PGF Math Function to compute cube root

Question

When I found that a simple x^(1.0/3.0) does not yield a graph in PGFplots for negative values of x, I attempted to define my own function for CubeRoot using pgfmathdeclarefunction as below. But, am not able to get them to work.

These are based on this example for a Gaussian distribution, and another one here.

This compiles as is, but has a problem if I set \DomainMin to a negative value. The CubeRootB function does not even compile, and I do not see what the problem with it is.

\documentclass{article}
\usepackage{tikz}

\newcommand*{\DomainMin}{0.1}
%\newcommand*{\DomainMin}{-2.0} % CubeRootA gives and error if DomainMin < 0

\tikzstyle{MyPlotStyle}=[domain=\DomainMin:2,samples=100,smooth]

\pgfmathdeclarefunction{gauss}{3}{% From links mentioned in the question
    \pgfmathparse{1/(#3*sqrt(2*pi))*exp(-((#1-#2)^2)/(2*#3^2))}%
}

\pgfmathdeclarefunction{CubeRootA}{1}{%
    \pgfmathparse{%
        ifthenelse(equal(#1,abs(#1)),%
            (#1)^(1.0/3.0),%            #1 >= 0
            -1.0*((abs(#1))^(1.0/3.0))% #1 is negative
        )%
    }%
}

\pgfmathdeclarefunction{CubeRootB}{1}{%
    \pgfmathparse{%
        \pgfmathifthenelse
            {\pgfmathequal{#1}{\pgfmathabs{#1}}}%
            {exp(ln(#1)/3.0)}%            #1 >= 0
            {-1.0*(exp(ln(abs(#1))/3.0))}% #1 is negative
    }%

    }

% Modified version of CubeRootB per Caramdir's suggestion
\pgfmathdeclarefunction{CubeRootC}{1}{%
        \pgfmathparse{%
            ifthenelse(
                equal(#1,abs(#1)),%
                exp(ln(#1)/3.0),%            #1 >= 0
                -1.0*(exp(ln(abs(#1))/3.0))% #1 is negative
            )
        }%
    }

    \begin{document}
\begin{tikzpicture}
\draw [help lines] (-2,-2) grid [step=0.5] (2,2);
\draw plot [MyPlotStyle] (\x,{gauss(\x,0,0.5)});
%
\draw plot [MyPlotStyle] (\x,{    CubeRootA(\x)});% Error if DomainMin < 0
%\draw plot [MyPlotStyle] (\x,{1.0+CubeRootB(\x)});% Does not compile
\draw plot [MyPlotStyle] (\x,{    CubeRootC(\x)});% Error if DomainMin < 0
\end{tikzpicture}
\end{document}

Answer 1

The problem with CubeRootC is the way pgf/tikz deals with ifthenelse. The error message I get shows that when dealing with an ifthenelse(test,A,B), both expressions A and B are evaluated, then only one is kept. Since both are evaluated, you get an error message. The way my example is set, this problem is avoided. If you want to keep your code, just replace #1 by abs(#1) in your true clause.

Here is my version of your code. The use of the exponential and logarithm functions is the best approach. The powfunction does not behave has described in the pgf/tikz manual. I shifted the plots so they don't overlap. Note also the declare function method. I find it is easier to read; the problem is that it is local to a tikzpicture. The code is

\documentclass{minimal}
\usepackage{tikz}

\pgfmathsetmacro{\inf}{-2}
\pgfmathsetmacro{\sup}{2}

\tikzstyle{MyPlotStyle}=[domain=\inf:\sup,samples=100,smooth]

\pgfmathdeclarefunction{CubeRootA}{1}{%
    \pgfmathparse{ifthenelse(#1<0,-1,1)*exp((ln(abs(#1)))/3)}
}

\pgfmathdeclarefunction{CubeRootB}{1}{%
    %\pgfmathparse{ifthenelse(#1<0,-1,1)*((abs(#1))^(1.0/3.0))}
    \pgfmathparse{ifthenelse(#1<0,-1,1)*pow(abs(#1),1/3)}
}

\begin{document}
\begin{tikzpicture}
[declare function={ CubeRootC(\t)=ifthenelse(\t<0,-1,1)*exp((ln(abs(\t)))/3);}]

\draw [help lines] (-2,-2) grid [step=0.5] (2,2);

\draw[red] plot [MyPlotStyle] (\x,{CubeRootA(\x)});
\draw[blue,shift={(0,0.2)}] plot [MyPlotStyle] (\x,{CubeRootB(\x)});
\draw[green,shift={(0,-0.2)}] plot [MyPlotStyle] (\x,{CubeRootC(\x)});

\end{tikzpicture}
\end{document}

The output is

Answer 2

run with lualatex

\documentclass{article}
\def\cubicRoot#1{%
  \directlua{%
    tex.print(#1^(1/3))}}

\begin{document}
\cubicRoot{3}\par
\cubicRoot{-3}

\cubicRoot{3e-30}\par
\cubicRoot{-3e-30}

\cubicRoot{3e30}\par
\cubicRoot{-3e30}
\end{document}

and the code for the complete plot:

\documentclass{minimal}
\usepackage{tikz}
\def\cubicRoot#1{%
  \directlua{%
    tex.print(#1^(1/3))}}
\tikzstyle{MyPlotStyle}=[domain=-2:2,samples=100,smooth]

\begin{document}
\begin{tikzpicture}
\draw [help lines] (-2,-2) grid [step=0.5] (2,2);
\draw[red] plot [MyPlotStyle] (\x,\cubicRoot{\x});
\draw[blue,shift={(0,0.2)}] plot [MyPlotStyle] (\x,\cubicRoot{\x});
\draw[green,shift={(0,-0.2)}] plot [MyPlotStyle] (\x,\cubicRoot{\x});
\end{tikzpicture}
\end{document}

Answer 3

Here's a workaround using the fact that \sqrt[3]{x} is the inverse function to x^3: A little dirty but works like a charm. Note the domain is actually the range because I've switched the x and y values.

\begin{tikzpicture}[xscale=.2,yscale=.5]
\draw[xstep=1, ystep=1, gray, very thin, dotted] (-30,-4) grid (30,4);
\draw[<->] (0,-4) -- (0,4) node[above]{$y$};
\draw[->] (-30,0) -- (30,0) node[right]{$x$};
\foreach \x in {-25,5,25} \draw(\x,1pt)--(\x,-1pt) node[below=2pt,fill=white]{\begin{tiny}$\x$\end{tiny}};
\foreach \y in {2} \draw(1pt,\y)--(-1pt,\y) node[left=2pt,fill=white]{\begin{tiny}$\y$\end{tiny}};
\draw[domain=-3.1:3.1,<->] plot ({\x^(3)},\x) node[right]{$cubert(x) = \sqrt[3]{x}$};
\draw[fill=black] (0,0) circle (2pt) node[below=10pt,right=2pt]{(0,0)};
\draw[fill=black] (1,1) circle (2pt) node[left]{(1,1)};
\draw[fill=black] (8,2) circle (2pt) node[below]{(8,2)};
\draw[fill=black] (27,3) circle (2pt) node[below]{(27,3)};
\draw[fill=black] (-1,-1) circle (2pt) node[left]{(-1,-1)};
\draw[fill=black] (-8,-2) circle (2pt) node[below]{(-8,-2)};
\draw[fill=black] (-27,-3) circle (2pt) node[below]{(-27,-3)};
\end{tikzpicture}

Answer 4

You are using \pgfmathifthenelse inside \pgfmathparse, which the math parser does not like. If you use \pgfmathparse, then you should use ifthenelse like in CubeRootA (and the same for \pgfmathequal and \pgfmathabs).

Answer 5

Here's some simple code for this!

\begin{tikzpicture}

\begin{axis}[axis lines=middle, axis line style={<->}, xmin=-10, xmax=10,ymin=-10, ymax=10, xtick = {-8,-6,-4,-2,0,2,4,6,8}, ytick = {-8,-6,-4,-2,0,2,4,6,8}, samples=200]

\addplot[domain=0:10, thick,->] {x^(1/3)};

\addplot[domain=-10:0, thick,<-] {-(abs(x)^(1/3))};

\end{axis}

\end{tikzpicture}