Generatingfunctionology (mostly combinatorics)

231 Pages·1.539 MB·English

by  Wilf.

generatingfunctionology
Herbert S. Wilf
Department of Mathematics
University of Pennsylvania
Philadelphia, Pennsylvania
Copyright 1990 and 1994 by Academic Press, Inc. All rights re-
served. This Internet Edition may be reproduced for any valid educational
purpose of an institution of higher learning, in which case only the reason-
able costs of reproduction may be charged. Reproduction for proflt or for
any commercial purposes is strictly prohibited.
vi
Preface
This book is about generating functions and some of their uses in
discrete mathematics. The subject is so vast that I have not attempted to
give a comprehensive discussion. Instead I have tried only to communicate
some of the main ideas.
Generating functions are a bridge between discrete mathematics, on
the one hand, and continuous analysis (particularly complex variable the-
ory) on the other. It is possible to study them solely as tools for solving
discrete problems. As such there is much that is powerful and magical in
the way generating functions give unifled methods for handling such prob-
lems. The reader who wished to omit the analytical parts of the subject
would skip chapter 5 and portions of the earlier material.
To omit those parts of the subject, however, is like listening to a stereo
broadcast of, say, Beethoven’s Ninth Symphony, using only the left audio
channel.
The full beauty of the subject of generating functions emerges only
from tuning in on both channels: the discrete and the continuous. See
how they make the solution of difierence equations into child’s play. Then
see how the theory of functions of a complex variable gives, virtually by
inspection, the approximate size of the solution. The interplay between the
two channels is vitally important for the appreciation of the music.
In recent years there has been a vigorous trend in the direction of
flnding bijective proofs of combinatorial theorems. That is, if we want to
prove that two sets have the same cardinality then we should be able to
do it by exhibiting an explicit bijection between the sets. In many cases
the fact that the two sets have the same cardinality was discovered in the
flrst place by generating function arguments. Also, even though bijective
arguments may be known, the generating function proofs may be shorter
or more elegant.
The bijective proofs give one a certain satisfying feeling that one ‘re-
ally’ understands why the theorem is true. The generating function argu-
ments often give satisfying feelings of naturalness, and ‘oh, I could have
thought of that,’ as well as usually ofiering the best route to flnding exact
or approximate formulas for the numbers in question.
This book was tested in a senior course in discrete mathematics at the
University of Pennsylvania. My thanks go to the students in that course for
helping me at least partially to debug the manuscript, and to a number of
my colleagues who have made many helpful suggestions. Any reader who
is kind enough to send me a correction will receive a then-current complete
errata sheet and many thanks.
Herbert S. Wilf
Philadelphia, PA
September 1, 1989
vii
Preface to the Second Edition
This edition contains several new areas of application, in chapter 4,
many new problems and solutions, a number of improvements in the pre-
sentation, and corrections. It also contains an Appendix that describes
some of the features of computer algebra programs that are of particular
importance in the study of generating functions.
I am indebted to many people for helping to make this a better book.
Bruce Sagan, in particular, made many helpful suggestions as a result of
a test run in his classroom. Many readers took up my ofier (which is now
repeated) to supply a current errata sheet and my thanks in return for any
errors discovered.
Herbert S. Wilf
Philadelphia, PA
May 21, 1992
viii
CONTENTS
Chapter 1: Introductory Ideas and Examples
1.1 An easy two term recurrence . . . . . . . . . . . . . . . . . 3
1.2 A slightly harder two term recurrence . . . . . . . . . . . . . 5
1.3 A three term recurrence . . . . . . . . . . . . . . . . . . . 8
1.4 A three term boundary value problem . . . . . . . . . . . . 10
1.5 Two independent variables . . . . . . . . . . . . . . . . . 11
1.6 Another 2-variable case . . . . . . . . . . . . . . . . . . 16
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . 24
Chapter 2: Series
2.1 Formal power series . . . . . . . . . . . . . . . . . . . . 30
2.2 The calculus of formal ordinary power series generating functions 33
2.3 The calculus of formal exponential generating functions . . . . 39
2.4 Power series, analytic theory . . . . . . . . . . . . . . . . 46
2.5 Some useful power series . . . . . . . . . . . . . . . . . . 52
2.6 Dirichlet series, formal theory . . . . . . . . . . . . . . . 56
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . 65
Chapter 3: Cards, Decks, and Hands: The Exponential Formula
3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . 73
3.2 Deflnitions and a question . . . . . . . . . . . . . . . . . 74
3.3 Examples of exponential families . . . . . . . . . . . . . . 76
3.4 The main counting theorems . . . . . . . . . . . . . . . . 78
3.5 Permutations and their cycles . . . . . . . . . . . . . . . 81
3.6 Set partitions . . . . . . . . . . . . . . . . . . . . . . . 83
3.7 A subclass of permutations . . . . . . . . . . . . . . . . . 84
3.8 Involutions, etc. . . . . . . . . . . . . . . . . . . . . . 84
3.9 2-regular graphs . . . . . . . . . . . . . . . . . . . . . 85
3.10 Counting connected graphs . . . . . . . . . . . . . . . . . 86
3.11 Counting labeled bipartite graphs . . . . . . . . . . . . . . 87
3.12 Counting labeled trees . . . . . . . . . . . . . . . . . . . 89
3.13 Exponential families and polynomials of ‘binomial type.’ . . . . 91
3.14 Unlabeled cards and hands . . . . . . . . . . . . . . . . . 92
3.15 The money changing problem . . . . . . . . . . . . . . . 96
3.16 Partitions of integers . . . . . . . . . . . . . . . . . . . 100
3.17 Rooted trees and forests . . . . . . . . . . . . . . . . . . 102
3.18 Historical notes . . . . . . . . . . . . . . . . . . . . . . 103
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . 104
vii
Chapter 4: Applications of generating functions
4.1 Generating functions flnd averages, etc. . . . . . . . . . . . 108
4.2 A generatingfunctionological view of the sieve method . . . . . 110
4.3 The ‘Snake Oil’ method for easier combinatorial identities . . . 118
4.4 WZ pairs prove harder identities . . . . . . . . . . . . . . 130
4.5 Generating functions and unimodality, convexity, etc. . . . . . 136
4.6 Generating functions prove congruences . . . . . . . . . . . 140
4.7 The cycle index of the symmetric group . . . . . . . . . . . 141
4.8 How many permutations have square roots? . . . . . . . . . 146
4.9 Counting polyominoes . . . . . . . . . . . . . . . . . . . 150
4.10 Exact covering sequences . . . . . . . . . . . . . . . . . . 154
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . 157
Chapter 5: Analytic and asymptotic methods
5.1 The Lagrange Inversion Formula . . . . . . . . . . . . . . 167
5.2 Analyticity and asymptotics (I): Poles . . . . . . . . . . . . 171
5.3 Analyticity and asymptotics (II): Algebraic singularities . . . . 177
5.4 Analyticity and asymptotics (III): Hayman’s method . . . . . 181
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . 188
Appendix: Using MapleTM and MathematicaTM . . . . . . . . 192
Solutions . . . . . . . . . . . . . . . . . . . . . . . . 197
References . . . . . . . . . . . . . . . . . . . . . . . 224
viii
Chapter 1
Introductory ideas and examples
A generating function is a clothesline on which we hang up a sequence
of numbers for display.
What that means is this: suppose we have a problem whose answer is
a sequence of numbers, a ;a ;a ;:::. We want to ‘know’ what the sequence
0 1 2
is. What kind of an answer might we expect?
A simple formula for a would be the best that could be hoped for. If
n
we flnd that a = n2+3 for each n = 0;1;2;:::, then there’s no doubt that
n
we have ‘answered’ the question.
But what if there isn’t any simple formula for the members of the
unknown sequence? After all, some sequences are complicated. To take
just one hair-raising example, suppose the unknown sequence is 2, 3, 5, 7,
11, 13, 17, 19, :::, where a is the nth prime number. Well then, it would
n
be just plain unreasonable to expect any kind of a simple formula.
Generating functions add another string to your bow. Although giv-
ing a simple formula for the members of the sequence may be out of the
question, we might be able to give a simple formula for the sum of a power
series, whose coe–cients are the sequence that we’re looking for.
For instance, suppose we want the Fibonacci numbers F ;F ;F ;:::,
0 1 2
and what we know about them is that they satisfy the recurrence relation
Fn+1 = Fn +Fn¡1 (n ‚ 1;F0 = 0; F1 = 1):
The sequence begins with 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, ::: There are
exact, not-very-complicatedformulasforF , aswewillseelater, inexample
n
2 of this chapter. But, just to get across the idea of a generating function,
here is how a generatingfunctionologist might answer the question: the
nth Fibonacci number, F , is the coe–cient of xn in the expansion of the
n
function x=(1¡x¡x2) as a power series about the origin.
You may concede that this is a kind of answer, but it leaves a certain
unsatisfled feeling. It isn’t really an answer, you might say, because we
don’t have that explicit formula. Is it a good answer?
In this book we hope to convince you that answers like this one are
often spectacularly good, in that they are themselves elegant, they allow
you to do almost anything you’d like to do with your sequence, and gener-
ating functions can be simple and easy to handle even in cases where exact
formulas might be stupendously complicated.
Here are some of the things that you’ll often be able to do with gener-
ating function answers:
(a) Find an exact formula for the members of your sequence.
Not always. Not always in a pleasant way, if your sequence is
1
2 1 Introductory ideas and examples
complicated. But at least you’ll have a good shot at flnding such
a formula.
(b) Find a recurrence formula. Most often generating functions
arise from recurrence formulas. Sometimes, however, from the
generating function you will flnd a new recurrence formula, not
the one you started with, that gives new insights into the nature
of your sequence.
(c) Find averages and other statistical properties of your se-
quence. Generating functions can give stunningly quick deriva-
tions of various probabilistic aspects of the problem that is repre-
sented by your unknown sequence.
(d) Find asymptotic formulas for your sequence. Some of the
deepest and most powerful applications of the theory lie here.
Typically, oneisdealingwithaverydi–cult sequence, andinstead
oflookingforanexactformula, whichmightbeoutofthequestion,
we look for an approximate formula. While we would not expect,
for example, to flnd an exact formula for the nth prime number,
it is a beautiful fact (the ‘Prime Number Theorem’) that the nth
prime is approximately nlogn when n is large, in a certain precise
sense. In chapter 5 we will discuss asymptotic problems.
(e) Prove unimodality, convexity, etc. A sequence is called uni-
modal if it increases steadily at flrst, and then decreases steadily.
Many combinatorial sequences are unimodal, and a variety of
methodsareavailableforprovingsuchtheorems. Generatingfunc-
tions can help. There are methods by which the analytic proper-
ties of the generating function can be translated into conclusions
about the rises and falls of the sequence of coe–cients. When
the method of generating functions works, it is often the simplest
method known.
(f) Prove identities. Many, many identities are known, in combina-
torics and elsewhere in mathematics. The identities that we refer
to are those in which a certain formula is asserted to be equal
to another formula for stated values of the free variable(s). For
example, it is well known that
(cid:181) ¶ (cid:181) ¶
Xn 2
n 2n
= (n = 0;1;2;:::):
j n
j=0
One way to prove such identities is to consider the generating
function whose coe–cients are the sequence shown on the left side
of the claimed identity, and to consider the generating function
formed from the sequence on the right side of the claimed identity,
and to show that these are the same function. This may sound
1.1 An easy two term recurrence 3
obvious, but it is quite remarkable how much simpler and more
transparent many of the derivations become when seen from the
point of view of the black belt generatingfunctionologist. The
‘Snake Oil’ method that we present in section 4.3, below, explores
some of these vistas. The method of rational functions, in section
4.4, is new, and does more and harder problems of this kind.
(g) Other. Is there something else you would like to know about
your sequence? A generating function may ofier hope. One ex-
ample might be the discovery of congruence relations. Another
possibility is that your generating function may bear a striking
resemblance to some other known generating function, and that
may lead you to the discovery that your problem is closely related
toanotherone, whichyouneversuspectedbefore. Itisnoteworthy
that in this way you may flnd out that the answer to your prob-
lem is simply related to the answer to another problem, without
knowing formulas for the answers to either one of the problems!
In the rest of this chapter we are going to give a number of examples
of problems that can be profltably thought about from the point of view
of generating functions. We hope that after studying these examples the
reader will be at least partly convinced of the power of the method, as well
as of the beauty of the unifled approach.
1.1 An easy two term recurrence
A certain sequence of numbers a ;a ;::: satisfles the conditions
0 1
a = 2a +1 (n ‚ 0;a = 0): (1:1:1)
n+1 n 0
Find the sequence.
First try computing a few members of the sequence to see what they
look like. It begins with 0, 1, 3, 7, 15, 31, ::: These numbers look sus-
piciously like 1 less than the powers of 2. So we could conjecture that
a = 2n ¡ 1 (n ‚ 0), and prove it quickly, by induction based on the
n
recurrence (1.1.1).
But this is a book about generating functions, so let’s forget all of
that, pretend we didn’t spot the formula, and use the generating function
method. Hence, insteadPof flnding the sequence fang, let’s flnd the gener-
ating function A(x) = a xn. Once we know what that function is,
n‚0 n
we will be able to read ofi the explicit formula for the a ’s by expanding
n
A(x) in a series.
To flnd A(x), multiply both sides of the recurrence relation (1.1.1) by
xn and sum over the values of n for which the recurrence is valid, namely,
over n ‚ 0. Then try to relate these sums to the unknown generating
function A(x).
4 1 Introductory ideas and examples
P
If we do this flrst to the left side of (1.1.1), there results a xn.
n‚0 n+1
How can we relate this to A(x)? It is almost the same as A(x). But the
subscript of the ‘a’ in each term is 1 unit larger than the power of x. But,
clearly,
X
a xn = a +a x+a x2 +a x3 +¢¢¢
n+1 1 2 3 4
n‚0
= f(a +a x+a x2 +a x3 +¢¢¢)¡a g=x
0 1 2 3 0
= A(x)=x
since a = 0 in this problem. Hence the result of so operating on the left
0
side of (1.1.1) is A(x)=x.
Next do the right side of (1.1.1). Multiply it by xn and sum over all
n ‚ 0. The result is
X X
(2a +1)xn = 2A(x)+ xn
n
n‚0 n‚0
1
= 2A(x)+ ;
1¡x
P
wherein we have used the familiar geometric series evaluation xn =
n‚0
1=(1¡x), which is valid for jxj < 1.
If we equate the results of operating on the two sides of (1.1.1), we flnd
that
A(x) 1
= 2A(x)+ ;
x 1¡x
which is trivial to solve for the unknown generating function A(x), in the
form
x
A(x) = :
(1¡x)(1¡2x)
Thisisthegeneratingfunctionfortheproblem. Theunknownnumbers
a are arranged neatly on this clothesline: a is the coe–cient of xn in the
n n
series expansion of the above A(x).
Suppose we want to flnd an explicit formula for the a ’s. Then we
n
would have to expand A(x) in a series. That isn’t hard in this example,
since the partial fraction expansion is
‰ (cid:190)
x 2 1
= x ¡
(1¡x)(1¡2x) 1¡2x 1¡x
= f2x+22x2 +23x3 +24x4 +¢¢¢g
¡fx+x2 +x3 +x4 +¢¢¢g
= (2¡1)x+(22 ¡1)x2 +(23 ¡1)x3 +(24 ¡1)x4 +¢¢¢
It is now clear that the coe–cient of xn, i.e. a , is equal to 2n¡1, for each
n
n ‚ 0.
1.2 A slightly harder two term recurrence 5
In this example, the heavy machinery wasn’t needed because we knew
the answer almost immediately, by inspection. The impressive thing about
generatingfunctionology is that even though the problems can get a lot
harder than this one, the method stays very much the same as it was here,
so the same heavy machinery may produce answers in cases where answers
are not a bit obvious.
1.2 A slightly harder two term recurrence
A certain sequence of numbers a ;a ;::: satisfles the conditions
0 1
a = 2a +n (n ‚ 0;a = 1): (1:2:1)
n+1 n 0
Find the sequence.
Asbefore, wemightcalculatetheflrstseveralmembersofthesequence,
to get 1, 2, 5, 12, 27, 58, 121, ::: A general formula does not seem to be
immediately in evidence in this case, so we use the method of generating
functions. That means that instead oPf looking for the sequence a0;a1;:::,
we will look for the function A(x) = a xj. Once we have found the
j‚0 j
function, the sequence will be identiflable as the sequence of power series
coe–cients of the function.*
As in example 1, the flrst step is to make sure that the recurrence
relation that we are trying to solve comes equipped with a clear indication
of the range of values of the subscript for which it is valid. In this case, the
recurrence (1.2.1) is clearly labeled in the parenthetical comment as being
valid for n = 0;1;2;::: Don’t settle for a recurrence that has an unqualifled
free variable.
Thenextstepistodeflnethegeneratingfunctionthatyouwilllookfor.
In this case, since we are looking forPa sequence a0;a1;a2;::: one natural
choice would be the function A(x) = a xj that we mentioned above.
j‚0 j
Next, take the recurrence relation (1.2.1), multiply both sides of it by
xn, and sum over all the values of n for which the relation is valid, which,
in this case, means sum from n = 0 to 1. Try to express the result of doing
that in terms of the function A(x) that you have just deflned.
If we do that to the left side of (1.2.1), the result is
a +a x+a x2 +a x3 +¢¢¢ = (A(x)¡a )=x
1 2 3 4 0
= (A(x)¡1)=x:
So much for the left side. What happens if we multiply the right side
of (1.2.1) byPxn and sum over nonnegative integers n? Evidently the result
is 2A(x)+ nxn. We need to identify the series
n‚0
X
nxn = x+2x2 +3x3 +4x4 +¢¢¢
n‚0
* If you are feeling rusty in the power series department, see chapter 2,
which contains a review of that subject.