# Declaremathoperator with language dependence

I would like to define a new operator for the range of a linear function. Since I am German I want to define it in a way that it changes depending on the used language.

\documentclass{scrartcl}
\usepackage[english]{babel}
\usepackage{amsmath}
\DeclareMathOperator{\range}{ran}
\DeclareMathOperator{\kernel}{ker}
% The following should be used if german or ngerman are defined for babel
% \DeclareMathOperator{\range}{Bild}
% \DeclareMathOperator{\kernel}{Kern}
\begin{document}
\begin{align}
T &: V \rightarrow W \\
\dim(V) &= \dim(\range(T)) + \dim(\kernel(T))
\end{align}
\end{document}


Is there an easy way to achieve this?

• do you just want to make a choice at definition time or within a document always pick up the language current outside this math expression? – David Carlisle Jan 14 '17 at 12:30
• Definition time is sufficient for this problem. – mcocdawc Jan 14 '17 at 12:31
• It turned out to be easier to do the other way:-) – David Carlisle Jan 14 '17 at 13:02

In my opinion the definition should be so that the operator doesn't change across a document even through language changes.

So if the main language is German, \ker should resolve to “Kern” also in English. However, it's easy to provide also a “variable” kind.

\documentclass{scrartcl}
\usepackage[english,ngerman]{babel}
\usepackage{amsmath}
\usepackage{xparse}

\ExplSyntaxOn
\NewDocumentCommand\DeclareBabelMathOperator{mmO{???}}
{
\DeclareMathOperator{#1}
{
\str_case:xnF { \use:c { bbl@main@language } } { #2 } { #3 }
}
}
\NewDocumentCommand\DeclareVariableBabelMathOperator{mmO{???}}
{
\DeclareMathOperator{#1}
{
\str_case:xnF { \languagename }{ #2 } { #3 }
}
}
\cs_generate_variant:Nn \str_case:nnF { x }
\ExplSyntaxOff

\DeclareBabelMathOperator{\range}{
{english}{ran}
{ngerman}{Bild}
}
\DeclareBabelMathOperator{\kernel}{
{english}{ker}
{ngerman}{Kern}
}[ker]
\DeclareVariableBabelMathOperator{\vrange}{
{english}{ran}
{ngerman}{Bild}
}
\DeclareVariableBabelMathOperator{\vkernel}{
{english}{ker}
{ngerman}{Kern}
}[ker]

\begin{document}

\section{Fixed names}
$\dim(V) = \dim(\range(T)) + \dim(\kernel(T))$
\begin{otherlanguage*}{english}
$\dim(V) = \dim(\range(T)) + \dim(\kernel(T))$
\end{otherlanguage*}

\section{Variable names}
$\dim(V) = \dim(\vrange(T)) + \dim(\vkernel(T))$
\begin{otherlanguage*}{english}
$\dim(V) = \dim(\vrange(T)) + \dim(\vkernel(T))$
\end{otherlanguage*}

\end{document}


The optional argument is for the string to be used with a language not included in the previous choices.

Here is the output if english and ngerman are swapped throughout.

• Thank you very much. That solves all possible cases, I could think of. – mcocdawc Jan 14 '17 at 17:09

\documentclass{scrartcl}
\usepackage[german,english]{babel}
\usepackage{amsmath}

\DeclareMathOperator{\range}{\rangename}
\DeclareMathOperator{\kernel}{\kernelname}

\def\rangename{range}%
\def\kernelname{ker}%
}

\def\rangename{Bild}%
\def\kernelname{Kern}%
}
\begin{document}

\selectlanguage{english}
english
\begin{align}
T &: V \rightarrow W \\
\dim(V) &= \dim(\range(T)) + \dim(\kernel(T))
\end{align}

\selectlanguage{german}
Und jetzt etwas auf Deutsch
\begin{align}
T &: V \rightarrow W \\
\dim(V) &= \dim(\range(T)) + \dim(\kernel(T))
\end{align}
\end{document}

• Thank you very much for the great answer. I accepted egreg's answer because it solves both cases, but if possible I would accept all three – mcocdawc Jan 14 '17 at 17:08
• @mcocdawc oh no! egreg will be unbearable in chat:-) – David Carlisle Jan 14 '17 at 17:11

I would stick to good old \operatorname and combine it with babel's \extrasLANGUAGE:

\documentclass{scrartcl}
\usepackage[german,english]{babel}
\usepackage{amsmath}

\def\range{\operatorname{ran}}%
\def\kernel{\operatorname{ker}}%
}

\def\range{\operatorname{Bild}}%
\def\kernel{\operatorname{Kern}}%
}
\begin{document}

English
\begin{align}
T &: V \rightarrow W \\
\dim(V) &= \dim(\range(T)) + \dim(\kernel(T))
\end{align}

\selectlanguage{german}
Und jetzt etwas auf Deutsch
\begin{align}
T &: V \rightarrow W \\
\dim(V) &= \dim(\range(T)) + \dim(\kernel(T))
\end{align}
\end{document}

• Thank you very much for your help. I finally accepted egreg's answer because it solves both cases, but if possible I would accept all three – mcocdawc Jan 14 '17 at 17:10