty}^{\infty} a^{n(n+1) / 2} b^{n(n-1) / 2}$

The triple product identity in form looks like

$\displaystyle f(a,b)=\sum_{n=-\infty}^{\infty} a^{n(n+1) / 2} b^{n(n-1) / 2} =(-a ; a b)_{\infty}(-b ; a b)_{\infty}(a b ; a b)_{\infty}$

where

$\displaystyle (a ; q)_{\infty} =\prod_{j=0}^{\infty}\left(1-a q^{j}\right)$

Watson quintuple product identity ;

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

Share this:

Related

Leave a comment Cancel reply