\documentclass[12pt]{article}
\usepackage[utf8]{inputenc}
\usepackage[letterpaper]{geometry}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{mathrsfs}
\usepackage{amsthm}
\usepackage{amssymb}
\usepackage{color}
\setlength{\topmargin}{-0.6 true in}
\setlength{\headsep}{0 true in}
\setlength{\topskip}{0 true in}
\setlength{\textwidth}{6.5 true in}
\setlength{\oddsidemargin}{0 true in}
\setlength{\evensidemargin}{0 true in}
\setlength{\textheight}{9.7 true in}
\renewcommand{\arraystretch}{1.2}
\newcommand{\eps}{\varepsilon}
\newcommand{\R}{{\mathbb R}}
\newcommand{\N}{{\mathbb N}}
\newcommand{\Q}{{\mathbb Q}}
\newcommand{\Z}{{\mathbb Z}}
\theoremstyle{definition}
\newtheorem{problem}{Problem}
\definecolor{darkblue}{RGB}{0,0,150}
\definecolor{darkred}{RGB}{180,0,0}
\definecolor{darkgreen}{RGB}{0,120,0}
\newenvironment{answer}{\par\medskip\bgroup\color{darkblue}}{\egroup\par\medskip}
\newenvironment{grading}{\par\medskip\bgroup\color{darkred}}{\egroup\par\bigskip}
\newenvironment{secondans}{\bgroup\color{darkgreen}}{\egroup}
\pagestyle{empty}
\allowdisplaybreaks


\begin{document}

\hbox to \hsize{\textbf{Math 331\hfill Homework 8}}
\nointerlineskip
\vskip 2pt
\hrule height 1pt

\medskip

\centerline{\textit{This homework is due by 11:59 PM on Tuesday, November 10.}}

\bigskip

\def\ds{\displaystyle}

\def\seq#1{\{#1_n\}_{n=1}^\infty}
\def\slim#1{{\ds\lim_{n\to\infty} #1_n}}

\begin{problem}  % problem 1
Prove that if $\seq x$ is an increasing sequence that is not bounded above,
then $\slim x = +\infty$.
\end{problem}

\begin{answer}
%You can put your answer here for problem 1, 3 points
\end{answer}




\begin{problem}[From Textbook problem 4.2.5]  % problem 2
Let $\seq a$ be defined inductively as follows: 
$$a_1=1,\qquad\qquad a_n = 1 + \frac{a_{n-1}}{4}\mbox{\ \ for }n>1$$
\begin{itemize}
\item[\bf(a)]
   Show by induction that $a_n$ is bounded above by $4/3$.
\item[\bf(b)]
    Show that $\seq a$ is convergent by showing that it is increasing.
\item[\bf(c)]
    Show that $\slim a = 4/3$. [Hint: Use the fact that
    $\slim a = \ds\lim_{n\to\infty}a_{n+1}$ and the recursive
    definition of $a_n$.]
\end{itemize}
\end{problem}

\begin{answer}
%You can put your answer here for problem 2, 3 points
\end{answer}




\begin{problem}[From Textbook problem 4.2.7]  % problem 3
Let $\seq a$ be defined inductively as follows:
$$a_1 = 1,\qquad\qquad a_n= 1 + \frac{1}{1+a_{n-1}}\mbox{\ \ for }n>1$$
\begin{itemize}
\item[(a)]
   Show that $\seq a$ converges by using the contraction principle.
   [Hint: First, show that $a_n\ge1$ for all $n$.]
\item[(b)]
   Show that $\slim a = \sqrt 2$. [Hint: Use the fact that
    $\slim a = \ds\lim_{n\to\infty}a_{n+1}$ and the recursive
    definition of $a_n$.]
\end{itemize}
\end{problem}

\begin{answer}
%You can put your answer here for problem 3, 3 points
\end{answer}




\begin{problem}  % problem 4
Suppose that $\seq a$ and $\seq b$ are sequences, and $\seq a$ is convergent with $\slim a = L$.
Suppose in addition that $\ds\lim_{n\to\infty}|a_n-b_n| = 0$.  Show that $\seq b$ is convergent
and $\slim b = L$.
\end{problem}

\begin{answer}
%You can put your answer here for problem 4, 3 points
\end{answer}




\begin{problem}  % problem 5
Suppose that the function $f:\R\to\R$ satisfies $|f(x)-f(y)|\le r|x-y|$ for all $x,y\in\R$,
where $r$ is a constant in the interval $0\le r<1$.  Such a function is said to be a 
\textbf{contraction} on $\R$.  Note that a contraction is simply a Lipschitz function with
Lipschitz constant strictly less than 1, so we already know that $f$ is continuous.
\begin{itemize}
\item[\bf(a)]
   Let $t$ be any real number.  Define a sequence $\{a_n\}_{n=0}^\infty$ by 
   $a_0 = t$, $a_n = f(a_{n-1})$ for $n>0$.   That is $a_0 = t, a_1 = f(t), a_2=f(f(t)), 
   a_3=f(f(f(t))), \dots, a_n = f^n(t),\dots$, where $f^n$ is the composition of
   $f$ with itself $n$ times.  Show that the sequence $\{a_n\}_{n=0}^\infty$ is contracting,
   and hence is convergent.
\item[\bf(b)]
   Let $z = \slim a$.  Show that $f(z) = z$, that is, $z$ is a fixed point of $f$.
   [Hint: Since $f$ is continuous, $\ds\lim_{n\to\infty}f(a_n) = f\big(\slim a\big)$.]
\end{itemize}
(Note: Recall that a \textbf{fixed point} of a function $f$ is a point $y$ such that $f(y) = y$.
It is clear that a contraction can have at most one fixed point.  This problem shows that
a contraction always does have a fixed point.  Furthermore, if $t$ is any real number, then
the sequence $\{f^n(t)\}_{n=0}^\infty$ converges to that unique fixed point.  This is the
\textbf{Contraction Mapping Theorem} for $\R$.)
\end{problem}

\begin{answer}
%You can put your answer here for problem 5, 3 points
\end{answer}






\end{document}
