fractionspunctuationvector
This is my first post so please go easy on me.
I am writing a vector using the bracket notation as shown below:
\[
\vec{u} = \left\langle \frac{1}{10} , \frac{1}{3} \right\rangle
\]
The comma is typeset in the center of the fractions in line with the Vinculum (division line), and I'm wondering if there is a way to put it at the bottom in line with the denominators?
Or, is this the proper typesetting notation, and I should just be okay with where it is at?
Note that you have extra space around your vector. You should probably using something like (pmatrix is part of the amsmath package)
\begin{pmatrix}a\\b\end{pmatrix}
The standard LaTeX \newcommand provides a way to have a single optional argument.
\newcommand*\colvec[3][]{
\begin{pmatrix}\ifx\relax#1\relax\else#1\\\fi#2\\#3\end{pmatrix}
}
Note that you have to use \colvec[a]{b}{c} if you want three elements or \colvec{a}{b} if you want two.
Update
As per your request in the comments, here's one that takes any number of elements based on the number passed in the first argument.
\newcount\colveccount
\newcommand*\colvec[1]{
\global\colveccount#1
\begin{pmatrix}
\colvecnext
}
\def\colvecnext#1{
#1
\global\advance\colveccount-1
\ifnum\colveccount>0
\\
\expandafter\colvecnext
\else
\end{pmatrix}
\fi
}
You use it exactly as you wanted, \colvec{5}{a}{b}{c}{d}{e}.
You have to collect the row beforehand, as doing it inside an array environment is quite difficult, as TeX ends a cell as soon as it sees an &. I prefer to do in with a token register and the \count255 scratch counter.
We collect two arguments: the first is the number of entries, the second the first entry, in order to initialize the token register with it. We set the counter to the number of requested element (since we have already collected one, we step it down immediately).
Then we start the recursion; the macro \rowvectnexta examines the value of the counter; if it's still greater than 0, it calls \rowvecnextb that gathers another argument, stores it into the token register (adding an &) and steps down the counter; otherwise it builds the pmatrix by releasing the contents of the register.
\newtoks\rowvectoks
\newcommand{\rowvec}[2]{%
\rowvectoks={#2}\count255=#1\relax
\advance\count255 by -1
\rowvecnexta}
\newcommand{\rowvecnexta}{%
\ifnum\count255>0
\expandafter\rowvecnextb
\else
\begin{pmatrix}\the\rowvectoks\end{pmatrix}
\fi}
\newcommand\rowvecnextb[1]{%
\rowvectoks=\expandafter{\the\rowvectoks}%
\advance\count255 by -1
\rowvecnexta}
It shouldn't be a big limitation the fact that you need at least one element:
\rowvec{1}{a}
\rowvec{2}{a}{b}
\rowvec{3}{a}{b}{c}
Notice: \usepackage{amsmath} is required. So a minimal example might be the following.
\documentclass{article}
\usepackage{amsmath}
\newtoks\rowvectoks
\newcommand{\rowvec}[2]{%
\rowvectoks={#2}\count255=#1\relax
\advance\count255 by -1
\rowvecnexta}
\newcommand{\rowvecnexta}{%
\ifnum\count255>0
\expandafter\rowvecnextb
\else
\begin{pmatrix}\the\rowvectoks\end{pmatrix}
\fi}
\newcommand\rowvecnextb[1]{%
\rowvectoks=\expandafter{\the\rowvectoks}%
\advance\count255 by -1
\rowvecnexta}
\begin{document}
$\rowvec{1}{a}$
$\rowvec{2}{a}{b}$
$\rowvec{3}{a}{b}{c}$
\end{document}
A completely different solution that can be generalized to column vectors uses xparse and LaTeX3.
\documentclass{article}
\usepackage{amsmath}
\usepackage{xparse}
\ExplSyntaxOn
\NewDocumentCommand{\Rowvec}{ O{,} m }
{
\vector_main:nnnn { p } { & } { #1 } { #2 }
}
\NewDocumentCommand{\Colvec}{ O{,} m }
{
\vector_main:nnnn { p } { \\ } { #1 } { #2 }
}
\seq_new:N \l__vector_arg_seq
\cs_new_protected:Npn \vector_main:nnnn #1 #2 #3 #4
{
\seq_set_split:Nnn \l__vector_arg_seq { #3 } { #4 }
\begin{#1matrix}
\seq_use:Nnnn \l__vector_arg_seq { #2 } { #2 } { #2 }
\end{#1matrix}
}
\ExplSyntaxOff
\begin{document}
$\Rowvec{a}\Rowvec{a,b}\Rowvec[;]{a;b;c}$
$\Colvec{a}\Colvec{a,b}\Colvec[;]{a;b;c}$
\end{document}
The optional argument is the delimiter to be used to separate entries in the list given as argument.
It's sufficient to change { p } into { b } in the two main definitions to get brackets instead of parentheses.
Note: this requires expl3 dated September 2012 or later.
Best Answer
As User Schrödinger's cat suggested, including
\includepackage{amsmath}in the preamble, the following code produces the desired result:\[\vec{u} = \left\langle \frac{1}{10} \genfrac{}{}{0pt}{}{}{,} \frac{1}{3} \right\rangle\]For my implementation, I've defined a new macro as a shortcut in the preamble as:
\newcommand{\fraccomma}{\genfrac{}{}{0pt}{}{}{,}}Then the final code produces:
Edit:
However, as User egreg also pointed out, when implementing this solution we should also consider that it might not be appropriate if not all of the terms are fractions. In that event, then a more appropriate typesetting would be to just leave the comma in it's default position:
\vec{u} = \left\langle \frac{1}{10} , 3 \right\rangle