# How to create the “typical arc or hat” that goes over the periodic numbers in LaTeX?

Hello! Anyone knows how to create the "typical arc or hat" that goes over the periodic numbers/repeating or recurring decimal in LaTeX? Thank you!!

• – egreg Sep 22 '18 at 16:49

EDIT: I was not aware of https://tex.stackexchange.com/a/451459/128553 when I made my first post. Here are other solutions.

I still think that the overline notation is the best for following reasons:

• it is more or less internationally known,
• it takes less horizontal space,
• some of the solutions are too bold to fit to the typeface and attract to much attention,
• if the number of digits is too long then the arc is not covering all digits, is ridiculously large, or is "too flat" to be easily distinguishable from a rotated bracket.

For some comparison: https://en.wikipedia.org/wiki/Repeating_decimal.

Variant A: Þe auld tikz: You can play around with the coordinates of wideArcAnchorA and wideArcAnchorB to get a shape of your taste.

\documentclass{article}

\usepackage{amsmath,tikz}
\usetikzlibrary{calc,topaths}

\newcommand\wideparen[1]{%
\tikz[baseline=(wideArcAnchor.base)]{
\node[inner sep=0] (wideArcAnchor) {$#1$};
\coordinate (wideArcAnchorA) at ($0.9*(wideArcAnchor.north west) + 0.1*(wideArcAnchor.north east)+(0.0em,0.75ex)$);
\coordinate (wideArcAnchorB) at ($0.1*(wideArcAnchor.north west) + 0.9*(wideArcAnchor.north east)+(0.0em,0.75ex)$);
%
\draw[line width=0.1ex,line cap=round]
($(wideArcAnchor.north west)+(0.0em,0.1ex)$)
.. controls (wideArcAnchorA) and (wideArcAnchorB) ..
($(wideArcAnchor.north east)+(0.0em,0.1ex)$)
;
}}
\newcommand\widearc[1]{%
\tikz[baseline=(wideArcAnchor.base)]{
\node[inner sep=0] (wideArcAnchor) {$#1$};
\coordinate (wideArcAnchorA) at ($(wideArcAnchor.north west) + (0.1em,0.0ex)$);
\coordinate (wideArcAnchorB) at ($(wideArcAnchor.north east) + (-0.1em,0.0ex)$);
%
\draw[line width=0.1ex,line cap=round,out=45,in=135] (wideArcAnchorA) to (wideArcAnchorB);
}}

\begin{document}
\centering{}Test foo bar.
\begin{align*}
\zeta
&=0.00\overline{1}\\
&=0.00\widehat{1}\\
&=0.00\wideparen{1}\\
&=0.00\widearc{1}\\
&\neq0.00\overline{12}\\
&=0.00\widehat{12}\\
&=0.00\wideparen{12}\\
&=0.00\widearc{12}\\
&\neq0.00\overline{1234567890}\\
&=0.00\widehat{1234567890}\\
&=0.00\wideparen{1234567890}\\
&=0.00\widearc{1234567890}.
\end{align*}
\end{document}


Variant B: Taken from https://ctan.org/pkg/comprehensive.

\documentclass{article}

\usepackage{amsmath}

\makeatletter
%Taken from https://ctan.org/pkg/comprehensive
\def\downparenthfill{$\m@th\braceld\leaders\vrule\hfill\bracerd$}
\def\wideparen#1{\mathop{\vbox{\ialign{##\crcr\noalign{\kern3\p@}
\downparenthfill\crcr\noalign{\kern3\p@\nointerlineskip}
$\hfil\displaystyle{#1}\hfil$\crcr}}}\limits}
\makeatother

\begin{document}
\centering{}Test foo bar.
\begin{align*}
\zeta
&=0.00\overline{12}\\
&=0.00\widehat{12}\\
&=0.00\wideparen{12}\\
&\neq0.00\overline{1234567890}\\
&=0.00\widehat{1234567890}\\
&=0.00\wideparen{1234567890}.\\
\end{align*}

\end{document}


Variant C:

\documentclass{article}

\usepackage{amsmath}
\usepackage{mathabx}\begin{document}
\centering{}Test foo bar.
\begin{align*}
\zeta
&=0.00\overline{12}\\
&=0.00\widehat{12}\\
&=0.00\wideparen{12}\\
&\neq0.00\overline{1234567890}\\
&=0.00\widehat{1234567890}\\
&=0.00\wideparen{1234567890}.\\
\end{align*}
\end{document}


Variant D: use Utopia which might be a restriction for your use:

\documentclass{article}

\usepackage{amsmath,fourier}
\begin{document}
\centering{}Test foo bar.
\begin{align*}
\zeta
&=0.00\overline{12}\\
&=0.00\widehat{12}\\
&=0.00\widearc{12}\\
&\neq0.00\overline{1234567890}\\
&=0.00\widehat{1234567890}\\
&=0.00\widearc{1234567890}.\\
\end{align*}
\end{document}


Variant E: The arcs package should provide a better solution but seems to be broken. The hack mentioned on http://tug.org/pipermail/xetex/2013-August/024674.html works on LuaLaTeX and XeLaTeX on my system.

If you want to stck to the default font, you hace the \wideparen from yhmath:

\documentclass{article}

\usepackage{amsmath, yhmath}

\begin{document}

$\frac{12}{35} = 0.3\wideparen{428\,571}$%

\end{document}