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author | Karl Berry <karl@freefriends.org> | 2010-06-08 13:40:50 +0000 |
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committer | Karl Berry <karl@freefriends.org> | 2010-06-08 13:40:50 +0000 |
commit | 6443467452320c296faa1f43f0606a9457bd4463 (patch) | |
tree | 4e906b09bab1664a6176f8c2655649e68dd23c47 /Master/texmf-dist/doc/latex/pstricks_calcnotes | |
parent | bf36456c4da174e063dd0b02cd0e0c0d263197aa (diff) |
new doc package pstricks_calcnotes (7jun10)
git-svn-id: svn://tug.org/texlive/trunk@18816 c570f23f-e606-0410-a88d-b1316a301751
Diffstat (limited to 'Master/texmf-dist/doc/latex/pstricks_calcnotes')
19 files changed, 918 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 new file mode 100644 index 00000000000..5ee3dcee8e9 --- /dev/null +++ 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 --- /dev/null +++ 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 Binary files differnew file mode 100644 index 00000000000..1bb82c71745 --- /dev/null +++ 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 Binary files differnew file mode 100644 index 00000000000..8efa93a02ca --- /dev/null +++ b/Master/texmf-dist/doc/latex/pstricks_calcnotes/For_Pdf_Output/Fig1-1.pdf 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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 Binary files differnew file mode 100644 index 00000000000..aacbb9457e0 --- /dev/null +++ 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 Binary files differnew file mode 100644 index 00000000000..cec79431078 --- /dev/null +++ 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 Binary files differnew file mode 100644 index 00000000000..68d3563a51a --- /dev/null +++ b/Master/texmf-dist/doc/latex/pstricks_calcnotes/two_apps.pdf |