9

The commath package use the TeX command \ifinner to select between inline math mode and display math mode. Because of this, it fail to work properly with some display math environments.

In the breqn package, we find:

The test \ifinner is unreliable for distinguishing whether we are in a displayed formula or an inline formula: (...) So we provide a more reliable test. But it might have been provided already by the amsmath package.

The more reliable test should be \if@display but only replace \ifinner with \if@display in commath.sty not work. Any suggestion?

4
  • 1
    There is no \if@display conditional. The commath package has many errors and I can't recommend using it.
    – egreg
    Oct 1, 2013 at 15:41
  • Did you recommend some replacement for commath? Oct 1, 2013 at 16:00
  • commath is just a collection of macros, some of them perhaps useful, but badly written in general. Define your own substitutes, which is not so difficult.
    – egreg
    Oct 1, 2013 at 16:05
  • Try the cool package instead.
    – jarmond
    Sep 15, 2015 at 6:22

1 Answer 1

25

Let's see how commath should define its macros.

\documentclass{article}
\usepackage[margin=1cm]{geometry}
\usepackage{amsmath}

%%% Here start the definitions
\newcommand{\dif}{\mathop{}\!\mathrm{d}}
\newcommand{\Dif}{\mathop{}\!\mathrm{D}}

\makeatletter
\newcommand{\spx}[1]{%
  \if\relax\detokenize{#1}\relax
    \expandafter\@gobble
  \else
    \expandafter\@firstofone
  \fi
  {^{#1}}%
}
\makeatother

\newcommand\pd[3][]{\frac{\partial\spx{#1}#2}{\partial#3\spx{#1}}}
\newcommand\tpd[3][]{\tfrac{\partial\spx{#1}#2}{\partial#3\spx{#1}}}
\newcommand\dpd[3][]{\dfrac{\partial\spx{#1}#2}{\partial#3\spx{#1}}}

\newcommand{\md}[6]{\frac{\partial\spx{#2}#1}{\partial#3\spx{#4}\partial#5\spx{#6}}}
\newcommand{\tmd}[6]{\tfrac{\partial\spx{#2}#1}{\partial#3\spx{#4}\partial#5\spx{#6}}}
\newcommand{\dmd}[6]{\dfrac{\partial\spx{#2}#1}{\partial#3\spx{#4}\partial#5\spx{#6}}}

\newcommand{\od}[3][]{\frac{\dif\spx{#1}#2}{\dif#3\spx{#1}}}
\newcommand{\tod}[3][]{\tfrac{\dif\spx{#1}#2}{\dif#3\spx{#1}}}
\newcommand{\dod}[3][]{\dfrac{\dif\spx{#1}#2}{\dif#3\spx{#1}}}

\newcommand{\genericdel}[4]{%
  \ifcase#3\relax
  \ifx#1.\else#1\fi#4\ifx#2.\else#2\fi\or
  \bigl#1#4\bigr#2\or
  \Bigl#1#4\Bigr#2\or
  \biggl#1#4\biggr#2\or
  \Biggl#1#4\Biggr#2\else
  \left#1#4\right#2\fi
}
\newcommand{\del}[2][-1]{\genericdel(){#1}{#2}}
\newcommand{\set}[2][-1]{\genericdel\{\}{#1}{#2}}
\let\cbr\set
\newcommand{\sbr}[2][-1]{\genericdel[]{#1}{#2}}
\let\intoo\del
\let\intcc\sbr
\newcommand{\intoc}[2][-1]{\genericdel(]{#1}{#2}}
\newcommand{\intco}[2][-1]{\genericdel[){#1}{#2}}
\newcommand{\eval}[2][-1]{\genericdel.|{#1}{#2}}
\newcommand{\envert}[2][-1]{\genericdel||{#1}{#2}}
\let\abs\envert
\newcommand{\sVert}[1][0]{%
  \ifcase#1\relax
  \rvert\or\bigr|\or\Bigr|\or\biggr|\or\Biggr
  \fi
}
\newcommand{\enVert}[2][-1]{\genericdel\|\|{#1}{#2}}
\let\norm\enVert
\newcommand{\fullfunction}[5]{%
  \begin{array}{@{}r@{}r@{}c@{}l@{}}
  #1 \colon & #2 & {}\longrightarrow{} & #3 \\
            & #4 & {}\longmapsto{}     & #5
  \end{array}
}
%%% end of the definitions

\linespread{2} % just for this test file

\begin{document}

$f(x)\dif x\quad f(x)\Dif x$

$\pd{f}{x}\displaystyle\pd[2]{f}{x}$

$\tpd{f}{x}\displaystyle\tpd[2]{f}{x}$

$\dpd{f}{x}\displaystyle\dpd[2]{f}{x}$

$\od{f}{x}\displaystyle\od[2]{f}{x}$

$\tod{f}{x}\displaystyle\tod[2]{f}{x}$

$\dod{f}{x}\displaystyle\dod[2]{f}{x}$

$\md{f}{5}{x}{2}{y}{3}\displaystyle\md{f}{5}{x}{2}{y}{3}$

$\tmd{f}{5}{x}{2}{y}{3}\displaystyle\tmd{f}{5}{x}{2}{y}{3}$

$\dmd{f}{5}{x}{2}{y}{3}\displaystyle\dmd{f}{5}{x}{2}{y}{3}$

$\del{\dfrac{1}{2}}\del[0]{x}\del[1]{x}\del[2]{x}\del[3]{x}\del[4]{x}$

$\sbr{\dfrac{1}{2}}\sbr[0]{x}\sbr[1]{x}\sbr[2]{x}\sbr[3]{x}\sbr[4]{x}$

$\set{\dfrac{1}{2}}\set[0]{x}\set[1]{x}\set[2]{x}\set[3]{x}\set[4]{x}$

$\intoo{a,b}\intoo[0]{a,b}\intoo[1]{a,b}\intoo[2]{a,b}\intoo[3]{a,b}\intoo[4]{a,b}$

$\intcc{a,b}\intcc[0]{a,b}\intcc[1]{a,b}\intcc[2]{a,b}\intcc[3]{a,b}\intcc[4]{a,b}$

$\intoc{a,b}\intoc[0]{a,b}\intoc[1]{a,b}\intoc[2]{a,b}\intoc[3]{a,b}\intoc[4]{a,b}$

$\intco{a,b}\intco[0]{a,b}\intco[1]{a,b}\intco[2]{a,b}\intco[3]{a,b}\intco[4]{a,b}$

$
\eval{f(x)}_a^b
\eval[0]{f(x)}_a^b
\eval[1]{f(x)}_a^b
\eval[2]{f(x)}_a^b
\eval[3]{f(x)}_a^b
\eval[4]{f(x)}_a^b
$

$\abs{x}\abs[0]{x}\abs[1]{x}\abs[2]{x}\abs[3]{x}\abs[4]{x}$

$\norm{x}\norm[0]{x}\norm[1]{x}\norm[2]{x}\norm[3]{x}\norm[4]{x}$

\linespread{1}\selectfont % return to normal

$\fullfunction{f}{A}{B}{x}{y}$
\end{document}

I have omitted the \...ref macros that are better managed with cleveref.

As you see, it's only a collection of dubiously useful macros.

Main errors in commath: the definitions of \dif and \Dif are plainly wrong. The usage of \ifinner is completely wrong; what the author intends to do by \ifinner with \tfrac and \dfrac is already done (better) by the standard \frac macro.

The "delimiters" macros are wrong in that they use by default \left and \right, which is disputable; I've left them as in the original, but defining them in terms of a generic macro.

enter image description here

7
  • Looking at the rendering of \pd{f}{x} or \od{f}{x}, there seems to be a bigger horizontal space between \partial and f in the numerator, than there is between \partial and x in the denominator. Is that due to the "empty" superscript?
    – Michael
    May 5, 2015 at 8:49
  • @Michael I added the check that the argument is empty
    – egreg
    May 5, 2015 at 11:10
  • I'm supposed to avoid such comments, but anyway thanks.
    – Michael
    May 5, 2015 at 11:20
  • There's a problem in the \eval (and in general, when someone uses . in \genericdel) in the “base” height. There should be a check to see if the #1 or #2 are equal to ., and, in that case, gobble it.
    – Manuel
    May 5, 2015 at 12:26
  • 1
    @HélitonMartins Actually, I'm not keen on packages that are just collections of macros, which are frequently just a cause of name conflicts.
    – egreg
    May 30, 2021 at 14:55

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