# homework in latex

When submitting homework as a .tex file please

• name the file yournamehmknumber.tex (no spaces or parentheses please);
• email the file as an attachment;
• include your name and homework number in the subject line;
• verify that the file compiles by running pdflatex or, if you are using sage math cloud or other online latex platform, by using the pdf option before submission;
• use the homework template below. Feel free to modify this template, except for the documentclass and the included packages and the macro \cc;
• for redoes,
• include the original submission with a \newpage separating the resubmission (first) from the original (second).
• Please also clearly mark the redo so that it is clear what is to be graded;
• add R1 to the end of the filename (so the new filename is yournamehmknumberR1.tex).
• Please no spaces or # (or other unusual characters) in the file name.

Latex homework template

\documentclass[12pt]{amsart}

\textwidth = 6.2 in
\textheight = 8.5 in
\oddsidemargin = 0.0 in
\evensidemargin = 0.0 in
\topmargin = 0.0 in
\parskip = 0.05 in
\parindent = 0.3 in

\usepackage{enumerate}
\usepackage{amsmath}
\usepackage{mathrsfs}
\usepackage{color}
\def\cc{\color{blue}}
\usepackage[normalem]{ulem}

\title{Homework number}

\begin{document}
\maketitle

Place solution here.

\end{document}

Sample homework (see Problem 1.2).

\documentclass[12pt]{amsart}

\textwidth = 6.2 in
\textheight = 8.5 in
\oddsidemargin = 0.0 in
\evensidemargin = 0.0 in
\topmargin = 0.0 in
\parskip = 0.05 in
\parindent = 0.3 in

\usepackage{enumerate}
\usepackage{amsmath}
\usepackage{color}
\def\cc{\color{blue}}
\usepackage[normalem]{ulem}

\title{Homework 1}
\author{Scott McCullough}

\begin{document}
\maketitle

Suppose $f:X\to S$. Show, if $C\subset X,$ then $f^{-1}(f(C))\supset C$ and give an example to show the inclusion can be strict.

\bigskip

Let $D=f(C)$ and suppose $x\in C$. It follows that $y=f(x)\in D$ and thus, by the definition of inverse image, $x\in f^{-1}(D)=f^{-1}(f(C))$ and the inclusion follows.

To see that the reverse inclusion does not hold in general, let $X=\{0,1\}$ and $S=\{0\}$ and define $f:X\to S$ by $f(x)=0$ for $x\in X$ (the only possible definition for $f$). Choose $C=\{0\}$ and note that $f(C)=\{0\}$ and hence $f^{-1}(f(C))=f^{-1}(\{0\}) =X\ne C$.
\newpage

Here are some other potentially useful typesetting commands.
\begin{aligned} \int_a^b f\, dx \\ \sum_{j=1}^\infty a_j \\ \lim_{n\to\infty} b_n \\ \|x\|\\ \dots\\ \cdots\\ \ldots\\ \frac{a}{b}\\ A\setminus B \\ \tilde{A}\\ \end{aligned}
\vskip 1 in
% the aligned environment aligns the material separated by the double backslash. The split environment, below, alings the lines separated by double backslash according to the ampersands.
$\begin{split} f(x)=& \lim \frac{g(x+h)-g(x)}{h}\\ =& g^\prime(x). \end{split}$
\bigskip
The {\it detexify} website offers a convenient way to search for latex commands.

\end{document}

Sample homework (problem 9.1) using the template.

\documentclass[12pt]{amsart}

\textwidth = 6.2 in
\textheight = 8.5 in
\oddsidemargin = 0.0 in
\evensidemargin = 0.0 in
\topmargin = 0.0 in
\parskip = 0.05 in
\parindent = 0.3 in

\usepackage{enumerate}
\usepackage{amsmath}
\usepackage{color}
\def\cc{\color{blue}}
\usepackage[normalem]{ulem}

\title{Homework 0}
\author{Scott McCullough}

\begin{document}
\maketitle

Suppose $X$ is a set and $g:X\to\mathbb R$. Define, for $n\in\mathbb N^+$, a sequence of functions $f_n:X\to\mathbb R$ by $f_n(x)=g(x)^n$. Show,
\begin{enumerate}[(i)]
\item $(f_n)$ converges pointwise if and only if the range of $g$ lies in $(-1,1]$; and
\item $(f_n)$ converges uniformly if and only if there exist an $0\le a<1$ such the range of $g$ lies in $[-a,a]\cup\{1\}$.
\end{enumerate}

\bigskip

To prove (i), first suppose that the range of $g$ lies in $(-1,1]$. Given $x\in X$, since $g(x)$ lies in $[-1,1],$ the sequence $a_n = g(x)^n$ converges. Hence $(f_n)$ converges pointwise. Conversely, if there exists an $x\in X$ such that $g(x)\notin (-1,1]$, then $(b_n = g(x)^n)$ does not converge and hence $(f_n)$ does not converge pointwise.

To prove (ii), suppose such an $a$ exists. To prove that the sequence $(f_n)$ converges uniformly to the function $f:X\to \mathbb R$ defined by

\label{eq:f}
f(x) =\begin{cases} 0 & \mbox{if} \ \ |g(x)|< b \\
1 & \mbox{if} \ \ g(x)=1, \end{cases}

where $a<b0$ be given. The sequence $(a^n)$ converges to $0$. Hence there is an $N$ such that if $n\ge N$, then
$0\le a^n<\epsilon$. Let $n\ge N$ and $x\in X$ be given. If $|g(x)|\le a$, then
\begin{equation*}
|f_n(x)-f(x)| = |f_n(x)-0| \le a^n <\epsilon
\end{equation*}
On the other hand, if $g(x)=1$, then
$|f_n(x)-f(x)| = 0 <\epsilon$
and the conclusion follows.

Now suppose no such $a$ exists. By part (i) we can assume that the range of $g$ lies in $(-1,1]$ and $(f_n)$ converges to the function $f$ defined in Equation \eqref{eq:f} with $b=1$. It suffices to show that $(f_n)$ does not converge uniformly to this $f$. To this end, fix $0<\eta|g(x_n)|> (\frac 12)^{\frac 1n}$. Hence,
$|f_n(x_n)-f(x_n)| = |g(x_n)|^n > \frac12 >\eta$
and the desired conclusion follows.

\end{document}