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\title{\bf The Hypercube and\\ Elementary Calculus}
\author{\large Roberto B. Salgado}
\maketitle
\abstract{We present two geometrical constructions involving hypercubes
to motivate the well-known formulae for the derivative and the
integral of $x^m$, where $m$ is a positive integer.  The novelty of these
constructions is that the coefficients that appear in these formulae
follow from simple counting arguments.}

\section{Introduction}
In elementary calculus, one learns that the derivative and the integral
of $x^m$, where $m$ is a positive
integer\footnote{Throughout this article, $m$ will be
                      taken to be a positive integer.},
are given by:
$$\frac{d}{dx}x^m=mx^{m-1},\qquad \int x^m\ dx=\frac{1}{m+1}x^{m+1}.$$
Without resorting to a thorough analysis,
how can one intuitively motivate these formulae?
Using ``dimensional analysis'', it easy to motivate that
the derivative of $x^m$ is proportional to $x^{m-1}$ and
that the integral of $x^m$ is proportional to $x^{m+1}$.
But how can one motivate the coefficients, $m$ and $1/(m+1)$,
respectively, in these formulae?

In this article, we present geometrical constructions involving
hypercubes.  In the \S 2, we motivate the coefficient $m$
in the formula for the derivative.  We interpret the expression
$(x+h)^m-x^m$ as ``difference of two hypercubes'' and show
that the leading-order term is equal to $mx^{m-1}h$.  From this,
the formula for the derivative follows.
In \S 3, we motivate the coefficient $1/(m+1)$
in the formula for the integral.  Using the Riemann definition for the
integral, one generally obtains an expression of the form
$\sum_{k=0}^{N-1} k^m$.
We observe that, for our purposes, we are interested in only the asymptotic
limit of this expression as $N\to\infty$.  We interpret this asymptotic
limit as ``the volume of a hyperpyramid of height $N$ and base $N^m$'',
which is equal to $N^{m+1}/(m+1)$.  From this, the formula for
the integral follows.



\section{The derivative}



For concreteness, we will motivate the formula for the derivative
of $x^m$ for the cases $m=2$ and $m=3$.  The generalization for
arbitrary $m$ is straightforward.

Applying the definition of the derivative,
$$\frac{d}{dx}(x^2)=\lim_{h\to 0}\frac{(x+h)^2-x^2}{h}=\lim_{h\to
0}\frac{2xh+O(h^2)}{h}=2x.$$
$$\frac{d}{dx}(x^3)=\lim_{h\to 0}\frac{(x+h)^3-x^3}{h}=\lim_{h\to
0}\frac{3x^2h+O(h^2)}{h}=3x^2,$$
where $O(h^2)$ represent higher-order terms in $h$.
Certainly, the algebra is straightforward.  But can this formula
be motivated geometrically?

The quantity
$(x+h)^2-x^2$ can be pictorially represented as
the ``difference of two squares''.  Similarly, the quantity
$(x+h)^3-x^3$ can be pictorially represented as
the ``difference of two cubes'':

 \bigskip
 \epsfxsize=3.0in
 \epsfysize=2.0in
 \centerline{\vbox{\epsfbox{cube-1a.eps}}}
 \epsfxsize=0pt
 \epsfysize=0pt
{\scriptsize {\bf Figure 1a.} The pictorial representation of
             $(x+h)^2-x^2$.}\\
 \bigskip

 \bigskip
 \epsfxsize=3.0in
 \epsfysize=2.0in
 \centerline{\vbox{\epsfbox{cube-1b.eps}}}
 \epsfxsize=0pt
 \epsfysize=0pt
{\scriptsize {\bf Figure 1b.} The pictorial representation of
             $(x+h)^3-x^3$.}\\
 \bigskip

\noindent In each case, the leading-order contribution of these
differences is the term linear in $h$.
For the $m=2$ case, the contribution is $2xh$ and can be pictorially
represented by the 2 ``rods'' of length $x$ and thickness $h$.
For the $m=3$ case, the contribution is $3x^2h$ and can be pictorially
represented by the 3 square ``plates'' of area $x^2$ and
thickness $h$.

It is obvious that, for arbitrary $m$, the leading-order
contribution in the analogous difference is $mx^{m-1}h$,
which is represented pictorially by $m$ ``plates'' of ``area''
$x^{m-1}$ and thickness $h$.  Hence, the coefficient $m$ appearing
in the derivative of $x^m$ follows from a simple counting argument.

\section{The integral}

Often, one motivates the coefficient $1/(m+1)$ using an antiderivative
of $x^m$, \ie, a function whose derivative is $x^m$.  But this
appears somewhat artificial. Moreover, it does not suggest the geometrical
notion as ``the area under a curve''. We now present a novel
way to motivate the coefficient $1/(m+1)$.

Using the Riemann method, we approximate the area under the graph of
$f(s)=s^m$ between $s=0$ and $s=x$ with a set of $N$ (inner) rectangles
with constant width $\Delta s=x/N$.

 \bigskip
 \epsfxsize=3.0in
 \epsfysize=2.0in
 \centerline{\vbox{\epsfbox{cube-2a.eps}}}
 \epsfxsize=0pt
 \epsfysize=0pt
{\scriptsize {\bf Figure 2a.} The Riemann integral
$\displaystyle \int_0^x f(s)\ ds
=\lim_{\Delta s \rightarrow 0} \sum_i f(s_i) \Delta s$.}\\
 \bigskip

\begin{eqnarray*}
\int_0^x s^m\ ds
               &=& \lim_{N \rightarrow \infty}
                       \left(
                         \left(\frac 0N x\right)^m\left(\frac xN\right)
                       + \left(\frac 1N x\right)^m\left(\frac xN\right)
                       + \left(\frac 2N x\right)^m\left(\frac xN\right)+
                       \right.\\
               & & \hspace{1.75in}\left.\ldots
                       + \left(\frac{N-1}{N} x\right)^m\left(\frac xN\right)
                       \right)\\
               &=& \lim_{N \rightarrow \infty} (0^m+1^m+2^m+\ldots+(N-1)^m)
                         \left(\frac xN\right)^{m+1}\\
               &=& \left(\lim_{N \rightarrow \infty}
                         \frac{1}{N^{m+1}} \sum_{k=0}^{N-1}k^m\right)
                                             x^{m+1}\\
\end{eqnarray*}

For $m=1$, the sum of the series is well-known and can be motivated many ways:
$$\sum_{k=0}^{N-1}k=\frac{(N-1)N}{2}.$$
Hence, the coefficient of $x^2$ is
$$\lim_{N \rightarrow\infty} \frac{1}{N^2}\sum_{k=0}^{N-1} k
=\lim_{N\to\infty}\frac{(N-1)N}{2N^2}
=\frac{1}{2}.$$

For integers $m$ greater than $1$, the corresponding summation formulae are
not as well-known. For example \cite{crc},
\begin{eqnarray*}
\sum_{k=1}^N k^{\hphantom{1}}&=& 1+2+3+\ldots+N=\frac{N(N+1)}{2}\\
     \sum_{k=1}^N k^2 &=& 1^2+2^2+3^2+\ldots+N^2=\frac{N(N+1)(2N+1)}{6}\\
     \sum_{k=1}^N k^3 &=& \frac{N^2(N+1)^2}{4}\\
     \sum_{k=1}^N k^4 &=& \frac{N}{30}(N+1)(2N+1)(3N^2+3N-1)\\
     \sum_{k=1}^N k^5 &=& \frac{N^2}{12}(N+1)^2(2N^2+2N-1)\\
     \sum_{k=1}^N k^6 &=& \frac{N}{42}(N+1)(2N+1)(3N^4+6N^3-3N+1)
\end{eqnarray*}
Without much of a pattern to follow, we feel a beginning calculus student
will have difficulty generating this list of summation formulae.
Although these formulae are interesting in their own right,
we feel that pursuing their derivation (as in \cite{calc})
would be a distraction from the task at hand.
However, without such a list, it appears difficult to motivate the
coefficient $1/(m+1)$.

Notice, however, that we are not really interested in the exact summation
formulae, but only in their asymptotic limits.  We now show how
to obtain these limits geometrically, completely avoiding the
explicit use of these summation formulae.

For the sum $0+1+2+\ldots+(N-1)$, consider the following
geometric figure.
We arrange the unit-squares within the shape of a triangle with a height $N$
and a base of length $N$.
Observe that this arrangement approximates
$(\mbox{area of a square with side $N$})/2$.
As $N$ tends to infinity, this approximation becomes exact.
Thus, intuitively, we conclude that
$$\lim_{N\rightarrow\infty} \sum_{k=0}^{N-1} k=\frac{1}{2} N^2.$$

 \bigskip
 \epsfxsize=3.0in
 \epsfysize=2.0in
 \centerline{\vbox{\epsfbox{cube-2b.eps}}}
 \epsfxsize=0pt
 \epsfysize=0pt
{\scriptsize {\bf Figure 2b.} The pictorial representation of
$0+1+2+3+4+5\approx\frac{1}{2} 6^2$.}\\
 \bigskip

For the sum $0^2+1^2+2^2+\ldots+(N-1)^2$, consider the analogous
geometric figure.
We arrange the unit-cubes within the shape of a pyramid with a height $N$
and a square base of area $N^2$.
Observe that this arrangement of cubes approximates
$(\mbox{volume of a cube with side $N$})/3$ \cite{cube}.
(Recall that the volume of a pyramid is $hA/3$,
where $h$ is the height of the pyramid and $A$ is the
area of the base of the pyramid.)
Thus, intuitively, we conclude that
$$\lim_{N \rightarrow\infty} \sum_{k=0}^{N-1} k^2=\frac{1}{3}N^3.$$

 \bigskip
 \epsfxsize=3.0in
 \epsfysize=2.0in
 \centerline{\vbox{\epsfbox{cube-2c.eps}}}
 \epsfxsize=0pt
 \epsfysize=0pt
{\scriptsize {\bf Figure 2c.} The pictorial representation of
$0^2+1^2+2^2+3^2+4^2+5^2\approx\frac{1}{3} 6^3$.}\\
 \bigskip

Generally, for the sum $0^m+1^m+2^m+\ldots+(N-1)^m$, we arrange the
unit-hypercubes (of dimension $m+1$) within
the shape of a hyperpyramid (of dimension $m+1$) with a height $N$ and a
``square base (of dimension $m$)'' of ``area'' $N^m$.
Observe that this arrangement of hypercubes approximates
$(\mbox{hypervolume of an $(m+1)$-cube with side $N$})/(m+1)$.
(The $m$-volume of an $m$-pyramid is $hA_m/m$,
where $h$ is the height of the pyramid and
      $A_m$ is the ``area'' [\ie, the $(m-1)$-volume]
      of the base \cite{ndim}.)
Thus, intuitively, we conclude that
$$\lim_{N\rightarrow\infty} \sum_{k=0}^{N-1} k^m=\frac{1}{m+1}N^{m+1}.$$
Hence, the coefficient $1/(m+1)$ appearing in the integral of $x^m$ also
follows from a simple counting argument.


\section{Conclusion}

We have presented geometrical motivations for the formula for
the derivative and the integral of $x^m$, where $m$ is a positive integer.
We have shown that these formulae can be interpreted as particular slicings
of a suitable (hyper)cube.  In particular, we showed that the coefficients
follow directly from simple counting arguments.




\begin{thebibliography}{9}

\bibitem{crc} {\it CRC Standard Mathematical Tables}, 26th-ed
(Boca Raton, Florida, CRC Press.) 1981.

\bibitem{calc} Chover, J.,
{\it The Green Book of Calculus},
(Menlo Park, California, W.A. Benjamin, Inc.) 1972.

\bibitem{cube} Ranucci, E.R. / Rollins, W.E.,
{\it Curiosities of the Cube},
(New York, Crowell) 1977.

\bibitem{ndim} Sommerville, D.M.Y.,
{\it An Introduction to the Geometry of n Dimensions}
(New York, Dover) 1958.

\end{thebibliography}

\ni{\em Department of Physics\\
        Syracuse University\\
        Syracuse NY 13244-1130\\
   {\tt salgado@suhep.phy.syr.edu}
   }

\end{document}
\end

