What is the difference between plotstyle=curve and plotpoints=1000?

Question

The first code:

\documentclass[pstricks]{standalone}
\usepackage{pst-plot}
\def\m{1/((x-1)^2)^(1/3)}

\begin{document}    
\begin{pspicture}(-1,-1)(3.5,3.5)
 \psaxes[labelFontSize=\scriptstyle]{->}(0,0)(3,3)[$x$,-90][$y$,0]
 \psplot[ algebraic,
         % plotstyle=curve,
            plotpoints=1000, <<-- notice
            yMaxValue=3,
            linewidth=1.5pt,
            linecolor=red]{0}{3}{\m}
\psline[linestyle=dashed](1,0)(1,3)
\end{pspicture} 
\end{document}

The second code:

\documentclass[pstricks]{standalone}
\usepackage{pst-plot}
\def\m{1/((x-1)^2)^(1/3)}

\begin{document}    
\begin{pspicture}(-1,-1)(3.5,3.5)
\psaxes[labelFontSize=\scriptstyle]{->}(0,0)(3,3)[$x$,-90][$y$,0]
\psplot[ algebraic,
         plotstyle=curve,
         %plotpoints=1000,
          yMaxValue=3,
         linewidth=1.5pt,
         linecolor=red]{0}{3}{\m}
\psline[linestyle=dashed](1,0)(1,3)
\end{pspicture} 
\end{document}

The result of compiling is frustrating.

1. What is the difference between them?
2. Can you fix the picture to make it become equal?

Answer 1

The best way to explain is showing the animations, right?

With curve plotstyle

It needs at least 3 points. When N=2 there is no graph.

\documentclass[12pt,pstricks]{standalone}
\usepackage{pst-plot}
\def\m{1/((x-1)^2)^(1/3)}
\def\xl{3 -1.5 exp neg 1 add}
\def\xr{3 -1.5 exp 1 add}
\begin{document}
\multido{\i=2+1}{20}{%
\begin{pspicture}[algebraic,showpoints,plotstyle=curve](-1,-1)(3.5,4)
    \psaxes[labelFontSize=\scriptstyle]{->}(0,0)(3,3.5)[$x$,-90][$y$,0]
    \psplot[linecolor=red,plotpoints=\i]{0}{\xl}{\m}
    \psplot[linecolor=red,plotpoints=\i]{\xr}{3}{\m}
    \psline[linestyle=dashed](1,0)(1,3)
    \rput[t](2,3){$N=\i$}
\end{pspicture}} 
\end{document}

With line plotstyle

It needs at least 2 points.

\documentclass[12pt,pstricks]{standalone}
\usepackage{pst-plot}
\def\m{1/((x-1)^2)^(1/3)}
\def\xl{3 -1.5 exp neg 1 add}
\def\xr{3 -1.5 exp 1 add}
\begin{document}
\multido{\i=2+1}{20}{%
\begin{pspicture}[algebraic,showpoints,plotstyle=line](-1,-1)(3.5,4)
    \psaxes[labelFontSize=\scriptstyle]{->}(0,0)(3,3.5)[$x$,-90][$y$,0]
    \psplot[linecolor=red,plotpoints=\i]{0}{\xl}{\m}
    \psplot[linecolor=red,plotpoints=\i]{\xr}{3}{\m}
    \psline[linestyle=dashed](1,0)(1,3)
    \rput[t](2,3){$N=\i$}
\end{pspicture}} 
\end{document}

Final output

\documentclass[12pt,pstricks]{standalone}
\usepackage{pst-plot}
\def\m{1/((x-1)^2)^(1/3)}
\def\xl{3 -1.5 exp neg 1 add}
\def\xr{3 -1.5 exp 1 add}
\begin{document}
\begin{pspicture}[algebraic](-.5,-.6)(4,4)
    \psaxes[labelFontSize=\scriptstyle]{->}(0,0)(3.5,3.5)[$x$,0][$y$,90]
    \psplot[linecolor=red]{0}{\xl}{\m}
    \psplot[linecolor=red]{\xr}{3}{\m}
    \psline[linestyle=dashed](1,0)(1,3)
\end{pspicture}
\end{document}

My best practices

• Split the graph into two or more invocation of \psplot when there are discontinuities in a single plot.
• Increasing plotpoints blindly will waste more storage because the size of PDF (or SVG) increases as well.
• You don't need to change plotstyle most of the time.

Explanation

• \def\xl{3 -1.5 exp neg 1 add} is the value of xl<1 such that f(xl)=3.

• \def\xr{3 -1.5 exp 1 add} is the value of xr>1 such that f(xr)=3.

• plotstyle represents the type of curves is used to connect points.

• plotpoints represents the number of points used to draw the curve.

Answer 2

\documentclass[pstricks]{standalone}
\usepackage{pst-plot}
\begin{document}    

\begin{pspicture}(-1,-1)(3.5,3.5)
\psaxes[labelFontSize=\scriptstyle]{->}(0,0)(3,3)[$x$,-90][$y$,0]
\psclip{\psframe[linestyle=none](3,3)}
  \psplot[ algebraic,linewidth=1.5pt,linecolor=red]{0}{3}{1/((x-1)^2)^(1/3)}
\endpsclip
\psline[linestyle=dashed](1,0)(1,3)
\end{pspicture} 

\end{document}