%  This is a LaTeX template for MCS 220 mastery problems.
% Tom LoFaro, Gustavus Adolphus College 2009.

% Anything after a percent sign is a comment.

\documentclass[12pt]{article} 
\usepackage{amssymb}

% These commands make the margins narrower
\setlength{\textwidth}{6.5in}
\setlength{\oddsidemargin}{0in}
\setlength{\evensidemargin}{\oddsidemargin}

% Make Theorems and Lemmas share the same numbering {theorem}, but
%  make definitions use their own numbers.
% \newtheorem{theorem}{Theorem}
% \newtheorem{lemma}{Lemma}[theorem]
% \newtheorem{definition}{Definition}

% The mastery problem environment definition

\newenvironment{mastery}[2]{\smallskip\noindent{\bf #1 #2:} \ \em }
{\smallskip}

% The lecture problem environment for the problems I demonstrate in
% class.  The argument is for the date


\newenvironment{lect}[1]{\smallskip\noindent {\bf Lecture #1:} \ \em }{\smallskip}

% This definitions make proofs the way I like them.
\def\QuadSpace{\vspace{0.25\baselineskip}}
\def\HalfSpace{\vspace{0.5\baselineskip}}
\def\FullSpace{\vspace{1.0\baselineskip}}
\def\qed{\enspace\vbox{\hrule\hbox{\vrule\kern4pt 
      \vbox{\kern4pt\kern4pt}\kern4pt\vrule}\hrule}}

\def\EndProof{ \quad \qed \newline }
\newenvironment{proof}{\QuadSpace\par\noindent{\bf Proof}:}{\EndProof\HalfSpace}
%\newenvironment{Proof}{\QuadSpace\par\noindent{\bf Proof}:}{\HalfSpace}

\def\pf{\smallskip\noindent{\sl Proof:}\quad} % the proof statement starting a proof

\title{MCS 220 Lecture Proofs}
\author{Tom LoFaro}

\begin{document}

\maketitle

\begin{lect}{9/15/09}
For all $\epsilon > 0$ there exists a $\delta >0$ such that if $|1-x|<\delta$ then $|2x+3-5|<\epsilon$.  
\end{lect}
\begin{proof} 
Fix $\epsilon > 0$.  Let $\delta \leq \epsilon/2$.  If
\[ 1- \delta < x < 1+\delta\]
then
\[1-\frac{\epsilon}{2} < x < 1+\frac{\epsilon}{2}.\]
If we multiply this inequality by 2 we get
\[2-\epsilon < 2x < 2+\epsilon.\]
Finally, add $3$ to get
\[5-\epsilon < 2x+3 < 5+\epsilon.\]
\end{proof}

\begin{lect}{9/18/09}
  For all $n \in \mathbb{N}, \, 7^n-6n-1$ is divisible by 36.
\end{lect}
\begin{proof}
  We will proceed by induction.  If $n=1$ then $7^n-6n-1=0$ which is
  divisible by 36 since 0 is divisible by all integers.

Assume that $7^n-6n-1$ is divisible by 36.  This implies that there
exists $k \in \mathbb{N}$ such that $7^n-6n-1=36k$.  To prove the
statement for the $n+1$ case we note that
\[ 7^{n+1} -6(n+1)-1 = 7^{n+1} -6n-7 = 7^{n+1} - 42n -7 +36n. \]
Factoring out a 7 from the first three terms we get
\[  7^{n+1} -6(n+1)-1 = 7(7^n-6n-1) + 36n = 7(36k)+36n = 36(7k+n).\]
Since $k$ and $n$ are both integers $7k+n$ is an integer and hence
$7^{n+1} -6n-7$ is divisible by 36 as desired.
\end{proof}

\begin{lect}{9/29/09}
$\displaystyle{\lim_{n \to \infty} {3n+10 \over n+2}=3.}$
\end{lect}
\begin{proof}
  Fix $\epsilon >0$ and let $N> 4/\epsilon - 3.$  If $n>N$ then
  $n>4/\epsilon - 2$ and hence
\[ {4 \over n+2} < \epsilon.\]
This implies that
\[  {4 \over n+2} + {3(n+2) \over n+2} - 3 = {3n+10 \over n+2} - 3<
  \epsilon.\]
Since the left hand side of this inequality is positive for $n>0$ we
get
\[  \left| {3n+10 \over n+2} - 3 \right|< \epsilon\]
as desired.
\end{proof}

\begin{lect}{10/22/09}
  The function $f(x)=2x^2+1$ is continuous.
\end{lect}
\begin{proof}
  Fix $x_0$ and let $(x_n)$ be any sequence such that $(x_n)$
  converges to $x_0$.  Then 
\[ \lim_{n \to \infty} f(x_n) =  \lim_{n \to \infty} 2x_n^2+1 = 
2x_0^2+1 = f(x_0). \]
The second equality follows from the properties of convergent
sequences.  Therefore $f(x)$ is continuous. 
\end{proof}

\begin{lect}{10/23/09}
  The function 
\[ f(x) = 
\left\{
  \begin{array}{rl}
    -1 & \mbox{{\rm if} $x < 0$} \\
    1 & \mbox{{\rm if} $x \geq 0$}
  \end{array}
\right. \]
is discontinuous at $x_0=0$.
\end{lect}
\begin{proof}
  Let $x_n = -1/n$ so that $\lim_{n \to \infty} x_n = 0.$  Then $f(x_n) = -1$ for all $n$.  Thus
  \[ \lim_{n \to \infty} f(x_n) = \lim_{n \to \infty} -1 = -1 \neq f(0). \]
  Therefore $f$ is discontinuous at $0$.
\end{proof}
\begin{lect}{10/23/09}
  The function 
\[ f(x) = 
\left\{
  \begin{array}{rl}
    x \sin \left( {1 \over x} \right) & \mbox{{\rm if} $x \neq 0$} \\
    0 & \mbox{{\rm if} $x = 0$}
  \end{array}
\right. \]
is continuous at $x_0=0$.

\end{lect}
\begin{proof}
Let $(x_n)$ be any sequence such that $(x_n)$ converges to 0.  Since $-1 \leq \sin(x) \leq 1$ for all $x$ it follows that 
\[ -x_n \leq x_n \sin\left({1 \over x_n}\right) \leq x_n. \]
Since both sequences $(x_n)$ and $(-x_n)$ converge to 0
\[ \lim_{n \to \infty} x_n \sin\left({1 \over x_n}\right) = 0 \]
and thus $f$ is continuous at $x=0.$
\end{proof}

\begin{lect}{11/1/09}
  The function $f(x)=x^{-2}$ is {\em not} uniformly continuous on the
  interval $(0,1)$.
\end{lect}

\begin{proof}
  We will show that for every $\delta >0$ there exists $x, \, y \in
  (0,1)$ such that if $|x-y|<\delta$ then $|f(x)-f(y)|>1$.

  Assume that $\delta < 1/2$ and let $x=\delta$ and $y=\delta +
  \delta/2 = 3\delta /2.$  Then
  \[ |f(x)-f(y)| = \frac{1}{\delta^2} - \frac{4}{9\delta^2} =
  \frac{5}{9\delta^2}. \]
  Since we have assumed that $\delta < 1/2$ we have that that
  \[ |f(x)-f(y)| = \frac{5}{9\delta^2} \geq \frac{5}{9({1 \over 2})^2}
  = {20 \over 9} > 1. \]
  Therefore $f(x)$ is not uniformly continous on the interval $(0,1).$
\end{proof}

\begin{lect}{11/9/09}
  If $(f_n)$ converges uniformly to a function $f$ on $S$ and $f_n$ is
  continous on S for all $n$ then $f$ is continuous on $S.$
\end{lect}

\begin{proof}
  Fix $\epsilon >0$ and $x_0 \in S.$  By the triangle inequality
  \[ |f(x)-f(x_0)| \leq |f(x)-f_n(x)|+|f_n(x)-f_n(x_0)| +
  |f_n(x_0)-f(x_0)|. \]
Since $f_n$ converges uniformly to $f$ on $S$ there exists $N>0$ such
that if $n>N$ then $|f(x)-f_n(x)| < \epsilon/3.$  In particular,
$|f(x_0)-f_n(x_0)| < \epsilon/3.$

Fix $n>N$.  Since $f_n$ is continuous on $S$ there exists $\delta >0$
such that if $|x-x_0|<\delta$ then  $|f_n(x)-f_n(x_0)| < \epsilon/3$.

Thus if $|x-x_0|<\delta$ and $n > N$
 \[ |f(x)-f(x_0)| \leq |f(x)-f_n(x)|+|f_n(x)-f_n(x_0)| +
  |f_n(x_0)-f(x_0)| < \epsilon. \]
Therefore $f(x)$ is continuous on $S$.
\end{proof}

\end{document}







