# How to do logarithmic shading with TikZ?

Using TikZ, the shade option is very convenient to produce color shadings, but it produces only linear color gradients. Is it possible to obtain other transitions, and especially a logarithmic one?

Thanks!

# Edit:

Here is an example for the linear case:

\documentclass[border=5mm]{standalone}
\usepackage{tikz}
\begin{document}

\begin{tikzpicture}
\shade[left color=black, right color=white] (0,0) rectangle (10,1);
\end{tikzpicture}

\end{document}


• You can with \pgfdeclarefunctionalshading. Check the manual or tex.stackexchange.com/questions/54193/… for examples. – percusse Apr 10 '15 at 9:54
• Thank you @percusse, I am pretty sure this is what I need, but the code I have to put into the function is puzzling to me! Any help would be appreciated. – Tobard Apr 10 '15 at 10:20
• Can you include a MWE and the details about the shading direction etc. ? Then we can look at it together. – percusse Apr 10 '15 at 10:30
• I have added a MWE. I would like something very simple : a horizontal band changing from black to white with a logarithmic progression. Thanks. – Tobard Apr 10 '15 at 11:42

Here are some examples.

• R=G=B=x
• R=G=B=x²
• R=G=B=√x
• R=G=B=log(1+x)
• R=G=B=log(1+9x)
• R=G=B= ... well ...

\documentclass{article}
\usepackage[a3paper]{geometry}
\usepackage{tikz}
\begin{document}

\begin{tikzpicture}
\fill(-11,0)|-(0,-2)|-(11,2)|-cycle;
\end{tikzpicture}

pop 50 div .5 sub % u
dup dup % u u u
}
\begin{tikzpicture}
\fill(-11,0)|-(0,-2)|-(11,2)|-cycle;
\end{tikzpicture}

pop 50 div .5 sub % u
dup mul % u²
dup dup % u² u² u²
}
\begin{tikzpicture}
\fill(-11,0)|-(0,-2)|-(11,2)|-cycle;
\end{tikzpicture}

pop 50 div .5 sub % u
sqrt % √u
dup dup % √u √u √u
}
\begin{tikzpicture}
\fill(-11,0)|-(0,-2)|-(11,2)|-cycle;
\end{tikzpicture}

pop 50 div .5 sub % u
dup dup % ㏒(1+u) ㏒(1+u) ㏒(1+u)
}
\begin{tikzpicture}
\fill(-11,0)|-(0,-2)|-(11,2)|-cycle;
\end{tikzpicture}

pop 50 div .5 sub % u
9 mul 1 add log % ㏒(1+9u)
dup dup % ㏒(1+9u) ㏒(1+9u) ㏒(1+9u)
}
\begin{tikzpicture}
\fill(-11,0)|-(0,-2)|-(11,2)|-cycle;
\end{tikzpicture}

50 div .5 sub exch % v U
50 div .5 sub 4 mul exch % 4u v
dup 1 exch sub % 4u v 1-v
2 index % 4u v (1-v) 4u
mul mul % u 4uv(1-v)
dup 1 exch sub 2 index mul mul
dup 1 exch sub 2 index mul mul
dup 1 exch sub 2 index mul mul
dup 1 exch sub 2 index mul mul
dup 1 exch sub 2 index mul mul
exch pop
}
\begin{tikzpicture}
\fill(-11,0)|-(0,-2)|-(11,2)|-cycle;
\end{tikzpicture}

\end{document}


### Remarks

• dup duplicates the topmost element.
• The dup dup at every very end is actually unnecessary.
• If you leave only a single number in the stack, PDF-renderer will treat it as the grayscale.
• If there are three, they are R, G, B respectively.
• If there are 0 or 2 or 4, 5, 6, ... I do not know.
• pop discards the topmost element.
• The pop at every very beginning throws away the y-coordinate, which is useless except the last case.
• Replacing pop by swap, you can leave the y-coordinate at the bottom alone. But then dup dup becomes necessary.
• Thank you! Could you just explain what pop mean, and why dup commands are needed? – Tobard Apr 15 '15 at 13:22