\documentclass[12pt]{article}
\usepackage{amsmath, amssymb}

\parskip=7pt

\newtheorem{theo}{Theorem}
\newtheorem{prop}[theo]{Proposition}
\newtheorem{coro}[theo]{Corollary}
\newtheorem{defi}[theo]{Definition}
\newtheorem{lem}[theo]{Lemma}

\makeatletter \@addtoreset{equation}{section}
\@addtoreset{theo}{section} \makeatother

\renewcommand{\theequation}{\arabic{section}.\arabic{equation}}
\renewcommand{\thesection}{\arabic{section}.}
\renewcommand{\thetheo}{\arabic{section}.\arabic{theo}}

\def\qed{\hfill \rule{4pt}{7pt}}
\def\pf{\noindent {\it Proof.} }

%%% Here's the symbols used in the article

\def\s{\sigma}
\def\e{{\rm exp\,}}
\def\P{\Psi}
\def\lg{{\rm log \,}}
\def\N{\mathbb{N}}
\def\qr{\beta_r}
\def\qs{\beta_s}
\def\a{\alpha}
\def\A{{\mathbb A}}
\def\B{{\mathbb B}}
\def\C{{\mathbb C}}

\newcommand\bx[1]{{\boxed{#1\vphantom{1}}}}
\def\moins{\raise 1pt\hbox{{$\scriptstyle -$}}}
\def\plus{\raise 1pt\hbox{{$\scriptstyle +$}} }

\begin{document}
\begin{center}
\Large Evaluation of Some Hankel Determinants
\end{center}

\begin{center}
Qing-Hu Hou\\
Center for Combinatorics, LPMC\\
Nankai University, Tianjin 300071, P. R. China\\
Email: hou@nankai.edu.cn

Alain Lascoux\\
CNRS, Institut Gaspard Monge, Universit\'e de Marne-la-Vall\'ee\\
77454 Marne-la-Vall\'ee Cedex, France\\[2pt]
 and \\[2pt]
Center for Combinatorics, LPMC\\
 Nankai University, Tianjin 300071, P.R. China\\
Alain.Lascoux@univ-mlv.fr

Yan-Ping Mu\\
Center for Combinatorics, LPMC\\
Nankai University, Tianjin 300071, P. R. China\\
Email: myphb@eyou.com

\end{center}

\begin{abstract}
We evaluate some Hankel determinants of Meixner polynomials, associated to the
series $\exp (\sum \a [i] z^i / i)$, where $[1],[2],\ldots$ are the
$q$-integers.
\end{abstract}

\section{Introduction}
Given a nonzero real number $\alpha$, let $\sum_{i \ge 0} a_i z^i$ be the
Taylor expansion of
\begin{equation}
f(z) = \exp \left( \sum_{i \ge 1} \alpha \frac{1-q^i}{1-q} \frac{z^i}{i}
\right) = 1 + z \alpha + \frac{z^2}{2}(q+\alpha+1)\alpha + \cdots \,.
\end{equation}
By using symmetric functions, we evaluate two families of Hankel determinants:
\begin{equation}
 H^1_n=\det( a_{i+j-1} )_{1 \le i,j \le n} \quad \mbox{and} \quad H^2_n = \det(
a_{i+j} )_{0 \le i,j \le n} \,.
\end{equation}
What is remarkable is that $H^1_n$ factors into linear factors of type
$(\alpha-r+rq)$ and that $H^2_n/H^1_n$ is a polynomial in $q$ and $\alpha$,
though the coefficients of the Taylor expansion of $f(z)$ are rather
complicated.

The explicit forms of $H^1_n$ and $H^2_n$ are given by the following theorem,
which is our main result.
\begin{theo} We have
\begin{equation}
H^1_n = w_n \prod_{n > |r|} (\alpha-r+rq)^{n-|r|} \,,
\end{equation}
and
\begin{equation}
H^2_n/H^1_n =  \sum_{k=0}^n (-1)^k {n \choose k}
                        \prod_{i=0}^{k-1} {\alpha+(1-q)(n-i) \over 2n-i} \,,
\end{equation}
with
\begin{equation}
w_n=\frac{(-1)^{n \choose 2}}{1 \cdot 2^2 \cdot 3^3 \cdots n^n (n+1)^{n-1}
\cdots (2n-1)^1} \,.
\end{equation}
\end{theo}


\section{Symmetric Functions}
Many properties of Hankel determinants are better understood when identifying
Hankel determinants with Schur functions indexed by rectangular partitions
\cite{Lascoux}. We first shall recall a few facts about symmetric functions. We
follow the convention of \cite{Lascoux} rather than \cite{Macdonald}. The basic
symmetric functions will be the power sums $p_i, i=1,2,\ldots$, and the
argument of a symmetric function will be a rational function with coefficients
in $\mathbb C$, of two types of variables: variables inside a box and variables
without a box. We define
\[
p_i(\alpha) = \alpha \quad \mbox{and} \quad p_i\left( \bx{x} \, \right) =
\bx{x}\,^i,
\]
and require that the $p_i$ be $\mathbb C$-ring homomorphisms satisfying
\[
\begin{cases}
p_i(\alpha A + \beta B) = \alpha p_i(A) + \beta p_i(B) \\
p_i(AB)=p_i(A) p_i(B) \\
p_i(p_j(A)) = p_{ij}(A).
\end{cases}
\]
Notice that
\[
p_i\left(3\, \bx{2}\,\right) = 3\, \bx{2}\,^i =
p_i\left(\bx{2}+\bx{2}+\bx{2}\,\right),
\]
but that
\[
p_i\left(\bx{2}\,\right) = \bx{2}\,^i \not
       = p_i\left(\bx{1}+\bx{1}\,\right) = \bx{1}\,^i+\bx{1}\,^i.
\]
One cannot expand the content of a box, inside the argument of a symmetric
function. Symmetric functions are considered here just as
a tool to write algebraic expressions, in a
simpler manner than having to manipulate coefficients of different generating
functions. Once they are evaluated in terms of their arguments, we can, of
course, erase boxes.

The usual power sums are recovered when taking as an argument a sum of letters
in boxes:
\[
p_i\left(\bx{a}+\bx{b}+\bx{c}+\cdots\right)
  = \bx{a}\,^i + \bx{b}\,^i + \bx{c}\,^i +
\cdots.
\]
By abuse of language, we shall still call the argument $A$ of a symmetric
function an {\it alphabet}.

The complete symmetric functions of $A$, $S^n(A)$,
are defined with the help of
the following generating function
\[
\sigma_z(A) = \sum_{n \ge 0} z^n S^n(A) = \exp \left( \sum_{i=1}^\infty z^i
p_i(A)/i \right).
\]
This definition implies in particular that
\begin{equation}
S^n(A+B) = \sum_{i=0}^n S^i(A) S^{n-i}(B).
\end{equation}

For any $n\in\N$ and $I=(i_1,\ldots,i_n) \in \N^n$, the \emph{Schur function}
$S_I(A)$ has a determinantal expression \cite[Section
1.3]{Macdonald}~:
$$
S_{I}(A) = \big| S^{i_s+s-r} (A) \big|_{1 \le r,s \le n}.
$$
If $I$ and $J$ are two conjugate partitions, then \cite[p.~8]{Lascoux}
\[
S_I(A) = (-1)^{|J|} S_J(-A).
\]

The determinantal expressions of Schur functions occur in many fields of
mathematics. For example, the notation $S_{n^n}(A-\bx{x}\,)$ encodes the
classical determinantal expressions of orthogonal polynomials in terms of
moments (cf. \cite[p.~273]{AAR}, \cite[p.117]{Lascoux}):
\begin{multline*}
S_{3^3} \left( A-\bx{x}\,\right) = \left|
\begin{array}{ccc}
S^3(A-\bx{x}\,) & S^4(A-\bx{x}\,) & S^5(A-\bx{x}\,) \\
S^2(A-\bx{x}\,) & S^3(A-\bx{x}\,) & S^4(A-\bx{x}\,) \\
S^1(A-\bx{x}\,) & S^2(A-\bx{x}\,) & S^3(A-\bx{x}\,)
\end{array}
\right| \\[5pt] = \left|
\begin{array}{cccc}
S^3(A) & S^4(A) & S^5(A) & x^3 \\
S^2(A) & S^3(A) & S^4(A) & x^2 \\
S^1(A) & S^2(A) & S^3(A) & x^1 \\
S^0(A) & S^1(A) & S^2(A) & x^0
\end{array}
\right|\ .
\end{multline*}
 More generally, given a non negative integer $k$ and a generic alphabet $A$
(by generic, we mean that for any $n \ge 0$, $S_{(n+k-1)^n}(A) \not= 0$), let
for any $n \in \N$,
\begin{multline*}
P_n \left( x \right) := S_{(n+k)^n} \left( A-\bx{x}\, \right)
  = S_{(n+k)^n}(A) - {x}\, S_{n+k-1,(n+k)^{n-1}}(A) \\
+ {x}^2 S_{(n+k-1)^2,(n+k)^{n-2}}(A) + \cdots +(-1)^n {x}^n S_{(n+k-1)^n}(A).
\end{multline*}
Then $\{P_0(x),P_1(x),\ldots\}$ is a family of orthogonal polynomials
associated to the functional $x^n \to \int x^n = S^{n+k}(A)$.

Notice that we have letters in boxes only as arguments of symmetric functions.

\section{A transformation of alphabets}
Given an alphabet $A$, we define a new alphabet $B$ by the following equality
of generating functions:
\begin{equation}
\label{tran} \sigma_z(B) = \sigma_{z/(1+z)}(A).
\end{equation}
We have the equivalence
\[
\sigma_z(B) = \sigma_{z/(1+z)}(A) \Longleftrightarrow \sigma_z(-B) =
\sigma_{z/(1+z)}(-A),
\]
and, by expansion of the generating functions,
\begin{equation}
\label{A-B}
\begin{cases}
 S^n(B) = \sum_{i=0}^{n-1} (-1)^i {n-1 \choose i} S^{n-i}(A)
= S^n(A-(n-1)), \\[10pt]
 S^n(A) = \sum_{i=0}^{n-1} {n-1 \choose i} S^{n-i}(B)
= S^n(B-(n-1)\bx{\moins 1}\, ).
\end{cases}
\end{equation}
This transformation has been used by Grothendieck to define Chern classes in
$K$-theory (cf. \cite[Exercise 2.13]{Lascoux}).

By row and column manipulations, we can easily transform the determinant
$S_{n^n}(A)$ into $S_{n^n}(B)$, i.e., one has
\begin{equation}
S_{n^n}(A) = S_{n^n}(B) = (-1)^{n^2} S_{n^n}(-B) = (-1)^{n^2}S_{n^n}(-A).
\end{equation}

Furthermore, we have
\begin{lem}
\label{main-lem}
\begin{equation}
\label{eqn-tran} S_{(n+1)^n}\left( -A -\bx{x \plus 1}\,\right) = S_{(n+1)^n}
\left( -B -\bx{x} \, \right).
\end{equation}
\end{lem}
\pf Define a linear functional $\int$ acting on the ring of polynomials in
$x$ by
\begin{equation}
\label{moment1} \int x^n = S^{n+1}(\moins B), \quad \forall\, n \ge 0.
\end{equation}
Then
\begin{equation}
\label{moment2} \int (x+1)^n = \sum_{i=0}^n {n \choose i}
S^{n-i+1}(\moins B) = S^{n+1}(\moins A)\, .
\end{equation}
From \eqref{moment1}, we have that $S_{(n+1)^n}(\moins B-\bx{x}\,)$ are
orthogonal polynomials associated to $\int$. Similarly, \eqref{moment2} implies
that $S_{(n+1)^n}(\moins A-\bx{x \plus 1}\,)$ are also orthogonal polynomials
associated to $\int$. Since their leading coefficients coincide,
\eqref{eqn-tran} holds. \qed

\begin{coro}
\label{n^(n+1)} Let $A$ and $B$ be alphabets such that
$\sigma_z(B)=\sigma_{z/(1+z)}(A)$. Then
\[   S_{(n+1)^n}(\moins A)=
S_{n^{n+1}}(A) = S_{n^{n+1}} \left(B+\bx{\moins 1}\,\right) = \sum_{i=0}^n
(-1)^{n-i} S_{i,n^n}(B).
\]
\end{coro}

\section{Evaluation of Hankel Determinants}
Given a nonzero real number $a$, let $\A$ be the alphabet defined by the
generating function
\begin{equation}
\label{defi} \s_z(\A) = \e \left( \sum_{i \ge 1} a {1-q^i \over 1-q} {z^i \over
i}\right).
\end{equation}
We shall now evaluate the determinants $S_{n^n}(\A)$ and $S_{n^{n+1}}(\A)$
(these are the determinants $H^1_n$ and $H^2_n$, mentioned in the introduction,
which are needed for writing $\sigma_z(\A)$ as a continued fraction).

Noticing that
\[
 \sum_{i \ge 1} {1-q^i \over 1-q} {z^i \over i}
 =  {1 \over 1-q} \sum_{i \ge 1} \left( {z^i \over i} - {(qz)^i
\over i} \right) = {1 \over 1-q} \log \left( {1-qz \over 1-z} \right),
\]
and taking the parameter $b=a/(1-q)$ instead of $a$, we derive that
\[
\sigma_z(\A) = \left( {1-qz \over 1-z}\right)^b
   = \sigma_z \left( b\, \left(1-\bx{q} \,\right)
\right).
\]
Let $\B=-b\, \bx{q \moins 1}\,$. Then
\[
\sigma_{z/(1+z)}(\A) = \big(1-(q-1)z \big)^b = \sigma_z (\B).
\]
and for any partition $\lambda=(\lambda_1,\ldots,\lambda_\ell)$
\[
S_\lambda(\B) = (q-1)^{\sum \lambda_i} \prod_{(i,j) \in \lambda} \frac{-b +
c(i,j)}{h(i,j)},
\]
where $h(i,j)$ and $c(i,j)$ are respectively the hook length and the content of
the box $(i,j)$ in the diagram of $\lambda$ (cf. \cite[p.~28]{Macdonald}).

Hence,
\begin{multline}
S_{n^n}(\A) = S_{n^n}(\B) =
   (q-1)^{n^n} \prod_{(i,j) \in n^n} {-b+c(i,j) \over h(i,j)}\\
    = \prod_{(i,j) \in n^n} {a + (q-1)c(i,j) \over h(i,j)}  \ .
\end{multline}
For example,
\[
S_{3^3}(\A) = (q-1)^9 \left. \boxed{\begin{array}{ccc}-b & -b+1 & -b+2 \\
-b-1 & -b & -b+1
\\ -b-2 & -b-1 & -b \end{array}} \right/ \boxed{\begin{array}{ccc}
5 & 4 & 3
\\ 4 & 3 & 2 \\ 3 & 2 & 1 \end{array}}\ ,
\]
(taking the product of all the elements in the arrays).

To evaluate $S_{n^{n+1}}(\A)$, we apply Corollary~\ref{n^(n+1)} and get
\begin{multline*}
S_{n^{n+1}}(\A) = \sum_{k=0}^n (-1)^{n-k} S_{k,n^n}(\B) \\
= \sum_{k=0}^n (-1)^{n-k} (q-1)^{n^n+k} \prod_{(i,j) \in (k,\,n^n)} {-b+c(i,j)
\over h(i,j)}.
\end{multline*}
Therefore,
\begin{eqnarray*}
S_{n^{n+1}}(\A)/S_{n^n}(\A) &=& (-1)^n \sum_{k=0}^n (1-q)^k  \prod_{i=1}^k
{-b+(-n+i-1) \over i} \cdot {n-k+i \over 2n-k+i} \nonumber \\
&=& (-1)^n {(n!)^2 \over
(2n)!} \sum_{k=0}^n (-1)^k (1-q)^k {b+n \choose k}
{2n-k \choose n-k} \, .
\end{eqnarray*}

Hence,
\begin{eqnarray}
S_{n^{n+1}}(\A)/S_{n^n}(\A) &=&
               (-1)^n  \sum_{k=0}^n (-1)^k {n \choose k}
                        \prod_{i=0}^{k-1} {a+(1-q)(n-i) \over 2n-i} \\
&=&  (-1)^n {(n!)^2 \over (2n)!} S^n\left((n+1) -(b+n)\bx{1 \moins q} \,\right)
\label{ra-(1-q)} \, .
\end{eqnarray}

One can write other expressions for \eqref{ra-(1-q)}. In terms of
hypergeometric function, it reads
\begin{equation}
\label{ra-Hyper} (-1)^n\ _2F_1 \left( \begin{array}{c} -b-n, -n \\ -2n
\end{array} ; 1-q \right).
\end{equation}
By \eqref{tran} and \eqref{A-B}, we have
\begin{multline}
\label{ra-q} S^n\left ((n+1) -(b+n)\bx{1 \moins q} \,\right) = S^n \left(
(b-n)\bx{\moins 1}-(b+n)\bx{\moins q} \,\right)
\\
= (-1)^n S^n \left( (b-n)-(b+n)\bx{q} \,\right).
\end{multline}
For example,
\begin{multline*}
S^2 \left( 3-(b+2)\bx{1\moins q} \,\right) =
S^2(3)-3(b+2)\bx{1\moins q}+{b+2 \choose 2}\bx{1\moins q}\,^2 \\
= S^2\left( (b-2)-(b+2)\bx{q} \,\right)  = {b-1 \choose 2} - (b-2)(b+2)\bx{q} +
{b+2 \choose 2} \bx{q}\,^2,
\end{multline*}
and thus
\begin{multline*}
{4 \choose 2}-3(b+2)(1-q)+{b+2 \choose 2}(1-q)^2 \\
= {b-1 \choose 2} - (b-2)(b+2)q + {b+2 \choose 2} q^2.
\end{multline*}

The referee has given a third expression
\begin{equation}
\label{referee}
\sum_{k=0}^n (-1)^k \frac{b\cdots (b\plus n)}{(b+k) k!(n-k)!}{(n+k)!
\over k!} \sum_{i=0}^{n-k} {{n \choose i} {n-k \choose i} q^i \over {i+k
\choose k}},
\end{equation}
which is no other than the expression of
\[
f(b)=(-1)^n n!\, S^n \left( (b-n)-(b+n)\bx{q} \,\right)
\]
in terms of its values at $b=0,-1,\ldots,-n$.

Recall (cf. \cite{I-S97}, see also \cite{AAR,I-S98}) that the Meixner
polynomials $M_k(x;\beta,c)$ have generating function
\[
\sum_{k=0}^\infty {M_k(x;\beta,c) \over k!} z^k = (1-z)^{-x-\beta} (1-z/c)^x.
\]
That is,
\begin{equation}
\label{M-S} M_n(x;\beta,c) = n!\, S^n\left( x+\beta -x \bx{1/c} \,\right).
\end{equation}
The referee pointed out that our coefficients $S^n(\A)$ are equal to
\[
{a \over n!}\, M_{n-1} \left( {a \over 1-q}-1; 2, 1/q \right) = {a \over n}\,
S^{n-1}\left( (b+1)-(b-1)\bx{q} \,\right) \ ,
\]
and that the value of $S_{n^n}(\A)$ results from the fact that
Meixner polynomials are moments associated to Jacobi polynomials
(cf. \cite{I-S97,I-S98}).




But from \eqref{M-S}, we recognize that \eqref{ra-(1-q)} and \eqref{ra-q}
themselves can be seen as specialization of Meixner polynomials:
\begin{align*}
& S^n\left ((n+1) -(b+n)\bx{1 \moins q} \,\right) = {1 \over n!}\, M_n(n+b;
1-b,1/(1-q)),
\\[5pt]
& S^n \left( (b-n)-(b+n)\bx{q} \,\right) = {1 \over n!}\, M_n(n+b; -2n,1/q).
\end{align*}

In fact, one can also obtain \eqref{ra-q} as a specialization of a Jacobi
polynomial. The homogeneous version $\widehat{P}^{\,\alpha,\beta}(\xi_1,\xi_2)$
of a Jacobi polynomial is (with $\xi_1=(x+1)/2$ and $\xi_2=(x-1)/2$ we get the
usual Jacobi polynomial):
\[
\widehat{P}^{\,\alpha,\beta}(\xi_1,\xi_2) = (\moins1)^n S^n\left( -(n\plus
\alpha)\,\bx{\xi_1} - (n\plus\beta)\,\bx{\xi_2} \,\right).
\]


Therefore, \eqref{ra-q} can be written as $\widehat{P}^{\,-b,b}(1,q)$.

\noindent {\bf Acknowledgments.} We thank the referee for having given us the
connection with Meixner polynomials, as well as formula \eqref{referee}. This
work was done under the auspices of the National ``973'' Project on
Mathematical Mechanization, the National Science Foundation, the Ministry of
Education, and the Ministry of Science and Technology of China.

\begin{thebibliography}{99}
\bibitem{AAR}
G.E.~Andrews, R.~Askey and R.~Roy, Special Functions, Encycl. Math. Appl. 71,
Cambridge University Press, Cambridge, 1999.

\bibitem{I-S97}
M.E.H.~Ismail and D.~Stanton, Classical orthogonal polynomials as moments,
Canad.~J.~Math. 49 (3) (1997), 520--542.

\bibitem{I-S98}
M.E.H.~Ismail and D.~Stanton, More orthogonal polynomials as moments, in:
B.E.~Sagan and R.P.~Stanley (Eds.), Mathematical Essays in Honor of Gian-Carlo
Rota, Birkhauser, 1998, pp. 377--396.

\bibitem{Lascoux}
A.~Lascoux, Symmetric Functions and Combinatorial Operators on Polynomials,
CBMS Regional Conference Series in Mathematics, Vol. 99, AMS, 2003.

\bibitem{Macdonald}
I.G.~Macdonald, Symmetric Functions and Hall Polynomials, The Clarendon Press,
Oxford University Press, New York, 1979.

\end{thebibliography}
\end{document}
