Stirling numbers and exponential generating functions

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The use of exponential generating functions or EGFs to study the properties of Stirling numbers is a classical exercise in combinatorics and possibly the canonical example of how symbolic combinatorics, the method that encapsulates the fundamental theorem of combinatorial enumeration, is used. It also illustrates the parallels in the construction of these two types of numbers, lending support to the binomial-style notation that is used for them.

This article uses the coefficient extraction operator $[z n]$ for formal power series, as well as the (labelled) operators $\mathfrak{C}$ (for cycles) and $\mathfrak{P}$ (for sets) on combinatorial classes, which are explained on the pages for symbolic combinatorics and the fundamental theorem of combinatorial enumeration. Quite simply, given a combinatorial class, the cycle operator creates the class obtained by placing objects from the source class along a cycle of some length, where cyclical symmetries are taken into account, and the set operator creates the class obtained by placing objects from the source class in a set (symmetries from the symmetric group, i.e. an "unstructured bag".) The two combinatorial classes (shown without additional markers) are

permutations (for unsigned Stirling numbers of the first kind):

$\mathcal{P} = \mathfrak{P}(\mathfrak{C}(\mathcal{Z})),$

and

set partitions into non-empty subsets (for Stirling numbers of the second kind):

$\mathcal{B} = \mathfrak{P}(\mathfrak{P}_{\ge 1}(\mathcal{Z})),$

where $\mathcal{Z}$ is the singleton class.

1 Stirling numbers of the first kind
2 Stirling numbers of the second kind
3 External links
4 References

[edit] Stirling numbers of the first kind

The unsigned Stirling numbers of the first kind count the number of permutations of [n] with k cycles. A permutation is a set of cycles, and hence the set $\mathcal{P}\,$ of permutations is given by

$\mathcal{P} = \mathfrak{P}(\mathcal{U} \mathfrak{C}(\mathcal{Z})),$

where the singletion $\mathcal{U}$ marks cycles. This decomposition is examined in some detail on the page on the statistics of random permutations.

Translating to generating functions we obtain the mixed generating function of the unsigned Stirling numbers of the first kind:

$G(z, u) = \exp \left( u \log \frac{1}{1-z} \right) = \left(\frac{1}{1-z} \right)^u = \sum_{n=0}^\infty \sum_{k=0}^n \left|\left[\begin{matrix} n \\ k \end{matrix}\right]\right| u^k \, \frac{z^n}{n!}.$

Now the signed Stirling numbers of the first kind are obtained from the unsigned ones through the relation

$\left[\begin{matrix} n \\ k \end{matrix}\right] = (-1)^{n-k} \left|\left[\begin{matrix} n \\ k \end{matrix}\right]\right|.$

Hence the generating function $H (z, u)$ of these numbers is

$H(z, u) = G(-z, -u) = \left(\frac{1}{1+z} \right)^{-u} = (1+z)^u = \sum_{n=0}^\infty \sum_{k=0}^n \left[\begin{matrix} n \\ k \end{matrix}\right] u^k \, \frac{z^n}{n!}.$

A variety of identities may be derived by manipulating this generating function:

$(1+z)^u = \sum_{n=0}^\infty {u \choose n} z^n = \sum_{n=0}^\infty \frac {z^n}{n!} \sum_{k=0}^n \left[\begin{matrix} n \\ k \end{matrix}\right] u^k = \sum_{k=0}^\infty u^k \sum_{n=k}^\infty \frac {z^n}{n!} \left[\begin{matrix} n \\ k \end{matrix}\right] = e^{u\log(1+z)}.$

In particular, the order of summation may be exchanged, and derivatives taken, and then z or u may be fixed.

[edit] Finite sums

A simple sum is

$\sum_{k=0}^n (-1)^k \left[\begin{matrix} n \\ k \end{matrix}\right] = (-1)^n n!.$

This formula holds because the exponential generating function of the sum is

$H(z, -1) = \frac{1}{1+z} \quad \mbox{and hence} \quad n! [z^n] H(z, -1) = (-1)^n n!.$

[edit] A convolution identity

The identity

$\sum_{p=k}^{n}{ \left|\left[\begin{matrix} n \\ p \end{matrix}\right]\right| \binom{p}{k}}= \left|\left[\begin{matrix} n+1 \\ k+1 \end{matrix}\right]\right|$

has a combinatorial proof as well as a proof using generating functions. The left side counts permutations of n elements, with at least k cycles, of which exactly k are marked. The right side counts permutations of $n + 1$ elements having exactly $k + 1$ cycles. It is easy to establish a bijection between these two, thereby showing that they have the same cardinality. To go from the left to the right, take the cycles that were not marked, which define a permutation, write down this permutation in table notation, i.e. as a sequence, and make it into a cycle by connecting the end of the table to the beginning, inserting $n + 1$ after the end, thereby obtaining a permutation on $n + 1$ elements having exactly $k + 1$ cycles. The other direction is as follows: mark all cycles except the one containing $n + 1$ , then take this cycle (the one that contains $n + 1$ ) and turn it into a sequence by taking it apart at the position of $n + 1$ , which you remove. Read this sequence as a permutation and factorize it into cycles. You now have a permutation on $[1, n]$ of at least k cycles, of which exactly k are marked. As an example, let $n = 8$ and suppose you have not marked $(18)(25)$ . Written in table notation, this reads $[8521]$ . Turn it into a cycle, inserting $n + 1 = 9$ , getting $(85219).$

The proof using generating functions follows directly from the generating function $G(z, u).\,$

We have

$\left|\left[\begin{matrix} n+1 \\ k+1 \end{matrix}\right]\right| = (n+1)! [z^{n+1}] [u^{k+1}] \left( \frac{1}{1-z} \right)^u.$

Differentiating with respect to z, we find

$\left|\left[\begin{matrix} n+1 \\ k+1 \end{matrix}\right]\right| = n! [z^n] [u^{k+1}] \frac{u}{1-z} \left( \frac{1}{1-z} \right)^u = n! [z^n] [u^k] \left( \frac{1}{1-z} \right)^{u+1}.$

Note that

$\left( \frac{1}{1-z} \right)^{u+1} = \sum_{p\ge 0} \frac{(u+1)^p}{p!} \left( \log \frac{1}{1-z} \right)^p,$

so that

$\left|\left[\begin{matrix} n+1 \\ k+1 \end{matrix}\right]\right| = n! [z^n] \sum_{p\ge k} \binom{p}{k} \frac{1}{p!} \left( \log \frac{1}{1-z} \right)^p,$

which is

$n! [z^n] \sum_{p\ge k} \binom{p}{k} [u^p] G(z, u) = \sum_{p\ge k} \binom{p}{k} n! [z^n] [u^p] G(z, u) = \sum_{p\ge k} \binom{p}{k} \left|\left[\begin{matrix} n \\ p \end{matrix}\right]\right|.$

But

$\left|\left[\begin{matrix} n \\ p \end{matrix}\right]\right| = 0 \quad \mbox{when} \quad p>n,$

so this simplifies to

$\left|\left[\begin{matrix} n+1 \\ k+1 \end{matrix}\right]\right| = \sum_{p=k}^n \left|\left[\begin{matrix} n \\ p \end{matrix}\right]\right| \binom{p}{k}, \quad \mbox{QED}.$

The references for this identity are from sci.math.research, more information may be found here.

[edit] Infinite sums

Some infinite sums include

$\sum_{n=k}^\infty \left[\begin{matrix} n \\ k \end{matrix}\right] \frac{z^n}{n!} = \frac {\left(\log (1+z)\right)^k}{k!}$

where $| z | < 1$ (the singularity nearest to $z = 0$ of $log(1 + z)$ is at $z = - 1.$ )

This relation holds because

$[u^k] H(z, u) = [u^k] \exp \left( u \log (1+z) \right) = \frac {\left(\log (1+z)\right)^k}{k!}.$

[edit] Stirling numbers of the second kind

These numbers count the number of partitions of $[n]$ into k nonempty subsets. First consider the total number of partitions, i.e. $B n$ where

$B_n = \sum_{k=1}^n \left\{\begin{matrix} n \\ k \end{matrix}\right\} \mbox{ and } B_0 = 1,$

i.e. the Bell numbers. The fundamental theorem of combinatorial enumeration applies (labelled case). The set $\mathcal{B}\,$ of partitions into non-empty subsets is given by ("set of non-empty sets of singletons")

$\mathcal{B} = \mathfrak{P}(\mathfrak{P}_{\ge 1}(\mathcal{Z})).$

This decomposition is entirely analogous to the construction of the set $\mathcal{P}\,$ of permutations from cycles, which is given by

$\mathcal{P} = \mathfrak{P}(\mathfrak{C}(\mathcal{Z})).$

and yields the Stirling numbers of the first kind. Hence the name "Stirling numbers of the second kind."

The decomposition is equivalent to the EGF

$B(z) = \exp \left(\exp z - 1\right).$

Differentiate to obtain

$\frac{d}{dz} B(z) = \exp \left(\exp z - 1\right) \exp z = B(z) \exp z,$

which implies that

$B_{n+1} = \sum_{k=0}^n {n \choose k} B_k,$

by convolution of exponential generating functions and because differentiating an EGF drops the first coefficient and shifts $B n + 1$ to $z n / n!.$

The EGF of the Stirling numbers of the second kind is obtained by marking every subset that goes into the partition with the term $\mathcal{U}\,$ , giving

$\mathcal{B} = \mathfrak{P}(\mathcal{U} \; \mathfrak{P}_{\ge 1}(\mathcal{Z})).$

Translating to generating functions, we obtain

$B(z, u) = \exp \left(u \left(\exp z - 1\right)\right).$

This EGF yields the formula for the Stirling numbers of the second kind:

$\left\{\begin{matrix} n \\ k \end{matrix}\right\} = n! [u^k] [z^n] B(z, u) = n! [z^n] \frac{(\exp z - 1)^k}{k!}$

$n! [z^n] \frac{1}{k!} \sum_{j=0}^k {k \choose j} \exp(jz) (-1)^{k-j}$

which simplifies to

$\frac{n!}{k!} \sum_{j=0}^k {k \choose j} (-1)^{k-j} \frac{j^n}{n!} = \frac{1}{k!} \sum_{j=0}^k {k \choose j} (-1)^{k-j} j^n.$

We can use $B (z, u)$ to evaluate the sum

$\sum_{k=0}^n \left\{\begin{matrix} n \\ k \end{matrix}\right\} (x)_k.$

This is equal to

$\sum_{k=0}^n n! [u^k] [z^n] B(z, u) (x)_k = n! [z^n] \sum_{k=0}^n [u^k] B(z, u) (x)_k$

$n! [z^n] \sum_{k=0}^n \frac{(\exp z - 1)^k}{k!} (x)_k = n! [z^n] \sum_{k=0}^\infty \frac{(\exp z - 1)^k}{k!} (x)_k$

where the last equality occurs because $[z^n] (\exp z - 1)^k = 0\,$ when $n<k\,.$

Now consider the Taylor series of $y x$ at $y = 1:$

$y^x = \sum_{k=0}^\infty \left( \left(\frac{d}{dy}\right)^k y^x \Bigg|_{y=1} \right) \frac{(y-1)^k}{k!} = \sum_{k=0}^\infty (x)_k \frac{(y-1)^k}{k!}.$

Hence

$\sum_{k=0}^n \left\{\begin{matrix} n \\ k \end{matrix}\right\} (x)_k = n! [z^n] \exp(xz) = n! \frac{x^n}{n!} = x^n.$

[edit] External links

Philippe Flajolet and Robert Sedgewick, Analytic combinatorics - Symbolic combinatorics.
Various, Identity for unsigned Stirling Numbers of the first kind, newsgroup sci.math.research

[edit] References

Ronald Graham, Donald Knuth, Oren Patashnik (1989): Concrete Mathematics, Addison-Wesley, ISBN 0-201-14236-8
D.S. Mitrinovic, Sur une classe de nombre relies aux nombres de Stirling, C. R. Acad. Sci. Paris 252 (1961), 2354--2356.
A.C.R. Belton, The monotone Poisson process, in: Quantum Probability (M. Bozejko, W. Mlotkowski and J. Wysoczanski, eds.), Banach Center Publications 73, Polish Academy of Sciences, Warsaw, 2006

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Category: Combinatorics