A circular version of \underbrace or \widehat

Question

I'm new to LaTeX and am searching for a function that acts like \underbrace or \widehat but is circular (I don't want the pointed middle part). Does such a function exist? I tried using Detexify but I didn't find what I was looking for. I would be using it for text.

Answer 1

Perhaps the macro \wideparen from yhmath package could be a solution, but it is not really circular and it does not scale very well with the fontsize, in my point of view.

\documentclass{article}


\usepackage{yhmath}


\begin{document}
\[ \wideparen{E = mc^2}\]
\end{document}

Answer 2

Here is a simple solution that uses pst-node. I define an \overarc and an underarc commands, both with an optional argument, the angle between the arc and the chord determined by the extremities of the arc.

Unfortunately, for some reason, it doesn't work with auto-pst-pdf, so that you have to compile either with the chain latex -> dvips -> pstopdf or with xelatex.

\documentclass[12pt]{article}

\usepackage{ifxetex, ifluatex}
\ifxetex\else \ifluatex\else\usepackage[utf8]{inputenc}\fi\fi
\usepackage{amsmath, amssymb, amsfonts}
\usepackage{fourier}
\usepackage{pst-node}
\usepackage{auto-pst-pdf}

\newcommand\overarc[2][30]{\psDefBoxNodes{F}{\mkern-2mu#2\mkern-2mu}\pcarc[offset=2pt, arcangle=#1]{c-c}(F:tl)(F:tr) }
\newcommand\underarc[2][30]{\psDefBoxNodes{F}{\mkern-2mu#2\mkern-2mu}\pcarc[offset=2pt, arcangle=#1]{c-c}(F:br)(F:bl) }

\begin{document}

\[ \overarc{E = mc²} \qquad \underarc{E = mc²} \]%

\end{document}