# How can I write this bar in LaTeX?

I want to write this in LaTeX:

However, $$\overline{S_n}^2 = \frac{1}{n}\sum\limits_{i=1}^{n}(X_i - M_n)^2$$ gives me:

whereas $$\bar{S_n}^2 = \frac{1}{n}\sum\limits_{i=1}^{n}(X_i - M_n)^2$$ gives me:

How can I fix this?

• Put the subscript outside the \overline: \overline{S}_n^2. BTW, use $...$ instead of $$...$$. However, my personal preference would be \bar{S}_n^2 (definitely not \bar{S_n}^2 with the subscript inside). – Ruixi Zhang Jan 22 at 13:06
• – Zarko Jan 22 at 13:47

Hope the below code may helps you...

\documentclass{book}

\begin{document}

$\overline{S}^{2}_{n}$

$\overline{S_{n}}^{2}$

$\bar{S_{n}}^{2}$
\end{document}


Output

Using Hendrik Vogt's code in his answer to this question, you have a \widebar command which takes into account the italic angle of the glyph:

    \documentclass{article}
\usepackage[utf8]{inputenc}%
\usepackage{mathtools, nccmath}

\makeatletter
\let\save@mathaccent\mathaccent
\newcommand*\if@single[3]{%
\setbox0\hbox{${\mathaccent"0362{#1}}^H$}%
\setbox2\hbox{${\mathaccent"0362{\kern0pt#1}}^H$}%
\ifdim\ht0=\ht2 #3\else #2\fi
}
%The bar will be moved to the right by a half of \macc@kerna, which is computed by amsmath:
\newcommand*\rel@kern[1]{\kern#1\dimexpr\macc@kerna}
%If there's a superscript following the bar, then no negative kern may follow the bar;
%an additional {} makes sure that the superscript is high enough in this case:
\DeclareRobustCommand\widebar[1]{\@ifnextchar^{{\wide@bar{#1}{0}}}{\wide@bar{#1}{1}}}
%Use a separate algorithm for single symbols:
\newcommand*\wide@bar[2]{\if@single{#1}{\wide@bar@{#1}{#2}{1}}{\wide@bar@{#1}{#2}{2}}}
\newcommand*\wide@bar@[3]{%
\begingroup
\def\mathaccent##1##2{%
%Enable nesting of accents:
\let\mathaccent\save@mathaccent
%If there's more than a single symbol, use the first character instead (see below):
\if#32 \let\macc@nucleus\first@char \fi
%Determine the italic correction:
\setbox\z@\hbox{$\macc@style{\macc@nucleus}_{}$}%
\setbox\tw@\hbox{$\macc@style{\macc@nucleus}{}_{}$}%
\dimen@\wd\tw@
%Now \dimen@ is the italic correction of the symbol.
\divide\dimen@ 3
\@tempdima\wd\tw@
%Now \@tempdima is the width of the symbol.
\divide\@tempdima 10
%Now \dimen@ = (italic correction / 3) - (Breite / 10)
\ifdim\dimen@>\z@ \dimen@0pt\fi
%The bar will be shortened in the case \dimen@<0 !
\rel@kern{0.6}\kern-\dimen@
\if#31
\overline{\rel@kern{-0.6}\kern\dimen@\macc@nucleus\rel@kern{0.4}\kern\dimen@}%
%Place the combined final kern (-\dimen@) if it is >0 or if a superscript follows:
\let\final@kern#2%
\ifdim\dimen@<\z@ \let\final@kern1\fi
\if\final@kern1 \kern-\dimen@\fi
\else
\overline{\rel@kern{-0.6}\kern\dimen@#1}%
\fi
}%
\macc@depth\@ne
\let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar
\mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax}%
\macc@set@skewchar\relax
\let\mathaccentV\macc@nested@a
%The following initialises \macc@kerna and calls \mathaccent:
\if#31
\macc@nested@a\relax111{#1}%
\else
%If the argument consists of more than one symbol, and if the first token is
%a letter, use that letter for the computations:
\def\gobble@till@marker##1\endmarker{}%
\futurelet\first@char\gobble@till@marker#1\endmarker
\ifcat\noexpand\first@char A\else
\def\first@char{}%
\fi
\macc@nested@a\relax111{\first@char}%
\fi
\endgroup
}
\makeatother

\begin{document}

$\widebar{S}_n^2 = \mfrac{1}{n}\sum_{i=1}^{n}(X_{i} - M_{n})^{2}$%

\end{document}


I would say

  \documentclass[]{article}

\begin{document}

$$\overline{S}_n^2 = \frac{1}{n}\sum\limits_{i=1}^{n}(X_i - M_n)^2$$

\end{document}