How to write Stirling numbers of the second kind?
The following is taken from amsmath
and uses \genfrac
- a generic fraction function:
\documentclass{article}
\usepackage{amsmath}% http://ctan.org/pkg/amsmath
\DeclareRobustCommand{\stirling}{\genfrac\{\}{0pt}{}}
\begin{document}
% Source: http://en.wikipedia.org/wiki/Stirling_numbers_of_the_second_kind
In mathematics, particularly in combinatorics, a Stirling number of the second kind
is the number of ways to partition a set of $n$~objects into $k$~non-empty subsets and
is denoted by~$S(n,k)$ or~$\stirling{a}{b}$. Stirling numbers of the second kind occur
in the field of mathematics called combinatorics and the study of partitions.
\end{document}
How does this work? \genfrac
takes five arguments to create a structure (from the amsmath
documentation; section 4.11.3 The \genfrac
command, p 14):
The last two correspond to
\frac
’s numerator and denominator; the first two are optional delimiters [...]; the third is a line thickness override [0 implies an invisible rule]; and the fourth argument is a mathstyle override: integer values 0-3 select respectively\displaystyle
,\textstyle
,\scriptstyle
, and\scriptscriptstyle
. If the third argument is left empty, the line thickness defaults to ‘normal’.
So \genfrac\{\}{0pt}{}
creates a fraction with an invisible horizontal rule (third argument is 0pt
), left and right delimiter given by \{
and \}
, respectively and no specific math style (an empty {}
fourth argument). \stirling
doesn't include a fifth and sixth argument for \genfrac
(numerator and denominator), because this is implicitly supplied by the user as the two "arguments" to \stirling
.
In a similar manner (perhaps for reference), amsmath
defines
\newcommand{\frac}[2]{\genfrac{}{}{}{}{#1}{#2}}
\newcommand{\tfrac}[2]{\genfrac{}{}{}{1}{#1}{#2}}
\newcommand{\binom}[2]{\genfrac{(}{)}{0pt}{}{#1}{#2}}
using \genfrac
.
Here is a plain pdfTeX solution, just to illustrate what type of typesetting Knuth used for such numbers (he wrote a few papers on this topic)
% ========= Fonts
\font\sc=cmcsc10
% ========== Heading macros
\magnification =\magstep 1
\overfullrule =0pt
%
\noindent 1. {\sc Stirling numbers} ---
Stirling cycle numbers ${ n\brack m}$ are defined by
$$ \ln^m(1+z) = m! \sum_n (-1)^{n+m} { n\brack m} {z^n\over n!}
\ .\leqno(1a) $$
The numbers $(-1)^{n+m}{n\brack m}$ are also called Stirling numbers of the first kind.
Stirling subset numbers ${n\brace m}$, also called Stirling numbers
of the second kind, are defined by
$$ \left( e^z-1\right)^m = m! \sum_n {n\brace m} {z^n\over n!}
\ ,\leqno(1b) $$
and 2-associated Stirling subset numbers ${n\brace m}_{\ge 2}$ are
defined by
$$ \left( e^z-1-z\right)^m = m!\sum_n {n\brace m}_{\!\ge 2} {z^n\over n!}
\ .\leqno(1c) $$
\bye
You can have a look at more examples at Knuth Papers
There is a Bmatrix
environment in amsmath
.
\documentclass{article}
\usepackage{amsmath}
\begin{document}
\begin{equation*}
\begin{Bmatrix}
x\\
y
\end{Bmatrix}
=\frac{1}{k!}\sum\limits_{j=0}^{k}(-1)^{k-j}\binom{k}{j}j^n
\end{equation*}
\end{document}