LaTeX/Math jokes [closed]

Question

Lighten up a bit guys. How about the following

\documentclass{article}
\usepackage{graphicx}
\begin{document}
If~~$\displaystyle\lim_{x\rightarrow8}\frac{1}{x{-}8}=\infty$
~~then~~$\displaystyle\lim_{x\rightarrow5}\frac{1}{x{-}5}=\rotatebox{90}{5}$
\end{document}

Which gives —

Can anyone give me some nice Tikz of PStricks code for the following

I am sure there there must be more Latex humor out there, even among the Germans ;-)

Answer 1

How about:

\font\donteveruseoutsidecartoons="Comic Sans MS" at 14pt
\font\donteveruseoutsidecartoonstwo="Comic Sans MS/IB:slant=.3pt" at 22pt
\input tikz
\usetikzlibrary{shapes.callouts,positioning}
\tikzpicture[mycallout/.style={draw,ellipse callout,inner sep=1.2ex,
    callout relative pointer={#1}}]
  \node[mycallout={(.5cm,-.5cm)}] (ico) {\donteveruseoutsidecartoons Be rational};
  \node[mycallout={(-.5cm,-.5cm)}, right=of ico] (pco) {\donteveruseoutsidecartoons Get real};
  \node[below=.1ex of ico.pointer] {\donteveruseoutsidecartoonstwo i};
  \node[below left=.1ex of pco.pointer] {\donteveruseoutsidecartoonstwo π\vphantom(};
\endtikzpicture
\bye

Answer 2

I was laughing about this one quite a bit. I had to recall it from German and translate it. Feel free to adjust the language.

\documentclass[a4paper,12pt]{article}
\usepackage[margin=1in]{geometry}

\usepackage{graphicx}
\usepackage{pgffor}
\pagestyle{empty}

\parindent=0pt
\begin{document}
\large\sffamily

\resizebox{\textwidth}{!}{The first 500 natural random numbers}

\vspace{1em}

\centerline{\LARGE For easy access listed in their natural order}
\bigskip

\par\noindent
\foreach \n in {1,...,500} {%
    \makebox[1.8em][r]{\n}
}

\end{document}

Answer 3

\documentclass{article}
\usepackage{amsmath}

\begin{document}

Expand $(a+b)^n$:
\begin{gather*}
  (a + b)^n\\
  (a\ + \ b)^n\\
  (a\quad + \quad b)^n\\
  (a\qquad + \qquad b)^n
\end{gather*}

\end{document}

EDIT: to Andrew:

\documentclass{article}

\newcount\mycntr

\begin{document}

Expand $(a+b)^n$:

\begin{center}
  \mycntr=0
  \loop\advance\mycntr by 1
  \ifnum\mycntr<40
    $(a\hskip\mycntr pt +\hskip\mycntr pt b)^n$\\
  \repeat
\end{center}

\end{document}

Answer 4

A silly one I found on a blog:

\documentclass{article}
\usepackage{graphicx}
\usepackage{cancel}

\begin{document}

\begin{eqnarray*}
    \frac{1}{n}\sin x & = & \mathrm{?} \\
    \frac{1}{\cancel{n}} \mathrm{si}\cancel{\mathrm{n}} ~x & = & \mathrm{?} \\
    \mathrm{six} & = & 6
\end{eqnarray*}

\end{document}

This one from xkcd is a classic:

\documentclass{article}
\usepackage{graphicx}
\usepackage{amsmath}

\begin{document}

$
\begin{bmatrix}
    \cos 90^{\circ} & \sin 90^{\circ}\\
   -\sin 90^{\circ} & \cos 90^{\circ}
\end{bmatrix}
\begin{bmatrix} a1 \\ a2 \end{bmatrix}
=
$
\rotatebox[origin=c]{270}{$\begin{bmatrix} a1 \\ a2 \end{bmatrix}$}

\end{document}

Answer 5

I never played with tikz before, so that's my first attempt. I simply love the following joke:

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{decorations.pathmorphing}
\usepackage{frcursive}
\begin{document}

Find $x$.

\begin{tikzpicture}
\draw (0,0) -- (4,0) node[midway,below] {4 cm}
   -- (4,3) node[midway,right] {3 cm}
   -- (0,0) node[midway,left,circle,draw=blue,decorate,decoration={random steps,segment length=1pt,amplitude=0.5pt}]{$x$}
   -- (4,0) rectangle (3.7,0.3)
   -- cycle;
\draw (0.4,0) arc (0:30:0.5);
\draw (4,2.6) arc (270:226:0.5);
\draw (1,2.1) node []{\color{blue}\fontfamily{frc}\selectfont{It's here!}};
\end{tikzpicture}

\end{document}

The result:

Answer 6

Mathematicians do it in \LaTeX.

I'll get my coat.

Answer 7

To all my Australian friends:

\documentclass{article}
\usepackage{graphicx}
\begin{document}
\noindent Are there any \raisebox{1ex}{\rotatebox[origin=B]{180}{Australians}}
here mate?\\ Hang on tight!
\end{document}

Answer 8

Here's another joke I remember:

\documentclass{article}

\usepackage[T1]{fontenc}
\usepackage[light,math]{anttor}

\begin{document}

Let $\displaystyle f(a) = \sqrt[n]{e^x}$.

\begin{minipage}{0.5\textwidth}
   \begin{eqnarray*}
      \lim_{t\rightarrow\infty} f(a) - \frac{i}{f(t)} & = & \frac{d}{dx} f(u) \\
      \lim_{t\rightarrow\infty} f(a) - \frac{i}{\infty} & = & \frac{d}{dx} f(u) \\
      \lim_{t\rightarrow\infty} f(a) - 0 & = & \frac{d}{dx} f(u)
   \end{eqnarray*}
\end{minipage}

\bigskip

Then

\begin{minipage}{0.5\textwidth}
   \begin{eqnarray*}
      \sqrt[n]{e^x} & = & \frac{d}{dx} f(u) \\
      (\sqrt[n]{e^x})^n & = & \frac{d}{dx} f(u)^n \\
      e^x & = & \frac{d}{dx} f(u)^n \\
      \int e^x & = & f(u)^n
   \end{eqnarray*}
\end{minipage}

\end{document}

The output:

Answer 9

Riffing on the "Australians" joke, borrowing from this answer.