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\begin{document}



\title{INTERPOLATION}
\author{Alain Lascoux}
\date{}
\maketitle


\part{Lagrange Interpolation}

\section{Introduction}
\vspace{3mm}
The original interpolation problem is to reconstruct a one-variable
function given its values at a finite number of points. This is
possible only for polynomials, in which case we have the well known
Lagrange Interpolation formula. In general, we 
shall have only an {\em approximation } of the function.

The problem can be restated as follows:

We favor a linear space $V$ of functions. Given $f$, one wants a function
\( p(f) \) in \( V \) which is as ``near" to \( f \) as possible.

The functional $p$ should satisfy:

Linearity: $ \lambda f + \mu g \rightarrow p( \lambda f + \mu g) = 
 \lambda p(f) + \mu p(g), \lambda , \mu \in {\bf C}. $
                                     
Idempotence: $p^{2} = p.$

In general, suppose that 
$\{L_{i}\}$ is a basis of the space dual to $V$; in particular, $n+1 = dim(V)$.
Given the values $L_{0}(f), \ldots , L_{n}(f)$ 
of linear functionals $L_{0}, \ldots L_{n}$ at $f$, we want to find $g$ in
$V$ such that
$$L_{0}(f) = L_{0}(g), \ldots , L_{n}(f) = L_{n}(g).$$


Other problem: find a basis $g_{i}$ of $V$ adapted to the $L_{i}$, for example, 
such that

$$L_{j}(g_{i}) = 1 \quad or \quad 0 \quad whether \quad i = j \quad or \quad not$$

Symmetrically, one can start from a basis of the space $V$ and look for the 
associated functionals. We shall give examples of these different points of view.
\pagebreak

\section{ Lagrange Interpolation}

\vspace{3 mm}

The simplest spaces of functions are the space of polynomials in one variable.
Take for example $V = {\sl Pol}(n) :=$ space of polynomials in $x$ of degree 
$ \leq n$.

At the same time, the simplest linear functinals are taking the value in a 
fixed set of points ${\bf A}$, with cardinal $({\bf A}) = n + 1$. 

The looked for interpolation polynomial $g$ is the (unique) polynomial
of degree $\leq n$ such that

\begin{equation}
\label{li-1} 
a \in {\bf A} \Rightarrow g(a) = f(a).
\end{equation}

Let
$$R({\bf A}, {\bf B})=\prod_{a \in {\bf A}, b \in {\bf B}}
(a-b)$$ be the resultant of $\bA$ and $\bB$,  
${\bf A} \setminus {\bf B}$ be the set difference.

We remark that for every $a \in {\bf A}$ the polynomial $R(x, {\bf A} \setminus a)$
vanishes in all the points of {\bf A} other than $a$. It is clear that by linear
combination of these $n + 1$ polynomials of degree $n$, one can express any 
element of $V$, and thus these polynomials constitute the adjoint basis of 
$L_{a} : f \rightarrow L_{a}(f) := f(a)$, modulo normalisation.

In other words, one has Lagrange formula (???J. Ecole Polyt., {\rm II}, p.277):

\begin{equation}
\label{li-2}
g(x) = \sum_{a \in {\bf A}}f(a) \frac{R(x, {\bf A} \setminus a)}{R(a, {\bf A}
\setminus a)}.
\end{equation}

The application $f \rightarrow g$ is a projector on the space ${\sl Pol}(n)$
of polynomials of degree no more than $n$. This projector is no other 
than the ``Remainder
modulo $R(x, {\bf A})$'' since it is the identity on ${\sl Pol}(n)$ and 
vanishes on every multiple of $R(x, {\bf A})$, for we have clearly that
$f(a)-g(a)=0, a\in {\bf A}$ and hence $R(x, {\bf A})|(f(x)-g(x))$.

Lagrange interpolation allows decomposition of rational fractions :\\
dividing both members of (\ref{li-2}) by $R(x, {\bf A})$, one gets the equivalent form :

\begin{equation}
\label{li-3}
\frac{g(x)}{R(x, {\bf A})} = \sum_{a \in {\bf A}} 
\frac{f(a)}{(x-a)R(a, {\bf A} \setminus a)}.
\end{equation}

Let $g(x)=f(x)=1$ in the equation (\ref{li-3}), we have the following identity.

\begin{equation}
\frac{1}{(x-a_1)\cdots (x-a_n)}=\sum_{i=1}^n 
\frac{1}{x-a_i} \prod_{j\neq i}\frac{1}{a_i-a_j}
\end{equation}

or equivalently, we have
\begin{equation}
\label{li-4}
\frac{1}{(1-a_1z)\cdots (1-a_nz)}=\sum_{i=1}^n 
\frac{1}{1-a_iz} \prod_{j\neq i}\frac{a_i}{a_i-a_j}
\end{equation}

Indeed, in the proceeding formula, the ``variable'' $x$ plays a role symmetrical
to the ``interpolation points'' $a \in {\bf A}$, i.e. it can be written as a 
symmetric summation on the alphabet ${\bf A} \bigcup \{ x \}$.

\vspace{2 mm}

Denote $x$ by $a_1$, one is therefore lead to replace
(\ref{li-3}) by the following operator on one variable-functions:

$${\sl Pol}(n) \ni f \rightarrow \sum_{a \in {\bf A}}f(a)/
R(a, {\bf A} \setminus a)=\frac{f(x)-g(x)}{R(x, {\bf A} \setminus x)} 
\in Sym({\bf A}).$$


This is not our last transformation of Lagrange formula. One notices that the
starting space does not need to be restricted to functions of one variable, 
but can be taken to be Sym(1$\mid$n), the space of functions of $n + 1$
variables which are symmetrical in the last $n$ ones. Put a total order on
the alphabet ${\bf A} = \{ a_{1}, a_{2}, \ldots , a_{n+1} \}$. Since the last 
century at least, one knows the following

\begin{thm}
Any function $f(a_{1}; a_{2}, \ldots , a_{n + 1})$, 
symmetrical in $a_{2}, \ldots , a_{n+1}$, can be expressed as functions of
$a_{1}$ only, with coefficients in $Sym({\bf A})$, the ring of functions which
are symmetrical in all the variables; on the other hand, the preceeding 
operator commutes with multiplication by every element of Sym({\bf A}), one
can thus extend it to the space Sym(1$\mid$n), considered as a module over
the ring $Sym({\bf A})$ generated by the powers of $a_{1}$.
\label{basis}
\end{thm}

Thus, {\em Lagrange operator} is the symmetrization operator:

\begin{equation}
\label{li-5}
Sym(1 \mid n) \ni f \rightarrow \sum_{a \in {\bf A}} f(a \mid {\bf A} \setminus a)/R(a, {\bf A} \setminus a) := L_{{\bf A}}(f) \in Sym({\bf A}).
\end{equation}

Here we give some propositions of $L_{\bf A}$.

1) $L_{\bf A}$ is linear, it means that

$$ L_{\bf A} (f_1+f_2)= L_{{\bf A}}(f_1) + L_{{\bf A}}(f_2)$$

2) $L_{\bf A}(\alpha f)=\alpha L_{{\bf A}}(f)$

where $\alpha$ has nothing to do with {\bf A}.

3) If $f \in Sym({\bf A})$, then

$$ L_{\bf A}=\sum_{a \in {\bf A}} f/R(a, {\bf A}-a)=f \sum_{a \in {\bf A}} 1/R(a, {\bf A}-a)=0.$$
It means that $L_{{\bf A}}(f)=0$ for $f \in Sym({\bf A})$.

4) We should give a proof of that:

Powers of $a_1$ generate Sym($a_{1} \mid a_{2}, \ldots , a_{n+1}$) as
a Sym({\bf A})-module.


Here's a proof of Theorem \ref{basis}:

Let $S_n, S'_n$ denote the symmetric group on $\{1, 2, \ldots, n \}$
and $\{0, 1, \ldots, n \}$ respectively.
Let $\lambda=(\lambda_1, \ldots, \lambda_k), 
\lambda_1 \geq \lambda_2 \geq \ldots \lambda_k >0$
be a partition, and $|\lambda|=k$ be its length.
For $\sigma \in S_n$ ( or $S'_n$ ), we denote
$x_{\sigma (1)}^{\lambda_1} x_{\sigma (2)}^{\lambda_2} \ldots x_{\sigma (k)}^{\lambda_k}$
by $\sigma ( x_1^{\lambda_1} x_2^{\lambda_2} \ldots x_k^{\lambda_k} )$, 
and let 
$$A(\lambda)=\{ \sigma( x^{\lambda_1}_1 \ldots x^{\lambda_k}_k )
| \sigma \in S_n \}, \ \ \ \ m_\lambda = \sum_{t \in A(\lambda)} t; $$
$$A'(\lambda)=\{ \sigma( x^{\lambda_1}_1 \ldots x^{\lambda_k}_k )
| \sigma \in S'_n \}, \ \ \ \ m'_\lambda = \sum_{t \in A'(\lambda)} t.$$
Clearly, we have $m'_\lambda \in Sym(x_0, x_1, \ldots, x_n)$.

Notice that for every $f \in Sym(x_0|x_1, \ldots, x_n)$, 
$f=\sum_\lambda a_\lambda m_\lambda x_0^{k(\lambda)}$, 
so we only need to prove: for every $\lambda$, 
$m_\lambda=\sum_\mu a'_\mu m'_\mu x_0^{k'(\mu)}$.

We proceed by induction on the length of $\lambda$.
If $|\lambda|=0$, there's nothing to prove.
Suppose the assertion is true for $k-1$, then for $|\lambda|=k$, 
let 
$$A_i= \{ \sigma(x_1^{\lambda_1} \ldots x_k^{\lambda_k}) |
\sigma \in S'_n, \sigma(i)=0 \}, $$
then we have $A'(\lambda)=\cup_{i=0}^n A_i.$

Let $\lambda^i=(\lambda_1, \ldots, \hat{\lambda_i}, \ldots, \lambda_k).\ (
\mbox{ if } i=0 \mbox{ or } i>k \mbox{ we consider } \lambda^i=\lambda )$
It's easy to verify that $A_i=\{ tx_0^{\lambda_i} | t \in A(\lambda^i) \}. 
\ (\mbox{ if } i=0 \mbox{ or } i > k \mbox{ we consider }\lambda_i=0 )$
In fact, given $\sigma \in S_n$, let $\mu=(\sigma(n), \sigma(1), \ldots, 
\sigma(i-1), 0, \sigma(i), \ldots, \sigma(n-1)) \in S'_n$;
given $\mu \in S'_n, \ \mu(i)=0$, 
let $\sigma=(\mu(1), \ldots, \mu(i-1), \mu(i+1), \ldots, \mu(n), \mu(0)) \in S_n$;
then we have
$$\sigma( x_1^{\lambda_1} \ldots x_{i-1}^{\lambda_{i-1}} x_i^{\lambda_{i+1}}
\ldots x_{k-1}^{\lambda_k} )x_0^{\lambda_i}=\mu( x_1^{\lambda_1} \ldots
x_k^{\lambda_k} ).$$

Suppose $\{ \lambda^i | i=0, \ldots, n \}
=\{ \lambda^{(0)}, \lambda^{(1)}, \ldots, \lambda^{(s)} \}$
where $\lambda^{(i)}$ are distinct, 
and $\lambda^{(0)}=\lambda$.
Clearly $|\lambda^{(i)}|=k-1$ for $1 \le i \le s$.

>From the definition of $A(\lambda)$, we know if $\lambda_1 \not= \lambda_2$, 
$A(\lambda_1) \cap A(\lambda_2) = \emptyset$.
Hence we have
\begin{eqnarray*}
m_\lambda & = & m_{\lambda^{(0)}} \\
& = & \sum_{i=0}^s x_0^{\lambda_i} m_{\lambda^{(i)}} -
\sum_{i=1}^s x_0^{\lambda_i}m_{\lambda^{(i)}} \\
& = & \sum_{t \in \cup_{i=0}^n A_i }t -
\sum_{i=1}^s x_0^{\lambda_i}m_{\lambda^{(i)}} \\
& = & m'_\lambda - \sum_{i=1}^s x_0^{\lambda_i}m_{\lambda^{(i)}}.
\end{eqnarray*}
Notice that $|\lambda^{(i)}|=k-1, \ i=1, \ldots, s$, by induction hypothesis
we know the assertion is true for $|\lambda|=k$. \qed

We may also prove the above theorem using some properties of
the Schur function.

5) For every $f \in Sym (1 \mid n)$, since powers of $a_{1}$ generate Sym($a_{1} \mid a_{2}, \ldots , a_{n+1}$) as
a Sym({\bf A})-module, that is to say:
$$f=\sum_{k}{a_1}^k T_k, T_k \in Sym({\bf A}), |{\bf A}|=n+1$$
then we have
$$L_{{\bf A}}(f)=L_{{\bf A}}(\sum_{k}{a_1}^k T_k)=\sum_{k}L_{{\bf A}}({a_1}^k T_k)$$
and
$$L_{{\bf A}}({a_1}^k T_k)=\sum_{a \in {\bf A}} a^k T_k/R(a, {\bf A}-a)= T_k \sum_{a \in {\bf A}} a^k /R(a, {\bf A}-a)=T^k L_{{\bf A}}(a_1^k)$$
it suffices to determine the values $L_{A}(a_{1}^{k}), 
k \geq 0$ to characterize Lagrange operator (we shall see later that 
$ \{ a_{1}^{0}, \ldots , a_{1}^{n} \}$ is a basis).



One can refer to Euler, who knew that 

\begin{equation}
\label{li-6}
\forall k \geq 0, \quad \sum_{a \in {\bf A}} a^{k}/R(a, {\bf A} \setminus a) = S_{k-n}({\bf A}), 
\end{equation}

where the $S_{k}({\bf A})$ are the {\em complete functions} of {\bf A}, i.e. $S_n({\bf A}):=[x^n]\frac {1}{\prod_{a \in {bf A}}(1-ax)}$

Here we shall give a proof of it. Indeed, let $f(x)=g(x)\equiv 1$, $x=1/z$ in (1.3), 
then we get 
$$\frac {1}{R(1/z, {\bf A})}=\sum_{a \in {\bf A}} \frac{1}{(\frac{1}{z}-a)R(a, {\bf A}\setminus a)}$$
here$|{\bf A}|= n+1$, thus we have


\begin{equation}
\label{li-7}
\sum_{a \in {\bf A}} \frac{1}{(1-za)R(a, {\bf A} \setminus a)} = \frac{z^{n}}{ \prod (1-za)}.
\end{equation}

Let's compare the coeffient of $z^k$ in both side of (\ref{li-7}), we see that the left is 
equeal to the left of (\ref{li-6}), and it is the same to the right!


More generally, given two alphabets {\bf A}, {\bf B}, with {\bf A} of cardinal 
$n + 1$, by linearity from (\ref{li-7}), one has 

\begin{equation}
\label{li-8}
L_{{\bf A}} \left( \frac{ \prod_{b \in {\bf B}}(1 - zb)}{(1 - za_1)} \right) = \frac{z^{n} \prod (1 - zb)}{ \prod (1 - za)} , 
\end{equation}

It is easy to understand from the second proposition of $L_{\bf A}$.
i.e. , for every $r$, writing $x$ for $a_{1}$, 

\begin{equation}
\label{li-9}
L_{{\bf A}}(S_{r}(x - {\bf B})) = S_{r - n}({\bf A} - {\bf B}).
\end{equation}

where $S_n({\bf A}-{\bf B}):=[z^n]\frac {\prod_{b \in {\bf B}}(1-zb)}{\prod_{a \in {\bf A}}(1-za)}$.

Lagrange operator given in (\ref{li-6}), sends polynomials in one variable of degree
$< n$ to $0$. In the general case, this is the operator which expresses the
{\em remainder} in the usual Lagrange formula. It can also be interpreted in
terms of the Euclidean division of polynomials, but this time it gives the 
{\em quotient} and not the remainder. 

Indeed, let {\bf B} be any alphabet, $m$ an integer (which does not need to
be the cardinal of {\bf B}), {\bf A} another alphabet of cardinal $n$, $x$ 
a variable. Using the Lagrange operator associated to the alphabet
$ \{ x, {\bf A} \} $, one can rewrite (\ref{li-9}) as

\begin{equation}
\label{li-10}
S_{m}(x - {\bf B})/R(x, {\bf A})- \sum_{a} \frac{S_{m}(a - {\bf B})}
{(x - a)R(a, {\bf A} \setminus a)}
= S_{m - n}(x + {\bf A} - {\bf B}), 
\end{equation}

which can be considered as a division

\begin{equation}
\label{li-11}
S_{m}(x - {\bf B}) = \sum_{a} \frac{S_{m}(a - {\bf B})R(x, {\bf A} \setminus a)}{R(a, {\bf A} \setminus a)} 
+ S_{m - n}(x + {\bf A} - {\bf B}) \cdot R(x, {\bf A}). 
\end{equation}


Showing that $S_{m-n}(x+{\bf A}-{\bf B})$ is the quotient in the Euclidean division, 
and that the remainder is equal to $L_{\bf A}(S_m(a_1-{\bf B})R(x, {\bf A}\setminus a_1))$.

At the end of this section, we will give some examples of $S_n({\bf A}):$

For example, 
Let ${\bf A}=\{1, 1, \ldots \}$ and ${\bf A}_n:=\{1, 1, \ldots, 1\}$, where$|{\bf A}_n|=n$, 
then we have 
$$ S_r({\bf A}_n)={n+r-1 \choose r}, \quad S_r({\bf A}_n)=(-1)^r {n\choose r}. $$


For example, 
Let ${\bf A} = \{ 1, q, \ldots , q^{n - 1}\}$, denote 
$${n\choose k}_q =\frac{(q^{n-1}+q^{n-2}+\ldots +1)\cdots 1}
{(q^{n-k-1}+q^{n-k-2}+\ldots +1)\cdots 1 \cdot 
(q^{k-1}+q^{k-2}\ldots +1)\cdots 1}, $$
then we have 
$$\sum S_r({\bf A})z^r = \prod ^{n - 1}_{i = 0}(1 - q^iz)^{-1} = \sum ^
{\infty}_{r = 0}
{n + r - 1 \choose r}_q z^r$$
and 
\begin{eqnarray*}
S_r({\bf A}) & = &
{n + r - 1 \choose r}_q \\
\sum S_r(-{\bf A})z^r & = & \sum ^{n - 1}_{i = 0}(1 - q^iz) = 
\sum^n_{r = 0}(-1)^rq^{r(r - 1)/2}
{n \choose r}_q z^r
\end{eqnarray*}


For example, 
Let ${\bf A} = \{ 1, 2, \ldots , n\}$

$$ S_r({\bf A}) = S(n + r, n)$$

where $S(n, k)$ is the Stirling number of the second kind.

For example, let ${\bf A} = \{\pm\pi , \pm 2\pi , \pm 3\pi , \ldots\}$\\
then we have

\begin{eqnarray*}
S_{2r}(-{\bf A}) & = & (-1)^r \frac{1}{(2r + 1)!}\\
S_{2r + 1}(-{\bf A}) & = & 0 \\
S_{2r}({\bf A}) & = & (-1)^{r - 1} \frac{(4^r - 2)B_{2r}}{(2r)!}, 
\end{eqnarray*}

where $\{ B_n\}$ are the Bernoulli nmubers, and they are defined by 
$$\frac{x}{e^x - 1} = \sum {\frac{B_nx^n}{n!}}.$$




{\bf Remark:}

Another aproach to Lagrange operator is to use Newton divided differences.
To illustrate this, we need the knowledge of permutations.

\pagebreak

\section{Permutations}

For each integer $n\geq 1$, let $G_n$ denote the symmetric group of degree $n$, 
that is to say the group of all permutations of the set $[1, n]=\{1, 2, \ldots, n\}$.
Each $w\in G_n$ is a mapping of $[1, n]$ onto itself. As is customary, we write all
mappings on the left of their arguments, so that the image of $i\in [1, n]$
under $w$ is $w(i)$. We shall sometimes denote $w$ by the sequence $(w(1), 
w(2), \ldots , w(n))$. Thus for example $(53214)$ is the element of $G_5$ that
sends 1 to 5, 2 to 3, 3 to 2, 4 to 1 and 5 to 4.

For $i=1, 2, \ldots , n-1$ let $s_i$ denote the transposition that interchanges 
$i$ and $i+1$, and fixes all other elements of $[1, n]$. We have

\begin{equation}
\label{s-1.1}
\left\{
\begin{array}{ll}
s_i^2=1 & \\
s_i s_j =s_j s_i & if |i-j|>1, \\
s_i s_{i+1} s_i =s_{i+1} s_i s_{i+1} & (1 \le i \le n-2).
\end{array}
\right.
\end{equation}

Also for each $w\in G_n$, let
$$ I(w)=\{(i, j) : 1\leq i <j \leq n , w(i)>w(j) \}. $$

We regard $I(w)$ as a subset of the square $\sum_n = [1, n]\times [1, n]$, 
and we shall adopt the convention of matrices, that in $\sum_n$ the first
coordinate increases from north to south, and the second coordinate from
west to east. The group $G_n \times G_n$ acts on $\sum_n: (u\times v)(i, j)
=(u(i), v(j))$. In particular, $G_n$ acts diagonally:
$w(i, j)=(w\times w)(i, j)=(w(i), w(j))$.

Let $w\in G_n, 1\leq r\leq n-1.$ Then $ws_r$ is the permutation
$$(w(1), \ldots , w(r+1), w(r), \ldots, w(n))$$
and it is clear that
\begin{equation}
\label{s-1.2}
I(ws_r)=\left\{ \begin{array}{ll}
        s_r I(w) \cup \{ (r, r+1) \} & if \; w(r)<w(r+1), \\
        s_r I(w) - \{ (r+1, r) \} & if \; w(r)>w(r+1).
        \end{array}
    \right.
\end{equation}

Let $\ell (w)=|I(w)|$. Then from (\ref{s-1.2}) we have
\begin{equation}
\label{s-1.3}
\ell (ws_r)=\left \{ \begin{array}{ll}
            \ell (w) +1 & if\ w(r)<w(r+1), \\
            \ell (w) -1 & if\ w(r)>w(r+1).
          \end{array}
      \right.
\end{equation}

\begin{thm}
\label{s-1.4}
$s_1, \ldots , s_{n-1}$ generate the group $G_n$.
\end{thm}
{\em Proof.} We shall show by induction on $\ell (w)$ that each $w\in S_{n}$ is
a product of $s$'s. If $\ell (w) = 0$, then $w=1$ and there is nothing to prove. 
If $\ell (w) > 0$ then $ w(r) > w(r+1) $ for some $r$, and hence $\ell (ws_r) = 
\ell (w) - 1$ by (\ref{s-1.3}). Hence $ws_r=s_{a_1} \ldots s_{a_p}$ say, 
and therefore (as $ s_r^2=1) w=s_{a_1} \ldots s_{a_p}s_r$.
\qed

For each $w \in G_n$, the {\sl length} of $w$ is the minimal length of a 
sequence $(a_1, \ldots, a_p)$ such that $w=s_{a_1} \ldots s_{a_p}$. 
 
\begin{thm} 
\label{s-1.5}
The length of $w \in G_n$ is equal to $\ell (w) = Card I(w)$.
\end{thm}
{\em Proof.} Let $\ell'(w)$ temporarily denote the length of $w$. The proof
of (\ref{s-1.4}) shows that $w$ can be written as a word of length $\ell (w)$ 
in the $s_i$, so that $\ell' (w) \leq \ell (w)$. Conversely, let $w=s_{a_1} \ldots
s_{a_p}$ be any expression of $w$ as a product of $s_i$. To show that 
$\ell (w) \leq \ell' (w)$ it is enough to show that $\ell(w) \leq p$. 
Let $w'=s_{a_1} \ldots s_{a_{p-1}}$; from (\ref{s-1.3}) we have 
$\ell (w) \leq \ell' (w)+1$ and hence 
$$\ell(w') \leq p-1 \Rightarrow \ell(w) \leq p.$$
Hence the proof is completed by induction on $p$.
\qed

\begin{thm}
\label{s-1.6}
Let $w \in G_n$. Then 
\begin{itemize}
\item[(i)] $\ell(w)=0 $ iff $ w=1.$
\item[(ii)] $\ell(w)=1$ iff $ w=s_r (1 \leq r \leq n-1).$
\item[(iii)] $\ell(w^{-1})=\ell(w).$ 
\item[(iv)] Let $ w_0=(n, n-1, \ldots, 2, 1) \in G_n$. Then
$\ell(w_0w)=\ell(ww_0)={1 \over 2}n(n-1)-\ell(w).$
\end{itemize}
\end{thm}
{\em Proof.} (i) and (ii) require no argument. Also (iii) is clear, since $w=s_{a_1}
\ldots s_{a_p}$ if and only if $w^{-1} = s_{a_p} \ldots s_{a_1}$.\\
(iv) The set $I(w_0)$ consists of all $(i, j) \in \Sigma_n$ such that 
$i < j$, so that $\ell(w_0)={1 \over 2}n(n-1)$. Next, we have 
$$ww_0=(w(n), w(n-1), \ldots, w(1))$$
so that $I(ww_0)$ is the complement of $I(w)$ in $I(w_0)$, and therefore
$$\ell(ww_0)={1 \over 2}n(n-1)-\ell(w).$$
Finally, since $w_0^2=1$ we have, by virtue of (iii) above.

\begin{eqnarray*}
\ell(ww_0) &=& \ell(w^{-1}w_0) \\
&=& {1 \over 2}n(n-1)-\ell(w^{-1}) \\
&=& {1 \over 2}n(n-1)-\ell(w).
\end{eqnarray*}
The element $w_0$ is called the {\sl longest element} of $G_n$.
\qed

For each $w \in G_n$ let $ R(w) $ denoted the set of all sequences $(a_1, 
\ldots, a_p)$ of length $p=\ell(w)$ such that $w=s_{a_1} \ldots s_{a_p}$. Such sequences
are called {\sl reduced words} for $w$. Clearly, 
$$(a_1, \ldots, a_p) \in R(w) \Leftrightarrow (a_p, \ldots, a_1) \in R(w^{-1}).$$


\begin{thm}
\label{s-1.7}
Let $(a_1, \ldots, a_p) \in R(w)$. Then 
$$I(w)=\{ s_{a_p} \ldots s_{a_{r+1}}(a_r, a_r+1):1 \le r \le p \}.$$
\end{thm}
{\em Proof.} Let $w'=ws_{a_p}=s_{a_1} \ldots s_{a_{p-1}}$. Then $\ell(w')=p-1$ 
and hence by (\ref{s-1.2}) and (\ref{s-1.3}) we have
$$I(w)=s_{a_p} I(w') \cup \{ (a_p, a_p+1) \}$$
from which (\ref{s-1.7}) follows by induction on $p$.
\qed

\begin{lem}[Exchange Lemma]
\label{s-1.8}
Let $(a_1, \ldots, a_p), (b_1, \ldots, b_p) \in R(w)$. Then \\
$(b_1, a_1, \ldots, \hat{a}_i, \ldots, a_p) \in R(w)$ for some $i=1, 2, \ldots, p$.
\end{lem}
{\em Proof.} By (\ref{s-1.7}), applied to $w^{-1}$, we have 
$(b_1, b_1+1) \in I(w^{-1})$ and hence
$$(b_1, b_1+1)=s_{a_1} \ldots s_{a_{i-1}}(a_i, a_i+1)$$
for some $i=1, \ldots, p$. It follows that
$$s_{b_1}=s_{a_1} \ldots s_{a_{i-1}}s_{a_i}(s_{a_1} \ldots s_{a_{i-1}})^{-1}, $$
so that $s_{b_1}s_{a_1} \ldots s_{a_{i-1}}=s_{a_1} \ldots s_{a_i}$, and therefore
$$s_{b_1}s_{a_1} \ldots \hat{s}_{a_i} \ldots s_{a_p}
 =s_{a_1} \ldots s_{a_p}=w.$$
\qed

A series of abstract operators $\{ D_i \}_{1 \le i}$ are said to fulfill
the Moore-Coxeter relation, if the following two equations hold.

\begin{equation}
\label{s-2.4-1}
\left\{ \begin{array}{ll}
        D_i D_j=D_j D_i & \mbox{\ \ if} |i-j|>1, \\
        D_i D_{i+1} D_i=D_{i+1} D_i D_{i+1}
     \end{array}
\right.
\end{equation}

For any sequence ${\bf a}=(a_1, \ldots, a_p)$ of positive integers, we define
$$D_{{\bf a}}=D_{a_1} \ldots D_{a_p}.$$

\begin{thm}
\label{s-2.5}
If there is a $w\in G_n$ such that ${\bf a, b} \in R(w)$, 
then $D_{\bf a}=D_{\bf b}.$
\end{thm}
{\em Proof.} We proceed by induction on $p=\ell(w)$. Let us write ${\bf a \equiv b}$ 
to mean that $D_{\bf a} = D_{\bf b}$. The inductive hypothesis then
implies that
\begin{equation}
\label{s-*}
{\bf a \equiv b} \mbox{ if either } a_1=b_1 \mbox{ or } a_p=b_p
\end{equation}
By the exchange lemma (\ref{s-1.8}) we have
$${\bf c}_i=(b_1, a_1, \ldots, \hat{a}_i, \dots, a_p) \in R(w)$$
for some $i=1, \ldots, p$. If $i \not= p$ then ${\bf b \equiv c}_i {\bf \equiv a}$ 
by virtue of (\ref{s-*}), so that ${\bf a \equiv b}$. If $i=p$ and $|b_1-a_1|>1$ 
then by (\ref{s-2.4-1}) and (\ref{s-1.1})
$${\bf c'}_p=(a_1, b_1, a_2, \ldots, a_{p-1}) \in R(w)$$
and ${\bf a \equiv c'}_p {\bf \equiv c}_p {\bf \equiv b}$, so that again ${\bf a \equiv b}$. 

Finally, if $i=p$ and $|b_1-a_1|=1$, we apply the exchange lemma again, 
this time to ${\bf c}_p$ and {\bf a}; this shows that
$${\bf d}_i=(a_1, b_1, a_1, \ldots, \hat{a}_i, \ldots, a_{p-1}) \in R(w)$$
for some $i=2, \dots, p-1$. But then by (\ref{s-2.4-1}) and (\ref{s-1.1}) we have
$${\bf d'}_i=(b_1, a_1, b_1, a_2, \ldots, \hat{a}_i, \ldots, a_{p-1}) \in R(w)$$
and ${\bf a \equiv d}_i {\bf \equiv d'}_i {\bf \equiv b}$. Hence
${\bf a \equiv b}$ in all cases.
\qed

So we can define $D_w=D_{\bf a}$ for any ${\bf a}\in R(w)$.

Another concern about permutations is the code of a permutation.
Let $w \in G_n$, and for each $i \ge 1$ denote 
$$c_i(w) = Card \{ j : j>i \mbox{ and } w(j)<w(i) \}.$$
Thus $c_i(w)$ is the number of points in the $i^{th}$ row of $I(w)$.
The vector $$c(w)=(c_1(w), \ldots, c_n(w)) \in N^n$$
is called the {\sl code} or {\sl inversion table} of $w$.

\begin{thm}
\label{code}
There is a bijection between the permutations of $G_n$ and the sequence
of integers $c_1, c_2, \ldots, c_n$ with $0\leq c_i \leq n-i$ for all
$1\leq i\leq n$.
\end{thm}
{\em Proof.}
Clearly, $c_i(w)$ denote the number of $j$ with $j>i$ which is before
$i$ in $w$, since the number of $j$ with $j>i$ is at most $n-i$, 
we have $c_i(w)\leq n-i$.
For any $w$ we have a unique $c(w)$ with $c_i(w)\leq n-i$.

For any sequence with $c_i\leq n-i$ we can construct $w\in G_n$ with
$c(w)=\{c_1, c_2, \ldots c_n\}$ as follows.
\begin{itemize}
\item Let $A=B=[1, n]$.

\item Find the smallest element $i$ in $A$, find the
$c_i+1$-th element $j$ in $B$, and we have $w(i)=j.$

\item Let $A:=A-\{i\}$ and $B=B-\{w(i)\}$, then repeat the above procedure
until $A=\phi $.
\end{itemize}

Since for each $i$, the $B$ becomes a set of $n-i+1$ elements, which is not
less than $c_i+1$, the above procedure gives a uniquepermutation.
\qed

The above bijection will be noted by $\psi (c(w))=w$. And if $c_i>c_{i+1}, $
it is clear that $w(i)>w(i+1)$ from the bijection then we have
\begin{equation}
\label{s_i}
c_i>c_{i+1}\Longrightarrow \left\{ 
\begin{array}{l}
c_j(ws_i)=c_j(w) \qquad if \quad j\neq i, i+1, \\
c_i(ws_i)=c_{i+1}(w), \\
c_{i+1}(ws_i)=c_i(w)-1, 
\end{array}
\right.
\end{equation}
and
\begin{equation}
c_i\leq c_{i+1}\Longrightarrow \left\{ 
\begin{array}{l}
c_j(ws_i)=c_j(w) \qquad if \quad j\neq i, i+1, \\
c_i(ws_i)=c_{i+1}(w)+1, \\
c_{i+1}(ws_i)=c_i(w).
\end{array}
\right.
\end{equation}

A permutation $w$ is said to be {\sl dominant} if its code is a partition, 
that is $c_1(w) \ge c_2(w) \ge \ldots \ge c_n(w) \ge 0$, and we say $w$ is
dominant of shape $\lambda=(c_1(w), c_2(w), \ldots, c_n(w))$


\pagebreak


\section{Schur Function}

Schur function has been defined by several methods in history. The famous
one is associated with Young Teablux, which need the concept of {\em
Partitions, Young Diagram.}


\subsection{Partitions}

Many of the objects we shall consider in this book will turn out to be
parametrized by partitions. The purpose of this section is to lay down some
notation and terminology which will be used throughout.

A partition is any finite sequence 

$$I = (i_1, i_2, \ldots , i_p)$$

of non-negative integers in increasing order:

$$i_1\leq i_2\leq\ldots , i_p\quad .$$

We shall find it convenient not to distinguish between two such sequences
which differ only by a string of zeros at the beginning.
Thus, for example, we regard $(1, 2)$, $(0, 1, 2)$, $(0, 0, 1, 2) \ldots$
as the same partition.

The weight of $I$ denoted by $|I|$, with

$$|I| = i_1 + i_2 + \ldots + i_p$$

If $|I| = n$, we say that $I$ is a partition of $n$.

The diagram of a partition $I$ may be formaly defined to be some squares, 
with the first row $i_1$ squars, the second row $i_2$ squares. $\ldots$
and the $l$-th row $i_l$ squares. 

The conjugate of a partition $I$ is the partition $I^*$ whose diagram is 
the transpose of the diagram $I$.

Suppose $i_p\leq q$, then we can have

$$i_s^* = card\{t : i_t \geq q + 1 - s\} \mbox{ and }
I^*=\{i_1^*, i_2^*, \ldots, i_q^* \}.$$

For example, the conjugate of $(1344)$ is $(2334)$, we can see this
in the following figue.

\begin{center}
\begin{picture}(30, 30)
\multiput(5, 5)(5, 0){4}{\framebox(5, 5)} 
\multiput(5, 10)(5, 0){4}{\framebox(5, 5)}
\multiput(5, 15)(5, 0){3}{\framebox(5, 5)}
\put(5, 20){\framebox(5, 5)}
\end{picture}
\end{center}


\begin{lem}
\label{conju}
Let $I$ be a partition contained in $(m^n)$, then the $m + n$ members
$$i_s + s - 1\quad (1\leq s\leq n), n + m - t - i^*_t\quad (1\leq t\leq m)$$
are a permutation of $\{ 0, 1, 2, \ldots , m + n - 1\}$.
\end{lem}

{\em Proof.}
The successive segments of the bandary line between the diagram of
$I$ and its complement
in $(m^n)$ (marked thickly in the picture) with the number $0, 1, \ldots , 
m + n - 1$. The number attached to the vertical segments are $i_s + s - 1$.
and by treansposition those attached to the horizontal segments are
$$(m + n - 1) - (i^*_t + t - 1) = m + n - i^*_t - t$$

\begin{picture}(70, 90)
\multiput(10, 10)(10, 0){7}{\framebox(10, 10)}
\multiput(10, 20)(10, 0){7}{\framebox(10, 10)}
\multiput(10, 30)(10, 0){7}{\framebox(10, 10)}
\multiput(10, 40)(10, 0){7}{\framebox(10, 10)} 
\multiput(10, 50)(10, 0){7}{\framebox(10, 10)} 
\thicklines{\put(10, 50){\line(0, 1){10}}}
\thicklines{\put(10, 20){\line(1, 0){10}}}
\thicklines{\put(20, 20){\line(0, 1){10}}}
\end{picture}
$I = (01344)\quad , \quad m=7 \quad , n=5$
\qed
%%%%%%

\subsection{Symmetric Functions}

For efficient recursions, one needs to consider symmetric functions as
``functors'' on alphabets. This essentially means that instead of defining
complete functions as coefficients in the expansion of a rational series
$\prod_{a \in \bA} 1/(1-za)$, we can take arbitrary families
$\{ S_k \}_{k \ge 0}$ of elements of a commutative ring, with the only
restriction that $S_0=1$. Indeed, to any formal series
$f = \sum _{ \geq 0}z^kS_k$, the $S_k$ belonging to a
commutative ring, Littlewood [ch. 6.4.] associated the infinite Hankel
matrix ${\bf S}(f) = (S_{(j - i)})_{j, i \geq 0}$, 
putting $S_i = 0$ if $i < 0$, and
defined {\em skew Schur functions} to be the minors of this matrix.

More precisely, given
$I = (i_1, \ldots , i_n) \in {\bf Z}^n, J \in {\bf Z}_n$
he defined 

\begin{equation}
\label{syf-1}
S_{J/I}(A)=\det (S_{j_s-i_t+s-t}(A))
\end{equation}

to be the minor of ${\bf S}({\bf A})$ taken on 
rows $i_1 + 1, 
i_2 + 2, \ldots , i_n + n$ and columns $j_1 +1, \ldots , j_n + n$ (the minor 
is $0$ if one of these numbers is $<0$). When $I = 0^n$, one writes $S_J$ 
instead of $S_{J/0^n}$, (indexing
corresponding to the convention that partition are decreasing).

\vspace{3mm}


The case of a rational series $f = \prod (1 - zb)/(1 - za)$ for two alphabets
${\bf A} = \{ a \}$ and ${\bf B} = \{ b \}$ corresponds by definition to the 
Schur functions of a difference of alphabets (``super Schur functons''), i.e. 
the coefficients of $f$ are the complete functions $S_J({\bf A} - {\bf B})$.
Let us denote ${\bf S}({\bf A} - {\bf B})$ the corresponding Hankel matrix.
This case covers in fact the case of general formal power series, since
when the cardinal of $\bA$ or $\bB$ is infinite, the $S_k(\bA)$
(resp. $S_k(\bB)$) are algebraically independent.

Now, addition of alphabets corresponds to multiplication of generating series
or multiplication of the associated Hankel matrices, and formal subtraction
corresponds to division of series or multiplication by the inverse of the 
Hankel matrix. In other words
\begin{eqnarray*}
&{\bf A} - {\bf B} + {\bf C} \Leftrightarrow {\bf S}({\bf A} - {\bf B} + {\bf C})
= {\bf S}({\bf A}){\bf S}({\bf B})^{-1}{\bf S}({\bf C})\\
\Leftrightarrow & \frac {\prod (1 - zb)}{\prod (1 - za) \prod (1 - zc)}
= \frac {1}{\prod (1 - za)} \prod (1 - zb) \frac{1}{\prod (1 - zc)}
\end{eqnarray*}

More explicitely, 

\begin{equation}
\label{syf-2}
S_k(\bA-\bB+\bC)=\sum_{i, j, h, i+j+h=k} S_i(\bA)S_j(-\bB)S_h(\bC)
\end{equation}

Erasing a common factor in the numerator and denominator of a rational
function corresponds to writing $(\bA+\bC)-(\bB+\bC)=\bA-\bB$.

Let us remark that, $\bA$ being finite of cardinal $n$, then the
$(-1)^kS_k(-\bA)$ are the {\em elementary symmetric functions}
in $\bA$, the $S_k(x-\bA), k < n$, are the {\em truncations} of the polynomial
$R(x, \bA)$, while $S_n(x-\bA)=R(x, \bA)$, and for $k>n$, 
$$S_k(-\bA)=0\ , \ S_k(x-\bA)=x^{k-n}S_n(x-\bA).$$

General identities on minors imply identities on Schur funcions.

For example, given two matrices $M, N$, with dimensions such that their product
$M\, N$ (using the above conventions for indexing: $I, J \in N^n$, $H$ runs
over all partitions in $N^n$ ):
$$[MN]_{I, J}=\sum_{H} [M]_{I, H} [N]_{H, J}.$$
For Schur funcions, this imply

\begin{equation}
\label{syf-3}
S_{J/I}({\bf A} \pm {\bf B})=\sum_H S_{H/I}({\bf A})S_{J/H}(\pm {\bf B}).
\end{equation}

Similarly, the correspondence, due to Jacobi, between the minors of a matrix
and those of its inverse reads

\begin{equation}
\label{syf-4}
S_{I/J}(\bA-\bB)=(-1)^{|J/I|} S_{J~/I~}(\bB-\bA), 
\end{equation}

for a pair of partitions $I, J$ and their conjugates $J~, I~$, where $|J|$
stands for $j_1+\cdots+j_n$ and $|J/I|$ for $j_1+\cdots+j_n-i_1-\cdots-i_n$.
We shall give a strong form of it in the last subsection following [Mcd1].

Given a finite alphabet ${\bf A}$ of cardinal $n$, one can also have recourse
to the alphabet of its inverses ${\bf A}^\star := \{1/a \}_{a\in{\bf A}}$.

The relation $S_k(-{\bf A}^\star)= S_{n-k}(-{\bf A}) \, / S_n(-{\bf A})$
extends to

\begin{equation}
\label{syf-5}
S_I(-{\bf A}^\star) = S_{n^r/I}(-{\bf A}) \, S_n(-{\bf A})^r \ 
\end{equation}

$I$ partition with parts $\leq n$, 
 $r$ sufficiently big so that $I\subseteq n^r$
\smallskip  We still need to enlarge the definition of a Schur function, 
this time allowing to play with different alphabets at the same time. 
In [L1], geometry imposed to take flags of alphabets
(i.e. ${\bf A}_1 \subseteq {\bf A}_2 \subseteq{\bf A}_3 \subseteq\cdots$), but 
the methods used thereby 
extend without pain, though not using the graphical display of partitions
({\it Ferrers' diagrams}) rendered in [L1] 
the involution ${\bf A} \rightarrow -{\bf A}$ more difficult to state.




Now we can give the definition of multi-Super Skew Schur Function.
Given $n$, given two sets of alphabets $\{ {\bf A}_1, {\bf A}_2, \cdots, {\bf A}_n\}$, 
$\{ {\bf B}_1, {\bf B}_2, \cdots, {\bf B}_n\}$, and $J, I \in {\cal N}^n$, 
we define that:
\begin{equation}
\label{syf-6}
S_{J/I}(A_1 - B_1, A_2 - B_2, \ldots , A_n - B_n):=
\det(S_{i_s-j_t-s+t}(A_t-B_t), 
\end{equation}


In the case where the alphabets are repeated, we indicate by a comma the 
corresponding block separation: given $H \in {\cal Z}^p$, $K \in {\cal Z}^q$, 
then $S_{H, K}({\bf A}-{\bf B}), {\bf C}-{\bf D})$ stands for the multi-schur 
function with index the concatenation of $H$ and $K$, and alphabets
${\bf A}_1=\cdots={\bf A}_p={\bf A}$, ${\bf B}_1=\cdots={\bf B}_p={\bf B}$, 
${\bf A}_{p+1}=\cdots={\bf A}_{p+q}={\bf C}$, ${\bf B}_{p+1}=\cdots={\bf B}_{p+q}={\bf D}$.

These functions are now sufficiently general to allow easy inductions, thanks to \
the following transformation lemma. 

\begin{thm}
\label{syf-7}
Let ${\bf D}_0, {\bf D}_1, \ldots {\bf D}_{n - 1}$ be $k$ sets of letters, 
and card$({\bf D}_i) \leq i$, then
\begin{eqnarray}
&&S_{J/0^n}({\bf A}_1 - {\bf B}_1, \ldots , {\bf A}_k - {\bf B}_n)
=\det(S_{j_s+s-t}({\bf A}_s-{\bf B}_s-{\bf D}_{n-t}))
\nonumber
\\
&=& \left|
\begin{array}{llll}
S_{j_1}({\bf A}_1-{\bf B}_1)& S_{j_2+1}({\bf A}_2-{\bf B}_2) &
\ldots & S_{j_n+n-1}({\bf A}_n-{\bf B}_n) \\
S_{j_1-1}({\bf A}_1-{\bf B}_1)& S_{j_2}({\bf A}_2-{\bf B}_2)&
\ldots & S_{j_k+n-2}({\bf A}_n-{\bf B}_n) \\
\ldots & \ldots &\ldots &\ldots \\
S_{j_1-n+1}({\bf A}_1-{\bf B}_1)& S_{j_2-n+2}({\bf A}_2-{\bf B}_2) &
\ldots & S_{j_n}({\bf A}_n-{\bf B}_n)
\end{array}
\right| 
\label{syf-8}
\\
&=& \left|
\begin{array}{llll}
S_{j_1}({\bf A}_1-{\bf B}_1-{\bf D}_{n-1})& S_{j_2+1}({\bf A}_2-{\bf B}_2-{\bf D}_{n-1}) &
\ldots & S_{j_n+n-1}({\bf A}_n-{\bf B}_n-{\bf D}_{n-1}) \\
S_{j_1-1}({\bf A}_1-{\bf B}_1-{\bf D}_{n-2})& S_{j_2}({\bf A}_2-{\bf B}_2-{\bf D}_{n-2})&
\ldots & S_{j_n+n-2}({\bf A}_n-{\bf B}_n-{\bf D}_{n-2}) \\
\ldots & \ldots &\ldots &\ldots \\
S_{j_1-n+1}({\bf A}_1-{\bf B}_1-{\bf D}_0)& S_{j_2-n+2}({\bf A}_2-{\bf B}_2-{\bf D}_0) &
\ldots & S_{j_n}({\bf A}_n-{\bf B}_n-{\bf D}_0)
\end{array}
\right|
\nonumber
\end{eqnarray}
In the (\ref{syf-8}), each row of the matrix minus
the set ${\bf D}_i$ in the corresponding rows.
\end{thm}

{\em Proof.} 


In other words, one does not change the value of a multi-Schur
function $S_J$ by replacing in row $h $ the difference 
${\bf A}-{\bf B}$ by ${\bf A}-{\bf B} - {\bf D}_{k-h}$. Indeed, thanks to 
the expansion ($\ref{syf-2}$) :
$$S_j({\bf A}-{\bf B}-{\bf D}_h) = S_j({\bf A}-{\bf B})+ S_1(-
{\bf D}_h)\, S_{j-1}({\bf A}-{\bf B}) 
+\cdots + S_h(-{\bf D}_h)\, S_{j-h}({\bf A}-{\bf B}) \ , $$
the sum terminating because the $S_k(-{\bf D}_h)$ are null for $k>h$, 
we see that the determinant has been transformed by multiplication
by a triangular matrix with 1's in the diagonal, and therefore has 
kept its value. Then the theorem follows.
\qed

\begin{cor}
\label{cosup}
If $|{\bf B}| = n$, we have
\begin{equation}
S_J({\bf A} - {\bf B}) = S_{J, 0^n}({\bf A}, {\bf B}).
\label{syf-9}
\end{equation}
\end{cor}

{\em Proof.}
For $S_{J, 0^n}({\bf A}, {\bf B}) $, simply let ${\bf D}_i=1, \; if \; i<n$
and ${\bf D}_i={\bf B}, \; if \; i\geq n$ in the theorem \ref{syf-7}.
\qed

For example, taking ${\bf D}_0=\emptyset$, ${\bf D}_1= \{x \}$, ${\bf D}_2=
\{y, z \}$, one has 
$$\left( \begin{array}{ccc}
        S_i({\bf A}_1-y-z) & S_{j+1}({\bf A}_2-y-z) & S_{h+2}({\bf A}_3-y-z) \\
        S_{i-1}({\bf A}_1-x) & S_{j}({\bf A}_2-x) & S_{h+1}({\bf A}_3-x) \\
        S_{i-2}({\bf A}_1) & S_{j-1}({\bf A}_2) & S_{h}({\bf A}_3) \\ 
    \end{array}
 \right)
$$
$$ = \left( \begin{array}{ccc}
        1 & -y-z & yz\\
        0 & 1 & -x\\
        0 &0 &1 \\ 
      \end{array} 
   \right) 
\cdot 
    \left( \begin{array}{ccc}
        S_i({\bf A}_1) & S_{j+1}({\bf A}_2) & S_{h+2}({\bf A}_3) \\
        S_{i-1}({\bf A}_1) & S_{j}({\bf A}_2) & S_{h+1}({\bf A}_3) \\
        S_{i-2}({\bf A}_1) & S_{j-1}({\bf A}_2) & S_{h}({\bf A}_3) \\ 
       \end{array}
    \right)
$$
and the determinant of the left matrix is equal to
$S_{ijh}({\bf A}_1, {\bf A}_2, {\bf A}_3)$. 


This lemma implies many factorization properties, e.g. for $r\geq0$, 

\begin{equation}
\label{syf-10}
S_J({\bf A}-{\bf B}-x)\, x^r = S_{J, r}({\bf A}-{\bf B}, x) 
\end{equation}

since taking ${\bf D}_1= {\bf D}_2=\cdots = \{x \}$ factorizes the determinant
$S_{J, r}({\bf A}-{\bf B}, x)$. 

More generally, for an alphabet ${\bf D}$ of cardinal $\leq r$ and $J \in {\bf N}^r$, 
one has 
\begin{equation}
\label{syf-11}
S_I({\bf A}-{\bf B}-{\bf D}) \, S_J({\bf D}) = S_{I, J}({\bf A}-
{\bf B}, \, {\bf D}) \ .
\end{equation}

In this case, let's consider $S_{I, J}({\bf A}-{\bf B}, {\bf D})$, for $|D| \le r$ 
and $J \in {\cal N}^r$, let ${\bf D}_1=\cdots={\bf D}_r=0$, 
${\bf D}_{r+1}={\bf D}_{r+2}={\bf D}$, then we get the previous eqution.




\smallskip
The resultant $R({\bf A}, {\bf B})$ is of fundamental importance in interpolation
theory, in elimination theory, in the theory of symmetric functions, 
in other words in premodern or postmodern algebra. 

There are many determinantal expressions of it, which all are related
to the fact that it is a special Schur function.

Indeed, let ${\bf A}$ and ${\bf B}$ be of cardinalities $m$ and $n$
respectively. 
Then
\begin{equation}
\label{syf-12}
R({\bf A}, {\bf B}) = S_{n^m}({\bf A}-{\bf B}) = (-
1)^{mn}S_{m^n}({\bf B}-{\bf A}) = 
S_{n^m, 0^n}({\bf A}, {\bf B}) \ .
\end{equation}

{\em Proof.} 
Take any $a$ in ${\bf A}$. Then $S_{n^m}({\bf A}-{\bf B}) =
S_{n, n^{m-1}} ({\bf A}-{\bf B}, {\bf A}-a-{\bf B}) $ according to ($\ref{syf-7}$), 
subtracting $a$ in all the columns except the first one 
(instead of subtracting in rows). Now subtract 
$\{ {\bf A}\backslash a\}$ in the first row of the new determinant. 
It factorizes into $S_n(a-{\bf B}) \, S_{n^{m-1}}( \{{\bf A}\backslash a\} -
{\bf B})$, 
which proves the result by induction on $m$. ($\ref{syf-11}$) gives 
the last expression where ${\bf A}$ and ${\bf B}$ have been separated. \quad 
\qed

For example, 
let $|A| = 3$ $A = \{ a_1, a_2, a_3 \}$, $|B| = 4$. we have
$$S_{444}(A - B) = \left|
\begin{array}{ccc}
S_4(A-B) & S_5(A-B) & S_6(A-B) \\
S_3(A-B) & S_4(A-B) & S_5(A-B) \\
S_2(A-B) & S_3(A-B) & S_4(A-B)
\end{array}
\right| $$
Using theorem \ref{syf-7}, let $C_0=1, C_1=1, C_2=\{a_3, a_2\}$
we have
$$S_{444}(A - B) = \left|
\begin{array}{ccc}
S_4(a_1-B) & S_5(a_1-B) & S_6(a_1-B) \\
S_3(A-B) & S_4(A-B) & S_5(A-B) \\
S_2(A-B) & S_3(A-B) & S_4(A-B)
\end{array}
\right|=R(a_1, B) \cdot |* |$$






\smallskip
Multi-Schur functions, being defined as determinants, 
are bound to techniques of minors.

For example, one can use Laplace's expansion along a subfamily of rows
or columns. Let $I\in{\bf N}^m$, $J\in{\bf N}^n$. Given $H\in{\bf N}^n$, 
denote
$H^\omega +J := (j_1+h_n, \ldots , j_n+h_1)$. Then
\begin{equation}
\label{syf-13}
S_{I, J}({\bf A}, {\bf B}) = \sum_H (-1)^{|H|} S_{I/\tilde{H}}({\bf A}) \, 
S_{J+H^\omega}({\bf B}) \, 
\end{equation}
sum on all partitions $H\in{\bf N}^n$, $H\subseteq m^n$.

In particular, for $n=1$, one has 
\begin{equation}
\label{syf-14}
S_{I, j}({\bf A}, {\bf B}) = \sum_{0\leq h\leq m} (-1)^h S_{I/1^h}({\bf A})\, 
S_{j+h}({\bf B}) \ .
\end{equation}

The fundamental involution ${\bf A} \rightarrow -{\bf A}$, which in the case ($\ref{syf-4}$) 
of a 
Schur function $S_{J/I}({\bf A})$ expresses it as a determinant of
elementary functions $S_k(-{\bf A})$, does not allow in general to
exchange the role of complete and elementary symmetric functions 
in a multi-Schur function. In [L1] are given conditions for having
two determinantal expressions of a multi-Schur function (see also
[Mcd2]). We skip this point at the moment, and shall handle it later
more easily with Schubert polynomials.

\subsection{Monomial}




Monomials themselves can be written as Schur functions. For an infinite
alphabet ${\bf A}$, let ${\bf A}_n$ be the initial sub-alphabet 
${\bf A}_n :=\{a_1, \ldots, a_n \}$. Let $I=(i_1, \ldots, i_n)\in{\bf N}^n$, 
and write $a^I$ for
$a_1^{i_1} \cdots a_n^{i_n}$, $I^\omega$ for $(i_n, \ldots , i_1)$.
Then one has 
\begin{equation}
\label{syf-15}
a^I = S_{I^\omega}( {\bf A}_n, \ldots, {\bf A}_1) \ , 
\end{equation}
as one can see by subtracting $0, {\bf A}_1, \ldots, {\bf A}_{n-1}$ in the successive 
rows, from the bottom.

For example, 
$$a^{742}=S_{247}({\bf A}_3, {\bf A}_2, {\bf A}_1)=
det \left| \begin{array}{ccc}
S_2({\bf A}_3) & S_5({\bf A}_2) & S_9({\bf A}_1) \\
S_1({\bf A}_3) & S_4({\bf A}_2) & S_8({\bf A}_1) \\
S_0({\bf A}_3) & S_3({\bf A}_2) & S_7({\bf A}_1) \\ 
\end{array}
\right|$$
$$= 
det\left| \begin{array}{ccc}
a_3^2 & 0 & 0 \\
a_2+a_3 & a_2^4 & 0 \\
1 & S_3(a_1+a_2) & a_1^7 \\ 
\end{array} 
\right|
$$

Any other ordering of the alphabet will lead to another 
determinantal expression of a monomial :
$$a^{742}= S_{742}( a_3+a_2+a_1, \,  a_3+a_2, \,  a_3) \ .$$

>From the above, we see that monomials in ${\bf A}_n$
can be expressed as linear combinations 
(with coefficients in $Sym({\bf A}_n)$ of monomials
$a^J : J\leq (n-1, \ldots , 1, 0)$, or monomials 
$a^I : I \leq(0, 1, \ldots , n-1) $.
Indeed, 

\begin{equation}
\label{syf-16}
a^H= S_H({\bf A}_n, \ldots , a_n+a_{n-1}, a_n)=
S_H({\bf A}_n, \ldots , {\bf A}_n - {\bf A}_{n-2}, {\bf A}_n- {\bf A}_{n-1}) 
\ . 
\end{equation}
Column $k$ is of degree 1 in the variables 
$a_1, \ldots , a_{k-1}$, and thus only monomials 
$a^J : J\leq (n-1, \ldots , 1, 0)$ appear in the expansion of the determinant.
It is very important for us!




\subsection{Duality}

Let $(X_n)_{n\in Z}$ be a sequence of alphabeta such that
$$X_{n-1}\subset X_n \qquad and \qquad |X_n-X_{n-1}|= 1$$
for all $n\in Z$.

\begin{lem}
Let $I$ be any interval in $Z$. Then the inverse of the matrix
$$H=(S_{i-j}(-X_i))_{i, j\in I}$$ is 
$H^{-1}=(S_{i-j}(X_{j+1}))_{i, j\in I}.$
\end{lem}
{\em Proof.}
Let $K$ denote the matrix $(S_{i-j}(X_{j+1}))$. The $(i, k)$ element of
$HK$ is then
$$\sum_{j}S_{i-j}(-X_i)S_{j-k}(X_{k+1})=S_{i-k}(-X_i+X_{k+1}).$$
If $i<k$ this is zero; if $i=k$ it is equal to 1; and if $i>k$ it is equal to 
$$S_{i-k}(-(X_i-X_{k+1}))$$ which is zero because $|X_i-X_{k+1}|\leq 
i-(k+1)<i-k.$ This complete the proof.
\qed

Then we have the following Duality Theorem.


\begin{thm}[Duality Theorem]
\label{dual}
Let $J > I$ be partitions contained in $(m^n)$. Then
$$S_{J/I}(-X_{j_1}, \ldots , -X_{j_{n + m - 1}}) = (-1)^{|J - I|}
S_{J^*/I^*}(X_{n + m - 1 - j_1^*}, \ldots , X_{m + n - j^*_m})$$
\end{thm}

{\em Proof.}
Let 
%\begin{center}
$$
\begin{array}{rll}
\xi _s = j_s - s, & \eta _s =i_s- s & (1 \leq s \leq n) \\
\xi _t^* = j_t^* - t, & \eta _t^* = i^*_t - t & (1\leq t\leq m)
\end{array}
$$
%\end{center}

Then by Lemma (\ref{conju}), the integers $\xi _s(1\leq s\leq n)$ and 
$-\xi^*_t - 1$ $(1\leq m)$ fill up the interval $[-m, n - 1]$, and so
do the $\eta _s$ and the $-\eta ^*_t - 1$.

The $(\xi _1, \ldots , \xi _n; \eta _1, \ldots , \eta _n)$ minor of the 
matrix $H$ is 
$$\det(h_{\xi_i - \eta _j}(-X_{\xi _i})) =
 S_{J/I}(-X_{\xi _i}, \ldots , -X_{\xi _n}).$$

Since $\det H = 1$, each minor of $H$ is equal to the complementary cofactor of 
$(H^{-1})'$.

and by \ref{conju}. which is just
$$(-1)^{J-I}S_{J^*/I^*}(X_{\xi ^*_1}, \ldots , X_{\xi ^*_m}).$$
\qed

Let $X_i=A$ for all integers $i$, then we can obtain the Jacobi-Trudi's
determinant expression


\begin{thm}
\label{xx}
We have $S_I(\bA)=(-1)^{|I|}S_{I^*}(-\bA)$, 
which can be writen as
\begin{equation}
S_I(\bA)=\left| \begin{array}{cccc}
e_{i^*_1}(\bA)) & e_{i^*_2+1}(\bA)) &\ldots & e_{i^*_l+l-1}(-\bA)) \\
e_{i^*_1-1}(\bA)) & e_{i^*_2}(\bA)) &\ldots & e_{i^*_l+l-2}(-\bA)) \\
\vdots & \vdots & \vdots & \vdots \\
e_{i^*_1-l+1}(\bA)) & e_{i^*_2-l+2}(\bA)) &\ldots & e_{i^*_l}(\bA))
\end{array} \right|
\end{equation}
when suppose $I$ is a partition of $k$ non zero components and with the
largest parts $l$.
\end{thm}

\vskip 5mm
{\bf Remark:}

Complete functions in the difference of two alphabets ${\bf A}$, ${\bf B}$, 
are defined by the generating function 
$$ \sum z^r S_r({\bf A}-{\bf B}) :=
\prod_{a\in{\bf A}, b\in{\bf B}} (1-zb)/(1-za) \ .$$
Multi-Schur functions are determinants in complete functions
$S_r({\bf A}^j-{\bf B}^j)$, with alphabets $A^j$, ${\bf B}^j$ fixed in each 
column
of the determinant.
\pagebreak

\section{Symmetrizing operators}

\vspace{3mm}

{\em Divided differences} $\partial _\mu$ have been introduced by Newton
to solve the interpolation problem in one variable. We shall also need the 
{\em isobaric divided differences $\pi _\mu$}.

In the case where $\mu$ is a simple transposition $\sigma _i$ exchanging
$a_i$ and $a_{i + 1}$, writing then $\partial _i, \pi _i, \sigma _i$, the
operators (acting on their left ) are: 

\begin{equation}
\label{e-2.1} 
\left\{ 
\begin{array}{rl}
\partial _i = & \frac{1}{(a_i - a_{i + 1})}(1 - \sigma _i) \\
\pi _i    = & \frac{1}{(1 - a_{i+1}/a_i)}(1 - \sigma _i) = a_i\partial _i 
\end{array}
\right.    
\end{equation}

More explicitly, $\partial _i$ and $\pi _i$ are the operators on functions
$f(a_1, \ldots , a_n)$ of several variables, acting only on the pair $(a_i, a_{i + 1})$:

\begin{eqnarray}
\label{e-2.2}
\left\{ 
\begin{array}{ll}
f\partial _i & = \frac{f( \ldots a_i, a_{i + 1} \ldots) - 
f( \ldots a_{i + 1}, a_i \ldots)}
{a_i - a_{i + 1}} \\
f \pi _i    & = \frac{a_if( \ldots a_i, a_{i +1} \ldots ) - a_{i + 1}
f( \ldots a_{i + 1}, a_i \ldots )}{a_i - a_{i + 1}}
\end{array}
\right.
\end{eqnarray}

One has to complete (\ref{e-2.1}) by the values for $\mu = identity$
which are

$$\partial _{id} = 1, \, \, \pi _{id} = 1$$

Both families of operators $\{ \sigma _i \}$, $\{ \partial _i\}$, 
$\{ \pi _i \}$
separately satisfy the Moore/Coxeter relations (writing $\{ D_i \}$ in each of 
these three cases).


\begin{equation}
\label{e-2.3}
D_iD_j = D_jD_i \quad if\quad \mid i - j \mid \geq 2, 
\end{equation}

\vspace{2 mm}

\begin{equation}
\label{e-2.4}
D_iD_{i + 1}D_i = D_{i + 1}D_iD_{i + 1}.
\end{equation}

Relations (\ref{e-2.3}) and (\ref{e-2.4}) imply that $\partial _\mu$, $\pi _\mu$
can be written as the product of elementary operators $\partial _i$, $\pi _i$
corresponding to any {\em reduced decomposition} of $\mu$ (i.e. any decomposition 
$\mu = \sigma _i \ldots \sigma _j$ of minimal length gives 
$\partial _\mu = \partial _i \ldots \partial_j, \pi _\mu = \pi _i \ldots \pi _j$). 

One can notice that

\begin{equation}
\label{e-2.5}
\left\{ 
\begin{array}{ll}
\partial _i \partial _i & = 0 \\
\pi _i \pi_i   & = \pi_i.
\end{array}
\right.
\end{equation}


Sepecially, $\partial _i$ satisfy the Braid relations.
\begin{equation}
\left\{ \begin{array}{l}
\partial _i \partial _i =0 \\
\partial _i \partial _j =\partial _j \partial _i , \quad |i-j|\geq 2 \\
\partial _i \partial _{i+1}\partial _i=
\partial _{i+1} \partial _i \partial _{i+1} 
\end{array}\right.
\end{equation}

{\em Proof.} The first and the second equation can be obtained immediately
from the definitions, and the third equation can be
verified by a direct calculation: each side is equal to
$$\frac{1}{(x_i-x_{i+1})(x_i-x_{i+2})(x_{i+1}-x_{i+2})}
\sum_{w \in S_3} {\varepsilon (w) w}. $$
where the symmetric group $S_3$ permutes $x_i, x_{i+1}, $ and $x_{i+2}$, 
 and $\varepsilon (w)$ is the sign of the permutation $w$.
\qed

For any sequence ${\bf a}=(a_1, \ldots, a_p)$ of positive integers, we define
$$\partial_{{\bf a}}=\partial_{a_1} \ldots \partial_{a_p}.$$

>From (\ref{s-2.5}) and (\ref{s-2.4-1}) it follows that we may define
$$\partial_w =\partial_{\bf a}$$
unambiguously, where {\bf a} is any reduced word for $w$.

\begin{thm}
\label{s-2.6}
For any ${\bf a} = (a_1, \ldots, a_p)$, if $\ell (s_{a_1}\cdots s_{a_p})<p$, 
then $\partial_{\bf a}=0$.
\end{thm}
{\em Proof.} By induction on $p$. If ${\bf a}'=(a_1, \ldots, a_{p-1})$ 
is not reduced, then $\partial_{{\bf a}'}=0$ and hence $\partial_{\bf a}=0$. 
So we may assume that {\bf a'} is reduced. Let $v=s_{a_1} \ldots s_{a_{p-1}}, $ 
$w=s_{a_1} \ldots s_{a_p}.$ We have $\ell(v)=p-1$ and $\ell(w) \le p-1$, 
hence by (\ref{s-1.3}) $\ell(w)=p-2$, so that 
$\ell(v)=\ell(ws_{a_p})=\ell(w)+1$. 
Consequently $\partial_v=\partial_w \partial_{a_p}$ and therefore 
$\partial_{\bf a}=\partial_v \partial{a_p}=\partial_w \partial_{a_p}^2=0.$
\qed

Then by theorem \ref{s-2.5} and \ref{s-2.6}, we have
\begin{thm}
\label{s-2.7}
Let $u, v$ be permutations. Then
$$\partial_u \partial_v=\left\{ \begin{array}{ll}
\partial_{uv} & if \quad \ell (uv)=\ell (u)+\ell (v), \\
0 & otherwise.
\end{array}
\right.$$
\end{thm}

In [{\bf L-S?}], one characterizes more general operators $D_i$ of the type

$$f \rightarrow f D_i = P(a_i, a_{i + 1})f + Q(a_i, a_{i + 1})f^{\sigma _i}, $$

(where $P$ and $Q$ are two rational functions of two variables) satisfying
Moore-Coxeter relations. It happens that they must also satisfy an Hecke
relation (with constants $q$ and $r$)

\begin{equation}
\label{e-2.6}
D_iD_i = qD_i + r\, .
\end{equation}

In other words, these operators $D_i$ furnish a representation of the Hecke
algebra of the symmetric group as an algebra of operators on the ring of 
polynomials.

The most useful of these operators are, apart from the two degenerate cases 
$\partial _i$ and $\pi _i$, 

$$D_i := c_1\partial _i + c_2\pi _i + c_3\sigma _i \, \, , $$

with constants $c_1, c_2, c_3$.

Let {\bf A} be of cardinal $n$, and $\omega$ be the maximal permutation of 
$S(A)$, i.e. $\omega$ sends $a_1, \ldots , a_n$ onto $a_n, \ldots, a_1$.
Then the operators $\partial _\omega$ and $\pi_\omega$ have the following global expression
(that we shall prove at the end of the section), apart from being (in different
ways) the products of the $\partial _i$, resp. $\pi _i$:

\begin{equation}
\label{e-2.7}
f\partial _\omega = \sum _{\mu \in S(A)}(f/ \Delta)^\mu \, \, , 
\end{equation}

\begin{equation}
\label{e-2.8}
f\pi _\omega = \sum _{\mu \in S(A)}(f \rho / \Delta)^\mu \, \, , 
\end{equation}
where $\Delta$ is the {\em Vandermonde} : $\Delta = \prod _{i < j}(a_i - a_j)$
and $\rho$ the monomial $a^n_1a^{n - 1}_2 \ldots a^0_{n + 1}$ (which corresponds 
to half the sum of positive roots, in the theory of classical groups).

Divided differences satisfy a Leibniz formula, as easily seen from the definition:

\begin{equation}
\label{e-2.9}
fg\partial _i = f(g\partial _i) + f\partial _ig^{\sigma _i} \, \, , 
\end{equation}
where $g^{\sigma_i}$ means exchange $a_i$ and $a_{i+1}$ in $g$.

In particular, symmetric functions in $a_i, a_{i + 1}$ are scalars for 
$\partial _i$ and $\pi _i$:

$$ g = g^{\sigma _i} \Rightarrow fg\partial = f\partial _ig\, \, and\, \, fg\pi _i = f\pi _ig\, \, .$$

Divided differences have a simple action on complete functions: let B be any
alphabet and $A_i$ denotes $\{ a_1, \ldots , a_i \}$ for any $i$ less than 
the cardinal of {\bf A}. Then

\begin{equation}
\label{e-2.10}
\frac{R(1, {\bf B})}{R(1, {\bf A}_i)}\partial _i
= \frac{R(1, {\bf B})}{R(1, {\bf A}_{i + 1})}\, , 
\end{equation}

\begin{equation}
\label{e-2.11}
\frac{R(1, {\bf B})}{R(1, {\bf A}_i)}\pi _i = \frac{R(1, {\bf B})}{R(1, {\bf A}_{i + 1})}\, , 
\end{equation}

because, writing $1/R(1, {\bf A}_i) = (1 - a_{i + 1})/R(1, {\bf A}_{i + 1})$, and 
using the fact that $R(1, {\bf A}_{i + 1})$ is a scalar for both $\partial _i$
and $\pi _i$, one is reduced to check that 

$$(1 - a_{i + 1})\partial _i = (1 - a_{i + 1})\pi _i =1\, .$$

More explicitely, (\ref{e-2.10}) and (\ref{e-2.11}) can be written

\begin{equation}
\label{e-2.12}
S_k({\bf A}_i - {\bf B})\partial _i = S_{k - 1}({\bf A}_{i + 1} - {\bf B}) \, , 
\end{equation}

and in the same way

\begin{equation}
\label{e-2.13}
S_k({\bf A}_i - {\bf B})\pi _i = S_k({\bf A}_{i + 1} - {\bf B})
\end{equation}

Here we give some computation:

\begin{eqnarray*}
&&(S_k(\bA_i-\bB)) \partial_i \\
& = & ([z^k] {\prod_{b \in \bB} (1-zb) \over \prod_{a \in A_i} (1-za) }) \partial_i \\
& = & ([z^k] {\prod_{b \in \bB} (1-zb) \over \prod_{a \in A_{i+1}} (1-za) }
(1-za_{i+1})) \partial_i \\
& = & ([z^k] {\prod_{b \in \bB} (1-zb) \over \prod_{a \in A_{i+1}} (1-za) }
-[z^{k-1}] {\prod_{b \in \bB} (1-zb) \over \prod_{a \in A_{i+1}} (1-za) } a_{i+1}) \partial\\
\end{eqnarray*}
Because ${\prod_{b \in \bB} (1-zb) \over \prod_{a \in A_{i+1}} (1-za) }$ is symmetric function in $a_i, a_{i+1}$, 
$$S_k(\bA_i-\bB) \partial_i=[z^{k-1}]{\prod_{b \in \bB} (1-zb) \over \prod_{a \in A_{i+1}} (1-za) }=S_{k-1}(\bA_{i+1}-\bB).$$

Divided differences have a simple action on (multi)-Schur functions, in the case 
that they operate only on the elements of one row or one column: take 
$S_{jk \ldots l}({\bf A}_i - {\bf B}_1, {\bf A}_p - {\bf B}_2, \ldots , {\bf A}_r - {\bf B}_q)$, 
and suppose that all the indices of the {\bf A}, apart from, say the first, 
column are different from $i$. Then all columns are scalars for $\partial _i$
and $\pi _i$, apart the first one, and from (\ref{e-2.12}) and (\ref{e-2.13})
one gets that the images of the proceeding function under $\partial _i$ and
$\pi _i$ are respectively

\begin{equation}
\label{e-2.14}
S_{j - 1 \, k \ldots l}({\bf A}_{i + 1} - {\bf B}_1, {\bf A}_p - {\bf B}_2, \ldots {\bf A}_r - {\bf B}_q)
\end{equation}

\begin{equation}
\label{e-2.15}
S_{jk \ldots l}({\bf A}_{i + 1} - {\bf B}_1, {\bf A}_p - {\bf B}_2, \ldots {\bf A}_r - {\bf B}_q)
\end{equation}

Let $\bA$ be of cardinal $n+1$ and let's denote

$$L'_{\bf A} = \partial _1\partial_2 \ldots \partial _n , $$
which is an $Sym(\bA)$ linear operator
and we shall prove that this is just
the Lagrange operator $L_A$ by using the forth proposition of $L_A$.

Here we will give some propositions of $L_{\bf A}$ first.

A natural result from the above identity is
\begin{equation}
\label{la1}
L'_{\bf A}\frac{1}{(1 - a_1)} = \frac{1}{\prod _{a \in {\bf A}}(1 - a)}
\end{equation}

A similar way we can get the following

\begin{eqnarray}
L'_{\bf A}\frac{1}{(1 -z a_1)} &=&
 \frac{z^n}{\prod _{a \in {\bf A}}(1 - za)} \\
\sum ^{\infty }_{k = 0}L'_{\bf A}(a^k_1)\cdot z^k &=&
 \sum S_k({\bf A})z^{n + k}
\end{eqnarray}

We also obtain $L'_{A}(a_1^k)=S_{k-n}(A)$ and more generally
\begin{equation}
\label{la2}
L'_{\bf A} S_k(a_1-{\bf B})=S_{k-n}({\bf A}-{\bf B})
\end{equation}

Proof of $L_A=L'_A$:

For the forth proposition of $L_A$, we know that for
every $f \in Sym(1 | n)$, we can say $f=\sum a_1^k \cdot T_k$, 
where $T_k \in Sym(A)$, for $L_A'$ is a linear operator, we can compute
$a_1^k \cdot T_k$ only. Hence
\begin{eqnarray*}
(a_1^k \cdot T_k)\partial_1 \partial_1 \ldots \partial_n
&=& (T_k \cdot (a_1^k \partial_1)) \partial_2 \ldots \partial_n \\
&=&\ldots \\
&=&T_k \cdot (a^k_1 \partial_1 \ldots \partial_n) \\
&=&T_k \cdot L'_A(a_1^k) = T_k \cdot S_{k-n}(A)
\end{eqnarray*}
and $L_A(a_1^k \cdot T_k) = T_k \cdot L_A(a_1^k) = S_{k-m}(A) T_k$.
Then we have the following conclusion: If $L'_A$ acts on $Sym(1 | n)$, then
it is the same operator as $L_A$.
Hence $L_A= \partial_1 \partial_2 \ldots \partial_n$.


Let us look now at the action of 

$$\partial _{\omega} = (\partial _n)(\partial _{n - 1}\partial _n)
\ldots (\partial _1 \ldots \partial _n).$$

Writing any monomial as a Schur function $S_I({\bf A}_n, 
\ldots , {\bf A}_2, {\bf A}_1)$, 
we see that at each step of applying the $\partial _i$, 
only one column is not
symmetrical in $a_i, a_{i + 1}$, and thus that the action increases alphabets 
and decreases degree; the final step gives a Schur function $S_J({\bf A})$.

Then we have the following theorem by $\ref{syf-15}$.
\begin{thm}
$\partial _w$ is a symtrizing operator on the polynomials of $\bA$, 
more explicitly, we have
\begin{equation}
\partial _w(a_{n+1}^{i_1}a_n^{i_2}\cdots a_1^{i_{n+1}})
=S_{i_1, i_2-1, \ldots , i_{n+1}-n}(\bA).
\end{equation}
\end{thm}

For example, $a^2_4a^5_3a^6_2a^9_1 = S_{2569}({\bf A}_4, {\bf A}_3, {\bf A}_2, {\bf A}_1)$, 
Our expression of $\partial _{\omega}$ is $\partial _{4321} = (\partial _3)
(\partial _2\partial _3)(\partial _1\partial _2\partial _3)$.

Under $\partial _3$ the monomial becomes $S_{2469}({\bf A}_4, {\bf A}_4, {\bf A}_2, {\bf A}_1)$.
then under $\partial _2\partial _3$, $S_{2449}({\bf A}_4, {\bf A}_4, {\bf A}_4, {\bf A}_1)$ 
and finaly, under $\partial _1\partial _2\partial _3$, 

$$S_{2446}({\bf A}_4, {\bf A}_4, {\bf A}_4, {\bf A}_4) = S_{2446}({\bf A}).$$

Now, the image of a monomial under the operator (\ref{e-2.7}) is an 
alternating sum of monomials, divided by the Vandermonde, which is Jacobi-Trudi expression of the same Schur function, and thus
$\partial _{\omega}$
coincide with (\ref{e-2.7}), because the images of any monomial under
both operators are the same. An identical proof works for $\pi _{\omega}$, 
this time the indices being preserved.

A simple proof of the validity of ($\ref{e-2.7}$) consists first in noticing that
both operators ($\partial_\omega$ and the summation) are $Sym(\bA)$-linear.
Therefore, we need only to check their action on a generating set, which from
($\ref{syf-16}$) can be taken as the monomials $\{ a^J : J \le \rho \}$. Now all these
monomials are sent to $0$ by both operators, except for $a^\rho$ which is sent
to $1$ by $f \longrightarrow \sum \pm f^\mu / \Delta(\bA)$, 
and to $S_{0\ldots0}=1$ by $\partial_\omega$

\qed

Multi-Schur functions can be obtained with the help of $\partial_\omega$.
For example, let $\{ \bB^1, \bB^2, \ldots, \bB^n \}$ be $n$ finite alphabets of
respective cardinals $i_1, i_2, \ldots, i_n$; we denote this sequence by $I$, 
and put $J=i_1-0, i_2-1, \ldots, i_n-(n-1)$. Then

\begin{equation}
\label{syo-30}
R(a_n, \bB^1) \cdots R(a_1, \bB^n) \partial = S_J(\bA_n-\bB^1, \ldots, \bA_n-\bB^n).
\end{equation}

{\em Proof.}
Write $R(a_n, \bB^1) \cdots R(a_1, \bB^n)$ as $S_I(\bA_n-\bB^1, \ldots, \bA_n-\bB^n)$, 
according to ($\ref{syf-7}$) : then determinant becomes triangular after subtraction of
$\{ 0, \bA_1, \ldots, \bA_{n-1} \}$, because in the upper part we are left with
elementary symmetric functions of degree strictly higher than the cardinals
of the alphabets involved. Now, the action of $(\partial_{n-1})
(\partial_{n-2} \partial_{n-1}) \cdots (\partial_1 \cdots \partial_n)$
consists, thanks to ($\ref{e-2.14}$), in transforming all the $\bA_i$ into $\bA_n$
while decreasing the indices. \qed

The Schur function $S_I(\bA-\bB)$ can be obtained by a double sommation on
$S(\bA)$ and $S(\bB)$ as stated by Sergeev-Pragacz formula.

{\bf Remark:}

For every permutation $\mu \in S(\bA)$, there exist a divided difference
$\partial_\mu$ and an isobaric divided defference $\pi_\mu$. The maximum divided
difference $\partial_\omega$ is the operator
$$f \longrightarrow \sum \pm f^\mu / \Delta(\bA).$$

\pagebreak

\section{The ring Sym($1|n$)}

We already used the fact that the ring
$$ \mbox{Sym}(1|n)=\mbox{Sym}(a_1) \otimes_\bZ \mbox{Sym}(a_2, \cdots, a_{n+1})$$
with coefficients in ${\bf Z}$ is a ${\bf Z}$-free module with basis
$\{ S_I (\bA-a_1) a_1^k, I \mbox{ partition } \in \bN^n, k \in \bN \}$. 
One of the essential properties of Lagrange opetator is its compatibility 
with Schur functions as shown by the following lemma:
\begin{lem}
\label{l-4.1}
Let $\bA$ be of cardinal $n+1$ and $L_\bA$ the correspongding Lagrange
operator. Then
\begin{itemize}
\item[1)] $L_\bA$ is equal to the product $\partial_1 \cdots \partial_n
(\mbox{ acting on its left })$.
\item[2)] $\forall I \in \bA^n, I \ge -(01 \cdots n-1), \forall k\ge 0, $
$$L_{\bA}(S_I(\bA-a_1)a_1^k)=S_{I\ k-n}(\bA).$$
\end{itemize}
\end{lem}
{\em Proof.} 
Here we shall give another proof for $L_\bA=\partial_1 \cdots \partial_n$.

First of all, we can write the product $S_I(\bA-a_1)a_1^k$ as the Schur function
$S_{I, k}(\bA, a_1)$ (for ($\ref{syf-10}$)). Now let's expand the determinant 
$S_{I, k}(\bA, a_1)$, the element of it should be the following form:
$$ S_h(a_1) \cdot f(\bA), \mbox{ where } f(\bA) \in \mbox{Sym}(\bA).$$
Then we can get the following from ($\ref{e-2.12}$)
$$(S_h(a_1)f(\bA)) \partial_1 \partial _2 \cdots \partial_n
= S_{h-n}(\bA) \cdot f(\bA), $$
and from ($\ref{li-9}$), we get
$$L_\bA(S_h(a_1) \cdot f(\bA))=S_{h-n}(\bA) \cdot f(\bA)$$
Thus $S_{I, k}(\bA, a_1)$ has the same images under the two
$$L_\bA(S_{I, k}(\bA, a_1))=S_{I, k}(\bA, a_1) \partial_1 \cdots \partial_n 
= S_{I, k-n}(\bA).$$
Because $\{ S_I(\bA-a_1)a_1^k \}$ is a basis of Sym$(1 | n)$ and 
$L_\bA, \partial_1 \cdots \partial_n$ have the same images acting on the 
basis, $L_\bA$ and $\partial_1 \cdots \partial_n$ are coincide for Sym$(1 | n)$.
\qed

The interest of taking $I \ge -(0\, 1\, \cdots \, n-1)$ instead 
of $I \in \bN^n$ will appear later.

Now we will give a more strong form of $L_{\bA}$'s second proposition. 

It was already clear in the preceeding proof that an essential property of 
Sym$(1 | n)$ is to be a Sym$(\bA)$-module. Indeed, the $\bA$-basis 
$S_I(\bA-a_1)a_1^k=S_{I, k}(\bA, a_1)$ can be writen in terms of terms of 
powers of $a_1$ only (with coefficients in Sym$(\bA)$); moreover, because the 
elementary symmetric functions of the alphabet $\{\bA \setminus a_1 \}$
vanish in degree $>n$, one has
$$a_1^k=S_k(a_1)=S_k(\bA+(a_1-\bA))=\sum_0^n S_{k-i}(\bA)S_i(a_1-\bA).$$
Thus, $\{a_1^0, \ldots, a_1^n \}$ is a generating system for our ring as a 
module over Sym($\bA$). To prove that it is a basis, i.e. that every element 
of Sym$(1 | n)$ can be uniquely written
$$f=c_0a_1^0+\cdots+c_n a_1^n, $$
with coefficients $c_i$ in Sym$(\bA)$, (in other words, the module is free), 
we shall first define on it a quadratic form with values in Sym$(\bA)$.
\begin{dfn}
\label{Pr2}
$\forall P, Q \in$ Sym$(1 | n)$, set $(P, Q):=L_{\bA}(PQ)$. 
\end{dfn}
\begin{pro}We have
\begin{itemize}
\item[1] Sym$(1 | n)$ is a Sym$(\bA)$-module of basis $\{1, a_1, \ldots, a_1^n \}$, 
as well as basis $\{ S_n(a_1-\bA), \ldots, S_0(a_1-\bA) \}$.
\item[2] The quadratic foem is onn-degenerate, i.e. there does not exist 
$P \not= 0$ such that for all $Q$ in Sym$(\bA)\ \ (P, Q)=0$. The two above bases 
are adjoint, i.e., for $0 \le h, k \le n$, $(a_1^k, S_{n-h}(a_1-\bA))=1$ or
$0$ according that $k=n$ or not.
\item[3] Every element $f$ of Sym$(1 | n)$ can be written
$$f=\sum_0^n L_\bA (fS_k(a_1-\bA)) a_1^{n-k}, $$
$$f=\sum_0^n L_\bA (fa_1^{n-k}) S_k(a_1-\bA)).$$
\end{itemize}
\end{pro}
{\em Proof.}
We have to write
$$S_h(a_1-\bA)a_1^k=(-1)^hS_{1^h}(\bA-a_1)a_1^k=(-1)^hS_{1^h, k}(\bA, a_1).$$
Now the image under $L_\bA=\partial_1 \cdots \partial_n$ is 
$(-1)^h S_{1^h, k-n} (\bA, \bA)$.
(The first equation can be obtained from duality theorem, which has the 
general form: $S_{J/I}(\bA-\bB)=(-1)^{|J/I|} S_{J~/I~}(\bB-\bA)$.)

The indices in the first row of this determinant are $1, 2, \ldots, h, k-n+h$. 
This determinant is determined by the first row. If there is some 
$i, 1 \le i \le h$, s.t. $i=k-n+h$, then the $i$th column is the same as the
$(h+1)$st column. This means that the determinant is $0$. If $k-n+h<0$, then 
the indices in the last column are all negative. The value of the determinant
is $0$ again. So we have to consider the special case: they are all different and oon negative. 
They are all different and non negative only when $k-n+h=0$, i.e. only when
the determinant is obtained from the determinant expressing $S_{0^{h+1}}(\bA)=1$
by permuting columns.

If $f \in$ Sym$(1 | n)$ can be written in two forms
$$f=c_0a_1^0+\cdots+c_na_1^n=d_0a_1^0+\cdots+d_na_1^n, \ \ c_i, d_i \in 
\mbox{Sym}(\bA)$$
We can prove $c_i=d_i$ for $0 \le i \le n$. Let's consider $(S_i(a_1-\bA), f)$, 
then 
$$c_i=L_\bA(S_i(a_1-\bA) \cdot f)=d_i$$
that means $c_i=d_i$ for $0 \le i \le n$, then $\{ a_1^0, \ldots, a_1^n \}$
is a basis. In the same way, we can prove 
$\{ S_n(a_1-\bA), \ldots, S_0(a_1-\bA) \}$ is a basis. The two bases are adjoint.
This pair of adjoint bases allows now to expand in two different ways any
element of the module by taking scalar products. \qed




\begin{thm}
\label{Jacobi-Trudi}
We have the following expression of the same Schur function.
\begin{equation}
S_I(A)=S_{I + 0, 1, \ldots n}(a_1, a_2, \ldots , a_{n + 1})/\Delta (A)
=\frac{\left| \begin{array}{cccc}
a_1^{i_1} & a_2^{i_1} &\cdots & a_{n+1}^{i_1} \\
a_1^{i_2+1} & a_2^{i_2+1} &\cdots & a_{n+1}^{i_2+1} \\
\cdots & \cdots & \cdots & \cdots \\
a_1^{i_{n+1}+n} & a_2^{i_{n+1}+n} &\cdots & a_{n+1}^{i_{n+1}+n}
\end{array} \right|}{\left| \begin{array}{cccc}
1 & 1 &\cdots &1 \\
a_1 & a_2 &\cdots & a_{n+1} \\
\cdots & \cdots & \cdots & \cdots \\
a_1^{n} & a_2^{n} &\cdots & a_{n+1}^{n}
\end{array} \right|}
\end{equation}
Where $\Delta(A)$ is the Vandermonde determinant.
\end{thm}

The Proof of this theorem is only a check of
$$S_I(A)\cdot S_{0^{n+1}}(-A)\cdot \Delta (A)
=S_{I+ 0, 1, \ldots n}(a_1, a_2, \ldots , a_{n + 1})$$
by means of matrix multiplication.



\begin{thm}
Suppose $A$ and $B$ are of the same cardinal $n$, then we have
\begin{eqnarray*}
&&\left| \frac{1}{1 - ab} \right| _{a \in A, b \in B} =
\frac{\Delta (A)\Delta (B)}{\prod _{a \in A, b \in B}(1 - ab)}
 \\
&\Leftrightarrow & \sum S_I(A)S_I(B) = \prod _{a \in A, b \in B}
\frac{1}{1 - ab}
\end{eqnarray*}
\end{thm}

The left side of $\Leftrightarrow$ is easy to compute, but the right side
is not easy, although it have been given a combinatorial proof.

{\em Proof.}
The proof is based on two methods of computing
$$\left|\begin{array}{cccc}
 1 & a_1 & a_1^2 & \ldots \\
 1 & a_2 & a_2^2 & \ldots \\
 \vdots & \vdots & \vdots & \vdots \\
 1 & a_n & a_n^2 & \ldots 
\end{array}
\right| \cdot 
\left|
\begin{array}{cccc}
1 & 1 &\ldots & 1\\
b_1 & b_2 &\ldots &b_n\\
b_1^2 & b_2^2 &\ldots &b_n^2\\
\vdots & \vdots & \vdots & \vdots
\end{array}
\right|. $$
The natural matrix multiplication shows that it is
$ \left|\frac{1}{1 - ab}\right| _{a \in A b\in B}$, 
but by the Cauchy's theorem, combined with the theorem \ref{Jacobi-Trudi}, 
it is $ \sum_I \Delta (A)S_I(A)\cdot \Delta (B)S_I(B)$, which is just
$ \Delta A\cdot \Delta B\cdot \sum_I S_I(A)S_I(B)$, the theorem then
follows.
\qed


The ring Sym$(1 | n)$ can be identified with the equivariant cohomology ring of 
a relative projective space of dimension $n$, and the operator $L_\bA$ to the 
{\em Gusin morphism} from Sym$(1 | n)$ to the cohomology of the base 
(identified with Sym$(\bA)$). The variable $a_1$ is the {\em first Chern class}
of the {\em tautological line bundle}, $(1+a_2) \cdots (1+a_{n+1})$ is the 
total Chern class of the tautological kernel (see [Ful])

{\bf Remark:}

The ring Sym$(1 | n)$ is a free module over Sym$(\bA)$, with basis 
$\{1, a_1, \ldots a_1^n \}$, and an explicit adjoint basis with respect to the scalar product 
$$(f, g):=L_\bA(fg).$$

\pagebreak

\section{The ring $Sym(r | m)$}

Lagrange operator gives a symmetrization of $Sym(1\mid n)$, 
it can be naturally
generalized to the ring of $Sym(r\mid m)$, where $r$, $m$ are two positive
integers. Let ${\bf A}$ be an alphabet of cardinal
$n = r+m$. Then the operator 
$$Sym(r)\times Sym(m)\ni f \rightarrow \sum _{\bA', 
card(\bA') = r}f(\bA', \bA\setminus \bA')/R(\bA', \bA\setminus \bA')$$

may be called the {\sl Sylvester operator of order $r$}, 
which is denoted by $L_{r|m}$. Indeed, Sylvester expressed
the discriminate of a polynomial, or the greatest common divisor of two 
polynomials using this operator.

Clearly, $L_{r\mid m}$ is $Sym(n)$ linear.

Let the alphabet be totally ordered, 
$L_{r\mid m}$ can also be expressed through divided differences.
Denote $\partial_{r | m}$ the product

$$\partial_{r\mid m} := (\partial _r \ldots \partial _{n-1})(\partial _{r - 1}
\ldots \partial _{n - 2}) \ldots (\partial _1 \ldots \partial_{m})$$

It is the divided difference indexed by the permutation
$$\mu = m + 1 \ldots n 1 \ldots m.$$
Moreover, Let $\bA_r$ denote the initial sub-alphabet
$\bA_r :=\{ a_1, \ldots , a_r \}$ and $Sym_{r\mid m}(\bA)$ the ring of
functions in $\bA$, 
Symmetrical in $\bA_r$ and symmetrical separately in $\bA\setminus \bA_r$.
Then we have the following theorem.

%%%%%%%%%%
\begin{thm}
\label{sylv}We have 
\begin{itemize}
\item[1)] $\{ S_J(\bA_r) \}, 0\leq J \leq m^r$ is a senerating set for $Sym(r\mid m)(\bA)$ as
as a $Sym(\bA)$-module, \\
\item[2)] $\partial_{r\mid m}(S_I(\bA-\bA_r)\cdot S_J(\bA_r)) = S_{(I, J)-(0^m, m^r)}(\bA)$
\end{itemize}
\end{thm}

{\em Proof.} 
1) Since $\{ S_I(\bA - \bA_r)S_J(\bA_r)\}$ is a basis of
$Sym(\bA_r\mid \bA - \bA_r)$
It is sufficient to express each $S_I(\bA - \bA_r)S_J(\bA_r)$
in terms of $\{ S_J(\bA_r) \} _{J \leq m^r}$.
We can write $S_I(\bA-\bA_r)S_J(\bA_r)$ as $S_{I, J}(\bA, \bA_r)$ by ($\ref{syf-11}$). The expansion of
this determinant is a linear combination of $S_k(\bA_r)$, $K$ is a partion: $K \le m^r$.

Therefore, when taking coefficients in $Sym(\bA)$, 
one can express any element of
$Sym(r\mid m)$ as a combination of $S_J(\bA_r)$, with $J$ partition contained 
in $m^r$.

2) One writes $S_I(\bA - \bA_r)S_J(\bA_r) = S_{I, J}(\bA, \bA_r)$. However, the action of
the divided difference $\partial_r$ is not easy, because the $r$
last columns of the determinant
are changed by it, and one would have to use Leibniz formula. Fortunately, 
one can overcome this by taking other products of divided differences.

Clearly, we have
\begin{eqnarray*}
&&S_{I, J}(\bA, \bA_r) \\
&=&S_{I, J+(0, 1, \ldots, r-1)}(\bA, \bA_r, \bA_{r-1}, \ldots, \bA_1)
\partial_{r-1}\partial_{r-2}\partial_{r-1}\cdots
\partial_{1}\cdots \partial_{r-1} \\
&=&S_{I, J+(0, 1, \ldots, r-1)}(\bA, \bA_r, \ldots, \bA_1)\partial_w , \\
\end{eqnarray*}
where $w={r, r-1, \ldots 1, r+1, \ldots , n}\, $,  and then 
$$S_{I, J}(\bA, \bA_r)\partial_{r\mid m}=S_{I, J}(\bA, \bA_r)\partial_\mu
=S_{I, J+0, 1, \ldots, r-1}(\bA, \bA_r, \ldots , \bA_1) \partial_w \partial_\mu , $$
where $\mu = m + 1, \ldots , n, 1, \ldots , m$ as above.
Clearly, $w \cdot \mu =n, n-1, \ldots, m+1, 1, \ldots, m$, and we have 
$$l(w)=r(r-1)/2, \; l(\mu)=rm, \; l(w\cdot \mu)=r+(r+1)+\cdots +n =l(w)+l(\mu)
.$$
On the other hand, we can find 
$$(r, \ldots , n-1, r-1, \ldots , n-1, \ldots , 1, \ldots, n-1)\in R(w\cdot \mu), $$
which exact mean to move $m+1$ to the right most step by step and then
$m+2, \ldots , n$. So by theorem \ref{s-2.7} and \ref{s-2.6}, we get
$$S_{I, J}(\bA, \bA_r)\partial_{r\mid m}=S_{I, J+0, 1, \ldots, r-1}(\bA, \bA_r, \ldots , \bA_1) 
\partial_r \cdots \partial_{n-1} \cdots \partial_1 \cdots \partial_{n-1} , $$
this time it is easy to compute, for each step only one column come into use.

Finally, we obtain the result
$$S_{I, J}(\bA, \bA_r)\partial_{r\mid m}=S_{I, J+0, 1, \ldots, r-1 -m, m+1, \ldots n-1}(\bA)
=S_{(I, J)-(0^m, m^r)}(\bA), $$
and the theorem followed.
\qed

For example, 
Let us take numerical values to illustrate how the proof of (2) works.
Let $r = 3$, 
$m = 2$, $I, J = 12467$. Instead of starting with $S_{12, 467}(\bA, \bA_3)$, 
we take
$S_{12479}(\bA, \bA, \bA_2, \bA_2, \bA_1)$ which gives under
$\partial _2\partial _1\partial _2$
the preceeding function.

Now we can compute the image of $S_{12479}(\bA, \bA, \bA_3, \bA_2, \bA_1)$ under \\
$(\partial _2\partial _1\partial _2)(\partial _3\partial _4\partial _2
\partial _3\partial _1\partial _2)$. But, thanks to the braid relations, this
operator can be written $(\partial _3\partial _4)(\partial _2\partial _3
\partial _4)(\partial _1\partial _2\partial _3\partial _4)$.

Of course, we could have proved the proceeding proposition by
mere expansions
of determinants, but using divided differences avoids the subtilities of 
determinantal calculus which have been lost since the time of Sylvester.

\begin{thm}
\begin{itemize}We have 
\item[1.] Sylvester operator $L_{r\mid m}$ is equal to $\partial_{r\mid m}$.
\item[2.] Sylvester operator is $Sym(\bA)$-linear and is the only $Sym(\bA)$-linear operator sending
the $S_J(\bA_r)$, $J$ partition: $J\subset m^r$ to zero and $S_{m^r}(\bA_r)$
to $1$.
\end{itemize}
\end{thm}

{\em Proof.}
Since $\{ S_J(\bA_r) \}, \ 0 \le J \le m^r$ is a generating set for $Sym(r|m)$ as
a $Sym(\bA)$-module and the two operators are both $Sym(\bA)$ linear, we should
compare the images of $\{S_j(a) \}$ under $L_{r | m}$ and $\partial_{r | m}$
for our proof. Now each operator $\partial_i$ decrease degree by $1$, then
$\partial_{r | m}$ decrease degree by $m \cdot r$. We can get the same conclusion
for the Sylvester poerator $L_{r|m}$ according to the definition. (for the degree of
$R(\bA', \bA \setminus \bA')$ is $m \cdot r$ ). Thus both of them send all $S_J(\bA_r), \ J \subset
m^r$, to zero and send $S_{m^r}(\bA_r)$ to constants.
When we act on $S_{m^r}(\bA_r)$ by $\partial_{r | m}$, we can apply the second part of
theorem (\ref{sylv}), let $I=0^m$ and $J=m^r$ then we get
$S_{m^r}(\bA_r) \partial_{r | m} = S_{0\cdots0}=1$. As for Sylvester summation, 
we should observe $R(\bA', \bA\setminus \bA_r)$ first. We know that
$$R(\bA', \bA\setminus \bA')=S_{m^r}(\bA'-(\bA\setminus \bA_r))$$
for the theorem (\ref{syf-12}) and
$$S_{m^r}(\bA'-(\bA \setminus \bA_r)) = \sum_{H \le m^r} S_H(\bA')
\cdot S_{m^r/H}(-(\bA\setminus \bA_r))$$
Now $\forall H < m^r$, we have
$$\sum_{\bA'} {S_H(\bA') \over R(\bA', \bA \setminus \bA')} \cdot S_{m^r/H}(-(\bA \setminus \bA'))
=S_{m^r/H}(-(\bA \setminus \bA_r)) \cdot L_{r | m}(S_H(\bA_r))=0, $$
$$\sum_{\bA'} {S_H(\bA') \over R(\bA', \bA \setminus \bA')} \cdot S_{m^r/m^r}(-(\bA \setminus \bA'))
=1 \cdot L_{r | m}(S_m^r(\bA_r)).$$
Then
\begin{eqnarray*}
&&\sum_{\bA'} {R(\bA', \bA \setminus \bA_r) \over R(\bA', \bA \setminus \bA')} \\
&=& \sum_{\bA'}{ \sum_{H \le m^r} S_H(\bA') \cdot S_{m^r/H} (-(\bA \setminus \bA_r))
    \over R(\bA', \bA \setminus \bA')} \\
&=& \sum_{H < m^r} (\sum_{\bA'} { S_H(\bA') \over R(\bA', \bA \setminus \bA')})
    \cdot S_{m^r/H(-\bA \setminus \bA_r)} +
  \sum_{\bA'} { S_{m^r}(\bA') \over R(\bA', \bA \setminus \bA') } \\
&=& \sum_{\bA'} { S_{m^r}(\bA') \over R(\bA', \bA \setminus \bA') }
\end{eqnarray*}
and clearly in $\sum_{\bA'} {R(\bA', \bA \setminus \bA_r) \over R(\bA', \bA \setminus \bA') }$, 
there is only the term $\bA'=\bA_r$ which is nonzero and equal to $1$, then
$$1=\sum_{\bA'} {S_{m^r}(\bA') \over R(\bA', \bA \setminus \bA')} = L_{r | m}(S_{m^r}(\bA_r))$$

The two operators, coinciding on a generating set, are equal, and thus the
theorem follows.
\qed


Having seen that we have to use the structure of $Sym(r\mid m)$ as a free
Sym(\bA)-module, we define on it a quadratic forms with the help of Sylvester
operator:

\begin{equation}
\forall f, g \in Sym(r\mid m), \ \ <f, g> := L_{r\mid m}(fg)
= \partial _{r\mid m}
(fg) \in Sym(\bA).
\end{equation}

\begin{thm}
\label{quadratic}
1) $Sym(r\mid m)$ is a free Sym(\bA)-module of basis $\{ S_I(\bA_r)\}$ and of basis
$\{ S_I(\bA_r - \bA)\}$, for all partitions $I: 0\leq I\leq m^r$.\\
2) The quadratic form is non degenerate, and the two bases $\{ S_I(\bA_r)\}$ 
and $\{S_I(\bA_r - \bA)\}$ are adjoint with respect to it. More precisely, one
has $<S_I(\bA_r - \bA), S_J(\bA_r)> = 1$ or $0$ according that
$I = (m - J_r, \ldots , m - J_1)$ or not.

3) Every element $f$ of $Sym(r\mid m)$ can be written as
$$f = \sum_{I \leq m^r}<f, S_{m^r-I}(\bA^r-\bA))S_I(\bA_r).$$
\end{thm}

{\em Proof.}
The proof is the same as in the case of $Sym(1 | m)$.
We have to compute all the $\partial _{r\mid m}(S_K(\bA - \bA_r)S_J(\bA_r))$.
Now, the theorem{sylv} gives us an image a determinant of complete 
functions in {\bf \bA}, with respective indices in the first row

$$k_1, k_2 + 1, \ldots , k_m + m-1, j_1, j_2 +1, \ldots , j_r + r - 1$$

Using the hypothesis that $0 \leq K \leq r^m$ and $0\leq J \leq M^r$, we 
see that these $n$ integers belongs to the set $\{ 0, 1, \ldots, n - 1 \}$.
Therefore, for every partition $J$, there exists one and only one partition
$K$ such that the determinant has all its columns distinct, and no column with
all indices negative. In this case, the determinant is obtained from
$S_{000\ldots } = 1$ by a permutation of columns. The conjugate of the 
partition $K$ gives the partition $I$ specified by the theorem \ref{xx}.

Thus the matrix $|\partial_{r\mid m}(S_I(\bA_r - \bA)S_J(\bA_r))|$ is the 
identity matrix, which proves that the quadratic form is non degenerate
and that the two adjoint bases, we can now expand any element of 
$Sym(r\mid m)$ by computing scalar products.

\qed

We have seen in the proceeding section that computing in the ring
Sym$(1|m)$ was related to finding the first remainder in the 
Euclidean division. Similarly, we can use Sym$(r|m)$ to express
the remainder of degree $r <n - 1$. This gives some determinantal
expressions of the remainder, as well as expressions using Sylvester
operator as a summation on subsets of given cardinal.


\smallskip
The ring $Sym(r|m)$ can be identified with the (equivariant) cohomology
ring of a relative Grassmann variety and $Sym({\bf A})$ to the cohomology
ring of the base. The variables 
$a_1, \ldots, a_r$ are the {\it Chern roots} of the 
{\it tautological quotient bundle} . The operator 
$\partial_{r|m}$ is the {\it Gysin morphism}. The Schur functions 
$S_I({\bf A}_r)$ are representatives of the Schubert subvarieties of 
the Grassamanniann. When computing modulo the ideal generated by
symmetric polynomials without constant term, the quadratic
form $\partial_{r|m}(f, g)$ becomes the {\it intersection form}. It was 
already well known at the end of the 19th century that Schubert
varieties in complementary degree intersect in 0 point or 1 point, 
and that the matrix representing the intersection form is the 
identity matrix. 


\vskip 5mm
{\bf Remark:}

The ring $Sym(r|m)$ is a free module over $Sym({\bf A})$, 
with basis $S_I(a_1, \ldots , a_r)$, 
and adjoint basis $S_{\tilde{I}} (a_{r+1}, \ldots , a_{r+m} )$, 
where the partition $I$ runs over all
partitions $I\leq r^m$.

\pagebreak

\section{ Large Rational interpolation }

\smallskip
The resultant $R(\bA, \bB)$ plays a fundamental role in the summations generalizing 
Lagrange interpolation, as well as in invariant theory or elimination theory. It can be itself expressed 
through summations implying other resultants, and we shall only present the following ``exchange 
lemma''.

\smallskip
\begin{lem}
\label{lri-1}
Let $\bA, \bB, \bC$ be three alphabets of respective cardinals
$m+n$, $n$, $m$. Then one has the following identity 
$$\sum_{\bA'+\bA''=\bA } { R(\bA', \bB) \over R(\bA', \bC) \,  
R(\bA', \bA'') }= { R(\bB, \bC) \over R(\bA, \bC) } $$
sum on all disjoint decomposition of $\bA$ in $\bA'+\bA''$, 
$card(\bA')=m$, $card(\bA'')=n$, as well as the identity 
$$ \sum_{\bA'+\bA''=\bA}
{ R(\bA', \bB) \, R(\bA'', \bC) \over R(\bA', \bA'') }= R(\bB, \bC).$$
\end{lem}

{\em Proof:} 

The two identities are equivalent up to multiplication by $R(\bA, \bC)= R(\bA', \bC) 
R(\bA'', \bC)$. 

The product of the two resultants can be written as the Schur function:
$$R(\bA', \bB)\, R(\bA'', \bC)=
S_{m^n, n^m }(\bA-\bC, \bA'-\bB) \ . $$

In fact, the expansion of $S_{m^n, n^m} (\bA-\bC, \bA'-\bB)$ is the following:
$$
\quad \begin{array}{r}
 n \left\{ \begin{array}{c}
       \\
       \\
       \\
     \end{array}
 \right. \\
\\
\\
\\
\end{array}
 \hspace{-0.6cm} \left| \begin{array}{cccccc}
        S_m(\bA-\bC) & \ldots & S_{n+m-1} (\bA-\bC) & S_{2n} (\bA'-\bB)
            & \ldots & S_{2n+m-1} (\bA'-\bB) \\
        & \ddots & \vdots & \vdots & \vdots & \vdots \\
        & & S_{m} (\bA-\bC) & S_{n+1} (\bA'-\bB) & \ldots & S_{n+m} (\bA'-\bB) \\
        & & & S_{n} (\bA'-\bB) & \ldots & S_{n+m-1} (\bA'-\bB) \\
        & & & & \ddots & \vdots \\
        & & & & & S_{n} (\bA'-\bB)
    \end{array}
 \right| \hspace{-0.6cm}
\begin{array}{c}
   \\
   \\
   \\
\left.  \begin{array}{c}
    \\
    \\
    \\
    \end{array}
\right\} m
\end{array}
$$

Because of $|\bA'|=m$, we can substract $\bA'$ in evert entry of the
first $n$ rows. The value of the determinant remains and it has the following form
$$
\left| \begin{array}{cccccc}
        S_m(\bA''-\bC) & \ldots & S_{n+m-1} (\bA''-\bC) & S_{2n} (-\bB)
            & \ldots & S_{2n+m-1} (-\bB) \\
        & \ddots & \vdots & \vdots & \vdots & \vdots \\
        & & S_{m} (\bA''-\bC) & S_{n+1} (-\bB) & \ldots & S_{n+m} (-\bB) \\
        & & & S_{n} (\bA'-\bB) & \ldots & S_{n+m-1} (\bA'-\bB) \\
        & & & & \ddots & \vdots \\
        & & & & & S_{n} (\bA'-\bB)
    \end{array}
\right|
$$
Because of $|\bB|=n$, $S_i(-\bB)=0$ for $i>n$. Hence
$$S_{m^n, n^m}(\bA-\bC, \bA'-\bB)$$
\begin{eqnarray*}
& = &
\left|
\begin{array}{cc}
    \begin{array}{ccc}
        S_m(\bA''-\bC) & \ldots & S_{n+m-1} (\bA''-\bC) \\
        & \ddots & \vdots \\
        & & S_{m} (\bA''-\bC) \\
    \end{array} & \mbox{\Large $0$} \\
   & \begin{array}{ccc}
        S_{n} (\bA'-\bB) & \ldots & S_{n+m-1} (\bA'-\bB) \\
        & \ddots & \vdots \\
        & & S_{n} (\bA'-\bB)
    \end{array}
\end{array}
\right| \\
& = & S_{m^n}(\bA''-\bC) \cdot S_{n^m}(\bA'-B) = R(\bA'', \bC) \cdot R(\bA', \bB).
\end{eqnarray*}

Using the factorization of Sylvester operator 
 $f\longrightarrow \sum f({\bf A }', {\bf A }'')/ R({\bf A }', {\bf A }'')$
as a product of divided differences, one obtains 
$$\sum R({\bf A }', {\bf B }) \, R({\bf A }'', {\bf C })/ R({\bf A }', {\bf A }'') =
          S_{m^n, 0^m}({\bf A }-{\bf C }, {\bf A }-{\bf B })  $$
Writing ${\bf A }-{\bf C } = ({\bf A }-{\bf B }) + ({\bf B }-{\bf C })$, one develops
\begin{eqnarray*}
&& S_{m^n\, 0^m} (\bA-\bC, \bA-\bB) \\
&=& \sum_H (-1)^H S_{m^n/\tilde{H}}(\bA-\bC) \cdot S_{H^\omega}(\bA-\bB)
    \hspace{5cm} (\mbox{ by ($\ref{syf-13}$) }) \\
&=& \sum_H (-1)^H (\sum_I S_{I/\tilde{H}}(\bA-\bB) \cdot S_{m^n/I}(\bB-\bC))
    \cdot S_{H^\omega}(\bA-\bB) \hspace{1.8cm} (\mbox{ by ($\ref{syf-3}$) }) \\ 
&=& \sum_I (\sum_H (-1)^H S_{I/\tilde{H}}(\bA-\bB) \cdot S_{H^\omega}(\bA-\bB))
    \cdot S_{m^n/I}(\bB-\bC) \\
&=& \sum_I S_{I\, 0^m}(\bA-\bB) \cdot S_{m^n/I}(\bB-\bC) \hspace{6.2cm} (\mbox{ by ($\ref{syf-13}$) }).
\end{eqnarray*}

All the $S_{I\, 0^m}$, $I\subseteq m^n$, are null
(having at least two identical columns), except 
for $I=0^n$; therefore 
$$\sum R({\bf A }', {\bf B }) \, R({\bf A }'', {\bf C })/ R({\bf A }', {\bf A }'') =
        S_{m^n}({\bf B }-{\bf C }) = R({\bf B }, {\bf C })$$
\qed 

A consequence of this exchange property is the determination of
the numerator and denominator of a rational function 
knowing sufficiently many of its values 
(Cauchy ?, Rosenhain (1846)):

\begin{thm}
\label{lri-2}
Let $P$ and $Q$ be two unitary polynomials of respective degrees
$n, m$, and $f= P/Q$. Let ${\bf A }$ be an alphabet
of cardinal $m+n+1$. Then, 
if ${\bf A }$ is generic enough, 
$$ f(x) = { \sum_{{\bf A }', card=n} R(x, {\bf A }') \prod f(a') \, /\, R({\bf A }', {\bf 
A }\backslash {\bf A }')
\over 
\sum_{{\bf A }', card=m}R(x, {\bf A }') \prod f(a') \, /\, R({\bf A }', {\bf A }\backslash {\bf A }')
} $$
\end{thm}

{\em Proof.}

Write $P(x)= R(x, {\bf B })$, $card({\bf B })=n$, 
$Q(x)= R(x, {\bf C })$, $card({\bf C })=m$. 

The rational function $f$ is therefore equal to 
$$f= R(x, {\bf B })/ R(x, {\bf C }) = (-1)^{mn} R(x+{\bf C }, {\bf B }) /R(x+{\bf B }, {\bf 
C }) $$
Using the proceeding lemma for the pair of alphabets $(x+{\bf B }, {\bf C })$, 
one has

\begin{eqnarray*}
{R(x+{\bf B }, {\bf C })\over R({\bf A }, {\bf C })} &=&
 \sum_{card({\bf A }')= m, {\bf A }''= {\bf A }-{\bf A }'}
 {R({\bf A }', x+{\bf B }) \over R({\bf A }', {\bf C }) R({\bf A }', 
{\bf A }'') } \\
& = & \sum_{{\bf A }'} {R({\bf A }', x) \over R({\bf A }', {\bf A }'') }
 {R({\bf A }', {\bf B })\over R({\bf A }', {\bf C })} 
= \sum_{{\bf A }'} \prod_{a'} f(a') {R({\bf A }', x)
 \over R({\bf A }', {\bf A }'') } 
\end{eqnarray*}
and similarly for the numerator.

Since we have introduced the extra factor
 $R({\bf B }, {\bf C })$, we must have a generosity
condition which ensures that it is non zero.
\qed

One can also give a determinantal form to the proceeding computation
 ( with $z=1/x$) :

\begin{pro}
\label{lri-3}
Let $f(z)= P(z)/Q(z)$ be a rational function, 
$d^o(P)\leq n$, $d^o(Q)\leq m$, and ${\bf A }$ be an alphabet of cardinal
 $m+n+1$.  
Then one has the identity
%\setbox2=\hbox{ ${\bf A }\quad $\left\{ \vbox to 1cm{} \right. }

$$  % \raise 3mm\box2 \quad 
\left| 
\begin{array}{cccccccc}
 \vdots &\vdots &    &\vdots &\vdots &   &\vdots&\vdots \\
1 &a &\cdots &a^n &a^m f(a) &\cdots & af(a) &f(a) \\
\vdots &\vdots &    &\vdots &\vdots &   &\vdots&\vdots \\
Q(z) & zQ(z) &\cdots &z^nQ(z) &z^mP(z) &\cdots &zP(z) &P(z)
\end{array}
\right|= 0$$
\end{pro}

{\em Proof:}
Multiply the row by the corresponding factor $Q(a)$, and we get a new determinant
\begin{equation}
\label{lri-4}
\left| \begin{array}{cccccccc}
    \vdots & \vdots & & \vdots & \vdots & & \vdots & \vdots \\
    Q(a) & aQ(a) & \ldots & a^nQ(a) & a^mP(a) & \ldots & aP(a) & P(a) \\
    \vdots & \vdots & & \vdots & \vdots & & \vdots & \vdots \\
    Q(z) & zQ(z) & \ldots & z^nQ(z) & z^mP(z) & \ldots & zP(z) & P(z) \\
     \end{array}
\right| 
\end{equation}

Here we must notice that each column has the same form, i.e. all the entries in one column
have the form $x^i \cdot Q(x)$ ( $x^i P(x)$ ), $x=a_1, \ldots, a_{n+m-1}, z$.
For every row, it has $n+m+2$ entries and its degree $\le n+m$.
Then we can let a column be zero by adding others on it. The determinant must
be null. \qed

Let's expand the preceeding determinant, we can get $f(z)$ as
Theorem (\ref{lri-2}). Now let's consider ($\ref{lri-4}$) and expand it
in the following form:

If we select an entry of the last row, for example $z \cdot Q(z)$, let's
consider the ??? of it. It is an $(m+n+1) \times (m+n+1)$ determinant.
Then the value of it have two possibilities. It may be null if
$S_1:=\{ Q(x), xQ(x), \ldots, x^nQ(x), x^mP(x), \ldots, xP(x), P(x) \}$
are linear dependent. Else, the elements in $S_1$ are linear independent, then we can get a
Vandermonde Determinant by some rearrangment of columns. That is to say, the
??? of $z \cdot Q(z)$ is $\epsilon \cdot T(\bA)$, where $\epsilon$ is a real
number and $T(\bA)=\prod_{i<j} (a_i-a_j)$. It is the same as $z \cdot Q(z)$
for other entries of the last row. Thus we get
$$(\sum_{i=0}^n \epsilon_i z^i) T(\bA) \cdot Q(z) -
 (\sum_{j=0}^n \delta_j z^j) T(\bA) \cdot P(z) = 0$$
and finally
$${P(z) \over Q(z)} =
{\sum_{i=0}^n \epsilon_i z^i \over \sum_{j=0}^n \delta_j z^j}$$
for $T(\bA) \not= 0$.

For example, for $f(z) =(1+2z)(1+3z)/(1+z) $ and
${\bf A }= \{0, 1, 2, 3 \}$, one has the determinant
$$\left| \begin{array}{ccccc}
1 &0 &0 &0 &1\\
1 &1 &1 &6 &6 \\
1 &2 &4 &70/3 &35/3 \\
1 &3 &9 &105/2 &35/2 \\
Q &zQ &z^2 Q & zP & P
\end{array}
 \right| $$
the nullity of which is equivalent to
$$(1+z)P = (1+2z) (1+3z) Q\ . $$


\vspace{3mm}
One can formulate differently the proceeding problem: 
instead of finding the numerator and denominator of a rational 
function from sufficiently many of its values, 
one looks for a rational approximation 
of an analytic function 
$$f(z) = 1 +z f_1+ z^2 f_2 + z^3 f_3 + \cdots\ , $$
i.e. one looks for a rational function 
$g(z) = \prod(1-zb)/ \prod(1-za)$, whose development in the neighbourhood of
$z=0$ coincide with the development of $f$ up to a certain 
power of $z$.

In other words, the first coefficients of $f$ are equal 
to the complete functions $S_j({\bf A }-{\bf B })$, and one has 
to find back the rational function 
 $g(z)$ from its 
 development in the neighbourhood of $z=0$.

Puting $x=1/z$, $card({\bf A })=m$, $card({\bf B })=n$, one can write 
$g(z) = x^{m-n} R(x, {\bf B })/ R(x, {\bf A })$, and transform this expression
by multiplying up and down by the factor $R({\bf A }, {\bf B })$ :
$$g(1/x) = \pm x^{m-n} R({\bf A }+x, {\bf B })\, /\, R({\bf B }+x, {\bf A }) $$

In other terms, the rational function is the quotient of two 
resultants. This is fortunate, because resultants can be written
as special Schur functions
and one has many determinantal expressions of them.

This gives without computation many expressions that one could
find in the flourishing 19th century litterature.

For example Sylvester gives 
\begin{equation}
\label{lri-5}
\pm x^{n-m} g(1/x) = S_{n^{m+1}} ({\bf A }+x-{\bf B }) / S_{{n+1}^m}( {\bf A }-{\bf 
B }-x) \ , 
\end{equation}

while Jacobi proposes
\begin{equation}
\label{lri-6}
\pm x^{n-m} g(1/x) = S_{n, n^{m}} ({\bf A }+x-{\bf B }, {\bf A }-{\bf B } ) / 
S_{m^{n+1}, 0 }( {\bf A }-{\bf B }, x) 
\end{equation}


One could also exchange the
roles of ${\bf A }$ and ${\bf B }$, or transforms the determinants 
by linear combination of rows and columns. 


For example, for $m=3$, $n=2$, 
writing ${\bf C } $ for ${\bf A }-{\bf B } $,  one has
$$-g(x)/x = \left| \begin{array}{cccc} 
 S_2(x+{\bf C }) & S_3(x+{\bf C }) & S_4(x+{\bf C }) & S_5(x+{\bf C }) \\
 S_1(x+{\bf C }) & S_2(x+{\bf C }) & S_3(x+{\bf C }) & S_4(x+{\bf C }) \\
 S_0(x+{\bf C }) & S_1(x+{\bf C }) & S_2(x+{\bf C }) & S_3(x+{\bf C }) \\
 0     & S_0(x+{\bf C }) & S_1(x+{\bf C }) & S_2(x+{\bf C }) 
\end{array}\right| \,  / \hfill $$

$$\hfill / \, \left| \begin{array}{ccc}
 S_3({\bf C }-x) & S_4({\bf C }-x) &S_5({\bf C }-x) \\
 S_2({\bf C }-x) & S_3({\bf C }-x) &S_4({\bf C }-x) \\
 S_1({\bf C }-x) & S_2({\bf C }-x) &S_3({\bf C }-x) 
\end{array}\right| $$

$$= \left| \begin{array}{cccc}
 S_2(x+{\bf C }) & S_3({\bf C }) & S_4({\bf C }) & S_5({\bf C }) \\
 S_1(x+{\bf C }) & S_2({\bf C }) & S_3({\bf C }) & S_4({\bf C }) \\
 S_0(x+{\bf C }) & S_1({\bf C }) & S_2({\bf C }) & S_3({\bf C }) \\
 0     & S_0({\bf C }) & S_1({\bf C }) & S_2({\bf C }) 
\end{array}\right| \, / \, 
\left| \begin{array}{cccc}
 S_3({\bf C }) & S_4({\bf C }) &S_5({\bf C })  &x^3 \\
 S_2({\bf C }) & S_3({\bf C }) &S_4({\bf C }) &x^2 \\
 S_1({\bf C }) & S_2({\bf C }) &S_3({\bf C }) &x^1 \\
 S_0({\bf C }) & S_1({\bf C }) &S_2({\bf C }) &x^0 
\end{array}\right| $$


\vspace{3mm}
The above expressions degenerate when the extra factor 
 $R({\bf A }, {\bf B }) = S_{n^m}({\bf A }-{\bf B })$ is null, i.e. 
when the determinant 
$$\left| \begin{array}{ccc}
f_n &\cdots & f_{m+n-1} \\
\vdots &  & \vdots \\
f_{n-m+1} &\cdots & f_n
\end{array}\right|$$        is null.

\smallskip
Qestions of convergence and algorithms to compute by recursion 
the different determinants corresponding to variable $m$ and $n$
have been studied by Pad\'e in his thesis, 
and have generated a numerous litterature which 
has not much relations with our algebraic manipulations.

{\bf Remark:}

Rational interpolation mostly reduces to using different determinantal expressions
of the resultant $R(\bA, \bB)$ coming from the fact that it is a special Schur function.

\pagebreak

\part{Newton Interpolation}
\vspace{3mm}

\section{Introduction to Newton interpolation}

Newton's approach to interpolation is different from Lagrange's Interpolation
in the sense that
the points in Newton's interpolation 
are totally ordered, and adding a new interpolation point has the
effect of adding an extra term to the interpolation polynomial. As for
Lagrange interpolation, we have to interpret Newton's interpolation as an 
identity on polynomials.

Given an infinite totally ordered alphabet $\bA$, we write $\bA_n$ for the 
subinitial alphabet $\bA_n := \{ a_1, \ldots , a_n\}$. Because $1$, 
$R(x, \bA_1)$is a linear basis of polynomials (triangulary equivalent to 
$\{ 1, x, x^2, \ldots \}$).
We can expand any polynomial $f(x)$ in this basis:

$$f(x) = f(a_1) + f^{\partial }R(x, \bA_1) +
f^{\partial \partial}R(x, \bA_2) + \ldots$$

the coefficients $f$, $f_{\partial}$, $f_{\partial \partial}$ $\ldots$
being uniquely determined.

Taking $x = a_1, a_2, \ldots $, we can easily determine these coefficients.
$$f^\partial = (f(a_1)-f(a_2)) /(a_1-a_2), $$
$$f^{\partial\, \partial} = (f(a_1)-f(a_2))/(a_1-a_2)(a_2-a_3)
                - (f(a_1)-f(a_2))/(a_1-a_3)(a_2-a_3), $$
$$\ldots$$
which infer that $f^{\partial \cdots \partial}$ is the image of $f(a_1)$
under a product of successive divided differences $\partial_1, \partial_2, \ldots$
acting on their left.

Thus, we have Newton's formula:

\begin{thm}
\label{inn-1}
For any polynomial $f(x)$, and infinite totally ordered alphabet $\bA$

$$f(x) = f(a_1) + f\partial _1\, R(x, \bA_1) +
f\partial_1 \partial _2 \, R(x, \bA_2)
 + \ldots , \quad \partial _i = \partial_{a_ia_{i + 1}}$$
\end{thm}

{\em Proof.} Since $\partial$ is a linear operator, it is sufficient to prove this formula for a linear basis of 
polynomials. Take a basis $1, R(x, B_1), R(x, B_2), \ldots $ associated to
another infinite alphabet $B$. Recall that $R(x, B_k) = S_k(x - B_k)$ and
$\partial _iS_k(\bA_i) = S_{k - 1}(\bA_{i + 1})$. We can only prove all the 
identities for $S_k(x-B_k)\, k = 0, 1, 2, \ldots$

$$S_k(x - B_k) = S_k(\bA_1 - B_k) + S_{k - 1}(\bA_2 - B_k)R(x, \bA_1) + \ldots +
S_0(\bA_{k + 1} - B_k)R(x, \bA_k)$$

But now, we could have taken any $\bB$ to have a linear basis of polynomials, 
for example $\bB=\bA$.

In this case, all the $S_{k - j + 1}(\bA_j - \bA_k)$, $j < k$, are null, because
they are, up to a sign, the $k - j + 1$ elementary symmetric function of 
an alphabet $(\bA_k\setminus \bA_j)$ of cardinal $k - j$. The above summation 
reduces to $R(x - \bA_k) = S_0(a_{k + 1})R(x - \bA_k)$, which checks.
 This completes the proof.
\qed

Here we give another proof:
We will prove
\begin{eqnarray}
\label{inn-2}
f(x) & = & f(a_1) + \sum_{k=1}^n
           f \partial_1 \ldots \partial_k (x-a_1) \ldots (x-a_k)\nonumber \\
&&   + f \partial_1 \ldots \partial_n \partial_{a_{n+1}x}
      (x-a_1) \ldots (x-a_{n+1}).
\end{eqnarray}
by induction on $n$.

When $n=0$, equation (\ref{inn-2}) becomes $f(x)=f(a_1)+f \partial_{a_1 x} (x-a_1)$, 
that is, $f\partial_{a_1 x}=(f(x)-f(a_1))/(x-a_1)$. It's just the definition
of $f \partial_{a_1 x}$.

Suppose $n=k$ has been proved, then notice that
$ f \partial_{ac} \partial_{bc} = f \partial_{ab} \partial_{bc} $
and so
\begin{eqnarray*}
f \partial_1 \ldots \partial_k \partial_{a_{k+1}x} & = &
   f \partial_1 \ldots \partial_k \partial_{a_{k+1}x}|_{x=a_{k+2}}
   +f \partial_1 \ldots \partial_k \partial_{a_{k+1}x}
    \partial_{a_{k+2}x}(x-a_{k+2}) \\
& = & f \partial_1 \ldots \partial_k \partial_{k+1}
   +f \partial_1 \ldots \partial_k \partial_{k+1}
   \partial_{a_{k+2}x}(x-a_{k+2}), 
\end{eqnarray*}
we know (\ref{inn-2}) is true for $n=k+1$.
\qed

A direct result of theorem \ref{inn-1} is
\begin{equation}
\label{inn-3}
\frac{1}{1-\alpha x}=\sum_{k=0}^\infty \frac{\alpha^{k} R(x, \bA_k)}
{\prod _{a\in \bA_{k+1}}(1-\alpha a)}
\end{equation}

When we let $\bA=\{1, 1, \ldots\}$ in (\ref{inn-3}), we obtain the Taylor
expansion of $\frac{1}{1-\alpha x}$ at $x=1$. In general, we have
the Taylor's formula as a corollary.

\begin{cor}
\label{inn-4}
We have the Taylor's expansion formula.
\begin{equation}
f(x) = \sum \frac{f^n(x_0)}{n!}(x - x_0)^n
\end{equation}
\end{cor}

{\em Proof.}
Simply take $\bA = \{ x_0, x_0\pm \epsilon, x_0\pm 2\epsilon, 
 x_0 \pm 3\epsilon, \ldots \}. \quad
\epsilon \rightarrow 0$.
\qed


\begin{lem}
\label{inn-5}
Let $\bA$ be an alphabeta, $\bA_k$ the initial $k$ letters, $B$ another finite
alphabeta. Then we have
\begin{equation}
S_m(B) = \sum _{k + i = n}S_k(B - \bA_k) S_i(\bA_{i + 1}).
\end{equation}
\end{lem}

{\em Proof.}
By the Newton interpolatin formula, we have
$$\frac{1}{1 - zx} = \sum _{k = 0}^{\infty}
\frac{z^kR(x, \bA_k)}{\prod_{a\in \bA_{k + 1}}(1 - za)}, $$

which is 
\begin{eqnarray*}
\sum ^{\infty}_{m = 0}x^mz^m & = \sum_{k = 0}z^kS_k(x - \bA_k)(\sum_{i = 0}^{\infty}S_i(\bA_{i + 1})z^i)\\
& = \sum ^{\infty}_{m = 0}(\sum_{k + i = m} S_k(x - \bA_k)S_i(\bA_{i + 1}))z^m
\end{eqnarray*}

Compare the coefficients of $z^m$, then we have

\begin{equation}
\label{inn-6}
x^m = \sum _{k + i = m} S_k(x - \bA_k)S_i(\bA_{i + 1})
\end{equation}

Suppose $|B| = l$, Let $b_1 = x$, operator on both sides of 
(\ref{inn-6}) by $L_B$, 
then we have 


$$ S_{m - l + 1}(B) = \sum _{k + i = m}S_{k - l + 1}(B - \bA_k)S_i(\bA_{i + 1}), $$
which is the same as (\ref{inn-5}).
\qed


For example

Let $\bA = \{ 1, q, q^2, \ldots , \}$, and denote
$$\sum\frac{z^k}{(k)_q!} = \exp_q(z).$$
Then we have the Jackson's formula.

\begin{thm}
Suppose $B$ is an alphabeta, then we have the identity
\begin{equation}
\label{inn-7}
\sum_{k=0}^\infty \frac{z^k}{( k)_q!} S_k(B)
=\exp _q(z)\sum_{k=0}^\infty \frac{z^k}{(k)_q!}\Delta_q^k(B), 
\end{equation}
where $\Delta_q^k(B)=S_k(B-\bA_k)$.
\end{thm}

{\em Proof.}
\begin{eqnarray*}
\sum ^{\infty}_{i = 0}S_i(\bA_{k + i})\frac{z^i}{(i + k)_q!} & = &
\sum\frac{(k + i)_q!}
{(k)_q!(i)_q!} = \frac{z^i}{(k + i)_q!}\\
& = & \frac{1}{(k)_q!} \exp_q(z).
\end{eqnarray*}

\begin{eqnarray*}
\exp_q(z)\sum_{k=0}^\infty \frac{z^k}{(k)_q!}\Delta_q^k(B) 
& = & \sum_{k=0}^{\infty}\left( \sum_{i=0}^{\infty}S_i(\bA_{k+1})\frac{z^i}
{(i+k)_q!} \right) S_k(B-\bA_k) z^k \\
& = & \sum_{m=0}^{\infty}\left( \sum_{k+i=m}S_i(\bA_{k+1})
 S_k(B-\bA_k) \right) \frac{z^m}{(m)_q!} \\
& = & \sum_{m=0}^{\infty}S_m(B)\frac{z^m}{(m)_q!}
\end{eqnarray*}

The last step is obtained by the Lemma \ref{inn-5}.
\qed



Newton's formula is closely connected with a discrete analog of the
Wronskian, and could be checked using it. 

Indeed, given $n$ one variable-functions $f_i$, we define their $\bA$-Wronskian
to be 

$$W_\bA(f_1, \ldots , f_n) :=
det|f_i(a_1)\partial _1\ldots \partial _{j - 1}|_{1 \leq i, j \leq n}$$

\begin{lem}
The $\bA$-Wronskian of $n$ functions $f_1, \ldots , f_n$ is equal to
the quotient $det|f_i(a_j)|/\Delta (\bA)$. It is also equal to the determinant
$det|f_i \partial_1 \cdots \partial_{j-1} \pi_j \cdots \pi_{n-1}|$
\end{lem}

{\em Proof.} Replace each $f(a_i)$ by its Newton's expansion: 
$f(a_i) = f(a_1) + \ldots +
f \partial _1 \cdots \partial _{i - 1}R(a_i, \bA_{i - 1})$.
Let's multiply $W_\bA(f_1, \ldots, f_n)$ and $T(\bA)$, where $T(\bA)=det|R(a_i, \bA_{j-1}|$
and $R(a_i, \bA_0)=1$.
\begin{eqnarray}
H & = & T(\bA) \cdot W_\bA(f_1, \ldots, f_n) \nonumber \\
&=& \left| \begin{array}{ccccc}
        R(a_1, \bA_0) & \ldots & R(a_1, \bA_{i-1}) & \ldots & R(a_1, \bA_{n-1}) \\
        R(a_2, \bA_0) & \ldots & R(a_2, \bA_{i-1}) & \ldots & R(a_2, \bA_{n-1}) \\
          \vdots & \vdots &   \vdots   & \vdots &   \vdots  \\
        R(a_i, \bA_0) & \ldots & R(a_i, \bA_{i-1}) & \ldots & R(a_i, \bA_{n-1}) \\
          \vdots & \vdots &   \vdots   & \vdots &   \vdots  \\
        R(a_n, \bA_0) & \ldots & R(a_n, \bA_{i-1}) & \ldots & R(a_n, \bA_{n-1})
     \end{array}
  \right| \times \nonumber \\
&& \left|
    \begin{array}{cccccc}
        f_1 & f_2 & \ldots & f_j & \ldots & f_n \\
        f_1 \partial_1 & f_2 \partial_1 & \ldots & f_j \partial_1
            & \ldots & f_n \partial_1 \\
        \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
        f_1 \partial_1 \cdots \partial_{j-1} & f_2 \partial_1 \cdots \partial_{j-1}
            & \ldots & f_j \partial_1 \cdots \partial_{j-1} & \ldots
            & f_n \partial_1 \cdots \partial_{j-1} \\
        \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
        f_1 \partial_1 \cdots \partial_{n-1} & f_2 \partial_1 \cdots \partial_{n-1}
            & \ldots & f_j \partial_1 \cdots \partial_{n-1} & \ldots
            & f_n \partial_1 \cdots \partial_{n-1} 
    \end{array}
 \right|
\end{eqnarray}
Then the $(i, j)$ entry of $H$ is
\begin{eqnarray}
h_{ij}&=&f_j(a_1)+f_j\partial_1 R(a_i, \bA_1)+f_j\partial_1\partial_2 R(a_i, \bA_2)
+\cdots+f_j\partial_1 \partial_2 \cdots \partial_{i-1}R(a_i, \bA_{i-1}) \nonumber \\
&& \quad f_j\partial_1 \partial_2 \cdots \partial_iR(a_i, \bA_i) +\cdots+
  \quad f_j\partial_1 \partial_2 \cdots \partial_{n-1}R(a_i, \bA_{n-1}) 
\nonumber
\end{eqnarray}
But obviously $R(a_i, \bA_k)=0, \, k \ge i$, thus we have $h_{ij}=f_j(a_i)$ which means
$det(f_j(a_i))=H$. At the same time, we know that $T(\bA)$ is an lower triangular 
matrix.
$$T(\bA)=R(a_2, \bA_1)R(a_3, \bA_2) \cdots R(a_n, \bA_{n-1})=\Delta(\bA).$$
Then 
$$W_\bA(f_1, \ldots, f_n)=det(f_i(a_j))/\Delta(\bA).$$
Because $det(f_i(a_j))/\Delta(\bA)$ is a symmetric function, one does not change
the value of the Wronskian by taking its image under
$$\pi_\omega=(\pi_{n-1})(\pi_{n-2}\pi_{n-1})\cdots(\pi_1\cdots\pi_{n-1}).$$
Now, the Wronskian has been written as a determinant such that the entries in
the successive rows are symmetric funcions of $\bA_1, \bA_2, \ldots, \bA_n$
respectively. The operator $\pi_{n-1}$ operates non trivially only on row
$n-1$, then the operator $\pi_{n-2}\pi_{n-1}$ operates on row $n-2$ of the
new determinant, and finally the image of the Wronskian under $\pi_\omega$
is the determinant where each of the operators
$(\pi_{n-1}), (\pi_{n-2}\pi_{n-1}), \cdots, (\pi_1\cdots\pi_{n-1})$
have been applied to the appropriate row. The entries of the final
determinant are now symetric functions in $\bA$.
\qed

In the special case where all the functions $f_j$ are powers of a variable, 
we see that the Wronskian interpolates betweenthe expression of a Schur function
as a determinant of complete funcions, and as a quotient of determinant of
powers by the Vandermonde. For example, with $\bA$ of cardinal $3$, one has
$$
W_{\bA}(x^2, x^5, x^9)=
\left| \begin{array}{ccc}
        S_2(\bA_1) & S_5(\bA_1) & S_9(\bA_1) \\
        S_1(\bA_2) & S_4(\bA_2) & S_8(\bA_2) \\
        S_0(\bA_3) & S_3(\bA_3) & S_7(\bA_3) \\
    \end{array}
\right|
=
\left| \begin{array}{ccc}
        a^2_1 & a^5_1 & a^9_1 \\
        a^1_2 & a^4_2 & a^8_2 \\
        a^0_3 & a^3_3 & a^7_3 \\
    \end{array}
\right| /\Delta(\bA)=
$$
$$
\left| \begin{array}{ccc}
        a^2_1\pi_1\pi_2 & a^5_1\pi_1\pi_2 & a^9_1\pi_1\pi_2 \\
        S_1(\bA_2)\pi_2 & S_4(\bA_2)\pi_2 & S_8(\bA_2)\pi_2 \\
        S_0(\bA_3) & S_3(\bA_3) & S_7(\bA_3) \\
    \end{array}
\right|
=\left|
    \begin{array}{ccc}
        S_2(\bA) & S_5(\bA) & S_9(\bA) \\
        S_1(\bA) & S_4(\bA) & S_8(\bA) \\
        S_0(\bA) & S_3(\bA) & S_7(\bA) \\
    \end{array}
 \right|
$$
Remark:

Given $\{a_1, a_2, \ldots\}$, Newton interpolation develops any polynomial
$f(x)$ in the basis $\{ (x-a_1)\cdots(x-a_k)\}$; the coefficients are
obtained by applying divided differences on $f$.

\pagebreak

\section{ Newton interpolation in several variables}

A linear basis of the ring of polynomials in an alphabet 
${\bf{A}}$ allows to expand any polynomial in ${\bf{A}}$, 
as we did for polynomials in one variable.
It is reasonable to ask that this
basis contains Newton's polynomials
$R(a_1, {\bf{B}}_k)$, and has simple vanishing properties. 

One may also start from the images 
of a polynomial in ${\bf{A}}$ under divided differences, evaluated 
in a second alphabet ${\bf{B}}$. It is easy to see that the
divided differences $\partial_\mu$, for all permutations
$\mu$, are linearly independent, and therefore there exist
 universal coefficients in ${\bf{A}}, {\bf{B}}$ which generalize Newton's polynomials.

We shall define recursively these polynomials, that have been called 
{\it Schubert polynomials} by M. P. Sch\"utzenberger and me.

For $K \in{\bf{N}}^\infty$, we denote $Y_K$ the Schubert polynomials
indexed by $K$.
\begin{itemize}
\item[1]
If $K_0=(n-1, \ldots, 1, 0, 0, \ldots)$, we define $Y_{K_0}(\bA, \bB)$ to be 
$$Y_{K_0}(\bA, \bB) := \prod_{(i+j \le n)} (a_i -b_j).$$
We will write $Y_k=Y_k(\bA, \bB)$ for short if no confusion.
\item[2]
If $Y_K$ is defined, and the $p$-th and $p+1$-th component of $K$ are $i, j$ 
respective, $i>j$
then we define $Y_{\ldots, j-1, i, \ldots}$ to be
$$Y_{\ldots, j-1, i, \ldots}:=Y_K \partial_p.$$
On the other hand, the nullity $\partial_p^2=0$ implies that
$i \le j \Rightarrow Y_{\cdots i j \cdots} \partial_p=0$. Here $\partial_p$
is the symmetric operator of $\bA$
\end{itemize}

The same $Y_I$ can be obtained in several manners from the 
$Y_{K_0}$. We shall prove that braid relations imply that:

The polynomials $Y_I$ are uniquely determined by the above recursions.

{\em Proof.} Let $c:G_n \rightarrow {\bf N}^\infty$ map a permutation to its code, 
$\phi:{\bf N}^\infty \rightarrow G_n$ map a code to the permutation whose code 
is just it.Then easily we know if the $p$-th component of I is larger than 
the $(p+1)$-th, we have 
$Y_I\partial_p=Y_{c(\phi(I))}\partial_p=Y_{c(\phi(I)s_p)}$. By induction, we 
get $Y_{K_0}\partial_w=Y_{c(w_0w)}$. If $c(w_0w_1)=c(w_0w_2)$, 
since $c$ is a bijection, then $w_1=w_2$ and so 
$Y_{K_0}\partial_{w_1}=Y_{K_0}\partial_{w_2}$. From this we know the polynomials $Y_I$ 
are uniquely determined by the above recursions. 

\begin{thm}
\label{s-**}
If $K$ is weakly decreasing, then
$$Y_K = \prod_{(i, j)} (a_i -b_j) \ , $$
product on all points $(i, j)$ of the Ferrers' diagram of $K$.
\end{thm}

For example 
$$Y_{221100\ldots}= \begin{array}{cccc}
(a_1-b_2) & (a_2 -b_2) & &\\
(a_1-b_1) & (a_2-b_1) &(a_3-b_1) &(a_4 -b_1)
\end{array} \ , $$
where the above matrix stands for the product of its entries.

Another example is the resultant $R(\bA, \bB)$, when $|\bA|=n, |\bB|=m$, 
we have $R(\bA, \bB)=Y_{m^n}$.

Now we shall give two proofs:

The first proof:

Let $K=(k_1, \ldots, k_n)_\ge$, let $m=max\{ k_i+i-1 \}_{1 \le i \le n}$, 
then $K \subset T=(m, m-1, \ldots, 2, 1)$. From the definition, we get
$$Y_T=\prod_{1 \le i \le m} R(a_i, \bB_{m-i+1}).$$
Here we have to pay attention to the following fact:
$$(R(a_1, \bB_k) \cdot R(a_{i+1}, \bB_{k-1})) \partial_i
 = R(a_i, \bB_{k-1}) \cdots R(a_{i+1}, \bB_{k-1}).$$
Then if $m=k_1$, we can do it by indution.

If $k_j+j-1=m$ and $k_i+i-1<m$ for $i<j$, we can consider
$Y_T \partial_1 \partial_2 \cdots \partial_{j-1}$, then we have
$$Y_T \partial_1 \partial_2 \cdots \partial_{j-1} =
    Y_{m-1 \cdots m-j+1 \, m-j+1 \, m-j+2 \cdots 2 \, 1} = Y_{T_1}, $$
by definition of $Y_K$. On the other hand, we have
$$
Y_T \partial_1 \partial_2 \cdots \partial_{j-1}
    = \prod_{1 \le i \le j-1} R(a_i, \bB_{m-i})
     \prod_{j \le i \le m} R(a_i, \bB_{m-i+1}).
$$
Let's compare $k_1 \ldots k_{j-1}$ and $m-1 \ldots m-j+1$. If $k_1=m-1$, 
we can proof the theorem by inducion. Else let $h$ be the smallest number
s.t. $k_n=m-n$. (If it doesn't exist, let $h=j$) Then let's consider
$Y_T \partial_1 \partial_2 \cdots \partial_{h-1}$.
This process will be end when the first index of the image of $Y_T$
is $k_1$. Then we can consider $k_2$ and it can be done by indution.

Following the above algorithm, we can get $Y_K$ easily and it has the
following expression:
$$R(a_1, \bB_{k_1}) \cdots R(a_n, \bB_{k_n}).$$
Thus we complete the proof.

The second proof:

Let $w$ denote the permutation corresponding to $K$ (i.e. the code of $w$ 
is $K$), and $\lambda$ denote the Ferrers' diagram of $K$, 
then $w$ is dominant of shape $\lambda$. 
We will proceed by descending induction on 
$\ell(w), \ w \in S_n$, and $w$ is dominant.
The result is true for $w=w_0$, since $K=(n-1, n-2, \ldots, 1, 0, 0, \ldots)$, 
and the definition of $Y_K$ 
is just the product of all $(a_i-b_j) \ (i, j) \in \delta$, 
where $\delta$ is the Ferrers' diagram of $K$.
Suppose $w \not= w_0$ is dominant of shape $\lambda$. Then 
$\lambda \subset \delta$ (and $\lambda \not= \delta$). 
Let $\alpha$ be the minimal integer such that 
$\lambda_\alpha=\lambda_{\alpha+1}$, i.e. the minimal interger such that
$w(\alpha)<w(\alpha+1)$, so the code of $w'=s_\alpha w$ is 
$K'=(k_1, \ldots, k_{\alpha-1}, k_\alpha+1, k_{\alpha+1}, \ldots, k_n)$. 
Then $w'$ is dominant, $\ell(w')=\ell(w)+1$, 
and $\lambda(w')=\lambda(w)+\varepsilon_\alpha$, 
( $\varepsilon_\alpha$ is the Ferrers' diagram 
whose $\alpha$-th column has $1$ square, and others have none. )
and therefore
\begin{eqnarray*}
Y_K & = & Y_{K'} \partial_\alpha \\
& = & ((a_\alpha-b_{r+1}) \prod_{(i, j) \in \lambda}
    (a_i-b_j))\partial_\alpha
\end{eqnarray*}
Since $\lambda_\alpha=\lambda_{\alpha+1}$ it follows that
\begin{eqnarray*}
Y_K & = & ((a_\alpha-b_{r+1}) \prod_{(i, j) \in \lambda}
        (a_i-b_j))\partial_\alpha\\
& = & \prod_{(i, j) \in \lambda}(a_i-b_j)
        [(a_\alpha-b_{r+1})\partial_\alpha]\\
& = & \prod_{(i, j) \in \lambda}(a_i-b_j).
\end{eqnarray*}

Remark:

The above induction cource can be exprees in the Ferrers' diagram: 
Suppose $K=(5, 2, 0, 0, \ldots)$, then we add squares to the Ferrers' diagram 
in the order marked below, at last we will reach $\delta$.


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\end{picture}

When $K$ is decreasing, the dominant Schubert polynomial $Y_K$ can be written
as a Schur function. Indeed, thanks to (Sf7), for $K=k_1 \ge k_2 \ge \cdots
\ge k_n \ge 00\ldots$, writing $K^\omega=k_n \ldots k_1$, one has
\begin{equation}
\label{Nis2}
Y_K=S_{K_\omega}(\bA_n-\bB_{k_n}, \ldots, \bA-\bB_{k_1}).
\end{equation}
Here's a short proof for it.

If $K=(k_1, \ldots, k_n)_{\ge}$, then we have
$$Y_K=R(a_1, \bB_{k_1})R(a_2, \bB_{k_2})\cdots R(a_n, \bB_{k_n}).$$
We know $R(a_i, \bB_{k_i})=S_{k_i}(a_i-\bB_{k_i})$ by (Sf12) for $1 \le i \le n$.
Thus
$$Y_K=S_{k_n}(a_n-\bB_{k_n})S_{k_{n-1}}(a_{n-1}-\bB_{k_{n-1}})
S_{k_1}(a_1-\bB_{k_1}).$$
Now let's consider the expansion of
$S_{K^\omega}(\bA_n-\bB_{k_n}, \ldots, \bA_1-\bB_{k_1})$, we have
\begin{eqnarray}
&&S_{K^\omega}(\bA_n-\bB_{k_n}, \ldots, \bA_1-\bB_{k_1}) \nonumber \\
 &\hspace{-1.7cm}=& \hspace{-1.7cm}\left|
    \begin{array}{cccc}
        S_{k_n}(\bA_n-\bB_{k_n}) & S_{k_{n-1}+1}(\bA_{n-1}-\bB_{k_{n-1}}) &
            \cdots & S_{k_1+n-1}(\bA-\bB_{k_1}) \\
        S_{k_n+1}(\bA_n-\bB_{k_n}) & S_{k_{n-1}}(\bA_{n-1}-\bB_{k_{n-1}}) &
            \cdots & S_{k_1+n-2}(\bA-\bB_{k_1}) \\
        \vdots & \vdots & \vdots & \vdots \\
        S_{k_n+n-1}(\bA_{n-1}-\bB_{k_{n-1}}) &
            S_{k_{n-1}+n-2}(\bA_{n-1}-\bB_{k_{n-1}}) & \cdots &
            S_{k_1}(\bA_1-\bB_{k_1})
    \end{array}
   \right|
\end{eqnarray}
For example, 
$$Y_{22110\ldots} = S_{1122}(\bA_4-\bB_1, \bA_3-\bB_1, \bA_2-\bB_2, \bA_1-\bB_2)=$$
$$
=\left|
    \begin{array}{cccc}
        S_1(\bA_4-\bB_1) & S_2(\bA_3-\bB_1) & S_4(\bA_2-\bB_2)
            & S_5(\bA_1-\bB_2) \\
        S_0(\bA_4-\bB_1) & S_1(\bA_3-\bB_1) & S_3(\bA_2-\bB_2)
            & S_4(\bA_1-\bB_2) \\
        0 & S_0(\bA_3-\bB_1) & S_2(\bA_2-\bB_2)
            & S_3(\bA_1-\bB_2) \\
        0 & 0 & S_1(\bA_2-\bB_2) & S_2(\bA_1-\bB_2) \\
    \end{array}
\right|
$$
In particular, $Y_{00\ldots}=1$.

\begin{lem}
Let $\bA=\{ a_1, \ldots, a_n \};w=(n, n-1, \ldots, 1), \mu(\not=w) \in G_n$, 
and $\bA^\mu$ denote the set $\{ a_{\mu(1)}, \ldots, a_{\mu(n)}) \}$, 
then we have
\begin{equation}
Y_{I_0}(\bA, \bA^\omega)=\Delta(\bA):=\prod_{i < j \le n}, 
\qquad Y_{I_0}(\bA, \bA^\mu)=0, 
\end{equation}
where $I_0=(n-1, n-2, \ldots, 1, 1, 0, 0, \ldots, )$.
\label{l-9.2}

\end{lem}

{\em Proof.}
if $Y_{I_0}(\bA, \bA^\mu) \not= 0$, that is to say $R(a_1, \bB_{n-1}) \not= 0$, 
where $\bB_{n-1}=\{ a_{\mu(1)}, a_{\mu(2)}, \ldots, a_{\mu(n-1)} \}$, then
$a_1$ has to be $a_{\mu(n)}$. Let's consider $R(a_2, \bB_{n-2})$, we can get
$a_2=a_{\mu(n-1)}$. Thus we have $\mu=\omega=(n, n-1, \ldots, 1)$ and
\begin{eqnarray}
Y_{I_0}(\bA, \bA^\omega) & = & \prod_{i=1}^{n-1} R(a_i, \bA_n \setminus \bA_i) \nonumber \\
&= & \prod_{i < j \le n} (a_i-a_j) = \Delta(\bA).
\end{eqnarray}
\qed

\begin{thm}
Let $\bA$ be an infinite alphabet, $f(a_1, a_2, \ldots)$ a polynomial in $\bA$, 
and $n \in \bN$ be such that all the exponents $I$ of the monomials appearing
in $f$ are (componentwise) majorized by $(n-1, \ldots, 1, 0)$. Let $\bB$ be an
alphabet of cardinal $n$. Then
$$f(\bA)=\sum_{\mu \in G(\bB)} f(\bB^\mu) Y_{I_0}(\bA, \bB_{\omega \mu})
        / \Delta(\bB^\mu).$$
\label{t-**}

\end{thm}
{\em Proof.}
For any permutation $\nu \in G(\bB)$, we have
$$f(\bB^\nu)= \sum_{\mu \in G(\bB)} f(\bB^\mu) Y_{I_0}(\bB^\nu, \bB_{\omega \mu})
        / \Delta(\bB^\mu).$$
By lemma(\ref{l-9.2}), these $n!$ equations on $f$ determine $f$ and force
the equality for general $\bA$.
\qed

\begin{lem}
\label{s-$}
Let ${\bA}=\{ a_1, \ldots, a_n \}$, then
$$Y_I({\bf{A}}, {\bf{A}})=\left \{
   \begin{array}{ll}
   1 & \ \ \ \mbox{ if }I=00\ldots, \\
   0 & \ \ \ \mbox{ otherwise. }
   \end{array}
   \right.
$$
\end{lem}
{\it Proof}: Suppose $Y_I=Y_{I_0} \partial_{j_1} \ldots \partial_{j_p}$, 
then compare the degree, we know $s_{j_1} \ldots s_{j_p}=w_0$ iff $I=00\ldots$. 
From the definition of $\partial_i$, we can easily know that 
$$Y_{I_0} \partial_{j_1} \ldots \partial_{j_p}=\sum_{(i_1, \ldots, i_m) 
\subset (j_1, \ldots, j_p)} P_{i_1 i_2 \ldots i_m}
Y_{I_0} \sigma_{i_1} \sigma_{i_2} \ldots \sigma_{i_m}, $$
where $P_{j_{i_1} j_{i_2} \ldots j_{i_m}}$ is some rational polynomial
of $a_1, \ldots, a_n$. From the above lemma, we get $Y_I({\bf{A}}, {\bf{A}})=0$
if $I \not=00\ldots$. It is easy to check that $Y_{00\ldots}=1$, 
so we finish the proof.
\qed

Remark: We shall prove stronger
vanishing properties at the end of the section.

We also need that the $Y_I$ are linearly independent, this coming from
the fact that their leading monomial (for the lexicographic order from
the right) is equal to $a^I = a_1^{I_1}\, a_2^{I_2} \, a_3^{I_3}\cdots$

It is remarkable that these simple properties are sufficient to obtain the 
multi-variables generalization of Newton's formula.

\begin{thm}
Let ${\bf{A}}$ and ${\bf{B}}$ be two infinite alphabets.
Then for any polynomial $f({\bf{A}})$,  one has
$$ f({\bf{A}}) = \sum_{I} f\partial_{[I]}({\bf{B}}) \ Y_I({\bf{A}}, {\bf{B}}) \ , $$
where $\partial_{[I]}$ is the divided difference which sends 
$Y_I$ onto $Y_{00\ldots}$. 
\end{thm}

{\em Proof.}
It is sufficient to check the theorem on a generating set
of the polynomials in ${\bf{A}}$, which can involve the elements of
${\bf{B}}$ which are scalar for the divided differences in ${\bf{A}}$.
Thus we take the $Y_J({\bf{A}}, {\bf{B}})$ as a generating set by
theorem\ref{t-**}. Now images of $Y_I$ under divided differences are either
$0$ or other Schubert polynomials. Moreover, the image of $Y_I$ under
$\partial_{[J]}$ cannot be $Y_{00\ldots}$ if $I \neq J$. Taking into account
that $Y_L({\bf{A}}, {\bf{A}}) =0$ for $L\neq 00\cdots$, we see that 
Newton's formula, in the case of the polynomial $f({\bf{A}}) = Y_J$
reduces to the irrefutable identity
$$ Y_J({\bf{A}}, {\bf{B}}) = Y_{00\ldots} Y_J({\bf{A}}, {\bf{B}}) $$
\qed

\pagebreak

\section{Interpolation of symmetric functions}

One can use Newton'formula to interpolate symmetric functions. Let
$f({\bf A_n})$ be a function symmetrical in $\{a_1, \ldots, a_n\}$. According to
the preceeding theorem, one can write this function
\begin{equation}
\label{isf-1}
f({\bf A}_n)=\sum_I c_I Y_I.
\end{equation}

We claim that all the $I$ corresponding to non zero coefficients $c_I$ are of
the type
$$I=0\leq i_1\leq i_2 \cdots \leq i_n00 \cdots$$

{\em Proof:}
Indeed, $f({\bf A}_n)$ depending only on $a_1, \cdots, a_n$is annilhilated by
all $\partial_k$, with $k>n$. Similarly, being symmetrical in $a_1, \cdots, a_n$, 
it is also annihilated by $\partial_1, \cdots, \partial_{n-1}.$

Therefore, all the Schubert polynomials appearing on the right
hand side must be
annihilated by the same divided differences, because otherwise we would have a sum
of Schubert polynomialss equal to 0. This is equivalent to say that $I$
can have only the above form.
\qed

Let us take for example a symmetric function of two variables, say
$Y_{25}({\bf A, C})$, where {\bf C}is another infinite alphabet. Write
$Y_IY_J$ for the product $Y_I({\bf B}, {\bf C})Y_J({\bf A}, {\bf B})$.
Then formula (\ref{isf-1}) reads
$$Y_{25}({\bf A, C})=Y_{25}Y_{00}+Y_{204}Y_{01}+Y_{014}Y_{11}+Y_{2003}Y_{02}$$
$$+Y_{0103}Y_{12}+Y_{20002}Y_{03}+Y_{0003}Y_{22}+Y_{01002}Y_{13}+Y_{200001}Y_{04}$$
$$+Y_{00002}Y_{23}+Y_{010001}Y_{14}+Y_{200000}Y_{05}+Y_{000001}Y_{24}
+Y_{010000}Y_{15}+Y_{000000}Y_{25}$$
In the case ${\bf B}=0$, the above expansion expresses the Schubert polynomial
$Y_{25}({\bf A, C})$ as a linear combination of all Schur functions
$S_I(a_1+a_2)=Y_I({\bf A}, 0)$, $I\subseteq 25$.

The symmetrical Schubert polynomials $Y_I({\bf A, B})$ are a natural generalization
of Schur functions:the leding term of $Y_I({\bf A, B})$(i.e. term of higher degher
degree in ${\bf A}$) is the Schur function $S_I({\bf A}_n)=Y_I({\bf A}, 0), $
and can be written as determinants.

Indeed, let $I= i_1\leq\cdots \leq i_n$, and $K= I^\omega+\rho :=
i_n+n-1, \ldots , \, i_2+1, i_1+0$. Then, still writing 
${{\bf B}}_k$ for $\{b_1, \ldots , b_k \}$, 
one has, according to (\ref{syf-16}) 
$$Y_K= R(a_n, {{\bf B}}_{k_n})  \cdots R(a_1, {{\bf B}}_{k_1 })=
S_{K^\omega}({{\bf A}}_n- {{\bf B}}_{k_n}, \ldots , {{\bf A}}_1-{{\bf B}}_{k_1} ) 
\ . $$
The image of $Y_K$ under $\partial_\omega$, with $\omega=n\cdots 1$, is 
by definition $Y_I$, but thanks to (\ref{syo-30}), it is also equal to
\begin{equation}
\label{isf-2}
Y_I = S_I( {{\bf A}}_n - {{\bf B}}_{k_1}, \ldots , {{\bf A}}_n-{{\bf B}}_{k_n} ) \ , 
\end{equation}
the leading term being evidently the Schur function $S_I({{\bf A}}_n)$.
It is a special case of the determinant used in [L1].

Expression (\ref{e-2.7}) shows that $\partial_\omega(Y_K)$ can be written as the 
quotient 
\begin{equation}
\label{isf-3}
Y_I = \left| R(a, {{\bf B}}_{k_1}), \, \ldots , \, R(a, {{\bf B}}_{k_n})
\right|_{a\in{{\bf A}}_n} \, / \, \Delta({{\bf A}}_n) \ , 
\end{equation}
i.e. 
\begin{equation}
\label{isf-4}
Y_I = \sum\nolimits_\mu \, (-1)^{\ell(\mu)}
\Bigl( R(a_1, {{\bf B}}_{k_1}) \cdots R(a_n, {{\bf B}}_{k_n}) \Bigr)^\mu
\, / \, \Delta( {{\bf A}}_n) \ .
\end{equation}

\smallskip
If $I$ has repeated components, one can reduce the preceeding
summation to a subset of the symmetric group.

For example, for any $m\geq0$, let $r= m+n-1$; then 
$$Y_{0^{n-1}m }= Y_{r\, 00\ldots} \partial_1\cdots \partial_{n-1} =
 L_{{{\bf A}}_n}\bigl( R(a_1, {{\bf B}}_r) \bigr) \ .  $$
Now Leibniz formula expresses the image of 
$R(a_1, {{\bf B}}_r)$ as a summation of products 
$\prod(a_i -b_j)$. 

This can be visualized as follows. A string of boxes, some of 
them occupied by a $\bullet$, means a product of linear factors
according to the following recepee~:

\noindent
" in an empty box at position $j$, put the factor $a_i -b_j$, 
where $i-1$ is equal to the number of dots on the left of the box.
Finally erase the $\bullet$'s."

\smallskip
For example, 
$$\carre\carre \bullet \bullet\, \carre\bullet\, \carre =
(a_1-b_1) (a_1-b_2)\ (a_3-b_5) \ (a_4-b_7) \ . $$
Now $Y_r$ is equal to a string of $r$ boxes. Its image under
$\partial_1$ is equal to the sum of all strings with one dot, 
$$Y_{0\, r-1} = \sum \carre\ldots \carre \bullet \carre\ldots \carre \ . $$
The next step, using $\partial_2$, puts one dot in a box right of 
the existing dot, and thus 
$Y_{0\, 0\, r-2}$ is the sum of all strings
of length $r$ with two dots~:
$$Y_{00\, r-2} = \sum \carre\ldots \carre \bullet\,  \carre\ldots \carre
       \bullet\, \carre\ldots \carre \ . $$


Finally, $Y_{0^{n-1} m }$ is equal to the sum of all strings 
of $r$ boxes with $n-1$ dots. In other words, 
\begin{equation}
\label{isf-5}
Y_{0^{n-1} m }= \sum (a_1-b_1)\cdots (a_1-b_i) \, 
(a_2-b_{i+2}) \cdots (a_2 - b_j) \cdots 
(a_n -b_h)\cdots (a_n-b_r) \ .
\end{equation}
Suppressing the void boxes, one could also write this summation as 
attached to a diagram of $m$ boxes, with shifts in the indices of
the $b$'s given by the {\it contents} of the boxes. 
For more general diagrams, with ${{\bf B}}=\{0, 1, 2, \ldots \}$, this
construction is due to Biedenharn-Louck. 


Indeed, "Stirling interpolation" 
 ${{\bf B}}={{\bf N}} := \{0, 1, 2, \ldots \}$ of symmetric functions 
involves the specialisation of Schubert polynomials 
$Y_I({{\bf A}}, {{\bf N}})$, $I= i_1\leq\cdots \leq i_n $, which 
are the {\it factorial Schur functions}
 of Biedenharn and Louck.

In that case, $R(a, {{\bf N}}_n) = a (a-1)\cdots (a-n+1)$ 
may be considered as a polarization of the $n$th-power
and may be conveniently represented by the symbol
$a^{[n]}$. Identity ($\ref{isf-3}$) reads now 

\begin{equation}
\label{isf-6}
Y_I({{\bf A}}, {{\bf N}}) = \left| a^{[k_1]}, \, \ldots , \, a^{[k_n]}
\right|_{a\in{{\bf A}}_n} \, / \, \Delta({{\bf A}}_n) \ , 
\end{equation}


while ($\ref{isf-5}$) becomes
\begin{equation}
\label{isf-7}
Y_{0^{n-1} m }= \sum (a_1)\cdots (a_1-i) \, 
(a_2-i-2) \cdots (a_2 -j) \cdots 
(a_n -h)\cdots (a_n -r+1) \ , 
\end{equation}

sum on all $-1\leq i\leq j \leq\cdots \leq h \leq r$.

Expression ($\ref{isf-6}$) proves that factorial Schur functions can be 
characterized by their vanishing properties, as noticed by
Okounkov. They are essential in the study of central
elements in the universal envelopping algebra $
U(Gl(n)) $ (Olshanski, Okounkov, Nazarov, Molev).


\smallskip 
One can add that factorial Schur functions can also be 
obtained as specializations of some Schubert polynomials in 
${\bf A}=\{1, 1, 1\ldots \}$ and ${\bf B}=\{0, 0, 0\ldots \}$, the variables 
being now the components of the index $I$
(and thus taking only integral values, but this sufficient to determine
them). This leads to the theory of binomial determinants, as a part 
of bijective combinatorics so notoriously illustrated 
by our friend Xavier Viennot. 

The relevant properties of factorial Schur functions are now related 
to dimensions of representations of the linear groups or unitary groups, 
as functions of the components of highest weight vectors.

\smallskip
Some operations on symmetric functions can be explicited using Newton's
interpolation. Let us describe, for example, the uniform translation 
${{\bf A}} \longrightarrow {{\bf A}}^+:= \{a_1+1, a_2+1, \ldots , \, a_n+1 \}$. Now, 
divided differences are 
invariant under this translation, and ($\ref{isf-1}$) becomes, for any 
$J = j_1\leq j2\leq\cdots \leq j_n \, 00\ldots$, 
denoting $Y_{J/I}$ the image of $Y_J$ under $\partial_{[I]}$~:
\begin{equation}
\label{isf-8}
Y_J({{\bf A}}^+) = \sum_I Y_{J/I}(\{1, \ldots, 1 \}, {0, \ldots 0}) \, 
 Y_J({{\bf A}}, \{0, \ldots , 0 \}  \ .
\end{equation}



{\bf Remark:}
Newton interpolation of functions symmetrical in $a_1, \ldots , a_n$, 
uses only Schubert polynomials which are symmetrical in
$a_1, \ldots , a_n$ (and have leading term a Schur function).
In the case where the interpolation alphabet $B$ is equal to
$\{0, 1, 2, \ldots \}$, these special Schubert polynomials are
called {\it factorial Schur functions}.

\pagebreak

\section{The ring of polynomials as a free module over {\rm Sym}}

We have seen that any polynomial in ${\bf A}$ can be uniquely 
expanded as a linear combination of Schubert polynomials. 
We skipped the proof of some simple properties of these polynomials.
We are going to reinterpret Newton's formula, and prove 
stronger properties that those left unproved.

First, we shall work with a finite alphabet ${\bf A}$ of cardinal $n$.
We shall describe the structure of the ring 
${\cal P}{\sl ol}({\bf A}):= {\bf Z}[{\bf B}][{\bf A}]$ of polynomials in ${\bf A}$, with 
coefficients 
involving a
second alphabet ${\bf B}$ of the same cardinal, 
 as a module over ${\rm Sym}({\bf A})$. 

We define a quadratic form :
\begin{equation}
\label{rp-1}
{\cal P}{\sl ol}({\bf A}) \ni f, g \Rightarrow (f, g) := fg\,  \partial_\omega \ , 
\end{equation}
where $\partial_\omega$ is the maximal divided difference relative to ${\bf A}$.

\begin{lem}
\label{rp-2}
The divided differences $\partial_i$, $i=1, \ldots, n-1$, are self-adjoint
with respect to $(\ , \ )$. Therefore, for 
 $\mu\in{\rm S}({{\bf A}})$,  $\partial_\mu$ is adjoint to
$\partial_{\mu^{-1} }$.

Moreover $\mu$ is adjoint to $(-1)^{\ell(\mu)} \mu^{-1}$.
\end{lem}

{\em proof:}

Given any $i <n$, factorize $\omega = \sigma_i\, \mu$.
Then $( f\,  \partial_i, g) = \bigl( f\partial_i\, g \,  \partial_i \bigr) \partial_\mu$.
Because $f\, \partial_i$ is a scalar for $\partial_i$, the preceeding expression
is equal to 
$$  \bigl( (f\partial_i) \, (g\partial_i) \bigr)\partial_\mu $$
and the symmetry between $f$ and $g$ implies that it is also
equal to $( f \, , \,  g\partial_i )$. By product, one gets the assertion
for every $\partial_\mu$. For$(f\, , \, g\partial_i)$=$(f\partial_i, , \, g)$, 
we have
$$(f\partial_{\mu}, g)=(f\partial_{i_1}\cdots \partial_{I_n}, g)
=(f\partial_{i_1}\cdots \partial_{i_{n-1}}, g\partial_{i_n})=\ldots
=(f, g\partial_{i_n}\cdots \partial_{i_1})=(f, g\partial_{\mu^{-1}})$$


Similarly $(f\mu \, , \, g) = f^\mu\, g\, \partial_\omega = f\, g^{\mu^{-1}} \, 
\mu\partial_\omega =
(-1)^{\ell(\mu)} f\, g^{\mu^{-1}} \, \partial_\omega =
(-1)^{\ell(\mu)} (f\, , \,  g\mu^{-1} ) $ .

{\em remark:}The third equation is by (\ref{e-2.7}), where
$$f\partial_{\omega}=\sum_{\mu \in G({\bf A})}(f/\Delta)^{\mu}$$
and we have
$$f\cdot g^{\mu^{-1}} \cdot {\mu} \cdot \partial_{\omega}
=\sum_{\nu \in G({\bf A})}(f \cdot g^{\mu^{-1}} \cdot \mu/\Delta)^{\nu}
=\sum_{\nu \in G({\bf A})}(f \cdot g^{\mu^{-1}} \cdot \mu/\Delta \mu)^{\nu}
(-1)^{l(\mu)}$$
$$=\sum_{\nu \in G({\bf A})}(f \cdot g^{\mu^{-1}}/\Delta)^{\nu \cdot \mu}
(-1)^{l(\mu)}$$
$$=(-1)^{l(\mu)}\cdot f \cdot g^{\mu^{-1}}\partial_{\omega}$$
 \quad\qed

\smallskip
The {\it code} of a permutation $\mu\in G(n)$ is the vector
$c[\mu]=[c_1, \ldots , c_n]$ such that $c_i:= card(j>i, \mu_j<\mu_i)$.
It is immediate that $\mu \longrightarrow c[\mu]$ is a bijection
between $G(n)$ and the vectors $c \in{{\bf N}}^n$ such that 
$c\leq\rho :=[n-1, \ldots , 0, 1]$
(i.e. $c_i \leq n-i$, $i=1, \ldots , n)$.

\vskip 3mm
Now we will give another definition of Schubert polynomial in terms
of permutation.
The {\it maximal} Schubert polynomial 
$X_\omega({{\bf A}}, {{\bf B}})$, that we choose to index by the permutation 
$\omega$
($:= n\ldots 1$) , is set to be 
\begin{equation}
\label{rp-3}
X_\omega({{\bf A}}, {{\bf B}}) := \prod_{i+j\leq n} \, (a_i-b_j) \ .
\end{equation}
For any permutation ${\mu}\in G(n)$, one defines the 
{\it Schubert polynomial } 
$X_\mu$ by 
\begin{equation}
\label{rp-4}
X_\mu({{\bf A}}, {{\bf B}}) := X_\omega({{\bf A}}, {{\bf B}})\, 
\partial_{\omega\mu}\, 
\end{equation}
recalling that divided differences act only the $a_i$'s.

A permutation is said to be {\it dominant} if its code $I$ is dominant, 
i.e. weakly decreasing. In that case, it is easy to see that there 
exists at least one sequence of operations of the type
$$ (\cdots j+1 \ j \cdots) \longrightarrow (\cdots j \ j \cdots)$$
which leads from $\rho$ to the code $I$.
This means that there exists at least a chain 
$$ \mu^0=\omega, \, \mu^1, \ldots , \,  \mu^r=\mu$$ 
of dominant permutations, together with a sequence of simple
transpositions $\sigma^1, \ldots , \sigma^r$, such that 
$$X_{\mu^{j-1}} \partial_{\sigma^j} = X_{\mu^j} \ , \quad j=1, \ldots , r \ .$$
Supposing that $X_{\mu^{j-1}}$ is equal to the polynomial
$Y_K$ defined in ($\ref{s-**}$), with 
$K = code(\mu^{j-1}) = [k_1, \ldots , k_n]$, let $p$
be the integer such that $\sigma^j= \sigma_p$. 
Now, because $k_p= k_{p+1}+1$, the polynomial $Y_K$ is
equal to a polynomial $F$ symmetrical in 
$a_p, a_{p+1}$, multiplied by a single factor 
$(a_p-b_m)$, for some $m$. Thus the image of
$Y_K$ under $\partial_p$ is equal to $F$, and finally 
$X_{\mu^j}=F$ is the dominant polynomial indexed by
the code of $\mu^j$.



\smallskip 
The following property shows that the maximal Schubert polynomial
is a "reproducing kernel", modulo the identification 
(that we note by "$\equiv $") of the symmetric
functions in ${\bf A}$ with the same 
 symmetric functions in ${\bf B}$.

\begin{pro}
\label{rp-5}
For any element $f\in{\cal P}{\sl ol}({\bf A})$, one has 
$$ \bigl( f({{\bf A}}) , X_\omega( {{\bf A}}^\omega, {{\bf B}}) \bigr) \equiv 
(-1)^{l(\omega)}f({{\bf B}})\ , $$
modulo the relation ${\rm Sym}({{\bf A}}) = {\rm Sym}({{\bf B}})$.
\end{pro}

{\em Proof.} The formula is linear. Writing any monomial 
 as a Schur function (according to (\ref{syf-16})):
$$S_{ij\ldots hk}({{\bf A}}, \, {{\bf A}}-a_1, \ldots , \,  {{\bf A}}-a_1-\cdots -a_{n-
2}, \, 
{{\bf A}}-a_1-\cdots -a_{n-1}) $$
we see that the expansion of such a function involves only 
monomials $a^J$ with $J\leq(n-1, \ldots , 1, 0):= \rho$, 
as we already have noticed in section 2.
In other words, any element of ${\cal P}{\sl ol}({\bf A})$ can be written as a linear 
combination of monomials $a^J : J\leq\rho$ with coefficients in
${\rm Sym}({{\bf A}})$. On the other hand, the expansion of $X_\omega({{\bf A}}^\omega, 
{{\bf B}})$
involves only monomials $a^I$: $I\leq(0, 1, \ldots , n-1)$. 
Therefore, the product $f({{\bf A}}) X_\omega({{\bf A}}^\omega, {{\bf B}})$ involves 
only 
monomials $a^H$: $H\leq(n-1, \ldots , n-1)$. Such a monomial 
has a non zero image under $\partial_\omega$ iff all the exponents are
different, (*) i.e. iff the exponent is a permutation of $\rho$; in this
case, the determinant is obtained by permutations of columns from
the determinant expressing 
 $S_{0\ldots 0}({{\bf A}})$(**) and thus is equal to $ \pm 1$. 

In any case, this image is independent of ${{\bf A}}$. Consequently, 
taking ${{\bf A}}={{\bf B}}$ in the evaluation of the quadratic form 
$\bigl( f({{\bf A}}), X_\omega({{\bf A}}^\omega, {{\bf B}}) \bigr)$ has the only effect 
of replacing 
symmetric functions in ${{\bf A}}$ by the same functions in ${{\bf B}}$. 
On the other hand, in the second expression (\ref{e-2.7}) 
of the scalar product 
$$\bigl( f({{\bf A}}), X_\omega({{\bf A}}^\omega, {{\bf B}}) \bigr)
= \sum\nolimits_\mu (-1)^{\ell(\mu)} f({{\bf A}}^\mu) 
X_\omega({{\bf A}}^{\omega\mu}, {{\bf B}})\,  /\, \Delta({{\bf A}}) $$
there remains only one term under the specialization
${{\bf A}}={{\bf B}}$, that is for $\mu= identity$, the term
$f({{\bf B}}) X_\omega({{\bf B}}^\omega, {{\bf B}}) / \Delta({\bf B}) = 
(-1)^{l(\omega)}f({{\bf B}})$, for $X_\omega({\bf B}^\omega, {\bf B})=
\Delta({\bf B}^{\omega})=(-1)^{l(\omega)}\cdot \Delta({\bf B}), $
which completes the proof. \quad \qed

{\em Remark:}
we wil give an explanation for (*) and (**).
$$(*)$$
We can get it by (\ref{e-2.7}).
$$a^H\partial_{\omega}=\sum_{{\mu}\in G({\bf A})}(a^H/\Delta)^{\mu}$$
if $h_i=h_j$, then $(a^H)^{{\mu}_0}=a^H$, where $\mu_0=(i, j)$.Now
$$(a^H/\Delta)^{\mu_0}=(a^H)^{\mu_0}/\Delta{\mu_0}=(-1)^{\mu_0}a^H/\Delta
=-a^H/\Delta.$$
Thus
$$a^H \partial_{\omega}=\sum_{\mu\in G({\bf A})}(a^H/\Delta)^{\mu_0\cdot\mu}
=-\sum_{\mu\in G({\bf A})}(a^H/\Delta)^{\mu}$$
then $a^H\partial_{omega}=0$.
$$(**)$$
For every $H=\rho\cdot\mu$, $\mu\in G(n)$, we have $a^H=S_{H^{\omega}}({\bf A}_n, 
{\bf A}_{n-1}, \cdots, {\bf A}_1).$ Thus we have the following expansion:

$$\left|
\begin{array}{cccc}
S_{h_n}({\bf A}_n) & S_{h_{n-1}-1}({\bf A}_{n-1}) & \cdots & S_{h_1+n-1}({\bf A}_1) \\
S_{h_n-1}({\bf A}_n) & S_{h_{n-1}}({\bf A}_{n-1}) & \cdots & S_{h_1+n-2}({\bf A}_1) \\
\vdots & \vdots & \vdots & \vdots \\
S_{h_n-n+1}({\bf A}_n) & S_{h_{n-1}-n+2}({\bf A}_{n-1}) & \cdots & S_{h_1}({\bf A}_1) \\
\end{array}
\right|$$


and we get another expansion after applying $\partial_{\omega}$ on it:
$$\left|
\begin{array}{cccc}
S_{h_n}({\bf A}_n) & S_{h_{n-1}}({\bf A}_n) & \cdots & S_{h_1}({\bf A}_n) \\
S_{h_n-1}({\bf A}_n) & S_{h_{n-1}-1}({\bf A}_n) & \cdots & S_{h_1-1}({\bf A}_n) \\
\vdots & \vdots & \vdots & \vdots \\
S_{h_n-n+1}({\bf A}_n) & S_{h_{n-1}-n+1}({\bf A}_n) & \cdots & S_{h_1-n+1}({\bf A}_n) \\
\end{array}
\right|$$

Thus the index in every row and every column has just one zero. Now we
can put all the $S_0({\bf A}_n)$ at the diagonal by permutation of column
from above. Then we get $S_{0^n}({\bf A}_n)$. Finally we know that $a^H\partial_{\omega}$
is equal to $\pm 1$.This means that $(f({\bf A})\cdot X_{\omega}({\bf A}^{\omega}, 
{\bf B}))\partial_{\omega}\in Sym({\bf A})$.

\vspace{5mm}
More generally, one can obtain divided differences by evaluating 
scalar products, as shown by the following proposition.

\begin{thm}
\label{rp-6}
For any element $f\in{\cal P}{\sl ol}({\bf A})$, any permutation $\mu$, 
one has 
$$ \bigl( f({{\bf A}}) , X_{\mu\omega}( {{\bf A}}^\omega, {{\bf B}}) \bigr) \equiv
(-1)^{l({\omega})+l({\mu})}f\, \partial_{\mu^{-1}}\, ({{\bf B}})\ $$
 \end{thm}

{\em Proof.}
One has 
$$X_{\mu\omega}( {{\bf A}}^\omega, {{\bf B}})= X_\omega({{\bf A}}, {{\bf B}})\,  
\partial_{\omega\mu\omega}\, \omega = 
X_\omega({{\bf A}}, {{\bf B}})\omega \, \bigl( \omega \partial_{\omega\mu\omega}\, 
\omega \bigr) =
X_\omega({{\bf A}}, {{\bf B}})\, \omega \, (-1)^{l(\mu)}\partial_\mu \ ,  $$
 and one can shift the divided difference
$\partial_\mu $ to the left of $(\, , \, ) $ thanks to ($\ref{rp-2}$).
 Now the formula becomes
the preceeding one for the function $f\, \partial_{\mu^{-1}}$ \quad \qed

\begin{cor}
\label{$Rp7$}

The quadratic form $(\, , \, )$ is non-degenerate. 
The space ${\cal P}{\sl ol}({\bf A}) $ is a free module over ${\rm Sym}$
with the two adjoint bases 
$\{ X_\nu({{\bf A}}, {{\bf B}}) \}$ and $\{(-1)^{l(\omega)+l(\nu)}X_{\nu\omega}({{\bf A}}^\omega, {{\bf B}}) 
\}$, for all permutations
$\mu, \nu$ in ${\rm Sym}({{\bf A}})$. 

More precisely 
$$ \bigl( X_\nu({{\bf A}}, {{\bf B}}) \, , \,  
(-1)^{l(\omega)+l(\mu)}X_{\mu\omega}({{\bf A}}^\omega, {{\bf B}}) \bigr) = \delta_{\mu, \nu} \ .$$ 
\end{cor}

\smallskip 
Having the basis adjoint to the Schubert polynomials, 
 one can now expand any polynomial $f$ in ${\cal P}{\sl ol}({\bf A})$
as a linear combination of 
Schubert polynomials :
\begin{equation} 
\label{rp-8}
f({{\bf A}}) \equiv \sum_\mu \bigl( f\, , \, 
(-1)^{l(\omega)+l(\mu)}X_{\mu\omega}({{\bf A}}^\omega, {{\bf B}})\bigr)
\, X_\mu({{\bf A}}, {{\bf B}}) 
\end{equation}
$$\qquad =  \sum_\mu f\, \partial_{\mu^{-1}}({{\bf B}}) \, 
X_\mu({{\bf A}}, {{\bf B}}) \ . $$

The preceeding formula uses the identification 
${\rm Sym}({{\bf A}}) ={\rm Sym}({{\bf B}})$. When $n$ is sufficiently big 
so that $f$ belongs to the space generated by monomials 
$a^I $, $I\leq (n-1, \ldots, 1, 0)$, then the symmetric functions
$\bigl( f\, , \, X_{\mu\omega}({{\bf A}}^\omega, {{\bf B}})\bigr)$ are of degree 0 in 
${{\bf A}}$, 
and thus the formula is exact in that case and coincide 
with Newton's interpolation. 

\smallskip
One can use ($\ref{rp-8}$) to develop a Schubert polynomial itself.
The following formula is a generalization of the expansion of the
resultant $R({{\bf A}}, {{\bf B}})$ as a sum of products 
$\pm S_I({{\bf A}}) S_{I'}({{\bf B}})$ due to Cauchy.

\begin{thm}
\label{rp-9}
Let ${{\bf A}}, {{\bf B}}, {{\bf C}}$ be three alphabets of cardinal $n$, and let
${\zeta}\in G(n)$. Then 
$$X_{\zeta}({\bf A}, {\bf C}) = \sum_{\mu, \eta} X_{\mu}({\bf A}, {\bf B})\, 
X_{\eta}({\bf B}, {\bf C}) \ , $$
sum on all reduced factorizations 
$\zeta= \mu\eta$ (i.e. factorizations such that 
$\ell(\mu) + \ell(\eta) = \ell(\zeta)$).
\end{thm}

{\em Proof.} To have the non nullity of $X_\zeta \partial_{\eta^{-1}}$ is the same 
as to require that the product $(\zeta \eta^{-1}) (\eta)$ be reduced. 
Taking into account that $X_\zeta({{\bf A}}, {{\bf C}})$ belong to the span
of monomials in ${{\bf A}}$ of degree $\leq\rho$, 
the theorem follows from ($\ref{rp-8}$) \quad\qed

All the operators that we have used are linear combinations of permutations, 
with coefficients which are rational functions of the $a_i$'s. 


For example, 
$$\partial_1 = {1\over a_1 -a_2} - \sigma_1 {1\over a_1-a_2}$$
$$\partial_1\partial_2 = ({1\over a_1 -a_2} - \sigma_1 {1\over a_1-a_2})
( {1\over a_2 -a_3} - \sigma_2 {1\over a_2-a_3} ) \qquad\qquad$$
$$\qquad \qquad = (1-\sigma_1) {1\over(a_1-a_2)(a_2-a_3)} +
( 1-\sigma_2) {1\over(a_1-a_3) (a_2-a_3) }  $$

All the computations shown in these notes could be performed
in the algebra of the symmetric group with rational coefficients
in ${{\bf A}}$. Let us call this algebra the {\it algebra of divided
differences}, not to confuse it with the group algebra with constant
coefficients. In it we have to use the commutations relations
($P$ denoting the operation "multiplication by $P$")~:
\begin{eqnarray*}
 P\partial_i &=\ &(P\partial_i) + \partial_i \, P^{\sigma_i} \\
     P\sigma_i &=\ &\sigma_i\, P^{\sigma_i} 
\end{eqnarray*}

The next proposition shows that we already know how to pass from permutations 
to divided differences, and conversely.

\begin{pro}

Let ${{\bf A}}=\{a_1, \ldots , a_n \}$, 
$\Delta:= \prod_{i<j} (a_i - a_j)$. then for any $\mu\in\S({{\bf A}})$, one has 
\begin{eqnarray*}
  \mu &= &\sum_{\nu} \partial_\nu X_{\nu^{-1}}({{\bf A}}^\mu{{\bf A}})  
\\
  \partial_\mu&= &\sum_{\nu}\nu X_{\omega\mu} 
({{\bf A}}, {{\bf A}}^{\omega\nu} ) 
\end{eqnarray*}
\label{rp-10}
\end{pro}

{\em Proof.} Newton's interpolation ($\ref{rp-8}$) states that for any polynomial $f$, 
any $\mu$, one has 
$$ f({{\bf A}}^\mu) \equiv f\partial_\nu({{\bf B}}) \, X_{\nu^{-1}}({{\bf A}}^\mu, 
{{\bf B}}) \ .$$
The left side can be interpreted as the operator 
$f({{\bf A}}) \longrightarrow f({{\bf A}}^\mu)$, that is $\mu$, while the right side, 
where, of course, we can put ${{\bf A}}={{\bf B}}$ because it does not affect the left 
side, 
is the operator $\sum \partial_\nu X_{\nu^{-1}}({{\bf A}}^\mu{{\bf A}}) $. This is 
exactly 
the first part of the claim.

Similarly, ($\ref{rp-6}$) states that $f\partial_\mu$, evaluated in $B$, is congruent to
$\sum f({{\bf A}}^\nu) X_{\mu^{-1} \omega}({{\bf A}}^{\omega\nu}, {{\bf B}}) / 
\Delta({{\bf A}}^\nu)$, and this is, 
for ${{\bf A}}={{\bf B}}$, the second part of the claim, taking into account that
$$ X_\nu({{\bf A}}, , {{\bf B}}) = (-1)^{\ell(\nu)} X_{\nu^{-1}}({{\bf B}}, {{\bf A}}) 
\ , $$
the sign being swallowed by 
$\Delta({{\bf A}}^\nu) = (-1)^{\ell(\nu)} \Delta({{\bf A}})$ \quad\qed

\smallskip
For example, for $\mu= \sigma_1 \sigma_2$, one has ${{\bf A}}^{\sigma_1 \sigma_2} =
\{a_3, a_2, a_1 \} :={{\bf A}}' $  , and 
$$ \sigma_1 \sigma_2 = 1 + \partial_1 X_{213}({{\bf A}}', {{\bf A}}) + \partial_2 
X_{132}({{\bf A}}', {{\bf A}})
+ \partial_1\partial_2 X_{312}({{\bf A}}', {{\bf A}}) $$
$$ = 1+ \partial_1(a_3-a_1) + \partial_2 (a_3+a_2 -a_1-a_2) + 
\partial_1\partial_2 (a_3-a_1) (a_3-a_2) \ .$$
Symmetrically, 
$$\partial_1 \partial_2 \Delta = 1 X_{213}({{\bf A}}, {{\bf A}}^\omega) + \sigma_1 
X_{213}({{\bf A}}, {{\bf A}}^{\omega\sigma_1 }) +
\sigma_2 X_{213}({{\bf A}}, {{\bf A}}^{\omega\sigma_2 }) +\sigma_1\sigma_2 
X_{213}({{\bf A}}, {{\bf A}}^{\omega\sigma_1\sigma_2 }) $$
$$= 1(a_1-a_3) + \sigma_1 (a_1-a_3) + \sigma_2 (a_1-a_2) 
+ \sigma_1 \sigma_2 (a_1-a_2) \ . $$

\smallskip
Without Schubert polynomials, it would have been more difficult
to obtain ($\ref{rp-10}$) in the algebra of divided differences
(cf. [KK ?]).

For example, to compute $\partial_\mu$, one has to start from a reduced decomposition
$\mu= \sigma_i \sigma_j \cdots \sigma_k$, and write accordingly 
$$ \partial_\mu = (1-\sigma_i) {1\over a_i -a_{i+1} } 
+(1-\sigma_j) {1\over a_j -a_{j+1} }+ \cdots +
(1-\sigma_k) {1\over a_k -a_{k+1} }  \ . $$
Now, one has to push all the coefficients either on the right, or on the left
according to one's preferences, and sum all terms
$\sigma_i^{\varepsilon_1} \sigma_j^{\varepsilon_2} \cdots$, $\varepsilon \in\{0, 1 \}$, 
which are decompositions
of the same permutation. 

With this method, it is not appearant that the entries of the inverse matrix
which expresses the $\mu$ in terms of the $\partial_\nu$ are the same, 
up to multiplication by $\Delta$ and twist of indices.
As a by-product however, one gets a relation between Schubert polynomials and 
the Ehresmann-Bruhat order on the symmetric group. 

\smallskip
This order is defined by :\hfill\break
$\nu\leq\mu$ iff, given an arbitrary reduced decomposition of 
$\mu $, $\mu= \sigma_i \sigma_j \cdots$, then there exists at least one choice of
$\varepsilon$ such that 
$\sigma_i^{\varepsilon_1} \sigma_j^{\varepsilon_2} \cdots = \nu$. 

\smallskip
In other words, we have just observed that in the expansion of 
$ \partial_\mu = (1-\sigma_i) {1\over a_i -a_{i+1} } 
 +(1-\sigma_j) {1\over a_j -a_{j+1} }+ \cdots $, and of 
$\mu= (1+\partial_i (x_{i+1}-x_i)) (1+\partial_j(x_{j+1}-x_j) \cdots  $, 
only permutations $\nu\leq\mu$ appear.

Comparing with ($\ref{rp-10}$), one obtains the implication
$\Leftarrow$ of the following lemma~:

\begin{lem}
\label{ref-11}
For any pair of permutations $\nu$, $\mu$, one has the equivalence
$$ \nu\leq\mu \ \Leftrightarrow \ X_{\nu^{-1}}( {{\bf A}}^\mu, {{\bf A}}) \neq0 
\ .$$
\end{lem} 


\vskip 5mm

Given two alphabets ${{\bf A}}$, ${{\bf B}}$ of cardinal $n$, 
the ring $K[{{\bf A}}]$, with $K={{\bf Z}}[{{\bf A}}]$, 
is a free-module over ${\rm Sym}({{\bf A}})$. Its Schubert basis 
$X_\mu({{\bf A}}, {{\bf B}})$ is in bijection with permutations $\mu$ in 
$G(n)$. The adjoint basis, with respect to the
scalar product $(f, g):= fg\partial_\omega$, 
is $\{(-1)^{l(\omega)+l(\mu)}X_{\mu\omega}({{\bf A}}^\omega, {{\bf B}}) \}$.

\pagebreak

\section{EXERCICES}

\begin{xex}  Suppose $m, n$ be two integers, $F$ be a symmetric function of 
$n$ variables which belong to the space generated by Schur functions $S_I$, $I\subseteq m^n$. 
${\bf{A}}$ be of cardinal $m+n$, and ${\bf{X}}$ be of cardinal $n$. 

Show that 
$$\sum_{{\bf{B}}\subset{\bf{A}}, card({\bf{B}})=n}  F({\bf{B}}) {R({\bf{X}}, {\bf{A}}-{\bf{B}}) \over R({\bf{B}}, {\bf{A}}-{\bf{B}}) } = F({\bf{X}})$$

\end{xex} \begin{xex}
Suppose ${\bf{A}}$ be of cardinal $n$. 
Show that $u_k := L_{\bf{A}}(x^k)$, $k\in{\bf{Z}}$ form a 
linear recurrent series of order $n$, i.e., if there exist 
constants $c_0=1, c_1, \ldots, c_n$ such that 
$$ \forall k\in{\bf{Z}}, \  c_0u_k + c_1 u_{k-1}+\cdots c_n u_{k-n} =0 \ .$$

What are these $c_i$?

It's enough to calculate the $n+1$ values $u_0, \ldots, u_{n-1}$. 
What are they? Calculate the function $\sum_{k\geq0} z^k u_{-k}$.


\end{xex} \begin{xex}
Let ${\bf{A}}$ be the isometric points: 
$0, 1/n, 2/n, \ldots, n/n$. 
Write down the corresponding Lagrange interpolation. 
% \sum f(i/n) {nx\choose i } {n-nx \choose n-i}

\end{xex} \begin{xex}
If we take ${\bf{A}}$ the $n$-th root set of unit. 
Show that every polynomial $f$ of degree $<n$ can be written as 
$$f(x) = \sum\nolimits_{a\in{\bf{A}}}f(a) (x^n-1)/(x/a-1) \ . $$
% PolyaII Ex 74 p.83 R(a, {\bf{A}}\setminus a) = n a^{n-1}

\end{xex} \begin{xex}
What form the Lagrange interpolation becomes when we 
specialize the interpolation points:

$a_2=a_1$, $a_4=a_3$, $\ldots$?

\end{xex} \begin{xex} Suppose $M$ be a matrix of order $n+1$
whose eigenvalues are distinct, and Suppose ${\bf{A}}$ be the set of 
these eigenvalues. Suppose $f$ a polynominal of degree $n+1$. 

With the aid of Lagrange interpolation, without using the theorem of Cayley-Hamilton, 
give a polynomial $g$ of degree $\leq n$
such that $g(M)= f(M)$. 

%%%%%%%%%%%%%%%%%%%%% 2eme serie
\end{xex} \begin{xex}  Suppose $n$ be an integer, 
$F(x, y)= \sum_{0\leq i, j\leq n-1} f_{ij}x^iy^j$ be a polynomial of two variables, 
 $\{y_1, \ldots, y_n \}$ be an alphabet of cardinal $n$. 
Let $\Delta $ denote the Vandermonde determinant, Show that the sum 
$$\Delta(y_1, \ldots, y_n))F(x, x) - \Delta(x, y_2, \ldots, y_n) F(x, y_1) +\cdots
+(-1)^n \Delta(x, y_1, \ldots, y_{n-1}) F(x, y_n)$$
is zero. 

%\ptisaut
% Muir(1909): Muir V p.198.  Let the alphabet 
%$A:= \{x, y_1, \ldots, y_n \}$. The sum en cause is the value for
%$z=x=a_1$ of 
%$$\Delta(A)  \sum_{a\in A} F(z, a)/R(a, A- a) = \Delta(A) L_A\bigl( F(z, a_1) \bigr) \ .$$
%This function is ???bien nulle puisque??? $F(z, y)$ is of degree $\leq n-1$ on $y$ 

%Ansi, for $n=3$, we have 

%$\Delta(y_1, y_2, y_3) F(x, x) - \Delta(x, y_2, y_3) F(x, y_1) +
%\Delta(x, y_1, y_3) F(x, y_2) \hfill $

%$\hfill - \Delta(x, y_1, y_2) F(x, y_3) =0 $ \quad $\Box$

\end{xex} \begin{xex}
Suppose ${\bf{A}}$ be of cardinal $n$,  $x=a_1$ and
$P$ be a polynomial of one variable.
Prove that the coefficient of $x^j$, $0\leq j\leq n-1$ of $P(x)$ modulo $R(x, {\bf{A}})$ 
is $$ |\,  1 \,  a \ldots a^{j-1} \, P(a)\, a^{j+1}\ldots a^{n-1}\, 
|\,  / \, \Delta({\bf{A}}) \ .$$

%\ptisaut
%Test the assertion with the base $\{x^0, x^1, \ldots, x^{n-1}, 
%R(x, {\bf{A}}), xR(x, {\bf{A}}), x^2R(x, {\bf{A}}), \ldots \}$

\end{xex} \begin{xex}
Suppose $P\in$ Sym$(1|n)$, ${\bf{A}}$ be of cardinal $n+1$. 
Show that 
$$\{u_k:= L_{\bf{A}}\bigl( x_1^k P(x_1, x_2, \ldots, x_{n+1}) \bigr) 
 \}_{k\in{\bf{N}}}$$
form a recurrent series with characteristic polynomial $R(x, {\bf{A}})$.

\setbox2=\hbox{the cardinal of ${\bf{B}}$ is arbitrary, }
Discuss the following particular cases:
$$\begin{array}{rlc}
P(x_1, x_2, \ldots, x_{n+1}) &= 1 \\
  &= x_1^{n+1} x_2 \cdots x_{n+1}   \\
  &= R(x, {\bf{B}}) & \box2 \\
 &= S_j(x-{\bf{A}}) \\  
 \end{array}$$

%\ptisaut
%$\{u_k \}$ is a recurrent series of characteristic polynomial
%$R(x, {\bf{A}})$ ?? ssi ?? 
%$$ k\geq0 \Rightarrow\ u_{k+n+1}S_0(-{\bf{A}})+\cdots + 4u_{k+1}S_{n}(-{\bf{A}})+
%u_k S_{n+1}(-{\bf{A}})= 0 \ .$$

%In the case where $P $ is an multiply of $x$, that is $P= x^j$, 
%$j\geq0$, the ?? propri\'et\'e ?? we wanted is
%$$L_{\bf{A}}\Bigl( x^{k+n}S_0(-{\bf{A}})+\cdots + x^{k+1}S_{n-1}(-{\bf{A}})+
%x^k S_k(-{\bf{A}}) \Bigr)= 0 \ . $$

%Since the initial sum is $S_{k+n}(x-{\bf{A}})$, 
%it's image of operator $L_{\bf{A}}$ is 
%$ S_{k+1}({\bf{A}}-{\bf{A}}) $ which is just zero when $k\geq0$.

%The nullity is still true for 
% all element $P$ of Sym$(1|n)$, because the ??ofrnier?? 
%can be written as a linear combination 
%(with coefficients in Sym$(n+1)$) of multiply of $x$.

%
%In the case of $P=1$, we find that $u_k= S_{k-n}({\bf{A}})$, 
%i.e. the series \hfill\break
%$\{0, \ldots, 0, 1, S_1({\bf{A}}), S_2({\bf{A}}), \ldots \}$
%??dont?? the first $n-1$ values are nulle, and the following
%is equal to 1.

%In the second case, we put a factor $x_1\cdots x_{n+1}$, 
%and then we find the same recurrent series, with movement of $n$
%in the {zhi, shu} and 
%?? au facteur $a_1\cdots a_{n+1}$ pr\`es ??.

%Rewrite $x^k R(x, {\bf{B}}) = S_{m+k}(x-{\bf{B}})$, where $m$ is the
%cardinal of ${\bf{B}}$, we obtain ?? in the case ??
%$u_k= S_{m+k-n}({\bf{A}}-{\bf{B}})$. 

%In the case where $m\leq n$, 
%the series $\sum_{k\geq0} u_k z^k$ is equal to
%$$ z^{n-m} \prod_{b\in{\bf{B}}}(1-zb) / \prod_{a\in{\bf{A}}}(1-za)  $$

%Finally, we brief the calculate of ?? ofrnier ?? case by write
%$x^k S_j(x-{\bf{A}})$ as the function of Schur
%$$ (-1)^j S_{1^j, k }({\bf{A}}, x)  $$
%so the image by $L_{\bf{A}}$ is the **shang san jiao** function 
%$$(-1)^j S_{1^j, k-n }({\bf{A}})\ .$$
%Beside this, we can give the ?? \'equerre ?? function
%compare the coefficients of restes dans la division euclidienne ofs $x^k$ par 
%le polyn\^ome $R(x, {\bf{A}})$.

\end{xex} \begin{xex} 
Suppose ${\bf{A}}^{-1}$ be the alphabet of the inverse 
of the letters in the alphabet ${\bf{A}}$ of cardinal $n+1 $. 
Show that the Lagrange operator associated with ${\bf{A}}^{-1}$ is the operator
$L_{{\bf{A}}^{-1} } : f \rightarrow (-1)^n L_{\bf{A}} (f\cdot a_1^n a_2\cdots a_{n+1} )$.

%\ptisaut
%L'op\'erateur $L_{{\bf{A}}^{-1}}$ est l'op\'erateur Sym$({\bf{A}})$-lin\'eaire
%qui envoie $\{a_1^0, \cdots , a_1^{n-1} \}$ sur 0 et
%$a_1^{-n}$ sur 1. Or 
%$L_{\bf{A}}(a_2\cdots a_{n+1}) = (-1)^n$  \quad$\Box$


%Remarque: \ L'op\'erateur $L_{{\bf{A}}^{-1} }$ est \'egal au produit
%$\partial'_1 \cdots \partial'_{n-1}$, o\`u $\partial'_i$ est la diff\'erence divis\'ee
%relative \`a la paire of variabthe $1/a_i, 1/a_{i+1}$, 
%d'o\`u $\partial'_i = -\partial_i (a_i a_{i+1})$.

\end{xex} \begin{xex}
Suppose ${\bf{A}}$ be of cardinal $n$. Show that 
$$L_{\bf{A}}\bigl({ x^n\over 1+x} R(1/x, A) \bigr) $$
 is equal to $(-1)^{n-1} (1- a_1\cdots a_n)$

%\ptisaut ( Polya II p.82 Ex 69) 
%La function initiale est 
%$\prod_{a\in{\bf{A}}} (1-xa)/(1+x)$. On peut calculer son image par $L_{\bf{A}}$
%en ordonnant l'alphabet et en 
%utilisant que $L_{\bf{A}} = \partial_1\cdots \partial_{n-1}$, apr\`es
%avoir pos\'e $x=a_1$. Seuls ofux termes du d\'eveloppement 
%of $(1-a_1)(1-a_1a_2)\cdots (1-a_1a_n)$ ont par 
%$L_{\bf{A}}$ une image non nulle, \`a savoir (pour n=6 par exemple)
%$$(-1)^3 a_3a_2 a_1^{3} \longrightarrow (-1)^3$$
%$$(-1)^6 a_6\cdots a_2 a_1^6 = (a_6\cdots a_1) a_1^5 \longrightarrow (a_6\cdots a_1) $$
%d'o\`u la formule cherch\'ee.

%Une autre m\'ethoof est d'utiliser l'alphabet 
%${\bf{B}} := \{1/a_1, \ldots , 1/a_n \}$. 
%D'apr\`es l'exercice pr\'ec\'eofnt, on a pour toute function $f$ d'une variable
%$$ L_{\bf{B}}( f(1/b_1) = (-1)^{n-1} a_1\cdots a_n 
%L_{\bf{A}}\bigl( a_1^{n-2} f(a_1) \bigr) \ .$$
%L'expression ofmand\'ee est donc 
%$$(-1)^{n-1} {1\over a_1\cdots a_n}  
%L_{\bf{B}}\Bigl( {R(b_1, {\bf{A}})\over b_1(1+b_1)  } \Bigr) \ . $$
%Or $(1/b_1) R(b_1, {\bf{A}}) = (1/b_1) S_n(-{\bf{A}}) + S_{n-1}(b_1-{\bf{A}})$ et
%$b_1^i R(b_1, {\bf{A}}) = S_{n+i}(b_1 -{\bf{A}})$ pour $i\geq0$. 

%L'image par $L_{\bf{B}}$ of la somme $S_{n-1}(b_1-{\bf{A}}) -S_{n}(b_1-{\bf{A}}) +\cdots$
%est \'egale \`a 
%$$S_0({\bf{B}}-{\bf{A}})- S_1({\bf{B}}-{\bf{A}}) +S_2({\bf{B}}-{\bf{A}}) -\cdots = 
%{\prod(1+a)\over \prod(1+b) }= {\prod(1+a) \over \prod(1+1/a) } = a_1\cdots a_n \ . $$
%Le terme manquant
%$L_{\bf{B}}\bigl( (1/b_1) S_n(-{\bf{A}})\bigr)$ est \'egal \`a
%$(-1)^{n-1} S_n(-{\bf{A}})/(b_1\cdots b_n) = -(a_1\cdots a_n)^2 $, 
%ce qui confirme le r\'esultat obtenu par the diff\'erences divis\'ees.

\end{xex} \begin{xex}
Suppose ${\bf{A}}$ and ${\bf{B}}$ be two alphabets of the same cardinal $n$.
Show that 
$$1- a_1\cdots a_n = \sum_j (1-a_j) \prod_{i, i\neq j} (b_j -a_ib_i)/(b_j -b_i) $$

%\ptisaut
%(Milne ) Pour pouvoir reconna\^\i tre la formule of Lagrange, 
%il faut introduire l'alphabet ${\bf{X}}:= \{a_1b_1, \ldots, a_nb_n \}$, 
%tout en gardant l'alphabet ${\bf{B}}$. 

%La somme \`a calculer est 
%$$\sum_j {1\over b_j} {R(b_j, {\bf{X}}) \over R(b_j, {\bf{B}}- b_j) }  $$
%Cette somme est donc une combinaison lin\'eaire ofs
%$L_{\bf{B}}(b_1^j) $, $j=-1, \ldots, n-1$. Seuthe the ofux puissances
%extr\`emes donnent une contribution non nulle:
%$L_{\bf{B}}(1/b_1) = (-1)^{n-1}/(b_1\cdots b_n)$ et
%$L_{\bf{B}}(b_1^{n-1}) =1$  \quad$\Box$

\end{xex} \begin{xex}
 Relations betwen the multiply section and formal series.

Let $f:= 1+xf_1+x^2f_2+\cdots$ be a formal series, 
$f|_n := 1+\cdots x^nf_n$ be the section of $f$ of order $n$.

For every $k\in{\bf{Z}}$, using functions $f^0|_n, \ldots, f^n|_n$
represent the multiply section $f^k|_n$ .

%\ptisaut
%Cayley, Quaterly {\bf 14}(1877) 83-84. \quad 
%Pour tout $j $, le coefficient of $x^j$ dans the $f^k|_n$
%est une function polyn\^omiale of $k$, of ofgr\'e $\leq n$.
%Prenant l'ensemble d'interpolation 
%$ {\bf{N}}_n := \{0, 1, \ldots, n \}$, on a donc d'apr\`es Lagrange, 
%pour $f^k|_n$ consid\'er\'ee comme une function of $k$

% $$\eqalign{
% f^k|_n &= \sum_i { R(k, {\bf{N}}_n-i)\over R(i, {\bf{N}}_n-i) } f^i|_n \\
% &=(k-0)\cdots (k-n) \sum_i {(-1)^{n-i}\over (k-i) i! (n-i)! } f^i|_n 
% \\}$$


\end{xex} \begin{xex}
Suppose ${\bf{A}}$ and ${\bf{B}}$ be two alphabets of the same cardinal $n$, 
$\partial_{\bf{B}}:= \partial_1\ldots \partial_{n-1}$ the divided difference 
relative to ${\bf{B}}$.

Show that 

$$ (1-a_1) (1+a_n + a_n a_{n-1}+ \ldots + a_n \cdots a_2) 
 = {1\over b_1} (b_1-b_1a_1)\cdots (b_1-b_na_n)\,  \partial_{\bf{B}}  \ . $$


%\ptisaut
%Pour raisonner par r\'ecurrence sur $n$ on d\'ecompose le
%membre of droite en
%$(b_1-b_1a_1)\cdots (b_1 - b_{n-1}a_{n-1}) \partial_{\bf{B}} + 
%-{1/b_1} (b_1-b_1a_1)\cdots (b_1-b_{n-1}a_{n-1})(b_na_n) \partial_{\bf{B}}   $.

%La premi\`ere partie a pour seule contribution non nulle
%$(b_1^{n-1} -b_1^{n-1}a_1 ) \partial_{\bf{B}} = 1-a_1$. Pour la seconof, on d\'efinit
%${\bf{B}}'= \{b_1, \ldots , b_{n-1} \}$, on utilise que 
%$\partial_{\bf{B}} = \partial_{{\bf{B}}'} \partial_{n-1}$, et que, par r\'ecurrrence, 
%$${1/b_1} (b_1-b_1a_1)\cdots (b_1-b_{n-1}a_{n-1}) \partial_{{\bf{B}}'}$$
%est un scalaire en ${\bf{B}}$. Comme en outre $ b_na_n \partial_{n-1}= -a_n $, 
%on a au total
%$$(1-a_1) + a_n (1-a_1)(1+a_{n-1}+\ldots+a_{n-1}\cdots a_2) $$
%ce qui est bien le r\'esultat voulu.

%On prendra garof que contrairement \`a l'exercice pr\'ec\'eofnt, 
%on ne peut introduire un alphabet 
%$\{a_1b_1, \ldots, a_nb_n \}$ invariant par le groupe 
%sym\'etrique ${{S}}({\bf{B}})$. 


%%%%%%%%%%% 3em serie d'exercices
 
\end{xex} \begin{xex} 
What form the Newton formule becomes for the alphabet
$${\bf{A}} = \{0, a_1, -a_1, a_2, -a_2, a_3, -a_3, \ldots \}$$
in the case of an odd function 
({\it Interpolation of Stirling}) ? 


The special case ${\bf{A}}=  \{0, \varepsilon, -\varepsilon, 2\varepsilon, -2\varepsilon, \ldots \}$ 
has been used by Gauss for the integrals.
 
%\ptisaut
%the diff\'erences divis\'ees $f\partial_1\partial_2\cdots \partial_{2n} $ sont nulthe , et
%donc une function impaire s'\'ecrit
%$$f(x) = x \Bigl( f\partial_1 + f\partial_1\partial_2\partial_3 (x-a_1^2) +\cdots
%+f\partial_1\cdots \partial_{2n+1}(x-a_1^2)\cdots (x- a_n^2) + \cdots \Bigr) $$

\end{xex} \begin{xex}
 Show that the normal derive, denoted by $D_x$, 
 acting on the functions of one variable $x$
can be rewritten as:
$$D_x = \partial_1 +\partial_1 \partial_2 (x-a_2) + 
\partial_1\partial_2\partial_3 (x-a_2) (x-a_3) + \cdots \ , $$
by given an infinite alphabet ${\bf{A}} := \{x, a_2, a_3, \ldots \}$
and the corresponding divided differences .

%\ptisaut
%Newton nous donne 
%$$f(y) = f(x) + f\partial_1 (y-x) + f\partial_1\partial_2 (y-x)(y-a_2) +\cdots$$
%d'o\`u l'expression of $f'(x)$ en faisant tendre $y$ vers $x$.

%Pour avoir the d\'eriv\'ees d'ordre sup\'erieur, on peut combiner
%formule of Newton et formule of Taylor, avec toujours le
%m\^eme alphabet ${\bf{A}}$ et $y= x+\varepsilon$:
%$$f(x+\varepsilon) = f(x) + f\partial_1\, \varepsilon + f\partial_1\partial_2 \, \varepsilon(x+\varepsilon-a_2) + \cdots $$
%Ainsi le coefficient of $\varepsilon^2/2$ est
%$${1\over 2} f" = f\partial_1\partial_2 + f\partial_1\partial_2\partial_3 (x+x-a_2-a_3)
%+f\partial_1\partial_2\partial_3\partial_4\,  S_2( x+x-a_2-a_3-a_4) +\cdots   $$

%Il est plus instructif d'utiliser que the d\'eriv\'ees successives sont
%ofs limites of diff\'erences divis\'ees. Cela revient \`a introduire
%un drapeau d'alphabets :
 %${\bf{X}}_n= \{ x_1, \ldots , x_n \}$, of diff\'erences divis\'ees
%associ\'ees $\Delta_1, \Delta_2, \ldots $
%simultan\'ement au drapeau 
%${\bf{A}}_n= \{a_1, \ldots , a_n \} $.

%L'image of
% l'expression of Newton pour $f(x_1)$ par le produit
%$\Delta_1\cdots \Delta_n$ donne
%$$\Delta_1\cdots \Delta_n = \partial_1\cdots \partial_n 
%+\partial_1\cdots \partial_{n+1} S_1({\bf{X}}_n -{\bf{A}}_{n+1})
% \partial_1\cdots \partial_{n+2} S_2({\bf{X}}_n -{\bf{A}}_{n+2}) +\cdots \ , $$
%d'o\`u $1/n! $ fois la $n$-i\`eme d\'eriv\'ee en faisant tendre 
%the $x_i$ vers $x_1$. Le cas pr\'ec\'eofnt s'obtient
%en posant $x_1=x=a_1$. 



\end{xex} \begin{xex} 
Suppose $T$ ( the {\it translation operator}) act on the series :
$$ (s_0, s_1, s_2, \ldots) \longrightarrow (s_1, s_2, s_3, \ldots)$$
Suppose $f$ be a polynomial of one variable and ${\bf{A}}$ be an alphabet.
Show that the operator $f(T)$ is given by the Newton formule :
$$f(T) = f(a_1) + f\partial_1 (T-a_1) + f\partial_1\partial_2 (T-a_1)(T-a_2) + \cdots $$

%\ptisaut
%On peut remarquer que the op\'erateurs en cause 
%sont d\'etermin\'es par leur action 
%sur la suite ofs puissances d'une variable 
%$(1, x, x^2, \ldots)$ qui est la base naturelle of
%l'anneau ofs polyn\^omes en $x$.
% L'op\'erateur $T$ s'interpr\`ete alors 
%comme l'op\'erateur of multiplication par $x$, 
%et $f(T)$ comme l'op\'erateur of multiplication
%par $f(x)$. Quelle que Supposela function $f$, 
%toute d\'ecomposition lin\'eaire of $f(x)$ s'interpr\`ete 
%comme une d\'ecomposition of l'op\'erateur 
%"multiplication par f(x)", i.e. comme une d\'ecomposition
%of $f(T)$  \quad $\Box$

%Le m\^eme raisonnement montre que la formule of Lagrange
%permet aussi d'interpoler the op\'erateurs sur the suites.

%%% donner combinaison of Lagrange et Newton
%Dans le cas particulier o\`u $a_1=a_2=\cdots = c$
%la formule of Newton ofvient
%$$f(T) = f(c) + f'(c) (T-c) + {1\over 2} f"(c) (T-c)^2 +\cdots $$



%%%%%%%%% Newton 
\end{xex} \begin{xex}
 Suppose ${\bf{A}}$ be an infinite alphabet, ${\bf{A}}_k := \{a_1, \ldots, a_k \}$, 
$k\geq1$ the first $k$ alphabets .

Show that, for every $j, k, n$ we have
$$S_{j+k}({\bf{A}}_n)= S_j({\bf{A}}_1) S_k({\bf{A}}_n) + S_{j-1}({\bf{A}}_2) S_{k+1}({\bf{A}}_n - {\bf{A}}_1)
+ S_{j-2}({\bf{A}}_3) S_{k+2}({\bf{A}}_n - {\bf{A}}_2)+ \cdots $$

%\ptisaut
%Donn\'e par Newton, pour la function $x^j= S_j(x)$ en $x=1$
%et l'alphabet $\{a_1, \ldots, a_n, a_{n+1}=1 \} $ :

%$S_j(1) = S_j({\bf{A}}_1) + S_{j-1}({\bf{A}}_2)\, R(1, {\bf{A}}_1) + 
% S_{j-2}({\bf{A}}_3) \, R(1, {\bf{A}}_2)+\cdots $ \hfill\break
%\hfill $S_{j-n+1}({\bf{A}}_n)R(1, {\bf{A}}_{n-1} +S_{j-n}({\bf{A}}_n+1) R(1, {\bf{A}}_n)  $

%On divise par $R(1, {\bf{A}}_n)$ et on iofntifie the termes 
%of ofgr\'e $j+k$.

\end{xex} \begin{xex}
Given the alphabet ${\bf{N}}_n=\{0, 1, 2\ldots , n\}$ ;
Suppose $L_n$ be the corresponding Lagrange operator .

Show that $L_n(y^x) = (y-1)^n/n!$ \ .

Suppose $f$ be a function of ${\bf{Z}}$ in a ${\bf{Q}} $-vector space .
Show that 
$$\sum f(n) x^n/n! = e^x \sum x^n\,  L_n(f)$$

%\ptisaut L'iofntit\'e 
% $$y^n= y^0 + {y-1\over 1!}(n-0) + {(y-1)^2\over 2! }(n-0)(n-1) +\cdots
%=\bigl( (y-1)+1 \bigr)^n $$
%que l'on iofntifie \`a la formule of Newton pour le d\'eveloppement
%of la function $y^x$ en $x=n$ permet of reconna\^\i tre
%the valeurs requises ofs diff\'erences divis\'ees.

\end{xex} \begin{xex}
 Show that 
$${1\over z-a}= {1\over z} + {a\over z(z-1)} + {a(a-1)\over z(z-1)(z-2)}
+ {a(a-1)(a-2)\over z(z-1)(z-2)(z-3)  }+\cdots $$

%\ptisaut
%C'est la formule of Newton pour la function $1/(1-x)$
%en $x=a$ et l'alphabet $\{0, 1, 2, \ldots \}$.

%Prenant un alphabet 
%${\bf{A}}= \{a_1=a, a_2, \ldots \} $ of cardinal $r+1$, 
%et the alphabets initiaux ${\bf{N}}_n= \{0, 1, \ldots, n \}$, 
%on obtient comme image of cette iofntit\'e 
%par the diff\'erences divis\'ees en ${\bf{A}} $ 
%$${1\over R(z, {\bf{A}}) } =
%\sum_n {1\over R(z, {\bf{N}}_n) }  S_{n-r}({\bf{A}}-{\bf{N}}_{n-1}) $$

\end{xex} \begin{xex}
Suppose ${\bf{A}}$ be of cardinal $n$, and $x, y$ be two variables . 
Show that 
$$ \Big( { 1\over R(x, {\bf{A}})} - {1\over R(y, {\bf{A}})}  \Big) {1\over x-y}=
\sum_{a\in{\bf{A}}} {1 \over R(a, {\bf{A}}-a+x+y)} $$

%\ptisaut 
%Regarofr l'image par $\partial_{xy}$ of l'iofntit\'e of Lagrange
%$$1/R(x, {\bf{A}}) = \sum\nolimits_a \, 1/R(a, {\bf{A}}-a+x)$$



\end{xex} \begin{xex}
Suppose ${\bf{A}}$ be of cardinal $n$ and $L_{\bf{A}}$ be the corresponding Lagrange operator (= $\partial_1\ldots \partial_{n-1} $). 
Show that 

$$L_{\bf{A}}\Big( (z-a_1)(z+a_2)\cdots (z+a_n) (a_1+a_2)\cdots (a_1+a_n)\Big) =$$

$$\left \{ \begin{array}{ll}
      (z-a_1)\cdots (z-a_n) & \ \ \mbox{ if } n \mbox{ odd }\\
      -2 \sum z^{2n-2j-1} \Lambda^{2j+1}({\bf{A}}) & \ \ \mbox{ if } n \mbox{ even.}
      \end{array}
  \right.
$$
%%%%% Emploi of \pi_\omega
\end{xex} \begin{xex} 
 Suppose ${\bf{A}}$ be a finite alphabet, $x, y$ be two variables, 
${\bf{A}}^\clubsuit$ be the alphabet 
$\{ {1\over a-x } \}_{a\in{\bf{A}}} $. 
By induction on the cardinal of ${\bf{A}}$ and using 
the divided differences, show that 
$$ R(x, {\bf{A}})/ R(y, {\bf{A}}) = \sum_{k\geq0} (y-k)^k S_k({\bf{A}}^\clubsuit) \ . $$

\end{xex} \begin{xex}
Suppose ${\bf{A}}$ be of cardinal $n\geq2$, and $\omega$ be the maximal permutation
of ${{S}}({\bf{A}})$. Show that the image of $\pi_\omega$ acting on 
$1/(1-a_1^2)(1-a_2^2)$ is equal to
$$ \prod_i 1/(1-a_i^2) \ .$$
By showing that $S_k( \psi_2({\bf{A}})) = \sum_{i\leq k}(-1)^i S_{i, 2k-i}({\bf{A}})$, 
where $\psi_2({\bf{A}}):= \{a^2 \}_{a\in{\bf{A}}}$.

%\ptisaut Posons $F(a_1, \ldots, a_k)=\prod_{i\leq k} 1/(1-a_i^2) $.
%Pour le cardinal 3, on \'ecrit la function of d\'epart
%$F(a_1, a_2) = (1-a_3^2)F(a_1, a_2, a_3)$.
%L'image par $\pi_1\pi_2\pi_1$ of $a_3^2$ \'etant nulle, on a donc 
%$F(a_1, a_2) \pi_1\pi_2\pi_1 = F(a_1, a_2, a_3)$. 

%On it\`ere le calcul sur the tripthe of variabthe cons\'ecutives, 
%en factorisant \`a gauche $\pi_\omega = \pi_k \pi_{k+1}\pi_k \pi_\omega $, 
%$k=1, \ldots, n-2$. 

%On obtient ainsi que l'image of $F(a_1, a_2)$ par $\pi_\omega$
%est \'egale \`a celle of $F(a_1, \ldots, a_n) $ par $\pi_\omega$, 
%i.e. \`a $F(a_1, \ldots, a_n)$.

%Or $F(a_1, a_2) = \sum_{0\leq i\leq k} (-1)^i S_{i, 2k-i}(a_1+a_2)$
%et chaque function of Schur $S_{ij}(a_1+a_2)$
%a pour image par $\pi_\omega$ la function of Schur $S_{ij}({\bf{A}})$
% \quad\Box 


\end{xex} \begin{xex}
 Suppose ${\bf{A}}$ be a finite alphabet. Show that 
$$\sum z^{col} S_I({\bf{A}})= \prod_i {1\over 1-za_i} \prod_{i<j} {1\over 1-a_i a_j} $$
by induction on the cardinal of ${\bf{A}}$, 
and using the isobaric symmetry $\pi_\omega$.
The sum runs all the partitions $I$ of lengh $<= card({\bf{A}})$, 
and $col$ is the number of odd colomns of diagramme of $I$, 
i.e. the number of odd parts of the conjugate partition of
$I$. 


\end{xex} \begin{xex}
Suppose ${\bf{A}}$ be an alphabet of cardinal $n+1$, 
$f$ be a function of one variable.
Suppose the isobaric operators $\Delta_i := \pi_i -1 = \partial_i\, a_{i+1}$.

Give the development of the image of $f(a_1)$ acted by $\pi_\omega$
with functions of the successive images of $f$ acted by the operators $\Delta_i$.

%\ptisaut
%Comme $f$ est une function d'une variable,  
%$f\pi_\omega = f\pi_1 \pi_2\cdots \pi_n$.
%Supposons par r\'ecurrence que
%$$f\pi_1 \pi_2\cdots \pi_{n-1} = f(a_1) + f\Delta_1 + f\Delta_1 \Delta_2 
%+ \ldots +f\Delta_1\cdots \Delta_{n-1} $$
%L'image of cette iofntit\'e par $\pi_n = 1 + \Delta_n$
%ne fait appara\^\i tre que le terme suppl\'ementaire 
%$ f\Delta_1\cdots \Delta_n$ car $\Delta_n$ annule tous the termes
%qui ne contiennent pas $a_n$, c'est-\`a-dire tous the 
%termes du membre of droite sauf le ofrnier \quad \Box

%Dans le cas of la function $f = 1/(1-x)$, 
%l'iofntit\'e d\'emontr\'ee est
%$${1\over (1-a_1} \pi_\omega =
%{1\over 1-a_1 } + {a_2\over (1-a_1)(1-a_2) } 
%+ {a_3\over (1-a_1)(1-a_2)(1-a_3) } + \cdots  $$

\end{xex} \begin{xex} 
Suppose ${\bf{A}}= \{a, b, \ldots, c, d \}$ be of cardinal $n$. 
By induction on $n$ and with the aid of maximal isobaric symmetry 
$\pi_\omega $, show that 
$$S_{12\ldots n}({\bf{A}})/(1-a^2)\cdots (1-d^2) 
= \sum S_{1\ldots n-1, n+2j}({\bf{A}})  $$
For exemple $S_{12}(a+b) /(1-a^2)(1-b^2) =
S_{12} + S_{14}+ S_{16} + \cdots  $

%\ptisaut 
%L'iofntit\'e r\'esulte of ce que 
%$d^2 S_{1\ldots n-1}(a+b+\cdots +c) \, \pi_\omega =0$.

%Une d\'emonstration plus \'el\'ementaire r\'esulte of l'expression 
% ofs fonctions of Schur comme quotient d'un d\'eterminant
%of puissances ofs \'el\'ements of ${\bf{A}}$ par le Vanofrmonof en ${\bf{A}}$
%(formule of Jacobi).

\end{xex} \begin{xex}
Suppose $\{S_k \}$ be a recurrent series of ordre $n$,  
i.e. a series of type $\{ S_k({\bf{A}}-{\bf{B}}) \}$,  
with $card({\bf{A}})=n$, $card({\bf{B}}) <n$.
Suppose $r$ be an integer $0\leq r\leq n$. 

Show that the series
$\{T_k := S_{k^r}({\bf{A}}-{\bf{B}}) \}$ is a recurrent series
of ordre ${n \choose r}$. 

\end{xex}

\pagebreak

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\end{document}


