\documentclass[12pt,a4paper]{article}
\usepackage{latexsym}
\newtheorem{thm}{Theorem}
\newcommand{\Om}{\Omega}
\newcommand{\epf}{\hfill$\Box$\smallskip}
\title{On a septuple product identity}
\author{Robin Chapman\\School of Mathematical Sciences
\\University of Exeter\\EX4 4QE
\\UK\\{\tt rjc@maths.exeter.ac.uk}}
\date{}
\begin{document}
\maketitle

For polynomials $f$ and $g$ in the variable $n$ with integer coefficients
let us define
$$\Om(f)=\sum_{n=-\infty}^\infty(-1)^nq^{f(n)}$$
and
$$\Om(f,g)=\sum_{n=-\infty}^\infty(-1)^nq^{f(n)}x^{g(n)}.$$
In \cite{FK} Farkas and Kra state and prove the following identity
\begin{thm}\label{main}
\begin{eqnarray*}
&&\prod_{n=1}^\infty(1-q^{2n})^2(1-xq^{2n-2})(1-x^{-1}q^{2n})(1-x^2q^{4n-2})\\
&\times&(1-x^{-2}q^{4n-2})(1-x^2q^{4n-4})(1-x^{-2}q^{4n})\\
=&&\Om(5n^2+n)(\Om(5n^2+3n,5n+3)+\Om(5n^2-3n,5n))\\
&-&\Om(5n^2+3n)(\Om(5n^2+n,5n+2)+\Om(5n^2-n,5n+1)).
\end{eqnarray*}
\end{thm}
\epf

Their proof exploits identities involving theta functions. Foata
and Han \cite{FH} gave a more elementary deduction of (1) from Jacobi's
triple product formula and Watson's quintuple product formula. Here
we give a still more elementary derivation just from Jacobi's triple
product formula.

Let $P$ denote the product on the of the identity in Theorem~\ref{main}.
First observe that we can rewrite $P$ as
$$\prod_{n=1}^\infty(1-q^{2n})^2(1-xq^{2n-2})(1-x^{-1}q^{2n})(1-x^2q^{2n-2})
(1-x^{-2}q^{2n}).\eqno(1)$$
We now use Jacobi's triple product formula in the form
$$\prod_{n=1}^\infty(1-q^{2n})(1-xq^{2n-2})(1-x^{-1}q^{2n})
=\sum_{k=-\infty}^\infty(-1)^kx^kq^{k^2-k}.\eqno(2)$$
We obtain (2) by replacing $q$ by $q^2$ and $k$ by $-k$ in formula
(1.1) of \cite{FH}. Replacing $x$ by $x^2$ in (2) gives
$$\prod_{n=1}^\infty(1-q^{2n})(1-x^2q^{2n-2})(1-x^{-2}q^{2n})
=\sum_{k=-\infty}^\infty(-1)^kx^{2k}q^{k^2-k}.\eqno(3)$$
Multiplying (2) by (3) and using (1) gives
$$P=\sum_{k,l=-\infty}^\infty(-1)^{k+l}x^{k+2l}q^{k^2-k+l^2-l}.$$
Thus
$$P=\sum_{r=-\infty}^\infty a_r(q)x^r$$
where
$$a_r(q)=\sum_{k+2l=r}(-1)^{k+l}q^{k^2-k+l^2-l}
=\sum_{l=-\infty}^\infty(-1)^{r-l}q^{5l^2-(4r-1)l+r^2-r}.$$

We now evaluate $a_r(q)$ by dividing into cases according to the
residue of $r$ modulo~5. If $r=5m$ then
$$5l^2-(4r-1)l+r^2-r=5(l-2m)^2+(l-2m)+5m^2-3m$$
and so
$$a_{5m}(q)=(-1)^mq^{5m^2-3m}\sum_{n=-\infty}^\infty(-1)^nq^{5n^2+n}
=(-1)^mq^{5m^2-3m}\Om(5n^2+n).$$
If $r=5m+1$ then
$$5l^2-(4r-1)l+r^2-r=5(2m-l)^2+3(2m-l)+5m^2-m$$
and so
$$a_{5m+1}(q)=
-(-1)^mq^{5m^2-m}\sum_{n=-\infty}^\infty(-1)^nq^{5n^2+3n}
=-(-1)^mq^{5m^2-m}\Om(5n^2+3n).$$
If $r=5m+2$ then
$$5l^2-(4r-1)l+r^2-r=5(l-2m-1)^2+3(l-2m-1)+5m^2+m$$
and so
$$a_{5m+2}(q)=-(-1)^mq^{5m^2+m}\sum_{n=-\infty}^\infty(-1)^nq^{5n^2+3n}
=-(-1)^mq^{5m^2+m}\Om(5n^2+3n).$$
If $r=5m+3$ then
$$5l^2-(4r-1)l+r^2-r=5(2m+1-l)^2+(2m+1-l)+5m^2+3m$$
and so
$$a_{5m+3}(q)=
(-1)^mq^{5m^2+3m}\sum_{n=-\infty}^\infty(-1)^nq^{5n^2+n}
=(-1)^mq^{5m^2+3m}\Om(5n^2+n).$$
If $r=5m+4$ then
$$5l^2-(4r-1)l+r^2-r=5(l-2m-2)^2+5(l-2m-2)+5m^2+5m+2$$
and so
$$a_{5m+4}(q)=(-1)^mq^{5m^2+5m+2}\sum_{n=-\infty}^\infty(-1)^nq^{5n^2+5n}.$$
But in this sum the terms for $n=j$ and $n=-j-1$ cancel and so
$a_{5m+4}(q)=0$. It follows that
\begin{eqnarray*}P&=&\Om(5n^2+n)
\sum_{m=-\infty}^\infty(-1)^m(q^{5m^2-3m}x^{5m}+q^{5m^2+3m}x^{5m+3})\\
&&-\Om(5n^2+3n)
\sum_{m=-\infty}^\infty(-1)^m(q^{5m^2-m}x^{5m+1}+q^{5m^2+m}x^{5m+2})\\
&=&\Om(5n^2+n)(\Om(5n^2-3n,5n)+\Om(5n^2+3n,5n+3))\\
&&-\Om(5n^2+3n)(\Om(5n^2-n,5n+1)+\Om(5n^2+n,5n+2))
\end{eqnarray*}
which establishes Theorem~\ref{main}.

\smallskip

\noindent\textbf{Remark} After submitting this manuscript, the author
received a copy of \cite{H} which contains a proof of an
assertion equivalent to Theorem~1, using a similar method.

\begin{thebibliography}{99}
\bibitem{FK} H.~M.~Farkas \& I.~Kra, `On the quintuple product identity',
\emph{Proc.\ Amer.\ Math.\ Soc.} \textbf{27} (1999), 771--778.
\bibitem{FH} D.~Foata \& G.-H.~Han, `The triple, quintuple and septuple
product identities revisited',
\emph{ S\'eminaire Lotharingien de Combinatoire}, B42o (1999), 12 pp.
\bibitem{H} M.~D.~Hirschhorn, `An identity of Ramanujan, and applications',
\emph{Contemp.\ Math.}, to appear.
\end{thebibliography}

\end{document}