diff --git a/Documents/LaTeX/Change of Variables in 2D.tex b/Documents/LaTeX/Change of Variables in 2D.tex new file mode 100644 index 0000000..74a8307 --- /dev/null +++ b/Documents/LaTeX/Change of Variables in 2D.tex @@ -0,0 +1,58 @@ +\documentclass[12pt,letterpaper]{article} +\usepackage[utf8]{inputenc} +\usepackage{pgfplots} +\usepackage[english]{babel} +\usepackage{amsthm} +\usepackage{cancel} +\usepackage{mathtools} +\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} +\newcommand{\limx}[2]{\displaystyle\lim\limits_{#1 \to #2}} +\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{\dotp}{\dot{\mathcal{P}}} +\newcommand{\dotq}{\dot{\mathcal{Q}}} +\newcommand{\dist}{\text{dist}} +\DeclareMathOperator{\sign}{sgn} +\newtheoremstyle{case}{}{}{}{}{}{:}{ }{} +\theoremstyle{case} +\newtheorem{case}{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*{lemma*}{Lemma} +\newtheorem*{remark}{Remark} +\setlist[enumerate]{font=\bfseries} +\renewcommand{\qedsymbol}{$\blacksquare$} +\author{Alexander J. Tusa} +\title{Change of Variables in 2D} +\begin{document} + \maketitle + \begin{enumerate} + \item Evaluate $\displaystyle\int \int_R\ dA$ where $R$ is the region bounded by the lines $x+y=1$, $x+y=2$, $2x-3y=2$, and $2x-3y=5$. Use the transformation $u=x+y$ and $v=2x-3y$. Draw the region $R$ in the $xy$-plane and the new region $S$ in the $uv$-plane. also, compute $x=x(u,v),\ y=y(u,v)$. + + \item Evaluate $\displaystyle\int\int_R \frac{2x-y}{2}\ dA$ where $R$ is the region bounded by the lines $y=2x$, $y=2x-2$, $y=0$, and $y=4$. Use the transformation $u=2x-y$ and $v=y$. Draw the region $R$ in the $xy$-plane and the new region $S$ in the $uv$-plane. Also, computer $x=x(u,v),$ and $y=y(u,v)$. + + \item Evaluate $\displaystyle\int\int_R 2(x-y)\ dA$ where $R$ is the region bounded by the lines $x=0$, $x=-3$, $y=x$ and $y=x+1$. Use the transformation $u=-x$ and $v=-x+y$. Draw the region $R$ in the $xy$-plane and the new region $S$ in the $uv$-plane. Also, compute $x=x(u,v)$, and $y=y(u,v)$. + + \item Evaluate $\displaystyle\int\int_R (2x^2-xy-y^2)\ dA$ where $R$ is the region bounded by the lines $y=-2x+4,\ y=-2x+7,\ y=x-2,$ and $y=x+1$ in the first quadrant. Use the transformation $u=x-y$, and $v=2x+y$. Draw the region $R$ in the $xy$-plane and the new region $S$ in the $uv$-plane. Also, compute $x=x(u,v)$ and $y=y(u,v)$. + + \item Evaluate $\displaystyle\int\int_R (3x^2+14xy+8y^2)\ dA$ where $R$ is the region bounded by the lines $y=-3/2x+1,\ y=-3/2x+3,\ y=-x/4,$ and $y=-x/4+1$. Use the transformation $u=3x+2y$ and $v=x+4y$. Draw the region $R$ in the $xy$-plane and the new region $S$ in the $uv$-plane. Also, compute $x=x(u,v)$, and $y=y(u,v)$. + \end{enumerate} +\end{document} diff --git a/Documents/LaTeX/Homework 1.pdf b/Documents/LaTeX/Homework 1.pdf new file mode 100644 index 0000000..2e140f1 Binary files /dev/null and b/Documents/LaTeX/Homework 1.pdf differ diff --git a/Documents/LaTeX/Homework 1.tex b/Documents/LaTeX/Homework 1.tex new file mode 100644 index 0000000..33b5218 --- /dev/null +++ b/Documents/LaTeX/Homework 1.tex @@ -0,0 +1,467 @@ +\documentclass[12pt,letterpaper]{article} +\usepackage[utf8]{inputenc} +\usepackage[english]{babel} +\usepackage{amsthm} +\usepackage{cancel} +\usepackage{mathtools} +\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} +\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{\dotp}{\dot{\mathcal{P}}} +\newcommand{\dotq}{\dot{\mathcal{Q}}} +\newcommand{\dist}{\text{dist}} +\DeclareMathOperator{\sign}{sgn} +\newtheoremstyle{case}{}{}{}{}{}{:}{ }{} +\theoremstyle{case} +\newtheorem{case}{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*{lemma*}{Lemma} +\newtheorem*{remark}{Remark} +\setlist[enumerate]{font=\bfseries} +\renewcommand{\qedsymbol}{$\blacksquare$} +\author{Alexander J. Tusa} +\title{Real Analysis II Homework 1} +\begin{document} + \maketitle + \textbf{Section 7.1 - The Riemann Integral} + \begin{enumerate} + \item + \begin{enumerate} +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +%%%%%%%%%%%%%%%%%%%%%%%% Section 7.1 Question 1 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% + \item[1.] If $I:=[0,4]$, calculate the norms of the following partitions: + \begin{enumerate} + \item[c.] $\mathcal{P}_3 := (0,1,1.5,2,3.4,4)$ + \\$||\mathcal{P}_3||=1.4$ + \item[d.] $\mathcal{P}_4 := (0,.5,2.5,3.5,4)$ + \\$||\mathcal{P}_4||=2$ + \end{enumerate} +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +%%%%%%%%%%%%%%%%%%%%%%%% Section 7.1 Question 2 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% + \item[2.] If $f(x):=x^2$ for $x \in [0,4]$, calculate the following Riemann sums, where $\dotp_i$ has the same partition points as in Exercise 1, and the tags are selected as indicated. + \\$\dotp_1 := (0,1,2,4)$ + \\$\dotp_2 := (0,2,3,4)$ + \begin{enumerate} + \item[(a)] $\dotp_1$ with the tags at the left endpoints of the subintervals. + \\The subintervals are: + \[I_1:=[0,1],\ I_2:=[1,2],\ I_3:=[2,4]\] + So the tags are: + \[t_1:=0,\ t_2:=1,\ t_3:=2\] + \begin{align*} + S(f,\dotp_1) &= \sum_{i=1}^{n} f(t_i)(x_i-x_{i-1}) \\ + &= f(0)(x_1-x_0)+f(1)(x_2-x_1)+f(2)(x_3-x_2) \\ + &= 0^2(1-0)+1^2(2-1)+2^2(4-2) \\ + &= 1+8 \\ + &= 9 + \end{align*} + \item[(b)] $\dotp_1$ with the tags at the right endpoints of the subintervals. + \\The subintervals are + \[I_1:=[0,1],\ I_2:=[1,2],\ I_3:=[2,4]\] + So the tags are: + \[t_1:=1,\ t_2:=2,\ t_3:=4\] + \begin{align*} + S(f,\dotp_1)&=\sum_{i=1}^{n}f(t_i)(x_i-x_{i-1}) \\ + &= f(1)(x_1-x_0)+f(2)(x_2-x_1)+f(4)(x_3-x_2) \\ + &= 1^2(1-0)+2^2(2-1)+4^2(4-2)\\ + &= 1+4+32\\ + &= 37 + \end{align*} + \item[(c)] $\dotp_2$ with the tags at the left endpoints of the subintervals. + \\The subintervals are: + \[I_1:=[0,2],\ I_2:=[2,3],\ I_3:=[3,4]\] + So the tags are: + \[t_1:=0,\ t_2:=2,\ t_3:=3\] + \begin{align*} + S(f,\dotp_2)&= \sum_{i=1}^{n} f(t_i)(x_i-x_{i-1})\\ + &= f(0)(x_1-x_0)+f(2)(x_2-x_1)+f(3)(x_3-x_2) \\ + &= 0^2(2-0)+2^2(3-2)+3^2(4-3) \\ + &= 4+9 \\ + &= 13 + \end{align*} + \item[(d)] $\dotp_2$ with the tags at the right endpoints of the subintervals. + \\So the subintervals are: + \[I_1:=[0,2],\ I_2:=[2,3],\ I_3:=[3,4]\] + So the tags are: + \[t_1:=2,\ t_2:=3,\ t_3:=4\] + \begin{align*} + S(f,\dotp_2)&= \sum_{i=1}^{n} f(t_i)(x_i-x_{i-1})\\ + &= f(2)(2-0)+f(3)(3-2)+f(4)(4-3) \\ + &= 2^2(2)+3^2(1)+4^2(1) \\ + &= 8+9+16 \\ + &= 33 + \end{align*} + \end{enumerate} +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +%%%%%%%%%%%%%%%%%%%%%%%% Section 7.1 Question 6 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% + \item[6.] + \begin{enumerate} + \item[(a)] Let $f(x):=2$ if $0 \leq x < 1$ and $f(x):= 1$ if $1 \leq x \leq 2$. Show that $f \in \mathcal{R}[0,2]$ and evaluate its integral. + \\\\We estimate by the graph of $f$ that the integral of $f$ is 3. We must now show by the definition of the integral that the integral of $f$ is 3. + \begin{proof} + Let $\dotp$ be a tagged partition of $[0,2]$. Let $\dotp_1 \subseteq \dotp$ with tags in $[0,1]$, and let $\dotp_2 \subseteq \dotp$ with tags in $[1,2]$. + \\\\We know that + \[[0,1-||\dotp||] \subseteq U_1 \subseteq [0,1+||\dotp||]\ \ \ \ (1)\] + and + \[[1+||\dotp||,2] \subseteq U_2 \subseteq [1-||\dotp||,2]\ \ \ \ (2)\] + where $U_1$ and $U_2$ are the union of the subintervals $\dotp_1$ and $\dotp_2$, respectively. + \\\\Now, we can calculate $S(f;\dotp_1)$ and $S(f;\dotp_2)$. + \begin{align*} + S(f;\dotp_1) &= \sum_{I_i \in \dotp_1} f(t_i)(x_i-x_{i-1}) \\ + &= \sum_{I_i \in \dotp_1} 2(x_i-x_{i-1}) \\ + &(I_i \in \dotp_1 \implies I_i \subseteq [0,1]\ \text{where the function value is }2)\\ + &=2\ \sum_{I_i \in \dotp_1} (x_i-x_{i-1}) \\ + &\in [2(1-||\dotp||), 2(1+||\dotp||)] = [2-2||\dotp||, 2+2||\dotp||] + \end{align*} + \begin{center} + (Because of (1) we know that the range of the subinterval lengths in $\dotp_1$) + \end{center} + \begin{align*} + S(f;\dotp_2)&=\sum_{I_i \in \dotp_2} f(t_i)(x_i-x_{i-1}) \\ + &= \sum_{I_i \in \dotp_2} 1 (x_i-x_{i-1}) \\ + &(I_i \in \dotp_2 \implies I_i \subseteq [1,2]\text{ where the function value is }1) \\ + &= \sum_{I_i \in \dotp_2} (x_i-x_{i-1}) \\ + &\in [1-||\dotp||, 1+||\dotp||] + \end{align*} + \begin{center} + (Because of (2), we know the range of the subinterval lengths in $\dotp_2$) + \end{center} + Therefore, + \[S(f;\dotp)=S(f;\dotp_1)+S(f;\dotp_2) \in [3(1-||\dotp||),3(1+||\dotp||)]\] + \[\Updownarrow\] + \[3-3||\dotp||\leq S(f;\dotp) \leq 3+3||\dotp||\] + \[\Updownarrow\] + \[|S(f;\dotp)-3| \leq 3||\dotp||\] + For arbitrary $\varepsilon >0$ we can pick a tagged partition $\dotp$ such that + \[||\dotp||<\frac{\varepsilon}{3}\] + Thus $f \in \mathcal{R}[0,2]$. + \end{proof} + \end{enumerate} +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +%%%%%%%%%%%%%%%%%%%%%%%% Section 7.1 Question 8 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% + \item[8.] If $f \in \mathcal{R}[a,b]$ and $|f(x)| \leq M$ for all $x\in [a,b]$, show that $\abs{\int_{a}^{b}f}\leq M (b-a)$. + \\Note that + \[-M \leq |f(x)| \leq M,\ \forall\ x \in [a,b]\] + By \textit{Theorem 7.1.5 c}, and since every constant function on $[a,b]$ is in $\mathcal{R}[a,b]$, we have that + \[-M(b-a) \leq \int_{a}^{b} (-M) \leq \int_{a}^{b} f \leq \int_{a}^{b} M = M(b-a)\] + Therefore, + \[\abs{\int_{a}^{b}f}\leq M(b-a)\] +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +%%%%%%%%%%%%%%%%%%%%%%%% Section 7.1 Question 12 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% + \item[12.] Consider the Dirichlet function, introduced in Example 5.1.6(g), defined by $f(x):=1$ for $x \in [0,1]$ rational and $f(x):=0$ for $x \in [0,1]$ irrational. Use the preceding exercise to show that $f$ is \textit{not} Riemann integrable on $[0,1]$.\\ + \\Let + \[\dotp_n:= \left\{\left[\frac{i-1}{n},\frac{i}{n}\right], \frac{i}{n}\right\}_{i=1}^n,\ n \geq 1\] + Then $||\dotp_n||=\frac{1}{n} \to 0$ as $n \to \infty$. + \\Then, + \[S(f;\dotp_n):=\sum_{i=1}^{n}f\left(\frac{i}{n}\right)\left(\frac{i}{n}-\frac{i-1}{n}\right)=\sum_{i=1}^{n}1 \cdot \frac{i}{n} = 1\] + because $\frac{i}{n}$ is rational. + \\\\Let + \[\dotq_n:=\left\{\left[\frac{i-1}{n},\frac{i}{n}\right],\alpha_i\right\}_{i=1}^n,\ n \geq 1\] + where $\alpha_i$ is an irrational number in the interval $\left[\frac{i-1}{n}, \frac{i}{n}\right]$, for $i=1,2,\dots,n$. + \\Then $||\dotq_n||=\frac{1}{n} \to 0$ as $n \to \infty$. + \\Then, + \[S(f;\dotq_n):=\sum_{i=1}^{n}f(\alpha_i)\left(\frac{i}{n}-\frac{i-1}{n}\right)=\sum_{i=1}^{n}0 \cdot \frac{i}{n}=0\] + because $\alpha_i$ is irrational. + \\\\Therefore, + \[\lim\limits_n S(f; \dotp_n)=1 \neq 0 = \lim\limits_n S(f;\dotq_n)\] + By the definition of a Riemann integrable function, for any $\varepsilon>0$, there exists $\delta > 0$ such that for all tagged partitions $\dotp$ with $||\dotp||<\delta$ we have + \[\abs{S(f;\dotp)-\int_{a}^{b}f}<\frac{\varepsilon}{2}\] + Because $||\dotp_n||\to 0$, there exists $n_1 \in \N$ such that + \[n>n_1 \implies ||\dotp_n||<\delta\] + Similarly, because $||\dotq_n|| \to 0$, there exists $n_2 \in \N$ such that + \[n>n_2 \implies ||\dotq_n||<\delta\] + Let $n_0:=\max \{n_1,n_2\}$. Then for all $n > n_0$ we have that + \[||\dotp_n||<\delta\ \&\ ||\dotq_n||<\delta\] + so we have + \[\abs{S(f;\dotp_n)-\int_{a}^{b}f}<\frac{\varepsilon}{2}\ \&\ \abs{S(f;\dotq_n)-\int_{a}^{b}f}<\frac{\varepsilon}{2}\] + Therefore, for all $n > n_0$, + \begin{align*} + \abs{S(f;\dotp_n)-S(f;\dotq_n)}&<\abs{S(f;\dotp_n)-\int_{a}^{b}f}+\abs{S(f;\dotq_n)-\int_{a}^{b}f} \\ + &< \frac{\varepsilon}{2}+\frac{\varepsilon}{2} \\ + &= \varepsilon + \end{align*} + By the definition of the limit of a sequence, + \[\lim\limits_n \left[S(f;\dotp_n)-S(f;\dotq_n)\right]=0,\] + that is, + \[\lim\limits_n S(f;\dotp_n)=\lim\limits_n S(f;\dotq_n)\] + which is a contradiction. Therefore $f \notin \mathcal{R}[a,b]$, and hence the Dirichlet function is not Riemann integrable. + \end{enumerate} + \item + \begin{enumerate} +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +%%%%%%%%%%%%%%%%%%%%%%%% Section 7.2 Question 8 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% + \item[8.] Suppose that $f$ is continuous on $[a,b]$, that $f(x) \geq 0$ for all $x \in [a,b]$ and that $\int_{a}^{b}f=0$. Prove that $f(x)=0$ for all $x \in [a,b]$. + \begin{proof} + Suppose there exists $c \in [a,b]$ such that $f(c)>0$. Since $f$ is continuous, there exists $\delta >0$ such that $f(x)>\frac{1}{2}f(c)$ for $x \in (c-\delta, c+ \delta) \subseteq [a,b]$. Then + \[\int_{a}^{b}f \geq \int_{c-\delta}^{c+\delta} f \geq \frac{1}{2} f(c) \cdot 2 \delta > 0\] + which contradicts the fact that $\int_{a}^{b}f=0$. If $c=a$, then there exists $\delta >0$ such that $f(x)>0$ for $x \in [a,a+\delta)$, and thus the same contradiction is present. The same applies for the case in which $a=b$. Therefore we have that $f(x)=0,\ \forall\ x \in [a,b]$. + \end{proof} +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +%%%%%%%%%%%%%%%%%%%%%%%% Section 7.2 Question 9 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% + \item[9.] Show that the continuity hypothesis in the preceding exercise cannot be dropped. + \\\\Consider the function $f:[0,1] \to \R$ given by + \[f(x):=\begin{cases} + 1, &x = 0\\ + 0, & x \neq 0 + \end{cases}\] + Then, $f$ has a discontinuity at the point $x=0$ and $\int_{0}^{1} f=0$, but $f$ is not zero everywhere. Therefore, continuity is a necessary part of the hypothesis.\\ +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +%%%%%%%%%%%%%%%%%%%%%%%% Section 7.2 Question 10 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% + \item[10.] If $f$ and $g$ are continuous on $[a,b]$ and if $\int_{a}^{b}f = \int_{a}^{b}g$, prove that there exists $c \in [a,b]$ such that $f(c)=g(c)$. + \begin{proof} + Let $f$ and $g$ be continuous functions on $[a,b]$ such that + \[\int_{a}^{b}f = \int_{a}^{b} g\] + Define $h:[a,b] \to \R$ as $h:=f-g$. Then, $h$ is continuous as a difference of continuous functions and + \[\int_{a}^{b}h=\int_{a}^{b} (f-g)=\int_{a}^{b}f-\int_{a}^{b}g=0\] + Suppose that there exists $c \in [a,b]$ such that $h(c)=0$ since $(f(c)=g(c))$. Then, since $h$ is continuous, it follows that $h(x)>0,\ \forall\ x \in [a,b]$. or $h(x)<0,\ \forall\ x \in [a,b]$ (recall \textit{Bolzano's Theorem}). + \\\\Suppose $h(x)>0,\ \forall\ x \in [a,b]$. Then because $h$ is a continuous function on a segment, by the \textit{Maximum-Minimum Theorem} there exists $m>0$ such that + \[h(x)\geq m > 0,\ \forall\ x \in [a,b]\] + Then we have + \[\int_{a}^{b} h \geq \int_{a}^{b} m=m(b-a)>0\] + This is a contradiction with the fact that $\int_{a}^{b} h=0$. + \\\\Now, for the case in which $h(x)<0,\ \forall\ x \in [a,b]$, by the \textit{Maximum-Minimum Theorem} we know that there exists $M<0$ such that + \[h(x)\leq M < 0\ \forall\ x \in [a,b]\] + and thus + \[\int_{a}^{b} h \leq \int_{a}^{b} M \leq M(b-a)<0\] + which again yields a contradiction. + \\\\Therefore, there exists $c \in [a,b]$ such that $h(c)=0$, that is, $f(c)=g(c)$. + \end{proof} +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +%%%%%%%%%%%%%%%%%%%%%%%% Section 7.2 Question 13 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% + \item[13.] Give an example of a function $f:[a,b] \to \R$ that is in $\mathcal{R}[c,b]$ for every $c \in (a,b)$ but which is not in $\mathcal{R}[a,b]$. + \\\\Define a function $f$ on $[0,1]$ by + \[f(x):=\begin{cases} + \frac{1}{x}, &x \in (0,1] \\ + 1, &x=0 + \end{cases}\] + For every $c>0,\ f \in \mathcal{R}[c,1]$ because $f$ is continuous on $[c,1]$. + \\\\Now, let's show that $f$ isn't Riemann integrable on $[0,1]$. + \\\\Define a tagged partition to be + \[\dotp := \left\{\left[\frac{i-1}{n},\frac{i}{n}\right],\frac{i}{n}\right\}_{i=1}^n\] + Then + \begin{align*} + S(f;\dotp) &= \sum_{i=1}^{n} f \left(\frac{i}{n}\right)\left(\frac{i}{n}-\frac{i-1}{n}\right) \\ + &= \sum_{i=1}^{n}\frac{1}{\frac{i}{n}} \cdot \frac{1}{n} \\ + &= \sum_{i=1}^{n} \frac{1}{i} + \end{align*} + As $n \to \infty$, $S(f;\dotp)$ diverges (since it is a harmonic series). Thus, $f$ is not Riemann integrable on $[0,1]$. + \end{enumerate} +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Question 3 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% + \item Use the right-endpoint Riemann sums to evaluate the following integrals: + \begin{enumerate} + \item $\int_{2}^{5}(3x-1)dx$ + \begin{align*} + \lim\limits_{n \to \infty} \sum_{i=1}^{n} f(t_i)\Delta x_i &= \sum_{i=1}^{n} f(a+i\Delta x) \Delta x \\ + &= \sum_{i=1}^{n} f\left(2+i\left(\frac{3}{n}\right)\right)\cdot \left(\frac{3}{n}\right) \\ + &= \sum_{i=1}^{n} \left(3 \cdot \left(2+\frac{3i}{n}\right)-1\right) \cdot \left(\frac{3}{n}\right) \\ + &= \sum_{i=1}^{n} \left(6+\frac{9i}{n}-1\right)\left(\frac{3}{n}\right) \\ + &= \sum_{i=1}^{n}\left(5+\frac{9i}{n}\right)\left(\frac{3}{n}\right) \\ + &= \sum_{i=1}^{n} \frac{15}{n} +\frac{27i}{n^2} \\ + &= \frac{15}{n} \sum_{i=1}^{n} 1 + \frac{27}{n^2} \sum_{i=1}^{n} i \\ + &= \frac{15}{\cancel{n}} \cdot \cancel{n} + \frac{27}{n^2} \cdot \frac{n(n+1)}{2} \\ + &= 15 + \frac{27}{n^2} \cdot \frac{n^2+n}{2} \\ + &= 15 + \frac{27n^2+27n}{2n^2} \\ + \int_{2}^{5} (3x-1)dx &= \lim\limits_{n \to \infty} 15+\frac{27n^2+27n}{2n^2}\\ + &= 15+\frac{27}{2}\\ + &=28.5 + \end{align*} + \item $\int_{0}^{4} (x^2+2x)dx$ + \begin{align*} + \int_{0}^{4} (x^2+2x)dx &= \lim\limits_{n \to \infty} \sum_{i=1}^{n} f(t_i)\Delta x \\ + &= \lim\limits_{n \to \infty} \sum_{i=1}^{n} f\left(a+i\Delta x\right) \Delta x \\ + &= \lim\limits_{n \to \infty} \sum_{i=1}^{n} f\left(0+i\left(\frac{4}{n}\right)\right)\cdot \left(\frac{4}{n}\right) \\ + &= \lim\limits_{n \to \infty} \sum_{i=1}^{n} \left[\left(\frac{4i}{n}\right)^2+2\left(\frac{4i}{n}\right) \cdot \right]\left(\frac{4}{n}\right) \\ + &= \lim\limits_{n \to \infty} \sum_{i=1}^{n} \left(\frac{16i^2}{n^2}+\frac{8i}{n}\right)\cdot \left(\frac{4}{n}\right) \\ + &= \lim\limits_{n \to \infty} \sum_{i=1}^{n} \left(\frac{64i^2}{n^3}+\frac{32i}{n^2}\right) \\ + &= \lim\limits_{n \to \infty} \left(\frac{64}{n^3}\ \sum_{i=1}^{n} i^2 + \frac{32}{n^2}\ \sum_{i=1}^{n} i\right) \\ + &= \lim\limits_{n \to \infty} \left(\frac{64}{n^3}\cdot\frac{n(n+1)(n+2)}{6}+\frac{32}{n^2}\cdot \frac{n(n+1)}{2}\right) \\ + &= \lim\limits_{n \to \infty} \left(\frac{64n^3+192n^2+128n}{6n^3}+\frac{32n^2+32n}{2n^2}\right) \\ + &= \frac{64}{6} + 16 \\ + &\approx 26.6667 + \end{align*} + \item $\int_{0}^{2} (2x^3+x)dx$ + \begin{align*} + \int_{0}^{2} (2x^3+x)dx &= \lim\limits_{n \to \infty} \sum_{i=1}^{n} f(t_i)\Delta x \\ + &= \lim\limits_{n \to \infty} \sum_{i=1}^{n} f(a+i\Delta x)\Delta x \\ + &=\lim\limits_{n \to \infty} \sum_{i=1}^{n} f\left(\frac{2i}{n}\right)\left(\frac{2}{n}\right) \\ + &= \lim\limits_{n \to \infty} \sum_{i=1}^{n} \left(2\left(\frac{2i}{n}\right)^3+\frac{2i}{n}\right)\left(\frac{2}{n}\right) \\ + &= \lim\limits_{n \to \infty} \sum_{i=1}^{n} \left(\frac{32i^3}{n^4}+\frac{4i}{n^2}\right) \\ + &= \lim\limits_{n \to \infty} \left[\frac{32}{n^4}\ \sum_{i=1}^{n}i^3+\frac{4}{n^2}\ \sum_{i=1}^{n}i\right] \\ + &= \lim\limits_{n \to \infty} \left[\frac{32}{n^4}\cdot\frac{n^2(n+1)^2}{4}+\frac{4}{n^2}\cdot\frac{n(n+1)}{2}\right] \\ + &= \lim\limits_{n \to \infty} \left[\frac{32n^4+64n^3+32n^2}{4n^4}+\frac{4n^2+4n}{2n^2}\right] \\ + &= 8+2 \\ + &= 10 + \end{align*} + \end{enumerate} +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Question 4 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% + \item Express each of the following as a definite integral. Then use calculus to evaluate the integral. + \begin{enumerate} + \item $\lim\limits_{|P| \to 0}\ \sum\limits_{i=1}^{n} \left(\frac{3}{w_i^2}\right) \Delta x_i$ where $P$ is a partition of $[1,3]$. + \begin{align*} + \lim\limits_{|P| \to 0}\ \sum\limits_{i=1}^{n} \left(\frac{3}{w_i^2}\right) \Delta x_i &= \int_{1}^{3} x dx \\ + &= \left.\frac{x^2}{2} \right|_1^3 \\ + &= \frac{3^2}{2} - \frac{1^2}{2} \\ + &= \frac{9}{2} - \frac{1}{2} \\ + &= \frac{8}{2} \\ + &= 4 + \end{align*} + \item $\lim\limits_{n \to \infty}\ \sum\limits_{i=1}^{n} \left(3 + \frac{2i}{n}\right)^2\cdot \left(\frac{2}{n}\right)$ + \begin{align*} + \lim\limits_{n \to \infty}\ \sum\limits_{i=1}^{n} \left(3 + \frac{2i}{n}\right)^2\cdot \left(\frac{2}{n}\right) &= \lim\limits_{n \to \infty} \sum_{i=1}^{n} \left(a+i\Delta x\right)^2\cdot \left(\Delta x\right) \\ + &= \lim\limits_{n \to \infty} \sum_{i=1}^{n} \left(a+i\left(\frac{b-a}{n}\right)\right)^2 \cdot \left(\frac{b-a}{n}\right) \\ + &= \lim\limits_{n \to \infty} \sum_{i=1}^{n} \left(3+i\left(\frac{2}{n}\right)\right)^2 \cdot \left(\frac{2}{n}\right) \\ + &= \lim\limits_{n \to \infty} \sum_{i=1}^{n} \left(3+i\left(\frac{5-3}{n}\right)\right)^2\cdot \left(\frac{5-3}{n}\right) \\ + &= \lim\limits_{n \to \infty} \sum_{i=1}^{n} f(t_i)\cdot \Delta x \\ + &= \int_{3}^{5} f(t_i) dx \\ + &= \int_{3}^{5} x^2 dx \\ + &= \left. \frac{x^3}{3} \right|_{3}^{5} \\ + &= \frac{5^3}{3}-\frac{3^3}{3} \\ + &= \frac{125}{3} - \frac{27}{3} \\ + &= \frac{125}{3} - 9 \\ + &\approx 32.66667 + \end{align*} + \item $\lim\limits_{n \to \infty}\ \sum\limits_{i=1}^{n} \left(1 + \frac{4(i-1)}{n}\right)^5\cdot \left(\frac{4}{n}\right)$ + \begin{align*} + \lim\limits_{n \to \infty}\ \sum\limits_{i=1}^{n} \left(1 + \frac{4(i-1)}{n}\right)^5\cdot \left(\frac{4}{n}\right) &= \lim\limits_{n \to \infty} \sum_{i=1}^{n} \left(a+i\Delta x\right)^5\cdot \left(\Delta x\right) \\ + &= \lim\limits_{n \to \infty} \sum_{i=1}^{n} \left(a+i\left(\frac{b-a}{n}\right)\right)^5 \cdot \left(\frac{b-a}{n}\right) \\ + &= \lim\limits_{n \to \infty} \sum_{i=1}^{n} \left(1+\frac{4i-4}{n}\right)^5\cdot\left(\frac{4}{n}\right)\\ + &=\int_{1}^{5} f(t_i)dx \\ + &= \int_{1}^{5} x^5 dx \\ + &= \left. \frac{x^6}{6} \right|_1^5 \\ + &= \frac{5^6}{6}-\frac{1^6}{6} \\ + &= \frac{15625}{6}-\frac{1}{6} \\ + &= \frac{15624}{6} \\ + &= \frac{7812}{3} \\ + &= 2604 + \end{align*} + \item $\lim\limits_{n \to \infty}\ \sum\limits_{i=1}^{n} \frac{i^2}{n^3}$ + \begin{align*} + \lim\limits_{n \to \infty}\ \sum\limits_{i=1}^{n} \frac{i^2}{n^3} &= \lim\limits_{n \to \infty} \sum_{i=1}^{n} (a+i\Delta x)\Delta x \\ + &= \lim\limits_{n \to \infty} \sum_{i=1}^{n} \left(a+i\left(\frac{b-a}{n}\right)\right)\cdot\left(\frac{b-a}{n}\right) \\ + &= \lim\limits_{n \to \infty} \sum_{i=1}^{n} \left(0+i\left(\frac{1-0}{n}\right)\right)^2 \cdot \left(\frac{1-0}{n}\right) \\ + &= \int_{0}^{1} f(t_i)dx \\ + &= \int_{0}^{1} x^2 dx \\ + &= \left.\frac{x^3}{3}\right|_0^1 \\ + &= \frac{1^3}{3} - \frac{0^3}{3} \\ + &= \frac{1}{3} + \end{align*} + \item Show that $\lim\limits_{n \to \infty}\ \sum\limits_{i=1}^{n} \frac{n}{n^2+i^2}=\frac{\pi}{4}$ + \begin{align*} + \lim\limits_{n \to \infty} \sum_{i=1}^{n} \frac{n}{n^2+i^2} &= \lim\limits_{n \to \infty} \sum_{i=1}^{n} \frac{n}{n^2+i^2}\cdot \frac{\frac{1}{n^2}}{\frac{1}{n^2}} \\ + &= \lim\limits_{n \to \infty} \sum_{i=1}^{n} \frac{1}{1+(\frac{i}{n})^2} \cdot \frac{1}{n} \\ + &= \int_{0}^{1} \frac{1}{x^2} dx \\ + &= \arctan(x) |_0^1 \\ + &= \arctan(1)-\arctan(0) \\ + &= \frac{\pi}{4} - 0 \\ + &= \frac{\pi}{4} + \end{align*} + \end{enumerate} +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Question 5 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% + \item Give examples of functions $f:[0,1] \to \R$ such that + \begin{enumerate} + \item $f \notin \mathcal{R}[0,1]$, but $|f|$ and $f^2$ are both in $\mathcal{R}[0,1]$. + \\\\Consider the function $f:[0,1] \to \R$ given by $f(x):=\begin{cases} + 1 &x \in \Q \\ + -1 &x \in \R\setminus\Q + \end{cases}$, and let $P:=\{x_0,x_1,\dots, x_n\}$ be any partition of $[0,1]$. Then $M_i=1$ and $m_i=-1$ for all $i=1,2,\dots,n$. thus $U(f,P)=1$ and $L(f,P)=-1$ for all $P$. Thus $U(f)=1$ and $L(f)=-1$. Thus $f$ is not integrable. + \\\\However, $|f|(x)=1$ for all $x \in [0,1]$. Since $|f|$ is a continuous function $|f|$ is integrable on $[0,1]$, and also since $f^2(x)=1$ is also a continuous function, we have that $f^2$ is also integrable on $[0,1]$.\\ + \item $f$ is bounded, but $f \notin \mathcal{R}[0,1]$. + \\\\Consider the function $f:[0,1] \to \R$ given by $f(x):=\begin{cases} + 1 &x \in \Q \\ + 0 &x \in \R\setminus\Q + \end{cases}$. Let $P$ be any partition of $[0,1]$, then + \[U(P;f):=\sum M_i \Delta x_i = \sum 1 \Delta x_i = b-a\] + and + \[L(P;f)=\sum m_i \Delta x_i = \sum 0 \Delta x_i = 0(b-a)=0\] + Hence + \[\overline{\int_{0}^{1}} fdx=b-a \neq 0 = \underline{\int_{0}^{1}} f dx \] + Hence $f$ is not Riemann integrable. + \item $f \in \mathcal{R}[0,1]$ and $f$ is not monotone. + \\\\Consider the function $f:[0,1] \to \R$ given by $f(x):=\begin{cases} + 3 & 0 \leq x < \frac{1}{3} \\ + 1 & \frac{1}{3} \leq x < \frac{2}{3} \\ + 3 & \frac{2}{3} \leq x \leq 1 + \end{cases}$. The function $f$ is non-monotonic on $[0,1]$. + \\\\The upper Riemann integral of $f$ is + \[\overline{\int_{[0,1]}}f=\inf \left\{\int_{[0,1]} g: g \text{ is a piecewise constant on }[0,1] \text{ and } g(x) \geq f(x)\ \forall\ x \in [0,1]\right\}\] + \[=\frac{7}{3}\] + Similarly, the lower integral of $f$ is given by + \[\underline{\int_{[0,1]}}f = \sup \left\{\int_{[0,1]} g: g \text{ is a piecewise constant on } [0,1] \text{ and } g(x) \leq f(x)\ \forall\ x \in [0,1]\right\}\] + \[=\frac{7}{3}\] + Since $\displaystyle\overline{\int_{[0,1]}}f=\underline{\int_{[0,1]}} f$, the function $f$ is Riemann integrable on $[0,1]$ and $\displaystyle\int_{[0,1]} f=\frac{7}{3}$. + \item $f \in \mathcal{R}[0,1]$ and $f$ is neither monotone nor continuous. + \\\\Consider the function $f:[0,1] \to \R$ given by $f(x):=\begin{cases} + 0 & x \in \{0,1\}\cup([0,1]\setminus\Q) \\ + \frac{1}{q} & x \in (0,1) \cap \Q,\ x=\frac{p}{q},\ p,q \in \N, \text{ and}\\ + & \text{$p,q$ are relatively prime} + \end{cases}$. We note that $f$ is known as the Riemann function. Thus it is well known that this function is not piecewise continuous nor is it monotone. + \end{enumerate} +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Question 6 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% + \item Prove or justify, if true or provide a counterexample, if false. + \begin{enumerate} + \item If $f(x) \leq g(x) \leq h(x)$ for all $x \in [a,b]$, and $f,h \in \mathcal{R}[a,b]$, then so is $g \in \mathcal{R}[a,b]$.\\\\ + This is true by the \textit{Squeeze Theorem}.\\ + \item If $f \in \mathcal{R}[a,b]$, then $f$ is continuous on $[a,b]$. + \\\\This is a false statement. Consider $f:[0,3] \to \R$ given by $f(x):=\begin{cases} + 2,\ &0 \leq x \leq 1 \\ + 3,\ &1 < x \leq 3 + \end{cases}$ + Then we have that $\int_{0}^{3} f(x)=8$, and thus $f \in \mathcal{R}[0,3]$, but $f$ is not continuous.\\ + \item If $|f| \in \mathcal{R}[a,b]$, then $f \in \mathcal{R}[a,b]$. + \\\\This is a false statement. Consider the function $f:[0,1] \to \R$ given by $f(x):=\begin{cases} + 1 &x \in \Q \\ + -1 &x \in \R\setminus\Q + \end{cases}$, and let $P:=\{x_0,x_1,\dots, x_n\}$ be any partition of $[0,1]$. Then $M_i=1$ and $m_i=-1$ for all $i=1,2,\dots,n$. thus $U(f,P)=1$ and $L(f,P)=-1$ for all $P$. Thus $U(f)=1$ and $L(f)=-1$. Thus $f$ is not integrable. + \\\\However, $|f|(x)=1$ for all $x \in [0,1]$. Since $|f|$ is a continuous function $|f|$ is integrable on $[0,1]$.\\ + \item Let $f$ be bounded on $[a,b]$. If $P$ and $Q$ are partitions of $[a,b]$, then $P \cup Q$ is a refinement of both $P$ and $Q$. + \\\\This is a true statement because this satisfies the definition of a refinement, since both $P \subseteq P \cup Q$ and $Q \subseteq P \cup Q$.\\ + \item If $f$ is continuous on $[a,b)$ and on $[b,c]$, then $f \in \mathcal{R}[a,c]$. + \\\\This is a false statement. Consider the function $f:[0,5] \to \R$ given by $f(x):= \begin{cases} + \frac{x}{x-2}, &0 \leq x < 2 \\ + 0, & 2 \leq x \leq 5 + \end{cases}$. Since an asymptote exists and is unbounded on $[0,5]$, we have that $f$ is not Riemann integrable. + \item If $f,g\in\mathcal{R}[a,b]$, then $f-g \in \mathcal{R}[a,b]$. + \\\\This is true by \textit{Theorem 7.1.5 c}, since it can be rewritten as $f+(-g)$.\\ + \item If $f$ is monotone on $[a,b]$, then $f \in \mathcal{R}[a,b]$. + \\\\This is a true statement by \textit{Theorem 7.2.6}: + \begin{theorem*} + If $f:[a,b]\to\R$ is monotone on $[a,b]$, then $f \in \mathcal{R}[a,b]$. + \end{theorem*} + \end{enumerate} + \end{enumerate} +\end{document} diff --git a/Documents/LaTeX/Homework 2.pdf b/Documents/LaTeX/Homework 2.pdf new file mode 100644 index 0000000..f4554a8 Binary files /dev/null and b/Documents/LaTeX/Homework 2.pdf differ diff --git a/Documents/LaTeX/Homework 2.tex b/Documents/LaTeX/Homework 2.tex new file mode 100644 index 0000000..db90539 --- /dev/null +++ b/Documents/LaTeX/Homework 2.tex @@ -0,0 +1,469 @@ +\documentclass[12pt,letterpaper]{article} +\usepackage[utf8]{inputenc} +\usepackage[english]{babel} +\usepackage{amsthm} +\usepackage{cancel} +\usepackage{mathtools} +\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} +\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{\dotp}{\dot{\mathcal{P}}} +\newcommand{\dotq}{\dot{\mathcal{Q}}} +\newcommand{\dist}{\text{dist}} +\DeclareMathOperator{\sign}{sgn} +\newtheoremstyle{case}{}{}{}{}{}{:}{ }{} +\theoremstyle{case} +\newtheorem{case}{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*{lemma*}{Lemma} +\newtheorem*{remark}{Remark} +\setlist[enumerate]{font=\bfseries} +\renewcommand{\qedsymbol}{$\blacksquare$} +\author{Alexander J. Tusa} +\title{Real Analysis II Homework 2} +\begin{document} + \maketitle + \begin{enumerate} + \item \textbf{Section 7.4} + \begin{enumerate} + \item[1.] Let $f(x):=|x|$ for $-1 \leq x \leq 2$. Calculate $L(f;P)$ and $U(f,P)$ for the following partitions: + \begin{enumerate} + \item[(a)] $\mathcal{P}_1 :=(-1,0,1,2)$ + \\\\Our terms are: + \[x_0:= -1,\ \ x_1:=0,\ \ x_2:=1,\ \ x_3:=2\] + and our intervals are: + \[I_1:=[-1,0],\ \ I_2:=[0,1],\ \ I_3:=[1,2]\] + thus $L(f,\mathcal{P}_1)$ is: + \begin{align*} + L(f,\mathcal{P}_1) &= \sum_{i=1}^{n} m_i(x_i-x_{i-1}) \\ + &= \sum_{i=1}^{n} \left(\inf\{f(x):x \in [x_{i-1},x_i]\}\right) (x_i-x_{i-1}) \\ + &= (\inf \{|x|:x \in [-1,0]\})(0-(-1))\\ + &+ (\inf \{|x|: x \in [0,1]\})(1-0)\\ + &+(\inf \{|x|: x\in [1,2]\})(2-1) \\ + &= 0 \cdot 1 + 0 \cdot 1 + 1 \cdot 1 \\ + &= 0+0+1 \\ + &= 1 + \end{align*} + and + \begin{align*} + U(f,\mathcal{P}_1) &= \sum_{i=1}^{n} M_i (x_i-x_{i-1}) \\ + &= \sum_{i=1}^{n} sup \{f(x): x \in [x_{i-1},x_i]\} (x_i-x_{i-1}) \\ + &= \sum_{i=1}^{n} \sup \{|x|: x \in [x_{i-1},x_i]\} (x_i-x_{i-1}) \\ + &= (\sup \{|x|:x \in [-1,0]\})(0-(-1)) \\ + &+ (\sup \{|x| : x \in [0,1]\})(1-0) \\ + &+ (\sup \{|x|: x \in [1,2]\})(2-1) \\ + &= 1 \cdot 1 + 1 \cdot 1 + 2 \cdot 1 \\ + &= 1 + 1 +2 \\ + &= 4 + \end{align*} + So, $L(f,\mathcal{P}_1)=1$ and $U(f,\mathcal{P}_1)=4$\\ + \item[(b)] $\mathcal{P}_2 := (-1,-1/2,0,1/2,1,3/2,2)$. + \\\\Our terms are: + \[x_0:=-1,\ \ x_1:=-\frac{1}{2},\ \ x_2:=0,\ \ x_3:=\frac{1}{2},\ \ x_4:= 1,\ \ x_5:=\frac{3}{2},\ \ x_6:=2\] + and our intervals are: + \[I_1:=\left[-1,-\frac{1}{2}\right],\ \ I_2:=\left[-\frac{1}{2},0\right],\ \ I_3:=\left[0,\frac{1}{2}\right],\ \ I_4:=\left[\frac{1}{2},1\right],\ \ I_5:=\left[1,\frac{3}{2}\right],\ \ I_6:=\left[\frac{3}{2}, 2\right]\] + So $L(f,\mathcal{P}_2)$ is + \begin{align*} + L(f,\mathcal{P}_2) &= \sum_{i=1}^{n} m_i(x_i-x_{i-1}) \\ + &= \sum_{i=1}^{n} \inf \{f(x): x \in [x_{i-1}, x_i]\}(x_i-x_{i-1}) \\ + &= \inf\left\{|x|: x \in \left[-1,-\frac{1}{2}\right]\right\}\left(-\frac{1}{2}-(-1)\right) \\ + &+ \inf\left\{|x|: x \in \left[-\frac{1}{2}, 0\right]\right\} \left(0-\left(-\frac{1}{2}\right)\right) \\ + &+ \inf\left\{|x|: x \in \left[0,\frac{1}{2}\right]\right\}\left(\frac{1}{2}-0\right) \\ + &+ \inf \left\{|x|: x \in \left[\frac{1}{2}, 1\right]\right\}\left(1-\frac{1}{2}\right) \\ + &+ \inf\left\{|x|: x \in \left[1,\frac{3}{2}\right]\right\}\left(\frac{3}{2}-1\right) \\ + &+ \inf \left\{|x|: x \in \left[\frac{3}{2}, 2\right]\right\}\left(2 - \frac{3}{2}\right) \\ + &= \frac{1}{2} \cdot \frac{1}{2} + 0 \cdot \frac{1}{2} + 0 \cdot \frac{1}{2} + \frac{1}{2} \cdot \frac{1}{2} + 1 \cdot \frac{1}{2} + \frac{3}{2} \cdot \frac{1}{2} \\ + &=\frac{1}{4} + 0 + 0+ \frac{1}{4} + \frac{1}{2} + \frac{3}{4} \\ + &= \frac{7}{4} + \end{align*} + and + \begin{align*} + U(f,\mathcal{P}_2) &= \sum_{i=1}^{n} M_i(x_i-x_{i-1}) \\ + &= \sum_{i=1}^{n} \sup \{f(x): x \in [x_{i-1},x_i]\}(x_i-x_{i-1}) \\ + &= \sup \left\{|x|: x \in \left[-1,-\frac{1}{2}\right]\right\}\left(-\frac{1}{2}-(-1)\right) \\ + &+ \sup\left\{|x|: x \in \left[-\frac{1}{2},0\right]\right\}\left(0-\left(-\frac{1}{2}\right)\right) \\ + &+ \sup\left\{|x|: x \in \left[0,\frac{1}{2}\right]\right\}\left(\frac{1}{2}-0\right) \\ + &+ \sup\left\{|x|: x \in \left[\frac{1}{2}, 1\right]\right\}\left(1-\frac{1}{2}\right) \\ + &+ \sup\left\{|x|: x \in \left[1, \frac{3}{2}\right]\right\}\left(\frac{3}{2}-1\right) \\ + &+ \sup\left\{|x|: x \in \left[\frac{3}{2}, 2\right]\right\}\left(2-\frac{3}{2}\right) \\ + &= 1 \cdot \frac{1}{2} + \frac{1}{2} \cdot \frac{1}{2} + \frac{1}{2} \cdot \frac{1}{2} + 1 \cdot \frac{1}{2} + \frac{3}{2} \cdot \frac{1}{2} + 2 \cdot \frac{1}{2} \\ + &= \frac{1}{2} + \frac{1}{4} + \frac{1}{4}+ \frac{1}{2} + \frac{3}{4} + 1 \\ + &= \frac{13}{4} + \end{align*} + So $L(f,\mathcal{P}_2) = \frac{7}{4}$ and $U(f,\mathcal{P}_2) = \frac{13}{4}$\\ + \end{enumerate} + \item[2.] Prove if $f(x):=c$ for $x \in [a,b]$, then its Darboux integral is equal to $c(b-a)$. + \begin{proof} + Let $\mathcal{P} := (x_0, x_1, \dots, x_n)$ be a partition of $[a,b]$ where + \[a=x_0 < x_1 < x_2 < \dots < x_n=b\] + then $M_i:= \sup f(x)=c$ since $f$ is constant, for all $x \in [x_{i-1}, x_i]$, and $m_i:= \inf f(x)=c$ again since $f$ is constant, for all $x \in [x_{i-1},x_i]$. + \\\\Then we have that + \begin{align*} + U(f, \mathcal{P}) &= \sum_{i=1}^{n} M_i (x_i-x_{i-1}) \\ + &= \sum_{i=1}^{n} c (x_i-x_{i-1}) \\ + &= c \sum_{i=1}^{n} (x_i-x_{i-1}) \\ + &= c (x_n-x_0) \\ + &= c (b-a), &\text{as both $b$ and $a$ were defined for $\mathcal{P}$} + \end{align*} + So $U(f,\mathcal{P}):= c(b-a)$. As for $L(f,\mathcal{P})$: + \begin{align*} + L(f,\mathcal{P}) &= \sum_{i=1}^{n} m_i (x_i-x_{i-1}) \\ + &= \sum_{i=1}^{n} c (x_i-x_{i-1}) \\ + &= c \sum_{i=1}^{n} (x_i-x_{i-1}) \\ + &= c (x_n-x_0) \\ + &= c(b-a), &\text{as both $b$ and $a$ were defined for $\mathcal{P}$} + \end{align*} + and thus $L(f,\mathcal{P}) = c(b-a)$. + \\\\Now we must find the Darboux integral of $f(x)$. So we have that the upper Darboux integral of $f(x)$ is + \begin{align*} + U(f) &= \inf \{U(f,\mathcal{P}): \mathcal{P} \in \mathscr{P}[a,b]\} \\ + &= \inf \{c(b-a): \mathcal{P} \in \mathscr{P}[a,b]\} \\ + &= c(b-a) + \end{align*} + and the lower Darboux integral is + \begin{align*} + L(f) &= \sup \{L(f, \mathcal{P}): \mathcal{P} \in \mathscr{P}[a,b]\} \\ + &= \sup \{c(b-a) : \mathcal{P} \in \mathscr{P}[a,b]\} \\ + &= c(b-a) + \end{align*} + Thus we have that $U(f)=L(f)$, which yields that $f$ is Darboux integrable on $[a,b]$ and the Darboux integral of $f$ is $c(b-a)$. + \end{proof} + \item[3.] Let $f$ and $g$ be bounded functions on $I:=[a,b]$. If $f(x) \leq g(x)$ for all $x \in I$, show that $L(f) \leq L(g)$ and $U(f) \leq U(g)$. + \begin{proof} + Let $f,g$ be bounded on $I:=[a,b]$ such that $f(x) \leq g(x)\ \forall\ x \in I$, and let $\mathcal{P}:=(x_0,x_1,\dots, x_n)$ be a partition of $[a,b]$ where + \[a=x_0 0$ be given, and let $\mathcal{P}:= (0, \frac{2}{n}, \frac{4}{n}, \dots, \frac{2n-1}{n}, 2)$, and we note that $\Delta x_i := \frac{2}{n}$. + \\\\Since $f$ is increasing on $[0,2]$, then on $[x_{i-1},x_i] = \left[\frac{2i-1}{n}, \frac{2i}{n}\right]$, $M_i$ occurs at the right endpoint, and is thus $f\left(\frac{2i}{n}\right)=\frac{4i^2}{n^2}+1$. Also, $m_i$ occurs at the left endpoint and is thus $f\left(\frac{2i-1}{n}\right)=\left(\frac{4i^2-4i+1}{n^2}+1\right)$. + \begin{align*} + U(f,\mathcal{P}_n) &= \sum_{i=1}^{n} M_i \Delta x_i \\ + &= \sum_{i=1}^{n} \left(\frac{4i^2}{n^2}+1\right) \cdot \left(\frac{2}{n}\right) \\ + &= \sum_{i=1}^{n} \frac{8i^2}{n^3} + \frac{2}{n} \\ + &= \frac{8}{n^3}\ \sum_{i=1}^{n} i^2 + \frac{2}{n}\ \sum_{i=1}^{n} 1 \\ + &= \frac{8}{n^3} \cdot \frac{n(n+1)(2n+1)}{6} + \frac{2 \cancel{n}}{\cancel{n}} \\ + &= \frac{8}{n^3} \cdot \frac{2n^3+3n^2+n}{6} + 2 \\ + &= \frac{16n^3+24n^2+8n}{6n^3} + 2 \\ + &= \lim\limits_{n \to \infty}\frac{16n^3+24n^2+8n}{6n^3} + 2 \\ + &= \frac{16}{6} + 2 \\ + &= \frac{8}{3} + 2 \\ + &= \frac{14}{3} \\ + &\geq U(f) + \end{align*} + and + \begin{align*} + L(f,\mathcal{P}_n) &= \sum_{i=1}^{n} m_i \Delta x_i \\ + &= \sum_{i=1}^{n} \left(\frac{4i^2-4i+1}{n^2}+1\right) \cdot \left(\frac{2}{n}\right) \\ + &= \sum_{i=1}^{n} \frac{8i^2-8i+2}{n^3} + \frac{2}{n} \\ + &= \sum_{i=1}^{n} \frac{2(2i-1)^2}{n^3} + \frac{2}{n}\ \sum_{i=1}^{n} 1 \\ + &= \frac{8}{3}-\frac{2}{3n^2} + \frac{2 \cancel{n}}{\cancel{n}} \\ + &= \lim\limits_{n \to \infty} \frac{8}{3} -\frac{2}{3n^2} + 2 \\ + &= \frac{8}{3} -0 + 2 \\ + &= \frac{8}{3} + 2 \\ + &= \frac{14}{3} \\ + &\leq L(f) + \end{align*} + So $\frac{14}{3} \leq L(f) \leq U(f) \leq \frac{14}{3}$. So $L(f)=U(f)=\frac{14}{3}$ + \end{enumerate} + \item + \begin{enumerate} + \item Prove that if $f,g:[a,b] \to \R$ are bounded, then $U(f+g, \mathcal{P}) \leq U(f,\mathcal{P})+U(g,\mathcal{P})$ for every partition $\mathcal{P}$ of $[a,b]$. + \begin{proof} + Since $f$ and $g$ are bounded, we know that $\sup(f+g) \leq \sup (f) + \sup(g)$. Then $M_{f+g,i}:=\sup \{f+g:x \in [x_{i-1},x_i]\}$. And so $M_{f+g,i} \leq M_{f,i} + M_{g,i}$, and thus + \[U(f+g,\mathcal{P}):=\sum_{i=1}^{n} M_{f+g,i}\Delta x_i \leq \sum_{i=1}^{n} M_{f,i}\Delta x_i +\sum_{i=1}^{n} M_{g,i}\Delta x_i=U(f,\mathcal{P})+U(g,\mathcal{P})\] + \end{proof} + \item Find examples of bounded functions $f,g:[a,b] \to \R$ such that $U(f+g,\mathcal{P}) < U(f, \mathcal{P}) + U(g,\mathcal{P})$ for some partition of $[a,b]$. + \\\\Consider the functions $f,g:[a,b] \to \R$ given by $f(x):=\begin{cases} + 1, &x \in \R\setminus\Q \\ + 0, &x \in \Q + \end{cases}$ and $g(x):=\begin{cases} + -1, &x \in \R\setminus\Q \\ + 0, &x \in \Q + \end{cases}$ + \\Thus, we have that $(f+g)(x) = 0\ \forall\ x$, and thus $U(f+g)=0$. However, we note that $U(f,\mathcal{P})+U(g,\mathcal{P})=1+0 = 1$. Thus, $0=U(f+g,\mathcal{P}) 0$, there exists a partition $P$ of $[a,b]$ such that $L(f,P) > U(f,P)-\varepsilon$. + \\\\This is a true statement. + \begin{proof} + Let $f \in \mathcal{R}[a,b]$. Recall the \textit{Equivalence Theorem}: + \begin{theorem*}{\textbf{Equivalence Theorem}} + A function $f$ on $I=[a,b]$ is Darboux integrable if and only if it is Riemann integrable. + \end{theorem*} + Thus by the \textit{Equivalence Theorem}, $f$ is also Darboux integrable. + \\\\Also, recall the \textit{Integrability Criterion}: + \begin{theorem*}{\textbf{Integrability Criterion}} + Let $I:=[a,b]$ and let $f:I \to \R$ be a bounded function on $I$. Then $f$ is Darboux integrable on $I$ if and only if for each $\varepsilon > 0$ there is a partition $\mathcal{P}_\varepsilon$ of $I$ such that + \[U(f;\mathcal{P}_\varepsilon)-L(f;\mathcal{P}_\varepsilon)<\varepsilon\] + \end{theorem*} + Note that we can rewrite the inequality as follows: + \begin{align*} + U(f,\mathcal{P}_\varepsilon) -L(f,\mathcal{P}_\varepsilon) < \varepsilon &\equiv -L(f,\mathcal{P}_\varepsilon) < \varepsilon-U(f,\mathcal{P}_\varepsilon) \\ + &\equiv L(f,\mathcal{P}\varepsilon) > U(f,\mathcal{P}_\varepsilon) -\varepsilon + \end{align*} + Thus by the \textit{Integrability Criterion}, we have that $L(f,\mathcal{P})>U(f,\mathcal{P})-\varepsilon$. + \end{proof} + \end{enumerate} + \end{enumerate} +\end{document} diff --git a/Documents/LaTeX/Homework 3.pdf b/Documents/LaTeX/Homework 3.pdf new file mode 100644 index 0000000..c4167e9 Binary files /dev/null and b/Documents/LaTeX/Homework 3.pdf differ diff --git a/Documents/LaTeX/Homework 3.tex b/Documents/LaTeX/Homework 3.tex new file mode 100644 index 0000000..be7ecf5 --- /dev/null +++ b/Documents/LaTeX/Homework 3.tex @@ -0,0 +1,521 @@ +\documentclass[12pt,letterpaper]{article} +\usepackage[utf8]{inputenc} +\usepackage{pgfplots} +\usepackage[english]{babel} +\usepackage{amsthm} +\usepackage{cancel} +\usepackage{mathtools} +\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} +\newcommand{\limx}[2]{\displaystyle\lim\limits_{#1 \to #2}} +\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{\dotp}{\dot{\mathcal{P}}} +\newcommand{\dotq}{\dot{\mathcal{Q}}} +\newcommand{\dist}{\text{dist}} +\DeclareMathOperator{\sign}{sgn} +\newtheoremstyle{case}{}{}{}{}{}{:}{ }{} +\theoremstyle{case} +\newtheorem{case}{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*{lemma*}{Lemma} +\newtheorem*{remark}{Remark} +\setlist[enumerate]{font=\bfseries} +\renewcommand{\qedsymbol}{$\blacksquare$} +\author{Alexander J. Tusa} +\title{Real Analysis II Homework 3} +\begin{document} + \maketitle +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +%%%%%%%%%%%%%%%%%%%%%%% Section 7.2 Question 16 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% + \begin{enumerate} + \item + \begin{enumerate} + \item[16.] If $f$ is continuous on $[a,b],a 0$, we have + \[m \leq \frac{\int_{a}^{b} f}{b-a} \leq M\] + By \textit{Bolzano's Theorem}, we can conclude that there exists $c \in [a,b] \st$ + \[f(c):=\frac{\int_{a}^{b}f}{b-a}\] + which can be equivalently written as + \[\int_{a}^{b} f = f(c)(b-a)\] + \end{proof} +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +%%%%%%%%%%%%%%%%%%%%%%% Section 7.2 Question 19 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% + \item[19.] Suppose that $a>0$ and that $f \in \mathcal{R}[-a,a]$. + \begin{enumerate} + \item[(a)] If $f$ is \textit{even} (that is, if $f(-x)=f(x)$ for all $x \in [0,a]$), show that $\displaystyle\int_{-a}^{a}f=2\int_{0}^{a}f$. + \begin{proof} + Since $f$ is even, we have + \begin{align*} + \int_{-a}^{b} f &= \int_{-a}^{0} f(x)\ dx + \int_{0}^{a} f(x)\ dx \\ + &= -\int_{a}^{0} f(-y)\ dy + \int_{0}^{a} f(x)\ dx + \end{align*} + where $y=-x$ for the first integral. Thus $x \mapsto -y,\ -a\mapsto a,\ 0 \mapsto 0$. + \begin{align*} + &= -\int_{a}^{0} f(y)\ dy + \int_{0}^{a} f(x)\ dx &\text{($f$ is even so $f(-y)=f(y)$} \\ + &= \int_{0}^{a} f(y)\ dy + \int_{0}^{a} f(x)\ dx + \end{align*} + since flipping the limits of integration changes the sign of the integral + \begin{align*} + &= 2 \int_{0}^{a} f(x) \\ + &= 2 \int_{0}^{a} f + \end{align*} + \end{proof} + \item[(b)] If $f$ is \textit{odd} (that is, if $f(-x)=-f(x)$ for all $x \in [0,a]$), show that $\int_{-a}^{a} f=0$. + \begin{proof} + Since $f$ is odd, we have + \begin{align*} + \int_{-a}^{a} f &= \int_{-a}^{0} f(x)\ dx + \int_{0}^{a} f(x)\ dx \\ + &= -\int_{a}^{0} f(-y)\ dy + \int_{0}^{a} f(x)\ dx + \end{align*} + where $y=-x$, thus giving us $x \mapsto -y,\ -a \mapsto a,\ 0 \mapsto 0$. + \begin{align*} + &= -\int_{a}^{0} (-f(y))\ dy + \int_{0}^{a} f(x)\ dx &\text{since $f$ is odd, $f(-x)=-f(x)$} \\ + &= \int_{0}^{a} (-f(y))\ dy + \int_{0}^{a} f(x)\ dx \\ + &= -\int_{0}^{a} f(y)\ dy + \int_{0}^{a} f(x)\ dx + \end{align*} + since flipping the limits of integration changes the sign of the integral, + \begin{align*} + &= 0 + \end{align*} + since both integrals cancel each other out. + \end{proof} + \end{enumerate} +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +%%%%%%%%%%%%%%%%%%%%%%% Section 7.2 Question 20 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% + \item[20.] If $f$ is continuous on $[-a,a]$, show that $\int_{-a}^{a}f(x^2)dx=2\int_{0}^{a}f(x^2)dx$. + \begin{proof} + Since $f$ is continuous on $[-a,a]$, and $x^2$ is a continuous function, we know that $f(x^2)$ is a continuous function since the composition of continuous functions is continuous. That is, $f(x^2)$ is a continuous function since it is the composition of the continuous functions$f(x)$, and $x \mapsto x^2$. + \\\\Let $g:[a,b] \to \R$ be given by $g(x):=f(x^2)$, then $g$ is also continuous as it is a composition of the continuous functions $x \mapsto f(x)$ and $x \mapsto x^2$. Notice, however, that $g(-x)=f((-x)^2)=f(x^2)=g(x)$. This means that $g$ is an even function. + \\\\So by the preceding problem, we know that + \[\int_{-a}^{a} g(x)\ dx = 2\int_{0}^{a} g(x)\ dx\] + Therefore + \[\int_{-a}^{a} f(x^2)\ dx = 2\int_{0}^{a} f(x^2)\ dx\] + \end{proof} + \end{enumerate} + \item + \begin{enumerate} +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +%%%%%%%%%%%%%%%%%%%%%%%% Section 7.3 Question 3 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% + \item[3.] If $g(x):=x$ for $|x| \geq 1$ and $g(x):=-x$ for $|x|<1$ and if $G(x):=\frac{1}{2}|x^2-1|$, show that $\int_{-2}^{3}g(x)dx=G(3)-G(-2)=\frac{5}{2}$. Also sketch the graphs of $g$ and $G$. + \begin{proof} + Let $g(x):=\begin{cases} + x, &|x| \geq 1 \\ + -x, &|x| < 1 + \end{cases}$, and let $G(x):=\frac{1}{2} \abs{x^2-1}$. We want to show that $\int_{-2}^{3} g(x)\ dx=G(3)-G(-2)=\frac{5}{2}$. + \\\\Notice that since $G$ is a composition of continuous functions, namely $|x^2-1|$ and $x^2$, we know that $G$ is also continuous. Also, notice that $g$ has a finite number of discontinuities, namely at $x=-1$ and $x=1$. Thus, $g \in \mathcal{R}[-2,3]$. + \\\\Lastly, note that + \[G'(x):=\begin{cases} + x, &|x| \geq 1 \\ + -x, &|x| < 1 + \end{cases}=g(x),\ \forall\ x \in [-2,3]\setminus\{-1,1\}\] + Thus, by the Fundamental Theorem of Calculus, we have + \[\int_{-2}^{3} g(x)\ dx = G(3)-G(-2) = \frac{1}{2} |9-1| - \frac{1}{2}|4-1| = \frac{8}{2}-\frac{3}{2}=\frac{5}{2}\] + \end{proof} + \begin{tikzpicture} + \begin{axis}[ + axis x line=middle, axis y line=middle, + ymin=-2, ymax=3, ylabel={$g(x)$}, + xmin=-2, xmax=3, xlabel={$x$}, + domain=-2:3 + ] + + \addplot[blue,thick][domain=-2:-1]{x}; + \addplot[blue,thick][domain=-1:1]{-x}; + \addplot[blue,thick][domain=1:3]{x}; + \end{axis} + \end{tikzpicture} + \begin{tikzpicture} + \begin{axis}[ + axis x line=middle, axis y line=middle, + ymin=-2, ymax=3, ylabel={$G(x)$}, + xmin=-2, xmax=3, xlabel={$x$}, + domain=-2:3 + ] + + \addplot[blue,thick][domain=-2:3, samples=150]{1/2 * abs(x^2-1)}; + \end{axis} + \end{tikzpicture} +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +%%%%%%%%%%%%%%%%%%%%%%%% Section 7.3 Question 9 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% + \item[9.] Let $f \in \mathcal{R}[a,b]$ and define $F(x):=\displaystyle\int_{a}^{x}f$ for $x \in [a,b]$. + \begin{enumerate} + \item[(a)] Evaluate $G(x):=\displaystyle\int_{c}^{x}f$ in terms of $F$, where $c \in [a,b]$. + \begin{align*} + G(x)&=\int_{c}^{x} f \\ + &=\int_{a}^{c} f + \int_{c}^{x} f - \int_{a}^{c} f \\ + &= \int_{a}^{x} f-\int_{a}^{c} f \\ + &= F(x)-F(c) + \end{align*} + \item[(b)] Evaluate $H(x):=\displaystyle\int_{x}^{b} f$ in terms of $F$. + \begin{align*} + H(x)&=\int_{x}^{b} f \\ + &= \int_{a}^{x} f + \int_{x}^{b} f- \int_{a}^{x} f \\ + &= F(b)-F(x) + \end{align*} + \item[(c)] Evaluate $S(x):=\displaystyle\int_{x}^{\sin x} f$ in terms of $F$. + \begin{align*} + S(x)&= \int_{x}^{\sin x} f \\ + &= \int_{a}^{x} f + \int_{x}^{\sin x} f - \int_{a}^{x} f \\ + &= \int_{a}^{\sin x} f - \int_{a}^{x} f \\ + &= F(\sin x) - F(x) + \end{align*} + \end{enumerate} +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +%%%%%%%%%%%%%%%%%%%%%%% Section 7.3 Question 11 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% + \item[11.] Find $F'(x)$ when $F$ is defined on $[0,1]$ by: + \begin{enumerate} + \item[(a)] $F(x):=\displaystyle\int_{0}^{x^2} (1+t^3)^{-1}\ dt$. + \\\\Since $x^2$ is continuous and differentiable on $[0,1]$, we can use \textit{Leibniz's Rule} to find $F'(x)$, where $f(x)=\frac{1}{1+x^3}$, $h(x):=x^2$, and $g(x):=0$. So, + \begin{align*} + F'(x)&= f(h(x))\cdot h'(x)-f(g(x))\cdot g'(x) \\ + &= \frac{1}{1+(x^2)^3}\cdot 2x - \frac{1}{1+(0)^3}\cdot 0 \\ + &= \frac{2x}{1+x^6}-0 \\ + &= \frac{2x}{1+x^6} + \end{align*} + \item[(b)] $F(x):=\displaystyle\int_{x^2}^{x} \sqrt{1+t^2}\ dt$. + \\\\Since both $x$ and $x^2$ are continuous and differentiable on $[0,1]$, we can use \textit{Leibniz's Rule} to find $F'(x)$, where $f(x):=\sqrt{1+x^2},\ h(x):=x,$ and $g(x):=x^2$. So, we must first rewrite $F(x)$ as + \[F(x):=\int_{x^2}^{x} \sqrt{1+x^2}\ dx = \int_{0}^{x} \sqrt{1+x^2}\ dx - \int_{0}^{x^2} \sqrt{1+x^2}\ dx\] + \begin{align*} + F'(x)&=f(h(x))\cdot h'(x) - f(g(x))\cdot g'(x) \\ + &= \sqrt{1+x^2} \cdot 1 - \sqrt{1+(x^2)^2} \cdot 2x \\ + &= \sqrt{1+x^2} - 2x \cdot \sqrt{1+x^4} + \end{align*} + \end{enumerate} +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +%%%%%%%%%%%%%%%%%%%%%%% Section 7.3 Question 12 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% + \item[12.] Let $f:[0,3] \to \R$ be defined by $f(x):=x$ for $0 \leq x < 1$, $f(x):=1$ for $1 \leq x < 2$ and $f(x):=x$ for $2 \leq x \leq 3$. Obtain formulas for $F(x):=\int_{0}^{x} f$ and sketch the graphs of $f$ and $F$. Where is $F$ differentiable? Evaluate $F'(x)$ at all such points. + \\\\Let $f(x):=\begin{cases} + x, &0 \leq x < 1 \\ + 1, &1 \leq x < 2 \\ + x, &2 \leq x \leq 3 + \end{cases}$ + \\Then we have the following: When $x \in [0,1)$: + \begin{align*} + F(x) &= \int_{0}^{x} f(t)\ dt \\ + &= \int_{0}^{x} t\ dt \\ + &= \frac{x^2}{2} + \end{align*} + When $x \in [1,2)$: + \begin{align*} + F(x) &= \int_{0}^{x} f(t)\ dt \\ + &= \int_{0}^{1} t\ dt + \int_{1}^{x} 1\ dt \\ + &= \frac{1}{2} + (x-1) \\ + &= x-\frac{1}{2} + \end{align*} + When $x \in [2,3]$: + \begin{align*} + F(x) &= \int_{0}^{x} t\ dt \\ + &= \int_{0}^{1} t\ dt + \int_{1}^{2} 1\ dt + \int_{2}^{3} t\ dt \\ + &= \frac{1}{2} + 1 + \left(\frac{x^2}{2}-\frac{2^2}{2}\right) \\ + &= \frac{x^2}{2} - \frac{1}{2} + \end{align*} + Therefore, we have + \[F(x)=\begin{cases} + \frac{x^2}{2}, &0 \leq x < 1 \\ + x-\frac{1}{2}, &1 \leq x < 2 \\ + \frac{x^2}{2} - \frac{1}{2}, &2 \leq x \leq 3 + \end{cases}\] + \begin{tikzpicture} + \begin{axis}[ + axis x line*=bottom, axis y line*=left, + ymin=0, ymax=4, ytick distance=1, ylabel={$F(x)$}, + xmin=0, xmax=3, xtick distance=1 + ] + + \addplot[blue,thick][domain=0:1]{x^2/2}; + \addplot[blue,thick][domain=1:2]{x-1/2}; + \addplot[blue,thick][domain=2:3]{x^2/2 - 1/2}; + \end{axis} + \end{tikzpicture}\\ + \begin{tikzpicture} + \begin{axis}[ + axis x line*=bottom, axis y line*=left, + ymin=0, ymax=4, ytick distance=1, ylabel={$f(x)$}, + xmin=0, xmax=3, xtick distance=1 + ] + + \addplot[blue,thick][domain=0:1]{x}; + \addplot[blue,thick][domain=1:2]{1}; + \addplot[blue,thick][domain=2:3]{x}; + \end{axis} + \end{tikzpicture} + \\$F$ is definitely differentiable at points $x \in (0,1) \cup (1,2) \cup (2,3)$ since at those points, $f$ is equal to a polynomial. So, we must now check for the differentiability of $f$ at $x=1$ and $x=2$. + \[\limx{x}{1^-}\frac{F(x)-F(1)}{x-1}=\limx{x}{1^-}\frac{\frac{x^2}{2}-\frac{1}{2}}{x-1}=\frac{1}{2} \limx{x}{1^-}\frac{\cancel{(x-1)}(x+1)}{\cancel{(x-1)}}=\frac{1}{2} \cdot (1+1) = 1\] + \[\limx{x}{1^+}\frac{F(x)-F(1)}{x-1}=\limx{x}{1^+}\frac{\left(x-\frac{1}{2}\right)-\frac{3}{2}}{x-1} = 1\] + Therefore, $F$ is differentiable at $x=1$ and $F'(x)=1$. As for when $x=2$, + \[\limx{x}{2^-}\frac{F(x)-F(2)}{x-1}=\limx{x}{2^-}\frac{\left(x-\frac{1}{2}\right)-\frac{3}{2}}{x-2}=1\] + \[\limx{x}{2^+}\frac{F(x)-F(2)}{x-2}=\limx{x}{2^+}\frac{\left(\frac{x^2}{2}-\frac{1}{2}\right)-\frac{1}{2}}{x-2}=2\] + Therefore, $F$ is not differentiable at $x=2$. Thus, + \[F'(x):=\begin{cases} + x, &0 \leq x < 1 \\ + 1, &1 \leq x < 2 \\ + x, &2 < x \leq 3 + \end{cases}\] + And notice that $F'(x)=f(x)$ for $x \in [0,1] \setminus \{2\}$ +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +%%%%%%%%%%%%%%%%%%%%%%% Section 7.3 Question 13 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% + \item[13.] The function $g$ is defined on $[0,3]$ by $g(x):=-1$ if $0 \leq x < 2$ and $g(x):=1$ if $2 \leq x \leq 3$. Find the indefinite integral $G(x)=\int_{0}^{x} g$ for $0 \leq x \leq 3$, and sketch the graphs of $g$ and $G$. Does $G'(x)=g(x)$ for all $x \in [0,3]$? + \[g(x):=\begin{cases} + -1, &0 \leq x < 2 \\ + 1, & 2 \leq x \leq 3 + \end{cases}\] + Then we have the following: when $x \in [0,2)$: + \begin{align*} + G(x) &= \int_{0}^{x} g(t)\ dt \\ + &= \int_{0}^{x} -1\ dt \\ + &= -x + \end{align*} + When $x \in [2,3]$: + \begin{align*} + G(x)&= \int_{0}^{x} g(t)\ dt \\ + &= \int_{0}^{2} -1\ dt + \int_{2}^{x} 1\ dt \\ + &= -2+x-2 \\ + &= x-4 + \end{align*} + Therefore, + \[F(x)=\begin{cases} + -x, &0 \leq x < 2 \\ + x-4, &2 \leq x \leq 3 + \end{cases}\] + \\\\\begin{tikzpicture} + \begin{axis}[ + axis x line=middle, axis y line=left, + ymin=-2, ymax=1, ytick distance=1, ylabel={$g(x)$}, + xmin=0, xmax=3, xtick distance=1 + ] + + \addplot[red,thick][domain=0:2]{-1}; + \addplot[red,thick][domain=2:3]{1}; + \end{axis} + \end{tikzpicture}\\ + \begin{tikzpicture} + \begin{axis}[ + axis x line=middle, axis y line=middle, + ymin=-3, ymax=3, ytick distance=1, ylabel={$G(x)$}, + xmin=0, xmax=3, xtick distance=1 + ] + + \addplot[blue,thick][domain=0:2]{-x}; + \addplot[blue,thick][domain=2:3]{x-4}; + \end{axis} + \end{tikzpicture} + \\\\Now, if it is possible, let $G'(x)=g(x)\ \forall\ x \in [0,3]$. Then + \[\limx{x}{2^-}\frac{G(x)-G(2)}{x-2}=\limx{x}{2^-}\frac{-x+2}{x-2} = -1\] + \[\limx{x}{2^+}\frac{G(x)-G(2)}{x-2}=\limx{x}{2^+}\frac{x-2}{x-2}=1\] + So, + \[\limx{x}{2^-} \frac{G(x)-G(2)}{x-2}\neq \limx{x}{2^+}\frac{G(x)-G(2)}{x-2}\] + Thus we have that the limit does not exist, and hence $G$ is not differentiable at $x=2$. Thus $G'(x) \neq g(x)$ for some $x \in [0,3]$.\\ +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +%%%%%%%%%%%%%%%%%%%%%%% Section 7.3 Question 16 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% + \item[16.] If $f:[0,1] \to \R$ is continuous and $\int_{0}^{x} f=\int_{x}^{1} f$ for all $x \in [0,1]$, show that $f(x)=0$ for all $x \in [0,1]$. + \begin{proof} + Let $F(x)=\displaystyle\int_{0}^{x} f(t)\ dt$ for $x \in [0,1]$. $F$ is well defined since $f$ is continuous on $[0,1]$, and thus is also integrable on $[0,1]$. $F$ is also differentiable since $f$ is continuous and $F'(x)=f(x)\ \forall\ x \in [0,1]$. So, + \begin{align*} + \int_{0}^{x} f = \int_{x}^{1} f &\Leftrightarrow \int_{0}^{x} f=\int_{0}^{1} f -\int_{0}^{x} f \\ + &\Leftrightarrow 2 \int_{0}^{x} f = \int_{0}^{1} f \\ + &\Leftrightarrow 2F(x)=F(1) + \end{align*} + And differentiating the last relation with respect to $x$, we get + \[2F'(x)=0 \Leftrightarrow F'(x)=0 \Leftrightarrow f(x)=0,\ \forall\ x \in [0,1]\] + \end{proof} +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +%%%%%%%%%%%%%%%%%%%%%%% Section 7.3 Question 18c %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% + \item[18.] Use the Substitution Theorem 7.3.8 to evaluate the following integral: + \begin{enumerate} + \item[(c)] $\int_{1}^{4} \frac{\sqrt{1+\sqrt{t}}}{\sqrt{t}} dt$ + \\\\Let $\phi(t)=1+\sqrt{t}$ for $t \in [1,4]$, and let $f(u)=\sqrt{u}$ for $u \in [2,3]$. $f$ is continuous on $[2,3]$ and $\phi$ has a continuous derivative (namely, $\phi'(t)=\frac{1}{2\sqrt{t}}$) on $[1,4]$. Thus we have + \begin{align*} + \int_{1}^{4} \frac{\sqrt{1+\sqrt{t}}}{\sqrt{t}}\ dt &= 2\int_{1}^{4} \frac{\sqrt{1+\sqrt{t}}}{\sqrt{t}}\ dt \\ + &= 2 \int_{1}^{4} \phi'(t)\cdot f(\phi(t))\ dt \\ + &= 2 \int_{\phi(1)}^{\phi(4)} f(u)\ dt &\text{by the \textit{Substitution Theorem}} \\ + &= 2 \int_{\phi(1)}^{\phi(4)} \sqrt{u}\ du \\ + &= 2 \int_{2}^{3} \sqrt{u}\ du \\ + &= 2 \cdot \left.\frac{2}{3}u^{\frac{3}{2}}\right|_{u=2}^{u=3} \\ + &= \frac{4}{3}\left(3^{\frac{3}{2}}-2^{\frac{3}{2}}\right) + \end{align*} + \end{enumerate} + \end{enumerate} +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Question 3 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% + \item Let $f:[0,1] \to \R$ given by $f(x)=\begin{cases} + \frac{1}{n}\ &\text{if } x \in \left(\frac{1}{n+1},\frac{1}{n}\right] \\ + 0\ &\text{if }x=0 + \end{cases}$. Sketch the graph of $f$ and show that $f \in \mathcal{R}[0,1]$.\\ + \begin{tikzpicture} + \begin{axis}[ + axis x line=bottom, axis y line=left, + ymin=0, ymax=4, ytick distance= 1, ytick={1,2,3,4}, yticklabels={$\dots$,$\frac{1}{3}$, $\frac{1}{2}$, $1$}, + xmin=0, xmax=4, xtick distance=1, xtick={1,2,3,4}, + xticklabels={$\dots$,$\frac{1}{3}$, $\frac{1}{2}$, $1$} + ] + + \addplot[blue,thick][domain=3:4]{4}; + \addplot[blue,thick][domain=2:3]{3}; + \addplot[blue,thick][domain=1:2]{2}; + \addplot[blue,thick][mark=*] coordinates{(4,4)}; + \addplot[blue,thick][mark=*,fill=white] coordinates{(3,4)}; + \addplot[blue,thick][mark=*] coordinates{(3,3)}; + \addplot[blue,thick][mark=*,fill=white] coordinates{(2,3)}; + \addplot[blue,thick][mark=*] coordinates{(2,2)}; + \addplot[blue,thick][mark=*,fill=white] coordinates{(1,2)}; + \addplot[blue,thick][mark=*] coordinates{(0,0)}; + \end{axis} + \end{tikzpicture} + \\Since $f$ is monotone, by \textit{Theorem 7.2.8}, $f \in \mathcal{R}[0,1]$. + +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Question 4 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% + \item + \begin{enumerate} + \item Give an example of two functions $f,g:[a,b] \to \R$ that are not Riemann integrable, but $fg \in \mathcal{R}[a,b]$. + \\\\Consider + \[f(x):=\begin{cases} + 1, &x \in \Q \\ + 0, &x \in \R\setminus\Q + \end{cases}\ \ \ \ \text{ and }\ \ \ \ g(x):=\begin{cases} + 0, &x \in \Q \\ + 1, &x \in \R\setminus\Q + \end{cases}\] + Since these are both Dirichlet and modified Dirichlet functions, we know that they are not Riemann integrable, however, + \[fg:=\begin{cases} + 0, x \in \Q \\ + 0, x \in \R\setminus\Q + \end{cases}=0\] + and thus $fg$ is a constant function, which is Riemann integrable. Thus we have that $f,g \notin \mathcal{R}[a,b]$, but $fg \in \mathcal{R}[a,b]$.\\ + \item Give an example of two functions $f,g:[a,b] \to \R$ where $f \in \mathcal{R}[a,b]$ and $g \notin \mathcal{R}[a,b]$, but $fg \in \mathcal{R}[a,b]$.\\\\ + Consider + \[f(x):=0,\ \forall\ x \in [a,b],\ \ \ \ \text{ and }\ \ \ \ g(x):=\begin{cases} + 1, &x \in \Q \\ + 0, &x \in \R\setminus\Q + \end{cases}\] + Then since $f$ is a constant function, $f \in \mathcal{R}[a,b]$, and since $g$ is the Dirichlet function, we know that $g \notin \mathcal{R}[a,b]$. However, + \[fg=\begin{cases} + 0, &x \in \Q \\ + 0, &x \in \R\setminus\Q + \end{cases}=0\] + And thus $fg$ is a constant function and is thus Riemann integrable. Thus we have that $f \in \mathcal{R}[a,b]$, $g \notin \mathcal{R}[a,b]$, and $fg \in \mathcal{R}[a,b]$.\\ + \item Let $f:[a,b] \to \R$, $f \in \mathcal{R}[a,b]$. Let $F:[a,b] \to \R$ be given by $F(x)=\displaystyle\int_{a}^{x} f(t)dt$. Prove that $F$ is Lipschitz. + \begin{proof} + Since $f \in \mathcal{R}[a,b]$, $f$ is bounded; that is, there is some $M \st |f(x)| \leq M\ \forall\ x \in [a,b]$. Now, if $y < x$, we have + \[|F(x)-F(y)|=\abs{\int_{y}^{x} f(t)\ dt} \leq \int_{y}^{x} |f(t)|\ dt \leq \int_{y}^{x} M\ dt = M(x-y)=M|x-y|\] + Similarly, $|F(x)-F(y)| \leq M(y-x)=M|x-y|$ if $y > x$. So, we see that for any $x,y \in [a,b]$, if we let $K=M$, we have + \[|F(x)-F(y)| \leq K|x-y|\] +% Thus, $F$ is Lipschitz since the above statement is the definition of a Lipschitz function. + \end{proof} + \end{enumerate} +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Question 5 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% + \item Let $f(t):=\begin{cases} + t\ &\text{for } 0 \leq t \leq 2 \\ + 3\ &\text{for } 2 < t \leq 4 + \end{cases}$ + \begin{enumerate} + \item Find an explicit expression for $F(x)=\int_{0}^{x}f(t)dt$. + \\\\When $x \in [0,2]$: + \begin{align*} + F(t)&= \int_{0}^{x} f(t)\ dt \\ + &= \int_{0}^{x} t\ dt \\ + &= \frac{x^2}{2} + \end{align*} + and when $x \in (2,4]$: + \begin{align*} + F(t)&= \int_{0}^{x} f(t)\ dt \\ + &= \int_{0}^{2} t\ dt + \int_{2}^{x} 3\ dt \\ + &= 2+3x-6 \\ + &= 3x-4 + \end{align*} + Thus, + \[F(x):=\begin{cases} + \frac{x^2}{2}, &0 \leq x \leq 2 \\ + 3x-4, &2 < t \leq 4 + \end{cases}\] + \item Sketch $F$ and determine where $F$ is differentiable.\\\\ + \begin{tikzpicture} + \begin{axis}[ + axis x line=bottom, axis y line=left, + ymin=0, ymax=8, ytick distance=1, + xmin=0, xmax=4, xtick distance=1 + ] + + \addplot[blue,thick][domain=0:2]{x^2/2}; + \addplot[blue,thick][domain=2:4]{3*x -4}; + \end{axis} + \end{tikzpicture} + \\\\Based on the graph, we can tell that the only place in which $F$ is not differentiable is at $x=2$, which we can see as follows: + \[\limx{x}{2^-} \frac{F(x)-F(x)}{x-2} = \limx{x}{2^-} \frac{\frac{x^2}{2}-2}{x-2}=2\] + \[\limx{x}{2^+}\frac{F(x)-F(2)}{x-2}=\limx{x}{2^+} \frac{3x-4-2}{x-2}=\limx{x}{2^+} \frac{3x-2}{x-2} = 3\] + Since $\limx{x}{2^-} \frac{F(x)-F(2)}{x-2} \neq \limx{x}{2^+} \frac{F(x)-F(2)}{x-2}$, we have that $F$ is not differentiable when $x=2$.\\ + \item Find formula for $F'(x)$ wherever $F$ is differentiable. + \\\\Since the only place in which $F$ is not differentiable is when $x=2$, we need only change one of the inequalities of $f$. So, + \[F'(x):=\begin{cases} + x, &0 \leq x < 2 \\ + 3, &2 0$. + \\\[\int_{1}^{e} f'(x)\cdot \ln (x)\ dx = \left.f(x)\ln(x)\right|_1^e - \int_{1}^{e} f(x) \cdot \frac{1}{x}\ dx\] + where $u=\ln (x)$, $dv=f'(x)\ dx$, $du=\frac{1}{x}\ dx$, $v=f(x)$. So + \[\left.\int_{1}^{e} f'(x)\cdot \ln (x)\ dx = f(x)\ln(x)\right|_1^e - \int_{1}^{e} f(x) \cdot \frac{1}{x}\ dx=f(e)-\int_{1}^{e} \frac{1}{x}\ dx > 0\] + Since $\int_{1}^{e} \frac{f(x)}{x}\ dx < f(e)$.\\ + \item If $f(0)=f(1)=0$, prove that $\displaystyle\int_{0}^{1} e^x [f(x)+f'(x)]\ dx =0$. + \begin{proof} + \[\int_{0}^{1}e^x \left[f(x)+f'(x)\right]\ dx = \int_{0}^{1} e^xf(x)\ dx +\int_{0}^{1} e^xf'(x)\ dx\] + Let us use \textit{Integration by Parts} on the second integral containing $f'(x)$. Let $u=e^x$. Then $du=e^x\ dx$, $dv=f'(x)\ dx$, and $v=f(x)$. Then we have the following: + \begin{align*} + \int_{0}^{1} e^xf(x)\ dx + \int_{0}^{1} f'(x)\ dx &= \left.\cancel{\int_{0}^{1} e^xf(x)\ dx} + e^xf(x)\right|_0^1 - \cancel{\int_{0}^{1} f(x)e^x\ dx} \\ + &= e^1f(1)-e^0f(0) \\ + &= e\cdot 0 - 1 \cdot 0 \\ + &= 0 - 0 \\ + &= 0 + \end{align*} + $\therefore\ \displaystyle\int_{0}^{1} e^x\left[f(x)+f'(x)\right]\ dx = 0$. + \end{proof} + \end{enumerate} + \item + \begin{enumerate} + \item Let $f:[0,b] \to \R,\ b>0$ be continuous and $f(x) \neq 0$ for all $x \in (0,b)$. Further, suppose $[f(x)]^2 = 2 \displaystyle\int_{0}^{x} f(t)\ dt$ for all $x \in [0,b]$. Prove that $f(x)=x$ for all $x \in [0,b]$.\\\\We have + \[2f(x)f'(x)=2f(x)\] + which implies $f(x)[f'(x)-1]=0$. Since $f(x) \neq 0$, then $f'(x)-1=0$. So $f'(x)=1$ and $f(x)=x+C$ for some arbitrary constant $C$. But $f(0)=0$ since $[f(0)]^2=0\implies f(0)=0$. So $f(x)=x$.\\ + \item Suppose that $f$ is defined on $[0,1]$ with $f(0)=0$ and $0 1$ 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, + \item (pr. 3a) $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2+3n+2}$ \begin{align*} - x&=2-\frac{1}{x} \\ - x^2 &= 2x -1 \\ - x^2 - 2x + 1 &= 0 \\ - (x-1)^2 &= 0 + \sum_{n=1}^{\infty} \frac{1}{n^2+3n+2} &= \sum_{n=1}^{\infty} \frac{1}{(n+1)(n+2)} \\ + &\Downarrow \\ + \frac{1}{n^2+3n+2}&=\frac{A}{n+1} + \frac{B}{n+2} \\ + 1&=A(n+2)+ B(n+2) \\ + 1&=An+2A+Bn+2B \\ + 1&=An+Bn+2A+2B \\ + 1&= (A+B)n+(A+B)2 \\ \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 + \[\begin{cases} + 0=A+B \\ + 1=A+B + \end{cases} \implies + \begin{cases} + A=1 \\ + B=-1 + \end{cases}\] \begin{align*} - \lim (x_n) &= \inf \{x_n: n \in \N\} \\ - &=\inf (1,2) \\ - &= 1 + &=\sum_{n=1}^{\infty} \frac{1}{n+1}-\frac{1}{n+2} \\ + &= \left(\frac{1}{2}-\cancel{\frac{1}{3}}\right) + \left(\cancel{\frac{1}{3}}-\cancel{\frac{1}{4}}\right) + \dots + \left(\cancel{\frac{1}{n}}-\cancel{\frac{1}{n+1}}\right)+ \left(\cancel{\frac{1}{n+1}}-\frac{1}{n+2}\right) \\ + &= \limx{n}{\infty} \frac{1}{2}+\frac{1}{n+2} \\ + &= \frac{1}{2} \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: + \item $\displaystyle\sum_{n=1}^{\infty} \frac{1}{(3n-2)(3n+1)}$ \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 + \frac{1}{(3n-2)(3n+1)} &= \frac{A}{3n-2} + \frac{B}{3n+1} \\ + 1&= A(3n+1)+B(3n-2) \\ + 1&= 3An+3Bn+A-2B \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, + \[\begin{cases} + 3A+3B=0 \\ + 1A-2B=1 + \end{cases} \implies \begin{cases} + A=\frac{1}{3} \\ + B=\frac{-1}{3} + \end{cases}\] \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 + \frac{1}{(3n-2)(3n+1)} &= \frac{1}{9n-6} - \frac{1}{9n+3} \\ + \sum_{n=1}^{\infty} \frac{1}{(3n-2)(3n+1)} &= \sum_{n=1}^{\infty} \frac{1}{9n-6} - \frac{1}{9n+3} \\ + &= \left(\frac{1}{3}-\cancel{\frac{1}{12}}\right) + \left(\cancel{\frac{1}{12}}-\cancel{\frac{1}{21}}\right) + \dots\\ + &+ \left(\cancel{\frac{1}{9n-15}}-\cancel{\frac{1}{9n-6}}\right) + \left(\cancel{\frac{1}{9n-6}}-\frac{1}{9n+3}\right) \\ + &= \limx{n}{\infty} \frac{1}{3}-\frac{1}{9n+3} \\ + &= \frac{1}{3} \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: + \item $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{e^{n-1}}$ \begin{align*} - \lim (x_n) &= \inf \{x_n:n \in \N\} \\ - &=2 + \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{e^{n-1}} &= \sum_{n=1}^{\infty} \frac{(-1)^n \cdot (-1)^1}{e^n\cdot e^{-1}} \\ + &=- \sum_{n=1}^{\infty} \frac{(-1)^n \cdot e}{e^n} \\ + &= - \sum_{n=1}^{\infty} (-1)^n \cdot \frac{e}{e^n} \\ + &= - \sum_{n=1}^{\infty} (-1)^n \cdot e^{1-n} \\ + &= \frac{e}{1+e} \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: + \item $\displaystyle\sum_{n=2}^{\infty} \frac{4^{n+1}}{9^{n-1}}$ \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 + \sum_{n=2}^{\infty} \frac{4^{n+1}}{9^{n-1}} &= \sum_{n=2}^{\infty} \frac{4^n \cdot 4^1}{9^n\cdot 9^{-1}} \\ + &= \sum_{n=2}^{\infty} \left(\frac{4}{9}\right)^n \cdot 4^1 \cdot 9^1 \\ + &= \sum_{n=2}^{\infty} \left(\frac{4}{9}\right)^n \cdot 36 \\ + &= \sum_{n=2}^{\infty} \left(\frac{4}{9}\right)^n \cdot 36 - 16-36 \\ + &= a\left(\frac{1}{1-r}\right) \\ + &= 36 \cdot \left(\frac{1}{1-\left(\frac{4}{9}\right)}\right) -52 \\ + &= \frac{36 \cdot 9}{5} - 52 \\ + &= \frac{324}{5} - 52 \\ + &= \frac{324}{5} - \frac{260}{5} \\ + &= \frac{64}{5} \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: + \item $\displaystyle\sum_{n=0}^{\infty} \frac{5^{n+1}+(-3)^n}{7^{n+2}}$ \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 + \sum_{n=0}^{\infty} \frac{5^{n+1}+(-3)^n}{7^{n+2}} &= \sum_{n=0}^{\infty} \frac{5^{n+1}}{7^{n+2}}+\frac{(-3)^n}{7^{n+2}} \\ + &= \sum_{n=0}^{\infty} \frac{5}{49} \cdot \left(\frac{5}{7}\right)^n + \frac{1}{49} \cdot \left(\frac{(-3)}{7}\right)^n \\ + &= \sum_{n=0}^{\infty} \frac{5}{49} \cdot \left(\frac{5}{7}\right)^n + \sum_{n=0}^{\infty} \frac{1}{49} \cdot \left(\frac{(-3)}{7}\right)^n \\ + &= \frac{5}{49} \cdot \frac{1}{1-\frac{5}{7}} + \frac{1}{49} \cdot \frac{1}{1-\frac{-3}{7}} \\ + &= \frac{5}{14} + \frac{1}{70} \\ + &= \frac{13}{35} + \end{align*} + \item $\displaystyle\sum_{n=2}^{\infty} \ln \frac{n^2-1}{n^2}$ + \begin{align*} + \sum_{n=2}^{\infty} \ln \left(\frac{n^2-1}{n^2}\right) &= \sum_{n=2}^{\infty} \ln \left(\frac{(n-1)(n+1)}{n^2}\right) \\ + &= \sum_{n=2}^{\infty} \ln \left(\frac{\frac{n-1}{n}}{\frac{n}{n+1}}\right) \\ + &= \sum_{n=2}^{\infty} \ln \left(\frac{n-1}{n}\right) - \ln\left(\frac{n}{n+1}\right) \\ + &= \left(\ln \frac{1}{2} - \cancel{\ln \frac{2}{3}}\right) + \left(\cancel{\ln \frac{2}{3}} - \cancel{\ln \frac{3}{4}}\right) +\\ + &\dots + \left(\cancel{\ln \frac{n-2}{n-1}}-\cancel{\ln \frac{n-1}{n}}\right) + \left(\cancel{\ln \frac{n-1}{n}}-\ln \frac{n}{n+1}\right) \\ + &= \limx{n}{\infty} \ln \left(\frac{1}{2}\right) - \ln \left(\frac{n}{n+1}\right) \\ + &= \ln \left(\frac{1}{2}\right) - \limx{n}{\infty} \frac{n+1}{n} \cdot \frac{n+1-n}{(n+1)^2}, &\text{by L'Hospital's Rule} \\ + &= \ln \left(\frac{1}{2}\right) - \limx{n}{\infty} \frac{1}{n(n+1)} \\ + &= \ln\left(\frac{1}{2}\right) - 0 \\ + &= \ln\left(\frac{1}{2}\right) \\ + &\approx -0.693147 + \end{align*} + \item $\displaystyle\sum_{n=2}^{\infty} \ln \frac{n(n+2)}{(n+1)^2}$ + \begin{align*} + \sum_{n=2}^{\infty} \ln\left(\frac{n(n+2)}{(n+1)^2}\right) &= \sum_{n=2}^{\infty} \ln \left(\frac{\frac{n}{n+1}}{\frac{n+1}{n+2}}\right) \\ + &= \sum_{n=2}^{\infty} \ln\left(\frac{n}{n+1}\right) - \ln\left(\frac{n+1}{n+2}\right) \\ + &= \left(\ln \frac{2}{3}-\cancel{\ln \frac{3}{4}}\right) + \left(\cancel{\ln \frac{3}{4}}-\cancel{\ln \frac{4}{5}}\right) + \dots \\ + &+ \left(\cancel{\ln \frac{n-1}{n}}-\cancel{\ln\frac{n}{n+1}}\right) + \left(\cancel{\ln \frac{n}{n+1}}-\ln\frac{n+1}{n+2}\right) \\ + &= \limx{n}{\infty}\ln \left(\frac{2}{3}\right) - \ln \left(\frac{n+1}{n+2}\right) \\ + &= \ln \left(\frac{2}{3}\right)-\limx{n}{\infty} \frac{n+2}{n+1} \cdot \frac{1\cdot (n+2) - (n+1)\cdot 1}{(n+2)^2}, &\text{by L'Hospital's Rule} \\ + &= \ln \left(\frac{2}{3}\right)- \limx{n}{\infty} \frac{1}{(n+1)(n+2)} \\ + &= \ln \left(\frac{2}{3}\right) - 0 \\ + &= \ln \left(\frac{2}{3}\right) \\ + &\approx -0.405465 + \end{align*} + \item (pr. 3c) $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n(n+1)(n+2)}$ + \begin{align*} + \frac{1}{n(n+1)(n+2)} &= \frac{A}{n} + \frac{B}{n+1} + \frac{C}{n+2} \\ + 1&=A(n+1)(n+2) + Bn(n+2)+Cn(n+1) \\ + 1&=An^2+3An+2A+Bn^2+2Bn+Cn^2+Cn \\ + 1&= An^2+Bn^2+Cn^2+3An+2Bn+Cn+2A + \end{align*} + \[\begin{cases} + An^2+Bn^2+Cn^2=0 \\ + 3An+2Bn+Cn=0 \\ + 2A=1 + \end{cases} = \begin{cases} + A=\frac{1}{2} \\ + B=-1 \\ + C=\frac{1}{2} + \end{cases}\] + \begin{align*} + \sum_{n=1}^{\infty} \frac{1}{n(n+1)(n+2)} &= \sum_{n=1}^{\infty} \frac{1}{2n} -\frac{1}{n+1} + \frac{1}{2n+4} \\ + &= \left(\frac{1}{2}-\frac{1}{2}+\frac{1}{6}\right) + \left(\frac{1}{4}-\frac{1}{3}+\frac{1}{8}\right) + \left(\frac{1}{6}-\frac{1}{4}+\frac{1}{10}\right) \\ + &+ \dots + \left(\frac{1}{2n-2}+\frac{1}{2n+2}-\frac{1}{n}\right) + \left(\frac{1}{2n}+\frac{1}{2n+4}-\frac{1}{n+1}\right) \\ + &= \limx{n}{\infty} \frac{1}{4} + \frac{1}{2(n+1)(n+2)} \\ + &= \frac{1}{4} + 0 \\ + &= \frac{1}{4} + \end{align*} + \item $\displaystyle\frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \dots$ + \\\\Notice that this is equal to the series $\displaystyle\sum_{n=1}^{\infty} \frac{1}{(2n-1)(2n+1)}$. So, + \begin{align*} + \sum_{n=1}^{\infty} \frac{1}{(2n-1)(2n+1)} &= \sum_{n=1}^{\infty} \frac{A}{2n-1} + \frac{B}{2n+1} \\ + &\Downarrow \\ + 1 &= 2An+2Bn+A-B \\ + \begin{cases} + 2An+2Bn=0 \\ + A-B=1 + \end{cases} + &\implies + \begin{cases} + A= \frac{1}{2} \\ + B= \frac{-1}{2} + \end{cases} \\ + &= \sum_{n=1}^{\infty} \frac{1}{4n-2} - \frac{1}{4n+2} \\ + &= \left(\frac{1}{2}-\cancel{\frac{1}{6}}\right) + \left(\cancel{\frac{1}{6}}-\cancel{\frac{1}{10}}\right) + \\ + &\dots + \left(\cancel{\frac{1}{4n-5}}-\cancel{\frac{1}{4n-2}}\right) + \left(\cancel{\frac{1}{4n-2}}-\frac{1}{4n+2}\right) \\ + &=\limx{n}{\infty} \frac{1}{2} -\frac{1}{4n+2} \\ + &= \frac{1}{2} - 0 \\ + &= \frac{1}{2} + \end{align*} + \item $\displaystyle\sum_{n=1}^{\infty} \frac{1}{1+2+3+\dots+n}$ + \begin{align*} + \sum_{n=1}^{\infty} \frac{1}{1+2+3+\dots+n} &= \sum_{n=1}^{\infty} \frac{1}{\sum\limits_{i=1}^{n} i} \\ + &= \sum_{n=1}^{\infty} \frac{2}{n(n+1)} \\ + &\Downarrow \\ + \frac{2}{n(n+1)} &= \frac{A}{n} + \frac{B}{n+1} \\ + 2&=An+A+Bn + \end{align*} + \[\begin{cases} + An+Bn=0 \\ + Bn=2 + \end{cases}=\begin{cases} + A=2 \\ + B=-2 + \end{cases}\] + \begin{align*} + \sum_{n=1}^{\infty} \frac{2}{n(n+1)} &= \sum_{n=1}^{\infty} \frac{2}{n}-\frac{2}{n+1} \\ + &= \left(\frac{2}{1}-\cancel{\frac{2}{2}}\right) + \left(\cancel{\frac{2}{2}}-\cancel{\frac{2}{3}}\right) + \left(\cancel{\frac{2}{3}}-\cancel{\frac{2}{4}}\right) + \\ + &\dots + \left(\cancel{\frac{2}{n-1}}-\cancel{\frac{2}{n}}\right) + \left(\cancel{\frac{2}{n}}-\frac{2}{n+1}\right) \\ + &= \limx{n}{\infty} 2-\frac{2}{n+1} \\ + &= 2-0 \\ + &= 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 Prove that each of the following series diverges. + \begin{enumerate} + \item $\displaystyle\sum_{n=1}^{\infty} \frac{n}{2n+1}$ + \begin{proof} + Recall \textit{Theorem 3.7.1 -- The $n^{\text{th}}$-Term Test}: + \begin{theorem*}[\textbf{The $n$th Term Test}] + If the series $\sum x_n$ converges, then $\lim (x_n) = 0$. + \end{theorem*} + Let $a_n$ be the sequence whose terms are obtained by $a_n:=\frac{n}{2n+1}$, for $n \in \N$. Then we have + \[\limx{n}{\infty} a_n = \limx{n}{\infty} \frac{n}{2n+1} = \limx{n}{\infty} \frac{1}{2} \text{ by L'Hospital's Rule} = \frac{1}{2} \neq 0\] + Thus since $\limx{n}{\infty} a_n \neq 0$, we know that by \textit{Theorem 3.7.1}, $\displaystyle\sum_{n=1}^{\infty}$ is divergent. + \end{proof} - \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 $\displaystyle\sum_{n=1}^{\infty} \cos \frac{1}{n^2}$ + \begin{proof} + Let $a_n$ be the sequence whose terms are obtained by $a_n:= \cos \left(\frac{1}{n^2}\right)$, for $n \in \N$. Then we have + \[\limx{n}{\infty} a_n = \limx{n}{\infty} \cos \left(\frac{1}{n^2}\right) = \cos (0) = 1\] + By \textit{The $n$th Term Test}, since $\limx{n}{\infty} a_n \neq 0$, the series $\displaystyle\sum_{n=1}^{\infty} \cos \left(\frac{1}{n^2}\right)$ is divergent. + \end{proof} - \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, + \item $\displaystyle\sum_{n=1}^{\infty} n \sin \frac{1}{n}$ + \begin{proof} + Let $a_n$ be the sequence whose terms are obtained by $a_n:=n\sin\left(\frac{1}{n}\right)$, for $n \in \N$. Then we have + \[\limx{n}{\infty} a_n = \limx{n}{\infty} n \sin \left(\frac{1}{n}\right)=\limx{n}{\infty} \frac{\sin\left(\frac{1}{n}\right)}{\frac{1}{n}}=\limx{n}{\infty} \frac{\frac{-\cos\left(\frac{1}{n}\right)}{n^2}}{-\frac{1}{n^2}}=\limx{n}{\infty} \cos\left(\frac{1}{n}\right) = \cos(0) = 1\] + By using \textit{L'Hospital's Rule}. Thus by the \textit{$n$th Term Test}, since $\limx{n}{\infty} a_n \neq 0$, the sum $\displaystyle\sum_{n=1}^{\infty} n\sin\left(\frac{1}{n}\right)$ is divergent. + \end{proof} + \item $\displaystyle\sum_{n=1}^{\infty} \left(1-\frac{1}{n}\right)^n$ + \begin{proof} + Let $a_n$ be the sequence whose terms are obtained by $a_n:=\left(1-\frac{1}{n}\right)^n$, for $n \in \N$. Then, we have + \begin{align*} + \limx{n}{\infty} a_n &= \limx{n}{\infty} \left(1-\frac{1}{n}\right)^n \\ + &= \limx{n}{\infty} e^{\ln \left(1-\frac{1}{n}\right)^n} \\ + &= \limx{n}{\infty} \exp\left\{\ln\left(1-\frac{1}{n}\right)^n\right\} \\ + &= \limx{n}{\infty} \exp \left\{n\ln\left(1-\frac{1}{n}\right)\right\} \\ + &= \limx{n}{\infty} \exp\left\{\frac{\ln\left(1-\frac{1}{n}\right)}{\frac{1}{n}}\right\} \\ + &= \limx{n}{\infty} \exp \left\{\frac{\frac{1}{1-\frac{1}{n}}\cdot \frac{1}{n^2}}{-\frac{1}{n^2}}\right\} \\ + &= \limx{n}{\infty} \exp \left\{\frac{\frac{1}{n^2-n}}{-\frac{1}{n^2}}\right\} \\ + &= \limx{n}{\infty} \exp\left\{-\frac{n^2}{n^2-n}\right\} \\ + &= \limx{n}{\infty} \exp \left\{-\frac{n}{n-1}\right\} \\ + &= \limx{n}{\infty} \exp \left\{-\frac{1}{1-\frac{1}{n}}\right\} \\ + &= \exp \left\{-\frac{1}{1-0}\right\} \\ + &= \exp (-1) \\ + &= e^{-1} \\ + &= \frac{1}{e} + \end{align*} + By using \textit{L'Hospital's Rule}. Thus by the \textit{$n$th Term Test}, since $\limx{n}{\infty} a_n \neq 0$, we have that the sum $\displaystyle\sum_{n=1}^{\infty} \left(1-\frac{1}{n}\right)^n$ diverges. + \end{proof} + \end{enumerate} + \item + \begin{enumerate} + \item Give an example of two series $\sum a_k$ and $\sum b_k$ that differ in the first five terms, yet converge to the same value. + \\\\Consider $\displaystyle\sum_{n=1}^{\infty} \left(\frac{1}{2}\right)^n$. This series converges to 2. Also notice that $1+2+3+4-10\displaystyle\sum_{n=1}^{\infty} \left(\frac{1}{2}\right)^n = 2$. Thus the first five terms are different but converge to the same value.\\ + \item Give an example of two series $\sum a_k$ and $\sum b_k$ that differ in infinitely many terms, yet converge to the same value. + \\\\Consider the sums + \[\sum_{n=0}^{\infty} \frac{15}{32} \left(\frac{1}{16}\right)^n\ \text{ and }\ \sum_{n=2}^{\infty} \frac{1}{n(n+1)}\] + Let $a_n$ and $b_n$ be the sequences whose terms are obtained by $a_n:=\frac{15}{32}\left(\frac{1}{16}\right)^n$, for $n =0,1,2,3,\dots$, and let $b_n:=\frac{1}{n(n+1)}$, for $n \in \N$. Then we have + \[a_n:=\left(\frac{15}{32}, \frac{15}{512}, \frac{15}{8192}, \frac{15}{131072}, \frac{15}{2097152}, \dots\right)\] + and + \[b_n:=\left(\frac{1}{6}, \frac{1}{12}, \frac{1}{20}, \frac{1}{30}, \frac{1}{42},\dots\right)\] + It is clear that since the numerator of each term of $a_n$ is always $15$, and that the numerator of $b_n$ is always $1$, these two sequences are different at every term and thus differ in infinitely many terms, and thus the terms of the sums $\sum a_n$ and $\sum b_n$ also differ in infinitely many terms. Thus the first five terms of $\sum a_n$ are + \[\frac{15}{32},\ \frac{255}{512},\ \frac{4095}{8192},\ \frac{65535}{131072},\ \frac{1048575}{2097152},\ \dots\] + and the first five terms of $\sum b_n$ are + \[\frac{1}{6},\ \frac{1}{4},\ \frac{3}{10},\ \frac{1}{3},\ \frac{5}{14},\ \dots\] + However, notice that $\sum a_n$ and $\sum b_n$ converge to the same value: \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} + \sum_{n=0}^{\infty} \frac{15}{32} \left(\frac{1}{16}\right)^n &= \frac{\frac{15}{32}}{1-\frac{1}{16}} \\ + &= \frac{\frac{15}{32}}{\frac{15}{16}} \\ + &= \frac{15 \cdot 16}{15 \cdot 32} \\ + &= \frac{240}{480} \\ + &= \frac{1}{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, + and \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} + \sum_{n=2}^{\infty} \frac{1}{n(n+1)} &= \sum_{n=2}^{\infty} \frac{A}{n} + \frac{B}{n+1} \\ + 1&=An+Bn+A \\ + \begin{cases} + An+Bn=0 \\ + A=1 + \end{cases} &\implies \begin{cases} + A=1 \\ + B=-1 + \end{cases} \\ + &\Downarrow \\ + \sum_{n=2}^{\infty} \frac{1}{n(n+1)}&= \sum_{n=2}^{\infty}\frac{1}{n}-\frac{1}{n+1} \\ + &= \left(\frac{1}{2}-\cancel{\frac{1}{3}}\right) + \left(\cancel{\frac{1}{3}}-\cancel{\frac{1}{4}}\right) + \left(\cancel{\frac{1}{4}}-\cancel{\frac{1}{5}}\right) + \\ + &\dots + \left(\cancel{\frac{1}{n-1}}-\cancel{\frac{1}{n}}\right) + \left(\cancel{\frac{1}{n}}-\frac{1}{n+1}\right) \\ + &= \limx{n}{\infty} \frac{1}{2} - \frac{1}{n+1} \\ + &= \frac{1}{2}-0 \\ + &= \frac{1}{2} \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} + Thus we have that + \[\sum_{n=0}^{\infty} \frac{15}{32} \left(\frac{1}{16}\right)^n = \sum_{n=2}^{\infty} \frac{1}{n(n+1)} = \frac{1}{2}\] + \item Give an example of two series $\sum a_k$ and $\sum b_k$ that converge to real numbers $A$ and $B$, respectively, but the series $\sum a_kb_k$ converges to a value different from $AB$. + \\\\Consider the series $\displaystyle\sum_{n=0}^{\infty} \left(\frac{1}{2}\right)^n$ and $\displaystyle\sum_{n=0}^{\infty} \left(\frac{1}{3}\right)^n$. Then we have + \[\sum_{n=0}^{\infty} \left(\frac{1}{2}\right)^n=\frac{1}{1-\frac{1}{2}}=\frac{2}{1}=2\] + and + \[\sum_{n=0}^{\infty} \left(\frac{1}{3}\right)^n = \frac{1}{1-\frac{1}{3}} = \frac{3}{2}\] + Note that $2 \cdot \frac{3}{2}=\frac{6}{2} = 3$. But the series + \[\sum_{n=0}^{\infty} \left(\frac{1}{2}\right)^n\left(\frac{1}{3}\right)^n = \sum_{n=0}^{\infty} \left(\frac{1}{2} \cdot \frac{1}{3}\right)^n = \sum_{n=0}^{\infty} \left(\frac{1}{6}\right)^n = \frac{1}{1-\frac{1}{6}}=\frac{6}{5}\] + Thus we have that the product of the sums, 3, is not equal to the sum of the products, $\frac{6}{5}$.\\ + \item Give an example of a series that diverges and whose sequence of partial sums is bounded. + \\\\Consider an alternating series, $\displaystyle\sum_{n=1}^{\infty} (-1)^n$. Then, note that $S_1:=-1,\ S_2:=-1+1=0,\ S_3:=-1+1-1=-1, S_4:=-1+1-1+1=0,\dots$. Then we have that the sequence of partial sums is bounded below by $-1$ and is bounded above by $1$. However, since this is an alternating series, we know that by the \textit{Geometric Series Test}, since $|r| = |-1|=1 \nless 1$, this series is divergent. + \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 $0 \leq A < 1/2$. - \\\\\textbf{TODO} - - \item Show $b_n = \frac{2 \cdot 4 \cdot 6 \dots (2n)}{3 \cdot 5 \cdot 7 \dots (2n+1)}$ converges to $B$ where $0 \leq B < 2/3$. - \\\\\textbf{TODO} + \item Prove or justify, if true. Provide a counterexample, if false. + \begin{enumerate} + \item If $a_n$ is strictly decreasing and $\limx{n}{\infty} a_n=0$, then $\sum a_n$ converges. + \\\\This is a false statement. Consider the series $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n}$. Then, we have that $\limx{n}{\infty} \frac{1}{n} =0$, however since this is a harmonic series, we know that it is divergent, thus $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n} = \infty$ is divergent.\\ + + \item If $a_n \neq b_n$ for all $n \in \N$ and if $\sum (a_n+b_n)$ converges, then either $\sum a_n$ converges or $\sum b_n$ converges. + \\\\This is a false statement. Consider the series $\displaystyle\sum_{n=1}^{\infty} (-1)^n$ and $\displaystyle\sum_{n=1}^{\infty} (-1)^{n+1}$. Then we have that $\displaystyle\sum_{n=1}^{\infty} (-1)^n = -1+1-1+1-1+\dots$, which is divergent by the \textit{Geometric Series Test}, and $\displaystyle\sum_{n=1}^{\infty} (-1)^{n+1} = 1-1+1-1+1-1+\dots$, which is also divergent by the \textit{Geometric Series Test}. However, $\displaystyle\sum_{n=1}^{\infty} (-1)^n+(-1)^{n+1} = 0+0+0+0+\dots = 0$ and thus converges. However, $a_n \neq b_n$ for all $n \in \N$ since + $a_n:= -1,1,-1,1,-1,1,\dots$ and $b_n:=1,-1,1,-1,1,-1,\dots$. Thus for all $n \in \N$, either $a_n=-1 \neq 1 = b_n$, or $a_n=1\neq -1=b_n$. Thus we have that $\sum (a_n+b_n)$ converges but neither $\sum a_n$ nor $\sum b_n$ converge.\\ + + \item Suppose $\sum (a_n+b_n)$ converges. Then $\sum a_n$ converges if and only if $\sum b_n$ converges. + \\\\This is a true statement since if $\sum (a_n+b_n)$ converges, then both $a_n+b_n$ converges, as was covered in our notes.\\ + + \item If $\limx{n}{\infty} a_n=A$, then $\displaystyle\sum_{n=1}^{\infty} (a_n-a_{n+2})=a_1+a_2-2A$. + \begin{proof} + Notice that we can rewrite the sum $\displaystyle\sum_{n=1}^{\infty} (a_n-a_{n+2})$ as $\displaystyle\sum_{n=1}^{\infty} \left((a_n-a_{n+1})+(a_{n+1}-a_{n+2})\right)$. Now, we have that the $n$th partial sum yields a telescoping series: + \[S_n:=[(a_1-\cancel{a_2})+(a_2-\cancel{a_3})]+ [(\cancel{a_2}-\cancel{a_3})+(\cancel{a_3}-\cancel{a_4})] + \dots + [(\cancel{a_n}-a_{n+1})+(\cancel{a_{n+1}}-a_{n+2})]\] + So $S_n=a_1+a_2-a_{n+1}-a_{n+2}$, which yields $\limx{n}{\infty} =a_1+a_2-A-A=a_1+a_2-2A$. + \end{proof} + + \item $\sum a_n$ converges if and only if $\limx{n}{\infty} a_n=0$. + \\\\This is a false statement. Consider the series $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n}$. This is the harmonic series, which we know diverges. However, $\limx{n}{\infty} \frac{1}{n} = 0$. Thus the limit is equal to 0, but the series does not converge.\\ + + \item Changing the first few terms in a series may affect the value of the sum of the series. + \begin{proof} + Suppose $x_n \to x$. Then for all $\varepsilon >0$, there exists some $N \in \N$ such that if $n \geq N$, then $|x_n-x| < \varepsilon$. Now, suppose $x'_n$ is a sequence such that for $n \geq M$, then $x'_n=x_n$. + \\\\Let $\varepsilon>0$ be given. Then there exists some $N \in \N$ such that for all $n \geq N$, $|x_n-x|<\varepsilon$. Let $N'=\max (N,M)$. Then if $n \geq N'$, we have that $|x'_n-x|<\varepsilon$. Hence $x'_n \to x$. + \\\\Consider a convergent series $\sum x_n$. If we let $s_n=x_1+x_2+\dots +x_n$, then we have that $s_n \to s$. + \\\\Consider the series $\sum x'_n$, where for $n \geq M$, then $x'_n=x_n$. Let $s'_n=x'_1+x'_2+\dots+x'_n$. Note that for $n \geq M$, we have $s'_n-s'_{M-1}=x'_M+\dots+x'_n=x_M+\dots + x_n$, and thus $s'_n-s'_{M-1}=s_n-s_{M-1}\to s-s_{M-1}$. Hence $s'_n \to (s-s_{M-1}+s'_{M-1})$. + \end{proof} + + \item Changing the first few terms in a series may affect whether or not the series converges. + \\\\This is a false statement. Consider the telescoping series used previously, given by $\displaystyle\sum_{n=2}^{\infty} \frac{1}{n(n+1)}$. We know this series converges to $\frac{1}{2}$. Consider changing the first few terms as follows: + \[\left(\frac{1}{2}-\cancel{\frac{1}{3}}\right) + \left(\cancel{\frac{1}{3}}-\cancel{\frac{1}{4}}\right) + \left(\cancel{\frac{1}{4}}-\cancel{\frac{1}{5}}\right) +\dots \to \left(1-\cancel{\frac{1}{2}}\right) + \left(\cancel{\frac{1}{2}}-\cancel{\frac{1}{4}}\right) + \left(\cancel{\frac{1}{4}}-\cancel{\frac{1}{5}}\right) + \dots\] + Now we have that this series converges to 1, not $\frac{1}{2}$. However, despite changing the first few terms of the series, we did not change whether or not it converges. This is because the first few terms of a series can only finitely affect the sum. Thus, if a series converges, a finite change to the terms will still create a finite sum. Likewise, if the series diverges, a finite change will not allow the series to converge to a finite sum.\\ + + \item If $\sum a_n$ converges and $\limx{n}{\infty} \frac{a_n}{b_n}=0$, then $\sum b_n$ converges. + \\\\This is a false statement. Consider the series $\displaystyle\sum_{n=1}^{\infty} \left(\frac{1}{2}\right)^n$ and $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n}$, which we note is the harmonic series. Then we have that $\displaystyle\sum_{n=1}^{\infty} \left(\frac{1}{2}\right)^n = \frac{1}{1-\frac{1}{2}} = 2$ since it is a geometric series. Thus $\sum a_n$ converges. Also, notice that + \[\limx{n}{\infty} \left(\frac{\left(\frac{1}{2}\right)^n}{\frac{1}{n}}\right)=0\] + However, since $\sum b_n = \displaystyle\sum_{n=1}^{\infty} \frac{1}{n}$, which is the harmonic series, we know that it is divergent, and thus if $\sum a_n$ converges and $\limx{n}{\infty} \frac{a_n}{b_n}=0$, $\sum b_n$ can still be divergent. + \end{enumerate} \end{enumerate} - - \item \textbf{Section 3.4} - \begin{enumerate} - \item[1)] Give an example of an unbounded sequence that has a convergent subsequence. - \\\\Let $a_n:=(-1)^n$. This sequence diverges since it oscillates between 1 and $-1$, however if we define $b_n:=(a_{2n})$; that is, let $(b_n)$ be the sequence of all terms of $(a_n)$ such that $n$ is even. Thus, $(b_n)$ is a subsequence of $(a_n)$, but $(b_n)$ converges since $a_2=1, a_4=1, \dots, a_{2n}=1$. Thus while $(a_n)$ is a divergent sequence, the subsequence $(b_n)$ of $(a_n)$ converges for all $n \in \N$.\\ - - \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*} - If $(x_n)$ is a bounded sequence of real numbers, then the following statements for a real number $x^*$ are equivalent: - \begin{enumerate} - \item $x^*= \lim \sup (x_n)$. - - \item If $\varepsilon > 0$, there are at most a finite number of $n \in \N$ such that $x^*+\varepsilon 1.99$. - - \item Compute the infimum, supremum, limit infimum, and limit supremum for $a_n = 3 - (-1)^n - (-1)^n/n$. - \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. - - \item If $a_n$ and $b_n$ are strictly increasing, then $a_n \cdot b_n$ is strictly increasing. - - \item If $a_n$ and $b_n$ are monotonic, then $a_n + b_n$ is monotonic. - - \item If $a_n$ and $b_n$ are monotonic, then $a_n \cdot b_n$ is monotonic. - - \item If a monotone sequence is bounded, then it is convergent. - - \item If a bounded sequence is monotone, then it is convergent. - - \item If a convergent sequence is monotone, then it is bounded. - - \item If a convergent sequence is bounded, then it is monotone. - \end{enumerate} - \end{enumerate} -\end{document} \ No newline at end of file +\end{document} diff --git a/Documents/LaTeX/Homework 6.pdf b/Documents/LaTeX/Homework 6.pdf new file mode 100644 index 0000000..274f404 Binary files /dev/null and b/Documents/LaTeX/Homework 6.pdf differ diff --git a/Documents/LaTeX/Homework 6.tex b/Documents/LaTeX/Homework 6.tex new file mode 100644 index 0000000..377dc2d --- /dev/null +++ b/Documents/LaTeX/Homework 6.tex @@ -0,0 +1,307 @@ +\documentclass[12pt,letterpaper]{article} +\usepackage[utf8]{inputenc} +\usepackage{pgfplots} +\usepackage[english]{babel} +\usepackage{amsthm} +\usepackage{cancel} +\usepackage{mathtools} +\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} +\newcommand{\limx}[2]{\displaystyle\lim\limits_{#1 \to #2}} +\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{\dotp}{\dot{\mathcal{P}}} +\newcommand{\dotq}{\dot{\mathcal{Q}}} +\newcommand{\dist}{\text{dist}} +\DeclareMathOperator{\sign}{sgn} +\newtheoremstyle{case}{}{}{}{}{}{:}{ }{} +\theoremstyle{case} +\newtheorem{case}{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*{lemma*}{Lemma} +\newtheorem*{remark}{Remark} +\setlist[enumerate]{font=\bfseries} +\renewcommand{\qedsymbol}{$\blacksquare$} +\author{Alexander J. Tusa} +\title{Real Analysis II Homework 6} +\begin{document} + \maketitle + \begin{enumerate} + \item \textbf{Section 3.7} + \begin{enumerate} + \item[6.] + \begin{enumerate} + \item Calculate the value of $\displaystyle\sum_{n=2}^{\infty} \left(\frac{2}{7}\right)^n$. (Note the series starts at $n=2$) + + \begin{align*} + \sum_{n=2}^{\infty} \left(\frac{2}{7}\right)^n &= \sum_{n=1}^{\infty} \left(\frac{2}{7}\right)^{n+2} \\ + &= \sum_{n=1}^{\infty} \left(\frac{2}{7}\right)^n-\frac{2}{7} - 1 \\ + &= \frac{1}{1-\frac{2}{7}} -\frac{2}{7} -1\\ + &= \frac{1}{\frac{5}{7}} - \frac{2}{7} -1\\ + &= \frac{7}{5} - \frac{2}{7} -1\\ + &= \frac{4}{35} + \end{align*} + + \item Calculate the value of $\displaystyle\sum_{n=1}^{\infty} \left(\frac{1}{3}\right)^{2n}$. (Note the series starts at $n=1$) + + \begin{align*} + \sum_{n=1}^{\infty} \left(\frac{1}{3}\right)^{2n} &= \sum_{n=1}^{\infty} \left(\frac{1}{9}\right)^n \\ + &= \sum_{n=1}^{\infty}\left(\frac{1}{9}\right)^n-1 \\ + &= \frac{1}{1-\frac{1}{9}}-1 \\ + &= \frac{1}{\frac{8}{9}} - 1 \\ + &= \frac{9}{8} - 1 \\ + &= \frac{1}{8} + \end{align*} + \end{enumerate} + \item[7.] Find a formula for the series $\displaystyle\sum_{n=1}^{\infty} r^{2n}$ when $|r| < 1$. + \\\\Let $S_n = r^2+r^4+r^6+\dots+r^{2(n-1)}+r^{2n}$. Then, we notice that $S_n=(r^2)^1+(r^2)^2+(r^2)^3+\dots + (r^2)^{n-1} + (r^2)^n$ which in turn yields $S_n=r^2[1+r^2+(r^2)^2+(r^2)^3+\dots+(r^2)^{n-1}]$ yielding $S_n=r^2 \cdot \frac{1-(r^2)^n}{1-r^2}=\frac{r^2}{1-r^2} \cdot (1-(r^2)^n)$. Since $|r|<1$, we know that $|r^2|<1$, and thus $\limx{n}{\infty} (r^2)^n=0$. So $\lim S_n=\frac{r^2}{1-r^2} \cdot (1-0)=\frac{r^2}{1-r^2}$. Thus $\displaystyle\sum_{n=1}^{\infty} r^{2n}$ converges and is equal to $\frac{r^2}{1-r^2}$. + \item[9.] + \begin{enumerate} + \item Show that the series $\displaystyle\sum_{n=1}^{\infty} \cos n$ is divergent. + \\\\We note that if the series converges, then $\limx{n}{\infty} \cos n = 0$. Consider the subsequence $n_k=2k\pi$, for some $k \in \N$. Then $\limx{k}{\infty} \cos (2k\pi)=1$. And for the subsequence $n_k=\frac{\pi}{2}+2k\pi$, for some $k \in \N$, then we have that $\limx{k}{\infty} \cos \left(\frac{\pi}{2}+2k\pi\right)=0$. And since $0 \neq 1$, we can conclude that $\cos n$ does not converge and that $\limx{n}{\infty} \cos n \neq 0$. Therefore, by the \textit{nth Term Test}, $\displaystyle\sum_{n=1}^{\infty} \cos n$ diverges. + + \item Show that the series $\displaystyle\sum_{n=1}^{\infty} \frac{\cos n}{n^2}$ is convergent. + \\\\We note that $\abs{\frac{\cos n}{n^2}} \leq \frac{1}{n^2}$, and since $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^2}$ is a convergent $p$-series, we have that $\displaystyle\sum_{n=1}^{\infty} \frac{\cos n}{n^2}$ is also convergent. + \end{enumerate} + \item[11.] If $\sum a_n$ with $a_n>0$ is convergent, then is $\sum a^2_n$ always convergent? Either prove it or give a counterexample. + \begin{proof} + Since $\sum a_n$ is convergent, we know that by the \textit{nth Term Test}, $\lim a_n=0$. This yields that for $\varepsilon=1>0$, there exists $N \in \N$ such that for all $n \geq N$, we have + \begin{align*} + |a_n-0| &<\varepsilon \\ + |a_n|&<\varepsilon \\ + |a_n|&<1 + \end{align*} + \[0 0,\ \forall n$. Since we know that $x^20$ is convergent, then is $\sum \sqrt{a_n}$ always convergent? Either prove it or give a counterexample. + \\\\This is a false statement. Consider $\sum a_n = \sum \frac{1}{n^2}$. Then we have that $\sum \sqrt{a_n} = \sum \frac{1}{n}$, which is a harmonic series, which we know diverges. Thus if $\sum a_n$ converges, then $\sum \sqrt{a_n}$ does not necessarily converge.\\ + + \item[13.] If $\sum a_n$ with $a_n>0$ is convergent, then is $\sum \sqrt{a_na_{n+1}}$ always convergent? Either prove it or give a counterexample. + \begin{proof} + We first notice that + \[\frac{a_n+a_{n+1}}{2} > \sqrt{a_na_{n+1}}\] + So, we then have + \[\frac{a_1}{2}+\sum_{n=1}^{\infty} \left(\frac{a_n+a_{n+1}}{2}\right)=\frac{a_1}{2}+\frac{a_1}{2}+\frac{a_2}{2}+\frac{a_2}{2} + \frac{a_3}{2} + \frac{a_3}{2} + \dots = a_1+a_2+a_3+\dots = \sum_{n=1}^{\infty}a_n\] + thus, since $\frac{a_1}{2}+ \displaystyle\sum_{n=1}^{\infty} \left(\frac{a_n+a_{n+1}}{2}\right)=\displaystyle\sum_{n=1}^{\infty}a_n$, and since we're given that $\sum a_n$ converges, then $\displaystyle\sum_{n=1}^{\infty} \left(\frac{a_n+a_{n+1}}{2}\right)$ converges. Thus by the \textit{Comparison Test}, we have that + \[a_n > \left(\frac{a_n+a_{n+1}}{2}\right)>\sqrt{a_na_{n+1}}\] + yields that $\displaystyle\sum_{n=1}^{\infty} \sqrt{a_na_{n+1}}$ also converges. + \end{proof} + + \item[16.] Use the \textit{Cauchy Condensation Test} to discuss the $p$-series $\displaystyle\sum_{n=1}^{\infty} (1/n^p)$ for $p>0$. + \\\\By the \textit{Cauchy Condensation Test}, we must show that the series $\displaystyle\sum_{n=0}^{\infty}2^n\cdot a(2^n)$ converges. So, + \begin{align*} + \sum_{n=0}^{\infty} 2^na(2^n) ^= \sum_{n=0}^{\infty} \left[2^n\frac{1}{(2^n)^p}\right] \\ + &= \sum_{n=0}^{\infty} 2^n \left[\frac{1}{2^{np}}\right] \\ + &= \sum_{n=0}^{\infty} 2^n (2^{-np}) \\ + &= \sum_{n=0}^{\infty} 2^{n(1-p)} \\ + &= \sum_{n=0}^{\infty} (2^{1-p})^n + \end{align*} + We notice that this is now a geometric series, with $|r|=|2^{1-p}|$. Then we note that if $|2^{1-p}| \leq 1$, the series converges, and otherwise the series diverges.\\\\ + + \item[17.] Use the \textit{Cauchy Condensation Test} to establish the divergence of the series: + \begin{enumerate} + \item $\displaystyle\sum \frac{1}{n \ln n}$ + \\\\By the \textit{Cauchy Condensation Test}, we have + \begin{align*} + \sum \frac{1}{n \ln n} &= \sum 2^n \cdot \frac{1}{2^n \cdot \ln 2^n} \\ + &= \sum \frac{1}{\ln 2^n} \\ + &= \sum \frac{1}{n \ln 2} \\ + &= \frac{1}{\ln 2} \sum \frac{1}{n} + \end{align*} + Since $\sum \frac{1}{n}$ is a harmonic series, we know that the series diverges. Thus we have that $\sum 2^n \cdot \frac{1}{2^n \cdot \ln 2^n}$ diverges as well. Thus by the \textit{Cauchy Condensation Test}, we have that $\sum \frac{1}{n \ln n}$ diverges also.\\ + + \item $\displaystyle\sum \frac{1}{n(\ln n)(\ln\ln n)}$ + \\\\By the \textit{Cauchy Condensation Test}, we have + \begin{align*} + \sum \frac{1}{n (\ln n)(\ln \ln n)} &= \sum 2^n \frac{1}{2^n\cdot (ln 2^n) \cdot (\ln (\ln 2^n))} \\ + &= \sum \frac{1}{\ln 2^n \cdot \ln \ln 2^n} \\ + &= \sum \frac{1}{(n \ln 2) \cdot (\ln (n \ln 2))} \\ + &= \sum \frac{1}{(n \ln 2) \cdot (\ln n+\ln\ln 2)} \\ + &> \sum \frac{1}{n \cdot (\ln 2 + \ln \ln 2} &(\ln 2 < 1) \\ + &> \sum \frac{1}{n \cdot \ln n} + \end{align*} + Since we showed in part (a) that $\sum 2^n \cdot \frac{1}{n \ln n}$ diverges, and thus by the \textit{Comparison Test}, since $0 \leq \frac{1}{n(\ln n)(\ln\ln n)} < \frac{1}{n \ln n}$, since $\sum \frac{1}{n \ln n}$ diverges, then so does $\sum \frac{1}{(\ln 2^n)(\ln\ln 2^n)}$.\\ + \end{enumerate} + \end{enumerate} + + \item Use the tests in section 3.7 to test the given series for convergence or divergence. State clearly which test is used. Also, for parts a-f, write out the first three terms of each series. + \begin{enumerate} + \item $\displaystyle\sum_{n=1}^{\infty} \frac{2n+5}{3n^2+2n-1}$ + \\\\First three terms: + \[\sum_{n=1}^{\infty} \frac{2n+5}{3n^2+2n-1} = \frac{7}{4}+\frac{3}{5}+\frac{11}{32}+\dots\] + Note that $2n<2n+5$ and that $3n^2+2n-1<4n^2 \implies \frac{1}{3n^2+2n-1} > \frac{1}{4n^2}$, and thus $\frac{2n}{4n^2} = \frac{1}{2n} < \frac{2n+5}{3n^2+2n-1}$. Since $\sum \frac{1}{2n} = \frac{1}{2} \sum \frac{1}{n}$ yields a half times the harmonic series, which we know is divergent, we have that by the \textit{Comparison Test}, $\sum \frac{2n+5}{3n^2+2n-1}$ also diverges.\\ + \item $\displaystyle\sum_{n=1}^{\infty} \frac{n-1}{n2^n}$ + \\\\The first three terms: + \[\sum_{n=1}^{\infty} \frac{n-1}{n2^n} = 0+ \frac{1}{8} + \frac{1}{12}+\dots\] + We note that $\displaystyle\sum_{n=1}^{\infty} \frac{n-1}{n2^n}$ looks similar to $\displaystyle\sum_{n=1}^{\infty} \left(\frac{1}{2}\right)^n$, which is a geometric series. We also note that the series $\displaystyle\sum_{n=1}^{\infty} \left(\frac{1}{2}\right)^n$ converges since $\abs{\frac{1}{2}} < 1$. Then, by the \textit{Limit Comparison Test}, we have $\limx{n}{\infty} \left(\frac{\frac{n-1}{n2^n}}{\left(\frac{1}{2}\right)^n}\right)=1 \neq 0$, and thus since $\displaystyle\sum_{n=1}^{\infty} \left(\frac{1}{2}\right)^n$ converges, then $\displaystyle\sum_{n=1}^{\infty} \frac{n-1}{n2^n}$ must also converge.\\ + + \item $\displaystyle\sum_{n=1}^{\infty} \frac{\sqrt{n} + \pi}{2+\sqrt[5]{n^8}}$ + \\\\The first three terms of this series are: + \[\sum_{n=1}^{\infty} \frac{\sqrt{n} + \pi}{2+\sqrt[5]{n^8}} = \frac{\pi + 1}{3} + \frac{\pi + \sqrt{2}}{2+2 \sqrt[5]{8}}+\frac{\pi + \sqrt{3}}{2+3\sqrt[5]{27}}+ \dots\] + We note that $\sqrt{n}+\pi < \sqrt{n}$ and that $2+\sqrt[5]{n^8} > n^{\frac{7}{4}} \implies \frac{1}{2+\sqrt[5]{n^8}} < \frac{1}{n^{\frac{7}{4}}}$. Thus we have that + \[\frac{\sqrt{n}+\pi}{2+\sqrt[5]{n^8}}< \frac{\sqrt{n}}{n^{\frac{7}{4}}}=\frac{1}{n^\frac{5}{4}}\] + We note that $\sum \frac{1}{n^\frac{5}{4}}$ converges because it is a $p$-series where $p > 1$. Thus by the \textit{Comparison Test}, we have that $\displaystyle\sum_{n=1}^{\infty} \frac{\sqrt{n}+\pi}{2+\sqrt[5]{n^8}}$ must also converge.\\ + \item $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^{\ln n}}$ + \\\\The first three terms are: + \[\sum_{n=1}^{\infty} \frac{1}{n^{\ln n}}=1+\frac{1}{e^{(\ln 2)^2}}+\frac{1}{e^{(\ln 3)^2}}+\dots\] + We note that $\frac{1}{n^{\ln n}}$ looks like $\frac{1}{n^{\ln 3}}$ and that $0 < \displaystyle\frac{1}{n^{\ln n}} \leq \frac{1}{n^{\ln 3}}$. By the \textit{Comparison Test}, since $\displaystyle\frac{1}{n^{\ln 3}}$ is a convergent $p$-series since $\ln 3 \geq 1$, $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^{\ln n}}$ must also be convergent. \\ + + \item $\displaystyle\sum_{n=2}^{\infty} \frac{1}{\ln n}$ + \\\\The first three terms of the series are: + \[\sum_{n=2}^{\infty} \frac{1}{\ln n}=\frac{1}{\ln (2)} + \frac{1}{\ln (3)} + \frac{1}{\ln (4)} + \dots\] + We note that $n > \ln n \implies \frac{1}{n} < \frac{1}{\ln n}$. Then by the \textit{Comparison Test}, since $\displaystyle\sum_{n=2}^{\infty} \frac{1}{n}$ is the Harmonic series, we know that it diverges, and thus by the \textit{Comparison Test}, since $\sum \frac{1}{n} \leq \frac{1}{\ln n}$, we have that $\displaystyle\sum_{n=2}^{\infty} \frac{1}{\ln n}$ must also diverge.\\ + \item $\displaystyle\sum_{n=2}^{\infty} \frac{n+\ln n}{n^2+1}$ + \\\\The first three terms of the series are: + \[\sum_{n=2}^{\infty} \frac{n+\ln n}{n^2+1}=\frac{\ln(2)+2}{5}+\frac{\ln(3)+3}{10} + \frac{\ln(4)+4}{17}+\dots\] + We notice that $\displaystyle\frac{n+\ln n}{n^2+1}$ looks like $\displaystyle\frac{1}{n}$. So, by the \textit{Limit Comparison Test}, we have + \begin{align*} + \limx{n}{\infty} \frac{\frac{n+\ln n}{n^2+1}}{\frac{1}{n}} &= \limx{n}{\infty} \frac{n^2+n\ln n}{n^2 +1} \\ + &= \limx{n}{\infty} \frac{1+\frac{\ln n}{n}}{1+\frac{1}{n^2}} \\ + &= \frac{1+0}{1+0} \\ + &= \frac{1}{1} \\ + &= 1 + \end{align*} + Thus by the \textit{Limit Comparison Test}, since $\displaystyle\sum_{n=2}^{\infty} \frac{1}{n}$ diverges, we have that $\displaystyle\sum_{n=2}^{\infty} \frac{n+\ln n}{n^2 + 1}$ diverges. \\ + + \item $\displaystyle\sum_{n=1}^{\infty} \frac{n+1}{n^2}$ + \\\\We note that $n < n+1$ and that $n^2 > \frac{n^2}{2} \implies \frac{1}{n^2}<\frac{2}{n^2}$ and gives us that + \[\frac{2n}{n^2}=\frac{2}{n}<\frac{n+1}{n^2}\] + Since $\displaystyle 2 \sum_{n=1}^{\infty} \frac{1}{n}$ is a harmonic series, it is divergent. Thus by the \textit{Comparison Test}, we have that $\displaystyle\sum_{n=1}^{\infty} \frac{n+1}{n^2}$ is also divergent.\\ + + \item $\displaystyle\sum_{n=2}^{\infty} \frac{1}{(\ln n)^n}$ + \\\\By the \textit{Cauchy Ratio Test}, we have the following: + \[\limx{n}{\infty} \abs{\frac{\frac{1}{\ln(n+1)^{n+1}}}{\frac{1}{\ln(n)^n}}}=\limx{n}{\infty} \abs{\frac{\ln(n)^n}{\ln(n+1)^{n+1}}}=0\] + Thus by the \textit{Cauchy Ratio Test}, since $L=0<1$, we have that $\displaystyle\sum_{n=1}^{\infty}\frac{1}{\ln (n)^n}$ is a convergent series. \\ + + \item $\displaystyle\sum_{n=2}^{\infty} \frac{\ln n}{n^2}$ + \\\\Notice that $\frac{\ln n}{n^2} < \frac{1}{n^2}$. So, we must note that $\displaystyle\sum_{n=2}^{\infty} \frac{1}{n^2}$ is a convergent series since it is a $p$-series with $p=2$ and $p>0$. Thus by the \textit{Comparison Test}, we have that the series $\displaystyle\sum_{n=2}^{\infty} \frac{\ln (n)}{n^2}$ is a convergent series.\\ + + \item $\displaystyle\sum_{n=1}^{\infty} \frac{\ln n}{n}$ + \\\\We notice that $\frac{\ln n}{n}$ looks like $\frac{1}{n}$. So, by the \textit{Limit Comparison Test}, we have + \[\limx{n}{\infty}\frac{\frac{\ln n}{n}}{\frac{1}{n}}= \limx{n}{\infty} \ln n = \infty\] + Since $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n}$ is the harmonic series, we know that it diverges, and thus by the \textit{Limit Comparison Test}, we have that $\displaystyle\sum_{n=1}^{\infty} \frac{\ln n}{n}$ diverges as well.\\ + + \item $\displaystyle\sum_{n=3}^{\infty} \frac{\sqrt{n}}{\sqrt{n^3}+1}$ + \\\\Notice that $\sqrt{n^3} < 2\sqrt{n^3} \implies \frac{1}{\sqrt{n^3}+1} > \frac{1}{2\sqrt{n^3}}$, which gives us that + \[\frac{\sqrt{n}}{2\sqrt{n^3}}<\frac{\sqrt{n}}{\sqrt{n^3}+1}\] + Since the series $\displaystyle \frac{1}{2}\sum_{n=3}^{\infty} \frac{1}{n}$ is a harmonic series ,we know that it diverges, and thus we know that by the \textit{Comparison Test} we have that the series $\displaystyle\sum_{n=3}^{\infty} \frac{\sqrt{n}}{\sqrt{n^3}+1}$ is also divergent.\\ + + \item $\displaystyle\sum_{n=1}^{\infty} \frac{n}{e^n}$ + \\\\By the \textit{McClaurin Integral Test}, we have + \[\int_{1}^{\infty}\frac{x}{e^x}\ dx = \int_{1}^{\infty}xe^{-x}\ dx\] + By \textit{Integration by Parts}, let $u=x,\ dv=e^{-x}dx,\ du=dx,\ v=-e^{-x}$. Then we have + \begin{align*} + \int_{1}^{\infty} xe^{-x}\ dx &= -xe^{-x}+\int_{1}^{\infty} e^{-x}\ dx \\ + &= \left.-xe^{-x}-e^{-x}\right|_1^\infty \\ + &= \frac{-\infty}{e^\infty}-\frac{1}{e^\infty} +\frac{1}{e}+\frac{1}{e} \\ + &= \frac{2}{e} + \end{align*} + Thus, since $\displaystyle\int_{1}^{\infty} \frac{x}{e^x}\ dx$ converges to $\frac{2}{e}$, we have that $\displaystyle\sum_{n=1}^{\infty} \frac{n}{e^n}$ converges. + \end{enumerate} + + \item Write the given expressions as a quotient of two integers. + \begin{enumerate} + \item $3+1+\frac{1}{2} + \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \dots $ + \begin{align*} + &= 3+1+\frac{1}{2}+\sum_{n=1}^{\infty} \frac{1}{3^n} \\ + &= 3+1+\frac{1}{2}+\frac{\frac{1}{3}}{1-\frac{1}{3}} \\ + &= 3+1+\frac{1}{2}+\frac{\frac{1}{3}}{\frac{2}{3}} \\ + &= 3+1+\frac{1}{2}+\frac{3}{6} \\ + &= 3+1+\frac{1}{2}+\frac{1}{2} \\ + &= 3+1+1 \\ + &= 5 + \end{align*} + + \item $3.2\overline{15}$ + \begin{align*} + 3.2\overline{15} &= 3+ \frac{2}{10} + \left(\frac{15}{1000}+\frac{15}{100000}+\frac{15}{10^7}+\dots\right) \\ + &= 3+\frac{2}{10}+\frac{15}{1000}\left(1+\frac{1}{10^2}+\frac{1}{10^3}+\dots\right)\\ + &= 3+\frac{2}{10} + \frac{\frac{15}{1000}}{1-\frac{1}{100}} \\ + &= 3+\frac{2}{10} + \frac{1}{66} \\ + &= \frac{1061}{330} + \end{align*} + \end{enumerate} + \item + \begin{enumerate} + \item Let $a_n$ be a sequence of real numbers and $b_n=a_n-a_{n-1}$ for all $n \in \N$. Prove: $\sum b_k$ converges if and only if the sequence $a_n$ converges. In this case, find the sum of $\sum b_k$. + \begin{proof} + Let $(a_n)\subseteq \R$ and let $b_n:=a_n-a_{n-1}\ \forall\ n \in \N$. + \begin{itemize} + \item[$(=>)$] Assume $\sum b_k$ converges. Then by the \textit{Cauchy Criterion for Series}, we have + \[\forall\ \varepsilon>0,\ \exists\ M(\varepsilon)\in \N \st \text{if}\ m>n\geq M(\varepsilon) \implies |s_m-s_n|<\varepsilon\] + This yields that + \begin{align*} + |s_m-s_n|&= |(\cancel{a_{n+1}}-a_n)+(\cancel{a_{n+2}}-\cancel{a_{n+1}})+\dots+(\cancel{a_{m-1}}-\cancel{a_{m-2}})+(a_m-\cancel{a_{m-1}})| \\ + &= |a_m-a_n| \\ + &<\varepsilon + \end{align*} + Thus by the definition of a \textit{Cauchy Sequence}, $a_n$ is a Cauchy sequence and thus by the \textit{Cauchy Convergence Criterion}, $a_n$ is a convergent sequence. + + \item[$(<=)$] Similarly, suppose $a_n$ is a convergent sequence. Then by the \textit{Cauchy Convergence Criterion}, $a_n$ is a Cauchy sequence. So $\forall\ \varepsilon>0,\ \exists\ H(\varepsilon) \in \N \st \forall\ m>n\geq H(\varepsilon),\ n,m \in \N,\ |a_m-a_n|<\varepsilon$. So, + \begin{align*} + |a_m-a_n| &= |(a_{n+1}-a_n)+(a_{n+2}-a_{n+1})+\dots+(a_{m-1}-a_{m-2})+(a_m-a_{m-1})| \\ + &= |(s_{m-1}+a_m)-(s_{n-1}+a_n)|, \\ + &\text{by the definition of the infinite series generated by $(a_n)$}\\ + &=|s_{m-1}+a_m-s_{n-1}-a_n|\\ + &= |s_m-s_n| \\ + &<\varepsilon + \end{align*} + Thus by the \textit{Cauchy Criterion for Series}, since $|s_m-s_n|<\varepsilon$ for all $m>n\geq H(\varepsilon), H(\varepsilon),m,n\in \N$, the series $\sum a_n-a_{n-1}$ converges, which implies that the series $\sum b_n$ converges. + \end{itemize} + \end{proof} + Thus, we have that $\sum b_n=\limx{n}{\infty} (b_n)=\limx{n}{\infty} (a_n-a_{n-1})$, by \textit{Theorem 3.7.3}, and thus by the \textit{$n$th Term Test}, since $\sum b_n$ converges, we have that $\limx{n}{\infty} (a_n-a_{n-1})=0$. Thus $\sum b_n=0$.\\ + + \item Let $a_n$ be a sequence of real numbers. If $\limx{n}{\infty} a_n=A$, find the sum of \[\displaystyle\sum_{n=1}^{\infty} (a_{n+1}-2a_n+a_{n-1})\] + By the \textit{Cauchy Convergence Criterion for Series}, we have that $\forall\ \varepsilon>0,\ \exists\ M(\varepsilon) \in \N \st \text{if } m,n \in \N,\ m>n\geq M(\varepsilon) \implies |s_m-s_n|=|a_{n+1}+a_{n+2}+\dots + a_m|<\varepsilon$. So, + \begin{align*} + |s_m-s_n|&=|(\cancel{a_{n+2}}-\cancel{2}a_{n+1}+a_n)+(\cancel{a_{n+3}}-\cancel{2a_{n+2}}+\cancel{a_{n+1}}) \\ + &+(\cancel{a_{n+4}}-\cancel{2a_{n+3}}+\cancel{a_{n+2}})+(\cancel{a_{n+5}}-\cancel{2a_{n+4}}+\cancel{a_{n+3}})\\ + &+\dots\\ + &+(\cancel{a_m}-\cancel{2a_{m-1}}+\cancel{a_{m-2}})+(a_{m+1}-\cancel{2}a_m+\cancel{a_{m-1}})| \\ + &= |a_n-a_{n+1}+a_{m+1}-a_m| \\ + &<\varepsilon + \end{align*} + This yields that the sequence of partial sums of $a_{n+1}-2a_n+a_{n-1}$ is bounded. Thus by \textit{Theorem 3.7.3}, $\displaystyle\sum_{n=1}^{\infty}(a_{n+1}-2a_n+a_{n-1})=\limx{n}{\infty} (a_{n+1}-2a_n+a_{n-1}) = A-2A+A = 0$. \\ + + \item If $\displaystyle\sum_{k=1}^{n} (k\ a_k)=\frac{n+1}{n+2}$ for $n \in \N$, show that $\displaystyle\sum_{k=1}^{\infty} a_k=\frac{3}{4}$. + \begin{proof} + If $n=1$, then $\displaystyle\sum_{k=1}^{1} 1\cdot a_1 = \frac{2}{3}$. + \\\\If $n \geq 2$, then + \begin{align*} + \sum_{k=1}^{n} ka_k &= a_1+a_2+\dots+(n-1)a_{n-1}+na_n \\ + &= \frac{n+1}{n+2} \\ + &\Downarrow \\ + na_n &= \frac{n+1}{n+2} - \frac{n}{n+1} \\ + &= \frac{1}{(n+1)(n+2)} + \end{align*} + So $a_n=\frac{1}{n(n+1)(n+2)}$, thus we have the following: + \begin{align*} + \sum_{n=2}^{\infty} &= \sum_{n=2}^{\infty} \frac{1}{n(n+1)(n+2)} \\ + &= \frac{1}{2} \sum_{n=2}^{\infty} \left[\frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)}\right] \\ + &= \frac{1}{12} &\text{as we did on the previous homework} + \end{align*} + So $\displaystyle\sum_{n=1}^{\infty} a_n=\frac{2}{3}+\frac{1}{12} = \frac{3}{4}$. + \end{proof} + \end{enumerate} + \end{enumerate} +\end{document} diff --git a/Documents/LaTeX/Homework 7.pdf b/Documents/LaTeX/Homework 7.pdf new file mode 100644 index 0000000..8cb95cc Binary files /dev/null and b/Documents/LaTeX/Homework 7.pdf differ diff --git a/Documents/LaTeX/Homework 7.tex b/Documents/LaTeX/Homework 7.tex new file mode 100644 index 0000000..8b2df1f --- /dev/null +++ b/Documents/LaTeX/Homework 7.tex @@ -0,0 +1,258 @@ +\documentclass[12pt,letterpaper]{article} +\usepackage[utf8]{inputenc} +\usepackage{pgfplots} +\usepackage[english]{babel} +\usepackage{amsthm} +\usepackage{cancel} +\usepackage{mathtools} +\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} +\newcommand{\limx}[2]{\displaystyle\lim\limits_{#1 \to #2}} +\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{\dotp}{\dot{\mathcal{P}}} +\newcommand{\dotq}{\dot{\mathcal{Q}}} +\newcommand{\dist}{\text{dist}} +\DeclareMathOperator{\sign}{sgn} +\newtheoremstyle{case}{}{}{}{}{}{:}{ }{} +\theoremstyle{case} +\newtheorem{case}{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*{lemma*}{Lemma} +\newtheorem*{remark}{Remark} +\setlist[enumerate]{font=\bfseries} +\renewcommand{\qedsymbol}{$\blacksquare$} +\author{Alexander J. Tusa} +\title{Real Analysis II Homework 7} +\begin{document} + \maketitle + \begin{enumerate} + \item \textbf{Section 9.1} + \begin{enumerate} + \item[7.] + \begin{enumerate} + \item If $\sum a_n$ is absolutely convergent and $(b_n)$ is a bounded sequence, show that $\sum a_nb_n$ is absolutely convergent. + \begin{proof} + We want to show that $\sum (a_nb_n)$ is also absolutely convergent.\\ + Since $(b_n)$ is bounded, we know that there is a $M > 0$ such that $|b_n| \leq M,\ \forall\ n$. Then we have that + \[|a_nb_n|=|a_n|\cdot|b_n|\leq M \cdot |a_n|\] + Since $\sum a_n$ is absolutely convergent, $M \cdot \sum |a_n|$ is also convergent. And since $|a_nb_n| \leq M \cdot |a_n|$ we know that $\sum |a_nb_n|$ is also convergent, and therefore $\sum (a_nb_n)$ is absolutely convergent. + \end{proof} + + \item Give an example to show that if the convergence of $\sum a_n$ is conditional and $(b_n)$ is a bounded sequence, then $\sum a_nb_n$ may diverge. + \\\\Consider the series $\sum a_n=\sum\displaystyle\frac{(-1)^n}{n}$ and the bounded sequence $(b_n)=(-1)^n$. We know that $\sum a_n$ is conditionally convergent and $(b_n)$ is bounded by 1. And since $\sum (a_nb_n) = \sum \displaystyle\frac{(-1)^n}{n}\cdot(-1)^n=\sum \frac{1}{n}$, we have that the product series is a harmonic series, and thus diverges.\\\ + \end{enumerate} + \item[8.] Give an example of a convergent series $\sum a_n$ such that $\sum a^2_n$ is not convergent. (Compare this with Exercise 3.7.11) + \\\\Consider the series $\sum a_n=\sum \displaystyle\frac{(-1)^n}{\sqrt{n}}$, which we know is convergent. But, $\sum (a_n)^2 = \sum \frac{1}{n}$, which is a harmonic series, and thus diverges.\\ + \item[9.] If $(a_n)$ is a decreasing sequence of strictly positive numbers and if $\sum a_n$ is convergent, show that $\lim (na_n)=0$. + \begin{proof} + Let $(a_n)$ be a sequence such that $a_1 \geq a_2 \geq \dots \geq a_n \geq 0$, and let $\sum a_n$ be convergent, and let $s_n=a_1+a_2+\dots+a_n$. + \\\\Let $\varepsilon>0$ be given. Since $\sum a_n$ is convergent, we know that $\limx{n}{\infty} s_n=0$ by the \textit{$n$th-Term Test}. By the \textit{Cauchy Criterion for Series}, $\exists\ M(\varepsilon) \in \N \st$ if $m > n \geq M(\varepsilon)$, then $|s_m-s_n| = |a_{n+1}+a_{n+2}+\dots+a_m|<\varepsilon$. Now, since $(a_n)$ is a decreasing sequence, we have + \begin{align*} + s_{2n}-s_n &= (a_1+\dots+a_{2n}) - (a_1+\dots+a_n) \\ + &= a_{n+1}+\dots + a_{2n} \\ + &\geq a_{2n}+\dots+a_{2n} \\ + &= n \cdot a_{2n}\\ + &> 0 + \end{align*} + Additionally, we note that + \begin{align*} + s_{2n-1}-s_n &= (a_1+\dots +a_{2n-1}) -(a_1+\dots+a_n) \\ + &= a_{n+1}+\dots + a_{2n-1} \\ + &\geq a_{2n-1}+\dots+a_{2n-1} \\ + &=(n-1)\cdot a_{2n} \\ + &> 0 + \end{align*} + Notice that if we let $n>M(\varepsilon)$, then $|s_{2n}-s_n| < \varepsilon$ and $|s_{2n-1}-s_n| <\varepsilon$. + \\\\Choose $\delta = 2M(\varepsilon)$ and for $n > \delta$, we get $n\cdot a_{2n} \cdot (n-1)\cdot a_{2n} = na_n < \varepsilon$, and thus by the definition of a limit, we have that + \[\limx{n}{\infty} na_n=0\] + \end{proof} + + \item[10.] Give an example of a divergent series $\sum a_n$ with $(a_n)$ decreasing and such that $\lim (na_n)=0$. + \\\\Consider the series $\sum a_n=\sum \frac{1}{n \ln n}$. We showed on the previous homework that this series diverges. And the sequence $(a_n)$ is decreasing since $n \ln n<(n+1)\ln(n+1)$. Thus, we have + \[\limx{n}{\infty} na_n = \limx{n}{\infty} n \cdot \frac{1}{n \ln n} = \limx{n}{\infty} \frac{1}{\ln n} = 0\] + + \item[11.] If $(a_n)$ is a sequence and if $\lim (n^2a_n)$ exists in $\R$, show that $\sum a_n$ is absolutely convergent. + \begin{proof} + Since $\lim na^2_n$ exists, we know that the sequence $(n^2a_n)$ is bounded. Thus we know that there exists $M>0$ such that + \[|n^2a_n| \leq M \implies |a_n| \leq \frac{M}{n^2},\ \forall\ n\] + Since we know that $\sum \frac{M}{n^2}$ is convergent, by the \textit{Comparison Test}, we know that $\sum |a_n|$ is also convergent, which yields that $\sum a_n$ is absolutely convergent. + \end{proof} + + \item[12.] Let $a > 0$. Show that the series $\sum (1+a^n)^{-1}$ is divergent if $01$. + \begin{proof} + Let $a > 0$. We want to show that the series $\displaystyle \sum a_n=\sum\frac{1}{1+a^n}$ is divergent when $0 < a \leq 1$ and is convergent when $a > 1$. + \begin{case}[$01$] + We notice that + \[\frac{1}{1+a^n}<\frac{1}{a^n}=\left(\frac{1}{a}\right)^n\] + which yields a geometric series, and since $a > 1 \implies \left(\frac{1}{a}\right) < 1$, by the \textit{Geometric Series Test}, the series $\sum \left(\frac{1}{a}\right)^n$ converges, and by the \textit{Comparison Test}, the series $\sum \frac{1}{1+a^n}$ must also converge. + \end{case} + \end{proof} + \item[13.] + \begin{enumerate} + \item Does the series $\displaystyle\sum_{n=1}^{\infty} \left(\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}}\right)$ converge? + \begin{align*} + \sum_{n=1}^{\infty} \left(\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}}\right) &= \sum_{n=1}^{\infty} \frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}} \cdot \frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}+\sqrt{n}} \\ + &= \sum_{n=1}^{\infty} \frac{(n+1)-n}{\sqrt{n}\cdot (\sqrt{n+1}+\sqrt{n})} \\ + &= \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}\cdot (\sqrt{n+1}+\sqrt{n})} \\ + &= \sum_{n=1}^{\infty} \frac{1}{\sqrt{n(n+1)}+n} + \end{align*} + We notice that $\displaystyle\frac{1}{\sqrt{n(n+1)}+n} \approx \frac{1}{2n}$ and that + \begin{align*} + \limx{n}{\infty} \frac{\frac{1}{\sqrt{n(n+1)}+n}}{\frac{1}{2n}} &= \limx{n}{\infty} \frac{2n}{\sqrt{n(n+1)}+n} \\ + &= \limx{n}{\infty} \frac{2n}{\sqrt{n(n+1)}+n} \cdot \frac{\frac{1}{n}}{\frac{1}{n}} \\ + &= \limx{n}{\infty} \frac{2}{\sqrt{1(1+\frac{1}{n})+1}} \\ + &= \frac{2}{\sqrt{1(1+0)}+1} \\ + &= 1 + \end{align*} + Thus we know that since $\sum \frac{1}{n}$ is a harmonic series, it diverges, and thus by the \textit{Comparison Test}, we have that $\displaystyle\sum_{n=1}^{\infty} \frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}}$ diverges.\\\\ + + \item Does the series $\displaystyle\sum_{n=1}^{\infty} \left(\frac{\sqrt{n+1}-\sqrt{n}}{n}\right)$ converge? + \begin{align*} + \sum_{n=1}^{\infty} \frac{\sqrt{n+1}-\sqrt{n}}{n} &= \sum_{n=1}^{\infty} \frac{\sqrt{n+1}-\sqrt{n}}{n} \cdot \frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}+\sqrt{n}} \\ + &= \sum_{n=1}^{\infty} \frac{(n+1)-n}{n\cdot (\sqrt{n+1}+\sqrt{n})} \\ + &= \sum_{n=1}^{\infty} \frac{1}{n\cdot(\sqrt{n+1}+\sqrt{n})} + \end{align*} + We notice that + \[\frac{1}{n \cdot (\sqrt{n+1}+\sqrt{n})} \leq \frac{1}{n \cdot \sqrt{n}}=\frac{1}{n^{\frac{3}{2}}}\] + And since we know that $\displaystyle\sum \frac{1}{n^{\frac{3}{2}}}$ is a convergent $p$-series, we have that by the \textit{Comparison Test}, the series $\displaystyle\sum_{n=1}^{\infty} \frac{\sqrt{n+1}-\sqrt{n}}{n}$ must also be convergent. + \end{enumerate} + \end{enumerate} + + \item + \begin{enumerate} + \item If $a_n \geq 0$ for all $n \in \N$ and $\displaystyle\sum_{n=1}^{\infty} a_n$ converges, prove that $\displaystyle\sum_{n=1}^{\infty}\frac{a_n}{n^p}$ converges for all $p \geq 0$. + \begin{proof} + We notice that the series $\displaystyle\sum \frac{a_n}{n^p}$ looks like $\displaystyle\sum \frac{1}{n^p}$, which is a convergent $p$-series when $p > 1$. So, by the \textit{Limit Comparison Test}, we have for $p >1$: + \[\limx{n}{\infty} \frac{\frac{a_n}{n^p}}{\frac{1}{n^p}}=\limx{n}{\infty} \frac{a_nn^p}{n^p} = \limx{n}{\infty} a_n = 0\] + Since by the \textit{$n$th Term Test}, since $\sum a_n$ converges, $\limx{n}{\infty} a_n = 0$. And since $\sum \frac{1}{n^p}$ is a convergent $p$ series, by the \textit{Limit Comparison Test}, the series $\displaystyle\sum_{n=1}^{\infty} \frac{a_n}{n^p}$ is convergent when $p > 1$. + \\\\As for the case in which $0 \leq p \leq 1$, consider the following subcases: + \begin{case}[$p=0$] + \[\sum_{n=1}^{\infty} \frac{a_n}{n^p} = \sum_{n=1}^{\infty} \frac{a_n}{n^0} = \sum_{n=1}^{\infty} \frac{a_n}{1} = \sum_{n=1}^{\infty} a_n\] + Which since we assumed that $\sum a_n$ converges, we have that when $p=0$, $\sum \frac{a_n}{n^p}$ converges. + \end{case} + \begin{case}[$p=1$] We notice that $\frac{a_n}{n} \leq \frac{na_n}{n}$, and thus we have that by the \textit{Comparison Test}, $\sum \frac{na_n}{n} = \sum a_n$, which we know converges, and thus by the \textit{Comparison Test}, $\sum \frac{a_n}{n}$ must also converge. + \end{case} + \begin{case}[$0 0$ such that $|b_n| \leq M,\ \forall\ n$. Then we have that + \[|a_nb_n|=|a_n|\cdot|b_n|\leq M \cdot |a_n|\] + Since $\sum a_n$ is absolutely convergent, $M \cdot \sum |a_n|$ is also convergent. And since $|a_nb_n| \leq M \cdot |a_n|$ we know that $\sum |a_nb_n|$ is also convergent, and therefore $\sum (a_nb_n)$ is absolutely convergent. + \\\\By \textit{Theorem 9.1.1}, since $\sum |a_nb_n|$ is absolutely convergent, then the series $\sum a_nb_n$ is also convergent. + \end{proof} + \end{enumerate} + + \item (pr. 18a, Sec. 3.7) Find the positive values of $p$ such that the logarithmic $p$-series $\displaystyle\sum_{k=2}^{\infty} \frac{1}{k(\ln k)^p}$ converges using a) the integral test and b) the Cauchy condensation test. + \\\\Since the terms are decreasing, we can use the \textit{Cauchy Condensation Test}. + \begin{align*} + \sum_{n=1}^{\infty} a_n &= \sum_{n=1}^{\infty}\frac{1}{n(\ln n)^p} \\ + \sum_{n=1}^{\infty} 2^na_{2n} &= \sum_{n=1}^{\infty} 2^n\cdot \frac{1}{2^n(\ln 2^n)^p} \\ + &= \sum_{n=1}^{\infty} \frac{1}{(n \ln 2)^p} &(\ln a^b = b \ln a) \\ + &= \sum_{n=1}^{\infty} \frac{1}{(\ln 2)^p} \cdot \frac{1}{n^p} \\ + &= \frac{1}{(\ln 2)^p} \sum_{n=1}^{\infty} \frac{1}{n^p} + \end{align*} + Now, since $\sum \frac{1}{n^p}$ is a $p$-series, the only way for the series to converge is if $p >1$ by the \textit{$p$-Series Test}, and thus we have that by the \textit{Cauchy Condensation Test} and by the \textit{Comparison Test}, $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n (\ln n)^p}$ converges if and only if $p > 1$. + \\\\So, by the \textit{Integral Test}, let $u=\ln k$ and $du = k\ dk$. Then we have: + \[\int_{2}^{\infty} \frac{1}{k(\ln k)^p}\ dk = \int_{2}^{\infty} \frac{1}{u^p}\ du = \int_{\ln 2}^{\infty} u^{-p}\ du = \limx{b}{\infty} \left.\frac{u^{-p+1}}{1-p}\right|_{\ln 2}^b=\limx{b}{\infty} \frac{b^{-p+1}}{1-p}-\frac{(\ln 2)^{-p+1}}{1-p}\] + which we have converges only when $p > 1$ since only then will the $b$ term go to 0. + \\\\And by the \textit{Cauchy Condensation Test}, we have $\displaystyle\sum_{n=2}^{\infty} \frac{2^n}{2^n(\ln 2^n)^p}=\sum_{n=2}^{\infty} \frac{1}{(\ln 2^n)^p}=\sum_{n=2}^{\infty} \frac{1}{(n \ln 2)^p} = \frac{1}{(\ln 2)^p} \sum_{n=2}^{\infty} \frac{1}{n^p}$, and by the definition of a $p$-series this converges when $p>1$. + + \item Give an example of two series $\sum a_n$ and $\sum b_n$ that converge but $\sum a_nb_n$ diverges. (Similar to pr. 8, Sec. 9.1) + \\\\Consider the series $\sum a_n=\sum \displaystyle\frac{(-1)^n}{\sqrt{n}}$ and $\sum a_n=\sum \displaystyle\frac{(-1)^{n+1}}{\sqrt{n}}$ which we know is convergent. But, $\sum (a_nb_n) = \sum -\frac{1}{n}$, which is a negative harmonic series, and thus diverges.\\ + + \item Prove or justify, if true; Provide a counterexample, if false. + \begin{enumerate} + \item Let $a_k$ and $b_k$ be sequences of positive real numbers. If $\sum a_n$ and $\sum b_n$ converge, then $\sum a_nb_n$ converges. + \\\\This is a true statement. + \begin{proof} + Since $\sum a_n$ converges, by the \textit{$n$th Term Test}, $\lim a_n = 0$. So there must exists some $N \in \N \st\ \forall\ n \geq N,\ a_n \leq 1$. So, $0 \leq a_n \leq 1$, which yields that $0 \leq a_nb_n \leq b_n\ \forall\ n \geq N$. Thus, since $\sum b_n$ converges, by the \textit{Comparison Test}, we have that $\sum a_nb_n$ converges. + \end{proof} + + \item Let $a_k$ and $b_k$ be sequences of positive real numbers. If $\sum a_nb_n$ converges, then $\sum a_n$ and $\sum b_n$ converge. + \\\\This is a false statement. Consider the series $\sum \frac{1}{n^3}$ and $\sum n$. Then we have that $\sum \frac{1}{n^3} \cdot n = \sum \frac{1}{n^2}$, which is a convergent $p$ series with $p=2>1$. However, we have that while $\sum \frac{1}{n^3}$ converges since it is a convergent $p$-series with $p=3>1$, but $\sum n$ diverges.\\ + + \item Let $a_k$ and $b_k$ be sequences of positive real numbers. If $\sum a_n$ and $\sum b_n$ converge, then $\sum\sqrt{(a_n)^2+(b_n)^2}$ converges. + \\\\This is a true statement. + \begin{proof} + Let $\sum a_n$ and $\sum b_n$ be convergent series, and let $a_n$ and $b_n$ be positive sequence of real numbers. Then by \textit{Theorem 3.7.3}, we know that the sequences of partial sums $(s_k)$ and $(t_k)$ of $\sum a_n$ and $\sum b_n$, respectively, must be bounded. And since the sequences of partial sums converges, we know that by the \textit{Cauchy Convergence Criterion}, $(s_k)$ and $(t_k)$ are Cauchy sequences. Additionally, by \textit{Theorem 3.2.3}, since both $(s_k)$ and $(t_k)$ converge, then $(s_k)+(t_k)$ must also converge. This yields that $\forall\ \varepsilon>0,\ \exists\ K(\varepsilon) \in \N\ \st \forall\ n>m\geq K(\varepsilon),$ by the \textit{Triangle Inequality}, we have + \[\sqrt{(s_n-s_m)^2+(t_n-t_m)^2} \leq \sqrt{(s_n-s_m)^2} + \sqrt{(t_n-t_m)^2} = |s_n-s_m|+|t_n-t_m| < \varepsilon\] + And thus since this is the definition of the \textit{Cauchy Criterion for Series}, we have that this is equal to the series $\sum \sqrt{(a_n)^2+(b_n)^2}$. Therefore the series $\sum \sqrt{(a_n)^2+(b_n)^2}$ converges. + \end{proof} + + \item Let $a_k$ and $b_k$ be sequences of positive real numbers. If $\sum \sqrt{(a_n)^2 + (b_n)^2}$ converges, then $\sum a_n$ and $\sum b_n$ converge. + \\\\This is a true statement. + \begin{proof} + By the \textit{Comparison Test}, since $0 \leq a_n^2 \leq a_n^2+b_n^2$, we have that $0 \leq a_n \leq \sqrt{a_n^2+b_n^2}$, and thus $\sum a_n$ converges. Similarly, by the \textit{Comparison Test}, we have that since $0 \leq b_n^2 \leq a_n^2+b_n^2$, we have that $0 \leq b_n \leq \sqrt{a_n^2+b_n^2}$, and thus $\sum b_n$ converges. + \end{proof} + + \item If $\sum a_n$ converges and $0 \leq b_n \leq a_n$ , then $\sum b_n$ converges. + \\\\This is true since it is the \textit{Comparison Test}.\\ + + \item If $\limx{n}{\infty} a_n=0,\ a_n \geq 0$ and $\sqrt{a_{n+1}} \leq a_n$ for all $n \in \N$, then $\sum a_n$ converges. + \begin{proof} + Since $\lim a_n = 0$, we know that $\exists\ N \in \N \st \forall\ n \geq N,\ |a_n|\leq \frac{1}{2}$. (That is, let $\varepsilon = \frac{1}{2}$). Also, since $\sqrt{a_{n+1}} \leq a_n$, we know that $a_{n+1} \leq a_N^2 \leq \frac{1}{4}$. So $a_{n+1} \leq \frac{1}{4}$. So, we have + \begin{align*} + a_{n+2} &\leq a_{n+1}^2\leq \frac{1}{16} = \frac{1}{4^2} \\ + &\dots \\ + a_{n+3} &\leq \frac{1}{4^4} + \end{align*} + So, we have that + \[|a_n| \leq \left(\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+\dots+\frac{1}{4^{n-N}}\right)\] + since $a_n=a_{N(n-N)}$. So, we have that $0 \leq |a_n| \leq \frac{1}{4}$, which is a convergent geometric series. Therefore by the \textit{Comparison Test}, we have that $\sum a_n$ is convergent. + \end{proof} + \end{enumerate} + \end{enumerate} +\end{document} diff --git a/Documents/LaTeX/Homework 8.pdf b/Documents/LaTeX/Homework 8.pdf new file mode 100644 index 0000000..ad9442e Binary files /dev/null and b/Documents/LaTeX/Homework 8.pdf differ diff --git a/Documents/LaTeX/Homework 8.tex b/Documents/LaTeX/Homework 8.tex new file mode 100644 index 0000000..5372f63 --- /dev/null +++ b/Documents/LaTeX/Homework 8.tex @@ -0,0 +1,395 @@ +\documentclass[12pt,letterpaper]{article} +\usepackage[utf8]{inputenc} +\usepackage{pgfplots} +\usepackage[english]{babel} +\usepackage{amsthm} +\usepackage{cancel} +\usepackage{mathtools} +\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} +\newcommand{\limx}[2]{\displaystyle\lim\limits_{#1 \to #2}} +\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{\dotp}{\dot{\mathcal{P}}} +\newcommand{\dotq}{\dot{\mathcal{Q}}} +\newcommand{\dist}{\text{dist}} +\DeclareMathOperator{\sign}{sgn} +\newtheoremstyle{case}{}{}{}{}{}{:}{ }{} +\theoremstyle{case} +\newtheorem{case}{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*{lemma*}{Lemma} +\newtheorem*{remark}{Remark} +\setlist[enumerate]{font=\bfseries} +\renewcommand{\qedsymbol}{$\blacksquare$} +\author{Alexander J. Tusa} +\title{Real Analysis II Homework 8} +\begin{document} + \maketitle + \begin{enumerate} + \item \textbf{Section 9.2} + \begin{enumerate} + \item[2. (c)] Establish the convergence or divergence of the series whose $n$th term is $n!/n^n$. + \\\\Since + \begin{align*} + \abs{\frac{a_{n+1}}{a_n}} &= \frac{\frac{(n+1)!}{(n+1)^{n+1}}}{\frac{n!}{n^n}} \\ + &= \frac{(n+1)!\cdot n^n}{(n+1)^{n+1}\cdot n!} \\ + &= \frac{n^n}{(n+1)^n} \\ + &\Downarrow \\ + \limx{n}{\infty} \abs{\frac{a_{n+1}}{a_n}} &= \frac{1}{\limx{n}{\infty} \left(1+\frac{1}{n}\right)^n} \\ + &= \frac{1}{e} \\ + &<1 + \end{align*} + By the \textit{Ratio Test}, and by \textit{Corollary 9.2.5}, we have that $\sum \frac{n!}{n^n}$ converges.\\ + + \item[5.] Show that the series $1/1^2+1/2^3+1/3^2+1/4^3+\dots$ is convergent, but that both the Ratio and the Root Tests fail to apply. + \\\\We notice that the specified series yields $S=\sum a_n$, where + \[a_{2n}=\frac{1}{(2n)^3},\ \text{ and }\ a_{2n-1}=\frac{1}{(2n-1)^2},\ \text{ for }\ n \in \N\] + \\\\First we show that the \textit{Ratio Test} fails. To do so, we must consider two cases: + \begin{align*} + \abs{\frac{a_{2n+1}}{a_{2n}}} &= \frac{\frac{1}{(2n+1)^2}}{\frac{1}{(2n)^3}} \\ + &= \frac{8n^3}{4n^2+4n+1} \cdot \frac{\frac{1}{n^3}}{\frac{1}{n^3}} \\ + &= \frac{8}{\frac{4}{n}+\frac{4}{n^2}+\frac{1}{n^3}} \\ + &\Downarrow \\ + \limx{n}{\infty} \frac{8}{\frac{4}{n}+\frac{4}{n^2}+\frac{1}{n^3}} = \infty + \end{align*} + and + \begin{align*} + \abs{\frac{a_{2n}}{a_{2n-1}}} &= \frac{\frac{1}{(2n)^3}}{\frac{1}{(2n-1)^2}} \\ + &= \frac{4n^2-4n+1}{8n^3} \cdot \frac{\frac{1}{n^3}}{\frac{1}{n^3}} \\ + &= \frac{\frac{4}{n}-\frac{4}{n^2}+\frac{1}{n^3}}{8} \\ + &\Downarrow \\ + \limx{n}{\infty} \frac{\frac{4}{n}-\frac{4}{n^2}+\frac{1}{n^3}}{8} &= 0 + \end{align*} + And thus we can see that the \textit{Ratio Test} is ineffective on this series. + \\\\As for the \textit{Root Test}, we must also consider two cases: + \[\abs{a_{2n}}^{\frac{1}{2n}}=\left(\frac{1}{8n^3}\right)^{\frac{1}{2n}}=\frac{1}{8^\frac{2}{n}}\cdot \left(\frac{1}{n^{\frac{2}{n}}}\right)^3\implies \limx{n}{\infty} \frac{1}{8^\frac{2}{n}}\cdot \left(\frac{1}{n^{\frac{2}{n}}}\right)^3 = \frac{1}{1} \cdot \left(\frac{1}{1}\right)^3=1\] + and + \[\abs{a_{2n-1}}^{\frac{1}{2n-1}}=\left(\frac{1}{(2n-1)^2}\right)^{\frac{1}{2n-1}}=\left(\frac{1}{2n-1}\right)^{\frac{2}{2n-1}} \implies \limx{n}{\infty} \left(\frac{1}{2n-1}\right)^{\frac{2}{2n-1}}=1\] + And thus we see that the \textit{Root Test} is ineffective on this series as well. + \\\\Now, we'll show that the series does converge by the \textit{Comparison Test}: + \\\\We notice that + \[a_{2n}=\frac{1}{8n^3}<\frac{1}{n^3}<\frac{1}{n^2}\] + \[a_{2n-1}=\frac{1}{(2n-1)^2}<\frac{1}{n^2}\] + and thus by the \textit{Comparison Test}, we have that since $\frac{1}{n^2}$ is a convergent $p$-series with $p=2>1$, the series $\sum a_n$ must also converge.\\ + + \item[7.] Discuss the series whose $n$th term is + \begin{enumerate} + \item[(a)] $\displaystyle\frac{n!}{3\cdot5\cdot7\cdot\dots\cdot(2n+1)}$ + \\\\By the \textit{Ratio Test}, we have: + \begin{align*} + \abs{\frac{a_{n+1}}{a_n}} &= \frac{\displaystyle\frac{(n+1)!}{3 \cdot 5 \cdot 7 \cdot \dots \cdot (2(n+1)+1)}}{\displaystyle\frac{n!}{3 \cdot 5 \cdot 7 \cdot \dots \cdot (2n+1)}} \\ + &= \frac{(n+1)!\cdot 3 \cdot 5 \cdot 7 \cdot \dots \cdot (2n+1)}{n! \cdot 3 \cdot 5 \cdot 7 \cdot \dots \cdot (2n+1)\cdot(2n+3)} \\ + &= \frac{n+1}{2n+3} \leq \frac{n+1}{2n+2} \\ + &= \frac{n+1}{2(n+1)} \\ + &= \frac{1}{2} \\ + &< 1 + \end{align*} + And thus by the \textit{Ratio Test}, we have that the series is absolutely convergent.\\ + + \item[(b)] $\displaystyle\frac{(n!)^2}{(2n)!}$ + \\\\By the \textit{Ratio test}, we have: + \begin{align*} + \abs{\frac{a_{n+1}}{a_n}} &= \frac{\displaystyle\frac{\big((n+1)!\big)^2}{\big(2(n+1)\big)!}}{\displaystyle\frac{(n!)^2}{(2n)!}} \\ + &= \frac{(n+1)! \cdot (n+1)! \cdot (2n)!}{(2n+2)!\cdot n! \cdot n!} \\ + &= \frac{(n+1)(n+1)}{(2n+1)(2n+2)} \\ + &= \frac{n+1}{2(2n+1)} \\ + &= \frac{n+1}{4n+2} \\ + &\leq \frac{n+1}{4n} \\ + &= \frac{1}{4} + \frac{1}{4n} \\ + &\leq \frac{1}{4}+\frac{1}{4} \\ + &= \frac{1}{2} \\ + &< 1 + \end{align*} + And thus by the \textit{Ratio Test}, we have that the series is absolutely convergent.\\ + + \item[(c)] $\displaystyle\frac{2\cdot4\cdot\dots\cdot(2n)}{3\cdot5\cdot\dots\cdot(2n+1)}$ + \\\\By the \textit{Ratio Test}, we have: + \begin{align*} + \abs{\frac{a_{n+1}}{a_n}} &= \frac{\displaystyle\frac{2 \cdot 4 \cdot \dots \cdot (2n) \cdot (2n+2)}{3 \cdot 5 \cdot \dots \cdot (2n+1) \cdot (2n+3)}}{\displaystyle\frac{2 \cdot 4 \cdot \dots \cdot (2n)}{3 \cdot 5 \cdot \dots \cdot (2n+1)}} \\ + &= \frac{2n+2}{2n+3} \\ + &\Downarrow \\ + \limx{n}{\infty} \abs{\frac{2n+2}{2n+3}} &= 1 + \end{align*} + Thus by \textit{Corollary 9.2.5}, the \textit{Ratio Test} is ineffective on this series. + \\\\By \textit{Raabe's Test}, we have: + \begin{align*} + \abs{\frac{a_{n+1}}{a_n}} &= \frac{2n+2}{2n+3} \\ + &= \frac{(2n+3)-1}{2n+3} \\ + &= 1-\frac{1}{2n+3} \\ + &\geq 1-\frac{1}{2n} \\ + &= 1-\frac{\frac{1}{2}}{n} + \end{align*} + Thus by \textit{Raabe's Test}, since $a=\frac{1}{2}$, we have that the series is divergent.\\ + + \item[(d)] $\displaystyle\frac{2\cdot4\cdot\dots\cdot(2n)}{5\cdot7\cdot\dots\cdot(2n+3)}$ + \\\\By the \textit{Ratio Test}, we have: + \begin{align*} + \abs{\frac{a_{n+1}}{a_n}} &= \frac{\displaystyle\frac{2 \cdot 4 \cdot \dots \cdot (2n)\cdot (2n+2)}{5 \cdot 7 \cdot \dots \cdot (2n+3) \cdot (2n+5)}}{\displaystyle\frac{2 \cdot 4 \cdot \dots \cdot (2n)}{5 \cdot 7 \cdot \dots \cdot (2n+3)}} \\ + &= \frac{2n+2}{2n+5} \\ + &\Downarrow \\ + \limx{n}{\infty} \frac{2n+2}{2n+5} &= 1 + \end{align*} + Thus by \textit{Corollary 9.2.5}, the \textit{Ratio Test} is ineffective on this series. + \\\\By \textit{Corollary 9.2.9}, we have: + \begin{align*} + a &= \limx{n}{\infty} \left(n \left(1-\abs{\frac{a_{n+1}}{a_n}}\right)\right) \\ + &= \limx{n}{\infty} \left(n \left(1-\frac{2n+2}{2n+5}\right)\right) \\ + &= \limx{n}{\infty} \left(n \cdot \frac{3}{2n+5}\right) \\ + &= \limx{n}{\infty} \left(\frac{3}{2+\frac{5}{n}}\right) \\ + &= \frac{3}{2} + \end{align*} + Thus by \textit{Corollary 9.2.9}, since $a=\frac{3}{2} > 1$, we have that the series is absolutely convergent. + \end{enumerate} + \end{enumerate} + + \item \textbf{Section 9.3} + \begin{enumerate} + \item[1.] Test the following series for convergence and for absolute convergence: + \begin{enumerate} + \item[(a)] $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2+1}$ + \\\\By the \textit{Alternating Series Test}, we have that $\limx{n}{\infty} \frac{1}{n^2+1}=0$. And by the \textit{Limit Comparison Test}, we notice that $\frac{1}{n^2+1}$ looks like $\frac{1}{n^2}$, which we note is a convergent $p$-series with $p=2>1$, which yields: + \[\limx{n}{\infty} \frac{\displaystyle\frac{1}{n^2+1}}{\displaystyle\frac{1}{n^2}}= \limx{n}{\infty} \frac{n^2}{n^2+1} = 1 \neq 0\] + And thus since $\sum \frac{1}{n^2}$ is convergent, by the \textit{Limit Comparison Test}, we have that $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2+1}$ is absolutely convergent.\\ + + \item[(b)] $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n+1}$ + \\\\By the \textit{Alternating Series Test}, we have that $\limx{n}{\infty} \frac{1}{n+1} = 0$, and thus the series is convergent. And by the \textit{Limit Comparison Test}, we note that the series looks like $\sum \frac{1}{n}$, which we note is a harmonic series and thus diverges, which yields + \[\limx{n}{\infty} \frac{\displaystyle\frac{1}{n+1}}{\frac{1}{n}} = \limx{n}{\infty} \frac{n}{n+1} = 1 \neq 0\] + And since $\sum \frac{1}{n}$ is a harmonic series and thus diverges, since the limit is not equal to 0, we have that by the \textit{Limit Comparison Test}, the series $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n+1}$ is conditionally convergent.\\ + + \item[(c)] $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}n}{n+2}$ + \\\\By the \textit{Alternating Series Test}, we have that $\limx{n}{\infty} \frac{n}{n+2} = 1 \neq 0$, and thus the series is divergent.\\ + + \item[(d)] $\displaystyle\sum_{n=1}^{\infty} (-1)^{n+1} \frac{\ln n}{n}$ + \\\\By the \textit{Alternating Series Test}, we have that $\limx{n}{\infty} \frac{\ln n}{n} = 0$, which yields that the series is convergent. And by the \textit{Integral Test}, we have that + \[\int_{1}^{\infty} \frac{\ln n}{n}\ dn = \int_{1}^{\infty} \frac{u}{n}\cdot n\ du = \int_{1}^{\infty} u\ du = \left.\frac{u^2}{2}\right|_1^\infty = \left.\frac{(\ln n)^2}{2}\right|_1^\infty = \frac{(\ln \infty)^2}{2} - \frac{(\ln 1)^2}{2}\] + \[= \infty - 0 = \infty\] + Thus by the \textit{Integral Test}, we have that this series is divergent. Thus the series is conditionally convergent.\\ + + \end{enumerate} + + \item[3.] Give an example to show that the Alternating Series Test 9.3.2 may fail if $(z_n)$ is not a decreasing sequence. + \\\\Let $\sum a_n$ be the series defined as + \[\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{6}+\dots+\frac{1}{n}-\frac{1}{2n}+\dots\right)\] + Let $A_n$ be the partial sums of the series $\sum a_n$. Then since $a_n$ is an alternating sequence that converges to 0 and isn't decreasing, we have + \[A_{2n}=\sum_{n=1}^{\infty} \left(\frac{1}{n}-\frac{1}{2n}\right) = \sum_{n=1}^{\infty} \left(\frac{1}{2n}\right)\] + As $A_{2n}$ diverges, we have that $A_n$ diverges and thus the series $\sum a_n$ is divergent.\\ + + \item[5.] Consider the series + \[1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{1}{6}-\frac{1}{7}++--\dots,\] + where the signs come in pairs. Does it converge? + \\\\We notice that $\frac{1}{n}$ is a monotone decreasing sequence that converges to 0. Then we notice that the series $\displaystyle\sum_{n=1}^{\infty} a_n$ where for every $n \in \N$, we have $a_1=1, a_{4n}=1, a_{4n-1}=-1=a_{4n+1}$. Now, let $s_n=a_1+a_2+\dots+a_n$. Then $s_{2n}=0$ and $s_{2n+1}=\pm 1$. This yields that $|s_n| \leq 1$. By \textit{Dirichlet's Test}, we have that $\displaystyle\sum_{n=1}^{\infty} a_n$ is convergent. Thus we have + \[\sum_{n=1}^{\infty} \frac{a_n}{n}=1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{1}{6}-\frac{1}{7}++--\dots\] + is also convergent. + \end{enumerate} + + \item Give an example of a series $\sum a_n$ that consists of nonzero terms with $\limx{n}{\infty} \abs{\frac{a_{n+1}}{a_n}}=1$ for each of the following conditions: + \begin{enumerate} + \item $\sum a_n$ converges absolutely + \\\\Consider Problem 7d from Section 9.2: $\displaystyle\sum_{n=1}^{\infty}\frac{2\cdot4\cdot\dots\cdot(2n)}{5\cdot7\cdot\dots\cdot(2n+3)}$ + \\\\By the \textit{Ratio Test}, we have: + \begin{align*} + \abs{\frac{a_{n+1}}{a_n}} &= \frac{\displaystyle\frac{2 \cdot 4 \cdot \dots \cdot (2n)\cdot (2n+2)}{5 \cdot 7 \cdot \dots \cdot (2n+3) \cdot (2n+5)}}{\displaystyle\frac{2 \cdot 4 \cdot \dots \cdot (2n)}{5 \cdot 7 \cdot \dots \cdot (2n+3)}} \\ + &= \frac{2n+2}{2n+5} \\ + &\Downarrow \\ + \limx{n}{\infty} \frac{2n+2}{2n+5} &= 1 + \end{align*} + Thus by \textit{Corollary 9.2.5}, the \textit{Ratio Test} is ineffective on this series. + \\\\By \textit{Corollary 9.2.9}, we have: + \begin{align*} + a &= \limx{n}{\infty} \left(n \left(1-\abs{\frac{a_{n+1}}{a_n}}\right)\right) \\ + &= \limx{n}{\infty} \left(n \left(1-\frac{2n+2}{2n+5}\right)\right) \\ + &= \limx{n}{\infty} \left(n \cdot \frac{3}{2n+5}\right) \\ + &= \limx{n}{\infty} \left(\frac{3}{2+\frac{5}{n}}\right) \\ + &= \frac{3}{2} + \end{align*} + Thus by \textit{Corollary 9.2.9}, since $a=\frac{3}{2} > 1$, we have that the series is absolutely convergent.\\ + + \item $\sum a_n$ converges conditionally + \\\\Consider problem 1b from Section 9.3: $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n+1}$ + \\\\By the \textit{Ratio Test}, we have: + \begin{align*} + \limx{n}{\infty} \abs{\frac{a_{n+1}}{a_n}} &= \limx{n}{\infty} \frac{\displaystyle\frac{1}{n+3}}{\displaystyle\frac{1}{n+1}} \\ + &= \limx{n}{\infty} \frac{n+1}{n+3} \\ + &= 1 + \end{align*} + Thus by \textit{Corollary 9.2.5}, the \textit{Ratio Test} is ineffective on this series. + \\\\By the \textit{Alternating Series Test}, we have that $\limx{n}{\infty} \frac{1}{n+1} = 0$, and thus the series is convergent. And by the \textit{Limit Comparison Test}, we note that the series looks like $\sum \frac{1}{n}$, which we note is a harmonic series and thus diverges, which yields + \[\limx{n}{\infty} \frac{\displaystyle\frac{1}{n+1}}{\frac{1}{n}} = \limx{n}{\infty} \frac{n}{n+1} = 1 \neq 0\] + And since $\sum \frac{1}{n}$ is a harmonic series and thus diverges, since the limit is not equal to 0, we have that by the \textit{Limit Comparison Test}, the series $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n+1}$ is conditionally convergent.\\ + + \item $\sum a_n$ diverges. + \\\\Consider Problem 7c from Section 9.2: $\displaystyle\sum_{n=1}^{\infty}\frac{2\cdot4\cdot\dots\cdot(2n)}{3\cdot5\cdot\dots\cdot(2n+1)}$ + \\\\By the \textit{Ratio Test}, we have: + \begin{align*} + \abs{\frac{a_{n+1}}{a_n}} &= \frac{\displaystyle\frac{2 \cdot 4 \cdot \dots \cdot (2n) \cdot (2n+2)}{3 \cdot 5 \cdot \dots \cdot (2n+1) \cdot (2n+3)}}{\displaystyle\frac{2 \cdot 4 \cdot \dots \cdot (2n)}{3 \cdot 5 \cdot \dots \cdot (2n+1)}} \\ + &= \frac{2n+2}{2n+3} \\ + &\Downarrow \\ + \limx{n}{\infty} \abs{\frac{2n+2}{2n+3}} &= 1 + \end{align*} + Thus by \textit{Corollary 9.2.5}, the \textit{Ratio Test} is ineffective on this series. + \\\\By \textit{Raabe's Test}, we have: + \begin{align*} + \abs{\frac{a_{n+1}}{a_n}} &= \frac{2n+2}{2n+3} \\ + &= \frac{(2n+3)-1}{2n+3} \\ + &= 1-\frac{1}{2n+3} \\ + &\geq 1-\frac{1}{2n} \\ + &= 1-\frac{\frac{1}{2}}{n} + \end{align*} + Thus by \textit{Raabe's Test}, since $a=\frac{1}{2}$, we have that the series is divergent.\\ + + \end{enumerate} + + \item Prove or justify, if true. Provide a counterexample, if false. + \begin{enumerate} + \item If $\sum |a_n|$ diverges, then $\sum a_n$ is conditionally convergent. + \\\\This is a false statement. Consider the sequence $a_n=(0,\frac{1}{2},0,\frac{1}{4},0,\dots)$. we note that all of the terms are greater than or equal to 0. We also note that the sequence can be defined piecewise as follows: + \[a_n:=\begin{cases} + 0, &n\text{ is odd} \\ + \frac{1}{n}, &n\text{ is even} + \end{cases}\] + for all $n \in \N$. By this definition, we have + \begin{align*} + \sum_{n=1}^{\infty} |a_n| &= \sum_{n=1}^{\infty} a_n \\ + &= 0+\frac{1}{2}+0+\frac{1}{4}+0+\frac{1}{6}+\dots \\ + &= \frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\dots \\ + &= \sum_{n=1}^{\infty} \frac{1}{2n} \\ + &= \infty + \end{align*} + Thus we have that $\sum |a_n|$ diverges since it yields a harmonic series, and since all terms of $a_n$ are greater than or equal to 0, we have that the series $\sum |a_n|=\sum a_n$. Thus $\sum a_n$ also diverges.\\ + + \item If $\sum |a_n|$ diverges, then $\sum |a_n|$ is conditionally convergent. + \\\\This is a false statement. Refer to the previous problem's counterexample: Consider the sequence $a_n=(0,\frac{1}{2},0,\frac{1}{4},0,\dots)$. we note that all of the terms are greater than or equal to 0. We also note that the sequence can be defined piecewise as follows: + \[a_n:=\begin{cases} + 0, &n\text{ is odd} \\ + \frac{1}{n}, &n\text{ is even} + \end{cases}\] + for all $n \in \N$. By this definition, we have + \begin{align*} + \sum_{n=1}^{\infty} |a_n| &= \sum_{n=1}^{\infty} a_n \\ + &= 0+\frac{1}{2}+0+\frac{1}{4}+0+\frac{1}{6}+\dots \\ + &= \frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\dots \\ + &= \sum_{n=1}^{\infty} \frac{1}{2n} \\ + &= \infty + \end{align*} + Thus we have that $\sum |a_n|$ diverges since it yields a harmonic series, and since all terms of $a_n$ are greater than or equal to 0, we have that the series $\sum |a_n|=\sum a_n$. Thus $\sum a_n$ also diverges.\\ + + \item If $\sum |a_n|$ diverges, then $\sum a_n$ diverges. + \\\\This is a false statement. Consider the series $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}$. Then, we have that $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} = \ln 2$, but $\displaystyle\sum_{n=1}^{\infty} \abs{\frac{(-1)^{n+1}}{n}} = \displaystyle\sum_{n=1}^{\infty} \frac{1}{n} = \infty$, since this is a harmonic series. Thus we have that $\sum |a_n|$ diverges since it yields the harmonic series, but $\sum a_n$ converges to $\ln 2$, hence $\sum a_n$ is conditionally convergent.\\ + + \item If $\sum |a_n|$ converges, then $\sum a_n$ is absolutely convergent. + \\\\This is true since it is the definition of \textit{Absolute Convergence}.\\ + + \item If $a_n \leq b_n$ for all $n \in \N$ and $\sum b_n$ is absolutely convergent, then $\sum a_n$ converges. + \\\\This is true by the \textit{Comparison Test}.\\ + + \item If $\sum a_n$ is absolutely convergent, then $\sum a_n^2$ is absolutely convergent. + \\\\This is a true statement. + \begin{proof} + Let $\sum a_n$ be an absolutely convergent series. Then $\limx{n}{\infty} a_n=0$ by the \textit{$n$th Term Test}. Thus, we know that $\exists\ N \in \N \st 0 < a_n < 1,\ \forall\ n \geq N$, and thus $0 < a_n^2