# [OS X TeX] Card Format

Alain Schremmer schremmer.alain at gmail.com
Sat May 9 01:11:28 EDT 2009

On May 6, 2009, at 4:42 PM, Alan Litchfield wrote:

> I think I would build a tabular framework for the page and set the
> width to linewidth, then I'd use minipages to construct each of the
> examples and solutions. You'd need to be fairly careful about
> spacing though to ensure that you don't end up with very large
> blank spots at the end of pages.
>
> Alternatively you could use frameboxes and minipages in them.

I just found out that one could use minipages within tables. Sigh.
>
> Be sure to let us know how you get on.

I bad-mouthed Companion 2ed as it has both Chapter 4 Layout of the
Page and an appendix on boxes and rules (p860). So, in preparation
for understanding the suggestions I got, both onlist and offlist, I
fiddled a bit and came up with the following. I still have to reduce
the margins so as to increase the working spaces, drop the answers to
the bottom of the boxes use parameters and, obviously the hard part,
"solidify" things.   I also just remembered about the package
boxedminipage.

Grateful regards
--schremmer

\documentclass[11pt]{book}
\usepackage{amsmath}
\usepackage{amssymb}
% ssssssssssssssssssssssssssssssssssssssssssssssssssssssss End PREAMBLE
\begin{document}
%ssssssssssssssssssssssssssssssss
\begin{center}
\begin{tabular}{@{} |l|l| @{}}
\hline
%Row0title
&	\textbf{Solution}
\\
\hline
%Row0text
\begin{minipage}[t]{50mm}
Given the function $f$ whose global input-output rule is
\begin{displaymath}
x \xrightarrow{\hspace{2mm} f\hspace{2mm}} f(x) =
\frac{\polynomial[reciprocal]{1,,,-8}}{\polynomial[reciprocal]{1,,-1}}
\end{displaymath}
find the sign of the \emph{concavity} of $f$ near $\infty$
\vspace{2mm}
\end{minipage}
&	\begin{minipage}[t]{110mm}
We approximate separately the numerator and the denominator and \emph
{then} divide:
\begin{displaymath}
x\; large \xrightarrow{\hspace{2mm} f_{\infty}\hspace{2mm}} f_{\infty}
(x) =
\frac{\polynomial[reciprocal]{1,,,}+[\dots]}{\polynomial[reciprocal]
{1,,}+[\dots]}
= x +[\dots]
\end{displaymath}
This, though, loses the \emph{concavity} in [\dots] so \emph{here} we
must approximate \emph{after} we divide (by stopping the division
when we reach a term with \emph{concavity}):
\begin{center}
\newdimen\digitwidth \settowidth\digitwidth{0}
\def~{\hspace{\digitwidth}}
\def\divrule#1#2{%
\noalign{\moveright#1\digitwidth%
\vbox{\hrule width#2\digitwidth}}}
%This is the divisor
$x^{2}$
%			
$-1$
\hspace{0mm}
\begin{tabular}[b]{@{}l@{}}
%This is the quotient
\hspace{3mm}
$x$
\hspace{5mm}
$+x^{-1}$
\\
\hline
\big)
\begin{tabular}[t]{@{}l@{}l@{}}
%This is the dividend (Its length controls the line.)
$x^{3}$
\hspace{4mm}

\hspace{4mm}

\hspace{4mm}
$-8$
\\
%This is the first product
%				\hspace{2mm}
$x^{3}$
\hspace{3mm}

\hspace{3mm}
$-x$
\\ %origin,length
\divrule{0}{17}
%This is the first remainder
\hspace{5mm}

\hspace{6mm}
$+x$
\hspace{1mm}
$-8$
\\
\end{tabular}
\end{tabular}
\end{center}
so that we have
\begin{displaymath}
x\; large \xrightarrow{\hspace{2mm} f_{\infty}\hspace{2mm}} f_{\infty}
(x) = x + x^{-1} + [\dots]
\end{displaymath}
where the $x^{-1}$ term gives $\left.\text{Concavity-sign }f\right|_ {\infty} = (\cup,\cap)$.
\vspace{1mm}
\end{minipage}
\\
\hline
\end{tabular}
\end{center}

%ssssssssssssssssssssssssssssssss
\vspace{-4.5mm}
%ssssssssssssssssssssssssssssssss
\begin{center}
\begin{tabular}{@{} |l|l| @{}}
\hline
\begin{minipage}[t]{145mm}
\textbf{Exercise 13.1.} Given the function $f$ whose global input-
output rule is
\begin{displaymath}
x \xrightarrow{\hspace{2mm} f\hspace{2mm}} f(x) =
\frac{\polynomial[reciprocal]{1,,,-8}}{\polynomial[reciprocal]{1,,-1}}
\end{displaymath}
find the sign of the \emph{slope} of $f$ near $\infty$.\vspace{22mm}
\end{minipage}
&	\begin{minipage}[b]{15mm}
$(\diagup,\diagup)$
\end{minipage}
\\
\hline
\begin{minipage}[t]{145mm}
\textbf{Exercise 13.2.} Given the function $f$ whose global input-
output rule is
\begin{displaymath}
x \xrightarrow{\hspace{2mm} f\hspace{2mm}} f(x) =
\frac{\polynomial[reciprocal]{1,,,-8}}{\polynomial[reciprocal]{1,,-1}}
\end{displaymath}
find the sign of the \emph{height} of $f$ near $\infty$.\vspace{22mm}
\end{minipage}
&	\begin{minipage}[b]{15mm}
$(+,-)$
\end{minipage}
\\
\hline
\end{tabular}
\end{center}

\end{document}