\documentclass[12pt]{amsart}
\usepackage[text={6.5in,9in},centering]{geometry}
\usepackage{amssymb}

\theoremstyle{plain}
\newtheorem{theorem}{Theorem}[section]

\newcommand{\NN}{{\mathbb N}}
\newcommand{\RR}{{\mathbb R}}
\newcommand{\ZZ}{{\mathbb Z}}

\begin{document}

\title{Some Mathematicul Results}
\author{Emmy Noether}
\address{Bryn Mawr College, Bryn Mawr, PA}

\maketitle

\section{Getting started}\label{start}
Here we go.  This is a sentence which just rambles on long enough so that you
can see what the shape of a paragraph is.

And here is the next paragraph.

\section{Simple math}\label{simple}
Finally, some math:

$$
\ell = r\theta
$$
\vskip.2cm
$$
A = \frac12r^2\theta
$$
\vskip.2cm
$$
A(\Delta) = \frac12ab\sin\theta = \sqrt{(s-a)(s-b)(s-c)}
$$
\vskip.2cm
$$
\frac{\cos A}{a} = \frac{\cos B}{b} = \frac{\cos C}{c}
$$
\vskip.2cm
$$
c^2 = a^2+b^2-2ab\sin C
$$
\vskip.2cm
$$
\text{truth} = \text{beauty}
$$
\vskip.2cm
$$
\text{brunch} = \text{breakfast} + \text{lunch} + \text{cantaloupe}
$$
\vskip.2cm
$$
a+ar+ar^2+\dots+ar^N=a\left(\frac{1-r^{N+1}}{1-r}\right)
$$
\section{Less simple math}\label{lesssimple}
Permutations are
$$
{}_nP_k=\frac{n!}{(n-k)!}
$$
while combinations are
$$
{}_nC_k=\begin{pmatrix}n\\k\end{pmatrix} = ...
$$
\vskip.2cm
\indent{\it The series $\sum_{n=1}^\infty 1/n^p$ converges if and only if
$p>1$.}
\vskip.2cm
For each $n\in\NN$, define a function $f_n:\RR\to\RR$ by
$$
f_n(x)=\begin{cases}1&\text{if $|x|\ge1/n$}\\
n|x|&\text{if $|x|<1/n$}
\end{cases}
$$
\vskip.2cm
The gradient is defined by
$$
\nabla f=\begin{pmatrix}\frac{\partial f}{\partial x}\\
\frac{\partial f}{\partial y}\\
\frac{\partial f}{\partial z}\end{pmatrix}=
\frac{\partial f}{\partial x}\vec{\imath} + \dots
$$
\vskip.2cm
Compute
$$
\lim_{x\to-\infty}\frac{\sqrt{4x^2+1}}{3x-1}\ .
$$
\vskip.2cm
What is $y^\prime$ if
$$
y = \left(x+\frac{1}{x^2}\right)^{\sqrt{7}}\ ?
$$
\vskip.2cm
\begin{equation}
\sum_{j=1}^nj=\frac{n(n+1)}{2}.
\end{equation}
\vskip.2cm
Say $T:\RR^2\to\RR^2$ is a linear transformation which satisfies
$$
T\left(\left[\begin{matrix}1\\1\end{matrix}\right]\right) =
\left[\begin{matrix}2\\3\end{matrix}\right] \qquad\text{and}\qquad
T\left(\left[\begin{matrix}\ 1\\-1\end{matrix}\right]\right) =
\left[\begin{matrix}0\\5\end{matrix}\right]\quad .
$$
Then $T\left(\left[\begin{matrix} x\\y\end{matrix}\right]\right) = ...$?

\section{Conclusion}
In ??? we gave simple equations, in ??? they were not so simple.
\end{document}