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%\section{preamble}


\title{\bf Schur's Determinants and Partition Theorems}

\author{Mourad E. H. Ismail\thanks{Research partially
supported by NSF grant DMS 99-70865}\, , Helmut Prodinger, and 
Dennis Stanton\thanks{Research partially
supported by NSF grant  DMS 99-70627}}
%\thanks{}

 \date{May 18, 2000}

% \keywords{Rogers--Ramanujan identities, Schur's determinant}
% \subjclass{05A30}
% \begin{document}

\maketitle

\begin{abstract}
Garrett, Ismail, and Stanton gave a general formula that
contains the Rogers--Ramanujan identities as special
cases. The theory of associated orthogonal polynomials is then
used to explain determinants that Schur introduced in 1917 and show  that the
Rogers--Ramanujan identities imply the Garrett, Ismail, and Stanton seemingly
more  general formula. Using a result of Slater a continued
fraction is explicitly evaluated.
\end{abstract}


\bigskip

{\bf Running Title}: Schur's Determinants

{\it Mathematics Subject Classification}. Primary 05A30.
 Secondary .

{\it Key words and phrases}. Rogers--Ramanujan identities, Schur's
determinant,
associated orthogonal polynomials, continued fractions.



\setcounter{section}{1}

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\setcounter{thm}{0}


\noindent{\bf 1. Introduction}.
In a recent paper
\cite{GaIsSt99} Garrett, Ismail, and Stanton
prove, amongst many other things, the following generalization
of the celebrated Rogers--Ramanujan identities:
\bea\label{stanton}
&{}& 1+ \sum_{n=1}^\infty \frac{q^{n^2+mn}}{(1-q)(1-q^2)\dots(1-q^n)} \\
&{}&\quad = (-1)^mq^{-\binom{m}{2}}E_{m-2}\prod_{n=0}^\infty
\frac{1}{(1-q^{5n+1})(1-q^{5n+4})}\nonumber \\
&{}& \qquad -(-1)^mq^{-\binom{m}{2}}D_{m-2}\prod_{n=0}^\infty
\frac{1}{(1-q^{5n+2})(1-q^{5n+3})},\nonumber
\eea
with the {\it Schur\/} polynomials, defined by
\bea      % \begin{align*}
D_{m}&=&D_{m-1}+q^mD_{m-2}, \quad  D_0=1, \ D_1=1+q,\\
E_{m}&=& E_{m-1}+q^mE_{m-2},  \quad  E_0=1, \ E_1=1.
\eea  %    \end{align*}
Another proof, based on generalized Engel expansions,
can be found in \cite{AnKnPa99}.

Schur \cite{Schur17} has computed the limits
\bea
D_\infty=\prod_{n=0}^\infty
\frac{1}{(1-q^{5n+1})(1-q^{5n+4})},
\qquad
E_\infty=\prod_{n=0}^\infty
\frac{1}{(1-q^{5n+2})(1-q^{5n+3})}.
\eea


The aim of this note is to explain the result (1.1) within the
context of associated orthogonal polynomials and their disguised
appearance in Schur's work in the form of determinants. One basic
ingredient is a sequence of orthogonal polynomials studied by
Al-Salam and Ismail in \cite{AlIs83}. In \S 2 we show how the 
polynomials
studied by Schur in \cite{Schur17} and the Rogers--Ramanujan 
identities can be
used to give a proof of (1.1).  In \S 3 the more general 
polynomials of
\cite{AlIs83} are used to formulate a more general identity 
than (1.1). An application to some of the Slater identities
\cite{Sl52}  is included in \S 4.



The Al-Salam and Ismail polynomials $\{U_n(x; a, b)\}$ are defined by
\bea
U_0(x; a, b) &=& 1, \quad U_1(x; a, b) = x(1+a), \\
U_{n+1}(x; a, b) &=& x(1+aq^n) U_n(x; a, b) - bq^{n-1} U_{n-1}(x; a, b),
\quad n \ge 1.
\eea
To indicate the dependence of $U_n(x; a, b)$ on $q$, when necessary we
will use the notation $U_n(x; a, b|q)$. In accordance with the theory of
orthogonal  polynomials \cite{AsIs},
the numerator  polynomials $\{U_n^{*}(x; a, b)\}$ satisfy the
recursion in (1.6) and the initial conditions
\bea
U_0^{*}(x; a, b) = 0, \quad U_1^{*}(x; a, b) = 1+a.
\eea
Therefore
\bea
U_n^{*}(x; a, b) = (1+a) U_{n-1} (x; qa, qb).
\eea
Schur \cite{Schur17} actually considered the polynomials $U_n(1; 0,-q)$ and
$U_n^{*}(1; 0,-q)$. In the  notation of (1.2) and (1.3) we have
\bea
D_n = U_{n+1}(1; 0, -q), \qquad  E_n = U_{n+1}^{*}(1; 0, -q) =
U_{n}(1; 0, -q^2).
\eea
Let
\bea
F(z; a, q)= \Sum \frac{(-1)^n z^n}{(q;q)_n (-a;q)_n} q^{n(n-1)}
\eea
where we have used the
standard notation for shifted factorials $(a;q)_n$ found in 
\cite{AnAsRo},
\cite{GaRa}.
Al-Salam and Ismail \cite{AlIs83} proved that the limiting relation
\bea
\lim_{n \to \infty} z^{-n}U_n(z; a, b) = (-a;q)_\infty \; F(b/z^2; a, q),
\eea
holds uniformly on compact subsets of the complex $z$-plane
which do not contain  $z = 0$. They also gave 
 the explicit  representation
\bea
U_{n}(x; a, b) = \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)}.
\eea

In \S 4 we recast (1.12) in the form of a generating function with 
generating function variables $a$ and $b$. In order to do so, we 
need the following form of  the
$q$-binomial theorem \cite{AnAsRo}, \cite{GaRa} 
\bea
(-u;q)_n = \sum_{k=0}^n\frac{(q;q)_nq^{k(k-1)/2}}
{(q;q)_k(q;q)_{n-k}}\;u^k.
\eea



\bigskip

\setcounter{section}{2}

 \setcounter{equation}{0}

\setcounter{thm}{0}


\noindent{\bf 2. Schur's determinants}. We first show how Schur would have
proved (1.1) in 1917. Consider
the following determinant of Schur:
%\begin{equation*}
\bea
\schur(b):=\left|
\begin{matrix}
1&bq^{1+m}&&&&\dots\\
-1&1&bq^{2+m}&&&\dots\\
&-1&1&bq^{3+m}&&\dots\\
&&-1&1&bq^{4+m}&\dots\\
&&&\ddots&\ddots&\ddots
\end{matrix}
\right|. \nonumber
%\end{equation*}
\eea

Expanding the determinant with respect to the first column
(``top--recursion'') we get
\bea
\schur(b)=\schur(bq)+bq^{1+m}\schur(bq^2). \nonumber
\eea
Setting
\bea               %\begin{equation*}
\schur(b)=\sum_{n=0}^\infty a_nb^n,\nonumber
\eea
%\end{equation*}
we get, upon comparing coefficients,
\bea     %\begin{equation*}
a_n=q^na_n+q^{1+m}q^{2n-2}a_{n-1},\nonumber
\eea    %\end{equation*}
or
\begin{equation*}
a_n=\frac{q^{2n-1+m}}{1-q^n}a_{n-1}.
\end{equation*}
Since $a_0=1$, iteration leads to
\begin{equation*}
a_n=\frac{q^{n^2+mn}}{(1-q)(1-q^2)\dots(1-q^n)},
\end{equation*}
and thus
the left hand side of (\ref{stanton})
can be expressed by $\schur(1)$.

On the other hand, $\schur(1)$ is the limit of the
{\it finite\/} determinants
\bea  %\begin{equation*}
\schur_n:=\left|
\begin{matrix}
1&q^{1+m}&&&&\dots\\
-1&1&q^{2+m}&&&\dots\\
&-1&1&q^{3+m}&&\dots\\
&&-1&1&q^{4+m}&\dots\\
\vdots&\vdots&\vdots&\ddots&\ddots&\ddots\\
&&&-1&1&q^{n+m}\\
&&&&-1&1\\
\end{matrix}
\right|.           \nonumber
\eea     % \end{equation*}
Expanding this determinant with respect to the last
row
(``bottom--recursion'') we get
\bea   % \begin{equation*}
\schur_n=\schur_{n-1}+q^{n+m}\schur_{n-2}.
\eea       % \end{equation*}
We see that the sequences ${\langle D_{n+m}\rangle}_n$
and ${\langle E_{n+m}\rangle}_n$ satisfy the  recursion (2.1) ,
and thus any linear combination will satisfy the same recurrence relation.
Set
\bea    %\begin{equation*}
\schur_n=\lambda D_{n+m}+\mu E_{n+m}.
\eea     %   \end{equation*}
We can determine the parameters $\lambda$ and $\mu$
using the initial conditions $\schur_0=1$,
$\schur_1=1+q^{1+m}$, which leads to the evaluations
\bea     %\begin{equation*}
\lambda &=& \frac{E_m-E_{m-1}}{D_{m-1}E_m-D_mE_{m-1}}=
\frac{q^mE_{m-2}}{D_{m-1}E_m-D_mE_{m-1}}, \\
    % \end{equation*}
            %  \begin{equation*}
\mu  &=&\frac{D_m-D_{m-1}}{D_{m}E_{m-1}-D_{m-1}E_{m}}=
\frac{q^mD_{m-2}}{D_{m}E_{m-1}-D_{m-1}E_{m}}.
\eea   %\end{equation*}
The denominators in (2.3) and (2.4) are Casorati 
determinants, the
discrete version of a Jacobian and  can be computed explicitly 
\cite{JoTh}.  Indeed
\bea
{D_{m-1}E_m-D_mE_{m-1}}=(-1)^mq^{\binom{m+1}{2}}.
\eea
The proof of (2.5) is by induction on $m$.  The beginning
$m=0$ is trivial; the induction step goes like this:
\bea   %   \begin{align*}
D_{m}E_{m+1}-D_{m+1}E_{m}&=&
D_{m}(E_{m}+q^{m+1}E_{m-1})-(D_{m}+q^{m+1}D_{m-1})E_{m} \nonumber      \\
&=& q^{m+1}(D_{m}E_{m-1}-D_{m-1}E_{m})\nonumber  \\
&=& -q^{m+1}(-1)^mq^{\binom{m+1}{2}}=(-1)^{m+1}q^{\binom{m+2}{2}}.
\nonumber
\eea   %  \end{align*}
This replaces (2.3)--(2.4) by the nicer forms
\bea   %   \begin{equation*}
\lambda     %\frac{q^mE_{m-2}}{D_{m-1}E_m-D_mE_{m-1}}
= (-1)^mq^{-\binom{m}{2}}E_{m-2}, \qquad \mu = (-1)^mq^{-\binom{m}{2}}D_{m-2}.
\eea    %   \end{equation*}
% which amounts to prove that
% \begin{equation*}
% {D_{m-1}E_m-D_mE_{m-1}}=(-1)^mq^{\binom{m+1}{2}}.
% \end{equation*}
% Andrews, Knopfmacher, and Paule \cite{AnKnPa99} give a nice
% formula for $D_iE_j-D_jE_i$. We only need a very particular case
% of that which can be easily proved by induction

Thus the above analysis has led to
\bea % \begin{align}
\label{avoid}
\schur_n=(-1)^mq^{-\binom{m}{2}}E_{m-2}D_{n+m}-
(-1)^mq^{-\binom{m}{2}}D_{m-2}E_{n+m}.
\eea  %\end{align}
Performing the limit $n\to\infty$ this turns into
\bea    % \begin{align*}
\schur(1) =(-1)^mq^{-\binom{m}{2}}E_{m-2}D_{\infty}-
(-1)^mq^{-\binom{m}{2}}D_{m-2}E_{\infty},
\eea    %   \end{align*}
 which is (\ref{stanton}).

\bigskip

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\setcounter{thm}{0}


\noindent{\bf 3. Associated orthogonal polynomials}.
The proof outlined in \S 2  can be considered in the context
of orthogonal polynomials which satisfy three term
recurrences such as (1.2)--(1.3). In this section we give
in Lemma 3.3 a result for general orthogonal polynomials which
specializes to the proof in \S 2. Some applications of Lemma 3.3
to the Al-Salam-Ismail polynomials are also given.

Any sequence of orthogonal polynomials $\{p_n(x)\}$
satisfies a three term recurrence relation
\bea
p_{n+1}(x) = (A_{n}x+B_n)p_n(x)+C_n p_{n-1}(x), \quad n\ge 1,
\eea
and we assume the initial conditions
\bea
 p_0(x)=1, \quad p_{1}(x)=A_0x+B_0.
\eea
The analogue  of Schur's finite determinant is the
well-known tridiagonal determinant
\newpage

\bea
p_n(x)= 
\eea
\bea =\left|
\begin{matrix}
A_0x+B_0&C_1&&&&\dots\\
-1&A_1x+B_1&C_2&&&\dots\\
&-1&A_2x+B_2&C_3&&\dots\\
&&-1&A_3x+B_3&C_4&\dots\\
\vdots&\vdots&\vdots&\ddots&\ddots&\ddots\\
&&&-1&A_{n-2}x+B_{n-2}&C_n\\
&&&&-1&A_{n-1}x+B_{n-1}\\
\end{matrix}
\right|.  \nonumber
\eea
If $A_n=1$, $B_n=0$, $C_n=q^{n+m}$, then
$p_{n+1}(1)=\schur_n$. To see that the polynomials defined by (3.3) satisfy 
(3.1) expand
 the determinant representing $p_{n+1}(x)$ about the
last row. We then verify that $p_1(x)$ and $p_2(x)$ of (3.3)
agree with $p_1(x)$ from (3.2) and $p_2(x)$ which arises from
(1.1) using the initial conditions (1.2).

Recall that the numerator polynomials $\{p_n^{*}(x)\}$ \cite{AsIs},
\cite{JoTh}
associated with $\{p_n(x)\}$ are
defined to be solutions of
\bea
z_{n+1}(x) = (A_{n}x + B_n)\, z_n(x) + C_n \, z_{n-1}(x), 
\quad n \ge1,
\eea
with the initial conditions
\bea
p_0^{*}(x) = 0, \quad p_1^{*}(x)= A_0.
\eea
The two sets of polynomials $\{p_n(x)\}$ and $\{p_{n}^*(x)\}$ form a
basis for solutions of the three-term recurrence (3.4).
One could also consider $\{p_{n+1}^*(x)\}$ as a solution to
the three-term recurrence relation which has the indices shifted up by one.
More generally, the $m$th associated polynomials are defined to be the
solution to
\bea
p_{n+1}^{(m)}(x)=(A_{n+m}x+B_{n+m})p_n^{(m)}(x)+C_{n+m} p_{n-1}^{(m)}(x),
\quad n\ge 1,
\eea
with
\bea
\ p_0^{(m)}(x)=1, \quad  p_{1}^{(m)}(x)=A_{m}x+B_{m}.
\eea

Thus we see that if $A_n=1$, $B_n=0$, $C_n=q^n$, then
\bea
p_{n+1}^{(m)}(1)=\schur_n, \quad D_n=p_{n+1}(1), \quad E_n=p_n^{(1)}(1).
\eea

\begin{thm}
The polynomials $U_n(x;a, -b|q)$ satisfy the polynomial  identity
\bea
 (-b)^{m-1} q^{\binom{m-1}{2}}U_n(x; aq^m, -bq^m|q) &=&
U_{m-1}(x; a, -b|q) U_{n+m-1}(x; aq, -bq|q)   \\
&{}& \quad - U_{m-2}(x; aq, -bq|q) U_{n+m}(x; a, -b|q),
\nonumber
\eea
for $m \ge 1, n \ge 0$, with $U_{-1}(x; a, b|q):= 0$.
\end{thm}


The relationship  (3.9) is an extension of (2.7). After applying
(1.11), the $n\rightarrow\infty$
limit of (3.9) becomes the following corollary.
\begin{cor}
We have the following generalization of (1.1)
\bea
&{}& (-b/x)^{m-1}q^{\binom{m-1}{2}}  F(-bq^m/x^2;aq^m, q)  \\
 &{}&  \quad
= \left[(-aq;q)_{m-1} U_{m-1}(x;a,b)F(-bq/x^2;aq, q)   \right. \nonumber \\
 &{}& \qquad \quad -\left.
(-a;q)_m x U_{m-2}(x,aq,-bq)F(-b/x^2;a,q) \right].\nonumber
\eea
\end{cor}

The proof of Theorem 3.1 depends on a Lemma well-known to
those who are familiar with the analytic theory of continued
fractions and orthogonal polynomials. We include its proof
only to make this work as self-contained as possible.
\begin{lem}
The associated polynomials $\{p_n^{(m)}(x)\}$ satisfy
\bea
p_{n}^{(m)}(x) &=& \frac{p_{m-1}^*(x)p_{n+m}(x) -
p_{m-1}(x)p_{n+m}^*(x)}{(-1)^mC_1C_2 \dots
C_{m-1}A_0}
\eea
\end{lem}
{\bf Proof}.
Fix $m\ge 2$. As a function of
$n$,
$\{p_{n-m}^{(m)}(x)\}_{n=m}^\infty$ also  satisfies (3.4), so it is a
linear combination of $\{p_n(x)\}$ and $\{p_{n}^{*}(x)\}$ with
coefficients that
are independent of $n$, but may depend upon $m$ and $x$.
Thus we use initial conditions (3.7) to find coefficients
$A_m(x)$ and $B_m(x)$ in
\bea
p_{n}^{(m)}(x)= A_m(x)p_{n+m}(x) + B_m(x) p_{n+m}^{*}(x).
\nonumber
\eea
The result is
\bea
A_m(x) = p_{m-1}^*(x)/\Delta_m(x), \quad B_m(x) =
p_{m-1}(x)/\Delta_m(x),
\eea
where
\bea
\Delta_m(x) &=& p_{m-1}^*(x)p_{m}(x) - p_{m-1}(x)p_{m}^*(x).
\eea
It is clear that $\Delta_m(x)$ is a discrete Wronskian
(Casorati Determinant) of $p_m(x)$ and $p_m^*(x)$ and can
be evaluated from
\bea
\Delta_{m+1}(x) = -C_m \Delta_m(x), \nonumber
\eea
which follows from (3.4).  Thus
\bea
\Delta_m(x) = (-1)^mC_1C_2 \dots C_{m-1}A_0, \nonumber
\eea
since $\Delta_0(x) = -A_0$.

Let
\bea
A_n=1+aq^n, \quad B_n=0, \quad C_n=bq^{n-1},
\eea
so that the $p_n(x) = U_n(x; a, -b)$, see (1.5) and (1.6). Thus the $m$th
associated polynomials are
\bea
p_{n}^{(m)}(x) = U_{n}(x;aq^m,-bq^m), \quad {\rm for}\quad m, n \ge 0.
\eea
In view of (1.9) and (3.8), it follows that  (3.9) and (3.10)
generalize  (2.7) and  (2.8), respectively.
In other words Schur's proof is the case $a=0$, $b=q$.




{\bf Remark}: The analogue of (1.1) for Rogers--Ramanujan identities of
any moduli have been found by Garrett. She also combinatorially proved
(1.1) by an involution, and
generalized (1.1) to partitions whose parts differ by at least $d$.

\bigskip 

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\setcounter{thm}{0}


\noindent{\bf 4. Further results}. The relationship  (1.1) is the case
$a = 0$, $x=1$ and $b=q$ of (3.10), if we assume the  Rogers--Ramanujan
identities, that is assume (1.4).  Another interesting result is found
by choosing
$a=-q^{1/2}$, $x = 1$, $b = q$, and then replacing $q$ by $q^2$.
 In our notation (38) and (39) in \cite{Sl52} are
\bea
F(-q^2;-q,q^2)&=& \sum_{n=0}^\infty \frac{q^{2n^2}}{(q;q)_{2n}}
= \prod_{n=0}^\infty \frac{(1+q^{3+8n})(1+q^{5+8n})(1-q^{8n+8})}
{(1-q^{2n+2})}, \\
 \frac{F(-q^4;-q^3, q^2)}{(1-q)} &=&\sum_{n=0}^\infty
\frac{q^{2n^2+2n}}{(q;q)_{2n+1}}=
 \prod_{n=0}^\infty \frac{(1+q^{1+8n})(1+q^{7+8n})(1-q^{8n+8})}
{(1-q^{2n+2})}.
\eea
These are Rogers--Ramanujan identities of order 8 as indicated in
\cite{Sl52}. It is interesting to note that with the choices 
$a=-q^{1/2}$, $x = 1$, $b = q$, formula  (3.10) is
\bea
&{}& \sum_{n=0}^\infty \frac{q^{2mn+n^2}}{(q;q)_n(q^{2n+1};q)_m} \\
&=& (-1)^mq^{\binom{m}{2}}\left[U_{m-2}(1;-q^3, -q^4|q^2)F(-q^2;-q,q^2)
\right. \nonumber\\
&{}& \qquad -  \left.U_{m-1}(1;-q,
-q^2|q^2)F(-q^4;-q^3,q^2)/(1-q)\right].
\nonumber
\eea


Al-Salam and Ismail \cite{AlIs83} established the continued fraction
representation 
\bea
\frac{F(-qbz^{-2}; qa, q)}{zF(-bz^{-2}; a,q)} 
= \frac{1+a}{(1+aq)z+}\; \frac{b}{(1+aq^2)z+}\;\frac{bq}{(1+aq^3)z+} \; 
\cdots
\eea
The special case $z=1, b =q, a=-\sqrt{q}$ gives, via (4.1)--(4.2),
\bea
\frac{1}{(1-q^3)+}\; \frac{q^2}{(1-q^5)+}\;\frac{q^4}{(1-q^7)+}
\dots
= \prod_{n=0}^\infty \frac{(1+q^{1+16n})(1+q^{14+16n})}
{(1+q^{6+16n})(1+q^{10+16n})}.
\eea 


Recall that a Gaussian (or $q$-)binomial and trinomial coefficients
are 
\bea
{\gauss{n}{k}} = \frac{(q;q)_n}{(q;q)_k(q;q)_{n-k}}, \quad 
{\gauss{n}{j,k}} = \frac{(q;q)_n}{(q;q)_j(q;q)_k(q;q)_{n-j-k}}.
\eea
The polynomials $\{U_n(x;a,b|q)\}$ contain a redundant parameter. In fact it
is clear from (1.12) that $x^{-n}U_n(x;a,bx^2|q)$ is independent of
$x$. As orthogonal polynomials the $x$ variable is important and we can
scale away the $b$ parameter. Set
\bea
V_n(a,b|q) = x^{-n}U_n(x;a, -bx^2|q) = 
\sum_{k=0}^{\lfloor n/2\rfloor}
{\gauss{n-k}{k}}\frac{(-a;q)_{n-k}}{(-a;q)_k}b^kq^{k(k-1)}.
\eea
Since $(-a;q)_{n-k}/(-a;q)_k = (-aq^k;q)_{n-2k}$ we can expand the quotient
using the $q$-binomial theorem (1.13) and obtain
\bea
V_n(a,b|q) =\sum_{k=0}^{\lfloor n/2\rfloor}\sum_{j=0}^{n-2k}
{\gauss{n-k}{j, k}} a^jb^kq^{k(k+j-1)+j(j-1)/2}.
\eea
For example the coefficient of $a^j b^k q^m$ in $V_n(a, b|q)$ has a 
combinatorial 
interpretation in terms of counting pairs of partitions. This combinatorial 
study is still in progress.    




\bibliographystyle{plain}
\begin{thebibliography}{1}

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{\it Orthogonal polynomials associated with the Rogers--Ramanujan continued
fraction}, Pacific J. Math. {\bf 104} (1983), 269--283.
%
% \bibitem{An} G. E. Andrews, {\it A polynomial identity which implies
%the Rogers--Ramanujan identities}, Scripta Math. {\bf 28} (1970),
%297--305.
%
\bibitem{AnAsRo}  G. E. Andrews, R. A. Askey and R. Roy, Special Functions,
Cambridge University Press, Cambridge, 1999.
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\bibitem{AnKnPa99}
G.~Andrews, A.~Knopfmacher, and P.~Paule, {\it
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\bibitem{AsIs} R.~A.~Askey and M.~E.~H.~Ismail,
{\it Recurrence relations, continued fractions and orthogonal
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%
\bibitem{GaRa}  G.~Gasper and M.~Rahman,  Basic Hypergeometric Series,
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\bibitem{JoTh}
W.~B. Jones and W.~Thron, Continued Fractions: Analytic Theory
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I.~Schur,
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\bibitem{Sl52} L. J.  Slater,
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\end{thebibliography}


\noindent  Department of Mathematics, University of South Florida, Tampa,
Florida  33620-5700. \\
email: {\tt ismail@math.usf.edu}


\noindent Department of Mathematics,
University of the Witwatersrand, P.~O. Wits,
2050 Johannesburg, South Africa. \\
 email: {\tt helmut@gauss.cam.wits.ac.za}\newline
homepage: {\tt http://www.wits.ac.za/helmut/index.htm}

\noindent School  of Mathematics, University of Minnesota, Minneapolis,
Minnesota 55455. \\
email: {\tt stanton@math.umn.edu}

\end{document}

