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\auteurcourant={DOMINIQUE FOATA AND GUO-NIU HAN}
\titrecourant={TRIPLE, QUINTUPLE AND SEPTUPLE IDENTITIES REVISITED}

\vglue2cm
\centerline{THE TRIPLE, QUINTUPLE AND SEPTUPLE PRODUCT}
\smallskip
\centerline{IDENTITIES REVISITED}
\bigskip
\centerline{\bf Dominique Foata and Guo-Niu Han}
\bigskip
\centerline{\sl Dedicated to George Andrews on the occasion
of his sixtieth birthday}
\bigskip
\abstract{This paper takes up again the study of 
the Jacobi triple and Watson quintuple identities that have been
derived combinatorially in several manners in the classical
literature. It also contains a proof of the recent Farkas-Kra
septuple product identity that makes use only of ``manipulatorics"
methods.}
\bigskip

\section 1. Introduction|In the classical literature the Jacobi triple
product appears in one of the following two forms
$$\displaylines{(1.1)\quad
\prod_{n=1}^\infty (1-x^{-1}q^{n-1})(1-xq^n)
=\prod_{i=1}^\infty {1\over (1-q^i)}
\sum_{k=-\infty}^{+\infty} (-1)^k\,x^k\,q^{k(k+1)/2},\hfill\cr
(1.2)\quad
\prod_{n=1}^\infty
(1-x^{-1}\,q^{2n-1})(1-x\,q^{2n-1})
=\prod_{i=1}^\infty {1\over (1-q^{2i})}
\sum_{k=-\infty}^{+\infty} (-1)^k\,x^{k}
q^{k^2},\hfill\cr
\noalign{\hbox{while the Watson quintuple product reads}}
(1.3)\quad
\prod_{n=1}^\infty 
(1-x^{-1}q^{n-1})(1-xq^n)
(1-x^{-2}\,q^{2n-1})(1-x^2\,q^{2n-1})
\hfill\cr
\hfill{}
=\prod_{i=1}^\infty {1\over (1-q^i)}
\sum_{k=-\infty}^{+\infty} q^{(3k^2+k)/2}
(x^{3k}-x^{-3k-1}).\quad\cr
}
$$
The letters $x$ and $q$ may be regarded as complex variables
with $|q|<1$ and $x\not=0$ or as simple indeterminates. In the
latter case consider the ring $\Omega[x,x^{-1}]$ of the
polynomials in the variables~$x$ and~$x^{-1}$ such that
$xx^{-1}=1$ with coefficients in a ring~$\Omega$.
Then the identities hold in the algebra of formal power series in
the variable~$q$ with coefficients in $\Omega[x,x^{-1}]$. 

As usual, let $(a;q)_n$ denote the $q$-ascending factorial 
$$\leqalignno{
(a;q)_n&=\cases{1,&if $n=0$;\cr
                 (1-a)(1-aq)\ldots (1-aq^{n-1}),&if $n\ge 1$;\cr}\cr
(a;q)_\infty&=\prod_{n\ge 0}(1-aq^n);\cr
\noalign{\hbox{and let the classical $q$-binomial coefficient
be denoted by:}}
{n\brack k}_q&={(q;q)_n\over  (q;q)_{n-k}(q;q)_k}\quad (0\le k\le
n).\cr}$$

The identities (1.1) and (1.2) have two finite
versions given by
$$
\leqalignno{\noalign{\vskip-5pt}
(x^{-1};q)_n\,(xq;q)_m&=\sum_{j=-n}^m{n+m\brack j+n}_q(-x)^j\,
q^{j(j+1)/2};&(1.4)\cr
(x^{-2};q^2)_n\,(x^2q;q^2)_m&=\sum_{j=-n}^m{n+m\brack
j+n}_{q^2}(-x^2)^j\, q^{j^2}.&(1.5)\cr
\noalign{\vskip-5pt}}
$$
Those two versions with $n$ and $m$ not necessarily equal are
apparently due to MacMahon ([Ma15], vol.~2, \S\kern2pt 323). He
proved (1.5) by using Sylvester's [Sy82] ``quasi-geometrical
method of demonstration" and notes that to obtain (1.4) the
variable $x$ is to be replaced by $xq$ and then $q^2$ by~$q$. With
similar substitutions (1.5) can be derived from (1.4). As those
substitutions are made within finite expressions the derivations
are straightforward.

Finally, as kindly mentioned to us by Garvan [Ga99], Farkas and Kra
[Fa99] derived a {\it septuple product identity} using the
algebra of $k$-order theta functions.
If $f$ (resp. $g$)
is a polynomial~$q$ (resp. in~$x$) with integral coefficients,
let
$$\eqalign{\Theta(f):&=\sum_{n\in {\bf Z}} q^{f(n)};\cr
\Theta(f,g):&=\sum_{n\in {\bf Z}} q^{f(n)}\,x^{g(n)};\cr
\Theta'(f):&=\sum_{n\in {\bf Z}} (-1)^n\, q^{f(n)};\cr
\Theta'(f,g):&=\sum_{n\in {\bf Z}} (-1)^n\, q^{f(n)}\,x^{g(n)}.\cr}
$$
Then Farkas and Kra [Fa99] imagined and proved the following
identity
$$\displaylines{(1.6)\quad
(q^2;q^2)_\infty^2\,
(x;q^2)_\infty\,
(x^{-1}q^2;q^2)_\infty\,
(x^2q^2;q^4)_\infty\hfill\cr
\qquad\qquad\qquad{}
\times(x^{-2}q^2;q^4)_\infty\,
(x^2;q^4)_\infty\,
(x^{-2}q^4;q^4)_\infty\hfill\cr
\qquad\qquad{}=
\Theta'(5n^2+n)\bigl(\Theta'(5n^2+3n,5n+3)
+\Theta'(5n^2-3n,5n)\bigr)\hfill\cr
\hfill{}
-\Theta'(5n^2+3n)\bigl(\Theta'(5n^2+n,5n+2)+\Theta'(5n^2-n,5n+1)\bigr).\cr}
$$
Notice that with the substitutions $x\leftarrow x^{-2}$ and
$q\leftarrow q^4$ the triple product identity (1.1) reads
$$
(x^2;q^4)_\infty\,(x^{-2}q^4;q^4)_\infty\, (q^4;q^4)_\infty
=\Theta'(2n^2+2n,-2n),
\leqno(1.7)$$
while the quintuple product identity (1.3) with the
substitutions
$q\leftarrow q^2$ and
$x\leftarrow x^{-1}$ takes the form:
$$\displaylines{(1.8)\quad
(q^2;q^2)_\infty\, (x;q^2)_\infty\, (x^{-1}q^2;q^2)_\infty\,
(x^2q^2;q^4)_\infty\, (x^{-2}q^2;q^4)_\infty\hfill\cr
\hfill{}
=\Theta(3n^2+n,-3n)-\Theta(3n^2+n,3n+1).\quad\cr}
$$

At the origin our intention was to give a combinatorial proof of
the quintuple product identity (1.3). A glance at the left-hand
sides of identities (1.1), (1.2), (1.3) shows that (1.3) must be
a consequence of (1.1) and (1.2) and the combinatorics
involved, once the products on the right-hand sides of the first
two identities are properly handled. This program was only
partially fulfilled, because (1.3) is an easy consequence of both
triple product identities and changing the ``manipulatorics"
needed into some combinatorial construction would have been a
useless task. As will be seen in section~3, besides the two triple
product identities, we only need the {\it Euler pentagonal number
formula} (see, e.g., [An76] p.~11), another special case of those
two identities, and a simple summation manipulation.

There remains to imagine the adequate bijections to prove 
(1.1) and (1.2). How can we dare construct such bijections, some
117 years after Sylverster [Sy82]? He already derived three
different combinatorial proofs, scholarly commented by
Joichi and Stanton~[Jo89]. We have to admit, indeed, that any kind
of new combinatorial construction for proving (1.1) and (1.2) can
only be a slight variation of Sylvester's method~[Sy82]. He had
been the source of a long tradition of combinatorial
construction makers. Even our ``rectangle-moving" method that we
were proud to discover did not escape his filiation. We have then
decided to leave our combinatorial construction on our own home
pages ([Fo99], [Ha99]) and, in the present paper, only provide
with straightforward proofs for all the identities above, i.e.
(1.1)---(1.6).

The first combinatorial proofs go back to Sylvester~[Sy82] and have
been the sources of inspiration of several subsequent ones, by
Wright~[Wr65], Sudler~[Su66], Ewell~[Ew81], Lewis~[Le84], Garvan
[Ga86] (see \S\kern2pt 3.2 in his Ph.D. thesis, as it was
mentioned to us by an anonymous referee). Joichi and Stanton~[Jo89]
discuss the various merits of those proofs. They are mostly
interested in building natural involutions for proving partition
identities; they also compare the approaches due to 
Zolnowsky~[79] and Cheema~[Ch64].

The other proofs are of formal nature, as in
MacMahon~([Ma15], vol.~2, \S\kern2pt 327), Bressoud~[Br97] or of
analytical nature, as in Andrews [An65], [An74], [An84], or in the
classical treatises by Hardy and Wright~[Ha38], Andrews~[An76],
Gupta~[Gu87]. A fairly complete bibliography can be found in Gasper and
Rahman~[Ga90].

\goodbreak
The quintuple product identity is originally due to 
Watson~[Wa29]. Other proofs were given by Gordon~[Go61],
Carlitz and Subbarao~[Ca72], Subbarao and Vidyasagar~[Su70].
Hirschhorn~[Hi88] proposes a generalisation of that identity
and claims that there are ``no fewer than twelve proofs of the
quintuple product identity," in particular by Bailey~[Ba51],
Sears~[Se52], Atkin and Swinnerton-Dyer~[At54],
Andrews~[An74] and more recently by Alladi~[Al96].

Finally, those identities are found in classical topics in Number
Theory or Lie Algebra, as in Adiga, Berndt, Bhargava and
Watson~[Ad85], Gustafson~[Gu87], Kac~[Ka78], [Ka85], Lepowsky
and Milne~[Le78], Macdonald~[Ma82], Menon~[Me65], Milne~[Mi85].

\smallskip
The paper is organized as follows. In the next section MacMahon's
finite versions (1.4) and (1.5) are derived and it is shown how
they imply (1.1) and (1.2). In section~3 we shall reprove (1.3)
using an argument very close to the one used by Carlitz and 
Subbarao~[Ca72]. In the final section we give our own proof of the
new elected {\it septuple product identity} obtained by Farkas and
Kra [Fa99]. We first make use of an extended Carlitz-Subbarao
trick (that was sufficient for the quintuple case), then introduce
two further specializations of both triple and quintuple product
identities to complete the calculation. It seems that Farkas-Kra's
identity is much deeper than its previous two sisters.
 
\section 2. The finite and infinite versions of the triple product|

As shown to us by Andrews [An98], and as it is
well-known in the case $m=n$, identity (1.4) can be proved by 
means of the $q$-binomial identity in its finite form. Proceed 
as follows:
$$
\leqalignno{
(x^{-1};q)_n\,(xq;q)_m\!
&=\!(-1)^nx^{-n}q^{n(n-1)/2}(xq^{1-n};q)_n\,(xq;q)_m\cr
&=\!(-1)^nx^{-n}q^{n(n-1)/2}(xq^{1-n};q)_{n+m}\cr
&=\!(-1)^nx^{-n}q^{n(n-1)/2}\sum_{j=0}^{n+m}{n+m\brack j}_q\!\!
(-xq^{1-n})^jq^{j(j-1)/2}\cr
\noalign{\vskip-10pt}
&=\!\sum_{j=0}^{n+m}{n+m\brack j}_q
(-x)^{j-n} q^{(j-n)(j-n+1)/2}\cr
&=\!\sum_{j=-n}^{m}{n+m\brack j+n}_q
(-x)^{j} q^{j(j+1)/2}.\cr
}
$$
Now to deduce the ``infinite" versions (1.1), (1.2) from the finite ones
we only have to let $n$ and $m$ tend to infinity.  Using (1.4) for $n=m$
the product $(x^{-1};q)_m\,(xq;q)_m\,(q;q)_\infty$
can be expressed as
$$
\sum_{j=-m}^m
(q^{m-j+1};q)_{m+j}\,(q^{m+j+1};q)_{\infty}\,(-x)^j\,
q^{j(j+1)/2}.
$$

\goodbreak
\noindent
In that sum the running term is equal to
$(-x)^j\,
q^{j(j+1)/2}(1-q^{m-|j|+1}a_j)$, with $a_j$ a series in~$q$, so that
$(x^{-1};q)_m\,(xq;q)_m\,(q;q)_\infty
=b_m+q^{m}c$, where $b_m$ is the series
$b_m=\sum\limits_{j=-m}^m (-x)^j\,q^{j(j+1)/2}$
and $c$ is a non-null series. Hence
$(x^{-1};q)_\infty\,(xq;q)_\infty\,(q;q)_\infty=\lim_mb_m
=\smash{\sum\limits_{j=-\infty}^{\infty} (-x)^j\,q^{j(j+1)/2}}$, which
is simply~(1.1).

Using the same method we can derive (1.5) that, in its turn,
implies~(1.2). The MacMahon finite versions (1.4) and (1.5) can be
regarded as the ``fundamental" triple product identities and, still,
they are derived by means of the $q$-binomial identity in its finite
form. Here we face one of the mysteries of mathematical tradition:
explain why  so many proofs of those identities can be found in the
literature.
\goodbreak

\section 3. The quintuple product identity|
To derive the quintuple product identity (1.3) it suffices to prove
$$\displaylines{\quad
\prod_{i=1}^\infty
{1\over 1-q^i}
\sum_{k\in {\bboard Z}} (-1)^k x^k
q^{k(k+1)/2}\times
\prod_{i=1}^\infty {1\over 1-q^{2i}}
\sum_{k\in {\bboard Z}} (-1)^k x^{2k} q^{k^2}\hfill\cr
\hfill{}
=\prod_{i=1}^\infty
{1\over 1-q^i}
\sum_{k\in {\bboard Z}} q^{(3k^2+k)/2}
(x^{3k}-x^{3k-1}),\quad\cr
\noalign{\hbox{or by using the Euler pentagonal number
identity (see [An76], p.~11)}}
\prod_{i\ge 1} (1-q^i)=\sum_{k\in {\bboard Z}}
(-1)^k\,q^{(3k^2-k)/2},\cr
\noalign{\hbox{to prove the identity}}
\sum_{k\in {\bboard Z}} (-1)^k x^k
q^{k(k+1)/2}\times
\sum_{l\in {\bboard Z}} (-1)^l x^{2l} q^{l^2}\hfill\cr
\hfill{}
=\sum_{n\in {\bboard Z}} (-1)^n\,q^{3n^2-n}\times
\sum_{m\in {\bboard Z}} q^{(3m^2+m)/2}
(x^{3m}-x^{-3m-1}).\quad\cr}
$$
Write the product of the two series of the left-hand side as the
sum of three series denoted by
$S_0$, $S_1$, $S_2$:
$$\eqalignno{
\sum_{k,l}(-1)^{k+l}\,x^{k+2l}q^{k(k+1)/2+l^2}
&=\sum_mx^{3m}\sum_{k+2l=3m}(-1)^{k+l}\,q^{k(k+1)/2+l^2}\cr
&\quad{}+\sum_mx^{3m-1}
\sum_{k+2l=3m-1}(-1)^{k+l}\,q^{k(k+1)/2+l^2}\cr
&\quad{}+\sum_mx^{3m-2}\sum_{k+2l=3m+2}
(-1)^{k+l}\,q^{k(k+1)/2+l^2}\cr
&=S_0+S_1 +S_2.\cr
}
$$
For $S_0$ notice that $k+2l=3m$ and $l-m=n$ imply: $k+l=2m-n$
and $k(k+1)/2+l^2=(3m^2+m)/2+3n^2-n$. Hence
$$
S_0=\sum_m q^{(3m^2+m)/2}\,x^{3m}
\sum_n(-1)^n\, q^{3n^2-n}.
$$
For $S_1$ the change of indices $k+2l=3m-1$ et $l-m=n$
imply:
$k+l=2m-n-1$ and
$k(k+1)/2+l^2=(3m^2-m)/2+3n^2+n$. Hence
$$\eqalign{
S_1&=-\sum_m q^{(3m^2-m)/2}\,x^{3m-1}
\sum_n(-1)^n\, q^{3n^2+n}\cr
&=-\sum_m  q^{(3m^2+m)/2}\,x^{-3m-1}\,
\sum_n(-1)^n\, q^{3n^2-n}.\cr}
$$
Finally, for $S_2$ make the change of indices
$k+2l=3m-2$ and $l-m=n$, so that
$k+l=2m-n-2$ et
$k(k+1)/2+l^2=(3m^2-3m+2)/2+3n^2+3n$. Hence
$$
S_2=\sum_m x^{3m-2}\,q^{(3m^2-3m+2)/2}
\sum_n(-1)^n\, q^{3n^2+3n}.
$$
But
$\sum\limits_{n\in {\bboard Z}}(-1)^n\, (q^3)^{n(n+1)}=0$,
and $S_2=0$. The sum $S_0+S_1$ is exactly the right-hand
side of the quintuple product identity~(1.3).\qed

\section 4. The septuple product identity|
Let $E$ be the left-hand side of identity (1.6), i.e., 
$$\displaylines{\quad 
E:=(q^2;q^2)_\infty^2\,(x;q^2)_\infty\,
(x^{-1}q^2;q^2)_\infty\,(x^2q^2;q^4)_\infty\hfill\cr
\hfill{}
\times(x^{-2}q^2;q^4)_\infty\,(x^2;q^4)_\infty\,
(x^{-2}q^4;q^4)_\infty.\quad\cr}
$$
Taking both identities (1.7) and (1.8) into account and using the
identity
$$\displaylines{
{(q^4;q^4)_\infty\over (q^2;q^2)_\infty}
=\prod_{m\ge 1} (1+q^{2m}),\cr
\noalign{\hbox{we may write:}}
\prod_{m\ge 1} (1+q^{2m})\,E
=\Theta'(2n^2+2n,-2n)\bigl(\Theta(3n^2+n,-3n)
-\Theta(3n^2+n,3n+1)\bigr).\cr}
$$
We now use the method of the previous section. However this time each
product of the right-hand side of the previous formula is
transformed into a sum of {\it five} products of two
$\Theta'$-series. We may write:
$$\displaylines{(4.1)\quad
\Theta'(2n^2+2n,-2n)\,\Theta(3n^2+n,-3n)
=S_0+S_1+S_2+S_3+S_4,\hfill\cr
\noalign{\hbox{where for each $k=0,1,2,3,4$ we let}}
S_k:=\sum_a z^{5a+k}\sum_{-2i-3j=5a+k}(-1)^i q^{2i^2+2i+3j^2+j}.\cr
}
$$

\goodbreak
In the second summation of each expression $S_k$ we make a change of
variables indicated in Table~1 below. For instance, as
shown in the second row of the table, when $-2i-3j=5a$
(first column), we let
$a+j=-2n$ (second column) noting that $a+j$ is necessarily even. Hence
$i\equiv a+n\pmod 2$ (third column). Finally, the exponent of~$q$ is
transformed into $30n^2+4n+5a^2-3a$ (fourth column).

\bigskip
\centerline{%
\vbox{
\halign{\vrule\strut\ \hfil$#$\hfil\ \vrule
&\ \hfil$#$\hfil\ \vrule 
&\ \hfil$#$\hfil\ \vrule  
&\ \hfil$#$\hfil\ \vrule \cr
\noalign{\hrule}
-2i-3j&a+j&i\equiv\kern-8pt\pmod 2&2i^2+2i+3j^2+j\cr
\noalign{\hrule}
5a&-2n&a+n&30n^2+4n+5a^2-3a\cr
5a+1&-2n-1&a+n+1&30n^2+28n+6+5a^2-a\cr
5a+2&2n&a+n+1&30n^2+8n+5a^2+a\cr
5a+3&-2n-1&a+n&30n^2+16n+2+5a^2+3a\cr
5a+4&2n&a+n&30n^2+20n+4+5a^2+5a\cr
\noalign{\hrule}
}}}
\smallskip
\centerline{Table 1}

\bigskip
With those changes of variables we get
$$\displaylines{(4.2)\quad
\Theta'(2n^2+2n,-2n)\,\Theta(3n^2+n,-3n)\hfill\cr
\kern3cm{}=\Theta'(5n^2-3n,5n)\,\Theta'(30n^2+4n)\hfill\cr
\kern4cm{}-\Theta'(5n^2-n,5n+1)\,\Theta'(30n^2+28n+6)\hfill\cr
\kern4cm{}-\Theta'(5n^2+n,5n+2)\,\Theta'(30n^2+8n)\hfill\cr
\kern4cm{}+\Theta'(5n^2+3n,5n+3)\,\Theta'(30n^2+16n+2)\hfill\cr
\kern4cm{}+\Theta'(5n^2+5n,5n+4)\,\Theta'(30n^2+20n+4).\hfill\cr   
}
$$
In the same manner, let
$$\displaylines{(4.3)\quad
\Theta'(2n^2+2n,-2n)\,\Theta(3n^2+n,3n+1)
=T_0+T_1+T_2+T_3+T_4,\hfill\cr
\noalign{\hbox{where for each $k=0,1,2,3,4$ we let}}
T_k:=\sum_a z^{5a+k}\sum_{-2i+3j+1=5a+k}(-1)^i q^{2i^2+2i+3j^2+j}.\cr
}
$$
Again, the changes of variables made in each $T_k$ are indicated in
Table~2.

\bigskip
\centerline{%
\vbox{
\halign{\vrule\strut\ \hfil$#$\hfil\ \vrule
&\ \hfil$#$\hfil\ \vrule 
&\ \hfil$#$\hfil\ \vrule  
&\ \hfil$#$\hfil\ \vrule \cr
\noalign{\hrule}
-2i-3j+1&a-j&i\equiv\kern-8pt\pmod 2&2i^2+2i+3j^2+j\cr
\noalign{\hrule}
5a&2n+1&a+n+1&30n^2+16n+2+5a^2-3a\cr
5a+1&-2n&a+n&30n^2+8n+5a^2-a\cr
5a+2&2n+1&a+n&30n^2+28n+6+5a^2+a\cr
5a+3&2n&a+n+1&30n^2+4n+5a^2+3a\cr
5a+4&-2n-1&a+n&30n^2+20n+4+5a^2+5a\cr
\noalign{\hrule}
}}}
\smallskip
\centerline{Table 2}

\bigskip
Those changes of variables yield:
$$\displaylines{(4.4)\quad
\Theta'(2n^2+2n,-2n)\,\Theta(3n^2+n,3n+1)\hfill\cr
\kern3cm{}=-\Theta'(5n^2-3n,5n)\,\Theta'(30n^2+16n+2)\hfill\cr
\kern4cm{}+\Theta'(5n^2-n,5n+1)\,\Theta'(30n^2+8n)\hfill\cr
\kern4cm{}+\Theta'(5n^2+n,5n+2)\,\Theta'(30n^2+28n+6)\hfill\cr
\kern4cm{}-\Theta'(5n^2+3n,5n+3)\,\Theta'(30n^2+4n)\hfill\cr
\kern4cm{}+\Theta'(5n^2+5n,5n+4)\,\Theta'(30n^2+20n+4).\hfill\cr   
}
$$
In particular, we notice that $S_4=T_4$. When taking 
the difference $(4.2)-(4.4)$ we simply get:
$$\displaylines{(4.5)\quad
\prod_{m\ge 1}(1+q^{2m})\,E
=
\bigl(\Theta'(5n^2+3n,5n+3)
+\Theta'(5n^2-3n,5n)\bigr)\hfill\cr
\noalign{\vskip-10pt}
\hfill{}\times\bigl(\Theta'(30n^2+4n)+\Theta'(30n^2+16n+2)\bigr)\cr
\kern3cm{}-\bigl(\Theta'(5n^2+n,5n+2)+\Theta'(5n^2-n,5n+1)\bigr)
\hfill\cr
\hfill{}\times
\bigl(\Theta'(30n^2+28n+6)+\Theta'(30n^2+8n)\bigr).\cr
}
$$
Now if we compare the last identity with the septuple
identity (1.6) we see that the latter is the consequence
of the next Lemma.

\th Lemma|We have:
$$
\leqalignno{\qquad
\prod_{m\ge 1}(1+q^{2m})\,\Theta'(5n^2+n)
&=\Theta'(30n^2+16n+2)+\Theta'(30n^2+4n);&(4.6)\cr
\qquad
\prod_{m\ge 1}(1+q^{2m})\,\Theta'(5n^2+3n)
&=\Theta'(30n^2+28n+6)+\Theta'(30n^2+8n).&(4.7)\cr
}$$
\finth

The proof of the Lemma has very much the flavor of the
proofs derived by our friend Mike Hirschhorn [Hi88] when
he masterly plays with the triple product identity for
various specializations. Here the quintuple identity also
gets into the picture. In the sequel, the range of
the index~$m$ (resp.~$n$) is ${\bboard N}\setminus\{0\}$
(resp. $\bboard Z$).

\medskip
{\it Proof of the Lemma}.\quad
The triple product (1.1) with $q:=q^{10}$ and $x:=q^{-4}$
reads
$$\leqalignno{
\prod_m (1-q^{10m})(1-q^{10m-4})(1-q^{10m-6}) 
&=\Theta'(5n^2+n),&(4.8)\cr
\noalign{\hbox{while with $q:=q^{10}$ and $x:=q^{-2}$
yields}}
\prod_m (1-q^{10m})(1-q^{10m-2})(1-q^{10m-8})
&=\Theta'(5n^2+3n).&(4.9)\cr
}
$$
Next the quintuple product (1.3) with $q:=q^{20}$ and
$x:=-q^{-2}$ reads
$$
\displaylines{(4.10)\ \prod_m
(1-q^{20m})(1+q^{20m-18})(1+q^{20m-2}) 
(1-q^{40m-16})(1-q^{40m-24}) \hfill\cr
\hfill{}=
\Theta'(30n^2+16n+2)+\Theta'(30n^2+4n),\cr
}
$$
that is, the right-hand side of (4.6). In the same manner
the quintuple product with $q:=q^{20}$ and $x:=-q^{-6}$
takes the form
$$
\displaylines{(4.11)\  
\prod_m (1-q^{20m})(1+q^{20m-14})(1+q^{20m-6}) (1-q^{40m-8})(1-q^{40m-32}) 
\hfill\cr
\hfill{}
=\Theta'(30n^2+28n+6)+\Theta'(30n^2+8n),\cr}
$$
which is the right-hand side of (4.7).

The two identities (4.6) and (4.7) may then rewritten as:
$$\leqalignno{
\prod_m (1+q^{2m})\times (\hbox{l.-h. s.  of
(4.8)})&=\hbox{l.-h. s.  of (4.10)},&(4.6')\cr  
\prod_m (1+q^{2m})\times (\hbox{l.-h. s.  of
(4.9)})&=\hbox{l.-h. s.  of (4.11)}.&(4.7')\cr  }
$$
As
$$
\prod_m (1+q^{2m})\!\! =\!\! \prod_m
(1+q^{10m})(1+q^{10m-2})(1+q^{10m-4})(1+q^{10m-6})(1+q^{10m-8}),
$$
the left-hand side of $(4.6')$ is equal to
$$
 \prod_m
(1-q^{20m})(1-q^{20m-8})(1-q^{20m-12})(1+q^{10m-2})
(1+q^{10m-8}).
$$
Now as
$$\displaylines{\quad
\prod_m (1+q^{10m-2})(1+q^{10m-8})\hfill\cr
\noalign{\vskip-10pt}
\hfill{}
= \prod_m
(1+q^{20m-2})(1+q^{20m-12})(1+q^{20m-8})(1+q^{20m-18}),\cr}
$$
we see that the left-hand side of $(4.6')$ is equal to
$$
\displaylines{\quad
 \prod_m
(1-q^{20m})(1-q^{20m-8})(1-q^{20m-12})\hfill\cr
\noalign{\vskip-10pt}
\hfill{}\times
(1+q^{20m-2})(1+q^{20m-12})(1+q^{20m-8})(1+q^{20m-18})\cr
\qquad{}= \prod_m
(1-q^{20m})(1+q^{20m-2})(1+q^{20m-18})\hfill\cr
\noalign{\vskip-10pt}
\hfill{}\times
(1-q^{20m-8})(1-q^{20m-12})(1+q^{20m-12})(1+q^{20m-8})\cr
\qquad{}= \prod_m
(1-q^{20m})(1+q^{20m-2})(1+q^{20m-18})  
(1-q^{40m-16})(1-q^{40m-24}),\cr
}$$
which is the left-hand side of (4.10). Hence (4.6) is
proved.

\goodbreak
In the same manner, the left-hand side of $(4.7')$
is equal to
$$\displaylines{
 \prod_m (1-q^{20m})(1-q^{20m-4})(1-q^{20m-16})
(1+q^{10m-4})(1+q^{10m-6}).\cr
\noalign{\hbox{As}}
\quad
\prod_m(1+q^{10m-4})(1+q^{10m-6})\hfill\cr
\noalign{\vskip -6pt}
\hfill{}= \prod_m
(1+q^{20m-4})(1+q^{20m-6})(1+q^{20m-14})(1+q^{20m-16}),\cr}
$$
we also see that the left-hand side of $(4.7')$ is equal to
$$
\displaylines{\quad
 \prod_m
(1-q^{20m})(1-q^{20m-4})(1-q^{20m-16})\hfill\cr
\noalign{\vskip -10pt}
\hfill{}\times 
(1+q^{20m-4})(1+q^{20m-6})(1+q^{20m-14})(1+q^{20m-16})\cr
\qquad{}= \prod_m (1-q^{20m})(1+q^{20m-6})(1+q^{20m-14})
\hfill\cr
\noalign{\vskip-10pt}
\hfill{}\times
(1-q^{20m-4})(1-q^{20m-16})(1+q^{20m-4})(1+q^{20m-16})\cr
\qquad{}= \prod_m (1-q^{20m})(1+q^{20m-6})(1+q^{20m-14})
(1-q^{40m-8})(1-q^{40m-32}),\hfill\cr }$$
which is the left-hand side of (4.11).
This achieves the proof of (4.7) and then of the
Lemma.\cqfd

\bigskip
{\bf Acknowledgements:} We should like to thank George
Andrews for the interest he has constantly showed in our
struggle with those very classical identities and Frank
Garvan for drawing our attention to the paper by Farkas
and Kra.




\vglue 2cm

\centerline{\bf References}

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}
\vskip 1cm
\line{\quad\vtop{\halign{#\hfil\cr
Dominique Foata\cr
D\'epartement de math\'ematique\cr
Universit\'e Louis Pasteur\cr
7, rue Ren\'e-Descartes\cr
F-67084 Strasbourg\cr
{\tt foata@math.u-strasbg.fr}\cr}}\hfil
\vtop{\halign{#\hfil\cr
Guo-Niu Han\cr
{\sevenrm I.R.M.A.} et {\sevenrm C.N.R.S.}\cr
Universit\'e Louis Pasteur\cr
7, rue Ren\'e-Descartes\cr
F-67084 Strasbourg\cr
{\tt guoniu@math.u-strasbg.fr}\cr}}\quad}
\bye




