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\begin{center}
\vskip 1cm{\LARGE\bf A Note on Cosine Power Sums} \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} Using the multisection series method, we establish formulas for various power sums of
cosine functions. As corollaries we derive several combinatorial identities.
\end{abstract}

\section{Introduction}
\label{intro}

In \cite{Mrk11,Mrk12}, we presented two open problems concerning the asymptotic behaviour of cosine power sums,
\begin{equation}
\sum_{k=1}^{\left\lfloor\frac{n-1}{2}\right\rfloor}\cos^p\left(\frac{k\pi}{n}\right)\ , 
\end{equation}
where $\lfloor x\rfloor$ denotes the largest integer not greater than $x$.
Under certain conditions, these cosine power sums can be determined exactly, without approximations \cite{Quo68}.

If
$$f(x)=a_0+a_1x+a_2x^2+\cdots+a_nx^n+\cdots$$
is a finite or convergent infinite series, then for $0\le r&lt;n$ the sum
$$a_rx^r+a_{r+n}x^{r+n}+a_{r+2n}x^{r+2n}+\cdots$$
is given by the multisection formula (see \cite[Ch. 16]{Hon85}, \cite[Ch. 4, S. 4.3]{Rio68} and \cite{Sim75})
\begin{equation}\label{eq1}
\sum_{k\ge0}a_{r+kn}x^{r+kn}=\frac{1}{n}\sum_{k=0}^{n-1}z^{-kr}f(z^kx)\ ,
\end{equation}
where $z=e^{\frac{2\pi i}{n}}$ is the $n$th root of $1$.

We shall use the multisection formula to prove:

\begin{theorem} \label{th:1}
Let $n$ and $p$ be two positive integers. Then
\begin{equation*}
\sum_{k=1}^{\left\lfloor \frac{n-1}{2}\right\rfloor} \cos^{2p}\left(\frac{k\pi}{n}\right) = -\frac{1}{2}+\frac{n}{2^{2p+1}}\sum_{k=-\left\lfloor \frac{p}{n}\right\rfloor}^{\left\lfloor \frac{p}{n}\right\rfloor} \binom{2p}{p+kn}\ .
\end{equation*}
\end{theorem}

\begin{theorem} \label{th:2}
Let $n$ and $p$ be two positive integers such that $n\equiv p \pmod 2$. Then
\begin{equation*}
\sum_{k=1}^{\left\lfloor\frac{n-1}{2}\right\rfloor} (-1)^k\cos^{p}\left(\frac{k\pi}{n}\right) = -\frac{1}{2}+\frac{n}{2^{p+1}} \sum_{k=-\left[\frac{p}{2n}\right]}^{\left[\frac{p}{2n}\right]}\binom{p}{\frac{p+n}{2}+kn}\ ,
\end{equation*}
where $[x]$ denotes the nearest integer to $x$, i.e., $[x]=\left\lfloor x+\frac{1}{2}\right\rfloor$.
\end{theorem}

\begin{theorem} \label{th:3}
Let $n$ and $p$ be two positive integers. Then
\begin{equation*}
\sum_{k=1}^{\left\lfloor\frac{n}{2}\right\rfloor} \cos^{2p}\left(\frac{2k-1}{n}\cdot\frac{\pi}{2}\right) = \frac{n}{2^{2p+1}} \sum_{k=-\left\lfloor\frac{p}{n}\right\rfloor}^{\left\lfloor\frac{p}{n}\right\rfloor}(-1)^k\binom{2p}{p+kn}\ .
\end{equation*}
\end{theorem}

We note that the formula proved in \cite{Quo68} is the case $p&lt;n$ of Theorem \ref{th:1}. 
The conjecture published in \cite{Mrk12} is solved by Theorem \ref{th:2}.
In the final section of this paper we present some relations that can be obtained immediately by the above theorems.


\section{Proof of Theorem \ref{th:1}}

We apply the multisection formula \eqref{eq1} for 
\begin{equation*}
r=p-n\left\lfloor\frac{p}{n}\right\rfloor \qquad\mbox{and}\qquad f(x)=(1+x)^{2p}\ .
\end{equation*}
Thus, when $x=1$, we get
\begin{equation}\label{eq2}
\sum_{k=-\left\lfloor\frac{p}{n}\right\rfloor}^{\left\lfloor\frac{p}{n}\right\rfloor}\binom{2p}{p+kn} =\frac{1}{n}\sum_{k=0}^{n-1}z^{-k\left( p-n\lfloor\frac{p}{n}\rfloor\right) }(1+z^k)^{2p}\ ,
\end{equation}
where $z=e^{\frac{2\pi i}{n}}$.
Having
\begin{eqnarray}
e^{-ir\varphi}(1+e^{i\varphi})^{p}&amp;=&amp;e^{-ir\varphi}\left(e^{\frac{i\varphi}{2}-\frac{i\varphi}{2}}+e^{\frac{i\varphi}{2}+\frac{i\varphi}{2}}\right)^{p}\nonumber\\
&amp;=&amp;e^{-ir\varphi}e^{\frac{ip\varphi}{2}}\left(e^{-\frac{i\varphi}{2}}+e^{\frac{i\varphi}{2}}\right)^{p}\nonumber\\
&amp;=&amp;2^{p}\cos^{p}\frac{\varphi}{2}\cdot e^{i\left(\frac{p}{2}-r\right) \varphi}\ ,\label{eq3}
\end{eqnarray}
we can write
\begin{equation*}
z^{-k\left( p-n\lfloor\frac{p}{n}\rfloor\right) }(1+z^k)^{2p} = 2^{2p}\cos^{2p}\left(\frac{k\pi}{n}\right)\cdot e^{i2k\left\lfloor\frac{p}{n}\right\rfloor\pi} \ .
\end{equation*}
So, we may rewrite \eqref{eq2} as
\begin{equation}\label{eq4}
\sum_{k=-\left\lfloor\frac{p}{n}\right\rfloor}^{\left\lfloor\frac{p}{n}\right\rfloor}\binom{2p}{p+kn}= \frac{2^{2p}}{n}\sum_{k=0}^{n-1}\cos^{2p}\left(\frac{k\pi}{n}\right)\ ,
\end{equation}
Taking into account that
$$\cos\left(\frac{(n-k)\pi}{n}\right) =-\cos\left(  \frac{k\pi}{n} \right)\ ,\quad k=0,1,\ldots,n\ ,$$
Theorem \ref{th:1} is proved.


\section{Proof of Theorem \ref{th:2}}

To prove the theorem we proceed as in Theorem \ref{th:1}. By the multisection formula \eqref{eq1}, with 
$$x=1\ ,\quad f(x)=(1+x)^{p}\quad\mbox{and}\quad r=\frac{p+n}{2}-n\left\lfloor\frac{p+n}{2n}\right\rfloor\ ,$$
we get
\begin{equation*}
\sum_{k=-\lfloor\frac{p+n}{2n}\rfloor}^{\lfloor\frac{p+n}{2n}\rfloor}\binom{p}{\frac{p+n}{2}+nk} 
= \frac{1}{n}\sum_{k=0}^{n-1}z^{-k\left(\frac{p+n}{2}-n\lfloor\frac{p+n}{2n}\rfloor \right) }(1+z^k)^{p}\ ,
\end{equation*}
where $z=e^{\frac{2\pi i}{n}}$.
By \eqref{eq3}, it immediately follows that
$$z^{-k\left(\frac{p+n}{2}-n\lfloor\frac{p+n}{2n}\rfloor \right) }(1+z^k)^{p}=2^{p}\cos^{p}\left(\frac{k\pi}{n} \right) \cdot e^{ik\pi\left(2\lfloor\frac{p+n}{2n}\rfloor-1 \right) }\ .$$
Thus we can write
\begin{equation}\label{eq5}
\sum_{k=-\lfloor\frac{p+n}{2n}\rfloor}^{\lfloor\frac{p+n}{2n}\rfloor}\binom{p}{\frac{p+n}{2}+kn} 
= \frac{2^{p}}{n}\sum_{k=0}^{n-1}(-1)^k\cos^{p}\left(\frac{k\pi}{n}\right) \ .
\end{equation}\label{eq6}
On the other hand, we have
\begin{eqnarray}
&amp;&amp;\sum_{k=0}^{n-1}(-1)^k\cos^{p}\left(\frac{k\pi}{n}\right)\nonumber\\
&amp;&amp;\qquad = 1+\sum_{k=1}^{\left\lfloor\frac{n-1}{2}\right\rfloor}(-1)^k\cos^{p}\left(\frac{k\pi}{n}\right) + \sum_{k=1}^{\left\lfloor\frac{n-1}{2}\right\rfloor}(-1)^{n-k}\cos^{p}\left(\frac{(n-k)\pi}{n}\right)\nonumber\\
&amp;&amp;\qquad = 1+\sum_{k=1}^{\left\lfloor\frac{n-1}{2}\right\rfloor}(-1)^k\cos^{p}\left(\frac{k\pi}{n}\right) + \sum_{k=1}^{\left\lfloor\frac{n-1}{2}\right\rfloor}(-1)^{n+p-k}\cos^{p}\left(\frac{k\pi}{n}\right)\nonumber\\
&amp;&amp;\qquad = 1+2\sum_{k=1}^{\left\lfloor\frac{n-1}{2}\right\rfloor}(-1)^k\cos^{p}\left(\frac{k\pi}{n}\right)\ .\label{eq7}
\end{eqnarray}

According to \eqref{eq6} and \eqref{eq7}, Theorem \ref{th:2} is proved.


\section{Proof of Theorem \ref{th:3}}

To prove the theorem, we use the formula 
\begin{equation}
\sum_{k=1}^{\left\lfloor\frac{n}{2}\right\rfloor}f(2k-1) = \frac{1}{2}\left(\sum_{k=1}^{n-1}f(k)-\sum_{k=1}^{n-1}(-1)^kf(k) \right) 
\end{equation}
for $f(k)=\cos^{2p}\left(\frac{k\pi}{2n} \right) $.
By Theorem \ref{th:1} (with $n$ replaced by $2n$) and Theorem \ref{th:2} (with $n$ and $p$ replaced by $2n$ and $2p$, respectively), we have
\begin{eqnarray*}
\sum_{k=1}^{\left\lfloor\frac{n}{2}\right\rfloor} f(2k-1) &amp;=&amp;\frac{1}{2}\left(-\frac{1}{2}+\frac{2n}{2^{2p+1}}\sum_{k=-\left\lfloor \frac{p}{2n}\right\rfloor}^{\left\lfloor \frac{p}{2n}\right\rfloor}\binom{2p}{p+2kn}\right. \\
&amp;&amp;\qquad\left. +\frac{1}{2}-\frac{2n}{2^{2p+1}}\sum_{k=-\left[ \frac{p}{2n}\right]}^{\left[ \frac{p}{2n}\right]}\binom{2p}{p+(2k+1)n}  \right) \\
&amp;=&amp; \frac{n}{2^{2p+1}}\sum_{k=-\left[ \frac{p}{2n}\right]}^{\left[ \frac{p}{2n}\right]}\left(  \binom{2p}{p+2kn}-\binom{2p}{p+(2k+1)n}\right) \\
&amp;=&amp;\frac{n}{2^{2p+1}}\sum_{k=-\left\lfloor \frac{p}{n}\right\rfloor}^{\left\lfloor \frac{p}{n}\right\rfloor}(-1)^k\binom{2p}{p+kn}\ .
\end{eqnarray*}
Theorem \ref{th:3} is proved.

\section{Special cases}

We begin this section with two immediate consequences of Theorems \ref{th:1}, \ref{th:2} and \ref{th:3}. 

\begin{corollary} \label{C1}
Let $n$ be a positive integer. Then
\begin{enumerate}
\item $\displaystyle{\sum_{k=1}^{\left\lfloor \frac{n-1}{2}\right\rfloor} \cos^{2n}\left(\frac{k\pi}{n}\right) = -\frac{1}{2}+\frac{n}{2^{2n}}+\frac{n}{2^{2n+1}}\binom{2n}{n}}$;
\item $\displaystyle{\sum_{k=1}^{\left\lfloor\frac{n-1}{2}\right\rfloor} (-1)^k\cos^{n}\left(\frac{k\pi}{n}\right) = -\frac{1}{2}+\frac{n}{2^n}}$;
\item $\displaystyle{\sum_{k=1}^{\left\lfloor\frac{n}{2}\right\rfloor} \cos^{2n}\left(\frac{2k-1}{n}\cdot\frac{\pi}{2}\right) = -\frac{n}{2^{2n}}+\frac{n}{2^{2n+1}}\binom{2n}{n}}$.
\end{enumerate}
\end{corollary}

\

\begin{corollary} \label{C2}
Let $n$ and $p$ be two positive integers such that $p&lt;n$. Then
\begin{enumerate}
\item $\displaystyle{\sum_{k=1}^{\left\lfloor \frac{n-1}{2}\right\rfloor} \cos^{2p}\left(\frac{k\pi}{n}\right) = -\frac{1}{2}+\frac{n}{2^{2p+1}}\binom{2p}{p}}$;
\item $\displaystyle{\sum_{k=1}^{\left\lfloor\frac{n-1}{2}\right\rfloor} (-1)^k\cos^{p}\left(\frac{k\pi}{n}\right) = -\frac{1}{2}}\ ,\quad n \equiv p \pmod 2$;
\item $\displaystyle{\sum_{k=1}^{\left\lfloor\frac{n}{2}\right\rfloor} \cos^{2p}\left(\frac{2k-1}{n}\cdot\frac{\pi}{2}\right) = \frac{n}{2^{2p+1}}\binom{2p}{p}}$.
\end{enumerate}
\end{corollary}

By Theorems \ref{th:1}, \ref{th:2} and \ref{th:3}, for fixed values of $n$, we can obtain some combinatorial identities. 

\begin{corollary}\label{C3}
Let $p$ be a positive integer. Then
\begin{enumerate}
	\item $\displaystyle{\sum_{k=0}^{p}\binom{2p}{p-k} = \frac{1}{2}\binom{2p}{p}+2^{2p-1}}\ ;$
	\item $\displaystyle{\sum_{k=0}^{\left\lfloor p/2\right\rfloor}\binom{2p}{p-2k} = \frac{1}{2}\binom{2p}{p}+2^{2p-2}}\ ;$
	\item $\displaystyle{\sum_{k=0}^{\left\lfloor p/3\right\rfloor}\binom{2p}{p-3k} = \frac{1}{2}\binom{2p}{p}+\frac{1+2^{2p-1}}{3}}\ ;$
	\item $\displaystyle{\sum_{k=0}^{\left\lfloor p/4\right\rfloor}\binom{2p}{p-4k} = \frac{1}{2}\binom{2p}{p}+2^{2p-3}+2^{p-2}}\ ;$
	\item $\displaystyle{\sum_{k=0}^{\left\lfloor p/5\right\rfloor}\binom{2p}{p-5k} = \frac{1}{2}\binom{2p}{p}+\frac{(3+\sqrt{5})^p+(3-\sqrt{5})^p+2^{3p-1}}{5\cdot2^p}}\ ;$
	\item $\displaystyle{\sum_{k=0}^{\left\lfloor p/6\right\rfloor}\binom{2p}{p-6k} = \frac{1}{2}\binom{2p}{p}+\frac{3^p+2^{2p-1}+1}{6}}\ .$
\end{enumerate}
\end{corollary}

\begin{corollary}\label{C4}
Let $p$ be a positive integer. Then
\begin{enumerate}
	\item $\displaystyle{\sum_{k=1}^{p}\binom{2p-1}{p-k} = 4^{p-1}}$ ;
	\item $\displaystyle{\sum_{k=1}^{\left\lfloor p/3\right\rfloor}\binom{2p-3}{p-3k} = \frac{4^{p-2}-1}{3}}\ ,\quad p&gt;1\ ;$
	\item $\displaystyle{\sum_{k=1}^{\left\lfloor p/5\right\rfloor}\binom{2p-5}{p-5k} = \frac{4^{p-3}}{5}-\frac{(\sqrt{5}+1)^{2p-5}-(\sqrt{5}-1)^{2p-5}}{5\cdot2^{2p-5}}}\ ,\quad p&gt;2\ .$
\end{enumerate}
\end{corollary}

\begin{corollary}\label{C5}
Let $p$ be a positive integer. Then
\begin{enumerate}
	\item $\displaystyle{\sum_{k=0}^{p}(-1)^k\binom{2p}{p-k} = \frac{1}{2}\binom{2p}{p}}\ ;$
	\item $\displaystyle{\sum_{k=0}^{\left\lfloor p/2\right\rfloor}(-1)^k\binom{2p}{p-2k} = \frac{1}{2}\binom{2p}{p}+2^{p-1}}\ ;$
	\item $\displaystyle{\sum_{k=0}^{\left\lfloor p/3\right\rfloor}(-1)^k\binom{2p}{p-3k} = \frac{1}{2}\binom{2p}{p}+3^{p-1}}\ ;$
	\item $\displaystyle{\sum_{k=0}^{\left\lfloor p/4\right\rfloor}(-1)^k\binom{2p}{p-4k} = \frac{1}{2}\binom{2p}{p}+\frac{(2+\sqrt{2})^p+(2-\sqrt{2})^p}{4}}\ ;$
	\item $\displaystyle{\sum_{k=0}^{\left\lfloor p/5\right\rfloor}(-1)^k\binom{2p}{p-5k} = \frac{1}{2}\binom{2p}{p}+\frac{(5+\sqrt{5})^p+(5-\sqrt{5})^p}{5\cdot2^p}}\ .$
\end{enumerate}
\end{corollary}

Note that Corollary \ref{C3} is related in \cite{Slo12} with the sequences $\seqnum{A032443}$, $\seqnum{A114121}$, $\seqnum{A007583}$,  $\seqnum{A007582}$, $\seqnum{A078789}$, $\seqnum{A085282}$, Corollary \ref{C4} with $\seqnum{A000302}$, $\seqnum{A002450}$, $\seqnum{A095931}$, Corollary \ref{C5} with $\seqnum{A088218}$, $\seqnum{A005317}$, $\seqnum{A191993}$, $\seqnum{A007052}$, $\seqnum{A081567}$, respectively.


\section{Acknowledgements}

The author would like to thank Professor Cecil C. Rousseau from the
University of Memphis for his comments on the exact formulas of these
cosine power sums.  The author expresses his gratitude to Oana Merca
for the careful reading of the manuscript and helpful remarks.


\begin{thebibliography}{10}


\bibitem{Hon85}
R. Honsberger, \textit{Mathematical Gems III}, Dolciani Math.\ 
Expositions No.\ 9, Mathematical Association of America, 1985.

\bibitem{Mrk11} 
M. Merca, Asymptotic behavior of cosine power sums, \textit{SIAM, Problems
and Solutions Online Archive}, 
\url{http://siam.org/journals/categories/11-002.php}, 2011.

\bibitem{Mrk12}
M. Merca, Problem 89, \textit{Eur. Math. Soc. Newsl.}, \textbf{81}
(2011), 59.

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

\bibitem{Quo68}
J. M. Quoniam, and M. G. Greening, A trigonometric summation,
\textit{Amer. Math. Monthly}, \textbf{75} (1968), 405--406.

\bibitem{Sim75}
T. Simpson, The invention of a general method for determining the sum
of every 2d, 3d, 4th, or 5th, \&amp;c. term of a series, taken in order;
The sum of the whole series being known, \textit{Philosophical
Transactions}, \textbf{50} (1757--1758), 757--769.  Available at
\url{http://www.jstor.org/stable/105328}.

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


\end{thebibliography}

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\noindent 2010 {\it Mathematics Subject Classification}:
Primary 33B10; Secondary 05A19, 11B65.

\noindent \emph{Keywords:} trigonometric power sum, multisection formula, combinatorial identity.

\bigskip
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\noindent (Concerned with sequences
\seqnum{A000302},
\seqnum{A002450},
\seqnum{A005317},
\seqnum{A007052},
\seqnum{A007582},
\seqnum{A007583},
\seqnum{A032443},
\seqnum{A078789},
\seqnum{A081567},
\seqnum{A085282},
\seqnum{A088218},
\seqnum{A095931},
\seqnum{A114121}, and
\seqnum{A191993}.)

\bigskip
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\noindent
Received February 15 2012;
revised version received May 8 2012.
Published in {\it Journal of Integer Sequences}, May 28 2012.

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