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\allowdisplaybreaks
\title[Engel expansions]{$\boldsymbol{ q}$--Engel
  Series Expansions and Slater's Identities}

\author[George E.~Andrews]{George E.~Andrews$^{\dag{}}$}
\thanks{$^\dag{}$Partially supported by National 
Science Foundation Grant DMS-9206993 }
\address[G.~E.~A.]{Department of Mathematics  \\
The Pennsylvania State University   \\
University Park, PA  16802   \\
U. S. A.}
\email{andrews@math.psu.edu}

\author[A. Knopfmacher]{Arnold Knopfmacher} 
\address[A. K.]{The John Knopfmacher Centre for Applicable Analysis and Number
  Theory\\
University of the Witwatersrand\\
Private Bag 3\\
WITS 2050, South Africa}
\email{arnoldk@cam.wits.ac.za}
\urladdr{http://www.wits.ac.za/science/number\underline{\ }theory/arnold.htm}



\author[Peter Paule]{Peter Paule$^{\ddag{}}$}
\thanks{$^\ddag{}$Partially supported by 
grant F1305   of the Austrian FWF}

\address[P. P.]{Research Institute for Symbolic Computation  \\
Johannes Kepler University Linz  \\
A-4040 Linz\\
Austria}
\email{Peter.Paule@risc.uni-linz.ac.at}
\urladdr{
http://www.risc.uni-linz.ac.at/research/combinat/}

\author[H. Prodinger]{Helmut Prodinger} 
\address[H. P.]{The John Knopfmacher Centre for Applicable Analysis and Number
  Theory\\
University of the Witwatersrand\\
Private Bag 3\\
WITS 2050, South Africa}
\email{helmut@cam.wits.ac.za}
\urladdr{http://www.wits.ac.za/helmut/index.htm}










\begin{document}

\keywords{Engel series, $q$--series, Al-Salam/Ismail polynomials, identities}
\subjclass{Primary: 11; Secondary: 11}
\date{\today}


\begin{abstract}
We describe the $q$--Engel series expansion for Laurent series 
discovered by John Knopfmacher and use this algorithm to shed
new light on partition identities related to two entries from
Slater's list. In our study Al-Salam/Ismail and Santos polynomials
play a crucial r{\^o}le.
\end{abstract}



\maketitle

\begin{center}{\it
Dedicated to the memory of John Knopfmacher 1937--1999}
\end{center}
\section{Introduction}

In 1987 John Knopfmacher conceived of the idea of representing formal Laurent series
as sums of reciprocals of polynomials. The initial motivation was some old results on
representations for real numbers by sums of reciprocals of integers, due originally to Lambert,
Engel and Sylvester. John together with Arnold had previously investigated various extensions of
the real number representations (see the article by Kalpazidou and Ganatsiou \cite{Kal} in this issue). However the development
of analogous expansions for Laurent Series turned out to have some unexpected benefits.

Around the time of publication of \cite{KK1} it was noticed that a number of
famous expansions including those of Euler and the Rogers--Ramanujan
identities were, in fact, special cases of the $q$--Engel expansion.  This led to the interesting project of using the
$q$--Engel algorithm (described below) to provide new proofs of these identities.  This gave rise to a first paper by George,
Arnold and John \cite{AnAKJK} in which the Rogers--Ramanujan 
identities
and some identities of Euler were given new inductive proofs using
the $q$--Engel algorithm discovered by John. Subsequently George, Arnold and Peter have continued these investigations
leading to the further publications \cite{AnAK,AnKnPaZi00,AnAKPP}.

To explain our new results we begin by recalling the $q$--Engel expansion \cite{KK1,KK2} for the field
$\EuScript{L}
 = \C(\!(q)\!)$ of formal Laurent series over the complex
numbers, $\C$.  If
$$
 A = \sum_{n=\nu}^{\infty} L_n q^n\quad\text{with\ \ $L_{\nu}\neq0$},
$$
we call $\nu = \nu(A)$ the {\sc order} of $A$ and we define the
{\sc norm} of $A$ to be
$$
 \|A\| = 2^{-\nu(A)}\,.
$$
In addition, we define the {\sc integral part} of $A$ by
\begin{equation}\label{eq:int}
 [A] = \sum_{\nu \leq n\leq 0} L_n q^n\,.
\end{equation}


Engel ({c.\;f.} \cite[$\S$34]{Perron}) originally defined a series expansion for
real numbers.  In \cite{KK1}, this concept was extended to 
$\EuScript{L}$
in the following way:


\begin{theorem}\label{thm1}{\rm [\/{$q$--Engel Expansion Theorem}
 (\cite[th. 1.\;4]{KK1}).]}   Every 
$A \in \EuScript{L}$ has a
finite or convergent (relative to the above norm) series
expansion of the form
\begin{equation}\label{eq:eng}
 A = a_0 + \sum_{n=1}^{\infty}  \frac1{a_1 a_2 \cdots a_n}\;,
\end{equation}

where $a_n \in \C [q^{-1}]$, $a_0 = [A]$,
\begin{equation}\label{eq:rec}
 \nu(a_n) \leq - n\,, \text{ and }\  \nu(a_{n+1}) \leq
 \nu(a_n) - 1\,.
\end{equation}


The series \eqref{eq:eng} is unique for $A$, 
and it is finite if and only if $A \in \C(q)$.  In addition,
if
$$
 a_0 + \sum_{j=1}^n \frac1{a_1 \cdots a_j} = 
 \frac{p_n}{q_n} \;, \quad \text{ where }\
 q_n = a_1 a_2 \cdots a_n\,,
$$
then
$$
 \left\| A - \frac{p_n}{q_n}\right\| \leq \frac1{2^{n+1}
 \| q_n\|}
$$
and
$$
 \nu\Big(A - \frac{p_n}{q_n}\Big) = - \nu(q_{n+1}) \geq
 \frac{(n+1)(n+2)}{2}\;.
$$

In fact, the $a_n$ (the ``digits'') are for $n\ge1$ given 
recursively by 
\begin{equation}\label{eq:alg}
 a_n = \left[ \frac1{A_n}\right]
\end{equation}
where $A_0 = A$, $a_0 = [A]$, $A_1=A-a_0$, and for $n\ge1$ 
\begin{equation}\label{eq:alg2}
 A_{n+1} = a_n A_n - 1.
\end{equation}\qed
\end{theorem}

To illustrate for example how the first Rogers--Ramanujan identity represents 
a \goodbreak {$q$--Engel expansion,} we write it as
\begin{align*}
  \prod_{n=0}^{\infty} \frac1{(1 - q^{5n+1})(1 - q^{5n+4})}
 &= 1 + \sum_{n=1}^{\infty} \frac{q^{n^2}}{(1 - q)(1 - q^2)
  \cdots (1 - q^n)}
 \\
& = 1 + \sum_{n=1}^{\infty} \frac1{a_1 a_2 \cdots a_n}
 \;,
\end{align*}
where $a_n=(1 - q^n)/q^{2n-1}$ for $n \geq 1$.


In Section 2, we will in fact make use of a slight variation of the 
$q$--Engel algorithm 
in which \eqref{eq:alg2}
is replaced by 
\begin{equation}\label{eq:alg3}
 A_{n+1} = q(a_n A_n - 1).
\end{equation}
Corresponding to \eqref{eq:eng} we have instead the modified expansion,
\begin{equation}\label{eq:engq}
 A = a_0 + \sum_{n=1}^{\infty}  \frac{q^{-n}}{a_1 a_2 \cdots a_n}\;.
\end{equation}
An explicit treatment of this modified $q$--Engel
expansion can be found in \cite{AnKnPaZi00}.

\medskip

In this paper we use the Engel approach
     e.~g. to derive in a new way the representation

  \[ \lim_{n\rightarrow \infty} U_n(0) =
          (q;q^2)_\infty \sum_{n\geq 0} \frac{q^{2 n^2}}{(q;q)_{2n}}
\]
with $ \qf xn:=(1-x)(1-xq)\dots(1-xq^{n-1})$
     for the limit of specialized Al-Salam/Ismail polynomials
     defined below.
 
On the other hand, it has
already been shown \cite{AnSa97} that the {\it Santos \/}
polynomials converge as follows
\begin{align*}
S_n&\to S_\infty:=\prod_{k\ge1,\ k\equiv \pm 2,\, \pm3,\, 
\pm4,\, \pm5
\pmod{16}}\frac{1}{1-q^k}. 
\end{align*}
We link the Al-Salam/Ismail polynomials to the Santos 
polynomials and shed in this way new light on formul{\ae}
related to two identities due to Slater one of which
being of the form
\begin{align*}
\sum_{n\ge0}\frac{q^{2n^2}}
{\qf q{2n}}&=\prod_{k\ge1,\ k\equiv \pm 2,\, \pm3,\, 
\pm4,\, \pm5
\pmod{16}}\frac{1}{1-q^k}.
\end{align*}


One of the main results is Theorem~2
     which embeds a new generating function relation for
     Al-Salam/Ismail polynomials into a $q$--Engel context.






\noindent
Finally, in Section~\ref{sec_paule}, we discuss Al-Salam/Ismail and 
Santos polynomials in the context of identities of
Garrett/Ismail/Stanton type.


\section{Al-Salam/Ismail polynomials and Slater's identities
(38) and (39)}

The Al-Salam and Ismail  polynomials $U_n(x; a, b\,|\,q)$ 
are defined by \cite{AlIs83}
\begin{align*}
U_{-1}(x; a, b\,|\,q) &= 0,\qquad
U_0(x; a, b\,|\,q) = 1,   \\
U_{n}(x; a, b\,|\,q) &= x(1+aq^{n-1}) U_{n-1}(x; a, b\,|\,q) 
- bq^{n-2} U_{n-2}(x; a, b\,|\,q),
\quad n \ge 1.
\end{align*}



Al-Salam and Ismail   gave 
 the explicit  representation
\begin{align*}U_{n}(x; a, b\,|\,q)
 &= \sum_{k =0}^{\lfloor n/2 \rfloor}
 \frac{(-a;q)_{n-k} (q;q)_{n-k}
x^{n-2k}}{(-a;q)_{k} (q;q)_{k}  (q;q)_{n-2k}} (-b)^kq^{k(k-
1)}\\
&= \sum_{k =0}^{\lfloor n/2 \rfloor}
 {(-aq^k;q)_{n-2k} 
x^{n-2k}}\gauss{n-k}{k} (-b)^kq^{k(k-1)};
\end{align*}
here we used the 
{\it Gaussian\/} polynomials
\begin{equation*}
\gauss nk=\frac{\qf qn}{\qf qk\qf q{n-k}}.
\end{equation*}








We specialize and consider
$U_n(1;-q^{2\alpha+1},-q^{2\alpha+2}\,|\,q^2)$,
but simply write $U_n(\alpha)$ for that. Let us
also rewrite the recursion


\begin{align*}
U_{-1}(\al) &= 0, \quad U_{0}(\al) = 1, \\
U_{n}(\al) &= (1-q^{2n+2\al-1}) U_{n-1}(\al) + q^{2n+2\al-
2} U_{n-2}(\al),
\quad n \ge 1 \end{align*} and the explicit formula



\begin{align*}
U_{n}(\al)
&= \sum_{k =0}^{\lfloor n/2 \rfloor}
 \gausss{n-k}{k}q^{2k^2+2k\al}{(q^{2k+1+2\al};q^2)_{n-
2k} }.
\end{align*}

In our treatment, $\al$ will be either 0 or 1, and we
try to treat both cases simultaneously whenever possible.



The recursion can be rewritten as 



\begin{equation*}
\frac{U_n(\al)}{\qff {q^{1+2\al}}{n}}
=\frac{U_{n-1}(\al)}{\qff {q^{1+2\al}}{n-1}}
+\frac{q^{2n-2+2\al}}{{\qff {q^{1+2\al}}{n}}
}U_{n-2}(\al)
\end{equation*}
and summed:




\begin{equation*}
\frac{U_n(\al)}{\qff {q^{1+2\al}}{n}}
=1+\sum_{k=1}^n\frac{q^{2k-2+2\al}}{{\qff 
{q^{1+2\al}}{k}}
}U_{k-2}(\al).
\end{equation*}
In the limit $n\to\infty$,
\begin{equation*}
\frac{U_\infty(\al)}{\qff {q^{1+2\al}}{\infty}}
=1+\sum_{k\ge1}\frac{q^{2k-2+2\al}}{{\qff 
{q^{1+2\al}}{k}}
}U_{k-2}(\al).
\end{equation*}


\begin{theorem}
If one applies the (modified) Engel algorithm to 
${U_\infty(\al)}\big/{\qff {q^{1+2\al}}{\infty}}$
the quantities $A_n(\al)$ are given by
\begin{equation*}   
A_n(\al)=\sum_{k\ge2}\frac{q^{2kn+2\al-1}U_{k-2}(\al)}
{\qff{q^{2n-1+2\al}}k}=
\sum_{j\ge0}\frac{q^{2j^2+2\al j+4nj+4n+2\al-1}}
{\qff{q^{2n-1+2\al}}{j+1}\qff{q^{2n}}{j+1}};
\end{equation*}
the digits $a_n(\al)$ are given by

\begin{equation*}   
a_n(\al)=\frac{(1-q^{2n-1+2\al})(1-q^{2n})}{q^{4n-
1+2\al}}\qquad\text{for $n\ge1$}
\end{equation*}
and $a_0(\al)=1$.
\end{theorem}


\begin{proof}
First, let us prove that the two expressions given
for $A_n(\al)$ are indeed equal:




\begin{align*}
\sum_{k\ge2}&\frac{q^{2kn+2\al-1}U_{k-2}(\al)}
{\qff{q^{2n-
1+2\al}}k}\\&=\sum_{k\ge0}\frac{q^{2(k+2)n+2\al-1}}
{\qff{q^{2n-1+2\al}}{k+2}}
\sum_{0\le 2j\le k}
 \gausss{k-j}{j}q^{2j^2+2j\al}{(q^{2j+1+2\al};q^2)_{k-2j} 
}\\
&=\sum_{j,k\ge0}\frac{q^{2(k+2j+2)n+2\al-1}}
{\qff{q^{2n-1+2\al}}{k+2j+2}}
 \gausss{k+j}{j}q^{2j^2+2j\al}{(q^{2j+1+2\al};q^2)_{k} 
}\\
&=\sum_{j,k\ge0}\frac{q^{2(2j+2)n+2\al-
1+2nk+{2j^2+2j\al}}}
{\qff{q^{2n-1+2\al}}{2j+2} \qff{q^{2n+4j+3+2\al}}{k} }
 \frac{\qff{q^{2j+2}}{k}}{\qff{q^{2}}{k}}
{(q^{2j+1+2\al};q^2)_{k} }\\
&=\sum_{j\ge0}
\frac{q^{2(2j+2)n+2\al-1+2j^2+2j\al}}
{\qff{q^{2n-1+2\al}}{2j+2}  }
\sum_{k\ge0}
\frac{\qff{q^{2j+2}}{k}{(q^{2j+1+2\al};q^2)_{k} 
q^{2nk}}} 
{\qff{q^{2}}{k}\qff{q^{2n+4j+3+2\al}}{k} }
\\
\intertext{(the inner sum can be computed by   
$q$--Gauss \cite[p.20]{Andrews76})}\\
&=\sum_{j\ge0}
\frac{q^{2(2j+2)n+2\al-1+2j^2+2j\al}}
{\qff{q^{2n-1+2\al}}{2j+2}  }
\frac{\qff{q^{2n+2\al+2j+1}}{\infty}\qff{q^{2n+2j+2}}
{\infty}  }
{\qff{q^{2n+2\al+4j+3}}{\infty}\qff{q^{2n}}{\infty}}\\
&=\sum_{j\ge0}
\frac{q^{4jn+4n+2\al-1+2j^2+2j\al}}
{\qff{q^{2n-1+2\al}}{2j+2}  }
\frac{\qff{q^{2n+2\al+2j+1}}{j+1}  }
{\qff{q^{2n}}{j+1}}\\
&=\sum_{j\ge0}
\frac{q^{4jn+4n+2\al-1+2j^2+2j\al}}
{\qff{q^{2n-1+2\al}}{j+1}  
\qff{q^{2n}}{j+1}},
\end{align*}
as desired. 

Let us now prove the announced formula
for the digits.

This is particularly easy since the $j=0$ term {\it is\/} the 
reciprocal
of $a_n(\al)$, and therefore

\begin{equation*}
\sum_{j\ge0}\frac{q^{2j^2+2\al j+4nj+4n+2\al-1}}
{\qff{q^{2n-1+2\al}}{j+1}\qff{q^{2n}}{j+1}}
=
\frac1{a_n}+O(q^{8n+1+4\al})=
\frac1{a_n}\Big(1+O(q^{4n+2\al+2})\Big).
\end{equation*}

The recursion is also  simple, since

\begin{align*}
&q\big( a_n(\al) A_n(\al) -1 \big)\\ 
&=q\bigg(a_n(\al)\sum_{j\ge0}\frac{q^{2j^2+2\al 
j+4nj+4n+2\al-1}}
{\qff{q^{2n-1+2\al}}{j+1}\qff{q^{2n}}{j+1}}-1\bigg)\\
&=qa_n(\al)\sum_{j\ge1}\frac{q^{2j^2+2\al j+4nj+4n+2\al-1}}
{\qff{q^{2n-1+2\al}}{j+1}\qff{q^{2n}}{j+1}}\\
&=\sum_{j\ge1}\frac{q^{2j^2+2\al j+4nj+1}}
{\qff{q^{2n+1+2\al}}{j}\qff{q^{2n+2}}{j}}\\
&=\sum_{j\ge0}\frac{q^{2(j+1)^2+2\al (j+1)+4n(j+1)+1}}
{\qff{q^{2n+1+2\al}}{j+1}\qff{q^{2n+2}}{j+1}}\\
&=\sum_{j\ge0}\frac{q^{2j^2+2\al j+4(n+1)j+4(n+1)+2\al-1}}
{\qff{q^{2n+1+2\al}}{j+1}\qff{q^{2n+2}}{j+1}}\\
&=A_{n+1}(\al),
\end{align*}
as desired.
\end{proof}


Thus our Engel proof established  the following relations.

\begin{theorem}
For $\al=0,1$ we have


\begin{equation*}
\lim_{n\to\infty}U_n(\al)
=\qff{q}{\infty}\sum_{n\ge0}\frac{q^{2n^2+2n\al}}
{\qf q{2n+\al}}.
\end{equation*}
\end{theorem}

This is a new proof of known results due to
Al-Salam and Ismail \cite{AlIs83}.
\medskip

Slater's identities (39) and (38)\footnote{Observe the order!} 
\cite{Slater52}
are the formul\ae
\begin{align*}
\sum_{n\ge0}\frac{q^{2n^2}}
{\qf q{2n}}&=\prod_{k\ge1,\ k\equiv \pm 2,\, \pm3,\, 
\pm4,\, \pm5
\pmod{16}}\frac{1}{1-q^k},\\
\sum_{n\ge0}\frac{q^{2n^2+2n}}
{\qf q{2n+1}}&=\prod_{k\ge1,\ k\equiv \pm 1,\, \pm4,\, 
\pm6,\, \pm7
\pmod{16}}\frac{1}{1-q^k}.
\end{align*}






Let us consider the Santos polynomials in the representation
\begin{align*} 
S_n&=\sum_{0\le 2j\le n} q^{2j^2}\gauss{n}{2j},\\
T_n&=\sum_{0\le 2j\le n-1}q^{2j^2+2j}\gauss{n}{2j+1}.
\end{align*}
Originally these polynomials were studied in \cite{Santos}
and were discussed further in \cite{AnSa97}. As pointed
out ibid.~they possess the following limit property,
\begin{align*}
S_n&\to S_\infty:=\prod_{k\ge1,\ k\equiv \pm 2,\, \pm 3,\, 
\pm 4,\, \pm 5
\pmod{16}}\frac{1}{1-q^k},\\ T_n&\to 
T_\infty:=\prod_{k\ge1,\ k\equiv \pm 
1,\, \pm 4,\, \pm 6,\, \pm 7 \pmod{16}}\frac{1}{1-q^k}. 
\end{align*} 
In \cite{AnKnPaZi00}, 
the quantities $S_\infty$ and $T_\infty$ were undergone
an Engel treatment; see also Section~\ref{sec_paule} below.

After observing these limits, Slater's identities
(39) and (38) are immediate by taking $n\rightarrow \infty$ in
the polynomial representations of $S_n$ and $T_n$ above.
However, the Engel approach encoded by Theorem 2 admits
another link to Slater's identities which is briefly
explained as follows.

Although it is not shown explicitly in \cite{AnSa97}, it is
easy to verify that the polynomials $S_n$
and $T_n$ satisfy the following
defining recurrences
\begin{align*}
S_n&=S_{n-1}+q^nT_{n-1},&\kern-2.5cm S_0&=1,\\
T_n&=T_{n-1}+q^{n-1}S_{n-1},&\kern-2.5cm T_0&=0.\\
\end{align*}
As a direct consequence of these recurrences we see that
\begin{equation*}
S_\infty=1+\sum_{j\ge1}q^jT_{j-1}
\quad\text{ and }\quad 
(1-q)\, T_\infty =1+\sum_{j\ge1}q^{2j}T_{j-1}.
\end{equation*}
So, by Theorem 2 we obtain an alternative statement
being equivalent to Slater's identities; namely,
\begin{equation*}
\frac{1}{q} A_1(\al)=\sum_{j\ge1}q^{(1+\al)j}T_{j-1} \qquad
(\al = 0,1) 
\end{equation*}
where $A_1(\al)$ is chosen as in Theorem 2.
As we shall see below, this fact is easily established, 
and---again as a by-product of the Engel context of  
Theorem 2---one obtains the more general
relation
\begin{equation*}
A_n(\al)=\sum_{j\ge1}q^{(2n-1+\al)j+1}T_{j-1}\qquad
 (\al = 0,1)  
\end{equation*}
where $A_n(\al)$ is chosen as in Theorem 2.
However, the proof for $n=1$ is at the same level of complexity 
as for  $n$, and so we do the computation for general $n$.

\noindent

\begin{align*}
\sum_{j\ge1}q^{(2n-1+\al)j+1}T_{j-1}
&=\sum_{j\ge1}q^{(2n-1+\al)j+1}
\sum_{0\le 2k\le j-2} q^{2k^2+2k}\gauss {j-1}{2k+1}\\
&=\sum_{j,k\ge0}q^{(2n-1+\al)(j+2k+2)+1+2k^2+2k}
\gauss {j+2k+1}{2k+1}\\
&=\sum_{k\ge0}q^{(2n-1+\al)(2k+2)+1+2k^2+2k}
\sum_{j\ge0}q^{(2n-1+\al)j}
\gauss {j+2k+1}{2k+1}\\
&=\sum_{k\ge0}\frac{q^{(2n-1+\al)(2k+2)+1+2k^2+2k}}
{\qf{q^{2n-1+\al}}{2k+2}}.
\end{align*}



For $\al=0$ we have


\begin{align*}
\sum_{k\ge0}\frac{q^{ 4nk +4n-1+2k^2}}
{\qf{q^{2n-
1}}{2k+2}}=\sum_{j\ge0}\frac{q^{2j^2+4nj+4n-1}}
{\qff{q^{2n-1}}{j+1}\qff{q^{2n}}{j+1}},
\end{align*}
and for $\al=1$

\begin{align*}
\sum_{k\ge0}\frac{q^{4nk+4n+1+2k^2+2k}}
{\qf{q^{2n}}{2k+2}}
=\sum_{j\ge0}\frac{q^{2j^2+2 j+4nj+4n+1}}
{\qff{q^{2n+1}}{j+1}\qff{q^{2n}}{j+1}},
\end{align*}
as it should.


(For other values of $\al$ it does not work!)

Finally we will establish that the 
Al-Salam/Ismail sequences
$U_n(\al)\big/\qff{q}n$ converge about
twice as fast to $S_\infty$ resp. $T_\infty$ as
$S_n$ resp. $T_n$.


For that we need two simple facts:

\begin{align*}
\gauss AB&=\frac{(1-q^{A})(1-q^{A-1})\dots(1-q^{A-B+1})}
{\qf qB}\\
&=\frac{1}{\qf qB}\Big(1-q^{A-B+1}+O(q^{A-B+2})\Big)
\end{align*}
and
\begin{equation*}   
\frac{1}{\qf {q^c}n}=1+q^c+O(q^{c+1}).
\end{equation*}

Therefore
\begin{align*}
S_n&=\sum_{j\ge0}\gauss{n}{2j}q^{2j^2}
=1+\sum_{j\ge1}\frac{q^{2j^2}}{\qf q{2j}}
\Big(1-q^{n-2j+1}+O(q^{n-2j+2})\Big)\\
&=\sum_{j\ge0}\frac{q^{2j^2}}{\qf q{2j}}
-q^{n+1}+O(q^{n+2}).
\end{align*}
Similarly
\begin{align*}
T_n&=\sum_{j\ge0}\gauss{n}{2j+1}q^{2j^2+2j}=\frac{1-q^n}{1-q}
+\sum_{j\ge1}\frac{q^{2j^2+2j}}{\qf q{2j+1}}
\Big(1-q^{n-2j}+O(q^{n-2j+1})\Big)\\
&=\sum_{j\ge0}\frac{q^{2j^2+2j}}{\qf q{2j+1}}
-q^{n}+O(q^{n+1}).
\end{align*}

On the other hand,
\begin{align*}
\frac{U_{n}(0)}{\qff qn}
&= \frac{1}{\qff qn}\sum_{j \ge0}
 \gausss{n-j}{j}q^{2j^2}{(q^{2j+1};q^2)_{n-2j}}\\
&= \sum_{j \ge0}
 \gausss{n-j}{j}q^{2j^2}\frac{1}{\qff qj\qff{q^{2n-
2j+1}}j}\\
&=1+\sum_{j \ge1}
 \frac1{\qff{q^2}j}\Big(1-q^{2n-4j+2}+O(q^{2n-4j+4})\Big)
\frac{q^{2j^2}}{\qff qj}
\\* &\qquad\qquad\qquad\qquad\qquad\qquad\times
\Big(1+q^{2n-2j+1}+O(q^{2n-2j+3})\Big)\\
&=\sum_{j \ge0}
 \frac{q^{2j^2}}{\qf q{2j}}-q^{2n}+O(q^{2n+1}).\\
\end{align*}
Similarly,
\begin{align*}
\frac{U_{n}(1)}{\qff qn}
&= \frac{1}{\qff qn}\sum_{j \ge0}
 \gausss{n-j}{j}q^{2j^2+2j}{(q^{2j+3};q^2)_{n-2j}}\\
&= \frac{1-q^{2n+1}}{1-q}+\sum_{j \ge1}
 \gausss{n-j}{j}q^{2j^2+2j}\frac{1}{\qff q{j+1}
\qff{q^{2n-2j+3}}{j-1}}\\
&=\frac{1-q^{2n+1}}{1-q}+\sum_{j \ge1}
 \frac1{\qff{q^2}j}\Big(1-q^{2n-4j+2}+O(q^{2n-4j+4})\Big)
\frac{q^{2j^2+2j}}{\qff q{j+1}}\\
&\qquad\qquad\qquad\qquad\qquad\qquad\times
\Big(1+q^{2n-2j+3}+O(q^{2n-2j+5})\Big)\\
&=\sum_{j \ge0}
 \frac{q^{2j^2+2j}}{\qf q{2j+1}}-
q^{2n+1}+O(q^{2n+2}).\\
\end{align*}




\section{Identities of  Garrett/Ismail/Stanton type}\label{sec_paule}

\medskip
\noindent
In \cite{GIS} Garrett et al.~presented a new 
parameterized generalization of the celebrated
Rogers--Ramanujan identities. As a by-product of an Engel
study, Andrews et al.~\cite{AnAKPP}
derived a polynomial version of it which in the limit
coincides with the Garrett/Ismail/ Stan\-ton result.
In \cite{IPS} Ismail et al.~have put the polynomial
version from \cite{AnAKPP} into the context of
orthogonal polynomials, in particular, of the
Al-Salam/Ismail polynomials $U_n$. 

It turned out that not only the Rogers--Ramanujan
identities but also other entries listed by Slater \cite{Slater52}
give rise to this type of generalization. For instance,
in \cite{IPS} it was shown that for $m \geq 0$,
\begin{equation}
\label{PP1}
(-1)^m q^{m^2+m} U_n(m+1) = 
U_{m}(0) \,U_{m+n}(1) - U_{m-1}(1) \,U_{m+n+1}(0).
\end{equation}
Above we have proved that
\begin{equation}
\label{PP2}
\frac{ U_{\infty}(0) }{ \qff{q}{\infty} }=
S_{\infty}
\hbox{\ \ \ and \ \ \ }
\frac{U_{\infty}(1)}{ \qff{q}{\infty} } =
T_{\infty}. 
\end{equation}  
So after dividing both sides of \eqref{PP1} by $\qff{q}{\infty}$ 
we obtain in the limit $n\rightarrow \infty$ 
an identity of Garrett/Ismail/Stanton type;
namely for $m\geq 0$,
\begin{equation}
\label{PP3}
(-1)^m q^{m^2+m} \sum_{k\geq 0}
\frac{ q^{2k^2+2(m+1)k} }{ \qf{q}{2k+1} \qff{q^{2k+3}}{m} } = 
U_m(0)\, T_{\infty}-U_{m-1}(1)\, S_{\infty}.
\end{equation}  

\medskip
The experimental use of \texttt{Engel}, a computer algebra implementation of 
the $q$--Engel expansion algorithm, led Andrews et al.~\cite{AnKnPaZi00} 
to the discovery of an identity similar to \eqref{PP3} but
using Santos polynomials instead; namely for $m\geq 0$,
\begin{align}
\label{PP4}
q^m (q;q^2)_m
\sum_{k \geq 0} \frac{q^{2k^2+ 2(m+1) k}}{(q;q)_{2k+1}}=
S_m \,T_\infty - T_m \,S_\infty.
\end{align}
The corresponding polynomial version, the counterpart to
\eqref{PP1}, reads as follows. For $m\geq 0$,
\begin{align}
\label{PP5}
q^m (q;q^2)_m
\sum_{k \geq 0} \gauss{n}{2k+1} q^{2 k^2 + 2(m+1) k} =
S_m\, T_{m+n} - T_m\, S_{m+n}.
\end{align}
Once found, such identities most often find quite 
elementary proofs. Nevertheless, the theme of this
section is to sketch a framework that helps to explain 
and to derive identities of this type. We also stress
the fact that this approach can be ideally supplemented 
by computer algebra packages like \texttt{Engel} and 
$q$--versions of Zeilberger's (``fast'') algorithm, as for 
instance \texttt{qZeil} \cite{PR}. Such packages
not only can help in proving, but also in {\it finding}
such identities.

\medskip 
All what follows is motivated by techniques from orthogonal
polynomials as used e.~g. in \cite{IPS}. However, we will 
not enter this theory (e.~g., numerator or associated polynomials)
but rather restrict ourselves to recall a few basic facts from the
general theory of difference equations.

Let $\mathbb{F}$ be a suitable field, as e.~g. 
$\mathbb{F}=\mathbb{C}(q)$ where $q$ is an indeterminate.
Let $\mathbb{F}^{\star}$ denote the non-zero elements
of $\mathbb{F}$. Let us fix two coefficient sequences
$\al = (\al_n)_{n\geq 0}$ and $\beta = (\beta_n)_{n\geq 0}$
with elements in $\mathbb{F}^{\star}$. Consider the
recurrence equation
\begin{align}
\label{star}
x_n = \al_{n-1} x_{n-1} + \beta_{n-2} x_{n-2} \hbox{\ \ \ } (n\geq 2).
\end{align}
Suppose the $\mathbb{F}$--sequences  $a = (a_n)_{n\geq 0}$ and 
$b = (b_n)_{n\geq 0}$ are solutions of \eqref{star}. For
$n\geq 0$ we define the discrete Wronskian as usual by
\begin{align*} 
W_n(a,b) :=
\left| \begin{matrix}
                  a_n & b_{n} \\
          a_{n+1} & b_{n+1}
          \end{matrix} \right|.
\end{align*}
As a matter of fact, the solutions $a$ and $b$ are linearly
independent over $\mathbb{F}$
if and only if $W_n(a,b)\neq 0$ for all $n\geq 0$.
But this is equivalent to $W_0(a,b)\neq 0$ since
\begin{align*}
W_n(a,b) = (-1)^n \beta_{n-1} \beta_{n-2} \dots \beta_{0}\, W_{0}(a,b) 
\hbox{\ \ \ } (n\geq 0).
\end{align*}
Combining these facts one can easily 
prove the following theorem. 
\begin{theorem}
\label{thm3}
Let $\al$, $\beta$, $a$, and $b$ be sequences as above where
$a$ and $b$ are linearly independent solutions of \eqref{star}.
Let $m$ be a non-negative integer and let 
$c(m)=\big( c_n(m) \big)_{n\geq 0}$ satisfy 
\begin{equation}
\label{z star}
z_n = \al_{m+n-1}\, z_{n-1} + 
\beta_{m+n-2}\, z_{n-2} \hbox{\ \ \ } (n\geq 2).
\end{equation}
Then 
\begin{equation*}
c_n(m) = u_{m}\, a_{m+n} + 
v_{m}\, b_{m+n} \hbox{\ \ \ } (n\geq 0)
\end{equation*}
where
\begin{equation*}
u_m = - \frac{c_1(m)\, b_{m} - c_0(m)\, b_{m+1}}{W_m(a,b)} 
\end{equation*}
and 
\begin{equation*}
v_m = \frac{c_1(m)\, a_{m} - c_0(m)\, a_{m+1}}{W_m(a,b)}. 
\end{equation*}
\end{theorem}

\medskip
Now we apply  Theorem~\ref{thm3} in order to prove the polynomial 
identities \eqref{PP1} and \eqref{PP4}. 

\medskip
\noindent
{\bf Example 1.} Let $m$ be a non-negative integer.  
The Al-Salam/Ismail polynomials $U_n(m)$
satisfy \eqref{z star} with $\al_n =1-q^n$ and
$\beta_n=q^n$. So in view of Theorem \ref{thm3}
we can take $a= (a_n)_{n\geq 0}$ with $a_n=U_n(0)$
and $c(m)= \big ( c_n(m) \big)_{n\geq 0} $ 
with $c_n(m)=U_n(m)$. But we also need a linearly
independent solution $b=(b_n)_{n\geq 0}$. Let us
try $b_n=U_{n-1}(1)$ since then $a$ and $b$ are both 
solutions of \eqref{star}. It turns out that $b$
chosen this way is also linearly independent since
$W_0(a,b)=U_0(0) U_0(1) - U_1(0) U_{-1}(1) = 1$;
therefore we can invoke Theorem \ref{thm3}.
It is easily checked that 
\begin{equation*}
W_m(a,b)= (-1)^m q^{2m} q^{2m-2} \dots q^2 W_0(a,b)=
(-1)^m q^{m^2 + m},
\end{equation*}
and 
\begin{equation*}
c_0(m)=U_0(m)=1 \hbox{\ \ \ and \ \ \ } c_1(m)=U_1(m)=1-q^{2m+1}.
\end{equation*}
For the coefficients we obtain
\begin{equation*}
u_m = -(-1)^m q^{-m^2-m} 
\big( (1-q^{2m+1}) U_{m-1}(1)-U_m(1) \big) = 
(-1)^m q^{-m^2+m} U_{m-2}(1)
\end{equation*}
where the last equation is by (\ref{z star}),
and similarly,
\begin{equation*}
v_m = (-1)^m q^{-m^2-m} 
\big( (1-q^{2m+1}) U_{m}(0)-U_{m+1}(0) \big) = 
-(-1)^m q^{-m^2+m} U_{m-1}(0).
\end{equation*}
Now Theorem \ref{thm3} yields identity \eqref{PP4}
with $m$ replaced by $m-1$.

\medskip
\noindent
{\bf Example 2.} As an easy consequence of the mixed 
recurrences for the Santos polynomials $S_n$ and $T_n$
one immediately derives that both polynomials are
solutions of 
\begin{align}
\label{santos rec}
x_n = (1+q)\, x_{n-1} - q(1-q^{2n-3})\, x_{n-2} 
\hbox{\ \ \ } (n\geq 2).
\end{align}
In view of \eqref{star} we have $\al_n=1+q$ and
$\beta_n=-q(1-q^{2n+1})$. Setting $a_n=S_n$ and
$b_n=T_n$ we see that the corresponding sequences
$a$ and $b$ are linearly independent since
$W_0(a,b)=S_0 T_1-T_0 S_1 = 1$. In addition, 
for any non-negative integer $m$ the sequence 
$c(m)= \big( c_n(m) \big)_{n\geq 0}$ with
\begin{equation*}
c_n(m)=\sum_{k \geq 0} \gauss{n}{2k+1} q^{2 k^2 + 2(m+1) k}
\end{equation*}
satisfies 
\begin{align}
\label{general santos}
z_n = (1+q)\, z_{n-1} - q(1-q^{2(m+n)-3})\, z_{n-2} 
\hbox{\ \ \ } (n\geq 2).
\end{align}
This e.~g.~can be proven automatically with the
package \texttt{qZeil}. Summarizing, we are in
the position to invoke Theorem \ref{thm3}. It is
easily checked that 
\begin{equation*}
W_m(a,b)= q^m (1-q^{2m-1})(1-q^{2m-3})\dots (1-q) W_0(a,b)=
q^m \qff{q}{m},
\end{equation*}
and 
\begin{equation*}
c_0(m)=0 \hbox{\ \ \ and \ \ \ } c_1(m)=1.
\end{equation*}
Hence by Theorem \ref{thm3}, identity \eqref{PP1} is proved.

\medskip
Finally we demonstrate the applicability of Theorem \ref{thm3}
by deriving a related identity which to our knowledge is new.

\noindent
{\bf Example 3.} By using the package \texttt{qZeil} one finds
that
\begin{equation*}
c_n(m)=\sum_{k \geq 0} \gauss{n}{2k} q^{2 k^2 + 2 m k}
\end{equation*}
is also a solution of \eqref{general santos};
this heuristic procedure will be described in a 
forthcoming paper.
 Thus we
can take $a$ and $b$ as in Example 2 which gives the
same expression for $W_m(a,b)$. The only difference
to the previous situation is that now
\begin{equation*}
c_0(m)=1 \hbox{\ \ \ and \ \ \ } c_1(m)=1.
\end{equation*}
We obtain for the coefficients 
\begin{equation*}
u_m = - \frac{q^{-m}}{\qff{q}{m}}(T_{m}-T_{m+1}) = 
\frac{1}{\qff{q}{m}}\, S_m
\end{equation*}
where the last equation is by the mixed Santos recurrence,
and similarly,
\begin{equation*}
v_m =  \frac{q^{-m}}{\qff{q}{m}}(S_{m}-S_{m+1}) = 
- \frac{q}{\qff{q}{m}}\, T_m.
\end{equation*}
Theorem 1 now implies that for $m\geq 0$,
\begin{align}
\label{PP new1}
(q;q^2)_m
\sum_{k \geq 0} \gauss{n}{2k} q^{2 k^2 + 2 m k} =
S_m\, S_{m+n} - q\, T_m\, T_{m+n}.
\end{align}
The corresponding limiting version, i.~e.~for
$n \rightarrow \infty$, reads as follows. For $m\geq 0$,
\begin{align}
\label{PP new2}
(q;q^2)_m
\sum_{k \geq 0} \frac{q^{2k^2+ 2 m k}}{(q;q)_{2k}}=
S_m \, S_\infty - q\, T_m \, T_\infty.
\end{align}



%\newpage





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\end{document}
