\documentclass[a4paper,11pt]{article}
\usepackage[T1]{fontenc}
\usepackage[latin1]{inputenc}
\usepackage{pamathx}
\usepackage{url}
\usepackage{amsmath}
\usepackage{tabularx}
\usepackage{longtable}
\usepackage{pstricks}
\usepackage{pst-func}
\let\pstFuncFV\fileversion
\usepackage{pst-math}
\usepackage{pstricks-add}% for the alg parser
\usepackage{showexpl}
\lstset{pos=t,wide=true}
%
\usepackage{xspace}
\def\PS{PostScript\xspace}
\def\CMD#1{{\ttfamily\textbackslash #1}}
\def\dt{\ensuremath{\,\mathrm{d}t}}
%
\def\pshlabel#1{\footnotesize#1}
\def\psvlabel#1{\footnotesize#1}
\usepackage[colorlinks,linktocpage]{hyperref}
%
\begin{document}
\title{\texttt{pst-func}\\plotting special mathematical functions\\
		  \small v.\pstFuncFV}
%\thanks{%
%		This document was written with \texttt{Kile: 1.6a (Qt: 3.1.1; KDE: 3.1.1;}
%		\protect\url{http://sourceforge.net/projects/kile/}) and the PDF output
%		  was build with VTeX/Free (\protect\url{http://www.micropress-inc.com/linux})}\\
\author{Herbert Vo\ss\thanks{%
%%JF
%Thanks to: Attila Gati  and to John Frampton.
Thanks to: Attila Gati, John Frampton and Lars Kotthoff, Jose-Emilio Vila-Forcen.
}}
\date{\today}

\maketitle

\tableofcontents

\clearpage

\section{\CMD{psPolynomial}}
The polynomial function is defined as
\begin{align}
f(x) &= a_0 + a_1x + a_2x^2 + a_3x^3 + \ldots +a_{n-1}x^{n-1} + a_nx^n\\
f^{\prime}(x) &= a_1 + 2a_2x + 3a_3x^2 + \ldots +(n-1)a_{n-1}x^{n-2} + na_nx^{n-1}\\
f^{\prime\prime}(x) &= 2a_2 + 6a_3x + \ldots +(n-1)(n-2)a_{n-1}x^{n-3} + n(n-1)a_nx^{n-2}
\end{align}

\noindent so \verb+pst-func+ needs only the coefficients of the
polynomial to calculate the function. The syntax is
\begin{verbatim}
\psPolynomial[<options>]{xStart}{xEnd}
\end{verbatim}

With the option \verb+xShift+ one can do a horizontal shift to the graph of the function. With another
than the predefined value the macro replaces $x$ by $x-x\mathrm{Shift}$; \verb+xShift=1+
moves the graph of the polynomial function one unit to the right.


\begin{center}
\bgroup
\psset{yunit=0.5cm,xunit=1cm}
\begin{pspicture*}(-3,-5)(5,10)
  \psaxes[Dy=2]{->}(0,0)(-3,-5)(5,10)
  \psset{linewidth=1.5pt}
  \psPolynomial[coeff=6 3 -1,linecolor=red]{-3}{5}
  \psPolynomial[coeff=2 -1 -1 .5 -.1 .025,linecolor=blue]{-2}{4}
  \psPolynomial[coeff=-2 1 -1 .5 .1 .025 .2 ,linecolor=magenta]{-2}{4}
  \psPolynomial[coeff=-2 1 -1 .5 .1 .025 .2 ,linecolor=magenta,xShift=1,linestyle=dashed]{-2}{4}
  \rput[lb](4,4){\textcolor{red}{$f(x)$}}
  \rput[lb](4,8){\textcolor{blue}{$g(x)$}}
  \rput[lb](2,4){\textcolor{magenta}{$h(x)$}}
\end{pspicture*}
\egroup
\end{center}


\begin{lstlisting}
\psset{yunit=0.5cm,xunit=1cm}
\begin{pspicture*}(-3,-5)(5,10)
  \psaxes[Dy=2]{->}(0,0)(-3,-5)(5,10)
  \psset{linewidth=1.5pt}
  \psPolynomial[coeff=6 3 -1,linecolor=red]{-3}{5}
  \psPolynomial[coeff=2 -1 -1 .5 -.1 .025,linecolor=blue]{-2}{4}
  \psPolynomial[coeff=-2 1 -1 .5 .1 .025 .2 ,linecolor=magenta]{-2}{4}
  \psPolynomial[coeff=-2 1 -1 .5 .1 .025 .2 ,linecolor=magenta,xShift=1,linestyle=dashed]{-2}{4}
  \rput[lb](4,4){\textcolor{red}{$f(x)$}}
  \rput[lb](4,8){\textcolor{blue}{$g(x)$}}
  \rput[lb](2,4){\textcolor{magenta}{$h(x)$}}
\end{pspicture*}
\end{lstlisting}


The plot is easily clipped using the star version of the
\verb+pspicture+ environment, so that points whose coordinates
are outside of the desired range are not plotted.
The plotted polynomials are:
\begin{align}
f(x) & = 6 + 3x -x^2 \\
g(x) & = 2 -x -x^2 +0.5x^3 -0.1x^4 +0.025x^5\\
h(x) & = -2 +x -x^2 +0.5x^3 +0.1x^4 +0.025x^5+0.2x^6\\
h^*(x) & = -2 +(x-1) -(x-1)^2 +0.5(x-1)^3 +\nonumber\\
       & \phantom{ = }+0.1(x-1)^4 +0.025(x-1)^5+0.2(x-1)^6
\end{align}




There are the following new options:

\noindent\medskip
{\tabcolsep=2pt
\begin{tabularx}{\linewidth}{@{}>{\ttfamily}l>{\ttfamily}l>{\ttfamily}lX@{}}
\textrm{Name} & \textrm{Value}  & \textrm{Default}\\\hline
coeff        & a0 a1 a2 ... & 0 0 1 & The coefficients must have the order $a_0\ a_1\ a_2 \ldots$ and
be separated by \textbf{spaces}. The number of coefficients
is limited only by the memory of the computer ... The default
value of the parameter \verb+coeff+ is \verb+0 0 1+, which gives
the parabola $y=a_0+a_1x+a_2x^2=x^2$.\\
xShift & <number>     & 0     & $(x-xShift)$ for the horizontal shift of the polynomial\\
Derivation & <number>     & 0     & the default is the function itself\\
markZeros    & false|true       & false & dotstyle can be changed\\
epsZero      & <value> &  0.1 & The distance between two zeros, important for
                                  the iteration function to test, if the zero value still
				  exists\\
dZero      & <value> &  0.1 & When searching for all zero values, the function is scanned
                              with this step\\
zeroLineTo    & <number>  & false & plots a line from the zero point to the value of the
                                    zeroLineTo's Derivation of the polynomial function\\
\end{tabularx}
}

\noindent
{\tabcolsep=2pt
\begin{tabularx}{\linewidth}{@{}>{\ttfamily}l>{\ttfamily}l>{\ttfamily}lX@{}}
\textrm{Name} & \textrm{Value}  & \textrm{Default}\\\hline
zeroLineStyle    & <line style>  & dashed & the style is one of the for PSTricks valid styles.\\
zeroLineColor  & <color>  & black & any valid xolor is possible\\
zeroLineWidth  & <width> & \rlap{0.5\textbackslash pslinewidth} & \\
\end{tabularx}
}



\bigskip
The above parameter are only 
valid for the \verb+\psPolynomial+
macro, except \verb+x0+, which can also be used for the Gauss function. All
options can be set in the usual way with \verb+\psset+. 



\bigskip
\begin{LTXexample}
\psset{yunit=0.5cm,xunit=2cm}
\begin{pspicture*}(-3,-5)(3,10)
  \psaxes[Dy=2]{->}(0,0)(-3,-5)(3,10)
  \psset{linewidth=1.5pt}
  \psPolynomial[coeff=-2 1 -1 .5 .1 .025 .2 ,linecolor=magenta]{-2}{4}
  \psPolynomial[coeff=-2 1 -1 .5 .1 .025 .2 ,linecolor=red,%
    linestyle=dashed,Derivation=1]{-2}{4}
  \psPolynomial[coeff=-2 1 -1 .5 .1 .025 .2 ,linecolor=blue,%
    linestyle=dotted,Derivation=2]{-2}{4}
  \rput[lb](2,4){\textcolor{magenta}{$h(x)$}}
  \rput[lb](1,1){\textcolor{red}{$h^{\prime}(x)$}}
  \rput[lb](-1,6){\textcolor{blue}{$h^{\prime\prime}(x)$}}
\end{pspicture*}
\end{LTXexample}
%$

\begin{LTXexample}
\psset{yunit=0.5cm,xunit=2cm}
\begin{pspicture*}(-3,-5)(3,10)
  \psaxes[Dy=2]{->}(0,0)(-3,-5)(3,10)
  \psset{linewidth=1.5pt}
  \psPolynomial[coeff=0 0 0 1,linecolor=blue]{-2}{4}
  \psPolynomial[coeff=0 0 0 1,linecolor=red,%
    linestyle=dashed,Derivation=2]{-2}{4}
  \psPolynomial[coeff=0 0 0 1,linecolor=cyan,%
    linestyle=dotted,Derivation=3]{-2}{4}
  \rput[lb](1.8,4){\textcolor{blue}{$f(x)=x^3$}}
  \rput[lb](0.2,8){\textcolor{red}{$f^{\prime\prime}(x)=6x$}}
  \rput[lb](-2,5.5){\textcolor{magenta}{$f^{\prime\prime\prime}(x)=6$}}
\end{pspicture*}
\end{LTXexample}
%$

\begin{LTXexample}
\begin{pspicture*}(-5,-5)(5,5)
  \psaxes{->}(0,0)(-5,-5)(5,5)%
  \psset{dotscale=2}
  \psPolynomial[markZeros,linecolor=red,linewidth=2pt,coeff=-1 1 -1 0 0.15]{-4}{3}%
  \psPolynomial[markZeros,linecolor=blue,linewidth=1pt,linestyle=dashed,%
    coeff=-1 1 -1 0 0.15,Derivation=1,zeroLineTo=0]{-4}{3}%
  \psPolynomial[markZeros,linecolor=magenta,linewidth=1pt,linestyle=dotted,%
    coeff=-1 1 -1 0 0.15,Derivation=2,zeroLineTo=0]{-4}{3}%
  \psPolynomial[markZeros,linecolor=magenta,linewidth=1pt,linestyle=dotted,%
    coeff=-1 1 -1 0 0.15,Derivation=2,zeroLineTo=1]{-4}{3}%
\end{pspicture*}
\end{LTXexample}

\begin{LTXexample}
\psset{xunit=1.5}
\begin{pspicture*}(-5,-5)(5,5)
  \psaxes{->}(0,0)(-5,-5)(5,5)%
  \psset{dotscale=2,dotstyle=x,zeroLineStyle=dotted,zeroLineWidth=1pt}
  \psPolynomial[markZeros,linecolor=red,linewidth=2pt,coeff=-1 1 -1 0 0.15]{-4}{3}%
  \psPolynomial[markZeros,linecolor=blue,linewidth=1pt,linestyle=dashed,%
    coeff=-1 1 -1 0 0.15,Derivation=1,zeroLineTo=0]{-4}{3}%
  \psPolynomial[markZeros,linecolor=magenta,linewidth=1pt,linestyle=dotted,%
    coeff=-1 1 -1 0 0.15,Derivation=2,zeroLineTo=0]{-4}{3}%
  \psPolynomial[markZeros,linecolor=magenta,linewidth=1pt,linestyle=dotted,%
    coeff=-1 1 -1 0 0.15,Derivation=2,zeroLineTo=1]{-4}{3}%
\end{pspicture*}
\end{LTXexample}


\clearpage
\section{\CMD{psFourier}}

A Fourier sum has the form:
\begin{align}
s(x) = \frac{a_0}{2} & + a_1\cos{\omega x} + a_2\cos{2\omega x} +
   a_3\cos{3\omega x} +
	\ldots + a_n\cos{n\omega x}\\
	& + b_1\sin{\omega x} + b_2\sin{2\omega x} + b_3\sin{3\omega x} +
	\ldots + b_m\sin{m\omega x}
\end{align}

\noindent The macro \verb+psFourier+ plots Fourier sums. The
syntax is similiar to \verb+psPolynomial+, except that there are
two kinds of coefficients:
\begin{verbatim}
\psPolynomial[cosCoeff=a0 a1 a2 ..., sinCoeff=b1 b2 ...]{xStart}{xEnd}
\end{verbatim}
The coefficients must have the orders $a_0\ a_1\ a_2\ \ldots$
and $b_1\ b_2\ b_3\ \ldots$ and be separated by
\textbf{spaces}. The default is \verb+cosCoeff=0,sinCoeff=1+,
which gives the standard \verb+sin+ function. Note that
%%JF, I think it is better without the angle brackets, but
%%you know the conventions used better than I do, so you
%%may disagree.
%the constant value can only be set with \verb+cosCoeff=<a0>+.
the constant value can only be set with \verb+cosCoeff=a0+.

\begin{LTXexample}
\begin{pspicture}(-5,-3)(5,5.5)
\psaxes{->}(0,0)(-5,-2)(5,4.5)
\psset{plotpoints=500,linewidth=1pt}
\psFourier[cosCoeff=2, linecolor=green]{-4.5}{4.5}
\psFourier[cosCoeff=0 0 2, linecolor=magenta]{-4.5}{4.5}
\psFourier[cosCoeff=2 0 2, linecolor=red]{-4.5}{4.5}
\end{pspicture}
\end{LTXexample}

\begin{LTXexample}
\psset{yunit=0.75}
\begin{pspicture}(-5,-6)(5,7)
\psaxes{->}(0,0)(-5,-6)(5,7)
\psset{plotpoints=500}
\psFourier[linecolor=red,linewidth=1pt]{-4.5}{4.5}
\psFourier[sinCoeff= -1 1 -1 1 -1 1 -1 1,%
	linecolor=blue,linewidth=1.5pt]{-4.5}{4.5}
\end{pspicture}
\end{LTXexample}

\begin{LTXexample}
\begin{pspicture}(-5,-5)(5,5.5)
\psaxes{->}(0,0)(-5,-5)(5,5)
\psset{plotpoints=500,linewidth=1.5pt}
\psFourier[sinCoeff=-.5 1 1 1 1 ,cosCoeff=-.5 1 1 1 1 1,%
	linecolor=blue]{-4.5}{4.5}
\end{pspicture}
\end{LTXexample}

\clearpage
\section{\CMD{psBessel}}
The Bessel function of order $n$ is defined as
\begin{align}
J_n(x) &=\frac{1}{\pi}\int_0^\pi\cos(x\sin t-nt)\dt\\
       &=\sum_{k=0}^{\infty}\frac{(-1)^k \left(\frac{x}{2}\right)^{n+2k}}{k!\Gamma(n+k+1)}
\end{align}

\noindent The syntax of the macro is
\begin{verbatim}
\psBessel[options]{order}{xStart}{xEnd}
\end{verbatim}

There are two special parameters for the Bessel function, and also the
settings of many \verb+pst-plot+ or \verb+pstricks+ parameters
affect the plot. 

\begin{verbatim}
\def\psset@constI#1{\edef\psk@constI{#1}}
\def\psset@constII#1{\edef\psk@constII{#1}}
\psset{constI=1,constII=0}
\end{verbatim}

These two "'constants"` have the following meaning:
\[
f(t) = constI \cdot J_n + constII
\]

\noindent
where $constI$ and $constII$ must be real PostScript expressions, e.g.:
\begin{verbatim}
\psset{constI=2.3,constII=t k sin 1.2 mul 0.37 add}
\end{verbatim}

The Bessel function is plotted with the parametricplot macro, this is the
reason why the variable is named \verb+t+. The internal procedure \verb+k+ 
converts the value t from radian into degrees. The above setting is
the same as
\[
f(t) = 2.3 \cdot J_n + 1.2\cdot \sin t + 0.37
\]
 

In particular, note that the default for
\verb+plotpoints+ is $500$. If the plotting computations are too
time consuming at this setting, it can be decreased in the usual
way, at the cost of some reduction in graphics resolution.

\begin{LTXexample}
{
\psset{xunit=0.25,yunit=5}
\begin{pspicture}(-13,-.85)(13,1.25)
\rput(13,0.8){%
	$\displaystyle J_n(x)=\frac{1}{\pi}\int_0^\pi\cos(x\sin t-nt)\dt$%
}
\psaxes[Dy=0.2,Dx=4]{->}(0,0)(-30,-.8)(30,1.2)
\psset{linewidth=1pt}
\psBessel[linecolor=red]{0}{-28}{28}%
\psBessel[linecolor=blue]{1}{-28}{28}%
\psBessel[linecolor=green]{2}{-28}{28}%
\psBessel[linecolor=magenta]{3}{-28}{28}%
\end{pspicture}
}
\end{LTXexample}


\begin{LTXexample}
{
\psset{xunit=0.25,yunit=2.5}
\begin{pspicture}(-13,-1.5)(13,3)
\rput(13,0.8){%
	$\displaystyle f(t) = 2.3 \cdot J_0 + 1.2\cdot \sin t + 0.37$%
}
\psaxes[Dy=0.8,dy=2cm,Dx=4]{->}(0,0)(-30,-1.5)(30,3)
\psset{linewidth=1pt}
\psBessel[linecolor=red,constI=2.3,constII={t k sin 1.2 mul 0.37 add}]{0}{-28}{28}%
\end{pspicture}
}
\end{LTXexample}

\clearpage
\section{\CMD{psGauss} and \CMD{psGaussI}}
The Gauss function  is defined as
%
\begin{align}
f(x) &= \dfrac{1}{\sigma\sqrt{2\pi}}\,e^{-\dfrac{\left(x-\mu\right)^2}{2\sigma{}^2}}
\end{align}
%
\noindent The syntax of the macros is
\begin{verbatim}
\psGauss[options]{xStart}{xEnd}
\psGaussI[options]{xStart}{xEnd}
\end{verbatim}

%%JF
%% comment, the angle brackets below, around "value", make sense
%% as a convention, so I left them in
%
%\noindent where the only new parameter is \verb+sigma=<value>+, with
%the default of \verb+0.5+ and can also be set in the usual way with
%\verb+\psset+. It is only valid for the \verb+psGauss+-macro.
\noindent where the only new parameter are \verb+sigma=<value>+ and  \verb+mue=<value>+ for the
horizontal shift,
which can also be set in the usual way with \verb+\psset+. It is
significant only for the \verb+psGauss+- and \verb+\psGaussI+-macro. The default is
\verb+sigma=0.5+ and \verb+mue=0+. The integral is caclulated wuth the Simson algorithm 
and has one special option, called \verb+Simpson+, which defines the number of intervalls per step
and is predefined with 5.

\bgroup
\psset{yunit=4cm,xunit=3}
\begin{pspicture}(-2,-0.2)(2,1.4)
%  \psgrid[griddots=10,gridlabels=0pt, subgriddiv=0]
  \psaxes[Dy=0.25]{->}(0,0)(-2,0)(2,1.25)
  \uput[-90](6,0){x}\uput[0](0,1){y}
  \rput[lb](1,0.75){\textcolor{red}{$\sigma =0.5$}}
  \rput[lb](1,0.5){\textcolor{blue}{$\sigma =1$}}
  \rput[lb](-2,0.5){$f(x)=\dfrac{1}{\sigma\sqrt{2\pi}}\,e^{-\dfrac{(x-\mu)^2}{2\sigma{}^2}}$}
  \psGauss[linecolor=red, linewidth=2pt]{-1.75}{1.75}%
  \psGaussI[linewidth=1pt,yunit=0.75]{-2}{2}%
  \psGauss[linecolor=cyan, mue=0.5, linewidth=2pt]{-1.75}{1.75}%
  \psGauss[sigma=1, linecolor=blue, linewidth=2pt]{-1.75}{1.75}
\end{pspicture}
\egroup


\begin{lstlisting}[xrightmargin=-2cm]
\psset{yunit=4cm,xunit=3}
\begin{pspicture}(-2,-0.5)(2,1.25)
%  \psgrid[griddots=10,gridlabels=0pt, subgriddiv=0]
  \psaxes[Dy=0.25]{->}(0,0)(-2,0)(2,1.25)
  \uput[-90](6,0){x}\uput[0](0,1){y}
  \rput[lb](1,0.75){\textcolor{red}{$\sigma =0.5$}}
  \rput[lb](1,0.5){\textcolor{blue}{$\sigma =1$}}
  \rput[lb](-2,0.5){$f(x)=\dfrac{1}{\sigma\sqrt{2\pi}}\,e^{-\dfrac{(x-x_0)^2}{2\sigma{}^2}}$}
  \psGauss[linecolor=red, linewidth=2pt]{-1.75}{1.75}%
  \psGaussI[linewidth=1pt,yunit=0.75cm]{-2}{2}%
  \psGauss[linecolor=cyan, mue=0.5, linewidth=2pt]{-1.75}{1.75}%
  \psGauss[sigma=1, linecolor=blue, linewidth=2pt]{-1.75}{1.75}
\end{pspicture}
\end{lstlisting}


\clearpage
\section{\CMD{psSi}, \CMD{pssi} and \CMD{psCi}}
The integral sin  and cosin are defined as
\begin{align}
\mathrm{Si}(x) &= \int_0^x\dfrac{\sin t}{t}\dt\\
\mathrm{si}(x) &= - \int_x^{\infty}\dfrac{\sin t}{t}\dt=\mathrm{Si}(x)-\frac{\pi}{2}\\
\mathrm{Ci}(x) &= -\int_x^{\infty}\dfrac{\cos t}{t}\dt=\gamma+\ln x +\int_0^{x}\dfrac{\cos t -1}{t}\dt
\end{align}
%
\noindent The syntax of the macros is
\begin{verbatim}
\psSi[options]{xStart}{xEnd}
\pssi[options]{xStart}{xEnd}
\psCi[options]{xStart}{xEnd}
\end{verbatim}


\begin{LTXexample}[pos=t]
\def\pshlabel#1{\footnotesize#1} \def\psvlabel#1{\footnotesize#1}
\psset{xunit=0.5}
\begin{pspicture}(-15,-4.5)(15,2)
  \psaxes[dx=1cm,Dx=2]{->}(0,0)(-15.1,-4)(15,2)
  \psplot[plotpoints=1000]{-14.5}{14.5}{ x RadtoDeg sin x div }
  \psSi[plotpoints=1500,linecolor=red,linewidth=1pt]{-14.5}{14.5}
  \pssi[plotpoints=1500,linecolor=blue,linewidth=1pt]{-14.5}{14.5}
  \rput(-5,1.5){\color{red}$Si(x)=\int\limits_{0}^x \frac{\sin(t)}{t}\dt$}  
  \rput(8,-1.5){\color{blue}$si(x)=-\int\limits_{x}^{\infty} \frac{\sin(t)}{t}\dt=Si(x)-\frac{\pi}{2}$}  
  \rput(8,.5){$f(x)= \frac{\sin(t)}{t}$}
\end{pspicture}
\end{LTXexample}



\begin{LTXexample}[pos=t]
\def\pshlabel#1{\footnotesize#1} \def\psvlabel#1{\footnotesize#1}
\psset{xunit=0.5}
\begin{pspicture*}(-15,-4.2)(15,4.2)
  \psaxes[dx=1cm,Dx=2]{->}(0,0)(-15.1,-4)(15,4)
  \psplot[plotpoints=1000]{-14.5}{14.5}{ x RadtoDeg cos x Div }
  \psCi[plotpoints=500,linecolor=red,linewidth=1pt]{-11.5}{11.5}
  \psci[plotpoints=500,linecolor=blue,linewidth=1pt]{-11.5}{11.5}
  \rput(-8,1.5){\color{red}$Ci(x)=-\int\limits_{x}^{\infty} \frac{\cos(t)}{t}\dt$}  
  \rput(8,1.5){\color{blue}$ci(x)=-Ci(x)+\ln(x)+\gamma$}  
\end{pspicture*}
\end{LTXexample}


\clearpage
\section{\CMD{psIntegral}, \CMD{psCumIntegral} and \CMD{psConv}}
These new macros\footnote{Created by Jose-Emilio Vila-Forcen}
allows to plot the result of an integral using the Simpson numerical integration rule. 
The first one is the result of the integral of a function with two variables, and 
the integral is performed over one of them. The second one is the cumulative 
integral of a function (similar to \verb+\psGaussI+ but valid for all functions). The third 
one is the result of a convolution. They are defined as:
\begin{align}
\text{psIntegral}(x) &= \int_a^b f(x,t)dt \\
\text{psCumIntegral}(x) &= \int_{\text{xStart}}^{x} f(t)dt \\
\text{psConv}(x) & = \int_a^b f(t)g(x-t)dt
\end{align}
In the first one, the integral is performed from $a$ to $b$ and the function $f$ depends 
on two parameters. In the second one, the function $f$ depends on only one parameter, and the 
integral is performed from the minimum value specified for $x$ (\verb|xStart|) and the current 
value of $x$ in the plot. The third one uses the \CMD{psIntegral} macro to perform an approximation 
to the convolution, where the integration is performed from $a$ to $b$.

The syntax of these macros is:
\begin{verbatim}
\psIntegral[<options>]{xStart}{xEnd}(a,b){ function }
\psCumIngegral[<options>]{xStart}{xEnd}{ function }
\psConv[<options>]{xStart}{xEnd}(a,b){ function f }{ function g }
\end{verbatim}

In the first macro, the function should be created such that it accepts two values: \verb|<x t function>| 
should be a value. For the second and the third functions, they only need to accept one 
parameter: \verb|<x function>| should be a value.

There are no new parameters for these functions. The two most important ones are \verb-plotpoints-, 
which controls the number of points of the plot (number of divisions on $x$ for the plot) and 
\verb-Simpson-, which controls the precision of the integration (a larger number means a smallest 
step). The precision and the smoothness of the plot depend strongly on these two parameters.

\bigskip
\begin{LTXexample}
%\usepackage{pst-math}
\psset{xunit=0.5cm,yunit=2cm}
\begin{pspicture}[linewidth=1pt](-10,-.5)(10,1.5)
  \psaxes[dx=1cm,Dx=2]{->}(0,0)(-10,0)(10,1.5)
  \psCumIntegral[plotpoints=200,Simpson=10]{-10}{10}{0 1 GAUSS}
  \psIntegral[plotpoints=200,Simpson=10,linecolor=red]{-10}{10}(-4,6){1 GAUSS}
  \psIntegral[plotpoints=200,Simpson=100,linecolor=green]{.1}{10}(-3,3){0 exch GAUSS}
\end{pspicture}
\end{LTXexample}

In the example, the cumulative integral of a Gaussian is presented in black. In red, a 
Gaussian is varying its mean from -10 to 10, and the result is the integral from -4 to 6. 
Finally, in green it is presented the integral of a Gaussian from -3 to 3, where the 
variance is varying from .1 to 10.

\begin{LTXexample}
\psset{xunit=1cm,yunit=4cm}
\begin{pspicture}[linewidth=1pt](-5,-.2)(5,0.75)
  \psaxes[dx=1cm,Dx=1,Dy=0.5]{->}(0,0)(-5,0)(5,0.75)
  \psplot[linecolor=blue,plotpoints=200]{-5}{5}{x abs 2 le {0.25}{0} ifelse}
  \psplot[linecolor=green,plotpoints=200]{-5}{5}{x abs 1 le {.5}{0} ifelse}
  \psConv[plotpoints=100,Simpson=1000,linecolor=red]{-5}{5}(-10,10)%
    {abs 2 le {0.25}{0} ifelse}{abs 1 le {.5} {0} ifelse}
\end{pspicture}
\end{LTXexample}

In the second example, a convolution is performed using two rectangle functions. 
The result (in red) is a trapezoid function.

\clearpage

\section{\CMD{psBinomial} and \CMD{psBinomialN}}

These two macros plot binomial distribution, \CMD{psBinomialN} the normalized one. It is always 
done in the $x$-Intervall $[0;1]$.
Rescaling to another one can be done by setting the \verb+xunit+ option
to any other value. 

The binomial distribution gives the discrete probability distribution $P_p(n|N)$ of obtaining
exactly $n$ successes out of $N$ Bernoulli trials (where the result of each 
Bernoulli trial is true with probability $p$ and false with probability
$q=1-p$.  The binomial distribution is therefore given by

\begin{align}
P_p(n|N) &= \binom{N}{n}p^nq^{N-n} \\
         &= \frac{N!}{n!(N-n)!}p^n(1-p)^{N-n},
\end{align}
where $(N; n)$ is a binomial coefficient and $P$ the probability. 

The syntax is quite easy:
\begin{verbatim}
\psBinomial[<options>]{N}{probability p}
\psBinomialN[<options>]{N}{probability p}
\end{verbatim}

There is a restriction in using the value for N. It depends to the probability, but in general
one should expect problems with $N>100$. PostScript cannot handle such small values and there will
be no graph printed. This happens on PostScript side, so \TeX\ doesn't report any problem in
the log file. The valid options for the macros are \verb+markZeros+ to draw rectangles instead
of a continous line and \verb+printValue+ for printing the $y$-values on top of the lines,
rotated by 90\textdegree. For this option all other options from section~\ref{sec:printValue}
for the macro \verb+\psPrintValue+ are valid, too. The only special option is \verb+barwidth+,
which is a factor (no dimension) and set by default to 1. This option is only valid for
the macro \CMD{psBinomial} and not for the normalized one!

\psset[pst-func]{barwidth=1}
\begin{LTXexample}[pos=t,preset=\centering]
\psset{xunit=1cm,yunit=5cm}%
\begin{pspicture}(-1,-0.15)(7,0.55)%
\psaxes[Dy=0.2,dy=0.2\psyunit]{->}(0,0)(-1,0)(7,0.5)
\uput[-90](7,0){$k$} \uput[90](0,0.5){$P(X=k)$}
\psBinomial[markZeros,printValue,fillstyle=vlines]{6}{0.4}
\end{pspicture}
\end{LTXexample}

\begin{LTXexample}[pos=t,preset=\centering]
\psset{xunit=1cm,yunit=10cm}%
\begin{pspicture}(-1,-0.1)(8,0.6)%
\psaxes[Dy=0.2,dy=0.2\psyunit]{->}(0,0)(-1,0)(8,0.5)
\uput[-90](8,0){$k$} \uput[90](0,0.5){$P(X=k)$}
\psBinomial[linecolor=red,markZeros,printValue,fillstyle=solid,
	fillcolor=blue,barwidth=0.2]{7}{0.6}
\end{pspicture}
\end{LTXexample}


\begin{LTXexample}[pos=t,preset=\centering]
\psset{xunit=0.25cm,yunit=10cm}
\begin{pspicture*}(-1,-0.1)(61,0.52)
\psaxes[Dx=5,dx=5\psxunit,Dy=0.2,dy=0.2\psyunit]{->}(60,0.5)
\uput[-90](60,0){$k$} \uput[0](0,0.5){$P(X=k)$}
\psBinomial[markZeros,linecolor=red]{4}{.5}
\psset{linewidth=1pt}
\psBinomial[linecolor=green]{5}{.5}
\psBinomial[linecolor=blue]{10}{.5}
\psBinomial[linecolor=red]{20}{.5}
\psBinomial[linecolor=magenta]{50}{.5}
\psBinomial[linecolor=cyan]{75}{.5}
\end{pspicture*}
\end{LTXexample}

The default binomial distribution has the mean of $\mu=E(X)=N\cdot p$ and a variant of $\sigma^2=\mu\cdot(1-p)$.
The normalized distribution has a mean of $0$. Instead of $P(X=k)$ we use $P(Z=z)$ with $Z=\dfrac{X-E(X)}{\sigma(X)}$
and $P\leftarrow P\cdot\sigma$.
The macros use the rekursive definition of the binomial distribution:
%
\begin{align}
P(k) &= P(k-1)\cdot\frac{N-k+1}{k}\cdot\frac{p}{1-p}
\end{align}


\begin{LTXexample}[pos=t,preset=\centering]
\psset{xunit=1cm,yunit=5cm}%
\begin{pspicture}(-3,-0.15)(4,0.55)%
\psaxes[Dy=0.2,dy=0.2\psyunit]{->}(0,0)(-3,0)(4,0.5)
\uput[-90](4,0){$z$} \uput[0](0,0.5){$P(Z=z)$}
\psBinomialN[markZeros,fillstyle=vlines]{6}{0.4}
\end{pspicture}
\end{LTXexample}


\begin{LTXexample}[pos=t,preset=\centering]
\psset{yunit=10}
\begin{pspicture*}(-8,-0.07)(8.1,0.55)
\psaxes[Dy=0.2,dy=0.2\psyunit]{->}(0,0)(-8,0)(8,0.5)
\uput[-90](8,0){$z$} \uput[0](0,0.5){$P(Z=z)$}
\psBinomialN{125}{.5}
\psBinomialN[markZeros,linewidth=1pt,linecolor=red]{4}{.5}
\end{pspicture*}
\end{LTXexample}

\begin{LTXexample}[pos=t,preset=\centering]
\psset{yunit=10}
\begin{pspicture*}(-8,-0.07)(8.1,0.52)
\psaxes[Dy=0.2,dy=0.2\psyunit]{->}(0,0)(-8,0)(8,0.5)
\uput[-90](8,0){$z$} \uput[0](0,0.5){$P(Z=z)$}
\psBinomialN[markZeros,linecolor=red]{4}{.5}
\psset{linewidth=1pt}
\psBinomialN[linecolor=green]{5}{.5}\psBinomialN[linecolor=blue]{10}{.5}
\psBinomialN[linecolor=red]{20}{.5} \psBinomialN[linecolor=gray]{50}{.5}
\end{pspicture*}
\end{LTXexample}

For the normalized distribution the plotstyle can be set to \verb+curve+ (\verb+plotstyle=curve+), 
then the binomial distribution looks like a normal distribution. This option is only
valid vor \CMD{psBinomialN}. The option \verb+showpoints+ is valid if \verb+curve+ was chosen.

\begin{LTXexample}[pos=t,preset=\centering]
\psset{xunit=1cm,yunit=10cm}%
\begin{pspicture*}(-4,-0.06)(4.1,0.57)%
\psaxes[Dy=0.2,dy=0.2\psyunit]{->}(0,0)(-4,0)(4,0.5)%
\uput[-90](4,0){$z$} \uput[90](0,0.5){$P(Z=z)$}%
\psBinomialN[linecolor=red,fillstyle=vlines,showpoints=true,markZeros]{36}{0.5}%
\psBinomialN[linecolor=blue,showpoints=true,plotstyle=curve]{36}{0.5}%
\end{pspicture*}
\end{LTXexample}

\begin{LTXexample}[pos=t,preset=\centering]
\psset{xunit=1cm,yunit=10cm}%
\begin{pspicture*}(-4,-0.06)(4.2,0.57)%
\psaxes[Dy=0.2,dy=0.2\psyunit]{->}(0,0)(-4,0)(4,0.5)%
\uput[-90](4,0){$z$} \uput[90](0,0.5){$P(Z=z)$}%
\psBinomialN[linecolor=red]{10}{0.6}%
\psBinomialN[linecolor=blue,showpoints=true,plotstyle=curve]{10}{0.6}%
\end{pspicture*}
\end{LTXexample}

\clearpage


\section{\CMD{psLame} -- Lamé Curve, a superellipse}
A superellipse is a curve with Cartesian equation
%	
\begin{align}
\left|\frac{x}{a}\right|^r + \left|\frac{y}{b}\right|^r & =1
\end{align}
%
first discussed in 1818 by Gabriel Lamé (1795--1870)%
\footnote{Lamé worked on a wide variety of different topics. 
His work on differential geometry and contributions to Fermat's Last Theorem 
are important. He proved the theorem for $n = 7$ in 1839.}. 
A superellipse may be described parametrically by
%
\begin{align}
x = a\cdot\cos^{\frac{2}{r}} t\\
y = b\cdot\sin^{\frac{2}{r}} t
\end{align}
%			
Superellipses with $a=b$ are also known as Lamé curves or Lamé ovals and
the restriction to $r>2$ is sometimes also made. The following 
table summarizes a few special cases. Piet Hein used $\frac{5}{2}$ with a number of different 
$\frac{a}{b}$ ratios for various of his projects. For example, he used $\frac{a}{b}=\frac{6}{5}$ 
for Sergels Torg 
(Sergel's Square) in Stockholm, and $\frac{a}{b}=\frac{3}{2}$ for his table.

\begin{center}
\begin{tabular}{@{}llm{1.5cm}@{}}
r & curve type & example\\\hline
$\frac{2}{3}$ &	(squashed) astroid 
  & \pspicture(-.5,-.5)(.5,.5)\psLame[radiusA=.5,radiusB=.5]{0.6667}\endpspicture\\
1             & (squashed) diamond
  & \pspicture(-.5,-.5)(.5,.5)\psLame[radiusA=.5,radiusB=.5]{1}\endpspicture\\
2	      & ellipse
  & \pspicture(-.5,-.5)(.5,.5)\psLame[radiusA=.5,radiusB=.5]{2}\endpspicture\\
$\frac{5}{2}$ & Piet Hein's ,,superellipse``
  & \pspicture(-.5,-.5)(.5,.5)\psLame[radiusA=.5,radiusB=.5]{2.5}\endpspicture
\end{tabular}
\end{center}

If is a rational, then a superellipse is algebraic. However, for irrational, 
it is transcendental. For even integers, the curve becomes closer to a 
rectangle as increases. The syntax of the \verb+\psLame+ macro is:

\begin{verbatim}
\psLame[settings]{r}
\end{verbatim}

It is internally ploted as a parametric plot with $0\le\alpha\le360$. Available keywords
are \verb+radiusA+ and \verb+radiusB+, both are preset to 1, but can have any valid value
and unit.

\bgroup
\begin{LTXexample}[pos=t,preset=\centering]
\definecolorseries{col}{rgb}{last}{red}{blue}
\resetcolorseries[41]{col}
\psset{unit=.5}
\pspicture(-9,-9)(9,9)
  \psaxes[Dx=2,Dy=2,tickstyle=bottom,ticksize=2pt]{->}(0,0)(-9,-9)(9,9)
  \multido{\rA=0.2+0.1,\iA=0+1}{40}{%
    \psLame[radiusA=8,radiusB=7,linecolor={col!![\iA]},linewidth=.5pt]{\rA}}
\endpspicture
\end{LTXexample}
\egroup

\clearpage


\section{\CMD{psplotImp} -- plotting implicit defined functions}
This macro is still experimental! For a given area, the macro calculates in a
first step row by row for every pixel (1pt) the function $f(x,y)$ and checks for an
changing of the value from $f(x,y)<0$ to $f(x,y)>0$ or vice versa. If this happens,
then the pixel must be a part of the curve of the function $f(x,y)=0$. In a second step the same is
done column by column. This will take some time because an area of $400\times 300$
pixel needs $120$ thousand calculations of the function value. The user still defines
this area in his own coordinates, the translation into pixel (pt) is done internally by the
macro.

\begin{verbatim}
\psplotImp[<options>](xMin,yMin)(xMax,yMax){<function f(x,y)>}
\end{verbatim}

The function must be of $f(x,y)=0$ and described in PostScript code, or alternatively with
the option \verb+algebraic+ (\verb+pstricks-add+) in an algebraic form. No other value names than $x$ and $y$
are possible. In general a starred \verb+pspicture+ environment maybe a good choice here.
The given area for \verb+\psplotImp+ should be \textbf{greater} than the given \verb+pspicture+ area.

\begin{LTXexample}[preset=\centering]
\begin{pspicture*}(-3,-3.2)(3.5,3.5)
\psaxes{->}(0,0)(-3,-3)(3.2,3)%
\psplotImp[linewidth=2pt,linecolor=red](-5,-2.1)(5,2.1){%
	 x dup mul y dup mul add 4 sub }% circle r=2
\uput[45](0,2){$x^2+y^2-4=0$}
\psplotImp[linewidth=2pt,linecolor=blue,algebraic]%
	(-5,-3)(4,2.4){ (x+1)^2+y^2-4 }% circle r=2
\end{pspicture*}
\end{LTXexample}

\begin{LTXexample}[preset=\centering]
\begin{pspicture*}(-3,-2.2)(3.5,2.5)
\psaxes{->}(0,0)(-3,-2)(3.2,2)%
\psplotImp[linewidth=2pt,linecolor=blue](-5,-2.2)(5,2.4){%
      /xqu x dup mul def
      /yqu y dup mul def
      xqu yqu add dup mul 2 dup add 2 mul xqu yqu sub mul sub }
\uput*[0](-3,2){$\left(x^2+y^2\right)^2-8(x^2-y^2)=0$}
\psplotImp[linewidth=1pt,linecolor=red,algebraic](-5,-2.2)(5,2.4){% Lemniskate a =2
    (x^2+y^2)^2-4*(x^2-y^2) }
\end{pspicture*}
\end{LTXexample}

\begin{LTXexample}[preset=\centering]
\begin{pspicture*}(-3,-3.2)(3.5,3.5)
\psaxes{->}(0,0)(-3,-3)(3.2,3)%
\psplotImp[linewidth=2pt,linecolor=green](-6,-6)(4,2.4){%
	 x 3 exp y 3 exp add 4 x y mul mul sub } 
\uput*[45](-2.5,2){$\left(x^3+y^3\right)-4xy=0$}
\end{pspicture*}
\end{LTXexample}


\begin{LTXexample}[preset=\centering]
\begin{pspicture*}(-5,-3.2)(5.5,4.5)
\psaxes{->}(0,0)(-5,-3)(5.2,4)%
\psplotImp[algebraic,linecolor=red](-6,-4)(5,4){ y*cos(x*y)-0.2 }
\psplotImp[algebraic,linecolor=blue](-6,-4)(5,4){ y*cos(x*y)-1.2 }
\end{pspicture*}
\end{LTXexample}



Using the \verb+polarplot+ option implies using the variables $r$ and $phi$ for describing
the function, $y$ and $x$ are not respected in this case. Using the \verb+algebraic+ option
for polar plots are also possible (see next example).

\begin{LTXexample}[preset=\centering]
\begin{pspicture*}(-3,-2.5)(3.75,2.75)\psaxes{->}(0,0)(-3,-2.5)(3.2,2.5)%
\psplotImp[linewidth=2pt,linecolor=cyan,polarplot](-6,-3)(4,2.4){ r 2 sub }% circle r=2
\uput*[45](0.25,2){$f(r,\phi)=r-2=0$}
\psplotImp[polarplot,algebraic](-6,-3)(4,2.4){ r-1 }% circle r=1
\end{pspicture*}
\end{LTXexample}

\begin{LTXexample}[preset=\centering]
\begin{pspicture*}(-5,-2.2)(5.5,3.5)
\pscircle(0,0){1}% 
\psaxes{->}(0,0)(-5,-2)(5.2,3)%
\multido{\rA=0.01+0.2}{5}{%
\psplotImp[linewidth=1pt,linecolor=blue,polarplot](-6,-6)(5,2.4){%
	 r dup mul 1.0 r div sub phi sin dup mul mul \rA\space sub }}%
\uput*[45](0,2){$f(r,\phi)=\left(r^2-\frac{1}{r}\right)\cdot\sin^2\phi=0$}
\end{pspicture*}
\end{LTXexample}

\begin{LTXexample}[preset=\centering]
\begin{pspicture*}(-4,-3.2)(4.5,4.5)
\psaxes{->}(0,0)(-4,-3)(4.2,4)%
\psplotImp[algebraic,polarplot,linecolor=red](-5,-4)(5,4){ r+cos(phi/r)-2 }
\end{pspicture*}
\end{LTXexample}

\clearpage

\section{\CMD{psPrintValue}}\label{sec:printValue}
This new macro allows to print single values of a math function. It has the syntax
\begin{verbatim}
\psPrintValue[<options>]{<PostScript code>}
\end{verbatim}

Important is the fact, that \CMD{psPrintValue} works on \PS\ side. For \TeX\ it is only a box of
zero dimension. This is the reason why you have to put it into a box, which reserves horizontal
space.

There are the following new options:

\noindent\medskip
\begin{tabularx}{\linewidth}{>{\ttfamily}l|>{\ttfamily}l>{\ttfamily}lX@{}}
\textrm{Name} & \textrm{Value}  & \textrm{Default}\\\hline
PSfont        & PS font name & Times & only valid \PS font names are possible, e.g. \texttt{Times-Roman}, \texttt{Helvetica}, \texttt{Courier}, \texttt{AvantGard}, \texttt{Bookman}\\
fontscale & <number>     & 10     & the font scale in pt\\
valuewidth & <number>     & 10     & the width of the string for the converted 
    real number; if it is too small, no value is printed\\
\end{tabularx}

\begin{center}
\psset{fontscale=12}
\makebox[2em]{x(deg)} \makebox[5em]{$\sin x$} \makebox[5em]{$\cos x$}
\makebox[5em]{$\sqrt x$}\makebox[7em]{$\sin x+\cos x$}\makebox[6em]{$\sin^2 x+\cos^2 x$}\\[3pt]
\multido{\iA=0+10}{18}{
  \makebox[1em]{\iA}
  \makebox[5em]{\psPrintValue[PSfont=Helvetica]{\iA\space sin}}
  \makebox[5em]{\psPrintValue[PSfont=Courier,fontscale=10]{\iA\space cos}}
  \makebox[5em]{\psPrintValue[valuewidth=15,linecolor=blue,PSfont=AvantGarde]{\iA\space sqrt}}
  \makebox[7em]{\psPrintValue[PSfont=Times-Italic]{\iA\space dup sin exch cos add}}
  \makebox[6em]{\psPrintValue[PSfont=Palatino-Roman]{\iA\space dup sin dup mul exch cos dup mul add}}\\}
\end{center}

\bigskip

\begin{lstlisting}
\psset{fontscale=12}
\makebox[2em]{x(deg)} \makebox[5em]{$\sin x$} \makebox[5em]{$\cos x$}
\makebox[5em]{$\sqrt x$}\makebox[7em]{$\sin x+\cos x$}\makebox[6em]{$\sin^2 x+\cos^2 x$}\\[3pt]
\multido{\iA=0+10}{18}{
  \makebox[1em]{\iA}
  \makebox[5em]{\psPrintValue[PSfont=Helvetica]{\iA\space sin}}
  \makebox[5em]{\psPrintValue[PSfont=Courier,fontscale=10]{\iA\space cos}}
  \makebox[5em]{\psPrintValue[valuewidth=15,linecolor=blue,PSfont=AvantGarde]{\iA\space sqrt}}
  \makebox[7em]{\psPrintValue[PSfont=Times-Italic]{\iA\space dup sin exch cos add}}
  \makebox[6em]{\psPrintValue[PSfont=Palatino-Roman]{\iA\space dup sin dup mul exch cos dup mul add}}\\}
\end{lstlisting}



\section{Credits}
Denis Girou | Manuel Luque | Timothy Van Zandt 


\nocite{*}
\bibliographystyle{plain}
\bibliography{pst-func-doc}

\end{document}


