# How to enclose theorems and definition in rectangles?

The following code

\documentclass{article}

\usepackage{amsthm}
\usepackage{amsmath}
\usepackage{mathtools}

\usepackage[left=1.5in, right=1.5in, top=0.5in]{geometry}

\newtheorem{definition}{Definition}
\newtheorem{theorem}{Theorem}

\begin{document}
\title{Extra Credit}
\maketitle

\begin{definition}
If f is analytic at $z_0$, then the series

$$f(z_0) + f'(z_0)(z-z_0) + \frac{f''(z_0)}{2!}(z-z_0)^2 + \cdots = \sum_{n=0}^{\infty} \frac{f^{(n)}(z_0)}{n!}(z-z_0)^n$$

is called the Taylor series for f around $z_0$.
\end{definition}

\begin{theorem}
If f is analytic inside and on the simple closed positively oriented contour $\Gamma$ and if $z_0$ is any point inside $\Gamma$, then
$$f^{(n)}(z_0) = \frac{n!}{2\pi i} \int_{\Gamma} \frac{f(\zeta)}{(\zeta - z_0)^{n+1}}d\zeta \hspace{1cm} (n=1,2,3, \cdots )$$
\end{theorem}

\begin{theorem}
(Cauchy's Integral Formula) Let $\Gamma$ be a simple closed positively oriented contour. If $f$ is analytic in some simply connected domain $D$ containing $\Gamma$ and $z_0$ is any point inside $\Gamma$, then
$$f(z_0)= \frac{1}{2\pi i} \int_{\Gamma} \frac{f(z)}{z-z_0} dz$$
\end{theorem}
\noindent \hrulefill

\begin{theorem}
If f is analytic in the disk $|z-z_0|<R'$, then the Taylor series $(1)$ converges to $f(z)$ for all $z$ in this disk.
\end{theorem}


produces the following image

How can I enclose Definition 1, Theorem 1, and Theorem 2 in separate rectangles. And have these rectangles separated by a space?

• Do you want all theorems/definition to be enclosed in a frame, or only some? – Bernard Apr 2 at 21:55
• I would like all theorems/definitions to be enclosed in a frame except for Theorem 3 – K.M Apr 2 at 21:57
• In this case you should take a look at the \newframedtheorem command in ntheorem. – Bernard Apr 2 at 22:06

## 2 Answers

You can try with shadethm package, it can do all you want and many more. In you example what you need is:

\documentclass{article}
\usepackage{shadethm}
\usepackage{mathtools}

\newshadetheorem{boxdef}{Definition}[section]
\newshadetheorem{boxtheorem}[boxdef]{Theorem}
\newtheorem{theorem}[boxdef]{Theorem}

\setlength{\shadeboxsep}{2pt}
\setlength{\shadeboxrule}{.4pt}
\setlength{\shadedtextwidth}{\textwidth}
\addtolength{\shadedtextwidth}{-2\shadeboxsep}
\addtolength{\shadedtextwidth}{-2\shadeboxrule}
\setlength{\shadeleftshift}{0pt}
\setlength{\shaderightshift}{0pt}
\definecolor{shadethmcolor}{cmyk}{0,0,0,0}
\definecolor{shaderulecolor}{cmyk}{0,0,0,1}

\begin{document}

\section{Boxed theorems}

\begin{boxdef}
If f is analytic at $z_0$, then the series

$$f(z_0) + f'(z_0)(z-z_0) + \frac{f''(z_0)}{2!}(z-z_0)^2 + \cdots = \sum_{n=0}^{\infty} \frac{f^{(n)}(z_0)}{n!}(z-z_0)^n$$

is called the Taylor series for f around $z_0$.
\end{boxdef}

\begin{boxtheorem}
If f is analytic inside and on the simple closed positively oriented contour $\Gamma$ and if $z_0$ is any point inside $\Gamma$, then
$$f^{(n)}(z_0) = \frac{n!}{2\pi i} \int_{\Gamma} \frac{f(\zeta)}{(\zeta - z_0)^{n+1}}d\zeta \hspace{1cm} (n=1,2,3, \cdots )$$
\end{boxtheorem}

\begin{boxtheorem}
(Cauchy's Integral Formula) Let $\Gamma$ be a simple closed positively oriented contour. If $f$ is analytic in some simply connected domain $D$ containing $\Gamma$ and $z_0$ is any point inside $\Gamma$, then
$$f(z_0)= \frac{1}{2\pi i} \int_{\Gamma} \frac{f(z)}{z-z_0} dz$$
\end{boxtheorem}
\noindent \hrulefill

\begin{theorem}
If f is analytic in the disk $|z-z_0|<R'$, then the Taylor series $(1)$ converges to $f(z)$ for all $z$ in this disk.
\end{theorem}

\end{document}


which produces the following:

• For \newshadetheorem{boxdef}{Definition}[section] \newshadetheorem{boxtheorem}[boxdef]{Theorem} \newtheorem{theorem}[boxdef]{Theorem}, why is boxdef in brackets? – K.M Apr 2 at 22:37
• In the first box, the space above the equation is larger than that below the equation The reason for this is the blank line above . Blank lines in that position should be avoided. – barbara beeton Apr 3 at 21:06
• @K.M the brackets [boxdef] is to enumerate all different kind of theorems with the same enumeration – Luis Turcio Apr 3 at 21:50
• @barbarabeeton the spacing is due to the original code written by K.M, it has a blank line before $$and one after$$. Removing or commenting this blank lines should be enough to correct spacing. – Luis Turcio Apr 3 at 21:54
• @LuisTurcio -- Indeed, commenting or removing the blank line is what is recommended. I really should have posted this comment to the original question. – barbara beeton Apr 4 at 0:12

Here is a solution with thmtools, which cooperates wit amsthm. Unrelated: you don't have to load amsmath if you load mathtools, as the latter does it for you:

\documentclass{article}
\usepackage{amsthm, thmtools}
\usepackage{mathtools}

\usepackage[left=1.5in, right=1.5in, top=0.5in]{geometry}

\newtheorem{definition}{Definition}
\newtheorem{theorem}{Theorem}

\declaretheorem[sibling=definition, shaded={rulecolor=black, rulewidth=0.6pt, bgcolor={rgb}{1,1,1}},name=Definition]{boxeddef}
\declaretheorem[sibling=theorem, shaded={rulecolor=black, rulewidth=0.6pt, bgcolor={rgb}{1,1,1}},name=Theorem]{boxedthm}

\begin{document}
\title{Extra Credit}
\author{}
\maketitle

\begin{boxeddef}
If f is analytic at $z_0$, then the series

$$f(z_0) + f'(z_0)(z-z_0) + \frac{f''(z_0)}{2!}(z-z_0)^2 + \cdots = \sum_{n=0}^{\infty} \frac{f^{(n)}(z_0)}{n!}(z-z_0)^n$$

is called the Taylor series for f around $z_0$.
\end{boxeddef}

\begin{boxedthm}
If f is analytic inside and on the simple closed positively oriented contour $\Gamma$ and if $z_0$ is any point inside $\Gamma$, then
$$f^{(n)}(z_0) = \frac{n!}{2\pi i} \int_{\Gamma} \frac{f(\zeta)}{(\zeta - z_0)^{n+1}}d\zeta \hspace{1cm} (n=1,2,3, \cdots )$$
\end{boxedthm}

\begin{boxedthm}
(Cauchy's Integral Formula) Let $\Gamma$ be a simple closed positively oriented contour. If $f$ is analytic in some simply connected domain $D$ containing $\Gamma$ and $z_0$ is any point inside $\Gamma$, then
$$f(z_0)= \frac{1}{2\pi i} \int_{\Gamma} \frac{f(z)}{z-z_0} dz$$
\end{boxedthm}

\noindent \hrulefill

\begin{theorem}
If f is analytic in the disk $|z-z_0|<R'$, then the Taylor series $(1)$ converges to $f(z)$ for all $z$ in this disk.
\end{theorem}

\end{document}