nLab double factorial

Definition

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,

$$(2n-1)!! = \frac1{2^n} \frac{(2n)!}{n!}$$

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

$$\binom{x}{n} = \frac{x(x-1)\ldots (x-n+1)}{n!}$$

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

$$\sum_{n \geq 0} \frac{a_n x^n}{n!} = \exp(x^2/2).$$

This is related to the fact that the double-factorials also crop up in calculations dealing with Gaussian integrals such as

$$\int_0^\infty p(x) e^{-x^2/2}\; d x$$

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.

Combinatorial interpretation

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.

Moments of the standard Gaussian distribution

The computation begins with a famous observation

$$\frac1{\sqrt{2\pi}} \int_{-\infty}^\infty e^{-x^2/2}\; d x = 1$$

which says that $x \mapsto \frac1{\sqrt{2\pi}} e^{-x^2/2}$ is a probability distribution on $\mathbb{R}$ with Lebesgue measure.

Proposition

For each $n \geq 0$,

$$\frac1{\sqrt{2\pi}} \int_{-\infty}^\infty x^n e^{-x^2/2}\; d x = a_n$$

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 $a_{2n} = (2n-1)!!$.

Proof

By matching MacLaurin coefficients, it is enough to show

$$\sum_{n \geq 0} \left(\frac1{\sqrt{2\pi}} \int_{-\infty}^\infty x^n e^{-x^2/2} \; d x\right) \frac{y^n}{n!} = e^{y^2/2}$$

However, the left side equals

$$\array{ \frac1{\sqrt{2\pi}} \displaystyle\int_{-\infty}^\infty \sum_{n \geq 0} \frac{(x y)^n}{n!} e^{-x^2/2}\; d x & = & \frac1{\sqrt{2\pi}} \displaystyle\int_{-\infty}^\infty e^{x y} e^{-x^2/2} \;d x \\ & = & e^{y^2/2} \cdot \frac1{\sqrt{2\pi}} \displaystyle\int_{-\infty}^\infty e^{-(x-y)^2/2} \;d x \\ & = & e^{y^2/2} }$$

where the last line results from the famous observation by substituting $x-y$ for $x$.

References

See also

• Wikipedia, Double factorial