Our last adventure in this section reveals another algebraic bridge to the Stirling numbers.
Let f .n/ be a function defined for integers n > 0. Define the difference operator by
f .n/WD f .n C 1/ f .n/:
The difference operator can be iterated as follows: kf .n/WD k 1f .n/
. For exam-ple,
2f .n/D f .n/
D f .n C 1/ f .n/
D f .n C 2/ f .nC 1/ f .nC 1/ f .n/ D f .n C 2/ 2f .nC 1/ C f .n/:
Since 3f .n/D 2f .n/
, it follows that
3f .n/D f .n C 2/ 2f .nC 1/ C f .n/ D f .n C 3/ 2f .nC 2/ C f .n C 1/
f .nC 2/ 2f .nC 1/ C f .n/ D f .n C 3/ 3f .nC 2/ C 3f .n C 1/ f .n/:
Indeed the pattern becomes obvious for 4f .n/:
4f .n/D f .n C 4/ 4f .nC 3/ C 6f .n C 2/ 4f .nC 1/ C f .n/:
Question 169 Verify this formula by calculating 3f .n/ .
The following theorem, which Exercise 14 asks you to prove by induction, shows how the binomial coefficients appear in the computation of mf .n/.
Theorem 4.3.9 Iff .n/ is a function defined for all integers n > 0, then
mf .n/D Xm kD0
. 1/k m k
!
f .nC m k/
for allm > 1.
In what we are about to derive, our main concern is when n D 0. In that case the formula of the theorem says
mf .0/D Xm kD0
. 1/k m k
!
f .m k/;
which can be rewritten (let j WD m kso that kD m j)
mf .0/D Xm jD0
. 1/m j m j
!
f .j /: (4.14)
This gives a formula for the m-th difference of f at 0 in terms of f .0/; f .1/; : : : ; f .m/.
The question is, can we invert this formula? That is, is it possible to get a formula for f .n/in terms of the differences kf .0/? The answer is yes. Exercise 10 asks for a proof of the following result.
Theorem 4.3.10 Iff .n/ is a function defined for all integers n > 0, then f .n/D
Xn kD0
n k
!
kf .0/:
The theorem says that if we know the differences of the function at 0, then we can recon-struct the function itself.
Example: a difference table
Given a function f .n/ defined on nonnegative integers n, the difference table for f at nD 0 is
f .0/ f .1/ f .2/ f .3/ f .4/
f .0/ f .1/ f .2/ f .3/
2f .0/ 2f .1/ 2f .2/ 2f .3/
3f .0/ 3f .1/ 3f .2/
4f .0/ 4f .1/ 4f .2/
: :: : :: : ::
The numbers f .0/; f .1/; f .2/; : : : go in the first row. To get the second row, we know
f .0/D f .1/ f .0/ so put that number directly below the space between f .0/ and f .1/.
In this way you can compute the rest of the row by taking the entry to the northeast and subtracting the entry to the northwest. Because of the definition of the difference operator, subsequent rows are computed in exactly the same way.
As an example, construct the difference table for f .n/D n3.
0 1 8 27 64 125
1 7 19 37 61
6 12 18 24
6 6 6
0 0
This shows that f .0/D 0, f .0/ D 1, 2f .0/ D 3f .0/ D 6, and mf .0/ D 0 for m > 3. These entries are printed in boldface in the difference table because they are the
4.3. Stirling numbers 173 ones appearing in Theorem 4.3.10. That theorem shows (those same entries are in boldface below)
Since this is true for infinitely many values of the nonnegative integer n, it is true as a polynomial equation when we replace n by an indeterminate x and use xk
D .x/kŠk. As
All of this shows that if f .x/D xn, then kf .0/has a combinatorial interpretation:
it equals the number of onto functions Œn ! Œk.
Summary
For fixed k > 0, the OGF of the numbers S.n; k/ is xk=.1 x/.1 2x/ .1 kx/. The EGF of the numbers B.n/ is eex 1and a beautiful formula for the n-th Bell number is
B.n/D 1
Algebraically, the Stirling numbers of the first and second kinds are the coefficients in certain polynomial expansions:
Although the Stirling numbers of the first kind alternate in sign, their absolute values have a combinatorial interpretation:ˇ
ˇs.n; k/ˇ
ˇ equals the number of permutations of Œn with exactly k cycles. The difference operator provides another link between polynomials and Stirling numbers of the second kind.
Exercises
1. Write the polynomial 3x4 x3C 4x C 10 as a linear combination of the polynomials .x/0; .x/1; .x/2; .x/3; .x/4.
2. Write 3.x/4 12.x/3 C 4.x/1 17 as a linear combination of the polynomials 1; x; x2; x3; x4.
3. Describe an algorithm that takes as its input a permutation of Œn, written as an n-list, and outputs the permutation written in cycle notation. The cycle notation should have the following properties: (1) the first cycle should begin with element 1; (2) each successive cycle should begin with the smallest element not belonging to any of the previous cycles.
4. (a) Give a combinatorial proof: for n > 1, c.n; n 1/D n2 . (b) Give a combinatorial proof: for n > 1, c.n; 1/D .n 1/Š.
(c) Give an algebraic proof: for n > 1, s.n; n 1/D n2
. (d) Give an algebraic proof: for n > 1, s.n; 1/D . 1/n 1.n 1/Š.
5. Let f be a continuous function. Prove that the general solution to the differential equation y0 D f .x/y is y D eF .x/CC where F is an antiderivative of f and C is a constant.
6. Prove: for any n > 0, . x/nD . 1/n.x/.n/. (The notation .x/.n/is “rising factorial”
notation. See Exercise 16 of Section 2.1.) 7. Prove: for any n > 0, .x/.n/ DP
k>0c.n; k/xk. (See previous exercise.)
8. Write the expansion of .1Cx/nas a linear combination of the polynomials .x/k. That is, determine the coefficients akso that .1C x/nDPn
kD0ak.x/k. 9. Let k > 0. In this section we derived the OGF of the sequence˚
S.n; k/
n>0. Show that the EGF of the same sequence is kŠ1.ex 1/k.
10. Prove Theorem 4.3.10.
11. Find and prove a formula for the number of partitions of Œn in which consecutive integers never appear in the same block.
12. Following the example for x3in this section, construct the difference table for f .n/D n4and then write x4as a linear combination of the polynomials xj
for 0 6 j 6 4.
13. Prove that, like the derivative, the difference operator satisfies f .n/C g.n/ D
f .n/C g.n/ and for any number c, cf .n/
D cf .n/.
14. Prove Theorem 4.3.9 by induction on m.
Travel Notes
When f .x/D xnthe appearance of the equation
f .x/D Xn kD0
kf .0/
kŠ .x/k
4.4. Integer partition numbers 175 may remind you of the Maclaurin series of an infinitely differentiable function, namely
g.x/DX
k>0
g.k/.0/
kŠ xk:
Both equations describe how to express a function (f .x/ or g.x/, respectively) as a linear combination of other functions (falling factorials .x/k or power functions xk) where the coefficients of that linear combination involve measurements of change at x D 0 (the k-th difference or the k-th derivative, respectively). Indeed there is a calculus of finite differencesthat is the discrete version of ordinary, continuous calculus.
4.4 Integer partition numbers
In Sections 2.4 and 3.4, we learned some facts about partitions of integers. Recall that P .n; k/equals the number of partitions of the integer n into k parts, and P .n/ equals the total number of partitions of the integer n. Among other things, the partition numbers satisfy the identity
P .n; k/D Xk jD1
P .n k; j /:
(This is Theorem 2.4.2 on page 79.) Simply put, this says that the number of partitions of ninto k parts equals the number of partitions of n kinto at most k parts. It is perhaps less cumbersome to write it like
P .n; k/D P .n k;at most k parts/: (4.15) We also learned that the OGF of˚
P .n/
n>0is 1
.1 x/.1 x2/.1 x3/ DY
j >1
1 1 xj:
(This is Theorem 3.4.3 on page 120.) In this section we will prove two more combinatorial theorems about partition numbers, find the OGF of P .n; k/ for fixed k, and investigate the prospect of formulas for partition numbers.