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\pdfminorversion=4 \documentclass[]{article} %%%%%%%%%%%%%%%%%%% % Packages/Macros % %%%%%%%%%%%%%%%%%%% \usepackage{amssymb,latexsym,amsmath} % Standard packages %%%%%%%%%%% % Margins % %%%%%%%%%%% \addtolength{\textwidth}{1.0in} \addtolength{\textheight}{1.00in} \addtolength{\evensidemargin}{-0.75in} \addtolength{\oddsidemargin}{-0.75in} \addtolength{\topmargin}{-.50in} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % Theorem/Proof Environments % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \newtheorem{theorem}{Theorem} \newenvironment{proof}{\noindent{\bf Proof:}}{$\hfill \Box$ \vspace{10pt}} %%%%%%%%%%%% % Document % %%%%%%%%%%%% \begin{document} \title{Sample \LaTeX ~File} \author{David P. Little} \maketitle \begin{abstract} This document represents the output from the file ``sample.tex" once compiled using your favorite \LaTeX compiler. This file should serve as a good example of the basic structure of a ``.tex" file as well as many of the most basic commands needed for typesetting documents involving mathematical symbols and expressions. For more of a description on how each command works, please consult the links found on our course webpage. \end{abstract} \section{Lists} %%%%%%%%%%%%%%% \begin{enumerate} \item {\bf First Point (Bold Face)} \item {\em Second Point (Italic)} \item {\Large Third Point (Large Font)} \begin{enumerate} \item {\small First Subpoint (Small Font)} \item {\tiny Second Subpoint (Tiny Font)} \item {\Huge Third Subpoint (Huge Font)} \end{enumerate} \item[$\bullet$] {\sf Bullet Point (Sans Serif)} \item[$\circ$] {\sc Circle Point (Small Caps)} \end{enumerate} \section{Equations} %%%%%%%%%%%%%%%%%%% \subsection{Binomial Theorem} \begin{theorem}[Binomial Theorem] For any nonnegative integer $n$, we have $$(1+x)^n = \sum_{i=0}^n {n \choose i} x^i$$ \end{theorem} \subsection{Taylor Series} The Taylor series expansion for the function $e^x$ is given by \begin{equation} e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots = \sum_{n\geq 0} \frac{x^n}{n!} \end{equation} \subsection{Sets} \begin{theorem} For any sets $A$, $B$ and $C$, we have $$ (A\cup B)-(C-A) = A \cup (B-C)$$ \end{theorem} \begin{proof} \begin{eqnarray*} (A\cup B)-(C-A) &=& (A\cup B) \cap (C-A)^c\\ &=& (A\cup B) \cap (C \cap A^c)^c \\ &=& (A\cup B) \cap (C^c \cup A) \\ &=& A \cup (B\cap C^c) \\ &=& A \cup (B-C) \end{eqnarray*} \end{proof} \section{Tables} %%%%%%%%%%%%%%%% \begin{center} \begin{tabular}{l||c|r} left justified & center & right justified \\ \hline 1 & 3.14159 & 5 \\ 2.4678 & 3 & 1234 \\ \hline \hline 3.4678 & 6.14159 & 1239 \end{tabular} \end{center} \section{A Picture} %%%%%%%%%%%%%%%%%%% \begin{center} \begin{picture}(100,100)(0,0) \setlength{\unitlength}{1pt} \put(20,70){\circle{30}} \put(20,70){\circle*{10}} % left eye \put(80,70){\circle{30}} \put(80,70){\circle*{10}} % right eye \put(40,40){\line(1,2){10}} \put(60,40){\line(-1,2){10}} \put(40,40){\line(1,0){20}} % nose \put(50,20){\oval(80,10)[b]} % mouth \multiput(0,90)(4,0){10}{\line(1,3){4}} % left eyebrow \multiput(100,90)(-4,0){10}{\line(-1,3){4}} % right eyebrow \end{picture} \end{center} \end{document}
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