transfinite arithmetic, cardinal arithmetic, ordinal arithmetic
prime field, p-adic integer, p-adic rational number, p-adic complex number
arithmetic geometry, function field analogy
For $n$ a natural number, the double factorial $(2n+1)!!$ is defined to be the product $(2n+1)(2n-1)\ldots 1$. Alternatively, in terms of the ordinary factorial,
so that in particular, $(-1)!!$ is defined to be $1$. Relatedly, we may define $(2n)!! = 2^n n!$.
Double-factorials have a number of applications in enumerative combinatorics. They are particularly prone to appear whenever dealing with binomial coefficients
in the case $x = 1/2$ or $x = -1/2$, or when dealing with middle binomial coefficients $\binom{2n}{n}$, or when dealing with the values of the Gamma function at half-integers.
According to the combinatorial interpretation below, the exponential generating function of the sequence $a_n$ defined by $a_{2n} = (2n-1)!!$ and $a_{2n+1} = 0$ is
This is related to the fact that the double-factorials also crop up in calculations dealing with Gaussian integrals such as
for even polynomials $p$, with consequent applications in quantum mechanics, for example calculations surrounding the quantum harmonic oscillator. See the section below on moments of Gaussian distributions.
The double-factorials $(2n-1)!!$ count the number of involutions without fixed points on a set with $2n$ elements, or the number of partitions of a $(2n)$-element set into $2$-element sets, or the number of isomorphism classes of rooted chord diagrams with $n$ chords. This follows readily from the exponential generating function expression above, and follows readily by considering the species composition of the exponential species $\exp$ (the terminal object in the category of species) with the species $x^2/2$, defined to be terminal at $2$-element sets and empty at others.
The computation begins with a famous observation
which says that $x \mapsto \frac1{\sqrt{2\pi}} e^{-x^2/2}$ is a probability distribution on $\mathbb{R}$ with Lebesgue measure.
For each $n \geq 0$,
where $a_n$ are the MacLaurin coefficients in $e^{y^2/2} = \sum_{n \geq 0} \frac{a_n y^n}{n!}$, namely, $a_n = 0$ if $n$ is odd, and $a_{2n} = (2n-1)!!$.
By matching MacLaurin coefficients, it is enough to show
However, the left side equals
where the last line results from the famous observation by substituting $x-y$ for $x$.
See also
Last revised on September 17, 2018 at 09:00:05. See the history of this page for a list of all contributions to it.