Two closed forms for the Apostol-Bernoulli polynomials
Su Hu
Min-Soo Kim
``In mathematics, a closed form is a mathematical expression that can be evaluated in a finite number of operations. It may contain constants, variables, four arithmetic operations, and elementary functions, but usually no limit." In this note, we shall obtain two closed forms for the Apostol-Bernoulli polynomials.
1509.04190.tex
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\begin{document}

\title[]
{Two closed forms for the Apostol-Bernoulli polynomials}

\author{Su Hu}
\address{Department of Mathematics, South China University of Technology, Guangzhou, Guangdong 510640, China}
\email{mahusu@scut.edu.cn}





\author{Min-Soo Kim}
\address{Center for General Education, Kyungnam University,
7(Woryeong-dong) kyungnamdaehak-ro, Masanhappo-gu, Changwon-si, Gyeongsangnam-do 631-701, Republic of Korea}
\email{mskim@kyungnam.ac.kr}

%\thanks{*Corresponding author}



\begin{abstract}
 ``In mathematics, a closed form is a mathematical expression that can be evaluated in a finite number of operations. It may contain constants, variables, four arithmetic operations, and elementary functions, but usually no limit." In this note, we shall obtain two closed forms for the Apostol-Bernoulli polynomials.
\end{abstract}



\subjclass[2000]{11B68,05A19}
\keywords{Stirling numbers, Apostol-Bernoulli polynomials, Closed forms}

%\thanks{Received May 24, 2009}

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%\begin{proof}[Proof of Lemma 4]
%The result follows in the same way as in the proof of Proposition 1 in \cite{AD2}.
%\end{proof}

\section{Introduction}
\label{Intro}
The Bernoulli numbers $B_{n}$ are defined by the generating function
\begin{equation}\label{Bn}
\frac{x}{e^x-1}=\sum_{n=0}^\infty B_n\frac{x^n}{n!}=1-\frac x2 +\sum_{n=1}^\infty B_{2n}\frac{x^{2n}}{(2n)!},\quad |x|<2 \pi
\end{equation}
and the Bernoulli polynomials $B_{n}(u)$ are defined by
\begin{equation}\label{Bp}
\frac{xe^{ux}}{e^x-1}=\sum_{n=0}^{\infty}B_{n}(u) \frac{x^n}{n!},\quad |x|<2 \pi.
\end{equation}
Because the function $\frac{x}{e^x-1}=1-\frac x2$ is odd in $x\in\mathbb R,$ all of the Bernoulli numbers $B_{2n+1}$ for $n\in\mathbb N$ equal to 0.
It is clear that $B_0=1$ and $B_1=-\frac 12$ (see \cite{Qi}).

The Apostol-Bernoulli polynomials $B_n(u,z)$ are natural generalizations of the Bernoulli polynomials, they were first introduced  by
Apostol \cite{Ap} in order to study the Lipschitz-Lerch zeta functions (also see \cite[Section 5]{Navas}). Their definitions are as follows,
\begin{equation}\label{AB-def}
\left(\frac{x}{z e^x-1}\right) e^{ux}=\sum_{n=0}^\infty B_n(u,z) \frac{x^n}{n!},
\end{equation}
where $|x|\leq 2\pi$ when $z=1$; $|x|\leq |\textrm{log}~z|$ when $z\neq 1$ (see ~\cite{LS}). In particular, $B_n(z)=B_n(0,z)$ are the
Apostol-Bernoulli numbers. Letting $z=1$ in (\ref{AB-def}), we
obtain the Bernoulli polynomials $B_n(u)$ and Bernoulli
numbers $B_n,$ respectively.

During recent years, the  Apostol-Bernoulli polynomials and numbers including their applications have been widely studied
  by many authors (see \cite{Ba,Bo,CJS,He1,He2,He3, KH,Kim, HB,Luof, LS, NFV, pre, Srivastava, Srivastava2, WZ,XC} and references therein).

Ten years ago, Jeong, Kim and Son~\cite{Jeong-Kim-Son-JNT-2005} proved the following explicit formula for Bernoulli numbers, which is equivalent to an old formula by Saalsch\"utz  in 1893 \cite{Sal} (also see the sentence above \cite[Theorem 3.1]{Jeong-Kim-Son-JNT-2005}).
\begin{theorem}[Jeong, Kim and Son, {\cite[Theorem 3.1]{Jeong-Kim-Son-JNT-2005}}]\label{th1} For $n\geq 1$, \begin{equation}\label{JKSf}
B_{n}=n!\sum_{j=1}^{n}(-1)^{j}\sum_{\substack{i_{1},\ldots,i_{n}\geq 0\\ i_{1}+\cdots+i_{n}=j\\
i_{1}+2i_{2}+\cdots+ni_{n}=n}}\binom{j}{i_1,\ldots,i_n}\frac{1}{2!^{i_{1}}3!^{i_{2}}\cdots (n+1)!^{i_{n}}}.
\end{equation}
\end{theorem}


The Stirling numbers $S(n,k)$ of the second kind for $n\ge k\ge1$ can be computed and generated by
$$S(n,k)=\frac1{k!}\sum_{\ell=1}^k(-1)^{k-\ell}\binom{k}{\ell}\ell^{n}$$
and
$$\frac{(e^x-1)^k}{k!}=\sum_{n=k}^\infty S(n,k)\frac{x^n}{n!}$$
respectively (see \cite[p.~206]{Comtet-Combinatorics-74} and \cite{Qi}).

Recently, Qi and Chapman~\cite{Qi} generalized the above formula to get a closed form for Bernoulli polynomials, which provide an explicit formula for computing these special polynomials in terms of Stirling numbers of the second kind $S(n,k)$.

\begin{theorem}[Qi and Chapman, {\cite[Theorem 1.1]{Qi}}]\label{th2}
The Bernoulli polynomials $B_n(u)$ for $n\in\mathbb{N}$ may be expressed as
\begin{equation}\label{BP-Stirl-form}
\begin{aligned}
B_n(u)&=\sum_{k=1}^nk! \sum_{r+s=k}\sum_{\ell+m=n}(-1)^m\binom{n}{\ell} \frac{\ell!}{(\ell+r)!}\frac{m!}{(m+s)!} \\
 &\quad\times\left[\sum_{i=0}^r\sum_{j=0}^s(-1)^{i+j}\binom{\ell+r}{r-i}\binom{m+s}{s-j}S(\ell+i,i)S(m+j,j)\right]\\
 &\quad\times u^{m+s}(1-u)^{\ell+r}.
 \end{aligned}
\end{equation}
Consequently, the Bernoulli numbers $B_n$ for $n\in\mathbb{N}$ can be represented as
\begin{equation}\label{Bernoulli-Stirling-formula}
B_n=\sum_{i=1}^n(-1)^{i}\frac{\binom{n+1}{i+1}}{\binom{n+i}{i}} S(n+i,i).
\end{equation}
\end{theorem}
\begin{remark}
Eq. (\ref{Bernoulli-Stirling-formula}) recovers Jeong, Kim and Son's formula (\ref{JKSf}) (see \cite[Remark 4.1]{Qi}).
\end{remark}
\begin{remark}Using an identity involving  the functions  $\frac{\pm1}{e^{\pm x}-1}$ and their derivatives, Guo and Qi ~\cite{GQc} also established two explicit formulas for computing Euler polynomials and two-parameter Euler polynomials in terms of Stirling numbers of the second kind.\end{remark}
\begin{remark} The formulas in the above theorem are named closed forms. As recalled by Qi and Chapman~\cite[p. 91]{Qi},

``In mathematics, a closed form is a mathematical expression that can be evaluated in a finite number of operations. It may contain constants, variables, four arithmetic operations, and elementary functions, but usually no limit."
\end{remark}


Furthermore, they also got another form for Bernoulli polynomials.
\begin{theorem}[Qi and Chapman, {\cite[Theorem 1.2]{Qi}}]\label{th3}
Under the conventions that $\binom{0}{0}=1$ and $\binom{p}{q}=0$ for $q>p\ge0$, the Bernoulli polynomials $B_n(u)$ for $n\in\mathbb{N}$ may be expressed as
\begin{equation}\label{Bern-Polyn-determ}
B_n(u)=(-1)^n\left|\frac1{\ell+1}\binom{\ell+1}{m} \left[(1-u)^{\ell-m+1}-(-u)^{\ell-m+1}\right]\right|_{1\le \ell\le n,0\le m\le n-1},
\end{equation}
where $|\cdot|_{1\le \ell\le n,0\le m\le n-1}$ denotes a $n\times n$ determinant.
Consequently, the Bernoulli numbers $B_k$ for $k\in\mathbb{N}$ can be represented as
\begin{equation}\label{Bern-No-determ}
B_n=(-1)^n\left|\frac1{\ell+1}\binom{\ell+1}{m}\right|_{1\le \ell\le n,0\le m\le n-1}.
\end{equation}
\end{theorem}



One of main tools used in ~\cite{Qi} to get the closed forms for the Bernoulli polynomials  is the following integral expression for the generating function    (\ref{Bp})
\begin{equation}\label{Bern-Polyn-Int-Expr}
\frac{xe^{ux}}{e^x-1}=\left[\frac{e^{(1-u)x}-e^{-ux}}x\right]^{-1}
=\frac1{\int_{-u}^{1-u}e^{xt} dt} =\frac{1}{\int_0^1 e^{x(t-u)} dt}
\end{equation}
(see \cite[p. 90, (1.1)]{Qi}).

In this note, we show that, different with~\cite{Qi}, if directly applying the  generating functions instead of their integral expressions, then the above theorems may be generalized to other special functions. As a result, we shall get the following two closed forms for Apostol-Bernoulli polynomials and numbers, which may be used to compute these special polynomials and numbers in a finite number of steps.

\begin{theorem}\label{th-main1}
Suppose that $z\not=1$. The Apostol-Bernoulli polynomials $B_n(u,z)$ for $n\in\mathbb N$ may be expressed as
\begin{equation}\label{AP-Stirl-form}
\begin{aligned}
B_n(u,z)
&=n\sum_{k=0}^{n-1}\frac{(-1)^{k}k!}{(z-1)^{k+1}}\sum_{r+s=k}\sum_{\ell+m=n-1}(-1)^{s+m}\binom{n-1}{\ell}\\
&\quad\times z^{r}(1-u)^{\ell}u^m S(\ell,r) S(m,s).
 \end{aligned}
\end{equation}
 Consequently, the Apostol-Bernoulli numbers $B_n(z)$ for $n\in\mathbb N$ can be represented as
\begin{equation}\label{APN-Stirl-form}
\begin{aligned}
B_n(z)=n\sum_{k=0}^{n-1}\frac{(-1)^{k}k!}{(z-1)^{k+1}}z^{k}S(n-1,k).
 \end{aligned}
\end{equation}
 \end{theorem}
 \begin{remark}Suppose that $z\not=1$. Letting $x=0$ in the both sides of (\ref{AB-def}), we get $B_{0}(u,z)=0$ and $B_{0}(z)=0.$
 Since by (\ref{AB-def}), we have
 $$ \frac{d}{dx}\left(\frac{xe^{ux}}{ze^{x}-1}\right)\biggl|_{x=0}=\sum_{m=1}^{\infty}B_{m}(u,z)\frac{x^{m-1}}{(m-1)!}\biggl|_{x=0}=B_{1}(u,z)$$
 and \begin{equation}\begin{aligned}
 \frac{d}{dx}\left(\frac{xe^{ux}}{ze^{x}-1}\right)&=\frac{e^{ux}}{ze^{x}-1}+x\frac{d}{dx}\left(\frac{e^{ux}}{ze^{x}-1}\right)\\
&\rightarrow \frac{1}{z-1}
\end{aligned}
\end{equation}
as $x\rightarrow 0$, so $B_{1}(u,z)=\frac{1}{z-1}$ and $B_{1}(z)=\frac{1}{z-1}.$
\end{remark}

\begin{remark}
There is another closed form for the Apostol-Bernoulli polynomials:
$$B_n(u,z)=\sum_{k=0}^n k \binom nk \sum_{j=0}^{k-1} (-1)^jz^j(z-1)^{-j-1}j! S(k-1,j)u^{n-k}.$$
(See \cite{Luo04} ).
\end{remark}

\begin{remark} Eq. (\ref{APN-Stirl-form}) recovers a formula by Apostol (see \cite[Eq. (3.7)]{Ap}), and recently, Xu and Chen~\cite{XC} provided another formula for the Apostol-Bernoulli numbers as follows,  $$B_n(z)=(-1)^{n-1}n\sum_{k=1}^n\frac{(k-1)!}{(z-1)^k}S(n,k)$$ (see \cite[Theorem 4.1]{XC}).
\end{remark}


\begin{theorem}\label{th-main2}
Suppose that $z\not=1$. Under the conventions that $\binom{0}{0}=1$ and $\binom{p}{q}=0$ for $q>p\ge0$, the Apostol-Bernoulli polynomials
$B_{n+1}(u,z)$ for $n\in\mathbb N$ may be expressed as
\begin{equation}\label{ABern-Polyn-determ}
B_{n+1}(u,z)=\frac{(-1)^{n}(n+1)}{(z-1)^{n+1}}\left|\binom{\ell}{m}\left[z(1-u)^{\ell-m}-(-u)^{\ell-m}\right]\right|_{1\le \ell\le n,0\le m\le n-1},
\end{equation}
where $|\cdot|_{1\le \ell\le n,0\le m\le n-1}$ denotes a $n\times n$ determinant.
Consequently, the Apostol-Bernoulli numbers $B_{n+1}(z)$ for $n\in\mathbb N$ can be represented as
\begin{equation}\label{ABern-No-determ}
B_{n+1}(z)=\frac{(-1)^{n}(n+1)}{(z-1)^{n+1}}\left|\binom{\ell}{m}(z-\delta_{\ell m})\right|_{1\le \ell\le n,0\le m\le n-1},
\end{equation}
where the Kronecker delta $\delta_{\ell m}$ is 1 if the variables are equal, and 0 otherwise.
\end{theorem}

\section{Bell polynomials}
As in \cite{Qi}, our proofs are also based on following properties of Bell polynomials.

The Bell polynomials of the second kind $\bell_{n,k}(x_1,x_2,\dotsc,x_{n-k+1})$ are defined by
\begin{multline}\label{Bd}
\bell_{n,k}(x_1,x_2,\dotsc,x_{n-k+1}) \\
=\sum
%_{\substack{\ell_1,\ldots,\ell_n\geq0\\ \ell_1+\ldots+\ell_{n-k+1}=k\\ \ell_1+2\ell_2+\cdots+(n-k+1)\ell_{n-k+1}=n}}
\frac{n!}{\ell_1!\cdots\ell_{n-k+1}!}\left(\frac{x_1}{1!}\right)^{\ell_1}\left(\frac{x_2}{2!}\right)^{\ell_2}\cdots
\left(\frac{x_{n-k+1}}{{n-k+1}!}\right)^{\ell_{n-k+1}},
\end{multline}
where the sum is taken over all sequences $\ell_1,\ldots,\ell_{n-k+1}$ of non-negative integers such that
$$\ell_1+\ldots+\ell_{n-k+1}=k\quad \text{ and }\quad \ell_1+2\ell_2+\cdots+(n-k+1)\ell_{n-k+1}=n$$
for $n\ge k\ge0$. See~\cite[p.~134, Theorem~A]{Comtet-Combinatorics-74}.

\begin{lemma}[{\cite[Example~2.6]{ABCLM-arXiv-2960},~\cite[Lemma 2.1]{Qi}}]
For $n\ge k\ge0$, the Bell polynomials of the second kind $\bell_{n,k}$ meets
\begin{multline}\label{Bell-times-eq}
\bell_{n,k}(x_1+y_1,x_2+y_2,\dotsc,x_{n-k+1}+y_{n-k+1})\\
=\sum_{r+s=k}\sum_{\ell+m=n}\binom{n}{\ell}\bell_{\ell,r}(x_1,x_2,\dotsc,x_{\ell-r+1})
\bell_{m,s}(y_1,y_2,\dotsc,y_{m-s+1}).
\end{multline}
\end{lemma}

\begin{lemma}[{\cite[p.~135]{Comtet-Combinatorics-74}, ~\cite[Lemma 2.2]{Qi}}]
For $n\ge k\ge0$, we have
\begin{equation}\label{Bell(n-k)}
\bell_{n,k}\left(abx_1,ab^2x_2,\dotsc,ab^{n-k+1}x_{n-k+1}\right) =a^kb^n\bell_{n,k}(x_1,x_2,\dotsc,x_{n-k+1}),
\end{equation}
where $a$ and $b$ are any complex numbers.
\end{lemma}

\begin{lemma}[{\cite[p.~135, Theorem B,~{[3g]}]{Comtet-Combinatorics-74}}]For $n\ge k\ge0$, we have
\begin{equation}\label{Bell3}\bell_{n,k}(\underbrace{1,1,\ldots,1}_{n-k+1})=S(n,k).\end{equation}
\end{lemma}

\section{Proof of Theorem~\ref{th-main1}}
Set
$$m(x)=x,\quad g(x)=\frac{z e^x-1}{e^{ux}}\quad\text{and}\quad f(y)=\frac{1}{y},$$
then we have
$$h(x)=m(x) (f\circ g)(x)=\frac{xe^{ux}}{z e^x-1}$$
which is the generating function of the Apostol-Bernoulli polynomials ~(\ref{AB-def}).
Thus by (\ref{AB-def}), we have
\begin{equation}\label{P1}
\begin{aligned}
\frac{\td^{n+1}}{\td x^{n+1}}h(x)\biggl|_{x=0}
&=B_{n+1}(u,z).
\end{aligned}
\end{equation}
On the other hand, since
\begin{equation}
\begin{aligned}
g(x)&=\frac{z e^x-1}{e^{ux}}=ze^{(1-u)x}-e^{-ux}\\
& = z\sum_{m=0}^{\infty}\frac{(1-u)^{m}}{m!}x^{m}-\sum_{m=0}^{\infty}\frac{(-u)^{m}}{m!}x^{m}\\
&\rightarrow z-1
\end{aligned}
\end{equation}
and
\begin{equation}\label{P1-1}
\begin{aligned}
g'(x)&=z\sum_{m=1}^{\infty}\frac{(1-u)^{m}}{(m-1)!}x^{m-1}-\sum_{m=1}^{\infty}\frac{(-u)^{m}}{(m-1)!} x^{m-1}\\
&\rightarrow z(1-u)-(-u),\\
 g''(x)&=z\sum_{m=2}^{\infty}\frac{(1-u)^{m}}{(m-2)!}x^{m-2}-\sum_{m=2}^{\infty}\frac{(-u)^{m}}{(m-2)!}x^{m-2} \\
&\rightarrow z(1-u)^{2}-(-u)^{2} ,\\
&\;\,\vdots\\
g^{(n-k+1)}(x)&=z\sum_{m=n-k+1}^{\infty}\frac{(1-u)^{m}}{(m-(n-k+1))!}x^{m-(n-k+1)} \\
&\quad-\sum_{m=n-k+1}^{\infty}\frac{(-u)^{m}}{(m-(n-k+1)!}x^{m-(n-k+1)} \\
&\rightarrow z(1-u)^{n-k+1}-(-u)^{n-k+1}
\end{aligned}
\end{equation}
as $x\rightarrow 0.$

In terms of the Bell polynomials of the second kind $B_{n,k},$ the  Fa\`a di Bruno formula for computing higher order derivatives of
composite functions is described in \cite[p.~139, Theorem C]{Comtet-Combinatorics-74} by
\begin{equation}\label{fabr}
\frac{\td^n}{\td x^n}(f\circ g)(x)=\sum_{k=0}^nf^{(k)}(g(x)) \bell_{n,k}\left(g'(x),g''(x),\dotsc,g^{(n-k+1)}(x)\right)
\end{equation}
(see also \cite[p.~93, (3.1)]{Qi}).


By the integral expression (\ref{AB-def}) and (\ref{P1-1}), applying the formula (\ref{fabr}) to the functions
$$f(y)=\frac1y \quad\text{and}\quad y=g(x)=\frac{ze^x-1}{e^{ux}},$$
we have
\begin{equation}\label{foru}
\begin{aligned}
\frac{\td^n}{\td x^n}(f\circ g)(x)&=\sum_{k=0}^nf^{(k)}(g(x)) \bell_{n,k}\left(g'(x),g''(x),\dotsc,g^{(n-k+1)}(x)\right)\\
&=\sum_{k=0}^{n}\frac{(-1)^{k}k!}{(g(x))^{k+1}}\bell_{n,k}\left(g'(x),g''(x),\dotsc,g^{(n-k+1)}(x)\right)\\
&\rightarrow \sum_{k=0}^{n}\frac{(-1)^{k}k!}{(z-1)^{k+1}}\\
&\quad \times\bell_{n,k}\left(z(1-u)-(-u),z(1-u)^{2}-(-u)^{2} ,\dots,z(1-u)^{n-k+1}-(-u)^{n-k+1}\right)
\end{aligned}
\end{equation}
as $x\rightarrow 0$ and
\begin{equation}\label{P2}
\begin{aligned}
\frac{\td^n}{\td x^n}(f\circ g)(x)\biggl|_{x=0}
&=\sum_{k=0}^n\frac{(-1)^{k}k!}{(z-1)^{k+1}} \sum_{r+s=k}\sum_{\ell+m=n}\binom{n}{\ell}\\
&\quad\times\bell_{\ell,r}\left(z(1-u), z(1-u)^{2}, \dotsc,z(1-u)^{\ell-r+1}\right)\\
&\quad\times\bell_{m,s}\left(-(-u),-(-u)^{2}, \dotsc,-(-u)^{m-s+1}\right)\\
&\quad\text{(by \eqref{Bell-times-eq})}
 \\&=\sum_{k=0}^n\frac{(-1)^{k}k!}{(z-1)^{k+1}}\sum_{r+s=k}\sum_{\ell+m=n}\binom{n}{\ell}\\
&\quad\times z^{r}(1-u)^{\ell} \bell_{\ell,r}(\underbrace{1,1,\ldots,1}_{\ell-r+1})\\
&\quad\times (-1)^{s}(-u)^m \bell_{m,s}(\underbrace{1, 1, \ldots,1}_{m-s+1})\\
&\quad\text{(by \eqref{Bell(n-k)})}
\\&=\sum_{k=0}^n\frac{(-1)^{k}k!}{(z-1)^{k+1}}\sum_{r+s=k}\sum_{\ell+m=n}\binom{n}{\ell}\\
&\quad\times z^{r}(1-u)^{\ell}S(\ell,r)
 (-1)^{s}(-u)^m S(m,s)\\
&\quad\text{(by \eqref{Bell3})}.
\end{aligned}
\end{equation}
Thus by Lebnitz's formula for the $n$th derivative of the product of two functions (see~\cite[p.~210, Example 24]{Zorich}), we have
\begin{equation}\label{PR}
\begin{aligned}
\frac{\td^n}{\td x^n}h(x)&=\frac{\td^n}{\td x^n}m(x)(f\circ g)(x)\\
&=\sum_{i=0}^{n}\binom{n}{i}(f\circ g)(x)^{(n-i)}m(x)^{(i)}\\
&=x\frac{\td^n}{\td x^n}(f\circ g)(x)+n\frac{\td^{n-1}}{\td x^{n-1}}(f\circ g)(x)\\
&\quad(\text{since}~m(x)=x)\\
&\rightarrow n\sum_{k=0}^{n-1}\frac{(-1)^{k}k!}{(z-1)^{k+1}}\sum_{r+s=k}\sum_{\ell+m=n-1}\binom{n-1}{\ell}\\
&\quad\times z^{r}(1-u)^{\ell}S(\ell,r)(-1)^s(-u)^m S(m,s)\\
&\quad\text{(by \eqref{P2})}
\end{aligned}
\end{equation}
as $x\rightarrow 0$.
Then by comparing (\ref{P1}) with (\ref{PR}), we get our result.

Finally, letting $u=0$ in (\ref{AP-Stirl-form}), we obtain (\ref{APN-Stirl-form}).

\section{Proof of Theorem~\ref{th-main2}}

Let $\mu=\mu(x)$ and $\nu=\nu(x)\neq0$ be differentiable functions.
Set
$$\frac{\td^n}{\td x^n}\left(\frac{\mu}{\nu}\right)=\frac{(-1)^nw_n}{\nu^{n+1}}$$
at every point $\nu(x)\neq0.$
By \cite[p.~40]{Bour}, we have
\begin{equation}\label{matrix}
w_n=
 \left|
   \begin{array}{cccccc}
     \mu         & \nu         & 0                       & 0                       & \cdots & 0                 \\
     \mu'        & \nu'        & \nu                     & 0                       & \cdots & 0                 \\
     \mu''       & \nu''       & 2\nu'                   & \nu                     & \cdots & 0                 \\
     \vdots      & \vdots      & \vdots                  & \vdots                  & \vdots & \vdots            \\
     \mu^{(n-1)} & \nu^{(n-1)} & \binom{n-1}1\nu^{(n-2)} & \binom{n-1}2\nu^{(n-3)} & \cdots & \nu               \\
     \mu^{(n)}   & \nu^{(n)}   & \binom{n}1\nu^{(n-1)}   & \binom{n}2\nu^{(n-2)}   & \cdots & \binom n{n-1}\nu' \\
   \end{array}
 \right|.
\end{equation}
As in \cite[p. 94--95, the~ first~ proof~ of~ Theorem 1.2]{Qi},
we may reformulate the formula (\ref{matrix}) as
\begin{equation}\label{reform-matrix}
\frac{\td^n}{\td x^n}\left(\frac{\mu}{\nu}\right)
=\frac{(-1)^n}{\nu^{n+1}}\begin{vmatrix} A_{(n+1)\times 1} & B_{(n+1)\times n}\end{vmatrix}_{(n+1)\times(n+1)} ,
\end{equation}
where the matrices
$$A_{(n+1)\times 1}=(a_{\ell,1})_{0\leq\ell\leq n}=
\left(
   \begin{array}{c}
     \mu         \\
     \mu'        \\
     \mu''       \\
     \vdots      \\
     \mu^{(n-1)} \\
     \mu^{(n)}   \\
   \end{array}
 \right)$$
and
$$\begin{aligned}
B_{(n+1)\times n}&=(b_{\ell,m})_{0\leq \ell\leq n,0\leq m\leq n-1} \\
&=
\left(
   \begin{array}{cccccc}
     \nu         & 0                       & 0                       & \cdots & 0                 \\
     \nu'        & \nu                     & 0                       & \cdots & 0                 \\
     \nu''       & 2\nu'                   & \nu                     & \cdots & 0                 \\
     \vdots      & \vdots                  & \vdots                  & \vdots & \vdots            \\
     \nu^{(n-1)} & \binom{n-1}1\nu^{(n-2)} & \binom{n-1}2\nu^{(n-3)} & \cdots & \nu               \\
     \nu^{(n)}   & \binom{n}1\nu^{(n-1)}   & \binom{n}2\nu^{(n-2)}   & \cdots & \binom n{n-1}\nu' \\
   \end{array}
 \right)
\end{aligned}
$$
satisfy
$$a_{\ell,1}=\mu^{(\ell)}(x)\quad\text{and}\quad b_{\ell,m}=\binom \ell m \nu^{(\ell-m)}(x)$$
under the conventions that $\nu^{(0)}(x)=\nu(x)$ and that $\binom \ell m=0$ and $\nu^{(\ell-m)}(x)\equiv0$ for $\ell< m$
(see \cite[(3.2) and (3.3)]{Qi}).
If we let
\begin{equation}
\mu(x)=1, \quad  \nu(x)=\frac{ze^{x}-1}{e^{ux}},
\end{equation}
then in this case,
the $a_{\ell,1}$ becomes
\begin{equation}\label{a ell}
a_{0,1}=1,\quad a_{\ell,1}=0 \quad\text{for }\ell\geq1
\end{equation}
and the $b_{\ell,m}$ becomes
\begin{equation}\label{P3}
\begin{aligned}
\binom{\ell}{m}\nu^{(\ell-m)}(x)&=\binom{\ell}{m}\frac{\td^{\ell-m}}{\td x^{\ell-m}}\left(\frac{ze^{x}-1}{e^{ux}}\right)\\
&=\binom{\ell}{m}\frac{\td^{\ell-m}}{\td x^{\ell-m}}\left(ze^{(1-u)x}-e^{-ux}\right)\\
&=\binom{\ell}{m}\frac{\td^{\ell-m}}{\td x^{\ell-m}}\left(z\sum_{k=0}^{\infty}\frac{(1-u)^{k}x^{k}}{k!}-\sum_{k=0}^{\infty}\frac{(-u)^{k}x^{k}}{k!}\right)\\
&=\binom{\ell}{m}\left(z\sum_{k=\ell-m}^{\infty}\frac{(1-u)^{k}x^{k-(\ell-m)}}{(k-(\ell-m))!}
-\sum_{k=l-m}^{\infty}\frac{(-u)^{k}x^{k-(\ell-m)}}{(k-(\ell-m))!}\right)\\
&\rightarrow\binom{\ell}{m} \left[z(1-u)^{\ell-m}-(-u)^{\ell-m}\right],\quad x\rightarrow 0
\end{aligned}
\end{equation}
for $0\leq\ell\leq n$ and $0\leq m\leq n-1$ with $\ell\geq m.$
Thus,
as in \cite[p. 95, the~ first~ proof~ of~ Theorem 1.2]{Qi}, by (\ref{reform-matrix}), (\ref{a ell}) and (\ref{P3}), we have
\begin{equation}\label{PRth2}
\begin{aligned}
\frac{\td^n}{\td x^n}\left(\frac{\mu}{\nu}\right)
&=\frac{(-1)^n}{\nu^{n+1}}
\left|
   \begin{array}{cccccc}
     1         & b_{0,0}         & 0                     & 0                     & \cdots & 0 \\
     0         & b_{1,0}         & b_{1,1}               & 0                     & \cdots & 0 \\
     0         & b_{2,0}         & b_{2,1}               & b_{2,2}               & \cdots & 0 \\
     \vdots    & \vdots          & \vdots                & \vdots                & \vdots & \vdots \\
     0         & b_{n-1,0}       & b_{n-1,1}             & b_{n-1,2}             & \cdots & b_{n-1,n-1} \\
     0         & b_{n,0}         & b_{n,1}               & b_{n,2}               & \cdots & b_{n,n-1} \\
   \end{array}
 \right| \\
&=\frac{(-1)^n}{\nu^{n+1}} \left|b_{\ell,m}\right|_{1\le \ell\le n,0\le m\le n-1}\\
&\to \frac{(-1)^n}{(z-1)^{n+1}}\left|\binom{\ell}{m}\left[z(1-u)^{\ell-m}-(-u)^{\ell-m}\right]\right|_{1\le \ell\le n,0\le m\le n-1}
\end{aligned}
\end{equation}
as $x\rightarrow 0.$
Then again by Lebnitz's formula for the $n$-th derivative of the product of two functions (see~\cite[p.~210, Example 24]{Zorich}), we have
\begin{equation}\label{PR2}
\begin{aligned}
\frac{\td^{n+1}}{\td x^{n+1}}\left(\frac{xe^{ux}}{ze^{x}-1}\right)&=\frac{\td^{n+1}}{\td x^{n+1}}\left(x\left(\frac{\mu}{\nu}\right)\right)\\
&=\sum_{i=0}^{n+1}\binom{n+1}{i}\left(\frac{\mu}{\nu}\right)^{(n+1-i)}x^{(i)}\\
&=x\frac{\td^{n+1}}{\td x^{n+1}}\left(\frac{\mu}{\nu}\right)+(n+1)\frac{\td^{n}}{\td x^{n}}\left(\frac{\mu}{\nu}\right)\\
&\rightarrow (n+1)\frac{\td^{n}}{\td x^{n}}\left(\frac{\mu}{\nu}\right) \\
&=\frac{(-1)^{n}(n+1)}{(z-1)^{n+1}}\left|\binom{\ell}{m}\left[z(1-u)^{\ell-m}-(-u)^{\ell-m}\right]\right|_{1\le \ell\le n,0\le m\le n-1} \\
&\quad\text{(by \eqref{PRth2})}
\end{aligned}
\end{equation}
as $x\rightarrow 0$.
Thus, by comparing (\ref{P1}) with (\ref{PR2}), we obtain (\ref{ABern-Polyn-determ}).

Finally, letting $u=0$ in (\ref{ABern-Polyn-determ}), we obtain (\ref{ABern-No-determ}).



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\end{thebibliography}

\end{document}







%%
%% End of file `elsarticle-template-num.tex'.



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{ABCLM-arXiv-2960 A. Aboud, J.-P. Bultel, A. Chouria, J.-G. Luque, O. Mallet, Bell polynomials in combinatorial Hopf algebras, http://arxiv.org/abs/1402.2960.
1509.04190.extracted_bib
{Ap T. M. Apostol, On the Lerch zeta function, Pacific J. Math. 1 (1951), 161--167.
1509.04190.extracted_bib
{Ba A. Bayad, Fourier expansions for Apostol-Bernoulli, Apostol-Euler and Apostol-Genocchi polynomials, Math. Comp. 80 (2011), 2219--2221.
1509.04190.extracted_bib
{Bo K.N. Boyadzhiev, Apostol-Bernoulli functions, derivative polynomials and Eulerian polynomials, Adv. Appl. Discrete Math. 1 (2) (2008), 109--122.
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{CJS J. Choi, D.S. Jang and H.M. Srivastava, A generalization of the Hurwitz-Lerch zeta function, Integral Transforms Spec. Funct. 19 (2008), no. 1-2, 65--79.
1509.04190.extracted_bib
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{He1Y. He, Some new results on products of Apostol-Bernoulli and Apostol-Euler polynomials, J. Math. Anal. Appl. 431 (2015), 34--46.
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{He3 Y. He, A new formula of products of the Apostol-Bernoulli and Apostol-Euler polynomials, Bulletin of the Malaysian Mathematical Sciences Society. 2015. http://link.springer.com/article/10.1007\%2Fs40840-015-0232-z
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{Jeong-Kim-Son-JNT-2005 S. Jeong, M.-S. Kim, and J.-W. Son, On explicit formulae for Bernoulli numbers and their counterparts in positive characteristic, J. Number Theory 113 (2005), 53--68.
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{KH M.-S. Kim and S. Hu, Sums of products of Apostol-Bernoulli numbers, Ramanujan J. 28 (2012), 113--123.
1509.04190.extracted_bib
{Kim M.-S. Kim and S. Hu, A note on the Apostol-Bernoulli and Apostol-Euler Polynomials, Publ. Math. Debrecen. 83 (2013), 449--464.
1509.04190.extracted_bib
{HB Y.L. Luke, The special functions and their applications, Volume I, Academic Press, New York, 1969.
1509.04190.extracted_bib
{Luo04 Q.-M. Luo, On the Apostol-Bernoulli polynomials, Cent. Eur. J. Math. 2 (2004), no. 4, 509--515.
1509.04190.extracted_bib
{Luof Q.-M. Luo, Fourier expansions and integral representations for the Apostol-Bernoulli and Apostol-Euler polynomials, Math. Comp. 78 (2009), 2193--2208.
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{LS Q.-M. Luo and H.M. Srivastava, Some generalizations of the Apostol-Bernoulli and Apostol-Euler polynomials, J. Math. Anal. Appl. 308 (1) (2005), 290--302.
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{Nie N. Nielson, Trait\'e \'el\'ementaire des nombers de Bernoulli, Gauthier-Villars, 1923.
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{NFVL.M. Navas, F.J. Ruiz and J.L. Varona, Asymptotic estimates for Apostol-Bernoulli and Apostol-Euler polynomials, Math. Comp. 81 (2012), 1707--1722.
1509.04190.extracted_bib
{Navas L.M. Navas, F.J. Ruiz and J.L. Varona, The Lerch transcendent from the point of view of Fourier analysis, J. Math. Anal. Appl. 431 (2015), 186--201.
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{preM. Pr\'evost, Pad\'e approximation and Apostol-Bernoulli and Apostol-Euler polynomials, J. Comput. Appl. Math. 233 (2010), 3005--3017.
1509.04190.extracted_bib
{Qi F. Qi and R.J. Chapman, Two closed forms for the Bernoulli polynomials, J. Number Theory 159 (2016), 89--100.
1509.04190.extracted_bib
{Sal L. Saalsch\"utz, Vorlesungen \"uber die Bernoullischen Zahlen, ihren Zusammenhang mit den Secanten-Coefficienten und ihre wichtigeren Anwendungen, Berlin, 1893 (Available since 1964 in Xerographed form from University Microfilms, Ann Arbor, Michigan, Order No. OP-17136).
1509.04190.extracted_bib
{Srivastava H. M. Srivastava and J. Choi, Zeta and $q$-Zeta Functions and Associated Series and Integrals, Elsevier Science Publishers, Amsterdam, London and New York, 2012.
1509.04190.extracted_bib
{Srivastava2 H. M. Srivastava, B. Kurt and Y. Simsek, Some families of Genocchi type polynomials and their interpolation functions, Integral Transforms Spec. Funct. 23 (2012), 919--938, see also Corrigendum, Integral Transforms Spec. Funct. 23 (2012), 939--940.
1509.04190.extracted_bib
{WZ K.S. Williams and N.Y. Zhang, Special values of the Lerch zeta function and the evaluation of certain integrals, Proc. Amer. Math. Soc. 119 (1993), 35--49.
1509.04190.extracted_bib
{XC A.-M. Xu and Z.-D. Cen, Some identities involving exponential functions and Stirling numbers and applications, J. Comput. Appl. Math. 260 (2014), 201--207.
1509.04190.extracted_bib
{Zhang-Yang-Oxford-Taiwan-12 Z.-Z. Zhang and J.-Z. Yang, Notes on some identities related to the partial Bell polynomials, Tamsui Oxf. J. Inf. Math. Sci. 28 (2012), 39--48.
1509.04190.extracted_bib
{Zorich V.A. Zorich, Mathematical analysis I, Translated from the 2002 fourth Russian edition by Roger Cooke, Universitext, Springer-Verlag, Berlin, 2009.