
\documentclass[12pt]{article}
\textwidth 14cm
\hoffset -.25in
\textheight 20cm
\begin{document}
\def\boxit#1{\vbox{\hrule\hbox{\vrule\kern3pt\vbox{\kern3pt#1\kern3pt}\kern3
pt\vrule}\hrule}}
\title{The number-theoretic content of the Jacobi triple product identity}
\author{Herbert S. Wilf\\\normalsize Department of Mathematics, University
of Pennsylvania\\\normalsize Philadelphia, PA 19104-6395}
\date{}
\maketitle
\vspace{.6in}
\begin{abstract}
We prove the Jacobi Triple Product Identity by exhibiting an elementary
number-theoretic proposition that is equivalent to it, and then proving
that the proposition is true.
\end{abstract}
\vspace{.9in}
\begin{center}
Dedicated to George Andrews on his sixtieth birthday:\\
\textit{``Unsolved problems tremble with fear as he approaches.''}
\end{center}
\vspace{.1in}
\newpage
The triple product identity of Jacobi asserts that
\begin{equation}
\label{eq:tpi}
\sum_{n=-\infty}^{\infty}z^nq^{n^2}=\prod_{n\ge
0}(1-q^{2n+2})(1+zq^{2n+1})(1+z^{-1}q^{2n+1}).
\end{equation}
Many proofs of the JTPI are known. There is a proof in Andrews's memoir
\cite{And2}, along with an extensive list of references, and another one by
George Andrews in
\cite{And}, which uses two identities of Euler. A proof of E.M. Wright
\cite{Wr} is combinatorial, and involves a direct bijection of bipartite
partitions. Proofs due to Cheema \cite{Ch} and to Sudler \cite{Su} are
variations of the one of Wright cited above. Here we give one which
proceeds by finding  an unusual proposition in elementary number theory to
which the Jacobi identity is equivalent, and proving it.

To discover the number-theoretic identity that is equivalent to
(\ref{eq:tpi}), do the following:
\begin{enumerate}
\item operate on the logarithm of both sides with $q\partial /\partial q$,
\item multiply by the left side of (\ref{eq:tpi}) to clear of fractions, and
\item equate the coefficients of $z^aq^b$ on both sides.
\end{enumerate}
The identity that results can be written as
\begin{eqnarray}
\label{eq:eqident}
&&\hspace{-.2in}\sum_{d\in S(n,1-2a)}(-1)^{d-1}\left\{{n\over
d}-d+2a\right\}+\sum_{d\in S(n,1+2a)}(-1)^{d-1}\left\{{n\over
d}-d-2a\right\}\nonumber\\
&&\qquad\qquad=\cases{2\sigma (n/2),&if $n\neq 0$;\cr
a^2,&if $n=0$,\cr}
\end{eqnarray}
where we have written $n=b-a^2$, and
\[S(n,x)=\bigl\{d: d\backslash n\ {\rm and}\ {n\over d}-d\ {\rm is\ odd\
and\ }{n\over d}-d\ge x\bigr\},\]
and $\sigma(N)$ is the sum of the divisors of $N$, if $N$ is a positive
integer, and is 0 otherwise.

Note that $b$ does not appear explicitly in (\ref{eq:eqident}). In fact,
\textit{we claim that (\ref{eq:eqident}) is true for all integers $n$ and
all real $a$}. We prove it for $n>0$ and all real $a$, the case $n< 0$
being similar, and $n=0$ being easy.

Indeed, if $n$ is odd then the left side of (\ref{eq:eqident}) vanishes and
the sums on the right side are empty, so the assertion is trivially true.
Hence we can suppose that $n$ is even.

To prove that (\ref{eq:eqident}) holds for all integers $a$ we will now
show that it holds for all sufficiently large $a$ and then that its truth
for $a$ implies the same for $a-1$.

First, for $n$ fixed, $\exists a_0$ such that $\forall a>a_0$ the second
sum on the right of (\ref{eq:eqident}) is empty, and for such $a$ we must
show that
\begin{equation}
\label{eq:(4)}
2\sigma(n/2)=\sum_{d\in S(n,-\infty)}(-1)^{d-1}\left\{{n\over
d}-d\right\}+2a\sum_{d\in S(n,-\infty)}(-1)^{d-1}.
\end{equation}
Let $n=2^mn'$, $n'$ odd, $m\ge 1$. Then the right side of (\ref{eq:(4)}) is
\begin{eqnarray*}
&&\sum_{d\backslash n'}\left\{2^m{{n'}\over d}-d\right\}-\sum_{d'\backslash
n'}\left\{{{n'}\over {d'}}-2^md'\right\}+(2a)\sum_{d\backslash
n'}1+(2a)\sum_{d'\backslash n'}(-1)\\
&&\qquad =2^m\sigma(n')-\sigma(n')-\sigma(n')+2^m\sigma(n')=2(2^m-1)\sigma
(n')=2\sigma(n/2),
\end{eqnarray*}
so (\ref{eq:eqident}) is proved for all large enough $a$.
Finally, if $f(a)$ denotes the right side of (\ref{eq:eqident}) then we
claim that $f$ is constant. Indeed, $f(a+1)-f(a)$ is equal to
\begin{eqnarray*}
&&\cases{(-1)^d(2a+1),&if $\exists\, d\backslash n:{n\over d}-d=2a+1$;\cr
0,&otherwise.\cr}\\
&&\quad+(2a+2)\sum_{d\in S(n,-2a-1)}(-1)^{d-1}-(2a)\sum_{d\in
S(n,1-2a)}(-1)^{d-1}\\
&&\quad-\cases{(-1)^d(2a+1)&if $\exists\, d\backslash n:{n\over
d}-d=2a+1$;\cr 0,&otherwise.\cr}\\
&&\quad-(2a+2)\sum_{d\in S(n,2a+3)}(-1)^{d-1}+(2a)\sum_{d\in
S(n,2a+1)}(-1)^{d-1}\\
&&=(2a+2)\left(\sum_{d\in S(n,-2a-1)}-\sum_{d\in S(n,2a+3)}\right)(-1)^{d-1}\\
&&\qquad +
  (2a)\left(\sum_{d\in S(n,2a+1)}-\sum_{d\in S(n,-2a+1)}\right)(-1)^{d-1}\\
&&=2a\left(\sum_{d\in S(n,-2a-1)}-\sum_{d\in S(n,-2a+1)}+\sum_{d\in
S(n,2a+1)}-\sum_{d\in S(n,-2a+1)}\right)(-1)^{d-1}\\
&&\qquad\qquad+2\left(\sum_{d\in S(n,-2a-1)}-\sum_{d\in
S(n,2a+3)}\right)(-1)^{d-1}.
\end{eqnarray*}
The terms in the four sums inside the first large parenthesis cancel
immediately except for two terms, and those are seen at once to cancel. The
terms in the second large parenthesis cancel in pairs, and the proof is
complete.

My thanks go to Robin Chapman for several helpful comments.


\newpage
\begin{thebibliography}{aaaa}
\bibitem{And}  George E. Andrews, A Simple Proof of Jacobi's Triple Product
Identity, Proceedings of the American Mathematical Society, Vol. 16, No. 2.
(Apr., 1965), pp. 333-334.
\bibitem{And2} George E. Andrews, Generalized Frobenius partitions,
\textit{Mem. Amer. Math. Soc.} \textbf{49} (1984), no. 301, iv+44 pp.
\bibitem{Ch} M. S. Cheema, Vector Partitions and Combinatorial Identities,
Math. Comp. \textbf{18} (1964), 414-420.
\bibitem{Su} C. Sudler, Two enumerative proofs of an identity of Jacobi,
Proc. Edinburgh Math. Soc. \textbf{15} (1966), 67--71.
\bibitem{Wr} E. M. Wright, An enumerative proof of an identity of Jacobi,
J. London Math. Soc. \textbf{40} (1965), 55--57.

\end{thebibliography}

\end{document}

