5

I'm looking for the best way to create the symbol very similar to \risingdotseq, just with circles instead of dots. Unfortunately, it looks like it's not in the standard set of symbols. Any help finding/creating it would be greatly appreciated.

equal sign with rising circles

  • Welcome to TeX.SE. There is a symbol \fallingdotseq but with the dots instead of the circles. Using a mirror (command \reflectbox) you can obtained your symbol but with the dots. – Sebastiano May 30 at 15:41
  • Yes, but I need circles, not dots. – wutek May 30 at 15:46
  • I think that this symbol not exist and it is important to built it. – Sebastiano May 30 at 15:47
  • What is the intended meaning? Is it to be used in contrast to, e.g., \fallingdotseq? Have you seen it in any other document? – barbara beeton May 30 at 23:39
  • @barbarabeeton: it's the symbol we use for Laplace transform. – wutek May 31 at 7:46
7

Here's a not-so-simple™ solution:

\documentclass{article}
\usepackage{amsmath,pict2e}

\makeatletter
\newcommand{\raisingcircleseq}{\mathrel{\mathpalette\raising@circles@eq\relax}}
\newcommand{\raising@circles@eq}[2]{%
  \vphantom{#1+}%
  \vbox{
    \settowidth\unitlength{$#1\mspace{2mu}$}%
    \offinterlineskip\m@th
    \ialign{##\cr
      \hfil\small@circle{#1}$#1\mspace{1.5mu}$\cr\noalign{\vskip0.5\unitlength}
      $#1=$\cr\noalign{\post@vskip{+}{#1}}
      $#1\mspace{1.5mu}$\small@circle{#1}\hfill\cr\noalign{\post@vskip{-}{#1}}
    }%
  }%
}
\newcommand{\small@circle}[1]{%
  \smash{%
    \begin{picture}(1,1)
    \small@linethickness{#1}
    \put(0.5,0.5){\circle{1}}
    \end{picture}%
  }%
}
\newcommand{\small@linethickness}[1]{%
  \linethickness{%
      \ifx#1\displaystyle 0.8\fontdimen8\textfont3\else
      \ifx#1\textstyle 0.8\fontdimen8\textfont3\else
      \ifx#1\scriptstyle0.8\fontdimen8\scriptfont3\else
      1\fontdimen8\scriptscriptfont3\fi\fi\fi
  }%
}
\newcommand{\post@vskip}[2]{%
  \expandafter\vskip\expanded{%
    #1\ifx#2\scriptscriptstyle0.9\else\ifx#2\scriptstyle0.6\else0.3\fi\fi\unitlength
  }%
}
\makeatother

\begin{document}

$a=\raisingcircleseq b$

$a\raisingcircleseq b$

${=\raisingcircleseq}{\scriptstyle=\raisingcircleseq}{\scriptscriptstyle=\raisingcircleseq}$

\end{document}

enter image description here

| improve this answer | |
4

Here's an option.

\documentclass{article}
\usepackage{amsmath, amssymb}
\usepackage{graphicx}
\newlength{\circheight}
\settoheight{\circheight}{\(=\)}
\addtolength{\circheight}{.5pt}
\newcommand*{\smallcirc}{\scalebox{.51}{\(\scriptscriptstyle\boldsymbol{\circ}\)}}
\newcommand*{\risingcircleseq}{%
    \mathrel{%
        \makebox[0pt][l]{\raisebox{-.5pt}{\smallcirc}}%
        \mbox{=}%
        \makebox[0pt][r]{\raisebox{\circheight}[0pt]{\smallcirc}}%
    }%
}
\begin{document}
\(A \risingcircleseq B\)
\end{document}
| improve this answer | |
3

Works in all math styles. Uses stackengine to build the macro and scalerel for automatic handling of math style.

\documentclass{article}
\usepackage{stackengine,graphicx,scalerel}
\newcommand\eqdots{\mathrel{\ensurestackMath{\ThisStyle{%
  \stackengine{-.5\LMpt}{\stackengine{.5\LMpt}{\SavedStyle=}%
  {\SavedStyle\scaleobj{.33}{\circ\,}}{O}{r}{F}{T}{S}}%
  {\SavedStyle\scaleobj{.33}{\,\circ}}{U}{l}{F}{T}{S}}}}}
\begin{document} 
$x\eqdots y$\par
$\scriptstyle x\eqdots y$\par
$\scriptscriptstyle x\eqdots y$\par
\end{document}

enter image description here

| improve this answer | |
2

A solution (in display style) with \makebox and  accents:

\documentclass{article}
\usepackage{makebox}
\usepackage{mathtools}
\usepackage{accents}
\usepackage{calc}

\newcommand{\risingcirceq}{\mathrel{%
\underaccent{\raisebox{-0.5ex}[0pt]{\makebox[\widthof{$=$}][l]{\scalebox{0.6}{$\circ$}}}}{%
\accentset{\raisebox{-0.15ex}[0pt][0pt]{\makebox[\widthof{$=$}][r]{\scalebox{0.6}{$\circ$}}}}{=}}
}}%

\begin{document}

    \[ a\risingcirceq b\]%

\end{document} 

enter image description here

| improve this answer | |

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