% file from FORM TEX
% written by David Radford 8/25/96, 8/29/96
\documentstyle[12pt]{article}
\pagestyle{empty} \textheight 7.5in
% macros etc
\newtheorem{Lemma}{Lemma}
\newcommand{\BF}[1]{\mbox{\boldmath $#1$}}
\newcommand{\pf}{\medskip\noindent{\sc Proof: }}
\newcommand{\qed}{$\Box$}
% this is file WIDEPAGE.STY to get a reasonable size page
with 12
\marginparwidth 0pt \oddsidemargin 0pt \evensidemargin 0pt
\marginparsep 0pt \topmargin 0pt \textwidth 6.4 in
\textheight
8.5 in
%
\begin{document}
\begin{center}
\bf Math 300, Spring
2003 \hfill Writing for Mathematics \hfill
Monday sections
\end{center}
\begin{center}
\bf {\huge The Mathematics for Essay 1}
\end{center}
\begin{center}
\rule[0in]{5.5in}{.1in}
\end{center}
\medskip
The purpose of these notes are two-fold: to explain the
mathematical background for Essay 1 and to show you some of
the
flexibility of TeX.
Your own essay should not repeat the
arguments here but should have a more geometric flavor. Write
about how one can physically place the blocks. You may
assume
basic facts about geometric sums and series for Essay 1. Let
$r$
be any real number and let $n$ be a non-negative integer.
The sum
\begin{equation}\label{EqGeoSum}
1 + r + r^2 + \cdots + r^n
\end{equation}
is a {\em geometric sum} and the infinite series
\begin{equation}\label{EqGeoSeries}
1 + r + r^2 + \cdots + r^n + \cdots
\end{equation}
is a {\em geometric series}.
Suppose further that $r \neq 1$. Then the geometric sum
(\ref{EqGeoSum}) can be computed by the formula
$$
1 + r + r^2 + \cdots + r^n
= \frac{1 - r^{n+1}}{1 - r}.
$$
This fact, which you may assume, is easily proved proved by
mathematical induction.
Now suppose that
$|r| < 1$. Then $\lim_{n \longrightarrow
\infty}r^n = 0$ which means the geometric series
(\ref{EqGeoSeries}) converges to $\displaystyle{\frac{1}{1 -
r}}$
by the preceding equation. We write
\begin{equation}\label{EqGeomSeriesSum}
1 + r + r^2 + \cdots + r^n + \cdots = \frac{1}{1 - r}
\end{equation}
to indicate that the series converges and to designate the
limit
of the sequence of partial sums.
Your essay will involve the geometric series
\begin{equation}\label{EqOneHalf}
1 + \frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2^n} +
\cdots .
\end{equation}
Since $\displaystyle{|\frac{1}{2}| < 1}$, it follows by
(\ref{EqGeomSeriesSum}) that (\ref{EqOneHalf}) converges and
$\displaystyle{ 1 + \frac{1}{2} + \frac{1}{4} + \cdots +
\frac{1}{2^n} + \cdots = 2. }$ The DeLux blocks are cubes
with
side lengths $\displaystyle{ 1, \quad \frac{1}{2}, \quad
\frac{1}{3}, \quad \frac{1}{5}, \quad \ldots }$ Your essay
involves analyzing the sum of their side lengths
$$
1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4}+ \frac{1}{5}+
\frac{1}{6}+ \frac{1}{7}+ \frac{1}{8} + \frac{1}{9} + \cdots +
\frac{1}{16}+ \cdots .
$$
The preceding series is called the {\em harmonic series}.
Think of
the terms of the geometric series (\ref{EqOneHalf}) as
markers for
grouping terms of the harmonic series as follows:
\begin{equation}\label{EqHarmonic}
{\BF 1} + \frac{\BF 1}{\BF 2} + (\frac{1}{3} + \frac{\BF
1}{\BF
4}) + (\frac{1}{5}+ \frac{1}{6}+ \frac{1}{7}+ \frac{\BF
1}{\BF 8})
+ (\frac{1}{9} + \cdots +
\frac{\BF 1}{\BF 16}) + \cdots .
\end{equation}
We will find an overestimate and an underestimate for the
sum of
the terms in each of the parenthesized groups. You will see
a
pattern emerging in our calculations:
$$
1 = \frac{1}{2} + \frac{1}{2} > \frac{1}{3} + \frac{1}{4}
>
\frac{1}{4} + \frac{1}{4} = \frac{1}{2},
$$
$$
1 = \frac{1}{4} + \frac{1}{4} + \frac{1}{4} + \frac{1}{4}
>
\frac{1}{5} + \frac{1}{6}+ \frac{1}{7}+ \frac{1}{8} >
\frac{1}{8}
+ \frac{1}{8} + \frac{1}{8} + \frac{1}{8} = \frac{1}{2},
$$
$$
1 = \frac{1}{8} + \cdots + \frac{1}{8} = 8(\frac{1}{8}) >
\frac{1}{9} + \cdots +
\frac{1}{16} > \frac{1}{16} + \cdots +
\frac{1}{16} = 8(\frac{1}{16}) = \frac{1}{2},
$$
$$
\vdots
$$
Using (\ref{EqHarmonic}) and our underestimates, we see that
\begin{eqnarray*}
\lefteqn{1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4}+ \frac{1}{5}+
\frac{1}{6}+ \frac{1}{7}+ \frac{1}{8} + \frac{1}{8} + \frac{1}{9} + \cdots
+ \frac{1}{16}+ \cdots} \\
& = & 1 + \frac{1}{2} + (\frac{1}{3} + \frac{1}{4})+
(\frac{1}{5}+ \frac{1}{6}+ \frac{1}{7}+ \frac{1}{8}) + (\frac{1}{9} + \cdots
+ \frac{1}{16}) + \cdots \\
& > & 1 + \frac{1}{2} + \frac{1}{2} +\frac{1}{2}
+\frac{1}{2} +
\cdots .
\end{eqnarray*}
Thus the partial sums of the harmonic series grow without
bound
which is expressed by
$$
1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4}+ \frac{1}{5}+
\cdots =
\infty.
$$
Below is a formal proof of the the fact that the sums of
terms in
parenthesized groupings lie between
$\displaystyle{\frac{1}{2}}$
and $1$. You should {\em not} include the proof in your
essay; the
mathematics of your essay is to be treated informally.
Observe
that the terms of a parenthesized group are given by
$\displaystyle{\frac{1}{2^n+1}, \ldots, \frac{1}{2^{n+1}}}$
for
some $n \geq 1$.
\begin{Lemma}\label{MemGroupSums}
Let $n$ be a positive integer. Then $\displaystyle{
\frac{1}{2} <
\frac{1}{2^n + 1} + \cdots + \frac{1}{2^{n+1}} < 1. }$
\end{Lemma}
\pf Since $2^{n+1} = 2^n + 2^n$ the sum in the statement of
the
lemma has $2^n$ terms. Each term has the form
$\displaystyle{\frac{1}{2^n + \ell}}$ for some $1 \leq \ell
\leq
2^n$ and thus
satisfies
$$
\frac{1}{2^{n+1}} \leq \frac{1}{2^n + \ell} \leq
\frac{1}{2^n}.
$$
At least one of the terms is larger than
$\displaystyle{\frac{1}{2}}$ and one at least one is smaller
then
$1$. Therefore
$$
\frac{1}{2} = 2^n(\frac{1}{2^{n+1}}) < \frac{1}{2^n + 1} + \cdots
+ \frac{1}{2^{n+1}} < 2^n(\frac{1}{2^n}) = 1.
$$
\qed \vfill
adapted from DER by JTB 1/24/03
\end{document}