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-rw-r--r--info/examples/Math/09-20-1.ltx48
1 files changed, 27 insertions, 21 deletions
diff --git a/info/examples/Math/09-20-1.ltx b/info/examples/Math/09-20-1.ltx
index 3642bc9c57..58c393d164 100644
--- a/info/examples/Math/09-20-1.ltx
+++ b/info/examples/Math/09-20-1.ltx
@@ -1,38 +1,44 @@
%%
%% Ein Beispiel der DANTE-Edition
%% Mathematiksatz mit LaTeX
-%% 2. Auflage
-%%
-%% Beispiel 09-20-1 auf Seite 208.
-%%
-%% Copyright (C) 2012 Herbert Voss
+%% 3. Auflage
+%% Beispiel 09-20-1 auf Seite 201.
+%% Copyright (C) 2018 Herbert Voss
%%
%% It may be distributed and/or modified under the conditions
%% of the LaTeX Project Public License, either version 1.3
%% of this license or (at your option) any later version.
-%%
%% See http://www.latex-project.org/lppl.txt for details.
%%
-%%
%% ====
% Show page(s) 1
%%
%%
-\documentclass[]{exaarticle}
+\documentclass[10pt]{exaartplain}
\pagestyle{empty}
-\setlength\textwidth{166.324pt}
+\setlength\textwidth{352.81416pt}
\AtBeginDocument{\setlength\parindent{0pt}}
-\setcounter{equation}{31}
-\renewcommand\theequation{9.\arabic{equation}}
-\StartShownPreambleCommands
-\usepackage{xypic}
-\StopShownPreambleCommands
+%StartShownPreambleCommands
+\usepackage{amsmath,mathastext} \MTDeclareVersion[n]{lmvtt}{T1}{lmvtt}{m}{n}
+%StopShownPreambleCommands
\begin{document}
-\begin{equation}
-\xymatrix{%
- A\POS[];[d]**\dir{~},[];[dr]**\dir{-} & B & C\\
- D & E\POS[];[l]**\dir{.},[];[r]**\dir{~}
- & F\POS[];[dl]**\dir{~}\\
- G & H & I}
-\end{equation}
+\MTversion{lmvtt}
+Let $(X,Y)$ be two functions of a variable $a$. If they obey the differential system $(VI_{\nu,n})$:
+\begin{align*} a\frac{d}{da} X &= \nu
+ X - (1 - X^2)\frac{2n a}{1 - a^2}\frac{aX+Y}{1+a XY} \\
+a\frac{d}{da} Y &= -(\nu+1) Y + (1-Y^2)\frac{2n a}{1 - a^2}\frac{X+aY}{1+a XY}
+\end{align*}
+then the quantity $q=a\frac{aX+Y}{X+aY}$ satisfies as function of $b=a^2$ the $P_{VI}$
+differential equation:
+\begin{equation*}
+\begin{split}
+ \frac{d^2 q}{db^2} = \frac12\left\{\frac1q+\frac1{q-1}
+ +\frac1{q-b}\right\}\left(\frac{dq}{db}\right)^2 - \left\{\frac1b+\frac1{b-1}
+ +\frac1{q-b}\right\}\frac{dq}{db}\\+\frac{q(q-1)(q-b)}{b^2(b-1)^2}\left\{\alpha+\frac{\beta
+ b}{q^2} + \frac{\gamma (b-1)}{(q-1)^2}+\frac{\delta
+ b(b-1)}{(q-b)^2}\right\}
+\end{split}
+\end{equation*}
+with parameters $(\alpha,\beta,\gamma,\delta) = (\frac{(\nu+n)^2}2,
+\frac{-(\nu+n+1)^2}2, \frac{n^2}2, \frac{1 - n^2}2)$.
\end{document}