Stirling Numbers and Exponential Generating Functions - Stirling Numbers of The Second Kind

Stirling Numbers of The Second Kind

These numbers count the number of partitions of into k nonempty subsets. First consider the total number of partitions, i.e. Bn 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 Symbolic combinatorics#The Flajolet–Sedgewick fundamental theorem applies (labelled case). The set of partitions into non-empty subsets is given by ("set of non-empty sets of singletons")

This decomposition is entirely analogous to the construction of the set of permutations from cycles, which is given by

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

Differentiate to obtain

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

which implies that

by convolution of exponential generating functions and because differentiating an EGF drops the first coefficient and shifts Bn+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, giving

Translating to generating functions, we obtain

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

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

or

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.

Read more about this topic:  Stirling Numbers And Exponential Generating Functions

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