Stirling numbers of the second kind

Stirling numbers of the second kind

Summary.

The

Stirling numbers

of the second kind,

S

⁢

(

n

,

k

)

=

{

n

k

}

,

k

,

n

∈

ℕ

,

1

≤

k

≤

n

,

are a doubly
indexed

sequence

of

natural numbers

, enjoying a wealth of interesting
combinatorial properties. There exist several

logically equivalent

characterizations

, but the starting point of the present entry will
be the following definition:

The Stirling number

S

⁢

(

n

,

k

)

is the number of ways to

partition

a set
of

n

objects into

k

groups.

For example,

S

⁢

(

4

,

2

)

=

7

because there are seven ways to partition 4
objects — call them a, b, c, d — into two groups, namely:

(

a

)

⁢

(

b

⁢

c

⁢

d

)

,

(

b

)

⁢

(

a

⁢

c

⁢

d

)

,

(

c

)

⁢

(

a

⁢

b

⁢

d

)

,

(

d

)

⁢

(

a

⁢

b

⁢

c

)

,

(

a

⁢

b

)

⁢

(

c

⁢

d

)

,

(

a

⁢

c

)

⁢

(

b

⁢

d

)

,

(

a

⁢

d

)

⁢

(

b

⁢

c

)

Four additional characterizations will be discussed in this entry:

•

a

recurrence relation

•

a

generating function

related to the

falling factorial

•

differential operators

•

a double-index generating function

Each of these will be discussed below, and shown to be

equivalent

.

A recurrence relation.

The Stirling numbers of the second kind can be characterized in terms of
the following recurrence relation:

S

⁢

(

n

,

k

)

=

k

⁢

S

⁢

(

n

-

1

,

k

)

+

S

⁢

(

n

-

1

,

k

-

1

)

,

1

≤

k

<

n

,

subject to the
following

initial conditions

:

S

⁢

(

n

,

n

)

=

S

⁢

(

n

,

1

)

=

1

.

Let us now show that the recurrence formula
follows from the enumerative definition. Evidently, there is only one
way to partition

n

objects into

1

group (everything is in that
group), and only one way to partition

n

objects into

n

groups
(every object is a group all by itself). Proceeding recursively, a
division of

n

objects

a

1

,

…

,

a

n

-

1

,

a

n

into

k

groups
can be achieved by only one of two basic maneuvers:

•

We could partition the first

n

-

1

objects into

k

groups, and
then add object

a

n

into one of those groups. There are

k

⁢

S

⁢

(

n

-

1

,

k

)

ways to do this.

•

We could partition the first

n

-

1

objects into

k

-

1

groups and
then add object

a

n

as a new, 1

element

group. This gives an
additional

S

⁢

(

n

-

1

,

k

-

1

)

ways to create the desired partition.

The

recursive

point of view, therefore explains the connection between
the recurrence formula, and the original definition.

Using the recurrence formula we can easily obtain a table of the
initial Stirling numbers:

n

\

k

1

2

3

4

5

1

1

2

1

1

3

1

3

1

4

1

7

6

1

5

1

15

25

10

1

Falling Factorials.

Consider the

vector space

of

polynomials

in indeterminate

x

. The
most obvious basis of this

infinite-dimensional

vector space is the
sequence of monomial powers:

x

n

,

n

∈

ℕ

.

However, the
sequence of falling factorials:

x

n

¯

=

x

⁢

(

x

-

1

)

⁢

(

x

-

2

)

⁢

…

⁢

(

x

-

n

+

1

)

,

n

∈

ℕ

is also a basis, and hence can be
used to

generate

the monomial basis. Indeed, the Stirling numbers of
the second kind can be characterized as the coefficients involved
in the corresponding

change of basis matrix

, i.e.

x

n

=

∑

k

=

1

n

S

⁢

(

n

,

k

)

⁢

x

k

¯

.

So, for example,

x

4

=

x

+

7

⁢

x

⁢

(

x

-

1

)

+

6

⁢

x

⁢

(

x

-

1

)

⁢

(

x

-

2

)

+

x

⁢

(

x

-

1

)

⁢

(

x

-

2

)

⁢

(

x

-

3

)

.

Arguing inductively, let us prove that this characterization follows
from the recurrence relation. Evidently the

formula

is true for

n

=

1

. Suppose then that the formula is true for a given

n

. We
have

x

⁢

x

k

¯

=

x

k

+

1

¯

+

k

⁢

x

k

¯

,

and hence using the recurrence relation we deduce that

x

n

+

1

=

∑

k

=

1

n

S

⁢

(

n

,

k

)

⁢

x

⁢

x

k

¯

=

∑

k

=

1

n

(

k

⁢

S

⁢

(

n

,

k

)

⁢

x

k

¯

+

S

⁢

(

n

,

k

+

1

)

)

⁢

x

k

+

1

¯

=

∑

k

=

1

n

+

1

S

⁢

(

n

,

k

)

⁢

x

k

¯

Differential operators.

Let

D

x

denote the ordinary

derivative

, applied to polynomials in
indeterminate

x

, and let

T

x

denote the differential operator

x

⁢

D

x

. We have the following characterization of the Stirling
numbers of the second kind in terms of these two

operators

:

(

T

x

)

n

=

∑

k

=

1

n

S

⁢

(

n

,

k

)

⁢

x

k

⁢

(

D

x

)

k

,

where an
exponentiated differential operator denotes the operator composed with
itself the indicated number of times. Let us show that this follows
from the recurrence relation. The proof is once again, inductive.
Suppose that the characterization is true for a given

n

. We have

T

x

⁢

(

x

k

⁢

(

D

x

)

k

)

=

k

⁢

x

k

⁢

(

D

x

)

k

+

x

k

+

1

⁢

(

D

x

)

k

+

1

,

and hence using the recurrence relation we deduce that

(

T

x

)

n

+

1

=

x

⁢

D

x

⁢

(

∑

k

=

1

n

S

⁢

(

n

,

k

)

⁢

x

k

⁢

(

D

x

)

k

)

=

∑

k

=

1

n

S

⁢

(

n

,

k

)

⁢

(

k

⁢

x

k

⁢

(

D

x

)

k

+

x

k

+

1

⁢

(

D

x

)

k

+

1

)

=

∑

k

=

1

n

+

1

S

⁢

(

n

+

1

,

k

)

⁢

x

k

⁢

(

D

x

)

k

Double index generating function.

One can also characterize the Stirling numbers of the second kind in
terms of the following generating function:

e

x

⁢

(

e

t

-

1

)

=

∑

n

=

1

∞

∑

k

=

1

n

S

⁢

(

n

,

k

)

⁢

x

k

⁢

t

n

n

!

.

Let us now prove this. Note that the

differential equation

d

⁢

ξ

d

⁢

t

=

ξ

,

admits the

general solution

ξ

=

e

t

⁢

x

.

It follows that for any polynomial

p

⁢

(

ξ

)

we have

exp

⁡

(

t

⁢

T

ξ

)

⁢

[

p

⁢

(

ξ

)

]

|

ξ

=

x

=

∑

n

=

0

∞

t

n

n

!

⁢

(

T

ξ

)

n

⁢

[

p

⁢

(

ξ

)

]

|

ξ

=

x

=

p

⁢

(

e

t

⁢

x

)

.

The proof is simple: just take

D

t

of both sides. To be more
explicit,

D

t

⁢

[

p

⁢

(

e

t

⁢

x

)

]

=

p

′

⁢

(

e

t

⁢

x

)

⁢

e

t

⁢

x

=

T

ξ

⁢

[

p

⁢

(

ξ

)

]

|

ξ

=

x

⁢

e

t

,

and that is exactly equal to

D

t

of the left-hand side. Since this

relation

holds for all polynomials, it also holds for all formal power
series. In particular if we apply the above relation to

e

ξ

,
use the result of the preceding

section

, and note that

D

ξ

⁢

[

e

ξ

]

=

e

ξ

,

we obtain

e

x

⁢

e

t

=

∑

n

=

1

∞

t

n

n

!

⁢

(

T

ξ

)

n

⁢

[

e

ξ

]

|

ξ

=

x

=

∑

n

=

1

∞

∑

k

=

1

n

S

⁢

(

n

,

k

)

⁢

t

n

n

!

⁢

ξ

k

⁢

(

D

ξ

)

k

⁢

[

e

ξ

]

|

ξ

=

x

=

e

x

⁢

∑

n

=

1

∞

∑

k

=

1

n

S

⁢

(

n

,

k

)

⁢

x

k

⁢

t

n

n

!

Dividing both sides by

e

x

we obtain the desired generating
function. Q.E.D.

Title

Stirling numbers of the second kind

Canonical name

StirlingNumbersOfTheSecondKind

Date of creation

2013-03-22 12:33:32

Last modified on

2013-03-22 12:33:32

Owner

rmilson (146)

Last modified by

rmilson (146)

Numerical id

11

Author

rmilson (146)

Entry type

Definition

Classification

msc 05A15

Related topic

StirlingNumbersOfTheFirstKind

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