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\begin{center}
\vskip 1cm{\LARGE\bf An Extended Version of Faulhaber's Formula
}
\vskip 1cm
\large
Raphael Schumacher\\
Department of Mathematics\\
ETH Zurich\\
R\"amistrasse 101 \\
8092 Zurich\\
Switzerland\\
\href{mailto:raphschu@ethz.ch}{\tt raphschu@ethz.ch}\\
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\begin{abstract}
This paper presents an extended version of the well-known Faulhaber
formula, which is used to compute the sum of the $m$-th powers of the
first $n$ natural numbers, where $m$ and $n$ are two natural numbers.
Our expression is analogous to Faulhaber's formula, but sums the $m$-th
powers of the natural numbers $\leq x$ for any non-negative real number
$x$.
\end{abstract}

\section{Introduction}
\label{sec:introduction}

For two natural numbers $m,n\in\mathbb{N}_{0}$, the Faulhaber formula
\cite{1}, which was found by Jacob Bernoulli around 1700, provides a
very efficient way to compute the sum of the $m$-th powers of the first
$n$ natural numbers. It is given by
\begin{displaymath}
\sum_{k=0}^{n}k^{m}=\frac{1}{m+1}\sum_{k=0}^{m}(-1)^{k}{m+1\choose k}B_{k}n^{m-k+1},
\end{displaymath}
where the $B_{k}$'s are the Bernoulli numbers.

In this paper, we will prove the analogous expression for the sum $\sum_{k=0}^{\lfloor x\rfloor}k^{m}$, where $x\in\mathbb{R}_{0}^{+}$ and $m\in\mathbb{N}_{0}$, in terms of Bernoulli polynomials $B_{k}(x)$ instead of Bernoulli numbers $B_{k}$. This expression is given in the following theorem:

\begin{theorem}(extended Faulhaber formula)

\noindent For any $x\in\mathbb{R}_{0}^{+}$ we have that
\begin{displaymath}
\begin{split}
\sum_{k=0}^{\lfloor x\rfloor}k^{m}=\frac{1}{m+1}x^{m+1}+(-1)^{m}\frac{B_{m+1}}{m+1}+\frac{1}{m+1}\sum_{k=1}^{m+1}(-1)^{k}{m+1\choose k}B_{k}\left(\left\{x\right\}\right)x^{m-k+1}.
\end{split}
\end{displaymath}
\end{theorem}

We have searched for this version of Faulhaber's formula in the
literature, but we have not found it and therefore we believe that this
result is new.

\section{Definitions}
\label{sec:Definitions}

As usual, we denote the floor of $x$ by $\lfloor x\rfloor$ and the
fractional part of $x$ by $\{x\}$.

\begin{definition}
For $k\in\mathbb{N}_{0}$ we define
the \emph{$k$-th Bernoulli polynomial} $B_{k}(x)$ via the following exponential generating function \cite{2}:
\begin{displaymath}
\frac{te^{xt}}{e^{t}-1}=\sum_{k=0}^{\infty}\frac{B_{k}(x)}{k!}t^{k}\;\;\forall t\in\mathbb{C}\;\text{with $|t|&lt;2\pi$}.
\end{displaymath}
\end{definition}

\begin{definition}
The \emph{$k$-th Bernoulli number} $B_{k}$ is defined as the value of the $k$-th Bernoulli polynomial $B_{k}(x)$ at $x=0$ \cite{2}, that is
\begin{displaymath}
B_{k}:=B_{k}(0).
\end{displaymath}
Moreover, we get from the definition of the Bernoulli polynomials \cite{1} that
\begin{displaymath}
\frac{t}{e^{t}-1}=\sum_{k=0}^{\infty}\frac{B_{k}}{k!}t^{k}\;\;\forall t\in\mathbb{C}\;\text{with $|t|&lt;2\pi$}.
\end{displaymath}
\end{definition}

\section{Proof of the extended Faulhaber formula}
\label{sec:Proof of the Extended Faulhaber Formula}

In this section we will prove our extended version of Faulhaber's formula.

\begin{proof}
Let $m,n\in\mathbb{N}_{0}$ be two natural numbers. Starting from \cite{1} the usual Faulhaber formula 
\begin{displaymath}
\sum_{k=0}^{n}k^{m}=\frac{1}{m+1}\sum_{k=0}^{m}(-1)^{k}B_{k}{m+1\choose k}n^{m-k+1},
\end{displaymath}
we obtain
\begin{displaymath}
\sum_{k=0}^{n}k^{m}=(-1)^{m}\frac{B_{m+1}}{m+1}+\frac{1}{m+1}\sum_{k=0}^{m+1}(-1)^{k}B_{k}{m+1\choose k}n^{m-k+1}.
\end{displaymath}
Setting here $n:=\lfloor x\rfloor=x-\{x\}$ for some $x\in\mathbb{R}_{0}^{+}$, we get

\begin{displaymath}
\begin{split}
\sum_{k=0}^{\lfloor x\rfloor}k^{m}&amp;=(-1)^{m}\frac{B_{m+1}}{m+1}+\frac{1}{m+1}\sum_{k=0}^{m+1}(-1)^{k}B_{k}{m+1\choose k}\left(x-\{x\}\right)^{m-k+1}\\
&amp;=(-1)^{m}\frac{B_{m+1}}{m+1}+\frac{1}{m+1}\sum_{k=0}^{m+1}(-1)^{k}B_{k}{m+1\choose k}\sum_{l=0}^{m-k+1}(-1)^{m-k-l+1}{m-k+1\choose l}x^{l}\{x\}^{m-k-l+1}\\
&amp;=(-1)^{m}\frac{B_{m+1}}{m+1}+\frac{1}{m+1}\sum_{k=0}^{m+1}B_{k}{m+1\choose k}\sum_{l=0}^{m-k+1}(-1)^{m-l+1}{m-k+1\choose l}x^{l}\{x\}^{m-k-l+1},
\end{split}
\end{displaymath}
where we have used the binomial theorem
\begin{displaymath}
(a+b)^{n}=\sum_{l=0}^{n}{n\choose l}a^{l}b^{n-l}
\end{displaymath}
for $a:=x$, $b:=\{x\}$ and $n:=m-k+1$.

We now interchange the order of summation and use the binomial identity
\begin{displaymath}
\begin{split}
{m+1\choose k}{m-k+1\choose l}&amp;=\frac{(m+1)!(m-k+1)!}{k!l!(m+1-k)!(m-k-l+1)!}\\
&amp;=\frac{(m+1)!}{k!l!(m-k-l+1)!}\\
&amp;=\frac{(m+1)!(m-l+1)!}{k!l!(m-l+1)!(m-k-l+1)!}\\
&amp;={m+1\choose l}{m-l+1\choose k}
\end{split}
\end{displaymath}
to obtain that
\begin{displaymath}
\begin{split}
\sum_{k=0}^{\lfloor x\rfloor}k^{m}&amp;=(-1)^{m}\frac{B_{m+1}}{m+1}+\frac{1}{m+1}\sum_{k=0}^{m+1}B_{k}{m+1\choose k}\sum_{l=0}^{m-k+1}(-1)^{m-l+1}{m-k+1\choose l}x^{l}\{x\}^{m-k-l+1}\\
&amp;=(-1)^{m}\frac{B_{m+1}}{m+1}+\frac{1}{m+1}\sum_{l=0}^{m+1}(-1)^{m-l+1}x^{l}\sum_{k=0}^{m-l+1}B_{k}{m+1\choose k}{m-k+1\choose l}\{x\}^{m-k-l+1}\\
&amp;=(-1)^{m}\frac{B_{m+1}}{m+1}+\frac{1}{m+1}\sum_{l=0}^{m+1}(-1)^{m-l+1}x^{l}\sum_{k=0}^{m-l+1}B_{k}{m+1\choose l}{m-l+1\choose k}\{x\}^{m-k-l+1}\\
\end{split}
\end{displaymath}
\begin{displaymath}
\begin{split}
\hspace{1.14cm}&amp;=(-1)^{m}\frac{B_{m+1}}{m+1}+\frac{1}{m+1}\sum_{l=0}^{m+1}(-1)^{m-l+1}{m+1\choose l}x^{l}\sum_{k=0}^{m-l+1}B_{k}{m-l+1\choose k}\{x\}^{m-k-l+1}.
\end{split}
\end{displaymath}
If we use now the following explicit formula \cite[\text{Proposition\ 23.2, p.\ 86}]{3} for the Bernoulli polynomials
\begin{displaymath}
B_{n}(x)=\sum_{k=0}^{n}B_{k}{n\choose k}x^{n-k}
\end{displaymath}
for $n:=m-l+1$ and $x:=\{x\}$, we get
\begin{displaymath}
\begin{split}
\sum_{k=0}^{\lfloor x\rfloor}k^{m}&amp;=(-1)^{m}\frac{B_{m+1}}{m+1}+\frac{1}{m+1}\sum_{l=0}^{m+1}(-1)^{m-l+1}{m+1\choose l}x^{l}\sum_{k=0}^{m-l+1}B_{k}{m-l+1\choose k}\{x\}^{m-k-l+1}\\
&amp;=(-1)^{m}\frac{B_{m+1}}{m+1}+\frac{1}{m+1}\sum_{l=0}^{m+1}(-1)^{m-l+1}{m+1\choose l}B_{m-l+1}\left(\{x\}\right)x^{l}.
\end{split}
\end{displaymath}
In the above formula we can change variables according
to $l:=m-k+1\Longleftrightarrow k=m-l+1$
and use the symmetry of the binomial coefficients
\begin{displaymath}
{m+1\choose m-k+1}={m+1\choose k},
\end{displaymath}
to conclude that
\begin{displaymath}
\begin{split}
\sum_{k=0}^{\lfloor x\rfloor}k^{m}&amp;=(-1)^{m}\frac{B_{m+1}}{m+1}+\frac{1}{m+1}\sum_{l=0}^{m+1}(-1)^{m-l+1}{m+1\choose l}B_{m-l+1}\left(\{x\}\right)x^{l}\\
&amp;=(-1)^{m}\frac{B_{m+1}}{m+1}+\frac{1}{m+1}\sum_{k=0}^{m+1}(-1)^{k}{m+1\choose m-k+1}B_{k}\left(\{x\}\right)x^{m-k+1}\\
&amp;=(-1)^{m}\frac{B_{m+1}}{m+1}+\frac{1}{m+1}\sum_{k=0}^{m+1}(-1)^{k}{m+1\choose k}B_{k}\left(\{x\}\right)x^{m-k+1}.
\end{split}
\end{displaymath}
Finally,
if we use the fact 
that $B_{0}(x)=1\;\forall x\in\mathbb{R}$, we get our claimed formula
\begin{displaymath}
\begin{split}
\sum_{k=0}^{\lfloor x\rfloor}k^{m}=\frac{1}{m+1}x^{m+1}+(-1)^{m}\frac{B_{m+1}}{m+1}+\frac{1}{m+1}\sum_{k=1}^{m+1}(-1)^{k}{m+1\choose k}B_{k}\left(\left\{x\right\}\right)x^{m-k+1}
\end{split}
\end{displaymath}
for all $x\in\mathbb{R}_{0}^{+}$.
\end{proof}

\begin{remark}
The ordinary Faulhaber formula follows by setting $x:=n\in\mathbb{N}_{0}$ in our developed extension, because
\begin{displaymath}
\begin{split}
\sum_{k=0}^{n}k^{m}&amp;=\sum_{k=0}^{\lfloor n\rfloor}k^{m}\\
&amp;=\frac{1}{m+1}n^{m+1}+(-1)^{m}\frac{B_{m+1}}{m+1}+\frac{1}{m+1}\sum_{k=1}^{m+1}(-1)^{k}{m+1\choose k}B_{k}\left(\left\{n\right\}\right)n^{m-k+1}\\
&amp;=\frac{1}{m+1}n^{m+1}+(-1)^{m}\frac{B_{m+1}}{m+1}+\frac{1}{m+1}\sum_{k=1}^{m+1}(-1)^{k}{m+1\choose k}B_{k}n^{m-k+1}\\
&amp;=(-1)^{m}\frac{B_{m+1}}{m+1}+\frac{1}{m+1}\sum_{k=0}^{m+1}(-1)^{k}{m+1\choose k}B_{k}n^{m-k+1}\\
&amp;=\frac{1}{m+1}\sum_{k=0}^{m}(-1)^{k}{m+1\choose k}B_{k}n^{m-k+1},
\end{split}
\end{displaymath}
where we have used that $B_{k}\left(\{n\}\right)=B_{k}(0)=B_{k}$ for all $k\in\mathbb{N}_{0}$.
\end{remark}


\section{Acknowledgment}
\label{sec:Acknowledgement}

The author thanks the referee for the valuable suggestions regarding
the improvement of this paper.

\begin{thebibliography}{9}

\bibitem{1}
Kevin J. McGown and Harold R. Parks, The generalization of Faulhaber's
formula to sums of non-integral powers, \emph{J. Math. Anal. Appl.}
{\bf330} (2007), 571--575.

\bibitem{2}
A. Bazs\'o, \'A. Pint\'er, and H. M. Srivastava, A refinement of
Faulhaber's theorem concerning sums of powers of natural numbers,
\emph{Appl. Math. Lett.} {\bf25} (2012), 486--489.

\bibitem{3}
Victor Kac and Pokman Cheung, \emph{Quantum Calculus}, Springer, 2002.

\end{thebibliography}

\bigskip
\hrule
\bigskip

\noindent 2010 {\it Mathematics Subject Classification}: Primary 11B68;
Secondary 34M30.

\noindent\emph{Keywords:} extended Faulhaber formula, floor function,
sum of powers of the first few natural numbers, 
Bernoulli polynomial, Bernoulli number.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences
\seqnum{A027641} and
\seqnum{A027642}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received November 7 2015;
revised versions received November 9 2015; March 19 2016.
Published in {\it Journal of Integer Sequences}, April 7 2016.

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