[Tex/LaTex] Big O notation – visual difference related to document configurations

amsmath

Please refer to the information present at the image attached.

The last image is the one I get when I type the code suggested in the same image, for which the author gets exactly the visual symbol I would like to get.

Important: I don´t want to mess up with previous document configurations, therefore I believe that what I need is to make a "local inibition" of the general document configuration, so that for this specific symbol I can get what I want.

I hope I could be understood. Not a native speaker, sorry for that.

Thanks a lot!
fskilnik.

Best Answer

Just revert \mathcal to using Computer Modern Symbols:

\documentclass{article}
\usepackage{fourier}

% add these two lines to your long preamble    
\DeclareMathAlphabet{\mathcal}{OMS}{cmsy}{m}{n}
\SetMathAlphabet{\mathcal}{bold}{OMS}{cmsy}{b}{n}

\begin{document}

$\mathcal{O}(n\log n)$

\end{document}

Actually I'd prefer to add also

\newcommand{\bigO}{\mathcal{O}}

so the formula can be written

$\bigO(n\log n)$

This has the advantage that you're free to redefine \bigO without having to change the document, except for one line.

If you just want that big O, but also want to keep the calligraphic letters from fourier, use

\documentclass{article}
\usepackage{amsmath}
\usepackage{fourier}

\DeclareRobustCommand{\bigO}{%
  \text{\usefont{OMS}{cmsy}{m}{n}O}%
}

\begin{document}

$\bigO(n\log n)$

\end{document}

Related Solutions

[Tex/LaTex] Hyperref jumps to the wrong equation if the equation has a \tag and cleveref is used

I have a solution. However this might not be the most elegant way.

You have to define your own numbering scheme for hyperref so that each number is unique:

\listfiles
\documentclass{book}

\usepackage{amsmath}
\usepackage{hyperref}
\usepackage{cleveref}

%%% define a new counter and use it for \theHequation which hyperref uses for the links
\newcounter{Hequation}
\renewcommand{\theHequation}{\thechapter.\arabic{Hequation}}
\makeatletter
\g@addto@macro\equation{\stepcounter{Hequation}}
\makeatother

\begin{document}
\begin{equation}\tag{first}\label{eq:first}
   \text{First equation}
\end{equation}

\begin{equation}\tag{second}\label{eq:second}
   \text{Second equation}
\end{equation}

\begin{equation}\label{eq:firstreal}
   \text{First equation without tag}
\end{equation}

\begin{equation}\tag{third}\label{eq:third}
   \text{Third equation}
\end{equation}

\begin{equation}\label{eq:secondreal}
   \text{Second equation without tag}
\end{equation}

\begin{itemize}
   \item A reference to~\eqref{eq:first}.
   \item A reference to~\eqref{eq:second}.
   \item A reference to~\eqref{eq:third}.
   \item A reference to~\eqref{eq:firstreal}.
   \item A reference to~\eqref{eq:secondreal}.
\end{itemize}

\newpage

For scrolling purposes.

\end{document}

As you can see \equation has to be redefined as well. Then equations with and without \tag can be mixed-up without any problems.

Best regards
Martin

[Tex/LaTex] Is eqnarray really obsolete

The output of eqnarray is wrong anyway and there's little to do about it. Consider also that it doesn't work with cross-references if hyperref or cleveref are involved.

No, there's no reason for using it.

The first example can be dealt with using array:

\documentclass{article}

\usepackage{amsmath,array}

\begin{document}

\begin{equation*}
\renewcommand{\arraystretch}{1.5}
\begin{array}{@{} r @{} >{{}} c <{{}} @{} l @{} }
2^n
 & =                   & \underbrace{2 \cdot 2 \cdot \ldots \cdot 2}_{n \text{ times}}\\
 & \overset{n>3}{=}    & 2 \cdot 2 \cdot 2 \cdot 2 \cdot \underbrace{2 \cdot 2 \cdot \ldots \cdot 2}_{n - 4 \text{ times}}\\
 & =                   & 16 \cdot \underbrace{2 \cdot 2 \cdot \ldots \cdot 2}_{n - 4 \text{ times}}\\
 & <                   & 24 \cdot \underbrace{2 \cdot 2 \cdot \ldots \cdot 2}_{n - 4 \text{ times}}\\
 & =                   & 1 \cdot 2 \cdot 3 \cdot 4 \cdot \underbrace{2 \cdot 2 \cdot \ldots \cdot 2}_{n - 4 \text{ times}}\\
 & \overset{n>3}{\leq} & 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot \ldots \cdot n\\
 & =                   & n!
\end{array}
\end{equation*}

A better proof, by induction, is
\begin{alignat*}{2}
2^4     &= 16 < 4! = 24  \\
2^{n+1} &= 2\cdot 2^n    &\quad&\text{for $n>3$} \\
        &< 2\cdot n!     &\quad&\text{induction hypothesis}\\
        &< (n+1)\cdot n! \\
        &= (n+1)!
\end{alignat*}
\end{document}

The second example is even easier: use a good alignment point:

\documentclass{article}

\usepackage{amsmath}

\begin{document}

\begin{align*}
\sigma =[ &x_1/t_1,x_2/t_2,\\
          & x_3/t_3,x_4/t_4]
\end{align*}

\end{document}

More lines can be accommodated:

\documentclass{article}

\usepackage{amsmath}

\begin{document}

\begin{align*}
\sigma = [ &x_1/t_1,x_2/t_2, \\
           & x_3/t_3,x_4/t_4] \\
\tau   = [ &y_1/t_1] \\
\psi   = [ &x_1/t_1,x_2/t_2,\\
           & x_3/t_3,x_4/t_4]
\end{align*}

\end{document}