\magnification=1200
\input a4new
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\auteurcourant={\hfill Zeng and Zhang}
\titrecourant={Zeng and Zhang\hfill}
\hautpagetitre={Results in Mathematics\hfil}
\baspagetitre={\hfil}
\font\aaa=cmr7 
\def\aab#1#2{#1{\aaa #2}}
\def\maj{\mathop {\rm maj}}
\def\Des{\mathop {\rm des}}
\def\proof{\aab P{ROOF.} --- }
\def\remark{\aab R{EMARK}: }
\def\vol{vol. }
\def\abstract{\bgroup\eightpoint\noindent Abstract\pointir} 
 \def\endabstract{\par\egroup}
\font\fourteenrm=cmbx10 scaled \magstep1
\
\vskip 2cm
\centerline{\fourteenrm
A q-ANALOG OF NEWTON'S SERIES, STIRLING }
\smallskip
\centerline{\fourteenrm
FUNCTIONS AND EULERIAN FUNCTIONS }
\vskip6mm
\centerline{Jiang ZENG and Changgui ZHANG} 
\vskip8mm
{\narrower\narrower
\abstract
Recently, Butzer et al. [BH1, BH2, BHS] have studied some
classical  combinatorial functions such as factorial functions,
Stirling numbers and Eulerian numbers of fractional orders. In the
present paper we show that much the same is true in the
case of the $q$-analogs. Meanwhile we give some
results for the convergence of a $q$-Newton interpolation
series.
\bigskip\noindent
Math. Subject Classification (1991) : 05A30, 33D05.
\endabstract}

\vskip 10mm

{\parindent 0mm
\sectiona
0. Introduction 
\endsection

Throughout this paper $q$ is a real number such
that $0<q<1$. For $x\in\bb C$ 
define 
$$
[x]={q^x-1\over q-1}\virg
$$
where and in the sequal, we take the {\sl principal value} of the
function $x\mapsto q^x$.
For $n\in\bb N$, let 
$$\eqalign{
[x]_0&=1,&
[x]_n&=[x][x-1]\cdots [x-n+1]\qquad(n\geq 1),\cr
[0]!&=1, &[n]!&=[n]_n=[n][n-1]\cdots [1]\qquad (n\geq
1),\cr} $$
and define the $q$-{\sl binomial coefficients} by
$$
{x\brack 0}=1,\qquad
{x\brack n}={[x]_n\over [n]!}\qquad(n\geq 1).
$$ 
Set also
$$ 
(x;q)_\infty
=\prod_{k=0}^{+\infty}(1-xq^k),\leqno(0.1) 
$$ 
and for $\alpha \in \bb C$
$$
(x;q)_\alpha ={(x;q)_\infty\over 
(xq^\alpha ;q)_\infty}\virg\leqno(0.2) 
$$
where $xq^\alpha\not=q^{-n}$ for any $n\in \bb N$. In particular,
we have the $q$-{\sl shifted factorials} :
$$
(x;q)_0=1,\qquad (x;q)_n=(1-x)(1-xq)\cdots (1-xq^{n-1}),
$$
and then the following expression [GR, p. 20] for $q$-binomial
coefficients :}
$$
{x\brack n}={(q^{-x};q)_n\over
(q;q)_n}(-q^x)^n q^{-{n\choose 2}}.
$$

Recall that the {\sl $q$-Gamma function} (\cf. [GR, p. 16]) is
defined for $\Re x>0$ by 
$$
\Gamma_q(x)={(q;q)_\infty\over (q^x;
q)_\infty}(1-q)^{1-x}\cdot 
\leqno (0.3)  
$$

Note that $\lim_{q\to 1^-}\Gamma_q(x)=\Gamma(x)$, 
the reader is refered to \pd Koorwinder [Ko] for a rigorous proof
of this formula.

We shall require the {\sl $q$-binomial formula} (\cf. [GR, p. 7]) :  
 $$ 
\sum_{n=0}^{+\infty}{(a;q)_n\over (q;q)_n}x^n
={(ax;q)_\infty\over (x;q)_\infty}
\qquad(|x|<1).\leqno (0.4) 
$$

There are two classical $q$-analogs of the exponential function
defined as follows :
$$
e_q(x)=\sum_{n=0}^{+\infty}{x^n\over[n]!}\qquad{\textstyle for}\qquad
|x|<{1\over 1-q}, 
$$
and 
$$
E_q(x)=\sum_{n=0}^{+\infty}{q^{{n\choose 2}}x^n\over[n]!}
\qquad{\textstyle for}\qquad x\in{\bb C}. 
$$ 
By the $q$-binomial formula (0.4) we may write
$$
e_q(x)={1\over ((1-q)x; q)_\infty},\qquad
E_q(x)=((q-1)x; q)_\infty. \leqno (0.5)
$$
Hence $e_q(x)\cdot E_q(-x)=1$. 

As usual, for any function $f(x)$ of $x$, we define
the {\sl shift operator} $E: \,Ef(x)=f(x+1)$, the 
{\sl identity operator}
$I: \, If(x)=f(x)$, and the {\sl $q$-difference operator} by means
of 
$$
\Delta_q^0=I, \quad \Delta_q^n=(E-q^{n-1}I)(E-q^{n-2}I)
\cdots (E-I).
$$
It is readily seen that
$$
\Delta_q^nf(x)=\sum_{k=0}^n(-1)^k{n\brack k}q^{{k\choose 2}}
f(x+n-k).\leqno(0.6)
$$

We call $q$-{\sl Newton's series} a series of the form 
$$
\sum_{n=0}^{+\infty}a_n{x-a \brack n},
\leqno(0.7) 
$$
where $a\in \bb C$. We remark that if series (0.7) converges to a
function $f(x)$ in an open region containing ${\bb N}$, then the
coefficients $a_n$ are given by (0.6) with $x=a$, \ie., 
$a_n=\Delta_q^nf(a)$. For this reason, we say also that
series (0.7) is the $q$-{\sl Newton's interpolation series}
associated to the function $f(x)$.

\proclaim Definition 1.
For $\alpha \in{\bb C}$ and $n\in{\bb N}$, 
the {\sl $q$-Stirling function of second kind} is defined by
$S_q(\alpha ,0)=\delta_{\alpha ,0} $ and for $n\ge 1$
$$\leqalignno{
S_q(\alpha ,n)&={1\over [n]!}\left.\Delta_q^n
[x]^\alpha \right|_{x=0}\cr
&={1\over [n]!}
\sum_{k=0}^{n-1}(-1)^k{n\brack k}q^{{k\choose 2}}
[n-k]^\alpha. &(0.8)\cr}
$$

\remark The $q$-Stirling function $S_q(\alpha ,n)$
was first introduced by \pc Carlitz| [Ca1] in the case
$\alpha \in \bb N$.

\proclaim Definition 2.
For $\alpha \in \bb C$ and $n\in \bb N$, the {\sl $q-$Eulerian
function} is defined by  $$
A_q(\alpha , n)=\sum_{k=0}^n(-1)^k{\alpha +1\brack k}
q^{{k\choose 2}} [n+1-k]^\alpha . \leqno (0.9)
$$

\remark The $q$-Eulerian function $A_q(\alpha , s)$ was
also first introduced by \pc Carlitz| [Ca2] in the case
$\alpha \in \bb N$. It may be interesting to recall here the 
combinatorial motivation of this
definition. Let $\goth S_n$ be the set of permutations of
 $\{1, 2,\cdots, n\}$. For $\sigma \in \goth S_n$ we say that $\sigma $
has a {\sl descent} at $i$, $1\le i\le n-1$, if $\sigma (i)>\sigma (i+1)$
and define
$$
\Des \sigma =\{i\,|\,\sigma (i)>\sigma (i+1)\}\quad{\textstyle and }\quad
\maj \sigma =\sum_{i\in \Des \sigma }i.
$$
\smallskip
For exemple, if $\sigma =4\,1\,3\,6\,5\,2\in \goth S_6$, then
$\Des \sigma =\{1, 4, 5\}$ and so $\maj \sigma =1+4+5=10$.
Let $\goth S_{n,k}$ be the set 
of permutations with $k$ descents. 
The $q$-Eulerian numbers $A_q(n,s)$ ($0\le s\le n$)
[Ca1, Ca2, Raw] is then the generating function of the $\maj$ 
on $\goth S_{n,k}$ :
$$
A_q(n,s)=\sum_{\sigma \in \goth S_{n,k}}q^{\maj \sigma }.
$$

\medskip
This paper was motivated by the recent works of \pc
Butzer| et al. [BH1, BH2, BHS], who investigated various
properties of some well-known combinatorial numbers
and functions such as {\sl factorial functions}, {\sl binomial
coefficients}, {\sl Stirling numbers} and {\sl Eulerian numbers} in
the case that the {\sl integer} $n$ is replaced by a {\sl real}
$\alpha$.  As is well-known (probably only for combinatorialists),
most combinatorial functions have natural ``$q$-analogues", so it
is natural to ask whether there are some ``$q$-analogues" of the
works of  \pc Butzer| et al. [{\sl loc. cit.}]. 

The purpose of this paper is to study the $q$-Stirling numbers
$S_q(n, k)$ of second kind and the $q$-Eulerian numbers
$A_q(n,k)$ when the integer $n\in \bb N$ is replaced by a complex
$\alpha \in \bb C$. As we shall see, most properties of these
numbers or functions remain true and can be proved
almost {\sl mutadis mutandis} in the same way as the restrictive
case $n\in \bb N$, except the two following formulae :
$$
\ldisplaylinesno{
\sum_{k=0}^{+\infty} S_q(\alpha , k)
 [x]_k=[x]^\alpha; &(0.10)\cr
\sum_{k=0}^{+\infty} A_q(\alpha ,k)
=\Gamma_q(\alpha+1).&(0.11)\cr 
} 
$$

Note that the proofs of the above two formulae are {\sl not easy}
in the case $q=1$. Indeed, for $q=1$, formula (0.10) for $\Re
x>-1$ was proved in [BHS] by a deep result on Newton's series,
while formula (0.11) was proved in [BH2] for $x\in(-1,+\infty)$ to
be equavelent to a  result of fractional calculus and also proved
in [Zh] in the half plane $\Re x>-1$ by combining a Tauberian
theorem and some fine analytic manipulations. In this paper, we
shall investigate the validity of the two formulae (0.10) and
(0.11).

This paper is organized as follows. We shall first 
introduce a {\sl $q-$analog of the Newton
interpolation series} in the first section, where we obtain a
{\sl $q$-analog of the formulae of \pd Cahen and \pd Pincherle for
the abscissa of convergence} (see Theorem 1.3). These results may
have some interests on themselves. 
In the second section, we review some results about the growth
of entire functions and then apply them to the study of the
convergence of $q$-Newton's series.
The third section will be devoted to the
$q-$Stirling functions, where we show especially that their 
generating function has $q$-{\sl exponential growth of order one
and of finite type} and give the {\sl convergence region}
of (0.10) (see Theorem 3.4 and Theorem 3.7). Finally,
the $q-$Eulerian functions will be studied in the last section,
where we shall determine the {\sl convergence region} of (0.11) in
particular (see Theorem 4.4).

{\parindent 0mm
\sectiona
1. A $q$-Newton interpolation series
\endsection

Consider the following $q$-Newton series : 
$$
a_0+a_1{x-1\brack 1}+
a_2{x-1\brack 2}+a_3{x-1\brack 3}+\cdots.
\leqno(1.1)   
$$ 
}

Note that, if $q\to1^-$, 
series (1.1) reduces {\sl formally}  to
$$
a_0+a_1{x-1\choose 1}+a_2{x-1\choose 2}
+a_3{x-1\choose 3}+\cdots,
$$
whose abscissa of convergence, say $\sigma$, is well-known (\cf.
[Ge, p. 133]) :  
$$ \leqalignno{
\sigma =\limsup_{n\to+\infty }{\log\left|\sum
_{k=0}^n(-1)^ka_k\right|\over \log n}
\qquad &\hbox{if }\quad \sigma \ge 0; &(\hbox{Cahen})\cr
\sigma =\limsup_{n\to+\infty }{\log\left|\sum _{k=n}^{+\infty}
(-1)^ka_k\right|\over \log n}\qquad &\hbox{if }\quad \sigma <
0. &(\hbox{Pincherle})\cr }
$$

For the latter use, set
$$\leqalignno{
b_n(x)&=a_n{x-1\brack n},&(1.2)\cr
c_n(x,y)&={b_n(x)\over b_n(y)}={[x-1]_n\over
[y-1]_n}\virg&(1.2a)\cr 
}$$
where $x, y\in{\bb C}$ and $y$ is a {\sl non positive integer}. 

\proclaim Lemma 1.1.
For given $x, y\in{\bb C}$ and $y\not=1$, 2, 3, $\ldots $, there
exist $A_1:=A_1(x,y)$, $A_2:=A_2(x,y)>0$ such that for
all $n\in{\bb N}$, 
$$
\leqalignno{
&\left|c_n(x,y)\right|\leq A_1q^{n\Re (x-y)},&(1.3)\cr
&\left|c_{n+1}(x,y)-c_n(x,y)\right|\leq A_2q^{n\Re
(x-y)}.
&(1.3a)} $$

\proof
 By definition (1.2a) we have 
$$
c_n(x,y)q^{-n(x-y)}=\prod_{j=1}^n{1-q^{j-x}\over
1-q^{j-y}}\virg
 $$
which is then {\sl bounded} by a positive function $A_1:=A_1(x,y)$
since it converges to ${(q^{1-x};q)_\infty/
(q^{1-y};q)_\infty}$ when $n\to +\infty $. Thus we have proved
(1.3). As to (1.3a), it suffices to take $A_2:=2A_1(x,y)$. \qed

\medskip
\proclaim Proposition 1.2.
If $q$-Newton's series (1.1) converges at $x_0
\not=1,2,\cdots $, then it converges for every $x$ such that
 $\Re x>\Re x_0$.  

\proof
By (1.2) and (1.2a), we may rewrite series (1.1) 
as $\sum_{n=0}^{+\infty}b_n(x_0)c_n(x,x_0)$. 
Since the series $\sum_{n\ge 0}b_n(x_0)$ is convergent,
so, by  du \pc Bois|-\pc Reymond|'s test (\cf. [Ge, p. 126] or [Kn,
p. 315]), it suffices to prove the convergence of the series
$$
\sum_{n=0}^{+\infty}|c_{n+1}(x,x_0)-c_n(x,x_0)|
$$ 
in the half plane $\Re x>\Re x_0$. Now, the latter is dominated by
the convergent geometric series $\sum_{n=0}^{+\infty}Aq^{n(\Re x-\Re
x_0)}$ in view of (1.3a). \qed 

\medskip
Therefore, we may define the {\sl convergence
abscissa} of $q$-Newton's series (1.1) by
$$
\sigma =\inf\{\sigma _0\ |\ \hbox{(1.1) converges 
for every $x$ such that $\Re x>\sigma _0$}\}. 
$$

The following theorem gives a $q$-analogue of the formulae of
\pd Cohen and \pd Pincherle.

\proclaim Theorem 1.3.
 Let $\sigma $ be the convergence abscissa of (1.1) and set
$$\leqalignno{
\alpha &=\limsup_{n\to+\infty }{\log\left|\sum
_{k=0}^n(-1)^ka_kq^{-{k+1\choose 2}}\right|\over n\log
q^{-1}};&(1.4)\cr
\beta &=\limsup_{n\to+\infty }{\log\left|\sum
_{k=n}^{+\infty }(-1)^ka_kq^{-{k+1\choose 2}}\right|\over 
n\log q^{-1}}\cdot&(1.5)\cr
}$$
Then $\sigma =\alpha $ if $\sigma \ge 0$,  
and $\sigma =\beta$ if $\sigma < 0$.

\proof
We shall only consider the 
$\sigma \ge 0$ case and leave the $\sigma < 0$ case 
to the interested reader, who would complete
the proof without difficulty on refering to
the counterpart of ordinary Newton's interpolation series
in [Mi, p. 281-283]. 

We divide the {\sl proof in two parts}. Suppose that series
(1.1) converges at a point $x_0$, where $x_0$ is a {\sl non
positive integer}. Let $\Re x_0=\sigma _0$.

For each $x\in{\bb C}$ and $n\geq 0$, set  
$$
\lambda _n(x)=\sum _{k=0}^na_k{x-1\brack k},\qquad
\mu_n(x)=\sum_{k=n}^{+\infty} a_k{x-1\brack k}.
$$
Note that
$$
\lambda _n(0)=\sum
_{k=0}^n(-1)^ka_kq^{-{k+1\choose 2}},\qquad
\mu _n(0)=\sum_{k=n}^{+\infty }(-1)^ka_kq^{-{k+1\choose
2}}.
$$

Using the notations in (1.2), (1.2a) and by Abel's Identity [Mi, p.
276-277], we may write $\lambda _n(x)-\lambda _l(x)$,
 $n>l\geq 0$, as follows :
$$
\leqalignno{
\sum_{k=l}^nb_k(x_0)c_k(x,x_0)&=\sum
_{k=l}^n(c_{k-1}(x,x_0)-c_k(x,x_0))\mu_k(x_0)&(1.6)\cr &
\hskip 2cm -c_{l-1}(x,x_0)\mu _l(x_0)+c_n(x,x_0)\mu _{n+1}
(x_0),\cr 
\sum _{k=l}^nb_k(x_0)c_k(x,x_0)&=\sum
_{k=l}^n(c_k(x,x_0)-c_{k+1}(x,x_0))\lambda_k(x_0)&(1.7)\cr
&\hskip 2cm -c_l(x,x_0)\lambda
_{l-1}(x_0)+c_{n+1}(x,x_0)\lambda _{n}(x_0),\cr}
$$
where $x_0\in{\bb C}$ and $x_0\not = 1, 2, 3,\cdots$. 

\medskip
(i) We prove that if $\sigma_0 >0$, then $\alpha \le\sigma_0$ ;
and consequently that $\alpha \le\sigma$, if $\sigma \ge0$. 

\medskip
Since (1.4) can be written as
$$
\alpha =\limsup_{n\to +\infty}{\log|\lambda _n(0)|\over n\log
q^{-1}}\virg 
$$
it is then sufficent (\cf. [Mi, p. 277-278]) to prove
$$
\lim_{n\to+\infty }q^{n\sigma _0}\lambda
_n(0)=0.\leqno(1.8)
$$ 

By hypothesis, the series $\sum b_n(x_0)$ converges, and hence
given $\varepsilon>0 $, we have $l\in\bb N$ such that  $|\mu_n
(x_0)|<\varepsilon$ if $n\ge l$. Thus
 we deduce from (1.6) and (1.3a) that
for $\Re x< \sigma _0$, 
$$\eqalign{
|\lambda _n(x)| &\leq |\lambda _l(x)|+A_2\varepsilon 
\left(\sum_{k=l+1}^n q^{k(\Re x-\sigma _0)}
+q^{l(\Re x-\sigma_0)}+q^{n(\Re x-\sigma_0)}\right)\cr
&\leq |\lambda_l(x)|+2 A_2\varepsilon 
{q^{(n+1)(\Re x-\sigma_0)}-q^{l(\Re x-\sigma _0)}
\over q^{\Re x-\sigma_0}-1}\cdot
\cr}\leqno(1.9)
$$

Noting that $\lim_{n\to+\infty}q^{n(\sigma_0 -\Re x)}=0$, 
therefore, if we multiply (1.9) by $q^{n(\sigma_0 -\Re x)}$
and let first $n\to+\infty$, and then 
$\varepsilon\to 0$, we get then
$$
\limsup_{n\to+\infty }q^{n(\sigma _0-\Re x)}\lambda_n(x)=0. 
$$
which yields clearly (1.8) by setting $x=0$. 

\medskip
(ii) Without the loss of generality, we suppose that $\alpha $
is finite. We prove that series (1.1)
converges at $\Re
x=\varepsilon+\alpha $ for any $\varepsilon >0$. This
implies that $\alpha \ge \sigma $.  

\medskip
Let $\Re x=\alpha +\varepsilon $ with $\varepsilon >0$. It
suffices to prove that the partial sum $\lambda_n(x)$ of (1.1)
converges when $n\to +\infty$. 

To prove this, setting $x_0=0$ and $l=1$ in (1.7) we have  
$$ 
\lambda_n(x)=\sum_{k=1}^n(c_k(x,0)-c_{k+1}
(x,0))\lambda_k(0)+(1-c_1(x,0))\lambda
_0(0)-c_{n+1}(x,0)\lambda_n(0). 
$$ 
Since $\alpha $ is finite, it follows from (1.4) that
for sufficiently large $n$, we have 
$$
|\lambda _n(0)|\leq
q^{-n(\alpha +\varepsilon/2)}.\leqno(1.10) 
$$
Combining (1.3) and (1.10), we have, for sufficiently
large $n$,
$$
\left|c_{n+1}(x,0)\lambda _n(0)\right|\leq A_1
q^{\alpha +(1+n/2)\epsilon}.
$$
Since the second member of the above inequality tends to
$0$ when $n\to +\infty$, it remains to prove the
convergence of the series  
$\sum_{k=0}^{+\infty} (c_k(x,0)-c_{k+1}(x,0))\lambda_k(0)$, 
but this is obvious in view of (1.3a) and (1.10). \qed

\medskip
\proclaim Proposition 1.4.
Let $\sigma$ be the convergence abscissa of (1.1).
If $\sigma $ is finite,
then series (1.1) defines a bounded regular
function in every $\delta $-half plane 
$\Re x\geq \sigma+\delta $, where
$\delta >0$. 

\proof
Let $x_0$ be a non integral point and
 $\Re x_0=\sigma _0=\sigma +\delta/2$ with $\delta >0$.
Set $M=\sup_{n\geq 0}\{|\mu
_n(x_0)|\}$, then $M$ is finite, since series (1.1) with $x=x_0$ is
convergent.  As in (1.9), it follows from (1.6) with $l=0$ and (1.3a)
that 
$$\eqalign{
|\lambda _n(x)|&<|a _0|+2A_2M{q^{(n+1)(\Re x-\sigma
_0)}- 1\over q^{\Re x-\sigma_0}-1}\cr
&<|a_0|+ {2A_2M\over 1-q^{\Re x-\sigma_0}}.\cr}
$$
where $\Re x> \sigma _0$ and $n\geq 0$.
The sum of series (1.1) is then bounded by
$|a_0|+ 2A_2M/(1-q^{\delta /2})$
in the half plane $\Re x\geq \sigma +\delta $. \qed

\medskip
In the same manner, we can define the {\sl absolute convergence
abscissa} of a $q$-Newton's series (1.1) by
$$
\bar\sigma=\inf\{\sigma _0\ |\ \hbox{(1.1) converges 
absolutely for every $x$ such that $\Re x>\sigma _0$}\}.
$$ 
It is obvious that $\bar\sigma\ge\sigma$. Similarly, we can prove
the following 

\proclaim Theorem 1.5.
The absolute convergence abscissa $\bar\sigma $ of (1.1) is given
by 
$$\leqalignno{
\bar\sigma &=\limsup_{n\to+\infty }{\log\sum
_{k=0}^n\left |a_kq^{-{k+1\choose 2}}\right|\over n\log
q^{-1}},
\qquad
\hbox{if } \bar\sigma \ge0;&(1.11)\cr
\bar\sigma &=\limsup_{n\to+\infty }{\log\sum
_{k=n}^{+\infty }\left |a_kq^{-{k+1\choose 2}}\right|\over 
n\log q^{-1}},
\qquad
\hbox{if }\bar\sigma <0.&(1.12)\cr
}$$

\medskip 
It should mention that there remains a lot to do with
$q$-Newton's series. Some natural questions are : What condition
should a function satisfy in order to be represented by a
$q$-Newton's series ? What is the relation between
the convergence abscissa and the absolute convergence abscissa
of a $q$-Newton's series ? 
Actually, there exist several $q$-analogues of Newton's series
(see, for exemple, [Wa]). It should be interesting to
compare these different $q$-Newton series and study their
relations.


{\parindent 0mm
\sectiona
2. Some lemmas about the growth of entire functions
\endsection

We shall first give some definitions about 
the $q$-{\sl exponential growth} of
entire functions following \pc Ramis| [Ram].}

\proclaim Definition 3. Let be given a non zero real
number $k$.  A power series   
$$  
a_0+a_1x+a_2x^2+a_3x^3+\cdots \leqno(2.1) 
$$ 
is said to be $q$-Gevrey of order $s=1/k$, if there exist real
numbers $A, K>0$ such that 
$$
\left|a_n\right|<KA^nq^{n(n+1)\over 2k}.\leqno(2.2)
$$

Thus formulae (1.11) and (1.12) imply immediately that
if the power series (2.1) is $q$-Gevrey of {\sl order one},
then the absolute convergence abscissa of the series (1.1) is
finite. The following lemma has been given in [Ram, Lemma 2.2, ii)].

\proclaim Lemma 2.1.
Let $f(x)=\sum _{n=0}^{+\infty}a_nx^n$ be an entire
function. If there exist real numbers $K, k>0$, $\mu $ such that 
$$
\vert f(x)\vert\leq Ke^{{k\over 2}(\log\vert x\vert)^2+\mu
\log\vert x\vert}\qquad {\textstyle when}\quad
\vert x\vert\to+\infty , 
$$
then
$$
\vert a_n\vert\leq Ke^{-{(n-\mu )^2\over 2k}}\qquad
{\textstyle for}\quad n\geq 0.
$$

\proclaim Definition 4. 
Let $f$ be an entire function. If there exist real numbers
$k\not = 0$ and $\mu $ such that, for a suitably chosen $K>0$,  
$$
|f(x)|<Ke^{-{1\over 2k}{\log^2 |x|\over \log q}+\mu 
\log|x|}\qquad {\textstyle when}\quad \vert x\vert\to+\infty,
\leqno(2.3)  
$$
we say that $f$ has $q$-exponential
growth of order $k$ and of finite type $\mu $.

\remark Comparing the above definition with that in
\pc Ramis| [Ram, p. 71], we have changed his $q^{-1}$ by $q$.
Note also that
for the same entire function $f(x)$, there may exist
several pairs of numbers $(k,\mu)$ satisfying (2.3). The largest
$k$ is called the {\it precise order of $q$-exponential growth}.


\medskip
The following result has been
given in [Ram, Prop. 5.5] and [Wa], but the proof given below
seems {\sl new}.

\proclaim Proposition 2.2.
The $q$-exponential function, $E_q(x)$, has $q$-exponetial
growth of order 1 and of finite type, more precisely 
$$
\left|E_q(x/(1-q))\right|=O(e^{-{\log^2|x|\over 2\log
q}+{\log |x|\over 2}})\qquad
\hbox{when}
\quad
|x|\to+\infty .\leqno(2.4) 
$$

\proof 
Since $|E_q(x)|\le E_q(|x|)$ in view of (0.5), and in particular
$|E_q(x)|=E_q(x)$ for real numbers $x>0$, we then need only to
consider the growth of the function $E_q(x)$ along the real
positive axe. For $x\in(0,+\infty )$, define the function
$$
f(x)=E_q(x/ (1-q))e^{{\log^2x\over 2\log
q}-{\log x\over 2}}\prod_{n=1}^{+\infty}
(1+q^n/x). \leqno(2.5)
$$
It is easy to check that 
$$
f(qx)=f(x)=(-x;q)_\infty(-q/x;q)_\infty{1\over
\sqrt{x}}e^{\log^2x\over 2\log q}, 
$$
so the function $f(x)$ is somewhat ``$q$-periodic".
Therefore, if we define 
$$
H=\min_{x\in[q,1]}\left \{f(x)\right \}
\qquad{\textstyle
and}\qquad 
K=\max_{x\in[q,1]}\left \{f(x)\right \},
$$
we should have
$$
0<H\leq f(x)\le K \qquad \hbox{for all} \quad x>0. 
 $$
On the other hand, noticing that
$$
\lim_{x\to +\infty }\prod_{n=1}^{+\infty} (1+q^n/x)= 1,
$$
 we derive from (2.5) that for sufficiently large $x$,
$$
{1\over 2}He^{-{\log^2x\over 2\log
q}+{\log x\over 2}}<E_q(x/ (1-q))<2Ke^{-{\log^2x\over 2\log
q}+{\log x\over 2}},\leqno(2.6)
$$
which is clearly equivalent to (2.4) for $x\in(0,+\infty )$ by
definition. \qed

\medskip
\proclaim Proposition 2.3. 
Let $f(x)$ be an entire function. If there exist real positive
numbers $\mu$, $k$, $K$, $d>0$ and $p>1$ 
such that for $|x|\in\{d,dp,dp^2,\cdots \}$
 $$
|f(x)|<Ke^{-{\log^2|x|\over 2k\log q}+\mu \log|x|}
\qquad {\textstyle when}\quad \vert x\vert\to+\infty,
$$
then $f(x)$ has 
$q$-exponential growth of order $k$ and of finite type 
$$
\mu ^\prime=\mu-{\log p\over k\log q}.\leqno (2.7) 
$$

\proof
Let $r_n=dp^n$ for $n\in \bb N$.
By hypothesis, for any $x\in \bb C$, there exists an
integer $n\geq 1$ such that 
$$
r_n\leq |x|<r_{n+1}.
$$
 Therefore, if $|x|$ is sufficiently large, 
by the maximum principle of analytic functions (\cf., for
exemple, [Ti, p. 165]),  we have 
$$\eqalign{
|f(x)|&<K
e^{-{\log^2|r_{n+1}|\over 2k\log q}+\mu \log|r_{n+1}|}\cr
&<Ke^{-{\log^2|xp|\over 2k\log q}+\mu \log|xp|}\cr
&\leq K'e^{-{\log^2|x|\over 2k\log q}+
(\mu -{\log p\over k\log q})\log|x|},\cr
}$$
where $K'=Ke^{-{\log^2p\over 2k\log q}}$. \qed

\medskip
Now we formulate a sufficient condition for the
absolute convergence abscissa of (1.1) to be {\sl finite} in terms
of the $q$-{\sl exponential growth} of functions.

\proclaim Proposition 2.4. 
 Let $\bar\sigma$ be the absolute convergence abscissa of series
(1.1). If the entire funcion $f(x)=\sum_{n=0}^{+\infty}a_nx^n$
has $q$-exponential growth of order 1 and of finite 
type $\mu $, then 
$$
\bar\sigma \leq \mu +{1\over 2}\cdot\leqno(2.8)
$$

\proof
By hypothesis and (2.3) we have 
$$
|f(x)|<Ke^{-{1\over 2}{\log^2 |x|\over \log q}+\mu 
\log|x|}\qquad {\textstyle when}\quad \vert x\vert\to+\infty.
$$
Applying Lemma 2.1 with $k=-1/\log q$ yields that 
there exists $K>0$ such that 
$$
\vert a_n\vert\leq Kq^{(n-\mu )^2\over 2}.
$$
We then get (2.8) from Theorem 1.5. \qed

\medskip
In order to illustrate the above proposition,
we end this section with the following example. 
Consider the $q$-Newton's interpolation series
associated to the $q$-exponential function $E_q(x)$ :
$$
\sum_{n=0}^{+\infty} {q^{n\choose 2}\over [n]!}{x-1\brack
n}\leqno(2.9) 
$$
By Proposition 2.2, we have
$$
\left|E_q(x)\right|=O(e^{-{\log^2|x|\over 2\log
q}+\mu \log |x|})\qquad (|x|\to +\infty),
$$
where $\mu ={1\over 2}- \log _q(1-q)$. Therefore,
in view of (2.8), the absolute convergence abscissa of (2.9) is
less than or equal to $1- \log_ q(1-q)$.

{\parindent 0mm
\sectiona 
3. $q$-Stirling function of second kind
\endsection

By definition 1 and (0.8), it is
readily seen that the $q-$Stirling functions satisfy the 
following recurrence : 
$$
S_q(\alpha +1, k+1)=q^kS_q(\alpha ,k)+[k+1]S_q(\alpha ,k+1),
$$
where $\alpha \in \bb C$ and $k\geq 1$.}

The first four values of $q-$Stirling functions are as follows :
$$\eqalign{
S_q(\alpha,0)&=\delta _{\alpha , 0}, \cr
S_q(\alpha ,1)&=1,\cr 
S_q(\alpha ,2)&=(1+q)^{\alpha -1}-1,\cr
S_q(\alpha ,3)&={1\over
1+q}((1+q+q^2)^{\alpha -1}-q^3(1+q)^\alpha +q).\cr} 
$$
\smallskip
For convenience, we set $[0]^\alpha=\delta
_{\alpha , 0}$ in what follows. 

\medskip
Let us introduce for $\alpha \in \bb C$, $|x|<1/(1-q)$ 
the generalized $q$-Stirling ``polynomials"
$$
f(\alpha ,x)={1\over
e_q(x)}\sum_{k=0}^{+\infty }{[k]^\alpha \over [k]!}x^k,\leqno
(3.1) $$
the radius of convergence of the series being $1/(1-q)$. 

\proclaim Proposition 3.1.
For $|x|<1/(1-q)$ and $\alpha \in{\bb C}$ we have
$$
f(\alpha ,x)=\sum_{n=0}^{+\infty } S_q(\alpha ,n)x^n. 
\leqno(3.2)
$$

\proof
This follows by comparing the
coefficients of $x^k$ of both sides in (3.2) and then the
definition of $q$-Stirling functions (0.8). \qed
\medskip
\proclaim Corollary 3.2.
For $\alpha \in{\bb C}$,  $|x|<1$ we have
$$\ldisplaylinesno{
\sum _{l=0}^k{k\brack l}[l]!S_q(\alpha, l)=[k]^\alpha
\qquad(k\ge 0),&(3.3)\cr
\sum_{k=0}^{+\infty}{S_q(\alpha, k)[k]!x^k\over (x;q)_{k+1}}
=\sum_{k=0}^{+\infty}x^k[k]^\alpha.&(3.4)\cr
}$$

\proof
Multiplying both sides of (3.2) by $e_q(x)$ yields
$$
e_q(x)\sum_{k=0}^{+\infty}S_q(\alpha ,k)x^k=
\sum_{k=0}^{+\infty }{[k]^\alpha \over [k]!}x^k,
$$
which is equivalent to (3.3) by identifying the coefficients of
$x^k$.
Next, note first that the series on the right side of (3.4) is 
convergent for $|x|<1$. By the
$q$-binomial formula (0.4), for $|x|<1$ we have  
$$
{1\over (x;q)_{k+1}}=
\sum _{l=0}^{+\infty }{k+l \brack l}x^l\cdot
$$
Substituting this into (3.4) and extracting the coefficients of
$x^k$ of both sides, we see that (3.4) is equivalent to (3.3).
\qed

\medskip
\remark If $x=1$, Proposition 3.1 reduces to  
$$
\sum_{k=0}^{+\infty}S_q(\alpha, k)={1\over
e_q(1)}\sum_{k=0}^{+\infty} {[k]^\alpha \over [k]!}\cdot
\leqno (3.5) 
$$
Note further that if $\alpha \in \bb N$ and $q=1$,
the left-hand side of (3.5) is actually the {\sl Bell numbers}
and (3.5) reduces to {\sl Dobinski's formula} (\cf., [Co, p. 45]).
In the case of $\alpha \in \bb N$, Proposition 3.2 has been 
proved in [Ze] as a $q$-analogue of Touchard's formula refining
{\sl Dobinski's formula}. 
Therefore, as suggested by the referer, 
(3.5) could be denoted as $q$-{\sl Bell numbers}.

\medskip
\proclaim Corollary 3.3.  
For all $\alpha \in \bb C$,  the function $f(\alpha ,x)$ is 
an entire function of $x$.

\proof
Proposition 3.1 shows that the function $f(\alpha ,x)$ is
analytic in the disc $|x|<1/(1-q)$. Therefore, in order
to prove the analyticity of $f(\alpha ,x)$ in the whole plane, it
suffices to do this for the right-hand side of (3.2).
Indeed, for any $n\in\bb N$, we may write  
$$
[k]^\alpha =\left({1-q^k\over 1-q}\right)^\alpha=
(1-q)^{-\alpha }\sum_{j=0}^n (-1)^j {\alpha \choose
j}q^{jk}+R(k,n)\quad(k\ge1),  
$$  
where, by Taylor's formula,  
$$
R(k,n)=(1-q)^{-\alpha}{\alpha \choose n+1}\theta
^{n+1}(1-\theta q^k)^{\alpha -n-1}q^{kn+k}
$$
for some $\theta\in (0,\, 1)$. Set  
$$
I_n(x)=\sum_{k=0}^n(-1)^k
{\alpha \choose k}{e_q(q^kx)\over
e_q(x)}\virg
\qquad
J_n(x)={1\over e_q(x)}
\sum_{k=0}^{+\infty }{R(k,n)\over
[k]!}x^k\virg\leqno(3.6)
$$
and write
$$
f(\alpha ,x)={1\over e_q(x)}\sum _{k=0}^{+\infty}{[k]^\alpha
\over [k]!}x^k =I_n(x)+J_n(x). \leqno(3.7)
$$

Note that  for $0<\theta <1$ and $k\ge1$, we have
$1>1-\theta q^k>1-q^k\geq 1-q$. Let
$T_0=\max\left\{(1-q)^{\Re\alpha},1\right\}$, a constant
independent of $k$ and $n$,  then 
$$ 
\left |(1-\theta q^k)^{\alpha
-n-1}\right | <T_0(1-q^k)^{-n-1}. 
$$
It follows that
$$
|R(k,n)|\le T_0(1-q)^{-\Re \alpha }\left({q^k\over
1-q^k}\right)^{n+1}\left |{\alpha \choose n+1}\right|
\qquad (k\geq 1),
\leqno(3.8)
$$
which implies the convergence of the series 
$\sum _{k\ge 0} R(k,n)x^k/[k]!$ for $|x|<1/q^{n+1}(1-q)$.
Besides, the functions $1/e_q(x)$, $e_q(q^jx)/e_q(x)$ 
($j\in{\bb N}$) are clearly regular for all $x\in{\bb C}$, 
so $I_n(x)$ and $J_n(x)$ are both analytic in the disc
$|x|<1/q^{n+1}(1-q)$ for every $n\in{\bb N}$, which 
implies the
analyticity of $f(\alpha , x)$ in the whole plane. \qed

\medskip
\proclaim Theorem 3.4. 
If $\Re\alpha >-1$, then $f(\alpha ,x)$
has $q$-exponential growth of order 1 and of type 
 $$
\mu ={3\over 2}+\log_q(1-q).\leqno(3.9)
$$ 

\proof
In view of (3.7) we have  
$$
|f(x,\alpha )|\le |I_n(x)|+|J_n(x)|.\leqno(3.10)
$$

Let us first estimate $|I_n(x)|$. By {\sl
Stirling's formula} about {\sl Gamma function} (\cf., [Mi, p. 254]),
we have
$$
\left |{\alpha \choose n}\right |=O(n^{-\Re\alpha -1})\qquad
{\textstyle when}\quad n\to+\infty .
$$
Hence, for $\Re \alpha >-1$, there exists a constant $K_1>0$
such that   
$$
\left |{\alpha \choose n}\right
|<K_1\qquad 
\forall n\in{\bb N}.\leqno(3.11)
$$
On the other hand, by (0.5) we have
$$
\left|{e_q(q^kx)\over e_q(x)}\right|=|(1-(1-q)x)\cdots 
(1-(1-q)q^{k-1}x^k)|\leq E_q(|x|).
$$
It then follows from
(3.6) and (3.11) that 
$$
|I_n(x)|\le\sum_{k=1}^n\left|{\alpha \choose
k}\right|\left|{e_q(q^kx)\over e_q(x)}\right|\leq nK_1 E_q(\vert
x\vert).\leqno(3.12)  
$$

Next, we shall estimate $|J_n(x)|$ on the circles :
$$
|x|=r_n={1\over 2}q^{-n-1}.\leqno (3.13)
$$
Applying
(3.11) to (3.8) implies that there exists a constant $K_2>0$,
independent of $k$ and $n$, such that    
$$
|R(k,n)|<K_2\left({q^k\over 1-q^k}\right)^{n+1}
\qquad (k\geq 1).
$$
Noting also that 
$1/|e_q(x)|=|E_q(-x)|\leq E_q(|x|)$, 
therefore, it follows from (3.6) that
$$
|J_n(x)|\le {K_2\over |e_q(x)|}\sum _{k=1}^{+\infty} 
{\left |q^{n+1}x\right |^k\over [k]!(1-q^k)^{n+1}}\le
K_2E_q(1/2)(1-q)^{-n-1}E_q(r_n),\leqno(3. 14)
$$
since $|q^{n+1}x|=1/2$ by (3.13). 
Combining (3.12) and (3.14) yields that for $|x|=r_n$,
$$
|f(\alpha,x)|\leq
\left(nK_1+K_2E_q(1/2)(1-q)^{-n-1}\right)
E_q(r_n).\leqno(3.15)
$$
Now, by (2.4), there exists a constant $C>0$ such that 
$$
|E_q(x)|<Ce^{-{\log^2|x|\over 2\log q}+({1\over
2}-\log_q(1-q))\log |x|}\leqno (3.16)
 $$
Note also that for sufficiently large $n$,
$$
n<(1-q)^{-n-1}=e^{2(\log_q(1-q))\log r_n}.\leqno (3.17)
$$
It then follows from (3.15), (3.16) and (3.17) that 
for sufficiently large $n$ and $|x|=r_n$, we have
$$
|f(\alpha ,x)|<Ke^{-{\log^2|x|\over 2\log q}+({1\over
2}+\log_q(1-q))\log |x|}.
$$
The theorem follows then 
according to Proposition 2.3 with $p=q^{-1}$. \qed

\medskip
\proclaim Corollary 3.5. If $\Re \alpha >-1$, then the series
$$
\sum _{n=0}^{+\infty}S_q(\alpha ,n)[x]_n\leqno(3.18)
$$
converges absolutely in the half plane $\Re
x>1+2\log_q(1-q)$.

\proof
It is easy to see that if $\Re \alpha >-1$, the
series 
$$
\sum _{n=0}^{+\infty} [n]!S_q(\alpha ,n)x^n
$$
represents an entire function having a $q$-exponential growth
of order 1 and of finite type $({3\over 2}+2\log_q(1-q))$.
 Hence, by Proposition 2.4
and noting that $[x]_n=[n]!{(x+1)-1\brack n}$, 
the proof is complete. \qed

\medskip
In the rest of this section we shall study the validity of (0.10),
\ie., determine the sum of series (3.18).
The following lemma is classical (\cf., [Ge, p.144]).

\proclaim Lemma 3.6.
Let $\beta>\sigma $, $0 <\eta <\pi $.
Let $f(x)$ be an analytic function 
in the half plane $\Re x>\sigma $. 
Assume that $f(\beta+n)=0$ for all 
$n\in{\bb N}$ and that there
exists $A>0$ satisfying  
$$
|f(\sigma +re^{i\theta })|<Ae^{\eta r}\qquad(r\to +\infty ),
$$
for all $\theta \in[-\pi /2,\pi /2]$. Then, the function $f$
is identically null. 


\proclaim Theorem 3.7.
If $\Re\alpha >-1$, then we have for $\Re
x>1+2\log_q(1-q)$
 $$
[x]^\alpha
=\sum _{n=0}^{+\infty }S_q(\alpha ,n)[x]_n. \leqno (3.19)
$$ 

\proof
In view of Proposition 1.4 and Corollary 3.5, we see that
if $\Re\alpha >-1$, series (3.15)
represents a bounded function in every $\delta $-half plane 
$$
 \Re x\geq 1+2\log_q(1-q)+\delta, \qquad \delta >0. 
$$  
Therefore, the difference
$$
[x]^\alpha -\sum _{n=0}^{+\infty }S_q(\alpha ,n)[x]_n \leqno
(3.20)
$$  
is bounded in every $\delta $-half plane. 
On the other hand, by (3.3) function (3.20) takes value zero at
every real integral point $x\in \bb N$. The
theorem follows then by Lemma 3.6. \qed

\medskip
Comparing (3.19) with its counterpart for $q=1$ [BH1], it is 
reasonable to conjecture that (0.10) has convergence abscissa
zero if $\Re\alpha >-1$. By (1.3) or (1.4) this would be proved if
the following identity 
$$
\sum_{n=0}^N(-1)^nS_q(\alpha ,n)[n]!q^{-{n+1\choose 2}}
=O(q^{-N})\qquad (N\to +\infty)
$$
is true.


{\parindent 0mm
\sectiona
4. $q$-Eulerian function
\endsection

By (0.9) of Defintion 2, it is easy to verify that the
$q$-Eulerian functions satisfy the following recurrence :
$$
A_q(\alpha +1, s)=[s+1]A_q(\alpha , s)+q^s[\alpha
+1-s]A_q(\alpha , s-1), 
$$
where $\alpha \in \bb C$ and $s\geq 1$.}

The first four $q$-Eulerian functions are
$$\eqalign{
A_q(\alpha ,0)&=\delta_{\alpha , 0} , \cr
A_q(\alpha ,1)&=[2]^\alpha -[\alpha +1],\cr 
A_q(\alpha ,2)&=[3]^\alpha -[\alpha +1][2]^\alpha +q{\alpha
+1\brack 2},\cr
 A_q(\alpha ,3)&=[4]^\alpha -[\alpha +1][3]^\alpha
+q{\alpha +1\brack 2}[2]^\alpha  -q^3{\alpha +1\brack 3}.\cr}
$$

\medskip
Let us introduce for $\Re \alpha >-1 $, $|x|<1$ the generalized
$q$-Eulerian ``polynomials"
$$
P_\alpha (x;q)={(x;q)_\infty\over
(q^{\alpha +1}x;q)_\infty}\sum_{n=0}^{+\infty}[n+1]^\alpha 
x^n,\leqno (4.1) $$
the radius of convergence of the series being obviously 1 since
$\lim_{n\to +\infty}\root n\of {|[n+1]^\alpha| }=1$.

\proclaim Lemma 4.1. 
For $\Re \alpha >-1$ and $|z|<1$,  the following identity holds 
$$
P_\alpha (z;q)=\sum_{s=0}^{+\infty}A_q(\alpha , s)z^s.
\leqno (4.2) 
$$

\proof By Definition 2, we have for $|z|<1$
$$
\sum_{s=0}^{+\infty}A_q(\alpha ,
s)z^s=\sum_{i=0}^{+\infty}(-1)^i{\alpha +1\brack i}
q^{{i\choose 2}}z^i\cdot \sum_{j= 0}^{+\infty}
[j+1]^\alpha  z^j,\leqno (4.3)
 $$
the two series of the second member being both convergent
in the open disc $|z|<1$. By the $q$-binomial formula (0.4), we
have for $|z|<1$
$$
\sum_{i=0}^{+\infty}(-1)^i{\alpha +1\brack i}
q^{{i\choose 2}}x^i
=\sum_{i=0}^{+\infty}{(q^{-\alpha -1};q)_i\over
(q;q)_i}(xq^{\alpha +1})^i ={(x;q)_\infty\over (xq^{\alpha
+1};q)_\infty}\cdot $$
The proof is thus complete by putting this into (4.3). \qed

\proclaim Proposition 4.2.
We have for $\alpha \in \bb C$ and $n\geq 0$
$$
[n+1]^\alpha =\sum_{k=0}^n{\alpha +k\brack k}
A_q(\alpha ,n-k),\leqno (4.4) 
$$
$$
\sum_{k=0}^nA_q(\alpha ,k)=\sum_{k=0}^n(-1)^k{\alpha \brack
k}q^{{k+1\choose2}}[n-k+1]^\alpha .\leqno (4.5)
$$

\proof 
By Lemma 4.1, we have, for $\Re \alpha >-1$ and $|x|<1$,
$$
{(q^{\alpha +1}x;q)_\infty\over (x;q)_\infty}
\sum_{s=0}^{+\infty}A_q(\alpha , s)x^s=\sum_{j=
0}^{+\infty}[j+1]^\alpha  x^j, 
$$
and also
$$
{1\over 1-x}\sum_{s=0}^{+\infty}A_q(\alpha , s)x^s=
{(qx;q)_\infty\over (q^{\alpha +1}x;q)_\infty}
\sum_{j=0}^{+\infty}[j+1]^\alpha  x^j. 
$$
We obtain (4.4) and (4.5) for $\Re \alpha >-1$ by identifying 
respectively the coefficients of $x^n$ in the above two
identities. Finally, the restriction $\Re \alpha >-1$ could be
released to $\alpha \in \bb C$ by analytic continuation. \qed

\medskip
\proclaim Proposition 4.3.
 For $\Re\alpha>-1$, $\Re \beta>0$, $z \in \bb C$ and $|x|<1$,
we have
$$
\sum_{n=0}^{+\infty}{P_{\alpha +n\beta }(x;q)\over
(x;q)_{\alpha +n\beta +1}}{z^n\over n!}
=\sum_{n=0}^{+\infty }[n+1]^\alpha \exp([n+1]^\beta z) x^n.\leqno (4.6)
$$

\proof 
By (0.2) and Lemma 4.1, the left side of (4.6) equals, after
substitution of (4.2), 
 $$
\sum_{n=0}^{+\infty}{P_{\alpha +n\beta }(x;q)\over
(x;q)_{\alpha +n\beta +1}}{z^n\over n!}=\sum_{n=0}^{+\infty}{z^n\over
n!} \sum_{j= 0}^{+\infty}[j+1]^{\alpha +n\beta }x^j\virg 
$$
which proves immediately (4.6) after an exchange of the order of
summation. \qed 
\medskip
We now come to the main result of this section.

\proclaim Theorem 4.4. 
We have
$$
\Gamma_q(x+1)=\sum_{s=0}^{+\infty}A_q(x, s) \qquad{\textstyle
for} \qquad \Re x>-1.\leqno(4.7)
$$

\proof
Observe that for $|z|<1$
$$
{(z;q)_\infty\over
(zq^{x+1};q)_\infty}={(zq;q)_\infty\over
(zq^{x+1};q)_\infty}(1-z)\cdot
$$

So, by Lemma 4.1 we have, for $|z|<1$,
$$
\sum_{s=0}^{+\infty}A_q(x, s)z^s
={(zq;q)_\infty\over
(zq^{x+1};q)_\infty}\Big(1+\sum_{j=1}^
{+\infty}\left([j+1]^x-[j]^x\right)z^j\Big).\leqno (4.8) 
$$

On the other hand, since the function ${(zq;q)_\infty/
(zq^{x+1};q)_\infty}$ is analytic for $z\in{\bb C}$ such that
$$
\vert zq\vert<1,\quad
\left\vert zq^{1+x}\right\vert<1,
$$
it is then regular at $z=1$ if $\Re x>-1$.

Furthermore, note that
$$
\lim_{j\to+\infty }\root {j} \of
{\left\vert[j+1]^x-[j]^x\right\vert} =q\ (<1), 
$$
hence the convergence radius of the power series
$
\sum_{j=1}^{+\infty}\left([j+1]^x-[j]^x\right)z^j $
is equal to $q^{-1}\ (>1)$. Therefore, the function defined
by (4.8) is actually regular at the point $z=1$. 
Finally, since
$$
\sum_{j=0}^{+\infty}\left([j+1]^x-[j]^x\right) 
=\lim_{j\to\infty }[j+1]^x=(1-q)^{-x},
$$
we obtain (4.7) by putting $z=1$ in (4.8) and referring to (0.3).
\qed
\smallskip
\remark
When $q\to 1^-$, formula (4.7) 
{\sl formally} reduces to 
$$
\Gamma(x+1)=\sum_{s=0}^{+\infty}\sum_{j=0}^s
(-1)^j(s+1-j)^x{x+1\choose j} \qquad (\Re x> -1),
$$
which has been respectively established by \pc Butzer| and \pc
Hauss| [Bu-Ha] and \pc Zhang| [Zh]. 
\medskip
The following proposition gives a relation between the
$q-$Stirling and the $q-$Eulerian functions.

\proclaim Proposition 4.5.
For $\alpha \in \bb C$, $|x|<1$ we have
$$
\sum_{k=0}^{+\infty}{S_q(\alpha ,k)[k]!x^k\over (x;q)_{k+1}}
=\sum_{k=0}^{+\infty}[k]^\alpha x^k =
{(xq^{\alpha +1}; q)_\infty \over (x;q)_\infty}
\sum_{k=0}^{+\infty}A_q(\alpha , k)x^{k+1}.
$$

\proof
It follows immediately from Lemma 4.1 and Corollary 3.2. \qed

\medskip
If $\alpha =n\in \bb N$, the above formula reduces to the 
$q$-{\sl Frobenius formula} (see [Ga]) :
$$
\sum_{k=0}^n{S_q(n,k)[k]!x^k\over (x;q)_{k+1}}
=\sum_{k=0}^{+\infty}[k]^nx^k=
{\sum_{\sigma \in {\goth S}_n}x^{\Des \sigma +1}
q^{\maj \sigma }\over (x;q)_{n+1}}\cdot\leqno (4.9)
$$
\medskip
\proclaim Proposition 4.6.
For $\alpha \in \bb C$, $|x|<1$ we have
$$\leqalignno{
A_q(\alpha , n)&=(-1)^{n+1}\sum_{k=1}^{n+1}
(-1)^k{\alpha -n\brack n+1-k}
q^{{n-k+2\choose 2}}[k]!S_q(\alpha ,k). &(4.10)\cr
S_q(\alpha ,n)&= {1\over
[n]!}\sum_{k=0}^{n-1}
q^{n(n-k-1)}{\alpha -k-1\brack \alpha -n} 
A_q(\alpha , k).&(4.11)\cr }$$

\proof
By Proposition 4.5 and the $q$-binomial formula (0.4), we have
for $|x|<1$
$$\eqalign{
\sum_{n=0}^{+\infty} A_q(\alpha
,k)x^{n+1}&=\sum_{k=1}^{+\infty}
 {(xq^{k+1}; q)_\infty\over
(xq^{\alpha +1};q)_\infty} S_q(\alpha ,k)[k]!x^k\cr
&=\sum_{k=1}^{+\infty}
S_q(\alpha ,k)[k]!x^k
\sum_{l=0}^{+\infty} {(q^{k-\alpha }; q)_l\over (q;
q)_l}(xq^{\alpha +1})^l.\cr  
}$$
Formula (4.10) follows then by equating the coefficients of 
$x^n$ of the above identity. 

Next, by (0.8) and (4.4), we have
$$\eqalign{
S_q(\alpha ,n)&={1\over [n]!}\sum_{k=0}^n(-1)^{n-k}{n\brack k}
q^{{n-k\choose 2}}[k]^\alpha \cr
&={1\over [n]!}\sum_{k=0}^n(-1)^{n-k}{n\brack k}
q^{{n-k\choose 2}}\sum_{l=0}^{k-1}{\alpha +l\brack l}
A_q(\alpha , k-l-1)\cr
&={1\over [n]!}\sum_{j=1}^n\left\{\sum_{k=j}^n
(-1)^{n-k}{n\brack k} q^{{n-k\choose 2}}
{\alpha +k-j\brack k-j}\right\}A_q(\alpha , j-1).\cr
}$$

We can evaluate the second summation by the 
$q$-{\sl Chu-Vandermonde formula} (see [GR, p. 236]) and get
$$
\sum_{k=j}^n
(-1)^{n-k}{n\brack k} q^{{n-k\choose 2}}
{\alpha +k-j\brack k-j}=q^{n(n-j)}{\alpha -j\brack n-j}.
$$
Substituting this in the above identity and rescaling $j-1$ to
$j$ yield (4.11). \qed

\medskip
We close this paper by pointing out that it is possible to
generalize the formulae of this section to the multiple
index's Eulerian functions as one did for the multiple index's
Eulerian numbers (see [Raw] and [FZ]). For exemple, we can
generalize (4.6) in the following manner.

Let ${\bf x}=(x_1, x_2, \ldots, x_m)$  be a sequence of complex
numbers. Define the {\sl multiple $q$-Eulerian functions} 
$A_q({\bf x}, s)$ by
$$
A_q({\bf x}, s)=\sum_{l=0}^s
(-1)^l{x_1+\cdots +x_m+1\brack l}q^{{l\choose
2}}\prod_{k=1}^m{s+x_k-l\brack
s-l}.\leqno (4.12)
$$

\proclaim Theorem 4.7.
If $\Re x_i>-1$ ($i=1, 2, \ldots, m)$ and
$\Re (x_1+\cdots +x_m)>-1$, then we have
$$
{\Gamma_q(x_1+x_2+\cdots +x_m+1)\over
\Gamma_q(x_1+1)\Gamma_q(x_2+1)\cdots\Gamma_q(x_m+1)}
=\sum_{s=0}^{+\infty}A_q({\bf x}, s).\leqno (4.13)
$$

\proof By definition (4.12) and the $q$-binomial formula, we have
$$\eqalign{
\sum_{s=0}^{+\infty}A_q({\bf x}, s)z^s&=\sum_{l=0}^{+\infty}
(-1)^l{x_1+\cdots + x_m+1\brack l}q^{{l\choose
2}}z^l\sum_{s=0}^{+\infty}\prod_{k=1}^m{s+x_k\brack s}z^s\cr
&={(z;q)_\infty\over (zq^{x_1+\cdots + x_m+1}; q)_\infty}
\sum_{s=0}^{+\infty}\prod_{k=1}^m{s+x_k\brack n_k}z^s\cr
&={(zq;q)_\infty\over (zq^{x_1+\cdots + x_m+1}; q)_\infty}
\sum_{s=0}^{+\infty}\left(\prod_{k=1}^m{s+1+x_k\brack
n_k}-\prod_{k=1}^m{s+x_k\brack n_k}\right) z^s.\cr
}$$

As in the proof of Theorem 4.4, the above series defines
a function which is regular at $z=1$. So
putting $z=1$ in the above identity yields
$$
\sum_{s=0}^{+\infty}A_q({\bf x}, s)=
{(q;q)_\infty\over (q^{x_1+\cdots + x_m+1};
q)_\infty} \prod_{k=1}^m{(q^{1+x_k};q)_\infty\over
(q;q)_\infty}, 
$$
which is clearly equivalent to (4.13). \qed


\goodbreak
{\bf Acknowledgement} : The authors are indebted to Professor \pc
Butzer| for sending them the preprint [BH2] before its publication
and the anonymous referer for the suggestions on the first
version of this paper.
\bigskip
\vfill

\ctitre
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\bigskip
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\vskip 40pt
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Institut de Recherche Math\'ematique Avanc\'ee
Universit\'e Louis Pasteur et CNRS
7, rue Ren\'e Descartes
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