new doc package pstricks_calcnotes (7jun10)

git-svn-id: svn://tug.org/texlive/trunk@18816 c570f23f-e606-0410-a88d-b1316a301751

21 files changed, 919 insertions, 0 deletions

diff --git a/Master/texmf-dist/doc/latex/pstricks_calcnotes/Convert_PstricsCode_To_Pdf/Readme.txt b/Master/texmf-dist/doc/latex/pstricks_calcnotes/Convert_PstricsCode_To_Pdf/Readme.txt
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+++ b/Master/texmf-dist/doc/latex/pstricks_calcnotes/Convert_PstricsCode_To_Pdf/Readme.txt
@@ -0,0 +1,16 @@
+Dear friends,
+
+To convert a PSTricks-code picture into a graphic file ".pdf", as doing an exercise, run 
+the file "convert.tex" by the following steps:
+
+  1) latex convert.tex (then you automatically get the output file "convert-fig1.tex").
+  2) latex convert-fig1.tex (then run divps program the dvi output file to get 
+             the file "convert-fig1.ps").
+  3) call the file "convert-fig1.ps" by the Ghostview progam to convert this file into its 
+   eps version (you can rename the result, say "fig1.eps").  
+  4) use the eps2pdf program to convert the file "fig1.eps" into the file "fig1.pdf". 
+
+
+The file "test.pdf" is the graph of the Max probability distribution function. 
+Check your file "fig1.pdf" to see this graph!
+
diff --git a/Master/texmf-dist/doc/latex/pstricks_calcnotes/Convert_PstricsCode_To_Pdf/convert.tex b/Master/texmf-dist/doc/latex/pstricks_calcnotes/Convert_PstricsCode_To_Pdf/convert.tex
new file mode 100644
index 00000000000..e72a3a2386d
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+++ b/Master/texmf-dist/doc/latex/pstricks_calcnotes/Convert_PstricsCode_To_Pdf/convert.tex
@@ -0,0 +1,42 @@
+\documentclass[11pt,a4paper,oneside]{article}
+\usepackage{pdftricks}
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+\def\RiemannSum#1#2#3#4#5#6#7#8#9{%
+\psplot[linecolor=blue]{#1}{#2}{#3}
+\pscustom[linecolor=red]{%
+\psline{-}(#1,0)(#1,0)
+\multido{\ni=#5,\ne=#6}{#4}
+{\psline(*{\ni} {#8})(*{\ne} {#9})}}
+\multido{\ne=#6,\nc=#7}{#4}
+{\psdot(*{\nc} {#3})
+\psline[linestyle=dotted,dotsep=1.5pt](\nc,0)(*{\nc} {#3})
+\psline[linecolor=red](\ne,0)(*{\ne} {#9})}
+}
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+\def\vecfld#1#2#3#4#5#6{%
+\multido{#2}{#4}
+{\multido{#1}{#3}
+{\parametricplot[algebraic,arrows=->,linecolor=red]{0}{1}
+{\nx+((#5)*t)*(1/sqrt(1+(#6)^2))|\ny+((#5)*t)*(1/sqrt(1+(#6)^2))*(#6)}}}}
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+\title{Making files \texttt{.ps} from \texttt{PSTricks}}
+\begin{document}
+\maketitle
+\begin{psinputs}
+\usepackage{pstricks}
+\usepackage[dvips]{geometry}
+\usepackage{pst-plot,pst-coil,multido,pst-xkey,pst-node}
+\usepackage{pst-func}
+\usepackage{pstricks-add}
+\end{psinputs}
+\section{Here is the Pstricks code to be converted}
+\begin{pdfpic}
+\begin{pspicture}(-1.2,-1)(5.5,5.2)
+\psset{xunit=0.5cm,yunit=5cm}
+\psplot[linecolor=blue,algebraic=true,plotpoints=500]
+{-2}{10}{IfTE(x<1.3,(ch(1)+sh(1))^(-0.5*(x-1)^2),IfTE(x<1.8,(ch(1)+sh(1))^(-0.5*(x-1.6)^2),%
+IfTE(x<2.25,(ch(1)+sh(1))^(-0.5*(x-2)^2),IfTE(x<2.75,(ch(1)+sh(1))^(-0.5*(x-2.5)^2),IfTE(x<5,(ch(1)+sh(1))^(-0.5*(x-3)^2),(ch(1)+sh(1))^(-0.5*(x-7)^2))))))}
+\psaxes[labelFontSize={$\footnotesize$},Dy=0.2,tickstyle=bottom]{->}(0,0)(-2.4,-0.1)(11,1.1)[$x$,-90][$y$,180]
+\end{pspicture}
+\end{pdfpic}
+\end{document}
diff --git a/Master/texmf-dist/doc/latex/pstricks_calcnotes/Convert_PstricsCode_To_Pdf/test.pdf b/Master/texmf-dist/doc/latex/pstricks_calcnotes/Convert_PstricsCode_To_Pdf/test.pdf
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+++ b/Master/texmf-dist/doc/latex/pstricks_calcnotes/Convert_PstricsCode_To_Pdf/test.pdf
diff --git a/Master/texmf-dist/doc/latex/pstricks_calcnotes/For_Pdf_Output/Fig1-1.pdf b/Master/texmf-dist/doc/latex/pstricks_calcnotes/For_Pdf_Output/Fig1-1.pdf
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index 00000000000..8efa93a02ca
--- /dev/null
+++ b/Master/texmf-dist/doc/latex/pstricks_calcnotes/For_Pdf_Output/Fig1-1.pdf
diff --git a/Master/texmf-dist/doc/latex/pstricks_calcnotes/For_Pdf_Output/Fig1-2.pdf b/Master/texmf-dist/doc/latex/pstricks_calcnotes/For_Pdf_Output/Fig1-2.pdf
new file mode 100644
index 00000000000..54c6972e706
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+++ b/Master/texmf-dist/doc/latex/pstricks_calcnotes/For_Pdf_Output/Fig1-2.pdf
diff --git a/Master/texmf-dist/doc/latex/pstricks_calcnotes/For_Pdf_Output/Fig1-3.pdf b/Master/texmf-dist/doc/latex/pstricks_calcnotes/For_Pdf_Output/Fig1-3.pdf
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index 00000000000..edae51373e5
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+++ b/Master/texmf-dist/doc/latex/pstricks_calcnotes/For_Pdf_Output/Fig1-3.pdf
diff --git a/Master/texmf-dist/doc/latex/pstricks_calcnotes/For_Pdf_Output/Fig2-1.pdf b/Master/texmf-dist/doc/latex/pstricks_calcnotes/For_Pdf_Output/Fig2-1.pdf
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index 00000000000..2be92abfbfd
--- /dev/null
+++ b/Master/texmf-dist/doc/latex/pstricks_calcnotes/For_Pdf_Output/Fig2-1.pdf
diff --git a/Master/texmf-dist/doc/latex/pstricks_calcnotes/For_Pdf_Output/Fig2-2.pdf b/Master/texmf-dist/doc/latex/pstricks_calcnotes/For_Pdf_Output/Fig2-2.pdf
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index 00000000000..ade96c92237
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+++ b/Master/texmf-dist/doc/latex/pstricks_calcnotes/For_Pdf_Output/Fig2-2.pdf
diff --git a/Master/texmf-dist/doc/latex/pstricks_calcnotes/For_Pdf_Output/Fig3.pdf b/Master/texmf-dist/doc/latex/pstricks_calcnotes/For_Pdf_Output/Fig3.pdf
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index 00000000000..0b6f0b825fc
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+++ b/Master/texmf-dist/doc/latex/pstricks_calcnotes/For_Pdf_Output/Fig3.pdf
diff --git a/Master/texmf-dist/doc/latex/pstricks_calcnotes/For_Pdf_Output/Fig4.pdf b/Master/texmf-dist/doc/latex/pstricks_calcnotes/For_Pdf_Output/Fig4.pdf
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index 00000000000..e1d694a316c
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+++ b/Master/texmf-dist/doc/latex/pstricks_calcnotes/For_Pdf_Output/Fig4.pdf
diff --git a/Master/texmf-dist/doc/latex/pstricks_calcnotes/For_Pdf_Output/Fig5.pdf b/Master/texmf-dist/doc/latex/pstricks_calcnotes/For_Pdf_Output/Fig5.pdf
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index 00000000000..96b8581d6c3
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+++ b/Master/texmf-dist/doc/latex/pstricks_calcnotes/For_Pdf_Output/Fig5.pdf
diff --git a/Master/texmf-dist/doc/latex/pstricks_calcnotes/For_Pdf_Output/Fig6.pdf b/Master/texmf-dist/doc/latex/pstricks_calcnotes/For_Pdf_Output/Fig6.pdf
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index 00000000000..d99748f5a33
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+++ b/Master/texmf-dist/doc/latex/pstricks_calcnotes/For_Pdf_Output/Fig6.pdf
diff --git a/Master/texmf-dist/doc/latex/pstricks_calcnotes/For_Pdf_Output/Fig7.pdf b/Master/texmf-dist/doc/latex/pstricks_calcnotes/For_Pdf_Output/Fig7.pdf
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index 00000000000..ddaa77ab5de
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+++ b/Master/texmf-dist/doc/latex/pstricks_calcnotes/For_Pdf_Output/Fig7.pdf
diff --git a/Master/texmf-dist/doc/latex/pstricks_calcnotes/For_Pdf_Output/Fig8.pdf b/Master/texmf-dist/doc/latex/pstricks_calcnotes/For_Pdf_Output/Fig8.pdf
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index 00000000000..aacbb9457e0
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+++ b/Master/texmf-dist/doc/latex/pstricks_calcnotes/For_Pdf_Output/Fig8.pdf
diff --git a/Master/texmf-dist/doc/latex/pstricks_calcnotes/For_Pdf_Output/Fig9.pdf b/Master/texmf-dist/doc/latex/pstricks_calcnotes/For_Pdf_Output/Fig9.pdf
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index 00000000000..cec79431078
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+++ b/Master/texmf-dist/doc/latex/pstricks_calcnotes/For_Pdf_Output/Fig9.pdf
diff --git a/Master/texmf-dist/doc/latex/pstricks_calcnotes/For_Pdf_Output/TwoApps_Pdf.tex b/Master/texmf-dist/doc/latex/pstricks_calcnotes/For_Pdf_Output/TwoApps_Pdf.tex
new file mode 100644
index 00000000000..fcdaf2e2688
--- /dev/null
+++ b/Master/texmf-dist/doc/latex/pstricks_calcnotes/For_Pdf_Output/TwoApps_Pdf.tex
@@ -0,0 +1,376 @@
+\documentclass[11pt,a4paper,oneside]{article}
+\usepackage{lmodern}
+\usepackage[T1,T5]{fontenc}
+\usepackage{amsthm}
+\usepackage{amsmath}
+\usepackage[dvips]{geometry}
+\usepackage{pstricks}
+\usepackage{graphicx}
+\usepackage{graphics}
+\usepackage{pst-plot}
+\usepackage{pst-node}
+\usepackage{multido}
+\usepackage{pst-xkey}
+\usepackage{pst-func}
+\usepackage[dvips,colorlinks,linktocpage]{hyperref}
+\usepackage{pstricks-add}
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+\def\RiemannSum#1#2#3#4#5#6#7#8#9{%
+\psplot[linecolor=blue]{#1}{#2}{#3}
+\pscustom[linecolor=red]{%
+\psline{-}(#1,0)(#1,0)
+\multido{\ni=#5,\ne=#6}{#4}
+{\psline(*{\ni} {#8})(*{\ne} {#9})}}
+\multido{\ne=#6,\nc=#7}{#4}
+{\psdot(*{\nc} {#3})
+\psline[linestyle=dotted,dotsep=1.5pt](\nc,0)(*{\nc} {#3})
+\psline[linecolor=red](\ne,0)(*{\ne} {#9})}
+}
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+\def\vecfld#1#2#3#4#5#6{%
+\multido{#2}{#4}
+{\multido{#1}{#3}
+{\parametricplot[algebraic,arrows=->,linecolor=red]{0}{1}
+{\nx+((#5)*t)*(1/sqrt(1+(#6)^2))|\ny+((#5)*t)*(1/sqrt(1+(#6)^2))*(#6)}}}}
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+\pagestyle{headings}
+\topmargin=-0.6cm
+\textwidth=16.7cm
+\textheight=23cm
+\headheight=2.5ex
+\headsep=0.6cm
+\oddsidemargin=.cm
+\evensidemargin=-.4cm
+\parskip=0.7ex plus0.5ex minus 0.5ex
+\baselineskip=17pt plus2pt minus2pt
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+\catcode`@=11
+\renewcommand\section{\@startsection {section}{1}{\z@}%
+                                   {-3.5ex \@plus -1ex \@minus -.2ex}%
+                                   {2.3ex \@plus.2ex}%
+                                   {\normalfont\large\bfseries}}
+\renewcommand\subsection{\@startsection{subsection}{2}{\z@}%
+                                     {-3.25ex\@plus -1ex \@minus -.2ex}%
+                                     {1.5ex \@plus .2ex}%
+                                     {\normalfont\normalsize\bfseries}}
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+\gdef\acknw{\section*{%
+{\acknwname}\markright{\protect\textsl{\acknwname}}}%
+\addcontentsline{toc}{section}{\acknwname}}
+\gdef\acknwname{Acknowledgment}
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+\renewcommand\sectionmark[1]{\markright{\thesection. #1}}
+\newcounter{lk}
+\newenvironment{listof}{\begin{list}{\rm(\roman{lk})}{\usecounter{lk}%
+\setlength{\topsep}{0ex plus0.1ex}%
+\setlength{\labelwidth}{1cm}%
+\setlength{\itemsep}{0ex plus0.1ex}%
+\setlength{\itemindent}{0.5cm}%
+}}{\end{list}}
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+\title{Two applications of macros in \texttt{PSTricks}\thanks{The author of this package
+is Timothy Van Zandt (email address: \texttt{tvz@econ.insead.fr}).}}
+\author{Le Phuong Quan\\
+\small{(Cantho University, Vietnam)}}
+\begin{document}
+\maketitle
+\tableofcontents
+\section{Drawing approximations to the area under a graph by rectangles}
+\subsection{Description}
+
+We recall here an application in Calculus. Let $f(x)$ be a function, defined and bounded on
+the interval $[a,b]$. If $f$ is integrable (in Riemann sense) on $[a,b]$, then its integration on this interval
+is
+$$\int_a^bf(x)dx=\lim_{\|P\|\to 0}\sum_{i=1}^nf(\xi_i)\Delta x_i,$$
+where $P\colon a=x_0<x_1<\cdots<x_n=b$, $\Delta x_i=x_i-x_{i-1}$, $\xi_i\in[x_{i-1},x_i]$, $i=1,2,\ldots,n$,
+and $\|P\|=\max\{\Delta x_i\colon i=1,2,\ldots,n\}$. Hence, when $\|P\|$ is small enough, we may have an
+approximation
+\begin{equation}\label{eqn1}
+I=\int_a^bf(x)dx\approx\sum_{i=1}^nf(\xi_i)\Delta x_i.
+\end{equation}
+Because $I$ is independent to the choice of the partition $P$ and of the $\xi_i$, we may
+divide $[a,b]$ into $n$ subintervals with equal length  and choose $\xi_i=(x_i+x_{i-1})/2$.
+Then, $I$ can be approximately seen as the sum of areas of the rectangles with sides
+$f(\xi_i)$ and $\Delta x_i$.
+
+We will make a drawing procedure to illustrate the approximation (\ref{eqn1}). Firstly, we establish
+commands to draw the \emph{sum\/} of rectangles, like the area under piecewise-constant functions
+(called \textsl{step shape\/}, for brevity). The choice here
+is a combination of the macros \texttt{\symbol{92}pscustom} (to \emph{join\/} horizontal segments, automatically)
+and \texttt{\symbol{92}multido}, of course. In particular, the horizontal segments are depicted within the loop
+\texttt{\symbol{92}multido} by
+$$\texttt{\symbol{92}psplot[{\it settings}]\{$x_{i-1}$\}\{$x_i$\}\{$f(\xi_i)$\}}$$
+The \texttt{\symbol{92}pscustom} will join these segments altogether with the end points
+$(a,0)$ and $(b,0)$, to make the boundary of the step shape. Then, we draw the points $(\xi_i,f(\xi_i))$, $i=1,2,\ldots,n$,
+and the dotted segments between these points and the points $(\xi_i,0)$, $i=1,2,\ldots,n$, by
+\begin{align*}
+&\texttt{\symbol{92}psdot[algebraic,\dots](*\{$\xi_i$\} \{$f(x)$\})},\\
+&\texttt{\symbol{92}psline[algebraic,linestyle=dotted,\dots]($\xi_i$,$0$)(*\{$\xi_i$\} \{$f(x)$\})},
+\end{align*}
+where we use the structure \texttt{(*\{{\it value}\} \{$f(x)$\})} to obtain the point $(\xi_i,f(\xi_i))$. Finally, we draw
+vertical segments to split the step shape into rectangular cells by
+$$\texttt{\symbol{92}psline[algebraic,\dots]($x_i$,$0$)(*\{$x_i$\} \{$f(x-\Delta x_i/2)$\})}$$
+\begin{figure}[htbp]
+\centering\begin{tabular}{cc}
+\includegraphics[height=5.5cm]{Fig1-1}
+&\includegraphics[height=5.5cm]{Fig1-2}
+\\
+\multicolumn{2}{c}{\includegraphics[height=5.5cm]{Fig1-3}}
+\end{tabular}
+\caption{Steps to make the drawing procedure.}\label{Fig1}
+\end{figure}
+
+We can combine the above steps to make a procedure whose calling sequence consists of main parameters
+$a$, $b$, $f$ and $n$, and dependent parameters $x_{i-1}$, $x_i$, $\xi_i$, $f(\xi_i)$ and
+$f(x\pm\Delta x_i/2)$. For instant, let us consider the approximations to the integration of $f(x)=\sin x-\cos x$
+on the interval $[-2,3]$ in the cases of $n=5$ and $n=20$. Those approximations are given in Figure \ref{Fig2}.
+
+\begin{figure}[htbp]
+\centering\includegraphics[width=6cm]{Fig2-1}
+\hskip4em
+\includegraphics[width=6cm]{Fig2-2}
+\caption{Approximations to the integration of $f(x)=\sin x-\cos x$ on $[-2,3]$.}\label{Fig2}
+\end{figure}
+
+In fact, we can make a procedure, say \texttt{RiemannSum}, whose calling sequence is of the form:
+$$\texttt{\symbol{92}RiemannSum\{$a$\}\{$b$\}\{$f(x)$\}\{$n$\}\{$x_{\rm ini}$\}\{$x_{\rm end}$\}\{$x_{\rm choice}$\}\{$f(x+\Delta x_i/2)$\}\{$f(x-\Delta x_i/2)$\}},$$
+where $x_0=a$ and for each $i=1,2\ldots,n$:
+\begin{align*}
+x_i&=a+\dfrac{b-a}{n}i,\quad\Delta x_i=x_i-x_{i-1}=\dfrac{b-a}{n},\\
+x_{\rm ini}&=x_0+\Delta x_i,\quad x_{\rm end}=x_1+\Delta x_i,\quad x_{\rm choice}=\dfrac{x_{\rm ini}+x_{\rm end}}{2}=\dfrac{x_0+x_1}{2}+\Delta x_i.
+\end{align*}
+Note that $x_{\rm ini}$, $x_{\rm end}$ and $x_{\rm choice}$ are given in such forms to be
+suitable to variable declaration in \texttt{\symbol{92}multido}. They are nothing but
+$x_{i-1}$, $x_i$ and $\xi_i$, respectively, at the step $i$-th in the loop.
+
+Tentatively, in \texttt{PSTricks} language, the definition of \texttt{RiemannSum} is suggested to be
+\bigskip\hrule
+\noindent\begin{tabular}{@{}l}
+\verb!\def\RiemannSum#1#2#3#4#5#6#7#8#9{%!\\
+\verb!\psplot[linecolor=blue]{#1}{#2}{#3}!\\
+\verb!\pscustom[linecolor=red]{%!\\
+\verb!\psline{-}(#1,0)(#1,0)!\\
+\verb!\multido{\ni=#5,\ne=#6}{#4}!\\
+\verb!{\psline(*{\ni} {#8})(*{\ne} {#9})}}!\\
+\verb!\multido{\ne=#6,\nc=#7}{#4}!\\
+\verb!{\psdot(*{\nc} {#3})!\\
+\verb!\psline[linestyle=dotted,dotsep=1.5pt](\nc,0)(*{\nc} {#3})!\\
+\verb!\psline[linecolor=red](\ne,0)(*{\ne} {#9})}}!
+\end{tabular}\smallskip\hrule
+\subsection{Examples}
+We just give here two more examples for using the drawing procedure with ease. In the first example, we approximate
+the area under the graph of the function $f(x)=x-(x/2)\cos x+2$ on the interval $[0,8]$. To draw the approximation, we try
+the case $n=16$; thus $x_0=0$ and for each $i=1,\ldots,16$, we have
+$x_i=0.5\,i$, $\Delta x_i=0.5$, $x_{\rm ini}=0.00+0.50$, $x_{\rm end}=0.50+0.50$ and $x_{\rm choice}=0.25+0.50$.
+\begin{figure}[htbp]
+\centering\includegraphics[width=5.1cm]{Fig3}
+\vskip0.5ex
+\caption{An approximation to the area under the graph of $f(x)=x-(x/2)\cos x+2$ on $[0,8]$.}\label{Fig3}
+\end{figure}
+
+To get Figure \ref{Fig3}, we have used the following \LaTeX\ code:
+\bigskip\hrule
+\noindent\begin{tabular}{@{}l}
+\verb!\begin{pspicture}(0,0)(4.125,5.5)!\\
+\verb!\psset{plotpoints=500,algebraic,dotsize=2.5pt,unit=0.5}!\\
+\verb!\RiemannSum{0}{8}{x-(x/2)*cos(x)+2}{16}{0.00+0.50}{0.50+0.50}{0.25+0.50}!\\
+\verb!{x+0.25-((x+0.25)/2)*cos(x+0.25)+2}{x-0.25-((x-0.25)/2)*cos(x-0.25)+2}!\\
+\verb!\psaxes[ticksize=2.2pt,labelsep=4pt]{->}(0,0)(8.5,11)!\\
+\verb!\end{pspicture}!
+\end{tabular}
+\smallskip\hrule\smallskip
+
+In the second example below, we will draw an approximation to the integration of $f(x)=x\sin x$ on $[1,9]$.
+Choosing $n=10$ and computing parameters needed, we get Figure \ref{Fig4}, mainly by
+the command
+\begin{align*}
+&\texttt{\symbol{92}RiemannSum\{$1$\}\{$9$\}\{$x\sin x$\}\{$10$\}\{$1.00+0.80$\}\{$1.80+0.80$\}\{$1.40+0.80$\}}\\
+&\texttt{\{$(x+0.4)\sin(x+0.4)$\}\{$(x-0.4)\sin(x-0.4)$\}}
+\end{align*}
+in the drawing procedure.
+\begin{figure}[htbp]
+\centering\includegraphics[width=4.75cm]{Fig4}
+\caption{An approximation to the integration of $f(x)=x\sin x$ on $[1,9]$.}\label{Fig4}
+\end{figure}
+\section{Drawing the vector field of an ordinary differential equation of order one}
+\subsection{Description}
+Let us consider the differential equation
+\begin{equation}\label{eqn2}
+\frac{dy}{dx}=f(x,y).
+\end{equation}
+At each point $(x_0,y_0)$ in the domain $D$ of $f$, we will put a vector $\mathbf{v}$ with slope
+$k=f(x_0,y_0)$. If $y(x_0)=y_0$, then $k$ is the slope of the tangent to the solution curve $y=y(x)$
+of (\ref{eqn2}) at $(x_0,y_0)$. The $\mathbf{v}$'s make a \textsl{vector field\/} and the picture
+of this field would give us information about the shape of solution curves of (\ref{eqn2}), even
+we have not found yet any solution of (\ref{eqn2}).
+
+The vector field of (\ref{eqn2}) will be depicted on a finite grid of points in $D$. This grid is made of
+lines, paralell to the axes $Ox$ and $Oy$. The intersectional points of those lines are called \textsl{grid points\/}
+and often indexed by $(x_i,y_j)$, $i=1,\ldots,m$, $j=1,\ldots,n$. For convenience, we will use
+polar coordinate to locate the terminal point $(x,y)$ of a field vector, with the initial point at
+the grid point $(x_i,y_j)$. Then, we can write
+\begin{align*}
+x&=x_i+r\cos\varphi,\\
+y&=y_j+r\sin\varphi.
+\end{align*}
+Because $k=f(x_i,y_j)=\tan\varphi$ is finite, we may take $-\pi/2<\varphi<\pi/2$.
+From $\sin^2\varphi+\cos^2\varphi=1$ and $\sin\varphi=k\cos\varphi$, we derive
+$$\cos\varphi=\frac{1}{\sqrt{1+k^2}},\quad\sin\varphi=\frac{k}{\sqrt{1+k^2}}.$$
+The field vectors should all have the same magnitude and we choose here that length to be
+$1/2$, that means $r=1/2$. Thus, vectors on the grid have their initial points and
+terminal ones as
+$$(x_i,y_j),\quad \Big(x_i+\frac{1}{2}\cos\varphi,y_j+\frac{1}{2}\sin\varphi\Big).$$
+
+Of macros in \texttt{PSTricks} to draw lines, we select \texttt{\symbol{92}parametricplot}\footnote{\footnotesize
+This macro is of ones, often added and updated in the package \texttt{pstricks-add}, the authors:
+Dominique Rodriguez (\texttt{dominique.rodriguez@waika9.com}), Herbert Vo\ss\ (\texttt{voss@pstricks.de}).}
+for its fitness. We immetiately have the simple parameterization of the vector at the grid point
+$(x_i,y_j)$ as
+\begin{align*}
+x&=x_i+\frac{t}{2}\cos\varphi=x_i+\frac{t}{2\sqrt{1+k^2}},\\
+y&=y_j+\frac{t}{2}\sin\varphi=y_j+\frac{tk}{2\sqrt{1+k^2}},
+\end{align*}
+where $t$ goes from $t=0$ to $t=1$, along the direction of the vector. The macro \texttt{\symbol{92}parametricplot}
+has the syntax as
+$$\texttt{\symbol{92}parametricplot[{\it settings}]\{$t_{\rm min}$\}\{$t_{\rm max}$\}\{$x(t)$|$y(t)$\}},$$
+where we should use the option \texttt{algebraic} to make the declaration of $x(t)$ and $y(t)$ simpler
+with \texttt{ASCII} code.
+\begin{figure}[htbp]
+\centering\includegraphics[width=5cm]{Fig5}
+\caption{Field vectors on a grid.}\label{Fig5}
+\end{figure}
+
+From the above description of one field vector, we go to the one of the whole vector field
+on the grid in the domain $R=\{(x,y)\colon a\le x\le b,\,c\le y\le d\}$. To determine the grid
+belonging to the interior of $R$, we confine grid points to the range
+\begin{equation}\label{eqn3}
+a+0.25\le x_i\le b-0.25,\quad c+0.25\le y_j\le d-0.25.
+\end{equation}
+With respect to the indices $i$ and $j$, we choose initial values as $x_1=a+0.25$ and
+$y_1=c+0.25$, with increments $\Delta x=\Delta y=0.5$, as corresponding to the length of vectors and the distance
+between grid points as indicated in Figure \ref{Fig5}. Thus, to draw vectors at grid points
+$(x_i,y_j)$, we need two loops for $i$ and $j$, with $0\le i\le [2m]$, $0\le j\le [2n]$, where
+$m=b-a$, $n=d-c$. Apparently, these two loops are nested \texttt{\symbol{92}multido}'s, with variable declaration
+for each loop as follows
+\begin{align*}
+\texttt{\symbol{92}nx}&=\text{initial value}+\text{increment}=x_1+\Delta x,\\
+\texttt{\symbol{92}ny}&=\text{initial value}+\text{increment}=y_1+\Delta y.
+\end{align*}
+Finally, we will replace \texttt{\symbol{92}nx}, \texttt{\symbol{92}ny} by $x_i$, $y_j$ in the
+below calling sequence for simplicity.
+
+Thus, the main procedure to draw the vector field of the equation (\ref{eqn2}) on the grid (\ref{eqn3})
+is
+\begin{align*}
+&\texttt{\symbol{92}multido\big\{$y_j=y_1+\Delta y$\big\}\big\{$[2n]$\big\}}\texttt{\bigg\{\symbol{92}multido\big\{$x_i=x_1+\Delta x$\big\}\big\{$[2m]$\big\}}\\
+&\quad\texttt{\Big\{\symbol{92}parametricplot[{\it settings}]\{$0$\}\{$1$\}\Big\{$x_i+\frac{t}{2\sqrt{1+\big[f(x_i,y_j)\big]^2}}$\Big|
+$y_j+\frac{tf(x_i,y_j)}{2\sqrt{1+\big[f(x_i,y_j)\big]^2}}$\Big\}\bigg\}}
+\end{align*}
+where we at least use \texttt{arrows=->} and \texttt{algebraic} for \textit{settings}.
+
+We can combine the steps mentioned above to define a drawing procedure, say \texttt{\symbol{92}vecfld},
+that consists of main parameters in the order as
+\texttt{\symbol{92}nx=}$x_1+\Delta x$, \texttt{\symbol{92}ny=}$y_1+\Delta y$, $[2m]$, $[2n]$, $r$
+and $f(\texttt{\symbol{92}nx},\texttt{\symbol{92}ny})$. We may change these values to modify
+the vector field or to avoid the vector intersection. However, we often take $\Delta x=\Delta y=r$.
+Such a definition is suggested to be
+\bigskip\hrule
+\noindent\begin{tabular}{@{}l}
+\verb!\def\vecfld#1#2#3#4#5#6{%!\\
+\verb!\multido{#2}{#4}{\multido{#1}{#3}!\\
+\verb!{\parametricplot[algebraic,arrows=->,linecolor=red]{0}{1}!\\
+\verb!{\nx+((#5)*t)*(1/sqrt(1+(#6)^2))|\ny+((#5)*t)*(1/sqrt(1+(#6)^2))*(#6)}}}}!
+\end{tabular}\smallskip\hrule
+\subsection{Examples}
+Firstly, we consider the equation that describes an object falling in a resistive medium:
+\begin{equation}\label{eqn4}
+\frac{dv}{dt}=9.8-\frac{v}{5},
+\end{equation}
+where $v=v(t)$ is the speed of the object in time $t$. In Figure \ref{Fig6}, the vector field of (\ref{eqn4}) is given
+on the grid $R=\{(t,y)\colon 0\le t\le 9,\,46\le v\le 52\}$, together with the graph of the equilibrium solution
+$v=49$.
+\begin{figure}[htbp]
+\centering\includegraphics[width=8.55cm]{Fig6}
+\caption{The vector field of (\ref{eqn4}).}\label{Fig6}
+\end{figure}
+
+Figure \ref{Fig6} is made of the following \LaTeX\ code:
+\bigskip\hrule
+\noindent\begin{tabular}{@{}l}
+\verb!\begin{pspicture}(0,46)(9.5,52.5)!\\
+\verb!\vecfld{\nx=0.25+0.50}{\ny=46.25+0.50}{18}{12}{0.5}{9.8-0.2*\ny}!\\
+\verb!\psplot[algebraic,linewidth=1.2pt]{0}{9}{49}!\\
+\verb!\psaxes[Dy=1,Dx=1,Oy=46]{->}(0,46)(0,46)(9.5,52.5)!\\
+\verb!\rput(9.5,45.8){$t$}\rput(-0.2,52.5){$y$}!\\
+\verb!\end{pspicture}!
+\end{tabular}
+\smallskip\hrule\smallskip
+Let us next consider the problem
+\begin{equation}\label{eqn5}
+\frac{dy}{dx}=x+y,\quad y(0)=0.
+\end{equation}
+It is easy to check that $y=e^x-x-1$ is the unique solution to the problem (\ref{eqn5}). We now draw
+the vector field of (\ref{eqn5}) and the solution curve\footnote{\footnotesize
+We have used ${\rm ch}(1)+{\rm sh}(1)$ for the declaration of $e$, natural base of logarithmic function.} on the grid $R=\{(x,y)\colon 0\le x\le 3,\,0\le y\le 5\}$ in
+Figure \ref{Fig7}.
+\begin{figure}[htbp]
+\centering\includegraphics[width=3.25cm]{Fig7}
+\caption{The vector field of (\ref{eqn5}).}\label{Fig7}
+\end{figure}
+
+We then go to the logistic equation, which is chosen to be a model for the dependence
+of the population size $P$ on time $t$ in Biology:
+\begin{equation}\label{eqn6}
+\frac{dP}{dt}=kP\Big(1-\frac{P}{M}\Big),
+\end{equation}
+where $k$ and $M$ are constants, respectively various to selected species and environment.
+For specification, we take, for instant, $k=0.5$ and $M=100$. The right hand side of
+(\ref{eqn6}) then becomes $f(t,P)=0.5\,P(1-0.01\,P)$. In Figure \ref{Fig8}, we draw the vector field
+of (\ref{eqn6}) on the grid $R=\{(t,P)\colon 0\le t\le 10,\,95\le P\le 100\}$ and the equilibrium
+solution curve $P=100$. Furthermore, with the initial condition $P(0)=95$, the equation (\ref{eqn6})
+has the unique solution $P=1900(e^{-0.5t}+19)^{-1}$. This solution curve is also given in Figure \ref{Fig8}.
+\begin{figure}[htbp]
+\centering\includegraphics[width=8.4cm]{Fig8}
+\caption{The vector field of (\ref{eqn6}) with $k=0.5$ and $M=100$.}\label{Fig8}
+\end{figure}
+
+The previous differential equations are all of seperated variable or linear cases that
+can be solved for closed-form solutions by some simple integration formulas. We will consider one more
+equation of the non-linear case whose solution can only be approximated by numerical methods.
+The vector field of such an equation is so useful and we will use the Runge-Kutta curves (of order $4$)
+to add more information about the behaviour of solution curve. Here, those Runge-Kutta curves are depicted by the procedure
+\texttt{\symbol{92}psplotDiffEqn}, also updated from the package \texttt{pstricks-add}.
+
+The vector field of the non-linear differential equation
+\begin{equation}\label{eqn7}
+\frac{dy}{dx}=y^2-xy+1
+\end{equation}
+will be depicted on the grid $R=\{(x,y)\colon -3\le x\le 3,\,-3\le y\le 3\}$ and the solutions
+of Cauchy problems for (\ref{eqn7}), corresponding to initial conditions
+\begin{listof}
+\item $y(-3)=-1$,
+\item $y(-2)=-3$,
+\item $y(-3)=-0.4$,
+\end{listof}
+will be approximated by the method of Runge-Kutta, with the grid size $h=0.2$. It is very easy
+to recognize approximation curves, respective to (i), (ii) and (iii) in Figure \ref{Fig9} below.
+\begin{figure}[htbp]
+\centering\includegraphics[width=8.5cm]{Fig9}
+\caption{The vector field of (\ref{eqn7}) and the Runge-Kutta curves.}\label{Fig9}
+\end{figure}
+\acknw
+I am very grateful to
+\begin{itemize}
+\item Timothy Van Zandt, Herbert Vo\ss\ and Dominique Rodriguez for helping me with
+their great works on \texttt{PSTricks}.
+\item H\`an Th\'\ecircumflex\ Th\`anh for helping me with his pdf\hskip.03em\LaTeX\ program.
+\end{itemize}
+\begin{thebibliography}{10}
+\bibitem{mot} Dominique Rodriguez \&\ Herbert Vo\ss. \textsl{PSTricks-add, additional macros for PSTricks\/}.
+Version 3.05, \url{http://ctan.org/tex-archive/graphics/pstricks/contrib}, 2008
+\bibitem{hai} Helmut Kopka \&\ Patrick W. Daly. \textsl{Guide to \LaTeX \/}.
+Addison-Wesley, Fourth Edition, 2004, ISBN 0321173856
+\bibitem{ba} Timothy Van Zandt. \textsl{User's Guide\/}. Version 1.5,\\
+\url{http://ctan.org/tex-archive/graphics/pstricks/base}, 2007
+\end{thebibliography}
+\end{document}
diff --git a/Master/texmf-dist/doc/latex/pstricks_calcnotes/For_Ps_Output/TwoApps_Ps.tex b/Master/texmf-dist/doc/latex/pstricks_calcnotes/For_Ps_Output/TwoApps_Ps.tex
new file mode 100644
index 00000000000..32241ca9a04
--- /dev/null
+++ b/Master/texmf-dist/doc/latex/pstricks_calcnotes/For_Ps_Output/TwoApps_Ps.tex
@@ -0,0 +1,479 @@
+\documentclass[11pt,a4paper,oneside]{article}
+\usepackage{lmodern}
+\usepackage[T1,T5]{fontenc}
+\usepackage{amsthm}
+\usepackage{amsmath}
+\usepackage[dvips]{geometry}
+\usepackage{pstricks}
+\usepackage{graphicx}
+\usepackage{graphics}
+\usepackage{pst-plot}
+\usepackage{pst-node}
+\usepackage{multido}
+\usepackage{pst-xkey}
+\usepackage{pst-func}
+\usepackage[dvips,colorlinks,linktocpage]{hyperref}
+\usepackage{pstricks-add}
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+\def\RiemannSum#1#2#3#4#5#6#7#8#9{%
+\psplot[linecolor=blue]{#1}{#2}{#3}
+\pscustom[linecolor=red]{%
+\psline{-}(#1,0)(#1,0)
+\multido{\ni=#5,\ne=#6}{#4}
+{\psline(*{\ni} {#8})(*{\ne} {#9})}}
+\multido{\ne=#6,\nc=#7}{#4}
+{\psdot(*{\nc} {#3})
+\psline[linestyle=dotted,dotsep=1.5pt](\nc,0)(*{\nc} {#3})
+\psline[linecolor=red](\ne,0)(*{\ne} {#9})}
+}
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+\def\vecfld#1#2#3#4#5#6{%
+\multido{#2}{#4}
+{\multido{#1}{#3}
+{\parametricplot[algebraic,arrows=->,linecolor=red]{0}{1}
+{\nx+((#5)*t)*(1/sqrt(1+(#6)^2))|\ny+((#5)*t)*(1/sqrt(1+(#6)^2))*(#6)}}}}
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+\pagestyle{headings}
+\topmargin=-0.6cm
+\textwidth=16.7cm
+\textheight=23cm
+\headheight=2.5ex
+\headsep=0.6cm
+\oddsidemargin=.cm
+\evensidemargin=-.4cm
+\parskip=0.7ex plus0.5ex minus 0.5ex
+\baselineskip=17pt plus2pt minus2pt
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+\catcode`@=11
+\renewcommand\section{\@startsection {section}{1}{\z@}%
+                                   {-3.5ex \@plus -1ex \@minus -.2ex}%
+                                   {2.3ex \@plus.2ex}%
+                                   {\normalfont\large\bfseries}}
+\renewcommand\subsection{\@startsection{subsection}{2}{\z@}%
+                                     {-3.25ex\@plus -1ex \@minus -.2ex}%
+                                     {1.5ex \@plus .2ex}%
+                                     {\normalfont\normalsize\bfseries}}
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+\gdef\acknw{\section*{%
+{\acknwname}\markright{\protect\textsl{\acknwname}}}%
+\addcontentsline{toc}{section}{\acknwname}}
+\gdef\acknwname{Acknowledgment}
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+\renewcommand\sectionmark[1]{\markright{\thesection. #1}}
+\newcounter{lk}
+\newenvironment{listof}{\begin{list}{\rm(\roman{lk})}{\usecounter{lk}%
+\setlength{\topsep}{0ex plus0.1ex}%
+\setlength{\labelwidth}{1cm}%
+\setlength{\itemsep}{0ex plus0.1ex}%
+\setlength{\itemindent}{0.5cm}%
+}}{\end{list}}
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+\title{Two applications of macros in \texttt{PSTricks}\thanks{The author of this package
+is Timothy Van Zandt (email address: \texttt{tvz@econ.insead.fr}).}}
+\author{Le Phuong Quan\\
+\small{(Cantho University, Vietnam)}}
+\begin{document}
+\maketitle
+\tableofcontents
+\section{Drawing approximations to the area under a graph by rectangles}
+\subsection{Description}
+
+We recall here an application in Calculus. Let $f(x)$ be a function, defined and bounded on
+the interval $[a,b]$. If $f$ is integrable (in Riemann sense) on $[a,b]$, then its integration on this interval
+is
+$$\int_a^bf(x)dx=\lim_{\|P\|\to 0}\sum_{i=1}^nf(\xi_i)\Delta x_i,$$
+where $P\colon a=x_0<x_1<\cdots<x_n=b$, $\Delta x_i=x_i-x_{i-1}$, $\xi_i\in[x_{i-1},x_i]$, $i=1,2,\ldots,n$,
+and $\|P\|=\max\{\Delta x_i\colon i=1,2,\ldots,n\}$. Hence, when $\|P\|$ is small enough, we may have an
+approximation
+\begin{equation}\label{eqn1}
+I=\int_a^bf(x)dx\approx\sum_{i=1}^nf(\xi_i)\Delta x_i.
+\end{equation}
+Because $I$ is independent to the choice of the partition $P$ and of the $\xi_i$, we may
+divide $[a,b]$ into $n$ subintervals with equal length  and choose $\xi_i=(x_i+x_{i-1})/2$.
+Then, $I$ can be approximately seen as the sum of areas of the rectangles with sides
+$f(\xi_i)$ and $\Delta x_i$.
+
+We will make a drawing procedure to illustrate the approximation (\ref{eqn1}). Firstly, we establish
+commands to draw the \emph{sum\/} of rectangles, like the area under piecewise-constant functions
+(called \textsl{step shape\/}, for brevity). The choice here
+is a combination of the macros \texttt{\symbol{92}pscustom} (to \emph{join\/} horizontal segments, automatically)
+and \texttt{\symbol{92}multido}, of course. In particular, the horizontal segments are depicted within the loop
+\texttt{\symbol{92}multido} by
+$$\texttt{\symbol{92}psplot[{\it settings}]\{$x_{i-1}$\}\{$x_i$\}\{$f(\xi_i)$\}}$$
+The \texttt{\symbol{92}pscustom} will join these segments altogether with the end points
+$(a,0)$ and $(b,0)$, to make the boundary of the step shape. Then, we draw the points $(\xi_i,f(\xi_i))$, $i=1,2,\ldots,n$,
+and the dotted segments between these points and the points $(\xi_i,0)$, $i=1,2,\ldots,n$, by
+\begin{align*}
+&\texttt{\symbol{92}psdot[algebraic,\dots](*\{$\xi_i$\} \{$f(x)$\})},\\
+&\texttt{\symbol{92}psline[algebraic,linestyle=dotted,\dots]($\xi_i$,$0$)(*\{$\xi_i$\} \{$f(x)$\})},
+\end{align*}
+where we use the structure \texttt{(*\{{\it value}\} \{$f(x)$\})} to obtain the point $(\xi_i,f(\xi_i))$. Finally, we draw
+vertical segments to split the step shape into rectangular cells by
+$$\texttt{\symbol{92}psline[algebraic,\dots]($x_i$,$0$)(*\{$x_i$\} \{$f(x-\Delta x_i/2)$\})}$$
+\begin{figure}[htbp]
+\centering\begin{pspicture}(-2,-2)(3,3.5)
+\psset{yunit=0.2}
+\psaxes[labelFontSize=$\footnotesize$,Dy=2,ticksize=2.2pt,labelsep=4pt]{->}(0,0)(-2.5,-10)(3.5,15.5)
+\psplot[plotpoints=500,algebraic,linecolor=red]{-2}{3}{x^3-2*x^2+6}
+\pscustom{
+\psline{-}(-2,0)(-2,0)
+\multido{\nx=-2.000+0.500,\ny=-1.750+0.500,\ni=-1.500+0.500}{10}
+{\psplot[algebraic]{\nx}{\ni}{(\ny)^3-2*(\ny)^2+6}}
+\psline{-}(3,0)}
+\end{pspicture}
+\hskip3em
+\begin{pspicture}(-2,-2)(3,3.5)
+\psset{yunit=0.2}
+\psaxes[labelFontSize=$\footnotesize$,Dy=2,ticksize=2.2pt,labelsep=4pt]{->}(0,0)(-2.5,-10)(3.5,15.5)
+\psplot[plotpoints=500,algebraic,linecolor=red]{-2}{3}{x^3-2*x^2+6}
+\pscustom{
+\psline{-}(-2,0)(-2,0)
+\multido{\nx=-2.000+0.500,\ny=-1.750+0.500,\ni=-1.500+0.500}{10}
+{\psplot[algebraic]{\nx}{\ni}{(\ny)^3-2*(\ny)^2+6}}
+\psline{-}(3,0)}
+\multido{\nx=-2.000+0.500,\ny=-1.750+0.500,\ni=-1.500+0.500}{10}
+{\psline[linestyle=dotted,dotsep=1.5pt,dotstyle=o,linecolor=gray](\ny,0)(*{\ny} {x^3-2*x^2+6})
+\psdot[dotsize=1.2pt 1,dotstyle=Bo](*{\ny} {x^3-2*x^2+6})}
+\end{pspicture}\\[3ex]
+\centering\begin{pspicture}(-2,-2)(3,3)
+\psset{yunit=0.2}
+\psaxes[labelFontSize=$\footnotesize$,Dy=2,ticksize=2.2pt,labelsep=4pt]{->}(0,0)(-2.5,-10)(3.5,15.5)
+\psplot[plotpoints=500,algebraic,linecolor=red]{-2}{3}{x^3-2*x^2+6}
+\pscustom{
+\psline{-}(-2,0)(-2,0)
+\multido{\nx=-2.000+0.500,\ny=-1.750+0.500,\ni=-1.500+0.500}{10}
+{\psplot[algebraic]{\nx}{\ni}{(\ny)^3-2*(\ny)^2+6}}
+\psline{-}(3,0)}
+\multido{\nx=-2.000+0.500,\ny=-1.750+0.500,\ni=-1.500+0.500}{10}
+{\psline[linestyle=dotted,dotsep=1.5pt,dotstyle=o,linecolor=gray](\ny,0)(*{\ny} {x^3-2*x^2+6})
+\psdot[dotsize=1.2pt 1,dotstyle=Bo](*{\ny} {x^3-2*x^2+6})}
+\multido{\nx=-2.000+0.500,\ny=-1.750+0.500,\ni=-1.500+0.500}{9}
+{\psline(\ni,0)(*{\ni} {(x-0.25)^3-2*(x-0.25)^2+6})}
+\end{pspicture}
+\caption{Steps to make the drawing procedure.}\label{Fig1}
+\end{figure}
+
+We can combine the above steps to make a procedure whose calling sequence consists of main parameters
+$a$, $b$, $f$ and $n$, and dependent parameters $x_{i-1}$, $x_i$, $\xi_i$, $f(\xi_i)$ and
+$f(x\pm\Delta x_i/2)$. For instant, let us consider the approximations to the integration of $f(x)=\sin x-\cos x$
+on the interval $[-2,3]$ in the cases of $n=5$ and $n=20$. Those approximations are given in Figure \ref{Fig2}.
+
+\begin{figure}[htbp]
+\centering\begin{pspicture}(-2.5,-3)(3.5,3.01)
+\psset{plotpoints=500,algebraic,dotsize=1pt 2,yunit=2}
+\psaxes[labelFontSize=$\footnotesize$,Dy=1]{->}(0,0)(-2.5,-1.5)(3.5,1.5)
+\RiemannSum{-2}{3}{sin(x)-cos(x)}{5}
+{-2.0+1.0}{-1.0+1.0}{-1.5+1.0}
+{sin(x+0.5)-cos(x+0.5)}{sin(x-0.5)-cos(x-0.5)}
+\end{pspicture}
+\hskip4em
+\begin{pspicture}(-2.5,-3)(3.5,3.01)
+\psset{plotpoints=500,algebraic,dotsize=1pt 2,yunit=2}
+\psaxes[labelFontSize=$\footnotesize$,Dy=1]{->}(0,0)(-2.5,-1.5)(3.5,1.5)
+\RiemannSum{-2}{3}{sin(x)-cos(x)}{20}
+{-2.000+0.250}{-1.750+0.250}{-1.875+0.250}
+{sin(x+0.125)-cos(x+0.125)}{sin(x-0.125)-cos(x-0.125)}
+\end{pspicture}
+\caption{Approximations to the integration of $f(x)=\sin x-\cos x$ on $[-2,3]$.}\label{Fig2}
+\end{figure}
+
+In fact, we can make a procedure, say \texttt{RiemannSum}, whose calling sequence is of the form:
+$$\texttt{\symbol{92}RiemannSum\{$a$\}\{$b$\}\{$f(x)$\}\{$n$\}\{$x_{\rm ini}$\}\{$x_{\rm end}$\}\{$x_{\rm choice}$\}\{$f(x+\Delta x_i/2)$\}\{$f(x-\Delta x_i/2)$\}},$$
+where $x_0=a$ and for each $i=1,2\ldots,n$:
+\begin{align*}
+x_i&=a+\dfrac{b-a}{n}i,\quad\Delta x_i=x_i-x_{i-1}=\dfrac{b-a}{n},\\
+x_{\rm ini}&=x_0+\Delta x_i,\quad x_{\rm end}=x_1+\Delta x_i,\quad x_{\rm choice}=\dfrac{x_{\rm ini}+x_{\rm end}}{2}=\dfrac{x_0+x_1}{2}+\Delta x_i.
+\end{align*}
+Note that $x_{\rm ini}$, $x_{\rm end}$ and $x_{\rm choice}$ are given in such forms to be
+suitable to variable declaration in \texttt{\symbol{92}multido}. They are nothing but
+$x_{i-1}$, $x_i$ and $\xi_i$, respectively, at the step $i$-th in the loop.
+
+Tentatively, in \texttt{PSTricks} language, the definition of \texttt{RiemannSum} is suggested to be
+\bigskip\hrule
+\noindent\begin{tabular}{@{}l}
+\verb!\def\RiemannSum#1#2#3#4#5#6#7#8#9{%!\\
+\verb!\psplot[linecolor=blue]{#1}{#2}{#3}!\\
+\verb!\pscustom[linecolor=red]{%!\\
+\verb!\psline{-}(#1,0)(#1,0)!\\
+\verb!\multido{\ni=#5,\ne=#6}{#4}!\\
+\verb!{\psline(*{\ni} {#8})(*{\ne} {#9})}}!\\
+\verb!\multido{\ne=#6,\nc=#7}{#4}!\\
+\verb!{\psdot(*{\nc} {#3})!\\
+\verb!\psline[linestyle=dotted,dotsep=1.5pt](\nc,0)(*{\nc} {#3})!\\
+\verb!\psline[linecolor=red](\ne,0)(*{\ne} {#9})}}!
+\end{tabular}\smallskip\hrule
+\subsection{Examples}
+We just give here two more examples for using the drawing procedure with ease. In the first example, we approximate
+the area under the graph of the function $f(x)=x-(x/2)\cos x+2$ on the interval $[0,8]$. To draw the approximation, we try
+the case $n=16$; thus $x_0=0$ and for each $i=1,\ldots,16$, we have
+$x_i=0.5\,i$, $\Delta x_i=0.5$, $x_{\rm ini}=0.00+0.50$, $x_{\rm end}=0.50+0.50$ and $x_{\rm choice}=0.25+0.50$.
+\begin{figure}[htbp]
+\centering\begin{pspicture}(0,0)(5.1,6.6)
+\psset{plotpoints=500,algebraic,dotsize=1pt 2,unit=0.6}
+\RiemannSum{0}{8}{x-(x/2)*cos(x)+2}{16}{0.00+0.50}{0.50+0.50}{0.25+0.50}
+{x+0.25-((x+0.25)/2)*cos(x+0.25)+2}{x-0.25-((x-0.25)/2)*cos(x-0.25)+2}
+\psaxes[labelFontSize=$\footnotesize$,Dy=1,ticksize=2.2pt,labelsep=4pt]{->}(0,0)(8.5,11)
+\end{pspicture}
+\vskip0.5ex
+\caption{An approximation to the area under the graph of $f(x)=x-(x/2)\cos x+2$ on $[0,8]$.}\label{Fig3}
+\end{figure}
+
+To get Figure \ref{Fig3}, we have used the following \LaTeX\ code:
+\bigskip\hrule
+\noindent\begin{tabular}{@{}l}
+\verb!\begin{pspicture}(0,0)(4.125,5.5)!\\
+\verb!\psset{plotpoints=500,algebraic,dotsize=2.5pt,unit=0.5}!\\
+\verb!\RiemannSum{0}{8}{x-(x/2)*cos(x)+2}{16}{0.00+0.50}{0.50+0.50}{0.25+0.50}!\\
+\verb!{x+0.25-((x+0.25)/2)*cos(x+0.25)+2}{x-0.25-((x-0.25)/2)*cos(x-0.25)+2}!\\
+\verb!\psaxes[ticksize=2.2pt,labelsep=4pt]{->}(0,0)(8.5,11)!\\
+\verb!\end{pspicture}!
+\end{tabular}
+\smallskip\hrule\smallskip
+
+In the second example below, we will draw an approximation to the integration of $f(x)=x\sin x$ on $[1,9]$.
+Choosing $n=10$ and computing parameters needed, we get Figure \ref{Fig4}, mainly by
+the command
+\begin{align*}
+&\texttt{\symbol{92}RiemannSum\{$1$\}\{$9$\}\{$x\sin x$\}\{$10$\}\{$1.00+0.80$\}\{$1.80+0.80$\}\{$1.40+0.80$\}}\\
+&\texttt{\{$(x+0.4)\sin(x+0.4)$\}\{$(x-0.4)\sin(x-0.4)$\}}
+\end{align*}
+in the drawing procedure.
+\begin{figure}[htbp]
+\centering\begin{pspicture}(0,-2.5)(4.75,4.25)
+\psset{plotpoints=500,algebraic,dotsize=1pt 2,unit=0.5}
+\RiemannSum{1}{9}{x*sin(x)}{10}%
+{1.00+0.80}{1.80+0.80}{1.40+0.80}%
+{(x+0.4)*sin(x+0.4)}{(x-0.4)*sin(x-0.4)}
+\psaxes[labelFontSize=$\footnotesize$,Dy=1,ticksize=2.2pt,labelsep=4pt]{->}(0,0)(0,-5)(9.5,8.5)
+\end{pspicture}
+\caption{An approximation to the integration of $f(x)=x\sin x$ on $[1,9]$.}\label{Fig4}
+\end{figure}
+\section{Drawing the vector field of an ordinary differential equation of order one}
+\subsection{Description}
+Let us consider the differential equation
+\begin{equation}\label{eqn2}
+\frac{dy}{dx}=f(x,y).
+\end{equation}
+At each point $(x_0,y_0)$ in the domain $D$ of $f$, we will put a vector $\mathbf{v}$ with slope
+$k=f(x_0,y_0)$. If $y(x_0)=y_0$, then $k$ is the slope of the tangent to the solution curve $y=y(x)$
+of (\ref{eqn2}) at $(x_0,y_0)$. The $\mathbf{v}$'s make a \textsl{vector field\/} and the picture
+of this field would give us information about the shape of solution curves of (\ref{eqn2}), even
+we have not found yet any solution of (\ref{eqn2}).
+
+The vector field of (\ref{eqn2}) will be depicted on a finite grid of points in $D$. This grid is made of
+lines, paralell to the axes $Ox$ and $Oy$. The intersectional points of those lines are called \textsl{grid points\/}
+and often indexed by $(x_i,y_j)$, $i=1,\ldots,m$, $j=1,\ldots,n$. For convenience, we will use
+polar coordinate to locate the terminal point $(x,y)$ of a field vector, with the initial point at
+the grid point $(x_i,y_j)$. Then, we can write
+\begin{align*}
+x&=x_i+r\cos\varphi,\\
+y&=y_j+r\sin\varphi.
+\end{align*}
+Because $k=f(x_i,y_j)=\tan\varphi$ is finite, we may take $-\pi/2<\varphi<\pi/2$.
+From $\sin^2\varphi+\cos^2\varphi=1$ and $\sin\varphi=k\cos\varphi$, we derive
+$$\cos\varphi=\frac{1}{\sqrt{1+k^2}},\quad\sin\varphi=\frac{k}{\sqrt{1+k^2}}.$$
+The field vectors should all have the same magnitude and we choose here that length to be
+$1/2$, that means $r=1/2$. Thus, vectors on the grid have their initial points and
+terminal ones as
+$$(x_i,y_j),\quad \Big(x_i+\frac{1}{2}\cos\varphi,y_j+\frac{1}{2}\sin\varphi\Big).$$
+
+Of macros in \texttt{PSTricks} to draw lines, we select \texttt{\symbol{92}parametricplot}\footnote{\footnotesize
+This macro is of ones, often added and updated in the package \texttt{pstricks-add}, the authors:
+Dominique Rodriguez (\texttt{dominique.rodriguez@waika9.com}), Herbert Vo\ss\ (\texttt{voss@pstricks.de}).}
+for its fitness. We immetiately have the simple parameterization of the vector at the grid point
+$(x_i,y_j)$ as
+\begin{align*}
+x&=x_i+\frac{t}{2}\cos\varphi=x_i+\frac{t}{2\sqrt{1+k^2}},\\
+y&=y_j+\frac{t}{2}\sin\varphi=y_j+\frac{tk}{2\sqrt{1+k^2}},
+\end{align*}
+where $t$ goes from $t=0$ to $t=1$, along the direction of the vector. The macro \texttt{\symbol{92}parametricplot}
+has the syntax as
+$$\texttt{\symbol{92}parametricplot[{\it settings}]\{$t_{\rm min}$\}\{$t_{\rm max}$\}\{$x(t)$|$y(t)$\}},$$
+where we should use the option \texttt{algebraic} to make the declaration of $x(t)$ and $y(t)$ simpler
+with \texttt{ASCII} code.
+\begin{figure}[htbp]
+\centering\begin{pspicture}(0,0)(5,5)
+\psset{unit=2}
+\psaxes[labelFontSize=$\footnotesize$,Dx=0.5,Dy=0.5,labels=none,ticksize=2pt,labelsep=2pt,linewidth=0.5pt]
+{->}(0,0)(2.5,2.5)
+\psdots[dotstyle=*,dotsize=3pt](1,1)(1.5,1)(1,1.5)
+\psline[linewidth=0.3pt](1.5,0.5)(1.5,2)(1,2)(1,0.5)(2,0.5)(2,1.5)
+\psline[linewidth=0.3pt](1,1.5)(2,1.5)\psline[linewidth=0.3pt](1,1)(2,1)
+\psarc[linewidth=0.5pt,linestyle=dotted,dotsep=1.5pt](1,1){0.5}{-90}{90}
+\psarc[linewidth=0.5pt,linestyle=dotted,dotsep=1.5pt](1.5,1){0.5}{-90}{90}
+\psarc[linewidth=0.5pt,linestyle=dotted,dotsep=1.5pt](1,1.5){0.5}{-90}{90}
+\psline[linewidth=0.8pt]{->}(1,1.5)(1.27,1.92)
+\psline[linewidth=0.8pt]{->}(1,1)(1.382,1.322)
+\psline[linewidth=0.8pt]{->}(1.5,1)(1.977,1.147)
+\rput(1,-0.1){$x_i$}\rput(1.5,-0.1){$x_{i+1}$}\rput(-0.1,1){$y_j$}\rput(-0.2,1.5){$y_{j+1}$}
+\end{pspicture}
+\caption{Field vectors on a grid.}\label{Fig5}
+\end{figure}
+
+From the above description of one field vector, we go to the one of the whole vector field
+on the grid in the domain $R=\{(x,y)\colon a\le x\le b,\,c\le y\le d\}$. To determine the grid
+belonging to the interior of $R$, we confine grid points to the range
+\begin{equation}\label{eqn3}
+a+0.25\le x_i\le b-0.25,\quad c+0.25\le y_j\le d-0.25.
+\end{equation}
+With respect to the indices $i$ and $j$, we choose initial values as $x_1=a+0.25$ and
+$y_1=c+0.25$, with increments $\Delta x=\Delta y=0.5$, as corresponding to the length of vectors and the distance
+between grid points as indicated in Figure \ref{Fig5}. Thus, to draw vectors at grid points
+$(x_i,y_j)$, we need two loops for $i$ and $j$, with $0\le i\le [2m]$, $0\le j\le [2n]$, where
+$m=b-a$, $n=d-c$. Apparently, these two loops are nested \texttt{\symbol{92}multido}'s, with variable declaration
+for each loop as follows
+\begin{align*}
+\texttt{\symbol{92}nx}&=\text{initial value}+\text{increment}=x_1+\Delta x,\\
+\texttt{\symbol{92}ny}&=\text{initial value}+\text{increment}=y_1+\Delta y.
+\end{align*}
+Finally, we will replace \texttt{\symbol{92}nx}, \texttt{\symbol{92}ny} by $x_i$, $y_j$ in the
+below calling sequence for simplicity.
+
+Thus, the main procedure to draw the vector field of the equation (\ref{eqn2}) on the grid (\ref{eqn3})
+is
+\begin{align*}
+&\texttt{\symbol{92}multido\big\{$y_j=y_1+\Delta y$\big\}\big\{$[2n]$\big\}}\texttt{\bigg\{\symbol{92}multido\big\{$x_i=x_1+\Delta x$\big\}\big\{$[2m]$\big\}}\\
+&\quad\texttt{\Big\{\symbol{92}parametricplot[{\it settings}]\{$0$\}\{$1$\}\Big\{$x_i+\frac{t}{2\sqrt{1+\big[f(x_i,y_j)\big]^2}}$\Big|
+$y_j+\frac{tf(x_i,y_j)}{2\sqrt{1+\big[f(x_i,y_j)\big]^2}}$\Big\}\bigg\}}
+\end{align*}
+where we at least use \texttt{arrows=->} and \texttt{algebraic} for \textit{settings}.
+
+We can combine the steps mentioned above to define a drawing procedure, say \texttt{\symbol{92}vecfld},
+that consists of main parameters in the order as
+\texttt{\symbol{92}nx=}$x_1+\Delta x$, \texttt{\symbol{92}ny=}$y_1+\Delta y$, $[2m]$, $[2n]$, $r$
+and $f(\texttt{\symbol{92}nx},\texttt{\symbol{92}ny})$. We may change these values to modify
+the vector field or to avoid the vector intersection. However, we often take $\Delta x=\Delta y=r$.
+Such a definition is suggested to be
+\bigskip\hrule
+\noindent\begin{tabular}{@{}l}
+\verb!\def\vecfld#1#2#3#4#5#6{%!\\
+\verb!\multido{#2}{#4}{\multido{#1}{#3}!\\
+\verb!{\parametricplot[algebraic,arrows=->,linecolor=red]{0}{1}!\\
+\verb!{\nx+((#5)*t)*(1/sqrt(1+(#6)^2))|\ny+((#5)*t)*(1/sqrt(1+(#6)^2))*(#6)}}}}!
+\end{tabular}\smallskip\hrule
+\subsection{Examples}
+Firstly, we consider the equation that describes an object falling in a resistive medium:
+\begin{equation}\label{eqn4}
+\frac{dv}{dt}=9.8-\frac{v}{5},
+\end{equation}
+where $v=v(t)$ is the speed of the object in time $t$. In Figure \ref{Fig6}, the vector field of (\ref{eqn4}) is given
+on the grid $R=\{(t,y)\colon 0\le t\le 9,\,46\le v\le 52\}$, together with the graph of the equilibrium solution
+$v=49$.
+\begin{figure}[htbp]
+\centering\begin{pspicture}(0,41.4)(8.55,47.25)
+\psset{xunit=0.9,yunit=0.9}
+\multido{\ny=46.25+0.50}{13}
+{\multido{\nx=0.25+0.50}{18}
+{\parametricplot[linewidth=0.5pt,algebraic,arrows=->,arrowinset=0.55,linecolor=red]{0}{1}{\nx+(t/2)*(1/sqrt(1+(9.8-0.2*(\ny))^2))|\ny+(t/2)*(1/sqrt(1+(9.8-(0.2)*(\ny))^2))*(9.8-(0.2)*(\ny))}}}
+\psplot[algebraic,linewidth=1.2pt]{0}{9.25}{49}
+\psaxes[labelFontSize=$\footnotesize$,ticksize=2.2pt,labelsep=4pt,Dy=1,Dx=1,Oy=46,linewidth=0.7pt]{->}(0,46)(0,46)(9.5,52.5)
+\rput(9.5,45.8){$t$}\rput(-0.2,52.5){$y$}
+\end{pspicture}
+\caption{The vector field of (\ref{eqn4}).}\label{Fig6}
+\end{figure}
+
+Figure \ref{Fig6} is made of the following \LaTeX\ code:
+\bigskip\hrule
+\noindent\begin{tabular}{@{}l}
+\verb!\begin{pspicture}(0,46)(9.5,52.5)!\\
+\verb!\vecfld{\nx=0.25+0.50}{\ny=46.25+0.50}{18}{12}{0.5}{9.8-0.2*\ny}!\\
+\verb!\psplot[algebraic,linewidth=1.2pt]{0}{9}{49}!\\
+\verb!\psaxes[Dy=1,Dx=1,Oy=46]{->}(0,46)(0,46)(9.5,52.5)!\\
+\verb!\rput(9.5,45.8){$t$}\rput(-0.2,52.5){$y$}!\\
+\verb!\end{pspicture}!
+\end{tabular}
+\smallskip\hrule\smallskip
+Let us next consider the problem
+\begin{equation}\label{eqn5}
+\frac{dy}{dx}=x+y,\quad y(0)=0.
+\end{equation}
+It is easy to check that $y=e^x-x-1$ is the unique solution to the problem (\ref{eqn5}). We now draw
+the vector field of (\ref{eqn5}) and the solution curve\footnote{\footnotesize
+We have used ${\rm ch}(1)+{\rm sh}(1)$ for the declaration of $e$, natural base of logarithmic function.} on the grid $R=\{(x,y)\colon 0\le x\le 3,\,0\le y\le 5\}$ in
+Figure \ref{Fig7}.
+\begin{figure}[htbp]
+\centering\begin{pspicture}(0,0)(3.25,5.5)
+\psset{unit=1}
+\multido{\ny=0.25+0.50}{10}
+{\multido{\nx=0.25+0.50}{6}
+{\parametricplot[linewidth=0.5pt,algebraic,arrows=->,arrowinset=0.55,linecolor=red]{0}{1}{\nx+(t/2)*(1/sqrt(1+(\nx+\ny)^2))|\ny+(t/2)*(1/sqrt(1+(\nx+\ny)^2))*(\nx+\ny)}}}
+\psplot[algebraic,linewidth=1.2pt]{0}{2.15}{(sh(1)+ch(1))^x-x-1}
+\psaxes[labelFontSize=$\footnotesize$,Dy=1,Dx=1,ticksize=2.2pt,labelsep=4pt,linewidth=0.7pt]{->}(0,0)(3.5,5.5)
+\rput(3.5,-0.2){$x$}\rput(-0.25,5.5){$y$}
+\end{pspicture}
+\caption{The vector field of (\ref{eqn5}).}\label{Fig7}
+\end{figure}
+
+We then go to the logistic equation, which is chosen to be a model for the dependence
+of the population size $P$ on time $t$ in Biology:
+\begin{equation}\label{eqn6}
+\frac{dP}{dt}=kP\Big(1-\frac{P}{M}\Big),
+\end{equation}
+where $k$ and $M$ are constants, respectively various to selected species and environment.
+For specification, we take, for instant, $k=0.5$ and $M=100$. The right hand side of
+(\ref{eqn6}) then becomes $f(t,P)=0.5\,P(1-0.01\,P)$. In Figure \ref{Fig8}, we draw the vector field
+of (\ref{eqn6}) on the grid $R=\{(t,P)\colon 0\le t\le 10,\,95\le P\le 100\}$ and the equilibrium
+solution curve $P=100$. Furthermore, with the initial condition $P(0)=95$, the equation (\ref{eqn6})
+has the unique solution $P=1900(e^{-0.5t}+19)^{-1}$. This solution curve is also given in Figure \ref{Fig8}.
+\begin{figure}[htbp]
+\centering\begin{pspicture}(0,76)(8.4,80.4)
+\psset{xunit=0.8,yunit=0.8}
+\multido{\ny=95.25+0.50}{10}
+{\multido{\nx=0.25+0.50}{20}
+{\parametricplot[linewidth=0.5pt,algebraic,arrows=->,arrowinset=0.55,linecolor=red]{0}{1}{\nx+(t/2)*(1/sqrt(1+((0.5)*(\ny)*(1-(0.01)*(\ny)))^2))|\ny+(t/2)*(1/sqrt(1+((0.5)*(\ny)*(1-(0.01)*(\ny)))^2))*((0.5)*(\ny)*(1-(0.01)*(\ny)))}}}
+\psplot[algebraic,linewidth=1.2pt]{0}{10.25}{100}
+\psplot[algebraic,linewidth=1.2pt]{0}{10.25}{1900/((ch(1)+sh(1))^(-0.5*x)+19)}
+\psaxes[labelFontSize=$\footnotesize$,Dy=1,Dx=1,Oy=95,ticksize=2pt,labelsep=4pt,linewidth=0.7pt]{->}(0,95)(0,95)(10.5,100.5)
+\rput(10.5,94.8){$t$}\rput(-0.25,100.5){$P$}
+\end{pspicture}
+\caption{The vector field of (\ref{eqn6}) with $k=0.5$ and $M=100$.}\label{Fig8}
+\end{figure}
+
+The previous differential equations are all of seperated variable or linear cases that
+can be solved for closed-form solutions by some simple integration formulas. We will consider one more
+equation of the non-linear case whose solution can only be approximated by numerical methods.
+The vector field of such an equation is so useful and we will use the Runge-Kutta curves (of order $4$)
+to add more information about the behaviour of solution curve. Here, those Runge-Kutta curves are depicted by the procedure
+\texttt{\symbol{92}psplotDiffEqn}, also updated from the package \texttt{pstricks-add}.
+
+The vector field of the non-linear differential equation
+\begin{equation}\label{eqn7}
+\frac{dy}{dx}=y^2-xy+1
+\end{equation}
+will be depicted on the grid $R=\{(x,y)\colon -3\le x\le 3,\,-3\le y\le 3\}$ and the solutions
+of Cauchy problems for (\ref{eqn7}), corresponding to initial conditions
+\begin{listof}
+\item $y(-3)=-1$,
+\item $y(-2)=-3$,
+\item $y(-3)=-0.4$,
+\end{listof}
+will be approximated by the method of Runge-Kutta, with the grid size $h=0.2$. It is very easy
+to recognize approximation curves, respective to (i), (ii) and (iii) in Figure \ref{Fig9} below.
+\begin{figure}[htbp]
+\centering\begin{pspicture}(-3.6,-3.6)(4.2,4.2)
+\psset{unit=1.2,dotsize=2.6pt}
+\vecfld{\nx=-3.00+0.4}{\ny=-3.00+0.4}{16}{16}{0.35}{(\ny)^2-(\nx)*(\ny)+1}
+\psplotDiffEqn[linewidth=1.2pt,algebraic,showpoints=true,plotpoints=24,method=rk4]{-3}{1.9}{-1}{(y[0])^2-x*y[0]+1}
+\psplotDiffEqn[linewidth=1.2pt,algebraic,showpoints=true,plotpoints=25,method=rk4]{-2}{3}{-3}{(y[0])^2-x*y[0]+1}
+\psplotDiffEqn[linewidth=1.2pt,algebraic,showpoints=true,plotpoints=10,method=rk4]{-3}{-0.96}{-0.4}{(y[0])^2-x*y[0]+1}
+\psaxes[labelFontSize=$\footnotesize$,Dy=1,Dx=1,ticksize=2.2pt,labelsep=4pt,linewidth=0.7pt]{->}(0,0)(-3,-3)(3.5,3.5)
+\rput(3.5,-0.2){$x$}\rput(-0.25,3.5){$y$}
+\end{pspicture}
+\caption{The vector field of (\ref{eqn7}) and the Runge-Kutta curves.}\label{Fig9}
+\end{figure}
+\acknw
+I am very grateful to
+\begin{itemize}
+\item Timothy Van Zandt, Herbert Vo\ss\ and Dominique Rodriguez for helping me with
+their great works on \texttt{PSTricks}.
+\item H\`an Th\'\ecircumflex\ Th\`anh for helping me with his pdf\hskip.03em\LaTeX\ program.
+\end{itemize}
+\begin{thebibliography}{10}
+\bibitem{mot} Dominique Rodriguez \&\ Herbert Vo\ss. \textsl{PSTricks-add, additional macros for PSTricks\/}.
+Version 3.05, \url{http://ctan.org/tex-archive/graphics/pstricks/contrib}, 2008
+\bibitem{hai} Helmut Kopka \&\ Patrick W. Daly. \textsl{Guide to \LaTeX \/}.
+Addison-Wesley, Fourth Edition, 2004, ISBN 0321173856
+\bibitem{ba} Timothy Van Zandt. \textsl{User's Guide\/}. Version 1.5,\\
+\url{http://ctan.org/tex-archive/graphics/pstricks/base}, 2007
+\end{thebibliography}
+\end{document}
diff --git a/Master/texmf-dist/doc/latex/pstricks_calcnotes/README b/Master/texmf-dist/doc/latex/pstricks_calcnotes/README
new file mode 100644
index 00000000000..957a13577ca
--- /dev/null
+++ b/Master/texmf-dist/doc/latex/pstricks_calcnotes/README
@@ -0,0 +1,5 @@
+This package illustrates using PSTricks to make LaTeX lecture notes for
+Calculus.
+
+Author: Le Phuong Quan
+License: lppl
\ No newline at end of file
diff --git a/Master/texmf-dist/doc/latex/pstricks_calcnotes/two_apps.pdf b/Master/texmf-dist/doc/latex/pstricks_calcnotes/two_apps.pdf
new file mode 100644
index 00000000000..68d3563a51a
--- /dev/null
+++ b/Master/texmf-dist/doc/latex/pstricks_calcnotes/two_apps.pdf
diff --git a/Master/tlpkg/tlpsrc/collection-pstricks.tlpsrc b/Master/tlpkg/tlpsrc/collection-pstricks.tlpsrc
index 83199663d0b..3bdb5da7903 100644
--- a/Master/tlpkg/tlpsrc/collection-pstricks.tlpsrc
+++ b/Master/tlpkg/tlpsrc/collection-pstricks.tlpsrc
@@ -71,6 +71,7 @@ depend pst-vue3d
 depend pst2pdf
 depend pstricks
 depend pstricks-add
+depend pstricks_calcnotes
 depend uml
 depend vaucanson-g
 depend collection-basic
diff --git a/Master/tlpkg/tlpsrc/pstricks_calcnotes.tlpsrc b/Master/tlpkg/tlpsrc/pstricks_calcnotes.tlpsrc
new file mode 100644
index 00000000000..e69de29bb2d
--- /dev/null
+++ b/Master/tlpkg/tlpsrc/pstricks_calcnotes.tlpsrc