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\begin{center}
\vskip 1cm{\LARGE\bf A Special Case of the Generalized \\
\vskip .1in
Girard-Waring Formula} \vskip 1cm \large
Mircea Merca\\
Department of Mathematics\\
University of Craiova\\
A. I. Cuza 13 \\
Craiova, 200585 \\
Romania \\
\href{mailto:mircea.merca@profinfo.edu.ro}{\tt mircea.merca@profinfo.edu.ro} \\
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\begin{abstract} 
In this note we introduce a new method to proving and discovering some
identities involving binomial coefficents and factorials.
\end{abstract}

\section{Introduction.}

Let $n$ be a positive integer. Being given a set of variables $\{x_1,x_2,\ldots,x_n\}$, the $k$th elementary symmetric function $e_k (x_1,x_2,\dots,x_n)$ on these variables is the sum of all possible products of $k$ of these $n$ variables, chosen without replacement
\begin{equation*}\label{ESP}
e_k (x_1,x_2,\ldots,x_n)=\sum _{1\le i_1 <i_2<\ldots <i_k\le n }x_{i_1} x_{i_2} \ldots x_{i_k}\ ,
\end{equation*}
for $k=1,2,\ldots,n$. We set $e_0 (x_1,x_2,\ldots,x_n)=1$ by convention (a single choice of the empty product, if you like that kind of thing).
For $k>n$ or $k<0$, we set $e_k (x_1,x_2,\ldots,x_n)=0$.

The starting point of this paper is the following result:

\begin{theorem}\label{T0}
Let $n$ be a positive integer and let $x_1,x_2,\ldots,x_n$ be $n$ independent variables. Then
\begin{equation}\label{gw}
e_k\left(x_1^2,\ldots,x_n^2 \right)=\sum_{i=-k}^{k}(-1)^ie_{k+i}\left(x_1,\ldots,x_n \right)e_{k-i}\left(x_1,\ldots,x_n \right)\ .
\end{equation}
\end{theorem}

\begin{proof}
Taking into account that
$$\prod_{i=1}^{n}\left(x+x_i \right) = \sum_{k=0}^{n}e_{n-k}\left(x_1,\ldots,x_n\right)x^k$$
and
$$e_k\left(-x_1,\ldots,-x_n\right)=(-1)^ke_k\left(x_1,\ldots,x_n\right)\ ,$$
we can write
\begin{eqnarray}
\prod_{i=1}^{n}\left(x^2-x_i^2 \right) &=& \sum_{k=0}^{n}e_{n-k}\left(-x_1^2,\ldots,-x_n^2\right)x^{2k}  \nonumber \\
&=& \sum_{k=0}^{n}(-1)^{n-k}e_{n-k}\left(x_1^2,\ldots,x_n^2\right)x^{2k} \ .\label{eq:T1a}
\end{eqnarray}
On the other hand, we have
\begin{eqnarray}
&&\prod_{i=1}^{n}\left(x^2-x_i^2 \right) =\nonumber \\
&&\qquad = \left( \prod_{i=1}^{n}\left(x-x_i \right)\right)\left( \prod_{i=1}^{n}\left(x+x_i \right)\right) \nonumber \\
&&\qquad = \left( \sum_{k=0}^{n}(-1)^{n-k}e_{n-k}\left(x_1,\ldots,x_n\right)x^{k}\right)
\left( \sum_{k=0}^{n}e_{n-k}\left(x_1,\ldots,x_n\right)x^{k}\right) \nonumber \\
&&\qquad = \sum_{k=0}^{n}\left( \sum_{i=0}^{2k}(-1)^{n-i}e_{n-i}\left(x_1,\ldots,x_n\right)e_{n-2k+i}\left(x_1,\ldots,x_n\right)\right) x^{2k}\ .\label{eq:T2}
\end{eqnarray}
By \eqref{eq:T1a} and \eqref{eq:T2}, we deduce the relation
$$(-1)^{n-k}e_{n-k}\left(x_1^2,\ldots,x_n^2\right)=\sum_{i=0}^{2k}(-1)^{n-i}e_{n-i}\left(x_1,\ldots,x_n\right)e_{n-2k+i}\left(x_1,\ldots,x_n\right)\ ,$$
that can be rewritten in the following way
\begin{eqnarray*}
(-1)^{k}e_{k}\left(x_1^2,\ldots,x_n^2\right)&=&\sum_{i=0}^{2(n-k)}(-1)^{n-i}e_{n-i}\left(x_1,\ldots,x_n\right)e_{2k-n+i}\left(x_1,\ldots,x_n\right)\\
&=&\sum_{i=k-n}^{n-k}(-1)^{k-i}e_{k-i}\left(x_1,\ldots,x_n\right)e_{k+i}\left(x_1,\ldots,x_n\right)\ .
\end{eqnarray*}
Since $e_k\left(x_1,\ldots,x_n\right)=0$ for $k<0$ or $k>n$, we have
\begin{eqnarray}
&&\sum_{i=k-n}^{n-k}(-1)^{i}e_{k-i}\left(x_1,\ldots,x_n\right)e_{k+i}\left(x_1,\ldots,x_n\right)\nonumber \\ 
&&\qquad\qquad=\sum_{i=-k}^{k}(-1)^{i}e_{k-i}\left(x_1,\ldots,x_n\right)e_{k+i}\left(x_1,\ldots,x_n\right)\ .\label{eq:T3}
\end{eqnarray}
The proof is finished.
\end{proof}

It is well-known that the power sum symmetric functions can be expressed in terms of elementary symmetric functions using Girard-Waring formula \cite[eq. 8]{Gou99}. 
In \cite{Kon96,Kon98,Zen97}, the Girard-Waring formula 
is generalised to monomial symmetric functions with equal exponents.
The relation \eqref{gw} is the case $n=2$ in the generalized Girard-Waring formula \cite[Eq.\ (3)]{Zen97} 
and can be used to proving and discovering some identities. 
To illustrate this we present two applications involving binomial coefficients and Stirling numbers of the first kind.


\section{Identities involving binomial coefficients}
%To begin, 
Let us consider the binomial coefficients
\begin{equation*}\label{eq:bc1}
\binom{n}{k}=e_k(\underbrace{1,\ldots,1}_n)\ .
\end{equation*}

The following identity is a direct consequence of Theorem \ref{T0}.

\begin{corollary} \label{c:1}
Let $k$ and $n$ be two nonnegative integers. Then
$$
\sum_{i=-k}^{k}(-1)^i\binom{n}{k+i}\binom{n}{k-i}=\binom{n}{k}\ .
$$
\end{corollary}

Taking into account that
\begin{equation*}\label{b1}
\sum_{k=0}^{n}\binom{n}{k}=2^n\qquad\mbox{and}\qquad\sum_{k=0}^{n}\binom{n}{k}^2=\binom{2n}{n}\ ,
\end{equation*}
by Corollary \ref{c:1}, we obtain a new identity:

\begin{corollary} 
Let $n$ be a positive integer. Then
\begin{equation*}\label{eq:1}
\sum_{0<i\le k<n}(-1)^{i}\binom{n}{k+i}\binom{n}{k-i}=2^{n-1}-\binom{2n-1}{n}\ .
\end{equation*}
\end{corollary}

This corollary is related in \cite{Slo12} with the sequences $\seqnum{A108958}$.
By Theorem \ref{T0}, we obtain the following result which is a generalization of Corollary \ref{c:1}.

\begin{corollary} \label{c:3a}
Let $k$ and $n$ be two positive integers, and let $p$ be a real number. Then
$$\sum_{i=-k}^{k}(-1)^i\left(1+\frac{(p-1)(k+i)}{n} \right)\left(1+\frac{(p-1)(k-i)}{n} \right)\binom{n}{k+i}\binom{n}{k-i}$$
$$= \left(1+\frac{(p^2-1)k}{n}\right)\binom{n}{k}\ .$$
\end{corollary}

\begin{proof}
Taking into account that
$$e_k\left(x_1,\ldots,x_n\right)=e_k\left(x_1,\ldots,x_{n-1}\right)+x_ne_{k-1}\left(x_1,\ldots,x_{n-1}\right)\ ,$$
we can write
\begin{eqnarray*}
e_k(\underbrace{1,\ldots,1}_{n-1},p)&=&\binom{n-1}{k}+p\binom{n-1}{k-1} \\
&=&\binom{n}{k}+(p-1)\frac{k}{n}\binom{n}{k}\\
&=&\left(1+\frac{(p-1)k}{n} \right)\binom{n}{k} \ .
\end{eqnarray*}
According to Theorem \ref{T0}, the corollary is proved.
\end{proof}

The following result is a consequence of Corollary \ref{c:3a}.

\begin{corollary}\label{c:3}
Let $k$ and $n$ be two positive integers. Then
$$\sum_{i=1}^{k}(-1)^{i+1}i^2\binom{n}{k+i}\binom{n}{k-i}=\frac{k(n-k)}{2}\binom{n}{k}\ .$$
\end{corollary}

\begin{proof}
Replacing $p$ by $2$ in Corollary \ref{c:3a}, we obtain
\begin{eqnarray*}
\left(1+\frac{3k}{n} \right) \binom{n}{k}
&=& \sum_{i=-k}^{k}(-1)^i\left(1+\frac{k-i}{n} \right)\left(1+\frac{k+i}{n} \right)\binom{n}{k-i}\binom{n}{k+i}\\
&=& \sum_{i=-k}^{k}(-1)^i\left(1+\frac{2k}{n}+\frac{k^2-i^2}{n^2}\right)  \binom{n}{k-i}\binom{n}{k+i}\\
&=& \left(1+\frac{k}{n}\right)^2\sum_{i=-k}^{k}(-1)^i \binom{n}{k-i}\binom{n}{k+i}\\ &&\qquad\qquad\qquad-\left( \frac{1}{n}\right)^2 \sum_{i=-k}^{k}(-1)^ii^2 \binom{n}{k-i}\binom{n}{k+i}
\end{eqnarray*}
Now, we use Corollary \ref{c:1} and, after some simple calculations, we obtain
$$\sum_{i=-k}^{k}(-1)^{i+1}i^2 \binom{n}{k-i}\binom{n}{k+i}=k(n-k)\binom{n}{k}\ .$$
The corollary is proved.
\end{proof}

\begin{remark}
To prove Corollary \ref{c:3} we use Corollary \ref{c:3a} with $p=2$. In fact, we could choose for $p$ any value with the exception of $1$. Corollary \ref{c:3} is related in \cite{Slo12} with the sequence $\seqnum{A094305}$.
\end{remark}

Taking into account the identities
$$\sum_{k=0}^{n}k\binom{n}{k}=n2^{n-1}\qquad\mbox{and}\qquad\sum_{k=0}^{n}k^2\binom{n}{k}=n(n+1)2^{n-2}\ ,$$
by Corollary \ref{c:3}, we get the following identity:

\begin{corollary}
Let $n$ be a nonnegative integer. Then
$$\sum_{0< i\le k< n}(-1)^{i+1}i^2\binom{n}{k+i}\binom{n}{k-i}=n(n-1)2^{n-3}\ .$$
\end{corollary}

This corollary is related in \cite{Slo12} with the sequence $\seqnum{A001788}$.

At the end of this section we propose the following two exercises:

\begin{exercise} 
Let $x_1,x_2,\ldots,x_n$ be the zeros of the polynomial
$$x^n+\sum_{k=1}^{n}(-1)^kk\binom{n}{k}x^{n-k}\ .$$
Show that
$$e_k\left( x_1^2,x_2^2,\ldots,x_n^2\right) =n^2\binom{n-1}{k-1}+(-1)^k4k\binom{n}{2k}\ .$$
\end{exercise}

\begin{exercise} 
Let $k$ and $n$ be two positive integers. Prove that
$$\sum_{i=1}^{k}(-1)^ii^4\binom{n}{k+i}\binom{n}{k-i}=\frac{k(n-k)(k(n-k)-n)}{2}\binom{n}{k}\ .$$
\end{exercise}


\section{Central factorial numbers of the first kind}

The numbers
\begin{equation}\label{s1}
s(n+1,n+1-k)=(-1)^ke_k(1,2,\ldots,n)
\end{equation}
are known as Stirling numbers of the first kind. They are the coefficients in the expansion
$$\left( x\right)_n = \sum_{k=0}^{n}s(n,k)x^k\ ,$$
where $\left( x\right)_n$ is the falling factorial, namely
$$\left( x\right)_n = \prod_{k=0}^{n-1}\left(x-k \right)\ $$
(see \cite[p. 278]{Cha02}).

Similarly, the central factorial numbers of the first kind are defined in Riordan's book \cite[p. 213-217]{Rio68} by
$$x^{[n]}=\sum_{k=0}^{n}t(n,k)x^k\ ,$$
where
$$x^{[n]}=x\left( x+\frac{n}{2}-1\right) _{n-1}\ .$$
It is clearly that the $t(n,k)$ are not always integers. For $n=2m$, we have
\begin{equation*}\label{cf1}
x^{[2m]}=\prod_{k=0}^{m-1}\left(x^2-k^2 \right) =\sum_{k=0}^{m}t(2m,2k)x^{2k}\ .
\end{equation*}
In \cite{Gel10} the central factorial numbers of the first kind with even indices are denoted by $u(n,k)=t(2n,2k)$. Thus, we can see that
\begin{equation}\label{cf}
u(n+1,n+1-k)=(-1)^k e_k(1^2,2^2,\ldots,n^2)\ .
\end{equation}

\begin{corollary}\label{T1}
Let $k$ and $n$ be two positive integers such that $k\le n$. Then %
\begin{equation*}\label{eq:T1}
u(n,k)=\sum_{i=-k}^{k}(-1)^{n-k+i}s(n,k+i)s(n,k-i)\ .
\end{equation*}
\end{corollary}

\begin{proof}
By \eqref{gw}, \eqref{s1} and \eqref{cf}, we deduce that
$$u(n,n-k)=\sum_{i=-k}^{k}(-1)^{k+i}s(n,n-k+i)s(n,n-k-i)\ .$$
According to \eqref{eq:T3}, the corollary is proved.
\end{proof}

Corollary \ref{T1} is related in \cite{Slo12} to the sequences
$\seqnum{A008955}$, $\seqnum{A000330}$, $\seqnum{A000596}$,
$\seqnum{A000597}$, $\seqnum{A001819}$, $\seqnum{A001820}$,
$\seqnum{A001821}$ and $\seqnum{A204579}$.

\section{Acknowledgement}

The author would like to thank Professor Jiang Zeng from Institut
Camille Jordan, Universit\'{e} Lyon 1  for his support. The author
expresses his gratitude to Oana Merca for the careful reading of the
manuscript and helpful remarks.



\begin{thebibliography}{10}


\bibitem{Cha02} Ch. A. Charalambides, \textit{Enumerative Combinatorics}, Chapman \& Hall/CRC Press, 2002.

\bibitem{Gel10} Y. Gelineau and J. Zeng, Combinatorial interpretations
of the Jacobi-Stirling numbers, \textit{Electron. J. Combin.}
\textbf{17} (2010), Paper \#R70.

\bibitem{Gou99} H. W. Gould, The Girard-Waring power sum formulas for
symmetric functions and Fibonacci sequences. \textit{Fibonacci Quart.}
\textbf{37} (2) (1999) 135--140.

\bibitem{Kon96} J. Konvalina, A generalization of Waring's formula,
\textit{J. Combin. Theory Ser. A} \textbf{75} (2) (1996)  281--294.

\bibitem{Kon98} J. Konvalina, A note on a generalization of Waring's
formula, \textit{Adv. in Appl. Math.} \textbf{20} (1998)  392--393.

\bibitem{Rio68} J. Riordan, \textit{Combinatorial Identities}, John
Wiley \& Sons, New York, 1968.

\bibitem{Slo12} N. J. A. Sloane, {\em The On-Line Encyclopedia of Integer Sequences}, Published electronically at \href{http://oeis.org} {\tt http://oeis.org}, 2012.

\bibitem{Zen97} J. Zeng, On a generalization of Waring's formula,
\textit{Adv. in Appl. Math.} \textbf{19} (1997) 450--452.

\end{thebibliography}

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\hrule
\bigskip

\noindent 2010 {\it Mathematics Subject Classification}:
Primary 05E05, 05A19;
Secondary 11B65, 11B73.

\noindent \emph{Keywords:} binomial coefficient, central factorial numbers,
Stirling number, symmetric function, generalized Girard-Waring formula.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences
\seqnum{A000330},
\seqnum{A000346},
\seqnum{A000596},
\seqnum{A000597},
\seqnum{A001788},
\seqnum{A001819},
\seqnum{A001820},
\seqnum{A001821},
\seqnum{A008955},
\seqnum{A094305},
\seqnum{A108958}, and
\seqnum{A135065}.)

\bigskip
\hrule
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\noindent
Received April 26 2012;
revised version received May 28 2012.
Published in {\it Journal of Integer Sequences},
June 12 2012.

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\noindent
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\htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}.
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