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\hyphenation{multi-convo-lu-tion}
\hyphenation{or-di-nary}
\hyphenation{con-vo-lu-tion}
\hyphenation{poly-no-mials}
\title{Multinomial convolution polynomials}
\author{\small\normalsize Jiang Zeng\\
\normalsize D\'epartement de Math\'ematique, Universit\'e Louis Pasteur,\\
\normalsize 7, rue Ren\'e Descartes, 67084 Strasbourg C\'edex, France}
\date{October 26, 1994}
\begin{document}
\maketitle
\begin{abstract}\small
In \cite{Knu} Knuth shows how to derive
the convolution formulas of Hagen, Rothe and Abel from
Vandermonde's convolution or binomial theorem for integer
exponents. In the present paper, we shall first present a short
and elementary proof of the multi-extension of the above
convolution formulas, due to Raney and Mohanty. 
In the second part we shall present a multi-version of Knuth's approach to
convolution polynomials and derive 
another short proof of the above formulas. 
\end{abstract}

\section{Introduction}
Recall that a family of 
polynomials $\{F_n(x)\}_{n\geq 0}$ is said to be
of convolution type if
$F_n(x)$ has degree $\leq n$ and satisfies the convolution
condition : 
$$
F_n(x+y)=\sum_{k=0}^nF_k(x)F_{n-k}(y).
$$
 
The two typical families of convolution polynomials are
the binomial theorem for integers
$$
{(x+y)^n\over n!}=\sum_{k=0}^n{x^k\over k!}{y^{n-k}\over (n-k)!} 
$$
and Vandermonde's convolution
$$
{x+y\choose n}=\sum_{k=0}^n{x\choose k}{y\choose n-k}.
$$
Some important extensions of the above identities have been given by
Abel, Rothe and Hagen in the last centry. More precisely,
for $a, b\in \bf C$ and $n\in \bf N$ let 
$$
A_n(a,b)={a\over a+bn}{a+bn\choose n}.
$$
In 1891 Hagen \cite{Hag,Goo,GK} proved the following identity :
$$
\sum_{k=0}^n(p+qk)A_k(a,b)A_{n-k}(c,b)={p(a+c)+aqn\over
a+c}A_n(a+c,b). 
$$
In the case $q=0$, this formula reduces to Rothe's
convolution formula \cite{Rot,Gou,Rio} dated 1793:
$$
\sum_{k=0}^nA_k(a,b)A_{n-k}(c,b)=A_n(a+c,b).
$$
Note that Hagen's formula implies also Abel's identity \cite{Abe} :
$$
(x+y)^n=\sum_{k=0}^n
{n\choose k}x(x-kz)^{k-1}(y+kz)^{n-k},\eqno 
$$
where $x, \ y,\  z\in \bf C$ and $n\in \bf N$.

In a {\sl refreshing} paper about {\it convolution polynomials}, 
Knuth \cite{Knu} shows how to derive these
seemingly non trivial formulas from
the basic binomial theorem for integers and Vandermonde's convolution.

Since the multi-extensions of the above identities are also known, it is natural
to give a multi-extension of Knuth's results in \cite{Knu}.
To express them more concisely, we shall use the vector notation 
as follows. Throughout this paper $m$ will be a fixed natural number. 
For ${\bf a}=(a_1,\cdots ,a_m)\in \bf
N^m$ and ${\bf b}=(b_1,\cdots, b_m)\in \bf N^m$ set
$|{\bf a}|=\sum_{i=1}^ma_i,\,\, {\bf a}!=a_1!\cdots a_m!$,
${\bf a}+{\bf b}=(a_1+b_1,\cdots ,a_m+b_m)$ and 
${\bf a}\cdot{\bf b}=\sum_{i=1}^ma_ib_i$. We order the elements of  $\bf N^m$
by {\it lexicographic order}, {\it i.e.}, ${\bf a}< {\bf a}$ if $a_1<b_1$ or
there is a $i>1$ such that $a_1=b_1$, $\ldots$, $a_{i-1}=b_{i-1}$ but 
$a_i<b_i$. Also for $1\le i\le m$ let
${\bf e}_i=(\delta_{i1}, \delta_{i2}, \cdots, \delta_{im})$.  Finally, 
for a complex number $x$ and ${\bf n}\in {\bf Z}^m$, we define 
$$
{x\choose \bf n}=\cases{
x(x-1)\cdots (x-|\bf n|+1)/\bf n!, 
& if ${\bf n}=(n_1,\cdots, n_m)\in \bf N^m$;\cr
0,& if some $n_i$ is negative.\cr
}$$ 

We now define the multi-analogue of $A_n(a,b)$ by
$$
A_{\bf n}(x,{\bf b})={x\over x+{\bf b}\cdot{\bf n}}{x+{\bf b}\cdot
{\bf n}\choose {\bf n}}. 
$$
The multi-extension of Rothe's convolution formula can then be stated as follows :
$$
\sum_{\bf k}A_{\bf k}(a,{\bf b})A_{{\bf n}-{\bf k}}
(c,{\bf b})=A_{\bf n}(a+c,{\bf b}).\eqno(1.1) 
$$

\noindent{\bf Remark:\/} The author first learned this identity in July 1988 at
Oberwolfach, where Louck \cite{Lou} communicated it as a conjecture. It was
shortly proved independently by Strehl, Paule and the
author by using different methods and reported at the 20th
session of  le séminaire Lotharingien de Combinatoire held at
Alghero, Italie, in September 1988. Shortly later, we noted that
formula (1.1) was already established by Raney  \cite[Th. 2.2, Th. 2.3]{Ran}  
in 1960. We refer the reader to the  recent paper of
Strehl \cite{Str} for a good account of the various aspects of this
formula. 

Curiously enough Raney's derivation of (1.1) was entirely
different from the three formentioned ones. Indeed, 
Strehl's proof is of combinatorial nature \cite{Str}, 
 Paule's proof is based on the one variable 
Lagrange inversion formula \cite{Pau} and the author's proof is inductive and will be 
produced below. Note that Chu \cite{Chu} quoted Raney's identity
explicitly in his paper to derive a combinatorial
interpretation of the generalized Catalan numbers.

A multi-extension of Hagen's identity has also been given by Mohanty \cite{Moh},
by applying the {\it multivariable Lagrange inversion formula}, in the following
form :  
$$
\sum_{\bf k}(p+{\bf q}\cdot {\bf k})A_{\bf k}(a,{\bf b})
A_{{\bf n}-{\bf k}}(c,{\bf b})
={p(a+c)+a({\bf q}\cdot {\bf n})\over a+c}A_{\bf n}(a+c,{\bf b}),
\eqno(1.2)
$$
where ${\bf q}=(q_1,\cdots ,q_m)\in {\bf C}^m$ and $a,\
c,\ p\in\bf C$.

The aim of this paper is two folds. 
First we present a very short and elementary
proof of (1.1) and (1.2) from scratch, which is similar in spirit to Good's short 
proof of Dyson's conjecture \cite{Goo}. In the the second part of this paper 
we shall present a natural multi-extension 
of Knuth's approach \cite{Knu} to ordinary convolution polynomials.
As a consequence of this generalization we see that (1.2) and (1.1) are
respectively equivalent to the following trivial convolution identities :
$$
{(x+y)^{|\bf n|}\over {\bf n}!}=\sum_{{\bf 0}\leq {\bf k}\leq {\bf n}}
{x^{|\bf k|}\over {\bf k}!}
{y^{|{\bf n}-{\bf k}|}\over ({\bf n}-{\bf k})!}\eqno (1.3)
$$
and 
$$
{x+y\choose {\bf n}}=\sum_{{\bf 0}\leq {\bf k}\leq {\bf n}}
{x\choose {\bf k}}{y\choose {\bf n}-{\bf k}}.\eqno (1.4)
$$ 

As remarked by  Knuth \cite{Knu}, in the special case that each
polynomial $F_n(x)$ has degree exactly $n$, the polynomials
$n!F_n(x)$ are said to be of {\it binomial type} \cite{MR}. We
note that in the later case Joni \cite{Jon} has given another
generalization of binomial polynomials. 

\section{ A short proof of formula (1.2)} Notice that formula (1.2) is eqivalent to (1.1) and 
the following identity :
$$
\sum_{\bf k}({\bf q}\cdot {\bf k})A_{\bf k}(a,{\bf b})A_{{\bf n}-{\bf k}}
(c,{\bf b})={a({\bf q}\cdot{\bf n})\over a+c}
A_{\bf n}(a+c,{\bf b})
\eqno(2.1) 
$$
A further look shows  that formula (1.3) is actually a consequence of (2.1).
Indeed, exchanging $a$ with $c$, and $\bf k$ with ${\bf n}-{\bf k}$ 
in (2.1) yields
$$
\sum_{\bf k}({\bf q}\cdot ({{\bf n}-{\bf k}}))A_{\bf k}(a,{\bf b})A_{{\bf n}-{\bf k}} (c,{\bf b})=
{c({\bf q}\cdot{\bf n})\over a+c}
A_{\bf n}(a+c,{\bf b})
\eqno(2.2) 
$$
Summing up the two above identities side by side and taking
${\bf q}={\bf n}$, we get (1.1). Therefore it suffices to prove (2.1). 
However it seems useful to first give an independent proof of (1.1)
to illustrate the method.  
In the sequel we assume $a$ to belong to $\bf N$. Let
$$
S_{\bf n}(a,c,{\bf b})=\sum_{\bf k}
A_{\bf k}(a,{\bf b})A_{{\bf n}-{\bf k}}(c,{\bf b}).
$$
Then, first
$$
S_{\bf n}(0,c,{\bf b})=A_{\bf n}(c,{\bf b}), \qquad 
S_{\bf 0}(a,c,{\bf b})=1. \eqno(2.3)
$$
Next, by the definition of $A_{\bf n}(a,{\bf b})$ we have
\begin{eqnarray*}
 A_{\bf n}(a,\bf b) 
&=&{1\over a-1+\bf b\cdot {\bf n}}\left(a-1+\sum_{i=1}^m{b_in_i\over a+{\bf b}\cdot 
{\bf n}}\right){a+{\bf b}\cdot {\bf n}\choose {\bf n}} \qquad (2.4)\\
&=&{1\over a-1+{\bf b}\cdot 
{\bf n}}\left[(a-1){a+{\bf b}\cdot {\bf n}\choose {\bf n}}+\sum_{i=1}^m
b_i{a-1+{\bf b}\cdot {\bf n}\choose {\bf n}-{\bf e}_i}\right] \\
            &=& A_{\bf n}(a-1,{\bf b})+\sum_{i=1}^mA_{{\bf n}-{\bf e_i}}(a+b_i-1,{\bf b}).
\end{eqnarray*}
The last line is due to ${x\choose {\bf n}}={x-1\choose {\bf n}}
+\sum_{i=1}^m{x-1\choose {\bf n}-{\bf e}_i}$. So that
\begin{eqnarray*}
S_{\bf n}(a,c,{\bf b}) &=&\sum_{\bf k}
\left(A_{\bf k}(a-1,{\bf b})+\sum_{i=1}^mA_{{\bf k}-{\bf e}_i}(a+b_i-1,{\bf b})\right)
A_{{\bf n}-{\bf k}}(c,{\bf b})\\
               &=&S_{\bf n}(a-1,c,{\bf b})
+\sum_{i=1}^mS_{{\bf n}-{\bf e}_i}(a+b_i-1,c,{\bf b}).\qquad (2.5)
\end{eqnarray*}
Eqs. (2.3)-(2.5) show that 
$S_{\bf n}(a,c,\bf b)$ and $A_{\bf n}(a,c, {\bf b})$
satisfy the same recurrence and the initial conditions.
Therefore they are equal. 

Now we turn to (2.1). Note that (2.1) is equivalent to the following $m$ identities: 
$$
\sum_{\bf k}k_jA_{\bf k}(a,{\bf b})A_{{\bf n}-{\bf k}}(c, {\bf b})
={a\,n_j\over a+c}A_{\bf n}(a+c,{\bf b}),
\qquad (1\le j\le m). \eqno(2.6)
$$
For $1\le j\le m$, let
$T_{\bf n}^{(j)}(a,c,{\bf b})$ and $S_{\bf n}^{(j)}(a,c,{\bf b})$ be 
respectively the left-hand and right-hand sides of (2.6).
By definition, we first have
$$
S_{\bf n}^{(j)}(a,c,{\bf b}) =(a-1){a-1+c+{\bf b}\cdot {\bf n}\choose {\bf
n}-{\bf e}_j} +{a-1+c+ {\bf b}\cdot {\bf n}\choose {\bf n}-{\bf e}_j}.
\eqno (2.7) 
$$
Now writing the first ${a-1+c+{\bf b}\cdot {\bf n}\choose {\bf n}-{\bf e}_j}$ as
$$
{a-2+c+{\bf b}\cdot {\bf n}\choose {\bf n}-{\bf e}_j}
+\sum_{i=1}^m{a-2+c+ {\bf b}\cdot {\bf n}\choose {\bf n}-{\bf e}_i}
$$
and the second one as
\begin{eqnarray*}
\left(\frac{{\bf b}\cdot({\bf n}-{\bf e}_j)}
{ a-1+c+{\bf b}\cdot{\bf n}}+ \frac{a-1+c+{b_j}}{ a-1+c+{\bf b}\cdot {\bf n}}\right)
{a-1+c+{\bf b}\cdot {\bf n} \choose {\bf n}-{\bf e}_j}\\
= \sum_{i=1}^m b_i{a-2+c+{\bf b}\cdot {\bf n}\choose {\bf n}-{\bf e}_j-{\bf e}_i}
+A_{{\bf n}-{\bf e}_j}(a-1+c+b_j, {\bf b}), 
\end{eqnarray*}
we see that
\begin{eqnarray*}
S_{\bf n}^{(j)}(a,c,{\bf b})=S_{\bf n}^{(j)}(a-1,c,{\bf b})
&+&\sum_{i=1}^mS_{{\bf n}-{\bf e}_i}^{(j)}(a-1+b_i,c, {\bf b})\qquad (2.8)\\
& +&A_{{\bf n}-{\bf e}_j}(a-1+c+b_j,{\bf b}).  
\end{eqnarray*}
Also we have the boundary conditions :
$$
S_{\bf n}^{(j)}(0,c,{\bf b})=0\qquad \hbox{and} \qquad S_{\bf 0}^{(j)}
(a,c,{\bf b})=0.\eqno (2.9)
$$ 
On the other hand, by (2.4) we verify easily that 
$T_{\bf n}^{(j)}(a,c,{\bf b})$ satisfies the same recurrence 
relation (2.8) and boundary
condition (2.9). Thus completes proof of (2.6). \qed

\section{Multi-convolution polynomials}
 A family of polynomials $\{F_{\bf n}(x)\}_{{\bf n}\geq {\bf 0}}$ 
forms a {\it multinomial convolution family} if $F_{\bf n}(x)$ has degree 
$\leq|\bf n|$ and if the convolution condition 
$$
F_{\bf n}(x+y)=\sum_{{\bf 0}\leq {\bf k}\leq {\bf n}}
F_{\bf k}(x)\ F_{{\bf n}-{\bf k}}(y)\eqno(3.1)
$$
holds for all $x$ and $y$ and for all ${\bf n}\geq {\bf 0}$.
In the case that $m=1$ we recover the monomial {\it convolution family} studied
by Knuth \cite{Knu}. We should note that this definition is different from the one
of ``higher dimentional polynomials of binomial type"  studied by Joni \cite{Jon}.
Many such families are known, and they appear frequently in applications.
For example, we can let $F_{\bf n}(x)=x^{|\bf n|}/{\bf n}!$, the condition
(1.7) is equivalent to the binomial theorem for integer exponents. Or we can
let  $F_{\bf n}(x)$ be the multinomial coefficient ${x\choose {\bf n}}$; the 
corresponding identity (1.8) may be called multi-Vandermonde's convolution.

Knuth showed that convolution polynomials arise 
as coefficients when a power series of {\it one variable} is raised to the
power $x$. Now we show that
multi-convolution polynomials arise 
as coefficients when a power series of {\it several variables} is raised to the
power $x$. Let ${\bf z}^{\bf n}=z_1^{n_1}\ldots z_m^{n_m}$. For a formal
series $F({\bf z})=\sum_{{\bf n}\geq {\bf 0}}F_{\bf n}{\bf z}^{\bf n}$ let
$[{\bf z}^{\bf n}]F({\bf z})$ be the coefficient of ${\bf z}^{\bf n}$ in
$F({\bf z})$.

\begin{thm} Let 
$$
F({\bf z})=1+\sum_{{\bf n}\not={\bf 0}}F_{\bf n}{\bf z}^{\bf n}\eqno(3.2)
$$
be any power series with $F({\bf 0})=1$. Then  the polynomials
$$
F_{\bf n}(x)=[{\bf z}^{\bf n}]F({\bf z})^x\eqno(3.3)
$$
form a multi-convolution family. 
Conversely, every convolution family arises 
in this way or is identically zero.
\end{thm}
\pf It is easy to verify that the coefficient of ${\bf z}^{\bf n}$
in $F({\bf z})^x$ is indeed a polynomial in $x$ of degree $\leq |\bf n|$
because
$$
F({\bf z})^x=1+\sum_{k\geq 1}{x\choose k}
(\sum_{{\bf n}\not={\bf 0}}F_{\bf n}{\bf
z}^{\bf n})^k  
$$  
This construction produces a convolution family because of
the rule for forming coefficients of the product 
$F({\bf z})^{x+y}=F({\bf z})^xF({\bf z})^y$.
Conversely, suppose the polynomials 
$$
F_{\bf n}(x)=(f_{{\bf n}0}+f_{{\bf n}1}x+
\ldots +f_{{\bf n}|{\bf n}|}x^{|{\bf n}|})/{\bf n}!
$$ 
form a convolution family. The condition $F_{\bf 0}(x)=F_{\bf 0}(x)^2$
can hold only if $F_{\bf 0}(x)=0$ or $F_{\bf 0}(x)=1$. In the former case it is
easy to prove by induction that $F_{\bf n}(x)=0$ for all $\bf n$ because
$F_{\bf n}(x)=\sum_{{\bf 0}<{\bf k}<{\bf n}}
F_{\bf k}(x)\ F_{{\bf n}-{\bf k}}(0)$. Otherwise, the condition 
$$
F_{\bf n}(2\ x)=2F_{\bf n}(x)+\sum_{{\bf 0}<{\bf k}<{\bf n}}
F_{\bf k}(x)\ F_{{\bf n}-{\bf k}}(x)\eqno(3.4)
$$
for ${\bf n}>{\bf 0}$ implies that
$F_{\bf n}(0)=f_{{\bf n}0}=0$ for ${\bf n}>{\bf 0}$ by induction. 
For $k>1$,
by equating coefficients
of $x^k$ on both sides of (3.4)
we find that the coefficient $f_{{\bf n}k}$ is forced to have 
certain values based on the coefficients of 
$F_{\bf i}(x)$ with ${\bf i}<{\bf n}$, 
 because $2^kf_{{\bf n}k}$
occurs on the left and $2f_{{\bf n}k}$ on the right. 
The coefficient $f_{{\bf n}1}$ can, however, be choosen freely. Any such choice
must make
$$
F_{\bf n}(x)=[{\bf z}^{\bf n}]
\exp(x\sum_{{\bf n}\not= {\bf 0}}f_{{\bf n}1}{\bf z}^{\bf n}), 
$$
by induction on $\bf n$. \qed \medskip
\noindent{\bf Examples:}
The first example (1.3), that is $F_{\bf n}(x)=x^{|\bf
n|}/{\bf n}!$, comes from the power series $F({\bf
z})=e^{z_1+z_2+\ldots +z_m}$; the second example (1.4), that is $F_{\bf
n}(x)={x\choose {\bf n}}$, comes from $F({\bf
z})=1+z_1+z_2+\ldots +z_m$.

\medskip 
As in the case of one-variable convolution polynomials \cite{Knu}, every
multi-convolution family $\{F_{\bf n}(x)\}$ satisfies another general
convolution formula in addition to the one we began with. In deed, we have the
following result.
 
\begin{lem}(A derived convolution). If $\{F_{\bf n}(x)\}$ is a set of
multi-convolution polynomilas, then
$$
\sum_{{\bf k}={\bf 0}}^{\bf n}k_iF_{\bf k}(x)
F_{{\bf n}-{\bf k}}(y)=xn_iF_{\bf n}(x+y)\qquad (1\leq i\leq m).
$$
or equivalently for all ${\bf q}\in {\bf C}^m$ we have
$$
(x+y)\sum_{{\bf k}={\bf 0}}^{\bf n}({\bf q}\cdot{\bf k})F_{\bf k}(x)
F_{{\bf n}-{\bf k}}(y)=x({\bf q}\cdot {\bf n})F_{\bf n}(x+y).
$$
\end{lem}
\pf The alternative convolution formula is proved by
differentiating the basic identity $F({\bf z})^x=\sum_{{\bf n}\geq {\bf
0}}F_{\bf n}(x){\bf z}^{\bf n}$ with respect to $z_i$ ($1\leq i\leq
m$) and  
$$
xz_i{\partial \over \partial z_i}F({\bf z}) F({\bf z})^{x-1}=
\sum_{\bf n}n_iF_{\bf n}(x){\bf z}^{\bf n}.
$$
Now observe that $\sum k_iF_{\bf k}(x)F_{{\bf n}-{\bf k}}(y)$ is the
coefficient of ${\bf z}^{\bf n}$ in the product
$xz_i(\partial F({\bf z})/\partial z_i) F({\bf z})^{x+y-1}$ while
$n_iF_{\bf n}(x+y)$ is the coefficient of ${\bf z}^{\bf n}$ in
$(x+y)z_i(\partial F({\bf z})/\partial z_i) F({\bf z})^{x+y-1}$. \qed
\medskip

\begin{lem} If $F_{\bf n}(x)=[{\bf z}^{\bf n}]F({\bf
z})^x$, then 
$$
{x\over x+n_i}F_{\bf n}(x+n_i)=[{\bf z}^{\bf n}]G_i({\bf
z})^x, \qquad 1\leq i\leq m
$$
where $G_i({\bf z})=F(z_1, \ldots, z_iG_i({\bf z}), \ldots,
z_m)$.
\end{lem}
\pf Since $F_{\bf n}(x)$ is a polynomial of degree
$\leq |{\bf n}|$, so it is sufficient to prove the above
identity for integer values $k$ of $x$. Let 
$w({\bf z})=z_i/F({\bf z})$, then $w(z_1, \ldots,
z_iG_i({\bf z}), \ldots, z_m)=z_i$. By Lagrange's inversion
formula (see [3, p. 159]), we have
$$
[z_i^{n_i}](z_i^kG_i^k({\bf z}))={k\over n_i}[z_i^{n_i-k}]
F({\bf z})^{n_i}, \qquad 1\leq i\leq m.
$$
Consequently,
$$
[z_i^{n_i}](G_i^k({\bf z}))={k\over n_i+k}[z_i^{n_i}]
F({\bf z})^{n_i+k},\qquad 1\leq i\leq m.
$$
It follows that for $1\leq i\leq m$ we have
$$
[{\bf z}^{\bf n}]G_i({\bf z})^x={k\over n_i+k}
[{\bf z}^{\bf n}]
F({\bf z})^{n_i+k}={k\over n_i+k}F_{\bf n}(n_i+k). 
$$
This completes the proof. \qed \medskip
 
\begin{thm} Let $F_{\bf n}(x)$ be any family
of polynomials in $x$ such that
$F_{\bf n}(x)$ has degree $\leq |\bf n|$. If
$$
F_{\bf n}(2\ x)=\sum_{{\bf k}={\bf 0}}^{\bf n}
F_{\bf k}(x)\ F_{{\bf n}-{\bf k}}(x)\eqno (3.5)
$$
holds for all $\bf n$ and $x$, 
then the following identities hold for all $\bf n$, $x$, $y$ and ${\bf t}$
\begin{eqnarray*}
&&{(x+y)\ F_{\bf n}(x+y+{\bf t}\cdot{\bf n})\over
x+y+{\bf t}\cdot{\bf n}}\\
&&=\sum_{{\bf k}={\bf 0}}^{\bf n}
{x\ F_{\bf k}(x+{\bf t}\cdot{\bf k})\over x+{\bf t}\cdot{\bf k}}
\cdot{y\ F_{{\bf n}-{\bf k}}(y+{\bf t}\cdot({\bf n}-{\bf k}))
\over y+{\bf t}\cdot({\bf n}-{\bf k})} \\
&&{({\bf q}\cdot{\bf n})\ F_{\bf n}(x+y+{\bf t}\cdot{\bf n})\over
x+y+{\bf t}\cdot{\bf n}}\\
&&=\sum_{{\bf k}={\bf 0}}^{\bf n}
{({\bf q}\cdot{\bf k})\ F_{\bf k}(x+{\bf t}\cdot{\bf k})\over x+{\bf t}\cdot{\bf k}}
\cdot {y\ F_{{\bf n}-{\bf k}}(y+{\bf t}\cdot({\bf n}-{\bf k}))
\over y+{\bf t}\cdot({\bf n}-{\bf k})}.
\end{eqnarray*}
\end{thm}
\pf The proof of Theorem 1 shows that the multinomial convolution
condition could be weakened to (3.4), or (3.5).
Hence (3.5) implies that polynomials $F_{\bf n}(x)$ form a set of convolution polynomials.
Now, for $1\leq i\leq m$, the process of going from
$F_{\bf n}(x)$ to $xF_{\bf n}(x+n_i)/(x+n_i)$ of Lemma 3 can be
iterated : another replacement gives $xF_{\bf
n}(x+2n_i)/(x+2n_i)$, and after $t_i$ iterations we discover
that the polynomials $xF_{\bf n}(x+t_in_i)/(x+t_in_i)$ also
form a convolution family. This holds for all positive
integers $t_i$ ($1\leq i\leq m$), and the convolution
condition is expressible as a set of polynomial relations in
$t_i$ ($1\leq i\leq m$); therefore 
$xF_{\bf n}(x+t_in_i)/(x+t_in_i)$ is a convolution
family for all complex numbers $t_i$ ($1\leq i\leq m$).
Thus by applying 
the above process we obtain succesfully a sequence of convolution polynomials families
$$
{x\over x+t_1n_1+\ldots +t_in_i}F_{\bf n}(x+t_1n_1+\ldots +t_in_i)
$$
for $i=1,\, 2,\ldots,\, m$ and the last family corresponds to the first identity.
The second one follows by
combining the first one with the derived convolution formula of Lemma 2.
\qed \medskip

\noindent{\bf Example 1}:
Let $F_{\bf n}(x)={x\choose {\bf n}}$, then it follows from 
(1.4) and Theorem 4 that
\begin{eqnarray*}
&&{x+y\over x+y+{\bf t}\cdot{\bf n}}{x+y+{\bf t}\cdot{\bf n}\choose {\bf n}}\\
&&=\sum_{{\bf k}={\bf 0}}^{\bf n}{x\over x+{\bf t}\cdot{\bf k}}{x+{\bf t}\cdot{\bf k}\choose {\bf k}}
{y\over y+{\bf t}\cdot({\bf n}-{\bf k})}
{y+{\bf t}\cdot({\bf n}-{\bf k})\choose {\bf n}-{\bf k}},\\
&&{({\bf q}\cdot {\bf n})\over x + y+{\bf t}\cdot {\bf n}}{x+y+{\bf t}\cdot{\bf n}\choose {\bf n}}\\
&& =\sum_{{\bf k}={\bf 0}}^{\bf n}
{{\bf q}\cdot {\bf k}\over x+{\bf t}\cdot{\bf k}}
{x+{\bf t}\cdot{\bf k}\choose {\bf k}}
{y\over y+{\bf t}\cdot({\bf n}-{\bf k})}
{y+{\bf t}\cdot({\bf n}-{\bf k})\choose {\bf n}-{\bf k}}.
\end{eqnarray*}
Combining the above two identies yields (1.2).
\medskip

\noindent{\bf Example 2}:
Let $F_{\bf n}(x)=x^{\bf n}/{\bf n}!$, then it follows from 
(1.3) and Theorem 4 that
\begin{eqnarray*}
&&{(x+y)(x+y+{\bf t}\cdot{\bf n})^{|\bf n|-1}\over {\bf n}!}\\
&&=\sum_{{\bf k}={\bf 0}}^{\bf n}{x (x+{\bf t}\cdot{\bf k})^{|\bf k|-1}\over {\bf k}!}
{y(y+{\bf t}\cdot ({\bf n}-{\bf k}))^{|{\bf n}-{\bf k}|-1}\over ({\bf n}-{\bf k})!},\\  
&&{({\bf q}\cdot {\bf n})(x+y+{\bf t}\cdot{\bf n})^{|\bf n|-1}\over  {\bf n}!}\\
&&=\sum_{{\bf k}={\bf 0}}^{\bf n}{({\bf q}\cdot {\bf k})(x+{\bf t}\cdot
{\bf k})^{|\bf k|-1}\over {\bf k}!}
{y(y+{\bf t}\cdot({\bf n}-{\bf k}))^{|{\bf n}-{\bf k}|-1}\over  ({\bf n}-{\bf k})!}.
\end{eqnarray*}
Combining the above two identies we obtain
$$
\sum_{\bf k}(p+{\bf q}\cdot {\bf k})B_{\bf k}(x,{\bf t})
B_{{\bf n}-
 {\bf k}}(y,{\bf t})
={p(x+y)+x({\bf q}\cdot {\bf n})\over x+y}B_{\bf n}(x+y,{\bf t}),
\eqno (3.6)
$$
where $B_{\bf n}(x,{\bf t})=x(x+{\bf t}\cdot{\bf n})^{|{\bf n}|-1}/{\bf n}!$.
If we take
${\bf q}={\bf t}$ and $p=1$ in the above formula we obtain a
multi-extension of Abel's identity :
 $$
(x+y+{\bf n}\cdot {\bf t})^{|\bf n|}=
\sum_{{\bf k}={\bf 0}}^{\bf n}{{\bf n}\choose {\bf k}}
(x+{\bf k}\cdot {\bf t})^{|\bf k|}y
(y+({\bf n}-{\bf k})\cdot {\bf t})^{|{\bf n}-{\bf k}|-1},
$$
where ${{\bf n}\choose {\bf k}}=\prod_{i=1}^m{n_i\choose k_i}$, or equivalently
$$
(x+y)^{|\bf n|}=\sum_{{\bf k}={\bf 0}}^{\bf n}
{{\bf n}\choose {\bf k}}(x-{\bf k}\cdot {\bf t})^{|{\bf n}-\bf k|}
y(y+{\bf k}\cdot {\bf t})^{|{\bf k}|-1}.
$$

We can also derive (3.6) from the first example as follows.
If we substitute $a$, $c$ and $\bf b$ by $xa$, $xc$ and $x{\bf b}=(xb_1,\cdots
,xb_m)$ respectively in (1.2), we get a polynomial identity in $x$. 
Equating the coefficients of $x^{|\bf n|}$ in both sides 
of this identity yields (3.6).

Note that formula (3.6) generalizes also Hurwitz's {\sl multinomial}
extension of Abel's identity \cite{Hur}. To see this, setting 
${\bf n}={\bf 1}=(1, \ldots, 1)$,  ${\bf q}={\bf 0}$ 
and $p=1$, then (3.6) reduces to 
$$
\sum_{{\bf k}\in \{0, 1\}^m}x(x+{\bf k}\cdot {\bf t})^{|{\bf k}|-1}
y(y+({\bf 1}-{\bf k})\cdot {\bf t})^{|{\bf 1}-{\bf k}|-1}
=(x+y)(x+y+|{\bf t}|)^{m-1}.
$$

\noindent{\bf Exemple 3}
Consider the set $[{\bf n}]=\{1\}\times [n_1]+
\{2\}\times [n_2]+\ldots +\{m\}\times [n_m]$.
For each element $\alpha=(i, j)$ of $[{\bf n}]$ we call $i$ the {\it color} of $\alpha$.
A partition $\pi$ of $[{\bf n}]$ 
is called {\it colored partition} (cf. \cite{Zen}) if each block of $\pi$ is only composed 
of elements with different colors. Let
$$
F({\bf z})=\exp({e_1({\bf z})+e_2({\bf z})+\ldots +e_m({\bf z})}),
$$
where $e_k({\bf z})$ is the k-th ($1\leq k\leq m$) 
elementary symmetric function of $z_1, z_2,\ldots, z_m$. 
Then the corresponding family of convolution polynomials is 
\begin{eqnarray*}
F_{\bf n}(x)
&=& [{\bf z}^{\bf n}]e^{x(e_1({\bf z})+e_2({\bf z})+\ldots +e_m({\bf z}))}\\
&=& \frac{
{{\bf n}\brace 0}+{{\bf n}\brace 1}x+\ldots +{{\bf n}\brace |{\bf n}|}x^{|{\bf n}|}}{{\bf n}!},
\end{eqnarray*}
where ${{\bf n}\brace k}$ is the number of colored partitions in 
$k$ blocks of $[{\bf n}]$ (see \cite{Zen}). 
We note that if ${\bf n}=(1,1,\ldots, 1)$, 
the number ${{\bf n}\brace k}$ is nothing but Stirling
number of second kind $S(m, k)$, that is, 
 the number of partitions in $k$ blocks of $[m]$.

We have shown how to create new multinomial convolution
families from given ones exactly as in the monomial case.
The addiional constructions mentioned by
Knuth \cite{Knu} can also be extended. For example, if $F_{\bf n}(x)$
and $G_{\bf n}(x)$ are convolution families, then so is the family
$H_{\bf n}(x)$ defined by 
$$
H_{\bf n}(x)=\sum_{{\bf k}
={\bf 0}}^{\bf n}F_{\bf k}(x)G_{{\bf n}-{\bf k}}(x).
$$
as did by Knuth \cite{Knu} in the case $m=1$.

\newpage
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\end{document}
\end
\vskip 40pt
\vbox{\sl
\leftline{D\'epartement de Math\'ematiques}
\leftline{Universit\'e Louis-Pasteur}
\leftline{7, rue Ren\'e Descartes}
\leftline{67084 Strasbourg Cedex, France}
\leftline{e-mail : jzeng@math.u-strasbg.fr}}
