# How do I display pi in LaTeX like Don?

Way back in 1979, Donald Knuth, tried to give the feeling that π is irrational using the idea of infinite sequence of smaller and smaller font sizes like what I have tried to do here using HTML and CSS. How do I do the same using LaTeX?

Now to the very difficult part, how do I some how create the feeling using typesetting that π is transcendental (not a solution of any finite polynomial having integer coefficients).

.pi sub { vertical-align: baseline; font-size: 96%; }

π = 3.14159265358979323846264338327950288419716939937510
• Thanks for all those interesting answers. I am humbled by all your answers. – Sukii Oct 23 '14 at 17:23
• 856142 repeats, so they usually indicate this repeat by $1/7=0.142\overline{856142}$. I think any rational number has these repeats. – Sukii Oct 24 '14 at 13:01
• You could think of transcendence as lack of rootedness. – Sukii Oct 24 '14 at 14:00
• Blabbering along the same vein, since polynomials with rational coefficients are countable, these rootless transcendental numbers are what fills the real line. Looks more like our modern society? – Sukii Oct 25 '14 at 7:45
• @Sukii: Or More directly as $1/7=0.\overline{142856}$. And yes, any rational number is of that form (if you count a repeating 0 the same way, although it would be highly unusual to set $1/4=0.25\overline{0}$). The same way, the continued fraction of any algebraic of degree 2 is eventually repeating. – Christopher Creutzig Oct 25 '14 at 13:07

EDITED to allow repeat invocations without having to reset parameters, using the syntax

\diminish[scale-down-ratio]{string}

The value of \defaultstartht sets the strut height of the first letter (default is set to \baselineskip), and the scale-down ratio defaults to 0.98, but may be supplied as an optional parameter.

Beware that arithmetic underflows are possible, if things get teeny enough.

See addendum for pseudo 3-D effect:

\documentclass{article}
\usepackage{scalerel}
\newlength\curht
\def\defaultdimfrac{.98}
\def\defaultstartht{\baselineskip}
\newcommand\diminish[2][\defaultdimfrac]{%
\curht=\defaultstartht\relax
\def\dimfrac{#1}%
\diminishhelpA{#2}%
}
\newcommand\diminishhelpA[1]{%
\expandafter\diminishhelpB#1\relax%
}
\def\diminishhelpB#1#2\relax{%
\scaleto{\strut#1}{\curht}%
\curht=\dimfrac\curht\relax%
\ifx\relax#2\relax\else\diminishhelpA{#2}\fi%
}
\begin{document}
\def\defaultstartht{14pt}
\def\pinum{3.14159265358979323846264338327950288419716939937510}
\diminish{\pinum}\par
\diminish[0.96]{\pinum}\par
\diminish[0.94]{\pinum}\par
\def\defaultstartht{38pt}
\diminish[0.92]{\pinum}
\end{document}


Based on Yori's desire to see something more "3-D", I have EDITED to incorporate Bruno's \slantbox at Shear transform a "box". Combining that with a \raisebox allows the effect which some may find more 3-D. The additional user parameters to control it are \slantvalue which is Bruno's model parameter and \zshft, which is an added vertical shift per letter.

\documentclass{article}
\usepackage{scalerel}
\newlength\curht
\newlength\zshft
\newcounter{letcount}
\def\defaultdimfrac{.98}
\def\slantvalue{0}
\zshft=0pt\relax
\def\defaultstartht{\baselineskip}
\newcommand\diminish[2][\defaultdimfrac]{%
\curht=\defaultstartht\relax
\def\dimfrac{#1}%
\setcounter{letcount}{0}
\diminishhelpA{#2}%
}
\newcommand\diminishhelpA[1]{%
\expandafter\diminishhelpB#1\relax%
}
\def\diminishhelpB#1#2\relax{%
\raisebox{\value{letcount}\zshft}{\scaleto{\strut\slantbox{#1}}{\curht}}%
\stepcounter{letcount}%
\curht=\dimfrac\curht\relax%
\ifx\relax#2\relax\else\diminishhelpA{#2}\fi%
}
\newsavebox{\foobox}
\newcommand{\slantbox}[2][\slantvalue]{\mbox{%
\sbox{\foobox}{#2}%
\hskip\wd\foobox
\pdfsave
\pdfsetmatrix{1 0 #1 1}%
\llap{\usebox{\foobox}}%
\pdfrestore
}}
\begin{document}
\def\pinum{3.14159265358979323846264338327950288419716939937510}
\def\defaultstartht{14pt}
\diminish{\pinum}\par
\def\slantvalue{.15}
\zshft=.1pt\relax
\diminish[0.96]{\pinum}\par
\diminish[0.94]{\pinum}\par
\def\slantvalue{.35}
\zshft=.4pt\relax
\def\defaultstartht{38pt}
\diminish[0.92]{\pinum}
\end{document}


While the curvy tail gives a certain stylistic fading, it might be preferable to follow the rules of perspective and have the numbers vanish along a linear path to the vanishing point. What this means is that rather than an additional \zshft added for each letter, the \zshft should also get smaller with each letter, so that the total shift (dZ) on the nth letter (after the 1st) should be

dZ = dz + k dz + k^2 dz + ... + k^(n-1) dz

where dz is the specified \zshft, and k is the scale-down parameter given by \dimfrac. Simple manipulation reveals that the shift for the nth letter (after the 1st) is:

dZ = dz (1 - k^n)/(1-k)

This can be calculated in LaTeX, but requires the much more computationally intensive fp package. So, at the expense of compilation time, we can achieve the following:

\documentclass{article}
\usepackage{scalerel,fp}
\newlength\curht
\newlength\zshft
\newcounter{letcount}
\def\defaultdimfrac{.98}
\def\slantvalue{0}
\zshft=0pt\relax
\def\defaultstartht{\baselineskip}
\newcommand\diminish[2][\defaultdimfrac]{%
\curht=\defaultstartht\relax
\def\dimfrac{#1}%
\setcounter{letcount}{0}
\diminishhelpA{#2}%
}
\newcommand\diminishhelpA[1]{%
\expandafter\diminishhelpB#1\relax%
}
\def\diminishhelpB#1#2\relax{%
\FPpow\localshift{\dimfrac}{\theletcount}\unskip%
\FPsub\localshift{1}{\localshift}%
\FPsub\localdenom{1}{\dimfrac}%
\FPdiv\localshift{\localshift}{\localdenom}%
\raisebox{\localshift\zshft}{\scaleto{\strut\slantbox{#1}}{\curht}}%
\stepcounter{letcount}%
\curht=\dimfrac\curht\relax%
\ifx\relax#2\relax\else\diminishhelpA{#2}\fi%
}
\newsavebox{\foobox}
\newcommand{\slantbox}[2][\slantvalue]{\mbox{%
\sbox{\foobox}{#2}%
\hskip\wd\foobox
\pdfsave
\pdfsetmatrix{1 0 #1 1}%
\llap{\usebox{\foobox}}%
\pdfrestore
}}
\begin{document}
\def\pinum{3.14159265358979323846264338327950288419716939937510}
\def\defaultstartht{14pt}
\diminish{\pinum}\par
\def\slantvalue{.15}
\zshft=.3pt\relax
\diminish[0.96]{\pinum}\par
\diminish[0.94]{\pinum}\par
\def\slantvalue{.35}
\zshft=1.7pt\relax
\def\defaultstartht{38pt}
\diminish[0.92]{\pinum}
\end{document}


Oops!! Spill in aisle 1! (coding left to the student)

• In some sense the "right" value for scale-down-ratio would be 0.1. – Nate Eldredge Oct 22 '14 at 21:57
• I wonder if there is a way to make this look more "3D", i.e., making it look like an infinite row of numbers vanishing at the horizon? – yori Oct 23 '14 at 1:25
• @Yori Does the addendum achieve what you desire? – Steven B. Segletes Oct 23 '14 at 10:43
• @StevenB.Segletes: Since we're working in base 10, the value of each successive decimal place is 1/10 the value of the one before it. So maybe one should make the heights diminish in the same proportion. (Or, as rob, suggests, the areas.) – Nate Eldredge Oct 24 '14 at 18:22
• @NateEldredge D'oh! So your comment contained both a joke and a greater mathematic truth! – Steven B. Segletes Oct 24 '14 at 19:38

Mark Wibrow enhanced the text decoration library of TikZ greatly and you can do all kinds of stuff based on formulas regarding the character number and total count of words and so on.

Here is a wacky take on it, (thus a simple monotone scaling along a path is very simple and is possible just by playing with the formula with scale instead of yscale)

\documentclass[tikz]{standalone}
\usetikzlibrary{decorations.text}
\tikzset{pi decoration/.style={
decoration={text effects along path,
text={3.14159265358979323846264338327950288419716939937510},
text align=center,
text effects/.cd,
character total=\n,character count=\i,
path from text,
characters={yscale=10*sin(deg(5*\i/\n))-\i/\n,inner sep=0,anchor=base}
}}}
\begin{document}
\begin{tikzpicture}
\path[pi decoration,decorate] (0,0);
\end{tikzpicture}
\end{document}


• Because, because, because .... because of the wonderful things TikZ does! (PS Thanks for the edit on my recent post.) – Loop Space Oct 22 '14 at 19:06
• @LoopSpace The pleasure is mine. – percusse Oct 22 '14 at 21:06
• Seeing this, I find myself wanting to ask the question "How do I display phi in LaTeX like Fibonacci?". I'd like to see a golden spiral with the digits of phi scaled so that when it reaches the second square then it is exactly 1/phi the size it was when it set out. – Loop Space Oct 22 '14 at 21:16
• @LoopSpace, please do! – Joe Oct 22 '14 at 21:24
• My phi code is now on github: github.com/loopspace/golden – Loop Space Oct 25 '14 at 21:33

A recursive macro:

\documentclass[border=2]{standalone}
\usepackage{mathptmx} % a scalable font is needed

\makeatletter
\def\makesmaller#1#2{%
\dimen0=\dimexpr\dimen0-.08pt\relax
\ifx#2\relax
\expandafter\@gobble
\else
{\fontsize{\dimen0}{0}\selectfont#1}%
\expandafter\@firstofone
\fi
{\makesmaller{#2}}%
}
\def\decreasingpi{%
{\dimen0=16pt\fontsize{\dimen0}{0}\selectfont
3.\makesmaller1415926535897932384626433832795028841971693993751058209749445923078164062862089986280348253
\relax}
}
\makeatletter

\begin{document}
\decreasingpi
\end{document}


One might first count the number of digits and decide the minimum size, so as to adjust the decrease step. I leave this as an exercise.

Just another way...

\documentclass[varwidth,border=5]{standalone}
\usepackage[T1]{fontenc}
\usepackage{lmodern}
\newdimen\dx
\def\dm#1#2{\dx=1em\relax\def\rt{#1}{\Dm#2\dm}}
\def\Dm#1{\ifx#1\dm\else\fontsize\dx\dx\selectfont#1\dx=\rt\dx\expandafter\Dm\fi}
\begin{document}
\dm{0.95}{3.14159265358979323846264338327950288419716939937510}
\end{document}