[pstricks] Dumb question -- asked before?

[Thomas Söll]

Hallo Hubert,

do you mean this? then you have to look for the negative roots.

\begin{psgraph}{->}(0,0)(-2,-2)(2,2){0.8\linewidth}{0.8\linewidth}
\parametricplot[linewidth=1.15pt,arrows=<-,plotpoints=200]{-8}{0}{t neg 1 3 div exp neg 1 t sub 1 3 div exp}
\parametricplot[linewidth=1.15pt,plotpoints=200]{0}{1}{t 1 3 div exp 1 t sub 1 3 div exp}
\parametricplot[linewidth=1.15pt,arrows=->,plotpoints=200]{1}{8}{t 1 3 div exp 1 t sub neg 1 3 div exp neg}
\end{psgraph}

\begin{psgraph}{->}(0,0)(-2,-2)(2,2){0.8\linewidth}{0.8\linewidth}
\psplot[linewidth=1.15pt,arrows=<-,plotpoints=200]{-2}{0}{x neg 1 3 div exp neg}
\psplot[linewidth=1.15pt,arrows=->,plotpoints=200]{0}{2}{x 1 3 div exp}
\end{psgraph}


Thomas


From: Hubert Lam 
Sent: Saturday, May 3, 2014 2:34 PM
To: pstricks at tug.org 
Subject: Re: [pstricks] Dumb question -- asked before?

Minimal example for x^\frac{1}{3} curve:

 

        \begin{psgraph}{->}(0,0)(-2,-2)(2,2){0.8\linewidth}{0.8\linewidth}

       \psplot[linewidth=1.15pt,arrows=<->]{-2}{2}{x 1 3 div exp}

        \end{psgraph}

 

 

Minimal example for parameterised curve:

 

\def\FuncA{t 1 3 div exp 1 t sub 1 3 div exp}

        \begin{psgraph}{->}(0,0)(-2,-2)(2,2){0.8\linewidth}{0.8\linewidth}

        \parametricplot[linewidth=1.15pt,arrows=<->]{-8}{-0.1}{\FuncA}

        \end{psgraph}

 

 

(by the way, it’s GhostScript that reports the error!)

 

From: Hubert Lam 
Sent: Saturday, May 3, 2014 10:33 PM
To: pstricks at tug.org
Subject: Dumb question -- asked before?

 

Hi all

 

I’m sure I’ve asked this before but I can’t seem to find it again in the archives.

 

I remember some years ago I asked about \psplot doing some cube root function, and it would fail at $x = 0$.

 

What was the workaround for this?

 

 

Reason:  I’m trying to sketch $x^3 + y^3 = 1$ but without \psplotimp which gives of course, a much poorer image than if I were to parameterise it to

 

$x = t^{\frac{1}{3}}$ and $y = (1- t)^{\frac{1}{3}}$.

 

Many thanks

 

Hubert



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