Added all Real Analysis I homeworks from Fall 2018
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Alex Tusa
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\documentclass[12pt,letterpaper]{article}
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\usepackage[utf8]{inputenc}
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\usepackage[english]{babel}
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\usepackage[normalem]{ulem}
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\usepackage{cancel}
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\usepackage{amsthm}
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\usepackage{amsmath}
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\usepackage{amsfonts}
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\usepackage{amssymb}
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\usepackage{graphicx}
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\usepackage{array}
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\usepackage[left=2cm, right=2.5cm, top=2.5cm, bottom=2.5cm]{geometry}
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\usepackage{enumitem}
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\usepackage{mathrsfs}
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\newcommand{\st}{\ \text{s.t.}\ }
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\newcommand{\abs}[1]{\left\lvert #1 \right\rvert}
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\newcommand{\R}{\mathbb{R}}
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\newcommand{\N}{\mathbb{N}}
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\newcommand{\Q}{\mathbb{Q}}
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\newcommand{\C}{\mathbb{C}}
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\newcommand{\Z}{\mathbb{Z}}
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\DeclareMathOperator{\sign}{sgn}
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\newtheoremstyle{case}{}{}{}{}{}{:}{ }{}
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\theoremstyle{case}
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\newtheorem{case}{Case}
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\theoremstyle{definition}
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\newtheorem{definition}{Definition}[section]
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\newtheorem{definition*}{Definition}
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\newtheorem{theorem}{Theorem}[section]
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\newtheorem*{theorem*}{Theorem}
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\newtheorem{corollary}{Corollary}[section]
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\newtheorem*{corollary*}{Corollary}
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\newtheorem{lemma}[theorem]{Lemma}
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\newtheorem*{lemma*}{Lemma}
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\newtheorem*{remark}{Remark}
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\setlist[enumerate]{font=\bfseries}
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\renewcommand{\qedsymbol}{$\blacksquare$}
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\author{Alexander J. Tusa}
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\title{Real Analysis Homework 10}
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\begin{document}
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\maketitle
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\begin{enumerate}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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%%%%%%%% Section 5.4 Questions %%%%%%%%
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\item \textbf{Section 5.4}
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\begin{enumerate}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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%%%%%%%% Section 5.4 #7 %%%%%%%%
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\item[7.] If $f(x):=x$ and $g(x):=\sin x$, show that both $f$ and $g$ are uniformly continuous on $\R$, but that their product $fg$ is not uniformly continuous on $\R$.
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\begin{proof}
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We want to show that $f(x):=x$ is uniformly continuous on $\R$.\\
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In the interest of being explicit, recall the definitions of continuity and uniform continuity, respectively, and note their differences if $f:A \subseteq \R \to \R$:\\\\
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\textit{Continuity:}
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\[\forall\ x \in A\ \forall\ \varepsilon>0\ \exists\ \delta > 0 \st \forall\ y \in A;\ |x-y|<\delta \implies |f(x)-f(y)|<\varepsilon\]
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\textit{Uniform Continuity:}
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\[\forall\ \varepsilon > 0\ \exists\ \delta >0 \st \forall\ x,y \in A;\ |x-y|<\delta \implies |f(x)-f(y)| < \varepsilon\]
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We note that the difference is that for continuity, one takes an arbitrary point $x \in A$, and thus there must exist a distance $\delta$, whereas for uniform continuity, we have that a single $\delta$ must work uniformly for all points $x$ and $y$.\\
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So, $\forall\ x,u \in \R$, we have the following:
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\[|f(x)-f(u)| = |x-u|<\varepsilon\]
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So, let $\delta = \varepsilon$.\\
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$\therefore\ f(x):=x$ is uniformly continuous on $\R$.
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\end{proof}
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\begin{proof}
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Now we want to show that $g(x):= \sin x$ is uniformly continuous.\\
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So, $\forall\ x,y \in \R$:
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\begin{align*}
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|g(x)-g(u)| &= |\sin x - \sin u|\\
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&=\abs{2 \cos \left(\frac{x+u}{2}\right) \sin \left(\frac{x-u}{2}\right)} \\
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&=2 \abs{\cos \left(\frac{x+u}{2}\right) \sin \left(\frac{x-u}{2}\right)} \\
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&\leq 2 \abs{\sin \left(\frac{x-u}{2}\right)} & \because\ |\cos (x)| \leq |x|\ \forall\ x \in \R \\
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&\leq 2 \abs{\frac{x-u}{2}} \\
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&= \frac{2}{2} |x-u| \\
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&= |x-u| \\
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&< \varepsilon
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\end{align*}
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So if we choose $\delta = \varepsilon$, we have that $g(x)$ is uniformly continuous.\\
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\end{proof}
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Now, we want to show that $fg$ is not uniformly continuous on $\R$. To do this, recall the \textit{Nonuniform Continuity Criteria}:
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\begin{theorem*}[\textbf{Nonuniform Continuity Criteria}]
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Let $A \subseteq \R$ and let $f:A \rightarrow \R$. Then the following statements are equivalent:
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\begin{enumerate}
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\item $f$ is not uniformly continuous on $A$.
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\item There exists an $\varepsilon_0 > 0$ such that for every $\delta > 0$ there are points $x_\delta, u_\delta$ in $A$ such that $|x_\delta - u_\delta|<\delta$ and $|f(x_\delta) - f(u_\delta)| \geq \varepsilon_0$.
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\item There exists an $\varepsilon_0 > 0$ and two sequences $(x_n)$ and $(u_n)$ in $A$ such that $\lim (x_n - u_n)=0$ and $|f(x_n)-f(u_n)|\geq \varepsilon_0=1$ for all $n \in \N$.
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\end{enumerate}
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\end{theorem*}
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So, let $(x_n), (u_n) \subseteq \R$ be given by $x_n:=2n\pi$, and $u_n:=2n\pi + \frac{1}{n}$ for $n \in \N$. Then we have that
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\[(x_n-u_n)=2n\pi - (2n\pi + \frac{1}{n})=-\frac{1}{n}\]
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and thus
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\[\lim\limits_{n \to \infty} (x_n-u_n)=\lim\limits_{n \to \infty} -\frac{1}{n} = 0\]
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Now we have the following:
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\begin{align*}
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|(fg)(x_n) - (fg)(u_n)| &= \abs{2n\pi \sin (2n\pi) - \left(2n\pi+\frac{1}{n}\right)\sin \left(2n\pi + \frac{1}{n}\right)} \\
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&=\abs{2n\pi \cdot 0 - \left(2n\pi + \frac{1}{n}\right) \sin \left(2n\pi + \frac{1}{n}\right)} \\
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&=\abs{-\left(2n\pi + \frac{1}{n}\right) \sin \left(2n\pi + \frac{1}{n}\right)} \\
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&= \left(2n\pi + \frac{1}{n}\right) \sin \left(2n\pi + \frac{1}{n}\right) \\
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&= \left(2n\pi + \frac{1}{n}\right) \sin \left(\frac{1}{n}\right) & \because\ \sin(2n\pi) = 0\ \forall\ n \in \N \\
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&= 2n\pi \sin \left(\frac{1}{n}\right) + \left(\frac{1}{n}\right) \sin \left(\frac{1}{n}\right)
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\end{align*}
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Note that $\lim\limits_{n \to \infty} \frac{1}{n}=0$ and $\left(\sin \left(\frac{1}{n}\right)\right)$ is a bounded sequence in $\R$. Then
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\[\lim\limits_{n \to \infty} \frac{1}{n} \sin \frac{1}{n} = 0\]
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and
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\[\lim\limits_{x \to 0} \frac{\sin (x)}{x}=1\ \forall\ x \in \R\]
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This yields
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\[\lim\limits_{n \to \infty} 2n\pi \sin \left(\frac{1}{n}\right) = 2\pi \lim\limits_{\frac{1}{n} \to 0} \frac{\sin \left(\frac{1}{n}\right)}{\frac{1}{n}}=2\pi\]
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Thus $\lim\limits_{n \to \infty} ((fg)(x_n) - (fg)(u_n)) = 2\pi$.\\
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Let $\varepsilon=\pi$. Then $\exists\ k \in \N \st \forall\ n \geq k$:
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\begin{align*}
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2\pi - \varepsilon < (fg)(x_n)-(fg)(u_n) &< 2\pi + \varepsilon \\
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(fg)(x_n)-(fg)(u_n) &>\pi \\
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|(fg)(x_n) - (fg)(u_n)| &> \pi \\
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|(fg)(x_{n+k}) -(fg)(u_{n+k})| &> \pi
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\end{align*}
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Now, for $\varepsilon_0 = \pi$ and two sequences $(x_{n+k}), (u_{n+k})$, by the \textit{Nonuniform Continuity Criteria}, $(fg)$ is not uniformly continuous on $\R$.\\
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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%%%%%%%% Section 5.4 #10 %%%%%%%%
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\item[10.] Prove that if $f$ is uniformly continuous on a bounded subset $A$ of $\R$, then $f$ is bounded on $A$.
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\begin{proof}
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Let $f$ be uniformly continuous on a bounded subset $A$ of $\R$. We want to show that $f$ is bounded on $A$.\\
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Since $f$ is uniformly continuous, we know that $\forall\ \varepsilon > 0,\ \exists\ \delta > 0 \st \forall\ x,y \in A;\ |x-y|<\delta \implies |f(x)-f(y)|<\varepsilon$. We must find a constant $M>0 \st |f(x)|\leq M\ \forall\ x \in A$.\\
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Let $\varepsilon>0$ be given, and let $\delta>0$ also be given.\\
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Recall the definitions of a cover, open cover, subcover, and compactness:
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\theoremstyle{definition}
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\begin{definition*}
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Let $A$ be a subset of a topological space $X$. Let $\mathscr{U}=\{{U_\alpha}\cup_{\alpha \in J} U_\alpha\}$ be a collection of subsets of $X$. We say that $\mathscr{U}$ is a \textbf{cover} of $A$ if $A \subseteq \cup_{\alpha \in J} U_\alpha$. $\mathscr{U}$ is called an \textbf{open cover} of $A$ if $\mathscr{U}$ is a cover of $A$ and each element of $\mathscr{U}$ is open. If $\mathscr{U}$ is a cover of $A$, then any subset of $\mathscr{U}$ that is also a cover of $A$ is called a \textbf{subcover} of $A$. A space $X$ is \textbf{compact} if every open cover of $X$ has a finite subcover.
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\end{definition*}
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Also recall the following theorem:
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\begin{theorem*}
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A compact subset of a metric space is bounded.
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\end{theorem*}
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Since $A$ is a subset of $\R$ under the usual metric, defined $d(x,y):=|x-y|$, we have that $A$ is compact. Thus by the definition of compactness, we know that every open cover of $A$ has a finite subcover. In particular, we know that there are finitely many open balls of radius $\delta$. Thus
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\[\exists n \in \N \st A \subseteq \cup_{i=0}^n B_d(x_i, \delta),\ x_i \in A\]
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Recall the definition of an open ball:
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\theoremstyle{definition}
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\begin{definition*}
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Let $(M,d)$ be a metric space. Given $x \in M$ and a positive real number $\varepsilon$, the \textbf{open ball} centered at $x$ with radius $\varepsilon$ is
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\[B_d(x,\varepsilon) := \{y \in M\ |\ d(x,y)<\varepsilon\}\]
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\end{definition*}
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Now, define $m$ and $M$ as follows:
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\[m:= \min\limits_{1 \leq i \leq n} f(x_i),\ M:=\max\limits_{1 \leq i \leq n} f(x_i)\]
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Let $x \in A$ be arbitrary. Since $A$ is covered with balls $B_d(x_i,\delta)$, we know that $\exists\ x_j \in A \st x \in B_d(x_j, \delta)$.
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So,
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\[x \in B_d(x_j, \delta) \implies |x-x_j|<\delta \implies |f(x)-f(x_j)|<\varepsilon \iff -\varepsilon < f(x)-f(x_j)<\varepsilon\]
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By the definition of $M$, we know that $f(x_j) \leq M$, and thus $f(x)\leq M+\varepsilon$.\\
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By the definition of $m$, we know that $f(x_j) \geq m$, and thus $f(x) \geq m-\varepsilon$.\\
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$\therefore\ \forall\ x \in A,$
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\[f(x) \in [m-\varepsilon, M+\varepsilon]\]
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$\therefore\ f$ is bounded on $A$.\\
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\end{proof}
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\textbf{Alternative Proof:}
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\begin{proof}
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By way of contradiction, assume that $f(A)$ is unbounded. Then we know that there exists a sequence $x_n \in A \st |f(x_n)| \geq n\ \forall\ n$.\\
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Since $(x_n)$ is bounded, we know that there exists a convergent subsequence $x_{n_k} \in A \st x_{n_k} \to x$. Now, since $f$ is uniformly continuous, we know that $f(x_{n_k}) \to f(x)$, which contradicts the fact that $f(A)$ is unbounded. Thus we have that if $f$ is uniformly continuous on $A$, then $f(A)$ is bounded as well.
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\end{proof}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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%%%%%%%% Section 5.4 #14 %%%%%%%%
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\item[14.] A function $f:\R \to \R$ is said to be \textbf{periodic} on $\R$ if there exists a number $p > 0$ such that $f(x+p)=f(x)$ for all $x \in \R$. Prove that a continuous periodic function on $\R$ is bounded and uniformly continuous on $\R$.\\
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\begin{proof}
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Let $f$ be a continuous periodic function on $\R$. Then we know that $\exists\ p>0 \st f(x+p)=f(x)\ \forall\ x \in \R$.
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Recall \textit{Theorem 5.3.9}:
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\begin{theorem*}
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Let $I$ be a closed bounded interval and let $f:I \rightarrow \R$ be continuous on $I$. Then the set $f(I):= \{f(x): x \in I\}$ is a closed bounded interval.
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\end{theorem*}
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Also recall the \textit{Uniform Continuity Theorem}:
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\begin{theorem*}[\textbf{Uniform Continuity Theorem}]
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Let $I$ be a closed bounded interval and let $f:I \rightarrow \R$ be continuous on $I$. Then $f$ is uniformly continuous on $I$.
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\end{theorem*}
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We must first show that $f$ is bounded. Let $f:[0,p] \to \R$ be given by $f([0,p]):= f([np,(n+1)p])\ \forall\ n \in \Z$.\\
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Since $[0,p]$ is a closed bounded interval, we know that by \textit{Theorem 5.3.9}, $f([0,p])$ is also a closed bounded interval. Thus we have that $f$ is bounded.\\
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Now we must show that $f$ is uniformly continuous on $\R$. So, by the \textit{Uniform Continuity Theorem}, since $f$ is continuous on $[0,p]$ and since $[0,p]$ is a closed bounded interval, we have that $f$ is uniformly continuous on $[0,p]$. Since $f([0,p])=f([np,(n+1)p])$, we have that $f$ is uniformly continuous on $[np, (n+1)p]$ and thus we have that $f$ is uniformly continuous on $\R$.\\
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\end{proof}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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%%%%%%%% Section 5.4 #15 %%%%%%%%
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\item[15.] Let $f$ and $g$ be Lipschitz functions on $A$.
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\begin{enumerate}
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\item[(a)] Show that the sum $f+g$ is also a Lipschitz function on $A$.
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\begin{proof}
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We want to show that $f+g$ is also a Lipschitz function on $A$.\\
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Recall the definition of a Lipschitz function:
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\theoremstyle{definition}
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\begin{definition*}
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Let $A \subseteq \R$ and let $f:A \rightarrow \R$. If there exists a constant $K > 0$ such that
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\[|f(x)-f(u)| \leq K|x-u|\]
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for all $x,u \in A$, then $f$ is said to be a \textbf{Lipschitz function} (or to satisfy a \textbf{Lipschitz condition}) on $A$.
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\end{definition*}
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Since $f$ and $g$ are Lipschitz functions on $A$, we know that $\forall\ x,y \in A$:
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\[\exists\ M_1 > 0 \st |f(x)-f(y)|<M_1|x-y|\]
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\[\exists\ M_2 > 0 \st |g(x)-g(y)|<M_2|x-y|\]
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In order to show that $f+g$ is also a Lipschitz function, we have the following:
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\begin{align*}
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|(f+g)(x)-(f+g)(y)| &= |f(x)+g(x)-(f(y)+g(y))| \\
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&=|f(x)+g(x)-f(y)-g(y)| \\
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&=|f(x)-f(y)+g(x)-g(y)| \\
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&\leq |f(x)-f(y)| + |g(x)-g(y)| \\
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&\leq M_1|x-y| + M_2|x-y| \\
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&= (M_1+M_2)|x-y|
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\end{align*}
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Thus, we have that $|(f+g)(x)-(f+g)(y)| \leq (M_1+M_2)|x-y|$. There by the definition of a Lipschitz function, we have that if $f$ and $g$ are Lipschitz, then $f+g$ is also Lipschitz.\\
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\end{proof}
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\item[(b)] Show that if $f$ and $g$ are bounded on $A$, then the product $fg$ is a Lipschitz function on $A$.
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\begin{proof}
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Let $f$ and $g$ be bounded on $A$. We want to show that the product $fg$ is a Lipschitz function on $A$.\\
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Recall the definition of a bounded function:
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\theoremstyle{definition}
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\begin{definition*}
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A function $f:A \rightarrow \R$ is said to be \textbf{bounded on} $A$ if there exists a constant $M > 0$ such that $|f(x)| \leq M$ for all $x \in A$.
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\end{definition*}
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So since $f$ and $g$ are bounded on $A$, we know
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\[\exists\ M_1 > 0 \st |f(x)| \leq M_1\ \forall\ x \in A\]
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\[\exists\ M_2 > 0 \st |g(x)| \leq M_2\ \forall\ x \in A\]
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And since $f(x)$ and $g(x)$ are Lipschitz, we know
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\[\exists\ K_1 \st |f(x)-f(u)|\leq K_1|x-u|\]
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\[\exists\ K_2 \st |g(x)-g(u) \leq K_2|x-u|\]
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We must show that $fg$ is Lipschitz. Let $x,y \in A$. Then
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\begin{align*}
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|(fg)(x)-(fg)(y)| &= |f(x)g(x)-f(y)g(y)| \\
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&= |f(x)g(x)-f(x)g(y)+f(x)g(y)-f(y)g(y)| \\
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&= |f(x)(g(x)-g(y))+ g(y)(f(x)-f(y))| \\
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& \leq |f(x)(g(x)-g(y))| + |g(y)(f(x)-f(y))| \\
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& = |f(x)| \cdot |g(x)-g(y)| + |g(y)| \cdot |f(x)-f(y)| \\
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&< M_1 \cdot |g(x)-g(y)| + M_2 \cdot |f(x)-f(y)| \\
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&< M_1K_1|x-y| + M_2K_2|x-y| \\
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&= (M_1K_2+M_2K_1)|x-y|
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\end{align*}
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Thus we have that $|(fg)(x)-(fg)(y)|\leq (M_1K_1+M_2K_2)|x-y|\ \forall\ x,y \in A$. Therefore by the definition of a Lipschitz function, we have that $fg$ is Lipschitz.\\
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\end{proof}
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\item[(c)] Give an example of a Lipschitz function $f$ on $[0, \infty)$ such that its square $f^2$ is \textit{not} a Lipschitz function.\\
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Consider the function $f(x):=x$, where $x \geq 0$. Then we have that $\forall\ x,y \geq 0$:
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\[|f(x)-f(y)|=|x-y| \leq 2|x-y|\]
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Thus we have that $f(x)$ is Lipschitz on the interval $[0, \infty)$. However, note that $f^2$ is not Lipschitz since $f^2$ is unbounded. Thus $f^2$ cannot be a Lipschitz function.
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\end{enumerate}
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\end{enumerate}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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%%%%%%%% Section 6.1 Questions %%%%%%%%
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\item \textbf{Section 6.1}
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\begin{enumerate}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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%%%%%%%% Section 6.1 #1ac %%%%%%%%
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\item[1.] Use the definition to find the derivative of each of the following functions:
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\begin{enumerate}
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\item[(a)] $f(x):=x^3$ for $x \in \R$.\\
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Recall the definition of the derivative:
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\theoremstyle{definition}
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\begin{definition*}
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Let $I \subseteq \R$ be an interval, let $f:I \rightarrow \R$, and let $ c \in I$. We say that a real number $L$ is the \textbf{derivative of $f$ at $c$} if given any $\varepsilon > 0$ there exists $\delta (\varepsilon) > 0$ such that if $x \in I$ satisfies $0 < |x-c|<\delta (\varepsilon)$, then
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\[\abs{\frac{f(x)-f(c)}{x-c}-L}<\varepsilon.\]
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In this case we say that $f$ is \textbf{differentiable} at $c$, and we write $f'(c)$ for $L$. In other words, the derivative of $f$ at $c$ is given by the limit
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\[f'(c) = \lim_{x\to c} \frac{f(x)-f(c)}{x-c}\]
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provided this limit exists. (We allow the possibility that $c$ may be the endpoint of the interval.)
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\end{definition*}
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Let $h:=x-c$. Then $x=c+h$:
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\begin{align*}
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\lim\limits_{h \to 0} \frac{f(x+h)-f(x)}{h} &= \lim\limits_{h \to 0} \frac{(x+h)^3-x^3}{h} \\
|
||||
&= \lim\limits_{h \to 0} \frac{h^3+3h^2x+3hx^2}{h} \\
|
||||
&=\lim\limits_{h \to 0} \frac{h(h^2+3hx+3x^2)}{h} \\
|
||||
&= \lim\limits_{h \to 0} (h^2+3hx+3x^2) \\
|
||||
&= 0^2 + 3 \cdot 0 \cdot x + 3x^2 \\
|
||||
&= 0+0+3x^2 \\
|
||||
&= 3x^2
|
||||
\end{align*}
|
||||
Thus $f'(x) = 3x^2$.\\
|
||||
\item[(c)] $h(x):=\sqrt{x}$ for $x>0$.\\
|
||||
|
||||
\begin{align*}
|
||||
\lim\limits_{h \to 0} \frac{f(x+h)-f(x)}{h} &= \lim\limits_{h \to 0} \frac{\sqrt{x+h}-\sqrt{x}}{h} \\
|
||||
&= \lim\limits_{h \to 0} \frac{\sqrt{x+h}-\sqrt{x}}{h} \cdot \frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}} \\
|
||||
&= \lim\limits_{h \to 0} \frac{x+h-x}{h(\sqrt{x+h}+\sqrt{x})} \\
|
||||
&= \lim\limits_{h \to 0} \frac{h}{h(\sqrt{x+h}+\sqrt{x})} \\
|
||||
&= \lim\limits_{h \to 0} \frac{1}{\sqrt{x+h}+\sqrt{x}} \\
|
||||
&= \frac{1}{\sqrt{x+0}+\sqrt{x}} \\
|
||||
&= \frac{1}{\sqrt{x}+\sqrt{x}} \\
|
||||
&= \frac{1}{2\sqrt{x}}
|
||||
\end{align*}
|
||||
Thus $h'(x)=\frac{1}{2\sqrt{x}}$.\\
|
||||
\end{enumerate}
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Section 6.1 #2 %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item[2.] Show that $f(x):=x^{1/3},\ x \in \R$ is not differentiable at $x=0$.\\
|
||||
|
||||
\begin{proof}
|
||||
We must show that $f(x)$ is not differentiable at $x=0$. By the definition of the derivative, the function $f(x)$ is differentiable at $x=0$ given that the limit exists. So, let's find the derivative at $x=0$:
|
||||
\begin{align*}
|
||||
\lim\limits_{x \to c} \frac{f(x)-f(c)}{x-c} &= \lim\limits_{x \to 0} \frac{f(x)-f(0)}{x-0} \\
|
||||
&= \lim\limits_{x \to 0} \frac{x^{\frac{1}{3}}-0^{\frac{1}{3}}}{x} \\
|
||||
&= \lim\limits_{x \to 0} \frac{x^{\frac{1}{3}}}{x} \\
|
||||
&= \lim\limits_{x \to 0} x^{\frac{1}{3}-1} \\
|
||||
&= \lim\limits_{x \to 0} x^{-\frac{2}{3}} \\
|
||||
&= \lim\limits_{x \to 0} \frac{1}{x^{\frac{2}{3}}} \\
|
||||
&= \frac{1}{0^{\frac{2}{3}}} \\
|
||||
&= \frac{1}{0} \\
|
||||
&= \text{undefined}
|
||||
\end{align*}
|
||||
Since the limit is undefined when $x=0$, we have that the limit does not exist at $x=0$, and thus $f(x)$ is not differentiable at $x=0$.\\
|
||||
\end{proof}
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Section 6.1 #8a %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item[8. (a)] Determine where $f:\R \to \R$ given by $f(x):=|x|+|x+1|$ is differentiable and find the derivative.\\
|
||||
|
||||
We can redefine $f(x)$ as a piecewise function:
|
||||
\[f(x):=\begin{cases}
|
||||
2x+1, & x \geq 0 \\
|
||||
1, & -1 \leq x < 0 \\
|
||||
-2x - 1, & x < -1
|
||||
\end{cases}\]
|
||||
|
||||
Thus to find where $f(x)$ is differentiable, we will find the derivatives of each of the three functions defined above in the piecewise definition of $f(x)$.\\
|
||||
|
||||
For $x \geq 0$, we have
|
||||
\begin{align*}
|
||||
\lim\limits_{h \to 0} \frac{f(x+h)-f(x)}{h} &= \lim\limits_{h \to 0} \frac{2(x+h)+1-(2x+1)}{h} \\
|
||||
&=\lim\limits_{h \to 0} \frac{2x+2h-2x}{h} \\
|
||||
&= \lim\limits_{h \to 0} \frac{2h}{h} \\
|
||||
&= \lim\limits_{h \to 0} 2 \\
|
||||
&= 2
|
||||
\end{align*}
|
||||
Thus $f'(x)=2$ when $x \geq 0$.\\
|
||||
|
||||
For $-1 \leq x < 0$:
|
||||
\begin{align*}
|
||||
\lim\limits_{x \to c} \frac{f(x)-f(c)}{x-c} &= \lim\limits_{x \to c} \frac{1-1}{x-c} \\
|
||||
&= \lim\limits_{x \to c} \frac{0}{x-c} \\
|
||||
&= \lim\limits_{x \to c} 0 \\
|
||||
&= 0
|
||||
\end{align*}
|
||||
Thus $f'(x)=0$ when $-1 \leq x < 0$.\\
|
||||
|
||||
For $x < -1$ we have:
|
||||
\begin{align*}
|
||||
\lim\limits_{h \to 0} \frac{f(x+h)-f(x)}{h} &= \lim\limits_{h \to 0} \frac{(-2(x+h)-1)-(-2x-1)}{h} \\
|
||||
&= \lim\limits_{h \to 0} \frac{-2x-2h-1+2x+1}{h} \\
|
||||
&= \lim\limits_{h \to 0} \frac{-2h}{h} \\
|
||||
&= \lim\limits_{h \to 0} -2 \\
|
||||
&= -2
|
||||
\end{align*}
|
||||
Thus $f'(x)=-2$ when $x < -1$.\\
|
||||
|
||||
Now, we must check for differentiability when $x=-1$ and when $x=0$.\\
|
||||
|
||||
So when $x=-1$:
|
||||
\begin{align*}
|
||||
\lim\limits_{x \to c} \frac{f(x)-f(c)}{x-c} &= \lim\limits_{x \to -1^+} \frac{f(x)-f(-1)}{x+1} \\
|
||||
&= \lim\limits_{x \to -1^+} \frac{1-1}{x+1} \\
|
||||
&= \lim\limits_{x \to -1^+} \frac{0}{x+1} \\
|
||||
&= \lim\limits_{x \to -1^+} 0 \\
|
||||
&= 0
|
||||
\end{align*}
|
||||
And
|
||||
\begin{align*}
|
||||
\lim\limits_{x \to c} \frac{f(x)-f(c)}{x-c} &= \lim\limits_{x \to -1^-} \frac{f(x)-f(-1)}{x+1} \\
|
||||
&= \lim\limits_{x \to -1^-} \frac{(-2x-1)-(2 \cdot (-1)-1)}{x+1} \\
|
||||
&= \lim\limits_{x \to -1^-} \frac{-2x-1+3}{x+1} \\
|
||||
&= \lim\limits_{x \to -1^-} \frac{-2x+2}{x+1} \\
|
||||
&= -2
|
||||
\end{align*}
|
||||
So we have that $f'(-1)$ does not exist since
|
||||
\[\lim\limits_{x \to -1^-} \frac{f(x)-f(-1)}{x+1}=0 \neq -2 = \lim\limits_{x \to -1^+} \frac{f(x)-f(-1)}{x+1}\]
|
||||
|
||||
Now for when $x=0$:
|
||||
\begin{align*}
|
||||
\lim\limits_{x \to c} \frac{f(x)-f(c)}{x-c} &= \lim\limits_{x \to 0^+} \frac{f(x)-f(0)}{x-0} \\
|
||||
&= \lim\limits_{x \to 0^+} \frac{(2x+1)-(2 \cdot 0 +1)}{x} \\
|
||||
&= \lim\limits_{x \to 0^+} \frac{2x}{x} \\
|
||||
&= \lim\limits_{x \to 0^+} 2 \\
|
||||
&= 2
|
||||
\end{align*}
|
||||
And
|
||||
\begin{align*}
|
||||
\lim\limits_{x \to c} \frac{f(x)-f(c)}{x-c} &= \lim\limits_{x \to 0^-} \frac{f(x)-f(0)}{x-0} \\
|
||||
&= \lim\limits_{x \to 0^-} \frac{1-1}{x} \\
|
||||
&= \lim\limits_{x \to 0^-} \frac{0}{x} \\
|
||||
&= \lim\limits_{x \to 0^-} 0 \\
|
||||
&= 0
|
||||
\end{align*}
|
||||
So we have that $f'(0)$ does not exist since
|
||||
\[\lim\limits_{x \to 0^-} \frac{f(x)-f(0)}{x-0} = 0 \neq -2 = \lim\limits_{x \to 0^+} \frac{f(x)-f(0)}{x}\]
|
||||
Thus we have that the function is not differentiable at $x=0$ or at $x=-1$; That is,
|
||||
\[f'(x):=\begin{cases}
|
||||
2, & x < 0 \\
|
||||
0, & -1 < x < 0 \\
|
||||
-2, & x < -1 \\
|
||||
\text{DNE} & x=1 \text{ or } x=0
|
||||
\end{cases}\]
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Section 6.1 #9 %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item[9.] Prove that if $f:\R \to \R$ is an \textbf{even function} [that is, $f(-x)=f(x)$ for all $x \in \R$] and has a derivative at every point, then the derivative $f'$ is an \textbf{odd function} [that is, $f'(-x)=-f'(x)$ for all $x \in \R$]. Also prove that if $g:\R \to \R$ is a differentiable odd function, then $g'$ is an even function.\\
|
||||
|
||||
\begin{proof}
|
||||
Let $f:\R \to \R$ be defined such that $f(-x)=f(x)\ \forall\ x \in \R$. Then we have
|
||||
\[f'(c) = \lim\limits_{x \to c} \frac{f(x)-f(c)}{x-c}\]
|
||||
\[f'(-c) = \lim\limits_{x \to -c} \frac{f(x)-f(-c)}{x-(-c)}\]
|
||||
Now if we change every $x$ for $-x$, then if $-x \to -c$ then $x \to c$. So
|
||||
\begin{align*}
|
||||
&= \lim\limits_{x \to c} \frac{f(-x)-f(c)}{-x+c} \\
|
||||
&= \lim\limits_{x \to c} \frac{f(x)-f(c)}{-(x-c)} &\because f(-x)=f(x)\ \forall\ x \in \R \\
|
||||
&= -\lim\limits_{x \to c} \frac{f(x)-f(c)}{x-c} \\
|
||||
&= -f'(c)
|
||||
\end{align*}
|
||||
Thus we have that $f'(-c)=-f'(c)$, an odd function.\\
|
||||
|
||||
$\therefore$ If $f$ is an even function, then $f'$ is an odd function.
|
||||
\end{proof}
|
||||
|
||||
\begin{proof}
|
||||
Let $g: \R \to \R$ be defined such that $g(-x)=-g(x)\ \forall\ x \in \R$. Then
|
||||
\[g'(c) = \lim\limits_{x \to c} \frac{g(x)-g(c)}{x-c}\]
|
||||
And
|
||||
\[g'(-c) = \lim\limits_{x \to -c} \frac{g(x)-g(-c)}{x-(-c)}\]
|
||||
Let us change every $x$ for $-x$ if $-x \to -c$ then $x \to c$. So
|
||||
\begin{align*}
|
||||
&=\lim\limits_{x \to c} \frac{g(-x) - (-g(c))}{-x+c} \\
|
||||
&=\lim\limits_{x \to c} \frac{-g(x)+g(c)}{-x+c} &\because g(-x)=-g(x)\ \forall\ x \in \R \\
|
||||
&=\lim\limits_{x \to c} \frac{-(g(x)-g(c))}{-(x-c)} \\
|
||||
&= \lim\limits_{x \to c} \frac{g(x)-g(c)}{x-c} \\
|
||||
&= g'(c)
|
||||
\end{align*}
|
||||
Thus we have that $g'(-c)=g'(c)$, an even function.\\
|
||||
|
||||
$\therefore$ If $g$ is an odd function, then $g'$ is an even function.
|
||||
\end{proof}
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Section 6.1 #11ac %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item[11.] Assume that there exists a function $L:(0,\infty) \to \R$ such that $L'(x)=1/x$ for $x>0$. Calculate the derivatives of the following functions:
|
||||
\begin{enumerate}
|
||||
\item[(a)] $f(x):=L(2x+3)$ for $x>0$\\
|
||||
|
||||
Recall the \textit{Chain Rule}:
|
||||
\begin{theorem*}[\textbf{Chain Rule}]
|
||||
Let $I, J$ be intervals in $\R$, let $g:I \rightarrow \R$ and $f:J \rightarrow \R$ be functions such that $f(J) \subseteq I$, and let $c \in J$. If $f$ is differentiable at $c$ and if $g$ is differentiable at $f(c)$, then the composite function $g \circ f$ is differentiable at $c$ and
|
||||
\[(g \circ f)'(c) = g'(f(c)) \cdot f'(c).\]
|
||||
\end{theorem*}
|
||||
So utilizing the \textit{Chain Rule}:
|
||||
\begin{align*}
|
||||
f'(x) &= (L(2x+3))' \\
|
||||
&= \frac{1}{2x+3} \cdot (2x+3)' \\
|
||||
&= \frac{2}{2x+3}
|
||||
\end{align*}
|
||||
So $f'(x) = \frac{2}{2x+3}$.\\
|
||||
|
||||
\item[(c)] $h(x):=L(ax)$ for $a >0,x>0$\\
|
||||
|
||||
Once again utilizing the \textit{Chain Rule}:
|
||||
\begin{align*}
|
||||
h'(x) &= (L(ax))' \\
|
||||
&= \frac{1}{ax} \cdot (ax)' \\
|
||||
&= \frac{1}{\cancel{a}x} \cdot \cancel{a} \\
|
||||
&= \frac{1}{x}
|
||||
\end{align*}
|
||||
So $h'(x)=\frac{1}{x}$.\\
|
||||
\end{enumerate}
|
||||
\end{enumerate}
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 3 %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item
|
||||
\begin{enumerate}
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 3 (a) %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item Show that $e^x=2\cos x+1$ for some $x \in [0, \pi]$\\
|
||||
|
||||
\begin{proof}
|
||||
Let $h(x):=e^x-2\cos(x)-1$. Then we have that $h(0)=2<0$ and $h(\pi)=e^\pi+1>0$. Since $h$ is continuous, by the \textit{Intermediate Value Theorem}, we know that there exists $c \in (0,\pi) \st h(c)=0$. Thus we have that $e^c=2\cos(c)+1$.\\
|
||||
\end{proof}
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 3 (b) %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item Let $h(x)=x^3+2x+1$. Compute $h(1),\ h'(1)$ and $[h^{-1}]'(1)$.\\
|
||||
|
||||
\[h(1) = 1^3+2(1)+1 = 1+2+1 = 4\]
|
||||
So $h(1)=4$.\\
|
||||
|
||||
Now, as for $h'(1)$:
|
||||
\begin{align*}
|
||||
h'(1) &= \lim\limits_{x \to 1} \frac{f(x)-f(1)}{x-1} \\
|
||||
&= \lim\limits_{x \to 1} \frac{x^3+2x+1-4}{x-1} \\
|
||||
&= \lim\limits_{x \to 1} \frac{x^3+2x-3}{x-1} \\
|
||||
&= \lim\limits_{x \to 1} \frac{\cancel{(x-1)}(x^2+x+3)}{\cancel{x-1}} \\
|
||||
&= \lim\limits_{x \to 1} (x^2+x+3) \\
|
||||
&= 1^2+1+3 \\
|
||||
&= 1 +1 +3 \\
|
||||
&= 5
|
||||
\end{align*}
|
||||
So $h'(1)=5$.\\
|
||||
|
||||
As for $[h^{-1}]'(1)$:\\
|
||||
|
||||
We first need to find $h^{-1}$. Since $h(1)=4$, we must solve $x^3+2x+1=1$. So
|
||||
\begin{align*}
|
||||
x^3+2x+1 &= 1 \\
|
||||
x^3+2x &= 0 \\
|
||||
x(x^2+1) &= 0 \\
|
||||
x(x+1)(x-1) &= 0
|
||||
\end{align*}
|
||||
Thus $x=0$.\\
|
||||
|
||||
Since $h'(1)$ exists and since $h'(1) \neq 0$, we have that
|
||||
\[[h^{-1}]'(1) = \frac{1}{h'(1)} = \frac{1}{5}\]
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 3 (c) %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item Show $f(x)=x^2$ for $x \in (-2,1]$ is a Lipschitz function.\\
|
||||
\begin{proof}
|
||||
To show that $f(x)$ is Lipschitz, we have the following:
|
||||
\begin{align*}
|
||||
\abs{\frac{f(x)-f(u)}{x-u}} &= \abs{\frac{x^2-u^2}{x-u}} \\
|
||||
&= \abs{\frac{\cancel{(x-u)}(x+u)}{\cancel{x-u}}} \\
|
||||
&= |(x+u)| \\
|
||||
&\leq 8
|
||||
\end{align*}
|
||||
\end{proof}
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 3 (d) %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item Suppose that $f:[a,b] \to \R$ and $g:[a,b] \to \R$ are continuous with $f(a) \leq g(a)$ and $f(b) \geq g(b)$. Prove that $f(c)=g(c)$ for some $c \in [a,b]$.\\
|
||||
|
||||
\begin{proof}
|
||||
Let $f:[a,b] \to \R$ and $g:[a,b] \to \R$ be continuous functions such that $f(a) \leq g(a)$, and $f(b) \geq g(b)$. We want to show that $f(c)=g(c)$ for some $c \in [a,b]$.\\
|
||||
|
||||
Let $h(x):=f(x)-g(x)$. Since both $f$ and $g$ are continuous, we know that $h$ is also continuous. And since $h$ is continuous, we know that $h(a)<0$, and $h(b)>0$.\\
|
||||
|
||||
Recall \textit{Bolzano's Intermediate Value Theorem}:
|
||||
\begin{theorem*}[\textbf{Bolzano's Intermediate Value Theorem}]
|
||||
Let $I$ be an interval and let $f:I \rightarrow \R$ be continuous on $I$. If $a,b \in I$ and if $k \in \R$ satisfies $f(a) < k<f(b)$, then there exists a point $c \in I$ between $a$ and $b$ such that $f(c) = k$.
|
||||
\end{theorem*}
|
||||
|
||||
Thus by \textit{Bolzano's Intermediate Value Theorem}, we have that $\exists\ c \in (a,b) \st h(c)=0$. Thus we have that $h(c)=0=f(c)-g(c) \implies g(c)=f(c)$. Thus $g(c)=f(c)$.\\
|
||||
\end{proof}
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 3 (e) %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item Prove: If $f$ is uniformly continuous on $(a,b)$, then for any sequence $x_n$ in $(a,b)$ that converges, then the sequence $(f(x_n))$ is Cauchy.\\
|
||||
|
||||
\begin{proof}
|
||||
Let $f$ be uniformly continuous on $(a,b)$, and let $(x_n)$ be a sequence in $(a,b)$ such that $(x_n)$ converges.\\
|
||||
|
||||
Recall the \textit{Cauchy Convergence Criterion}:
|
||||
\begin{theorem*}
|
||||
A sequence of real numbers is convergent if and only if it is a Cauchy sequence.
|
||||
\end{theorem*}
|
||||
Since we have that $(x_n)$ is a convergent sequence, we know that $(x_n)$ is a Cauchy sequence by the \textit{Cauchy Convergence Criterion}.\\
|
||||
|
||||
Also recall \textit{Theorem 5.4.7}:
|
||||
\begin{theorem*}
|
||||
If $f:A \rightarrow \R$ is uniformly continuous on a subset $A$ of $\R$ and if $(x_n)$ is a Cauchy sequence in $A$, then $(f(x_n))$ is a Cauchy sequence in $\R$.
|
||||
\end{theorem*}
|
||||
Since $(x_n)$ is a sequence in $(a,b)$ and since $f$ is uniformly continuous on $(a,b)$, we have that by \textit{Theorem 5.4.7}, the sequence $(f(x_n))$ is Cauchy.\\
|
||||
\end{proof}
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 3 (f) %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item Prove: If $f:[a,b] \to [a,b]$ is a contraction, then $f$ has a unique fixed point $c$ satisfying $f(c)=c$ for some $c \in [a,b]$.\\
|
||||
|
||||
\begin{proof}
|
||||
Since $f$ is a contraction, we know that $f$ is Lipschitz with $k \in (0,1)$. This in turn implies that $f$ is continuous.\\
|
||||
|
||||
By the \textit{Brower Fixed Point Theorem}, there exists $c \st f(c)=c$. We want to show the uniqueness of this point.\\
|
||||
|
||||
Assume that there exists $c_1,c_2 \st f(c_1)=c_1$ and $f(c_2)=c_2$, and $c_1 \neq c_2$.\\
|
||||
|
||||
Assume without loss of generality that $c_1<c_2$.\\
|
||||
|
||||
Consider the interval $[c_1,c_2]\leq[a,b]$. Since $f$ is a contraction, we have that
|
||||
\[|f(c_1)-f(c_2)|\leq K|c_1-c_2|<|c_1-c_2|\]
|
||||
Thus we have that $|c_1-c_2|\leq|c_1-c_2|$, which is a contradiction. Thus if $f$ is a contraction, then $f$ has a unique fixed point $c$ satisfying $f(c)=c$.\\
|
||||
\end{proof}
|
||||
\end{enumerate}
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 4 %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item Prove or justify, if true; Provide a counterexample, if false. For all parts, assume $f$ is a function defined on the given interval or set.
|
||||
\begin{enumerate}
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 4 (a) %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item If $f$ is bounded and continuous on $A$, then $f$ is uniformly continuous on $A$.\\
|
||||
|
||||
This is a false statement. Consider $f:(0,2) \to \R$ given by $f(x):=\frac{1}{x}$. It is clear that $f$ is continuous in $(0,2)$ as it is the quotient of two polynomials and the denominator is never equal to 0. Let $x,u \in (0,2)$. First, suppose $x > \frac{u}{2}$. So,
|
||||
\begin{align*}
|
||||
|f(x)-f(u)| &= \abs{\frac{1}{x}-\frac{1}{u}} \\
|
||||
&= \abs{\frac{x-u}{xu}} \\
|
||||
&= \frac{|x-u|}{xu} &\because\ x > 0,\ u>0,\ \forall\ x,u \in (0,2) \\
|
||||
&< \frac{2|x-u|}{u \cdot u} &\because\ x > \frac{u}{2} \implies \frac{1}{x} < \frac{2}{u} \\
|
||||
&= \frac{2|x-u|}{u^2} \\
|
||||
&= \frac{2}{u^2}|x-u| \\
|
||||
&< \varepsilon
|
||||
\end{align*}
|
||||
So $|x-u| < \frac{u^2\varepsilon}{2}$. Thus, let $\delta=\min\{\frac{u^2\varepsilon}{2},\frac{u}{2}\}$. Hence $f(x)$ is continuous on $(0,2)$.\\
|
||||
|
||||
Recall that in order for $f(x)$ to be considered uniformly continuous, $\forall\ \delta>0$ must always satisfy $|x-u|<\delta\ \forall\ \varepsilon > 0$. However, if we let $\varepsilon=1$, then we have that $\forall\ \delta>0$:
|
||||
\[x:=\min\{\delta, 1\},\ u:=\frac{x}{2} \implies |x-u|=\frac{x}{2}<\delta\]
|
||||
but
|
||||
\[\abs{\frac{1}{x}-\frac{1}{u}} = \abs{\frac{1}{x}-\frac{2}{x}}=\abs{\frac{1}{x}} \geq 1 = \varepsilon\]
|
||||
Thus $f(x)$ is both bounded and continuous on $A:=(0,2)$, but $f(x)$ is not uniformly continuous.\\
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 4 (b) %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item If $f$ is uniformly continuous on $A$, then $f$ is bounded on $A$.\\
|
||||
|
||||
This is a false statement. Consider $f:\R \to \R$ given by $f(x):=3x+7$. Then if we let $x,u \in \R$:
|
||||
\begin{align*}
|
||||
|3x+7-(3u+7)| &= |3x+7-3u-7| \\
|
||||
&=|3x-3u| \\
|
||||
&= 3|x-u| \\
|
||||
&< \varepsilon
|
||||
\end{align*}
|
||||
So, let $\delta := \frac{\varepsilon}{3}$. Thus we have that $f(x)$ is uniformly continuous. However, since $\R$ is unbounded, we have that $f$ is not bounded on $A:=\R$ since $\nexists\ M>0 \st |f(x)|\leq M\ \forall\ x \in \R$.\\
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 4 (c) %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item If $f$ is uniformly continuous on $(a,b)$, then $f$ is bounded on $(a,b)$.\\
|
||||
|
||||
This is a true statement. We prove it by way of contradiction:
|
||||
\begin{proof}
|
||||
Assume that $f$ is uniformly continuous on $(a,b)$, and suppose by way of contradiction that $f$ is not bounded on $(a,b)$. Then we have that $\forall\ n \in \N$, there exists a corresponding $f(x_n) \st |f(x_n)|>n$, where $x_n \in (a,b)$. By the \textit{Bolzano-Weirstrass Theorem}, there exists a convergent subsequence $(x_{n_k}) \subseteq (x_n)$.\\
|
||||
|
||||
Recall \textit{Lemma 3.5.2}:
|
||||
\begin{lemma*}
|
||||
If $X=(x_n)$ is a convergent sequence of real numbers, then $X$ is a Cauchy sequence.
|
||||
\end{lemma*}
|
||||
Thus by \textit{Lemma 3.5.2}, $(x_{n_k})$ is a Cauchy sequence.\\
|
||||
|
||||
Since $f$ is uniformly continuous on $(a,b)$, we know that $f(x_{n_k})$ is also Cauchy. However, this is a contradiction since $f(x_{n_k})$ is clearly divergent. Thus, we have that $f$ is bounded on $(a,b)$.\\
|
||||
\end{proof}
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 4 (d) %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item If $f$ is bounded on $A$, then $f$ is uniformly continuous on $A$.\\
|
||||
|
||||
This is a false statement. Consider $f:(0,2) \to \R$ given by $f(x):=\frac{1}{x}$. It is clear that $f$ is continuous in $(0,2)$ as it is the quotient of two polynomials and the denominator is never equal to 0. Let $x,u \in (0,2)$. First, suppose $x > \frac{u}{2}$. So,
|
||||
\begin{align*}
|
||||
|f(x)-f(u)| &= \abs{\frac{1}{x}-\frac{1}{u}} \\
|
||||
&= \abs{\frac{x-u}{xu}} \\
|
||||
&= \frac{|x-u|}{xu} &\because\ x > 0,\ u>0,\ \forall\ x,u \in (0,2) \\
|
||||
&< \frac{2|x-u|}{u \cdot u} &\because\ x > \frac{u}{2} \implies \frac{1}{x} < \frac{2}{u} \\
|
||||
&= \frac{2|x-u|}{u^2} \\
|
||||
&= \frac{2}{u^2}|x-u| \\
|
||||
&< \varepsilon
|
||||
\end{align*}
|
||||
So $|x-u| < \frac{u^2\varepsilon}{2}$. Thus, let $\delta=\min\{\frac{u^2\varepsilon}{2},\frac{u}{2}\}$. Hence $f(x)$ is continuous on $(0,2)$.\\
|
||||
|
||||
Recall that in order for $f(x)$ to be considered uniformly continuous, $\forall\ \delta>0$ must always satisfy $|x-u|<\delta\ \forall\ \varepsilon > 0$. However, if we let $\varepsilon=1$, then we have that $\forall\ \delta>0$:
|
||||
\[x:=\min\{\delta, 1\},\ u:=\frac{x}{2} \implies |x-u|=\frac{x}{2}<\delta\]
|
||||
but
|
||||
\[\abs{\frac{1}{x}-\frac{1}{u}} = \abs{\frac{1}{x}-\frac{2}{x}}=\abs{\frac{1}{x}} \geq 1 = \varepsilon\]
|
||||
Thus $f(x)$ is bounded on $A:=(0,2)$, but $f(x)$ is not uniformly continuous on $A$.\\
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 4 (e) %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item The derivative of $f$ at $x=c$ is defined by $f'(c)=\lim\limits_{x \to c}\frac{f(x)-f(c)}{x-c}$ provided the limits exists.\\
|
||||
|
||||
This is true since this is the definition of the derivative.\\
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 4 (f) %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item If $f$ is continuous at $c$, then $f$ is differentiable at $c$.\\
|
||||
|
||||
This is a false statement. Consider $f:\R \to \R$ given by
|
||||
\[f(x):=\begin{cases}
|
||||
\frac{1}{q} & x=\frac{p}{q},\ x \in \Q,\ p \in \Z,\ q \in \N \st \gcd (p,q)=1 \\
|
||||
0 & x \in \R \setminus \Q
|
||||
\end{cases}\]
|
||||
This is Thomae's function, which we know is continuous in the irrationals, and discontinuous in the rationals. So, we know that $f(c)$ is continuous at $c=\sqrt{2}$, but notice that $f$ is not differentiable at $c=\sqrt{2}$. In fact, $f$ is not differentiable for any values of $c$. We know that it's not differentiable when $x \in \Q$ since $f$ is not continuous for any value $x \in \Q$. As for our particular case where $c=\sqrt{2}$:
|
||||
\begin{align*}
|
||||
\lim\limits_{x \to c} \frac{f(x)-f(c)}{x-c} &= \lim\limits_{x \to \sqrt{2}} \frac{f(x)-f(\sqrt{2})}{x-\sqrt{2}} \\
|
||||
&\Downarrow \\
|
||||
x \in \Q &\implies x=\frac{p}{q},\ p \in \Z,\ q \in \N,\ \gcd (p,q)=1 \\
|
||||
&\Downarrow \\
|
||||
&= \lim\limits_{x \to \sqrt{2}} \frac{\frac{1}{q}-0}{x-0} \\
|
||||
&= \lim\limits_{x \to \sqrt{2}} \frac{\frac{1}{q}}{x} \\
|
||||
&= \lim\limits_{x \to \sqrt{2}} \frac{\frac{1}{q}}{\frac{p}{q}} &\because\ x=\frac{p}{q} \\
|
||||
&= \lim\limits_{x \to \sqrt{2}} \frac{q}{qp} \\
|
||||
&= \lim\limits_{x \to \sqrt{2}} \frac{1}{p} \\
|
||||
&= \text{DNE}
|
||||
\end{align*}
|
||||
Thus the limit does not exist since it is a contradiction that as $x \to \sqrt{2},\ x \in \Q$, and since $\sqrt{2} \notin \Q$, it's impossible to substitute $\sqrt{2}$ for $x$. Additionally, since $p \in \Z$, $p$ can equal 0, and thus the function is undefined at this point, again rendering the limit non-existent. Thus $f$ is continuous at $c=\sqrt{2}$, but $f$ is not differentiable at $c=\sqrt{2}$.\\
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 4 (g) %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item If $f$ is differentiable at $c$, then $f$ is continuous at $c$.\\
|
||||
|
||||
This is true by \textit{Theorem 6.1.2}:
|
||||
\begin{theorem*}
|
||||
If $f:I \rightarrow \R$ has a derivative at $c \in I$, then $f$ is continuous at $c$.\\
|
||||
\end{theorem*}
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 4 (h) %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item If $f$ is differentiable on $[a,b]$, then $f$ is uniformly continuous on $[a,b]$.\\
|
||||
|
||||
This is a true statement.
|
||||
\begin{proof}
|
||||
Let $f: [a,b] \to [a,b]$ be a function such that $f$ is differentiable $\forall\ x \in A$. Let $x,c \in A$, and without loss of generality, let $x \neq c$ (since $f$ is differentiable everywhere, $f$ is also differentiable at $c=0$.).\\
|
||||
|
||||
Recall \textit{Carathéodory's Theorem}:
|
||||
\begin{theorem*}[\textbf{Carathéodory's Theorem}]
|
||||
Let $f$ be defined on an interval $I$ containing the point $c$. Then $f$ is differentiable at $c$ if and only if there exists a function $\phi$ on $I$ that is continuous at $c$ and satisfies
|
||||
\[f(x)-f(c)=\phi (x)(x-c)\ \ \ \ \text{for}\ \ \ \ x \in I\]
|
||||
In this case, we have $\phi (c)=f'(c)$.
|
||||
\end{theorem*}
|
||||
Since $f$ is differentiable at all points, we know that $\exists\ \phi \st f(x)-f(c)=\phi(x)(x-c)$. Let $\phi$ be given by $\phi(x):= f'(c)$.\\
|
||||
|
||||
Also, recall the \textit{Mean Value Theorem}:
|
||||
\begin{theorem*}[\textbf{Mean Value Theorem}]
|
||||
Suppose that $f$ is continuous on a closed interval $I:=[a,b]$, and that $f$ has a derivative in the open interval $(a,b)$. Then there exists at least one point $c$ in $(a,b)$ such that
|
||||
\[f(b)-f(a)=f'(c)(b-a)\]
|
||||
\end{theorem*}
|
||||
Thus we have that $f'(c)=\frac{f(x)-f(c)}{x-c} \leq M$ since $f'(x) \leq M\ \forall\ x \in \R$. Thus $|x-c|<\frac{\varepsilon}{M}|f(x)-f(c)|<\varepsilon$.
|
||||
\begin{align*}
|
||||
|f(x)-f(c)|&=\lim\limits_{x \to c} \frac{f(x)-f(c)}{x-c} \cdot (x-c) \\
|
||||
&= \lim\limits_{x \to c} f(x)-f(c) \\
|
||||
&< M \cdot \varepsilon
|
||||
\end{align*}
|
||||
Thus let $\delta = \frac{\varepsilon}{M}$.\\
|
||||
|
||||
Recall the definition of a \textit{Lipschitz Function}:
|
||||
\theoremstyle{definition}
|
||||
\begin{definition*}
|
||||
Let $A \subseteq \R$ and let $f:A \rightarrow \R$. If there exists a constant $K > 0$ such that
|
||||
\[(4)\ \ \ \ \ \ \ \ |f(x)-f(u)| \leq K|x-u|\]
|
||||
for all $x,u \in A$, then $f$ is said to be a \textbf{Lipschitz function} (or to satisfy a \textbf{Lipschitz condition}) on $A$.\\
|
||||
|
||||
The condition $(4)$ that a function $f: I \to \R$ on an interval $I$ is a Lipschitz function can be interpreted geometrically as follows. If we write the condition as
|
||||
\[\abs{\frac{f(x)-f(u)}{x-u}}\leq K,\ x,u \in I,\ x \neq u,\]
|
||||
then the quantity inside the absolute values is the slope of a line segment joining the points $(x,f(x))$ and $(u,f(u))$. Thus a function $f$ satisfies a Lipschitz condition if and only if the slopes of all line segments joining two points on the graph of $y=f(x)$ over $I$ are bounded by some number $K$.
|
||||
\end{definition*}
|
||||
If we let $M \delta=K$, we have that $f$ is a Lipschitz function.\\
|
||||
|
||||
Now, recall \textit{Theorem 5.4.3}:
|
||||
\begin{theorem}
|
||||
If $f:A \rightarrow \R$ is a Lipschitz function, then $f$ is uniformly continuous on $A$.
|
||||
\end{theorem}
|
||||
Also, recall the \textit{Continuous Extension Theorem}:
|
||||
\begin{theorem*}[\textbf{Continuous Extension Theorem}]
|
||||
A function $f$ is uniformly continuous on the interval $(a,b)$ if and only if it can be defined at the endpoints $a$ and $b$ such that the extended function is continuous on $[a,b]$.
|
||||
\end{theorem*}
|
||||
And recall \textit{Theorem 6.1.1}:
|
||||
\begin{theorem*}
|
||||
If $f:I \to \R$ has a derivative at $c \in I$, then $f$ is continuous at $c$.
|
||||
\end{theorem*}
|
||||
So, by \textit{Theorem 5.4.3} we have that $f$ is uniformly continuous.\\
|
||||
|
||||
By \textit{Theorem 6.1.1}, since $f$ is differentiable $\forall\ c \in [a,b]$, $f$ is continuous on $[a,b]$. \\
|
||||
|
||||
Thus by the \textit{Continuous Extension Theorem}, since $f$ is continuous on $[a,b]$, we have that $f$ is uniformly continuous on $(a,b)$.\\
|
||||
|
||||
Recall the \textit{Uniform Continuity Theorem}:
|
||||
\begin{theorem*}[\textbf{Uniform Continuity Theorem}]
|
||||
Let $I$ be a closed bounded interval and let $f:I \rightarrow \R$ be continuous on $I$. Then $f$ is uniformly continuous on $I$.
|
||||
\end{theorem*}
|
||||
Since $[a,b]$ is a closed interval, and since $f$ is continuous on $[a,b]$, we have that by the \textit{Uniform Continuity Theorem}, $f$ is uniformly continuous on $[a,b]$.\\
|
||||
|
||||
$\therefore$ If $f$ is differentiable on $[a,b]$, then $f$ is uniformly continuous on $[a,b]$.\\
|
||||
\end{proof}
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 4 (i) %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item If $f$ is differentiable on $(a,b)$ and $f(a)=f(b)=0$, then $f$ is uniformly continuous at $[a,b]$.\\
|
||||
|
||||
This is a false statement. Consider $f:(0,\pi) \to \R$ given by $f(x):=x\sin (x)$. Then we have that
|
||||
\[f(a)=f(0)=0 \sin (0) = 0 \cdot 0 = 0 = \pi \cdot 0 = \pi \sin (\pi) = f(\pi)=f(b)\]
|
||||
Recall \textit{Theorem 6.1.2}:
|
||||
\begin{theorem*}
|
||||
Let $I \subseteq \R$ be an interval, let $c \in I$ , and let $f:I \rightarrow \R$ and $g:I \rightarrow \R$ be functions that are differentiable at $c$. Then:
|
||||
\begin{enumerate}
|
||||
\item If $\alpha \in \R$, then the function $\alpha f$ is differentiable at $c$, and \[(\alpha f)'(c) = \alpha f'(c)\]
|
||||
|
||||
\item The function $f+g$ is differentiable at $c$, and
|
||||
\[(f+g)'(c) = f'(c)+g'(c)\]
|
||||
|
||||
\item (Product Rule) The function $fg$ is differentiable at $c$, and
|
||||
\[(fg)'(c) = f'(c)g(c) + f(c)g'(c).\]
|
||||
|
||||
\item (Quotient Rule) If $g(c) \neq 0$, then the function $f/g$ is differentiable at $c$, and
|
||||
\[\left( \frac{f}{g}\right)'(c) = \frac{f'(c)g(c)-f(c)g'(c)}{(g(c))^2}\]
|
||||
\end{enumerate}
|
||||
\end{theorem*}
|
||||
So, if we let $g(x):=x$, and $h(x):=\sin (x)$, we then have
|
||||
\begin{align*}
|
||||
\lim\limits_{c \to 0} \frac{f(x+c)-f(x)}{c} &= \lim\limits_{c \to 0} \frac{g(x+c)-g(x)}{c} \cdot h(x) + g(x) \cdot \lim\limits_{c \to 0} \frac{h(x+c)-h(x)}{c} \\
|
||||
&= \lim\limits_{c \to 0} \frac{x+c-x}{c} \cdot \sin (x) + x \cdot \lim\limits_{c \to 0} \frac{\sin (x+c)-\sin (x)}{c} \\
|
||||
&= \lim\limits_{c \to 0} \frac{c}{c} \cdot \sin(x) + x \cdot \lim\limits_{c \to 0} \frac{(\sin (x)\cos (c) + \sin(c)\cos(x))-\sin(x)}{c} \\
|
||||
& (\because\ \sin(a+b)=\sin(a)\cos(b)+\sin(b)\cos(a)) \\
|
||||
&= \lim\limits_{c \to 0} 1 \cdot \sin(x) + x \cdot\lim\limits_{c \to 0} \left(\cos(x)\cdot\frac{\sin(c)}{c}+\sin(x)\cdot\frac{\cos(c)-1}{c}\right) \\
|
||||
&=\sin(x)+x\cdot\lim\limits_{c \to 0} \left(\cos(x) \cdot \frac{\sin(c)}{c} + \sin(x) \cdot \frac{\cos^2(c)-1}{(\cos(c)+1)c}\right) \\
|
||||
&=\sin(x)+x\cdot\lim\limits_{c \to 0} \left(\cos(x)\cdot\frac{\sin(c)}{c}-\sin(x)\cdot\frac{\sin^2(c)}{(\cos(c)+1)c}\right) \\
|
||||
&(\because\ \sin^2(c)+\cos^2(c)=1) \\
|
||||
&= \sin(x)+x\cdot\lim\limits_{c \to 0} \left(\left(\cos(x)-\frac{\sin(x)\sin(c)}{\cos(c)+1}\right)\frac{\sin(c)}{c}\right) \\
|
||||
&= \sin(x)+x\cdot\left(\lim\limits_{c \to 0} \left(\cos(x)-\frac{\sin(x)\sin(c)}{\cos(c)+1}\right)\right)\cdot\left(\lim\limits_{c \to 0} \frac{\sin(c)}{c} \right) \\
|
||||
&=\sin(x)+ x\cdot\cos(x)\left(\lim\limits_{c \to 0} \frac{\sin(c)}{c}\right) \\
|
||||
&(\because\ \lim\limits_{c \to 0} \left(\cos(x)-\frac{\sin(x)\sin(c)}{\cos(c)+1}\right)=\cos(x)-\frac{\sin(x)\sin(0)}{\cos(0)+1}=\cos(x) \\
|
||||
&\text{by continuity} )\\
|
||||
&= \sin(x)+x\cos(x) \\
|
||||
&(\because\ \lim\limits_{c \to 0} \frac{\sin(c)}{c}=1)
|
||||
\end{align*}
|
||||
Thus $x\sin(x)$ is differentiable on $(0,\pi)$.\\
|
||||
|
||||
Now, let $(x_n),(u_n) \subseteq (0,\pi)$ be given by $x_n:=\pi$, and $u_n:=\pi+\frac{1}{n}$ for $n \in \N$. Then we have
|
||||
\[(x_n-u_n)=\pi-(\pi+\frac{1}{n})=-\frac{1}{n}\]
|
||||
and thus
|
||||
\[\lim\limits_{n \to \infty} (x_n-u_n)=\lim\limits_{n \to \infty} -\frac{1}{n}=0\]
|
||||
Now we have
|
||||
\begin{align*}
|
||||
|f(x_n)-f(u_n)| &= \abs{\pi\sin(\pi)-\left(\pi+\frac{1}{n}\right)\sin(\pi+\frac{1}{n})} \\
|
||||
&= \abs{\pi \cdot 0-\left(\pi+\frac{1}{n}\right)\sin\left(\pi+\frac{1}{n}\right)} \\
|
||||
&= \abs{-\left(\pi+\frac{1}{n}\right)\sin\left(\pi+\frac{1}{n}\right)} \\
|
||||
&= \left(\pi+\frac{1}{n}\right)\sin\left(\pi+\frac{1}{n}\right) \\
|
||||
&= \left(\pi+\frac{1}{n}\right)\sin(\frac{1}{n}) &\because\ \sin(\pi)=0 \\
|
||||
&= \pi\sin\left(\frac{1}{n}\right)+\left(\frac{1}{n}\right)\sin\left(\frac{1}{n}\right)
|
||||
\end{align*}
|
||||
We notice that $\lim\limits_{n \to \infty} \frac{1}{n}=0$ and $(\sin\left(\frac{1}{n}\right))$ is a bounded sequence in $(0,\pi)$. Thus
|
||||
\[\lim\limits_{n \to \infty} \left(\frac{1}{n}\sin\frac{1}{n}\right)=0\]
|
||||
and
|
||||
\[\lim\limits_{x \to 0} \frac{\sin(x)}{x}=1\ \forall\ x \in (0,\pi)\]
|
||||
This yields
|
||||
\[\lim\limits_{n \to \infty} \pi\sin\left(\frac{1}{n}\right)=\pi\lim\limits_{\frac{1}{n}\to 0} \left(\frac{\sin\left(\frac{1}{n}\right)}{\frac{1}{n}}\right)=\pi\]
|
||||
Thus $\lim\limits_{n \to \infty} (f(x_n)-f(u_n))=\pi$.\\
|
||||
|
||||
Let $\varepsilon=\pi$. Then $\exists\ k \in \N \st \forall\ n \geq k$:
|
||||
\begin{align*}
|
||||
\pi-\varepsilon<f(x_n)-f(u_n)&<\pi+\varepsilon \\
|
||||
f(x_n)-f(u_n)&>\pi \\
|
||||
|f(x_n)-f(u_n)|&>\pi \\
|
||||
|f(x_{n+k})-f(u_{n+k})|&>\pi
|
||||
\end{align*}
|
||||
Thus, for $\varepsilon_0=\pi$, and the two sequences $(x_{n+k}),\ (u_{n+k})$, by the \textit{Nonuniform Continuity Criteria}, $f$ is not uniformly continuous on $(0,\pi)$\\
|
||||
|
||||
$\therefore$ If $f$ is differentiable on $(a,b)$, and $f(a)=f(b)=0$, then $f$ is \textbf{not} uniformly continuous on $[a,b]$.
|
||||
\end{enumerate}
|
||||
\end{enumerate}
|
||||
\end{document}
|
||||
Binary file not shown.
+428
@@ -0,0 +1,428 @@
|
||||
\documentclass[12pt,letterpaper]{article}
|
||||
\usepackage[utf8]{inputenc}
|
||||
\usepackage[english]{babel}
|
||||
\usepackage[normalem]{ulem}
|
||||
\usepackage{cancel}
|
||||
\usepackage{amsthm}
|
||||
\usepackage{amsmath}
|
||||
\usepackage{amsfonts}
|
||||
\usepackage{amssymb}
|
||||
\usepackage{graphicx}
|
||||
\usepackage{array}
|
||||
\usepackage[left=2cm, right=2.5cm, top=2.5cm, bottom=2.5cm]{geometry}
|
||||
\usepackage{enumitem}
|
||||
\newcommand{\st}{\ \text{s.t.}\ }
|
||||
\newcommand{\abs}[1]{\left\lvert #1 \right\rvert}
|
||||
\newcommand{\R}{\mathbb{R}}
|
||||
\newcommand{\N}{\mathbb{N}}
|
||||
\newcommand{\Q}{\mathbb{Q}}
|
||||
\newcommand{\C}{\mathbb{C}}
|
||||
\newcommand{\Z}{\mathbb{Z}}
|
||||
\DeclareMathOperator{\sign}{sgn}
|
||||
\newtheoremstyle{case}{}{}{}{}{}{:}{ }{}
|
||||
\theoremstyle{case}
|
||||
\newtheorem{case}{Case}
|
||||
\theoremstyle{definition}
|
||||
\newtheorem{definition}{Definition}[section]
|
||||
\newtheorem{definition*}{Definition}
|
||||
\newtheorem{theorem}{Theorem}[section]
|
||||
\newtheorem{theorem*}{Theorem}
|
||||
\newtheorem{corollary}{Corollary}[section]
|
||||
\newtheorem{corollary*}{Corollary}
|
||||
\newtheorem{lemma}[theorem]{Lemma}
|
||||
\newtheorem{remark}{Remark}
|
||||
\setlist[enumerate]{font=\bfseries}
|
||||
\renewcommand{\qedsymbol}{$\blacksquare$}
|
||||
\author{Alexander J. Tusa}
|
||||
\title{Real Analysis Homework 11}
|
||||
\begin{document}
|
||||
\maketitle
|
||||
\begin{enumerate}
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 1 %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item Give an example of nonconstant functions $f$ and $g$ such that $(fg)'=f'g'$.\\
|
||||
|
||||
Let $f(x)=e^{2x}=g(x)$. Then
|
||||
\begin{align*}
|
||||
(fg)'&=f'g' \\
|
||||
f'(x)g(x)+f(x)g'(x) &= f'(x)g'(x) \\
|
||||
2e^{2x} \cdot e^{2x} + e^{2x} \cdot 2e^{2x} &= 2e^{2x} \cdot 2e^{2x} \\
|
||||
2e^{4x}+2e^{4x} &= 4e^{4x} \\
|
||||
4e^{4x} &= 4e^{4x}
|
||||
\end{align*}
|
||||
Thus we have that $(fg)'=f'g'$.\\
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 2 %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item Suppose that $f$ is differentiable at 2 and 4 with $f(2)=2,\ f(4)=3,\ f'(2)=\pi$, and $f'(4)=e$.
|
||||
\begin{enumerate}
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 2 (a) %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item If $g(x)=xf(x^2)$, find $g'(2)$.\\
|
||||
|
||||
\begin{align*}
|
||||
g'(x) &= (xf(x^2))' \\
|
||||
&= x' \cdot f(x^2)+x \cdot f'(x^2) \cdot 2x &\text{by the } \textit{Product Rule} \text{ and the } \textit{Chain Rule} \\
|
||||
&= 1 \cdot f(x^2) + x \cdot f'(x^2) \cdot 2x \\
|
||||
&\Downarrow \\
|
||||
g'(2) &= f(x^2) + x \cdot f'(x^2) \cdot 2x \\
|
||||
&= f(2^2) + 2 \cdot f'(2^2) \cdot 2(2)\\
|
||||
&= f(4)+2 \cdot f'(4) \cdot 2(2) \\
|
||||
&= 3+2 \cdot e \cdot 4 \\
|
||||
&= 3+8e
|
||||
\end{align*}
|
||||
So $g'(2) = 3+8e$.\\
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 2 (b) %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item If $g(x)=f^2(\sqrt{x})$, find $g'(4)$.\\
|
||||
|
||||
\begin{align*}
|
||||
g'(x) &= ([f(\sqrt{x})]^2)' \\
|
||||
&= 2[f(\sqrt{x})] \cdot f'(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}} &\text{by the } \textit{Chain Rule} \\
|
||||
&\Downarrow \\
|
||||
g'(4) &= 2[f(\sqrt{4})] \cdot f'(\sqrt{4}) \cdot \frac{1}{2 \sqrt{4}} \\
|
||||
&= 2 \cdot f(2) \cdot f'(2) \cdot \frac{1}{2 \cdot 2} \\
|
||||
&= 2 \cdot 2 \cdot \pi \cdot \frac{1}{4} \\
|
||||
&= \frac{4\pi}{4} \\
|
||||
&= \pi
|
||||
\end{align*}
|
||||
So $g'(4)=\pi$.\\
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 2 (c) %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item If $g(x)=x/f(x^3)$, find $g'(\sqrt[3]{2})$. \\
|
||||
|
||||
\begin{align*}
|
||||
g'(x) &= \left(\frac{x}{f(x^3)}\right)' \\
|
||||
&= \frac{x' \cdot f(x^3) - x \cdot f'(x^3) \cdot 3x^2}{\left(f(x^3)\right)^2} &\text{by the } \textit{Quotient Rule} \\
|
||||
&= \frac{1 \cdot f(x^3)-x \cdot f'(x^3) \cdot 3x^2}{(f(x^3))^2} \\
|
||||
&\Downarrow \\
|
||||
g'(\sqrt[3]{2}) &= \frac{f((\sqrt[3]{2})^3)-\sqrt[3]{2} \cdot f'((\sqrt[3]{2})^3) \cdot 3(\sqrt[3]{2})^2}{(f((\sqrt[3]{2})^3))^2} \\
|
||||
&= \frac{f(2)-3(\sqrt[3]{2})^2\sqrt[3]{2} \cdot f'(2)}{(f(2))^2} \\
|
||||
&= \frac{2-3\cdot 2 \cdot \pi}{2^2} \\
|
||||
&= \frac{2-6 \cdot \pi}{4} \\
|
||||
&= \frac{1-3 \cdot \pi}{2}
|
||||
\end{align*}
|
||||
So $g'(\sqrt[3]{2})=\frac{1-3\pi}{2}$.\\
|
||||
|
||||
\end{enumerate}
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 3 %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item Determine if each function is differentiable at the given point. If so, find its derivative. If not, explain why not.
|
||||
\begin{enumerate}
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 3 (a) %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item At $x=1$ for $f(x)=\begin{cases}
|
||||
3x-2 &\text{if } x<1 \\
|
||||
x^3 &\text{if } x \geq 1
|
||||
\end{cases}$\\
|
||||
|
||||
We note that $f(1)=1^3=1$. First, let us find $f'(x)$ for each part of the function. So when $x<1$:
|
||||
\begin{align*}
|
||||
\lim\limits_{h \to 0} \frac{f(x+h)-f(x)}{h} &= \lim\limits_{h \to 0} \frac{3(x+h)-2-(3x-2)}{h} \\
|
||||
&= \lim\limits_{h \to 0} \frac{3x+3h-2-3x+2}{h} \\
|
||||
&= \lim\limits_{h \to 0} \frac{3h}{h} \\
|
||||
&= \lim\limits_{h \to 0} 3 \\
|
||||
&= 3
|
||||
\end{align*}
|
||||
So $f'(x)=3$ when $x <1$
|
||||
And when $x \geq 1$:
|
||||
\begin{align*}
|
||||
\lim\limits_{h \to 0} \frac{f(x+h)-f(x)}{h} &= \lim\limits_{h \to 0} \frac{(x+h)^3-x^3}{h} \\
|
||||
&= \lim\limits_{h \to 0} \frac{h^3-3h^2x+3hx^2+x^3-x^3}{h} \\
|
||||
&= \lim\limits_{h \to 0} \frac{h^3-3h^2x+3hx^2}{h} \\
|
||||
&= \lim\limits_{h \to 0} \frac{h(h^2-3hx+3x^2)}{h} \\
|
||||
&= \lim\limits_{h \to 0} h^2-3hx+3x^2 \\
|
||||
&= 0^2+3(0)x+3x^2 \\
|
||||
&= 0+0+3x^2 \\
|
||||
&= 3x^2
|
||||
\end{align*}
|
||||
So $f'(x)=3x^2$ when $x \geq 1$.\\
|
||||
|
||||
Now we must check for differentiability when $x=1$. So
|
||||
\begin{align*}
|
||||
\lim\limits_{x \to c} \frac{f(x)-f(c)}{x-c} &= \lim\limits_{x \to 1^+} \frac{x^3-1}{x-1} \\
|
||||
&= \lim\limits_{x \to 1^+} \frac{\cancel{(x-1)}(x^2+x+1)}{\cancel{(x-1)}} \\
|
||||
&= \lim\limits_{x \to 1^+} (x^2+x+1) \\
|
||||
&= 1^2+1+1 \\
|
||||
&= 1 + 1 + 1 \\
|
||||
&= 3
|
||||
\end{align*}
|
||||
And
|
||||
\begin{align*}
|
||||
\lim\limits_{x \to c} \frac{f(x)-f(c)}{x-c} &= \lim\limits_{x \to 1^-} \frac{3x-2-1}{x-1} \\
|
||||
&= \lim\limits_{x \to 1^-} \frac{3x-3}{x-1} \\
|
||||
&= \lim\limits_{x \to 1^-} \frac{3\cancel{(x-1)}}{\cancel{(x-1)}} \\
|
||||
&= \lim\limits_{x \to 1^-} 3 \\
|
||||
&= 3
|
||||
\end{align*}
|
||||
So since $\lim\limits_{x \to 1^-} \frac{f(x)-f(c)}{x-c} = 3 = \lim\limits_{x \to 1^+} \frac{f(x)-f(c)}{x-c}$, we have that the limits are equal, and thus the limit exists which yields that $f$ is differentiable at $x=1$.\\
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 3 (b) %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item At $x=1$ for $f(x):=\begin{cases}
|
||||
2x+1 &\text{if } x < 1 \\
|
||||
x^2 &\text{if } x \geq 1
|
||||
\end{cases}$\\
|
||||
|
||||
In order for $f$ to be differentiable, it must also be continuous. So let us first ensure that $f$ is continuous.\\
|
||||
|
||||
So we have
|
||||
\[\lim\limits_{x \to 1^-} f(x)=\lim\limits_{x \to 1^-} 2x+1 =2(1)+1=3\]
|
||||
and
|
||||
\[\lim\limits_{x \to 1^+} f(x)=\lim\limits_{x \to 1^+} x^2 = 1^2=1\]
|
||||
Since $\lim\limits_{x \to 1^-} f(x) = 3 \neq 1 = \lim\limits_{x \to 1^+} f(x)$, we have that $f$ is not continuous, and thus $f$ is not differentiable at $x=1$.\\
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 3 (c) %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item At $x=1$ for $f(x):=\begin{cases}
|
||||
3x-2 &\text{if } x<1 \\
|
||||
x^2 &\text{if } x \geq 1
|
||||
\end{cases}$\\
|
||||
|
||||
We will show that $f$ is \textbf{not} differentiable at $x=1$. We first note that $f(1)=1^2=1$. Now let us find the derivatives of $f$ for each part of the piece wise function. So for $x<1$:
|
||||
\begin{align*}
|
||||
\lim\limits_{h \to 0} \frac{f(x+h)-f(x)}{h} &= \lim\limits_{h \to 0} \frac{3(x+h)-2-(3x-2)}{h} \\
|
||||
&= \lim\limits_{h \to 0} \frac{3x+3h-2-3x+2}{h} \\
|
||||
&= \lim\limits_{h \to 0} \frac{3h}{h} \\
|
||||
&= \lim\limits_{h \to 0} 3 \\
|
||||
&= 3
|
||||
\end{align*}
|
||||
So $f'(x)=3$ when $x<1$.\\
|
||||
|
||||
And for the case where $x \geq 1$:
|
||||
\begin{align*}
|
||||
\lim\limits_{h \to 0} \frac{f(x+h)-f(x)}{h} &= \lim\limits_{h \to 0} \frac{(x+h)^2-x^2}{h} \\
|
||||
&= \lim\limits_{h \to 0} \frac{h^2+2hx+x^2-x^2}{h} \\
|
||||
&= \lim\limits_{h \to 0} \frac{h(h+2x)}{h} \\
|
||||
&= \lim\limits_{h \to 0} (h+2x) \\
|
||||
&= 0+2x \\
|
||||
&=2x
|
||||
\end{align*}
|
||||
So $f'(x)=2x$ for $x \geq 1$.\\
|
||||
|
||||
Now we must show that $f$ is \textbf{not} differentiable at $x=1$. So
|
||||
\begin{align*}
|
||||
\lim\limits_{x \to c} \frac{f(x)-f(c)}{x-c} &= \lim\limits_{x \to 1^-} \frac{3x-2-1}{x-1} \\
|
||||
&= \lim\limits_{x \to 1^-} \frac{3x-3}{x-1} \\
|
||||
&= \lim\limits_{x \to 1^-} \frac{3\cancel{(x-1)}}{\cancel{x-1}} \\
|
||||
&= \lim\limits_{x \to 1^-} 3 \\
|
||||
&= 3
|
||||
\end{align*}
|
||||
And
|
||||
\begin{align*}
|
||||
\lim\limits_{x \to c} \frac{f(x)-f(c)}{x-c} &= \lim\limits_{x \to 1^+} \frac{x^2-1}{x-1} \\
|
||||
&= \lim\limits_{x \to 1^+} \frac{\cancel{(x-1)}(x+1)}{\cancel{x-1}} \\
|
||||
&= \lim\limits_{x \to 1^+} (x+1) \\
|
||||
&= 1+1 \\
|
||||
&= 2
|
||||
\end{align*}
|
||||
So since $\lim\limits_{x \to 1^-} f(x) = 3 \neq 2 = \lim\limits_{x \to 1^+} f(x)$, we have that the limit does not exist at $x=1$, and thus $f$ is not differentiable at $x=1$, but is continuous.
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 3 (d) %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item (Sec. 6.1, pr. 4) At $x=0$ for $f(x):=\begin{cases}
|
||||
x^2 &\text{if } x \text{ is rational} \\
|
||||
0 &\text{if } x \text{ is irrational}
|
||||
\end{cases}$\\
|
||||
|
||||
First, we note that $0 \in \Q$ and $f(0)=0^2=0$.\\
|
||||
|
||||
Let $g(x):=\frac{f(x)-f(0)}{x-0}=\frac{f(x)}{x}$. Then for $x \neq 0$:
|
||||
\[g(x):=\begin{cases}
|
||||
x, & x \in \Q \\
|
||||
0, & x \in \R \setminus \Q
|
||||
\end{cases}\]
|
||||
So, we have that the given function is differentiable at 0 if and only if $\lim\limits_{x \to 0} g(x)$ exists. In this case, the limit is $f'(0)$.\\
|
||||
|
||||
We note that $-|x| \leq g(x) \leq |x|$, and that $\lim\limits_{x \to 0} -|x|=\lim\limits_{x \to 0} |x|= 0$.\\
|
||||
|
||||
Recall the \textit{Squeeze Theorem}:
|
||||
\begin{theorem*}[\textbf{Squeeze Theorem}]
|
||||
Let $A \subseteq \R$, let $f,g,h:A \rightarrow \R$, and let $c \in \R$ be a cluster point of $A$. If
|
||||
\[f(x) \leq g(x) \leq h(x)\ \ \ \ \text{for all}\ \ \ \ x \in A,\ x \neq c,\]
|
||||
and if $\lim\limits_{x\to c} f = L = \lim\limits_{x\to c} h$, then $\lim\limits_{x\to c} g =L$.
|
||||
\end{theorem*}
|
||||
By the \textit{Squeeze Theorem} $\lim\limits_{x \to 0} g(x)=0$. Thus we have that $f$ is differentiable at $x=0$ and $f'(0)=0$.\\
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 3 (e) %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item At $x=0$ for $f(x):=\begin{cases}
|
||||
x &\text{if } x \text{ is rational} \\
|
||||
0 &\text{if } x \text{ is irrational}
|
||||
\end{cases}$ \\
|
||||
|
||||
We want to show that $f(x)$ is \textbf{not} differentiable at $x=0$. First we note that $f(0)=0$. So let us first find the derivatives of $f(x)$ for each piece of the function. So for $x \in \Q$:
|
||||
\begin{align*}
|
||||
\lim\limits_{h \to 0} \frac{f(x+h)-f(x)}{h} &= \lim\limits_{h \to 0} \frac{x+h-x}{h} \\
|
||||
&= \lim\limits_{h \to 0} \frac{h}{h} \\
|
||||
&= \lim\limits_{h \to 0} 1 \\
|
||||
&= 1
|
||||
\end{align*}
|
||||
So $f'(x)=1$ when $x \in \Q$.\\
|
||||
|
||||
Now, for $x \in \R \setminus \Q$:
|
||||
\begin{align*}
|
||||
\lim\limits_{h \to 0} \frac{f(x+h)-f(x)}{h} &= \lim\limits_{h \to 0} \frac{0-0}{h} \\
|
||||
&= \lim\limits_{h \to 0} 0 \\
|
||||
&= 0
|
||||
\end{align*}
|
||||
So $f'(x)=0$ when $x \in \R\setminus\Q$.\\
|
||||
|
||||
Now we want to show that $f$ is \textbf{not} differentiable at $x=0$. So there's four cases we must consider: $x$ approaches zero from the left along a sequence of rational numbers, $x$ approaches zero from the right along a sequence of rational numbers, $x$ approaches zero from the left along a sequence of irrational numbers, and $x$ approaches zero from the right along a sequence of irrational numbers.\\
|
||||
|
||||
For the case where $x$ approaches 0 from the left along a sequence of rational numbers:
|
||||
\begin{align*}
|
||||
\lim\limits_{x \to c^-} \frac{f(x)-f(c)}{x-c} &= \lim\limits_{x \in \Q \to 0^-} \frac{f(x)-f(0)}{x-0} \\
|
||||
&=\lim\limits_{x \in \Q \to 0^-} \frac{x-0}{x} \\
|
||||
&= \lim\limits_{x \in \Q \to 0^-} \frac{x}{x} \\
|
||||
&= \lim\limits_{x \in \Q \to 0^-} 1 \\
|
||||
&= 1
|
||||
\end{align*}
|
||||
And for $x$ approaches 0 from the right along a sequence of rational numbers:
|
||||
\begin{align*}
|
||||
\lim\limits_{x \to c^+} \frac{f(x)-f(c)}{x-c} &= \lim\limits_{x \in \Q \to 0^+} \frac{f(x)-f(0)}{x-0} \\
|
||||
&= \lim\limits_{x \in \Q \to 0^+} \frac{x-0}{x} \\
|
||||
&= \lim\limits_{x \in \Q \to 0^+} \frac{x}{x} \\
|
||||
&= \lim\limits_{x \in \Q \to 0+} 1 \\
|
||||
&= 1
|
||||
\end{align*}
|
||||
For the case where $x$ approaches 0 from the left along a sequence of irrational numbers:
|
||||
\begin{align*}
|
||||
\lim\limits_{x \to c^-} \frac{f(x)-f(c)}{x-c} &= \lim\limits_{x \in \R \setminus \Q \to 0^-} \frac{f(x)-f(0)}{x-0} \\
|
||||
&= \lim\limits_{x \in \R \setminus \Q \to 0^-} \frac{0-0}{x} \\
|
||||
&= \lim\limits_{x \in \R \setminus \Q \to 0^-} \frac{0}{x} \\
|
||||
&= \lim\limits_{x \in \R \setminus \Q \to 0^-} 0 \\
|
||||
&= 0
|
||||
\end{align*}
|
||||
And for $x$ approaching 0 from the right along a sequence of irrational numbers:
|
||||
\begin{align*}
|
||||
\lim\limits_{x \to c^+} \frac{f(x)-f(c)}{x-c} &= \lim\limits_{x \in \R \setminus \Q \to 0^+} \frac{f(x)-f(0)}{x-0} \\
|
||||
&= \lim\limits_{x \in \R \setminus \Q \to 0^+} \frac{0-0}{x} \\
|
||||
&= \lim\limits_{x \in \R \setminus \Q \to 0^+} \frac{0}{x} \\
|
||||
&= \lim\limits_{x \in \R \setminus \Q \to 0^+} 0 \\
|
||||
&= 0
|
||||
\end{align*}
|
||||
So we have that
|
||||
\[\lim\limits_{x \in \Q \to 0^-} \frac{f(x)-f(0)}{x}=\lim\limits_{x \in \Q \to 0^+} \frac{f(x)-f(0)}{x}=1\]
|
||||
and
|
||||
\[0 = \lim\limits_{x \in \R \setminus \Q \to 0^+} \frac{f(x)-f(0)}{x} = \lim\limits_{x \in \R \setminus \Q \to 0^-} \frac{f(x)-f(0)}{x}\]
|
||||
However, since these limits are not equal to each other since $0 \neq 1$, we have that the limit does not exist at $x=0$, and thus $f$ is \textbf{not} differentiable at $x=0$ but is continuous.\\
|
||||
\end{enumerate}
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 4 %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item
|
||||
\begin{enumerate}
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 4 (a) %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item Prove: If $f$ is differentiable on $\R$, then $f'(x)=\lim\limits_{h \to 0} \frac{f(x)-f(x-h)}{h}$\\
|
||||
|
||||
\begin{proof}
|
||||
Assume that $f:A \subseteq \R \to \R$ is differentiable on $\R$. Then we know that
|
||||
\[f'(x):=\lim\limits_{x \to c} \frac{f(x)-f(c)}{x-c} = \lim\limits_{h \to 0} \frac{f(x+h)-f(x)}{h}\]
|
||||
Recall the definition of the derivative:
|
||||
\theoremstyle{definition}
|
||||
\begin{definition*}
|
||||
Let $I \subseteq \R$ be an interval, let $f:I \rightarrow \R$, and let $ c \in I$. We say that a real number $L$ is the \textbf{derivative of $f$ at $c$} if given any $\varepsilon > 0$ there exists $\delta (\varepsilon) > 0$ such that if $x \in I$ satisfies $0 < |x-c|<\delta (\varepsilon)$, then
|
||||
\[\abs{\frac{f(x)-f(c)}{x-c}-L}<\varepsilon.\]
|
||||
In this case we say that $f$ is \textbf{differentiable} at $c$, and we write $f'(c)$ for $L$. In other words, the derivative of $f$ at $c$ is given by the limit
|
||||
\[f'(c) = \lim_{x\to c} \frac{f(x)-f(c)}{x-c}\]
|
||||
provided this limit exists. (We allow the possibility that $c$ may be the endpoint of the interval.)
|
||||
\end{definition*}
|
||||
Assume $f'(x)=L$ for some $L \in \R$. Then
|
||||
\[\forall\ \varepsilon > 0,\ \exists\ \delta(\varepsilon) > 0 \st [x \in A \st 0<|x-c|<\delta(\varepsilon)] \implies \abs{\frac{f(x)-f(c)}{x-c}-L}<\varepsilon\]
|
||||
Now, let $h:=x-c$. Then $c=x-h$. So
|
||||
\[\forall\ \varepsilon > 0,\ \exists\ \delta(\varepsilon) > 0 \st [x \in A \st 0<|h|<\delta(\varepsilon)] \implies \abs{\frac{f(x)-f(x-h)}{h}-L}<\varepsilon\]
|
||||
\[\equiv\]
|
||||
\[\lim\limits_{h \to 0} \frac{f(x)-f(x-h)}{h}=L\]
|
||||
And thus
|
||||
\[f'(x)=\lim\limits_{x \to c} \frac{f(x)-f(c)}{x-c}=\lim\limits_{h \to 0} \frac{f(x+h)-f(x)}{h}=\lim\limits_{h \to 0} \frac{f(x)-f(x-h)}{h}\]
|
||||
\end{proof}
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 4 (b) %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item Show $f'(x)=\lim\limits_{h \to 0} \frac{3f(x+h)-f(x)-2f(x-h)}{5h}$\\
|
||||
|
||||
\begin{align*}
|
||||
f'(x) &= \lim\limits_{h \to 0} \frac{3f(x+h)-f(x)-2f(x-h)}{5h} \\
|
||||
&= \frac{3}{5}\lim\limits_{h \to 0} \frac{f(x+h)}{h} - \frac{1}{5} \lim\limits_{h \to 0} \frac{f(x)}{h} - \frac{2}{5} \lim\limits_{h \to 0} \frac{f(x-h)}{h} \\
|
||||
&= \frac{3}{5}\lim\limits_{h \to 0} \frac{f(x+h)}{h} - \frac{3}{5} \lim\limits_{h \to 0} \frac{f(x)}{h} + \frac{2}{5} \lim\limits_{h \to 0} \frac{f(x)}{h} - \frac{2}{5} \lim\limits_{h \to 0} \frac{f(x-h)}{h} \\
|
||||
&= \frac{3}{5} \lim\limits_{h \to 0} \frac{f(x+h)-f(x)}{h} + \frac{2}{5} \lim\limits_{h \to 0} \frac{f(x)-f(x-h)}{h} \\
|
||||
&= \frac{3}{5} \cdot f'(x)+\frac{2}{5} \cdot f'(x) \\
|
||||
&= f'(x)
|
||||
\end{align*}
|
||||
Thus $f'(x)=\lim\limits_{h \to 0} \frac{3f(x+h)-f(x)-2f(x-h)}{5h}$.\\
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 4 (c) %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item Find constant $c$ such that $f'(x)=\lim\limits_{h \to 0} \frac{5f(x+h)-f(x)-4f(x-h)}{ch}$\\
|
||||
|
||||
Let $c=9$. Then we want to show that
|
||||
\[f'(x)=\lim\limits_{h \to 0}\frac{5f(x+h)-f(x)-4f(x-h)}{9h}\]
|
||||
So
|
||||
\begin{align*}
|
||||
f'(x)&= \lim\limits_{h \to 0}\frac{5f(x+h)-f(x)-4f(x-h)}{9h} \\
|
||||
&= \frac{5}{9} \lim\limits_{h \to 0} \frac{f(x+h)}{h} - \frac{1}{9} \lim\limits_{h \to 0} \frac{f(x)}{h} - \frac{4}{9} \lim\limits_{h \to 0} \frac{f(x-h)}{h} \\
|
||||
&= \frac{5}{9} \lim\limits_{h \to 0} \frac{f(x+h)}{h} - \frac{5}{9} \lim\limits_{h \to 0} \frac{f(x)}{h} + \frac{4}{9} \lim\limits_{h \to 0} \frac{f(x)}{h} - \frac{4}{9} \lim\limits_{h \to 0} \frac{f(x-h)}{h} \\
|
||||
&= \frac{5}{9} \lim\limits_{h \to 0} \frac{f(x+h)-f(x)}{h} + \frac{4}{9} \lim\limits_{h \to 0} \frac{f(x)-f(x-h)}{h} \\
|
||||
&= \frac{5}{9} \cdot f'(x) + \frac{4}{9} \cdot f'(x) \\
|
||||
&= f'(x)
|
||||
\end{align*}
|
||||
Thus $f'(x)=\lim\limits_{h \to 0}\frac{5f(x+h)-f(x)-4f(x-h)}{ch}$ for $c=9$.\\
|
||||
\end{enumerate}
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 5 %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item Prove, if true or provide a counterexample, if false.\\
|
||||
\textbf{For all parts:} Let $A$ be an interval, $c \in A$ and $f,g:A \to \R$.
|
||||
\begin{enumerate}
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 5 (a) %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item If $f$ and $g$ are differentiable at $c$, then $f+g$ is differentiable at $c$.\\
|
||||
|
||||
This is true by \textit{Theorem 6.1.3}:
|
||||
\begin{theorem*}
|
||||
Let $I \subseteq \R$ be an interval, let $c \in I$ , and let $f:I \rightarrow \R$ and $g:I \rightarrow \R$ be functions that are differentiable at $c$. Then:
|
||||
\begin{enumerate}
|
||||
\item If $\alpha \in \R$, then the function $\alpha f$ is differentiable at $c$, and \[(\alpha f)'(c) = \alpha f'(c)\]
|
||||
|
||||
\item The function $f+g$ is differentiable at $c$, and
|
||||
\[(f+g)'(c) = f'(c)+g'(c)\]
|
||||
|
||||
\item (Product Rule) The function $fg$ is differentiable at $c$, and
|
||||
\[(fg)'(c) = f'(c)g(c) + f(c)g'(c).\]
|
||||
|
||||
\item (Quotient Rule) If $g(c) \neq 0$, then the function $f/g$ is differentiable at $c$, and
|
||||
\[\left( \frac{f}{g}\right)'(c) = \frac{f'(c)g(c)-f(c)g'(c)}{(g(c))^2}\]
|
||||
\end{enumerate}
|
||||
\end{theorem*}
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 5 (b) %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item If $f$ and $g$ are differentiable at $c$, then $g \circ f$ is differentiable at $c$.\\
|
||||
|
||||
This is a false statement. Consider $f(x):=x+1$. Then $f(c)=c+1$.\\
|
||||
|
||||
Let $g(x):=\begin{cases}
|
||||
1, &x\leq c+1 \\
|
||||
4, &x > c+1
|
||||
\end{cases}$\\
|
||||
|
||||
Then we have that $g'(c+1)$ does not exist, and thus $g$ is not differentiable at $f(c)$.\\
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 5 (c) %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item If $f=g^2$ and $f$ is differentiable on $(a,b)$, then $g$ is differentiable on $(a,b)$.\\
|
||||
|
||||
This is a false statement. Consider $g^2:(0,1) \to \R$ given by $g(x):=-1$. Then $f:(0,1) \to \R$ is $f(x):=-1$, which is differentiable on $(0,1)$. However, $g$ is not differentiable on $(0,1)$ since $g(x):=\sqrt{-1}=i \notin \R$, and thus $g$ is not differentiable on $(0,1)$ since $(0,1) \subset \R$, and $i \notin (0,1)$.
|
||||
\end{enumerate}
|
||||
\end{enumerate}
|
||||
\end{document}
|
||||
Binary file not shown.
+551
@@ -0,0 +1,551 @@
|
||||
\documentclass[12pt,letterpaper]{article}
|
||||
\usepackage[utf8]{inputenc}
|
||||
\usepackage[english]{babel}
|
||||
\usepackage{amsthm}
|
||||
\usepackage{amsmath}
|
||||
\usepackage{amsfonts}
|
||||
\usepackage{amssymb}
|
||||
\usepackage{graphicx}
|
||||
\usepackage{array}
|
||||
\usepackage[left=2cm, right=2.5cm, top=2.5cm, bottom=2.5cm]{geometry}
|
||||
\usepackage{enumitem}
|
||||
\usepackage{mathrsfs}
|
||||
\usepackage{hyperref}
|
||||
\hypersetup{
|
||||
colorlinks=true,
|
||||
linkcolor=blue,
|
||||
filecolor=magenta,
|
||||
urlcolor=cyan,
|
||||
}
|
||||
\newcommand{\st}{\ \text{s.t.}\ }
|
||||
\newcommand{\abs}[1]{\left\lvert #1 \right\rvert}
|
||||
\newcommand{\R}{\mathbb{R}}
|
||||
\newcommand{\N}{\mathbb{N}}
|
||||
\newcommand{\Q}{\mathbb{Q}}
|
||||
\newcommand{\C}{\mathbb{C}}
|
||||
\newcommand{\Z}{\mathbb{Z}}
|
||||
\DeclareMathOperator{\sign}{sgn}
|
||||
\newtheoremstyle{case}{}{}{}{}{}{:}{ }{}
|
||||
\theoremstyle{case}
|
||||
\newtheorem{case}{Case}
|
||||
\theoremstyle{definition}
|
||||
\newtheorem{definition}{Definition}[section]
|
||||
\newtheorem{definition*}{Definition}
|
||||
\newtheorem{theorem}{Theorem}[section]
|
||||
\newtheorem{theorem*}{Theorem}
|
||||
\newtheorem{corollary}{Corollary}[section]
|
||||
\newtheorem*{corollary*}{Corollary}
|
||||
\newtheorem{lemma}[theorem]{Lemma}
|
||||
\newtheorem{lemma*}{Lemma}
|
||||
\newtheorem{remark}{Remark}
|
||||
\setlist[enumerate]{font=\bfseries}
|
||||
\renewcommand{\qedsymbol}{$\blacksquare$}
|
||||
\author{Alexander J. Tusa}
|
||||
\title{Real Analysis Homework 12}
|
||||
\begin{document}
|
||||
\maketitle
|
||||
\begin{enumerate}
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 1 %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item
|
||||
\begin{enumerate}
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 1 (a) %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item \textbf{Section 6.2 Problem 15} Let $I$ be an interval. Prove that if $f$ is differentiable on $I$ and if the derivative $f'$ is bounded on $I$, then $f$ satisfies a Lipschitz condition on $I$. (See Definition 5.4.4)
|
||||
|
||||
\begin{proof}
|
||||
Let $M>0$ be such that $|f'(c)| \leq M\ \forall\ c \in I$. This follows from the fact that $f'$ is bounded on $I$. For $x,y \in I$ such that $x<y$, we know that by the mean value theorem, $\exists\ c \in (x,y) \subseteq I \st f(y)-f(x)=f'(c)(y-x)$. This yields the following:
|
||||
\begin{align*}
|
||||
|f(x)-f(y)| &= |f'(c)||y-x| \leq M|x-y| \dots (\because\ c \in I) \\
|
||||
\implies |f(x)-f(y)| &\leq M|x-y|\ \forall\ x,y \in I
|
||||
\end{align*}
|
||||
This holds since $M>0$ is true regardless of $x,y$. Thus we have that by the definition of a Lipschitz condition, $f$ satisfies a Lipshitz condition on $I$.\\
|
||||
\end{proof}
|
||||
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 1 (b) %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item Suppose $f$ and $g$ are differentiable functions on $(a,b)$. Show that between two consecutive roots of $f$ there exists a root $f'+fg'$. (Hint: Apply Rolle's Theorem to the function $h(x)=f(x)e^{g(x)}$) \\
|
||||
|
||||
\begin{proof}
|
||||
Let $x_1,x_2 \in (a,b)$ with $x_1<x_2$, and $f(x_1)=f(x_2)=0$.\\
|
||||
|
||||
Let $h(x)=f(x)^{g(x)}$.\\
|
||||
|
||||
Then we have $h(x_1)=h(x_2)=0$ and $h$ is differentiable on $(a,b)$, and $h$ is continuous on $[x_1,x_2]$.\\
|
||||
|
||||
Recall \textit{Rolle's Theorem}:
|
||||
\begin{theorem*}[\textbf{Rolle's Theorem}]
|
||||
Suppose that $f$ is continuous on a closed interval $I:= [a,b]$, that the derivative $f'$ exists at every point of the open interval $(a,b)$, and that $f(a)=f(b)=0$. Then there exists at least one point $c$ in $(a,b)$ such that $f'(c)=0$.
|
||||
\end{theorem*}
|
||||
Thus by \textit{Rolle's Theorem}, we know that there exists some $c \in (x_1,x_2) \st h'(c)=0$ and
|
||||
\[h'(x)=f'(x)e^{g(x)}+f(x)\cdot g'(x)e^{g(x)}=[f'(x)+f(x)g'(x)]e^{g(x)}\]
|
||||
So $h'(c)=[f'(c)+f(c)g'(c)]\cdot e^{g(c)}=0$ \\
|
||||
|
||||
So $f'(c)+f(c)g'(c)=0$.
|
||||
\end{proof}
|
||||
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 1 (c) %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item Suppose that $f$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $f(a)=f(b)=0$. Prove that for each real number $\alpha$, there exists some $c \in (a,b)$ such that $f'(c)=\alpha f(c)$. (Hint: Apply Rolle's Theorem to the function $g(x)=e^{-\alpha x} f(x)$)
|
||||
|
||||
\begin{proof}
|
||||
Consider the function $g(x)=e^{-\alpha x} f(x)$. Since $f(a)=f(b)=0$, we know that $g(a)=e^{-\alpha(a)}f(a)=e^{-\alpha(b)}f(b)=g(b)$, and thus $g(a)= e^{-\alpha(a)}\cdot 0 = 0 = g(b)$. By \textit{Rolle's Theorem}, we know that since $f(a)=f(b)=0$, $f$ is continuous on $[a,b]$, and since $f$ is differentiable on $(a,b)$, then there exists some $c \in (a,b)$, such that $f'(c)=0$. Thus there exists one point $c \in (a,b)$, such that $g'(c)=0$, and
|
||||
\begin{align*}
|
||||
g'(x)&=-\alpha e^{-\alpha x}f(x)+e^{-\alpha x}f'(x) \\
|
||||
0 &= e^{-\alpha x} \left(-\alpha f(x)+f'(x)\right) \\
|
||||
0 &= -\alpha f(x)+f'(x) \\
|
||||
\alpha f(x) &= f'(x)
|
||||
\end{align*}
|
||||
Thus we have that there exists some $c \in (a,b)$ such that $f'(c) = \alpha f(c)$.
|
||||
\end{proof}
|
||||
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 1 (d) %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item Suppose that $f$ is differentiable on $(a,b)$ and $f'$ is bounded on $(a,b)$. Show that $f$ is uniformly continuous.
|
||||
|
||||
\begin{proof}
|
||||
Let $M>0$ satisfy $|f'(x)| \leq M\ \forall\ x \in (a,b)$. Recall the definition of differntiability:
|
||||
\theoremstyle{definition}
|
||||
\begin{definition*}
|
||||
Let $I \subseteq \R$ be an interval, let $f:I \rightarrow \R$, and let $ c \in I$. We say that a real number $L$ is the \textbf{derivative of $f$ at $c$} if given any $\varepsilon > 0$ there exists $\delta (\varepsilon) > 0$ such that if $x \in I$ satisfies $0 < |x-c|<\delta (\varepsilon)$, then
|
||||
\[\abs{\frac{f(x)-f(c)}{x-c}-L}<\varepsilon.\]
|
||||
In this case we say that $f$ is \textbf{differentiable} at $c$, and we write $f'(c)$ for $L$. In other words, the derivative of $f$ at $c$ is given by the limit
|
||||
\[f'(c) = \lim\limits_{x\to c} \frac{f(x)-f(c)}{x-c}\]
|
||||
provided this limit exists. (We allow the possibility that $c$ may be the endpoint of the interval.)
|
||||
\end{definition*}
|
||||
Thus since $f$ is differentiable on $(a,b)$, we know that given any $\varepsilon>0$, there exits $\delta(\varepsilon) >0 \st$ if $x \in (a,b)$ satisfies $0<|x-c|<\delta(\varepsilon)$, then $\lim\limits_{x \to c} \frac{f(x)-f(c)}{x-c} = f'(c)$, for any $c \in (a,b)$.\\
|
||||
|
||||
By result of $1\ (a)$, we know that since $f$ is differentiable on $(a,b)$, and since $f'$ is bounded on $(a,b)$, then $f'$ satisfies a Lipschitz condition on $(a,b)$.\\
|
||||
|
||||
Recall \textit{Theorem 5.4.3}:
|
||||
\begin{theorem*}
|
||||
If $f:A \rightarrow \R$ is a Lipschitz function, then $f$ is uniformly continuous on $A$.
|
||||
\end{theorem*}
|
||||
Thus we have that by \textit{Theorem 5.4.3}, since $f$ is a Lipschitz function, $f$ is uniformly continuous on $(a,b)$.
|
||||
\end{proof}
|
||||
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 1 (e) %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item Give an example of a function $f$ that is differentiable, uniformly continuous on $(a,b)$, but $f'$ is not bounded.\\
|
||||
|
||||
Consider the function $f:(0,\infty) \to (-\infty,\infty)$ given by $f(x)=x\sin\left(\frac{1}{x}\right)$. First, we must show that $f$ is uniformly continuous, as follows:
|
||||
\begin{align*}
|
||||
|f(x)-f(c)| &= |x\sin\left(\frac{1}{x}\right) - c\sin\left(\frac{1}{c}\right)| \\
|
||||
&\leq |x-c| &\because\ \max \abs{\sin\left(\frac{1}{x}\right)}=1,\text{ and } \max\abs{\sin\left(\frac{1}{c}\right)} = 1 \\
|
||||
&<\varepsilon
|
||||
\end{align*}
|
||||
Thus let $\delta(\varepsilon)=\varepsilon$. Hence $f$ is uniformly continuous on $(-\infty, \infty)$. We also know that $f$ is differentiable on $(-\infty,\infty)$ since
|
||||
\[f'(x)=1\cdot\sin\left(\frac{1}{x}\right)+x\cdot\cos\left(\frac{1}{x}\right)\cdot -\frac{1}{x^2}= \sin\left(\frac{1}{x}\right)-\frac{x}{x^2}\cos\left(\frac{1}{x}\right)=\sin\left(\frac{1}{x}\right)-\frac{1}{x}\cos\left(\frac{1}{x}\right)\]
|
||||
However, since the maximum value of $\sin$ and $\cos$ is 1, and since $x \in (1,\infty)$, the derivative $f'(x)$ is unbounded, since $\sup \{(0,\infty)\} =\infty$, and $\max\{(0,\infty)\}=$ DNE, and $\lim\limits_{x \to 0^+} \frac{1}{x}=\infty$. Thus $\frac{1}{x}$ can be infinitely large, meaning $f'(x)$ is unbounded.\\
|
||||
\end{enumerate}
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 2 %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item Prove the given inequalities.
|
||||
\begin{enumerate}
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 2 (a) %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item $1+x \leq e^x$ for all $x \in \R$.
|
||||
|
||||
\begin{proof}
|
||||
Let $f:(0, \infty) \to \R$ given by $f(x):=e^x$. Then we know that $f'(x)=e^x$, which is greater than or equal to 1 since $f$ is defined on $\R^+$. Then we know by the \textit{Mean Value Theorem} that $\exists\ c \in (0,\infty) \st f'(c)=\frac{f(x)-f(0)}{x-0}=\frac{e^x-1}{x} \implies e^x-1 = f'(c) \cdot x \geq x$. Since $e^x \geq 1\ \forall\ x \in (0,\infty)$, we have that $e^x \geq x+1\ \forall\ x \in (0,\infty)$.\\
|
||||
|
||||
For the case where $f:(-\infty,0) \to \R$, we have that $f'(x)=e^x \leq 1\ \forall\ x \in (-\infty, 0)$. Now, by the \textit{Mean Value Theorem}, we know that $\exists\ c \in (-\infty, 0) \st f'(c)=\frac{f(0)-f(x)}{0-x} = \frac{1-e^x}{-x}$. Thus we have that $1-e^x = f'(c) \cdot -x \leq 1$, which implies that $e^x-1=x\cdot-f'(c) \geq -1$. Thus $1+ x \leq e^x$.\\
|
||||
\end{proof}
|
||||
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 2 (b) %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item $2x+0.7<e^x$ for all $x \geq 1$. \\
|
||||
|
||||
\begin{proof}
|
||||
Let $f(x)=e^x-2x-0.7$. Then $f'(x)=e^x-2>0\ \forall\ x \geq 1$. So $f$ is increasing on $[1,\infty)$. In particular, $f(x)=e^x-2x-0.7 \geq f(1) = e-2.7 > 0$.\\
|
||||
|
||||
So $e^x>2x+0.7$.
|
||||
\end{proof}
|
||||
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 2 (c) %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item $x^e \leq e^x$ for all $x > 0$.\\
|
||||
|
||||
\begin{proof}
|
||||
Let $f(x)=x^{\frac{1}{x}}=y$. Then we have
|
||||
\begin{align*}
|
||||
\ln(y) &= \frac{\ln(x)}{x} \\
|
||||
\frac{1}{y}\cdot y' &= \frac{x\left(\frac{1}{x}-\ln(x)\right)}{x^2} \\
|
||||
&=\frac{1-\ln(x)}{x^2} \\
|
||||
&\Downarrow \\
|
||||
y' &= \frac{y(1-\ln(x))}{x^2} \\
|
||||
&= \frac{x^{\frac{1}{x}}(1-\ln(x))}{x^2}
|
||||
\end{align*}
|
||||
We note that $x=e$ is a critical point. Thus by the first derivative test, we know that $f$ is increasing at $e$.\\
|
||||
|
||||
So $f(x)=x^{\frac{1}{x}} \leq e^\frac{1}{x}\ \forall\ x \in (0,\infty)$. Thus
|
||||
\begin{align*}
|
||||
\frac{ln(x)}{x} &\leq \frac{1}{e} \\
|
||||
\ln(x) &\leq \frac{x}{e} \\
|
||||
&(x \leq e^\frac{x}{e})^e \\
|
||||
x^e &\leq e^x\ \forall\ x >0
|
||||
\end{align*}
|
||||
\end{proof}
|
||||
\end{enumerate}
|
||||
\textbf{Section 6.3 - L'Hôpital's Rule}
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 3 Section 6.3 %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item \textbf{Section 6.3}
|
||||
\begin{enumerate}
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Section 6.3 - 6 (a) %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item[6.] \textbf{(a)} Evaluate $\lim\limits_{x \to 0} \frac{e^x+e^{-x}-2}{1-\cos x}$\\
|
||||
|
||||
\begin{align*}
|
||||
\lim\limits_{x \to 0} \frac{e^x+e^{-x}-2}{1-\cos (x)} &= \frac{e^0+e^{-0}-2}{1-\cos(0)} \\
|
||||
&= \frac{1+1-2}{1-1} \\
|
||||
&= \frac{0}{0}
|
||||
\end{align*}
|
||||
We note that this is in one of the indeterminate forms that L'Hopital's Rule accounts for.\\
|
||||
|
||||
Recall L'Hopital's rules:
|
||||
\begin{theorem*}[\textbf{L'Hopital's Rule, I}]
|
||||
Let $-\infty \leq a < b \leq \infty$ and let $f,g$ be differentiable on $(a,b)$ such that $g'(x) \neq 0$ for all $x \in (a,b)$. Suppose that
|
||||
\[\lim\limits_{x\to a+} f(x) = 0 = \lim\limits_{x\to a+} g(x)\]
|
||||
|
||||
\begin{enumerate}
|
||||
\item If $\lim\limits_{x\to a+} \frac{f'(x)}{g'(x)}=L \in \R$, then $\lim\limits_{x\to a+} \frac{f(x)}{g(x)}=L$.
|
||||
|
||||
\item If $\lim\limits_{x\to a+} \frac{f'(x)}{g'(x)}=L \in \{-\infty, \infty\}$, then $\lim\limits_{x\to a+} \frac{f(x)}{g(x)}=L$.
|
||||
\end{enumerate}
|
||||
\end{theorem*}
|
||||
|
||||
\begin{theorem*}[\textbf{L'Hopital's Rule, II}]
|
||||
Let $-\infty \leq a < b \leq \infty$ and let $f,g$ be differentiable on $(a,b)$ such that $g'(x) \neq 0$ for all $x \in (a,b)$. Suppose that
|
||||
\[\lim\limits_{x\to a+} g(x) = \pm \infty\]
|
||||
\begin{enumerate}
|
||||
\item If $\lim\limits_{x\to a+} \frac{f'(x)}{g'(x)}=L \in \R$, then $\lim\limits_{x\to a+} \frac{f(x)}{g(x)}=L$.
|
||||
|
||||
\item If $\lim\limits_{x\to a+} \frac{f'(x)}{g'(x)}=L \in \{-\infty, \infty\}$, then $\lim\limits_{x\to a+} \frac{f(x)}{g(x)}=L$.
|
||||
\end{enumerate}
|
||||
\end{theorem*}
|
||||
Thus by utilizing L'Hopital's rule, we have
|
||||
\begin{align*}
|
||||
\lim\limits_{x \to 0} \frac{e^x+e^{-x}-2}{1-\cos(x)} &= \lim\limits_{x \to 0^+} \frac{\frac{d}{dx} (e^x+e^{-x}-2)}{\frac{d}{dx}(1-\cos (x))} \\
|
||||
&= \lim\limits_{x \to 0^+} \frac{e^x-e^{-x}}{\sin (x)} \\
|
||||
&= \frac{e^0-e^{-0}}{\sin(0)} \\
|
||||
&= \frac{1-1}{0} \\
|
||||
&= \frac{0}{0}
|
||||
\end{align*}
|
||||
And since this is once again in one of the indeterminate forms that L'Hopital's rule accounts for, we perform the rule again, thus
|
||||
\begin{align*}
|
||||
\lim\limits_{x \to 0^+} \frac{e^x-e^{-x}}{\sin (x)} &= \lim\limits_{x \to 0^+} \frac{\frac{d}{dx} (e^x-e^{-x})}{\frac{d}{dx} \sin(x)} \\
|
||||
&= \lim\limits_{x \to 0^+} \frac{e^x+e^{-x}}{\cos (x)} \\
|
||||
&= \frac{e^0+e^{-0}}{\cos (0)} \\
|
||||
&= \frac{1+1}{1} \\
|
||||
&= \frac{2}{1} \\
|
||||
&= 2
|
||||
\end{align*}
|
||||
Thus we have
|
||||
\[\lim\limits_{x \to 0} \frac{e^x+e^{-x}-2}{1-\cos(x)}=\lim\limits_{x \to 0^+} \frac{e^x-e^{-x}}{\sin (x)} = \lim\limits_{x \to 0^+} \frac{e^x+e^{-x}}{\cos (x)}=2\]
|
||||
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Section 6.3 - 7 (a) %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item[7.] \textbf{(a)} Evaluate $\lim\limits_{x \to 0^+} \frac{\ln (x+1)}{\sin x}$ where the domain of the quotient is $(0, \pi/2)$.\\
|
||||
|
||||
\begin{align*}
|
||||
\lim\limits_{x \to 0^+} \frac{\ln (x+1)}{\sin (x)} &= \frac{\ln (0+1)}{\sin (0)} \\
|
||||
&= \frac{\ln (1)}{0} \\
|
||||
&= \frac{0}{0}
|
||||
\end{align*}
|
||||
Since this is now in one of the indeterminate forms that L'Hopital's rule accounts for, we apply the rule as follows:
|
||||
\begin{align*}
|
||||
\lim\limits_{x \to 0^+} \frac{\ln(x+1)}{\sin (x)} &= \lim\limits_{x \to 0^+} \frac{\frac{d}{dx} \ln(x+1)}{\frac{d}{dx} \sin(x)} \\
|
||||
&= \lim\limits_{x \to 0^+} \frac{\frac{1}{x+1} \cdot 1}{\cos (x)} \\
|
||||
&= \lim\limits_{x \to 0^+} \frac{\frac{1}{x+1}}{\cos(x)} \\
|
||||
&= \frac{\frac{1}{0+1}}{\cos(0)} \\
|
||||
&= \frac{1}{1} \\
|
||||
&= 1
|
||||
\end{align*}
|
||||
Thus we have that
|
||||
\[\lim\limits_{x \to 0^+} \frac{\ln(x+1)}{\sin(x)}=\lim\limits_{x \to 0^+} \frac{\frac{1}{x+1}}{\cos(x)}=1\]
|
||||
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Section 6.3 - 10 %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item[10.] Evaluate the following limits:
|
||||
\begin{enumerate}
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Section 6.3 - 10 (b) %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item[(b)] $\lim\limits_{x \to 0} (1+3/x)^x$ $(0,\infty)$ \\
|
||||
|
||||
\begin{align*}
|
||||
\lim\limits_{x \to 0} \left(1+\frac{3}{x}\right)^x &= \left(1+\frac{3}{0}\right)^0 \\
|
||||
&= \left(1+\infty\right)^0 \\
|
||||
&= \infty^0
|
||||
\end{align*}
|
||||
Since this is in indeterminate form, we know that we need to somehow get it into the form in which it can be solved using L'Hopital's rule; So,
|
||||
\begin{align*}
|
||||
\lim\limits_{x \to 0} \left(1+\frac{3}{x}\right)^x &=\lim\limits_{x \to 0} e^{\ln \left(\left(1+\frac{3}{x}\right)^x\right)} \\
|
||||
&= \lim\limits_{x \to 0} e^{x \cdot \ln \left(1+\frac{3}{x}\right)} \\
|
||||
&= e^{0 \cdot \ln \left(1+\frac{3}{0}\right)} \\
|
||||
&= 0 \cdot 0
|
||||
\end{align*}
|
||||
and thus we can now use L'Hopital's rule on the exponent; So
|
||||
\begin{align*}
|
||||
&\lim\limits_{x \to 0} \ln \left(1+\frac{3}{x}\right) \\
|
||||
&= \lim\limits_{x \to 0} \left[\frac{\ln(1+\frac{3}{x})}{\left(\frac{1}{x}\right)}\right] \\
|
||||
&= \lim\limits_{x \to 0} \left[\frac{\left(1+\frac{3}{x}\right)^{-1}\cdot \left(\frac{-3}{x^2}\right)}{\left(-\frac{1}{x^2}\right)}\right] \\
|
||||
&= \lim\limits_{x \to 0} \left[\frac{3}{1+\frac{3}{x}}\right] \\
|
||||
&= 0
|
||||
\end{align*}
|
||||
And thus
|
||||
\[\lim\limits_{x \to 0} \left(1+\frac{3}{x}\right)^x=\lim\limits_{x \to 0} \left[e^{x \ln \left(1+\frac{3}{x}\right)}\right] = e^0=1\]
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Section 6.3 - 10 (c) %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item[(c)] $\lim\limits_{x \to \infty} (1+3/x)^x$ $(0, \infty)$ \\
|
||||
|
||||
We know from the previous problem that this will be in the indeterminate form of $1^\infty$, and thus we apply the same logic from step 2, we know that
|
||||
\[\lim\limits_{x \to \infty} \left(1+\frac{3}{x}\right)^x = \lim\limits_{x \to 0} \left[e^{x \cdot \ln \left(1+\frac{3}{x}\right)}\right]\]
|
||||
So
|
||||
\begin{align*}
|
||||
&\lim\limits_{x \to \infty} \ln \left(1+\frac{3}{x}\right) \\
|
||||
&= \lim\limits_{x \to \infty} \left[\frac{\ln(1+\frac{3}{x})}{\left(\frac{1}{x}\right)}\right] \\
|
||||
&= \lim\limits_{x \to \infty} \left[\frac{\left(1+\frac{3}{x}\right)^{-1}\cdot \left(\frac{-3}{x^2}\right)}{\left(-\frac{1}{x^2}\right)}\right] \\
|
||||
&= \lim\limits_{x \to \infty} \left[\frac{3}{1+\frac{1}{x}}\right] \\
|
||||
&= \frac{3}{1} \\
|
||||
&= 3
|
||||
\end{align*}
|
||||
and thus
|
||||
\[\lim\limits_{x \to \infty} \left(1+\frac{3}{x}\right)^x = \lim\limits_{x \to \infty} \left[e^{x \cdot \ln \left(1+\frac{3}{x}\right)}\right]=e^3\]
|
||||
\end{enumerate}
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Section 6.3 - 11 %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item[11.] Evaluate the following limits:
|
||||
\begin{enumerate}
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Section 6.3 - 11 (b) %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item[(b)] $\lim\limits_{x \to 0^+} (\sin x)^x$ $(0, \pi)$\\
|
||||
|
||||
We first note that
|
||||
\[\lim\limits_{x \to 0^+} (\sin (x))^x=(\sin(0))^0 = 0^0\]
|
||||
So we know that we need to get this into a form that can utilize L'Hopital's rule, so we consider
|
||||
\[(\sin(x))^x=e^{\ln((\sin(x))^x)}=e^{x\cdot\ln(\sin(x))}\]
|
||||
This yields an indeterminate form of $0 \cdot -\infty$, thus we can use L'Hopital's rule on the exponent; thus
|
||||
\begin{align*}
|
||||
&\lim\limits_{x \to 0^+} \left[\frac{\ln(\sin(x))}{\frac{1}{x}}\right] \\
|
||||
&= \lim\limits_{x \to 0^+} \left[\frac{\frac{1}{\sin(x)}\cdot \cos(x)}{-\frac{1}{x^2}}\right] \\
|
||||
&= \lim\limits_{x \to 0^+} \left[\frac{-x^2}{\tan(x)}\right] \\
|
||||
&= \lim\limits_{x \to 0^+} \left[\frac{-2x}{\sec^2(x)}\right] \\
|
||||
&= \frac{0}{1} \\
|
||||
&= 0
|
||||
\end{align*}
|
||||
Which thus yields
|
||||
\[\lim\limits_{x \to 0} (\sin(x))^x=\lim\limits_{x \to 0^+} e^{x \cdot \ln(\sin(x))}= e^0=1\]
|
||||
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Section 6.3 - 11 (c) %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item[(c)] $\lim\limits_{x \to 0^+} x^{\sin x}$ $(0, \infty)$\\
|
||||
|
||||
We first note that
|
||||
\[\lim\limits_{x \to 0} x^{\sin(x)}=0^{\sin(0)}=0^0\]
|
||||
which is in indeterminate form, thus we know that we need to somehow get this into an indeterminate form that can utilize L'Hopital's rule. Thus we consider
|
||||
\[x^{\sin(x)}=e^{\ln\left(x^{\sin(x)}\right)}=e^{\sin(x)\cdot\ln(x)}\]
|
||||
Thus we can use L'Hopital's rule on the exponent as follows:
|
||||
\begin{align*}
|
||||
&\lim\limits_{x \to 0^+} \left[\frac{\ln(x)}{\csc(x)}\right] \\
|
||||
&=\lim\limits_{x \to 0^+} \left[\frac{\frac{1}{x}}{-\csc(x)\cot(x)}\right] \\
|
||||
&= \lim\limits_{x \to 0^+} \left[\frac{-\cos(x)\tan(x)-\sin(x)\sec^2(x)}{1}\right] \\
|
||||
&= 0
|
||||
\end{align*}
|
||||
And thus we have that
|
||||
\[\lim\limits_{x \to 0^+} x^{\sin(x)}=\lim\limits_{x \to 0^+} e^{\sin(x)\ln(x)}=e^0=1\]
|
||||
\end{enumerate}
|
||||
\end{enumerate}
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 4 %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item Suppose that the function $f$ is twice differentiable.
|
||||
\begin{enumerate}
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 4 (a) %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item Prove $f''(x)=\lim\limits_{h \to 0} \frac{f(x+3h)-3f(x+h)+2f(x)}{3h^2}$
|
||||
|
||||
\begin{proof}
|
||||
\begin{align*}
|
||||
\lim\limits_{h \to 0} \frac{f(x+3h)-3f(x+h)+2f(x)}{3h^2} &= \frac{f(x+3(0))-3f(x+0)+2f(x)}{3(0)^2} \\
|
||||
&= \frac{f(x)-3f(x)+2f(x)}{0} \\
|
||||
&= \frac{0}{0} \implies \text{ Use L'Hopital's Rule} \\
|
||||
&\Downarrow \\
|
||||
&= \lim\limits_{h \to 0} \frac{3f'(x+3h)-3f'(x+h)}{6h} \\
|
||||
&= \frac{3f'(x+3(0))-3f'(x+0)}{6(0)} \\
|
||||
&= \frac{3f'(x)-3f'(x)}{0} \\
|
||||
&= \frac{0}{0} \implies \text{ Use L'Hopital's Rule} \\
|
||||
&\Downarrow \\
|
||||
&= \lim\limits_{h \to 0} \frac{9f''(x+3h)-3f''(x+h)}{6} \\
|
||||
&= \frac{9f''(x+3(0))-3f''(x+(0))}{6} \\
|
||||
&= \frac{9f''(x)-3f''(x)}{6} \\
|
||||
&= \frac{6f''(x)}{6} \\
|
||||
&= f''(x)
|
||||
\end{align*}
|
||||
Thus $f''(x)=\lim\limits_{h \to 0} \frac{f(x+3h)-3f(x+h)+2f(x)}{3h^2}$
|
||||
\end{proof}
|
||||
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 4 (b) %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item Prove $f''(x)=\lim\limits_{h \to 0} \frac{2f(x+3h)-3f(x+2h)+f(x)}{3h^2}$ \\
|
||||
|
||||
\begin{align*}
|
||||
\lim\limits_{h \to 0} \frac{2f(x+3h)-3f(x+2h)+f(x)}{3h^2} &= \frac{2f(x+3(0))-3f(x+2(0))+f(x)}{3(0)^2} \\
|
||||
&= \frac{2f(x)-3f(x)+f(x)}{0} \\
|
||||
&= \frac{0}{0} \implies \text{ Use L'Hopital's Rule} \\
|
||||
&\Downarrow \\
|
||||
&= \lim\limits_{h \to 0} \frac{6f'(x+3h)-6f'(x+2h)}{6h} \\
|
||||
&= \lim\limits_{h \to 0} \frac{f'(x+3h)-f'(x+2h)}{h} \\
|
||||
&= \frac{f'(x+3(0))-f'(x+2(0))}{0} \\
|
||||
&= \frac{f'(x)-f'(x)}{0} \\
|
||||
&= \frac{0}{0} \implies \text{ Use L'Hopital's Rule} \\
|
||||
&\Downarrow \\
|
||||
&= \lim\limits_{h \to 0} \frac{3f''(x+3h)-2f''(x+2h)}{1} \\
|
||||
&= 3f''(x+3(0))-2f''(x+2(0)) \\
|
||||
&= 3f''(x)-2f''(x) \\
|
||||
&= f''(x)
|
||||
\end{align*}
|
||||
Thus $f''(x)=\lim\limits_{h \to 0} \frac{2f(x+3h)-3f(x+2h)+f(x)}{3h^2}$
|
||||
\end{enumerate}
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 5 %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item Prove, if true or provide a counterexample, if false.
|
||||
\begin{enumerate}
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 5 (a) %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item If $f$ and $g$ are increasing on $[a,b]$, then $f+g$ is increasing on $[a,b]$.\\
|
||||
|
||||
This is a true statement.
|
||||
\begin{proof}
|
||||
Recall \textit{Theorem 6.2.5}:
|
||||
\begin{theorem*}
|
||||
Let $f:I \rightarrow \R$ be differentiable on the interval $I$. Then:
|
||||
\begin{enumerate}
|
||||
\item $f$ is increasing on $I$ if and only if $f'(x) \geq 0$ for all $x \in I$.
|
||||
\item $f$ is decreasing on $I$ if and only if $f'(x) \leq 0$ for all $x \in I$.
|
||||
\end{enumerate}
|
||||
\end{theorem*}
|
||||
So we know that $f'(x) \geq 0\ \forall\ x \in [a,b]$, and $g'(x) \geq 0\ \forall\ x \in [a,b]$. Thus $f'(x)+g'(x)\geq 0\ \forall\ x \in [a,b]$, which implies that $f+g$ is increasing, by \textit{Theorem 6.2.5}.\\
|
||||
\end{proof}
|
||||
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 5 (b) %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item If $f$ and $g$ are increasing on $[a,b]$, then $fg$ is increasing on $[a,b]$.\\
|
||||
|
||||
This is a false statement. Consider $f,g:[-6,-4] \to \R$ given by $f(x)=2x$, and $g(x)=3x$. Then both $f'(x)$ and $g'(x)$ are greater than 0 for any $x \in [-6,-4]$, since $f'(x)=2 \geq 0$ and $g'(x)=3 \geq 0$. However, their product $fg=2x\cdot3x=6x^2$, yet $fg'(x) \ngeq 0\ \forall\ x \in [-6,-4]$, since $fg'(x)=12x$, and $fg'(-6)=12(-6)=-72 \ngeq 0$. Thus $fg$ is not increasing.\\
|
||||
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 5 (c) %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item If $f$ and $g$ are differentiable on $[a,b]$ and $|f'(x)| \leq 1 \leq |g'(x)|$ for all $x \in (a,b)$, then $|f(x)-f(a)| \leq |g(x)-g(a)|$ for all $x \in [a,b]$.\\
|
||||
|
||||
This is a true statement.
|
||||
\begin{proof}
|
||||
There's two cases that we must consider: $f'(x)=g'(x)$, and $f'(x)\neq g'(x)$. \\
|
||||
\begin{case}
|
||||
Let $f'(x)=g'(x)$ such that $|f'(x)|=1=|g'(x)|$. Recall \textit{Corollary 6.2.2}:
|
||||
\begin{corollary*}
|
||||
Suppose that $f$ and $g$ are continuous on $I:=[a,b]$, that they are differentiable on $(a,b)$, and that $f'(x)=g'(x)$ for all $x \in (a,b)$. Then there exists a constant $C$ such that $f=g+C$ on $I$.
|
||||
\end{corollary*}
|
||||
Thus we have that $|f'(x)| \leq 1 \leq |g'(x)|$, and by \textit{Corollary 6.2.2}, we know that there exists some constant $C$ such that $g=f+C$. Thus
|
||||
\begin{align*}
|
||||
|f(x)-f(a)| &\leq |g(x)-g(a)| \\
|
||||
&\leq |(f(x)+C)-(f(a)+C)| &\text{By \textit{Corollary 6.2.2}} \\
|
||||
&=|f(x)-f(a)|
|
||||
\end{align*}
|
||||
Thus if $f'(x)=g'(x)$, and $|f'(x)| \leq 1 \leq |g'(x)|\ \forall\ x \in (a,b)$, then $|f(x)-f(a)| \leq |g(x)-g(a)|$ for all $x \in [a,b]$.
|
||||
\end{case}
|
||||
\begin{case}
|
||||
Now, let $f'(x) \neq g'(x)$ such that $|f'(x)| < 1 < |g'(x)|$. Since $f$ is differentiable on $[a,b]$, we know that the \textit{Mean Value Theorem} holds true. Thus we have
|
||||
\begin{align*}
|
||||
|f'(x)|&<1 \\
|
||||
\abs{\frac{f(x)-f(c)}{x-c}} &<1 \\
|
||||
\frac{|f(x)-f(c)|}{|x-c|} &< 1 \\
|
||||
|f(x)-f(c)|&<|x-c|
|
||||
\end{align*}
|
||||
and
|
||||
\begin{align*}
|
||||
1 &< |g'(x)| \\
|
||||
1 &< \abs{\frac{g(x)-g(c)}{x-c}} \\
|
||||
1 &< \frac{|g(x)-g(c)}{|x-c|} \\
|
||||
|x-c| &< |g(x)-g(c)|
|
||||
\end{align*}
|
||||
Let $c=a$; then we have that
|
||||
\[|f(x)-f(a)|<|x-a|<|g(x)-g(a)|\]
|
||||
and thus
|
||||
\[|f(x)-f(a)|<|g(x)-g(a)|\]
|
||||
for all $x \in [a,b]$.\\
|
||||
|
||||
Hence we have that if $f'(x) \neq g'(x)$ satisfying $|f'(x)|<1<|g'(x)|$ for all $x \in (a,b)$, then $|f(x)-f(a)|<|g(x)-g(a)|$ for all $x \in [a,b]$.
|
||||
\end{case}
|
||||
Since both cases account for all possible outcomes of functions $f$ and $g$ whose derivatives satisfy the inequality $|f'(x)|\leq 1 \leq |g'(x)|\ \forall\ x \in (a,b)$, we have that this statement holds true.
|
||||
\end{proof}
|
||||
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 5 (d) %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item A continuous function defined on a bounded interval assumes its maximum and minimum values.\\
|
||||
|
||||
This is a false statement. Consider the function $f:(-2,2) \to (-6,6)$ given by $f(x)=3x$. Then we have that $f$ is bounded on $(-2,2)$, since $|f(x)|< 2\ \forall\ x \in (-2,2)$. However, notice that $(f(-2),f(2))=(-6,6)$ is a bounded interval, yet $\min\{(-6,6)\}=$ DNE, and $\max \{(-6,6)\}=$ DNE, but $\inf \{(-6,6)\}=-6$, and $\sup \{(-6,6)\}=6$. Thus $f$ does not assume its minimum and maximum values since the minimum and maximum values do not exist.\\
|
||||
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 5 (e) %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item If $f$ is continuous on $[a,b]$, then there exists a point $c \in (a,b)$ such that $f'(c)=\frac{f(b)-f(a)}{b-a}$.\\
|
||||
|
||||
This is a false statement. Consider the function $f:\R \to (-1,1)$ given by $f(x):=\sin\left(\frac{1}{x}\right)$. We know that $f$ is not uniformly continuous, which thus means that $f$ cannot be differentiable, hence it does not satisfy the \textit{Mean Value Theorem}.\\
|
||||
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 5 (f) %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item Suppose $f$ is differentiable on $(a,b)$. If $c \in (a,b)$ and $f'(c)=0$, then $f(c)$ is either the maximum or the minimum value of $f$ on $(a,b)$.\\
|
||||
|
||||
This is false. Consider the function $f:(-\infty,\infty) \to \R$ given by $f(x)=x^3$. Then we note that the derivative of $x^3 = 3x^2$. If we let $3x^2=0$, then we have that $x=0$. However, the slope at $x=0$ is $0$, and thus we have that $x$ is not a maximum or a minimum of $f$.\\
|
||||
\end{enumerate}
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 6 %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item \textbf{Not collected} The following is an outline of a proof that $e$ is irrational:
|
||||
\begin{enumerate}
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 6 (a) %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item Show that $f(x)=e^x$ is strictly increasing on $\R$.
|
||||
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 6 (b) %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item Use Taylor's Theorem about $x=0$ and the estimate $e < 3$ to show for all $n \in \N$,
|
||||
\[0<e-\left(1+1+\frac{1}{2!}+\frac{1}{3!}+ \dots + \frac{1}{n!}\right)< \frac{3}{(n+1)!}\]
|
||||
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 6 (c) %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item Suppose that $e$ is rational. Then $e=a/b$ for some $a,b \in \N$. Choose $n > \max \{b,3\}$. Substitute into part b) and show that this leads to the existence of an integer between 0 and 3/4.
|
||||
\end{enumerate}
|
||||
\end{enumerate}
|
||||
\end{document}
|
||||
Binary file not shown.
+208
@@ -0,0 +1,208 @@
|
||||
\documentclass[12pt,letterpaper]{article}
|
||||
\usepackage{amsthm}
|
||||
\usepackage[latin1]{inputenc}
|
||||
\usepackage{amsmath}
|
||||
\usepackage{amsfonts}
|
||||
\usepackage{amssymb}
|
||||
\usepackage{graphicx}
|
||||
\usepackage[left=2cm, right=2.5cm, top=2.5cm, bottom=2.5cm]{geometry}
|
||||
\newtheoremstyle{case}{}{}{}{}{}{:}{ }{}
|
||||
\theoremstyle{case}
|
||||
\newtheorem{case}{Case}
|
||||
\renewcommand{\qedsymbol}{$\blacksquare$}
|
||||
\author{Alexander J. Tusa}
|
||||
\title{Real Analysis Homework 3}
|
||||
\begin{document}
|
||||
\maketitle
|
||||
\begin{enumerate}
|
||||
\item Find the infimum and supremum, if they exist.
|
||||
\begin{enumerate}
|
||||
\item Section 2.3
|
||||
\begin{enumerate}
|
||||
\item[4)] Let $S_4 := \{1-\frac{(-1)^n}{n}: n \in \mathbb{N}\}$.
|
||||
\\$\inf S_4 = \frac{1}{2}$, $\sup S_4 = 2$
|
||||
\item[5)]
|
||||
\begin{enumerate}
|
||||
\item[a)]
|
||||
\begin{align*}
|
||||
A :&= \{x \in \mathbb{R}: 2x + 5 > 0\}
|
||||
\\ &= \{x \in \mathbb{R}: 2x > -5\}
|
||||
\\ &= \{x \in \mathbb{R}: x > \frac{-5}{2}\}
|
||||
\end{align*}
|
||||
So $\inf A$ exists. So $\inf A = \frac{-5}{2}$. But since $\nexists$ an upper bound or the upper bound of $A = \infty$, then either $\sup A = \infty$, or $\sup A = DNE$.
|
||||
\item[b)]
|
||||
\begin{align*}
|
||||
B :&= \{x \in \mathbb{R}: x + 2 \geq x^2\}
|
||||
\\ &= \{x \in \mathbb{R}: 0 \geq x^2-x-2\}
|
||||
\\ &= \{x \in \mathbb{R}: 0 \geq (x-2)(x+1)\}
|
||||
\\ &= [-1, 0]\cup[0,2]
|
||||
\end{align*}
|
||||
So the infimum and supremum exist. So $\inf B = -1$, and $\sup B = 2$.
|
||||
\item[d)]
|
||||
\begin{align*}
|
||||
D :&= \{x \in \mathbb{R}: x ^2-2x-5 < 0\}
|
||||
\\ &= \{x \in \mathbb{R}: (x-(1 + \sqrt{6}))(x-(1-\sqrt{6}))\}
|
||||
\\ &= \{x \in \mathbb{R}: 1 -\sqrt{6} < x < 1 + \sqrt{6}\}
|
||||
\\ &= (1-\sqrt{6}, 1+\sqrt{6})
|
||||
\end{align*}
|
||||
So both the $\inf D$ and $\sup D$ exist. So $\inf D = 1 - \sqrt{6}$ and $\sup D = 1 + \sqrt{6}$.
|
||||
\end{enumerate}
|
||||
\end{enumerate}
|
||||
\item $A=\{x \in \mathbb{R}: x = \frac{1}{n} + (-1)^n$ for $n \in \mathbb{N}\}$
|
||||
\\ $\Rightarrow \inf A = -1$, and $\sup A = \frac{3}{2}$.
|
||||
\item $B = \{x \in \mathbb{R}: x = 2 - \dfrac{(-1)^n}{n^2}$ for $n \in \mathbb{N}\}$
|
||||
\\ $\Rightarrow \inf B = \frac{7}{4}$, $\sup B = 3$
|
||||
\end{enumerate}
|
||||
\item Section 2.3
|
||||
\begin{enumerate}
|
||||
\item[9)] Let $S \subseteq \mathbb{R}$ be nonempty. Show that if $u = \sup S$, then for every number $n \in \mathbb{N}$, the number $\frac{u-1}{n}$ is not an upper bound of $S$, but the number $\frac{u+1}{n}$ is an upper bound of $S$. (The converse is also true; see Exercise 2.4.3)
|
||||
\begin{proof}
|
||||
Let $S \subseteq \mathbb{R}$ be nonempty. We want to show that if $u = \sup S$, then for every number $n \in \mathbb{N}$, the number $\frac{u-1}{n}$ is not an upper bound of $S$, but the number $\frac{u+1}{n}$ is an upper bound of $S$.
|
||||
\\\\Let $u = \sup S$. Recall the definition of the supremum:
|
||||
$$\alpha = \sup S \iff (i)\ x \leq \alpha\ \forall x \in S,\ \wedge\ (ii)\ \forall \epsilon \in S, \exists x \in S\ \text{s.t.}\ x > \alpha - \epsilon$$
|
||||
$u$ is by definition an upper bound of $S$, and thus by the definition of $u$, $u+ \frac{1}{n}>u$, thus $u +\frac{1}{n}$ is also an upper bound of $S$, since $u+\frac{1}{n} > u\ \forall n \in \mathbb{N}$.
|
||||
\\\\ Now, let $\epsilon = \frac{1}{n}$. By Lemma 2.3.4, we have that $\exists s_\epsilon \in S\ \text{s.t.}\ \sup S - \epsilon < s_\epsilon < \sup S$, so
|
||||
$$u-\frac{1}{n} = u-\epsilon < s_\epsilon$$
|
||||
$\therefore u-\frac{1}{n}$ is not an upper bound of $S$.
|
||||
\end{proof}
|
||||
\item[10)] Show that if $A$ and $B$ are bounded subsets of $\mathbb{R}$, then $A \cup B$ is a bounded set. Show that $\sup(A \cup B) = \sup \{\sup A, \sup B\}$.
|
||||
\begin{proof}
|
||||
Let $A, B \subseteq \mathbb{R}$ such that $A, B$ are bounded. We want to show that $A \cup B$ is a bounded set, and that $\sup(A \cup B) = \sup\{\sup A, \sup B\}$.
|
||||
\\\\Since $A$ is bounded, we have that $$\inf A \leq A \leq \sup A,$$ and since $B$ is bounded, we have that $$\inf B \leq B \leq \sup B$$
|
||||
Let $s = \max\{|\inf A|, |\sup A|\}$, and let $t = \max\{|\inf B|, |\sup B|\}$. Let $x \in A \cup B$. Then, by the definition of union, $x \in A$ or $x \in B$.
|
||||
\\\\If $x \in A$, then $|x| \leq s$.
|
||||
\\If $x \in B$, then $|x| \leq t$.
|
||||
\\\\Let $r = \max \{s,t\}$.
|
||||
\\\\Then if $x \in A \cup B$, then $|x| \leq r$.
|
||||
\\\\ $\therefore A \cup B$ is bounded. $\square$
|
||||
\\\\Now, we want to show that $$\sup (A \cup B) = \sup \{\sup A, \sup B\}$$
|
||||
Since $A$ is bounded, $\sup A$ exists by the completeness axiom. Since $B$ is bounded, $\sup B$ exists by the completeness axiom.
|
||||
\\\\Let $w = \sup \{\sup A, \sup B\} = \max \{\sup A, \sup B\}$. Then $w$ is an upper bound for $A \cup B$ since $w \geq |\sup A|$ and $w \geq |\sup B|$. By completeness, $\sup(A \cup B)$ exists. And $\sup (A \cup B) \leq w = \sup \{\sup A, \sup B\}$.
|
||||
\\\\Let $z$ be any upper bound for $A \cup B$. Then $z$ is an upper bound for $A$ and for $B$. So $x \leq \sup A \leq z,\ \forall a \in A$ and $x \leq \sup B \leq z,\ \forall b \in B$. So $\sup \{\sup A, \sup B\} \leq z$.
|
||||
\\\\$\therefore z$ is an upper bound for $A \cup B$, choose $z = \sup (A \cup B)$. So $\sup \{\sup A, \sup B\} \leq \sup (A \cup B)$.
|
||||
\\\\Then $\sup\{\sup A, \sup B\} = \sup (A \cup B)$.
|
||||
\end{proof}
|
||||
\end{enumerate}
|
||||
\item Section 2.4\\
|
||||
\begin{enumerate}
|
||||
\item[4a)]
|
||||
\begin{proof}
|
||||
Let $S$ be a nonempty bounded set in $\mathbb{R}$. Let $a > 0$, and let $aS = \{as: s \in S\}$. We want to show that $$\inf (aS) = a\inf S,\ \text{and}\ \sup (aS) = a\sup S$$
|
||||
\begin{align*}
|
||||
\because \inf S &\leq s,\ \forall s \in S,
|
||||
\\ \Rightarrow a\inf S &\leq aS,\ \forall as \in aS
|
||||
\end{align*}
|
||||
For any $\epsilon > 0, \frac{\epsilon}{s} > 0$. Then we have that $\exists s_0 \in S\ \text{s.t.}\ s_0 \leq \inf S + \frac{\epsilon}{a} \Rightarrow as_0 \leq a\inf S + \epsilon$, where $as_0 \in aS$.
|
||||
\\\\$\therefore \inf (aS) = a\inf S$. $\square$
|
||||
\\\\Now, we want to show that $\sup (aS) = a\sup S$. By the definition of the supremum, we have that $s \leq \sup S,\ \forall s \in S \Rightarrow as \leq a\sup S,\ \forall as \in aS$. So for any $\epsilon > 0, \frac{\epsilon}{a} > 0$, we have that $\exists s' \in S\ \text{s.t.}\ s' = \sup S - \frac{\epsilon}{a}$.
|
||||
\\\\$\therefore \sup (aS) = a\sup S$.
|
||||
\end{proof}
|
||||
\item[5)] Let $S$ be a set of nonnegative real numbers that is bounded above, and let $T := \{x^2: x \in S\}$. Prove that if $u = \sup S$, then $u^2 = \sup T$. Give an example that shows the conclusion may be false if the restriction against negative numbers is removed.
|
||||
\begin{proof}
|
||||
Let $S$ be a set of nonnegative real numbers that is bounded above, and let $T:=\{x^2:x \in S\}$. We want to show that if $u = \sup S$, then $u^2 = \sup T$.
|
||||
\\\\Suppose $u = \sup S$. Then $s \leq u,\ \forall s \in S$.
|
||||
\\$\Rightarrow 0 \leq s \leq u$
|
||||
\\$\Rightarrow 0 \leq s^2 \leq u^2$, because if $a,b \geq 0\ \text{s.t.}\ a \leq b$, then $a^2 \leq b^2$.
|
||||
\\\\So $s^2 \leq u^2, \forall s \in S \Rightarrow t \leq u^2 \ \forall t \in T$.
|
||||
\\\\$\therefore T$ is bounded above, where $u^2$ is an upper bound of $T$. $\square$
|
||||
\\\\Thus we've satisfied one property of the supremum. Now, for the other, suppose $w$ is an upper bound of $T$ and $w \leq u^2$. Then $w \geq 0$, and $\sqrt{w} \leq u$, by the definition of $T$.
|
||||
\\\\Since $u = \sup S$, we have that $\exists s_0 \in S\ \text{s.t.}\ \sqrt{w} \leq s_0$.
|
||||
\\$\Rightarrow w < s^2_0$, which contradicts the fact that $w$ is an upper bound of $T$.
|
||||
\\\\$\therefore \sup T = u^2$.
|
||||
\end{proof}
|
||||
\textbf{Example:} Let $S:= (-2,1)$. Then $\sup S = 1$. Then $T:=(1,4)$, which yields $\sup T = 4$, and $4 \neq 1$.
|
||||
\item [8)] Let $X$ be a nonempty set, and let $f$ and $g$ be defined on $X$ and have bounded ranges in $\mathbb{R}$. Show that $$\sup\{f(x)\ +\ g(x): x \in X\} \leq \sup\{f(x): x \in X\}\ + \ \sup\{g(x): x \in X\}$$
|
||||
\begin{proof}
|
||||
Let $A=\{f(x): x \in X\}, B=\{g(x): x \in X\}$, where $A$ and $B$ are bounded above. Let $C=\{a+b:a \in A, b \in B\}$. Since $A$ and $B$ are bounded, we have that $a \leq \sup A\ \forall\ a \in A$, and $b \leq \sup B\ \forall\ b \in B$. Thus we have that $a + b \in C$, by the definition of $C$. So $a+b\leq \sup A + \sup B \ \forall a \in A$, and $\forall b \in B$. Thus we also have that $a\ +\ b \in C$. Since $a+b \leq \sup A + \sup B \Rightarrow \sup A \ + \sup B$ is an upper bound for $C$. Thus by completeness and the definition of $C$, $\sup C \leq \sup A + \sup B$.
|
||||
\end{proof}
|
||||
\textbf{Example:} Let $X=[-1,1]$ and let $f(x)=x$ and $g(x)=-x$. Then we have $\sup \{f(x):x \in X\} = 1$ and $\sup \{g(x): x \in X\} = 1$. But $\{f(x) + g(x): x \in X\} = \{x - x: x \in X\} = \{0\}$.\\
|
||||
\begin{align*}
|
||||
\therefore \sup\{f(x) + g(x): x \in X\} &\leq \sup \{f(x):x \in X\} + \sup \{g(x): x \in X\} \\ &= 2
|
||||
\end{align*}
|
||||
\item [9a)] Let $X=Y\ :=\{x \in \mathbb{R}: 0 < x < 1\}$. Define $h:X \times Y \rightarrow \mathbb{R}$ by $h(x,y) := 2x + y$. For each $x \in X$, find $f(x) := \sup \{h(x,y): y \in Y \}$; then find $\inf \{f(x): x \in X\}$.\\
|
||||
If $X$ and $Y$ are between 0 and 1, then the range of $f(x) = (0,3)$, thus $\inf(f(x)) = 0$.
|
||||
\item[14)] If $y > 0$, show that $\exists n \in \mathbb{N}$ such that $\frac{1}{2^n} < y$.
|
||||
\begin{proof}
|
||||
Let $y > 0$. By Corollary 2.4.5, $\exists n \in \mathbb{N}$ such that $0 < \frac{1}{n} < y$.\\ Since $n < 2^n$, we have $$0 < \frac{1}{2^n} < \frac{1}{n} < y$$
|
||||
\end{proof}
|
||||
\end{enumerate}
|
||||
\item Section 2.5
|
||||
\begin{enumerate}
|
||||
\item[2)] If $S\subseteq\mathbb{R}$ is nonempty, show that $S$ is bounded if and only if there exists a closed bounded interval $I$ such that $S \subseteq I$.
|
||||
\begin{proof}
|
||||
Let $S \subseteq \mathbb{R}$ be nonempty. We want to show that $S$ is bounded if and only if there exists a closed, bounded interval $I$ such that $S \subseteq I$. We prove it by cases, one for each direction of the "if and only if" condition.
|
||||
\begin{case} ($\Leftarrow$)
|
||||
Assume that there exists a closed, bounded interval $I$ such that $S \subseteq I$; that is, define $I := [a, b],\ \text{where}\ a,b \in \mathbb{R}$.
|
||||
\\\\Then $\min I = a,\ \text{and}\ \max I = b$. Thus we have that $\forall x \in I, a \leq x$, and so $a$ is a lower bound of $I$. Also, $\forall x \in I, x \leq b$, and so $b$ is an upper bound of $I$. By completeness, we have that $\inf I\ \text{and}\ \sup I$ exist. Specifically, we have that $\min I = \inf I = a,\ \text{and}\ \max I = \sup I = b$.
|
||||
\\\\Since $S \subseteq I$, we know that $\forall s \in S, a \leq s \leq b$. Thus by transitivity, we have that $\because \sup I = b \Rightarrow \sup S = b,\ \text{and}\ \because \inf I = a \Rightarrow \inf S = a$.
|
||||
\\\\$\therefore$ If there exists a closed, bounded interval $I$ such that $S \subseteq I$, then $S$ is bounded. $\square$
|
||||
\end{case}
|
||||
\begin{case} ($\Rightarrow$)
|
||||
Conversely, Assume that $S$ is bounded. Then we have that $\exists x \in S\ \text{s.t.}\ x \leq s,\ \forall s \in S$, and that $\exists y \in S\ \text{s.t.}\ s \leq y,\ \forall s \in S$. Thus by completeness, $\inf S$ and $\sup S$ exist.
|
||||
\\\\Let $\inf S = a$, and let $\sup S = b$. Since this holds, we can explicitly define $S := (a, b)$.
|
||||
\\\\By the Archimedian property, we have that $\forall s \in S,\ \exists n \in \mathbb{N},\ \text{s.t.}\ n \leq s < n+1$.
|
||||
\\\\Define an interval $I := [\lfloor a \rfloor, \lceil b \rceil]$. Thus we now have that $\lfloor a \rfloor \leq \inf S$, and that $\sup S \leq \lceil b \rceil$. Hence $S \subseteq I$.
|
||||
\\\\$\therefore\ I$ is a closed, bounded interval by construction, such that $S \subseteq I$.
|
||||
\end{case}
|
||||
\end{proof}
|
||||
\item[3)] If $S \subseteq \mathbb{R}$ is a nonempty bounded set, and $I_s := [\inf S, \sup S]$, show that $S\subseteq I_s$. Moreover, if $J$ is any closed bounded interval containing $S$, show that $I_s \subseteq J$.
|
||||
\begin{proof}
|
||||
Let $S \subseteq \mathbb{R}$ be a nonempty, bounded set, and let $I_s := [\inf S, \sup S]$. We want to show that $S \subseteq I_s$, and that if $J$ is any closed, bounded interval that contains $S$, then $I_s \subseteq J$.
|
||||
\\\\Let $\inf S = a$ and $\sup S = b$.
|
||||
\\\\First, assume that $\nexists \min S, \max S$. Then we have that $I_s \supset S$. Since $\sup \notin S$ and $\inf \notin S$, by the definition of infimum and supremum, respectively. We know this to be the case since the only time $\inf S \in S$ is if $\exists \min S$, and also $\sup S \in S$ if $\exists \max S$. But since $\inf S \in I_s$, and $\sup S \in I_s$, by the definition of $I_s$, we have that $S \subset I_s$.
|
||||
\\\\Now suppose that $\sup S, \inf S \in S$. Then $I_s = S$, since the bounds are the same. That is, let $\inf S = \alpha$, and let $\sup S = \beta$. Then $S = I_s \iff S:= [\alpha, \beta]$. This is because $\min S = \inf S = \alpha$ and $\max S = \sup S = \beta$.
|
||||
\\\\$\therefore\ S \subseteq I_s$.
|
||||
\\\\Now, let $J$ be a nonempty, bounded, closed set such that $S \subseteq J$. We want to show that $I_s \subseteq J$. Since $J$ is bounded, we can define $J := [a,b]$, where $a,b \in \mathbb{R}$. Similarly as was to be shown above, we know that $\min J = a$, and $\max J = b$. So, we know that $\inf J = \min J = a$, and $\sup J = \max J = b$. Since $S \subseteq J$, we know that if $S \subsetneq J$,
|
||||
\begin{enumerate}
|
||||
\item $\inf S \notin S\ \text{but}\ \inf S \in J$, and
|
||||
\item $\sup S \notin S\ \text{but}\ \sup S \in J$
|
||||
\end{enumerate}
|
||||
Thus since $\inf S, \sup S \in J, I_s \subseteq J$, since $\inf S, \sup S \in I_s$. Also, if $S = I_s$, then clearly $I_s \subseteq J$.
|
||||
\end{proof}
|
||||
\end{enumerate}
|
||||
\item Prove that for every $x \in \mathbb{R}$ and for each $n \in \mathbb{N}$, there exists a rational number $r_n$ such that $|x-r_n| < \frac{1}{n}$.
|
||||
\begin{proof}
|
||||
Let $x \in \mathbb{R}$, and let $n \in \mathbb{N}$. Then we have $x - \frac{1}{n} < x + \frac{1}{n}$. So $x-\frac{1}{n}, x+\frac{1}{n} \in \mathbb{R}$. By Theorem 2.4.8, we have that $\exists r_n \in \mathbb{Q}\ \text{s.t.}\ x-\frac{1}{n} < r_n < \frac{1}{n} \Rightarrow \frac{-1}{n} < r_n - x < \frac{1}{n}$.
|
||||
\\\\ So $|r_n -x| < \frac{1}{n}$ and $|x - r_n| < \frac{1}{n}$.
|
||||
\end{proof}
|
||||
\item A \textit{dyadic rational} is a number of the form $\frac{k}{2^n}$ for some $k,n \in \mathbb{Z}$. Prove that if $a,b \in \mathbb{R}$ and $a < b$, then there exists a dyadic rational $q$ such that $a < q < b$.
|
||||
\begin{proof}
|
||||
Let $a,b \in \mathbb{R}$ such that $a < b$. We want to show that $\exists q = \frac{k}{n}\ \text{s.t.}\ a < q < b$.
|
||||
\\\\By question 14 from Section 2.4, we know that $\forall y > 0, \exists n\ \text{s.t.}\ \frac{1}{2^n} < y$. By the Archimedian property, we have $0 < \frac{1}{2^n} < \frac{1}{n} < y$.
|
||||
\\\\
|
||||
Case 1: Let $a > 0$. So $0 < a < b$. By the Archimedian property again, $\exists n \in \mathbb{N}\ \text{s.t.}\ 0 < \frac{1}{2^n} < \frac{1}{n} < b-a$. So $\frac{1}{2^n} < b-a$. So $1+a*2^n < b*2^n$. By the Archimedian property again, since $a*2^n > 0,\ \exists m \in \mathbb{N}\ \text{s.t.}\ m-1 \leq a*2^n < m$. So $m \leq a*2^n+1 < m+1$.
|
||||
\\\\Now, combine $a*2^n < m \leq a*2^n+1 < b*2^n$. So $a < \frac{m}{2^n} < b$, and $q=\frac{m}{2^n}$. $\square$
|
||||
\\\\
|
||||
Case 2: If $a \leq 0$, choose $p \in \mathbb{Z}\ \text{s.t.}\ p \geq |a|$. Apply Case 1 to $0 < a + p < b +p$ to get $a+p < \frac{m}{2^n} < b+p$. So $a < \frac{m}{2^n}-p < b$. So $a<q<b,\ \text{where}\ q=\frac{m-2^n*p}{2^n} = \frac{k}{2^n}$.
|
||||
\end{proof}
|
||||
\item Prove, if true. Provide a counterexample if false.
|
||||
\begin{enumerate}
|
||||
\item If $A$ and $B$ are nonempty, bounded subsets of $\mathbb{R}$, then $\sup(A \cap B) \leq \sup A$.
|
||||
\begin{proof}
|
||||
Let $A$ and $B$ be nonempty, bounded subsets of $\mathbb{R}$. We want to show that $\sup (A \cap B) \leq \sup A$.
|
||||
\\\\ Consider the case where $A \cap B = \emptyset$. Then $\sup (\emptyset) = -\infty$. Since $A,B \subseteq \mathbb{R}$ and $A,B \nsubseteq \overline{\mathbb{R}}$, we have that since $A$ is nonempty, $A \neq \{-\infty\} \Rightarrow \sup A \neq -\infty$. Thus if $A \cap B = \emptyset \Rightarrow \sup (A \cap B) < \sup A$.
|
||||
\\\\Now, consider the case where $A \cap B \neq \emptyset$.
|
||||
\\\\By the definition of intersection, $\sup(A \cap B) = \sup A \iff A \cap B = A$. Also by the definition of intersection, we have that if $A \cap B \neq A$, then $A \cap B \subset A$ and $A \cap B \subset B$. This means that it's impossible to have a set after the intersection that is larger than both $A$ and $B$. This implies that the resulting set will yield $\sup (A \cap B) < \sup A$, since $\sup (A \cap B) = \sup A \Rightarrow A \cap B = A$.
|
||||
\\\\$\therefore \sup (A \cap B) \leq \sup A$.
|
||||
\end{proof}
|
||||
\item If $A + B=\{a + b: a\in A, b \in B\}$, where $A$ and $B$ are nonempty, bounded subsets of $\mathbb{R}$, then $\sup(A + B) = \sup A + \sup B$.
|
||||
\begin{proof}
|
||||
Let $A$ and $B$ be nonempty bounded subsets of $\mathbb{R}$, and let $A+B = \{a + b: a \in A,b \in B\}$. We want to show that $\sup (A+B) = \sup A + \sup B$.
|
||||
\\\\Since $A$ and $B$ are bounded, we know that $\sup A$ and $\sup B$ exist, and that $x \leq \sup A,\ \forall x \in A$, and $y \leq \sup B,\ \forall y \in B$. So $x+y \in A+B$ and $x+y \leq \sup A + \sup B,\ \forall x \in A,\forall y \in B$. Then by completeness, $\sup (A+B) \leq \sup A + \sup B$. $\square$
|
||||
\\\\Now we must show that $\sup A + \sup B \leq \sup (A+B)$.
|
||||
\\\\Let $y \in B$ be fixed. Since $x+y \leq \sup (A+B)$, then $x \leq \sup (A+B)-y,\ \forall x \in A$. So $\sup (A+B)-y$ is an upper bound for $A$. By completeness, we have that $\sup A \leq \sup (A+B)-y$. Then $y \leq \sup (A+B)-\sup A$. This is true for all $y \in B$.
|
||||
\\\\So $\sup (A+B)-\sup A$ is an upper bound for B, and $\sup B \leq \sup (A+B)- \sup A$.
|
||||
\\\\$\therefore \sup A + \sup B \leq \sup (A+B)$.
|
||||
\end{proof}
|
||||
\item If $A - B=\{a-b:a \in A, b \in B\}$, where $A$ and $B$ are nonempty, bounded subsets of $\mathbb{R}$, then $\sup(A - B)=\sup A - \sup B$.
|
||||
\\\\ \textbf{Counterexample:} Let $A := [-2, 0]$ and let $B := [1,4]$. Then we have that $A-B := [-4, -3]$. Then we have the following:
|
||||
\begin{align*}
|
||||
\sup (A-B) &= \sup A - \sup B
|
||||
\\ \sup ([-4,-3]) &= \sup ([-2, 0]) - \sup ([1,4])
|
||||
\\ -3 &= 0 - 4
|
||||
\\ -3 &\neq -4
|
||||
\end{align*}
|
||||
Thus $\sup (A-B) \neq \sup A - \sup B$.
|
||||
\end{enumerate}
|
||||
\end{enumerate}
|
||||
\end{document}
|
||||
Binary file not shown.
+345
@@ -0,0 +1,345 @@
|
||||
\documentclass[12pt,letterpaper]{article}
|
||||
\usepackage{amsthm}
|
||||
\usepackage[latin1]{inputenc}
|
||||
\usepackage{amsmath}
|
||||
\usepackage{amsfonts}
|
||||
\usepackage{amssymb}
|
||||
\usepackage{graphicx}
|
||||
\usepackage{array}
|
||||
\usepackage[left=2cm, right=2.5cm, top=2.5cm, bottom=2.5cm]{geometry}
|
||||
\newcommand{\st}{\ \text{s.t.}\ }
|
||||
\newcommand{\abs}[1]{\left\lvert #1 \right\rvert}
|
||||
\newcommand{\R}{\mathbb{R}}
|
||||
\newcommand{\N}{\mathbb{N}}
|
||||
\newtheorem*{3.1.10}{Theorem 3.1.10}
|
||||
\newtheorem*{3.2.3}{Theorem 3.2.3}
|
||||
\newtheorem*{3.2.10}{Theorem 3.2.10}
|
||||
\newtheorem*{3.2.7}{Squeeze Theorem}
|
||||
\newtheorem*{3.2.11}{Theorem 3.2.11}
|
||||
\newtheoremstyle{case}{}{}{}{}{}{:}{ }{}
|
||||
\theoremstyle{case}
|
||||
\newtheorem{case}{Case}
|
||||
\renewcommand{\qedsymbol}{$\blacksquare$}
|
||||
\author{Alexander J. Tusa}
|
||||
\title{Real Analysis Homework 4}
|
||||
\begin{document}
|
||||
\maketitle
|
||||
\begin{enumerate}
|
||||
\item Section 3.1
|
||||
\begin{enumerate}
|
||||
\item[5)] Use the definition of the limit of a sequence to establish the following limits.
|
||||
\begin{enumerate}
|
||||
\item[(a)] $\lim (\frac{n}{n^2+1})=0$
|
||||
\\\\Recall the definition of the limit of a sequence:
|
||||
$$\text{A sequence converges to a limit}\ A\ \text{if}\ \forall \ \varepsilon > 0,\ \exists N_\varepsilon\ \text{s.t.}\ |a_n-A|<\varepsilon\ \forall \ n \geq N_\varepsilon,$$
|
||||
$$n \in \mathbb{N},\ N_\varepsilon \in \mathbb{N},\ \text{written}\ \lim_{n\to\infty} a_n = A$$
|
||||
Thus, what we ultimately want to show here is that $|\frac{n}{n^2+1}-0| < \varepsilon, \forall \ n \geq N_\varepsilon$.
|
||||
\\\\Let's first take care of the denominator. We want to maximize the size of the denominator. So, we have
|
||||
$$n^2+1 > n^2\ \forall\ n \in \mathbb{N},\ \text{thus we have that}\ \frac{1}{n^2+1} < \frac{1}{n^2}\ \forall \ n \in \mathbb{N}$$
|
||||
Since there's no way to maximize the size of the numerator from what it currently is, combining both the numerator and denominator, we have
|
||||
$$\frac{n}{n^2+1} < \frac{n}{n^2}=\frac{1}{n}\ \forall \ n \in \mathbb{N}$$
|
||||
Now, given that $\varepsilon > 0$, we know that by Corollary 2.4.5 (If $t > 0$, then $\exists n_t \in \mathbb{N}\ \text{s.t.}\ 0 < \frac{1}{n_t} < t$), we know that $\exists N_\varepsilon \in \mathbb{N} \st 0 < \frac{1}{N_\varepsilon} < \varepsilon$, where $n=\varepsilon$. Let $N_0$ be the smallest of these numbers with this property. Then if $n \geq N_0,\ \frac{1}{n} < \frac{1}{N_0} < \varepsilon$. Thus we have
|
||||
$$\abs{\frac{n}{n^2 + 1}-0}=\abs{\frac{n}{n^2+1}} = \frac{n}{n^2+1} < \frac{1}{n} < \varepsilon$$
|
||||
$\therefore \lim_{n\to\infty} (\frac{n}{n^2+1}) = 0$.
|
||||
|
||||
\item[(c)] $\lim (\frac{3n+1}{2n+5})=\frac{3}{2}$
|
||||
\\\\We want to show that \(\abs{\frac{3n+1}{2n+5}-\frac{3}{2}} < \varepsilon,\ \forall\ n \geq N_\varepsilon,\ n \in \N,\ N_\varepsilon \in \N,\ \forall\ \varepsilon > 0\)
|
||||
\\\\Since \(n \in \N\), we know that \(\abs{\frac{3n+1}{2n+5}-\frac{3}{2}} = \frac{3n+1}{2n+5}-\frac{3}{2}\). So,
|
||||
\[\frac{3n+1}{2n+5}-\frac{3}{2}=\frac{6n+2}{4n+10}-\frac{6n+15}{4n+10}=-\frac{13}{4n+10} < 0\]
|
||||
and
|
||||
\[\frac{13}{4n+10} < \frac{13}{4n} < \frac{16}{4n} < \frac{4}{n}\ \forall\ n \in \N\]
|
||||
Given $\varepsilon > 0$, and since $\frac{4}{n} < \varepsilon \implies \frac{n}{4} > \varepsilon$, we know that by Corollary 2.4.5 ($t > 0 \implies \exists n_t \in \N \st 0 < \frac{1}{n_t} < t$), then $\exists N_\varepsilon \in \N \st 0 < \frac{1}{N_\varepsilon} < \frac{\varepsilon}{4}$, where $n = \varepsilon$. Let $N_0 \in \N$ be the smallest of these numbers with this property. Thus, if $n \geq N_0$, we have that $\frac{1}{n} < \frac{1}{N_0} < \frac{\varepsilon}{4}$. This gives us that $\frac{1}{n} < \frac{\varepsilon}{4}\ =\ \frac{4}{n} < \varepsilon$. So
|
||||
\[\abs{\frac{3n+1}{2n+5}-\frac{3}{2}}=\abs{-\frac{13}{4n+10}}=\frac{13}{4n+10}<\frac{4}{n}<\varepsilon\]
|
||||
$\therefore \lim (\frac{3n+1}{2n+5})=\frac{3}{2}$.
|
||||
|
||||
\item[(d)] $\lim (\frac{n^2-1}{2n^2 + 3})= \frac{1}{2}$
|
||||
\\\\We want to show that $\abs{\frac{n^2-1}{2n^2+3}-\frac{1}{2}} < \varepsilon,\ \forall\ \varepsilon>0,\ \forall\ n \geq N_\varepsilon$, where $n,N_\varepsilon \in \N$. Since
|
||||
\[\frac{n^2-1}{2n^2+3}-\frac{1}{2}=\frac{2n^2-2}{4n^2+6}-\frac{2n^2+3}{4n^2+6}=\frac{-5}{4n^2+6}<0\]
|
||||
So,
|
||||
\[\abs{\frac{n^2-1}{2n^2+3}-\frac{1}{2}}=\abs{\frac{2n^2-2}{4n^2+6}-\frac{2n^2+3}{4n^2+6}}=\abs{\frac{-5}{4n^2+6}}=\frac{5}{4n^2+6}<\frac{5}{4n^2}<\frac{5}{4n}<\frac{5}{n}\]
|
||||
$\forall n \in \N$
|
||||
\\\\Given $\varepsilon >0$, and since $\frac{5}{n},\varepsilon \implies \frac{n}{5} > \varepsilon$, we know that by Corollary 2.4.5 ($t>0 \implies \exists n_t \in \N \st 0<\frac{1}{n_t}<t$), then $\exists N_\varepsilon \in \N \st 0<\frac{1}{N_\varepsilon}<\frac{\varepsilon}{5}$, where $n=\varepsilon$. Let $N_0$ be the smallest of these numbers with this property. Then if $n \geq N_0$, we have that $\frac{1}{n}<\frac{1}{n_0}<\frac{\varepsilon}{5}$. This yields $\frac{1}{n}<\frac{\varepsilon}{5}=\frac{5}{n}<\varepsilon$. So
|
||||
\[\abs{\frac{n^2-1}{2n^2+3}-\frac{1}{2}}=\abs{-\frac{5}{4n^2+6}}=\frac{5}{4n^2+6}<\frac{5}{n}<\varepsilon\]
|
||||
$\therefore \lim (\frac{n^2-1}{2n^2+3})=\frac{1}{2}$
|
||||
\end{enumerate}
|
||||
\item[6)] Show that
|
||||
\begin{enumerate}
|
||||
\item[(a)] $\lim (\frac{1}{\sqrt{n+7}})=0$
|
||||
\\\\We want to show that $\abs{\frac{1}{\sqrt{n+7}}-0}<\varepsilon,\ \forall\ \varepsilon>0,\ \forall\ n \geq N_\varepsilon$, where $n,N_\varepsilon \in \N$.
|
||||
Since $n+7 > n\ \forall\ n \in \N$, we have that $\sqrt{n+7}>\sqrt{n}\ \forall\ n \in \N$ and therefore, $\frac{1}{\sqrt{n+7}}<\frac{1}{\sqrt{n}}\ \forall\ n \in \N$.
|
||||
\\\\Given $\varepsilon>0$, we know by Corollary 2.4.5 ($t>0 \implies \exists n_t \st 0 < \frac{1}{n_t}<t$), then $\exists N_\varepsilon \in \N \st 0 < \frac{1}{N_\varepsilon}<\varepsilon^2$. Thus we have that $\frac{1}{\sqrt{N_\varepsilon}}<\varepsilon$.
|
||||
\\\\Thus if $n \geq N_\varepsilon$, we have that $\sqrt{N_\varepsilon}\leq \sqrt{n}$ which gives us that $\frac{1}{\sqrt{n+7}}<\frac{1}{\sqrt{n}}\leq\frac{1}{\sqrt{N_\varepsilon}}<\varepsilon$. Therefore we have $\abs{\frac{1}{\sqrt{n+7}}-0}=\abs{\frac{1}{n+7}}=\frac{1}{\sqrt{n+7}}<\varepsilon\ \forall\ n \geq N_\varepsilon$.
|
||||
\\\\$\therefore\ \lim (\frac{1}{\sqrt{n+7}})=0$
|
||||
\end{enumerate}
|
||||
\item[9)] Show that if $x_n \geq 0\ \forall\ n \in \mathbb{N}\ \text{and}\ \lim (x_n),\ \text{then}\ \lim (\sqrt{x_n})=0$.
|
||||
\\\\Let $(x_n)$ be a sequence such that $\forall\ n \in \N,\ x_n \geq 0$ and $\lim (x_n)$. We want to show that $\lim (\sqrt{x_n})=0$.
|
||||
\\\\Let $\varepsilon >0$. By the definition of the limit of a sequence, we know that $\exists N_\varepsilon \in \N \st \forall\ n \geq N_\varepsilon$, the following inequality holds:
|
||||
\[\abs{x_n-0}=\abs{x_n}=x_n \geq 0 = x_n < \varepsilon^2\]
|
||||
Thus, if $n \geq N_\varepsilon$, we have that
|
||||
\[\abs{\sqrt{x_n}-0}=\abs{\sqrt{x_n}}=\sqrt{x_n}>\sqrt{\varepsilon^2}=\varepsilon\]
|
||||
and by the definition of the limit of a sequence, again, we have that $\lim (\sqrt{x_n})=0$.
|
||||
|
||||
\item[11)] Show that $\lim (\frac{1}{n}-\frac{1}{n+1})=0$.
|
||||
\\\\We want to show the following:
|
||||
\[\abs{\left(\frac{1}{n}-\frac{1}{n+1}\right)-0}<\varepsilon,\ \forall\ \varepsilon > 0,\ \forall\ n \geq N_\varepsilon,\ \text{for}\ n,N_\varepsilon \in \N\]
|
||||
Recall Theorem 3.1.10:
|
||||
\begin{3.1.10}
|
||||
Let $(x_n)$ be a sequence of real numbers and let $x \in \R$. If $(a_n)$ is a sequence of positive real numbers with $\lim (a_n) = 0$ and if for some constant $C>0$ and some $m \in \N$ we have
|
||||
\[\abs{x_n-x}\leq Ca_n\ \forall\ n \geq m,\]
|
||||
then it follows that $\lim (x_n)=x$.
|
||||
\end{3.1.10}
|
||||
We want to find a constant $C>0$ and a sequence $(a_n)$ such that $a_n>0,\ \lim (a_n)=0$, and
|
||||
\[\abs{\left(\frac{1}{n}-\frac{1}{n+1}\right)-0}\leq Ca_n\ \forall\ n \geq m,\ \text{for some}\ m \in \N\]
|
||||
Let's first find $C$ and $a_n$.
|
||||
\begin{align*}
|
||||
\abs{\left(\frac{1}{n}-\frac{1}{n+1}\right)-0} &= \abs{\frac{1}{n}-\frac{1}{n+1}}
|
||||
\\ &=\frac{1}{n}-\frac{1}{n+1} &\left(\frac{1}{n}>\frac{1}{n+1}\right)
|
||||
\\ &=\frac{(n+1)-n}{n(n+1)}
|
||||
\\ &=\frac{1}{n^2+n}
|
||||
\end{align*}
|
||||
Since $n < n+n^2\ \forall\ n$, we have that $\frac{1}{n^2+n}<\frac{1}{n}$. Thus, choose $C=1$, $(a_n)=\frac{1}{n}$ and $m=1$.
|
||||
\\\\Furthermore, we proved in class that $\lim (\frac{1}{n})=0$.
|
||||
\\\\Now that all of the conditions of Theorem 3.1.10 have been satisfied, apply it to our original sequence of $x_n = \frac{1}{n}-\frac{1}{n+1}$, which yields:
|
||||
\[\lim \left(\frac{1}{n}-\frac{1}{n+1}\right)=0\]
|
||||
|
||||
\item[12)] Show that $\lim (\sqrt{n^2+1}-n)=0$.
|
||||
\\\\Let $\varepsilon > 0$ be given. Then, we have the following:
|
||||
\begin{align*}
|
||||
\abs{(\sqrt{n^2+1}-n)-0} &= \abs{\sqrt{n^2+1}-n}
|
||||
\\ &= \abs{(\sqrt{n^2+1}-n)*\frac{\sqrt{n^2+1}+n}{\sqrt{n^2+1}+n}}
|
||||
\\ &= \abs{\frac{1}{\sqrt{n^2+1}+n}}
|
||||
\end{align*}
|
||||
Now, let's look at the denominator, $\sqrt{n^2+1}+n$. Then, we know that $n^2+1 > n^2$. So, we have
|
||||
\begin{align*}
|
||||
n^2+1 &> n^2
|
||||
\\ \sqrt{n^2+1}&> n
|
||||
\\ \sqrt{n^2+1}+n &\geq 2n
|
||||
\\ \frac{1}{\sqrt{n^2 + 1}+n} &\leq \frac{1}{2n} = \varepsilon
|
||||
\end{align*}
|
||||
Choose $N \geq \frac{1}{2\varepsilon}$.
|
||||
\\\\Thus we have that
|
||||
\[\forall\ \varepsilon > 0, \text{ choose } N \geq \frac{1}{2\varepsilon} \text{ then }\abs{a_n-A} < \varepsilon\ \forall\ n \geq N\]
|
||||
|
||||
\item[13)] Show that $\lim (\frac{1}{3^n})=0$.
|
||||
\\\\Since $n \leq 3n \iff \frac{1}{3^n} \leq \frac{1}{n}$, we have
|
||||
\[\abs{\frac{1}{3^n}-0}\leq \frac{1}{n}\]
|
||||
Using Theorem 3.1.10 and $\lim \frac{1}{n}=0$ we get
|
||||
\[\lim \frac{1}{3^n}=0\]
|
||||
\end{enumerate}
|
||||
\item Section 3.2
|
||||
\begin{enumerate}
|
||||
\item[1)] For $x_n$ given by the following formulas, establish either the convergence or the divergence of the sequence $X = (x_n)$.
|
||||
|
||||
\begin{enumerate}
|
||||
\item[(a)] $x_n := \frac{n}{n+1}$
|
||||
\[x_n=\frac{1}{n+1}=\frac{1}{1+\frac{1}{n}}\]
|
||||
\[\lim_{n\to\infty}x_n=\lim_{n\to\infty} \frac{1}{1+\frac{1}{n}}=\frac{1}{1+\lim_{n\to\infty}\frac{1}{n}}=1\]
|
||||
therefore we have that the sequence $\{x_n\}$ converges to 1.
|
||||
\end{enumerate}
|
||||
\item[2)] Give an example of two divergent sequences $X$ and $Y$ such that:
|
||||
\begin{enumerate}
|
||||
\item[(a)] their sum $X + Y$ converges,
|
||||
\\\\Let $X=(0,1,0,1,0,1,...)$ and let $Y=(1,0,1,0,1,0,...)$. Clearly $X$ and $Y$ are divergent because the difference of two consecutive terms is equal to 1.
|
||||
\\Thus, $X+Y=(1,1,1,1,1,....)$, and thus the sequence converges as it's a constant sequence.
|
||||
|
||||
\item[(b)] their product $XY$ converges.
|
||||
\\\\Let $X=(0,1,0,1,0,1,...)$ and let $Y=(1,0,1,0,1,0,...)$. Once more, these two sequences clearly diverge as the difference of two consecutive terms is equal to 1.
|
||||
\\Thus, $XY=(0,0,0,0,0,...)$, which clearly converges.
|
||||
\end{enumerate}
|
||||
\item[3)] Show that if $X$ and $Y$ are sequences such that $X$ and $X+Y$ are convergent, then $Y$ is convergent.
|
||||
\\\\Recall Theorem 3.2.3:
|
||||
\begin{3.2.3}
|
||||
\begin{enumerate}
|
||||
\item[\textbf{(a)}] Let $X=(x_n)$ and let $Y=(y_n)$ be sequences of real numbers that converge to $x$ and $y$, respectively, and let $c \in \R$. Then the sequences $X+Y,\ X-Y,\ X*Y,$ and $cX$ converge to $x+y,\ x-y,\ xy,$ and $cx$, respectively.
|
||||
\item[\textbf{(b)}] If $X=(x_n)$ converges to $x$ and $Z=(z_n)$ is a sequence of nonzero real numbers that converges to $z$ and $z \neq 0$, then the quotient sequence $X/Z$ converges to $x/z$
|
||||
\end{enumerate}
|
||||
\end{3.2.3}
|
||||
If $X$ and $X+Y$ are convergent, then by Theorem 3.2.3, $Y=(X+Y)-X$. is also convergent.
|
||||
|
||||
\item[5)] Show that the following sequence is not convergent.
|
||||
\begin{enumerate}
|
||||
\item[(a)] $(2^n)$
|
||||
\\\\$2^n > n$, and $(n)$ is an unbounded sequence. Therefore, $(2^n)$ is also unbounded.
|
||||
\\\\Every convergent sequence must be bounded so we can conclude that $(2^n)$ is unbounded.
|
||||
\end{enumerate}
|
||||
\item[6)] Find the limits of the following sequence:
|
||||
\begin{enumerate}
|
||||
\item[(a)] $\lim ((\frac{2+1}{n})^2)$
|
||||
\begin{align*}
|
||||
\lim \left(\left(2+\frac{1}{n}\right)^2\right) &= \lim \left(\left(2+\frac{1}{n}\right)*\left(2+\frac{1}{n}\right)\right)
|
||||
\\ &= \text{By Theorem 3.2.3 (a): Limit of a product = Product of limits}
|
||||
\\ &= \left(\lim\left(2+\frac{1}{n}\right)\right) * \left(\lim\left(2+\frac{1}{n}\right)\right)
|
||||
\\ &= \text{By Theorem 3.2.3 (a): Limit of a sum = Sum of Limits}
|
||||
\\ &= \left(\lim (2)+\lim\left(\frac{1}{n}\right)\right)*\left(\lim (2) + \lim \left(\frac{1}{n}\right)\right)
|
||||
\\ &= (2+0)*(2+0)
|
||||
\\ &=4
|
||||
\end{align*}
|
||||
\end{enumerate}
|
||||
\item[9)] Let $y_n := \sqrt{n_1}-\sqrt{n}\ \text{for}\ n \in \mathbb{N}$. Show that $(\sqrt{n}y_n)$ converges. Find the limit.
|
||||
\\\\To show that the sequences $(y_n)$ and $(\sqrt{n}y_n)$ converge, we first need to recall Theorem 3.2.10:
|
||||
|
||||
\begin{3.2.10}
|
||||
Let $X=(x_n)$ be a sequence of real numbers that converges to $x$ and suppose that $x_n \geq 0$. Then the sequence $(\sqrt{x_n})$ of positive square roots converges and $\lim (\sqrt{x_n})=\sqrt{x}$.
|
||||
\end{3.2.10}
|
||||
|
||||
\begin{align*}
|
||||
y_n=\sqrt{n+1}-\sqrt{n} &= \frac{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{(\sqrt{n+1}+\sqrt{n})}
|
||||
\\ &= \frac{(n+1)-n}{\sqrt{n+1}+\sqrt{n}}
|
||||
\\ &= \frac{1}{\sqrt{n+1}+\sqrt{n}}
|
||||
\end{align*}
|
||||
$\implies\ |y_n-0|=\frac{1}{\sqrt{n+1}+\sqrt{n}} < \frac{1}{2\sqrt{n}}$. Now we have that $\lim (\frac{1}{n})=0$ implies $\lim (\frac{1}{\sqrt{n}})=0$ by Theorem 3.2.10.
|
||||
\\\\Now, we have
|
||||
\[\sqrt{n}y_n=\frac{\sqrt{n}}{\sqrt{n}+\sqrt{n+1}}=\frac{1}{1+\sqrt{\frac{n+1}{n}}}=\frac{1}{1+\sqrt{1+\frac{1}{n}}}\]
|
||||
Now, by the algebra of limits and of convergent sequences, we have that as $\lim (\frac{1}{n})=0 \implies \lim (1+\frac{1}{n})=1 \implies \lim \sqrt{1+\frac{1}{n}}=\sqrt{1}=1$ and thus $\lim \left(\frac{1}{1+\sqrt{1+\frac{1}{n}}}\right)=\frac{1}{1+\sqrt{1+0}}=\frac{1}{2}$. Therefore $\lim (\sqrt{n}y_n)$ exists and is equal to $\frac{1}{2}$. \\
|
||||
\item[14)] Use the Squeeze Theorem 3.2.7 to determine the limits of the following,
|
||||
\begin{enumerate}
|
||||
\item[(a)] $(n^{1/n^2})$.
|
||||
\\\\Recall Theorem 3.2.7 Squeeze Theorem:
|
||||
|
||||
\begin{3.2.7}
|
||||
Suppose that $X=(x_n),\ Y=(y_n),$ and $Z=(z_n)$ are sequences of real numbers such that
|
||||
\[x_n \leq y_n \leq z_n\ \forall\ n \in \N\]
|
||||
and that $\lim (x_n)= \lim (z_n)$. Then $Y=(y_n)$ is convergent and
|
||||
\[\lim(x_n)=\lim(y_n)=\lim(z_n)\]
|
||||
\end{3.2.7}
|
||||
|
||||
So, notice that
|
||||
\[1 \leq n^{\frac{1}{n^2}} \leq n^{\frac{1}{n}}\]
|
||||
and $\lim (n^{\frac{1}{n}})=1$. By the Squeeze Theorem,
|
||||
\[1 \leq \lim (n^{\frac{1}{n^2}}) \leq \lim (n^{\frac{1}{n}})=1\]
|
||||
Therefore we have that $\lim (n^{\frac{1}{n^2}}=1)$. \\
|
||||
\end{enumerate}
|
||||
\item[16)] Apply Theorem 3.2.11 to the following sequences, where $a, b$ satisfy $0 < a < 1, b > 1$.
|
||||
\\\\Recall Theorem 3.2.11:
|
||||
\begin{3.2.11}
|
||||
Let $(x_n)$ be a sequence of positive real numbers such that $L := \lim (x_{n+1}/x_n)$ exists. If $L < 1$, then $(x_n)$ converges and $\lim (x_n)=0$.
|
||||
\\
|
||||
\end{3.2.11}
|
||||
\begin{enumerate}
|
||||
\item[(c)] $(\frac{n}{b^n})$
|
||||
\\Since $\frac{n}{b^n}>0\ \forall\ n$, we have that
|
||||
\[\lim \left(\frac{\frac{n+1}{b^{n+1}}}{\frac{n}{b^n}}\right)=\frac{1}{b} < 1\]
|
||||
Thus, let $(x_n)$ be a sequence of positive real numbers such $L:=\lim \left(\frac{x_{n+1}}{x_n}\right)$ exists. If $L<1$, then $(x_n)$ converges and $\lim (x_n)=0$. Therefore we have that $\lim (\frac{n}{b^n}) = 0$.
|
||||
\\
|
||||
|
||||
\item[(d)] $(2^{3n}/3^{2n})$
|
||||
\\Since $\frac{2^{3n}}{3^{2n}}>0\ \forall\ n \in \N$, we have that
|
||||
\[\lim \left(\frac{\frac{2^{3(n+1)}}{3^{2(n+1)}}}{\frac{2^{3n}}{3^{2n}}}\right)=\frac{8}{9} < 1\]
|
||||
Thus, let $(x_n)$ be a sequence of positive real numbers such that $L:= \lim (\frac{x_{n+1}}{x_n})$ exists. If $L < 1$, then $(x_n)$ converges and $\lim (x_n) = 0$. Therefore $\lim (\frac{2^{3n}}{3^{2n}})=0$.
|
||||
\end{enumerate}
|
||||
\item[17)]
|
||||
\begin{enumerate}
|
||||
\item[(a)] Give an example of a convergent sequence $(x_n)$ of positive numbers with $\lim (x_{n+1}/x_n)=1$.
|
||||
\\\\Let $(x_n)$ be a sequence such that $x_n=1\ \forall\ n \in \N$. This is a constant sequence and is thus convergent and $\lim (\frac{x_{n+1}}{x_n})=1$. \\
|
||||
|
||||
\item[(b)] Give an example of a divergent sequence with this property. (Thus, this property cannot be used as a test for convergence.)
|
||||
\\\\Let $(x_n)$ be a sequence such that $(x_n)=(n)$. This sequence is divergent because it's not bounded, however $\lim (\frac{x_{n+1}}{x_n})=1$.
|
||||
\end{enumerate}
|
||||
\item[22)] Suppose that if $(x_n)$ is a convergent sequence and $(y_n)$ is such that for any $\varepsilon > 0$ there exists $M$ such that $|x_n-y_n| < \varepsilon$ for all $n \geq M$. Does it follow that $(y_n)$ is convergent?
|
||||
\\\\It does follow that $(y_n)$ is convergent. In fact, $\lim (y_n) = \lim (x_n)$. To show this, let $x=\lim (x_n)$.
|
||||
\\\\For any $\varepsilon > 0$, choose $M_1,M_2>0 \st$
|
||||
\begin{align*}
|
||||
|x_n-x|<\frac{\varepsilon}{2},\ &\forall\ n\geq M_1 &(a)
|
||||
\\\text{and } |x_n-y_n|< \frac{\varepsilon}{2},\ &\forall\ n \geq M_2 &(b)
|
||||
\end{align*}
|
||||
Choose $M=\max \{M_1, M_2\}$. Then $\forall\ n \geq M$,
|
||||
\begin{align*}
|
||||
|y_n-x| &= |y_n-x_n+x_n-x|
|
||||
\\ &\leq |y_n-x_n|+|x_n-x|
|
||||
\\ &< \frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon &\text{ from (a) and (b)}
|
||||
\end{align*}
|
||||
Thus we have that $\lim (y_n)=x$.
|
||||
\end{enumerate}
|
||||
\item Give an example of each of the following:
|
||||
\begin{enumerate}
|
||||
\item A convergent sequence of rational numbers having an irrational limit.
|
||||
\\\\Let $a_n=\left(1 + \frac{1}{n}\right)^n$ be a sequence of rationals. Then, notice that $a_n \in \mathbb{Q}$, but notice also that $\lim_{n\to\infty} a_n = e$.
|
||||
|
||||
\item A convergent sequence of irrational numbers having a rational limit.
|
||||
\\\\Let $a_n=\frac{\sqrt{2}}{n}$. Then we have that $a_n \notin \mathbb{Q}$, but we also have that $\lim_{n\to\infty} a_n = 0$, and $0 \in \mathbb{Q}$
|
||||
\end{enumerate}
|
||||
\item Prove: Let $a_n$ and $b_n$ be sequences of real numbers and $A \in \mathbb{R}$. If for some $k > 0$ and some $m \in \mathbb{N}$, we have $|a_n-A|\leq k|b_n|$ for all $n > m$, and if $\lim_{n\to\infty} b_n = 0$, then $\lim_{n\to\infty} a_n = A$.
|
||||
\begin{proof}
|
||||
Let $a_n$ and $b_n$ be sequences of real numbers and let $A \in \R$. Let $k >0$, and some $m \in \N$. We have $\abs{a_n-A}\leq k|b_n|\ \forall\ n > m$, and $\lim_{n\to\infty} b_n = 0$. We want to show that $\lim_{n\to\infty} a_n = A$.
|
||||
\\\\Let $\varepsilon>0$ be given. Since $\lim b_n=0$, we know that $b_n$ converges. So by the definition of the limit of a sequence, we know that $\exists\ N_1 \st |b_n-0| =|b_n| < \varepsilon\ \forall\ n \geq N_1$. This also means that since $\lim b_n = 0$, we have $k|b_n| < \varepsilon \implies |b_n| < \frac{\varepsilon}{k}$, by algebra.
|
||||
\\\\Recall that we have $|a_n-A|\leq k|b_n|\ \forall\ n > m$. So, let $N=\max \{m,N_1\}$. Then we have that $\forall\ n \geq N,\ |a_n-A|\leq k|b_n| \leq k\left(\frac{\varepsilon}{k}\right)=\varepsilon$. Thus we have that $a_n$ converges since it is also always less than $\varepsilon$. This is by the definition of the limit of a sequence.
|
||||
\\\\$\therefore\ \lim_{n\to\infty} a_n = A$; That is, the sequence $a_n$ converges to $A$.
|
||||
\end{proof}
|
||||
|
||||
\item (Similar to Section 3.2, problem 7)
|
||||
\begin{enumerate}
|
||||
\item Suppose that $\lim_{n\to\infty} a_n=0$. If $b_n$ is a bounded sequence, prove that $\lim_{n\to\infty} a_nb_n=0$.
|
||||
\begin{proof}
|
||||
Let $b_n$ be a bounded sequence. Then by the definition of a bounded sequence we know that $\exists\ M>0,\ M \in \R \st |b_n| \leq M\ \forall\ n \in \N$.
|
||||
\\\\Let $\varepsilon >0$ be given. Since $a_n$ converges, we know that by the definition of the limit of a sequence, $\exists\ N \st |a_n-0|=|a_n|<\frac{\varepsilon}{M}\ \forall\ n \geq N$, since we know that by the definition of bounded, $|b_n| \leq M$. Thus, by the utilization of the triangle inequality, we have that $|a_nb_n| \leq |a_n||b_n|\leq M\left(\frac{\varepsilon}{M}\right)=\varepsilon,\ \forall\ n \geq N$.
|
||||
\\\\$\therefore$ we have that $a_nb_n$ converges to 0.
|
||||
\end{proof}
|
||||
|
||||
\item Show by counterexample that the boundedness of $b_n$ is a necessary condition for part (a).
|
||||
\\\\Let $a_n=\frac{1}{n}$, and let $b_n=n^3$. Then we have
|
||||
\[\lim_{n\to\infty} a_n = 0, \text{ and } \lim_{n\to\infty} b_n = \infty\]
|
||||
, since $n^3$ is an unbounded sequence since $n^3$ is not bounded above, and in order to be bounded, the sequence must be both bounded above and bounded below. So, we have the following:
|
||||
\[\lim_{n\to\infty}a_nb_n = \lim_{n\to\infty} \left(\frac{1}{n}\right)(n^3) = \lim_{n\to\infty} \left(\frac{n^3}{n}\right) = \lim_{n\to\infty} n^2 = \infty \neq 0\]
|
||||
$\therefore$ the boundedness of $b_n$ is a necessary condition for part (a) to be true.
|
||||
\end{enumerate}
|
||||
|
||||
\item Prove or justify, if true. Provide a counterexample, if false.
|
||||
\begin{enumerate}
|
||||
\item If $a_n$ converges, then $a_n/n$ also converges.
|
||||
\\\\This is true.
|
||||
\begin{proof}
|
||||
Let $\lim a_n = A$. Let $b_n = n$, and let $\lim b_n = B$. Recall Theorem 3.2.3. Since $n \in \N$, we know that $n \neq 0$, and thus we know that $0 \notin b_n$. Thus by Theorem 3.2.3, we have that $\lim \frac{a_n}{b_n} = \frac{A}{B}$. And thus since the limit exists, we have that $\frac{a_n}{n}$ also converges.
|
||||
\end{proof}
|
||||
|
||||
\item If $a_n$ does not converge, then $a_n/n$ does not converge.
|
||||
\\\\This is false.
|
||||
\\\\Counterexample: Let $a_n = n$. Then we know that $a_n$ does not converge, however $\frac{a_n}{n} = \frac{n}{n} = 1$, which converges since $\lim 1 = 1$.
|
||||
|
||||
\item If $a_n$ converges and $b_n$ is bounded, then $a_nb_n$ converges.
|
||||
\\\\This is false.
|
||||
\\\\Counterexample: Let $a_n=(\frac{1}{n} + 1)$, and let $b_n = (-1)^n$. Then we have that $\lim a_n = 1$, and $b_n$ is bounded both above and below, by $-1$ and $1$, respectively. Thus $(-1)^n$ is bounded. So if we multiply the two together, we have
|
||||
\[a_nb_n = (\frac{1}{n}+1)(-1)^n\]
|
||||
which is not convergent since this sequence oscillates. Thus by the fact that a convergent sequence multiplied by a divergent sequence diverges, we have that even though $a_n$ is convergent and $b_n$ is bounded, $a_nb_n$ does not converge.
|
||||
|
||||
\item If $a_n$ converges to zero and $b_n > 0$ for all $n \in \mathbb{N}$, then $a_nb_n$ converges.
|
||||
\\\\This is false.
|
||||
\\\\Counterexample: Let $a_n=\frac{1}{n}$, and let $b_n = n^3$. Then we have that $\lim a_n = 0$, and that $b_n >0\ \forall\ n \in \N$. Then, we have that $a_nb_n=(\frac{1}{n})(n^3) = \frac{n^3}{n} = n^2$. And thus we have that $\lim a_nb_n = \lim n^2 = \infty$, thus $a_nb_n$ diverges.
|
||||
|
||||
\item If $a_n \rightarrow A$ and $b_n \rightarrow A$ as $n \rightarrow \infty$, then $a_n = b_n$ for all $n \in \mathbb{N}$.
|
||||
\\\\This is false.
|
||||
\\\\Counterexample: Let $a_n = \frac{n}{3n+1}$, and let $b_n = \frac{1}{n} + 1$. Then we have $\lim a_n = 1$, and $\lim b_n = 1$, thus $\lim a_n = \lim b_n = 1$. However, $a_n \neq b_n$, since $\frac{n}{3n+1} \neq \frac{1}{n} + 1 = \frac{n}{n + 1} \neq \frac{n}{3n+1}$.
|
||||
|
||||
\item Every convergent sequence is bounded.
|
||||
\\\\This is true.
|
||||
\begin{proof}
|
||||
Refer to proof of Theorem 3.2.2.
|
||||
\end{proof}
|
||||
|
||||
\item Every bounded sequence is convergent.
|
||||
\\\\This is false.
|
||||
\\\\Counterexample: Let $a_n = (-1)^n$. Then we have that $a_n$ is bounded both above and below, and thus $a_n$ is bounded. However since $a_n$ oscillates, we know that $a_n$ is not convergent.
|
||||
|
||||
\item If $a_n \rightarrow 0$ as $n \rightarrow \infty$, then for every $\epsilon > 0$ there exists $N \in \mathbb{N}$ such that if $n >N$, then $a_n < \epsilon$.
|
||||
\\\\This is true.
|
||||
\begin{proof}
|
||||
Since we have that $\lim_{n\to\infty} a_n = 0$, then by the definition of the limit of a sequence, we have
|
||||
\[\forall\ \varepsilon > 0,\ \exists\ N \in \N \st |a_n-A|<\varepsilon,\ \forall\ n \geq N\]
|
||||
Thus we have that
|
||||
\begin{align*}
|
||||
|a_n-0| &<\varepsilon
|
||||
\\ |a_n| &< \varepsilon
|
||||
\\ -\varepsilon &< a_n < \varepsilon
|
||||
\end{align*}
|
||||
Thus we have that if $n > N$, then $a_n < \varepsilon$, by the definition of the limit of a sequence.
|
||||
\end{proof}
|
||||
|
||||
\item If for every $\epsilon >0 $ there exists $N \in \mathbb{N}$ such that $n > N$ implies $a_n < \epsilon$, then $a_n \rightarrow 0$ as $n \rightarrow \infty$.
|
||||
\\\\This is false.
|
||||
\\\\Counterexample: Let $a_n = -1,\ \forall\ n \in \N$. Then we have that by the definition of the limit of a sequence, for every $\varepsilon > 0$, there exists $N \in \N$ such that $n > N$ implies that $a_n < \varepsilon$, but $a_n$ does not converge to 0 as $n \rightarrow \infty$. Rather, we have that $a_n \rightarrow -1$.
|
||||
\item Given sequences $a_n$ and $b_n$, if for some $A \in \mathbb{R}, k > 0$ and $m \in \mathbb{N}$ we have $|a_n-A| \leq k|b_n|$ for all $n >m$, then $a_n \rightarrow A$ as $n \rightarrow \infty$.
|
||||
\\\\This is false.
|
||||
\\\\Counterexample: Note the proof of question 4: We showed that the above statement is true if $\lim b_n = 0$ is also given. However, given that this limit is omitted, we have that we can provide a counterexample to disprove said statement.
|
||||
\\\\Suppose $b_n=\frac{1}{n}+1$. Then we have that $\lim b_n=1$. Let $a_n=-2$. Assume for some $A \in \R, k>0, m \in \N$, we have $|-2 - (-2)| \leq k|1|\ \forall\ n > m = 2 \leq k$. However, note that there doesn't exist an $m$ such that $2 \leq k\ \forall\ n > m$, since $m=n\ \forall\ m \in \N,\ n \in \N$.
|
||||
\end{enumerate}
|
||||
\end{enumerate}
|
||||
\end{document}
|
||||
Binary file not shown.
+496
@@ -0,0 +1,496 @@
|
||||
\documentclass[12pt,letterpaper]{article}
|
||||
\usepackage[utf8]{inputenc}
|
||||
\usepackage[english]{babel}
|
||||
\usepackage{amsthm}
|
||||
\usepackage{amsmath}
|
||||
\usepackage{amsfonts}
|
||||
\usepackage{amssymb}
|
||||
\usepackage{graphicx}
|
||||
\usepackage{array}
|
||||
\usepackage[left=2cm, right=2.5cm, top=2.5cm, bottom=2.5cm]{geometry}
|
||||
\usepackage{enumitem}
|
||||
\newcommand{\st}{\ \text{s.t.}\ }
|
||||
\newcommand{\abs}[1]{\left\lvert #1 \right\rvert}
|
||||
\newcommand{\R}{\mathbb{R}}
|
||||
\newcommand{\N}{\mathbb{N}}
|
||||
\newcommand{\Q}{\mathbb{Q}}
|
||||
\newcommand{\C}{\mathbb{C}}
|
||||
\newcommand{\Z}{\mathbb{Z}}
|
||||
\DeclareMathOperator{\sign}{sgn}
|
||||
\newtheoremstyle{case}{}{}{}{}{}{:}{ }{}
|
||||
\theoremstyle{case}
|
||||
\newtheorem{case}{Case}
|
||||
\theoremstyle{definition}
|
||||
\newtheorem{definition}{Definition}[section]
|
||||
\newtheorem{theorem}{Theorem}[section]
|
||||
\newtheorem*{theorem*}{Theorem}
|
||||
\newtheorem{corollary}{Corollary}[section]
|
||||
\newtheorem{lemma}[theorem]{Lemma}
|
||||
\newtheorem*{remark}{Remark}
|
||||
\setlist[enumerate]{font=\bfseries}
|
||||
\renewcommand{\qedsymbol}{$\blacksquare$}
|
||||
\author{Alexander J. Tusa}
|
||||
\title{Real Analysis Homework 5}
|
||||
\begin{document}
|
||||
\maketitle
|
||||
\begin{enumerate}
|
||||
\item For the following sequences, i) write out the first 5 terms, ii) Use the Monotone Sequence Property to show that the sequences converges.
|
||||
\begin{enumerate}
|
||||
\item \textbf{Section 3.3}
|
||||
\begin{enumerate}
|
||||
\item[2)] Let $x_1 > 1$ and $x_{n+1} := 2-1/x_n$ for $n \in \N$. Show that $(x_n)$ is bounded and monotone. Find the limit.
|
||||
\\\\ The first five terms of this sequence are $x_1 \geq 2,x_2 \geq \frac{3}{2}, x_3 \geq \frac{4}{3}, x_4 \geq \frac{5}{4}, x_5 \geq \frac{6}{5}, \dots \approx x_1 \ge 2, x_2 \geq 1.5, x_3 \geq 1.3333, x_4 \geq 1.25, x_5 \geq 1.2, \dots$. This sequence appears to be decreasing.
|
||||
\\\\Recall the Monotone Sequence Property:
|
||||
\begin{theorem*}{Monotone Sequence Property}
|
||||
A monotone sequence of real numbers is convergent if and only if it is bounded. Further,
|
||||
\begin{enumerate}
|
||||
\item If $X=(x_n)$ is a bounded increasing sequence, then
|
||||
\[\lim (x_n) = \sup \{x_n:n \in \N\}\]
|
||||
|
||||
\item If $Y=(y_n)$ is a bounded decreasing sequence, then
|
||||
\[\lim (y_n) = \inf \{y_n : n \in \N \}\]
|
||||
\end{enumerate}
|
||||
\end{theorem*}
|
||||
|
||||
To show that this sequence converges, we must first find the possible limit points (fixed points) of this sequence. So,
|
||||
\begin{align*}
|
||||
x&=2-\frac{1}{x} \\
|
||||
x^2 &= 2x -1 \\
|
||||
x^2 - 2x + 1 &= 0 \\
|
||||
(x-1)^2 &= 0
|
||||
\end{align*}
|
||||
Thus, $x=1$ is a possible limit of this sequence.
|
||||
\\\\Now, we will prove that $(x_n)$ is bounded by $1$, and since we hypothesized that $(x_n)$ is decreasing, we say that $(x_n)$ is bounded below by 1.
|
||||
\begin{proof}
|
||||
We want to show that the sequence $(x_n)$ is bounded below by 1; that is, we want to show that $1 \leq x_n,\ \forall\ n \in \N$. We prove it by method of mathematical induction.
|
||||
\\\\\textbf{Basis Step:} Let $n=1$. Then
|
||||
\begin{align*}
|
||||
x_n &\geq x_{n+1}, &\text{by the definition of decreasing,} \\
|
||||
x_1 &\geq x_{1+1} \\
|
||||
x_1 &\geq x_2
|
||||
\end{align*}
|
||||
Since $x_1>1 \Rightarrow \frac{1}{x_1} < 1$, we have
|
||||
\[x_2 = 2-\frac{1}{x_1} > 1\]
|
||||
\[\Rightarrow 1 < x_2 < 2.\]
|
||||
Since $x_1>1$ and because $1 < x_2 < 2$, we have that $x_1 \geq x_2$.
|
||||
\\\\\textbf{Inductive Step:} Assume $1 < x_n < 2,\ \forall\ n \in \N$.
|
||||
\\\\\textbf{Show:} Now we want to show that $x_n \leq x_{n+1}$.
|
||||
\\So,
|
||||
\[1 < x_n <2\]
|
||||
\[1 > \frac{1}{x_n} > \frac{1}{2}\]
|
||||
\[-1 < -\frac{1}{x_n} < -\frac{1}{2}\]
|
||||
\[1 < 2-\frac{1}{x_n} < 2-\frac{1}{2} < 2\]
|
||||
\[1 < x_{n+1} < 2\]
|
||||
Thus we have that $(x_n)$ is bounded between 1 and 2.
|
||||
\end{proof}
|
||||
|
||||
Now we need to show that $(x_n)$ is monotone decreasing; that is, we must show that $x_1 \geq x_2 \geq \dots \geq x_n$.
|
||||
|
||||
\begin{proof}
|
||||
We want to show that $x_1 \geq x_2 \geq \dots \geq x_n,\ \forall\ n \in \N$. We prove it by method of mathematical induction.
|
||||
\\\\\textbf{Basis Step:} Let $n=1$. Then since $x_1 >1$ is given, we have that $\frac{1}{x_1} < 1$. This yields $x_2=2-\frac{1}{x_1}>1$, as was determined for the boundedness proof, and thus we have that $1 < x_2 < 2$. This means that $1 > \frac{1}{x_2} > \frac{1}{2}$, and since $\frac{1}{2} \leq \frac{1}{x_n}$, we have $x_2 \geq x_1$.
|
||||
\\\\\textbf{Inductive Step:} Assume $x_n \geq x_{n+1}\ \forall\ n \in \N$.
|
||||
\\\\\textbf{Show:} We now want to show that $x_{n+2} \leq x_{n+1}$.
|
||||
\\So,
|
||||
\[x_{n+2}=2-\frac{1}{x_{x+1}}\]
|
||||
Recall the inductive hypothesis, in that $x_n \geq x_{n+1} \Rightarrow \frac{1}{x_n} \leq \frac{1}{x_{n+1}}$. Thus,
|
||||
\[-\frac{1}{x_n} \geq -\frac{1}{x_{n+1}}\]
|
||||
\[\Rightarrow 2-\frac{1}{x_n} \leq 2 -\frac{1}{x_{n+1}}\]
|
||||
\[x_{n+1} \leq x_{n+2}\]
|
||||
$\therefore$ we have that $x_1 \geq x_2 \geq \dots \geq x_n,\ \forall\ n \in \N$.
|
||||
\end{proof}
|
||||
Thus $(x_n)$ is monotone decreasing.
|
||||
\\\\By the \textit{Monotone Sequence Property}, since we have shown that $(x_n)$ is both bounded (and thus converges), and that $(x_n)$ is monotone decreasing, we have that
|
||||
\begin{align*}
|
||||
\lim (x_n) &= \inf \{x_n: n \in \N\} \\
|
||||
&=\inf (1,2) \\
|
||||
&= 1
|
||||
\end{align*}
|
||||
Hence the sequence converges to the previously found possible limit of 1. \\
|
||||
|
||||
\item[3)] Let $x_1 > 1$ and $x_{n+1} := 1 + \sqrt{x_n - 1}$ for $n \in \N$. Show that $(x_n)$ is decreasing and bounded below by $2$. Find the limit.
|
||||
\\\\The first 5 terms of this sequence are $x_1 \geq 2, x_2 \geq 2, x_3 \geq 2, x_4 \geq 2, x_5 \geq 2, \dots$. Notice the following, however:
|
||||
\begin{align*}
|
||||
x_{n+1} \leq x_n &\iff 1+\sqrt{x_n - 1} \leq x_n \\
|
||||
&\iff \sqrt{x_n -1} \leq x_n -1
|
||||
\end{align*}
|
||||
which we know is always true since the square root function is a decreasing function.
|
||||
\\\\Now we must find the possible limit points (fixed points) of this sequence. So,
|
||||
\begin{align*}
|
||||
x &= 1 + \sqrt{x-1} \\
|
||||
x-1 &= \sqrt{x-1} \\
|
||||
x-1 &= (x-1)^2 \\
|
||||
x-1 &= x^2 -2x +1 \\
|
||||
(x-1)-(x^2-2x+1) &= 0 \\
|
||||
-x^2+3x-2 &=0 \\
|
||||
-(x^2-3x+2) &= 0 \\
|
||||
-(x-1)(x-2) &= 0 \\
|
||||
(x-1)(x-2) &= 0
|
||||
\end{align*}
|
||||
Thus $x=1$, or $x=2$. These are the possible limits of $(x_n)$. Since we hypothesized that $(x_n)$ is decreasing, then we say that $(x_n)$ is bounded below by 2, since we are given that $x_1 > 1$.
|
||||
\\\\Now we will prove that $(x_n)$ is bounded below by 2.\\
|
||||
\begin{proof}
|
||||
We want to show that $(x_n)$ is bounded below by 1; that is, we want to show that $1 \leq x_n,\ \forall\ n \in \N$. We prove it by method of mathematical induction.
|
||||
\\\\\textbf{Basis Step:} Let $n=1$. Then we are given that $x_1 \geq 2$.
|
||||
\\\\\textbf{Inductive Step:} Assume that $x_n \geq 2,\ \forall\ n \in \N$.
|
||||
\\\\\textbf{Show:} We now want to show that $x_{n+1} \geq 2,\ \forall\ n \in \N$.
|
||||
\\\\So,
|
||||
\begin{align*}
|
||||
x_{n+1} &= 1+\sqrt{x_n -1} \\
|
||||
&\geq 1+\sqrt{2 -1} \\
|
||||
&=1 + 1 \\
|
||||
&= 2
|
||||
\end{align*}
|
||||
Thus, $x_n \geq 2,\ \forall\ n \in \N$. By the definition of boundedness, we have that $(x_n)$ is bounded below by 2.
|
||||
\end{proof}
|
||||
|
||||
Since we have also shown earlier that $(x_n)$ is monotone decreasing, we have that by the monotone sequence property, since $(x_n)$ is bounded, $(x_n)$ converges, and since $(x_n)$ is monotone decreasing, we have:
|
||||
\begin{align*}
|
||||
\lim (x_n) &= \inf \{x_n:n \in \N\} \\
|
||||
&=2
|
||||
\end{align*}
|
||||
|
||||
\item[7)] Let $x_1 := a>0$ and $x_{n+1} := x_n+1/x_n$ for $n \in \N$. Determine whether $(x_n)$ converges or diverges.
|
||||
\\\\The first 5 terms of this sequence are $x_1 \geq 1, x_2 \geq 2, x_3 \geq \frac{5}{2}, x_4 \geq \frac{29}{10}, x_5 \geq \frac{941}{290}, \dots \approx x_1 \geq 1, x_2 \geq 2, x_3 \geq 2.5, x_4 \geq 2.9, x_5 \geq 3.244828, \dots$. This sequence appears to be increasing. We show this to be true as follows:
|
||||
\begin{align*}
|
||||
x_{n+1} \geq x_n &\iff x_n + \frac{1}{x_n} \geq x_n \\
|
||||
&\iff x_n^2 + 1 \geq x_n^2 \\
|
||||
&\iff 1 \geq 0
|
||||
\end{align*}
|
||||
which is true.
|
||||
However, notice that one of the terms of the sequence is $x_n$. We know that $x_n$ is an unbounded sequence. Thus, we can infer that $(x_n)$ is unbounded above. We show this as follows:
|
||||
\begin{align*}
|
||||
x_{n+1}^2 &= \left(x_n + \frac{1}{x_n} \right)^2 \\
|
||||
&= x_n^2+2+\frac{1}{x_n^2} \\
|
||||
&> x_n^2 +2
|
||||
\end{align*}
|
||||
Since:
|
||||
\[x_{n+1}^2 > x_n^2+2 > x_{n-1}^2 +4 > \dots > x_1^2+2 \cdot n = a^2+2 \cdot n\]
|
||||
\[\Downarrow\]
|
||||
\[x_n > \sqrt{a^2 + 2 \cdot (n-1)}\]
|
||||
Since the right hand side of this inequality is unbounded, the left hand side is also unbounded.
|
||||
\\\\Thus we have that this sequence $(x_n)$ is unbounded above.
|
||||
\\\\Since this sequence is increasing and unbounded above, we have that the sequence is divergent.\\
|
||||
|
||||
\item[8)] Let $(a_n)$ be an increasing sequence, $(b_n)$ be a decreasing sequence, and assume that $a_n \leq b_n$ for all $n \in \N$. Show that $\lim (a_n) \leq \lim (b_n)$, and thereby deduce the Nested Intervals Property 2.5.2 from the Monotone Convergence Theorem 3.3.2.
|
||||
\\\\Since $(a_n)$ is an increasing sequence, we know that $(a_1 \leq a_2 \leq \dots \leq a_n)$, and since $(b_n)$ is a decreasing sequence, we know that $(b_1 \geq b_2 \geq \dots \geq b_n)$. Also, since we have that $a_n \leq b_n,\ \forall\ n \in \N$, we know that $(a_n)$ is bounded above by $(b_1)$. Thus, by the \textit{Monotone Convergence Theorem}, we know that
|
||||
\[\lim (a_n) = \sup \{a_n: n \in \N\}\]
|
||||
Also, since $(b_n)$ is a decreasing sequence such that it is bounded below by $(a_1)$, by the \textit{Monotone Convergence Theorem}, we have
|
||||
\[\lim (b_n) = \inf \{b_n: n \in \N\}\]
|
||||
Recall Theorem 3.2.5:
|
||||
\begin{theorem*}
|
||||
If $X=(x_n)$ and $Y=(y_n)$ are convergent sequences of real numbers and if $x_n \leq y_n\ \forall\ n \in \N$, then $\lim (x_n) \leq \lim (y_n)$.
|
||||
\end{theorem*}
|
||||
Also, recall the \textit{Nested Intervals Property}:
|
||||
\begin{theorem*}
|
||||
If $I_n=[a_n,b_n],\ n \in \N$, is a nested sequence of closed bounded intervals, then there exists a number $\xi \in \R \st \xi \in I_n\ \forall\ n \in \N$.
|
||||
\end{theorem*}
|
||||
Note that we have a nested sequence of closed, bounded intervals: $[a_n, b_n],\ n \in \N$. Since we showed that $\lim (a_n) \leq \lim (b_n)$, (and we are given that $(a_n)$ is increasing and $(b_n)$ is decreasing), we know that there exists $\xi$ such that
|
||||
\[\lim (a_n) \leq \xi \leq \lim (b_n)\]
|
||||
which means that $\xi \in [a_n, b_n],\ \forall\ n \in \N$.
|
||||
\end{enumerate}
|
||||
|
||||
\item $a_1 = 1,\ a_{n+1}=\frac{a_n^2+5}{2a_n}$
|
||||
\\\\The first 5 terms of this sequence are $1, 3, \frac{7}{3}, \frac{47}{21}, \frac{2207}{987}, \dots \approx 1, 3, 2.3333, 2.2381, 2.2361, \dots$. This is a decreasing sequence.
|
||||
\\\\First, we must find the possible limits (fixed points) of the sequence. So,
|
||||
\begin{align*}
|
||||
a&=\frac{a^2+5}{2a} \\
|
||||
2a^2 &= a^2+5 \\
|
||||
a^2 &= 5 \\
|
||||
a &= \pm \sqrt{5}
|
||||
\end{align*}
|
||||
Since we're given that $a_1=1$, we know that the most likely lower bound will be $\sqrt{5}$.
|
||||
\\\\Now we want to show that $(a_n)$ is bounded below by $\sqrt{5}$.
|
||||
\begin{proof}
|
||||
We want to show that $a_n \geq \sqrt{5},\ \forall\ n \in \N$. We prove it by method of mathematical induction.
|
||||
\\\\\textbf{Basis Step:} Since $1 \geq \sqrt{5}$, we have that $a_1 \geq \sqrt{5}$
|
||||
\\\\\textbf{Inductive Step:} Assume that $a_n \geq \sqrt{5}\ \forall\ n \in \N$.
|
||||
\\\\\textbf{Show:} We want to show that $a_{n+1} \geq \sqrt{5}\ \forall\ n \in \N$. So,
|
||||
\[a_{n+1} = \frac{a_n^2 + 5}{2a_n}\]
|
||||
\begin{align*}
|
||||
(a_n-\sqrt{5})^2 &\geq 0 \\
|
||||
a_n^2 -2\sqrt{5}a_n +5 &\geq 0 \\
|
||||
a_n^2 +5 &\geq 2\sqrt{5}a_n \\
|
||||
\Downarrow \\
|
||||
\frac{a_n^2+5}{2a_n} &\geq \frac{2\sqrt{5}a_n}{2a_n} \\
|
||||
\frac{a_n^2+5}{2a_n} &\geq \sqrt{5} \\
|
||||
a_{n+1} \geq \sqrt{5}
|
||||
\end{align*}
|
||||
Thus we have that $(a_n)$ is bounded below by $\sqrt{5}$.
|
||||
\end{proof}
|
||||
Now we must show that $(a_n)$ is monotone decreasing.\\
|
||||
\begin{proof}
|
||||
We want to show that $(a_n)$ is monotone decreasing; that is, we want to show that $(a_2 \geq a_3 \geq \dots \geq a_n),\ \forall\ n \geq 2$. We prove it by method of mathematical induction.
|
||||
\\\\\textbf{Basis Step:} Since $3 \geq \frac{7}{3}$, we have that $a_2 \geq a_3$.
|
||||
\\\\\textbf{Inductive Step:} Assume that $a_n \geq a_{n+1},\ \forall\ n \geq 2$.
|
||||
\\\\\textbf{Show:} We want to show that $a_{n+2} \leq a_{n+1},\ \forall\ n \geq 2$.
|
||||
\\So,
|
||||
\[a_{n+2} = \frac{a_{n+1}^2+5}{2a_{n+1}} \leq \frac{a_n^2+5}{2a_n}\]
|
||||
Since we have:
|
||||
\begin{align*}
|
||||
a_{n+1} &\geq \sqrt{5}, &\text{by the previous proof of boundedness} \\
|
||||
a_{n+1}^2 &\geq 5
|
||||
\end{align*}
|
||||
We can equivalently write the inequality as
|
||||
\[\frac{a_{n+1}^2+5}{2a_{n+1}} \leq \frac{a_{n+1}^2+a_{n+1}^2}{2a_{n+1}}=a_{n+1}\]
|
||||
Thus we have that $(a_n)$ is monotone decreasing.
|
||||
\end{proof}
|
||||
Since $(a_n)$ is both monotone decreasing and bounded, we have
|
||||
\begin{align*}
|
||||
\lim (a_n) &= \inf \{a_n:n \in \N\} \\
|
||||
&= \sqrt{5}
|
||||
\end{align*}
|
||||
|
||||
\item $a_1 = 5,\ a_{n+1}=\sqrt{4+a_n}$
|
||||
\\\\The first 5 terms of this sequence are 5, 3, $\sqrt{7}$, $ \frac{\sqrt{14}}{2} + \frac{\sqrt{2}}{2} $, $ \frac{\sqrt{2 \cdot (\sqrt{14}+\sqrt{2}+8)}}{2} $, $\dots$, $\approx$ 5, 3, 2.64575131106, 2.57793547457, 2.5647486182, $\dots$. This sequence is decreasing.
|
||||
\\\\First, we must find the possible limits (fixed points) of the sequence. So,
|
||||
\begin{align*}
|
||||
a&=\sqrt{4+a} \\
|
||||
\sqrt{4+a} &= a \\
|
||||
4+a &= a^2 \\
|
||||
-a^2+a+4 &= 0 \\
|
||||
a^2-a-4 &= 0 \\
|
||||
a^2-a&=4 \\
|
||||
a^2-a+\frac{1}{4}&=4+\frac{1}{4} \\
|
||||
a^2-a+\frac{1}{4}&=\frac{17}{4} \\
|
||||
(a-\frac{1}{2})^2&=\frac{17}{4} \\
|
||||
a-\frac{1}{2}&=\pm \frac{\sqrt{17}}{2}
|
||||
\end{align*}
|
||||
So we have that $a=\frac{1}{2} + \frac{\sqrt{17}}{2}$, or $a=\frac{1}{2}-\frac{\sqrt{17}}{2}$. We must now check these solutions for correctness; so,
|
||||
\begin{align*}
|
||||
a \Rightarrow \frac{1}{2}-\frac{\sqrt{17}}{2} &=\frac{1}{2} \left(1-\sqrt{17}\right) \\
|
||||
&\approx -1.56155 \\\\
|
||||
\sqrt{a+4} &= \sqrt{\left(\frac{1}{2}-\frac{\sqrt{17}}{2}\right)+4} \\
|
||||
&=\frac{\sqrt{9-\sqrt{17}}}{\sqrt{2}} \\
|
||||
&\approx 1.56155
|
||||
\end{align*}
|
||||
Thus, this solution is incorrect. Now we must validate that $a=\frac{1}{2}+\frac{\sqrt{17}}{2}$ is correct. So,
|
||||
\begin{align*}
|
||||
a \Rightarrow \frac{1}{2}+\frac{\sqrt{17}}{2} &= \frac{1}{2}\left(1+\sqrt{17}\right) \\
|
||||
&\approx 2.56155 \\\\
|
||||
\sqrt{a+4} &= \sqrt{\left(\frac{\sqrt{17}}{2}+\frac{1}{2}\right)+4} \\
|
||||
&= \frac{\sqrt{9+\sqrt{17}}}{\sqrt{2}} \\
|
||||
&\approx 2.56155
|
||||
\end{align*}
|
||||
Thus $a=\frac{1}{2}+\frac{\sqrt{17}}{2}$ is a correct solution.
|
||||
\\\\Now we want to show that $(a_n)$ is bounded below by $\frac{1}{2}+\sqrt{17}$.
|
||||
\begin{proof}
|
||||
We want to show that $a_n \geq \frac{1}{2}+\frac{\sqrt{17}}{2},\ \forall\ n \in \N$, by the definition of a lower bound. We prove this by method of mathematical induction.
|
||||
\\\\\textbf{Basis Step:} Since $5 \geq \frac{1}{2}+\frac{\sqrt{17}}{2}$, we have that $a_1 \geq \frac{1}{2}+\frac{\sqrt{17}}{2}$.
|
||||
\\\\\textbf{Inductive Step:} Assume $a_n \geq \frac{1}{2}+\frac{\sqrt{17}}{2},\ \forall\ n \in \N$.
|
||||
\\\\\textbf{Show:} We now want to show that $ a_{n+1} \geq \frac{1}{2}+\frac{\sqrt{17}}{2}\ \forall\ n \in \N$. So,
|
||||
\begin{align*}
|
||||
a_{n+1} &= \sqrt{4+a_n}, &\text{by the definition of the sequence} \\
|
||||
&\geq \sqrt{4+\left(\frac{1}{2}+\frac{\sqrt{17}}{2}\right)}, &\text{by the inductive hypothesis} \\
|
||||
&\geq \sqrt{\frac{8}{2}+\frac{1}{2}+\frac{\sqrt{17}}{2}} \\
|
||||
&\geq \sqrt{\frac{9+\sqrt{17}}{2}} \\
|
||||
&\geq \sqrt{\frac{1}{2}\left(9+\sqrt{17}\right)} \\
|
||||
&\geq \sqrt{\frac{1}{4}+\frac{\sqrt{17}}{2}+\frac{17}{4}}, &\text{by expressing } \frac{9+\sqrt{17}}{2} \text{ as a square} \\
|
||||
&\geq \sqrt{\frac{1+2\sqrt{17}+17}{4}} \\
|
||||
&\geq \sqrt{\frac{1+2\sqrt{17}+(\sqrt{17})^2}{4}} \\
|
||||
&\geq \sqrt{\frac{(\sqrt{17}+1)^2}{4}} \\
|
||||
&\geq \sqrt{\frac{1}{4}\left(1+\sqrt{17}\right)^2} \\
|
||||
&\geq \frac{\sqrt{(1+\sqrt{17})^2}}{\sqrt{4}} \\
|
||||
&\geq \frac{\sqrt{17}+1}{2} \\
|
||||
&\geq \frac{1}{2} + \frac{\sqrt{17}}{2}
|
||||
\end{align*}
|
||||
Thus we have that $ a_{n+1} \geq \frac{1}{2} + \frac{\sqrt{17}}{2}\ \forall\ n \in \N$.
|
||||
\end{proof}
|
||||
Now, we want to show that $ (a_n) $ is monotone decreasing; that is, we want to show that $(a_1 \geq a_2 \geq \dots \geq a_n)$.
|
||||
\begin{proof}
|
||||
We want to show that $(a_1 \geq a_2 \geq \dots \geq a_n),\ \forall\ n \in \N$. We prove this by method of mathematical induction.
|
||||
\\\\\textbf{Basis Step:} Since $5 \geq 3$, we have that $a_1 \geq a_2$.
|
||||
\\\\\textbf{Inductive Step:} Assume $a_n \geq a_{n+1}\ \forall\ n \in \N$.
|
||||
\\\\\textbf{Show:} We want to show that $a_{n+1} \geq a_{n+2}\ \forall\ n \in \N$. So,
|
||||
\begin{align*}
|
||||
a_{n+2} &= \sqrt{4+a_{n+1}} &\text{by the definition of the sequence} \\
|
||||
&\leq \sqrt{4+a_n} &\text{by the inductive hypothesis} \\
|
||||
&=a_{n+1}
|
||||
\end{align*}
|
||||
Thus we have that $a_{n+1} \geq a_{n+2}\ \forall\ n \in \N$.
|
||||
\end{proof}
|
||||
Since $(a_n)$ is both bounded and monotone decreasing, by the \textit{Monotone Convergence Theorem}, we have that $(a_n)$ converges. Also by the \textit{Monotone Sequence Property}, we have that $(a_n)$ converges to the following:
|
||||
\begin{align*}
|
||||
\lim (a_n) &= \inf \{a_n: n \in \N\} \\
|
||||
&= \frac{1}{2}+\frac{\sqrt{17}}{2} \approx 2.56155281281
|
||||
\end{align*}
|
||||
\end{enumerate}
|
||||
|
||||
\item
|
||||
\begin{enumerate}
|
||||
\item Show $a_n=\frac{3 \cdot 5 \cdot 7 \cdot \dots (2n-1)}{2 \cdot 4 \cdot 6 \dots (2n)}$ converges to $A$ where $0 \leq A < 1/2$.
|
||||
\\\\First, we note the first few terms of this sequence: $\frac{3}{2}, \frac{15}{8}, \frac{105}{48}, \dots$. Now, since we have that
|
||||
\[0 <\frac{1 \cdot 3 \cdot 5 \cdot 7 \dots (2n-1)}{2 \cdot 4 \cdot 6 \cdot \dots 2n}<\frac{1}{2} \cdot 1 \cdot 1 \dots \cdot 1=\frac{1}{2}\]
|
||||
We have that $a_n$ is bounded. Also, we note that $a_n$ is strictly decreasing since $\frac{a_{n+1}}{a_n}=\frac{2n-1}{2n}<1$ (i.e. $a_{n+1}<a_n$) Thus by the \textit{Monotone Sequence Property}, we have that $a_n$ converges to $A$ where $0 \leq A < \frac{1}{2}$.
|
||||
|
||||
\item Show $b_n = \frac{2 \cdot 4 \cdot 6 \dots (2n)}{3 \cdot 5 \cdot 7 \dots (2n+1)}$ converges to $B$ where $0 \leq B < 2/3$.
|
||||
\\\\To begin, note that the first few terms of this sequence are $\frac{2}{3}, \frac{8}{15}, \frac{48}{105}, \dots$. Now, since we have that
|
||||
\[0 < \frac{2 \cdot 4 \cdot 6 \cdot \dots (2n)}{3 \cdot 5 \cdot 7 \cdot \dots (2n-1)}<\frac{2}{3}\cdot 1 \cdot 1 \dots \cdot 1 = \frac{2}{3}\]
|
||||
Thus we have that $b_n$ is bounded, and that it's strictly decreasing since $\frac{b_{n+}1}{b_n}=\frac{2n}{2n+1}<1$. That is, $b_{n+1} < b_n$. Hence by the \textit{Monotone Sequence Property}, we have that $b_n$ converges to $B$ where $0 \leq B < \frac{2}{3}$.
|
||||
\end{enumerate}
|
||||
|
||||
\item \textbf{Section 3.4}
|
||||
\begin{enumerate}
|
||||
\item[1)] Give an example of an unbounded sequence that has a convergent subsequence.
|
||||
\\\\Consider the sequence $(a_n)=(1,1,2,1,3,1,4,1, \dots)$ Clearly, this sequence is bounded below by 1 but is unbounded above, and thus this sequence is unbounded. However, consider the subsequence $(a_{2n-1})$. Then the resulting sequence is $(a_{2n-1})=(1,1,1,1,1, \dots)$. Since this is a constant sequence, we have that $(a_{2n-1})$ converges to 1. And hence we have an unbounded sequence that has a convergent subsequence.
|
||||
|
||||
\item[3)] Let $(f_n)$ be the Fibonacci sequence of Example 3.1.2(d), and let $x_n := f_{n+1}/f_n$. Given that $\lim (x_n) =L$ exists, determine the value of $L$.
|
||||
\\\\We can rewrite $(x_n)$ as follows:
|
||||
\begin{align*}
|
||||
x_n &= \frac{f_{n+1}}{f_n} \\
|
||||
&=\frac{f_n+f_{n-1}}{f_n} \\
|
||||
&= 1+\frac{f_{n-1}}{f_n} \\
|
||||
&= 1+\frac{\frac{1}{f_n}}{f_{n-1}} \\
|
||||
&= 1+\frac{1}{x_{n-1}}
|
||||
\end{align*}
|
||||
Since we're given that $L=\lim (x_n)$ exists and since we just showed that it's equal to $\lim (x_{n-1})$, we get the following:
|
||||
\begin{align*}
|
||||
x_n &= \left. 1+\frac{1}{x_{n-1}}\ \ \ \ \ \right| \lim \\
|
||||
\lim (x_n) &= 1+\frac{1}{\lim(x_{n-1})} \\
|
||||
L &= \left. 1+\frac{1}{L}\ \ \ \ \ \right| \cdot L \\
|
||||
L^2 &= L + 1 \\
|
||||
L^2-L-1 &= 0 \\
|
||||
L_{1,2} &= \frac{1 \pm \sqrt{1-4 \cdot 1 \cdot (-1)}}{2} \\
|
||||
L_1 &= \frac{1-\sqrt{5}}{2} <0 \\
|
||||
L_2 &= \frac{1+\sqrt{5}}{2}>0
|
||||
\end{align*}
|
||||
Now, since $f_n >0 \Rightarrow x_n >0 \Rightarrow L>0$, we can infer that the proper limit is
|
||||
\[L=\frac{1+\sqrt{5}}{2}\]
|
||||
\\
|
||||
|
||||
\item[4a)] Show that the sequence $(1-(-1)^n+1/n)$ converges.
|
||||
\\\\ Let $(x_n):=(1-(-1)^n+1/n)$. Let $(z_n)=(x_{2n})$, and $(w_n)=(x_{2n-1})$ be subsequence of $(x_n)$. Then $(z_n)$ is the subsequence of all terms of $(x_n)$ such that $n$ is even, and $(w_n)$ is the subsequence of all terms of $(x_n)$ such that $n$ is odd.
|
||||
\\\\These subsequences yield the following:
|
||||
\[z_n = x_{2n} = 1-(-1)^{2n}+\frac{1}{2n}=1-1+\frac{1}{2n}=\frac{1}{2n}\]
|
||||
\[w_n=x_{2n-1}=1-(-1)^{2n-1}+\frac{1}{2n-1}=1+1+\frac{1}{2n-1}=2+\frac{1}{2n-1}\]
|
||||
Now, if we take the limit of each sequence as $n \rightarrow \infty$ yields
|
||||
\[\lim_{n \to \infty} (z_n) = 0 \neq 2 = \lim_{n \to \infty} (w_n)\]
|
||||
Recall Theorem 3.4.5 \textit{Divergence Criteria}:
|
||||
\begin{theorem*}
|
||||
If a sequence $X=(x_n)$ of real numbers has either of the following properties, then $X$ is divergent.
|
||||
\begin{enumerate}
|
||||
\item $X$ has two convergent subsequences $X'=(x_{n_k})$ and $X''=(x_{r_k})$ whose limits are not equal.
|
||||
|
||||
\item $X$ is unbounded
|
||||
\end{enumerate}
|
||||
\end{theorem*}
|
||||
Thus by the \textit{Divergence Criteria}, we have that since $(z_n)$ and $(w_n)$ satisfy the first property of the \textit{Divergence Criteria}, we can conclude that the sequence $(x_n)$ is divergent. \\
|
||||
|
||||
\item[16)] Give an example to show that Theorem 3.4.9 fails if the hypothesis that $X$ is a bounded sequences is dropped.
|
||||
\\\\Recall \textit{Theorem 3.4.9}:
|
||||
\begin{theorem*}
|
||||
Let $X=(x_n)$ be a bounded sequence of real numbers and let $x \in \R$ have the property that every convergent subsequence of $X$ converges to $x$. Then the sequence $X$ converges to $x$.
|
||||
\end{theorem*}
|
||||
Consider the sequence $a_n=(0,-1,0,-2,0,-3,0,-4, \dots)$. Note that the subsequence $a_{2n-1}=(0,0,0,0,0,\dots)$. Thus, $\lim (a_{2n-1}=0)$. However, since $(a_n)$ is not bounded, we have also that $\lim (a_n) \neq 0$, since $(a_n)$ is divergent. It is only bounded above by $0$, but it is not bounded below.
|
||||
|
||||
\item[18)] Show that if $(x_n)$ is a bounded sequence, then $(x_n)$ converges if and only if $\lim \sup (x_n) = \lim \inf (x_n)$.
|
||||
\begin{proof}
|
||||
Let $(x_n)$ be a bounded sequence. We want to show that $(x_n)$ converges if and only if $\lim \sup (x_n) = \lim \inf (x_n)$.
|
||||
\\\\$(\Rightarrow)$ Suppose that $(x_n)$ is a bounded sequence, and suppose that $(x_n)$ converges. We want to show that $\lim \sup (x_n)=\lim \inf (x_n)$.
|
||||
\\\\Recall \textit{Theorem 3.4.2}:
|
||||
\begin{theorem*}
|
||||
If a sequence $X=(x_n)$ of real numbers converges to a real number $x$, then any subsequence $X'=(x_{n_k})$ of $X$ also converges to $x$.
|
||||
\end{theorem*}
|
||||
By \textit{Theorem 3.4.2}, we have that $(x_n)$ has one and only one limit, $x$. Thus we have that $\lim \sup (x_n)=\lim \inf (x_n)$.
|
||||
\\\\$(\Leftarrow)$ Conversely, suppose that $\lim \sup (a_n)=\lim \inf (a_n)$.
|
||||
Recall the definitions of the supremum and infimum for some nonempty subset $S$ of the real numbers:
|
||||
\[\sup(S)=u \Leftrightarrow\ i)\ \ \forall\ s \in S,\ u \geq s,\ \land\ \ ii)\ \forall\ \varepsilon >0,\ \exists s_\varepsilon \in S \st u-\varepsilon<s_\varepsilon\]
|
||||
and
|
||||
\[\inf (S)=w \Leftrightarrow\ \ i)\ \forall\ s \in S,\ w \leq s,\ \land\ \ ii)\ \forall\ \varepsilon>0,\ \exists\ s_\varepsilon \in S \st s_\varepsilon<w+\varepsilon\]
|
||||
Also, recall the definition of the limit of a sequence for some sequence $(a_n)$:
|
||||
\[\lim (a_n)=A \implies\ \forall\ \varepsilon>0,\ \exists\ K(\varepsilon) \in \N \st |a_n-A|<\varepsilon,\ \forall\ n \geq K(\varepsilon)\]
|
||||
So, if we let $\lim \sup x_n=a$, then we know that there exists a natural number $K(\varepsilon_1)$ such that $x_n<a+\varepsilon,\ \forall\ n \geq K(\varepsilon_1)$. Also, for $\lim \inf (x_n)=a$ yields that $\exists\ K(\varepsilon_2) \in \N \st x_n>a-\varepsilon\ \forall\ n \geq K(\varepsilon_2)$. Now, let $K(\varepsilon)=\max \{K(\varepsilon_1), K(\varepsilon_2)\}$. Then $a-\varepsilon<x_n<a+\varepsilon\ \forall\ n \geq K(\varepsilon)\ \Rightarrow\ |x_n-a|<\varepsilon\ \forall\ n \geq K(\varepsilon)$. Thus, by definition we have that $\lim x_n=a$.
|
||||
$\therefore$ we have that $x_n$ is convergent.
|
||||
\end{proof}
|
||||
|
||||
\item[19)] Show that if $(x_n)$ and $(y_n)$ are bounded sequences, then
|
||||
\[\lim \sup (x_n + y_n) \leq \lim \sup (x_n) + \lim \sup (y_n).\]
|
||||
Give an example in which the two sides are not equal.
|
||||
\begin{proof}
|
||||
We first note that since $(x_n)$ and $(y_n)$ are bounded sequences of real numbers, we have that $\{x_n+y_n\}$ is also bounded. Let $\lim \sup x_n=a_1, \lim \sup (y_n)=a_2$, and $\lim \sup (x_n+y_n)=p$. Let $\varepsilon > 0$. Since $\lim \sup (x_n)=a_1$, we know that there exists $K(\varepsilon_1) \in \N \st x_n<a_1+\frac{\varepsilon}{2},\ \forall\ n \geq K(\varepsilon_1)$. We also know that since $\lim \sup (y_n)=a_2,\ \exists\ K(\varepsilon_2) \in \N \st y_n<a_2+\frac{\varepsilon}{2},\ \forall\ n \geq K(\varepsilon_2)$.
|
||||
\\Let $K(\varepsilon)=\max \{K(\varepsilon_1), K(\varepsilon_2)\}$. Then we have that $x_n<a_1+\frac{\varepsilon}{2},\ y_n<a_2+\frac{\varepsilon}{2}\ \ \forall\ n \geq K(\varepsilon)\ \Rightarrow\ x_n+y_n<a_1+a_2+\varepsilon,\ \forall\ n \geq K(\varepsilon)$. Hence no subsequential limit of $(x_n+y_n)$ can be greater than $a_1+a_2+\varepsilon$.
|
||||
\\\\Thus $p \leq a_1+a_2+\varepsilon$. Now since $\varepsilon>0$ is arbitrary, we have that $p \leq a_1+a_2$.
|
||||
\\So $\lim \sup (x_n+y_n) \leq \lim \sup (x_n)+\lim \sup (y_n)$.
|
||||
\end{proof}
|
||||
Example: Let $x_n=\sin (\frac{n\pi}{2})$, and let $y_n=\cos(\frac{n\pi}{2})$, for $n \in \N$. Then we have that $(x_n)$ is a sequence of 0, 1, and $-1$. Additionally, $x_{4n+1}=\sin(\frac{(4n+1)\pi}{2})=1,\ \forall\ n \in \N$. Thus $\lim (x_{4n+1})=1$. Therefore we have that 1 is a subsequential limit of $(x_n)$. Also, we have that 1 is the greatest subsequential limit since $1>0$ and $1>-1$. So, $\lim \sup (x_n)=1$ and by similar logic we have that $\lim \sup (y_n)=1$, and $\lim \sup (x_n+y_n)=1$ because $(x_n+y_n)$ and $(y_n)$ are also sequence of $-1, 0, 1$, and $\lim (y_{4n})=\lim \cos 2n\pi=1$, $\lim (x_{4n+1}+y_{4n+1})=1$. Then $\lim \sup (x_n+y_n)=1<2=\lim \sup (x_n)+\lim \sup (y_n)$.
|
||||
\end{enumerate}
|
||||
\item
|
||||
\begin{enumerate}
|
||||
\item Show that $x_n=e^{\sin (5n)}$ has a convergent subsequence.
|
||||
\\\\Let $y_n=x_{\frac{n \pi}{10n}+\frac{2 \pi n}{5}}$.
|
||||
\\\\Then we have that the first 5 terms of is subsequence of $(x_n)$ are $e^{\sin(\frac{5\pi}{2})}$, $e^{\sin(\frac{9\pi}{2})}$, $e^{\sin(\frac{13\pi}{2})}$, $e^{\sin(\frac{17\pi}{2})}$, $e^{\sin(\frac{21\pi}{2})}$, $\dots = e^1, e^1, e^1, e^1, e^1, \dots = e,e,e,e,e,\dots$. Thus we have that the subsequence $(y_n)$ converges to $e$. \\
|
||||
|
||||
\item Give an example of a bounded sequence with three subsequences converging to three different numbers.
|
||||
\\\\Let $a_n=(n \mod 3 + 1)$ be a bounded sequence. We have that $(a_n)$ is bounded above by $\frac{10}{3}$ and is bounded below by 2. The first five terms of this sequence are $\frac{5}{2}$, $\frac{10}{3}$, 2, $\frac{5}{2}$, $\frac{10}{3}$, $\dots$. Now, let $x_n=a_{3n+1}$, $y_n=a_{3n+2}$, and $z_n=a_{3n+3}$. Thus for each of the sequences, we have the following:
|
||||
\[x_n=(\frac{5}{2}, \frac{5}{2}, \frac{5}{2}, \frac{5}{2}, \frac{5}{2}, \dots)\]
|
||||
\[y_n=(\frac{10}{3}, \frac{10}{3}, \frac{10}{3}, \frac{10}{3}, \frac{10}{3}, \dots)\]
|
||||
\[z_n=(2,2,2,2,2,\dots)\]
|
||||
By this, we can conclude that $(x_n)$ converges to $\frac{5}{2}$, $(y_n)$ converges to $\frac{10}{2}$, and that $(z_n)$ converges to $2$. Hence we have that there exist three different subsequences that converge to three different numbers. \\
|
||||
|
||||
\item Give an example of a sequence $x_n$ with $\lim \sup x_n = 5$ and $\lim \inf x_n = -3$.
|
||||
\\\\Consider the sequence $a_n=(-3,5,-3,5,-3,5,-3,5,\dots)$. For the subsequences $b_n=a_{2n-1}=(-3,-3,-3,-3,-3,\dots)$, and $c_n=a_{2n}=(5,5,5,5,5,\dots)$. Then we have the following:
|
||||
\begin{align*}
|
||||
\inf (a_n)=-3\ \ \ \ \ \ &\sup (a_n)=5 &\lim \inf (a_n)=-3\ \ \ \ \ \ &\lim \sup (a_n)=5 \\
|
||||
\inf (b_n)=-3\ \ \ \ \ \ &\sup (b_n)=-3 &\lim \inf (b_n)=-3\ \ \ \ \ \ &\lim \sup (b_n)=-3 \\
|
||||
\inf (c_n)=5\ \ \ \ \ \ &\sup (c_n)=5 &\lim \inf (c_n)=5\ \ \ \ \ \ &\lim \sup (c_n)=5
|
||||
\end{align*}
|
||||
|
||||
\item Let $\lim \sup x_n = 2$. True or False: if $n$ is sufficiently large, then $x_n > 1.99$.
|
||||
\\\\This statement is true. Let $(x_{n_k})$ be a subsequence of $(x_n)$ such that $\lim (x_{n_k})=2$. Then we have that if $(x_{n_k})$ decreases to 2, then any element of $(x_{n_k})$ is going to be greater than 1.99. If $(x_{n_k})$ is a constant sequence, then the same is also true. Lastly, if $(x_{n_k})$ is increasing to 2, then we have that elements of $(x_{n_k})$ must have an arbitrary distance $\varepsilon$ between the elements themselves and the limit of 2. Thus, we have that $\forall\ \varepsilon>0,\ \exists\ K(\varepsilon) \in \N \st |x_{n_k}-2|<\varepsilon,\ \forall\ n_k \geq K(\varepsilon)$, by the definition of the limit of a sequence.
|
||||
|
||||
\item Compute the infimum, supremum, limit infimum, and limit supremum for $a_n = 3 - (-1)^n - (-1)^n/n$.
|
||||
\[\inf (a_n)=1.5\]
|
||||
\[\sup (a_n)=5\]
|
||||
\[\lim \inf (a_n)=2\]
|
||||
\[\lim \sup (a_n)=4\]
|
||||
\end{enumerate}
|
||||
|
||||
\item Prove or justify, if true. Provide a counterexample, if false.
|
||||
\begin{enumerate}
|
||||
\item If $a_n$ and $b_n$ are strictly increasing, then $a_n + b_n$ is strictly increasing.
|
||||
\begin{proof}
|
||||
Since $(a_n)$ and $(b_n)$ are strictly increasing, we have the following:
|
||||
\[a_1<a_2<a_3<\dots a_n\]
|
||||
and
|
||||
\[b_1<b_2<b_3<\dots <b_n\]
|
||||
Thus, the sum of the sequences $(a_n+b_n)$ is
|
||||
\[a_1+b_1<a_2+b_2<a_3+b_3< \dots < a_n+b_n\]
|
||||
\end{proof}
|
||||
|
||||
\item If $a_n$ and $b_n$ are strictly increasing, then $a_n \cdot b_n$ is strictly increasing.
|
||||
\\\\Let $(a_n)=\frac{-1}{n}$ and let $(b_n)=n$. Then we have the following:
|
||||
\[a_n=-1,\frac{-1}{2},\frac{-1}{3},\frac{-1}{4}, \frac{-1}{5}, \dots \]
|
||||
\[b_n=1,2,3,4,5,\dots \]
|
||||
However, since the product of these two strictly increasing sequences is
|
||||
\[a_n \cdot b_n = -1,-1,-1,-1,-1, \dots \]
|
||||
we have that the product is not strictly increasing and thus this statement is false.
|
||||
|
||||
\item If $a_n$ and $b_n$ are monotonic, then $a_n + b_n$ is monotonic.
|
||||
\\\\This is a false statement. Consider the following monotonic sequences:
|
||||
\[a_n=(1,2,2,3,3,\dots)\]
|
||||
\[b_n=(-1,-1,-2,-2,-3, \dots)\]
|
||||
Their sum is the sequence $(a_n+b_n)=(0,1,0,1,0,\dots)$. This sequence is not monotonic since it oscillates between 0 and 1.
|
||||
|
||||
\item If $a_n$ and $b_n$ are monotonic, then $a_n \cdot b_n$ is monotonic.
|
||||
\\\\This statement is false. Consider the following monotonic sequences:
|
||||
\[a_n=(1,1,2, \dots)\]
|
||||
\[b_n=(1, \frac{1}{2}, \frac{1}{2}, \dots)\]
|
||||
Then we have that the product of these two monotonic sequences is $a_n \cdot b_n=(1,\frac{1}{2},1,\dots)$, which is an oscillating sequence. Thus we have that the product of two monotonic sequences is not monotonic.
|
||||
|
||||
\item If a monotone sequence is bounded, then it is convergent.
|
||||
\\\\This statement is true. For proof, consult the proof of the \textit{Monotone Convergence Theorem} (Monotone Sequence Property).
|
||||
|
||||
\item If a bounded sequence is monotone, then it is convergent.
|
||||
\\\\This statement is true. For proof, consult the proof of the \textit{Monotone Convergence Theorem} (Monotone Sequence Property).
|
||||
|
||||
\item If a convergent sequence is monotone, then it is bounded.
|
||||
\\\\This statement is true. For proof, consult the proof of \textit{Theorem 3.2.2}:
|
||||
\begin{theorem*}
|
||||
A convergent sequence of real numbers is bounded.
|
||||
\end{theorem*}
|
||||
|
||||
\item If a convergent sequence is bounded, then it is monotone.
|
||||
\\\\This statement is false. Consider the sequence generated by $a_n=(-1)^n\frac{1}{n}=(-1,\frac{1}{2},\frac{-1}{3},\frac{1}{4},\frac{-1}{5}, j\dots)$. This sequence converges to 0, however since it does oscillate as it converges, we have that $a_n$ is not monotone, yet it is bounded.
|
||||
\end{enumerate}
|
||||
\end{enumerate}
|
||||
\end{document}
|
||||
Binary file not shown.
+332
@@ -0,0 +1,332 @@
|
||||
\documentclass[12pt,letterpaper]{article}
|
||||
\usepackage[utf8]{inputenc}
|
||||
\usepackage[english]{babel}
|
||||
\usepackage{amsthm}
|
||||
\usepackage{amsmath}
|
||||
\usepackage{amsfonts}
|
||||
\usepackage{amssymb}
|
||||
\usepackage{graphicx}
|
||||
\usepackage{array}
|
||||
\usepackage[left=2cm, right=2.5cm, top=2.5cm, bottom=2.5cm]{geometry}
|
||||
\usepackage{enumitem}
|
||||
\newcommand{\st}{\ \text{s.t.}\ }
|
||||
\newcommand{\abs}[1]{\left\lvert #1 \right\rvert}
|
||||
\newcommand{\R}{\mathbb{R}}
|
||||
\newcommand{\N}{\mathbb{N}}
|
||||
\newcommand{\Q}{\mathbb{Q}}
|
||||
\newcommand{\C}{\mathbb{C}}
|
||||
\newcommand{\Z}{\mathbb{Z}}
|
||||
\DeclareMathOperator{\sign}{sgn}
|
||||
\newtheoremstyle{case}{}{}{}{}{}{:}{ }{}
|
||||
\theoremstyle{case}
|
||||
\newtheorem{case}{Case}
|
||||
\theoremstyle{definition}
|
||||
\newtheorem{definition}{Definition}[section]
|
||||
\newtheorem*{definition*}{Definition}
|
||||
\newtheorem{theorem}{Theorem}[section]
|
||||
\newtheorem*{theorem*}{Theorem}
|
||||
\newtheorem{corollary}{Corollary}[section]
|
||||
\newtheorem*{lemma*}{Lemma}
|
||||
\newtheorem{lemma}[theorem]{Lemma}
|
||||
\newtheorem*{remark}{Remark}
|
||||
\setlist[enumerate]{font=\bfseries}
|
||||
\renewcommand{\qedsymbol}{$\blacksquare$}
|
||||
\author{Alexander J. Tusa}
|
||||
\title{Real Analysis Homework 6}
|
||||
\begin{document}
|
||||
\maketitle
|
||||
\begin{enumerate}
|
||||
\item \textbf{Section 3.5}
|
||||
\begin{enumerate}
|
||||
\item[2)a)] Show directly from the definition that $\left(\frac{n+1}{n}\right)$ is a Cauchy sequence.
|
||||
|
||||
\begin{proof}
|
||||
Let $a_n:=(\frac{n+1}{n})=(1+\frac{1}{n})$ be a sequence. We want to show that $\forall\ \varepsilon > 0,\ \exists\ H(\varepsilon) \in \N \st \forall\ x_n,x_m \in a_n,\ |x_n-x_m|<\varepsilon,\ \text{for } m,n \geq H(\varepsilon)$.
|
||||
\\\\Recall that if $\varepsilon > 0,\ \exists\ n_\varepsilon \in \N \st 0<\frac{1}{n_\varepsilon}<\varepsilon$. So, we want to show that if we let $n_\varepsilon=H(\varepsilon)$, $\frac{1}{H(\varepsilon)}<\frac{\varepsilon}{2}$.
|
||||
\\\\Let $m>n \geq H(\varepsilon)$.
|
||||
\\\\So,
|
||||
\begin{align*}
|
||||
\abs{\frac{m+1}{m}-\frac{n+1}{n}}&=\abs{\frac{1}{m}-\frac{1}{n}} \\
|
||||
&\leq \frac{1}{m}+\frac{1}{n} &\text{by the Triangle Inequality} \\
|
||||
&\leq \frac{2}{n} &\text{since } m>n \Rightarrow \frac{1}{m}<\frac{1}{n} \\
|
||||
&\leq \frac{2}{H(\varepsilon)} &\text{since } n \geq H(\varepsilon) \\
|
||||
&< \varepsilon &\text{since } \frac{1}{H(\varepsilon)} < \frac{\varepsilon}{2}
|
||||
\end{align*}
|
||||
Thus, $a_n=(\frac{n+1}{n})$ is a Cauchy sequence.
|
||||
\end{proof}
|
||||
|
||||
\item[3)b)] Show directly from the definition that $\left(n+\frac{(-1)^n}{n}\right)$ is not a Cauchy sequence.
|
||||
\\\\Let $x_n:=n+\frac{(-1)^n}{n}$, for $n \geq 1$.
|
||||
\\\\Then we have the following:
|
||||
\begin{align*}
|
||||
x_{n+1}-x_n &= (n+1)+\frac{(-1)^{n+1}}{n+1}-n-\frac{(-1)^n}{n} \\
|
||||
&=1+\frac{(-1)^{n+1}}{n+1}-\frac{(-1)^n}{n} \\
|
||||
&=1+(-1)^{n+1}\left(\frac{1}{n+1}+\frac{1}{n}\right)
|
||||
\end{align*}
|
||||
So,
|
||||
\begin{align*}
|
||||
x_{2m+2}-x_{2m+1} &=1+(-1)^{2m+1+1}\left(\frac{1}{2m+2}+\frac{1}{2m+1}\right) \\
|
||||
&=1 +\frac{1}{2m+2}+\frac{1}{2m+1}
|
||||
\end{align*}
|
||||
Thus, $|x_{2m+2}-x_{2m+1}|>1$. Hence, if we let $\varepsilon=\frac{1}{2}$, there doesn't exist $H(\varepsilon) \in \N \st |x_n-x_m|<\frac{1}{2},\ \forall\ m,n \geq H(\varepsilon)$. This is because
|
||||
\[\abs{x_{2H(\varepsilon)+2}-x_{2H(\varepsilon)+1}} = 1 + \frac{1}{2H(\varepsilon)+2}+\frac{1}{2H(\varepsilon)+1}>1>\frac{1}{2}\]
|
||||
Thus, $\left(n+\frac{(-1)^n}{n}\right)$ is not a Cauchy sequence.\\
|
||||
|
||||
\item[4)] Show directly from the definition that if $(x_n)$ and $(y_n)$ are Cauchy sequences, then $(x_n+y_n)$ and $(x_ny_n)$ are Cauchy sequences.
|
||||
\begin{proof}
|
||||
Recall \textit{Lemma 3.5.4}:
|
||||
\begin{lemma*}
|
||||
A Cauchy sequence of real numbers is bounded.
|
||||
\end{lemma*}
|
||||
Thus, we have that $(x_n)$ and $(y_n)$ are bounded. By the definition of boundedness, we know that there exists $M_1,M_2 \in \R$ such that the following hold:
|
||||
\[\exists M_1>0 \st |x_n|\leq M_1\ \forall\ n \in \N\]
|
||||
and
|
||||
\[\exists M_2>0 \st |y_n| \leq M_2 \forall\ n \in \N\]
|
||||
Let $M=\max \{M_1, M_2\}$.
|
||||
\\\\Since $(x_n)$ and $(y_n)$ are Cauchy sequences, we have the following:
|
||||
\[\forall\ \varepsilon>0,\ \exists\ H_1(\varepsilon) \in \N \st |x_n-x_m|<\frac{\varepsilon}{2},\ \forall\ m,n\geq H_1(\varepsilon)\]
|
||||
\[\forall\ \varepsilon>0,\ \exists\ H_2(\varepsilon) \in \N \st |y_n-y_m|<\frac{\varepsilon}{2},\ \forall\ m,n\geq H_2(\varepsilon)\]
|
||||
Thus, by the Triangle Inequality, we have
|
||||
\[|(x_n+y_n)-(x_m+y_m)|\leq|x_n-x_m|+|y_n-y_m|\]
|
||||
Let $H(\varepsilon) = \max \{H_1(\varepsilon), H_2(\varepsilon)\}$. Then we have that $|x_n-x_m| \leq \frac{\varepsilon}{2}$ and $|y_n-y_m| \leq \frac{\varepsilon}{2},\ \forall\ m,n \geq H(\varepsilon)$. Thus,
|
||||
\[|(x_n+y_n)-(x_m-y_m)| \leq \frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon,\ \forall\ m,n \geq H(\varepsilon)\]
|
||||
$\therefore\ (x_n+y_n)$ is Cauchy.
|
||||
\\\\Similarly, to show that $x_ny_n$ is Cauchy, we have
|
||||
\begin{align*}
|
||||
|x_ny_n-x_my_m| &= |x_ny_n-y_nx_m+x_my_n-x_my_m| \\
|
||||
&\leq |y_n||x_n-x_m|+|x_m||y_n-y_m| \\
|
||||
&\leq M(|x_n-x_m|+|y_n-y_m|) \\
|
||||
&\leq M \left(\frac{\varepsilon}{2}+\frac{\varepsilon}{2}\right), &\forall\ n,m\geq H(\varepsilon) \\
|
||||
&= M_\varepsilon
|
||||
\end{align*}
|
||||
Note that we can initially replace $\varepsilon$ by $\frac{\varepsilon}{M}$ to get $|x_ny_n-x_my_m|\leq \varepsilon,\ \forall\ n,m \geq H(\varepsilon)$.
|
||||
\\\\Thus we have that $(x_ny_n)$ is also Cauchy.
|
||||
|
||||
\end{proof}
|
||||
|
||||
\item[5)] If $x_n:= \sqrt{n}$, show that $(x_n)$ satisfies $\lim |x_{n+1}-x_n|=0$, but that it is not a Cauchy sequence.
|
||||
\\\\Note that $(x_n)$ is an unbounded sequence. By the \textit{Cauchy Convergence Criterion}, since $\sqrt{n}$ is not bounded, $(x_n)$ is not a Cauchy sequence. Now, we only must show that $|x_{n+1}-x_n|$ converges to 0.
|
||||
\\So,
|
||||
\begin{align*}
|
||||
|x_{n+1}-x_n| &= \sqrt{n+1}-\sqrt{n} \\
|
||||
&= \frac{1}{\sqrt{n+1}+\sqrt{n}} \\
|
||||
\Rightarrow \lim\limits_{n \to \infty} |x_{n+1}-x_n| &= \lim\limits_{n \to \infty} \frac{1}{\sqrt{n+1}+\sqrt{n}} \\
|
||||
&=0
|
||||
\end{align*}
|
||||
Thus, we have that $\lim |x_{n+1}-x_n|=0$, but $(x_n)$ is not Cauchy.\\
|
||||
|
||||
\item[12)] If $x_1>0$ and $x_{n+1}:=(2+x_n)^{-1}$ for $n \geq 1$, show that $(x_n)$ is a contractive sequence. Find the limit.
|
||||
\begin{proof}
|
||||
Recall the definition of a contractive sequence:
|
||||
\theoremstyle{definition}
|
||||
\begin{definition*}
|
||||
We say that a sequence $(x_n)$ of real numbers is \textbf{contractive} if there exists a constant $C$, $0<C<1$, such that
|
||||
\[|x_{n+2}-x_{n+1}| \leq C|x_{n+1}-x_n|\]
|
||||
for all $n \in \N$. The number $C$ is called the \textbf{constant} of the contractive sequence.
|
||||
\end{definition*}
|
||||
For $(x_n)$, we have the following:
|
||||
\begin{align*}
|
||||
|x_{n+2}-x_{n+1}| &= |(2+x_{n+1})^{-1}-(2+x_n)^{-1}| \\
|
||||
&= \abs{\frac{1}{2+x_{n+1}}-\frac{1}{2+x_n}} \\
|
||||
&= \abs{\frac{2-x_n-(2+x_{n+1})}{(2+x_{n+1})(2+x_n)}} \\
|
||||
&= \frac{|x_n-x_{n+1}|}{(2+x_{n+1})(2+x_n)} \\
|
||||
&\leq \frac{|x_n-x_{n+1}|}{(2+0)(2+0)} \\
|
||||
&= \frac{1}{4} \cdot |x_{n+1}-x_n|
|
||||
\end{align*}
|
||||
Thus, if we let $C=\frac{1}{4}$, then $(x_n)$ is a contractive sequence.
|
||||
\end{proof}
|
||||
Now we want to find the limit of $(x_n)$.
|
||||
\\\\Recall \textit{Theorem 3.5.8}:
|
||||
\begin{theorem*}
|
||||
Every contractive sequence is a Cauchy sequence, and therefore is convergent.
|
||||
\end{theorem*}
|
||||
By \textit{Theorem 3.5.8}, we have that $\lim (x_n)=x$ exists, and since we know that $\lim (x_{n+1})=\lim (x_n)=x$, we have the following:
|
||||
\begin{align*}
|
||||
x_{n+1} &= (2+x_n)^{-1} \\
|
||||
x_{n+1} &= \left.\frac{1}{2+x_n}\ \ \ \ \ \right| \lim \\
|
||||
x &= \frac{1}{2+x}\ \ \ \ \ \left.\right| \cdot (2+x) \\
|
||||
x^2+2x &=1 \\
|
||||
x^2+2x-1 &= 0 \\
|
||||
x &= \frac{-2 \pm \sqrt{4-4 \cdot 1 \cdot (-1)}}{2} \\
|
||||
x_a &= -1-\sqrt{2} < 0 \\
|
||||
x_b &= -1 + \sqrt{2}>0
|
||||
\end{align*}
|
||||
Since $x_n>0,\ \forall\ n$, we know that $x_a<0$ cannot be the limit, and thus we can conclude that $\lim (x_n)=-1+\sqrt{2}$.\\
|
||||
|
||||
\item[13)] If $x_1:=2$ and $x_{n+1}:=2+1/x_n$ for $n \geq 1$, show that $(x_n)$ is a contractive sequence. What is the limit?
|
||||
\begin{lemma}
|
||||
We want to show that $x_n\geq 2\ \forall\ n \in \N$. We prove this by method of mathematical induction.
|
||||
\\\\\textbf{Basis Step:} Let $n=1$. Then we have that $x_1=2 \geq 2$.
|
||||
\\\\\textbf{Inductive Step:} Assume that $x_n \geq 2$ for arbitrary $n \in \N$.
|
||||
\\\\\textbf{Show:} We want to now show that $x_{n+1} \geq 2,\ \forall\ n \in \N$. So we have the following:
|
||||
\begin{align*}
|
||||
x_{n+1} &= 2 + \frac{1}{x_n} \\
|
||||
&\geq 2+0 \\
|
||||
&=2
|
||||
\end{align*}
|
||||
$\therefore$ by mathematical induction, we have that $x_n\geq 2,\ \forall\ n \in \N$
|
||||
\end{lemma}
|
||||
|
||||
\begin{proof}
|
||||
By the definition of a contractive sequence, we have the following:
|
||||
\begin{align*}
|
||||
|x_{n+2}-x_{n+1}| &= \abs{\left(2+\frac{1}{x_{n+1}}\right)- \left(2+\frac{1}{x_n}\right)} \\
|
||||
&=\abs{\frac{1}{x_n+1}-\frac{1}{x_n}} \\
|
||||
&= \abs{\frac{x_n-x_{n+1}}{x_{n+1} \cdot x_n}}
|
||||
&\leq \abs{\frac{x_n-x_{n+1}}{2 \cdot 2}} &\text{by Lemma 0.1} \\
|
||||
&= \frac{1}{4}|x_n-x_{n+1}|
|
||||
\end{align*}
|
||||
So, if we let $C=\frac{1}{4}$, then we have shown that by the definition of a contractive sequence, $(x_n)$ is contractive.
|
||||
|
||||
\end{proof}
|
||||
Now we want to find the limit of $(x_n)$. By \textit{Theorem 3.5.8}, we have that since $(x_n)$ is contractive, it is also convergent. Thus, we know that $\lim (x_n)=x$ exists, and since we also know that $\lim (x_{n+1})=\lim (x_n)=x$, we have the following:
|
||||
\begin{align*}
|
||||
x_{n+1} &=2 + \frac{1}{x_n}\ \ \ \ \ \left.\right| \lim \\
|
||||
x &=2+\frac{1}{x}\ \ \ \ \ \left.\right| \cdot x \\
|
||||
x^2 &= 2x + 1 \\
|
||||
x^2-2x-1 &= 0 \\
|
||||
x &= \frac{2 \pm \sqrt{4 - 4 \cdot 1 \cdot (-1)}}{2} \\
|
||||
x_a &= 1-\sqrt{2} < 2 \\
|
||||
x_b &= 1+\sqrt{2}>2
|
||||
\end{align*}
|
||||
By \textit{Lemma 0.1}, we know that $x_n \geq 2,\ \forall\ n \in \N$, and thus we know that $x_a<2$ can't be the limit. Thus, we can conclude that $\lim (x_n)=1 +\sqrt{2}$.
|
||||
\end{enumerate}
|
||||
|
||||
\item Find examples of sequences of real numbers satisfying each set of properties:
|
||||
\begin{enumerate}
|
||||
\item Cauchy, but not monotone
|
||||
\\\\Consider the sequence $a_n:=(1,\frac{1}{3},\frac{1}{2},\frac{1}{5},\frac{1}{4}, \dots)$. Notice that $\forall\ \varepsilon > 0,\ \exists\ H(\varepsilon) \in \N \st |a_n-a_m|<\varepsilon,\ \text{for}\ m,n \in \N,\ \forall\ m,n \geq H(\varepsilon)$. Thus, this sequence satisfies the definition of a Cauchy sequence, and thus it is a Cauchy sequence. However, notice that while it is decreasing overall, it is not decreasing in a way that coincides with the definition of monotone decreasing.
|
||||
\\\\That is, in order for this sequence to be considered monotone decreasing, we must have the following:
|
||||
\[a_1 \geq a_2 \geq a_3 \geq a_4 \geq a_5 \geq \dots \geq a_n,\ \forall\ n \in \N\]
|
||||
But, given the first five terms of this sequence, we have
|
||||
\[1 \geq \frac{1}{3} \leq \frac{1}{2} \geq \frac{1}{5} \leq \frac{1}{4} \dots\]
|
||||
Hence this sequence is Cauchy but not monotone.\\
|
||||
|
||||
\item Monotone, but not Cauchy
|
||||
\\\\Consider the sequence $a_n:=(1,4,9,16,25,\dots)=n^2$. This sequence is monotone, since it is an increasing sequence, however it is not Cauchy. Consider $\varepsilon=1$. Since $(a_n)$ is an increasing sequence, we have that for any two $m,n \in \N$, $|a_n-a_m| > 1$. Thus, $\nexists\ H(\varepsilon) \in \N \st \forall\ \varepsilon > 0,\ |a_n-a_m|<\varepsilon,\ \text{for}\ m,n\in \N,\ \forall\ m,n \geq H(\varepsilon)$. Thus $(a_n)$ is a monotone sequence but it is not a Cauchy sequence.\\
|
||||
|
||||
\item (Section 3.5, Problem 1) Bounded, but not Cauchy
|
||||
\\\\Consider the sequence $a_n:=(-1,1,-1,1,-1, \dots)=(-1)^n$. This sequence is bounded above by $1$ and is bounded below by $-1$, and thus this sequence is bounded. However, since $\forall\ m,n \in \N,\ |a_n-a_m|=|a_m-a_n|=2$, we have that the sequence is not Cauchy. Consider $\varepsilon = 1$. Then we have that $\nexists\ H(\varepsilon) \in \N \st |a_n-a_m|<\varepsilon,\ \text{for}\ m,n \in \N,\ \forall\ m,n\geq H(\varepsilon)$. Thus, we have a sequence that is bounded, but is not a Cauchy sequence.\\
|
||||
\end{enumerate}
|
||||
|
||||
\item
|
||||
\begin{enumerate}
|
||||
\item Let $a_n=\frac{1}{2}+\frac{1}{6}+ \dots + \frac{1}{n(n+1)}$. Show $a_n$ is Cauchy.
|
||||
\begin{proof}
|
||||
Let $a_n=\frac{1}{2}+\frac{1}{6}+ \dots + \frac{1}{n(n+1)}$. We want to show that $a_n$ is Cauchy. That is, we want to show that $\forall\ \varepsilon>0,\ \exists\ H(\varepsilon) \in \N, \st |a_m-a_n|<\varepsilon,\ \forall\ m,n\geq H(\varepsilon)$.
|
||||
\\\\Let $\varepsilon >0$ be given. Without loss of generality, let $m \geq n$. Then we have that $0<a_m-a_n$, which yields the following:
|
||||
\begin{align*}
|
||||
&= \left(\frac{1}{2}+\frac{1}{6}+ \dots + \frac{1}{n(n+1)}+\frac{1}{(n+1)(n+2)}+ \dots + \frac{1}{m(m+1)}\right) - \left(\frac{1}{2} + \dots + \frac{1}{n(n+1)}\right) \\
|
||||
&= \frac{1}{(n+1)(n+2)}+ \dots + \frac{1}{m(m+1)} \\
|
||||
&= \left(\frac{1}{n+1}-\frac{1}{n+2}\right)+ \left(\frac{1}{n+2}-\frac{1}{n+3}\right) + \dots + \left(\frac{1}{m}-\frac{1}{m+1}\right) \\
|
||||
&= \frac{1}{n+1}-\frac{1}{m+1} < \frac{1}{n+1} < \frac{1}{n}=\varepsilon
|
||||
\end{align*}
|
||||
So, if we let $H(\varepsilon)\geq \frac{1}{\varepsilon}$, we then have that $|a_m-a_n|<\varepsilon,\ \forall\ m,n\geq H(\varepsilon)$.
|
||||
\\\\$\therefore\ a_n$ is Cauchy.
|
||||
\end{proof}
|
||||
|
||||
\item Let $a_n$ satisfy $|a_n-a_{n+1}| \leq 1/3^n$. Show $a_n$ converges.
|
||||
\begin{proof}
|
||||
First note that if we let $b_n:=\frac{1}{3^n}$, then we have the sequence $b_n=(\frac{1}{3}, \frac{1}{9}, \frac{1}{27})$. Notice that this sequence is a Cauchy sequence. This is because if we have the following:
|
||||
\begin{align*}
|
||||
3^n &< 3^{2n} \\
|
||||
\Rightarrow \frac{1}{3^n} &> \frac{1}{3^{2n}}
|
||||
\end{align*}
|
||||
Thus we have that if we let $\varepsilon = \frac{1}{3^n}$, then $|a_n-a_{n+1}|<\frac{1}{3^2n}<\frac{1}{3^n}=\varepsilon$. So, if we let $H(\varepsilon)=\frac{\log (\frac{1}{\varepsilon})}{2 \log (3)}$, then $\forall\ n \geq H(\varepsilon),\ |a_n-a_{n+1}|<\varepsilon$. Consider if we let $m=n+1$. Then we can rewrite this as follows:
|
||||
\[\forall\ \varepsilon>0,\ \exists\ H(\varepsilon) \in \N \st |a_n-a_m|<\varepsilon,\ \forall\ m,n \geq H(\varepsilon),\ \text{where}\ H(\varepsilon)=\frac{\log (\frac{1}{\varepsilon})}{2 \log (3)}\]
|
||||
Thus we have that any sequence $a_n$ that satisfies this property must be a Cauchy Sequence. Thus by the \textit{Cauchy Convergence Criteria}, we have that $a_n$ is convergent since it is also Cauchy. Hence $\lim (a_n) = A$ exists.
|
||||
\end{proof}
|
||||
|
||||
\item Prove that if $a_n$ converges, then $\lim\limits_{n \to \infty} |a_{n+1}-a_n|=0$.
|
||||
\begin{proof}
|
||||
Suppose that $(a_n)$ is a convergent sequence. Then we have that by the \textit{Cauchy Convergence Criterion}, $(a_n)$ is a Cauchy sequence.
|
||||
\\\\Recall \textit{Theorem 3.1.3}:
|
||||
\begin{theorem*}
|
||||
Let $X=(x_n:n \in \N)$ be a sequence of real numbers and let $m \in \N$. Then the $m$-tail $X_m=(x_{m+n}:n \in \N)$ of $X$ converges if and only if $X$ converges. In this case, $\lim X_m = \lim X$.
|
||||
\end{theorem*}
|
||||
Thus by \textit{Theorem 3.1.3}, we have that if we let $m=1$, then $a_{m+n}=a_{n+1}$ also converges since $a_n$ converges.
|
||||
\\\\Also, recall \textit{Theorem 3.4.2}:
|
||||
\begin{theorem*}
|
||||
If a sequence $X=(x_n)$ of real numbers converges to a real number $x$, then any subsequence $X'=(x_{n_k})$ of $X$ also converges to $x$.
|
||||
\end{theorem*}
|
||||
Thus we have that since the sequence $(a_{n+1})$ is a subsequence of $(a_n)$, by \textit{Theorem 3.4.2}, the subsequence $(a_{n+1})$ also converges to $x$.
|
||||
\\\\Let $\lim\limits_{n \to \infty} (a_n)=A$ and let $\lim\limits_{n \to \infty} (a_{n+1}) = B$, for $A,B \in \R$.
|
||||
\\\\Consider first the sequence generated by $(a_{n+1}-a_n)$.
|
||||
\\\\Recall \textit{Theorem 3.2.3}:
|
||||
\begin{theorem*}
|
||||
\begin{enumerate}
|
||||
\item Let $X=(x_n)$ and $Y=(y_n)$ be sequences of real numbers that converge to $x$ and $y$, respectively, and let $c \in \R$. Then t he sequences $X+Y, X-Y, X \cdot Y$, and $cX$ converge to $x+y, x-y, xy$, and $cx$, respectively.
|
||||
|
||||
\item If $X=(x_n)$ converges to $x$ and $Z=(z_n)$ is a sequence of nonzero real numbers that converges to $z$ and if $z \neq 0$, then the quotient sequence $X/Z$ converges to $x/z$.
|
||||
\end{enumerate}
|
||||
\end{theorem*}
|
||||
Thus, by \textit{Theorem 3.2.3}, we have
|
||||
\[\lim\limits_{n \to \infty} (a_{n+1}-a_n)=B-A\]
|
||||
However, since we have that $\lim\limits_{n \to \infty} (a_n)=x$ and by \textit{Theorem 3.4.2}, since $(a_{n+1})$ is a subsequence of $(a_n)$, we know that
|
||||
\[\lim\limits_{n \to \infty} (a_{n+1}) = \lim\limits_{n \to \infty} (a_n) = x\]
|
||||
Which yields that $A=x=B$. So, we have
|
||||
\[\lim\limits_{n \to \infty} (a_{n+1}-a_n)=B-A=A-A=B-B=x-x=0\]
|
||||
Hence,
|
||||
\[\lim\limits_{n \to \infty} |a_{n+1}-a_n|=|0|=0\]
|
||||
$\therefore$ if $(a_n)$ converges, then $\lim\limits_{n \to \infty} |a_{n+1}-a_n| = 0$.
|
||||
\end{proof}
|
||||
|
||||
\item Give an example of a sequence $a_n$ where $\lim\limits_{n \to \infty}|a_{n+1}-a_n|=0$, but $a_n$ diverges.
|
||||
\\\\Consider the sequence $a_n:=(\frac{1}{2}, 1, \frac{4}{3}, \frac{19}{12}, \frac{107}{60}, \frac{38}{20}, \dots)$ $\approx$ (0.5, 1, 1.3333, 1.583333, 1.783333, 1.9, $\dots$). Thus we note that this sequence is monotone increasing, and also is unbounded. Thus this is a divergent sequence. However, note the resulting sequence of $|a_{n+1}-a_n|$:
|
||||
\[|a_{n+1}-a_n| = (\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}, \frac{1}{6}, \dots)\]
|
||||
We can note that the resulting sequence of $|a_{n+1}-a_n|$ is not only monotone decreasing, but also converges to 0. Thus we have defined a sequence $(a_n)$ such that $\lim\limits_{n \to \infty} (a_n) = \infty$, however $\lim\limits_{n \to \infty} |a_{n+1}-a_n| = 0$.\\
|
||||
|
||||
\item Let $a_n$ satisfy $a_{n+1}=a_n^2$ for all $n \in \N$ where $0 < a_1 \leq 1/3$. Show $a_n$ is contractive.
|
||||
\\\\To begin, we want to show that $a_{n+1} < \frac{1}{3}$. We prove it by method of mathematical induction.
|
||||
\begin{lemma}
|
||||
\textbf{Basis Step:} Let $n=1$. Then we have that $a_1 < \frac{1}{3}$
|
||||
\\\\\textbf{Inductive Step:} Assume that $a_2 < \frac{1}{3}$ for arbitrary $n \in \N$.
|
||||
\\\\\textbf{Show:} We want to now show that $a_{n+1} < \frac{1}{3},\ \forall\ n \in \N$. So,
|
||||
\begin{align*}
|
||||
a_{n+1} &= a_n^2 < \left(\frac{1}{3}\right)^2 &\text{by definition of } a_{n+1} \\
|
||||
&= a_n^2 < \frac{1}{9} < \frac{1}{3}
|
||||
\end{align*}
|
||||
Thus, by mathematical induction, we have that $a_n < \frac{1}{3},\ \forall\ n \in \N$.
|
||||
\end{lemma}
|
||||
Now, we want to show that $a_n$ is contractive.
|
||||
\begin{proof}
|
||||
By the definition of a contractive sequence, we have the following:
|
||||
\begin{align*}
|
||||
|a_{n+2}-a_{n+1}| &= a_{n+1}^2-a_n^2| \\
|
||||
&= |a_{n+1}-a_n||a_{n+1}+a_n| \\
|
||||
&\leq |a_{n+1}-a_n|(|a_{n+1}|+|a_n|) \\
|
||||
&\leq \frac{2}{3}|a_{n+1}-a_n|
|
||||
\end{align*}
|
||||
So, if we let $C=\frac{2}{3}$, then we have shown that by the definition of a contractive sequence, $(a_n)$ is contractive.
|
||||
\end{proof}
|
||||
\end{enumerate}
|
||||
|
||||
\item Show that the following sequences are not Cauchy.
|
||||
\begin{enumerate}
|
||||
\item $a_n=n^2$.
|
||||
\\\\Consider $\varepsilon=1$. Since $(a_n)$ is an increasing sequence, we have that for any two $m,n \in \N$, $|a_n-a_m| > 1$. Thus, $\nexists\ H(\varepsilon) \in \N \st \forall\ \varepsilon > 0,\ |a_n-a_m|<\varepsilon,\ \text{for}\ m,n\in \N,\ \forall\ m,n \geq H(\varepsilon)$. Thus $(a_n)$ is not a Cauchy sequence.\\
|
||||
|
||||
\item (Section 3.5, Problem 2b) $a_n=n+\frac{(-1)^n}{n}$.
|
||||
\\\\If we let $\varepsilon=\frac{1}{2}$, there doesn't exist $H(\varepsilon) \in \N \st |x_n-x_m|<\frac{1}{2},\ \forall\ m,n \geq H(\varepsilon)$. This is because
|
||||
\[\abs{x_{2H(\varepsilon)+2}-x_{2H(\varepsilon)+1}} = 1 + \frac{1}{2H(\varepsilon)+2}+\frac{1}{2H(\varepsilon)+1}>1>\frac{1}{2}\]
|
||||
Thus, $\left(n+\frac{(-1)^n}{n}\right)$ is not a Cauchy sequence.\\
|
||||
\end{enumerate}
|
||||
|
||||
\item Prove or justify, if true. Provide a counterexample, if false.
|
||||
\begin{enumerate}
|
||||
\item If $a_n$ is Cauchy and $b_n$ is bounded, then $a_n \cdot b_n$ is Cauchy.
|
||||
\\\\This is a false statement. Consider the sequences $(a_n):=1$, and $b_n:=(-1)^n$. We have that $a_n$ is a Cauchy sequence and we also have that $(b_n)$ is a bounded sequence since it is bounded below by $-1$ and it is bounded above by $1$. Then we have that the resulting sequence of $a_nb_n$ is
|
||||
\[a_n \cdot b_n = (-1,1,-1,1,-1, \dots)\]
|
||||
Thus since this sequence oscillates between $-1$ and $1$, it is not a Cauchy sequence since by the \textit{Cauchy Convergence Criterion}, since this sequence doesn't converge, it isn't Cauchy.
|
||||
|
||||
\item If $a_n$ is a monotone increasing sequence such that $a_{n+1}-a_n \leq 1/n$, then $a_n$ converges.
|
||||
\\\\This is a false statement. Consider the following sequence: $a_n:=(-1, -\frac{1}{2}, -\frac{1}{3}, -\frac{1}{4}, \dots, -\frac{1}{n})$. Thus we have that this sequence is both monotone increasing and is a harmonic sequence. Also, note for $n=1$, we have
|
||||
\[a_{n+1}-a_n = -\frac{1}{2}-(-1)= -\frac{1}{2}+1=\frac{1}{2} \leq \frac{1}{1}=1\]
|
||||
However, if we let $\varepsilon=\frac{1}{3}$, then we have that $|a_2-a_1| \nless \varepsilon$, since $|a_2-a_1|=\frac{1}{2} \nless \frac{1}{3}$. Hence this sequence is not true for all $\varepsilon$.
|
||||
|
||||
\item The Cauchy convergence criteria holds in $\Q$.
|
||||
\\\\This is a false statement. Consider the sequence $a_n:= (1.4,1.41,1.414, \dots)$. We have that this sequence is indeed a Cauchy sequence, however this sequence does not converge to a value in $\Q$. Rather, this sequence converges to $\sqrt{2} \notin \Q$. Thus the Cauchy convergence criteria does not hold in $\Q$.
|
||||
\end{enumerate}
|
||||
|
||||
\end{enumerate}
|
||||
\end{document}
|
||||
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|
||||
\documentclass[12pt,letterpaper]{article}
|
||||
\usepackage[utf8]{inputenc}
|
||||
\usepackage[english]{babel}
|
||||
\usepackage{amsthm}
|
||||
\usepackage{amsmath}
|
||||
\usepackage{amsfonts}
|
||||
\usepackage{amssymb}
|
||||
\usepackage{graphicx}
|
||||
\usepackage{array}
|
||||
\usepackage[left=2cm, right=2.5cm, top=2.5cm, bottom=2.5cm]{geometry}
|
||||
\usepackage{enumitem}
|
||||
\newcommand{\st}{\ \text{s.t.}\ }
|
||||
\newcommand{\abs}[1]{\left\lvert #1 \right\rvert}
|
||||
\newcommand{\R}{\mathbb{R}}
|
||||
\newcommand{\N}{\mathbb{N}}
|
||||
\newcommand{\Q}{\mathbb{Q}}
|
||||
\newcommand{\C}{\mathbb{C}}
|
||||
\newcommand{\Z}{\mathbb{Z}}
|
||||
\newcommand{\sgn}{\text{sgn}}
|
||||
\DeclareMathOperator{\sign}{sgn}
|
||||
\newtheoremstyle{case}{}{}{}{}{}{:}{ }{}
|
||||
\theoremstyle{case}
|
||||
\newtheorem{case}{Case}
|
||||
\theoremstyle{definition}
|
||||
\newtheorem{definition}{Definition}[section]
|
||||
\newtheorem{theorem}{Theorem}[section]
|
||||
\newtheorem*{theorem*}{Theorem}
|
||||
\newtheorem{corollary}{Corollary}[section]
|
||||
\newtheorem{lemma}[theorem]{Lemma}
|
||||
\newtheorem*{remark}{Remark}
|
||||
\setlist[enumerate]{font=\bfseries}
|
||||
\renewcommand{\qedsymbol}{$\blacksquare$}
|
||||
\author{Alexander J. Tusa}
|
||||
\title{Real Analysis Homework 7}
|
||||
\begin{document}
|
||||
\maketitle
|
||||
\begin{enumerate}
|
||||
\item \textbf{Section 4.1}
|
||||
\begin{enumerate}
|
||||
\item[1.] Determine a condition on $|x-1|$ that will assure that:
|
||||
\begin{enumerate}
|
||||
\item[(a)] $|x^2-1|<\frac{1}{2}$
|
||||
|
||||
We notice first that $|x^2-1|=|x-1||x+1|$.\\
|
||||
|
||||
Consider the case where $|x-1|<1$. Then we have that $-1 < x-1<1$ and thus $-3<1<x+1<3$ and thus $|x+1|<3$.\\
|
||||
|
||||
Hence we have that $|x^2-1|=|(x-1)(x+1)|=|x-1||x+1|<1 \cdot 3 = 3$.\\
|
||||
|
||||
Now, since $\frac{1}{6}<1$, if $|x-1|<\frac{1}{6}$, then $|x+1|<3$ and thus $|x^2-1|=|(x-1)(x+1)|=|x-1||x+1|<\frac{1}{6} \cdot 3 = \frac{1}{2}$.\\
|
||||
|
||||
$\therefore$ If we let $|x-1|<\frac{1}{6}$ then $|x^2-1|<\frac{1}{2}$.\\
|
||||
|
||||
\item[(c)] $|x^2-1|<1/n$ for a given $n \in \N$.\\
|
||||
|
||||
Notice that $0<\frac{1}{n}\leq 1$, and thus $0<\frac{1}{3n}\leq \frac{1}{3}<1$.\\
|
||||
|
||||
So, by \textit{Part (a)}, we have that if $|x-1|<\frac{1}{3n}$, then $|x+1|< 3$. Hence, $|x^2-1|=|(x-1)(x+1)|=|x-1||x+1|<\frac{1}{3n} \cdot 3 = \frac{1}{n}$.\\
|
||||
|
||||
$\therefore$ If $|x-1|<\frac{1}{3n}$, then $|x^2-1|<\frac{1}{n}$.
|
||||
\end{enumerate}
|
||||
|
||||
\item[7.] Show that $\lim\limits_{x \to c} x^3=c^3$ for any $c \in \R$. (Hint: Use $c=2$)\\
|
||||
|
||||
Suppose $|x-2|<1$.
|
||||
\begin{align*}
|
||||
|x^3-8| &= |x-2||x^2+2x+4|
|
||||
\end{align*}
|
||||
So, we have
|
||||
\begin{align*}
|
||||
|x^2+2x+4|=|(x^2-4x+4)+6x| \\
|
||||
&=|(x-2)^2+6(x-2)+12| \\
|
||||
&\leq |x-2|^2 + 6|x-2|+12 \\
|
||||
&< 1 +6 + 12 = 19
|
||||
\end{align*}
|
||||
So we have that if we let $\delta=\min \{1, \frac{\varepsilon}{19}\}$, then we have that $\lim\limits_{x \to c}x^3 = c^3$, for $c=2$.\\
|
||||
|
||||
More generally, we have the following:\\
|
||||
|
||||
$\forall\ c \in \R$, let $b=|c|+1$. Then we have that if $|x|<b$, then
|
||||
\[|x^2+cx+c^2| \leq |x|^2 + |c||x|+|c|^2 \leq 3b^2\]
|
||||
Now, if we let $\varepsilon>0$ be arbitrary, we have that for $\delta = \frac{\varepsilon}{3b^2}$ and $x$ such that $|x-c|<\delta$, then
|
||||
\begin{align*}
|
||||
|x^3-c^3| &= |(x-c)(x^2+cx+c^2)| \\
|
||||
&\leq |x-c||x^2+cx+c^2| \\
|
||||
&\leq 3b^2|x-c| \\
|
||||
&<3b^2 \frac{\varepsilon}{3b^2} \\
|
||||
&= \varepsilon
|
||||
\end{align*}
|
||||
Hence we have that $\lim\limits_{x \to c} x^3 = c^3,\ \forall\ c \in \R$.
|
||||
|
||||
\item[9.] Use either the $\varepsilon-\delta$ definition of the limit or the Sequential Criterion for limits, to establish the following limit:
|
||||
\begin{enumerate}
|
||||
\item[(b)] $\lim\limits_{x \to 1} \frac{x}{1+x}=\frac{1}{2}$\\
|
||||
|
||||
Let $\varepsilon>0,\ \delta=\min \{\frac{1}{2},\varepsilon\}$. Now for $0<|x-1|<\delta\leq \frac{1}{2}$, we have:
|
||||
\begin{align*}
|
||||
\abs{\frac{x}{1+x}-\frac{1}{2}} &= \abs{\frac{2x-(1+x)}{2(1+x)}} \\
|
||||
&=\abs{\frac{x-1}{2(1+x)}} \\
|
||||
&< |x-1| &(|x-1|<\frac{1}{2} \implies x+1>\frac{3}{2}) \\
|
||||
&<\delta \\
|
||||
&<\varepsilon
|
||||
\end{align*}
|
||||
And thus by the $\varepsilon-\delta$ definition of the limit, we have
|
||||
\[\lim\limits_{x \to 1} \frac{x}{1+x}=\frac{1}{2}\]
|
||||
\end{enumerate}
|
||||
|
||||
\item[10.] Use the definition of a limit to show that
|
||||
\begin{enumerate}
|
||||
\item[(a)] $\lim\limits_{x \to 2} (x^2+4x)=12$\\
|
||||
|
||||
\begin{align*}
|
||||
|x-2|<1 \implies |x^2+4x-12| &\leq |(x+6)(x-2)| \\
|
||||
&\leq |x+6||x-2| \\
|
||||
&= |x-2+8||x-2| \\
|
||||
&\leq (|x-2|+8)|x-2| \\
|
||||
&\leq (1+8)|x-2| &(|x-2|<1) \\
|
||||
&=9|x-2|
|
||||
\end{align*}
|
||||
For $\delta=\min \{1,\frac{\varepsilon}{9}\}$ and $x$ such that $|x-2|<\delta$ we have that $|x^2+4x-12|<\varepsilon$.\\
|
||||
|
||||
$\therefore\ \lim\limits_{x \to 2} x^2+4x=12$.
|
||||
|
||||
\item[(b)] $\lim\limits_{x \to -1} \frac{x+5}{2x+3} = 4$
|
||||
|
||||
\begin{align*}
|
||||
\abs{\frac{x+5}{2x+3}-4} &= \abs{\frac{x+5-4(2x+3)}{2x+3}} \\
|
||||
&=\abs{\frac{-7x-7}{2x+3}} \\
|
||||
&= 7 \abs{\frac{x+1}{2x+3}} \\
|
||||
&=\frac{7|x+1|}{|2x+3|}
|
||||
\end{align*}
|
||||
Now, if $|x+1|<\frac{1}{4}$, then
|
||||
\begin{align*}
|
||||
-\frac{5}{4}<x<-\frac{3}{4} &\implies \frac{1}{2}<2x+3<\frac{3}{2} \\
|
||||
&\implies 0<\frac{1}{2x+3}<2 \\
|
||||
\implies \abs{\frac{x+5}{2x+3}-4} &\leq \frac{7|x+1|}{|2x+3|} \\
|
||||
&< 14|x+1|
|
||||
\end{align*}
|
||||
Thus, let $\delta:=\min \{\frac{1}{4}, \frac{\varepsilon}{14}\}$.\\
|
||||
|
||||
$\therefore\ \lim\limits_{x \to -1} \frac{x+5}{2x+3}=4$
|
||||
\end{enumerate}
|
||||
|
||||
\item[15.] Let $f:\R \to \R$ be defined by the setting $f(x):=x$ if $x$ is rational, and $f(x)=0$ if $x$ is irrational.
|
||||
\begin{enumerate}
|
||||
\item[(a)] Show that $f$ has a limit at $x=0$.\\
|
||||
|
||||
$\forall\ \varepsilon > 0$, choose $\delta = \varepsilon$. Then we have that $\forall\ x \in \R$ with $|x|<\delta$, we have
|
||||
\[|f(x)-0|=|f(x)=\begin{cases}
|
||||
|x|, & \text{if } x \text{ is rational} \\
|
||||
0, & \text{if } x \text{ is irrational}
|
||||
\end{cases}\]
|
||||
\[\implies |f(x)|<\varepsilon\]
|
||||
$\therefore$ We have that $f$ has a limit of $0$ at $x=0$.
|
||||
|
||||
\item[(b)] Use a sequential argument to show that if $c \neq 0$, then $f$ does not have a limit at $c$.\\
|
||||
|
||||
Let $\{x_n\}$ and $\{y_n\}$ be two sequences, both converging to $c \st x_n \in \R \setminus \Q$ and since the rationals and irrationals are dense in the set of real numbers, $y_n \in \Q\ \forall\ n \in \N$.\\
|
||||
|
||||
Thus, $f(x_n)=0\ \forall\ n$ and $f(y_n)=y_n\ \forall\ n$. This yields that $f(x_n) \to 0$, but $f(y_n)=y_n \to c$. Thus we now have that $\lim\limits_{n \to \infty} f(x_n) \neq \lim\limits_{n \to \infty} f(y_n)$. Hence we have that $f$ does not have a limit at $x=c$.
|
||||
\end{enumerate}
|
||||
\end{enumerate}
|
||||
\item Use the definition of a limit to establish the following limits:
|
||||
\begin{enumerate}
|
||||
\item $\lim\limits_{x \to 1} \frac{x^2-x-2}{2x-3}=2$\\
|
||||
|
||||
We want to show the following $\forall\ \varepsilon > 0$:
|
||||
\begin{align*}
|
||||
\abs{\frac{x^2-x-2}{2x-3}-2} &= \abs{\frac{x^2-x-2-4x+6}{2x-3}} \\
|
||||
&= \abs{\frac{x^2-5x+4}{2x-3}} \\
|
||||
&= \frac{|x-4||x-1|}{|2x-3|} \\
|
||||
&<\varepsilon
|
||||
\end{align*}
|
||||
So, we have for the numerator:
|
||||
\begin{align*}
|
||||
|x-4| &= |(x-1)-3| \\
|
||||
&= |x-1|-3 \\
|
||||
&\leq 1-3 \\
|
||||
&= -2
|
||||
\end{align*}
|
||||
And for the denominator we have for $|x-1|$:
|
||||
\begin{align*}
|
||||
-1 &< x-1 < 1 \\
|
||||
0 &< x < 2 \\
|
||||
0 &< 2x < 4 \\
|
||||
-3 &< 2x-3 < 1 \\
|
||||
-\frac{1}{3} &> \frac{1}{2x-3} \\
|
||||
\frac{1}{2x-3} &< -\frac{1}{3}
|
||||
\end{align*}
|
||||
This then yields that
|
||||
\[\frac{|x-4||x-1|}{|2x-3|}<(-2)\cdot\left(-\frac{1}{3}\right)|x-1|<\varepsilon\]
|
||||
Which then gives us that
|
||||
\[|x-1|<\frac{3\varepsilon}{2}\]
|
||||
Hence if we let $\delta = \min \{1, \frac{3\varepsilon}{2}\}$, and x such that $|x-1|<\delta$ gives us that $\abs{\frac{x^2-x-2}{2x-3}-2}=\frac{|x-4||x-1|}{|2x-3|}< \varepsilon,\ \forall\ \varepsilon>0$.\\
|
||||
|
||||
$\therefore\ \lim\limits_{x \to 1} \frac{x^2-x-2}{2x-3}=2$
|
||||
|
||||
\item $\lim\limits_{x \to 5} (x^2-3x+1)=11$ \\
|
||||
|
||||
We want to show the following $\forall\ \varepsilon>0$:
|
||||
\begin{align*}
|
||||
|x^2-3x+1-11| &= |x^2-3x-10| \\
|
||||
&=|x-5||x+2| \\
|
||||
&<\varepsilon
|
||||
\end{align*}
|
||||
So we have the following for $|x+2|$:
|
||||
\begin{align*}
|
||||
|x+2| &= |(x-5)+7| \\
|
||||
&= |x-5|+7 \\
|
||||
&\leq 1+7 \\
|
||||
&= 8
|
||||
\end{align*}
|
||||
This then yields that $|x-5||x+2|<8|x-5|=\varepsilon$, and thus $|x-5|<\frac{\varepsilon}{8}$.\\
|
||||
|
||||
So if we let $\delta=\min \{1, \frac{\varepsilon}{8}\}$, and $x$ such that $|x-5|<\delta$, we have that $|x^2-3x-1|=|x-5||x+2|<\varepsilon$.\\
|
||||
|
||||
$\therefore\ \lim\limits_{x \to 5} (x^2-3x+1) = 11$
|
||||
\end{enumerate}
|
||||
|
||||
\item Find a $\delta > 0$ so that $|x-2|<\delta$ implies that
|
||||
\begin{enumerate}
|
||||
\item $|x^2+x-6|<1$\\
|
||||
|
||||
Note that $|x^2+x-6|=|x-2||x+3|$. So, we have that $|x+3|=|(x-2)+5|=|x-2|+5\leq 1+5=6$. Thus we have that $6|x-2|<1 \implies |x-2| < \frac{1}{6}$. So if we let $\delta = \min \{1, \frac{1}{6}\}=\frac{1}{6}$ gives us that $|x-2|<\delta$.\\
|
||||
|
||||
\item $|x^2+x-6|<1/n$ for a given $n \in \N$\\
|
||||
|
||||
By similar logic as the previous problem, we have that if we let $\delta=\min \{1,\frac{1}{6n}\}=\frac{1}{6n}$, we have that $|x-2|<\delta$.
|
||||
|
||||
\item $|x^2+x-6|<\varepsilon$\\
|
||||
|
||||
Suppose that $|x-2|<1$. Then we have that $|x+3|=|(x-2)+5|\leq |x-2|+5 < 6$. So we have that if we let $\delta=\min \{1,\frac{\varepsilon}{6}\}$, we have that $|x-2|<\delta$.
|
||||
\end{enumerate}
|
||||
|
||||
\item Show that for any $a$, $\lim\limits_{x \to a} F(x)$ does not exist for $F(x) = \begin{cases}
|
||||
1, & \text{if } x \text{ is rational} \\
|
||||
0, & \text{if } x \text{ is irrational}
|
||||
\end{cases}$\\
|
||||
|
||||
Let $x_n=$ sequence of rationals $\rightarrow\ a$, and let $y_n=$ sequence of irrationals such that $\rightarrow\ a$. Then we have that $\lim f(x_n)=1$ and $\lim f(y_n)=0$. Thus by \textit{Theorem 4.1.9}, we have that $\lim f(x)$ does not exist.
|
||||
|
||||
\item Let $f(x)=\begin{cases}
|
||||
3x+2 & \text{if } x \text{ is rational} \\
|
||||
6-x & \text{if } x \text{ is irrational}
|
||||
\end{cases}$
|
||||
\begin{enumerate}
|
||||
\item Show $\lim\limits_{x \to 1} f(x)=5$\\
|
||||
|
||||
Let $\varepsilon>0$ be given. Then we have the following for $|f(x)-L|$:
|
||||
\begin{align*}
|
||||
|f(x)-5| &= \begin{cases}
|
||||
|(3x+2)-5| \\
|
||||
|(6-x)-5|
|
||||
\end{cases} \\
|
||||
&= \begin{cases}
|
||||
|3x-3| \\
|
||||
|1-x|
|
||||
\end{cases}\\
|
||||
&= \begin{cases}
|
||||
3|x-1| \\
|
||||
|x-1|
|
||||
\end{cases} \\
|
||||
&= \varepsilon
|
||||
\end{align*}
|
||||
|
||||
Thus, let $\delta=\min \{\varepsilon, \frac{\varepsilon}{3}\}$, and we have that $\lim\limits_{x \to 1} f(x)=5$.
|
||||
|
||||
\item Show $\lim\limits_{x \to a} f(x)$ does not exist if $a \neq 1$.\\
|
||||
|
||||
Let $x_n$ be a sequence of rational numbers such that $x_n \neq a,\ \forall\ n \in \N$, and $\lim (x_n) =a$. Let $y_n$ be a sequence of irrational numbers such that $y_n \neq a,\ \forall\ n \in \N$, and $\lim (y_n)=a$. Then we have that $\lim\limits_{x \to a} f(x_n) = 3a+2$, and that $\lim\limits_{x \to a} f(y_n) = 6-a$. Thus, we have that if the limit exists, then $\lim\limits{x \to a} f(x_n) = \lim\limits_{x \to a} f(y_n)$. So,
|
||||
\begin{align*}
|
||||
3a+2 &= 6-a \\
|
||||
4a+2 &= 6 \\
|
||||
4a &= 4 \\
|
||||
a &= 1
|
||||
\end{align*}
|
||||
However, since we have that $a \neq 1$, we have that by \textit{Theorem 4.1.8} (\textit{The Divergence Criteria}), we know that $3a+2 \neq 6-a \implies \lim\limits_{x \to a} f(x_n) \neq \lim\limits_{x \to a} f(y_n)$, and thus this limit does not exists if $a \neq 1$.
|
||||
\end{enumerate}
|
||||
|
||||
\item Let $f:[-1,1] \to \R$ be given by, $f(x)=\begin{cases}
|
||||
0, & \text{if } x=\pm\frac{1}{n} \text{ where } n \in \N \\
|
||||
1, & \text{otherwise}
|
||||
\end{cases}$\\
|
||||
Find the limit, if it exists, using the definition of a limit.
|
||||
\begin{enumerate}
|
||||
\item $\lim\limits_{x \to \frac{3}{8}} f(x)$\\
|
||||
|
||||
Notice first that the range of $f$ is $R(f):= \{0,1\}$. This yields that $\varepsilon = 0$, or $\varepsilon=1$. However, by the definition of a limit, we have that $\varepsilon>0$, and thus $\varepsilon=1$. This gives us a guarantee that $\varepsilon$ will always equal 1. Thus all we need to worry about is finding a $\delta$, that doesn't need to be in terms of $\varepsilon$.\\
|
||||
|
||||
First, recall the definition of a limit when a function $f:A \to \R$ for $A \subseteq \R$, and $c$ is a cluster point of $A$:
|
||||
\[\lim\limits_{x \to c} f(x)=L \implies \forall\ \varepsilon>0\ \exists\ \delta >0 \st x \in A \wedge 0<|x-c|<\delta \implies |f(x)-L|<\varepsilon\]
|
||||
However, also recall the definition of a cluster point of $A$ for $c \in \R$:
|
||||
\[\forall\ \delta > 0\ \exists\ x \in V_\delta (c):=\{x \in A:|x-c|<\delta\}=(c-\delta, c+\delta) \st x \neq c\]
|
||||
So we have now that if we observe the different parts of the piecewise function that create the domain are the following sets: $B:=\{-1, -\frac{1}{2}, -\frac{1}{3}, \dots, \frac{1}{3}, \frac{1}{2}, 1\}$ and $C:= [-1,1] \setminus B$.\\
|
||||
|
||||
We now note that since $c = \frac{3}{8}$, we know by evaluating the limit through substitution, $\lim\limits_{x \to \frac{3}{8}}=1$. Thus we also know that since $\frac{3}{8}$ is not of the form $\pm \frac{1}{n}$ for $n \in \N$, that by the corollaries to \textit{The Archimedian Property}, there exists a rational number between any two real numbers. Thus we have that there must exist two rational numbers $p,q \in \Q \st p < \frac{3}{8}<q$.\\
|
||||
|
||||
We note by the elements of $B$ that if we let $p=\frac{1}{3}$ and let $q=\frac{1}{2}$ we have that $\frac{1}{3}<\frac{3}{8}<\frac{1}{2}$. So we now have that since $\delta$ must be defined such that $0<|x-c|<\delta$, we can solve for $\delta$ as being $\frac{3}{8} - \frac{1}{3}=\frac{1}{24}$, since $\frac{1}{3}<\frac{1}{2}$, and thus we have that if we let $\delta=\frac{1}{24}$, then $(c-\delta, c+\delta)$ is such that $|x-c|<\delta$.\\
|
||||
|
||||
$\therefore\ \lim\limits_{x \to \frac{3}{8}} f(x)=1$ when $|x-\frac{3}{8}|<\delta$ for $\delta=\frac{1}{24}$
|
||||
|
||||
\item $\lim\limits_{x \to -\frac{1}{3}} f(x)$\\
|
||||
|
||||
By applying similar logic as the previous problem, we notice that since $-\frac{1}{2}<-\frac{1}{3}<-\frac{1}{4}$, since $\delta$ does not need to be defined in terms of $\varepsilon$ since $\varepsilon=1$, we have that we can let $\delta=|-\frac{1}{3}+\frac{1}{4}|=\frac{1}{12}$, since $\delta>0$. Thus, we have that the cluster point $-\frac{1}{3}$ is defined such that $|x+\frac{1}{3}|<\frac{1}{12}$. Hence $\lim\limits_{x \to -\frac{1}{3}} f(x)=0$.
|
||||
|
||||
\item $\lim\limits_{x \to 0} f(x)$\\
|
||||
|
||||
We have that the limit does not exist when $x=0$. We can see this if we let $x_n=\frac{1}{n} \to 0$, and $y_n=\frac{\sqrt{2}}{n} \to 0$. Then we have that $f(x_n)=0$ and $f(y_n)=1$. Thus we have that $\lim\limits_{x \to 0} f(x)$ does not exist.\\
|
||||
|
||||
\end{enumerate}
|
||||
|
||||
\item \textbf{Section 4.2}
|
||||
\begin{enumerate}
|
||||
\item[3.] Find $\lim\limits_{x \to 0} \frac{\sqrt{1+2x}-\sqrt{1+3x}}{x+2x^2}$ where $x>0$.\\
|
||||
|
||||
Recall \textit{Theorem 4.2.4}:
|
||||
\begin{theorem*}
|
||||
let $A \subseteq \R$, let $f$ and $g$ be functions on $A$ to $\R$, and let $c \in \R$ be a cluster point of $A$. Further, let $b \in \R$.
|
||||
\begin{enumerate}
|
||||
\item If $\lim\limits_{x\to c} f = L$ and $\lim\limits_{x\to c} g = M$, then
|
||||
\[\lim\limits_{x\to c} (f+g) = L+M,\]
|
||||
\[\lim\limits_{x\to c} (f-g)=L-M,\]
|
||||
\[\lim\limits_{x\to c} (fg) = LM,\]
|
||||
\[\lim\limits_{x\to c} (bf) = bL.\]
|
||||
|
||||
\item If $h: A \rightarrow \R$, if $h(x) \neq 0$ for all $x \in A$, and if $\lim\limits_{x\to c} h = H \neq 0$, then
|
||||
\[\lim\limits_{x\to c} \left( \frac{f}{h} \right)= \frac{L}{H}\]
|
||||
\end{enumerate}
|
||||
\end{theorem*}
|
||||
This yields the following:
|
||||
\begin{align*}
|
||||
\lim\limits_{x \to 0} \frac{\sqrt{1+2x}-\sqrt{1+3x}}{x+2x^2} &= \lim\limits_{x \to 0} \frac{(1+2x)-(1+3x)}{(x+2x^2)(\sqrt{1+2x}+\sqrt{1+3x})} \\
|
||||
&= \lim\limits_{x \to 0} \frac{-x}{x(1+2x)(\sqrt{1+2x}+\sqrt{1+3x})} \\
|
||||
&= \lim\limits_{x \to 0} \frac{-1}{(1+2x)(\sqrt{1+2x}+\sqrt{1+3x})} \\
|
||||
&= \frac{-1}{(1+0)(\sqrt{1+0}+\sqrt{1+0})} &\text{by } \textit{Theorem 4.2.4} \\
|
||||
&= \frac{-1}{1(2)} \\
|
||||
&= \frac{-1}{2}
|
||||
\end{align*}
|
||||
Thus we have that $\lim\limits_{x \to 0} \frac{\sqrt{1+2x}-\sqrt{1+3x}}{x+2x^2}=-\frac{1}{2}$
|
||||
|
||||
\item[4.] Prove that $\lim\limits_{x \to 0} \cos (1/x)$ does not exist but that $\lim\limits_{x \to 0} x \cos (1/x)=0$\\
|
||||
|
||||
\begin{proof}
|
||||
Let $x_n:=\left((2n+1)\frac{\pi}{2}\right)^{-1} \implies x_n \to 0$. And also let $y_n:=(2n\pi)^{-1}$. Then we have that $y_n \to 0$. We can now note that $\cos \left(\frac{1}{x_n}\right)=\cos \left((2n+1)\frac{\pi}{2}\right)=0$, and that $\cos \left(\frac{1}{y_n}\right) = \cos (2n\pi) = 1$. This yields that $\cos \left(\frac{1}{x_n}\right) \to 0$ and $\cos \left(\frac{1}{y_n}\right) \to 1$.\\
|
||||
|
||||
Recall the \textit{Sequential Criterion}:
|
||||
\begin{theorem*}[Sequential Criterion]
|
||||
Let $f:A \rightarrow \R$ and let $c$ be a cluster point of $A$. Then the following are equivalent.
|
||||
\begin{enumerate}
|
||||
\item $\lim\limits_{x\to c} f=L$.
|
||||
\item For every sequence $(x_n)$ in $A$ that converges to $c$ such that $x_n\neq c$ for all $n \in \N$, the sequence $(f(x_n))$ converges to $L$.
|
||||
\end{enumerate}
|
||||
\end{theorem*}
|
||||
So we have that $\lim\limits_{x \to 0} \cos \left(\frac{1}{x}\right)$ doesn't exist. However, $\lim\limits_{x \to 0}x \cos \left(\frac{1}{x}\right)=0$. This is given to us by the fact that $\abs{x \cos \left(\frac{1}{x}\right)}\leq |x|$. Thus, if we let $\delta = \varepsilon$, we have that $\lim\limits_{x \to 0} x \cos \left(\frac{1}{x}\right)=0$.
|
||||
\end{proof}
|
||||
|
||||
\item[6.] Use the definition of the limit to prove the first assertion in Theorem 4.2.4(a).\\
|
||||
|
||||
\begin{proof}
|
||||
Let $\varepsilon>0$ be given.\\
|
||||
|
||||
Since we have that $\lim\limits_{x \to c} f(x) = L$, by the definition of the limit, we know that $\exists\ \delta_f > 0 \st |x-c|<\delta_f \implies |f(x)-L|<\frac{\varepsilon}{2}$.\\
|
||||
|
||||
Since $\lim\limits_{x \to c} g(x)=M$, then we have that by the definition of the limit, we know that $\exists\ \delta_g > 0 \st |x-c|<\delta_g \implies |g(x)-M|<\frac{\varepsilon}{2}$.\\
|
||||
|
||||
Now, define $\delta=\max \{\delta_f, \delta_g\}$. Then we have that for $|x-c|<\delta$:
|
||||
\begin{align*}
|
||||
|(f(x)+g(x))-(L+M)| &= |(f(x)-L)+(g(x)-M)| \\
|
||||
&\leq |f(x)-L| + |g(x)-M| \\
|
||||
&< \frac{\varepsilon}{2} + \frac{\varepsilon}{2} \\
|
||||
&= \varepsilon
|
||||
\end{align*}
|
||||
This yields that by the definition of the limit again, we have that $\lim\limits_{x \to c} (f+g)(x) = L+M$.\\
|
||||
|
||||
$\therefore$ We have that if $\lim\limits_{x \to c} f(x) = L$ and $\lim\limits_{x \to c} g(x) = M$, then $\lim\limits_{x \to c} (f+g)(x) = L + M$.\\
|
||||
|
||||
A similar argument can be used to show the case for subtraction.
|
||||
\end{proof}
|
||||
|
||||
\item[9.] Let $f,g$ be defined on $A$ to $\R$ and let $c$ be a cluster point of $A$.
|
||||
\begin{enumerate}
|
||||
\item[(a)] Show that if both $\lim\limits_{x \to c} f$ and $\lim\limits_{x \to c}(f+g)$ exist, then $\lim\limits_{x \to c}$ exists.\\
|
||||
|
||||
\begin{proof}
|
||||
Since we have that $\lim\limits_{x \to c} f$ and $\lim\limits_{x \to c} (f+g)$ exist, by \textit{Theorem 4.2.4} we have that $\lim\limits_{x \to c} g=\lim\limits_{x \to c} ((f + g)-f)$ also exists.\\
|
||||
|
||||
$\therefore\ \exists\ \lim\limits_{x \to c} f \wedge \exists\ \lim\limits_{x \to c} (f+g) \implies \exists\ \lim\limits_{x \to c} g$.
|
||||
\end{proof}
|
||||
|
||||
\item[(b)] If $\lim\limits_{x \to c} f$ and $\lim\limits_{x \to c} fg$ exist, does it follow that $\lim\limits_{x \to c} g$ exists?\\
|
||||
|
||||
If we have that $\lim\limits_{x \to c} f$ and $\lim\limits_{x \to c} fg$ exist, then we have that $\lim\limits_{x \to c}g$ doesn't necessarily have to exist. Consider the following:\\
|
||||
|
||||
Let $c=0, f(x)=x,g(x)=\frac{1}{x}$. Then we have that $\lim\limits_{x \to c} f(x)=\lim\limits_{x \to 0} x = 0$, and that $\lim\limits_{x \to c}fg(x) = \lim\limits_{x \to 0} x \cdot \frac{1}{x} = \lim\limits_{x \to 0} 1 = 1$. However, we have that $\lim\limits_{x \to c} g(x) = \lim\limits_{x \to 0} \frac{1}{x}$ does not exist.\\
|
||||
\end{enumerate}
|
||||
|
||||
\item[10.] Give examples of functions $f$ and $g$ such that $f$ and $g$ do not have limits at a point $c$, but such that both $f+g$ and $fg$ have limits at $c$.\\
|
||||
|
||||
Consider the following: Let $c=0, f(x)=\sgn (x), g(x)=-\sgn(x)$. By the definition of the signum function, we know that $\lim\limits_{x \to 0} \sgn$ does not exist. This yields that $\lim\limits_{x \to 0} -\sgn$ also does not exit. However, we have that $(f+g)(x)=0$ and $(f \cdot g)(x)=-1$ for $x\neq 0$, and $(f \cdot g)(0)=0$. This gives us that $\lim\limits_{x \to 0} (f+g)(x)=0$ and that $\lim\limits_{x \to 0} (f \cdot g)(x) = -1$.\\
|
||||
|
||||
\item[11.] Determine whether the follow limits exist in $\R$.
|
||||
\begin{enumerate}
|
||||
\item[(a)] $\lim\limits_{x \to 0} \sin (1/x^2)\ \ \ (x \neq 0)$\\
|
||||
|
||||
$\lim\limits_{x \to 0} \sin \frac{1}{x^2}$ for $x \neq 0$ does not exist.\\
|
||||
|
||||
Consider $f(x) = \sin \frac{1}{x^2}$, and $x \neq 0$. Also, let $x_n = \frac{1}{\sqrt{n\pi}}$ for $n \in \N$. Then we have that $\lim x_n = \frac{1}{\pi}\cdot \lim \frac{1}{\sqrt{n}}=0$. Thus $f(x_n) = \sin \frac{1}{x_n^2}=\sin \frac{1}{\frac{1}{n\pi}}=\sin n \pi = 0$. Which yields $\lim f(x_n)=0$.\\
|
||||
|
||||
Now, let $y_n = \frac{1}{\sqrt{(4n+1)\frac{\pi}{2}}}=\sin (4n+1)\frac{\pi}{2}=1$. Thus $\lim f(y_n)=1$.\\
|
||||
|
||||
Hence we have that $(x_n)$ and $(y_n)$ both converge to $0$, however the sequences $(f(x_n))$ and $(f(y_n))$ converge to two complete separate limits.\\
|
||||
|
||||
$\therefore\ \lim\limits_{x \to 0} f(x) = \lim\limits_{x \to 0} \sin \frac{1}{x^2}$ does not exist.\\
|
||||
|
||||
\item[(b)] $\lim\limits_{x \to 0} x \sin (1/x^2)\ \ \ (x \neq 0)$\\
|
||||
|
||||
$\lim\limits_{x \to 0} x \sin \frac{1}{x^2}=0$.\\
|
||||
|
||||
\begin{lemma}
|
||||
Let $A \subseteq \R$, let $f,g:A \to \R$, and let $c$ be a cluster point of $A$. Suppose that $\lim\limits_{x \to c} g(x)=0$ and that $f$ is bounded on some neighborhood of $c$. We want to show that $\lim\limits_{x \to c}fg(x)=0$.\\
|
||||
|
||||
Since $f$ is bounded on some neighborhood of $c$, we know that $\exists\ \delta_0>0$ and $M>0$ such that $|x-c|<\delta_0 \implies |f(x)|<M$.\\
|
||||
|
||||
Let $\varepsilon >0$ be given. Since $\lim\limits_{x \to c} g(x)=0$, we know that $\exists\ \delta_1 >0$ such that $|x-c|<\delta_1 \implies |g(x)|=|g(x)-0|<\frac{\varepsilon}{M}$.\\
|
||||
|
||||
Choose $\delta := \min \{\delta_0, \delta_1\}$. Thus we now have
|
||||
\[|x-c|<\delta \implies |f(x)|<M \text{ and } |g(x)|<\frac{\varepsilon}{M} \implies |f(x)g(x)|<\varepsilon\]
|
||||
Thus we have that by the definition of the limit, $\lim\limits_{x \to c} fg(x)=0$.\\
|
||||
$\square$
|
||||
\end{lemma}
|
||||
\begin{proof}
|
||||
Let $f(x)=\begin{cases}
|
||||
\sin \frac{1}{x^2}, &x \neq 0\\
|
||||
0, &x=0
|
||||
\end{cases}$ \\
|
||||
|
||||
and let $g(x)=x$ for $x \in \R$.\\
|
||||
|
||||
Then we have that $|f(x)|\leq 1\ \forall\ x \in \R$. This gives us that $f$ is bounded in $\R$. Hence $\lim\limits_{x \to 0} g(x)=\lim\limits_{x \to 0} x = 0$. Thus we have that by \textit{Lemma 0.1}, we have that $\lim\limits_{x \to 0} f(x)=0 \implies \lim\limits_{x \to 0} x \sin \frac{1}{x^2}=0$.
|
||||
\end{proof}
|
||||
\end{enumerate}
|
||||
|
||||
\item[13.] Functions $f$ and $g$ are defined on $R$ by $f(x):=x+1$ and $g(x):=2$ if $x \neq 1$ and $g(1):=0$.
|
||||
\begin{enumerate}
|
||||
\item[(a)] Find $\lim\limits_{x \to 1}g(f(x))$ and compare with the value of $g(\lim\limits_{x \to 1} f(x))$.\\
|
||||
|
||||
We first note that $\lim\limits_{x \to 1} f(x) = \lim\limits_{x \to 1} (x+1)=2$. Thus we have that $g(\lim\limits_{x \to 1} f(x))=g(2)=2$.\\
|
||||
|
||||
Now, $g(f(x))=g(x+1)=\begin{cases}
|
||||
2 & x \neq 0 \\
|
||||
0 & x=0
|
||||
\end{cases}$\\
|
||||
Thus we have that $\lim\limits_{x \to 1} g(f(x))=2$. Thus we have that $\lim\limits_{x \to 1} g(f(x))=2=g(\lim\limits_{x \to 1} f(x))$.
|
||||
|
||||
\item[(b)] Find $\lim\limits_{x \to 1} f(g(x))$ and compare with the value of $f(\lim\limits_{x \to 1} g(x))$.\\
|
||||
|
||||
$f(g(x))=\begin{cases}
|
||||
f(2) & x \neq 1 \\
|
||||
f(0) & x = 1
|
||||
\end{cases} = \begin{cases}
|
||||
3 & x \neq 1 \\
|
||||
1 & x = 1
|
||||
\end{cases}$\\
|
||||
Thus we have that $\lim\limits_{x \to 1}f(g(x))=3$. Now $\lim\limits_{x \to 1} g(x)=2$, we have that $f(\lim\limits_{x \to 1} g(x))=f(2)=3$.\\
|
||||
|
||||
$\therefore\ \lim\limits_{x \to 1}f(g(x))=f(\lim\limits_{x \to 1}g(x))$.
|
||||
\end{enumerate}
|
||||
\end{enumerate}
|
||||
|
||||
\item Prove or justify if true. Provide a counterexample if false.
|
||||
\begin{enumerate}
|
||||
\item $\lim\limits_{x \to 3a} f(x) = 3 \lim\limits_{x \to a} f(x)$\\
|
||||
|
||||
This is a false statement. Consider the function $f(x):=\frac{1}{x}$. Then we have that if $a=4$:
|
||||
\[\lim\limits_{x \to 3(4)} f(x) = \lim\limits_{x \to 12} \frac{1}{x} = \frac{1}{12}\]
|
||||
However, we also have the following:
|
||||
\[3\lim\limits_{x \to 4} f(x) = 3\lim\limits_{x \to 4} \frac{1}{x} = \frac{3}{4}\]
|
||||
Thus, we have that
|
||||
\[\lim\limits_{x \to 3(4)} \frac{1}{x} = \frac{1}{12} \neq \frac{3}{4} = 3\lim\limits_{x \to 4} \frac{1}{x}\]
|
||||
And thus
|
||||
\[\lim\limits_{x \to 3a} f(x) \neq 3\lim\limits_{x \to a} f(x)\]
|
||||
|
||||
\item $\lim\limits_{x \to a} f(3x) = 3 \lim\limits_{x \to a} f(x)$\\
|
||||
|
||||
This is also a false statement. Consider $f(x)=\frac{1}{x}$ and $a=5$. Then we have
|
||||
\[\lim\limits_{x \to 5} f(3x) = \lim\limits_{x \to 5} \frac{1}{3x} = \frac{1}{15}\]
|
||||
But
|
||||
\[3\lim\limits_{x \to 5} f(x) = 3\lim\limits_{x \to 5} \frac{1}{x} = \frac{3}{5}\]
|
||||
Thus we have that
|
||||
\[\lim\limits_{x \to 5} \frac{1}{3x} = \frac{1}{15} \neq \frac{3}{5} = 3\lim\limits_{x \to 5} \frac{1}{x}\]
|
||||
Hence
|
||||
\[\lim\limits_{x \to a} f(3x) \neq 3\lim\limits_{x \to a} f(x)\]
|
||||
|
||||
\item $\lim\limits_{x \to 3a} f(x) = \lim\limits_{x \to a} f(3x)$\\
|
||||
|
||||
This statement is true.
|
||||
\begin{proof}
|
||||
Let $y=\frac{x}{3}$. Then we have that $x = 3y$. Thus as $x \to 3a$, we get that $y \to a$. Thus $\lim\limits_{x \to 3a} f(x)=\lim\limits_{y \to a} f(3y) = \lim\limits_{x \to a} f(3a)$.
|
||||
\end{proof}
|
||||
\end{enumerate}
|
||||
\end{enumerate}
|
||||
\end{document}
|
||||
Binary file not shown.
+503
@@ -0,0 +1,503 @@
|
||||
\documentclass[12pt,letterpaper]{article}
|
||||
\usepackage[utf8]{inputenc}
|
||||
\usepackage[english]{babel}
|
||||
\usepackage[normalem]{ulem}
|
||||
\usepackage{cancel}
|
||||
\usepackage{amsthm}
|
||||
\usepackage{amsmath}
|
||||
\usepackage{amsfonts}
|
||||
\usepackage{amssymb}
|
||||
\usepackage{graphicx}
|
||||
\usepackage{array}
|
||||
\usepackage[left=2cm, right=2.5cm, top=2.5cm, bottom=2.5cm]{geometry}
|
||||
\usepackage{enumitem}
|
||||
\newcommand{\st}{\ \text{s.t.}\ }
|
||||
\newcommand{\abs}[1]{\left\lvert #1 \right\rvert}
|
||||
\newcommand{\R}{\mathbb{R}}
|
||||
\newcommand{\N}{\mathbb{N}}
|
||||
\newcommand{\Q}{\mathbb{Q}}
|
||||
\newcommand{\C}{\mathbb{C}}
|
||||
\newcommand{\Z}{\mathbb{Z}}
|
||||
\DeclareMathOperator{\sign}{sgn}
|
||||
\newtheoremstyle{case}{}{}{}{}{}{:}{ }{}
|
||||
\theoremstyle{case}
|
||||
\newtheorem{case}{Case}
|
||||
\theoremstyle{definition}
|
||||
\newtheorem{definition}{Definition}[section]
|
||||
\newtheorem{theorem}{Theorem}[section]
|
||||
\newtheorem*{theorem*}{Theorem}
|
||||
\newtheorem{corollary}{Corollary}[section]
|
||||
\newtheorem*{corollary*}{Corollary}
|
||||
\newtheorem{lemma}[theorem]{Lemma}
|
||||
\newtheorem*{remark}{Remark}
|
||||
\setlist[enumerate]{font=\bfseries}
|
||||
\renewcommand{\qedsymbol}{$\blacksquare$}
|
||||
\author{Alexander J. Tusa}
|
||||
\title{Real Analysis Homework 8}
|
||||
\begin{document}
|
||||
\maketitle
|
||||
\begin{enumerate}
|
||||
\item \textbf{Section 5.1}
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Section 5.1 Questions %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\begin{enumerate}
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 1 - 5.1 %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item[1.] Prove the Sequential Criterion 5.1.3.\\
|
||||
|
||||
Recall the \textit{Sequential Criterion}:
|
||||
\begin{theorem*}
|
||||
A function $f:A \to \R$ is continuous at the point $c \in A$ if and only if for every sequence $(x_n)$ in $A$ that converges to $c$, the sequence $(f(x_n))$ converges to $f(c)$.
|
||||
\end{theorem*}
|
||||
|
||||
Let $\varepsilon > 0$ be given. Since we're given that $f$ is continuous, we know that by the definition of continuity, there exists a $\delta >0$ such that if $x \in A$ satisfies $|x-c|<\delta$, then $|f(x)-f(c)|<\varepsilon$. Thus, for that $\delta$, since $x_n \to c,\ \exists\ n_0 \in \N \st n \geq n_0 \implies |x_n-c|<\delta$. Thus we have
|
||||
\begin{align*}
|
||||
n \geq n_0 &\implies |x_n - c|<\delta \\
|
||||
&\implies |f(x_n)-f(c)|<\varepsilon \\
|
||||
&\implies \lim\limits_{n \to \infty} f(x_n)=f(c)
|
||||
\end{align*}
|
||||
Now, suppose that for every sequence $(x_n)$ in $A$ that converges to $c$, $f(x_n)$ converges to $f(c)$. We will now show that the $f$ is continuous at $c$.
|
||||
\begin{proof}
|
||||
We want to show that $f$ is continuous at $c$.\\
|
||||
|
||||
By way of contradiction, suppose that $f$ is discontinuous at $c$. Then we know that $\exists\ \varepsilon >0\ \forall\ \delta > 0\ |x-c|<\delta$ and $|f(x)-f(c)| \geq \varepsilon$.\\
|
||||
|
||||
Let $\delta = \frac{1}{n}$. Then we have that $x_n$ is such that $|x-c|<\frac{1}{n}$ and that $|f(x_n)-f(c)| \geq \varepsilon$.\\
|
||||
|
||||
$\therefore$ We have a sequence $(x_n)$ such that $\lim\limits_{n \to \infty} x_n = c$ and $\lim\limits_{n \to \infty} f(x_n) \neq f(c)$. Notice however that we have a contradiction here, and thus we have that $f$ is continuous at $c$.
|
||||
\end{proof}
|
||||
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 5 - 5.1 %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item[5.] Let $f$ be defined for all $x \in \R, x \neq 2$, by $f(x)=(x^2+x-6)/(x-2)$. Can $f$ be defined at $x=2$ in such a way that $f$ is continuous at this point?\\
|
||||
|
||||
Since we have that $f$ is not defined at $x=2$, we know that we have to factor the numerator to evaluate the limit.\\
|
||||
|
||||
So, for $x \neq 2$:
|
||||
\[f(x)=\frac{x^2+x-6}{x-2}=\frac{\cancel{(x-2)}(x+3)}{\cancel{x-2}}=x+3\]
|
||||
Thus, if we let $f(2):=5$, then $f$ is continuous:
|
||||
|
||||
\[\lim\limits_{x \to 2} f(x) = \lim\limits_{x \to 2} (x+3) = 5\]
|
||||
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 7 - 5.1 %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item[7.] Let $f:\R \to \R$ be continuous at $c$ and let $f(c)>0$. Show that there exists a neighborhood $V_\delta (c)$ of $c$ such that if $x \in V_\delta (c)$, then $f(x)>0$.\\
|
||||
|
||||
\begin{proof}
|
||||
Let $\varepsilon = \frac{f(c)}{2} > 0$. Since we are given that $f$ is continuous at $c$, we know that there exists $\delta > 0$ such that
|
||||
\[|x-c|<\delta \implies |f(x)-f(c)|<\varepsilon = \frac{f(c)}{2}\]
|
||||
Thus, if we define $V_\delta (c) := (c-\delta, c+\delta)$ and if we let $x \in V_\delta (c)$, then we have
|
||||
\begin{align*}
|
||||
x \in V_\delta (c) &\implies |x-c|<\delta \\
|
||||
&\implies |f(x)-f(c)|<\varepsilon \\
|
||||
&\implies -\varepsilon < f(x)-f(c)<\varepsilon \\
|
||||
&\implies f(c)-\frac{f(c)}{2} < f(x) &(\varepsilon = \frac{f(c)}{2}) \\
|
||||
&\implies \frac{1}{2}f(c) < f(x) \\
|
||||
&\implies f(x)>0 &(f(c)>0)
|
||||
\end{align*}
|
||||
Thus we have that there exists a neighborhood $V_\delta(c)$ such that if $x \in V_\delta (c)$, then $f(x)>0$.
|
||||
\end{proof}
|
||||
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 11 - 5.1 %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item[11.] Let $K>0$ and let $f:\R \to \R$ satisfy the condition $|f(x)-f(y)|\leq K|x-y|$ for all $x,y \in \R$. Show that $f$ is continuous at every point $c \in \R$.\\
|
||||
|
||||
\begin{proof}
|
||||
We want to show that $f$ is continuous at every point $c \in \R$.\\
|
||||
|
||||
Let $\varepsilon >0$ be given, and let $\delta = \frac{\varepsilon}{k}$. Then, for any $c \in \R$,
|
||||
\begin{align*}
|
||||
|x-c|<\delta &\implies |x-c|<\frac{\varepsilon}{k} \\
|
||||
&\implies k|x-c| < \varepsilon \\
|
||||
&\implies |f(x)-f(c)| \leq k|x-c|<\varepsilon \\
|
||||
&\implies |f(x)-f(c)|<\varepsilon
|
||||
\end{align*}
|
||||
Thus we have that $f$ is continuous for $c \in \R$. Since $c$ is arbitrary, this also shows that $f$ is continuous for all $c \in \R$.
|
||||
\end{proof}
|
||||
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 12 - 5.1 %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item[12.] Suppose that $f:\R \to \R$ is continuous on $\R$ and that $f(r)=0$ for every rational number $r$. Prove that $f(x)=0$ for all $x \in \R$.\\
|
||||
|
||||
\begin{proof}
|
||||
We want to show that $f(x)=0,\ \forall\ x \in \R$.\\
|
||||
|
||||
Let $x \in \R$. Since we know that $\Q$ is dense in $\R$, we know that we can find a sequence $(x_n)$ of rational numbers such that $(x_n)$ converges to $x$. Thus, by continuity of $f$ at $x$, $(f(x_n))$ converges to $f(x)$. Since $x_n$ is rational, we have that $f(x_n)=0\ \forall\ n \in \N$. Thus we have that $f(x) = \lim f(x_n) = \lim 0 = 0$.\\
|
||||
|
||||
$\therefore$ We have that $f(x)=0$ for all $x \in \R$.
|
||||
\end{proof}
|
||||
\end{enumerate}
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 2 %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item Use the definition of continuity to show that the given $f$ is continuous.
|
||||
\begin{enumerate}
|
||||
\item Let $f:\R \to \R$ be given by $f(x)=x^2$.\\
|
||||
|
||||
Recall the definition of continuity:
|
||||
\theoremstyle{definition}
|
||||
\begin{definition}
|
||||
Let $A \subseteq \R$, let $f:A \to \R$, and let $c \in A$. We say that $f$ is \textbf{continuous at} $c$ if, given any number $\varepsilon > 0$, there exists $\delta >0$ such that if $x$ is any point of $A$ satisfying $|x-c|<\delta$, then $|f(x)-f(c)|<\varepsilon$. If $f$ fails to be continuous at $c$, then we say that $f$ is \textbf{discontinuous at} $c$.
|
||||
\end{definition}
|
||||
|
||||
\begin{proof}
|
||||
In order to show that $f$ is continuous for all $c \in \R$, we must consider two cases: $c\neq 0$, and $c=0$.\\
|
||||
|
||||
Suppose $c \neq 0$. Then we have the following:
|
||||
\begin{align*}
|
||||
|f(x)-f(c)| &= |x^2 - c^2| \\
|
||||
&=|(x-c)(x+c)| \\
|
||||
&=|x-c||x+c| \\
|
||||
&< |x-c|(|x|+|c|) \\
|
||||
&<\varepsilon
|
||||
\end{align*}
|
||||
Which yields that
|
||||
\[|x-c|<\frac{\varepsilon}{|x|+|c|}< \frac{\varepsilon}{|x|}\]
|
||||
So, for the case when $c \neq 0$, let $\delta=\frac{\varepsilon}{|x|}$.\\
|
||||
|
||||
Now, suppose that $c=0$. Then we have the following:
|
||||
\begin{align*}
|
||||
|f(x)-f(c)| &= |x^2-0^2| \\
|
||||
&= |x^2| \\
|
||||
&=(|x|)^2 \\
|
||||
&< \varepsilon
|
||||
\end{align*}
|
||||
So, we have now that
|
||||
\[(|x|)^2 < \varepsilon \implies |x|<\sqrt{\varepsilon}\]
|
||||
Thus, let $\delta=\sqrt{\varepsilon}$.\\
|
||||
|
||||
$\therefore$ Since we have found definitions of $\delta$ that satisfy the definition of continuity, we have that $f$ is continuous $\forall c \in \R$.
|
||||
\end{proof}
|
||||
|
||||
\item Let $f:(0,\infty) \to \R$ be given by $f(x)=1/x$. \\
|
||||
|
||||
\begin{proof}
|
||||
Suppose $x>\frac{c}{2}$, since $c>0$. Then we have that
|
||||
\[\abs{f(x)-f(c)}=\abs{\frac{1}{x}-\frac{1}{c}}=\frac{|x-c|}{xc}<\frac{2}{c^2}|x-c|<\varepsilon \implies |x-c|<\frac{c^2\varepsilon}{2}\]
|
||||
So, if we let $\delta < \min \{\frac{c^2\varepsilon}{2}, \frac{c}{2}\}$, then we have that if $|x-c|<\delta,\ x>\frac{c}{2}$, and $\abs{\frac{1}{x}-\frac{1}{c}}<\frac{2}{c^2}\delta \leq \varepsilon$.\\
|
||||
|
||||
$\therefore$ We have that $f$ is continuous $\forall\ c \in (0,\infty)$.
|
||||
\end{proof}
|
||||
|
||||
\item Let $f:\R \to \R$ be given by $f(x)=|x|$ (This is Problem 10 in Section 5.1)\\
|
||||
|
||||
\begin{proof}
|
||||
We must first note the following:
|
||||
\begin{align*}
|
||||
|a| &= |a-b+b| \\
|
||||
&\leq |a-b|+|b| \\
|
||||
&\Downarrow \\
|
||||
|a|-|b| &\leq |a-b|
|
||||
\end{align*}
|
||||
$\implies \pm(|a|-|b|) \leq |a-b|$
|
||||
We also note:
|
||||
\begin{align*}
|
||||
|b| &= |b-a+a| \\
|
||||
&\leq |b-a| + |a| \\
|
||||
&=|a-b|+|a| \\
|
||||
&\Downarrow \\
|
||||
|b|-|a| &\leq |a-b|
|
||||
\end{align*}
|
||||
$\implies ||a|-|b|| \leq |a-b|$
|
||||
\begin{align*}
|
||||
|f(x)-f(c)| &= ||x|-|c|| \\
|
||||
&\leq |x-c| \\
|
||||
&<\varepsilon
|
||||
\end{align*}
|
||||
Thus if we let $\varepsilon = \delta$, then we have that $|x-c|<\delta$.\\
|
||||
|
||||
$\therefore$ We have that $f(x)$ is continuous $\forall\ c \in \R$.
|
||||
\end{proof}
|
||||
\end{enumerate}
|
||||
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 3 %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item Let $f(x)=\begin{cases}
|
||||
3x+2 & \text{if } x \text{ is rational} \\
|
||||
6-x & \text{if } x \text{ is irrational}
|
||||
\end{cases}$
|
||||
\begin{enumerate}
|
||||
\item Determine whether or not $f$ is continuous at $x=1$.\\
|
||||
|
||||
$f$ is continuous at $x=1$ since we proved in $HW 7$ that $\lim\limits_{x \to 1} f(x)$ exists, $\lim\limits_{x \to 1} f(x)=f(1)=5$.
|
||||
|
||||
\item Determine whether or not $f$ is continuous at $x=0$.\\
|
||||
|
||||
$f$ is not continuous for $x \neq 1$ (and not $x = 0$) since from Homework 7 again, $\lim\limits_{x \to c} f(x)$ does not exists for $x \neq 1$.
|
||||
\end{enumerate}
|
||||
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 4 %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item Let $g(x)=\begin{cases}
|
||||
2x & \text{if } x \text{ is rational} \\
|
||||
x+3 & \text{if } x \text{ is irrational}
|
||||
\end{cases}$ \\
|
||||
Find all points where $g$ is continuous (This is Problem 13 of Section 5.1)\\
|
||||
|
||||
First, let $c$ be some point of continuity of $g$. Then since both $\Q$ and $\R \setminus \Q$ are dense in $\R$, we can find sequences $(x_n) \subset \Q$ and $(y_n) \subset \R \setminus \Q$ such that $(x_n) \to c$ and $(y_n) \to c$. Then we have the following:
|
||||
\begin{align*}
|
||||
g(c) &= \lim\limits_{n \to \infty} g(x_n) \\
|
||||
&= \lim\limits_{x \to \infty} 2x_n \\
|
||||
&= 2c &(x_n \to c) \\
|
||||
\\
|
||||
g(c) &= \lim\limits_{n \to \infty} g(y_n) \\
|
||||
&= \lim\limits_{n \to \infty} y_n + 3 \\
|
||||
&= c+3 &(y_n \to c) \\
|
||||
\\
|
||||
\implies 2c &= c+3 \\
|
||||
&\Updownarrow \\
|
||||
c &= 3
|
||||
\end{align*}
|
||||
Thus, we have that if $g$ is continuous at a point $c$, then it must be $c=3$. Thus we conjecture that $g$ is continuous only at $3$.\\
|
||||
|
||||
\begin{proof}
|
||||
\begin{align*}
|
||||
|g(x)-g(3)| &= |g(x)-6| &(3 \in \Q) \\
|
||||
&\leq \sup \{|2x-6|,|(x+3)-6|\} &(x \text{ is either rational or irrational}) \\
|
||||
&= \sup \{2|x-3|,|x-3|\} \\
|
||||
&=2|x-3|
|
||||
\end{align*}
|
||||
Thus $\forall\ \varepsilon >0$ let $\delta=\frac{\varepsilon}{2}$. Then we have that
|
||||
\[|x-3| < \delta \implies |g(x)-g(3)|<\varepsilon\]
|
||||
$\therefore$ We have that $g$ is only continuous at $c=3$.
|
||||
\end{proof}
|
||||
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 5 %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item Give an example of the following, if possible.
|
||||
\begin{enumerate}
|
||||
\item A function $f$ defined on $\R$ such that it is not continuous at any point in $\R$.\\
|
||||
|
||||
Consider the Dirichlet function. This function is not continuous at any points.
|
||||
|
||||
\item A function $f$ defined on $\R$ such that it is continuous at exactly one point in $\R$.\\
|
||||
|
||||
Consider the function $f(x):= \begin{cases}
|
||||
x, & \text{if } x \in \Q \\
|
||||
0, & \text{if } x \in \R \setminus \Q
|
||||
\end{cases}$ \\
|
||||
Notice that this function is continuous only at $x=0$.
|
||||
|
||||
\item A function $f$ defined on $\R$ such that it is continuous at exactly two points in $\R$.\\
|
||||
|
||||
Consider the function $f(x):=\begin{cases}
|
||||
x^2, & \text{if } x \in \Q \\
|
||||
1, &\text{if } x \notin \Q
|
||||
\end{cases}$ \\
|
||||
We notice that this function is continuous only at the points $-1$ and $1$.
|
||||
\end{enumerate}
|
||||
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Section 5.2 %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item \textbf{Section 5.2}
|
||||
\begin{enumerate}
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 2 - 5.2 %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item[2.] Show that if $f:A \to \R$ is continuous on $A \subseteq \R$ and if $n \in \N$, then the function $f^n$ defined by $f^n(x)=(f(x))^n$, for $x \in A$, is continuous on $A$.\\
|
||||
|
||||
\begin{proof}
|
||||
We want to show that $f^n$ is continuous. We prove it by method of mathematical induction.\\
|
||||
|
||||
\textbf{Basis Step:} Let $n=1$. Then we have that $f^1(x)=(f(x))^1=f(x) \implies f^1=f$. Since we are given that $f$ is continuous, we know that $f^1$ is continuous as well.\\
|
||||
|
||||
\textbf{Inductive Step:} Assume that $f^n(x)=(f(x))^n$ is continuous $\forall\ n \in \N$.\\
|
||||
|
||||
\textbf{Show:} We want to show that $f^{n+1}(x)=(f(x))^{n+1}$ is continuous $\forall\ n \in \N$. Then we have
|
||||
\[f^{n+1}=f^n \cdot f\]
|
||||
Recall \textit{Theorem 5.2.1}:
|
||||
\begin{theorem*}
|
||||
Let $A \subseteq \R$, let $f$ and $g$ be functions on $A$ to $\R$, and let $b \in \R$. Suppose that $c \in A$ and that $f$ and $g$ are continuous at $c$.
|
||||
\begin{enumerate}
|
||||
\item Then $f+g,\ f-g,\ fg$, and $bf$ are continuous at $c$.
|
||||
|
||||
\item If $h:A \rightarrow \R$ is continuous at $c \in A$ and if $h(x) \neq 0$ for all $x \in A$, then the quotient $f/h$ is continuous at $c$.
|
||||
\end{enumerate}
|
||||
\end{theorem*}
|
||||
|
||||
Since both $f^n$ and $f$ are continuous functions, by \textit{Theorem 5.2.1}, we have that $f^n\cdot f=f^{n+1}$ is a continuous function on $A$ since it is the product of continuous functions on $A$.\\
|
||||
|
||||
$\therefore$ By the Principle of Mathematical Induction, we have that $f^n$ is continuous $\forall\ n \in \N$ on $A$.
|
||||
\end{proof}
|
||||
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 3 - 5.2 %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item[3.] Give an example of functions $f$ and $g$ that are both discontinuous at a point $c$ in $\R$ such that $(a)$ the sum $f+g$ is continuous at $c$ and $(b)$ the product $fg$ is continuous at $c$.\\
|
||||
|
||||
Consider the functions $f(x):=\begin{cases}
|
||||
1, & x=0 \\
|
||||
0, & x \neq 0
|
||||
\end{cases}$, and $g(x):=\begin{cases}
|
||||
0, & x=0 \\
|
||||
1, & x \neq 0
|
||||
\end{cases}$ \\
|
||||
We first note that both $f(x)$ and $g(x)$ are discontinuous at $x=0$.
|
||||
\begin{enumerate}
|
||||
\item[(a)] Note that the sum
|
||||
\[f(x)+g(x):=\begin{cases}
|
||||
1+0, & x=0 \\
|
||||
0+1, & x \neq 0
|
||||
\end{cases} = \begin{cases}
|
||||
1, & x=0 \\
|
||||
1, & x \neq 0
|
||||
\end{cases}=1\].
|
||||
Thus, we have that the sum $f(x)+g(x)$ is a constant function, and thus is defined $\forall\ x \in \R$, and thus $f(x)+g(x)$ is continuous at $c$, where $c=0$.
|
||||
|
||||
\item[(b)] Now, the product $f(x)\cdot g(x)$ is
|
||||
\[f(x) \cdot g(x):= \begin{cases}
|
||||
1 \cdot 0, & x=0 \\
|
||||
0 \cdot 1, & x \neq 0
|
||||
\end{cases} = \begin{cases}
|
||||
0, & x=0 \\
|
||||
0, & x \neq 0
|
||||
\end{cases}=0\]
|
||||
Thus we have that the product of $f(x) \cdot g(x)$ is also a continuous function since $f(x) \cdot g(x)=0$. Thus, we have that $f(x) \cdot g(x)$ is continuous $\forall\ x \in \R$. Hence $f(x) \cdot g(x)$ is continuous at $c$, where $c=0$.
|
||||
\end{enumerate}
|
||||
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 5 - 5.2 %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item[5.] Let $g$ be defined on $\R$ by $g(1):=0$, and $g(x):=2$ if $x \neq 1$, and let $f(x):=x+1$ for all $x \in \R$. Show that $\lim\limits_{x \to 0} g \circ f \neq (g \circ f)(0)$. Why doesn't this contradict Theorem 5.2.6?\\
|
||||
|
||||
Recall \textit{Theorem 5.2.6}:
|
||||
\begin{theorem*}
|
||||
Let $A,B \subseteq \R$ and let $f:A \rightarrow \R$ and $g:B \rightarrow \R$ be functions such that $f(A) \subseteq B$. If $f$ is continuous at a point $c \in A$ and g is continuous at $b= f(c) \in B$, then the composition $g \circ f:A \rightarrow \R$ is continuous $c$.
|
||||
\end{theorem*}
|
||||
|
||||
Note that we have $g(x):=\begin{cases}
|
||||
0, & x=1 \\
|
||||
2, & x \neq 1
|
||||
\end{cases}$, and $f(x):=x+1$.\\
|
||||
|
||||
Thus $(g \circ f)(x)=g(f(x))=g(x+1)=\begin{cases}
|
||||
0, & x+1=1 \\
|
||||
2, & x+1 \neq 1
|
||||
\end{cases} = \begin{cases}
|
||||
0, & x=0 \\
|
||||
2, & x \neq 0
|
||||
\end{cases}$\\
|
||||
|
||||
This gives us that $\lim\limits_{x \to 0} (g \circ f)(x) = \lim\limits_{x \to 0} 2 = 2$, since $x \to 0 \implies x \neq 0$\\
|
||||
|
||||
Thus we have that $\lim\limits_{x \to 0} (g \circ f)(x) \neq (g \circ f)(0)$ since $2 \neq 0$.\\
|
||||
|
||||
The reason that this does not violate \textit{Theorem 5.2.6} is because $g$ is discontinuous at $f(0)$, since $f(0)=1$. \\
|
||||
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 7 - 5.2 %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item[7.] Give an example of a function $f:[0,1] \to \R$ that is discontinuous at every point of $[0,1]$ but such that $|f|$ is continuous on $[0,1]$.\\
|
||||
|
||||
Let $f(x):=\begin{cases}
|
||||
-1, & x \in \Q \\
|
||||
1, & x \in \R \setminus \Q
|
||||
\end{cases}$\\
|
||||
|
||||
Recall \textit{Theorem 2.4.8 - The Density Theorem}:
|
||||
\begin{theorem*}[\textbf{The Density Theorem}]
|
||||
If $x$ and $y$ are any real numbers with $x<y$, then there exists a rational number $r \in \Q$ such that $x < r < y$.
|
||||
\end{theorem*}
|
||||
And also recall \textit{Corollary 2.4.9}:
|
||||
\begin{corollary*}
|
||||
If $x$ and $y$ are real numbers with $x < y$, then there exists an irrational number $z$ such that $x < z < y$.
|
||||
\end{corollary*}
|
||||
|
||||
Recall the \textit{Discontinuity Criterion}:
|
||||
\begin{theorem*}[\textbf{Discontinuity Criterion}]
|
||||
Let $A \subseteq \R$, let $F:A \rightarrow \R$, and let $c \in A$. Then $f$ is discontinuous at $c$ if and only if there exists a sequence $(x_n)$ in $A$ such that $(x_n)$ converges to $c$, but the sequence $(f(x_n))$ does not converge to $f(c)$.
|
||||
\end{theorem*}
|
||||
|
||||
Consider the sequence $x_n:=(\frac{1}{3}, \frac{1}{3}, \dots, \frac{1}{3})$. Clearly $x_n \in [0,1]\ \forall\ n \in \N$. However, note that $\lim\limits_{x \to \frac{1}{3}} (x_n) = \frac{1}{3}$, but $\lim\limits_{x \to \frac{1}{3}} (f(x_n)) = 1$ but $f(\frac{1}{3})=-1$. Thus we have that $\lim\limits_{x \to \frac{1}{3}} (f(x_n)) \neq f(c)$, where $c=\frac{1}{3}$. Thus by the \textit{Discontinuity Criterion}, since there exists a sequence $(x_n) \in [0,1]$ such that $(x_n) \to c \in [0,1]$ but $(f(x_n)) \nrightarrow f(c)$. Thus we have that $f$ is discontinuous at $c=\frac{1}{3}$. However, note that by the \textit{Density Theorem} and by \textit{Corollary 2.4.9}, we have that $f$ is discontinuous at $c\ \forall c \in [0,1]$.\\
|
||||
|
||||
However, note that $|f(x)|=1\ \forall\ x \in [0,1]$. Thus we have that $|f|$ is continuous $\forall\ c \in [0,1]$.\\
|
||||
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 8 - 5.2 %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item[8.] Let $f,g$ be continuous from $\R$ to $\R$ and suppose that $f(r)=g(r)$ for all rational numbers $r$. Is it true that $f(x)=g(x)$ for all $x \in \R$?\\
|
||||
|
||||
This is actually true.
|
||||
\begin{proof}
|
||||
Let $x \in \R$ be arbitrary. By the \textit{Density Theorem}, we know that we can find a sequence $(x_n) \subset \Q \st (x_n) \to x$.\\
|
||||
|
||||
Then we have:
|
||||
\begin{align*}
|
||||
f(x) &= \lim\limits_{n \to \infty} f(x_n) &(f\text{ is continuous}) \\
|
||||
&= \lim\limits_{n \to \infty} g(x_n) &(x_n \in \Q) \\
|
||||
&= g(x) &(g\text{ is continuous}) \\
|
||||
\implies f(x) &= g(x)\ \forall\ x \in \R
|
||||
\end{align*}
|
||||
\end{proof}
|
||||
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 14 - 5.2 %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item[14.] Let $g:\R \to \R$ satisfy the relation $g(x+y)=g(x)g(y)$ for all $x,y \in \R$. Show that if $g$ is continuous at $x=0$, then $g$ is continuous at every point of $\R$. Also if we have $g(a)=0$ for some $a \in \R$, then $g(x)=0$ for all $x \in \R$.\\
|
||||
|
||||
\begin{proof}
|
||||
We first note that since we're given that $g$ is continuous at $x=0$, we have
|
||||
\begin{align*}
|
||||
g(0) &= g(0+0) \\
|
||||
&=g(0) \cdot g(0) &(g(x+y)=g(x)g(y),\ \forall x,y \in \R) \\
|
||||
&=g(0)^2
|
||||
\end{align*}
|
||||
This tells us that we have two cases to consider for $g(0)$: $g(0)=0$ or $g(0)=1$.\\
|
||||
|
||||
\begin{case}{$g(0)=0$}
|
||||
Let $g(a)=0$ for some $a \in \R$. Then we have that $\forall\ x \in \R$,
|
||||
\[g(x) = g((x-a)+a)=g(x-a) \cdot g(a)=0\]
|
||||
Thus, if $g(a)=0\ \forall\ a \in \R$, then $g \equiv 0$ and thus $g$ is constant. Thus, $g$ is continuous $\forall\ x \in \R$.
|
||||
\end{case}
|
||||
|
||||
\begin{case}{$g(0)=1$}
|
||||
Let $g(c) \neq 0\ \forall\ c \in \R$. Otherwise we have that $g$ is a zero function, which was shown to be continuous as the result of \textit{Case 1}, and this contradicts the fact that we let $g(0)=1 \neq 0$.\\
|
||||
|
||||
Let $\varepsilon > 0$ be given, and let $c \in \R$ be arbitrary such that $c \neq 0$. Since $g$ is continuous at 0 and $g(c) \neq 0$, we know that $\exists\ \delta >0 \st \forall\ x \in \R$, we have:
|
||||
\[|x-0|<\delta \implies |g(x)-g(0)|<\frac{\varepsilon}{|g(c)|}\]
|
||||
that is, since $g(0)=1$,
|
||||
\[|x| < \delta \implies |g(x)-1|<\frac{\varepsilon}{|g(c)|}\]
|
||||
Thus, for $x \in \R \st |x-c|<\delta$, we have
|
||||
\begin{align*}
|
||||
|g(x)-g(c)| &= |g((x-c)+c)-g(c)| \\
|
||||
&=|g(x-c)g(c)-g(c)| &(g(a+b)=g(a)g(b)) \\
|
||||
&= |g(c)(g(x-c)-1)| \\
|
||||
&=|g(c)||g(x-c)-1| \\
|
||||
&= |g(c)||g(h)-1| &(\text{Substitue } h:=x-c) \\
|
||||
&<|g(c)|\frac{\varepsilon}{|g(c)|} &(|h|=|x-c|<\delta) \\
|
||||
&= \varepsilon
|
||||
\end{align*}
|
||||
\end{case}
|
||||
$\therefore$ We have that $g$ is continuous at $c$. Since $c \neq 0$ was arbitrary and $g$ is continuous at 0, we have that $g$ is continuous on $\R$.
|
||||
\end{proof}
|
||||
\end{enumerate}
|
||||
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 7 %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item Prove or justify if true. Provide a counterexample if false.
|
||||
\begin{enumerate}
|
||||
\item Let $f$ be defined on $[a,b]$. Let $x_n$ be any sequence in $[a,b]$. The sequence $\{f(x_n)\}$ converges to $f(c)$ for $x_n$ converging to $c \in [a,b]$ implies that $f$ is continuous at $x=c$. \\
|
||||
|
||||
This is true. Recall the \textit{Sequential Criterion for Continuity}:
|
||||
\begin{theorem*}[\textbf{Sequential Criterion for Continuity}]
|
||||
A function $f:A \rightarrow \R$ is continuous at the point $c \in A$ if and only if for every sequence $(x_n)$ in $A$ that converges to $c$, the sequence $(f(x_n))$ converges to $f(c)$.
|
||||
\end{theorem*}
|
||||
This is true by the \textit{Sequential Criterion for Continuity}.\\
|
||||
|
||||
\item If $f$ is continuous on $D$ and the sequence $x_n$ in $D$ is a converging sequence, then the sequence $\{f(x_n)\}$ converges.\\
|
||||
|
||||
This statement is false. Consider the function $f:(0,1) \to \Q$ given by $f(x)=\frac{1}{x}$, and consider the sequence $x_n:= (0.1, 0.01,0.001,0.0001, \dots),\ \forall\ n \in \N$. Then we have that $x_n \to 0$, but since $0 \notin (0,1)$ and since $(0,1)$ is not a compact set, this violates the \textit{Sequential Criterion for Continuity} and thus this statement is false.
|
||||
\end{enumerate}
|
||||
\end{enumerate}
|
||||
\end{document}
|
||||
Binary file not shown.
+547
@@ -0,0 +1,547 @@
|
||||
\documentclass[12pt,letterpaper]{article}
|
||||
\usepackage[utf8]{inputenc}
|
||||
\usepackage[english]{babel}
|
||||
\usepackage[normalem]{ulem}
|
||||
\usepackage{cancel}
|
||||
\usepackage{amsthm}
|
||||
\usepackage{amsmath}
|
||||
\usepackage{amsfonts}
|
||||
\usepackage{amssymb}
|
||||
\usepackage{graphicx}
|
||||
\usepackage{array}
|
||||
\usepackage[left=2cm, right=2.5cm, top=2.5cm, bottom=2.5cm]{geometry}
|
||||
\usepackage{enumitem}
|
||||
\newcommand{\st}{\ \text{s.t.}\ }
|
||||
\newcommand{\abs}[1]{\left\lvert #1 \right\rvert}
|
||||
\newcommand{\R}{\mathbb{R}}
|
||||
\newcommand{\N}{\mathbb{N}}
|
||||
\newcommand{\Q}{\mathbb{Q}}
|
||||
\newcommand{\C}{\mathbb{C}}
|
||||
\newcommand{\Z}{\mathbb{Z}}
|
||||
\DeclareMathOperator{\sign}{sgn}
|
||||
\newtheoremstyle{case}{}{}{}{}{}{:}{ }{}
|
||||
\theoremstyle{case}
|
||||
\newtheorem{case}{Case}
|
||||
\theoremstyle{definition}
|
||||
\newtheorem{definition}{Definition}[section]
|
||||
\newtheorem{theorem}{Theorem}[section]
|
||||
\newtheorem*{theorem*}{Theorem}
|
||||
\newtheorem{corollary}{Corollary}[section]
|
||||
\newtheorem*{corollary*}{Corollary}
|
||||
\newtheorem{lemma}[theorem]{Lemma}
|
||||
\newtheorem*{remark}{Remark}
|
||||
\setlist[enumerate]{font=\bfseries}
|
||||
\renewcommand{\qedsymbol}{$\blacksquare$}
|
||||
\author{Alexander J. Tusa}
|
||||
\title{Real Analysis Homework 9}
|
||||
\begin{document}
|
||||
\maketitle
|
||||
\begin{enumerate}
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 1 %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item \textbf{Section 5.3}
|
||||
\begin{enumerate}
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Section 5.3 #1 %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item[1.] Let $I:=[a,b]$ and let $f:I \to \R$ be a continuous function such that $f(x)>0$ for each $x$ in $I$. Prove that there exists a number $\alpha >0$ such that $f(x) \geq \alpha$ for all $x \in I$.
|
||||
|
||||
\begin{proof}
|
||||
Recall the \textit{Maximum-Minimum Theorem}:
|
||||
\begin{theorem*}
|
||||
Let $I := [a,b]$ be a closed bounded interval and let $f: I \rightarrow \R$ be continuous on $I$. Then $f$ has an absolute maximum and an absolute minimum on $I$.
|
||||
\end{theorem*}
|
||||
Since $f$ is continuous on the closed bounded interval $I$, by the \textit{Maximum-Minimum Theorem}, there exists $x_* \in I$ such that
|
||||
\[f(x_*) \leq f(x),\ \forall\ x \in I\]
|
||||
Since $f(x)>0$, for all $x \in I$, we have that $f(x_*)>0$. Thus, let $\alpha:=f(x_*)$.\\
|
||||
|
||||
$\therefore f(x)\geq \alpha,\ \forall\ x \in I$.
|
||||
\end{proof}
|
||||
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Section 5.3 #3 %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item[3.] Let $I:=[a,b]$ and let $f:I \to \R$ be a continuous function on $I$ such that for each $x$ in $I$ there exists $y$ in $I$ such that $|f(y)| \leq \frac{1}{2}|f(x)|$. Prove that there exists a point $c$ in $I$ such that $f(c)=0$.
|
||||
|
||||
\begin{proof}
|
||||
Let $x_1 \in I$ be arbitrary. For $x_1,\ \exists\ x_2 \in I \st$
|
||||
\[|f(x_2)|\leq \frac{1}{2}f(x_1)\]
|
||||
By method of Mathematical Induction, we find a sequence $(x_n) \subseteq I$ such that
|
||||
\[|f(x_{n+1})| \leq \frac{1}{2}|f(x_n)|,\ \forall\ n \in \N\]
|
||||
Thus,
|
||||
\[|f(x_{n+1})| \leq \frac{1}{2}|f(x_n)| \leq \left(\frac{1}{2}\right)^2 \leq \dots \leq \left(\frac{1}{2}\right)^n|f(x_1)|\ \ (1)\]
|
||||
Recall the \textit{Bolzano-Weirstrass Theorem}:
|
||||
\begin{theorem*}
|
||||
A bounded sequence of real numbers has a convergent subsequence.
|
||||
\end{theorem*}
|
||||
Then, by the \textit{Bolzano-Wierstrass Theorem} there exists a subsequence $(x_{p(n)})$ of $(x_n)$ which converges to $c \in I$. Since $f$ is continuous, then $(f(x_{p(n)}))$ is also convergent and converges to $f(c)$. Because of $(1)$ we have
|
||||
\[\lim\limits_{n \to \infty} f(x_{p(n)})=0\]
|
||||
$\therefore f(c)=0\ \& \ c \in I$.
|
||||
\end{proof}
|
||||
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Section 5.3 #6 %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item[6.] Let $f$ be continuous on the interval $[0,1]$ to $\R$ and such that $f(0)=f(1)$. Prove that there exists a point $c$ in $[0, \frac{1}{2}]$ such that $f(c)=f(c+\frac{1}{2})$. (Hint: Consider $g(x)=f(x)-f(x+\frac{1}{2})$.) Conclude that there are, at any time, antipodal points on the earth's equator that have the same temperature.
|
||||
|
||||
\begin{proof}
|
||||
Let $g(x):=f(x)-f(x+\frac{1}{2})$. We note that $g$ is continuous, and thus yields
|
||||
\begin{align*}
|
||||
g(0) &= f(0)-f\left(\frac{1}{2}\right) \\
|
||||
g\left(\frac{1}{2}\right) &= f\left(\frac{1}{2}\right) - f(1) \\
|
||||
&= f\left(\frac{1}{2}\right)-f(0)
|
||||
\end{align*}
|
||||
We must now consider three separate cases: \textbf{1.} if $f(0) = f(\frac{1}{2})$, \textbf{2.} if $f(0)\neq f(\frac{1}{2})$ such that $f(0)>f(\frac{1}{2})$, and \textbf{3.} if $f(0) \neq f(\frac{1}{2})$ such that $f(0)<f(\frac{1}{2})$.\\
|
||||
|
||||
\begin{case}
|
||||
Let $f(0)=f(\frac{1}{2})$. Then if we let $c=0$, we are done.
|
||||
\end{case}
|
||||
|
||||
\begin{case}
|
||||
Let $f(0) \neq f(\frac{1}{2})$ such that $f(0)>f(\frac{1}{2})$. Then we have
|
||||
\[g(0)>0\ \text{and}\ g\left(\frac{1}{2}\right)<0\]
|
||||
Recall the \textit{Location of Roots Theorem (Theorem 5.3.5)}:
|
||||
\begin{theorem*}
|
||||
Let $I=[a,b]$ and let $f:I \rightarrow \R$ be continuous on $I$. If $f(a) < 0 < f(b)$, or if $f(a) > 0 > f(b)$, then there exists a number $c \in (a,b)$ such that $f(c)=0$.
|
||||
\end{theorem*}
|
||||
By the \textit{Location of Roots Theorem} we know that $\exists\ c \in [0,\frac{1}{2}] \st g(c)=0$. Thus $f(c)=f(c+\frac{1}{2})$.
|
||||
\end{case}
|
||||
|
||||
\begin{case}
|
||||
Let $f(0) \neq f(\frac{1}{2})$ such that $f(0)<f(\frac{1}{2})$. Then we have
|
||||
\[g(0) < 0 \text{ and } g\left(\frac{1}{2}\right)>0\]
|
||||
By the \textit{Location of Roots Theorem}, we know that $\exists\ c \in [0, \frac{1}{2}]$ such that $g(c)=0$. Thus $f(c)=f(c+\frac{1}{2})$.
|
||||
\end{case}
|
||||
We have thus proven that in any case there exists a point $c \in [0,\frac{1}{2}]$ such that $f(c)=f(c+\frac{1}{2})$.\\
|
||||
\end{proof}
|
||||
Now, we want to show that there are, at any time, antipodal points on the earth's equator that have the exact same temperature.\\
|
||||
|
||||
\begin{proof}
|
||||
Let $t:[0,2\pi] \to \R$ be the temperature at a point on the equator with angle $\rho$ from some predetermined point on the equator. We notice that $t$ is continuous in $\rho$ on $[0,2\pi]$ and that $t$ is a periodic function with a regular interval of $2\pi$.\\
|
||||
|
||||
Now, let us define antipodal difference in temperatures as:
|
||||
\[d:[0,2\pi] \to \R,\ \ d(\rho):=(\rho + \pi)-t(\rho)\]
|
||||
Also, note that $d$ is also continuous on $[0,2\pi]$ and:
|
||||
\[d(0)=t(\pi)-t(0) \text{ and } d(\pi)=t(2\pi)-t(\pi)\]
|
||||
Hence since $t$ is $2\pi$ periodic:
|
||||
\[d(0)=-d(\pi)\]
|
||||
\begin{case}
|
||||
Let $d(0)=0$. Then we have that we are done. We have thus found antipodal points with equal temperature.
|
||||
\end{case}
|
||||
|
||||
\begin{case}
|
||||
Let $d(0) \neq 0$. Then without loss of generality, assume that $d(0)<0<d(\pi)$. Then since $d$ is continuous on $[0, \pi]$, we have that $[0,\pi] \subset [0,2\pi],\ \exists\ \delta \in [0, \pi] \st d(\delta)=0$.\\
|
||||
|
||||
Thus, $d(\delta)=d(\delta + \pi)$, which means that we have found our antipodal points with equal temperature.
|
||||
\end{case}
|
||||
$\therefore$ We have shown that at any time, there are two points on the earth's equator that have the exact same temperature.
|
||||
\end{proof}
|
||||
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Section 5.3 #8 %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item[8.] Show that the function $f(x):=2\ln x+\sqrt{x}-2$ has root in the interval $[1,2]$, Use the Bisection Method and a calculator to find the root with error less than $10^{-2}$.\\
|
||||
|
||||
If we let $f(x)=2\ln x + \sqrt{x}-2$ over the interval $[1,2]$, then $f(x)$ is an increasing function taking on values $[-1,0.8005\dots]$. The following table shows the steps taken to calculate the root using the Bisection Method to successively narrow the interval containing the root. Note that after 7 iterations, the root is contained in the interval $(1.469, 1.484)$ and the error is less than $10^{-2}$.\\
|
||||
|
||||
\[\begin{array}{|l|l|l|l|l|l|}
|
||||
\hline
|
||||
n & a_n & b_n & c_n &f(c_n) & \frac{1}{2}(b_n-a_n) \\
|
||||
\hline
|
||||
1 & 1 & 2 & 1.5 & 0.035675088 & 0.5 \\
|
||||
\hline
|
||||
2 & 1 & 1.5 & 1.25 & -0.435678909 & 0.25 \\
|
||||
\hline
|
||||
3 & 1.25 & 1.5 & 1.375 & -0.190488598 & 0.125 \\
|
||||
\hline
|
||||
4 & 1.375 & 1.5 & 1.4375 & -0.075231132 & 0.0625 \\
|
||||
\hline
|
||||
5 & 1.4375 & 1.5 & 1.46875 & -0.019256638 & 0.03125 \\
|
||||
\hline
|
||||
6 & 1.46875 & 1.5 & 1.484375 & 0.00833691 & 0.015625 \\
|
||||
\hline
|
||||
7 & 1.48675 & 1.484375 & 1.4765625 & -0.005427617 & 0.0078125\\
|
||||
\hline
|
||||
\end{array}\]
|
||||
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Section 5.3 #15 %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item[15.] Examine which open [respectively, closed] intervals are mapped by $f(x):=x^2$ for $x \in \R$ onto open [respectively, closed] intervals.\\
|
||||
|
||||
We first observe the open intervals. Take any open interval $I=(a,b)$ where $a<b$.\\
|
||||
|
||||
If $0 \in I$, then $f(x)=x^2$ attains the absolute minimum value of $f(0)=0$. For any $x \in I,\ f(x) \geq 0\ f(I)=[0,\max \{a^2, b^2\}]$. Thus the image of $f$ on $(a,b)$ is not open at the end point $x=0$.\\
|
||||
|
||||
Now, if $0 \notin I$, then both $a$ and $b$ are positive or negative.
|
||||
\begin{itemize}
|
||||
\item In the case where $0<a<b$, then $0 < a^2 < b^2$. Thus, $f((a,b))=(a^2,b^2)$ which is an open interval.
|
||||
|
||||
\item In the case where $a < b < 0$, then $0 < b^2 < a^2$. Thus, $f((a,b))=(b^2,a^2)$ which is also an open interval.
|
||||
\end{itemize}
|
||||
Therefore open intervals that do not contain $0$ map to an open interval by the given function $f$.\\
|
||||
|
||||
Now, consider the closed intervals. Let $I=[a,b]$ be a closed interval where $a<b$.\\
|
||||
|
||||
If $0 \in I$, then $f(x) \geq 0\ \forall\ x \in I$ and $f(0)=0$.
|
||||
\begin{align*}
|
||||
&0 \leq x \leq b &a \leq x \leq 0 \\
|
||||
&0 \leq x^2 \leq b^2 &0 \leq x^2 \leq a^2
|
||||
\end{align*}
|
||||
\[0\leq x^2 \leq \max \{a^2, b^2\}\]
|
||||
\[f([a,b])=[0, \max \{a^2, b^2\}]\]
|
||||
Now, if $0 \notin I$, then both $a$ and $b$ are positive or negative.
|
||||
\begin{itemize}
|
||||
\item In the case where $a>0,\ b>0$, then $0<a^2<b^2$. Thus $f([a,b])=[a^2,b^2]$.
|
||||
|
||||
\item In the case where $a < b < 0$ then for all $x$ such that $a \leq x \leq b \implies b^2 \leq x^2 \leq a^2$. Thus $f([a,b])=[b^2,a^2]$
|
||||
\end{itemize}
|
||||
Therefore, in any case, closed intervals map to a closed interval by a given function $f$.
|
||||
\end{enumerate}
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 2 %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item \textbf{Section 5.4}
|
||||
\begin{enumerate}
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Section 5.4 #2 %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item[2.] Show that the function $f(x):= 1/x^2$ is uniformly continuous on $A:=[1, \infty)$, but that it is not uniformly continuous on $B:=(0,\infty)$.\\
|
||||
|
||||
\begin{proof}
|
||||
Let $x,y \in A$. Then
|
||||
\begin{align*}
|
||||
\abs{\frac{1}{x^2}-\frac{1}{y^2}} &= \frac{|x^2-y^2|}{xy} \\
|
||||
&= \frac{|x-y||x+y|}{xy} \\
|
||||
&= \left(\frac{1}{x}+\frac{1}{y}\right) \cdot |x-y|
|
||||
\end{align*}
|
||||
Since $x,y \in [1, \infty)$, it follows that $\frac{1}{x} \leq 1$ and $\frac{1}{y} \leq 1$ which means that
|
||||
\[\left(\frac{1}{x}+\frac{1}{y}\right) \leq (1+1) \implies \left(\frac{1}{x}+\frac{1}{y}\right) \cdot |x-y| \leq (1+1) \cdot |x-y|\]
|
||||
Now, we can write:
|
||||
\[\abs{\frac{1}{x^2}-\frac{1}{y^2}}=\left(\frac{1}{x}+\frac{1}{y}\right) \cdot |x-y| \leq (1+1) \cdot |x-y|\]
|
||||
\[\Downarrow\]
|
||||
\[\abs{\frac{1}{x^2}-\frac{1}{y^2}} \leq 2 \cdot |x-y|\ \forall\ x,y \in A\]
|
||||
Recall the definition of a Lipschitz function:
|
||||
\theoremstyle{definition}
|
||||
\begin{definition}
|
||||
Let $A \subseteq \R$ and let $f:A \rightarrow \R$. If there exists a constant $K > 0$ such that
|
||||
\[(4)\ \ \ \ \ \ \ |f(x)-f(u)| \leq K|x-u|\]
|
||||
for all $x,u \in A$, then $f$ is said to be a \textbf{Lipschitz function} (or to satisfy a \textbf{Lipschitz condition}) on $A$.\\
|
||||
|
||||
The condition $(4)$ that a function $f:I \to \R$ on an interval $I$ is a Lipschitz function can be interpreted geometrically as follows. If we write the condition as
|
||||
\[\abs{\frac{f(x)-f(u)}{x-u}}\leq K,\ \ x,u \in I,\ x \neq u,\]
|
||||
then the quantity inside the absolute values is the slope of a line segment joining the points $(x,f(x))$ and $(u,f(u))$. Thus a function $f$ satisfies a Lipschitz condition if and only if the slopes of all line segments joining two points on the graph of $y=f(x)$ over $I$ are bounded by some number $K$.
|
||||
\end{definition}
|
||||
Also, recall \textit{Theorem 5.4.5}:
|
||||
\begin{theorem*}
|
||||
If $f:A \rightarrow \R$ is a Lipschitz function, then $f$ is uniformly continuous on $A$.
|
||||
\end{theorem*}
|
||||
So we have proven that by the definition of a Lipschitz function that the given function $f$ is a Lipschitz function, and thus by \textit{Theorem 5.4.5}, we can conclude that $f$ is uniformly continuous on $A$.
|
||||
\end{proof}
|
||||
Now, we will show that $f$ is not uniformly continuous on $(0, \infty)$.
|
||||
\begin{proof}
|
||||
Let $x_n,y_n \in B \st x_n=\frac{1}{\sqrt{n}} \text{ and } y_n=\frac{1}{\sqrt{n+1}}$. We note that $\lim\limits_{n \to \infty} (x_n-y_n)=0$.
|
||||
\begin{align*}
|
||||
\abs{\frac{1}{x^2}-\frac{1}{y^2}} &= |(\sqrt{n})^2 - (\sqrt{n+1})^2| \\
|
||||
&=|n+1-n| \\
|
||||
&=1
|
||||
\end{align*}
|
||||
Recall the \textit{Nonuniform Continuity Criteria (Theorem 5.4.2)}:
|
||||
\begin{theorem*}
|
||||
Let $A \subseteq \R$ and let $f:A \rightarrow \R$. Then the following statements are equivalent:
|
||||
\begin{enumerate}
|
||||
\item $f$ is not uniformly continuous on $A$.
|
||||
|
||||
\item There exists an $\varepsilon_0 > 0$ such that for every $\delta > 0$ there are points $x_\delta, u_\delta$ in $A$ such that $|x_\delta - u_\delta|<\delta$ and $|f(x_\delta) - f(u_\delta)| \geq \varepsilon_0$.
|
||||
|
||||
\item There exists an $\varepsilon_0 > 0$ and two sequences $(x_n)$ and $(u_n)$ in $A$ such that $\lim (x_n - u_n)=0$ and $|f(x_n)-f(u_n)|\geq \varepsilon_0=1$ for all $n \in \N$.
|
||||
\end{enumerate}
|
||||
\end{theorem*}
|
||||
Thus we have shown that for $x_n, y_n \in B$, it's true that $\lim\limits_{n \to \infty} (x_n-y_n)=0$ but $|f(x_n)-f(y_n)|=1$ and thus by the \textit{Nonuniform Continuity Criteria}, we can conclude that $f$ is not uniformly continuous on $B$.
|
||||
\end{proof}
|
||||
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Section 5.4 #3b %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item[3. (b)] Use the Nonuniform Continuity Criterion 5.4.2 to show that $g(x):=\sin (1/x)$ is not uniformly continuous on $B:=(0, \infty)$.\\
|
||||
|
||||
\begin{proof}
|
||||
Let $(a_n)$ be the sequence defined by $a_n=(\frac{\pi}{2}+2n\pi)^{-1}$, and let $(b_n)$ be the sequence defined by $b_n=(2n\pi)^{-1}$. We notice that $(a_n) \subseteq B$ and $(b_n) \subseteq B$.\\
|
||||
|
||||
We now want to show that $\lim\limits_{n \to \infty} (a_n-b_n)=0$. So,
|
||||
\begin{align*}
|
||||
\lim\limits_{n \to \infty} (a_n - b_n) &= \lim\limits_{n \to \infty} \left(\frac{1}{\frac{\pi}{2}+2n\pi}-\frac{1}{2n\pi}\right) \\
|
||||
&= \lim\limits_{n \to \infty} \left(\frac{1}{\frac{\pi}{2}+2n\pi}\right)-\lim\limits_{n \to \infty} \left(\frac{1}{2n\pi}\right) \\
|
||||
&= \frac{1}{\infty} - \frac{1}{\infty} \\
|
||||
&= 0-0 \\
|
||||
&= 0
|
||||
\end{align*}
|
||||
Now, we want to find the value of $|g(a_n)-g(b_n)|$ for the given function $g(x)=\sin \left(\frac{1}{x}\right)$
|
||||
\begin{align*}
|
||||
|g(a_n-b_n)| &= \abs{g\left(\frac{1}{\frac{\pi}{2}+2n\pi}\right)-g\left(\frac{1}{2n\pi}\right)} \\
|
||||
&= |\sin(\frac{\pi}{2}+2n\pi)-\sin(2n\pi)| \\
|
||||
&= |\cos (2n\pi) - \sin (2n\pi)| \\
|
||||
&=1-0 \\
|
||||
&= 1
|
||||
\end{align*}
|
||||
Thus we have that $\lim\limits_{n \to \infty} (a_n-b_n)=0$, but $|f(a_n)-f(b_n)|=1\ \forall\ n \in \N$ and therefore by the \textit{Nonuniform Continuity Criterion}, we have that $g$ is not uniformly continuous.
|
||||
\end{proof}
|
||||
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Section 5.4 #5 %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item[5.] Show that if $f$ and $g$ are uniformly continuous on a subset $A$ of $\R$, then $f+g$ is uniformly continuous on $A$.\\
|
||||
|
||||
\begin{proof}
|
||||
Recall the definition of uniform continuity:\\
|
||||
|
||||
We say a function $f$ is continuous if
|
||||
\[\forall\ \varepsilon > 0,\ \exists\ \delta > 0 \st |f(x)-f(y)|<\varepsilon\ \forall\ x,y \in \R \st |x-y|<\delta\]
|
||||
Let $\varepsilon > 0$ be given, and choose $\delta_1,\ \delta_2 >0$ such that:
|
||||
\begin{align*}
|
||||
&|f(x)-f(y)|<\frac{\varepsilon}{2} &\text{for any } |x-y|<\delta_1 \\
|
||||
&|g(x)-g(y)|<\frac{\varepsilon}{2} &\text{for any } |x-y|<\delta_2
|
||||
\end{align*}
|
||||
Choose $\delta=\min \{\delta_1, \delta_2\}$. Then $\forall\ x,y \in \R$:
|
||||
\[|x-y|<\delta \implies |f(x)+g(x)-f(y)-g(y)|<|f(x)-f(y)|+|g(x)-g(y)|<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon\]
|
||||
Therefore we have shown that $\forall\ |x-y|<\delta\ |(f+g)(x)-(f+g)(y)|<\varepsilon$.\\
|
||||
|
||||
$\therefore\ f+g$ is uniformly continuous.
|
||||
\end{proof}
|
||||
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Section 5.4 #6 %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item[6.] Show that if $f$ and $g$ are uniformly continuous on $A \subseteq \R$ and if they are \textit{both} bounded on $A$, then their product $fg$ is uniformly continuous on $A$.\\
|
||||
|
||||
\begin{proof}
|
||||
Functions $f$ and $g$ are both bounded on $A \subseteq \R$.\\
|
||||
|
||||
Recall the definition of boundedness (\textit{Definition 5.3.1}):
|
||||
\theoremstyle{definition}
|
||||
\begin{definition}
|
||||
A function $f:A \rightarrow \R$ is said to be \textbf{bounded on} $A$ if there exists a constant $M > 0$ such that $|f(x)| \leq M$ for all $x \in A$.
|
||||
\end{definition}
|
||||
Let $M$ be the maximum value obtained by either $|f|$ or $|g|$. That is to say that $\exists\ M > 0 \st |f(x)| < M$ and $|g(x)|<M\ \forall\ x \in A$.\\
|
||||
|
||||
We have defined $M$ as a positive number, but what happens if $M=0$? For example, the constant function $c(x)=0\ \forall\ x$ is bounded by 0.\\
|
||||
|
||||
If $M=0$, then $f=g=0$ and the function $f(x)g(x)=0$ is uniformly continuous. Now, assuming that $M>0$ we can divide by $M$.\\
|
||||
|
||||
Let $\varepsilon>0$ be given. Then there exists $\delta_1, \delta_2 >0$ such that :
|
||||
\begin{align*}
|
||||
&|f(x)-f(y)| < \frac{\varepsilon}{2M} &\text{for any } |x-y|<\delta_1 \\
|
||||
&|g(x)-g(y)| < \frac{\varepsilon}{2M} &\text{for any } |x-y|<\delta_2
|
||||
\end{align*}
|
||||
Choose $\delta=\min \{\delta_1, \delta_2\}$. Then $\forall\ x,y \in \R$:
|
||||
\begin{align*}
|
||||
|x-y|<\delta \implies |f(x)g(x)-f(y)g(y)| &= |f(x)g(x)-f(y)g(x)+f(y)g(x)-f(y)g(y)| \\
|
||||
&\leq |g(x)||f(x)-f(y)| + |f(y)||g(x)-g(y)| \\
|
||||
&\leq M|f(x)-f(y)| + M|g(x)-g(y)| \\
|
||||
&\leq M\frac{\varepsilon}{2M} + M \frac{\varepsilon}{2M} \\
|
||||
&= \varepsilon
|
||||
\end{align*}
|
||||
Thus we have shown that $\forall\ |x-y|<\delta,\ |(f \cdot g)(x) - (f \cdot g)(y)|<\varepsilon$.\\
|
||||
|
||||
$\therefore\ fg$ is uniformly continuous by the definition of uniform continuity.
|
||||
\end{proof}
|
||||
\end{enumerate}
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 3 %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item[3.] Prove or justify, if true; provide a counterexample, if false. For all parts, assume $f$ and $g$ are functions defined on the given intervals or sets.
|
||||
\begin{enumerate}
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 3 (a) %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item If $f$ is bounded on $[a,b]$, then $f$ is continuous on $[a,b]$.\\
|
||||
|
||||
This is a false statement. Consider the case $f:[-1,1] \to \R$ be given by $f(x):=\frac{1}{x}$. Then we have that $f$ is bounded on $[-1,1]$, but notice that $f$ is discontinuous at $x=0$, since $f$ is undefined at $x=0$. That is, $\frac{1}{0}$ is undefined.\\
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 3 (b) %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item If $f$ is continuous on $(a,b)$, then $f$ is bounded on $(a,b)$.\\
|
||||
|
||||
This is a false statement. Consider $f:(-\infty, \infty) \to \R$ given by $f(x):=x$. Then we have that $f$ is continuous, but since $(-\infty, \infty)$ is an unbounded set, we have that this is a false statement.\\
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 3 (c) %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item If $[f(x)]^2$ is continuous on $(a,b)$, then $f$ is continuous on $(a,b)$.\\
|
||||
|
||||
This is a false statement. Consider the case where $f^2:(-\infty, \infty) \to (-\infty, \infty)$ given by $[f(x)]^2:=-1$. Then we have that since $[f(x)]^2$ is a constant function, and thus it is continuous, however since $f(x)=\sqrt{-1}$, since both our domain and range are the real numbers, we have that this function is undefined in the real numbers since $\sqrt{-1} \in \C$. Thus $f$ is not continuous on $(-\infty, \infty)$.\\
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 3 (d) %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item If $f$ and $g$ are not continuous on $(a,b)$, then $f+g$ is not continuous on $(a,b)$.\\
|
||||
|
||||
This is a false statement. Consider $f,g:(-5,5) \to (-5,5)$ given by $f(x):=\begin{cases}
|
||||
1, & x \in \R \setminus \Q \\
|
||||
0, & x \in \Q
|
||||
\end{cases}$ and $g(x):=\begin{cases}
|
||||
0, & x \in \R \setminus \Q \\
|
||||
1, & x \in \Q
|
||||
\end{cases}$\\
|
||||
Since these are both different variant forms of the Dirichlet function, we know that they're both discontinuous, however note that their sum yields
|
||||
\[f(x)+g(x)=\begin{cases}
|
||||
1+0, & x \in \R \setminus \Q \\
|
||||
0+1, & x \in \Q
|
||||
\end{cases}\ \ =\begin{cases}
|
||||
1, & x \in \R \setminus \Q \\
|
||||
1, & x \in \Q
|
||||
\end{cases}\ \ = 1\ \forall\ x \in \R\]
|
||||
which is a constant function, and thus is continuous.\\
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 3 (e) %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item If $f$ and $g$ are not continuous on $(a,b)$, then $fg$ is not continuous on $(a,b)$.\\
|
||||
|
||||
This is a false statement. Consider $f,g:(-\infty,\infty) \to (-\infty,\infty)$ given by $f(x):=\begin{cases}
|
||||
x, & x \neq \pi \\
|
||||
0, & x = \pi
|
||||
\end{cases}$ and $g(x):=\begin{cases}
|
||||
x, & x = \pi \\
|
||||
0, & x \neq \pi
|
||||
\end{cases}$\\
|
||||
Then we have that both of these functions are discontinuous, but their product yields
|
||||
\[f(x)g(x)=\begin{cases}
|
||||
x \cdot 0 \\
|
||||
0 \cdot x
|
||||
\end{cases} = \begin{cases}
|
||||
0 \\
|
||||
0
|
||||
\end{cases} = 0\]
|
||||
which is a constant function, which is always continuous.\\
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 3 (f) %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item If $f$ and $g$ are not continuous on $(a,b)$, then $f \circ g$ is not continuous on $(a,b)$.\\
|
||||
|
||||
This is a false statement. Consider $f,g:(-\infty, \infty) \to (-\infty, \infty)$ given by $f(x)=\begin{cases}
|
||||
1, & x \in \Q \\
|
||||
-1, & x \in \R \setminus \Q
|
||||
\end{cases}$ and $g(x):=\begin{cases}
|
||||
-1, & x \in \Q \\
|
||||
1, & x \in \R \setminus \Q
|
||||
\end{cases}$\\
|
||||
Then we have that both $f$ and $g$ are discontinuous, however $f \circ g$ is continuous everywhere. \\
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 3 (g) %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item If $fg$ and $f$ are continuous on $(a,b)$, then $g$ is continuous on $(a,b)$.\\
|
||||
|
||||
This is a false statement. Consider $f,g:(-\infty, \infty) \to (-\infty, \infty)$ given by $f(x):=0$, $g(x):=\begin{cases}
|
||||
1, & x \in \Q \\
|
||||
0, & x \in \R \setminus \Q
|
||||
\end{cases}$\\
|
||||
Then we have that $fg=0$, which is a continuous function and $f(x)$ is continuous as well since they are both constant functions, however since $g$ is discontinuous $\forall x \in \R$, we have that this is a false statement.\\
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 3 (h) %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item If $f+g$ and $f$ are continuous on $(a,b)$, then $g$ is continuous on $(a,b)$.\\
|
||||
|
||||
This is true. Recall \textit{Theorem 5.2.1}:
|
||||
\begin{theorem*}
|
||||
Let $A \subseteq \R$, let $f$ and $g$ be functions on $A$ to $\R$, and let $b \in \R$. Suppose that $c \in A$ and that $f$ and $g$ are continuous at $c$.
|
||||
\begin{enumerate}
|
||||
\item Then $f+g,\ f-g,\ fg$, and $bf$ are continuous at $c$.
|
||||
|
||||
\item If $h:A \rightarrow \R$ is continuous at $c \in A$ and if $h(x) \neq 0$ for all $x \in A$, then the quotient $f/h$ is continuous at $c$.
|
||||
\end{enumerate}
|
||||
\end{theorem*}
|
||||
|
||||
Thus, if we let $g=(f+g)-f$, we have that $g$ is also continuous by \textit{Theorem 5.2.1}.\\
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 3 (i) %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item If $|f|$ is continuous on $(a,b)$, then $f$ is continuous on $(a,b)$.\\
|
||||
|
||||
This is a false statement. Consider $f:(-\infty, \infty) \to (-\infty, \infty)$ given by $f(x):=\begin{cases}
|
||||
1, & x > 0 \\
|
||||
-1, & x \leq 0
|
||||
\end{cases}$\\
|
||||
Then we note that $|f(x)|=1$ is continuous, but $f(x)$ is not continuous.\\
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 3 (j) %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item If $f(x)=\frac{\sin x}{x}$ and $g(x)=\frac{1}{x}$, then $f/g$ is continuous on $\R$.\\
|
||||
|
||||
This is a false statement. Note that $f,g$ have a domain $\R \setminus \{0\}$. So, if we use $f,g$ to form $\frac{f}{g}$, we have that the domain of the quotient is also $\R \setminus \{0\}$, then the quotient is also not continuous on $\R$.\\
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 3 (k) %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item $f$ is continuous at $c \in A$ if and only if for all $\varepsilon > 0$, there exists a $\delta >0$ such that if $|x-c|<\delta$ and $x \in A$, then $|f(x)-f(c)|<\varepsilon$.\\
|
||||
|
||||
This is a true statement, since this is the definition of continuity at a point $c$.\\
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 3 (l) %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item If $f(A)$ is bounded, then $f$ is continuous on $A$.\\
|
||||
|
||||
This is a false statement. Consider $f:(-5,5) \to (-1,1)$ given by $f(x):=\begin{cases}
|
||||
1, & x > 0 \\
|
||||
-1, & x \leq 0
|
||||
\end{cases}$\\
|
||||
Then we have that $f((-5,5))$ is bounded between $(-1,1)$. However, $f$ is not continuous. \\
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 3 (m) %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item If $f$ is continuous at $c \in A$, and $x_n$ is a sequence in $A$, then $x_n \to c$ whenever $f(x_n) \to f(c)$.\\
|
||||
|
||||
This is a false statement. Consider $f(x):=x^2$. Then note that $f$ is continuous at $x=2$. Also, let $(x_n):=(-1)^n \cdot 2=-2,2,-2,2, \dots$. Then we have that $f(x_n)=4\ \forall\ n \in \N$. Thus, $f(x_n) \to f(c)=4$ but $x_n$ is divergent.\\
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 3 (n) %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item If $x_n$ is a Cauchy sequence in $A$, then $f(x_n)$ converges.\\
|
||||
|
||||
This is a false statement. Consider the sequence $x_n=1, \frac{1}{2}, \frac{1}{3}, \dots$. Then, if we let $f: \R \to \R$ given by $f(x):=\frac{1}{x}$. Then we have that $f(x_n)$ diverges since $(f(x_n)) = 1, 2, 3, 4, \dots$.\\
|
||||
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 3 (o) %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item If $f:\R \to \R$ is continuous at each irrational number, then $f$ is continuous on $\R$.\\
|
||||
|
||||
This is a false statement. Consider the Thomae function, defined by \[f(x):= \begin{cases}
|
||||
\frac{1}{q}, & x=\frac{p}{q},\ x \in \Q,\ p \in \Z,\ q \in \N \\
|
||||
0, & x \in \R \setminus \Q
|
||||
\end{cases}\]
|
||||
Then we have that we know that the Thomae function is only continuous on the irrationals, but it is discontinuous on all of the rational numbers, thus the Thomae function is not continuous on $\R$.\\
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 3 (p) %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item There exists $x_1 \in A$ such that $f(x_1) \geq f(x)$ for all $x \in A$.\\
|
||||
|
||||
This is a false statement. Consider $f:\R \to \R$ given by $f(x):=x$. Then $f(x)$ is not bounded. Thus by the \textit{Archimedian Property}, we have that there does not exist an $x_1 \in \R \st f(x_1) \geq f(x)\ \forall\ x \in \R$.\\
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
%%%%%%%% Question 3 (q) %%%%%%%%
|
||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||
\item Let $A$ be a bounded subset of $\R$. Then $f(A)$ is bounded.\\
|
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This is a false statement. Consider $f:(0,1) \to \R$ given by $f(x):=\frac{1}{x}$. Then we have that $\lim\limits_{x \to 0^+} f(x) = \infty$, and thus the image $f((0,1))$ is unbounded.\\
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%%%%%%%% Question 3 (r) %%%%%%%%
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\item Let $A=[a,b]$ and $f(a)<0<f(b)$. Then there exists $c \in (a,b)$ such that $f(c)=0$.\\
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This is a false statement. Consider $f:[-1,1] \to \R$ given by $f(x):=\frac{1}{x}$. Then we have that $-1 < 0 < 1$, which satisfies the problem statement. However, since $f$ is discontinuous at $x=0$, we have that there does not exist $c \in (-1,1)$ such that $f(c)=0$, because $f$ is undefined at $x=0$.\\
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%%%%%%%% Question 3 (s) %%%%%%%%
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\item Let $A=[a,b]$ and $f(a)<k<f(b)$. Then there exists $c \in [a,b]$ such that $f(c)=k$.\\
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This is a false statement. Consider $f:[-1,1] \to \R$ given by $f(x):=\frac{1}{x}$, and let $k=0$. Then we have that $-1 < 0 < 1$, which satisfies the problem statement. However, since $f$ is discontinuous at $x=0$, we have that there does not exist $c \in (-1,1)$ such that $f(c)=k$, because $f$ is undefined at $x=0$.\\
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%%%%%%%% Question 3 (t) %%%%%%%%
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\item Let $A$ be bounded. Then $f$ assumes maximum and minimum values on $A$.\\
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This is a false statement. Consider $f:(0,1) \to \R$ given by $f(x):=\frac{1}{x}$. Then we have that since there exists an uncountably infinite number of values of $x \in (0,1)$, then it is not possible to choose an $x$ such that $f(x) \leq f(y)\ \forall\ y \in (0,1)$, and it is also not possible to find an $x$ such that $f(x) \geq f(y)\ \forall\ y \in (0,1)$.\\
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%%%%%%%% Question 3 (u) %%%%%%%%
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\item If $A$ is unbounded, then $f(A)$ is unbounded.\\
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This is a false statement. Consider $f:\R \to \R$ given by $f(x):=1$. Then we have that $\R$ is unbounded, however $f(\R)=1\ \forall\ x \in \R$. Thus $f(\R)$ is bounded between $[1,1]=[1]$.
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\end{enumerate}
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\end{enumerate}
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\end{document}
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Reference in New Issue
Block a user