[Tex/LaTex] Absolute positioning of wrapfig

floatspositioningwrapfigure

In Microsoft Word, it is common to create a floating text box, separate from the normal body text, that can be placed anywhere on the page. The body text will then wrap around this text box so everything is readable.

In (La)TeX, the wrapfigure package allows figures to be created that the body text will wrap around, but they may only be placed at the beginning of paragraphs and on the edges of the page. The package doesn't seem to allow absolute placement independent of the body text and margins. Using the absolute-placement package textpos in conjunction with wrapfig will simply break the wrapping.

So how can I have the best of both worlds—text wrapping and absolute placement (at the very least, vertical placement)?

Something like the diagram in this question, or this image from Word.

(Edited for generality and clarity.)

Best Answer

Placing a picture somewhere on the page is easy. The problem is the wrapping. It is easy to say "indent line 5-10 of the next paragraph". But due to the asynchronic page builder there is in tex no way (unless your document is very simple) to say "indent line 5-10 of the next page". Perhaps it is possible with luatex. But with pdflatex you will have to do it manually. See also on CTAN wrapfig/multiple-span.txt.

Don't

use capital letters (like R) for float specifiers. Capital letters mean the figure can float.
don't use a group to wrap the wrapfigure environment.

Do

put sufficient text in the paragraph following wrapfigure.
specify the number of lines to be cut (so as to accomodate wrapfigure) as tols by olga.saucedo.

Here is the commented code.

\documentclass[11pt]{article}
\usepackage{graphicx}
\usepackage{amsmath}
\usepackage[margin={2.2cm}]{geometry}%
\usepackage{wrapfig, blindtext}

\begin{document}
\section{The problem}
\begin{wrapfigure}[12]{r}{0.3\textwidth}
\centering
\rule{0.9\linewidth}{0.75\linewidth}
\caption{\label{fig:d1}Integrating a function}
\end{wrapfigure}
%
To find a numerical value for the integral of a real valued function of a real
variable over a specific range over the real line. This means to
evaluate
\[
I={\displaystyle\int\limits_{a}^{b}}f\left(  x\right)  \ dx
\]
Geometrically, this integral represents the area under $f(x)$
from $a$ to $b$.%

%% uncomment following dummy text to see the undesired effect
This is some dummy text. This is some dummy text. This is some dummy text. This is some dummy text.
This is some dummy text. This is some dummy text. This is some dummy text. This is some dummy text.
This is some dummy text. This is some dummy text. This is some dummy text. This is some dummy text.
This is some dummy text. This is some dummy text. This is some dummy text. This is some dummy text.
This is some dummy text. This is some dummy text. This is some dummy text. This is some dummy text.
This is some dummy text. This is some dummy text. This is some dummy text. This is some dummy text.
This is some dummy text. This is some dummy text. This is some dummy text. This is some dummy text.

\section{Solution}
%\begingroup   %% why group?
\begin{wrapfigure}[12]{r}{0.3\textwidth}
\centering
\rule{0.9\linewidth}{0.75\linewidth}
\caption{\label{fig:d2}Numerical integration}
\end{wrapfigure}
%\endgroup

We can always approximate the area by dividing it in equal width strips and
then sum the areas of all the strips.

enter image description here

In general, there will always be an error in the estimate of the area using
this method. The error will become smaller the more strips
we use (which implies a smaller strip width). Hence we can write%

\[
{\displaystyle\int\limits_{a}^{b}}
f\left(  x\right)  \ dx=\left(
{\displaystyle\sum\limits_{i=1}^{N}}
\Delta x\ f\left(  x_{i}\right)  \right)  +E
\]

Where $E$ is the error between the actual area and the approximated area using the
above method of numerical integration. $N$ above is the number of strips or
can also be refereed to as the number of integration points.

Instead of keep referring to the 'width of the strip' all the time, we will
call this quantity the weight $w_{i}$ that we will multiply the value of the
function with to obtain the area. Hence the above becomes

\[%
{\displaystyle\int\limits_{a}^{b}}
f\left(  x\right)  \ dx=\left(
{\displaystyle\sum\limits_{i=1}^{N}}
w_{i}\ f\left(  x_{i}\right)  \right)  +E
\]

Using implied summation on indices the above becomes

\[%
{\displaystyle\int\limits_{a}^{b}}
f\left(  x\right)  \ dx=w_{i}\ f\left(  x_{i}\right)  +E
\]

In the above we divided the range of the integration (the distance between the
upper and lower limits of integration) into equal intervals. We can decide to
evaluate $f(x_{i})$ at the middle of the strip or at the start
of the strip or at the end of the strip. In the diagram above we evaluated the
$f(x)$ at the left end of the strip.

Our goal is to evaluate this integral such as the error $E$ is minimum and using
the smallest number of integration points. In a sense this can be considered
an optimization problem with constraints: minimize the error of integration
using the smallest possible number of points.

\end{document}

Best Answer

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Don't

Do

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