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wilf on generating functions

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Textbook by Herbert S. Wilf (University of Pennsylvania), second edition with 1990 and 1994 copyright, apparently kept in Phil's math miscellany folder as a reference. Chapters cover recurrences, formal power series, exponential families and counting, applications such as the sieve method, the Snake Oil method, WZ pairs, and cycle indices, and analytic and asymptotic methods including Lagrange inversion. It also has exercises with solutions and a Maple/Mathematica appendix.

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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 proﬁt 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 inthe way generating functions give uniﬁed 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 diﬀerence equations into child’s play. Thensee 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 ﬁnding 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 ﬁrst place by generating function arguments. Also, even though bijectivearguments 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 oﬀering the best route to ﬁnding 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 whois 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 oﬀer (which is nowrepeated) 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 . . . . . . . . . . . . . 51.3 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 E x e r c i s e s ......................... 2 4 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 . . . . . . . . . . . . . . . . . . 522.6 Dirichlet series, formal theory . . . . . . . . . . . . . . . 56 E x e r c i s e s ......................... 6 5 Chapter 3: Cards, Decks, and Hands: The Exponential Formula 3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . 73 3.2 Deﬁnitions 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 . . . . . . . . . . . . . . . . . 863.11 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 E x e r c i s e s ......................... 1 0 4 vii Chapter 4: Applications of generating functions 4.1 Generating functions ﬁnd 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. . . . . . 1364.6 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 E x e r c i s e s ......................... 1 5 7 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 E x e r c i s e s ......................... 1 8 8 Appendix: Using Maple TMandMathematicaTM........ 1 9 2 Solutions ........................ 1 9 7 References ....................... 2 2 4 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, a0,a1,a2,.... We want to ‘know’ what the sequence is. What kind of an answer might we expect? A simple formula for anwould be the best that could be hoped for. If we ﬁnd that an=n2+ 3 for each n=0,1,2,..., then there’s no doubt that 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, ..., whereanis thenth prime number. Well then, it would 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 thequestion, we might be able to give a simple formula for the sum of a power series, whose coeﬃcients are the sequence that we’re looking for . For instance, suppose we want the Fibonacci numbers F 0,F1,F2,..., 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-complicated formulas for Fn, as we will see later, in example 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, Fn, is the coeﬃcient of xnin the expansion of the functionx/(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 unsatisﬁed 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 exactformulas 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 ﬁnding such a formula. (b)Find a recurrence formula. Most often generating functions arise from recurrence formulas. Sometimes, however, from the generating function you will ﬁnd a new recurrence formula, notthe 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, one is dealing with a very diﬃcult sequence, and instead of looking for an exact formula, which might be out of the question, we look for an approximate formula. While we would not expect, for example, to ﬁnd an exact formula for the nth prime number, it is a beautiful fact (the ‘Prime Number Theorem’) that the nth prime is approximately nlognwhennis 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 ﬁrst, and then decreases steadily. Many combinatorial sequences are unimodal, and a variety of methods are available for proving such theorems. Generating func-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 coeﬃcients. 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 referto 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 n/summationdisplay j=0/parenleftbiggn j/parenrightbigg2 =/parenleftbigg2n n/parenrightbigg (n=0,1,2,...). One way to prove such identities is to consider the generating function whose coeﬃcients 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 oﬀer 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 relatedto another one, which you never suspected before. It is noteworthy that in this way you may ﬁnd 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 proﬁtably thought about from the point of view of generating functions. We hope that after studying these examples thereader will be at least partly convinced of the power of the method, as well as of the beauty of the uniﬁed approach. 1.1 An easy two term recurrence A certain sequence of numbers a 0,a1,...satisﬁes the conditions an+1=2an+1 (n≥0;a0=0 ). (1.1.1) 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 an=2n−1(n≥0), and prove it quickly, by induction based on the 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, instead of ﬁnding the sequence {an}, let’s ﬁnd the gener- ating function A(x)=/summationtext n≥0anxn. Once we know what that function is, we will be able to read oﬀ the explicit formula for the an’s by expanding A(x) in a series. To ﬁndA(x), multiply both sides of the recurrence relation (1.1.1) by xnand sum over the values of nfor which the recurrence is valid, namely, overn≥0. Then try to relate these sums to the unknown generating functionA(x). 4 1 Introductory ideas and examples If we do this ﬁrst to the left side of (1.1.1), there results/summationtext n≥0an+1xn. 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, /summationdisplay n≥0an+1xn=a1+a2x+a3x2+a4x3+··· ={(a0+a1x+a2x2+a3x3+···)−a0}/x =A(x)/x sincea0= 0 in this problem. Hence the result of so operating on the left side of (1.1.1) is A(x)/x. Next do the right side of (1.1.1). Multiply it by xnand sum over all n≥0. The result is /summationdisplay n≥0(2an+1 )xn=2A(x)+/summationdisplay n≥0xn =2A(x)+1 1−x, wherein we have used the familiar geometric series evaluation/summationtext n≥0xn= 1/(1−x), which is valid for |x|<1. If we equate the results of operating on the two sides of (1.1.1), we ﬁnd that A(x) x=2A(x)+1 1−x, which is trivial to solve for the unknown generating function A(x), in the form A(x)=x (1−x)(1−2x). This is the generating function for the problem. The unknown numbers anare arranged neatly on this clothesline: anis the coeﬃcient of xnin the series expansion of the above A(x). Suppose we want to ﬁnd an explicit formula for the an’s. Then we would have to expand A(x) in a series. That isn’t hard in this example, since the partial fraction expansion is x (1−x)(1−2x)=x/braceleftbigg2 1−2x−1 1−x/bracerightbigg ={2x+22x2+23x3+24x4+··· } −{x+x2+x3+x4+··· } =( 2−1)x+( 22−1)x2+( 23−1)x3+( 24−1)x4+··· It is now clear that the coeﬃcient of xn, i.e.an, is equal to 2n−1, for each 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 a0,a1,...satisﬁes the conditions an+1=2an+n (n≥0;a0=1 ). (1.2.1) Find the sequence. As before, we might calculate the ﬁrst several members of the sequence, 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 generatingfunctions. That means that instead of looking for the sequencea 0,a1,..., we will look for the functionA(x)=/summationtext j≥0ajxj. Once we have found the function, the sequence will be identiﬁable as the sequence of power series coeﬃcients of the function.* As in example 1, the ﬁrst 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 forn=0,1,2,...Don’t settle for a recurrence that has an unqualiﬁed free variable. The next step is to deﬁne the generating function that you will look for. In this case, since we are looking for a sequence a0,a1,a2,...one natural choice would be the function A(x)=/summationtext j≥0ajxjthat we mentioned above. Next, take the recurrence relation (1.2.1), multiply both sides of it by xn, and sum over all the values of nfor which the relation is valid , which, in this case, means sum from n=0t o ∞. Try to express the result of doing that in terms of the function A(x) that you have just deﬁned. If we do that to the left side of (1.2.1), the result is a1+a2x+a3x2+a4x3+···=(A(x)−a0)/x =(A(x)−1)/x. So much for the left side. What happens if we multiply the right side of (1.2.1) by xnand sum over nonnegative integers n? Evidently the result is 2A(x)+/summationtext n≥0nxn. We need to identify the series /summationdisplay n≥0nxn=x+2x2+3x3+4x4+··· * If you are feeling rusty in the power series department, see chapter 2, which contains a review of that subject. 6 1 Introductory ideas and examples There are two ways to proceed: (a) look it up (b) work it out. To work it out we use the following stunt, which seems artiﬁcial if you haven’t seen it before, but after using it 4993 times it will seem quite routine: /summationdisplay n≥0nxn=/summationdisplay n≥0x(d dx)xn=x(d dx)/summationdisplay n≥0xn=x(d dx)1 1−x=x (1−x)2. (1.2.2) In other words, the series that we are interested in is essentially the derivative of the geometric series, so its sum is essentially the derivative of the sum of the geometric series. This raises some nettlesome questions, which we will mention here and deal with later. For what values of xis (1.2.2) valid? The geometric series converges only for |x|<1, so the ana- lytic manipulation of functions in (1.2.2) is legal only for those x. However, often the analytic nature of the generating function doesn’t interest us; we love it only for its role as a clothesline on which our sequence is hanging out to dry. In such cases we can think of a generating function as only a formal power series, i.e., as an algebraic object rather than as an analytic one. Then (1.2.2) would be valid as an identity in the ring of formal power series, which we will discuss later, and the variable xwouldn’t need to be qualiﬁed at all. Anyway, the result of multiplying the right hand side of (1.2.1) by xn and summing over n≥0i s2A(x)+x/(1−x)2, and if we equate this with our earlier result from the left side of (1.2.1), we ﬁnd that (A(x)−1) x=2A(x)+x (1−x)2, (1.2.3) and we’re ready for the easy part, which is to solve (1.2.3) for the unknown A(x), getting A(x)=1−2x+2x2 (1−x)2(1−2x). (1.2.4) Exactly what have we learned? The original problem was to ‘ﬁnd’ the numbers {an}that are determined by the recurrence (1.2.1). We have, in a certain sense, ‘found’ them: the number anis the coeﬃcient of xnin the power series expansion of the function (1.2.4). This is the end of the ‘ﬁnd-the-generating-function’ part of the method. We have it. What we do with it depends on exactly why we wanted to know the solution of (1.2.1) in the ﬁrst place. Suppose, for example, that we want an exact, simple formula for the membersanof the unknown sequence. Then the method of partial fractions will work here, just as it did in the ﬁrst example, but its application is now a little bit trickier. Let’s try it and see. The ﬁrst step is to expand the right side of (1.2.4) in partial fractions. Such a fraction is guaranteed to be expandable in partial fractions in the 1.2 A slightly harder two term recurrence 7 form 1−2x+2x2 (1−x)2(1−2x)=A (1−x)2+B 1−x+C 1−2x, (1.2.5) and the only problem is how to ﬁnd the constants A,B,C . Here’s the quick way. First multiply both sides of (1.2.5) by (1 −x)2, and then let x= 1. The instant result is that A=−1 (don’t take my word for it, try it for yourself!). Next multiply (1.2.5) through by 1 −2xand letx=1/2. The instant result is that C= 2. The hard one to ﬁnd is B, so let’s do that one by cheating. Since we know that (1.2.5) is an identity,i.e., is true for all values of x, let’s choose an easy value of x,s a yx=0 , and substitute that value of xinto (1.2.5). Since we now know AandC, we ﬁnd at once that B=0 . We return now to (1.2.5) and insert the values of A,B,C that we just found. The result is the relation A(x)=1−2x+2x 2 (1−x)2(1−2x)=(−1) (1−x)2+2 1−2x. (1.2.6) What we are trying to do is to ﬁnd an explicit formula for the coeﬃcient ofxnin the left side of (1.2.6). We are trading that in for two easier problems, namely ﬁnding the coeﬃcient of xnin each of the summands on the right side of (1.2.6). Why are they easier? The term 2 /(1−2x), for instance, expands as a geometric series. The coeﬃcient of xnthere is just 2·2n=2n+1. The series ( −1)/(1−x)2was handled in (1.2.2) above, and its coeﬃcient of xnis−(n+ 1). If we combine these results we see that our unknown sequence is an=2n+1−n−1(n=0,1,2,...). Having done all of that work, it’s time to confess that there are better ways to deal with recurrences of the type (1.2.1), without using generating functions.* However, the problem remains a good example of how gener- ating functions can be used, and it underlines the fact that a single uniﬁed method can replace a lot of individual special techniques in problems about sequences. Anyway, it won’t be long before we’re into some problems that essentially cannot be handled without generating functions. It’s time to introduce some notation that will save a lot of words in the sequel. Deﬁnition. Letf(x)be a series in powers of x. Then by the symbol [xn]f(x)we will mean the coeﬃcient of xnin the series f(x). Here are some examples of the use of this notation. [xn]ex=1/n!; [tr]{1/(1−3t)}=3r;[um](1 +u)s=/parenleftbiggs m/parenrightbigg . * See, for instance, chapter 1 of my book [Wi2]. 8 1 Introductory ideas and examples A perfectly obvious property of this symbol, that we will use repeatedly, is [xn]{xaf(x)}=[xn−a]f(x). (1.2.7) Another property of this symbol is the convention that if βis any real number, then [βxn]f(x)=( 1/β)[xn]f(x), (1.2.8) so, for instance, [ xn/n!]ex= 1 for alln≥0. Before we move on to the next example, here is a summary of the method of generating functions as we have used it so far. THE METHOD Given: a recurrence formula that is to be solved by the method of generating functions. 1. Make sure that the set of values of the free variable (say n) for which the given recurrence relation is true, is clearly delineated. 2. Give a name to the generating function that you will look for, and write out that function in terms of the unknown sequence (e.g., call itA(x), and deﬁne it to be/summationtext n≥0anxn). 3. Multiply both sides of the recurrence by xn, and sum over all values ofnfor which the recurrence holds. 4. Express both sides of the resulting equation explicitly in terms of your generating function A(x). 5. Solve the resulting equation for the unknown generating function A(x). 6. If you want an exact formula for the sequence that is deﬁned by the given recurrence relation, then attempt to get such a formula by expanding A(x) into a power series by any method you can think of. In particular, if A(x) is a rational function (quotient of two polynomials), then success will result from expanding in partial fractions and then handling each of the resulting termsseparately. 1.3 A three term recurrence Now let’s do the Fibonacci recurrence F n+1=Fn+Fn−1. (n≥1;F0=0 ;F1=1 ). (1.3.1) Following ‘The Method,’ we will solve for the generating function F(x)=/summationdisplay n≥0Fnxn. 1.3 A three term recurrence 9 To do that, multiply (1.3.1) by xn, and sum over n≥1. We ﬁnd on the left side F2x+F3x2+F4x3+···=F(x)−x x, and on the right side we ﬁnd {F1x+F2x2+F3x3+··· }+{F0x+F1x2+F2x3+··· }={F(x)}+{xF(x)}. (Important: Try to do the above yourself, without peeking, and see if you get the same answer.) It follows that ( F−x)/x=F+xF, and therefore that the unknown generating function is now known, and it is F(x)=x 1−x−x2. Now we will ﬁnd some formulas for the Fibonacci numbers by expand- ingx/(1−x−x2) in partial fractions. The success of the partial fraction method is greatly enhanced by having only linear (ﬁrst degree) factors in the denominator, whereas what we now have is a quadratic factor. So let’s factor it further. We ﬁnd that 1−x−x2=( 1−xr+)(1−xr−)(r±=( 1±√ 5)/2) and sox 1−x−x2=x (1−xr+)(1−xr−) =1 (r+−r−)/parenleftbigg1 1−xr+−1 1−xr−/parenrightbigg =1√ 5/braceleftbigg/summationdisplay j≥0rj +xj−/summationdisplay j≥0rj −xj/bracerightbigg , thanks to the magic of the geometric series. It is easy to pick out the coeﬃcient of xnand ﬁnd Fn=1√ 5(rn +−rn −)(n=0,1,2,...)( 1 .3.3) as an explicit formula for the Fibonacci numbers Fn. This example oﬀers us a chance to edge a little further into what gen- erating functions can tell us about sequences, in that we can get not only the exact answer, but also an approximate answer, valid when nis large. Indeed, when nis large, since r+>1 and |r−|<1, the second term in (1.3.3) will be minuscule compared to the ﬁrst, so an extremely good ap- proximation to Fnwill be Fn∼1√ 5/parenleftBigg 1+√ 5 2/parenrightBiggn . (1.3.4) 10 1 Introductory ideas and examples But, you may ask, why would anyone want an approximate formula when an exact one is available? One answer, of course, is that sometimes exact answers are fearfully complicated, and approximate ones are more revealing. Even in this case, where the exact answer isn’t very complex, we can still learn something from the approximation. The reader should take a few moments to verify that, by neglecting the second term in (1.3.3), we neglect a quantity that is never as large as 0.5 in magnitude, and conse- quently not only is Fnapproximately given by (1.3.4), it is exactly equal to the integer nearest to the right side of (1.3.4). Thus consideration of anapproximate formula has found us a simpler exact formula! 1.4 A three term boundary value problem This example will diﬀer from the previous ones in that the recurrence relation involved does not permit the direct calculation of the members of the sequence, although it does determine the sequence uniquely. The situation is similar to the following: suppose we imagine the Fibonacci re- currence, together with the additional data F 0= 1 andF735= 1. Well then, the sequence {Fn}would be uniquely determined, but you wouldn’t be able to compute it directly by recurrence because you would not be in possession of the two consecutive values that are needed to get the recurrence started. We will consider a slightly more general situation. It consists of the recurrence aun+1+bun+cun−1=dn (n=1,2,...,N −1;u0=uN= 0) (1.4.1) where the positive integer N, the constants a,b,cand the sequence {dn}N−1 n=1 are given in advance. The equations (1.4.1) determine the sequence {ui}N 0 uniquely, as we will see, and the method of generating functions gives us a powerful way to attack such boundary value problems as this, which arise in numerous applications, such as the theory of interpolation by spline func-tions. To begin with, we will deﬁne two generating functions. One of them is our unknown U(x)=/summationtext N j=0ujxj, and the second one is D(x)=/summationtextN−1 j=1djxj, and it is regarded as a known function (did we omit any given values of the dj’s, liked0?o rdN? Why?). Next, following the usual recipe, we multiply the recurrence (1.4.1) by xnand sum over the values of nfor which the recurrence is true, which in this case means that we sum from n=1t oN−1. This yields aN−1/summationdisplay n=1un+1xn+bN−1/summationdisplay n=1unxn+cN−1/summationdisplay n=1un−1xn=N−1/summationdisplay n=1dnxn. If we now express this equation in terms of our previously deﬁned generating functions, it takes the form a x{U(x)−u1x}+bU(x)+cx{U(x)−uN−1xN−1}=D(x). (1.4.2) 1.4 A three term boundary value problem 11 Next, with only a nagging doubt because u1anduN−1are unknown, we press on with the recipe, whose next step asks us to solve (1.4.2) for the unknown generating function U(x). Now that isn’t too hard, and we ﬁnd at once that {a+bx+cx2}U(x)=x{D(x)+au1+cuN−1xN}. (1.4.3) The unknown generating function U(x) is now known except for the two still-unknown constants u1anduN−1, but (1.4.3) suggests a way to ﬁnd them, too. There are two values of x, call them r+andr−, at which the quadratic polynomial on the left side of (1.4.3) vanishes. Let us suppose thatrN +/negationslash=rN −, for the moment. If we let x=r+in (1.4.3), we obtain one equation in the two unknowns u1,uN−1, and if we let x=r−, we get another. The two equations are au1+(crN +)uN−1=−D(r+) au1+(crN −)uN−1=−D(r−).(1.4.4) Once these have been solved for u1anduN−1, equation (1.4.3) then gives U(x) quite explicitly and completely. We leave the exceptional case where rN +=rN −to the reader. Here is an application∗of these results to the theory of spline interpo- lation. Suppose we are given a table of values y0,y1,...,y nof some function y(x), at a set of equally spaced points ti=t0+ih(0≤i≤n). We want to construct a smooth function S(x) that ﬁts the data, subject to the following conditions: (i) Within each interval ( ti,ti+1)(i=0,...,n −1) our function S(x)i s to be a cubic polynomial (a diﬀerent one in each interval!); (ii) The functions S(x),S/prime(x) andS/prime/prime(x) are to be continuous on the whole interval [ t0,tn]; (iii)S(ti)=yifori=0,...,n . A function S(x) that satisﬁes these conditions is called a cubic spline . Suppose our unknown spline S(x) is given by S0(x), ifx∈[t0,t1],S1(x), if x∈[t1,t2],...,Sn−1(x), ifx∈[tn−1,tn], and we want now to determine all of the cubic polynomials S0,...,S n−1. To do this we have 2 ninterpolatory conditions Si−1(ti)=yi=Si(ti)(i=1,...,n −1);S0(t0)=y0;Sn−1(tn)=yn (1.4.5) along with 2 n−2 continuity conditions S/prime i−1(ti)=S/prime i(ti);S/prime/prime i−1(ti)=S/prime/prime i(ti)(i=1,...,n −1). (1.4.6) ∗This application is somewhat specialized, and may be omitted at a ﬁrst reading. 12 1 Introductory ideas and examples There are altogether 4 n−2 conditions to satisfy. We have ncubic polyno- mials to be determined, each of which has 4 coeﬃcients, for a total of 4 n unknown parameters. Since the conditions are linear, such a spline S(x) surely exists and we can expect it to have two free parameters. It is con- ventional to choose these so that S(x) has a point of inﬂection at t0and at tn. Now here is the solution. The functions Si(x) are given by Si(x)=1 6h/parenleftbig zi(ti+1−x)3+zi+1(x−ti)3+( 6yi+1−h2zi+1)(x−ti) +( 6yi−h2zi)(ti+1−x)/parenrightbig (i=0,1,...,n −1), (1.4.7) provided that the numbers z1,...,z n−1satisfy the simultaneous equations zi−1+4zi+zi+1=6 h2(yi+1−2yi+yi−1)(i=1,2,...,n −1) (1.4.8) in whichz0=zn= 0. It is easy to check this, by substituting x=tiand x=ti+1into (1.4.7) to verify that (1.4.5) and (1.4.6) are satisﬁed. Hence it remains only to solve the equations (1.4.8). The system of equations (1.4.8) is of the form (1.4.1), hence we can ﬁnd the solutions from (1.4.3), (1.4.4). To do this, begin with the given set of points {(ti,yi)}n i=0, through which we wish to interpolate. Use them to write down D(x)=6 h2n−1/summationdisplay i=1(yi+1−2yi+yi−1)xi. (1.4.9) Since (a,b,c )=( 1,4,1) in this example, we have r±=−2±√ 3. Now our unknown generating function U(x) is given by (1.4.3), which reads as U(x)=x(D(x)+z1+zn−1xn) (1 + 4x+x2), (1.4.10) in which the unknown numbers z1,zn−1are determined by the requirement that the right side of (1.4.10) be a polynomial, or equivalently by the two equations (1.4.4), which become z1+(√ 3−2)nzn−1=−D(√ 3−2) z1+(−√ 3−2)nzn−1=−D(−√ 3−2).(1.4.11) When we know U(x), which is, after all,/summationtextn−1 i=1zixi, we can read oﬀ its coeﬃcients to ﬁnd the z’s, and use them in (1.4.7) to ﬁnd the interpolating spline. 1.4 A three term boundary value problem 13 Example. Now let’s try an example with real live numbers in it. Suppose we are trying to ﬁt the powers of 2 by a cubic spline on the interval [0 ,5]. Our input data are yi=2ifori=0,1,..., 5,h= 1, andn= 5. From (1.4.9) we ﬁnd that D(x)=6x(1 + 2x+4x2+8x3). Then we solve (1.4.11) to ﬁnd thatz1= 204/209 andz4= 2370/209. Next (1.4.10) tells us that U(x)=204x 209+438x2 209+552x3 209+2370x4 209, and now we know all of the zi’s. Finally, (1.4.7) tells us the exact cubic polynomials that form the spline. For example, S0(x), which lives on the subinterval [0 ,1], is S0(x)=1+175 209x+34 209x3. Note thatS0(0) = 1 and S0(1) = 2, so it correctly ﬁts the data at the endpoints of its subinterval, and that S/prime/prime 0(0) = 0, so the ﬁt will have an inﬂection point at the origin. The reader is invited to ﬁnd all of the Si(x) (i=0,1,..., 5), in this example, and check that they smoothly ﬁt into each other at the points 1 ,2,3,4, in the sense that the functions and their ﬁrst two derivatives are continuous. One reason why you might like to ﬁt some numerical data with a spline is because you want to integrate the function that the data represent. Integration of (1.4.7) from ti=ihtoti+1=(i+1 )hshows that /integraldisplayti+1 tiSi(x)dx=h 2(yi+yi+1)−h3 24(zi+zi+1). (1.4.12) Thus, ﬁtting some data by a spline and integrating the spline amounts to numerical integration by the trapezoidal rule with a third order correction term. If we sum (1.4.12) over i=0,...,n −1 we get for the overall integral, /integraldisplaynh 0S(x)dx= trap −h 12(y0−y1−yn−1+yn)−h3 72(z1+zn−1)( 1.4.13) in which ‘trap’ is the trapezoidal rule, and z1,zn−1satisfy (1.4.11). Interpolation by spline functions is an important subject. It occurs in the storage of computer fonts, such as the one that you are now reading. Did you ever wonder how the shapes of the letters in the fonts are actually stored in a computer? One way is by storing the parameters of spline functions that ﬁt the contours of the letters in the font. 14 1 Introductory ideas and examples 1.5 Two independent variables In this section we will see how generating functions can be helpful in problems that involve functions of two discrete variables. We will use the opportunity also to introduce the binomial coeﬃcients, since they are surely one of the most important combinatorial counting sequences. Letnandkbe integers such that 0 ≤k≤n. In how many ways can we choose a subset of kobjects from the set {1,2,...,n }? Let’s pretend that we don’t know how this will turn out, and allow generating functions to help us ﬁnd the answer. Supposef(n,k) is the answer to the question. We imagine that the collection of all possible subsets of kof thesenobjects are in front of us, and we will divide them into two piles. In the ﬁrst pile we put all of those subsets that docontain the object ‘ n’, and into the second pile we put all subsets that do not contain ‘n’. The ﬁrst of these piles obviously contains f(n−1,k−1) subsets. The second pile contains f(n−1,k) subsets. The two piles together originally contained f(n,k) subsets. So it must be that our unknown numbers f(n,k) satisfy the recurrence f(n,k)=f(n−1,k)+f(n−1,k−1) (f(n,0) = 1). (1.5.1) To ﬁnd formulas for these numbers we use (what else?) generating functions. For each n=0,1,2,...deﬁne the generating function Bn(x)=/summationdisplay k≥0f(n,k)xk. Now multiply (1.5.1) throughout by xkand sum over k≥1. The result is thatBn(x)−1=(Bn−1(x)−1) +xBn−1(x), forn≥1, withB0(x)=1 . Hence Bn(x)=( 1+x)Bn−1(x)(n≥1;B0(x)=1 ). (1.5.2) ThusBn(x)=( 1+x)n, for alln≥0.Our unknown number f(n,k) is revealed to be the coeﬃcient of xkin the polynomial (1 +x)n. To ﬁnd a formula for f(n,k) we might, for example, use Taylor’s formula, which would tell us that f(n,k) is thekth derivative of (1 + x)n, evaluated at x= 0, all divided by k!. The diﬀerentiation is simple to do. Indeed, the kth derivative of (1 + x)nisn(n−1)···(n−k+ 1)(1 +x)n−k. If we put x= 0 and divide by k! we quickly discover that f(n,k), the number of k-subsets of nthings, is given by /parenleftbiggn k/parenrightbigg =n! k!(n−k)!=n(n−1)(n−2)···(n−k+1 ) k!(1.5.3) for integers n,kwith 0 ≤k≤n. 1.5 Two independent variables 15 That pretty well takes care of the binomial coeﬃcients/parenleftbign k/parenrightbig when 0 ≤ k≤nandn,kare integers. When kis a negative integer the binomial coeﬃcient/parenleftbign k/parenrightbig =0 . Although the second member of equation (1.5.3) is diﬃcult to decipher ifnis not a nonnegative integer, the third member isn’t a bit hard to understand, even if nis a complex number, so long as kis a nonnegative integer. So that gives us an extension of the deﬁnition of the binomial coeﬃcients to arbitrary complex numbers n, namely, /parenleftbiggn k/parenrightbigg =n(n−1)(n−2)···(n−k+1 ) k!(integerk≥0). (1.5.4) Thus/parenleftbig−3 3/parenrightbig =(−3)(−4)(−5)/6=−10, and/parenleftbigi 2/parenrightbig =i(i−1)/2=(−1−i)/2, etc. The generating function Bn(x)=( 1+x)n=/summationdisplay k≥0/parenleftbiggn k/parenrightbigg xk=1+nx+n(n−1) 2x2+··· remains valid for all complex numbers n: the series terminates if nis a nonnegative integer, and it converges for |x|<1 in any case. What is the support of/parenleftbign k/parenrightbig ? That is, for which values of n,kis it true that/parenleftbign k/parenrightbig /negationslash= 0? First, kmust be a nonnegative integer. If nis not a nonnegative integer then/parenleftbign k/parenrightbig is surely nonzero, by (1.5.4). If nis a nonnegative integer then (1.5.4) shows that/parenleftbign k/parenrightbig /negationslash=0i ﬀ0 ≤k≤n(in accordance with the combinatorial deﬁnition!). A very important consequence of these facts is that if nis a nonnegative integer, instead of writing something like/summationtextn k=0/parenleftbign k/parenrightbig xk, we can equally well write/summationtext∞ k=−∞/parenleftbign k/parenrightbig xk, because the binomial coeﬃcients vanish on all of the seemingly extra values of kthat appear in the second form of the sum. The binomial coeﬃcients “cut oﬀ” the sum by themselves, so there is no need to do it again with the range of summation. That being the case, we introduce two conventions that we will adhere to throughout the book. Convention 1. When the range of a variable that is being summed over is not speciﬁed, it is to be understood that the range of summation is from −∞ to+∞. Convention 2. When the range of a free variable in an equation is not speciﬁed, it is to be understood that the equation holds for all integer values of that variable. For example, we will write/summationtext k/parenleftbign k/parenrightbig xk=( 1+x)n. These conventions will save us an enormous amount of work in the sequel, mainly in that we won’t have to worry about changing the limits of summation when we change the variable of summation by a constant shift. 16 1 Introductory ideas and examples Let’s look at generating functions of some other kinds. If we multiply Bn(x)b yynand sum only over n≥0, we ﬁnd that /summationdisplay n≥0Bn(x)yn=/summationdisplay n≥0/summationdisplay k/parenleftbiggn k/parenrightbigg xkyn=/summationdisplay n≥0(1 +x)nyn=1 1−y(1 +x). Thus for integer n≥0,/parenleftbign k/parenrightbig =[xkyn](1−y(1 +x))−1. For another exercise, let’s evaluate, for nonnegative integer k, the sum/summationtext n/parenleftbign k/parenrightbig yn. Note that the index nruns over all integers, but that the sum- mand vanishes unless n≥k. That sum is clearly [xk]/summationdisplay n≥0/summationdisplay k/parenleftbiggn k/parenrightbigg xkyn=[xk]1 1−y(1 +x)=1 1−y[xk]1 1−(y 1−y)x =1 1−y/parenleftbigy 1−y/parenrightbigk=yk (1−y)k+1. For future reference we will place side-by-side these two power series gen- erating functions of the binomial coeﬃcients: /summationdisplay k/parenleftbiggn k/parenrightbigg xk=( 1+x)n;/summationdisplay n/parenleftbiggn k/parenrightbigg yn=yk (1−y)k+1. (1.5.5) 1.6 Another 2-variable case. This example will have a stronger combinatorial ﬂavor than the pre- ceding ones. It concerns the partitions of a set. By a partition of a set S we will mean a collection of nonempty, pairwise disjoint sets whose union is S. Another name for a partition of Sis an equivalence relation onS. The sets into which Sis partitioned are called the classes of the partition. For instance, we can partition [5]∗in several ways. One of them is as {123}{4}{5}. In this partition there are three classes, one of which contains 1 and 2 and 3, another of which contains only 4, while the other contains only 5. No signiﬁcance attaches to the order of the elements within the classes, nor to the order of the classes. All that matters is ‘who is together and who is apart.’ Here is a list of allof the partitions of [4] into 2 classes: {12}{34};{13}{24};{14}{23};{123}{4};{124}{3};{134}{2};{1}{234}. (1.6.1) There are exactly 7 partitions of [4] into 2 classes. The problem that we will address in this example is to discover how many partitions of [ n] intokclasses there are. Let/braceleftbign k/bracerightbig denote this number. ∗Recall that [ n] is the set {1,2,...,n } 1.6 Another 2-variable case. 17 It is called the Stirling number of the second kind. Our list above shows that/braceleftbig4 2/bracerightbig =7 . To ﬁnd out more about these numbers we will follow the method of generating functions. First we will ﬁnd a recurrence relation, then a few generating functions, then some exact formulas, etc. We begin with a recurrence formula for/braceleftbign k/bracerightbig , and the derivation will be quite similar to the one used in the previous example. Let positive integers n,kbe given. Imagine that in front of you is the collection of all possible partitions of [ n] intokclasses. There are exactly/braceleftbign k/bracerightbig of them. As in the binomial coeﬃcient example, we will carve up this collection into two piles; into the ﬁrst pile go all of those partitions of [ n] intokclasses in which the letter nlives in a class all by itself. Into the second pile go all other partitions, i.e., those in which the highest letter n lives in a class with other letters. The question is, how many partitions are there in each of these two piles (expressed in terms of the Stirling numbers)? Consider the ﬁrst pile. There, every partition has nliving alone. Imag- ine marching through that pile and erasing the class ‘( n)’ that appears in every single partition in the pile. If that were done, then what would re- main after the erasures is exactly the complete collection of all partitions of [n−1] intok−1 classes. There are/braceleftbign−1 k−1/bracerightbig of these, so there must have been/braceleftbign−1 k−1/bracerightbig partitions in the ﬁrst pile. That was the easy one, but now consider the second pile. There the letternalways lives in a class with other letters. Therefore, if we march through that pile and erase the letter nwherever it appears, we won’t aﬀect the numbers of classes; we’ll still be looking at partitions with kclasses. After erasing the letter nfrom everything, our pile now contains partitions ofn−1 letters into kclasses. However, each one of these partitions appears not just once, but several times. For example, in the list (1.6.1), the second pile contains the partitions {12}{34};{13}{24};{14}{23};{124}{3};{134}{2};{1}{234},(1.6.2) and after we delete ‘4’ from every one of them we get the list {12}{3};{13}{2};{1}{23};{12}{3};{13}{2};{1}{23}. What we are looking at is the list of all partitions of [3] into 2 classes, where each partition has been written down twice. Hence this list contains exactly 2/braceleftbig3 2/bracerightbig partitions. In the general case, after erasing nfrom everything in the second pile, we will be looking at the list of all partitions of [ n−1] intokclasses, where every such partition will have been written down ktimes. Hence that list will contain exactly k/braceleftbign−1 k/bracerightbig partitions. 18 1 Introductory ideas and examples Therefore the second pile must have also contained k/braceleftbign−1 k/bracerightbig partitions before the erasure of n. The original list of/braceleftbign k/bracerightbig partitions was therefore split into two piles, the ﬁrst of which contained/braceleftbign−1 k−1/bracerightbig partitions and the second of which contained k/braceleftbign−1 k/bracerightbig partitions. It must therefore be true that /braceleftbiggn k/bracerightbigg =/braceleftbiggn−1 k−1/bracerightbigg +k/braceleftbiggn−1 k/bracerightbigg ((n,k) =????). To determine the range of nandk, let’s extend the deﬁnition of/braceleftbign k/bracerightbig to all pairs of integers. We put/braceleftbign k/bracerightbig =0i fk>n orn< 0o rk< 0. Further,/braceleftbign 0/bracerightbig =0i fn/negationslash= 0, and we will take/braceleftbig0 0/bracerightbig = 1. With those conventions, the recurrence above is valid for all ( n,k) other than (0 ,0), and we have /braceleftbiggn k/bracerightbigg =/braceleftbiggn−1 k−1/bracerightbigg +k/braceleftbiggn−1 k/bracerightbigg ((n,k)/negationslash=( 0,0);/braceleftbigg0 0/bracerightbigg =1 ).(1.6.3) The stage is now set for ﬁnding the generating functions. Again there are three natural candidates for generating functions that might be ﬁnd- able, namely An(y)=/summationdisplay k/braceleftbiggn k/bracerightbigg yk Bk(x)=/summationdisplay n/braceleftbiggn k/bracerightbigg xn C(x,y)=/summationdisplay n,k/braceleftbiggn k/bracerightbigg xnyk.(1.6.4) Before we plunge into the calculations, let’s pause for a moment to develop some intuition about which of these choices is likely to succeed. To ﬁndAn(y) will involve multiplying (1.6.3) by ykand summing over k.D o you see any problems with that? Well, there are some, and they arise from the factor of kin the second term on the right. Indeed, after multiplying by ykand summing over kwe will have to deal with something like/summationtext kk/braceleftbign k/bracerightbig yk. This is certainly possible to think about, since it is related to the derivativeofA n(y), but we do have a complication here. If instead we choose to ﬁnd Bk(x), we multiply (1.6.3) by xnand sum onn. Then the factor of kthat seemed to be troublesome is not involved in the sum, and we can take that koutside of the sum as a multiplicative factor. Comparing these, let’s vote for the latter approach, and try to ﬁnd the functionsBk(x)(k≥0). Hence, multiply (1.6.3) by xnand sum over n, to get Bk(x)=xBk−1(x)+kxB k(x)(k≥1;B0(x)=1 ). 1.6 Another 2-variable case. 19 This leads to Bk(x)=x 1−kxBk−1(x)(k≥1;B0(x)=1 ) and ﬁnally to the evaluation Bk(x)=/summationdisplay n/braceleftbiggn k/bracerightbigg xn=xk (1−x)(1−2x)(1−3x)···(1−kx)(k≥0). (1.6.5) The problem of ﬁnding an explicit formula for the Stirling numbers could therefore be solved if we could ﬁnd the power series expansion of the function that appears in (1.6.5). That, in turn, calls for a dose of partial fractions, notof Taylor’s formula! The partial fraction expansion in question has the form 1 (1−x)(1−2x)···(1−kx)=k/summationdisplay j=1αj (1−jx). To ﬁnd the α’s, ﬁxr,1≤r≤k, multiply both sides by 1 −rx, and let x=1/r. The result is that αr=1 (1−1/r)(1−2/r)···(1−(r−1)/r)(1−(r+1 )/r)···(1−k/r) =(−1)k−rrk−1 (r−1)!(k−r)!(1≤r≤k). (1.6.6) From (1.6.5) and (1.6.6) we obtain, for n≥k, /braceleftbiggn k/bracerightbigg =[xn]/braceleftbiggxk (1−x)(1−2x)···(1−kx)/bracerightbigg =[xn−k]/braceleftbigg1 (1−x)(1−2x)···(1−kx)/bracerightbigg =[xn−k]k/summationdisplay r=1αr 1−rx(k≥1) =k/summationdisplay r=1αr[xn−k]1 1−rx =k/summationdisplay r=1αrrn−k =k/summationdisplay r=1(−1)k−rrk−1 (r−1)!(k−r)!rn−k =k/summationdisplay r=1(−1)k−rrn r!(k−r)!(n,k≥0),(1.6.7) 20 1 Introductory ideas and examples which is just what we wanted: an explicit formula for/braceleftbign k/bracerightbig . Do check that this formula yields/braceleftbig4 2/bracerightbig = 7, which we knew already. At the same time, note that the formula says that/braceleftbign 2/bracerightbig =2n−1−1(n>0). Can you give an independent proof of that fact? Next, we’re going to try one of the other approaches to solving the recurrence for the Stirling numbers, namely that of studying the functions An(y) in (1.6.4). This method is much harder to carry out to completion than the one we just used, but it turns out that these generating functions have other uses that are quite important from a theoretical point of view. Therefore, let’s ﬁx n>0, multiply (1.6.3) by ykand sum over k. The result is An(y)=/summationdisplay k/braceleftbiggn−1 k−1/bracerightbigg yk+/summationdisplay kk/braceleftbiggn−1 k/bracerightbigg yk =yAn−1(y)+(yd dy)An−1(y) ={y(1 +Dy)}An−1(y)(n>0;A0(y)=1 ).(1.6.8) The novel feature is the appearance of the diﬀerentiation operator d/dy that was necessitated by the factor kin the recurrence relation. Hence each function Anis obtained from its predecessor by applying the operator y(1 +Dy). Beginning with A0= 1, we obtain successively y,y+y2,y+3y2+y3,..., but as far as an explicit formula is concerned, we ﬁnd only that An(y)={y+yDy}n1(n≥0) (1 .6.9) by this approach. There are, however, one or two things that can be seen more clearly from these generating functions than from the Bn(x)’s. One of these will be discussed in section 4.5, and concerns the shape of the sequence/braceleftbign k/bracerightbig for ﬁxedn,a skruns from 1 to n. It turns out that the sequence increases for a while and then it decreases. That is, it has just one maximum. Many combinatorial sequences are unimodal , like this one, but in some cases it can be very hard to prove such things. In this case, thanks to the formula (1.6.9), we will see that it’s not hard at all. For an application of (1.6.7), recall that the Stirling number/braceleftbign k/bracerightbig is the number of ways of partitioning a set of nelements into kclasses. Suppose we don’t particularly care how many classes there are, but we want to know the number of ways to partition a set of nelements. Let these numbers be{b(n)}∞ 0. They are called the Bell numbers. It is conventional to take b(0) = 1. The sequence of Bell numbers begins 1, 1, 2, 5, 15, 52, ... Can we ﬁnd an explicit formula for the Bell numbers? Nothing to it. In (1.6.7) we have an explicit formula for/braceleftbign k/bracerightbig . If we sum that formula from k=1t onwe will have an explicit formula for b(n).However , there’s one 1.6 Another 2-variable case. 21 more thing that it is quite proﬁtable to notice. The formula (1.6.7) is valid forallpositive integer values of nandk. In particular, it is valid if k>n . But/braceleftbign k/bracerightbig =0i fk>n . This means that the formula (1.6.7) doesn’t have to betoldthat/braceleftbig13 19/bracerightbig = 0; it knows it; i.e., if we blissfully insert n= 13,k=1 9 into the monster sum and work it all out, we will get 0. Hence, to calculate the Bell numbers, we can sum the last member of (1.6.7) from k=1t oM, whereMis any number you please that is ≥n. Let’s do it. The result is that b(n)=M/summationdisplay k=1k/summationdisplay r=1(−1)k−rrn−1 (r−1)!(k−r)! =M/summationdisplay r=1rn−1 (r−1)!M/summationdisplay k=r(−1)k−r (k−r)! =M/summationdisplay r=1rn−1 (r−1)!/braceleftBiggM−r/summationdisplay s=0(−1)s s!/bracerightBigg . But now the number Mis arbitrary, except that M≥n. Since the partial sum of the exponential series in the curly braces above is so inviting, let’s keepnandrﬁxed, and let M→∞ . This gives the following remarkable formula for the Bell numbers (check it yourself for n= 1): b(n)=1 e/summationdisplay r≥0rn r!(n≥0). (1.6.10) This formula for the Bell numbers, although it has a certain charm, doesn’t lend itself to computation. From it, however, we can derive a generating function for the Bell numbers that is unexpectedly simple and elegant. We will look for the generating function in the form B(x)=/summationdisplay n≥0b(n) n!xn. (1.6.11) This is the ﬁrst time we have found it necessary to introduce an extra factor of 1/n! into the coeﬃcients of a generating function. That kind of thing happens frequently, however, and we will discuss in chapter 2 how to recognize when extra factors like these will be useful. A generating function of the form (1.6.11), with the 1 /n!’s thrown into the coeﬃcients, is called anexponential generating function . We would say, for instance, that ‘ B(x) is the exponential generating function of the Bell numbers.’ When we wish to distinguish the various kinds of generating functions, we may use the phrase the ordinary power series generating function of the sequence {an}is/summationtext nanxnorthe exponential generating function of the sequence {an}is/summationtext nanxn/n!. 22 1 Introductory ideas and examples To ﬁndB(x) explicitly, take the formula (1.6.10), which is valid for n≥1, multiply it by xn/n! (don’t forget the n!), and sum over all n≥1. This gives B(x)−1=(1 e)/summationdisplay n≥1xn n!/summationdisplay r≥1rn−1 (r−1)! =(1 e)/summationdisplay r≥11 r!/summationdisplay n≥1(rx)n n! =(1 e)/summationdisplay r≥11 r!(erx−1) =(1 e){eex−e} =eex−1−1. We have therefore shown Theorem 1.6.1. The exponential generating function of the Bell numbers iseex−1, i.e., the coeﬃcient of xn/n!in the power series expansion of eex−1 is the number of partitions of a set of nelements. This result is surely an outstanding example of the power of the gen- erating function approach. The Bell numbers themselves are complicated, but the generating function is simple and easy to remember. The next novel element of this story is the fact that we can go from generating functions torecurrence formulas, although in all our examples to date the motion has been in the other direction. We propose now to derive from Theorem 1.6.1 a recurrence formula for the Bell numbers, one that will make it easy to compute as many of them as we might wish to look at. First, the theorem tells us that /summationdisplay n≥0b(n) n!xn=eex−1. (1.6.12) We are going to carry out a very standard operation on this equation, but the ﬁrst time this operation appears it seems to be anything but standard. Thex(d/dx ) log operation (1) Take the logarithm of both sides of the equation. (2) Diﬀerentiate both sides and multiply through by x. (3) Clear the equation of fractions.(4) For each n, ﬁnd the coeﬃcients of x non both sides of the equation and equate them. Although the best motivation for the above program is the fact that it works, let’s pause for a moment before doing it, to see why it is likely to 1.6 Another 2-variable case. 23 work. The point of taking logarithms is to simplify the function eex−1, whose power series coeﬃcients are quite mysterious before taking loga- rithms, and are quite obvious after doing so. The price for that simpli- ﬁcation is that on the left side we have the log of a sum, which is an awesome thing to have. The next step, the diﬀerentiation, changes the log of the sum into a ratio of two sums, which is much nicer. The reason for multiplying through by xis that the diﬀerentiation dropped the power of x by 1 and it’s handy to restore it. After clearing of fractions we will simply be looking at two sums that are equal to each other, and the work will beover. In this case, after step 1 is applied to (1.6.12), we have log/braceleftbigg/summationdisplay n≥0b(n) n!xn/bracerightbigg =ex−1. Step 2 gives/summationtext nnb(n)xn n!/summationtext nb(n)xn n!=xex. To clear of fractions, multiply both sides by the denominator on the left, obtaining/summationdisplay nnb(n)xn n!=(xex)/summationdisplay nb(n)xn n!. Finally, we have to identify the coeﬃcients of xnon both sides of this equation. On the left it’s easy. On the right we have to multiply two power series together ﬁrst, and then identify the coeﬃcient. Since in chapter 2 we will work out a general and quite easy-to-use rule for doing things like this, let’s postpone this calculation until then, and merely quote the result here. It is that the Bell numbers satisfy the recurrence b(n)=/summationdisplay k/parenleftbiggn−1 k/parenrightbigg b(k)(n≥1;b(0) = 1). (1.6.13) We have now seen several examples of how generating functions can be used to ﬁnd recurrence relations. It often happens that the method of generating functions ﬁnds a recurrence, and only later are we able to give a direct, combinatorial interpretation of the recurrence. In some cases, re- currences are known that look like they ought to have simple combinatorialexplanations, but none have yet been found. 24 1 Introductory ideas and examples Exercises 1. Find the ordinary power series generating functions of each of the follow- ing sequences, in simple, closed form. In each case the sequence is deﬁned for alln≥0. (a)an=n (b)an=αn+β (c)an=n2 (d)an=αn2+βn+γ (e)an=P(n), wherePis a given polynomial, of degree m. (f)an=3n (g)an=5·7n−3·4n 2. For each of the sequences given in part 1, ﬁnd the exponential generating function of the sequence in simple, closed form. 3. Iff(x) is the ordinary power series generating function of the sequence {an}n≥0, then express simply, in terms of f(x), the ordinary power series generating functions of the following sequences. In each case the range of nis 0,1,2,... (a){an+c} (b){αan+c} (c){nan} (d){P(n)an}, wherePis a given polynomial. (e) 0,a1,a2,a3,... (f) 0,0,1,a3,a4,a5,... (g)a0,0,a2,0,a4,0,a6,0,a8,0,... (h)a1,a2,a3,... (i){an+h} (ha given constant) (j){an+2+3an+1+an} (k){an+2−an+1−an} 4. Letf(x) be the exponential generating function of a sequence {an}.F o r each of the sequences in exercise 3, ﬁnd the exponential generating function simply, in terms of f(x). Exercises 25 5. Find (a) [xn]e2x (b) [xn/n!]eαx (c) [xn/n!] sinx (d) [xn]{1/((1−ax)(1−bx))} (a/negationslash=b) (e) [xn](1 +x2)m 6. In each part, a sequence {an}n≥0satisﬁes the given recurrence relation. Find the ordinary power series generating function of the sequence. (a)an+1=3an+2 (n≥0;a0=0 ) (b)an+1=αan+β (n≥0;a0=0 ) (c)an+2=2an+1−an (n≥0;a0=0 ;a1=1 ) (d)an+1=an/3+1 (n≥0;a0=0 ) 7. Give a direct combinatorial proof of the recurrence (1.6.13), as follows: givenn; consider the collection of all partitions of the set [ n]. There are b(n) of them. Sort out this collection into piles numbered k=0,1,...,n −1, where thekth pile consists of all partitions of [ n] in which the class that contains the letter ‘ n’ contains exactly kother letters. Count the partitions in thekth pile, and you’ll be all ﬁnished. 8. In each part of problem 6, ﬁnd the exponential generating function of the sequence (you may have to solve a diﬀerential equation to do so!). 9. A function fis deﬁned for all n≥1 by the relations (a) f(1) = 1 and (b)f(2n)=f(n) and (c)f(2n+1 )=f(n)+f(n+ 1). Let F(x)=/summationdisplay n≥1f(n)xn−1 be the generating function of the sequence. Show that F(x)=( 1+x+x2)F(x2), and therefore that F(x)=∞/productdisplay j≥0/braceleftBig 1+x2j+x2j+1/bracerightBig . 10. LetXbe a random variable that takes the values 0 ,1,2,...with re- spective probabilities p0,p1,p2,..., where the p’s are given nonnegative real numbers whose sum is 1. Let P(x) be the opsgf of {pn}. (a) Express the mean µand standard deviation σofXdirectly in terms ofP(x). 26 1 Introductory ideas and examples (b) Two values of Xare sampled independently. What is the probability p(2) nthat their sum is n? Express the opsgf P2(x)o f{p(2) n}in terms ofP(x). (c)kvalues ofXare sampled independently. Let p(k) nbe the probability that their sum is equal to n. Express the opsgf Pk(x)o f{p(k) n}n≥0 in terms of P(x). (d) Use the results of parts (a) and (c) to ﬁnd the mean and standard deviation of the sum of kindependently chosen values of X, in terms ofµandσ. (e) LetA(x) be a power series with A(0) = 1, and let B(x)=A(x)k.I t is desired to compute the coeﬃcients of B(x), without raising A(x) to any powers at all. Use the ‘ xDlog ’ method to derive a recurrence formula that is satisﬁed by the coeﬃcients of B(x). (f) A loaded die has probabilities .1, .2, .1, .2, .2, .2 of turning up with, respectively, 1, 2, 3, 4, 5, or 6 spots showing. The die is then thrown 100 times, and we want to calculate the probability p∗that the total number of spots on all 100 throws is ≤300. Identify p∗ as the coeﬃcient of x300in the power series expansion of a certain function. Say exactly what the function is (you are notbeing asked to calculate p∗). Use the result of part (e) to say exactly how you would calculate p∗if you had to. (g) A random variable Xassumes each of the values 1 ,2,...,m with probability 1 /m. LetSnbe the result of sampling nvalues ofX independently and summing them. Show that for n=1,2,..., Prob{Sn≤j}=1 mn/summationdisplay r(−1)r/parenleftbiggn r/parenrightbigg/parenleftbiggj−mr n/parenrightbigg . 11. Letf(n) be the number of subsets of [ n] that contain no two consecu- tive elements, for integer n. Find the recurrence that is satisﬁed by these numbers, and then ‘ﬁnd’ the numbers themselves. 12. For given integers n,k, letf(n,k) be the number of k-subsets of [ n] that contain no two consecutive elements. Find the recurrence that is satisﬁed by these numbers, ﬁnd a suitable generating function and ﬁnd the num- bers themselves. Show the numerical values of f(n,k) in a Pascal triangle arrangement, for n≤6. 13. By comparing the results of the above two problems, deduce an identity. Draw a picture of the elements of Pascal’s triangle that are involved in this identity. 14. Let the integers 1 ,2,...,n be arranged consecutively around a circle, and letg(n) be the number of ways of choosing a subset of these, no two Exercises 27 consecutive on the circle. That is, gdiﬀers from the fof problem 11 in thatnand 1 are now regarded as consecutive. Find g(n). 15. As in the previous problem, ﬁnd, analogously to problem 12 above, the numberg(n,k) of ways of choosing kelements from narranged on a circle, such that no two chosen elements are adjacent on the circle. 16. Find the coeﬃcient of xnin the power series for f(x)=1 (1−x2)2, ﬁrst by the method of partial fractions, and second, give a much simpler derivation by being sneaky. 17. An inversion of a permutation σof [n] is a pair of letters i,jsuch that i<j andσ(i)>σ(j). In the 2-line form of writing the permutation, an inversion shows up as a pair that is ‘in the wrong order’ in the second line. The permutation σ=/parenleftbigg 123456789 492581673/parenrightbigg of [9] has 19 inversions, namely the pairs (4,2), (4,1), ..., (7,3). Let b(n,k) be the number of permutations of nletters that have exactly kinversions. Find a ‘simple’ formula for the generating function Bn(x)=/summationtext kb(n,k)xk. Make a table of values of b(n,k) forn≤5. 18. (a) Givenn,k. For how many of the permutations of nletters is it true that their ﬁrst kvalues decrease? (b) What is the average length of the decreasing sequence with which the values of a random n-permutation begin? (c) Iff(n,k) is the number of permutations that have exactly k ascending runs, ﬁnd the Pascal-triangle-type recurrence sat- isﬁed byf(n,k). They are called the Euler numbers. As an example, the permutation /parenleftbigg 123456789 416925837/parenrightbigg has 4 such runs, namely 4, 1 6 9, 2 5 8, and 3 7. 19. Consider the 256 possible sums of the form /epsilon11+/epsilon12+2/epsilon13+5/epsilon14+1 0/epsilon15+1 0/epsilon16+2 0/epsilon17+5 0/epsilon18 (1) where each /epsilon1is 0 or 1. 28 1 Introductory ideas and examples (a) For each integer n, letCnbe the number of diﬀerent sums that representn. Write the generating polynomial C0+C1x+C2x2+C3x3+···+C99x99 as a product. (b) Next, consider all of the possible sums that are formed as in (1), where now the /epsilon1’s can have any of the three values −1,0,1. For each integern, letDnbe the number of diﬀerent sums that represent n. Show that some integer nis representable in at least 33 diﬀerent ways. Then write the generating function 99/summationdisplay n=−99Dnxn as a product. (c) Generalize the results of parts (a) and (b) of this problem by re- placing the particular set of weights by a general set. Factor the polynomial that occurs. (d) In the general case of part (d) of this problem, state precisely what all of the zeros of the generating polynomial are, and state precisely what the multiplicity of each of the zeros is, in terms of the set of weights. 20. Letf(n,m,k ) be the number of strings of n0’s and 1’s that contain exactlym1’s, nokof which are consecutive. (a) Find a recurrence formula for f. It should have f(n,m,k ) on the left side, and exactly three terms on the right . (b) Find, in simple closed form, the generating functions Fk(x,y)=/summationdisplay n,m≥0f(n,m,k )xnym(k=1,2,...). (c) Find an explicit formula for f(n,m,k ) from the generat- ing function (this should involve only a single summation, of an expression that involves a few factorials). 21. (a) We want to ﬁnd a formula for the nth derivative of the function eex. Diﬀerentiate it a few times, study the pattern, and conjecture the form of the answer for general n, including some constants to be determined. Then ﬁnd a recurrence formula for the constants in question, and identify them as some ‘famous’ numbers that we have studied. 1.6 Another 2-variable case. 29 (b) Next let f(x1,...,x n) be some function of nvariables. Find a formula for the mixed partial derivative ∂n ∂x1∂x2···∂xnef that expresses it in terms of various partial derivatives of fitself. 30 2 Series Chapter 2 Series This chapter is devoted to a study of the diﬀerent kinds of series that are widely used as generating functions. 2.1 Formal power series To discuss the formal theory of power series, as opposed to their an- alytic theory, is to discuss these series as purely algebraic objects, in their roles as clotheslines, without using any of the function-theoretic properties of the function that may be represented by the series or, indeed, without knowing whether such a function exists. We study formal series because it often happens in the theory of gen- erating functions that we are trying to solve a recurrence relation, so we introduce a generating function, and then we go through the various ma- nipulations that follow, but with a guilty conscience because we aren’t sure whether the various series that we’re working with will converge. Also, we might ﬁnd ourselves working with the derivatives of a generating function, still without having any idea if the series converges to a function at all. The point of this section is that there’s no need for the guilt, because the various manipulations can be carried out in the ring of formal power series, where questions of convergence are nonexistent. We may execute the whole method and end up with the generating series, and only then discover whether it converges and thereby represents a real honest function or not. If not, we may still get lots of information from the formal series, but maybe we won’t be able to get analytic information, such as asymptotic formulas for the sizes of the coeﬃcients. Exact formulas for the sequences in question, however, might very well still result, even though the method rests, in those cases, on a purely algebraic, formal foundation. The series f=1+x+2x2+6x3+2 4x4+ 120x5+···+n!xn+···, (2.1.1) for instance, has a perfectly ﬁne existence as a formal power series, despite the fact that it converges for no value of xother than x= 0, and therefore oﬀers no possibilities for investigation by analytic methods. Not only that, but this series plays an important role in some natural counting problems. Aformal power series is an expression of the form a0+a1x+a2x2+··· where the sequence {an}∞ 0is called the sequence of coeﬃcients . To say that two series are equal is to say that their coeﬃcient sequences are the same. 2.1 Formal power series 31 We can do certain kinds of operations with formal power series. We canaddorsubtract them, for example. This is done according to the rules /summationdisplay nanxn±/summationdisplay nbnxn=/summationdisplay n(an±bn)xn. Power series can be multiplied by the usual Cauchy product rule, /summationdisplay nanxn/summationdisplay nbnxn=/summationdisplay ncnxn(cn=/summationdisplay kakbn−k). (2.1.2) It is certainly this product rule that accounts for the wide applicability of series methods in combinatorial problems. This is because frequently we can construct all anof the objects of type nin some family by choosing an object of type kand an object of type n−kand stitching them together to make the object of type n. The number of ways of doing that will be akan−k, and if we sum on kwe ﬁnd that the Cauchy product of two formal series is directly relevant to the problem that we are studying. If we follow the multiplication rule we obtain, for instance, (1−x)(1 +x+x2+x3+···)=1. Thus we can say that the series (1 −x) has a reciprocal, and that reciprocal is 1 +x+x2+···(and the other way around, too). Proposition. A formal power series f=/summationtext n≥0anxnhas a reciprocal if and only if a0/negationslash=0. In that case the reciprocal is unique. Proof. Letfhave a reciprocal, namely 1 /f=/summationtext n≥0bnxn. Thenf·(1/f)= 1 and according to (2.1.2), c0=1=a0b0,s oa0/negationslash= 0. Further, in this case (2.1.2) tells us that for n≥1,cn=0=/summationtext kakbn−k, from which we ﬁnd bn=(−1/a0)/summationdisplay k≥1akbn−k (n≥1). (2.1.3) This determines b1,b2,...uniquely, as claimed. Conversely, suppose a0/negationslash= 0. Then we can determine b0,b1,...from (2.1.3), and the resulting series/summationtext nbnxnis the reciprocal of f. The collection of formal power series under the rules of arithmetic that we have just described forms a ring, in which the invertible elements are the series with nonvanishing constant term. The above idea of a reciprocal of a formal power series is not to be confused with the subtler notion of the inverse of such a series. The inverse of a series f, if it exists, is a series gsuch thatf(g(x)) =g(f(x)) =x. When can such an inverse exist? First we need to be able to deﬁne the symbolf(g(x)), then we can worry about whether or not it is equal to x. 32 2 Series Iff=/summationtext nanxn, thenf(g(x)) means f(g(x)) =/summationdisplay nang(x)n. (2.1.4) If the series g(x) has a nonzero constant term, g0, then every term of the series (2.1.4) may contribute to the coeﬃcient of each power of x. On the other hand, if g0= 0, then we will be able to compute the coeﬃcient of, say,x57in (2.1.4) from just the ﬁrst 58 terms of the series shown. Indeed, notice that every single term ang(x)n=an(g1x+g2x2+...)n =anxn(g1+g2x+...)n withn>57 will contain only powers of xhigher than the 57th, and there- fore we won’t need to look at those terms to ﬁnd the coeﬃcient of x57. Thus ifg0= 0 then the computation of each one of the coeﬃcients of the seriesf(g(x)) is a ﬁnite process, and therefore all of those coeﬃcients are well deﬁned, and so is the series. If g0/negationslash= 0, though, the computation of each coeﬃcient of f(g(x)) is an inﬁnite process unless fis a polynomial, and therefore it will make sense only if the series ‘converge.’ In a formal, algebraic theory, however, ideas of convergence have no place. Thus the composition f(g(x))of two formal power series is deﬁned if and only if g0=0orfis a polynomial . For instance, the series eex−1is a well deﬁned formal series, whereas the serieseexis not deﬁned, at least from the general deﬁnition of composition of functions. To return to the question of ﬁnding a series inverse of a given series f, we see that if such an inverse series gexists, then f(g(x)) =g(f(x)) =x (2.1.5) must both make sense and be true. We claim that if f(0) = 0 the inverse series exists if and only if the coeﬃcient of xis nonzero in the series f. Proposition. Let the formal power series f,gsatisfy (2.1.5) and f(0) = 0 . Thenf=f1x+f2x2+··· (f1/negationslash=0), andg=g1x+g2x2+··· (g1/negationslash=0). Proof. Suppose that f=frxr+···andg=gsxs+···, wherer,s≥0 and frgs/negationslash= 0. Thenf(g(x)) =x=frgr sxrs+···, whencers= 1, andr=s=1 , as claimed. In the ring of formal power series there are other operations deﬁned, which mirror the corresponding operations of function calculus, but which make no use of limiting operations. The derivative of the formal power series f=/summationtext nanxnis the series f/prime=/summationtext nnanxn−1. Diﬀerentiation follows the usual rules of calculus, such as the sum, product, and quotient rules. Many of these properties are even easier to prove for formal series than they are for the functions of calculus. For example: 2.2 The calculus of formal ordinary power series generating functions 33 Proposition. Iff/prime=0thenf=a0is constant. Proof. Take another look at the ‘=’ sign in the hypothesis f/prime= 0. It means that the formal power series f/primeis identical to the formal power series 0, and that means that each and every coeﬃcient of the formal series f/primeis 0. But the coeﬃcients of f/primearea1,2a2,3a3,..., so each of these is 0, and therefore aj= 0 for allj≥1, which is to say that fis constant. Next, try this one: Proposition. Iff/prime=fthenf=cex. Proof. Sincef/prime=f, the coeﬃcient of xnmust be the same in fas inf/prime, for alln≥0. Hence ( n+1 )an+1=anfor alln≥0, whence an+1=an/(n+1 ) (n≥0). By induction on n,an=a0/n! for alln≥0, and sof=a0ex. 2.2 The calculus of formal ordinary power series generating func- tions Operations on formal series involve corresponding operations on their coeﬃcients. If the series actually converge and represent functions, then operations on those functions correspond to certain operations on the power series coeﬃcients of the expansions of those functions. In this section we will explore some of these relationships. They are of great importance in helping to spot which kind of generating function is appropriate for which kind of recurrence relation or other combinatorial situation. Deﬁnition. The symbol fops ←→{an}∞ 0means that the series fis the ordi- nary power series (‘ops’) generating function for the sequence {an}∞ 0. That is, it means that f=/summationtext nanxn. Supposefops ←→{an}∞ 0. Then what generates {an+1}∞ 0? To answer that we do a little calculation: /summationdisplay n≥0an+1xn=1 x/summationdisplay m≥1amxm=(f(x)−f(0)) x. Therefore fops ←→{an}∞ 0⇒((f−a0)/x)ops ←→{an+1}∞ 0. (2.2.1) Thus a shift of the subscript by 1 unit changes the series represented to the diﬀerence quotient ( f−a0)/x. If we shift by 2 units, of course, we just iterate the diﬀerence quotient operation, and ﬁnd that {an+2}∞ 0ops ←→((f−a0)/x)−a1 x =f−a0−a1x x2. 34 2 Series Note how this point of view allows us to see ‘at a glance’ that the Fibonacci recurrence relation Fn+2=Fn+1+Fn(n≥0;F0=0 ;F1= 1) translates directly into the ordinary power series generating function relationf−x x2=f x+f. Indeed, the purpose of this section is to develop this facility for passing from sequence relations to series relations quickly and conveniently. Rule 1. Iffops ←→{an}∞ 0, then, for integer h>0, {an+h}∞ 0ops ←→f−a0−···−ah−1xh−1 xh. Next let’s look into the eﬀect of multiplying the sequence by powers of n. Again, suppose that fops ←→{an}∞ 0. Then what generates the sequence {nan}∞ 0? The question means this: can we express the series/summationtext nnanxn in some simple way in terms of the series f=/summationtext nanxn? The answer is easy, because the former series is exactly xf/prime. Therefore, to multiply the nth member of a sequence by ncauses its ops generating function to be ‘multiplied’ by x(d/dx ), which we will write as xD. In symbols: fops ←→{an}∞ 0⇒(xDf)ops ←→{nan}∞ 0. (2.2.2) As an example, consider the recurrence (n+1 )an+1=3an+1 (n≥0;a0=1 ). Iffis the opsgf of the sequence {an}∞ 0, then from Rule 1 and (2.2.2), f/prime=3f+1 1−x, which is a ﬁrst order diﬀerential equation in the unknown generating func- tion, and it can be solved by standard methods. Next suppose fops ←→{an}∞ 0. Then what generates the sequence {n2an}∞ 0? Obviously we re-apply the multiply-by- noperatorxD, so the answer is (xD)2f. In general, (xD)kfops ←→{nkan}n≥0. OK, what generates {(3−7n2)an}n≥0? Again obviously, we do the same thing to xDthat is done to n, i.e., (3 −7(xD)2)fis the answer. The general prescription is: 2.2 The calculus of formal ordinary power series generating functions 35 Rule 2. Iffops ←→{an}∞ 0, andPis a polynomial, then P(xD)fops ←→{P(n)an}n≥0. Example 1. Find a closed formula for the sum of the series/summationtext n≥0(n2+4n+5 )/n!. According to the rule, the answer is the value at x= 1 of the series {(xD)2+4 (xD)+5}ex={x2+x}ex+4xex+5ex =(x2+5x+5 )ex. Therefore the answer to the question is 11 e. But we cheated. Did you catch the illegal move? We took our gen- erating function and evaluated it at x= 1, didn’t we? Such an operation doesn’t exist in the ring of formal series. There, series don’t have ‘values’ at particular values of x. The letter xis purely a formal symbol whose powers mark the clothespins on the line. What canbe evaluated at a particular numerical value of xis a power series that converges at that x, which is an analytic idea rather than a formal one. The way we make peace with our consciences in such situations, which occur frequently, is this: if, after writing out the recurrence relation and solving it by means of a formal power series generating function, we ﬁnd that the series so obtained converges to an analytic function inside a certain disk in the complex plane, then the whole derivation that we did formally is actually valid analytically for all complex xin that disk. Therefore we can shift gears and regard the series as a convergent analytic creature if itpleases us to do so. Example 2. Find a closed formula for the sum of the squares of the ﬁrst Npositive integers. To do that, begin with the fact that N/summationdisplay n=0xn=xN+1−1 x−1, and notice that if we apply ( xD)2to both sides of this relation and then setx= 1, the left side will be the sum of squares that we seek, and the right side will be the answer! Hence N/summationdisplay n=1n2=(xD)2/braceleftbiggxN+1−1 x−1/bracerightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle x=1. 36 2 Series After doing the two diﬀerentiations and lots of algebra, the answer emerges as N/summationdisplay n=1n2=N(N+ 1)(2N+1 ) 6(N=1,2,...), which you no doubt knew already. Do notice, however, that the generating function machine is capable of doing, quite mechanically, many formidable- looking problems involving sums. Our third rule will be a restatement of the way that two opsgf’s are multiplied. Rule 3. Iffops ←→{an}∞ 0andgops ←→{bn}∞ 0, then fgops ←→/braceleftbiggn/summationdisplay r=0arbn−r/bracerightbigg∞ n=0. (2.2.3) Now consider the product of more than two series. For instance, in the case of three series, if f,g,h are the series, and if they generate sequences a,bandc, respectively, then a brief computation shows that fghgenerates the sequence/braceleftbigg/summationdisplay r+s+t=narbsct/bracerightbigg∞ n=0. (2.2.4) A comparison with Rule 3 above will suggest the general formulas that apply to products of any number of power series. One case of this is worth writing down, namely the expressions for the kth power of a series. Rule 4. Letfops ←→{an}∞ 0, and letkbe a positive integer. Then fkops ←→/braceleftBigg/summationdisplay n1+n2+···+nk=nan1an2···ank/bracerightBigg∞ n=0. (2.2.5) Example 3. Letf(n,k) denote the number of ways that the nonnegative integer n can be written as an ordered sum of knonnegative integers. Find f(n,k). For instance, f(4,2) = 5 because 4=4+0=3+1=2+2=1+3=0+4. To ﬁndf, consider the power series 1 /(1−x)k. Since 1/(1−x)ops ←→{1}, by (2.2.5) we have 1/(1−x)kops ←→{f(n,k)}∞ n=0. By (1.5.5), f(n,k)=/parenleftbign+k−1 n/parenrightbig , and we are ﬁnished. Next consider the eﬀect of multiplying a power series by 1 /(1−x). Supposefops ←→{an}∞ 0. Then what sequence does f(x)/(1−x) generate? 2.2 The calculus of formal ordinary power series generating functions 37 To ﬁnd out, we have f(x) (1−x)=(a0+a1x+a2x2+···)(1 +x+x2+···) =a0+(a0+a1)x+(a0+a1+a2)x2 +(a0+a1+a2+a3)x3+··· which clearly leads us to: Rule 5. Iffops ←→{an}∞ 0then f (1−x)ops ←→/braceleftbiggn/summationdisplay j=0aj/bracerightbigg n≥0. That is, the eﬀect of dividing an opsgf by (1−x)is to replace the sequence that is generated by the sequence of its partial sums. Example 4. Here is another derivation of the formula for the sum of the squares of the ﬁrstnwhole numbers. Since 1 /(1−x)ops ←→{1}n≥0, we have by Rule 2, (xD)2(1/(1−x))ops ←→{n2}n≥0, and by Rule 5, 1 1−x(xD)21 1−xops ←→/braceleftbiggn/summationdisplay j=0j2/bracerightbigg n≥0. That is, the sum of the squares of the ﬁrst npositive integers is the coeﬃ- cient ofxnin the series 1 1−x(xD)21 1−x=x(1 +x) (1−x)4. However, by (1.5.5) with k=3 , [xn]/parenleftbigg1 (1−x)4/parenrightbigg =/parenleftbiggn+3 3/parenrightbigg . Hence, by (1.2.7), [xn]x(1 +x) (1−x)4=/parenleftbiggn+2 3/parenrightbigg +/parenleftbiggn+1 3/parenrightbigg =n(n+ 1)(2n+1 ) 6, so this must be the sum of the squares of the ﬁrst npositive integers. 38 2 Series Fig. 2.1: A (28,12)fountain Example 5. The harmonic numbers {Hn}∞ 1are deﬁned by Hn=1+1 2+1 3+···+1 n(n≥1). How can we ﬁnd their ops generating function? By Rule 5, that function is 1/(1−x) times the opsgf of the sequence {1/n}∞ 1of reciprocals of the positive integers. So what is f=/summationtext n≥1xn/n? Well, its derivative is 1 /(1− x), so it must be −log (1 −x). That means that the opsgf of the harmonic numbers is∞/summationdisplay n=1Hnxn=1 1−xlog/parenleftbigg1 1−x/parenrightbigg . Example 6. Prove that the Fibonacci numbers satisfy F0+F1+F2+···+Fn=Fn+2−1(n≥0). By Rule 5, the opsgf of the sequence on the left side is F/(1−x), whereF is the opsgf of the Fibonacci numbers, which we found in section 1.3 to be x/(1−x−x2). By Rule 1, the opsgf of the sequence on the right hand side isF−x x2−1 1−x, and it is the work of just a moment to check that these are equal. Example 7. By a fountain of coins we mean an arrangement of ncoins in rows such that the coins in the ﬁrst row form a single contiguous block, and that in all higher rows each coin touches exactly two coins from the row beneath it. If the ﬁrst row contains kcoins, we will speak of an ( n,k)-fountain. In Fig. 2.1 we show a (28 ,12) fountain. Among all possible fountains we distinguish a special type: those in which every row consists of just a single contiguous block of coins. Let’s call these block fountains . 2.3 The calculus of formal exponential generating functions 39 The question here is this: how many block fountains have a ﬁrst row that consists of exactly kcoins? Letf(k) be that number, for k=0,1,2,...If we strip oﬀ the ﬁrst row from such a block fountain, then we are looking at another block fountain that haskfewer coins in it. Conversely, if we wish to form all possible block fountains whose ﬁrst row has kcoins, then begin by laying down that row. Then choose a number j,0≤j≤k−1. Above the row of kcoins we will place a block fountain whose ﬁrst row has jcoins. Ifj= 0 there is just one way to do that. Otherwise there are k−jways to do it, depending on how far in we indent the row of jover the row of kcoins. It follows that f(0) = 1 and f(k)=k/summationdisplay j=1(k−j)f(j)+1 (k=1,2,...). (2.2.6) Deﬁne the opsgf F(x)=/summationtext j≥0f(j)xj. The appearance, under the summation sign in (2.2.6), of a function of k−jtimes a function of jshould trigger a reﬂex reaction that Rule 3, above, applies, and that the product of two ordinary power series generating functions is involved. The two series in question are the opsgf’s of the integers {j}∞ 1and of the unknowns {f(j)}∞ 1, respectively. However the former opsgf is x/(1−x)2, and the latter is F(x)−1. Hence, after multiplying equation (2.2.6) by xkand summing over k≥1 we obtain F(x)−1=x (1−x)2(F(x)−1) +x 1−x, and therefore F(x)=1−2x 1−3x+x2. (2.2.7) The sequence {f(k)}∞ 0begins with 1 ,1,2,5,13,34,89,...If these num- bers look suspiciously like Fibonacci numbers, then see exercise 19. 2.3 The calculus of formal exponential generating functions In this section we will investigate the analogues of the rules in the preceding section, which applied to ordinary power series, in the case of exponential generating functions. Deﬁnition. The symbol fegf ←→{an}∞ 0means that the series fis the ex- ponential generating function of the sequence {an}∞ 0, i.e., that f=/summationdisplay n≥0an n!xn. 40 2 Series Let’s ask the same questions as in the previous section. Suppose fegf ←→{an}∞ 0. Then what is the egf of the sequence {an+1}∞ 0? We claim that the answer is f/prime, because f/prime=∞/summationdisplay n=1nanxn−1 n! =∞/summationdisplay n=1anxn−1 (n−1)! =∞/summationdisplay n=0an+1xn n! which is exactly equivalent to the assertion that f/primeegf ←→{an+1}∞ 0. Hence the situation with exponential generating functions is just a triﬂe simpler, in this respect, than the corresponding situation for ordinary power series. Displacement of the subscript by 1 unit in a sequence is equivalent to action of the operator Don the generating function, as opposed to the operator (f(x)−f(0))/x, in the case of opsgf’s. Therefore we have, by induction: Rule 1/prime.Iffegf ←→{an}∞ 0then, for integer h≥0, {an+h}∞ 0egf ←→Dhf. (2.3.1) The reader is invited to compare this Rule 1/primewith Rule 1, stated above. Example 1. To get a hint of the strength of this point of view in problem solving, let’s ﬁnd the egf of the Fibonacci numbers. Now, with just a glance at the recurrence Fn+2=Fn+1+Fn (n≥0) we see from Rule 1/primethat the egf satisﬁes the diﬀerential equation f/prime/prime=f/prime+f. At the corresponding stage in the solution for the ops version of this prob- lem, we had an equation to solve for fthat did not involve any derivatives. We solved it and then had to deal with a partial fraction expansion in or- der to ﬁnd an exact formula for the Fibonacci numbers. In this version, we solve the diﬀerential equation, getting f(x)=c1er+x+c2er−x(r±=( 1±√ 5)/2) wherec1andc2are to be determined by the initial conditions (which haven’t been used yet!) f(0) = 0;f/prime(0) = 1. After applying these two 2.3 The calculus of formal exponential generating functions 41 conditions, we ﬁnd that c1=1/√ 5 andc2=−1/√ 5, from which the egf of the Fibonacci sequence is f=(er+x−er−x)/√ 5. (2.3.2) Now it’s easier to get the exact formula, because no partial fraction expan- sion is necessary. Just apply the operator [ xn/n!] to both sides of (2.3.2) and the formula (1.3.3) materializes. To compare, then, the ops method in this case involves an easier func- tional equation to solve for the generating function: it’s algebraic instead of diﬀerential. The egf method involves an easier trip from there to the exact formula, because the partial fraction expansion is unnecessary. Both methods work, which is, after all, the primary desideratum . To continue, we discuss next the analogue of Rule 2 for egf’s, and that one is easy: it’s the same. That is, multiplication of the members of a sequence by a polynomial in nis equivalent to acting on the egf with the same polynomial in the operator xD, and we have: Rule 2/prime.Iffegf ←→{an}∞ 0, andPis a given polynomial, then P(xD)fegf ←→{P(n)an}n≥0. Next let’s think about the analogue of Rule 3, i.e., about what happens to sequences when their egf’s are multiplied together. Precisely, supposef egf ←→{an}∞ 0andgegf ←→{bn}∞ 0. The question is, of what sequence is fgthe egf? This turns out to have a pretty, and uncommonly useful, answer. To ﬁnd it, we carry out the multiplication fgand try to identify the coeﬃcient ofxn/n!. We obtain fg=/braceleftbigg∞/summationdisplay r=0arxr r!/bracerightbigg/braceleftbigg∞/summationdisplay s=0bsxs s!/bracerightbigg =/summationdisplay r,s≥0arbs r!s!xr+s =/summationdisplay n≥0xn/braceleftbigg/summationdisplay r+s=narbs r!s!/bracerightbigg . The coeﬃcient of xn/n! is evidently /bracketleftbiggxn n!/bracketrightbigg (fg)=/summationdisplay r+s=nn!arbs r!s! =/summationdisplay r/parenleftbiggn r/parenrightbigg arbn−r. We state this result as: 42 2 Series Rule 3/prime.Iffegf ←→{an}∞ 0andgegf ←→{bn}∞ 0, thenfggenerates the sequence /braceleftBigg/summationdisplay r/parenleftbiggn r/parenrightbigg arbn−r/bracerightBigg∞ n=0. (2.3.3) This rule should be contrasted with Rule 3, the corresponding rule for multiplication of opsgf’s, the result of which is to generate the sequence /braceleftBigg/summationdisplay rarbn−r/bracerightBigg∞ n=0. (2.3.4) We remarked earlier that the convolution of sequences that is shown in (2.3.4) is useful in counting problems where structures of size nare ob- tained by stitching together structures of sizes randn−rin all possible ways. Correspondingly, the convolution (2.3.3) is useful in combinatorial situations where we not only stitch together two such structures, but we alsorelabel the structures. For then, roughly speaking, there are/parenleftbign r/parenrightbig ways to choose the new labels of the elements of the structure of size r, as well asarways to choose that structure and bn−rways to choose the other one. Since this no doubt all seems to be very abstract, let’s try to make it concrete with a few examples. Example 2. In (1.6.13) we found the recurrence formula for the Bell numbers, which we may write in the form b(n+1 )=/summationdisplay k/parenleftbiggn k/parenrightbigg b(k)(n≥0;b(0) = 1). (2.3.5) We will now apply the methods of this section to ﬁnd the egf of the Bell numbers. This will give an independent proof of Theorem 1.6.1, since (2.3.5) can be derived directly, as described in exercise 7 of chapter 1. LetBbe the required egf. The egf of the left side of (2.3.5) is, by Rule 1/prime,B/prime. If we compare the right side of (2.3.5) with (2.3.3) we see that the egf of the sequence on the right of (2.3.5) is the product of Band the egf of the sequence whose entries are all 1’s. This latter egf is evidently ex, and so we have B/prime=exB as the equation that we must solve in order to ﬁnd the unknown egf. But obviously the solution is B=cexp (ex), and since B(0) = 1, we must havec=e−1, from which B(x) = exp (ex−1), completing the re-proof of Theorem 1.6.1. 2.3 The calculus of formal exponential generating functions 43 Example 3. In order to highlight the strengths of ordinary vs. exponential gen- erating functions, let’s do a problem where the form of the convolution of sequences that occurs suggests the ops form of generating function. We will count the ways of arranging npairs of parentheses, each pair consisting of a left and a right parenthesis, into a legal string. A legal string of parentheses is one with the property that, as we scan the string from left to right we never will have seen more right parentheses than left. There are exactly 5 legal strings of 3 pairs of parentheses, namely: ((())); (()()); (())(); ()()(); ()(()) . (2.3.6) Letf(n) be the number of legal strings of npairs of parentheses ( f(0) = 1), forn≥0. With each legal string we associate a unique nonnegative integer k,a s follows: as we scan the string from left to right, certainly after we have seen allnpairs of parentheses, the number of lefts will equal the number of rights. However, these two numbers may be equal even earlier than that. In the last string in (2.3.6), for instance, after just k= 1 pairs have been scanned, we ﬁnd that all parentheses that have been opened have also been closed. In general, for any legal string, the integer kthat we associate with it is the smallest positive integer such that the ﬁrst 2 kcharacters of the string do themselves form a legal string. The values of kthat are associated with each of the strings in (2.3.6) are 3, 3, 2, 1, 1. We will say that a legal string of 2nparentheses is primitive if it hask=n. The ﬁrst two strings in (2.3.6) are primitive. How many legal strings of 2 nparentheses will have a given value of k? Letwbe such a string. The ﬁrst 2 kcharacters of ware a primitive string, and the last 2 n−2kcharacters of ware an arbitrary legal string. There are exactly f(n−k) ways to choose the last 2 n−2kcharacters, but in how many ways can we choose the ﬁrst 2 k? That is, how many primitive strings of length 2kare there? Lemma 2.3.1. Ifk≥1andg(k)is the number of primitive legal strings, andf(k)is the number of all legal strings of 2kparentheses, then g(k)=f(k−1). Proof. Given any legal string of k−1 pairs of parentheses, make a primitive one of length 2 kby adding an initial left parenthesis and a terminal right parenthesis to it. Conversely, given a primitive string of length 2 k, if its initial left and terminal right parentheses are deleted, what remains is an arbitrary legal string of length 2 k−2. Hence there are as many primitive strings of length 2 kas there are all legal strings of length 2 k−2, i.e., there aref(k−1) of them. 44 2 Series Hence the number of legal strings of length 2 nthat have a given value ofkisf(k−1)f(n−k). Since every legal string has a unique value of k,i t must be that f(n)=/summationdisplay kf(k−1)f(n−k)(n/negationslash=0 ;f(0) = 1) (2 .3.7) with the convention that f= 0 at all negative arguments. The recurrence easily allows us to compute the values 1 ,1,2,5,14,... Now let’s ﬁnd a generating function for these numbers. The clue as to which kind of generating function is appropriate comes from the form of the recurrence (2.3.7). The sum on the right is obviously related to the coeﬃcients of the product of two ordinary power series generating functions, so that is the species that we will use. LetF=/summationtext kf(k)xkbe the opsgf of {f(n)}n≥0. Then the right side of (2.3.7) is almost the coeﬃcient of xnin the series F2. What is it exactly ? It is the coeﬃcient of xnin the product of the series Fand the series/summationtext kf(k−1)xk. How is this latter series related to F? It is just xF. Therefore, if we multiply the right side of (2.3.7) by xnand sum over n/negationslash=0 , we getxF2. If we multiply the left side by xnand sum over n/negationslash= 0, we get F−1. Therefore our unknown generating function satisﬁes the equation F(x)−1=xF(x)2. (2.3.8) Here we have a new wrinkle. We are accustomed to going from recur- rence relations on a sequence to functional equations that have to be solved for generating functions. In previous examples, those functional equations have either been simple linear equations or diﬀerential equations. In (2.3.8) we have a generating function that satisﬁes a quadratic equation. When we solve it, we get F(x)=1±√1−4x 2x. Which sign do we want? If we choose the ‘+’ then the numerator will approach 2 as x→0, so the ratio will become inﬁnite at 0. But our generating function takes the value 1 at 0, so that can’t be right. If we choose the ‘ −’ sign, then a dose of L’Hospital’s rule shows that we will indeed have F(0) = 1. Hence our generating function is F(x)=1−√1−4x 2x. (2.3.9) This is surely one of the most celebrated generating functions in com- binatorics. The numbers f(n) are the Catalan numbers , and in (2.5.10) there is an explicit formula for them. For the moment, we declare that this exercise, which was intended to show how the form of a recurrence can guide the choice of generating function, is over. 2.3 The calculus of formal exponential generating functions 45 Example 4. By a derangement ofnletters we mean a permutation of them that has no ﬁxed points. Let Dndenote the number of derangements of nletters, and letD(x)egf ←→{Dn}∞ 0. We will ﬁnd a recurrence for the sequence, then D(x), then an explicit formula for the members of the sequence. The number of permutations of nletters that have a particular set of k≤nletters as their set of ﬁxed points is clearly Dn−k. There are/parenleftbign k/parenrightbig ways to choose the set of kﬁxed points, and so there are exactly/parenleftbign k/parenrightbig Dn−k permutations of nletters that have exactly kﬁxed points. Since every permutation has some set of ﬁxed points, it must be that n!=/summationdisplay k/parenleftbiggn k/parenrightbigg Dn−k (n≥0). If we take the egf of both sides we get, by Rule 3/prime, 1 1−x=exD(x) (see how easy that was?), from which D(x)=e−x/(1−x). Next, by Rule 5, if we take [ xn] on both sides, we ﬁnd that Dn n!=1−1+1 2!−1 3!+···+(−1)n1 n!, and we are ﬁnished. Just as in the case of ordinary power series generating functions, pleas- ant and useful things happen when we consider products of more than two exponential generating functions. For instance, if we multiply three of them, f,g, andh, which generate a,b, and c, respectively, then we ﬁnd that fghegf ←→/braceleftBigg/summationdisplay r+s+t=nn! r!s!t!arbsct/bracerightBigg∞ n=0, (2.3.10) and therefore such operations can be expected to be helpful in dealing with sums that involve multinomial coeﬃcients. Iffegf ←→{an}∞ 0then fkegf ←→/braceleftBigg/summationdisplay r1+···+rk=nn! r1!r2!···rk!ar1ar2···ark/bracerightBigg∞ n=0. (2.3.11) 46 2 Series 2.4 Power series, analytic theory The formal theory of power series shows us that we can manipulate recurrences and solve functional equations, such as diﬀerential equations, for power series without necessarily worrying about whether the resulting series converge. If they do converge though, and they represent functions, that’s a big advantage, for then we may be in a position to ﬁnd analytic information about the recurrence relation that might not otherwise be easily obtainable. In this section we will review the basic analytic properties of power series and their coeﬃcient sequences. First, suppose we are given a power series f=/summationdisplay n≥0anzn, where we now use the letter zto encourage thinking about complex vari- ables. Question: for exactly what set of complex values of zdoes the series fconverge? We want to give a fairly complete answer to this question, and express it in terms of the coeﬃcient sequence {an}∞ 0. Theorem 2.4.1. There exists a number R,0≤R≤+∞, called the radius of convergence of the series f, such that the series converges for all values ofzwith|z|<R and diverges for all zsuch that |z|>R. The number R is expressed in terms of the sequence {an}∞ 0of coeﬃcients of the series by means of R=1 lim supn→∞|an|1/n(1/0=∞;1/∞=0 ). (2.4.1) Before proving the theorem, we recall the deﬁnition of the limit superior of a sequence. Let {xn}∞ 0be a sequence of real numbers, and let Lbe a real number (possibly = ±∞). Deﬁnition. We say that Lis the limit superior (‘upper limit’) of the sequence {xn}if (a)Lis ﬁnite and (i) for every /epsilon1> 0all but ﬁnitely many members of the sequence satisfy xn<L+/epsilon1, and (ii) for every /epsilon1>0, inﬁnitely many members of the sequence satisfyxn>L−/epsilon1,o r (b)L=+∞and for every M> 0, there is an nsuch thatxn>M , or (c)L=−∞ and for every x, there are only ﬁnitely many nsuch that xn>x. IfLis the limit superior of the sequence {xn}∞ 0, then we write L= lim supn→∞{xn}, or perhaps just L= lim sup {xn}, if the context is clear enough. 2.4 Power series, analytic theory 47 The limit superior has the following properties: •Every sequence of real numbers has one and only one limit superior in the extended real number system (i.e., including ±∞). •If a sequence has a limitL, thenLis also the limit superior of the sequence. •IfSis the set of cluster points of the sequence {xn}∞ 0, then lim sup {xn}is the least upper bound of the numbers in S. Proof of theorem 2.4.1. LetRbe the number shown in (2.4.1), and suppose ﬁrst that 0 <R< ∞. Choosezsuch that |z|<R. We will show that the series converges at z. For the given z, we can ﬁnd /epsilon1>0 such that |z|<R 1+/epsilon1R. Now, by the deﬁnition of the lim sup, there exists Nsuch that for all n>N we have |an|1/n<1 R+/epsilon1. Hence, for these same n, |an||z|n</braceleftbigg |z|(1 R+/epsilon1)/bracerightbiggn . Letαdenote the number in the curly brace. Then by our choice of/epsilon1, we haveα< 1. Hence the series/summationtextanznconverges absolutely, by comparison with the terms of a convergent geometric series. Therefore our series converges absolutely at z, and hence it does so for all |z|<R. Next we claim the series diverges if |z|>R. Indeed, we will show that for such z, the sequence of terms of the series does not approach zero. Since |z|>R, we can choose /epsilon1>0 such that if θ=|(z/R)−/epsilon1z|, then θ>1. By deﬁnition of the lim sup, for inﬁnitely many values of nwe have |an|1/n>(1/R)−/epsilon1. Hence, for those values of n, |anzn|>/vextendsingle/vextendsingle/vextendsingle/vextendsingle/parenleftbigg1 R−/epsilon1/parenrightbigg z/vextendsingle/vextendsingle/vextendsingle/vextendsinglen =θn which increases without bound since θ> 1. Hence, that subsequence of terms of the power series does not approach zero and the series diverges. This completes the proof of the theorem in the case that 0 <R< ∞. The cases where R=0o rR=+∞are similar, and are left to the reader. Theorem 2.4.2. Suppose the power series/summationtextanznconverges for all zin |z|<R, and letf(z)denote its sum. Then f(z)is an analytic function in 48 2 Series |z|<R. If furthermore the series diverges for |z|>R, then the function f(z)must have at least one singularity on the circle of convergence |z|=R. In other words: a power series keeps on converging until something stops it, namely a singularity of the function that is being represented. Proof. Iffhas no singularity on its circle of convergence |z|=R, then about each point of that circle we can draw an open disk in which fremains analytic. By the Heine-Borel theorem, a ﬁnite number of these disks cover the circle |z|=R, and therefore fmust remain analytic in some larger disk |z|<R+/epsilon1. By Cauchy’s inequality, the Taylor coeﬃcients of the series for fsatisfy |an|≤M/(R+/epsilon1)n, for alln, and so the series must converge in a larger disk, a contradiction. Example 1. The series/summationtextznconverges if |z|<1 and diverges if |z|>1. Hence the function that is represented must have a singularity somewhere on the circle |z|= 1. That function is 1 /(1−z), and sure enough it has a singularity at z=1 . Example 2. Take the function f(z)=1/(2−ez). Suppose we expand f(z)i na power series about z= 0. What will be the radius of convergence of the series? According to the theorem, the series will converge in the largest disk |z|<R in whichfis analytic. The function fails to be analytic only at the pointszwhereez= 2. Those points are of the form z= log 2 + 2kπi, for integer k, and the nearest one to the origin is log 2. Therefore f(z)i s analytic in the disk |z|<log 2 and in no larger disk. Hence the radius of convergence of the series will be R= log 2. Remember that, if f(z) is given, the best way to ﬁnd the radius of convergence of its power series expansion about the origin may well be to look for its singularity that is nearest to the origin. Example 3. Take the function f(z)=z/(ez−1) (f(0) = 1). Estimate the size of the coeﬃcients of its power series about the origin directly from the analyticity properties of the function. This is where things start getting more interesting. This f(z) is ana- lytic except possibly at points zwhereez= 1, i.e., except possibly at the pointsz=2kπifor integerk. The nearest of these to the origin is the origin itself (k= 0). However, fis not singular at z= 0 because even though the denominator of fis 0 there, the numerator is also, and L’Hospital’s rule, or whatever, reveals that the value f(0) = 1 removes the singularity. Hence the singularity of this function that is nearest to the origin is at z=2πi. 2.4 Power series, analytic theory 49 The power series z ez−1=∞/summationdisplay n=0anzn therefore has radius of convergence R=2π. The problem asks for estimates of the sizes of the coeﬃcients {an}∞ 0. But since the radius of convergence is 2 π, we have, from theorem 2.4.1, lim sup |an|1/n=1 2π. It follows that, ﬁrst of all, for all suﬃciently large values of nwe have |an|1/n<1 2π+/epsilon1, and for inﬁnitely many values of nwe have |an|1/n>1 2π−/epsilon1. Therefore, what we ﬁnd out about the coeﬃcients is that for each /epsilon1>0, there exists Nsuch that |an|</parenleftbigg1 2π+/epsilon1/parenrightbiggn (n>N ) and further, for inﬁnitely many values of n, |an|>/parenleftbigg1 2π−/epsilon1/parenrightbiggn . Therefore the coeﬃcients of this series decrease to zero exponentially fast, at roughly the rate of 1 /(2π)n, for large n. This is quite a lot to have found out about the sizes of the coeﬃcients without having calculated any of them! We state, for future reference, a general proposition that summarizes what we learned in this example. Theorem 2.4.3. Letf(z)=/summationtextanznbe analytic in some region containing the origin, let a singularity of f(z)of smallest modulus be at a point z0/negationslash=0, and let/epsilon1>0be given. Then there exists Nsuch that for all n>N we have |an|</parenleftbigg1 |z0|+/epsilon1/parenrightbiggn . Further, for inﬁnitely many nwe have |an|>/parenleftbigg1 |z0|−/epsilon1/parenrightbiggn . 50 2 Series In chapter 5 we will learn how to make much more precise estimates of the sizes of the coeﬃcients of power series based on the analyticity, or lack thereof, of the function that is represented by the series. For instance, the method of Darboux (Theorem 5.3.1) is a powerful technique for asymptotic analysis of coeﬃcient sequences of generating functions. The existence of such methods is an excellent reason why we should be knowledgeable about the analytic, as well as the formal, side of the subject of generating func- tions. Another path to the asymptotic analysis of coeﬃcient sequences ﬂows from Cauchy’s formula an=1 2πi/integraldisplayf(z)dz zn+1(n=0,1,2,...)( 2 .4.2) that expresses the nth coeﬃcient of the Taylor’s series expansion f(z)=/summationtextanznas a contour integral involving the function f. The contour can be any simple, closed curve that encloses the origin and that lies entirely inside a region in which fis analytic. One has immediately, from (2.4.2), Cauchy’s inequality, which states that |an|≤M(r) rn, and which holds for all n≥0 and all 0 <r<R , whereRis the radius of convergence of the series, and M(r) = max |z|≤r|f(z)|= max |z|=r|f(z)|. (2.4.3) Just as the analysis of Example 3 above is reﬁned by the method of Dar- boux to a much more precise method of estimating the growth of coeﬃcientsequences, so is Cauchy’s inequality reﬁned by the method of Hayman (The- orem 5.4.1) to another very precise tool for the same purpose. If a power series actually converges to a function, then we can use roots of unity to pick out a progression of terms from a series. For instance, how can we select just the even powers out of a power series? If the series represents a function f, then, as is well known, ( f(x)+f(−x))/2 has just the terms that involve even powers of xfrom the series for f(x), and (f(x)− f(−x))/2 has just the odd ones. But suppose, instead of wanting to keep every second term of the series, we want to keep only every third term? For instance, what function do we get if we take the exponential series and keep just the terms where the powers ofxare multiples of 3? In other words, who is g(x)=/summationdisplay n≥0x3n (3n)!?( 2 .4.4) 2.4 Power series, analytic theory 51 Well, what makes the ( f(x)+f(−x))/2 thing work is that the two square roots of unity , namely ±1, have the property that 1n+(−1)n 2=/braceleftbigg 1,ifnis even; 0,ifnis odd. Now here is a correspondingly helpful property of the three cube roots of unity 1,ω1,ω2: (1n+ωn 1+ωn 2) 3=/braceleftbigg 1,if 3\n; 0,else.(2.4.5) Since that is the case, we have, for any convergent power series f=/summationtext rarxr, f(x)+f(ω1x)+f(ω2x) 3=/summationdisplay ra3rx3r. (2.4.6) Sinceω1=e(2πi)/3andω2=e(4πi)/3, we can unmask the mystery functiong(x) in (2.4.4) as g(x)=1 3(ex+eω1x+eω2x) =1 3/parenleftBigg ex+2e−x/2cos (√ 3x 2)/parenrightBigg .(2.4.7) Example 4. For ﬁxedn, ﬁnd λn=/summationdisplay k(−1)k/parenleftbiggn 3k/parenrightbigg . We could do this one if we knew the function f(x)=/summationdisplay k/parenleftbiggn 3k/parenrightbigg x3k, becauseλn=f(−1). Butf(x) picks out every third term from the series F(x)=( 1+x)n, and so f(x)=(F(x)+F(ω1x)+F(ω2x))/3 ={(1 +x)n+( 1+ω1x)n+( 1+ω2x)n}/3. Thus the numbers that we are asked to ﬁnd are, for n>0, λn=f(−1) ={(1−ω1)n+( 1−ω2)n}/3 =1 3/braceleftBigg/parenleftBigg 3−√ 3i 2/parenrightBiggn +/parenleftBigg 3+√ 3i 2/parenrightBiggn/bracerightBigg =2·3(n/2−1)cos (nπ 6).(2.4.8) 52 2 Series The ﬁrst few values of the {λn}n≥0are 1, 1, 1, 0, −3,−9,−18,.... To complete this example we want to prove the helpful property (2.4.5) of the cube roots of unity. But for every r>1, therth roots of unity do the same sort of thing, namely 1 r/summationdisplay ωr=1ωn=/braceleftBig1i fr\n 0 else.(2.4.9) Indeed, the left side is 1 rr−1/summationdisplay j=0e(2πijn )/r, which is a ﬁnite geometric series whose sum is easy to ﬁnd, and is as stated in (2.4.9). So, with more or less diﬃculty, it is always possible to select a subset of the terms of a convergent series in which the exponents form an arithmetic progression. See exercise 25. 2.5 Some useful power series Generatingfunctionologists need reference lists of known power series and other series that occur frequently in applications of the theory. Here is such a list. For each series we show the series and its sum. The radius of the largest open disk, centered at the origin, in which convergence takes place will be, of course, the modulus of the singularity of the function that is nearest to the origin. Considering the relatively simple forms of the functions, the locations of those singularities will be suﬃciently obvious that the radii of convergence are not explicitly shown in the table below. 1 1−x=/summationdisplay n≥0xn(2.5.1) log1 1−x=/summationdisplay n≥1xn n(2.5.2) ex=/summationdisplay n≥0xn n!(2.5.3) sinx=/summationdisplay n≥0(−1)nx2n+1 (2n+ 1)!(2.5.4) cosx=/summationdisplay n≥0(−1)nx2n (2n)!(2.5.5) 2.5 Some useful power series 53 (1 +x)α=/summationdisplay k/parenleftbiggα k/parenrightbigg xk(2.5.6) 1 (1−x)k+1=/summationdisplay n/parenleftbiggn+k n/parenrightbigg xn(2.5.7) x ex−1=/summationdisplay n≥0Bnxn n!(2.5.8) tan−1x=/summationdisplay n≥0(−1)nx2n+1 2n+1(2.5.9) 1 2x(1−√ 1−4x)=/summationdisplay n1 n+1/parenleftbigg2n n/parenrightbigg xn(2.5.10) =1+x+2x2+5x3+1 4x4+4 2x5+ 132x6 + 429x7+ 1430x8+ 4862x9+··· 1√1−4x=/summationdisplay k/parenleftbigg2k k/parenrightbigg xk(2.5.11) =1+2x+6x2+2 0x3+7 0x4+ 252x5+ 924x6 + 3432x7+ 12870x8+ 48620x9+··· xcotx=/summationdisplay k≥0(−4)kB2k (2k)!x2k(2.5.12) =1−x2 3−x4 45−2x6 945−x8 4725−2x10 93555−··· tanx=/summationdisplay r≥1(−1)r−122r(22r−1)B2r (2r)!x2r−1(2.5.13) =x+x3 3+2x5 15+17x7 315+62x9 2835+1382x11 155925+··· +21844x13 6081075+929569x15 638512875+··· 54 2 Series x sinx=/summationdisplay r≥0(−1)r−1(4r−2)B2r (2r)!x2r =1+x2 6+7x4 360+31x6 15120+··· (2.5.14) 1√1−4x/parenleftbigg1−√1−4x 2x/parenrightbiggk =/summationdisplay n/parenleftbigg2n+k n/parenrightbigg xn(2.5.15) /parenleftbigg1−√1−4x 2x/parenrightbiggk =/summationdisplay n≥0k(2n+k−1)! n!(n+k)!xn(k≥1) (2 .5.16) sin−1(x)=x+1 2x3 3+1·3 2·4x5 5+1·3·5 2·4·6x7 7+··· (2.5.17) exsinx=/summationdisplay n≥12n 2sinnπ 4 n!xn(2.5.18) =x+x2+x3 3−x5 30−x6 90−x7 630+··· 1 2tan−1(x) log (1 +x2)=/summationdisplay r≥1(−1)r−1H2rx2r+1 2r+1(2.5.19) =x3 2−5x5 12+7x7 20−761x9 2520+··· 1 4tan−1(x) log1+x 1−x=/summationdisplay r≥0x4r+2 4r+2/parenleftbigg 1−1 3+1 5−··· +1 4r+1/parenrightbigg =x2 2+13x6 90+263x10 3150+··· (2.5.20) 1 2/braceleftbigg log1 1−x/bracerightbigg2 =/summationdisplay r≥2Hr−1 rxr(2.5.21) 2.5 Some useful power series 55 =x2 2+x3 2+11x4 24+5x5 12+137x6 360+7x7 20+··· /radicalBigg 1−√1−x x=∞/summationdisplay k=0(4k)! 16k√ 2(2k)!(2k+ 1)!xk(2.5.22) =1√ 2/parenleftbigg 1+x 8+7x2 128+33x3 1024+715x4 32768+4199x5 262144 +52003x6 4194304+334305x7 33554432+17678835x8 2147483648 +119409675x9 17179869184+1641030105 x10 274877906944+···/parenrightbigg earcsin x=∞/summationdisplay k=0/producttextk−1 j=0(4j2+1 ) (2k)!x2k+∞/summationdisplay k=04k/producttextk j=1(1 2−j+j2) (2k+ 1)!x2k+1 =1+x+x2 2+x3 3+5x4 24+x5 6+17x6 144+13x7 126 +629x8 8064+325x9 4536+8177x10 145152+··· (2.5.23) /parenleftbiggarcsinx x/parenrightbigg2 =∞/summationdisplay k=04kk!2 (k+ 1)(2k+ 1)!x2k(2.5.24) =1+x2 3+8x4 45+4x6 35+128x8 1575+128x10 2079+··· (x+/radicalbig 1+x2)a=∞/summationdisplay k=02k·(a 2−k 2+1 )k (1 +k/a)k!xk(2.5.25) =1+ax+a2x2 2+/parenleftbigg−a 6+a3 6/parenrightbigg x3+/parenleftbigg−a2 6+a4 24/parenrightbigg x4 +a/parenleftbig 9−10a2+a4/parenrightbig x5 120+a2/parenleftbig 64−20a2+a4/parenrightbig x6 720 +a/parenleftbig −225 + 259a2−35a4+a6/parenrightbig x7 5040 +a2/parenleftbig −2304 + 784 a2−56a4+a6/parenrightbig x8 40320+··· 56 2 Series In the above, the {Bn}are the Bernoulli numbers , and they are deﬁned by (2.5.8). The Bernoulli numbers {Bn}16 0have the values 1,−1/2,1 6,0,−1 30,0,1 42,0,−1 30,0,5 66,0,−691 2730,0,7 6,0,−3617 510. The{Hn}are the harmonic numbers that were deﬁned in section 2.2. The symbol mk, in (2.5.25), means m(m+1)···(m+k−1). The expansions (2.5.22)-(2.5.25) are taken from [Ko]. 2.6 Dirichlet series, formal theory We have already discussed two slightly diﬀerent forms of generating functions of sequences, namely the ordinary power series form and the ex- ponential generating function form. We remarked that when, in a particular problem, one has to decide which of these forms to use, the choice is mostoften dictated by the form of the multiplicative convolution of the two se- quences that occurs in the problem. If the form is as in Rule 3 /primeand (2.3.3), then we choose the egf, whereas if it is of the form (2.3.4), the opsgf may well be preferred. To help highlight the basis for this kind of choice, we will now discuss yet another kind of generating function that matches yet another kind ofconvolution of two sequences, a kind that also occurs naturally in many problems in combinatorics and number theory. Deﬁnition. Given a sequence {a n}∞ 1; we say that a formal series f(s)=∞/summationdisplay n=1an ns =a1+a2 2s+a3 3s+a4 4s+···(2.6.1) is the Dirichlet series generating function (Dsgf) of the sequence, and we write f(s)Dir ←→{an}∞ 1. The importance of Dirichlet series stems directly from their multipli- cation rule. Suppose f(s)Dir ←→{an}∞ 1andg(s)Dir ←→{bn}∞ 1. The question is, what sequence is generated by f(s)g(s)? To ﬁnd out, consider the product of these series, fg=(a1+a22−s+a33−s+···)(b1+b22−s+b33−s+···) =(a1b1)+(a1b2+a2b1)2−s+(a1b3+a3b1)3−s +(a1b4+a2b2+a4b1)4−s+··· 2.6 Dirichlet series, formal theory 57 What is the general rule? In the product fg, what is the coeﬃcient of n−s? It is the sum of all products of a’s andb’s where the product of their subscripts is n, i.e., it is/summationdisplay rs=narbs. Now ifrs=nthenrandsare divisors of n, so the above sum can also be written as /summationdisplay d\nadbn d, in which the symbol ‘ d\n’ is read ‘ddividesn.’ We state this formally as: Rule 1/prime/prime.Iff(s)Dir ←→{an}∞ 1andg(s)Dir ←→{bn}∞ 1, then f(s)g(s)Dir ←→  /summationdisplay d\nadbn d  ∞ n=1. (2.6.2) Let’s hasten to say what kind of a problem gives rise to this kind of a convolution of sequences. It is, roughly, a situation in which all objects of sizenare obtained by stitching together dobjects of size n/d, wheredis some divisor of n. Before we get to examples of this sort of thing, since the multiplication is so important, let’s look at a few more of its properties. What happens to the sequence generated if we take the kth power of a Dirichlet series? Let’s work it out, as follows: f(s)k= /summationdisplay n≥1ann−s k =/summationdisplay n1,...,n k≥1an1···ank(n1n2···nk)−s =/summationdisplay n≥1n−s/braceleftBigg/summationdisplay n1···nk=nan1···ank/bracerightBigg . This shows: Rule 2/prime/prime.Iff(s)Dir ←→{an}∞ 1thenf(s)kDir ←→a sequence whose nth member is the sum, extended over all ordered factorizations of nintokfactors, of the products of the members of the sequence whose subscripts are the factors in that factorization. What series fgenerates the sequence of all 1/primes:{1}∞ 1? When we asked that question in the cases of the opsgf and the egf, the answers turned out to be ‘famous’ functions. For opsgf’s it was 1 /(1−x) and for egf’s it was ex. In the present case, the formal Dirichlet series whose coeﬃcients are all 1’s 58 2 Series is not related to any simple function of analysis, it is a new creature, and it gets a new name: the Riemann zeta function . It is the Dirichlet series ζ(s)=∞/summationdisplay n=11 ns =1−s+2−s+3−s+4−s+···,(2.6.3) and it is one of the most important functions in analysis. Now, sinceζ(s)Dir ←→{1}∞ 1, what sequence does ζ2(s) generate? Directly from (2.6.2), [n−s]ζ2(s)=/summationdisplay d\n1·1=d(n), whered(n) is the number of divisors of the integer n. The sequence d(n)i s quite irregular, and begins with 1,2,2,3,2,4,2,4,3,4,2,... Nevertheless, its Dirichlet series generating function is ζ2(s), by Rule 2/prime/prime. Likewise,ζ(s)kgenerates the number of ordered factorizations of n intokfactors. If the factor 1 is regarded as inadmissible, then ( ζ(s)−1)k generates the number of ordered factorizations of nin which there are k factors, all ≥2. One can go on and study further examples of interesting number- theoretic sequences that are generated by relatives of the Riemann zeta function, but there is a somewhat breathtaking generalization that takes in all of these at a single swoop, so let’s prepare the groundwork for that. Deﬁnition. A number-theoretic function is a function whose domain is the set of positive integers. A number-theoretic function fis said to be multiplicative if it has the property that f(mn)=f(m)f(n)for all pairs of relatively prime positive integers mandn. Since every positive integer nis uniquely, apart from order, a product of powers of distinct primes, n=pa1 1pa2 2···par r, (2.6.4) it follows that a multiplicative number-theoretic function is completely de- termined by its values on all powers of primes . Indeed, f(n)=f(pa1 1)f(pa2 2)···f(par r). (2.6.5) For instance, suppose that I have a certain function fin mind. It is multiplicative and, further, for every prime pand positive integer mwe 2.6 Dirichlet series, formal theory 59 havef(pm)=p2m. Well then, it must be that f(n)=n2for alln, because ifnis as shown in (2.6.4), then f(n)=f(/productdisplay pai i)=/productdisplay if(pai i) =/productdisplay ip2ai i=/braceleftBigg/productdisplay ipai i/bracerightBigg2 =n2, as claimed. Another, less obvious, example of a multiplicative function is d(n), the number of divisors of n. For instance, 6=d(12) =d(3·4) =d(3)d(4) = 2 ·3=6. To see that d(n) is multiplicative in general, let mandnbe relatively prime positive integers. Then every divisor dofmnisuniquely the product of a divisord/primeofmand a divisor d/prime/primeofn. Indeed, we can take d/prime=gcd(d,m) andd/prime/prime=gcd(d,n). Therefore the number of divisors of mnis the product of the number of divisors of mand the number of divisors of n, which was to be shown. It is quite easy, therefore, to dream up examples of multiplicative func- tions: let your function fdo anything it likes on the powers of primes, then declare it to be multiplicative, and walk away. Multiplicative number-theoretic functions satisfy an amazing identity, which we will state, then prove, and then use. Theorem 2.6.1. Letfbe a multiplicative number-theoretic function. Then we have the formal identity ∞/summationdisplay n=1f(n) ns=/productdisplay p/braceleftbig 1+f(p)p−s+f(p2)p−2s+f(p3)p−3s+···/bracerightbig (2.6.6) in which the product on the right extends over all prime numbers p. Proof. Imagine, if you will, multiplying out the product that appears on the right side of (2.6.6). Each factor in that product is an inﬁnite series. The product looks like this, when spread out in detail: (1 +f(2)2−s+f(22)2−2s+f(23)2−3s+···)× (1 +f(3)3−s+f(32)3−2s+f(33)3−3s+···)× (1 +f(5)5−s+f(52)5−2s+f(53)5−3s+···)× (1 +f(7)7−s+f(72)7−2s+f(73)7−3s+···)×···(2.6.7) 60 2 Series To multiply out a bunch of formal inﬁnite series like this, we reach into the ﬁrst parenthesis and pull out one term, for instance f(23)2−3s. Then we reach into the second parenthesis, pull out one term, say f(3)3−sand multiply it by the one we got earlier. This gives us an accumulated product (so far) of f(23)f(3)2−3s3−s=f(23)f(3) (24)s. (2.6.8) Suppose, just as an example, that in all of the following parentheses we exercise our choice of one term by pulling out the term ‘1.’ Then, as a result of having made all of those choices, one out of each parenthesis, the contribution to the answer would be the single term shown in (2.6.8) above. Now here’s the interesting part. That particular set of choices has produced a term that involves (24)−s. What other sequence of choices of a single term out of each parenthesis would alsolead to a net contribution that involves (24)−s? The answer: no other set of choices can do that . Indeed, if from the ﬁrst parenthesis we choose any term other than f(23)2−3s, then no matter what terms we pull out of all following paren- theses, there is no way we will ever ﬁnd the power of 2, namely 2−3s, that occurs in (24)−s. We need three 2’s, and no other parenthesis has any 2’s at all to oﬀer, so we’d better take them when we have the chance. Similarly, we need a factor of 3−sin order to complete the formation of the term (24)−s. There are no 3’s available in any parenthesis other than the second one, and there, to get the right number of 3’s, namely one, we had better take the term f(3)3−sthat we actually chose. Thus, the coeﬃcient of (24)−son the right side of (2.6.6) is just what we found, namely f(23)f(3). Now since fis multiplicative, that’s the same asf(24). Hence the coeﬃcient of (24)−sisf(24). But that is just what the left side of (2.6.6) claims. Let’s say that again, using ‘ n’ instead of ‘24.’ Let nbe some ﬁxed integer, and let (2.6.4) be its factorization into prime powers. In order to obtain a term that involves n−s, i.e., that involves/producttextp−ais i, we are forced to choose the ‘1’ term in every parenthesis on the right side of (2.6.7), except for those parentheses that involve the primes pithat actually occur in n. Inside a parenthesis that belongs to pi, we must choose the one and only term in which piis raised to the power with which it actually occurs in n, else we won’t have a chance of getting n−s. Thus we are forced to choose the termf(pai i)p−ais i out of the parenthesis that belongs to pi. That means that the coeﬃcient of n−sin the end will be /productdisplay if(pai i)=f(n), by (2.6.5). Let’s look again at (2.6.6). One thing that is very apparent is that a multiplicative function is completely determined by its values on all prime 2.6 Dirichlet series, formal theory 61 powers. Indeed, on the right side of (2.6.6) we see only the values of fat prime powers, but on the left, all values appear. Try an example of the theorem. Take the multiplicative function f(n) = 1 (alln). Then (2.6.6) says that ζ(s)=/productdisplay p/braceleftbig 1+p−s+p−2s+···/bracerightbig =/productdisplay p/braceleftbigg1 1−p−s/bracerightbigg =1/producttext p(1−p−s),(2.6.9) which is a fundamental factorization of the zeta function. For another example, take the multiplicative function µ(n) whose val- ues on prime powers are µ(pa)=/braceleftBigg+1,ifa=0 ; −1,ifa=1 ; 0,ifa≥2. With this function substituted for fin (2.6.6), one sees that the once formidable series in the braces now has only two terms, and (2.6.6) reads /summationdisplay n≥1µ(n) ns=/productdisplay p{1−p−s}. (2.6.10) An important fact emerges by comparison of (2.6.9) with (2.6.10): the seriesζ(s) and the series on the left side of (2.6.10) are reciprocals of each other. Hence, 1 ζ(s)=/summationdisplay n≥1µ(n) ns, or, what amounts to the same thing, 1 /ζ(s)Dir ←→{µ(n)}∞ 1. The function µ(n) is the M¨ obius function , and it plays a central role in the analytic theory of numbers, because of the fact that it is generated by the reciprocal of the Riemann zeta function. For instance, watch this: Suppose we have two sequences {an}∞ 1and{bn}∞ 1, and suppose that these two sequences are connected by the following equations- an=/summationdisplay d\nbd (n≥1). (2.6.11) The question is, how can we invert these equations, and solve for the b’s in terms of the a’s? 62 2 Series Nothing to it. Let the Dsgf’s of the two sequences be A(s) andB(s). Then, if we take a step into Generatingfunctionland, we see that (2.6.11) means A(s)=B(s)ζ(s) by Rule 1/prime/prime. HenceB(s)=A(s)/ζ(s), and then from Rule 1/prime/primeagain, bn=/summationdisplay d\nµ/parenleftBign d/parenrightBig ad (n=1,2,3,...)( 2 .6.12) This is the celebrated M¨ obius Inversion Formula. The reciprocal relation- ships (2.6.11) and (2.6.12) of the sequences mirror the reciprocal relation- ships of their Dsgf’s ζ(s) and 1/ζ(s). Example 1. Primitive bit strings. How many strings of n0’s and 1’s are primitive , in the sense that such a string is notexpressible as a concatenation of several identical smaller strings? For instance, 100100100 is not primitive, but 1101 is. There are a total of 2nstrings of length n. Supposef(n) of these are primitive. Every string of length nisuniquely expressible as a concatenation of some number, n/d, of identical primitive strings of length d, wheredis a divisor of n. Thus we have 2n=/summationdisplay d\nf(d)(n=1,2,...) By (2.6.12) we have f(n)=/summationdisplay d\nµ(n d)2d(n=1,2,...)( 2 .6.13) for the required number. Example 2. Cyclotomic polynomials Among the nroots of the equation xn= 1, the primitiventh roots of unity are those that are not also mth roots of unity for some m<n . Thus the 4th roots of unity are ±1,±i, but±1 are roots of x2= 1, so they aren’t primitive 4th roots. In general, the nth roots of unity are {e2πir/n}n−1 r=0, and the primitive ones are {e2πir/n} (0≤r≤n−1; gcd(r,n)=1 ). So for each nthere are exactly φ(n) primitive nth roots of unity. The equation whose roots are allnof thenroots of unity is obviously the equation xn−1 = 0. The question is this: what is the polynomial 2.6 Dirichlet series, formal theory 63 Φn(x) of degree φ(n) whose roots are exactly the set of primitiventh roots of unity? In other words, what can be said about the polynomial Φn(x)=/productdisplay 0≤r≤n−1 gcd(r,n)=1(x−e2πir/n)(n=1,2,3,...)? The polynomials Φ n(x) are called the cyclotomic (“circle-cutting”) polyno- mials. The important fact for answering this question is that /productdisplay d\nΦd(x)=1−xn(n=1,2,3,...). (2.6.14) Indeed, the right side of the equation is the product of all possible factors (ω−x) whereωis annth root of unity, primitive or not. But every nth root of unity is a primitivedth root of unity for exactly one d≤n, and thatdis a divisor of n. In detail, if we have some nth rootω=e2πir/n, then letg= gcd(r,n), d=n/g, andr/prime=r/g. Sinceω=e2πir/prime/dwe see that ωis a primitive dth root of unity and that d\n. Thus every linear factor on the right side of (2.6.14) occurs in one and only one of the cyclotomic polynomials on the left side of (2.6.14), which proves the assertion. From (2.6.14) we will obtain a fairly explicit formula for the Φ n(x), by inverting the equation to solve for the Φ’s. The form of the equation reminds us of the setup (2.6.11) for the M¨ obius inversion formula, but we have a product over divisors instead of a sum over divisors. A small dose of logarithms will convert products to sums, however, so we take the logarithm of both sides of (2.6.14), to get /summationdisplay d\nlog Φ d(x) = log (1 −xn)(n=1,2,3,...). This is now precisely in the form (2.6.11), so we can use (2.6.12) to invert it, the result being log Φ n(x)=/summationdisplay d\nµ(n d) log (1 −xd)(n=1,2,3,...). Finally, we exponentiate both sides to obtain our “fairly explicit formula,” Φn(x)=/productdisplay d\n(1−xd)µ(n/d)(n=1,2,3,...). (2.6.15) This is a good time to remember that the values of the M¨ obius function µcan only be ±1 or 0. So the exponents on the right side of (2.6.15) tell 64 2 Series us whether to omit a certain factor, which we do if µ= 0, to put it in the numerator (if µ= 1), or to put it in the denominator (if µ=−1). For instance, Φ 12(x)i s (1−x)µ(12)(1−x2)µ(6)(1−x3)µ(4)(1−x4)µ(3)(1−x6)µ(2)(1−x12)µ(1) =( 1−x)0(1−x2)1(1−x3)0(1−x4)−1(1−x6)−1(1−x12)1 =(1−x2)(1−x12) (1−x4)(1−x6) =1−x2+x4, which didn’t look much like a polynomial at all until the very last step! An important and beautiful fact about these polynomials is that the equation Φ n(z) = 0 can always be solved by radicals. That is, the solutions can always be obtained by a ﬁnite number of root extractions and rationaloperations. This is certainly not the case for general polynomial equations. As an example of this property we note the splendid fact that cos2π 17=1 16/braceleftbigg −1+√ 17 +/radicalBig 2(17−√ 17) +2/radicalbigg 17 + 3√ 17−/radicalBig 2(17−√ 17)−2/radicalBig 2(17 +√ 17)/bracerightbigg . The proof is fairly diﬃcult, and can be found in Rademacher [Ra]. Some applications of cyclotomic polynomials will appear in section 4.10. Exercises 65 Exercises 1. Calculate the ﬁrst three coeﬃcients of the reciprocals of the power series of the functions: (a) cosx (b) (1 +x)m (c) 1 +t2+t3+t5+t7+t11+··· 2. Calculate the ﬁrst three coeﬃcients of the inverses of the power series for the functions: (a) sinx (b) tanx (c)x+x2√1+x (d)x+x3 (e) log (1 −x) 3. Letfbe a formal power series such that f/prime/prime+f= 0. Give a careful proof thatf=Asinx+Bcosx. 4. Find simple closed formulas for the opsgf’s of the following sequences: (a){n+7}∞ 0 (b){1}∞ 4 (c){1,0,1,0,1,0,1,0,...} (d){1/(n+1 )}∞ 2 (e){1/(n+ 5)!}∞ 0 (f)F1,2F2,3F3,4F4,...(theF’s are the Fibonacci numbers) (g){(n2+n+1 )/n!}∞ 1 5. Use generating functions to prove that/summationtext k/parenleftbign k/parenrightbig =2n. 6. Given positive integers n,k; deﬁnef(n,k) as follows: for each way of writingnas an ordered sum of exactly knonnegative integers, let Sbe the product of those kintegers. Then f(n,k) is the sum of all of the S’s that are obtained in this way. Find the opsgf of fand an explicit, simple formula for it. 7. Letf(n,k,h ) be the number of ordered representations of nas a sum of exactlykintegers, each of which is ≥h. Find/summationtext nf(n,k,h )xn. 8. Find the limit superior of each of the following sequences. In each case give a careful proof that your answer is correct. (a) 1,0,1,0,1,0,... (b){(−1)n}∞ 0 66 2 Series (c){cos (nπ/k )}n≥0 (k/negationslash= 0 is a ﬁxed integer) (d){1+( ( −1)n/n)}n≥1 (e){n1/n}n≥1 9. Prove that if a sequence has a limit then its limit superior is equal to that limit. 10. Prove that a sequence cannot have two distinct limits superior. 11. Find the radius of convergence of each of the following power series: (a)/summationtext n≥1xn/(n2) (b) 1 +x3+x6+x9+x12+··· (c) 1 + 5x2+2 5x4+ 125x6+··· (d) 1 + 2!x2+4 !x4+6 !x6+··· (e)/summationtext n≥0xn! 12. Finish the proof of theorem 2.4.1 in the cases where R= 0 andR=∞. 13. Show that if {f(n)}∞ 1is a multiplicative function, then so is g(n)=/summationdisplay d\nf(d)(n=1,2,...) 14. Euler’s function φ(n) is the number of integers 1 ≤m≤nsuch thatm is relatively prime to n. Show by a direct counting argument that /summationdisplay d\nφ(d)=n (n=1,2,...). 15. Show that each of the following functions is multiplicative. In each case ﬁnd the value of the function when nis a prime power, and thereby ﬁnd a formula for its value on any integer n. (a) Euler’s function φ(n) (use the results of problems 13, 14 above). (b)σ(n), which is the sum of the divisors of n. (c) The function |µ(n)|, which is 1 if nis not divisible by a square and 0 otherwise. 16. For each of the functions deﬁned in problem 15 above, ﬁnd its Dirichlet series generating function by using Theorem 2.6.1. First substitute into (2.6.6) the values of the function at prime powers. Then try to sum the power series that occurs, in closed form. Finally, by comparing the product that results with (2.6.9), try to express your answer simply in terms of the Riemann zeta function. In each case the Dsgf can be simply expressed interms ofζ(s), or small variations thereof. 17. Find the Dsgf of each of the following sequences: Exercises 67 (a){n}∞ 1 (b){nα}∞ 1 (c){logn}∞ 1 (d){/summationtext d\ndq}∞ n=1 18. For each of the following identities: ﬁrst check that the identity is sometimes correct by calculating both sides of the alleged equation when n=1,2,3,4,5,6,7,8; next ﬁnd the Dsgf’s of the sequences on both sides of the claimed identity, observe that they are the same, and thereby prove the identity. Use the results of exercise 16 above. (a)/summationtext d\nφ(d)=n (n≥1) (b)/summationtext d\nµ(d)=0i fn≥2 and =1 if n=1 (c)/summationtext δ\nµ(δ)d(n/δ)=1 (n≥1) 19. If {f(k)}is the sequence in example 7 of section 2.2, show that f(k)= F2k−1fork≥1, where the {Fk}are the Fibonacci numbers. 20. Prove the binomial theorem (x+y)n=/summationdisplay k/parenleftbiggn k/parenrightbigg xkyn−k by comparing the coeﬃcient of tn/n! on both sides of the equation et(x+y)= etxety. Prove the multinomial theorem (x1+···+xk)n=/summationdisplay r1+···+rk=nn! r1!···rk!xr1 1···xrk k by a similar device. 21. (a) LetTbe a ﬁxed set of nonnegative integers. Let f(n,k,T )b e the number of ordered representations of nas a sum of kintegers chosen from T. Find/summationtext nf(n,k,T )xn. (b) Letg(n,k,T ) be the number of ordered representations of nas a sum ofkdistinct integers chosen from T. Find/summationtext ng(n,k,T )xn. (c) Finally, let S,T be two ﬁxed sets of nonnegative integers. Let f(n,k,S,T ) be the number of ordered representations of nas a sum ofkintegers chosen from T, each being chosen with a multi- plicity that belongs to S. Find/summationtext nf(n,k,S,T )xn. 22. Letf(n) be the excess of the number of ordered representations of n as the sum of an even number of positive integers over those as a sum of an odd number of them. Find f(n) by ﬁnding/summationtext nf(n)xnand reading oﬀ its coeﬃcients. 68 2 Series 23. Let {Bn}be the sequence of Bernoulli numbers deﬁned by (2.5.8), and letmbe a positive integer. By considering the generating function x(emx−1) ex−1 in two ways, ﬁnd an evaluation of the sum of the rth powers of the ﬁrst N positive integers as a polynomial of degree r+1i nN, whose coeﬃcients are given quite explicitly in terms of the Bernoulli numbers. 24. (a) Make a table of values of the classical M¨ obius function µ(n) for n=1,2,..., 30. (b) Make a table of the values of the function f(n) of (2.6.13) for n=1,2,..., 12. (c) Make a list of the primitive strings of length 6, and verify your value off(6). 25. This problem is intended to show how generating functions occur in coding theory. An important question in coding theory is the following: for ﬁxed integers nandd, what is the length A(n,d) of the longest list of n-bit strings of 0’s and 1’s ( codewords ) such that two distinct codewords always diﬀer in at least dbit positions? (a) Assign to each of the 2ncodewords ( /epsilon11,...,/epsilon1 n)acolor , as follows: the color of /epsilon1is/summationtext jj/epsilon1jmodulo 2n. Show that if two codewords diﬀer in just 1 or 2 coordinates, then they are assigned distinct colors in this scheme. (b) From part (a), show that A(n,3)≥2n/(2n). (c) Letajbe the number of codewords for which/summationtext rr/epsilon1r=j, for each j. Find the opsgf f(z)=/summationtext jajzjexplicitly as a product. (d) Ifβris the number of codewords of color r, then express βrin terms of the aj’s above. Then use the roots of unity method of section 2.4 to ﬁnd that βr=1 2nn/prime/prime/summationdisplay j=122a(j,n/prime/prime)e−2πirj n/prime/prime for eachr=0,1,..., 2n−1. Hereaandn/prime/primeare deﬁned by n=2an/prime/prime wheren/prime/primeis odd, and ( b,c) denotes the g.c.d. of bandc. (e) Deduce that β0is the largest of the βr’s, and therefore ﬁnd the stronger bound A(n,3)≥1 2nn/prime/prime/summationdisplay j=122a(j,n/prime/prime)≥2n 2n. Exercises 69 (f) Use Parseval’s identity and the result of part (d) to ﬁnd the vari- ance of the occupancy numbers β0,...,β 2n−1. Make an estimate that shows that the variance is in some sense very small, so that this coloring scheme is shown to distribute codewords into color classes very uniformly. 26. Derive (2.5.7) from (2.5.6). That is, show that /parenleftbigg−n k/parenrightbigg =(−1)k/parenleftbiggn+k−1 k/parenrightbigg . 27. LetD(n) be the number of derangements of nletters, discussed in Example 4. (a) Find, in simple explicit form, the egf of {D(n)}∞ 0. (b) Prove, by any method, that D(n+1 )=(n+1 )D(n)+(−1)n+1(n≥0;D(0) = 1) (c) Prove, by any method, that D(n+1 )=n(D(n)+D(n−1)) (n≥1;D(0) = 1;D(1) = 0). (c) Show that the number of permutations of nletters that have exactly 1 ﬁxed point diﬀers from the number with no ﬁxed points by ±1. (d) LetDk(n) be the number of permutations of nletters that have exactlykﬁxed points. Show that /summationdisplay k,n≥0Dk(n)xnyk n!=e−x(1−y) 1−x. 28. Prove the following variation of the M¨ obius inversion formula. Let {an(x)}and{bn(x)}be two sequences of functions that are connected by the relation an(x)=/summationdisplay d\nbn d(xd)(n=1,2,3,...). Then we have bn(x)=/summationdisplay d\nµ(n d)ad(xn/d)(n=1,2,3,...). 29. (a) Make a table of the values of φ(n) for 1 ≤n≤25. 70 2 Series (b) As far as your table goes, verify that φis a multiplicative function, by actual computation. Then check by actual computation from your table, that the result stated in exercise 14 above is true when n=2 0 andn= 24. (c) Letn=pawherepis a prime number. What is φ(n)? (d) Use the results above to ﬁnd a general formula for φ(n) in terms of the prime factorization n=pa1 1pa2 2···pak kofn. Use your result to calculate φ(2592). (e) Find the Dirichlet series generating function of φ(n), using (2.6.6), and express it in terms of the Riemann zeta function. (f) Apply the M¨ obius inversion formula to the result of exercise 18(a), and thereby “solve” 18(a) for φ(n), to get an explicit formula for φ(n) that involves a sum of various values of the M¨ obius function. (g) Show that your answers to parts (d) and (f) of this problem are iden- tical, even though they look diﬀerent. 30. Find the Dirichlet series generating functions for the sequences (a)an=√n (b)an=|µ(n)|, whereµis the M¨ obius function. (c) A number-theoretic function f(n)i sstrongly multiplicative if it is true thatf(mn)=f(m)f(n) for all pairs m,n of positive integers. Letλ(n) be the strongly multiplicative function that takes the value −1 on every prime, and λ(1) = 1. Find its Dsgf, and then prove that/summationdisplay d\nλ(d)=/braceleftBig1i fnis a square; 0 otherwise. 31. A Lambert series is a series of the form f(x)=/summationdisplay n≥1anxn 1−xn, and we then say that fis the Lambert series gf for the sequence {an}. (Lambert series are only rarely used because they’re hard to analyze.) (a) Suppose fis the Lambert series gf of a sequence {an}∞ 1, and the samefis the opsgf of a sequence {bn}∞ 1. Find the b’s in terms of thea’s. Exercises 71 (b) Thus prove the amazing identity /summationdisplay n≥1µ(n)xn 1−xn=x, where again µis the M¨ obius function. (c) Find the Lambert series generating function of Euler’s φfunction. 32. Let a={an}n≥0be a given sequence. Let Sbe the operator that transforms ainto its sequence of partial sums: ( Sa)n=a0+···+an, for n≥0. (a) Iffis the opsgf of a, what is the opsgf of Sa? (b) Iffis the opsgf of aandr≥0 what is the opsgf of Sra? (c) What is Sraifais the sequence of all 1’s? (d) For a general sequence a, ﬁnd an explicit formula, involving a single summation sign, for the nth member of the sequence Sra. (e) An unknown sequence ahas the following property: if, beginning withawe iteratertimes the operation S, of replacing the sequence by its sequence of partial sums, we obtain the sequence {1,0,0,...}. Finda. 33. (a) Write out the ﬁrst twelve cyclotomic polynomials. (b) Ifn=pais a prime power, what is Φ n(x)? (c) Show that for n≥1, Φn(1) =/braceleftBigg1,ifn>1 is not a prime power; p,ifn=pkis a prime power; 0,ifn=1 . 34. Consider the following sequence of polynomials. ψn(x)=/summationdisplay 1≤m≤n gcd(m,n)=1xm(n=1,2,...). Thusψ1=x,ψ2=x,ψ3=x+x2,ψ4=x+x3, etc. (a) Show that /summationdisplay d\nψn d(xd)=x(1−xn) 1−x(n=1,2,...). (b) Use the result of exercise 28 to show that ψn(x)=( 1 −xn)/summationdisplay d\nµ(d)xd 1−xd(n=1,2,...). 72 2 Series (c) Show that at every primitive nth root of unity ωwe haveψn(ω)= µ(n), and therefore the polynomial ψn(x)−µ(n) is divisible by the nth cyclotomic polynomial Φ n(x). 3.1 Introduction 73 Chapter 3 Cards, Decks, and Hands: The Exponential Formula 3.1 Introduction In this chapter we will discuss a particularly rich vein of applications of the theory of generating functions to counting problems. The exponential formula, which is our main goal here, is a cornerstone of the art of count-ing. It deals with the question of counting structures that are built out of connected pieces. The structures themselves need not be connected, but their pieces always are. The question is, if we know how many pieces of each size there are, how many structures of each size can we build out of those pieces? We begin with a little example. There is 1 connected labeled graph that has 1 vertex, there is 1 connected labeled graph that has 2 vertices, and there are 4 connected labeled graphs that have 3 vertices. These creatures are all shown in Fig. 3.1 below. 321213123132112 Fig. 3.1: The six labeled, connected graphs of ≤3 vertices. Now think of graphs that have exactly 3 labeled vertices but are not necessarily connected. There are 8 of them, as shown in Fig. 3.2. The question is, how can we develop a theory that will show us the connection between the number 8, of allgraphs of ≤3 vertices, and the numbers 1, 1, 4 of connected labeled graphs of 1, 2, and 3 vertices? After all, such a theory should exist, because the connected graphs are the building blocks out of which all graphs are constructed. How, exactly, are those building blocks used? Suppose we want to construct a graph Gofnvertices and kconnected components. We can ﬁrst choose which kconnected graphs to use for the connected components, subject only to the condition that the sum of their numbers of vertices must be n. Second, after deciding which connected graphs to use, we need to re- label all of their vertices. That is because the connected graphs that we 74 3 Cards, Decks, and Hands: The Exponential Formula 321213123132123132231231 Fig. 3.2: The eight not-necessarily-connected labeled graphs of 3 vertices. use, as in Fig. 3.1, each have their own private sets of vertex labels. A connected graph of 5 vertices will have labels 1, 2, 3, 4, 5 on its vertices, etc. However, the ﬁnal assembled graph G, that we are manufacturing out of those connected pieces, will use each vertex label 1 ,2,3,...,n exactly once, as in Fig. 3.2. Note, for instance, that the connected graph of 1 vertex appears three times in the ﬁrst graph of Fig. 3.2 with 3 diﬀerent vertex labels. So our counting theory will have to take into account the choices of the connected graphs that are used as building blocks, as well as the number of ways to relabel the vertices of those connected graphs to obtain the ﬁnal product. Now we’re going to raise the ante. Instead of going ahead and an- swering these counting questions in the case of graphs, it turns out to be better to be a bit more general right from the start, because a lot of niceapplications don’t quite fall under the heading of graphs. So we are going to develop the theory in a context of ‘playing cards’ and ‘hands,’ instead of ‘connected graphs’ and ‘all graphs.’ Next you will see a number of deﬁnitions of the basic terminology. After all of those deﬁnitions, a few examples will no doubt be welcome, and will be immediately forthcoming. Then we will get on with the development of the theory and its numerous applications. 3.2 Deﬁnitions and a question We suppose that there is given an abstract set Pof ‘pictures.’ Deﬁnition. AcardC(S,p) is a pair consisting of a ﬁnite set S(the ‘label set’) of positive integers, and a picture p∈P. The weight ofCisn=|S|. A card of weight nis called standard if its label set is [ n].* * Recall that [ n] is the set {1,2,...,n }. 3.3 Examples of exponential families 75 Deﬁnition. AhandHis a set of cards whose label sets form a partition of [n], for some n. This means that if ndenotes the sum of the weights of the cards in the hand, then the label sets of the cards in Hare pairwise disjoint, nonempty, and their union is [ n]. Deﬁnition. The weight of a hand is the sum of the weights of the cards in the hand. Deﬁnition. Arelabeling of a card C(S,p) with a set S/primeis deﬁned if |S|= |S/prime|, and it is the card C(S/prime,p). IfS/prime=[|S|] then we have the standard relabeling of the card. Deﬁnition. AdeckDis a ﬁnite set of standard cards whose weights are all the same and whose pictures are all diﬀerent. The weight of the deck is the common weight of all of the cards in the deck. Deﬁnition. Anexponential family Fis a collection of decks D1,D2,... where for each n=1,2,..., the deck Dnis of weight n. IfFis an exponential family, we will write dnfor the number of cards in deck Dn, and we will call D(x), the egf of the sequence {dn}∞ 1, the deck enumerator of the family. Question: Given an exponential family F. For each n≥0andk≥1, let h(n,k)denote the number of hands Hof weightnthat consist of kcards, and are such that each card in the hand is a relabeling of some card in some deck in F. Repetitions are allowed. That is, we are permitted to take several copies of the same card from one deck, and to relabel those copies with diﬀerent label sets. How can we express h(n,k)in terms of d1,d2,d3,..., wherediis the number of diﬀerent cards in deck Di(i≥1)? Ifh(n,k) is the number of hands Hof weightnthat have exactly k cards, then we introduce the 2-variable generating function H(x,y)=/summationdisplay n,k≥0h(n,k)xn n!yk. (3.2.1) This is a generator of mixed type; it is an opsgf with respect to the y variable and an egfwith respect to x. We will call it the 2-variable hand enumerator of the family. Ifh(n)=/summationtext kh(n,k) is the number of hands of weight nwithout regard to the number of cards in it, then we write H(x) for the egf of {h(n)}, instead of H(x,1). It is the 1-variable hand enumerator of F. One way to answer the question raised above would be to exhibit a simple relationship between the generating functions H(x,y) and D(x), and that, of course, is exactly what we are about to do (see (3.4.4) below for a look at the answer). 76 3 Cards, Decks, and Hands: The Exponential Formula 3.3 Examples of exponential families Before we get on with the business of answering the question that was raised in the previous section, here are a few examples of exponential families that have important roles in combinatorial theory. Example 1. The ﬁrst exponential family that we will describe is the family of all vertex-labeled, undirected graphs. We will call this family F1. A graphGis a set of vertices some pairs of which are designated as edges. A labeled graph is a graph that has a positive integer associated with each vertex. The integers (‘labels’) are all diﬀerent. The graph has thestandard labeling if the set of its vertex labels is [ n], wherenis the number of vertices of G. There are/parenleftbign 2/parenrightbig possible edges in graphs of nvertices, so there are 2(n 2) labeled graphs of nvertices. For instance, there are 8 labeled graphs of 3 vertices, and these are shown in Fig. 3.2 (graphs are drawn by ﬁrst drawing thenvertices and then, between each pair of vertices that is designated as an edge, drawing a line). Some graphs are connected and some are disconnected . A graph is con- nected if, given any pair of vertices, we can walk from one to the other along edges in the drawing of the graph. Otherwise, the graph is disconnected. Of the 8 graphs of 3 vertices, shown in Fig. 3.2, 4 are connected, namely the last 4 that are pictured there. Now let’s describe our exponential family. First, we describe a card C(S,p). There is a card corresponding to every connected labeled graph G. The setSis the set of vertex labels that is used in the graph. Before we can describe the ‘picture’ on the card we need to say what a standard relabeling of a graph is. Let Gbe a graph of nvertices that are labeled with a set Sof labels. Then relabel the vertices with [ n],preserving the order of the labels . That is, the vertex that had the smallest label in Swill then get label 1, etc. Therefore the standard relabeling is uniquely deﬁned. Now, ifGis a labeled, connected graph, the picture pon the card C(S,p) that corresponds to Gis the standard relabeling of G. Hence, on a card Cwe see two things: a picture of a connected graph with standard labels, and another set of labels, of equal cardinality. For instance, one card of weight 3 might be (S,p)=/parenleftbig {5,9,11},132/parenrightbig which would correspond to the connected labeled graph 5119 3.3 Examples of exponential families 77 So cards correspond to connected graphs with not-necessarily-standard label sets. What is a hand? A hand is a collection of cards whose label sets partition [n], wherenis the weight of the hand, which is to say, it is the total number of vertices in all of the connected graphs on all of the cards of the hand. But that is something very useful; a hand Hcorresponds to a not-necessarily-connected graph with standard labels! Its individual connected components may have nonstandard labels, but the graph itself uses exactly the labels 1 ,2,...,n , wherenis its number of vertices. In summary then, the set of all vertex labeled graphs forms an expo- nential family. Each card is a labeled connected graph, each deck Dnis the set of all connected standard labeled graphs of nvertices, each hand is a standard (not-necessarily-connected) labeled graph. The number dnof cards in the nth deck is the number of standard connected labeled graphs ofnvertices, and the number h(n,k) of hands of weight nwithkcards is the number of standard labeled graphs of nvertices with kconnected components. The question posed at the end of the last section in this case asks for the relationship between the numbers of alllabeled graphs and all connected labeled graphs of all sizes. Example 2. In this example we will ﬁnd that the set of all permutations can be thought of as an exponential family. First let’s say what the cards are. On a card, the picture will show n points arranged in a circle, the points being labeled with the set [ n], in some order, and there will be arrowheads around the circle, all pointing clockwise, to tell us that the points are arranged in clockwise circular sequence. So much for the ‘picture’ part of the card. Additionally, there is a set Sofnpositive integers on the card. The reader will recognize that such a card corresponds to a cyclic permutation of the elements of S, i.e., a permutation of Sthat has a single cycle. For instance, the card whose picture is shown in Fig. 3.3 5 2 3 14 Fig. 3.3: A cyclic permutation is in the cards. and whose set is S={2,4,7,9,10}represents the cyclic permutation 2−→7−→4−→10−→9−→2 of the setS. 78 3 Cards, Decks, and Hands: The Exponential Formula Now what is a deck of these cards? The cards in a deck are standard cards, and they consist of one sample of every distinct standard card of a given weight. In this case the nth deck Dncontains exactly ( n−1)! cards, one for each cyclic permutation of [ n]. So far we have accounted for the permutations with one cycle. They are the building blocks out of which all permutations are constructed, using hands of cards. So what is a hand , in this example? A hand is a collection of cards, and on each card there are two things: a cyclic permutation and a label set. The label sets are pairwise disjoint and their union is {1,2,...,n }. The cardinality of the label set on each card matches that of the cyclic permutation that is shown there. The collection of all of the cards in the hand represents a permutation of nletters. The cycles of this permutation are the ones shown on the individual cards of the hand after the cycle on each card has been relabeled, in an order-preserving way, with the elementsof the label set on the card. Since every permutation of nletters has a unique decomposition into cycles, we see that hands of weight ncorrespond exactly to permutations of nletters . Hence the set of all permutations is an exponential family. We call it F 2. How many cards are in deck Dn? There are dn=(n−1)! of them. The question raised at the end of the last section asks for the number h(n,k) of hands of weight nandkcards. Such a hand represents a permutation ofnletters that has kcycles. Hence in this case h(n,k) is the number of permutations of nletters that have kcycles. When we have our general theorems in place, the ones that give the relationships between the dn’s and theh(n,k)’s, we’ll learn a lot about permutations of various kinds with given numbers and sizes of cycles. Later, in chapter 5, we’ll return to this subject and re-use these generating functions to get asymptotic information about permutations and their cycles. 3.4 The main counting theorems In this section we will state and prove various forms of the exponential formula. The next section contains 109applications of the method. First, let two exponential families be given. We will say what it means tomerge them. Roughly, it means to form a new family whose decks of each weight are the unions of the decks of those weights in the two given families. Some care is necessary, however, to insure that the two decks haveall diﬀerent cards, so we will now give a precise deﬁnition. LetF /primeandF/prime/primebe two exponential families whose picture sets P/prime,P/prime/prime are disjoint. We form a third family F, and write F=F/prime⊕F/prime/prime, as follows: 3.4 The main counting theorems 79 ﬁxn≥1. From F/primewe take all of the d/prime ncards of deck D/prime nand put them in a new pile. Then from F/prime/primewe take all d/prime/prime nof its cards from deck D/prime/prime nand add thesed/prime/prime ncards to the pile, which now contains dn=d/prime n+d/prime/prime ndiﬀerent cards. Repeat this for each n≥1. The Fundamental Lemma of Labeled Counting. LetF/prime,F/prime/primebe two exponential families, and let F=F/prime⊕F/prime/primebe their merger. Further, let H/prime(x,y),H/prime/prime(x,y),H(x,y)be the respective 2-variable hand enumerators of these families. Then H(x,y)=H/prime(x,y)H/prime/prime(x,y). Proof. Consider a hand Hin the merged family F. Some of its cards came from F/primeand some came from F/prime/prime. The collection of cards that came fromF/primeforms a sub-hand H/primeof weight, say, n/prime, and having k/primecards, that has been relabeled, in an order-preserving way, with a certain label set S⊂[n]. All hands Hin the merged family are uniquely determined by a particular hand H/primefromF/prime, the choice of new labels Swith which that hand is to be relabeled, and the remaining subhand H/prime/primefromF/prime/prime, which must be relabeled, again preserving the order of the labels, with [ n]−S. Consequently the number of hands in the merged family that have weightnand have exactly kcards is h(n,k)=/summationdisplay n/prime,k/prime/parenleftbiggn n/prime/parenrightbigg h/prime(n/prime,k/prime)h/prime/prime(n−n/prime,k−k/prime) =/bracketleftbiggxn n!yk/bracketrightbigg H/prime(x,y)H/prime/prime(x,y),(3.4.1) and we are ﬁnished. The main idea is that the processes of merging families and of mul- tiplying egf’s correspond exactly. The fact that in equation (3.4.1) the n/prime variable in the sum carries a binomial coeﬃcient along in its wake, while thek/primedoes not, accounts for the mixed nature of the generating function that was chosen, with the ‘ x’ variable being egf-like and the ‘ y’ variable ops-like. The Fundamental Lemma will allow us to build up the general rela- tionship between deck and hand enumerators very easily, in a ‘Sorcerer’s Apprentice’ fashion, beginning with a trickle and ending with a ﬂood. We begin with a starkly simple exponential family that consists of exactly one nonempty deck that has just one card in it. The hand enumerator there will be obvious. Then we consider a family that has a number of cards in one deck, and no other decks. Finally we jump to the general situation, at each stage using the Fundamental Lemma, because we will be carrying out a merging operation. 80 3 Cards, Decks, and Hands: The Exponential Formula Step 1: The trickle. Fix a positive integer r. Let therth deck, Dr, contain exactly one card, and let all other decks be empty. The deck counts are dr= 1 and all other dj= 0. The deck enumerator is D(x)=xr/r!. A handHconsists of some number, say s, of copies of the one card that exists. The weight of Hisrs. Therefore the number of hands of kcards and of weight nish(n,k)=0 unlessn=kr.I fn=kr, then how many hands of weight nare there? We can choose the labels for the ﬁrst card in/parenleftbign r/parenrightbig ways, for the second in/parenleftbign−r r/parenrightbig ways, etc, for the kth card in/parenleftbign−(k−1)r r/parenrightbig = 1 way. Since the order of the labeled cards is immaterial, the number of hands is therefore h(kr,k)=1 k!n! r!k. The hand enumerator of this elementary family is therefore H(x,y)=/summationdisplay n,kh(n,k)xnyk/n! =/summationdisplay kxkryk k!r!k = exp/braceleftbiggyxr r!/bracerightbigg .(3.4.2) We won’t have to do any more computation to get the general result; the Fundamental Lemma will do it for us. Step 2: The ﬂow Fix positive integers randdr, and consider an exponential family F that hasdrcards in its rth deck Dr, and has no other nonempty decks. We claim that the hand enumerator of this family is H(x,y) = exp/braceleftbiggydrxr r!/bracerightbigg . (3.4.3) The proof is by induction on dr. The claim is correct when dr= 1, for that is (3.4.2). Suppose the claim is true for dr=1,2,...,m −1, and let the family Fhavemcards in its rth deck. Then Fis the result of merging a family with m−1 cards in the rth deck and a family with 1 card in that deck. By the inductive hypothesis and the Fundamental Lemma, the hand enumerator is the product exp{y(m−1)xr/r!}exp{yxr/r!}= exp {ymxr/r!}, and the claim is proved. Step 3: The ﬂood. We are now ready to prove the main counting theorem. 3.5 Permutations and their cycles 81 Theorem 3.4.1 (The exponential formula). LetFbe an exponential family whose deck and hand enumerators are D(x)andH(x,y), respec- tively. Then H(x,y)=eyD(x). (3.4.4) In detail, the number of hands of weight nandkcards is h(n,k)=/bracketleftbiggxn n!/bracketrightbigg/braceleftbiggD(x)k k!/bracerightbigg . (3.4.5) Proof. In (3.4.3) we have proved this result in the special case where there is only one nonempty deck. But a general exponential family with a full sequence of nonempty decks D1,D2,...is the merger of the special families Fr(r=1,2,...), each of which has just a single nonempty deck Dr. By the Fundamental Lemma, the hand enumerator of the general family is the product of the hand enumerators of the special families. But the generating function (3.4.4) claimed in the theorem is indeed the product of the enumerators (3.4.3) of the special families Fr, and the proof is ﬁnished. By summing (3.4.5) over all kwe obtain the following: Corollary 3.4.1. LetFbe an exponential family, let D(x)be the egf of the sequence {dn}∞ 1of sizes of the decks, and let H(x)egf ←→{hn}∞ 0, where hnis the number of hands of weight n. Then H(x)=eD(x). (3.4.6) By summing (3.4.5) over just those kthat lie in a given set T,w e obtain Corollary 3.4.2 (The exponential formula with numbers of cards restricted). LetTbe a set of positive integers, let eT(x)=/summationtext n∈Txn/n!, and lethn(T)be the number of hands whose weight is nand whose number of cards belongs to the allowable set T. Then {hn(T)}∞ 0egf ←→eT(D(x)). (3.4.7) The next several sections of this chapter will contain applications of the exponential formula. 3.5 Permutations and their cycles We apply the theorems to the exponential family F2of permutations, that was described in example 2 of section 3.3. There we observed that the 82 3 Cards, Decks, and Hands: The Exponential Formula deckDncontainsdn=(n−1)! cards. The exponential generating function of the sequence {(n−1)!}∞ 1is D(x)=/summationdisplay n≥1(n−1)!xn n! =/summationdisplay n≥1xn n = log1 1−x. Now from theorem 3.4.1 we have H(x,y) = exp/braceleftbigg ylog1 1−x/bracerightbigg =1 (1−x)y.(3.5.1) In this exponential family, h(n,k) is the number of permutations of n letters that have kcycles, and it is called the Stirling number of the ﬁrst kind. We will use one of the standard notations,/bracketleftbign k/bracketrightbig *, for these numbers, and will reserve the h(n,k) for the general situation. Now, /summationdisplay k/bracketleftbiggn k/bracketrightbigg yk=/bracketleftbiggxn n!/bracketrightbigg (1−x)−y =n!/parenleftbiggy+n−1 n/parenrightbigg (by (2.5.7)) =y(y+1 )···(y+n−1),(3.5.2) so the numbers of permutations of nletters with various numbers of cycles are the coeﬃcients in the expansion of the ‘rising factorial’ function y(y+ 1)···(y+n−1). The enumerator of hands of kcards is obviously 1 k!/braceleftbigg log1 1−x/bracerightbiggk (k=1,2,...), which tells us that the Stirling number is also given by /bracketleftbiggn k/bracketrightbigg =/bracketleftbiggxn n!/bracketrightbigg1 k!/braceleftbigg log1 1−x/bracerightbiggk . (3.5.3) * There are as many notations for/bracketleftbign k/bracketrightbig as there are books on combina- torics. It is called ( −1)ks(n,k)o rs1(n,k), ors(n,k), orc(n,k), or several other things. Similarly the/braceleftbign k/bracerightbig are calleds2(n,k)o rS(n,k), etc. 3.7 A subclass of permutations 83 One thing that we don’t ﬁnd is a simple little formula for these Stirling numbers. One can ﬁnd formulas for them, but they’re fairly unpleasant, involving double sums of summands with sign alternations, etc. But with the generating function apparatus we can do just about whatever we want to without such a formula. To calculate numerical values of the/bracketleftbign k/bracketrightbig , for in- stance, one can use the very simple recurrence relations that can be derived from these generating functions (see Exercise 8). 3.6 Set partitions We introduce a new exponential family F3, as follows: ﬁrst, for each n≥1, in the deck Dnthere is just onecard of weight n. On that card there is a picture of a smiling rabbit,* and there is the label set [ n]. What is a hand? There is a hand Hcorresponding to every partition of the set [ n]. Indeed, given such a partition, take the sets in it and let them relabel the label sets on the cards in the hand. Then the cards are otherwise uniquely determined since there’s only one card of each weight. So in this exponential family the number of hands of weight nthat have kcards is equal to the number of partitions of the set [ n] intokclasses. But that is something we’ve met before, in example 6 of chapter 1, where we called those numbers/braceleftbign k/bracerightbig , the Stirling numbers of the second kind. To apply the exponential formula we ﬁrst compute the egf of the num- bersdnof cards in each deck. But these numbers are all 1, if n≥1, and are 0 else, so D(x)=/summationdisplay ndnxn n!=/summationdisplay n≥1xn n!=ex−1. Now by the exponential formula the enumerator of hands is H(x,y)=ey(ex−1), (3.6.1) and in particular/braceleftbiggn k/bracerightbigg =/bracketleftbiggxn n!/bracketrightbigg/braceleftbigg(ex−1)k k!/bracerightbigg . (3.6.2) Compare this result with the generating function (1.6.12) of the Bell num- bers and ﬁnd that we have here a reﬁnement of that generating function. Not only does eex−1generate the numbers of partitions of n-sets, but each term of the expansion eex−1=/summationdisplay k≥0(ex−1)k k! has signiﬁcance with respect to the numbers of classes in the partitions. * Why not? Since there’s only one card the picture is immaterial, so it might as well be cheerful. 84 3 Cards, Decks, and Hands: The Exponential Formula 3.7 A subclass of permutations How many permutations σofnletters have the property that σhas an even number of cycles and all of them are of odd lengths? This problem takes place in an exponential family that is like the family F2of permutations, except that it contains only the decks of odd weights, D1,D3,.... The numbers {dn}∞ 1that count the cards in the decks are now 1, 0, 2, 0, 24, 0, 720, .... The egf of the deck counts is D(x)=/summationdisplay nodd(n−1)!xn n! =/summationdisplay r≥0x2r+1 2r+1 = log/radicalbigg 1+x 1−x by (2.5.2). Since the number of cycles is required to be even, the allowable numbers of cards in a hand are the set T=the even numbers. By (3.4.7), the egf of the answer is cosh/braceleftBigg log/radicalbigg 1+x 1−x/bracerightBigg =1√ 1−x2 =/summationdisplay m≥0/parenleftbigg2m m/parenrightbigg (x/2)2m. The number of permutations that meet the conditions of the problem is the coeﬃcient of xn/n! here, namely /parenleftbiggn n 2/parenrightbiggn! 2n. That’s one way to answer the question, but the answer can be restated in quite a striking form, like this- Theorem 3.7.1. Let a positive integer nbe ﬁxed. The probabilities of the following two events are equal: (a) a permutation is chosen at random from among those of nletters, and it has an even number of cycles, all of whose lengths are odd (b) a coin is tossed ntimes and exactly n/2heads occur. 3.8 Involutions, etc. Fix positive integers m,n. How many permutations σ,o fnletters, satisfyσm= 1, where ‘1’ is the identity permutation? To do this problem, we need the following: 3.9 2-regular Graphs 85 Lemma. Forσm=1 it is necessary and suﬃcient that all of the cycle lengths ofσbe divisors of m. Proof. Consider a cycle Cofσ, of length r. Letibe some letter that is inC. Then, by deﬁnition of a cycle, σm(i) is the letter on Cthat we encounter by beginning at iand moving msteps around the cycle, namely the letter that is mmodrsteps around Cfromi. Butσm(i)=i. Therefore mmodr= 0, i.e.,rdividesm. Therefore mis a multiple of the length of every cycle of C. The converse is clear, and the proof is ﬁnished. Now back to the problem. Consider the exponential family F4in which the cards are the usual ones for cycles of permutations, but in which the only decks that occur are those whose weights are divisors of m. Then dr=(r−1)! ifr\m, and is 0 else. Hence D(x)=/summationdisplay r≥1drxr/r!=/summationdisplay d\mxd d. (3.8.1) By the exponential formula (theorem 3.4.1) we have the following elegant result: Theorem 3.8.1. Fixm> 0. The numbers of permutations of nletters whosemth power is the identity permutation have the generating function exp/parenleftbigg/summationdisplay d\m(xd/d)/parenrightbigg . (3.8.2) Let’s try a special case of this theorem. Take m= 2. Then we are talking about permutations whose square is 1. These are called involutions . Involutions can have cycles of lengths 1 or 2 only, by the lemma above. If tnis the number of involutions of nletters, then by (3.8.2) we have /summationdisplay n≥0tn n!xn=ex+1 2x2. (3.8.3) 3.9 2-regular Graphs How many undirected, labeled graphs are there on nvertices, in which every vertex is of degree 2 (such graphs are called 2-regular )? Such a graph is a disjoint union of undirected cycles, so we have an ex- ponential family F5in which the cards stand for undirected cycles, instead of directed ones, as in the case of permutations. For ﬁxedn≤2 there are no undirected cycles at all. For n≥3, the numberdnof cards in the nth deck is the number of undirected circular 86 3 Cards, Decks, and Hands: The Exponential Formula arrangements of nletters, and that number is ( n−1)!/2. Therefore the generating function of the deck sizes is D(x)=/summationdisplay n≥3(n−1)! 2n!xn =1 2/summationdisplay n≥3xn/n =1 2/braceleftbigg log1 1−x−x−x2 2/bracerightbigg . By the exponential formula (3.4.4), the exponential generating function of the number g(n) of undirected 2-regular labeled graphs is /summationdisplay n≥0g(n)xn n!= exp/braceleftbigg1 2log1 1−x−x 2−x2 4/bracerightbigg =e−1 2x−1 4x2 √1−x.(3.9.1) This answer is a sparkling example of the ability of the generating function method to produce answers to diﬃcult counting problems with minimal eﬀort. 3.10 Counting connected graphs How many labeled, connected graphs of nvertices are there? Now we’re back in the exponential family F1of labeled graphs, but there are one or two little twists. The exponential formula can tell you the number of all gadgets of each size if you know the number of connected ones, or vice versa. This problem is ‘vice versa.’ The number of all labeled graphs ofnvertices is 2(n 2), so in the equation ‘Hands = eDecks’ we know ‘Hands’ and we want to ﬁnd ‘Decks,’ rather than the other way around. There’s one more twist. Let D(x) and H(x) be the egf’s of the decks and the hands, respectively. Then H(x)=/summationdisplay n≥02(n 2) n!xn, and this series does not converge for any x/negationslash= 0. So this is a formal power series generating function only, and we should not expect analytic functions at the end of the road. Having said all of that, the machinery still works very nicely. We will now ﬁnd a recurrence formula for the number of connected graphs by the ‘xDlog ’ method of section 1.6. It isn’t any harder to ﬁnd a general recurrence relation than for this special case, however, so let’s do it in general. 3.11 Counting labeled bipartite graphs 87 Theorem 3.10.1. The counting sequences {dn}and{hn}, of decks and hands in an exponential family satisfy the recurrence nhn=/summationdisplay k/parenleftbiggn k/parenrightbigg kdkhn−k (n≥1;h0=1 ). (3.10.1) Proof. Apply the ‘xDlog ’ method of section 1.6 to the exponential formula (3.4.6). It follows that the numbers dnof connected labeled graphs of nvertices satisfy the recurrence n2(n 2)=/summationdisplay k/parenleftbiggn k/parenrightbigg kdk2(n−k 2)(n≥1). (3.10.2) From this formula we are able, for example, to compute the dn’s for small n.F o rn=1,..., 6 we ﬁnd the values 1, 1, 4, 38, 728, 26704. 3.11 Counting labeled bipartite graphs How many bipartite vertex-labeled graphs of nvertices are there? The exponential formula can handle even this problem with just a little bit of coaxing. A bipartite graph Gis a graph whose vertex set V(G) can be partitioned into V=A∪Bsuch that every edge of Gis of the form (a,b), wherea∈Aandb∈B. A bipartite graph of 10 vertices is shown in Fig. 3.4. 9641 1087532 Fig. 3.4: A bipartite graph Now, of the 2(n 2)labeled graphs of nvertices, how many are bipartite? Well, there’s a little problem. The exponential formula can count the hands if you can count the decks, or it can count the decks if you can count the hands. But it can’t do both, and in this problem it isn’t immediately clear how many connected bipartite graphs there are orhow many there are altogether. A thought might be to choose the sets A,Bof the partition [ n]=A∪B, and then count the bipartite graphs that have that partition. The latter is easy; since there are |A||B|possible edges, there must be 2|A||B|ways to exercise the freedom to draw or not to draw all of those edges. 88 3 Cards, Decks, and Hands: The Exponential Formula The problem is that a ﬁxed bipartite graph might get counted several times in the process. In other words, there may be several ways to exhibit a partition of the vertex set with all edges running between vertices in diﬀerent classes. For instance, the graph Gof Fig. 3.4 would turn up several times: once with A={1,4,6,9}, again with A={2,3,5,7,8,10}, again with A={1,4,5,9}, etc. In general, a bipartite graph that has cconnected components would be created 2ctimes by the construction that we are considering, the reason being that for each connected component GiofGwe can choose which of the two sets in its vertex partition, AiorBi, will get put on the left hand side, inA, and which on the right hand side, in B. To get around this conundrum we use slightly diﬀerent playing cards. By a 2-colored bipartite graph we mean a vertex-labeled bipartite graph G together with a coloring of the vertices of Gin two colors (‘Red,’ ‘Green’), such that whenever ( v,w) is an edge of G, thenvandwhave diﬀerent colors. Aconnected bipartite graph, for instance, creates two 2-colored graphs. A bipartite graph with cconnected components creates 2csuch 2-colored graphs. In the exponential family F6that we are making, there will be a card Ccorresponding to each 2-colored connected labeled bipartite graph. Im- printed on the card there will be, as always, S, the set of vertex labels that are used, and a picture of a 2-colored, connected bipartite graph of |S| vertices with standard vertex labels. What have we gained by coloring the cards? Just this: we now know how many hands of weight nthere are. That number is γn=/summationdisplay k/parenleftbiggn k/parenrightbigg 2k(n−k), (3.11.2) because each and every hand arises exactly once from the following con- struction: (i) ﬁx an integer k,0≤k≤n. (ii) choose kof the elements of [ n] and color them ‘Red.’ (iii) color the remaining elements of [ n] ‘Green.’ (iv) decide independently for each vertex pair ( ρ,γ), whereρis Red andγis Green, whether or not to make ( ρ,γ) an edge. It is obvious that (3.11.2) counts the possible outcomes of the con- struction. So, even though we are in the wrong exponential family, because things are colored that we wish weren’t, at least we know how many hands there are! Next, let’s use the exponential formula to ﬁnd the egf for the decks, which correspond to connected 2-colored bipartite graphs. It tells us in- 3.12 Counting labeled trees 89 stantly that D(x) = log/braceleftbigg/summationdisplay n≥0γn n!xn/bracerightbigg , (3.11.3) whereγnis deﬁned by (3.11.2). Now that we have the connected colored graphs counted, is it hard to count the connected uncolored graphs? Not at all, because there are just half as many uncolored and connected as there are colored and connected. So the egf of ordinary, uncolored connected bipartite graphs is D(x)/2, where D(x) is given by (3.11.3). But now we have achieved, in the correct exponential family, the ob- jective that we had not reached before: we know how many cards there are in each deck. So we know one of the two items that the exponential formula relates, and therefore we can ﬁnd the other one. Since D(x)/2 generates the deck counts, it must be that eD(x)/2= exp/braceleftbigg1 2log/braceleftbigg/summationdisplay n≥0γn n!xn/bracerightbigg/bracerightbigg =/radicalBigg/summationdisplay n≥0γn n!xn(3.11.4) generates the hand counts, and we have: Theorem 3.11.1. Letβ(n)denote the number of vertex labeled bipartite graphs ofnvertices. Then /summationdisplay n≥0β(n) n!xn=/radicalBigg/summationdisplay n≥0γn n!xn, (3.11.5) where theγnare given by (3.11.2). So all of the complications about multiple counting were resolved by taking the square root of the generating function that we started with! 3.12 Counting labeled trees A tree is a connected graph that has no cycles. How many (standard) labeled trees of nvertices are there? In this example we will derive the answer to that question in the form of one of the most famous results in combinatorics, namely: Theorem 3.12.1. For eachn≥1there are exactly nn−2labeled trees of nvertices. Although many proofs are known, the one by generating functions, which uses the exponential formula, is particularly enchanting, and here it is: 90 3 Cards, Decks, and Hands: The Exponential Formula Arooted tree is a tree that has a distinguished vertex called the root. There are obviously ntimes as many labeled rooted trees of nvertices as there are trees, so we will be ﬁnished if we can count the rooted ones. Lettnbe the number of rooted trees (with standard labels) of nvertices forn≥1. We deﬁne an exponential family F7as follows. The cards correspond to rooted labeled trees. On a card C(S,p),pis a picture of a standard rooted tree of |S|vertices, and Sis a set of labels. InF7, what is a hand? A hand Hcorresponds to a rooted labeled forest , which is a labeled graph each of whose connected components is a rooted tree. The exponential formula will tell us how many forests there are if we know how many trees there are, or vice versa. But this is one of those unsettling situations where we know neither. The solution? Press on, and keep the faith. By the exponential formula, H(x)=eD(x), (3.12.1) where H(x)egf ←→{fn},D(x)egf ←→{tn}andfnis the number of rooted forests ofnvertices. Now (3.12.1) is one equation in two unknown functions. To get another one we use a fact that was discovered by P´ olya, namely that tn+1=(n+1 )fn (n≥0). (3.12.2) To prove (3.12.2), let Fbe a rooted labeled forest of nvertices. In- troduce a new vertex v, and assign to it a label j, where 1 ≤j≤n+1 . RelabelFwith the set 1 ,2,...,j −1,j+1,...,n + 1, preserving the or- der of the labels. Then draw edges between vand all of the roots of the components of F, and root the resulting tree at v. The result is a rooted labeled tree of n+ 1 vertices. As we vary the label j, we construct n+1 rooted trees corresponding to each rooted forest F. The construction is easily reversible, so every rooted tree of n+ 1 vertices occurs exactly once, which proves (3.12.2). The sequence fn=tn+1/(n+ 1) has the egf H(x)=/summationdisplay n≥0fn n!xn =/summationdisplay n≥0tn+1 (n+ 1)!xn =1 xD(x). If we combine this with (3.12.1) we get D(x)=xeD(x). (3.12.3) 3.13 Exponential families and polynomials of ‘binomial type.’ 91 Now, in previous problems where there was an unknown generating function it has always happened that we obtained some sort of functional equation that had to be solved in order to ﬁnd the function. We have seen situations where the equation was a diﬀerential equation, and others where it was a quadratic equation. In (3.12.3) we have a functional equation that is to be solved for D(x), which in fact determines D(x) uniquely, but which is not a diﬀerential equation or an algebraic equation, and whose solution isn’t obvious at all. There is a powerful tool for dealing with this kind of a functional equation, called the Lagrange Inversion Formula, which will be discussed in section 5.1. There we will ﬁnish the enumeration of trees as an illustration of the use of the Lagrange formula. 3.13 Exponential families and polynomials of ‘binomial type.’ Associated with each exponential family there is a sequence of polyno- mials φn(y)=/summationdisplay kh(n,k)yk(n=0,1,2,...), (3.13.1) whereh(n,k) is the number of hands of weight nandkcards. In view of the exponential formula (3.4.4) these polynomials satisfy the generating relation eyD(x)=/summationdisplay n≥0φn(y) n!xn. (3.13.2) Polynomial sequences that satisfy (3.13.2) have been called polynomials of binomial type by Rota and Mullin [RM]. The reason for the name is that since euD(x)egf ←→{φn(u)};evD(x)egf ←→{φn(v)}; it follows that φn(u+v)=/summationdisplay r/parenleftbiggn r/parenrightbigg φr(u)φn−r(v)(n≥0), which is reminiscent of the binomial theorem. Although various authors have given combinatorial interpretations for such polynomial sequences, the very natural interpretation that appears above seems not to have been discussed. That interpretation is: when the coeﬃcients of polynomials {φn(y)}of binomial type are nonnegative, then there exists an exponential family Fsuch that for each n≥0,φn(y) generates the hands of weight n, by numbers of cards. Conversely, every exponential family has a family of polynomials of binomial type associated with it. 92 3 Cards, Decks, and Hands: The Exponential Formula 3.14 Unlabeled cards and hands In the remainder of this chapter we will consider the same kinds of problems, except that there will be no label sets to worry about. This would seem to simplify things, and it does in some respects, but not in all. We will be concerned with how many structures (hands) can be built out of given building blocks (cards). A cardC=C(n,p) now has only its weight nand its picture p.F o r eachn=1,2,...there is a deck Dnthat contains dncards, all of weight n. A hand is a multiset of cards. That is, we may reach into one of the decks Drand pull out of it some number of copies of a single card C(r,p/prime), then a number of copies of C(r,p/prime/prime), and so forth, then from another deck we can take more cards, etc. No signiﬁcance attaches to the sequence of cards in the hand. What matters is which cards have been selected and with which multiplicities. The weight of a hand is the sum of the weights of the cards in the hand, taking account of their multiplicities. As before, we let h(n,k) be the number of hands of weight nthat contain exactly kcards, and we let H(x,y)=/summationdisplay n,kh(n,k)xnyk. (3.14.1) Notice that the ‘ n!’ is missing in the assumed form of the generating func- tion. Instead of the mixed egf-ops that was appropriate for labeled counting, a pure ops is the way to go for unlabeled counting. We need a generic name for the systems that we are constructing. We will call them prefabs (instead of exponential families, which applies in the labeled case), and will use letters like Pto represent them. Thus a prefab Pconsists of a sequence of decks D1,D2,...from which we can form hands, as described above. In Pwe let D(x)ops ←→{dn}∞ 1. The main problem is to ﬁnd the functional relationship between H(x,y) andD(x), so let’s do that now. We will use the Sorcerer’s Apprentice method once more. For the trickle, consider a prefab Pthat consists of just one nonempty deck, Dr, and suppose that Drcontains only a single card. In this prefab, a hand His a fairly simple-minded thing. It consists of some number, ksay, of copies of the one and only card that there is, and its weight will be n=rk. Hence in this prefab the number h(n,k) of hands of weightnthat have exactly kcards is 1 if n=rkand is 0 else. Thus H(x,y)=/summationdisplay n,kh(n,k)xnyk =/summationdisplay k≥01·xrkyk =1 1−yxr.(3.14.2) 3.14 Unlabeled cards and hands 93 Next, just as in section 3.4, we deﬁne the merge operation. If P/primeand P/prime/primeare prefabs whose picture sets are disjoint, then by their merger P= P/prime⊕P/prime/primewe mean the prefab whose deck Dn, for eachn, is the union of the corresponding decks of P/primeandP/prime/prime. If there were d/prime n,d/prime/prime ncards, respectively, in those two decks, then there are dn=d/prime n+d/prime/prime ncards in Dn. Fundamental lemma of unlabeled counting. LetH/prime(x,y),H/prime/prime(x,y) andH(x,y)be the hand enumerators of prefabs P/prime,P/prime/primeandP=P/prime⊕P/prime/prime, respectively. Then H=H/primeH/prime/prime. Proof. Consider a hand H∈P, of weight n, and containing exactly k cards. Some k/primeof those cards come from P/prime, and their total weight is, say, n/prime, while the remaining k−k/primecards come from P/prime/prime, and their total weight must ben−n/prime. Thus h(n,k)=/summationdisplay k/prime,n/primeh/prime(n/prime,k/prime)h/prime/prime(n−n/prime,k−k/prime), but, by a strange coincidence, that is exactly the relationship which holds between the coeﬃcients of the power series H,H/primeandH/prime/prime. Armed with the fundamental lemma, we can now consider a slightly more complicated prefab Pr, which still contains just one nonempty deck Dr, but now that deck contains drdiﬀerent cards. By induction on dr= 1,2,..., we see at once that the hand enumerator of this prefab is H(x,y)=1 (1−yxr)dr. (3.14.3) Finally ( d´ej´ a vu anybody?), in a general prefab Pin which there are dncards in deck Dn, for eachn=1,2,3,..., we observe that P=⊕∞ n=1Pn, where the Pnare as deﬁned in the previous paragraph. We obtain at once: Theorem 3.14.1. In a prefab Pwhose hand enumerator is H(x,y)we have H(x,y)=∞/productdisplay n=11 (1−yxn)dn, (3.14.4) wherednis the number of cards in the nth deck (n≥1). This is the analogue of the exponential formula in the case where there are no labels. Like the exponential formula, this one too has an astounding number of elegant applications, and we will discuss a number of them in the sequel. Before we get to that, let’s convert (3.14.4) into a formula from which we could actually compute the h’s from the d’s, using the ‘ yDlog ’ method of section 1.6. 94 3 Cards, Decks, and Hands: The Exponential Formula If we take the logarithm of both sides of (3.14.4), logH(x,y)=∞/summationdisplay s=1log1 (1−yxs)ds =/summationdisplay s≥1dslog1 (1−yxs) =/summationdisplay s≥1ds/summationdisplay m≥1ymxsm m =/summationdisplay n,m≥1dn mxnym m, wheredjis to be interpreted as 0 if its subscript is not a positive integer. Next we diﬀerentiate with respect to yand multiply by yH, getting y∂H(x,y) ∂y=H(x,y)/summationdisplay n,m≥1xnymdn m. Finally, we take [ xnym] of both sides, which yields mh(n,m)=/summationdisplay r,m/prime≥1h(n−rm/prime,m−m/prime)dr(n,m≥1;h(n,0) =δn,0). (3.14.5) This recurrence holds in any prefab, and permits the numerical computation of the hand counts from the deck counts. Often the 2-variable deck enumerators H(x,y)o r{h(n,k)}n,k≥0give more detail than is necessary. If hn=/summationtext kh(n,k) is the number of hands of weightn, however many cards they contain, and if H(x)ops ←→{hn}∞ 0, then, since we obtain H(x) from H(x,y) by formally replacing yby 1, the general counting theorem (3.14.4) becomes H(x)=∞/productdisplay r=11 (1−xr)dr. (3.14.6) The recurrence (3.14.5) can be replaced by nhn=/summationdisplay m≥1Dmhn−m (n≥1;h0=1 ), (3.14.7) whereDm=/summationtext r\mrdr(m=1,2,...). 3.15 The money changing problem 95 Considerably more detailed information can be obtained with just a little more eﬀort. Suppose we restrict the multiplicities with which the cards can be used in hands. For instance, suppose we decree that every card that appears in a hand must appear there with multiplicity that is divisible by 3, etc. Then what can be said about the number of hands? LetWbe a ﬁxed set of nonnegative integers, containing 0. For each nandkwe leth(n,k;W) be the number of hands of weight nthat have exactlykcards (counting multiplicities!), each appearing with a multiplicity that belongs to W. Let H(x,y;W)=/summationdisplay n,kh(n,k;W)xnyk. Finally, let w(t)=/summationdisplay k∈Wtk. (3.14.8) The generating functions are again multiplicative under merger of pre- fabs with disjoint picture sets. Consider a prefab with just 1 card of weight r, and no other decks. Then h(n,k;W)=1i fk∈Wandn=kr, and is 0 otherwise, and so H(x,y;W)=/summationdisplay k∈Wxkryk=w(yxr). If there are drcards in the rth deck, and no other cards, then H(x,y;W)= w(yxr)dr, and ﬁnally we obtain: Theorem 3.14.2. Let the prefab Pcontain decks of sizes d1,d2,..., and letWbe a set of nonnegative integers, 0∈W.I fh(n,k;W)is the number of hands ofkcards of weight n, such that each card appears with a multiplicity that belongs to W, then H(x,y;W)=/summationdisplay n,kh(n,k;W)xnyk=/productdisplay r≥1w(yxr)dr, (3.14.9) wherew(t)is given by (3.14.8). Observe that the theorem reduces to theorem 3.14.1 in the case where W=Z+, the set of all nonnegative integers. A noteworthy special case is W={0,1}, which means that we can choose a card for our hand or not, but we can’t take more than one copy of it. In that case (3.14.9) gives H(x,y;{0,1})=/productdisplay r≥1(1 +yxr)dr =1 H(x,−y;Z+).(3.14.10) 96 3 Cards, Decks, and Hands: The Exponential Formula We proceed with several examples of the use of these formulas. 3.15 The money changing problem Suppose that in the coinage of a certain country there are 5-cent coins, 11-cent coins, and 37-cent coins. In how many ways can we make change for $17.19? In general terms, we are given Mpositive integers 1≤a1<a 2<···<aM, and we ask the following question: for each positive integer n, in how many ways can we write n=x1a1+x2a2+···+xMaM (∀i:xi≥0), (3.15.1) where thex’s are integers? This problem is of great importance in a number of areas, both pure and applied, and it has a very beautiful theory, some of which we will give here. For givena1,...,a Mwe write S=S(a1,...,a M) for the set of all n that can be written in the form (3.15.1). Sis a semigroup of nonnegative integers. First let’s identify the prefab Pin which everything will be happening. The decks are almost all empty. The only decks that are not empty are the Mdecks Da1,...,DaM. Each of these contains just a single card. Hence the deck enumerating sequence is dn=/braceleftBig1i fn=a1,...,a M 0 else. In a sense, then, the problem is all over. If h(n,k) denotes the number of ways of making change that use exactly kcoins, i.e., the number of representations (3.15.1) in which/summationtext ixi=k, then according to the main counting theorem (eq. (3.14.4)) we have H(x,y)=1 (1−yxa1)(1−yxa2)···(1−yxaM). (3.15.2) Ifhnis the number of ways of representing nwithout regard to the number of coins, then from the cruder formula (3.14.6) H(x)=1 (1−xa1)(1−xa2)···(1−xaM). (3.15.3) Even though the generating functions are known, substantial questions remain. Here are a few of them. 3.15 The money changing problem 97 How can we describe the set S? That is, which sums of money can be changed? Given 8-cent and 12-cent coins only, it wouldn’t be reasonable to expect to make change for 53 cents. In general, if the greatest common divisor of the set {a1,...,a M}isg> 1, then only multiples of gcan be represented. But suppose that g= 1, i.e., that the ai’s are relatively prime . Then which integers are representable? The central result of this subject is due to I. Schur. It states that Sthen contains all suﬃciently large integers, i.e. there exists an integer Nsuch that every integer n≥Nis representable in the form (3.15.1) . The smallest integer Nthat has the property stated in the theorem will be called the conductor of the set S={a1,...,a M}, and will be denoted by the symbol κ=κ(S). For instance, every integer ≥8 can be represented as a nonnegative integer linear combination of 3 and 5, and 7 cannot be so represented, so κ({3,5})=8 . The problem of determining the conductor of a set Sexactly seems to be of enormous diﬃculty. There are no general ‘formulas’ for the conductor ifM≥3, and no good algorithms for calculating it if M≥4. The case M= 2 is already very pretty, and the answers are known, so here they are: Theorem 3.15.1. Letaandbbe relatively prime positive integers. Then (a) every integer n≥κ=(a−1)(b−1)is of the form n=xa+yb, x,y≥0, and (b) the integer κ−1is not of that form, and (c) of the integers 0,1,2,...,κ −1, exactly half are representable and half are not. Proof. (Our proof follows [NW]) Since gcd(a,b) = 1, we can certainly write every integer masxa+ybifx,ycan have either sign. The representation is unique if we require that 0 ≤x<b . Thenm∈Sify≥0, andm/∈Sif y<0. The largest integer that is not representable is therefore obtained by choosingx=b−1,y=−1. Henceκ(S) is one unit larger than ( b−1)a−b, and parts (a) and (b) of the theorem are proved. To prove (c), let 0 ≤m<κ (S), and again consider the unique way of writingm=xa+yb, with 0 ≤x<b . Then m/prime=κ−1−m=(b−1−x)a+(−1−y)b. Now 0 ≤b−1−x<b , so ify≥0 thenmis representable and m/primeis not, while ify<0 thenm/primeis representable and mis not. Hence exactly half of the numbers 0 ,1,...,κ −1 are representable. Now we’re going to prove Schur’s theorem. The idea of the proof is that we will consider (without ever writing it down) the partial fraction expansion of the right side of (3.15.3). Among the multitude of terms that occur there we will identify one term whose power series coeﬃcients grow more rapidly than any other, and this will give the desired result. 98 3 Cards, Decks, and Hands: The Exponential Formula The generating function H(x) in (3.15.3) is a rational function whose poles all lie on the unit circle |x|= 1. In fact, the poles are at various roots of unity. What are the multiplicities of these poles? The point x=1i sap o l eo f multiplicity M, because the denominator of H(x) is divisible by (1 −x)M. Letω=e2πir/sbe a primitive (i.e., gcd(r,s)=1 )sth root of 1. What is the multiplicity with which this point x=ωoccurs as a pole of H(x)? It is equal to the number of ai’s that are divisible by s. Since the ai’s are relatively prime, it cannot be that allof them are divisible by s. Thereforex=1 is a pole of order MofH(x), and every other pole has multiplicity <M . Supposeωis a pole of order r. Then the portion of the partial fraction expansion of Hthat comes from ωis of the form c1 (1−x/ω)r+c2 (1−x/ω)r−1+···. Now refer to the power series expansion (2.5.7), which we repeat here: 1 (1−x)k+1=/summationdisplay n≥0/parenleftbiggn+k k/parenrightbigg xn. Ifk= 1, the coeﬃcients of this expansion are linear functions of n.I f k= 2 they are quadratic functions of n. In general, the coeﬃcients of xn are growing, as n→∞ , likenk/k!. The contribution of one ﬁxed pole of order rto the coeﬃcient sequence ofH(x) therefore grows like cnr−1. There is one pole, at x= 1, of order M. Its portion of the partial fraction expansion contributes ∼cnM−1to thenth coeﬃcient of H(x). Since all other poles have strictly lower multiplicities, none of them can alter the asymptotic rate of growth that is contributed by the principal pole at x= 1. Hence, for n→∞ we havehn∼cnM−1. That certainly implies that for all large enough values of nwe will have hn/negationslash=0 , and that ﬁnishes the proof. However, as long as we’re here, why not ﬁnd out the value of calso? The partial fraction expansion of H(x) is of the form H(x)=1 (1−xa1)(1−xa2)···(1−xaM) =c (1−x)M+O((1−x)−M+1). To calculate c, multiply both sides by (1 −x)Mand letx→1. This gives c=1/(a1···aM). Thus we get a growth estimate along with the proof of the theorem. 3.15 The money changing problem 99 Theorem 3.15.2 (Schur’s theorem). Ifhndenotes the number of rep- resentations of nas a nonnegative integer linear combination of a1,...,a M, these being a relatively prime set of positive integers, then hn∼nM−1 (M−1)!a1a2···aM(n→∞ ). (3.15.4) In particular, there exists an integer Nsuch that every n≥Nis so repre- sentable in at least one way. Example 1. Given two relatively prime integers a,b. Find an explicit formula for f(n), the number of ways to change ncents using those coins. From (3.15.3) we have /summationdisplay nf(n)xn=1 (1−xa)(1−xb), (3.15.5) so what remains is a partial fraction expansion. We ﬁnd 1 (1−xa)(1−xb)=A (1−x)2+B (1−x)+/summationdisplay ωa=1 ω/negationslash=1Cω 1−x/ω+/summationdisplay ζb=1 ζ/negationslash=1Dζ 1−x/ζ. (3.15.6) As regards the constants, we already know that A=1/(ab), from (3.15.4). To ﬁnd B, multiply (3.15.6) by (1 −x)2, diﬀerentiate, and let x= 1. This gives B=(a+b−2)/(2ab). To ﬁndCω, multiply by (1 −x/ω) and letx=ω. The result is that Cω=1/(a(1−ωb)), and similarly for Dζ. Finally we take the coeﬃcient of xnthroughout (3.15.6) to get the formula f(n)=n ab+a+b 2ab+/summationdisplay ωa=1 ω/negationslash=1Cω ωn+/summationdisplay ζb=1 ζ/negationslash=1Dζ ζn. (3.15.7) If we examine the two sums that appear in (3.15.7) as functions of n,w e see that each of them is a periodic function of n. The ﬁrst sum is periodic of periodaand the second is periodic of period b. The sum of these two sums is therefore periodic of period ab. We have therefore found that the number of ways to change ncents into coins of a- andb-cent denominations is f(n)=n ab+a+b 2ab+per(n), (3.15.8) whereper(n)is periodic of period ab, and is on the average 0. We might like to see this periodicity in action, so let’s take a= 3 and b= 5. A good way to compute the numbers f(n) is to use the recurrence 100 3 Cards, Decks, and Hands: The Exponential Formula formula that is implicit in the generating function (3.15.5). If we use the xDlog method on (3.15.5), we ﬁnd the recurrence in the form nf(n)=3/summationdisplay j≥1f(n−3j)+5/summationdisplay j≥1f(n−5j)(n≥1;f(0) = 1),(3.15.9) with the understanding that f(m)=0i fm< 0. Table 3.1 shows n,f(n), and 15(f(n)−(n/15)−(4/15)) (which is periodic of period 15, according to (3.15.8)). 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 100 10 1 1 0 1 1 1 1111 11−5−68 −865 −11 3 2 1 0 −1−2−3 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 211 21 2 2 1 2 2 2 2222 11−5−68 −865 −1 13210 −1−2−3 Table 3.1 3.16 Partitions of integers A partition of a positive integer nis a representation n=r1+r2+···+rk (r1≥r2≥···≥rk≥1). (3.16.1) The numbers r1,...,r kare the parts of the partition. Hence (3.16.1) is a partition of nintokparts. There are 7 partitions of 5, namely 5=5, =4+1, =3+2, =3+1+1, =2+2+1, =2+1+1+1, =1+1+1+1+1. The number of partitions of nis de- noted byp(n), andp(n,k) is the number of partitions of nintokparts.The investigation of the deeper properties of p(n) was one of the jewels of 20th century analysis, involving researches of Hardy and Ramanujan and further work by Rademacher, the result of which was an exact closed formula for p(n) that was at the same time a complete asymptotic series. The whole story can be found in Andrews [An]. From our point of view, the theory of partitions is the case of the money-changing problem where coins of every positive integer size are avail- able. Thus, theorem 3.14.1 gives us immediately an opsgf of the partitionfunction in the form of the reciprocal of an inﬁnite product, /summationdisplay n,k≥0p(n,k)xnyk=1 (1−yx)(1−yx2)(1−yx3)(1−yx4)··· (p(0,k)=δ0,k).(3.16.2) 3.16 Partitions of integers 101 Withy=1w eﬁ n d /summationdisplay n≥0p(n)xn=1 (1−x)(1−x2)(1−x3)(1−x4)···(p(0) = 1) (3 .16.3) as the generating function for {p(n)}itself. At the other extreme, we can think about partitions with constrained parts and constrained multiplicities of parts. Let two sets W, of nonnegative integers, and R, of positive integers, be given, with 0 ∈W. Letp(n,k;W,R ) be the number of partitions of nintokparts such that all of the parts lie inR, and all of their multiplicities lie in W. Then from (3.14.9) /summationdisplay n,kp(n,k;W,R )xnyk=/productdisplay r∈R/parenleftBigg/summationdisplay k∈Wykxkr/parenrightBigg . (3.16.4) From this generating function we can prove many theorems about parti- tions. Example 1. LetW={0,1},R={1,2,...}. Thenp(n,k;W,R ) is the number of partitions of nintokdistinct parts, and we have /summationdisplay n,kp(n,k;{0,1},Z+)xnyk=/productdisplay r≥1(1 +yxr). (3.16.5) If we lety= 1 we obtain /summationdisplay np(n;{0,1},Z+)xn=/productdisplay r≥1(1 +xr) =/productdisplay r≥11−x2r 1−xr =(1−x2)(1−x4)··· (1−x)(1−x2)(1−x3)(1−x4)··· =1 (1−x)(1−x3)(1−x5)(1−x7)···. The last member, however, generates the partitions of ninto odd parts, and we have a generating function proof of: Theorem 3.16.1. For eachn=1,2,3,..., the number of partitions of n into odd parts is equal to the number of partitions of ninto distinct parts. For instance, the partitions of 5 into odd parts are 5, 3+1+1, and 1+1+1+1+1, while its partitions into distinct parts are 5, 4+1, and 3+2. 102 3 Cards, Decks, and Hands: The Exponential Formula Theorem 3.16.1 was discovered by Euler. A great many proofs of it have been given. Some of the most interesting proofs are bijective ; that is, they give explicit constructions that match each partition into odd parts with a partition into distinct parts. Example 2. Now letW={0,1,...,q }andR=Z+. The right side of (3.16.4) becomes/productdisplay r≥1(1 +tr+···+tqr)=/productdisplay r≥1/parenleftbigg1−tr(q+1) 1−tr/parenrightbigg . Each factor in the numerator of this product cancels one in the denominator, leaving in the denominator only those factors in which ris not divisible by q+ 1. This proves the following result, which reduces to theorem 3.16.1 whenq=1 . Theorem 3.16.2. Fixq≥1. For each n≥1, the number of partitions ofninto parts that are not divisible by q+1 is equal to the number of partitions of nin which no part appears more than qtimes. 3.17 Rooted trees and forests A rooted tree is a tree whose vertices are unlabeled, except that one of them is distinguished as ‘the root.’ In section 3.12 we counted labeled trees, using the exponential formula. Here we will count unlabeled, rooted trees. On each card in a deck Dnthere is now the integer nand a picture of a rooted tree of nvertices. The deck D4is shown in Fig. 3.5. 4444RRRR Fig. 3.5: The rooted trees of 4 vertices A hand of weight nandkcards is, in this case, a rooted forest of n vertices and kconnected components (rooted trees). If h(n,k) is the number of these and if h(n)=/summationtext kh(n,k) is the number of all rooted forests of n vertices, then by (3.14.6) /summationdisplay nh(n)xn=/productdisplay n≥11 (1−xn)t(n), (3.17.1) 3.18 Historical notes 103 wheret(n)=h(n,1) is the number of rooted trees of nvertices. Next, just as we found in the labeled case (see 3.12.2), there is a simple relationship between the number of rooted forests of nvertices and of rooted trees ofn+1 vertices: they are equal. Just add a new vertex rto the forest, call it the new root, connect it to all of the former roots of the trees in the forest, and there is the rooted tree that corresponds to the given forest. Henceh(n)=t(n+1 ) (n≥0). Then (3.17.1) takes the form /summationdisplay nt(n+1 )xn=/productdisplay n≥11 (1−xn)t(n). (3.17.2) This equation in fact determines all of the numbers {t(n)}. It is, however, a fairly formidable equation, and we should not expect simple formulas for these numbers. 3.18 Historical notes The exponential formula ﬁrst appeared in the thesis of Riddell [RU], in the form of counting connected labeled graphs from a knowledge of the number of all labeled graphs. Since then the idea has been generalized andextended by several researchers. In [BG], and at about the same time in [FS], signiﬁcant extensions of the idea were made to very general labeled and unlabeled applications, by Bender and Goldman and by Foata and Sch¨ utzenberger. The former introduced ‘prefabs’ and the latter used the ‘ compos´ e partitionnel .’ Further developments of the method can be found in Stanley ([St1], [St2]) who worked with a partition-based approach, in Joyal [Jo] who used functorial methods in his theory of ‘species,’ in Beissinger [Bei], and in Garsia and Joni [GaJ]. The approach taken in this book is most closely akin to the compos´ e partitionnel . The suggestion to cast the discussion in terms of cards, decks, and hands was made to me in private conversation by Adriano Garsia, when I showed him a set of lecture notes of mine that were based on the graph- theoretical point of view. I think that his suggestion aﬀords maximum clarity of the ideas along with maximum generality of applications. 104 3 Cards, Decks, and Hands: The Exponential Formula Exercises 1. Give an explicit 1-1 correspondence between partitions of ninto distinct parts and partitions of ninto odd parts. 2. Fix integers n,k. Letf(n,k) be the number of permutations of nletters whose cycle lengths are all divisible by k. Find a simple, explicit egf for {f(n,k)}n≥0. Find a simple, explicit formula for f(n,k). (Hint: You might need the discussion at the end of section 3.4.) 3. Find the egf for the partitions of the set [ n], all of whose classes have a prime number of elements. 4. In a group Γ, the order of an element gis the least positive integer ρ such thatgρ=1 Γ. (a) In the group of all permutations of nletters, express the order of a permutation σin terms of the lengths of its cycles. (b) Letg(n,k) be the number of permutations of nletters whose or- der isk. Expressg(n,k) in terms of the number ˜ g(n,m)o fn- permutations whose cycle lengths all divide m. 5. LetTnbe the number of involutions of nletters. (a) Find a recurrence formula that is satisﬁed by these numbers. (b) Compute T1,...,T 6. (c) Give a combinatorial and constructive interpretation of the re- currence. That is, after having derived it from the generatingfunction, re-derive it without the generating function. 6. Find, in simple form, the egf of the sequence of numbers of permutations ofnletters that have no cycles of lengths ≤3. Your answer should not contain any inﬁnite series. 7. Find the generating function for labeled graphs with all vertices of degrees 1 or 2, and an odd number of connected components. Find arecurrence formula for these numbers, calculate the ﬁrst few, and draw the graphs involved. 8. From (3.5.2) ﬁnd a three term recurrence relation that is satisﬁed by the Stirling numbers of the ﬁrst kind. Give a direct combinatorial proof ofthis recurrence relation. That is, reprove it, without using any generating functions. 9. As in section 3.7, ﬁnd the egf of the numbers {g(n)} ∞ 0of permutations ofnletters that have both of the following two properties: (a) they have an odd number of cycles and (b) the lengths of all of their cycles are even. Find a simple, explicit formula for these numbers. 10. Find an explicit formula for/braceleftbign k/bracerightbig , the Stirling number of the second kind by expanding the kth power that appears in (3.6.2) by the binomial Exercises 105 theorem. Your formula should be in the form of a single ﬁnite sum. 11. LetS,Tbe ﬁxed sets of positive integers. Let f(n;S,T) be the number of partitions of [ n] whose class sizes all lie in Sand whose number of classes lies inT. Show that {f(n;S,T)}n≥0has the egf eT(eS(x)), whereeS(x)=/summationtext s∈Sxs/s!. 12. Fixk> 0. Letf(n,k) be the number of permutations of nletters whose longest cycle has length k. Find the egf of {f(n,k)}n≥0, forkﬁxed. 13. IfT(x) and G(x) denote, respectively, the egf’s of involutions, in (3.8.3), and of 2-regular graphs, in (3.9.1), then observe that T(x)G(x)2=1 1−x. (a) Write out the identity between the sequences {g(n)},{tn}that is implied by the above generating function relation. (b) Show that for each ﬁxed n≥1 there are exactly the same numbers of (i) permutations of nletters and of (ii) triples ( τ,G 1,G2), whereτis an involution of a set R,G1 is a 2-regular graph on a vertex set S,G2is a 2-regular graph on a vertex set T, andR,S,T partition [n]. (c) Find, explicitly, a 1-1 correspondence such as is described in part (b) above. 14. Let Fbe an exponential family with associated polynomials {φn(x)} of binomial type, and with deck enumerator D(x). (a) IfDydenotes the diﬀerential operator ∂/∂y , then show that D(−1)(Dy)φn(y)=nφn−1(y)(n≥0) by directly applying the operator to the egf of the polynomial sequence (here D(−1)denotes the inverse function in the sense of functional composition). (b) In the case of the exponential family of permutations by cycles, ﬁnd the associated polynomials of binomial type, and verify the identity proved in part (a) by direct computation with those poly- nomials. 15. In an exponential family F, let˜h(n) be the number of hands of weight nwhose cards have all diﬀerent weights. (a) Show that /summationdisplay n≥0˜h(n) n!xn=∞/productdisplay k=1/braceleftbigg 1+dk k!xk/bracerightbigg . 106 3 Cards, Decks, and Hands: The Exponential Formula (b) Letpnbe the probability that a permutation of nletters has cycles whose lengths are all diﬀerent. Then {pn}ops ←→/productdisplay k≥1/braceleftbigg 1+xk k/bracerightbigg . (c) Ifp(x) denotes the generating function in part (b) above, deter- mine the growth of p(x)a sx→1−. Do this by inserting additional factors ofe−xk/kin the product. 16. Let numbers {cn}be deﬁned by xx=1+/summationdisplay n≥1cn n!(x−1)n. Show that each cnis an integer multiple of n, and in fact is a multiple of n(n−1) if and only if n−1 divides (n−2)!. 17. Here we want to show that the Stirling numbers of the ﬁrst and second kinds are inverse to each other, in a certain sense. In the generating function (3.5.2) for the former, replace xby 1/xand compare with the generating function (1.6.5) for the latter. Multiply the functions together so that the hard part cancels out. Read oﬀ the coeﬃcient of xnin what remains, and state it as an assertion that a certain pair of matrices, each involving Stirling numbers, are inverses of each other. 18. Letanbe the number of unlabeled graphs ofnvertices each of whose connected components is a path or a cycle. Let F(x) be the opsgf of the sequence {an}. FindF(x) and express it in terms of Euler’s opsgf for the sequence {p(n)}of the numbers of partitions of integers n. 19. Letanbe the number of unlabeled rooted trees of nvertices in which the degree of the root is 2. That is, there are exactly 2 edges incident at the root. Let T(x) be the opsgf of the sequence {tn}that counts allrooted trees ofnvertices. Show that /summationdisplay nanxn−1=1 2/parenleftbigg T(x)2+T(x2)/parenrightbigg . 20. Find the largest integer that is notof the form 6 x+1 0y+1 5zwhere x,y,z are nonnegative integers. Prove that your answer is correct, i.e., that your integer is not so representable, and that every integer larger than it is so representable. 21. In a country that has 1-cent, 2-cent, and 3-cent coins only, the number of ways of changing ncents is exactly the integer nearest to (n+3 )2/12. 22. This exercise develops a considerable sharpening of the exponential formula, that will be used again in section 4.7. 3.18 Historical notes 107 (a) In an exponential family F, the number of hands of weight nthat contain exactly a1cards of weight 1 and a2cards of weight 2 and a3of weight 3 and ..., wherea1+2a2+···=n, is the coeﬃcient of (tnxa1 1xa2 2···)/n! in the expansion of exp/braceleftbig/summationdisplay i≥1xiditi i!/bracerightbig . (b) Letf(n,r,s ) be the number of partitions of the set [ n] that have exactlyrclasses of size 1 and exactly sclasses of size 2 (however many classes of other sizes they may have). Then /summationdisplay n,r,sf(n,r,s )xrystn n!= exp/parenleftbig xt+yt2 2+et−1−t−t2 2/parenrightbig . 108 4 Applications of generating functions Chapter 4 Applications of generating functions 4.1 Generating functions ﬁnd averages, etc. Power series generating functions are exceptionally well adapted to ﬁnding means, standard deviations, and other moments of distributions, with minimum work. Suppose f(n) is the number of objects, in a certain setSofNobjects, that have exactly nproperties, for each n=0,1,2,..., with/summationtext nf(n)=N. What is the average number of properties that an object inShas? Evidently it is µ=1 N/summationdisplay nnf(n). (4.1.1) Suppose we happen to be fortunate enough to be in possession of the opsgf of the sequence {f(n)},s a yF(x)ops ←→{f(n)}. Is there some convenient way to express the mean µof (4.1.1) in terms of F? But of course. Clearly, µ=F/prime(1)/F(1). So averages can be computed directly from generating functions. Let’s go to the next moment, the standard deviation σ, of the distri- bution. This is deﬁned as follows: σ2=1 N/summationdisplay ω∈S(n(ω)−µ)2, (4.1.2) whereωrepresents an object in the set S, andn(ω) is the number of prop- erties that ωhas.σ2, which is known as the variance of the distribution, is therefore the mean square of the diﬀerence between the number of prop- erties that each object has and the mean number of properties µ. Every one of the f(n) objectsωthat has exactly nproperties will contribute ( n−µ)2to the sum in (4.1.2), and therefore σ2=1 N/summationdisplay n(n−µ)2f(n) =1 N/summationdisplay n(n2−2µn+µ2)f(n) =1 N{(xD)2−2µ(xD)+µ2}F(x)|x=1 =(F/prime/prime(1) + (1 −2µ)F/prime(1) +µ2F(1))/F(1) =F/prime/prime(1)/F(1) +F/prime(1)/F(1)−(F/prime(1)/F(1))2 ={(logF)/prime+ (logF)/prime/prime}x=1.(4.1.3) 4.1 Generating functions ﬁnd averages, etc. 109 So the standard deviation can also be calculated in terms of the values of Fand its ﬁrst two derivatives at x=1 . Let’s work this out in exponential families. In an exponential family F, what is the average number, µ(n), of cards in a hand of weight n? Ifh(n,k) is the number of hands of weight nthat havekcards, then the average is µ(n)=1 h(n)/summationdisplay kkh(n,k). (4.1.4) Now if we begin with the exponential formula /summationdisplay n,kh(n,k)xn n!yk=eyD(x) the thing to do is to apply the operator ∂/∂y and then set y= 1. The result is that /summationdisplay nxn n!/summationdisplay kkh(n,k)=D(x)eD(x)=D(x)H(x). (4.1.5) Theorem 4.1.1. In an exponential family F, the average number of cards in hands of weight nis µ(n)=/bracketleftbiggh(n)xn n!/bracketrightbigg D(x)H(x) =1 h(n)/summationdisplay r/parenleftbiggn r/parenrightbigg drh(n−r).(4.1.6) Example 1. Cycles of permutations The averaging relations (4.1.6) are particularly happy if h(n)=n!, as in the family of all permutations. There, (4.1.6) becomes µ(n)=1 n!/summationdisplay r/parenleftbiggn r/parenrightbigg (r−1)!(n−r)! =1+1 2+1 3+···+1 n. Consequently, the average number of cycles in a permutation of nletters is the harmonic number Hn. What is the standard deviation? The function F(x) that appears in (4.1.3), in the case of permutations, is, for nﬁxed, F(x)=/summationdisplay kh(n,k)xk=x(x+ 1)(x+2 )···(x+n−1), 110 4 Applications of generating functions by (3.5.2). After taking logarithms and diﬀerentiating, following (4.1.3), we ﬁndF(1) =n!, (logF)/prime(1) =Hn, and (logF)/prime/prime(1) = −1−1/4−1/9−1/16−···− 1/n2. If we substitute this into (4.1.3), we ﬁnd that the variance of the distribution of cycles over permutations of nletters is σ2=Hn−1−1/4−1/9−···− 1/n2 = logn+γ−π2/6+o(1). whereγis Euler’s constant. Hence the average number of cycles is ∼lognwith a standard deviation σ∼√logn. 4.2 A generatingfunctionological view of the sieve method The sieve method* is one of the most powerful general tools in com- binatorics. It is explained in most texts in discrete mathematics, however it most often appears as a sequence of manipulations of alternating sums of binomial coeﬃcients. Here we will emphasize the fact that generating functions can greatly simplify the lives of users of the method. We are given a ﬁnite set Ω of objects and a set Pof properties that the objects may or may not possess.** In this context, we want to answerquestions of the following kind: how many objects have no properties at all? how many have exactly rproperties? what is the average number of properties that objects have? etc., etc. The characteristic ﬂavor of problems that the sieve method can handle is that, although it is hard to see how many objects have exactlyrproper- ties, for instance, it is relatively easy to see how many objects have at least a certain set of properties and maybe more. What the method does is to convert the ‘at least’ information into the ‘exactly’ information. To see how this works, if S⊆Pis a set of properties, let N(⊇S) be the number of objects that have at least the properties in S. That is, N(⊇S) is the number of objects whose set of properties contains S. For ﬁxedr≥0, consider the sum N r=/summationdisplay |S|=rN(⊇S). (4.2.1) * A.k.a. ‘the principle of inclusion-exclusion,’ and often abbreviated as ‘p.i.e.’ ** Strictly speaking, a property is just a subset of the objects, but in practice we will usually have simple verbal descriptions of the properties. 4.2 A generatingfunctionological view of the sieve method 111 Introduce the symbol P(ω) for the set of properties that ωhas. Then we can writeNras follows: Nr=/summationdisplay |S|=rN(⊇S) =/summationdisplay |S|=r/summationdisplay ω∈Ω S⊆P(ω)1 =/summationdisplay ω∈Ω  /summationdisplay |S|=r S⊆P(ω)1   =/summationdisplay ω∈Ω/parenleftbigg|P(ω)| r/parenrightbigg .(4.2.2) Therefore every object that has exactly tproperties contributes/parenleftbigt r/parenrightbig to Nr. If there are etobjects that have exactly tproperties, then (4.2.2) simpliﬁes to Nr=/summationdisplay t≥0/parenleftbiggt r/parenrightbigg et (r=0,1,2,...). (4.2.3) Recall the philosophy of the method: the Nr’s are easier to calculate than theer’s because they can be found from (4.2.1). However, the er’s are what we want. Therefore it is desirable to be able to solve the equations (4.2.3) for the e’s in terms of the N’s. But how can we do that? After all, (4.2.3) is a set of simultaneous equations. At ﬁrst glance that might seem to be a tall order, but with a friendly generating function at your side, it’s easy. Let N(x) andE(x) denote* the opsgf’s of the sequences {Nr},{er}, respectively. What relation between the two generating functions is implied by the equations (4.2.3)? Multiply (4.2.3) by xrand sum on r. We then get N(x)=/summationdisplay r/summationdisplay t/parenleftbiggt r/parenrightbigg etxr =/summationdisplay tet/braceleftBigg/summationdisplay r/parenleftbiggt r/parenrightbigg xr/bracerightBigg =/summationdisplay tet(x+1 )t =E(x+1 ).(4.2.4) * The letters ‘N’ and ‘E’ are intended to suggest the Nr’s and the word ‘Exactly.’ 112 4 Applications of generating functions In the language of generating functions, the set of equations (4.2.3) boils down to the fact that N(x)=E(x+ 1). Now the problem of solving for thee’s in terms of the N’s is a triviality, and the solution is obviously E(x)=N(x−1) (4.2.5) This is the sieve method. The act of replacing the variable xbyx−1in the generating function N(x)replaces the unﬁltered data {Nr}by the sieved quantities {er}. If theN’s are known, then in principle we can read oﬀ the e’s as the coeﬃcients of N(x−1). For example, e0is the number of objects that have no properties at all. By (4.2.5), e0=E(0) =N(−1) =/summationdisplay t(−1)tNt. (4.2.6) It’s easy to ﬁnd explicit formulas for all of the ej’s by looking at the coef- ﬁcient ofxjon both sides of (4.2.5). The result is ej=/summationdisplay t(−1)t−j/parenleftbiggt j/parenrightbigg Nt. (4.2.7) But (4.2.5) says it all, in a much cleaner fashion. We will now summarize the sieve method, and then give a number of examples of its use. The Sieve Method (A) ( Find ΩandP) Given an enumeration problem, ﬁnd a set of objects and properties such that the problem would be solved if we knew the number of objects with each number of properties. (B) ( Find the unﬁltered counts N(⊇S)) For each set Sof properties, ﬁndN(⊇S), the number of objects whose set of properties contains S. (C) ( Find the coeﬃcients Nr) For eachr≥0, calculate the Nrby sum- ming theN(⊇S) over all sets Sofrproperties, as in (4.2.1). (D) ( The answer is here. ) The numbers erare the coeﬃcients of the powers ofxin the polynomial N(x−1). Before we get to some examples, we would like to point out that the numberN1has a special role to play. According to (4.2.3), N1=/summationtext ttet. That, however, is what you would want to know if you were trying to 4.2 A generatingfunctionological view of the sieve method 113 calculate the average number of properties that objects have. Hence it is good to remember that when using the sieve method on a set of Nobjects, the average number of properties that an object has is N1/N. Example 1. The ﬁxed points of permutations. Of then! permutations of nletters, how many have exactly rﬁxed points? Step (A) of the sieve method asks us to say what the set of objects is and what the set of properties is. It is almost always worthwhile to be quite explicit about these. In the case at hand, the set Ω of objects is the set of all permutations of nletters. There are nproperties: for each i=1,...,n, a permutation τhas property iifiis a ﬁxed point of τ, i.e., ifτ(i)=i. With those deﬁnitions of Ω and P, it is indeed true that we would like to know the numbers of objects that have exactly rproperties, for each r. In step (B) we must ﬁnd the N(⊇S). Hence let Sbe a set of properties. ThenS⊆[n] is a set of letters, and we want to know the number of permutations of nletters that leave at least the letters in Sﬁxed. If a permutation leaves the letters in Sﬁxed, then it can act freely on only the remaining n−|S|letters, and so there are ( n−|S|)! such permutations. Hence N(⊇S)=(n−|S|)!. For step (C) we calculate the Nr’s. But, for each r=0,...,n , Nr=/summationdisplay |S|=rN(⊇S)=/summationdisplay |S|=r(n−|S|)! =/parenleftbiggn r/parenrightbigg (n−r)! =n! r!. In step (D) we’re ready for the answers. It will save some writing if we introduce the abbreviation exp|αfor the truncated exponential series exp|α(x)=/summationdisplay 0≤r≤αxr r!. (4.2.8) Now we form the opsgf N(x) from theNr’s that we just found: N(x)=n/summationdisplay r=0n! r!xr=n!n/summationdisplay r=0xr r!. Thenetis the coeﬃcient of xtinN(x−1), i.e., E(x)=/summationdisplay tetxt=n!n/summationdisplay r=0(x−1)r r!=n! exp|n(x−1). (4.2.9) As an extra dividend, the average number of ﬁxed points that permu- tations ofnletters have is N1 N=n! n!=1. 114 4 Applications of generating functions On the average, a permutation has 1 ﬁxed point. The number of permutations that have no ﬁxed points at all is e0=E(0) =N(−1) =n! exp|n(−1)∼n! e. (4.2.10) Finally, if we really want a formula for the et’s, it’s quite easy to ﬁnd from (4.2.9) that et=n! t!exp|(n−t)(−1) ∼e−1n! t!(n→∞ ).(4.2.11) Example 2. The number of k-cycles in permutations. Fix positive integers n,k, andr≥0. How many permutations of n letters have exactly rcycles of length k? Whatever the answer is, it should at least have the good manners to reduce to the answer of the previous example when k= 1, since a ﬁxed point is a cycle of length 1. What are the objects and the properties? Evidently Ω is the set of all permutations of nletters. Further, the set Pof properties is the set of all possiblek-cycles chosen from nletters. How many such k-cycles are there? Thekletters can be chosen in/parenleftbign k/parenrightbig ways, and they can be arranged around a cycle in (k−1)! ways, so we are facing a list of/parenleftbign k/parenrightbig (k−1)! properties. Choose a set Sofk-cycles from P. How many permutations have at least the set Sof properties? None at all, unless the sets of letters in those cycles are pairwise disjoint. If the sets are pairwise disjoint, then there are N(⊇S)=(n−k|S|)! permutations that have at least all of those k-cycles. Next we calculate Nr, the sum of N(⊇S) over all sets of rproperties. The terms in this sum are either 0 or ( n−kr)!. So we really need to know only how many of them are not 0, that is, in how many ways we can choose a set ofrk-cycles from nletters in such a way that the cycles operate on disjoint sets of letters. The letters for the ﬁrst cycle can be chosen in/parenleftbign k/parenrightbig ways, and they can be ordered around the cycle in ( k−1)! ways. The letters for the second cycle can then be chosen in/parenleftbign−k k/parenrightbig ways, and ordered in ( k−1)! ways, etc. Finally, since the sequence in which the cycles are constructed is of no signiﬁcance, we divide by r!. Hence Nr=(n−kr)! r!n!(k−1)!r (k!)r(n−kr)! =n! krr!(0≤r≤n/k).(4.2.12) 4.2 A generatingfunctionological view of the sieve method 115 We can get a little piece of the solution right here, with no more work: the average number ofk-cycles that permutations of nletters have isN1/n!=1/k. The opsgf of {Nr}is N(x)=n!/summationdisplay 0≤r≤n/kxr krr! =n! exp|(n/k)(x k).(4.2.13) Finally, in the sieving step, we convert this to exact information by replacingxbyx−1, to obtain E(x)=n! exp|(n/k)/parenleftbiggx−1 k/parenrightbigg . (4.2.14) Example 3. Stirling numbers of the second kind. The Stirling numbers/braceleftbign k/bracerightbig , which we studied in section 1.6, are the numbers of partitions of a set of nelements into kclasses. We can ﬁnd out about them with the sieve method if we can invent a suitable collection of objects and properties. For the set Ω of objects we take the collection of all knways of arranging nlabeled balls in klabeled boxes. Further, such an arrangement will have property Piif boxiis empty (i=1,...,k ). Then k!/braceleftbign k/bracerightbig is the number of objects that have exactly no properties. LetSbe some set of properties. How many arrangements of balls in boxes have at least the set Sof properties? If N(⊇S) is that number, then N(⊇S) counts the arrangements of nlabeled balls into just k−|S|labeled boxes, because all of the boxes that are labeled by Smust be empty. There are obviously ( k−|S|)nsuch arrangements. Hence N(⊇S)=/braceleftbigg (k−|S|)nif|S|≤k, 0, else. If we now sum over all sets Sofrproperties, we obtain for r≤k, Nr=/parenleftbiggk r/parenrightbigg (k−r)n, whose opsgf is N(x)=/summationdisplay 0≤r≤k/parenleftbiggk r/parenrightbigg (k−r)nxr. We can now invoke the sieve to ﬁnd that the number of arrangements that have exactly tempty cells is the coeﬃcient of xtinN(x−1). On the 116 4 Applications of generating functions other hand, the number of arrangements that have exactly tempty cells is clearly/parenleftbiggk t/parenrightbigg (k−t)!/braceleftbiggn k−t/bracerightbigg =k! t!/braceleftbiggn k−t/bracerightbigg . The result is the identity /summationdisplay 0≤r≤k/parenleftbiggk r/parenrightbigg (k−r)n(x−1)r=k!/summationdisplay 0≤t≤k/braceleftbiggn k−t/bracerightbiggxt t!. (4.2.15) If we putx= 0, we ﬁnd the explicit formula (1.6.7) again. If, on the other hand, we compare (4.2.15) with the rule (2.3.3) for ﬁnding the coeﬃcients of the product of two egf’s, we discover the following remarkable identity: /summationdisplay 1≤k≤n/braceleftbiggn k/bracerightbigg yk=e−y/summationdisplay r≥1rn r!yr. (4.2.16) This shows that e−ytimes the inﬁnite series is a polynomial! The special casey= 1 has been previously noted in (1.6.10). Example 4. Rooks on chessboards Fornﬁxed, a chessboard Cis a subset of [ n]×[n]. We are given C, and we deﬁne a sequence {rk}as follows:rkis the number of ways we can placeknonattacking (i.e., no two in the same row or column) rooks on C. Next, letσbe a permutation of nletters. For each jwe letejdenote the number of permutations that ‘meet the chessboard Cin exactlyjsquares,’ i.e., if the event ( i,σ(i))∈Coccurs for exactly jvalues ofi,1≤i≤n. The question is, how can we ﬁnd the ej’s in terms of the rk’s? Let the objects Ω be the n! permutations of [ n]. There will be a propertyP(s) corresponding to each square s∈C. A permutation σhas propertyP(s)i fσmeets the mini-chessboard that consists of the single cell s. LetSbe a set of properties, i.e., of cells in C, and consider the sum Nk=/summationtext |S|=kN(⊇S). Each arrangement of knonattacking rooks on C contributes ( n−k)! to this sum. Indeed, when the set Scorresponds to the cells on which those rooks can be placed, then we are looking at kof then values of a permutation that hits Cin at leastksquares. The permutation can be completed, in the remaining n−krows, in (n−k)! ways. HenceNk=rk(n−k)!, for each k,0≤k≤n. Therefore N(x)=/summationdisplay k(n−k)!rkxk, and immediately we ﬁnd that the number of n-permutations that hit Cin exactlyjcells is [xj]/summationdisplay k(n−k)!rk(x−1)k. (4.2.17) 4.3 The ‘Snake Oil’ method for easier combinatorial identities 117 Example 5. A problem on subsets. This example is more cute than profound, but we will at least ﬁnish with a combinatorial proof of an interesting identity, as well as illustrating the generating function aspect of the sieve method. For a ﬁxed positive n, take as our set Ω of objects the/parenleftbig2n n/parenrightbig ways of choosing an n-subset of [2 n]. For the set Pof properties we take the following list of n(not 2n) properties: an n-subsetQhas property iif i/∈Q, for eachi=1,2,...,n (note that we are working with only the ﬁrst half of the possible elements of S). IfSis a set of properties (i.e., is a set of letters chosen from [ n]), then the number of ‘objects’ Qthat have at least that set of properties (i.e., are missing at least all of the i∈S) is clearly N(⊇S)=/parenleftbigg2n−|S| n/parenrightbigg . Hence Nr=/summationdisplay |S|=rN(⊇S)=/parenleftbiggn r/parenrightbigg/parenleftbigg2n−r n/parenrightbigg . If we substitute these N’s into the sieve (4.2.5) we ﬁnd that /summationdisplay jejtj=/summationdisplay r/parenleftbiggn r/parenrightbigg/parenleftbigg2n−r n/parenrightbigg (t−1)r. (4.2.18) This formula tells us the number ejof objects that have exactlyjproperties, for eachj. But we didn’t need to be told that!An object that has exactly jof these properties is a subset Qof [2n] that is missing exactly jof the elements 1 ,2,...,n . Obviously there are just/parenleftbig n j/parenrightbig2such subsets Q, because we can choose the jelements that they are missing in/parenleftbign j/parenrightbig ways, and we can then choose the other jelements that are needed to ﬁll the subset from n+1,..., 2nin/parenleftbign j/parenrightbig ways also. Thus, with no assistance from the sieve method, we already knew that ej=/parenleftbign j/parenrightbig2, for allj. Hence, according to (4.2.18), it must be true that /summationdisplay j/parenleftbiggn j/parenrightbigg2 tj=/summationdisplay r/parenleftbiggn r/parenrightbigg/parenleftbigg2n−r n/parenrightbigg (t−1)r. (4.2.19) We therefore have an odd kind of a combinatorial proof of the identity (4.2.19). The reader should suspect that something of this sort is going on whenever an identity involves an expansion around the origin on one side, and an expansion around t= 1 on the other side. 118 4 Applications of generating functions 4.3 The ‘Snake Oil’ method for easier combinatorial identities Combinatorial mathematics is full of dazzling identities. Legions of them involving binomial coeﬃcients alone ﬁll text- and reference books (see below for some references). It is a ﬁne skill for a working discrete mathematician to have if he/she is able to evaluate or simplify complicated looking sums that involve combinatorial numbers, because they have a wayof turning up in connection with problems in graphs, algorithms, enumer- ation, etc. (they’re fun, too!). In the past, one had to have built up a certain arsenal of special devices, the more the better, in order to be able to trot out the correct one forthe correct occasion. Recently, however, a good deal of quite dramatic systematization has taken place, and there are uniﬁed methods for handling vast sub-legions of the legions referred to above. In this section we are going to do two things. First we will give a single method (the Snake Oil * method) that uses generating functions to deal with the evaluation of combinatorial sums. That one method is capableof handling a great variety of sums involving binomial coeﬃcients, but there’s nothing special about binomial coeﬃcients in this respect. The method also works beautifully, within its limitations, on sums involving other combinatorial numbers. The philosophy is roughly this: don’t try to evaluate the sum that you’re looking at. Instead, ﬁnd the generating function for the whole parameterized family of them, then read oﬀ the coeﬃcients. Second, we will confess that Snake Oil doesn’t cure them all. Some combinatorial sums are really hard. Many of the very hardest binomial coeﬃcient sums can now be proved by computers using the method of rational functions, which we will discuss next. Not only that, but use of the computer has resulted in some new proofs of classical identities. The hallmarks of these proofs are that (a) they are very short comparedto the previously known proofs, (b) they seem extremely unmotivated to the reader, but (c) nothing is left out, and they really are proofs. The computerized proof techniques rely on a very simple-looking observation, which we will describe and illustrate. Therefore, in this section you can expect to see one uniﬁed method that works on a lot of relatively easy sums, and one other uniﬁed methodthat works on many more kinds of binomial coeﬃcient sums, including some ﬁendishly diﬃcult ones. First let’s talk about the Snake Oil method. The basic idea is what I might call the external approach to identities * The Random House Dictionary of the English Language deﬁnes ‘snake oil’ as a purported cure for everything, and gives the example The governor promised to lower taxes, but it was the same old snake oil. The date of the expression is given as ‘1925-30, Amer.’ 4.3 The ‘Snake Oil’ method for easier combinatorial identities 119 rather than the usual internal method. To explain the diﬀerence between these two points of view, suppose we want to prove some identity that involves binomial coeﬃcients. Typically such a thing would assert that some fairly intimidating-looking sum is in fact equal to such-and-such a simple function of n. One approach that is now customary, thanks to the skillful exposition and deft handling by Knuth in [Kn], and by Graham, Knuth and Patashnik in [GKP], consists primarily of looking inside the summation sign (‘inter- nally’), and using binomial coeﬃcient identities or other manipulations of indices inside the summations to bring the sum to manageable form. The method that we are about to discuss is complementary to the in- ternal approach. In the external , or generatingfunctionological, approach that we are selling here, one begins by giving a quick glance at the expres- sion that is inside the summation sign, just long enough to spot the ‘free variables,’ i.e., what it is that the sum depends on after the dummy vari- ables have been summed over. Suppose that such a free variable is called n. Then instead of trying to grapple with the sum, just sweep it all under the rug, as follows: The Snake Oil Method for Doing Combinatorial Sums (a) Identify the free variable, say n, that the sum depends on. Give a name to the sum that you are working on; call it f(n). (b) LetF(x) be the opsgf whose [ xn]i sf(n), the sum that you’d love to evaluate. (c) Multiply the sum by xn, and sum on n. Your generating func- tion is now expressed as a double sum over n, and over whatever variable was ﬁrst used as a dummy summation variable. (d) Interchange the order of the two summations that you are now looking at, and perform the inner one in simple closed form. For this purpose it will be helpful to have a catalogue of series whose sums are known, such as the list in section 2.5 of this book. (e) Try to identify the coeﬃcients of the generating function of the answer, because those coeﬃcients are what you want to ﬁnd. If that seems complicated, just wait till you see the next seven exam- ples. By then it will seem quite routine. The success of the method depends on favorable outcomes of steps (d) and (e). What is surprising is the high success rate. It also has the‘advantage’ of requiring hardly any thought at all; when it works, you know it, and when it doesn’t, that’s obvious too. We will adhere strictly to the customary conventions about binomial 120 4 Applications of generating functions coeﬃcients and the ranges of summation variables. These are: ﬁrst that the binomial coeﬃcient/parenleftbigx m/parenrightbig vanishes if m< 0 or ifxis a nonnegative integer that is smaller than m. Second, a summation variable whose range is not otherwise explicitly restricted is understood to be summed from −∞to∞. Thus we have, for integer n≥0, /summationdisplay k/parenleftbiggn k/parenrightbigg =2n, in the sense that the sum ranges over all positive and negative and 0 values ofk, the summand vanishes unless 0 ≤k≤n, and the sum has the value advertised. These conventions will save endless fussing over changing limitsof summation when the dummy variables of summation get changed. For example, we ﬁnd that /summationdisplay k/parenleftbiggn r+k/parenrightbigg xk=x−r/summationdisplay k/parenleftbiggn r+k/parenrightbigg xr+k=x−r/summationdisplay s/parenleftbiggn s/parenrightbigg xs=x−r(1 +x)n, for nonnegative integer nand integer r, without ever even thinking about the ranges of the summation variables. The series evaluations that are most helpful in the examples that follow are, ﬁrst and foremost, /summationdisplay r≥0/parenleftbiggr k/parenrightbigg xr=xk (1−x)k+1(k≥0), (4.3.1) which is basically a rewrite of (2.5.7). Also useful are the binomial theorem /summationdisplay r/parenleftbiggn r/parenrightbigg xr=( 1+x)n(4.3.2) and (2.5.11), which we repeat here for easy reference: /summationdisplay n1 n+1/parenleftbigg2n n/parenrightbigg xn=1 2x(1−√ 1−4x). (4.3.3) Example 1. Openers Consider the sum /summationdisplay k≥0/parenleftbiggk n−k/parenrightbigg (n=0,1,2,...). The free variable is n, so let’s call the sum f(n). Write it out like this: f(n)=/summationdisplay k≥0/parenleftbiggk n−k/parenrightbigg . 4.3 The ‘Snake Oil’ method for easier combinatorial identities 121 OK, now multiply both sides by xnand sum over n. You have now arrived at step (c) of the general method, and you are looking at F(x)=/summationdisplay nxn/summationdisplay k≥0/parenleftbiggk n−k/parenrightbigg . Ready for step (d)? Interchange the sums, to get F(x)=/summationdisplay k≥0/summationdisplay n/parenleftbiggk n−k/parenrightbigg xn. We would like to ‘do’ the inner sum, the one over n. The trick is to get the exponent of xto be exactly the same as the index that appears in the binomial coeﬃcient. In this example the exponent of xisn, andnis involved in the downstairs part of the binomial coeﬃcient in the form n−k. To make those the same, the correct medicine is to multiply inside the sum byx−kand outside the inner sum by xk, to compensate. The result is F(x)=/summationdisplay k≥0xk/summationdisplay n/parenleftbiggk n−k/parenrightbigg xn−k. Now the exponent of xis the same as what appears downstairs in the binomial coeﬃcient. Hence take r=n−kas the new dummy variable of summation in the inner sum. We ﬁnd then F(x)=/summationdisplay k≥0xk/summationdisplay r/parenleftbiggk r/parenrightbigg xr. We recognize the inner sum immediately, as (1 + x)k. Hence F(x)=/summationdisplay k≥0xk(1 +x)k=/summationdisplay k≥0(x+x2)k=1 1−x−x2. The generating function on the right is an old friend; it generates the Fi- bonacci numbers (see Example 1.3 of chapter 1). Hence f(n)=Fn+1, and we have discovered that /summationdisplay k≥0/parenleftbiggk n−k/parenrightbigg =Fn+1 (n=0,1,2,...). 122 4 Applications of generating functions Example 2. Another one Consider the sum /summationdisplay k/parenleftbiggn+k m+2k/parenrightbigg/parenleftbigg2k k/parenrightbigg(−1)k k+1(m,n≥0). (4.3.4) Can it be that the same method will do this sum, without any further infusion of ingenuity? Indeed; just pour enough Snake Oil on it and it will be cured. Let f(n) denote the sum in question, and let F(x) be its opsgf. Dive in immediately by multiplying by xnand summing over n≥0, to get F(x)=/summationdisplay n≥0xn/summationdisplay k/parenleftbiggn+k m+2k/parenrightbigg/parenleftbigg2k k/parenrightbigg(−1)k k+1 =/summationdisplay k/parenleftbigg2k k/parenrightbigg(−1)k k+1x−k/summationdisplay n≥0/parenleftbiggn+k m+2k/parenrightbigg xn+k =/summationdisplay k/parenleftbigg2k k/parenrightbigg(−1)k k+1x−k/summationdisplay r≥k/parenleftbiggr m+2k/parenrightbigg xr =/summationdisplay k/parenleftbigg2k k/parenrightbigg(−1)k k+1x−kxm+2k (1−x)m+2k+1(by (4.3.1)) =xm (1−x)m+1/summationdisplay k/parenleftbigg2k k/parenrightbigg1 k+1/braceleftbigg−x (1−x)2/bracerightbiggk =−xm−1 2(1−x)m−1/braceleftBigg 1−/radicalBigg 1+4x (1−x)2/bracerightBigg =−xm−1 2(1−x)m−1/braceleftbigg 1−1+x 1−x/bracerightbigg =xm (1−x)m. The original sum is now unmasked: it is the coeﬃcient of xnin the last member above. But that is/parenleftbign−1 m−1/parenrightbig , by (4.3.1) again, and we have our answer. See exercise 16 for a generalization of this sum. If the train of manipulations seemed long, consider that at least it’s always the same train of manipulations, whenever the method is used, and also that with some eﬀort a computer could be trained to do it! Example 3. A discovery Is it possible to write the sum fn=/summationdisplay k≤n 2(−1)k/parenleftbiggn−k k/parenrightbigg yn−2k(n≥0) (4 .3.5) 4.3 The ‘Snake Oil’ method for easier combinatorial identities 123 in a simpler closed form? This example shows the whole machine at work again, along with a few new wrinkles. The ﬁrst step is to let Fops ←→{fn}, and try to ﬁnd the generating function Finstead of the sequence {fn}. To do that we multiply (4.3.5) on both sides by xnand sum over n≥0 to obtain F(x)=/summationdisplay n≥0xn/summationdisplay k≤n 2(−1)k/parenleftbiggn−k k/parenrightbigg yn−2k. The next step is invariably to interchange the summations and hope. To try to make the innermost summation as clean looking as possible, be sure to take to the outer sum any factors that depend only on k. This yields F(x)=/summationdisplay k(−1)ky−2k/summationdisplay n≥2k/parenleftbiggn−k k/parenrightbigg xnyn. Now focus on (4.3.1), and try to make the inner sum look like that. If in our inner sum the powers of xandywerexn−kyn−k, then those exponents would match exactly the upper story of the binomial coeﬃcient/parenleftbign−k k/parenrightbig , and so after a change of dummy variable of summation we would be looking exactly at the left side of (4.3.1). Hence we next multiply inside the inner sum by x−ky−k, and outside the inner sum by xkyk. Now we have F(x)=/summationdisplay k(−1)ky−2kxkyk/summationdisplay n≥2k/parenleftbiggn−k k/parenrightbigg xn−kyn−k =/summationdisplay k(−1)kxky−k/summationdisplay a≥k/parenleftbigga k/parenrightbigg (xy)a =/summationdisplay k≥0(−1)kxky−k(xy)k (1−xy)k+1(by (4.3.1)) =1 1−xy/summationdisplay k≥0/braceleftbigg−x2 1−xy/bracerightbiggk =1 1−xy1 1+x2 1−xy =1 1−xy+x2.(4.3.6) (Question: Why, after the third equals sign above, did the range of kget restricted to ‘ k≥0’?) 124 4 Applications of generating functions We now expand (4.3.6) in partial fractions to obtain a closed form for the sum (4.3.5). This gives F(x)=1 (1−xx+)(1−xx−) =x+ (x+−x−)(1−xx+)−x− (x+−x−)(1−xx−), where x±=y±/radicalbig y2−4 2. Hence, forn≥0 the coeﬃcient of xnis fn=1/radicalbig y2−4  /parenleftBigg y+/radicalbig y2−4 2/parenrightBiggn+1 −/parenleftBigg y−/radicalbig y2−4 2/parenrightBiggn+1  . We now have our answer, but just for a demonstration of the eﬀec- tiveness of cleanup operations, let’s invest a little more time in making the answer look as neat as possible. Because of the ubiquitous appearance of/radicalbig y2−4 in the answer, we replace yformally by x+( 1/x). Then /radicalbig y2−4=x−1 x, and our formula becomes /summationdisplay k≤n 2(−1)k/parenleftbiggn−k k/parenrightbigg (x2+1 )n−2kx2k=x2n+2−1 x2−1(n≥0). Finally we write t=x2to obtain the pretty evaluation /summationdisplay k≤n 2(−1)k/parenleftbiggn−k k/parenrightbigg (t+1 )n−2ktk=1−tn+1 1−t(n≥0). (4.3.7) For instance, the value t= 1 gives /summationdisplay k≤n 2(−1)k/parenleftbiggn−k k/parenrightbigg 2n−2k=n+1 (n≥0). (4.3.8) As a ﬁnal touch, we can read oﬀ the coeﬃcient of tmin (4.3.7) to discover the interesting fact that /summationdisplay k≤n 2(−1)k/parenleftbiggn−k k/parenrightbigg/parenleftbiggn−2k m−k/parenrightbigg =/braceleftbigg 1,if 0≤m≤n; 0,otherwise.(4.3.9) 4.3 The ‘Snake Oil’ method for easier combinatorial identities 125 Try this identity with n= 2 and watch what happens. Here is another example of the same technique. Example 4. Evaluate the sums fn=/summationdisplay k/parenleftbiggn+k 2k/parenrightbigg 2n−k(n≥0). (4.3.10) Without stopping to think, let Fbe the opsgf of the sequence, multiply both sides of (4.3.10) by xn, sum overn≥0, and interchange the two sums on the right. This produces F=/summationdisplay k2−k/summationdisplay n≥0/parenleftbiggn+k 2k/parenrightbigg 2nxn =/summationdisplay k2−k(2x)−k/summationdisplay n≥0/parenleftbiggn+k 2k/parenrightbigg (2x)n+k =/summationdisplay k≥02−k(2x)−k(2x)2k (1−2x)2k+1(by (4.3.1)) =1 1−2x/summationdisplay k≥0/braceleftbiggx (1−2x)2/bracerightbiggk =1 1−2x1 1−x (1−2x)2 =1−2x (1−4x)(1−x) =2 3(1−4x)+1 3(1−x). It is now a triviality to read oﬀ the coeﬃcient of xnon both sides and discover the answer: /summationdisplay k/parenleftbiggn+k 2k/parenrightbigg 2n−k=22n+1+1 3(n≥0). (4.3.11) Example 5. Our next example will be of a sum that we won’t succeed in evaluating in a neat, closed form. However, the generating function that we obtain will be rather tidy, and that is about the most that can be expected from this family of sums. 126 4 Applications of generating functions The sum is fn(y)=/summationdisplay k/parenleftbiggn k/parenrightbigg/parenleftbigg2k k/parenrightbigg yk(n≥0). (4.3.12) Follow the usual prescription. Deﬁne F(x,y)=/summationtext n≥0fn(y)xn. To ﬁnd F, multiply (4.3.12) by xn, sum over n≥0 and interchange the inner and outer sums, to obtain F(x,y)=/summationdisplay k/parenleftbigg2k k/parenrightbigg yk/summationdisplay n≥0/parenleftbiggn k/parenrightbigg xn =/summationdisplay k/parenleftbigg2k k/parenrightbigg ykxk (1−x)k+1 =1 1−x/summationdisplay k/parenleftbigg2k k/parenrightbigg/parenleftbiggxy 1−x/parenrightbiggk .(4.3.13) Now since/summationdisplay k/parenleftbigg2k k/parenrightbigg zk=1√1−4z, (4.3.14) by (2.5.11), we obtain F(x,y)=1 (1−x)/radicalBig 1−4xy 1−x =1/radicalbig (1−x)(1−x(1 + 4y)).(4.3.15) For general values of y, that’s about all we can expect. There are two special values of yfor which we can go further. If y=−1/4, we ﬁnd that /summationdisplay k/parenleftbigg2k k/parenrightbigg/parenleftbiggn k/parenrightbigg (−1 4)k=2−2n/parenleftbigg2n n/parenrightbigg (n≥0). (4.3.16) Ify=−1/2, then F(x,−1/2) = 1//radicalbig 1−x2 =/summationdisplay m/parenleftbigg2m m/parenrightbigg (x/2)2m(by (2.5.11)). Hence we have Reed Dawson’s identity /summationdisplay k/parenleftbigg2k k/parenrightbigg/parenleftbiggn k/parenrightbigg (−1)k2−k=/braceleftbigg/parenleftbign n/2/parenrightbig 2−nifn≥0 is even, 0i f n≥0 is odd,(4.3.17) 4.3 The ‘Snake Oil’ method for easier combinatorial identities 127 and Snake Oil triumphs again. Example 6. Suppose we have two complicated sums and we want to show that they’re the same. Then the generating function method, if it works, should be very easy to carry out. Indeed, one might just ﬁnd the generating functions of each of the two sums independently and observe that they are the same. Suppose we want to prove that /summationdisplay k/parenleftbiggm k/parenrightbigg/parenleftbiggn+k m/parenrightbigg =/summationdisplay k/parenleftbiggm k/parenrightbigg/parenleftbiggn k/parenrightbigg 2k(m,n≥0) without evaluating either of the two sums. Multiply on the left by xn, sum onn≥0 and interchange the summa- tions, to arrive at /summationdisplay k/parenleftbiggm k/parenrightbigg x−k/summationdisplay n≥0/parenleftbiggn+k m/parenrightbigg xn+k=/summationdisplay k/parenleftbiggm k/parenrightbigg x−kxm (1−x)m+1 =xm (1−x)m+1/parenleftbigg 1+1 x/parenrightbiggm =(1 +x)m (1−x)m+1. If we multiply on the right by xn, etc., we ﬁnd /summationdisplay k/parenleftbiggm k/parenrightbigg 2k/summationdisplay n≥0/parenleftbiggn k/parenrightbigg xn=1 (1−x)/summationdisplay k/parenleftbiggm k/parenrightbigg/parenleftbigg2x (1−x)/parenrightbiggk =1 (1−x)/parenleftbigg 1+2x 1−x/parenrightbiggm =(1 +x)m (1−x)m+1. Hence the two sums are equal, even if we don’t know what they are! Example 7. There are, in combinatorics, a number of inversion formulas , and gen- erating functions give an easy way to prove many of those. An inversion formula in general is a relationship that expresses one sequence in terms of another, along with the inverse relation, which recovers the original se- quence from the constructed one. We have already seen a couple of famous examples of these. One is the M¨ obius inversion formula, which is the pair (2.6.11), (2.6.12). Another 128 4 Applications of generating functions is the pair (4.2.3), (4.2.7) that occurred in the sieve method. We repeat that pair here, for ready reference. It states that if we compute a sequence {Nr}from a sequence {er}by the relations Nr=/summationdisplay t≥0/parenleftbiggt r/parenrightbigg et (r=0,1,2,...), (4.3.18) then we can recover the original sequence (‘invert’) by means of et=/summationdisplay j(−1)j−t/parenleftbiggj t/parenrightbigg Nt (t≥0). To give just one more example of such a pair of formulas, consider the relation ar=/summationdisplay s/parenleftbiggr s/parenrightbigg bs (r≥0), (4.3.19) which diﬀers from the previous pair in that the summation is over the lower index in the binomial coeﬃcient. How can we ﬁnd the relations that are inverse to (4.3.19)? That is, how can we solve for the b’s in terms of the a’s? The answer is that we convert the relation (4.3.18) between two se- quences into a relation between their exponential generating functions, which we then invert. By (2.3.3) we have A(x)=exB(x), whereAand Bare the egf’s. Hence B(x)=e−xA(x), and therefore bn=/summationdisplay m/parenleftbiggn m/parenrightbigg (−1)n−mam (n≥0). (4.3.20) An inversion formula of a somewhat deeper kind appears in (5.1.5), (5.1.6). Example 8. Snake Oil vs. hypergeometric functions. Many combinatorial identities are special cases of identities in the the- ory of hypergeometric series (we’ll explain that remark, brieﬂy, in a mo-ment). However, the Snake Oil method can cheerfully deal with all sorts of identities that are not basically about hypergeometric functions. So the approaches are complementary. A hypergeometric series is a series /summationdisplay kTk 4.3 The ‘Snake Oil’ method for easier combinatorial identities 129 in which the ratio of every two consecutive terms is a rational function of the summation variable k. That means that Tk+1 Tk=P(k) Q(k), wherePandQare polynomials, and it takes in a lot of territory. Many binomial coeﬃcient identities, including all of the examples in this chapter so far, are of this type. There are some general tools for dealing with such sums, and these are very important considering how frequently they occur in practice. For a discussion of some of these tools, see, for example, the article by Roy [Ro]. In this example we want to emphasize that the scope of the Snake Oil method includes a lot of sums that are not hypergeometric. Consider, for instance, the following sum- f(n)=/summationdisplay k/bracketleftbiggn k/bracketrightbigg Bk, where the/bracketleftbig/bracketrightbig ’s are the Stirling numbers of the ﬁrst kind, and the B’s are the Bernoulli numbers. Now one thing, at least, is clear from looking at this sum: it is not hypergeometric. The ratio of two consecutive terms is certainly not a ra- tional function of k. The Snake Oil method is, however, unfazed by this turn of events. If you follow the method exactly as before, you could deﬁne F(x) to be the egf of the sequence {f(n)}, multiply by xn/n!, sum onn, interchange the indices, etc., and obtain F(x)=/summationdisplay nf(n)xn n! =/summationdisplay nxn n!/summationdisplay k/bracketleftbiggn k/bracketrightbigg Bk =/summationdisplay kBk/summationdisplay n/bracketleftbiggn k/bracketrightbiggxn n! =/summationdisplay kBk/braceleftBigg 1 k!/parenleftbigg log1 1−x/parenrightbiggk/bracerightBigg (by (3.5.3)) =/summationdisplay kBk k!uk(u= log1 1−x) =u eu−1(by (2.5.8)) =1−x xlog1 1−x. 130 4 Applications of generating functions If we now read oﬀ the coeﬃcient of xn/n! on both sides, we ﬁnd that the unknown sum is /summationdisplay k/bracketleftbiggn k/bracketrightbigg Bk=−(n−1)! n+1(n≥1). (4.3.21) Example 9. The scope of the Snake Oil method The success of the Snake Oil method depends upon being given a sum to evaluate in which there is a free variable that appears in only one place. Then, after interchanging the order of the summations, one ﬁnds one of the basic power series (4.3.1) or (4.3.2) to sum. At the risk of diminishing the charm of the method somewhat by adding gimmicks to it, one must remark that in many important cases this limitation on the scope is easy to overcome. This is because it fre- quently happens that when an identity is presented that has a free variablerepeated several times, that identity turns out to be a special case of a more general identity in which each of the repeated appearances of the free vari- able is replaced by a diﬀerent free variable. Before abandoning the method on some given problem, this possibility should be explored. Consider the identity /summationdisplay i/parenleftbiggn i/parenrightbigg/parenleftbigg2n n−i/parenrightbigg =/parenleftbigg3n n/parenrightbigg . At ﬁrst glance the possibilities for successful Snake Oil therapy seem dim because of the multiple appearances of nin the summand. However, if we generalize the identity by splitting the appearances of ninto diﬀerent free variables, we might be led to consider the sum /summationdisplay i/parenleftbiggn i/parenrightbigg/parenleftbiggm r−i/parenrightbigg , which is readily evaluated by the Snake Oil method. It is characteristic of the subject of identities that it is usually harder to prove special cases than general theorems. Multiple appearances of a free variable are often a hint that one should try to ﬁnd a suitable generalization. 4.4 WZ pairs prove harder identities Computers can now ﬁnd proofs of combinatorial identities, including most of the identities that we did by the Snake Oil method in the previous section, as well as many, many more. In this section we will say how that is done. Although ﬁnding proofs this way requires more work than a human would care to do, the result, after the computer is ﬁnished, is a neat and compact proof that a human can often easily check, and can always check 4.4 WZ pairs prove harder identities 131 with the assistance of one of the numerous symbolic manipulation programs that are now available on personal computers. Hence there is no need for blind trust in the computer. One can ask it to ﬁnd a proof of an identity, and one can readily check that the proof is correct. These developments are quite recent, and they will surely change our attitudes towards, for instance, binomial coeﬃcient identities. Instead of regarding each one as a challenge to our ingenuity, we can instead ask our computer to ﬁnd a proof. It will not always succeed, but in the vast majority of cases it will. In fact, even more powerful methods are nowbecoming available, which promise a 100% success rate in certain classes of identities. This doesn’t mean that it was a waste of time to have learned the Snake Oil method. There were identities that Snake Oil handled that the method of [WZ] (which we’re about to discuss) cannot deal with, like the fact that/summationtext k≥0/parenleftbigk n−k/parenrightbig =Fnforn≥0, which was the ﬁrst example in the previous section. Another one that oﬀers no hope to the WZ method is (4.3.21), which involves Stirling and Bernoulli numbers. Also, to use the Snake Oil method, one doesn’t need to know the right hand side of theidentity in advance; the method will ﬁnd it. The method that we are about to describe will prove a given identity, but it won’t discover the identity for itself. With those disclaimers, however, it is fair to say that the method is quite versatile, and seems able to handle in a uniﬁed way some of the knottiest identities that have ever been discovered. It stems from some totally obvious facts, concatenated in a slightly un-obvious way. Suppose we want to prove that an identity /summationdisplay kU(n,k)=rhs(n)(n=0,1,2,...) is true. The ﬁrst thing we do is to divide by the right hand side to get the standard form/summationdisplay kF(n,k)=1 (n=0,1,2,...). (4.4.1) So, in standard form, we are trying to prove that a certain sum is indepen- dent ofn, forn≥0. To do that, rewrite (4.4.1) with nreplaced by n+ 1 and then subtract (4.4.1), to get /summationdisplay k{F(n+1,k)−F(n,k)}=0 (n=0,1,...)( 4 .4.2) Wouldn’t it be helpful if there were a nice function G(n,k) such that F(n+1,k)−F(n,k)=G(n,k+1 )−G(n,k), (4.4.3) 132 4 Applications of generating functions for then the sum (4.4.2) would telescope? In detail, that would mean that k=K/summationdisplay k=−L{F(n+1,k)−F(n,k)}=k=K/summationdisplay k=−L{G(n,k+1 )−G(n,k)} =G(n,K +1 )−G(n,−L). Well, as long as we’re wishing, why not wish for G(n,±∞) = 0 too, for then, by letting K,L→∞ we would ﬁnd that (4.4.2) is indeed true! This line of reasoning leads quickly to the following: Theorem 4.4.1. (Wilf, Zeilberger [WZ]) Let (F,G)satisfy (4.4.3), and suppose lim k→±∞G(n,k)=0 (n=0,1,2,...). (4.4.4) Then the identity /summationdisplay kF(n,k)=const. (n=0,1,2,...) holds. Example 1. Suppose we want to prove the identity /summationdisplay k/parenleftbiggn k/parenrightbigg =2n(n≥0). (4.4.5) If we divide by the right hand side we ﬁnd that the function F(n,k)o f (4.4.1) is F(n,k)=/parenleftbiggn k/parenrightbigg /2n(n≥0). (4.4.6) Now we need to ﬁnd the mate G(n,k) of thisF. Well, it is G(n,k)=−/parenleftbign k−1/parenrightbig 2n+1. (4.4.7) That raises two questions. Does this Greally work, and where did it come from? Let’s take the easy one ﬁrst. To check that it works we need to check ﬁrst that (4.4.3) holds, which in this case says that 2−n−1/parenleftbiggn+1 k/parenrightbigg −2−n/parenleftbiggn k/parenrightbigg =−2−n−1/parenleftbiggn k/parenrightbigg +2−n−1/parenleftbiggn k−1/parenrightbigg . 4.4 WZ pairs prove harder identities 133 After a few moments of work we can satisfy ourselves that this is true. Further, the boundary conditions (4.4.4) are easy to verify, and we are all ﬁnished. Deﬁnition. We say that an identity (4.4.1) is certiﬁed by a pair (F,G) (‘WZ pair’) if the conditions (4.4.3), (4.4.4) hold. Hence, the simple identity (4.4.5) is certiﬁed by the pair ( F,G)o f (4.4.6), (4.4.7). We can cut down still further on the complexity of the apparatus, as follows. It will turn out in the class of identities that we are discussing here that the mate Gwill always be of the form G(n,k)=R(n,k)F(n,k−1), whereR(n,k) is a rational function of nandk. Hence, instead of describing the pair (F,G), we need to give only FandR. ButFcomes directly from the identity that we’re trying to prove, just by dividing the summand by the right hand side. So if we have the identity in front of us, then the rational function Ris the only extra certiﬁcation that we need. The class of identities for which the above simpliﬁcation is true is the class of those that are of the form (4.4.1) in which the function Fhas the property that both F(n+1,k) F(n,k)andF(n,k−1) F(n,k) are rational functions of nandk. This class includes just about every binomial coeﬃcient identity that we have encountered or will encounter. Let’s recapitulate the complete proof procedure for an identity that is certiﬁed by a single rational function R(n,k): (a) Given an identity/summationtext kU(n,k)=rhs(n)(n=0,1,2,...), and also given a rational function proof certiﬁcate R(n,k). (b) Deﬁne F(n,k)=U(n,k)/rhs(n), forn≥0 and integer k. (c) DeﬁneG(n,k)=R(n,k)F(n,k−1). (d) Check that the conditions (4.4.3) and (4.4.4) are satisﬁed. (e) Check that the identity is true when n=0 . (f) The proof of the identity is now complete. Now here are some more examples of the technique in action. Do take the time to check one or more of these using the full proof technique (a)-(f) given above. Theorem./summationtext k(−1)k/parenleftbign k/parenrightbig/parenleftbig2k k/parenrightbig 4n−k=/parenleftbig2n n/parenrightbig (n≥0) Proof: Take R(n,k)=( 2k−1)/(2n+ 1). Let’s check that statement, one step at a time, following the proof procedure above. From step (b) we ﬁnd that F(n,k)=(−1)k/parenleftbign k/parenrightbig/parenleftbig2k k/parenrightbig 4n−k /parenleftbig2n n/parenrightbig. 134 4 Applications of generating functions From step (c) we ﬁnd that G(n,k)=(−1)k−12k−1 2n+1/parenleftbiggn k−1/parenrightbigg/parenleftbigg2k−2 k−1/parenrightbigg4n−k+1 /parenleftbig2n n/parenrightbig. Now we know the pair ( F,G). In step (d) we must ﬁrst check that the conditionF(n+1,k)−F(n,k)=G(n,k+1 )−G(n,k) is satisﬁed. At this point it would be very helpful to have a symbolic manipulation program available, for then one would simply type in the functions FandG, and ask it to verify that the condition holds. Otherwise, it’s a rather dull pencil and paper computation of ﬁve minutes’ length, and we will omit it. The second part of step (d) is the check of the boundary condition lim k→±∞G(n,k)=0 (n=0,1,2,...). That, however, is a triviality, because in this case not only are the limits 0, but the function G(n,k) is 0 for every single value of k>n + 1 and for all values ofk<1. In step (e) we quickly check and ﬁnd that 1=1, and the proof is com- plete. Theorem./summationtext k(−1)k/parenleftbign k/parenrightbig //parenleftbigk+a k/parenrightbig =a/(n+a)(n≥0) Proof: Take R(n,k)=k/(n+a). Theorem./summationtext k(−1)n−k/parenleftbig2n k/parenrightbig2=/parenleftbig2n n/parenrightbig (n≥0) Proof: Take R(n,k)=−(10n2−6kn+1 7n+k2−5k+7 )/(2(2n−k+2 )2). After the last three examples the reader will probably be wondering how to ﬁndtheR(n,k)’s, instead of just checking that an R(n,k) snatched out of the blue sky seems to work. So we are going to tell that story too, because it’s a very important one, not just for these purposes but for symbolic manipulation in general. The problem is this: the function F(n,k) is known, and we want to ﬁndG(n,k) so that the identity (4.4.3) is true. Since Fis known, so is F(n+1,k)−F(n,k), so we might as well call it f(n,k). Next, observe that at this moment the index nis a silent partner. That is, we are looking for Gso thatf(n,k)=G(n,k+1 )−G(n,k), and we see that kis an active index butnis just a parameter, since nhas the same value throughout. So we might as well suppress the appearance of naltogether, and state the problem this way: if fkis a given function of k(and other parameters), how can we ﬁnd gkso thatfk=gk+1−gkfor all integers k? We still don’t quite have the right question, but we’re getting there. The question as asked is a triviality. There is always such a gkand it is 4.4 WZ pairs prove harder identities 135 just/summationtext j<kfj. Take the function fk=k, for instance. Then we can take gk to be/summationtext 0≤j<kj. To ask the right question we have to add the condition that the sum that represents gkcan be done in closed form . This whole problem is about closed forms. What is closed form? Well roughly it means that the answer should be pleasant to look at and have no summation signs left in it. That idea is too nebulous to work with, so we will use one way of making it precise that has proved to be productive. Deﬁnition. A function fkof the integer kis a hypergeometric term if fk+1/fkis a rational function of k. Thusk! is a hypergeometric term. So is (3 k+ 2)!/(5k−6)!, and so is (−1)k4n−k/parenleftbig2k k/parenrightbig /parenleftbign+k k/parenrightbig. The function kkis not a hypergeometric term, nor is e√ k. The function fk=/summationdisplay 0≤j≤k/parenleftbiggn j/parenrightbigg is not obviously a hypergeometric term, nor is it obviously not a hyperge- ometric term. The expression serves to deﬁne the function but does not immediately reveal the nature of the beast. Now we’re ready for the right question. Letfkbe deﬁned for integer kand be a hypergeometric term. Does there exist a hypergeometric term gksuch thatfk=gk+1−gkfor all integers k? An algorithm that is due to R. W. Gosper, Jr. [Gos] gives a complete algorithmic answer to this question. That is to say, if we input fto Gosper’s algorithm it will then either return a function gwith the desired properties, or it will return a guarantee that no such function exists. It will notever return a statement ‘I don’t know.’ We will not describe Gosper’s algorithm here because that would take us rather far aﬁeld from generating functions. However, the reader is urged to consult either the original reference [Gos]or the lucid explanation in [GKP]. Gosper’s algorithm is built in to some of the commercially available symbolic manipulation packages. At this writing (June, 1989) the algorithm is fully implemented in Macsyma, where it can be invoked with the nusum command, and it is partially implemented in Mathematica. No doubt it will be more widely available as its usefulness becomes recognized. Now here are the statements and proofs of two more general and dif- ﬁcult identities, using the [WZ] method of proof by rational function certi- ﬁcation. 136 4 Applications of generating functions Theorem 4.4.2. (The Pfaﬀ-Saalsch¨ utz identity) /summationdisplay k(a+k)!(b+k)!(c−a−b+n−1−k)! (k+ 1)!(n−k)!(c+k)!= (a−1)!(b−1)!(c−a−b−1)!(c−a+n)!(c−b+n)! (c−a−1)!(c−b−1)!(n+ 1)!(c+n)!. Proof: Take R(n,k)=−(b+k)(a+k) (c−b+n+ 1)(c−a+n+1 ). Theorem 4.4.3. (Dixon’s identity) /summationdisplay k(−1)k/parenleftbiggn+b n+k/parenrightbigg/parenleftbiggn+c c+k/parenrightbigg/parenleftbiggb+c b+k/parenrightbigg =(n+b+c)! n!b!c!. Proof: TakeR(n,k)=(c+1−k)(b+1−k)/(2(n+k)(n+b+c+ 1)). 4.5 Generating functions and unimodality, convexity, etc. The binomial coeﬃcients are the prototype of unimodal sequences. A sequence is unimodal if its entries rise to a maximum and then decrease. The binomial coeﬃcients {/parenleftbign k/parenrightbig }n k=0do just that. The maximum (‘mode’) of the binomial coeﬃcient sequence occurs at k=n/2i fnis even, and at k=(n±1)/2i fnis odd. In general, a sequence c0,c1,...,c nis unimodal if there exist indices r,ssuch that c0≤c1≤c2≤···≤cr=cr+1=···=cr+s≥cr+s+1≥···≥cn.(4.5.1) Many of the sequences that occur in combinatorics are unimodal. Sometimes it is easy and sometimes it can be very hard to prove that a given sequence is unimodal. Generating functions can help with this kind of a problem, though they are far from a panacæa. A stronger property than unimodality is logarithmic concavity . First recall that a function fon the real line is concave if whenever x<y we havef((x+y)/2)≥(f(x)+f(y))/2. This means that the graph of the function bulges up over every one of its chords. Similarly, a sequence c0,c1,...,c nof positive numbers is log concave if logcµis a concave function of µ, which is to say that (logcµ−1+ logcµ+1)/2≤logcµ. If we exponentiate both sides of the above, to eliminate all of the logarithms, we ﬁnd that the sequence is log concave if cµ−1cµ+1≤c2 µ (µ=1,2,...,n −1). (4.5.2) If, in (4.5.2) we can replace the ‘ ≤’b y‘<’, then we will say that the sequence isstrictly log concave. 4.5 Generating functions and unimodality, convexity, etc. 137 Proposition. Let{cr}n 0be a log concave sequence of positive numbers. Then the sequence is unimodal. Proof. If the sequence is not unimodal then it has three consecutive mem- bers that satisfy cr−1>cr<c r+1which contradicts the assumed log concavity. In many cases of interest, generating functions can help to prove log concavity of a sequence, and therefore unimodality too. The source of such results is usually some variant of the following: Theorem 4.5.2. Letp(x)=c0+c1x+c2x2+···+cnxnbe a polynomial all of whose zeros are real and negative. Then the coeﬃcient sequence {cr}n 0 is strictly log concave. To prove the theorem we need to recall Rolle’s theorem of elementary calculus. It holds that if f(x) is continuously diﬀerentiable in ( a,b), and iff(a)=f(b), then somewhere between aandbthe derivative f/primemust vanish. Iffis a polynomial this can be considerably strengthened. Let uand vbe two consecutive distinct zeros of f. Then by Rolle’s theorem there is a zero off/primein (u,v). Suppose fis of degree n, has only real zeros, and has exactlyrdistinct real zeros. Then Rolle’s theorem accounts for r−1 of the zeros off/prime, because we ﬁnd one between each pair of consecutive distinct zeros off. The remaining n−rzeros offare copies of the distinct zeros. But ifx0is a root of fof multiplicity m> 1, then (x−x0)mis a factor of f, and so (x−x0)m−1is a factor of f/prime. Thusx0is a zero of multiplicity m−1o ff/prime. This accounts for the other n−1−(r−1) =n−rzeros off/prime. In particular, the zeros of f/primeare all real if the zeros of fare.F o r maximum utility in our present discussion, we summarize this discussion in the following way: Lemma 4.5.1. Let f(x,y)=c0xn+c1xn−1y+···+cnyn(4.5.3) be a polynomial all of whose roots x/yare real. Let g(x,y)be the result of diﬀerentiating fsome number of times with respect to xandy.I fgis not identically zero, then all of its zeros are real. Proof of theorem 4.5.2 : Since the zeros of fare all negative, we have f(x)=c0+c1x+···+cnxn=n/productdisplay j=1(x+xj), (4.5.4) where thexj’s are positive real numbers. Hence none of the ci’s can vanish. Now apply the diﬀerential operator Dm xDn−m−2 y to the polynomial f(x,y) of (4.5.3). Then only three terms survive, viz.: cn−m−2(m+2 ) n−m−1x2+2cn−m−1xy+(n−m)cn−m m+1y2. (4.5.5) 138 4 Applications of generating functions We can put this in a cleaner form by writing cj=/parenleftbign j/parenrightbig pj, in which case the result (4.5.5) becomes /parenleftbiggn m+1/parenrightbigg (pn−m−2x2+2pn−m−1xy+pn−my2). But this quadratic polynomial, according to lemma 4.5.1 above, must have two real roots, and so its discriminant must be nonnegative, i.e., p2 n−m−1≥pn−m−2pn−m, and the sequence of p’s is log concave. If we substitute back the c’s, we ﬁnd that c2 n−m−1≥(m+ 2)(n−m) (m+ 1)(n−m−1)cn−m−2cn−m >cn−m−2cn−m, and the strict log concavity is established. Corollary 4.5.1. The binomial coeﬃcient sequence/braceleftbig/parenleftbign k/parenrightbig/bracerightbign k=0is log con- cave, and therefore unimodal. Proof. The zeros of the generating polynomial (1 + x)nare evidently real and negative. Corollary 4.5.2. The sequence of Stirling numbers of the ﬁrst kind {/bracketleftbiggn k/bracketrightbigg }n k=1 is log concave, and therefore unimodal. Proof. According to (3.5.2), the opsgf of these Stirling numbers is the polynomial /summationdisplay j/bracketleftbiggn j/bracketrightbigg xj−1=(x+ 1)(x+2 )···(x+n−1), whose zeros are clearly real and negative. Corollary 4.5.3. The sequence of Stirling numbers of the second kind {/braceleftbign k/bracerightbig }n k=1is log concave, and therefore unimodal. Proof. We’ll have to work just a little harder for this one, because the zeros of the polynomial An(x)=/summationdisplay j/braceleftbiggn j/bracerightbigg xj 4.5 Generating functions and unimodality, convexity, etc. 139 are not easy to ﬁnd. They are, however, real and negative, and here is one way to see that: by (1.6.8) we have the recurrence formula An(y)={y(1 +Dy)}An−1(y)(n>0;A0=1 ), which can be rewritten in the form eyAn(y)=y(eyAn−1(y))/prime(n>0;A0=1 ). (4.5.6) We claim that for each n=0,1,2,..., the function eyAn(y) has exactly nzeros, which are real, distinct, and negative except for the one at y=0 . This is true for n= 0, and if it is true for 0 ,1,...,n −1, then (4.5.6) and Rolle’s theorem guarantee that ( eyAn−1(y))/primehasn−2 negative, distinct zeros, one between each pair of zeros of An−1(y). After multiplying by y, as in (4.5.6), we have n−1 negative, distinct zeros for eyAn(y), but we need to ﬁnd still one more. But eyAn−1(y) obviously approaches zero as y→− ∞ . Hence its derivative must have one more zero to the left of the leftmost zero of An−1(y), and we are ﬁnished. The theorem is very strong, but one must not be left with the im- pression that unimodality or log concavity has something essential to do with reality of the zeros of the generating polynomials. Many sequences are known that are unimodal, and have generating polynomials whose zeros alllie on the unit circle, and are quite uniformly distributed, in angle, around the circle. In such cases our theorem will be of no help. For example, an inversion of a permutation σofnletters is a pair (i,j) for which 1 ≤i<j ≤n, butσ(i)>σ(j). A permutation may have between 0 and/parenleftbig n 2/parenrightbig inversions. It is well known that if b(n,k) is the number of permutations of nletters that have exactly kinversions, then {b(n,k)}k≥0ops ←→(1 +x)(1 +x+x2)···(1 +x+x2+···+xn−1).(4.5.7) The zeros of the generating polynomial are very uniformly sprinkled around the unit circle, so the hypotheses of theorem 4.5.2 are extravagantly vio- lated. Nonetheless, the sequence is unimodal; it rises steadily for k≤/parenleftbign 2/parenrightbig /2, and falls steadily thereafter. 140 4 Applications of generating functions 4.6 Generating functions prove congruences In this section we give one or two examples of the power of the generat- ing function method in proving congruences among combinatorial numbers. A congruence between two generating functions means that the congruence holds between every pair of their corresponding coeﬃcients. Example 1. Stirling numbers of the ﬁrst kind We found, in chapter 3, that the Stirling numbers of the ﬁrst kind/bracketleftbign k/bracketrightbig have the generating function /summationdisplay k/bracketleftbiggn k/bracketrightbigg xk=x(x+ 1)(x+2 )···(x+n−1). (4.6.1) Suppose we are interested in ﬁnding some criterion for deciding the evenness or oddness of these numbers. If we read (4.6.1) modulo 2, it becomes /summationdisplay k/bracketleftbiggn k/bracketrightbigg xk≡x(x+1 )x(x+1 )··· (mod 2) =x⌈n/2⌉(x+1 )⌊n/2⌋.(4.6.2) Now take the coeﬃcient of xkon both sides, and ﬁnd that /bracketleftbiggn k/bracketrightbigg ≡[xk]x⌈n/2⌉(x+1 )⌊n/2⌋(mod 2) =/bracketleftBig xk−⌈n/2⌉/bracketrightBig (1 +x)⌊n/2⌋ =/parenleftbigg⌊n/2⌋ k−⌈n/2⌉/parenrightbigg .(4.6.3) Theorem 4.6.1. The Stirling number/bracketleftbign k/bracketrightbig has the same parity as the bino- mial coeﬃcient/parenleftbig⌊n/2⌋ k−⌈n/2⌉/parenrightbig . In particular,/bracketleftbign k/bracketrightbig is an even number if k<⌈n/2⌉. Example 2. The other Stirling numbers In the case of the Stirling numbers that count set partitions, the/braceleftbign k/bracerightbig ’s, we found in (1.6.5) that they have the ops generating function /summationdisplay n/braceleftbiggn k/bracerightbigg xn=xk (1−x)(1−2x)···(1−kx). Again, suppose we read the equation modulo 2. Then we would ﬁnd that/summationdisplay n/braceleftbiggn k/bracerightbigg xn≡xk (1−x)⌈k/2⌉(mod 2) =xk/summationdisplay h/parenleftbigg⌈k/2⌉+h−1 h/parenrightbigg xh. 4.7 The cycle index of the symmetric group 141 Now take the coeﬃcient of xnthroughout. The result is: Theorem 4.6.2. The Stirling number/braceleftbign k/bracerightbig has the same parity as the binomial coeﬃcient/parenleftbigg⌈k/2⌉+n−k−1 n−k/parenrightbigg . (4.6.4) 4.7 The cycle index of the symmetric group We have already studied the Stirling numbers of the ﬁrst kind, which give the number of permutations of nletters that have exactly kcycles. Now we’ll look for much more detailed information about the cycles of permutations. Instead of considering only the number of cycles that a permutation has, we will be interested in the numbers of cycles that it has of each length . So let a={a1,a2,a3,...}be a given sequence of nonnegative integers for whichn=a1+2a2+3a3+···is ﬁnite. How many permutations of n letters have exactly a1cycles of length 1 and exactly a2cycles of length 2 and etc.? For a given permutation σ, we will call the vector a=a(σ) the cycle type ofσ. It tells us the numbers of cycles of each length that σhas. Letc(a) denote the required number of permutations, and write φn(x)=/summationdisplay a1+2a2+···=n a1≥0,a2≥0,...c(a)xa1 1xa2 2···. (4.7.1) Thenφn(x) is called the cycle index of the symmetric group Sn. If we can somehow ﬁnd φn(x) then the coeﬃcient of each monomial xais the number of permutations of nletters whose cycle type is a. We are going to ﬁnd the “grand” generating function C(x,t)=∞/summationdisplay n=1φn(x)tn n!(4.7.2) which will turn out to have a surprisingly elegant form (see (4.7.5) below), considering the large amount of information that it contains. The derivation will be unusual in at least one respect. Most often a generating function is a way-station on the road to ﬁnding an exact formula for something. But in this problem we will begin by ﬁnding an exact formula forc(a). It will then be easy to check that its generating function really generates the sequence. Well, for a given a, how many permutations σhaveafor their cycle type? We will ﬁrst prove a lemma, and then give the answer. 142 4 Applications of generating functions Lemma A. Given integers m,a,k . The number of ways of choosing ka letters from m(distinct) given letters, and arranging them into acycles of lengthkis f(m,a,k )=m! (m−ka)!kaa!. (4.7.3) Proof. First choose an ordered ka-tuple of the letters, which can be done inm!/(m−ka)! ways. Then arrange each consecutive block of kletters in a cycle, which gives us our set of acycles. However, we claim that every ﬁxed set ofacycles of length kwill arise exactly kaa! times in this construction. Indeed, that set will occur in every ordering of the list of cycles ( a! such). Furthermore, the same set of acycles results from each of the kpossible circular permutations of elements within blocks of kconsecutive entries in the original ka-tuple, i.e., katimes. Hence if we are given nletters, and a sequence of nonnegative integers a1,a2,...such that a1+2a2+3a3+···=n, then the number of ways of forming these letters into a11-cycles and a2 2-cycles, and ..., is evidently f(n,a 1,1)f(n−a1,a2,2)f(n−a1−2a2,a3,3)··· =/parenleftbiggn! (n−a1)!1a1a1!/parenrightbigg/parenleftbigg(n−a1)! (n−a1−2a2)!2a2a2!/parenrightbigg ··· =n! a1!a2!···1a12a23a3···,(4.7.4) where Lemma A was used. This yields the following explicit formula for the number of permutations of each given cycle type. Theorem 4.7.1. Letabe nonnegative integers for which/summationtext jjaj=n. Then the number of permutations of nletters that have afor their cycle type is exactly c(a)=n!/producttext j≥1(aj!jaj). Next we’ll look for the generating function of the quantities c(a). Since there are inﬁnitely many variables awe shouldn’t be surprised by the need 4.7 The cycle index of the symmetric group 143 for a generating function in inﬁnitely many variables. We calculate C(x,t)=/summationdisplay n≥0φn(x) n!tn =/summationdisplay n≥0tn n!/summationdisplay a1+2a2+···=n a1≥0,a2≥0,...c(a)xa =/summationdisplay n≥0tn n!/summationdisplay a1+2a2+···=n a1≥0,a2≥0,...c(a)xa1 1xa2 2··· =/parenleftbigg/summationdisplay a1≥0(tx1)a1 1a1a1!/parenrightbigg/parenleftbigg/summationdisplay a2≥0(t2x2)a2 2a2a2!/parenrightbigg ··· =etx1et2x2/2et3x3/3··· = exp/parenleftbig/summationdisplay j≥1xjtj j/parenrightbig . We have proved the following result. Theorem 4.7.2. The coeﬃcient of tn/n!in C(x,t) = exp/parenleftbig/summationdisplay j≥1xjtj j/parenrightbig (4.7.5) is the cycle index of Sn, i.e., the generating function φn(x)in (4.7.1) above, of the numbers of permutations of nletters that have each possible cycle type. In more detail, the coeﬃcient of xatn/n!is the number of permuta- tions ofnletters whose cycle type is a. If this result rather reminds you of the exponential formula, and if you suspect that there must be some connection, you are quite correct. The result of exercise 22 of the previous chapter is a generalization of theorem 4.7.2 to exponential families. Indeed, the theorem is an immediate special case of the result of that exercise, but we thought it might be interestingto give an elementary proof also. Thus the generating function C(x,t) in (4.7.5) generates the cycle in- dexes of all of the symmetric groups. We will now give some of its applica- tions to the probabilistic theory of permutations. The polynomials φ n(x) of (4.7.1) have coeﬃcients that give the number of permutations of nletters with given cycle type. If we divide them by n!, as in (4.7.2), then since n! is the total number of permutations of nletters, we will then be ﬁnding the probabilities that a permutation has various 144 4 Applications of generating functions cycle type vectors. Thus C(x,t)=/summationdisplay nφn(x) n!tn =/summationdisplay npn(x)tn(4.7.6) where pn(x)=/summationdisplay a1+2a2+···=n a1≥0,a2≥0,...Prob( a,n)xa(4.7.7) and Prob( a,n) is the probability that a permutation of nletters has the cycle type a. Just ahead of us now there lie some very pretty theorems. They are theorems that give very quantitative answers to questions that don’t seem to have any quantities in them. For instance, take this question, which is a simple illustration of the genre: what is the probability that a permutation has no ﬁxed points? Notice that the question doesn’t tell us how many letters the permu- tation permutes. There is no ‘ n’ in the question. But it has a nontrivial answer: 1/e, as we discovered in (4.2.10). To interpret a question like this, one proceeds as follows. Let f(n) be the number of permutations of nletters that have no ﬁxed points. Then f(n)/n! is the probability that a permu- tation ofnletters has no ﬁxed points. Since, in this case, lim n→∞f(n)/n! exists and is equal to 1 /e, we can then say that, in this precise sense, the probability that a permutation has no ﬁxed points is 1/e. There are many lovely questions about permutations that sound like what is the probability that a permutation has ...? , in which the number of letters that the permutations act upon is not even mentioned, and to which the answers are nontrivial numbers, like the 1 /eabove. We are about to derive handfuls of them at once. But ﬁrst we need a lemma. Lemma B. Let/summationtext jbjbe a convergent series. Then in the power series expansion of the function 1 1−t/summationdisplay jbjtj=/summationdisplay nαntn, we have that limnαnexists and is equal to/summationtext jbj. Proof. By Rule 5 of section 2.2, αnis the sum of those bjfor whichj≤n. The latter sum clearly approaches the limit stated. Now letSbe a (ﬁnite or inﬁnite) set of positive integers, with the property that/summationdisplay s∈S1 s<∞. 4.7 The cycle index of the symmetric group 145 In the generating function C(x,t) we set all xi= 1 fori/∈S. That means that we are declaring ourselves to have no interest in any cycle lengths other than those in S. The others can be whatever they please. Then Cbecomes C(x,t) = exp/parenleftbigg/summationdisplay i∈Sxiti i+/summationdisplay i/∈Sti i/parenrightbigg = exp/parenleftbigg/summationdisplay i∈S(xi−1)ti i+ log1 1−t/parenrightbigg =1 1−texp/parenleftbigg/summationdisplay i∈S(xi−1)ti i/parenrightbigg . By Lemma B above, the coeﬃcient of tnin this last expression approaches the limit exp/parenleftbigg/summationdisplay i∈S(xi−1) i/parenrightbigg asn→∞ , which proves the following. Theorem 4.7.3. LetSbe a set of positive integers for which/summationtext s∈S1/s converges, and let abe a ﬁxed cycle type vector. The probability that the cycle type vector of a random permutation agrees with ain all of its components whose subscripts lie in Sexists and is equal to e−/parenleftbig/summationtext s∈S1/s/parenrightbig [xa] exp/parenleftbig/summationdisplay s∈Sxs s/parenrightbig =1/producttext s∈S/parenleftbig e1 ssasas!/parenrightbig. (4.7.8) As a ﬁrst example, take S={1}, so we are interested only in ﬁxed points. Then from (4.7.8), the probability that a random permutation hasexactlyaﬁxed points is 1 /(a!e), fora≥0. TakeS={r}. Then we see at once that the probability that a random permutation has exactly ar-cycles is 1/(e 1 rraa!), for each a=0,1,.... LetS={r,s}. The probability that a random permutation has exactly arr-cycles and exactly ass-cycles is therefore e−1/r−1/s rarsasr!s!. OK, now blindfold yourself, reach into a bag that contains every per- mutation in the world, and pull one out. What is the probability that none of its cycle lengths is the square of an integer? We claim that the probabil- ity ise−π2/6=.193025..., so the odds are about 5 to 1 against this event. Indeed, this is just the case where we take Sto be the set of all squares of integers, and use the fact that 1 + 1 /22+1/32+···=π2/6. For a ﬁnal example, what is the probability that a randomly chosen permutation contains equal numbers of 1-cycles and 2-cycles? Well, if that 146 4 Applications of generating functions number isj, then the probability is e−3/2/(2jj!2), for eachj=0,1,2,.... If we sum over all of these j, we ﬁnd that the required probability is e−3/2∞/summationdisplay j=01 2jj!2=0.34944033.... We can recast the result in theorem 4.7.3 in the language of the Poisson distribution. The Poisson distribution is a probability distribution on the nonnegative integers j=0,1,2,...that occurs very naturally in a number of areas of application, such as in the theory of waiting lines. It is given by Prob(j)=e−MMj j!(j=0,1,2,...) whereMis the mean. Theorem 4.7.3 then asserts the following: If a set Sis ﬁxed, for which/summationtext s∈S1/s < ∞, then for a randomly chosen permutation the numbers of cycles of each length s∈Shave asymptotically independent Poisson distributions in which the mean number of s-cycles is 1/sfor eachs∈S. 4.8 How many permutations have square roots? Letσbe a permutation. There may or may not be a permutation τ such thatσ=τ2. We want to describe and to count the σ’s that do have square roots, in this sense. More generally, σhas akth root if there is aτsuch thatσ=τk, and again, we would like to know the number of permutations of nletters that have kth roots. The answers to these questions involve some fairly spectacular generat- ing functions, and the methods will lean strongly on the cycle index results of the previous section, and in particular on theorem 4.7.2. We begin with the square root problem. So let τbe a permutation, and consider a single cycle of τ, say this one: 8→3→13→19→7→12→8. What happens to that cycle when we square τ? The mapping τ2executes the permutation τtwice, so it carries 8 into 13 and 13 into 7, etc. Thus we go hopping around the cycle of τ, visiting every second member, until we return to our starting point. A cycle of even length therefore falls apart into two cycles of half the length, while an odd cycle remains a cycle of the same length, although it becomes a diﬀerent cycle of that length. In the example above, the cycle shown breaks into two cycles, like this: 8→13→7→8 and 3 →19→12→3. In general, every cycle of τwhose length is 2 mwill contribute two cycles of lengthmtoτ2. 4.8 How many permutations have square roots? 147 A cycle of even length in τ2, therefore, can only be the result of splitting a cycle of twice its length in τinto two cycles. Hence, if σhas a square root, then the number of cycles that it has of each even length must be even . Conversely, let σbe any permutation that has this property. Then we claim thatσ=τ2for at least one τ. In fact we can construct such a τ(in how many ways?). To do that, pick up a pair of cycles of the same even length and thread them together into a single cycle of twice that length, as in the two examples above, by taking alternately a letter from one of the cycles and a letter from the other. These threaded cycles are all parts of the permutation τthat is being constructed. What do we do with the odd cycles of σ? If we are given a cycle of odd length 2 m+1i nσwe convert it into a cycle of the same odd length inτas follows. Let the letters in the given cycle of σbe a1→a2→a3→···→a2m→a2m+1→a1. Then intoτwe put the cycle a1→am+2→a2→am+3→a3→am+4→···→a2m+1→am+1. This cycle clearly has the property that if we square it then it will be back in the original order as in σ, and completes the proof of the following theorem (as well as giving us an algorithm for ﬁnding the square root of a permutation!). Theorem 4.8.1. A permutation σhas a square root if and only if the numbers of cycles of σthat have each even length are even numbers. Now letf(n,2) be the number of permutations of nletters that have square roots. We seek the generating function of the sequence. Consider the cycle type vector aof such a permutation. The even-indexed components must be even numbers, and the odd-indexed components are arbitrary. According to theorem 4.7.2, the coeﬃcient of xatn/n! in the product ex1tex2t2/2ex3t3/3··· is the number of permutations of nletters whose cycle type is a. The sum of these coeﬃcients over all of the cycle types that we are considering, namely a’s for which the even-indexed entries are even, is obtained as follows. In the product of the exponential functions above, put x1= 1, because all values ofa1are admissible. In the second exponential factor, ex2t2/2, don’t use the whole exponential series. Since only even powers of x2are admissible, use the subseries of even powers of the exponential series, namely the cosh series, and then put x2= 1. Then put x3= 1. Then use the cosh ( x4t4/4) series and put x4= 1, and so forth. 148 4 Applications of generating functions The result will be that the number f(n,2) of permutations of nletters that have square roots satisﬁes /summationdisplay n≥0f(n,2)tn n!=etcosh (t2/2)et3/3cosh (t4/4)et5/5··· = exp (t+t3/3+t5/5+···)/productdisplay m≥1cosh (t2m 2m) =/radicalbigg 1+t 1−t/productdisplay m≥1cosh (t2m 2m) =1+t+t2 2!+3t3 3!+1 2t4 24+6 0t5 5!+···.(4.8.1) Hence the sequence {f(n,2)}begins as 1,1,1,3,12,60,270,1890,14280,.... Corollary. Letp(n)be the probability that a permutation of nletters has a square root. Then for each n=0,1,2,..., we havep(2n)=p(2n+1 ). Proof. The generating function in (4.8.1) is of the form 1 /(1−t) times an even function of t. Hencef(n,2)/n! is thenth partial sum of the coeﬃcient sequence of an even function. More generally, the sequence of probabilities is actually weakly de- creasing, i.e., p(0) =p(1)≥p(2) =p(3)≥p(4) =p(5)≥···. Bijective proofs of these facts have been found by Dennis White (p.c.). Now let’s try the question of kth roots. We let f(n,k) be the number of permutations of nletters that have kth roots, and we seek the egf of {f(n,k)}n≥0. Some additional notation will be helpful here. For a prime p and an integer nwe will write e(p,n) for the highest power of pthat divides n. Next, for a pair m,k of positive integers, we deﬁne ( ( m,k)) t o b e ((m,k)) =/productdisplay p\mpe(p,k). First, ifkis given, for which permutations σis it true that there exists a permutation τsuch thatσ=τk? The generalization of theorem 4.8.1 is the following. Theorem 4.8.2. A permutation σhas akth root if and only if for every m=1,2,...it is true that the number of m-cycles that σhas is a multiple of((m,k)). To prove this, let σ=τkbe a permutation of nletters, and suppose σhas exactly νmcycles of length m, for eachm=1,2,.... Consider a 4.8 How many permutations have square roots? 149 cycle of length rinτ.I nτkthis contributes ( r,k) cycles of lengths r/(r,k). Hence theνmcycles of length minσmust come from cycles of length rin τwherer/(r,k)=m. From this equation r=m(r,k) it is easy to see that rmust be a multiple of m((m,k) ). Hence the cycles of length minσall come from cycles of lengths that are multiples of m((m,k)) i nτ. But every such cycle in τcontributes a multiple of ( ( m,k))m-cycles inσ. Hence the number ofm-cycles inσmust be a multiple of ( ( m,k)) . To show that it is also suﬃcient, let σbe a permutation that satisﬁes the condition. We will construct a kth root,τ,o fσ. Fixm, and write g=( (m,k) ). Then the number of m-cycles ofσis a multiple of g, so we can tie them up into bundles of gm-cycles each, and then for each bundle we can construct a single new cycle of length mgas follows: construct a circle withmgplaces marked consecutively around it. Take the ﬁrst m-cycle in the bundle and arrange its elements in the marked places, consecutive elements being spaced apart by gplaces. Then do the same for the second m-cycle in the bundle, etc. Repeat for each mto complete the proof. Theorem 4.8.2 is due to Arnold Knopfmacher and Richard Warlimont, and it corrects an error that appeared in this discussion in the previous printing of this book. To obtain the egf of the sequence {f(n,k)}we proceed as in (4.8.1) above. It will be convenient to have a name for the subseries of the expo- nential series that occur. So let us write expq(x) for the subseries of the exponential series exthat is obtained by choosing only the powers of xthat are divisible by q. That is expq(x)=/summationdisplay j≥0xjq (jq)!(q=1,2,3,...). Thus exp1(x)=ex, exp2(x) = coshx, exp3(x) is explicitly shown in eq. (2.4.7) of chapter 2, etc. Now following the argument that led to (4.8.1) we obtain this generalization. Theorem 4.8.3. Letf(n,k)be the number of permutations of nletters that have a kth root. Then we have ∞/summationdisplay n=0f(n,k)xn n!=∞/productdisplay m=1exp((m,k))/parenleftbigxm m/parenrightbig (k=1,2,3,...). (4.8.2) The reader is invited to check that this reduces to a triviality when k= 1, and to (4.8.1) when k= 2. A short table of f(n,k)( 1≤n≤10; 2≤ k≤7) is shown below. 150 4 Applications of generating functions k= 2 : 1 1 3 12 60 270 1890 14280 128520 1096200 k= 3 : 1 2 4 16 80 400 2800 22400 181440 1814400 k= 4 : 1 1 3 9 45 225 1575 11130 100170 897750 k= 5 : 1 2 6 24 96 576 4032 32256 290304 2612736 k= 6 : 1 1 1 4 40 190 1330 8680 52920 340200 k= 7 : 1 2 6 24 120 720 4320 34560 311040 3110400 4.9 Counting polyominoes By a cellwe will mean the interior and boundary of a unit square in the x-yplane, if the vertices of the square are at lattice points (points whose coordinates are both integers). Let Pbe a collection of cells. We associate withPa graph, whose vertices correspond to the cells of P, and in which two vertices are joined by an edge in the graph if the two cells to which they correspond intersect in a line segment (rather than in a vertex, or not at all). We say that Pis a connected collection of cells if the graph associated withPis a connected graph. A collection Pof cells is in standard position if all of its cells lie in the ﬁrst quadrant, and at least one of them intersects the yaxis and at least one of them intersects the xaxis. Apolyomino is a connected collection of cells that is in standard posi- tion. Here are all of the polyominoes that have one, two, or three cells: Sometimes polyominoes are called animals . This is because one can imagine a single cell that ‘grows’ by sprouting a new cell along one of its edges. Then that two-celled animal would grow a new cell along one of its edges, etc. If f(n) is the number of n-celled polyominoes, then from the picture above we see that {f(n)}=1,2,6,.... It would be good to be able to say that in this section we are going to derive the generating function etc. for the sequence f(n). We aren’t going to do that, though. The sequence and its generating function are unknown, despite a great deal of eﬀort that has been invested in the problem. Various special kinds of polyominoes, however, have been counted, with respect to various properties of the polyomino. For instance, among the properties that a polyomino has, one might mention its area, or number of cells, and its perimeter . So one might ask for the number of polyominoes of some special kind whose area is n, or the number whose perimeter is m, 4.9 Counting polyominoes 151 or the number whose area is nand whose perimeter is m, or the generating functions of any of these, etc. For a survey of recent progress in such questions see [De1] and [De2]. What we will do in this section will be to count a special kind of polyomino that is called horizontally convex (HC). An HC-polyomino is one in which every row is a single contiguous block of cells. The picture below shows a typical HC-polyomino. Another special kind of polyomino is called convex . A polyomino is convex if it is both vertically and horizontally convex. One of the striking results in the theory of polyominoes is the fact that there are exactly (2n+ 11)4n−4(2n+1 )/parenleftbigg2n n/parenrightbigg (4.9.1) convex polyominoes of perimeter 2 n+ 8. The number of area nhas been found by M. Bousquet-Melou [Bo]. There are interesting problems involved in counting HC-polyominoes either by area or by perimeter. We are going to count them here by area. It is worth noting that the question of enumerating them by perimeter has also been solved [De2], and the solution involves a remarkable generating function, which looks like this: if cnis the number of HC-polyominoes whose perimeter is 2 n+ 2 then /summationdisplay n≥0cntn=/radicalbig −(AC1/3+D+EC−1/3)−F 2√ AH−H 2√ 2A−G in whichA,B,...,H are certain speciﬁc functions. For instance A= 18t4(2t3−23t2+3 8t−18)2. For the complete list see [De2]. We return to the problem of counting HC-polyominoes by area, which is similar to the enumeration of ‘fountains’ in section 2.2, and our methodof attack will be similar. Letf(n,k,t ) be the number of HC-polyominoes of ncells, having k rows, of which tare in the top row. If we strip oﬀ the top row of one of these polyominoes, what will remain will have n−tcells, arranged in k−1 rows, with some number r≥1 in the top row. Hence after removing the top row, there are f(n−t,k−1,r) possibilities for what remains, for some r. However, each one of those possibilities generates r+t−1 of the original 152 4 Applications of generating functions (n,k,t ) HC-polyominoes, by adjoining a top row of tcells, and sliding it left and right through all legal positions atop the second row. Hence we have f(n,k,t )=/summationdisplay r≥1f(n−t,k−1,r)(r+t−1) (k≥2;f(n,1,t)=δt,n). (4.9.2) If we deﬁne the generating functions Fk,t(x)=/summationtext nf(n,k,t )xn, then we haveF1,t(x)=xt, fort≥1 and after multiplying (4.9.2) by xnand sum- ming overn, we obtain Fk,t(x)=xt/summationdisplay r≥1(r+t−1)Fk−1,r(x)(k≥2). (4.9.3) Now letUk(x)=/summationtext r≥1Fk,r(x) andVk(x)=/summationtext r≥1rFk,r(x). Then U1(x)=x/(1−x) andV1(x)=x/(1−x)2. Further, from (4.9.3), Fk,t(x)=xt(Vk−1(x)+(t−1)Uk−1(x)) (k≥2), (4.9.4) and if we sum on twe ﬁnd that Uk(x)=x 1−xVk−1(x)+x2 (1−x)2Uk−1(x)(k≥2). (4.9.5) If we ﬁrst multiply (4.9.4) by tand then sum on twe ﬁnd Vk(x)=x (1−x)2Vk−1(x)+2x2 (1−x)3Uk−1(x)(k≥2). (4.9.6) We now have two simultaneous recurrences to solve for the sequences UkandVk. To do that we eliminate the Vksequence as follows: solve (4.9.5) forVk−1in terms ofUkandUk−1, and substitute the result in (4.9.6). After simpliﬁcation we obtain a single three term recurrence for the U’s, viz. 1−x xUk+1(x)−x+1 1−xUk(x)−x2 (1−x)3Uk−1(x)=0 (k≥1),(4.9.7) along with the initial data U0(x) = 0 andU1(x)=x/(1−x). Finally, to solve (4.9.7) we introduce the generating function φ(x,y)=/summationtext k≥0Uk(x)yk. Then, if we multiply (4.9.7) by ykand sum over k≥1w e get 1−x xy/braceleftbigg φ(x,y)−U1(x)y/bracerightbigg −x+1 1−xφ(x,y)−x2y (1−x)3φ(x,y)=0. 4.10 Exact covering sequences 153 If we use the initial conditions and solve for φ, the result is that φ(x,y)=/summationdisplay k≥0Uk(x)yk=/summationdisplay n,k,rf(n,k,r )xnyk =xy(1−x)3 (1−x)4−xy(1−x−x2+x3+x2y).(4.9.8) Notice that the sum over rhas no variable attached to it; it acts directly on f(n,k,r ) and yields the number of HC-polyominoes of ncells andkrows, without regard to how many cells are in the top row. Thus if g(n,k) is that number, then /summationdisplay n,kg(n,k)xnyk=xy(1−x)3 (1−x)4−xy(1−x−x2+x3+x2y). (4.9.9) For the complete 3-variable generating function of the sequence {f(n,k,r )}, see exercise 21 at the end of this chapter. Perhaps we are interested only in the total number of HC-polyominoes, and we don’t need to know the number of rows. In that case we let y=1 in (4.9.8) and we ﬁnd the following result, which is due to D. Klarner, who used diﬀerent methods. Theorem 4.9.1. Letf(n)be the number of n-celled HC-polyominoes. Then /summationdisplay n≥1f(n)xn=x(1−x)3 1−5x+7x2−4x3 =x+2x2+6x3+1 9x4+6 1x5+ 196x6+ 629x7+ 2017x8 + 6466x9+ 20727x10+ 66441x11+ 212980x12+···. (4.9.9) We now will give a preview of the material in chapter 5, by working out anasymptotic formula forf(n). To do this we take the generating function in (4.9.9) and expand it in partial fractions. This gives x(1−x)3 1−5x+7x2−4x3=−5 16+x 4+c1 1−ξ1x+c2 1−ξ2x+c3 1−ξ3x, in whichξ1=3.20556943...andξ2,3=0.897215 ±.665457i. Thus the number of HC-polyominoes is, for n≥2, f(n)=c1ξn 1+c2ξn 2+c3ξn 3 =c1ξn 1+O(|ξ2|n) =0.1809155018 ...(3.2055694304 ...)n+O(1.1171n). 154 4 Applications of generating functions The ﬁrst term of this formula gives, for example, f(12) = 212979 .61, compared to 212980, the exact value, as shown in (4.9.9). 4.10 Exact covering sequences Every positive integer nis either 1 mod 2 or 0 mod 4 or 2 mod 4, as a moment’s reﬂection will conﬁrm. So the three pairs ( a1,b1)=( 1,2), (a2,b2)=( 0,4) and (a3,b3)=( 2,4) of residues and moduli exactly cover the positive integers. Anexact covering sequence (ECS) is a set ( ai,bi)(i=1,...,k )o f ordered pairs of nonnegative integers with the property that for every non- negative integer nthere is one and only one isuch that 1 ≤i≤kand n≡aimodbi. In this section we will give the basic theory of such sequences and deal, in two or three diﬀerent ways, with the question of how we can tell if a given sequence of pairs is or is not an exact covering sequence. Here is what generating functions have to contribute to this subject. Suppose (ai,bi)(i=1,...,k ) is an exact covering sequence. Then in the series k/summationdisplay i=1/summationdisplay t≥0xai+tbi every nonnegative integer noccurs exactly once as an exponent of x, so it must be true that the series shown is equal to 1 /(1−x). If we perform the summation over t, we ﬁnd that k/summationdisplay i=1xai 1−xbi=1 1−x. (4.10.1) For example x 1−x2+1 1−x4+x2 1−x4=1 1−x. Theorem 4.10.1. For a set of pairs (ai,bi)(i=1,...,k )to be an exact covering sequence it is necessary and suﬃcient that the relation (4.10.1) hold. One conclusion that we can draw immediately is that in an ECS we must have/summationtext i1/bi= 1. To see that, just multiply (4.10.1) by 1 −xand letx→1. But we can learn much more by comparing the partial fraction expansions of the left and right sides of (4.10.1). For the left side, we have k/summationdisplay i=1xai 1−xbi=/summationdisplay ω:ωN=1A(ω) ω−x 4.10 Exact covering sequences 155 where A(ω) = lim x→ω(ω−x)k/summationdisplay i=1xai 1−xbi =/summationdisplay j:ωbj=1ωaj+1 bj, andN=l.c.m.{bj}. If we compare with the right side of (4.10.1) we see that theA(ω)’s must all vanish except that A(1) = 1. Hence we must have /summationdisplay j:ωbj=1ωaj bj=/braceleftbigg 1,ifω=1 ; 0,otherwise. NowA(ω) surely vanishes unless ωis a root of unity. So let ω=e2πir/s, wheres≥1 and (r,s) = 1, be a primitive sth root of unity. Then our conditions take the form /summationdisplay j:s\bjωaj bj=/braceleftBig1i fs=1 ; 0 otherwise.(4.10.2) Hence, associated with any sequence of pairs ( ai,bi)|k i=1we can deﬁne polynomials ψs(z)=/summationdisplay j:s\bjzaj bj(s=1,2,3,...). (4.10.3) In terms of these polynomials we can restate our conditions (4.10.2) as follows: necessary and suﬃcient for the given set of pairs to constitute an exact covering sequence is that for each s>1 the polynomial ψsshould vanish at the primitive sth rots of unity, and ψ1(1) = 1. However, any polynomial that vanishes at all of the primitive sth roots of unity must be divisible by the cyclotomic polynomial (see section 2.6, example 2) Φs(z)=/productdisplay r:(r,s)=1;0 <r<s(z−e2πir/s)(s=1,2,3,...) since Φ s(z) has only those roots. The ﬁrst few cyclotomic polynomials are 1−z,1+z,1+z+z2,1+z2,1+z+z2+z3+z4,1−z+z2,.... If we put all of this together, we obtain the following result. 156 4 Applications of generating functions Theorem 4.10.2. A set of pairs of integers (a1,b1),..., (ak,bk), in which thea’s are nonnegative and the b’s are positive, is an exact covering se- quence if and only if/summationtext j1/bj=1and for each s>1, the polynomial ψs(z), of (4.10.3), is divisible by the cyclotomic polynomial Φs(z). For an example, take the pairs (0,4),(2,4),(1,6),(3,6),(5,12),(11,12). Then/summationtext j1/bj=1/4+1/4+1/6+1/6+1/12+1/12 = 1, and the divisibility conditions of the theorem look like this: Φ2(z)=( 1+z) divides1 4+z2/4+z/6+z3/6+z5/12 +z11/12 Φ3(z)=( 1+z+z2) divides z/6+z3/6+z5/12 +z11/12 Φ4(z)=( 1+z2) divides1 4+z2/4+z5/12 +z11/12 Φ6(z)=( 1 −z+z2) divides z/6+z3/6+z5/12 +z11/12 Φ12(z)=( 1 −z2+z4) divides z5/12 +z11/12. These are all readily checked, and so the given pairs are an ECS. Theorem 4.10.2 has a number of corollaries, some of which are left as exercises. One of them, however, is quite clear. If B= max {bj}, thenB cannot occur just once among the moduli {bj}. Indeed, if we take s=B in the theorem, we discover that ψB(z) must have enough monomials in it to allow it to be divisible by Φ B(z), so it must surely have at least two monomials. Exercises 157 Exercises 1. Given a coin whose probability of turning up ‘heads’ is p, letpnbe the probability that the ﬁrst occurrence of ‘heads’ is at the nth toss of the coin. Evaluatepnand the opsgf of the sequence {pn}. Use that opsgf to ﬁnd the mean of the number of trials till the ﬁrst ‘heads’ and the standard deviation of that number. 2. In the coupon collector’s problem we imagine that we would like to get a complete collection of photos of movie stars, where each time we buy abox of cereal we acquire one such photo, which may of course duplicate one that is already in our collection. Suppose there are ddiﬀerent photos in a complete collection. Let p nbe the probability that exactly ntrials are needed in order, for the ﬁrst time, to have a complete collection. (a) Show that pn=d!/braceleftbign−1 d−1/bracerightbig dn, where/braceleftbign k/bracerightbig is the Stirling number of the second kind (see section 1.6). (b) Letp(x)ops ←→{pn}. Show that p(x)=(d−1)!xd (d−x)(d−2x)···(d−(d−1)x). (c) Find, directly from the generating function p(x), the average num- ber of trials that are needed to get a complete collection of all d coupons. (d) Similarly, using p(x), ﬁnd the standard deviation of that number of trials. (e) In the case d= 10, how many boxes of cereal would you expect to have to buy in order to collect all 10 diﬀerent kinds of pictures? 3. (First return times on trees )B ya random walk on a graph we mean a walk among the vertices of the graph, which, having arrived at some vertexv, goes next to a vertex wthat is chosen uniformly from among the neighbors of vin the graph. IfTis a tree, and vis a vertex of T, letp(j;v;T) denote the probability that a random walk on Twhich starts at vertex v, returns to vfor the ﬁrst time after exactly jsteps. Now letT1,T2be trees, let vibe a vertex of Tifori=1,2, and let Tbe the tree that is formed from these two by adding edge ( v1,v2). Finally, let F1(x;v1),F2(x;v2),F(x;v1) be the opsgf’s of the sequences {p(j;v1;T1)}j≥0,{p(j;v2;T2)}j≥0, and {p(j;v1;T)}j≥0, respectively. 158 4 Applications of generating functions (a) Show that F(x;v1)=1 d1+1/braceleftbigg d1F1(x;v1)+x2 d2+1−d2F2(x;v2)/bracerightbigg , wherediis the degree of vertex viin the treeTi, fori=1,2. (b) LetµT(a) be the average number of steps in a random walk that starts at vertex a∈Tand stops when it returns to afor the ﬁrst time. Show, by diﬀerentiating the answer to part (a), that µT(v1)=1 d1+1/parenleftbig 2+d1µT1(v1)+d2µT2(v2)/parenrightbig . (c) Let the tree Tbe a path of n+ 1 vertices. Show that the mean return time of a walk that begins at vertex vis 2nifvis one of the two endpoints, and is nfor all other v(surprisingly?). (d) Again, if Tis a path of nvertices, and if Pn(x) denotes the gener- ating function F(x;v1) of part (a), where v1is an endpoint of T, then ﬁnd an explicit formula for Pn(t). 4. Find, in terms of N(x), the opsgf of the sequences {e≤m}(resp. {e≥m}) which count the objects that have at mostmproperties (resp. at least m properties). 5. What chessboard would you use to derive the number of permutations that have no ﬁxed points? Rederive the formula for this number using the chessboard method. 6. (Bonferroni’s inequalities ) In the sieve method, eq. (4.2.6) computes the number of objects that have no properties at all. Suppose the alternating series on the right were cut oﬀ after a certain value t=m, say. Show that the result would overestimate e0ifmwere even, and underestimate it for modd. To do this, show that the sequence αm=/summationdisplay r≥m(−1)r−mNr (m=0,1,2,...) has the opsgf e0+x/summationtext r≥0er+1(x+1 )r, whose coeﬃcients are obviously nonnegative. 7. (Bonferroni’s inequalities, cont. ) Not only is e0alternately under- and overestimated by the successive partial sums of its sieve formula, the same is true of every ek, the number of objects that have exactly kproperties. To show this generatingfunctionologically, deﬁne, for each k,t≥0, γ(k,t)=(−1)t+1  ek−/summationdisplay j≤t(−1)j/parenleftbiggk+j j/parenrightbigg Nk+j  . Exercises 159 Then the problem is to show that all γ(k,t)≥0. To do this, (a) let Γ(x,y) be the 2-variable opsgf of the γ’s. Then multiply the deﬁnition of the γ’s byxkyt, sum overk,t≥0, and show that Γ(x,y)=E(x+( 1+y))−E(x) (1 +y). (b) It now follows that the γ’s are nonnegative, and in fact that γ(k,t)=/summationdisplay r>k/parenleftbiggr k/parenrightbigg/parenleftbiggr−k−1 t/parenrightbigg er (k,t≥0). 8. Show that/summationdisplay r/parenleftbiggn/floorleftbigr 2/floorrightbig/parenrightbigg xr=( 1+x)(1 +x2)n. Then use Snake Oil to evaluate /summationdisplay k/parenleftbiggn k/parenrightbigg/parenleftbiggn−k/floorleftbigm−k 2/floorrightbig/parenrightbigg yk explicitly, when y=±2 (due to D. E. Knuth). Find the generating function of these sums, whatever the value of y. 9. LetGbe a graph of nvertices, and let positive integers x,λbe given. LetP(λ;x;G) denote the number of ways of assigning one of λgiven colors to each of the vertices of Gin such a way that exactly xedges ofGhave both endpoints of the same color. Formulate the question of determining Pas a sieve problem with a suitable set of objects and properties. Find a formula for P(λ;x;G), and observe that it is a polynomial in the two variables λandx. The chromatic poly- nomial ofGisP(λ;0 ;G). 10. (a) Letwbe a word of mletters over an alphabet of kletters. Suppose that no ﬁnal substring of wis also an initial string of w. Use the sieve method to count the words of nletters, over that alphabet ofkletters, that do not contain the substring w. (b) Use the Snake Oil method on the sum that you got for the answer in part (a). 11. Use the Snake Oil method to do all of the following: (a) Find an explicit formula, not involving sums, for the polynomial /summationdisplay k≥0/parenleftbiggk n−k/parenrightbigg tk. 160 4 Applications of generating functions (b) Invent a really nasty looking sum involving binomial coeﬃcients that isn’t any of the ones that we did in this chapter, and evaluate it in simple form. (c) Evaluate /summationdisplay k/parenleftbigg2n+1 2p+2k+1/parenrightbigg/parenleftbiggp+k k/parenrightbigg , and thereby obtain a ‘Moriarty identity.’ (d) Show that /summationdisplay m/parenleftbiggr m/parenrightbigg/parenleftbiggs t−m/parenrightbigg =/parenleftbiggr+s t/parenrightbigg . Then evaluate /summationdisplay k/parenleftbiggn k/parenrightbigg2 . (e) Show that (Graham and Riordan) /summationdisplay k/parenleftbigg2n+1 2k/parenrightbigg/parenleftbiggm+k 2n/parenrightbigg =/parenleftbigg2m+1 2n/parenrightbigg . (f) Show that for all n≥0 /summationdisplay k/parenleftbiggn k/parenrightbigg/parenleftbiggk j/parenrightbigg xk=/parenleftbiggn j/parenrightbigg xj(1 +x)n−j. (g) Show that for all n≥0 x/summationdisplay k/parenleftbiggn+k 2k/parenrightbigg/parenleftbiggx2−1 4/parenrightbiggn−k =/parenleftbiggx−1 2/parenrightbigg2n+1 +/parenleftbiggx+1 2/parenrightbigg2n+1 . (h) Show that for n≥1 /summationdisplay k≥1/parenleftbiggn+k−1 2k−1/parenrightbigg(x−1)2kxn−k k=(xn−1)2 n. 12. The Snake Oil Method works not only on sums that involve binomial coeﬃcients, but on all sorts of counting numbers, as this exercise shows. (a) Let {an}and{bn}be two sequences whose egf’s are, respectively, A(x),B(x). Suppose that the sequences are connected by bn=/summationdisplay k/bracketleftbiggn k/bracketrightbigg ak (n≥0), Exercises 161 where the/bracketleftbig/bracketrightbig ’s are the Stirling numbers of the ﬁrst kind. Show that their egf’s are connected by B(x)=A/parenleftbigg log1 (1−x)/parenrightbigg . (b) Let ˜bnbe the number of ordered partitions of [ n] (see (5.2.7)). Show that/summationdisplay k/bracketleftbiggn k/bracketrightbigg ˜bk=n!2n−1(n≥1). (c) Let {an}be the numbers of derangements (= ﬁxed point free permutations) of nletters, and let {bn}be deﬁned as in part (a). Show that {bn}egf ←→1−x 1 + log (1 −x). (d) Repeat parts (a)-(c) on the Stirling numbers of the second kind, and discover a few identities of your own that involve them. (e) Generalize parts (a)-(d) to exponential families. 13. Prove that /summationdisplay k(−1)n−k/parenleftbigg2n k/parenrightbigg2 =/parenleftbigg2n n/parenrightbigg by exhibiting this sum as a special case of a sum with two free parameters, and by using Snake Oil on the latter. 14. To do a sum that is of the form S(n)=/summationdisplay kf(k)g(n−k), the natural method is to recognize S(n)a s[xn]{F(x)G(x)}, whereFand Gare the opsgf’s of {fn}and{gn}. Use this method to evaluate S(n)=/summationdisplay k1 k+1/parenleftbigg2k k/parenrightbigg1 n−k+1/parenleftbigg2n−2k n−k/parenrightbigg . 15. (a) Prove the following generalization of (4.2.19), and show that it is indeed a generalization. For all m,n,q ≥0, we have /summationdisplay r/parenleftbiggm r/parenrightbigg/parenleftbiggn−r n−r−q/parenrightbigg (t−1)r=/summationdisplay r/parenleftbiggm r/parenrightbigg/parenleftbiggn−m n−r−q/parenrightbigg tr. 162 4 Applications of generating functions (b) The Jacobi polynomials may be deﬁned, for n≥0, by P(a,b) n(x)=/summationdisplay k/parenleftbiggn+a k/parenrightbigg/parenleftbiggn+b n−k/parenrightbigg/parenleftbiggx−1 2/parenrightbiggn−k/parenleftbiggx+1 2/parenrightbiggk . Use the result of part (a) to show also that P(a,b) n(x)=/summationdisplay j/parenleftbiggn+a+b+j j/parenrightbigg/parenleftbiggn+a j+a/parenrightbigg/parenleftbiggx−1 2/parenrightbiggj . (c) Use the result of part (b) and a dash of Snake Oil to show that P(a,b) n(x)=2−n(x−1)−a/bracketleftbig tn+a+b/bracketrightbig/braceleftbigg(1 +x−2t)n+a (1−t)n+1/bracerightbigg . 16. Prove the following generalization of the sum in example 2: if two sequences {fn}and{ck}are connected by the equations fn=/summationdisplay k/parenleftbiggn+k m+2k/parenrightbigg ck (n≥0), wherem≥0 is ﬁxed, then their opsgf’s are connected by F(x)=xm (1−x)m+1C/parenleftbiggx (1−x)2/parenrightbigg . Say exactly what was special about the sequence {ck}that was used in example 2 that made the result turn out to be so neat in that case. 17. The purpose of this problem is to show the similarity of the method of [WZ] to some well known continuous phenomena. (a) LetF(x,y),G(x,y) be diﬀerentiable functions that satisfy the conditions that Fx=Gyand lim y→±∞G(x,y) = 0, for all xin a certain interval a<x<b . Show that we have the ‘identity’ /integraldisplay∞ −∞F(x,y)dy=const. (a<x<b ). (b) Show, using the result of part (a), that if f(z) is analytic in the strip−∞<a< /Rfracturz<b< ∞, and iff→0 on all vertical lines in that strip, then the conclusion of part (a) holds, where F(x,y)i s the real part of f(z). (c) Show that the result stated in part (b) is true without using the result of part (a), but using instead the Cauchy integral theorem applied to a suitable rectangle. Exercises 163 (d) Apply these results to f(z)=ez2and thereby discover the ‘iden- tity’/integraldisplay∞ −∞e−y2cos (2xy)dy=ce−x2(xreal), which states that the function e−y2is its own Fourier transform. Findc. 18. This problem gives a neat proof of Cayley’s formula for the number of trees ofnvertices, by showing more, namely that there is a pretty formula for the number of such trees even if the degrees of all vertices are speciﬁed. (a) Letd1,...,d nbe positive integers whose sum is 2 n−2. Show that the number of vertex-labeled trees Tofnvertices, in which for all i=1,...,n it is true that diis the degree of vertex iofT, is exactly fn(d1,...,d n)=(n−2)! (d1−1)!(d2−1)!···(dn−1)!. (Do this by induction on n. Show that one of the di’s, at least, must be =1, and go from there.) (b) Find the generating function Fn(x1,...,x n)=/summationdisplay d1+···+dn=2n−2 d1,...,d n≥1fn(d1,...,d n)xd1 1···xdnn in a pleasant, explicit form, involving no summation signs. (c) Letx1=x2=···=xn= 1 in your answer to part (b), and thereby prove Cayley’s result that there are exactly nn−2labeled trees of n vertices. (d) Use the sieve method to show that if ekis the number of vertex labeled trees on nvertices of which kare endpoints (vertices of degree 1), then /summationdisplay kekxk=/summationdisplay r/parenleftbiggn r/parenrightbigg rn−2(x−1)n−r. (e) Show that the average number of endpoints that trees of nvertices have is n/parenleftbigg 1−1 n/parenrightbiggn−2 ∼n e(n→∞ ), i.e.,the probability that a random vertex of a tree is an endpoint is about 1/e. 164 4 Applications of generating functions 19. (a) IfN(⊆S) is the number of objects whose set of properties is con- tained inS, then for all sets T, the number of objects whose set of properties is preciselyTis N(=T)=/summationdisplay S⊆T(−1)|S|−|T|N(⊆S). (b) LetSbe a ﬁxed set of positive integers, and let hn(S) be the number of hands of weight n, in a certain labeled exponential family, whose card sizes all belong to S. Find the egf of {hn(S)}. (c) Multiply by ( −1)|S|−|T|and sum over S⊆Tto ﬁnd the egf of {ψn(T)}n≥0, the number of hands whose set of distinct card sizes is exactlyT, in the form (the d’s are the deck sizes) /summationdisplay n≥0ψn(T) n!xn=/productdisplay t∈T/parenleftbigg edtxt t!−1/parenrightbigg . (d) Letρ(n,k) be the number of hands of weight nthat have exactly k diﬀerent sizes of cards (however many cards they might have!). Sum the result of (c) over all |T|=k, etc., to ﬁnd that /summationdisplay n,k≥0ρ(n,k) n!xnyk=/productdisplay t≥1/braceleftbigg 1+y/parenleftbig edtxt/t!−1/parenrightbig/bracerightbigg . (e) Letcnbe the average number of diﬀerent sizes of cycles that occur in permutations of nletters. Show that the opsgf of {cn}is 1 1−x/summationdisplay t≥1/parenleftbigg 1−e−xt/t/parenrightbigg , and ﬁnd an explicit formula for cn. 20. Begin with the set {1,2,...,n }. Toss a coin ntimes, once for each member of the set. Keep the elements that scored ‘Heads’ and discard the elements that got ‘Tails’. You now have a certain subset Sof the original set. Call this whole process a ‘step’. Now take a step from S. That is, toss a coin for each element of S, and keep those that get ‘Heads’, getting a sub-subset S/prime, etc. The game halts when the empty set is reached. Let f(n,k,r ) be the probability that after ksteps, exactly r objects remain. (a) Find a recurrence relation for f, ﬁnd the generating function for f, and ﬁndfitself. (b) What is the average number of steps in a complete game? Exercises 165 (c) What is the standard deviation of the number of steps in the game? 21. As in section 4.7, let f(n,k,t ) be the number of HC-polyominoes that havencells, inklayers, the highest layer consisting of exactly tcells. Show that the ‘grand’ three-variable generating function is /summationdisplay n,k,tf(n,k,t )xnykzt=xyz(1−x)2((1−xz)(1−x)2+x2y(z−1)) (1−xz)2((1−x)4−xy(1−x−x2+x3+x2y)). Note that you do not have to start from scratch here, but instead you can use the results of section 4.7. 22. Let (a1,b1),..., (ak,bk) be an exact covering sequence. Show that /summationdisplay j:ajis even1 bj=1/2=/summationdisplay j:ajis odd1 bj. Generalize this result to other residue classes for the aj. 23. What is the probability that a random permutation has equal numbers ofr-cycles and s-cycles? Express your answer in terms of Bessel functions (see chapter 2). Make a table of your answer, as a function of rands, for 1<r<s ≤6. 24. Find a three term recurrence relation, whose coeﬃcients are polyno- mials inn, that is satisﬁed by the quantity shown in (4.8.1), which is the number of convex polyominoes of perimeter 2 n+8 . 25. For (ai,bi)|k i=1to be an exact covering sequence it is necessary and suﬃcient that for all nsuch that 0 ≤n≤N,nis congruent to aimodbi for exactly one i, whereNis the least common multiple of b1,...,b k. 26. (a) Develop the generalization of the exponential formula that we were really using in section 4.7. Precisely, suppose that for each i= 1,2,3,...we are given a set Siof positive integers. Let h(n) be the number of hands of weight nthat can be formed from a given collec- tion of decks if our choices of cards are restricted by the condition that for each i=1,2,3,..., the number of cards of weight ithat are chosen for the hand must lie in the set Si. Then show that /summationdisplay n≥0h(n)tn n!=∞/productdisplay i=1expSi/parenleftbigditi i!/parenrightbig , where expS(x) is the subseries of the exponential series whose indices lie in the set Sand, as in chapter 3, diis the number of cards in the ith deck. 166 4 Applications of generating functions (b) Find the egf of {f(n)}, wheref(n) is the number of partitions of the set [n] in which the number of classes of size 2 is divisible by 2 and the number of classes of size 3 is divisible by 3, etc. 27. In order that ( ai,bi)|k i=1be an exact covering sequence of residues and moduli, it is necessary and suﬃcient that [Fr] k/summationdisplay i=1bn−1 iBn(ai bi)=Bn (n=0,1,2,...) where the {Bn}are the Bernoulli numbers (deﬁned by (2.5.8)), and the Bn(x) are the Bernoulli polynomials , deﬁned by text et−1=∞/summationdisplay n=0Bn(x)tn n!. 28. Find a formula for the number of square roots that a permutation has. What kind of a permutation has a unique square root? 5.1 The Lagrange Inversion Formula 167 Chapter 5 Analytic and asymptotic methods In the preceding chapters we have emphasized the formal aspects of the theory of generating functions, as opposed to the analytic theory. One of the attractions of the subject, however, is how easily one can shift gearsfrom thinking of generating functions as mere clotheslines for sequences to regarding them as analytic functions of complex variables. In the latter state of mind, one can deduce many properties of the sequences that are generated that would be inaccessible to purely formal approaches. Notable among these properties are the asymptotic growth rates of the sequences, which are probably the main focus of the analytic side of the theory. Hence in this chapter we will develop some of the analytic machinery that is invalu- able for such studies. For an introduction to asymptotics and deﬁnitions of all of the symbols of asymptotics, see, for example, chapter 4 of [Wi1]. 5.1 The Lagrange Inversion Formula The Lagrange Inversion Formula is a remarkable tool for solving certain kinds of functional equations, and at its best it can give explicit formulas where other approaches run into stone walls. The form of the functional equation that the LIF can help with is u=tφ(u). (5.1.1) Hereφis a given function of u, and we are thinking of the equation as determining uas a function of t. That is, we are ‘solving for uin terms of t.’ We found one example of such an equation in section 3.12. There we saw that if T(x) is the egf for the numbers of rooted labeled trees of each number of vertices, then T(x) satisﬁes the equation T=xe T, which is indeed of the form (5.1.1), with φ(u)=eu. The general problem is this: suppose we are given the power series expansion of the function φ=φ(u), convergent in some neighborhood of the origin (of the u-plane). How can we ﬁnd the power series expansion of the solution of (5.1.1), u=u(t), in some neighborhood of the origin (in the t-plane)? The answer is surprisingly explicit, and it even allows us to ﬁnd the expansion of some function of the solution u(t). Theorem 5.1.1. (The Lagrange Inversion Formula) Let f(u)andφ(u)be formal power series in u, withφ(0) = 1 . Then there is a unique formal power series u=u(t)that satisﬁes (5.1.1). Further, the value f(u(t))off at that root u=u(t), when expanded in a power series in taboutt=0, satisﬁes [tn]{f(u(t))}=1 n/bracketleftbig un−1/bracketrightbig {f/prime(u)φ(u)n}. (5.1.2) 168 5 Analytic and asymptotic methods Proof. First we note that it suﬃces to prove the theorem in the case wherefandφare polynomials. Indeed, if nis ﬁxed, and if fandφare full formal power series, then suppose that we truncate both of those series by discarding all terms that involve powers ukfork>n . If the result is true for these polynomials then it remains true for the original untruncated series, since the higher order terms that were discarded do not aﬀect (5.1.2), for the ﬁxed n, at all. Therefore we now suppose that fandφare polynomials. We will ﬁrst make a formal computation, and then discuss the range of validity of the results. We have /bracketleftbig un−1/bracketrightbig {f/prime(u)φ(u)n}=/bracketleftbig un−1/bracketrightbig {f/prime(u)(u/t)n} =/bracketleftbig u−1/bracketrightbig {f/prime(u)/tn} =1 2πi/integraldisplayf/prime(u) t(u)ndu =1 2πi/integraldisplay t−nf/prime(u(t))u/prime(t)dt =[tn]{t(d/dt)f(u(t))} =n[tn]f(u(t)).(5.1.3) In the above, the ﬁrst equality comes from (5.1.1), the second is trivial, and the third is the residue theorem of complex integration, in which theintegrand is a function of uand the contour is, say, a small circle enclosing 0. The fourth equality needs some discussion. Consider the behavior of the function g(u)=u/φ(u) near the origin. Since φ(0) = 1, the function φ remains nonzero in a neighborhood of 0. Hence gis analytic there, and it has a power series development of the form u+cu 2+···. It follows that g is a 1-1 conformal map near 0. Hence it has a well deﬁned inverse mapping that is itself analytic near 0. Thus (5.1.1) has a unique solution u=u(t) in some neighborhood /Rfracturoft= 0, anduis an analytic function of tthere. If the contour of integration in the integral that appears after the fourthequals sign above is a circle around t= 0 that lies in /Rfractur, then the sign of equality is valid simply as a change of variable from utotin the integral. The remaining equalities are trivialities, and the proof is complete. Example 1. In section 3.12 we embarked on proving theorem 3.12.1, to the eﬀect that there are exactly nn−2labeled trees of nvertices. We found that if D(x)egf ←→{tn}, wheretnis the number of rooted labeled trees of nvertices, thenD(x) satisﬁes the functional equation D(x)=xeD(x). (5.1.4) 5.1 The Lagrange Inversion Formula 169 Now, with the Lagrange Inversion Formula, we can actually solve (5.1.4), because it is the case φ(u)=euof (5.1.1). If we take f(u)=u, then according to (5.1.2), [xn]D(x)=( 1/n)/bracketleftbig un−1/bracketrightbig {φ(u)n} =( 1/n)/bracketleftbig un−1/bracketrightbig {enu} =( 1/n)nn−1 (n−1)! =nn−1 n!. Hence, tn n!=nn−1 n!, andtn=nn−1. Buttnis the number of rooted trees of nvertices, and every labeled tree contributes nrooted labeled trees, so the number of labeled trees ofnvertices isnn−2, which completes the proof of theorem 3.12.1. Example 2. At the end of chapter 4 we discussed inverse pairs of summation for- mulas and gave some examples beyond the M¨ obius inversion formula. Here we’ll derive a much fancier example with the help of the LIF. We propose to show that if two sequences {an}and{bn}are related by bn=/summationdisplay k/parenleftbiggk n−k/parenrightbigg ak, (5.1.5) then we have the inversion nan=/summationdisplay k/parenleftbigg2n−k−1 n−k/parenrightbigg (−1)n−kkbk. (5.1.6) Indeed, if (5.1.5) holds, then, as we did so often in section 4.3, multiply byxn, sum onn, and interchange the kandnsummations, to get B(x)=/summationdisplay kakxk/summationdisplay n/parenleftbiggk n−k/parenrightbigg xn−k =/summationdisplay kakxk(1 +x)k =A(x(1 +x)),(5.1.7) whereA,Bare the opsgf’s of the sequences. Now we can solve for Ain terms ofBby settingy=x+x2. Then if we read (5.1.7) backwards, we ﬁnd that A(y)=B(x(y)), (5.1.8) 170 5 Analytic and asymptotic methods wherex(y) is the solution of the quadratic equation y=x+x2that vanishes wheny=0 . Now we could , of course, solve the quadratic explicitly. If we were to do that (which we won’t) we would ﬁnd from (5.1.8) that A(y)=B/parenleftbigg√1+4y−1 2/parenrightbigg , and then we would want to ﬁnd a nice formula for the coeﬃcient of ynon the right hand side for every n. That, in turn, would require a nice formula for [yn]/braceleftbigg√1+4y−1 2/bracerightbiggk (5.1.9) for everynandk. Instead of trying to deal with (5.1.9) by explicitly raising the quantity in braces to the kth power and working with the square root, it is a lot more elegant to let the LIF do the hard work. We begin by rephrasing the question (5.1.9) implicitly , rather than explicitly. What we want is [yn]{x(y)k}, wherex=x(y) is the solution of y=x+x2that is 0 at y= 0. In terms of the LIF, we write the equation as x=y 1+x, which is of the form (5.1.1) with φ(u)=1/(1 +u). Further, since we want the coeﬃcients of the kth power of x(y), the function f(u) in the LIF is nowf(u)=uk. The conclusion (5.1.2) of the LIF tells us that [yn]{x(y)}k=( 1/n)[xn−1]/braceleftbiggkxk−1 (1 +x)n/bracerightbigg =(k/n)[xn−k]1 (1 +x)n =(k/n)(−1)n−k/parenleftbigg2n−k−1 n−k/parenrightbigg , and we are all ﬁnished with the proof of (5.1.6). It should be particularly noted that the LIF is as adept at computing the coeﬃcients of the kth power of the unknown function as those of the unknown function itself. The function fin the statement of the LIF simply speciﬁes the function of the unknown function whose coeﬃcients we would like to know. The LIF then hands us those coeﬃcients. 5.2 Analyticity and asymptotics (I): Poles 171 5.2 Analyticity and asymptotics (I): Poles Suppose we have found the generating function f(z) for a certain se- quence of combinatorial numbers that interests us. Next we might want to ﬁnd the asymptotic behavior of the sequence, i.e., to ﬁnd a simple function ofnthat aﬀords a good approximation to the values of our sequence when nis large. The ﬁrst law of doing asymptotics is: look for the singularity or sin- gularities of f(z)that are nearest to the origin . The reason is that f(z)i s analytic precisely in the largest circle centered at the origin that contains no singularities, and we ﬁnd the radius of that circle by ﬁnding the singu- larities nearest to the origin. Once we have that radius, we have also theradius of convergence of the power series f(z). Once we have the radius of convergence we know something about the sizes of the coeﬃcients when n is large, as in theorem 2.4.3. By various reﬁnements of this process we can discover more detailed information. Therefore, in this and the following sections we will study the inﬂuence of the singularities of analytic functions on the asymptotic behavior of their coeﬃcients. These methods, taken together, provide a powerful technique for obtaining the asymptotics of combinatorial sequences, and provide yet another justiﬁcation, if one were needed, of the generating function ap- proach. In this section we will concentrate on functions whose only singularities are poles. Letf(z) be analytic in some region of the complex plane that includes the origin, with the exception of a ﬁnite number of singularities. If Ris the smallest of the moduli of these singularities, then fis analytic in the disk|z|<R, so this will be precisely the disk in which its power series expansion about the origin converges. Conversely, if the power series expansion of a certain generating func- tionfconverges in the disk |z|<R but in no larger disk centered at the origin, then there are one or more singularities of the function fon the circumference |z|=R. A number of methods for dealing with questions of asymptotic growth of coeﬃcient sequences rely on the following strategy: ﬁnd a simple function gthat has the same singularities that fhas on the circle |z|=R. Then f−gis analytic in some larger disk, of radius R /prime>R,s a y . Then, according to theorem 2.4.3 (q.v.), the power series coeﬃcients off−gwill be<(1 R/prime+/epsilon1)nfor largen, and therefore they will be much smaller than the coeﬃcients of fitself. The latter, according to the same theorem 2.4.3, will inﬁnitely often be as large as (1 R−/epsilon1)n. Therefore we will be able to ﬁnd the most important aspects of the growth of the coeﬃcients of fby looking at the growth of the coeﬃcients ofg. The strategy that wins, therefore, is that of ﬁnding a simple function 172 5 Analytic and asymptotic methods that mimics the singularities of the function that one is interested in, and then of using the growth of the coeﬃcients of the simple function for the estimate. These considerations come through most clearly in the case of a mero- morphic functionf(z), i.e., one that is analytic in the region with the exception of a ﬁnite number of poles , and so we will study such functions ﬁrst. The idea is that near a pole z0, a meromorphic function is well ap- proximated by the principal part of its Laurent expansion, i.e., by the ﬁnite number of terms of the series that contain ( z−z0) raised to negative powers. Example 1. The function f(z)=ez/(z−1) is meromorphic in the whole ﬁnite plane. Its only singularity is at z0= 1, and the principal part at that singularity is e/(z−1). Hence the function f(z)−e/(z−1) is analytic in the whole plane. Thus if {cn}are the power series coeﬃcients of fabout the origin, and {dn}are the same for the function e/(z−1), the diﬀerence cn−dnis small when nis large. In fact, theorem 2.4.3 guarantees that for every/epsilon1>0 we have |cn−dn|</epsilon1nfor all large enough n. Therefore the ‘unknown’ coeﬃcients of f(z) are very well approximated by those of the simple function e/(z−1). It is easy to work out this particular example completely to see just how the machine works. The expansion of ez/(z−1) about the origin is (see Rule 5 of section 2.2) ez (z−1)=−/summationdisplay n≥0{1+1+1/2! +···+1/n!}zn. On the other hand, the expansion of e/(z−1) is e (z−1)=−e−ez−ez2−ez3−···. Hence the act of replacing the function by its principal part yields in one step the approximation of the true coeﬃcient of zn, which is the nth partial sum of the power series for −e,b y−eitself, which is a smashingly good approximation indeed. The reason for the great success in this case is that merely subtracting oﬀ one principal part from the function expands its disk of analyticity from radius =1 to radius = ∞. Here’s another way to look at this example, without mentioning any of the heavy machinery. Consider the innocent fact that f(z)=ez z−1=e z−1+ez−e z−1. In the ﬁrst term, the power series coeﬃcients are all equal to −e. The second term has no singularities at all in the ﬁnite plane, i.e., it is an entire 5.2 Analyticity and asymptotics (I): Poles 173 function ofz. By theorem 2.4.3, its coeﬃcients are O(/epsilon1n) for every positive /epsilon1. Therefore the coeﬃcients of f(z) are =−e+O(/epsilon1n)(n→∞ ) for every/epsilon1>0. In less favorable cases one may have to subtract oﬀ several principal parts in order to increase the size of the disk of analyticity at all (i.e., if there are several poles on the circumference), and even then it may increase only a little bit if there are other singularities on a slightly larger circle. Iffis meromorphic in /Rfractur, letz0be a pole of fof orderr,1≤r<∞. Then in some punctured disk centered at z0,fhas an expansion f(z)=r/summationdisplay j=1a−j (z−z0)j+∞/summationdisplay j=0aj(z−z0)j. (5.2.1) The ﬁrst one of the two sums above, the one containing the negative powers of (z−z0), is called the principal part of the expansion of faround the singu- larityz0, and we will denote it by PP(f;z0). The function f−PP(f;z0)i s analytic at z0. That is to say, we can remove the singularity by subtracting oﬀ the principal part. If, besidesz0, there are other poles of fon the same circle |z|=R= |z0|, then letz1,...,z sbe all such poles. The function h(z)=f(z)−PP(f;z0)−PP(f;z1)−···−PP(f;zs)( 5.2.2) is regular (analytic) at every one of the points {zj}s 0. Butfhad no other singularities on that circle, so his analytic in a circle centered at the origin that has radius R/prime, whereR/prime>R. That means, by theorem 2.4.3 again, that the power series coeﬃcients ofh, about the origin, cannot grow faster than /parenleftbigg1 R/prime+/epsilon1/parenrightbiggn for all large n. Thus, iffops ←→{an}, and if g(z)=PP(f;z0)+PP(f;z1)+···+PP(f;zs)ops ←→{bn}, (5.2.3) then an=bn+O/parenleftbigg/parenleftbigg1 R/prime+/epsilon1/parenrightbiggn/parenrightbigg (n→∞ ). We may then be well on our way towards ﬁnding the asymptotic behavior of the coeﬃcients of f. 174 5 Analytic and asymptotic methods Indeed, let us now study the power series coeﬃcients, about the origin, of the sum of the principal parts that are shown in (5.2.3). We have, if z0 is a pole of multiplicity r, PP(f;z0)=r/summationdisplay j=1a−j (z−z0)j =r/summationdisplay j=1(−1)ja−j zj 0(1−(z/z0))j =r/summationdisplay j=1(−1)ja−j zj 0/summationdisplay n≥0/parenleftbiggn+j−1 n/parenrightbigg (z/z0)n =/summationdisplay n≥0zn/braceleftbiggr/summationdisplay j=1(−1)ja−j zn+j 0/parenleftbiggn+j−1 j−1/parenrightbigg/bracerightbigg .(5.2.4) We see, therefore, that a pole of order ratz0, of a function f, con- tributes r/summationdisplay j=1(−1)ja−j zn+j 0/parenleftbiggn+j−1 j−1/parenrightbigg (5.2.5) to the coeﬃcient of zninf. The basic theorem, which asserts that we can well approximate the coeﬃcients of a meromorphic function by the coeﬃcients of the principal parts at its poles of smallest modulus, can be stated as follows: Theorem 5.2.1. Letfbe analytic in a region /Rfracturcontaining the origin, except for ﬁnitely many poles. Let R> 0be the modulus of the pole(s) of smallest modulus, and let z0,...,z sbe all of the poles of f(z)whose modulus is R. Further, let R/prime>R be the modulus of the pole(s) of next- smallest modulus of f, and let/epsilon1>0be given. Then [zn]f(z)=[zn]/braceleftbiggs/summationdisplay j=0PP(f;zj)/bracerightbigg +O/parenleftbigg/parenleftbigg1 R/prime+/epsilon1/parenrightbiggn/parenrightbigg . (5.2.6) Proof. By theorem 2.4.3, this theorem will be proved as soon as we estab- lish that if we subtract from f(z) the sum of all of its principal parts from singularities on the circle |z|=R, then the resulting function is analytic in the larger disk |z|<R/prime. Consider the moment when we subtract PP(f;z0) fromf(z). Certainly the resulting function, g, say, is analytic at z0. Next, however, we subtract PP(f;z1) fromginstead of subtracting PP(g;z1) fromg. We claim that this doesn’t matter, i.e., that PP(f−PP(f;z0);z1)=PP(f;z1). 5.2 Analyticity and asymptotics (I): Poles 175 To see this, observe that PP(f−PP(f;z0);z1)=PP(f;z1)−PP(PP(f;z0);z1). But the second term on the right vanishes because PP(f;z0) is analytic at z1. By induction on s, the result follows. Example 1. Ordered Bell numbers We now investigate the asymptotic behavior of the ‘ordered Bell num- bers.’ These are deﬁned as follows: a set of nelements has/braceleftbign k/bracerightbig partitions intokclasses. If we now regard the order of the classes as important, but not the order of the elements within the classes, then we see that [ n] has k!/braceleftbign k/bracerightbig ordered partitions into kclasses . The ordered Bell number ˜b(n)i s the total number of ordered partitions of [ n], i.e., it is/summationtext kk!/braceleftbign k/bracerightbig . Our question concerns the growth of {˜b(n)}whennis large. To ﬁnd a nice formula for these numbers, multiply both sides of the identity (4.2.16) bye−yand integrate from 0 to ∞. This gives the neat result that ˜b(n)=/summationdisplay r≥0rn 2r+1. (5.2.6) Then the exponential generating function of the ordered Bell numbers is* f(z)=/summationdisplay n≥0˜b(n) n!zn=1 2−ez. (5.2.7) We’re in luck! The generating function f(z) has only simple poles, namely at the points log 2 ±2kπifor all integer k. The principal part at the polez0= log 2 is ( −1/2)/(z−log 2). That principal part all by itself contributes1 2(log 2)n+1 to the coeﬃcient of zn. There are no other singularities of f(z) on the circle of radius log 2 centered at the origin. Hence h(z)=f(z)−(−1/2) (z−log 2) is analytic in the larger circle that extends from the origin to log 2 + 2 πi. The radius of that circle is ρ=/radicalbig (log 2)2+4π2=6.321.... * Be sure to work this out for yourself. 176 5 Analytic and asymptotic methods Hence the coeﬃcients of h(z) areO((.16)n). Altogether, we have shown that the ordered Bell numbers ˜b(n)are of the form ˜b(n)=1 2(log 2)n+1n!+O((.16)nn!), (5.2.8) which is not bad for so little eﬀort invested. More terms of the asymptotic expansion can be produced as desired from the principal parts of f(z)a t its remaining poles, taken in nondescending order of their absolute values. The reader should look into the contribution of the next two poles together, which are complex conjugates of each other. Below we show a table of some values of n,˜b(n), andn!/(2(log 2)n+1). n 12 3 5 1 0 ˜b(n) 1 3 13 541 102247563 n!/(2(log 2)n+1)1.04 3.002 12.997 541.002 102247563 The agreement is astonishingly close. Basically all we have done is to use the Taylor coeﬃcients of the series for 1 /(2(log 2 −z)) as approximations to the coeﬃcients of the series for 1 /(2−ez). Yet we are rewarded with a superb approximation. Example 2. Permutations with no small cycles Fix a positive integer q. Letf(n,q) be the number of permutations ofnletters whose cycles all have lengths >q. We want the asymptotic behavior of f(n,q). By exercise 11 of chapter 3, the egf of {f(n,q)}∞ 0is fq(z) = exp/summationdisplay n>qzn n = exp  log1 1−z−/summationdisplay 1≤n≤qzn n   =1 1−ze−{z+···+zq/q}.(5.2.9) The only singularity of fq(z) in the ﬁnite plane is a pole of order 1 at z= 1 with principal part e−Hq/(1−z), where Hq=1+1 2+1 3+···+1 q is theqth harmonic number. 5.3 Analyticity and asymptotics (II): Algebraic singularities 177 This is the kind of situation where we get very accurate asymptotic estimates, because the diﬀerence between the function and its principal part atz=1i s h(z)=fq(z)−e−Hq 1−z =e−{z+···+zq/q}−e−Hq 1−z, and is analytic in the whole plane, i.e., is an entire function. Again, by theorem 2.4.3, the nth coeﬃcient of h(z)i sO(/epsilon1n)a sn→∞ for every /epsilon1>0. Therefore, f(n,q) n!=e−Hq+O(/epsilon1n)(n→∞ ). (5.2.10) The strikingly small error term in this estimate suggests that for each ﬁxedqthe probability that an n-permutation has no cycles of length ≤q should be very nearly independent of n. Consider the case q= 1 to get some of the ﬂavor of what is going on here. Then f(n,1)/n! is the probability that ann-permutation has no ﬁxed point. But we saw in (4.2.10) that f(n,1)/n!=e−1 |n=1−1+1/2−··· +(−1)n/n! =e−1+O(1/n!). Indeed, the probability is very nearly independent of n, and the error in- volved in using the principal part is O(/epsilon1n) for every positive /epsilon1. The two examples above have shown the method at work in situations where it was atypically accurate. More commonly one ﬁnds not just one pole of order 1 in the entire plane, but many poles of various multiplicities.The method remains the same in such cases, but a lot more work may be necessary in order to get estimates of reasonable accuracy. An example that shows this kind of phenomenon was worked out in section 3.15, in connection with the money-changing problem. In fact, the proof of Schur’s theorem (Theorem 3.15.2) was an exercise in the use of principal parts at the poles of a meromorphic function. The importance of the single dominant singularity was, in that case, much less, though it was enough to get the theorem proved! 5.3 Analyticity and asymptotics (II): Algebraic singularities Again, letf(z) be analytic in some region that contains the origin, but now suppose that the singularity z 0offthat is nearest to 0 is not a pole, but is an algebraic singularity ( branch point ). What that means is that f(z)=(z0−z)αg(z), wheregis analytic at z0andαis not an integer, but is a real number. 178 5 Analytic and asymptotic methods A case in point was given in (3.9.1), where we found that the egf for the numbers of graphs of nvertices whose vertex-degrees are all equal to 2 is f(z)=e−z/2−z2/4 √1−z. (5.3.1) In this section we will derive the theorem (‘Darboux’s lemma’) that allows us to deduce the asymptotics of sequences with this kind of a gener- ating function. What it all boils down to is that one should do exactly the same thing in this case as in the case of meromorphic functions, and the right answer will fall out. The proof that this is indeed valid, however, is more demanding in the present case. We follow the proof in [KnW]. By considering f(zz0) instead of f(z), if necessary, we see that we can assume without loss of generality that z0= 1. Hence we are dealing with a functionfthat is analytic in the unit disk, and which has a branch point atz= 1. We will also assume, until further notice, that z0= 1 is the only singularity that fhas in some disk |z|<1+η, whereη>0. After the lessons of the previous section on meromorphic functions, here’s how we might proceed in this case. First we have f(z)=( 1 −z)αg(z), wheregis analytic at z= 1. That being the case, we can expand gin a power series g(z)=/summationdisplay k≥0gk(1−z)k that converges in a neighborhood of z= 1. Hence fitself has an expansion f(z)=/summationdisplay k≥0gk(1−z)k+α. (5.3.2) By analogy with the procedure for meromorphic functions, we might expect that each successive term in the above series expansion generates the next term of the asymptotic expansion of the coeﬃcients of f. That is in fact true. The dominant behavior of the coeﬃcient of zninf(z) comes from the ﬁrst term in (5.3.2). That is, the simple function g0(1−z)αhas, for its coeﬃcient of zn, the main contribution to that coeﬃcient of f, etc. We will now prove all of these things. Lemma 5.3.1. Let{an},{bn}be two sequences that satisfy (a) an= O(n−γ)and (b)bn=O(θn)(0<θ< 1). Then /summationdisplay kakbn−k=O(n−γ). Proof. We have ﬁrst (the C’s are not all the same constant)/vextendsingle/vextendsingle/vextendsingle/vextendsingle/summationdisplay 0≤k≤n/2akbn−k/vextendsingle/vextendsingle/vextendsingle/vextendsingle≤/braceleftbigg max 0≤k≤n/2|ak|/bracerightbigg/braceleftbigg/summationdisplay 0≤k≤n/2Cθn−k/bracerightbigg ≤max{C,Cn−γ}{Cθn/2} ≤C˜θn(0<˜θ<1). 5.3 Analyticity and asymptotics (II): Algebraic singularities 179 Further, /vextendsingle/vextendsingle/vextendsingle/vextendsingle/summationdisplay n/2<k≤nakbn−k/vextendsingle/vextendsingle/vextendsingle/vextendsingle≤/braceleftbigg max n/2<k≤n|ak|/bracerightbigg  /summationdisplay n/2<k≤nθn−k   ≤Cn−γ. Lemma 5.3.2. Ifβ/∈{0,1,2,...}, then [zn](1−z)β∼n−β−1 Γ(−β). (5.3.3) Proof. We have [zn](1−z)β=/parenleftbiggβ n/parenrightbigg (−1)n =/parenleftbiggn−β−1 n/parenrightbigg =Γ(n−β) Γ(−β)Γ(n+1 ), and the result follows from Stirling’s formula, which is Γ(n+1 )=n!∼/parenleftBign e/parenrightBign√ 2πn (n→∞ ). Lemma 5.3.3. Letu(z)=( 1 −z)γv(z), wherev(z)is analytic in some disk|z|<1+η,(η>0). Then [zn]u(z)=O(n−γ−1). (5.3.4) Proof. Apply lemma 5.3.1 with an=[zn](1−z)γandbn=[zn]v(z). Since vis analytic in a disk |z|<1+η, we havebn=O(θn). The result follows by lemma 5.3.2. Theorem 5.3.1. (Darboux) Let v(z)be analytic in some disk |z|<1+η, and suppose that in a neighborhood of z=1 it has the expansion v(z)=/summationtextvj(1−z)j. Letβ/∈{0,1,2,...}. Then [zn]/braceleftbig (1−z)βv(z)/bracerightbig =[zn]  m/summationdisplay j=0vj(1−z)β+j  +O(n−m−β−2) =m/summationdisplay j=0vj/parenleftbiggn−β−j−1 n/parenrightbigg +O(n−m−β−2).(5.3.5) 180 5 Analytic and asymptotic methods Proof. We have (1−z)βv(z)−m/summationdisplay j=0vj(1−z)β+j=/summationdisplay j>mvj(1−z)β+j =( 1−z)β+m+1˜v(z), where the regions of analyticity of ˜ vand ofvare the same. The result now follows from lemma 5.3.3. Example 1. 2-regular graphs For the exponential generating function f(z), in (5.3.1), of the number of 2-regular graphs of nvertices, we have f(z)=( 1 −z)βv(z) withβ=−1/2 andv(z) = exp {−z/2−z2/4}. The ﬁrst few terms of the expansion of v(z) aboutz= 1 are e−z/2−z2/4=e−3/4+e−3/4(1−z)+1 4e−3/4(1−z)2+··· Then according to (5.3.5) this expansion of v(z) aroundz= 1 ‘lifts’ to an asymptotic formula for the coeﬃcients of f(z), which are in this case γ(n)/n!, whereγ(n) is the number of 2-regular graphs of nvertices. If we use (5.3.5) with m= 2, we obtain γ(n) n!=e−3/4/parenleftbiggn−1/2 n/parenrightbigg +e−3/4/parenleftbiggn−3/2 n/parenrightbigg +1 4e−3/4/parenleftbiggn−5/2 n/parenrightbigg +O(n−7/2).(5.3.6) If we like, we can further simplify the answer by using the known asymptotic expansion of the binomial coeﬃcient /parenleftbiggn−α−1 n/parenrightbigg ≈n−α−1 Γ(−α)/bracketleftbigg 1+α(α+1 ) 2n+α(α+ 1)(α+ 2)(3α+1 ) 24n2+···/bracketrightbigg . (5.3.7) If this be substituted into (5.3.6), the result is γ(n)≈n!e−3/4 √nπ/braceleftbigg 1−5 8n+1 128n2+···/bracerightbigg . (5.3.8) The form of Darboux’s method that we have proved applies when there is just one algebraic singularity on the circle of convergence. The method can be extended to several such singularities. We quote without proof a more general result of this kind ([Sz], thm. 8.4): 5.4 Analyticity and asymptotics (III): Hayman’s method 181 Theorem 5.3.2. (Szeg¨ o) Let h(w)be analytic in |w|<1and suppose it has a ﬁnite number of singularities {eiφk}r 1on|w|=1. Suppose that in the neighborhood of each singularity eiφkthere is an expansion h(w)=/summationdisplay ν≥0c(k) ν(1−we−iφk)αk+νβk, whereβk>0. Then the following is a complete asymptotic series for the coeﬃcients of h(w): [wn]h(w)≈/summationdisplay ν≥0r/summationdisplay k=1c(k) ν/parenleftbiggαk+νβk n/parenrightbigg (−eiφk)n. 5.4 Analyticity and asymptotics (III): Hayman’s method In the previous two sections we have seen how to handle the asymp- totics of sequences whose generating functions have singularities in the ﬁnite plane. Essentially, one looks for the singularity(ies) nearest the origin, ﬁnds simple functions whose behavior near the singularities is the same as thatof the generating function in question, and then proves that the asymptotic behavior of the coeﬃcients of that generating function is the same as that of the coeﬃcients of the simple functions that behave the same way near the singularities. But what shall we do if the generating function doesn’t have any sin- gularities, i.e., if it is an entire function ? Example 1. The coeﬃcients of e z Consider the function ez. The coeﬃcient of zninezis 1/n!. Can we think of some fairly general method for handling the asymptotics of entire functions, which in this case will derive Stirling’s formula for us? Here’s how we might begin. By Cauchy’s formula we have 1 n!=1 2πi/integraldisplayezdz zn+1, where the contour of integration is some simple closed curve that encloses the origin. If we use for the contour a circle of radius rcentered at the origin, then by taking absolute values we ﬁnd that 1 n!≤1 2πmax |z|=r/braceleftbigg|ez| |z|n+1/bracerightbigg (2πr) =er rn. Sinceezis an entire function the value of r>0 is entirely up to us, so we might as well choose it to minimize the upper bound that we willobtain. But min r>0er rn(5.4.1) 182 5 Analytic and asymptotic methods is attained at r=n, so the best possible estimate that we can get from this argument is that 1 n!≤(e n)n. If we compare this with Stirling’s formula 1 n!∼1√ 2nπ(e n)n, we see that we haven’t done too badly, since our rather crude estimate diﬀers from the ‘truth’ by only a factor of about 1 /√ 2nπ. To do better than this we are going to have to treat the variation of ezaround the contour of integration with a little more respect, and not just replace it by the maximum absolute value that it attains. Indeed, formost of the way around the circumference |z|=r, the absolute value of e z is considerably smaller than er. It is only in a small neighborhood of the pointz=rthat it is nearly that large. In his 1956 paper A generalisation of Stirling’s formula , W. K. Hayman [Ha] developed machinery of considerable power for dealing more precisely with this kind of situation. Further, his method is uncommonly useful for generating functions that arise in combinatorial theory, because these tend to have nonnegative real coeﬃcients. For that reason, on any circle centered at the origin, such a function will be largest in modulus at the positive real point on that circle. Hayman’s method is strongest on just such functions. Hayman’s machinery applies not only to entire functions, but to all analytic functions, even those with singularities in the ﬁnite plane. In practice it has most often been used on entire functions, mainly because, as we have seen, other methods are available when singularities exist in the ﬁnite plane. Letf(z) be analytic in a disk |z|<R in the complex plane, where 0<R≤∞ . Suppose further that f(z) is an admissible function for the method. Operationally, that simply means that f(z) is a function on which the method works. We will give some suﬃcient conditions for admissibility below. Deﬁne M(r) = max |z|=r{|f(z)|}. (5.4.2) It will be a consequence of the admissibility conditions that M(r)=f(r)( 5 .4.3) for all large enough r. This is because, as we remarked above, the method is aimed at functions that take their largest values in the direction of the positive real axis. 5.4 Analyticity and asymptotics (III): Hayman’s method 183 Next deﬁne two auxiliary functions, a(r)=rf/prime(r) f(r)(5.4.4) and b(r)=ra/prime(r)=rf/prime(r) f(r)+r2f/prime/prime(r) f(r)−r2/parenleftbiggf/prime(r) f(r)/parenrightbigg2 . (5.4.5) The main result is the following: Theorem 5.4.1. (Hayman) Let f(z)=/summationtextanznbe an admissible function. Letrnbe the positive real root of the equation a(rn)=n, for eachn= 1,2,..., wherea(r)is given by eq. (5.4.4) above. Then an∼f(rn) rnn/radicalbig 2πb(rn)asn→+∞, (5.4.6) whereb(r)is given by (5.4.5) above. It will be noted that the recipe itself is quite straightforward to apply. What is often diﬃcult is determining whether the function f(z) is admis- sible for the method or not. Before we explain that notion, let’s apply the theorem to f(z)=ez, taking on faith, for now, the fact that it is admissible. Example 1. (continued) The function ez First we would calculate a(r)=r, in this case, from (5.4.4). Then the equationa(rn)=n, which determines {rn}, becomes just rn=n. These numbersrnare the same as those that we found earlier from the condition (5.4.1). They are simply the values of rat which the minimum of f(r)/rn occurs. Next we ﬁnd b(r)=rfrom (5.4.5), and Hayman’s result (5.4.6) reads as1 n!∼en nn√ 2nπ, which is Stirling’s formula again, but this time in its exact form. Next let’s give a precise deﬁnition of the class of admissible functions. Letf(z)=/summationtext n≥0anznbe regular in |z|<R, where 0<R≤∞ . Suppose that (a) there exists an R0<R such that f(r)>0(R0<r<R ), and (b) there exists a function δ(r) deﬁned for R0<r<R such that 0<δ(r)<π for thoser, and such that as r→Runiformly for |θ|≤δ(r), we have f(reiθ)∼f(r)eiθa(r)−1 2θ2b(r), 184 5 Analytic and asymptotic methods and (c) uniformly for δ(r)≤|θ|≤πwe have f(reiθ)=o(f(r))/radicalbig b(r)(r→R), and (d) asr→Rwe haveb(r)→+∞, wherea(r),b(r) are deﬁned by (5.4.4), (5.4.5). Then we will say that f(z)i sadmissible , and the apparatus of theorem 5.4.1 above is available for determining the asymptotic growth of the coeﬃcients {an}. However, it isn’t always necessary to appeal directly to the deﬁnition of admissibility in order to be sure that a certain function is admissible. Here are some theorems that give suﬃcient conditions for admissibility, conditions that are much easier to verify than the formal deﬁnition above. (A) Iff(z) is admissible, then so is ef(z). (B) Iffandgare admissible in |z|<R, then so is fg. (C) Letfbe admissible in |z|<R. LetPbe a polynomial with real coeﬃcients which satisﬁes P(R)>0, ifR<∞, and which has a positive highest coeﬃcient, if R=+∞. Thenf(z)P(z)i s admissible in |z|<R. (D) LetPbe a polynomial with real coeﬃcients, and let fbe admis- sible in |z|<R. Thenf+Pis admissible, and if the highest coeﬃcient of Pis positive, then P[f(z)] is also admissible. (E) LetP(z) be a nonconstant polynomial with real coeﬃcients, and letf(z)=eP(z).I f [zn]f(z)>0 for all suﬃciently large n, then f(z) is admissible in the plane. Example 2. Lettnbe the number of involutions of nletters, i.e., the number of permutations of nletters whose cycles have lengths ≤2. We will ﬁnd the asymptotic behavior of {tn}. By (3.8.3), the egf of the sequence {tn}is f(z)=/summationdisplay n≥0tn n!zn=ez+1 2z2. By criterion (E) above, f(z) is clearly Hayman admissible in the whole plane. Hence theorem 5.4.1 applies. To use it, we ﬁrst calculate the func- tionsa(r),b(r) of (5.4.4) and (5.4.5). We ﬁnd that a(r)=rf/prime(r) f(r)=r+r2 and b(r)=ra/prime(r)=r+2r2. 5.4 Analyticity and asymptotics (III): Hayman’s method 185 Next we let rnbe the positive real solution of the equation a(rn)=n, which in this case is the equation rn+r2 n=n. (5.4.7) Evidently rn=/radicalbigg n+1 4−1 2 =√n/braceleftbigg 1+1 4n/bracerightbigg1 2 −1 2 =√n/braceleftbigg 1+1 8n−1 128n2+···/bracerightbigg −1 2(by (2.5.6)) =√n−1 2+1 8√n−1 128n3/2+···,(5.4.8) so we have a very good ﬁx on where rnis, in this case. Now, it would seem, all we have to do is to plug things into Hayman’s estimate (5.4.6), and that’s true, but there will be one little subtlety that will require a bit of explanation. If we take a look at (5.4.6) we see that we will need asymptotic estimates of the ‘ ∼’ kind forf(rn),b(rn), andrn n. Let’s take them one at a time. First, f(rn)=ern+1 2r2 n=e1 2(rn+n)=en/2ern/2, where (5.4.7) was used again. But in view of (5.4.8), ern/2= exp/braceleftbigg√n 2−1 4+O(n−1/2)/bracerightbigg ∼e1 2√n−1 4 (n→∞ ), and so f(rn)∼exp/braceleftbiggn 2+1 2√n−1 4/bracerightbigg (n→∞ ). (5.4.9) So far, so good. Next on the list is b(rn), and that one is easy since b(rn)=rn+2r2 n∼2r2 n∼2n (n→∞ ). (5.4.10) The last one is the hardest, and it is rn n=/braceleftbigg√n−1 2+1 8√n−···/bracerightbiggn =nn 2/braceleftbigg 1−1 2√n+1 8n−···/bracerightbiggn .(5.4.11) 186 5 Analytic and asymptotic methods What we want to do now is to ﬁnd just one term of the asymptotic behavior of the large curly brace to the nth power, and of course, it’s that nth power that causes the diﬃculty. To illustrate the method in a simpler context, consider (1+1 n)n. What does this behave like for large n? Does it approach 1? We know that it doesn’t; in fact it approaches e. So the correct asymptotic relation is /parenleftbigg 1+1 n/parenrightbiggn ∼e (n→∞ ). Hence, although 1 +1 n∼1, (1 +1 n)n∼e. In general, one cannot raise both sides of an asymptotic equality to the nth power and expect it still to be true. In exercise 7 below there are a number of situations of this kind to think about. To take a slightly harder example, how would we deal with /parenleftbigg 1+1√n/parenrightbiggn ?( 5 .4.12) What does it behave like when nis large? The way to deal with all of these questions is ﬁrst to replace (1 + ···)nby exp {nlog (1 + ···)}. Next, the logarithm should be expanded by using the power series (2.5.2), to get (1 +···)n= exp {nlog (1 + ···)} = exp/parenleftbig n{(···)−(···)2/2+(···)3/3−··· }/parenrightbig . The inﬁnite series in the argument of the exponential must now be broken oﬀ at exactly the right place. Terms can be ignored beginning with the ﬁrst one which, when multiplied by n, still approaches 0. More brieﬂy, we can ignore all terms of that inﬁnite series which are o(n−1). In the example (5.4.12) we have /parenleftbigg 1+1√n/parenrightbiggn = exp/braceleftbigg nlog/parenleftbigg 1+1√n/parenrightbigg/bracerightbigg = exp/braceleftbigg n/parenleftbigg1√n−1 2n+O(n−3/2)/parenrightbigg/bracerightbigg ∼exp/braceleftbigg n/parenleftbigg1√n−1 2n/parenrightbigg/bracerightbigg = exp/braceleftbigg√n−1 2/bracerightbigg . Now that we have that subject under our belts, we can return to the 5.4 Analyticity and asymptotics (III): Hayman’s method 187 real problem, which is (5.4.11). We now ﬁnd that /braceleftbigg 1−1 2√n+1 8n−···/bracerightbiggn = exp/braceleftbigg nlog/parenleftbigg 1−1 2√n+1 8n−···/parenrightbigg/bracerightbigg = exp/braceleftBigg n/parenleftBigg/parenleftbigg −1 2√n+1 8n/parenrightbigg −1 2/parenleftbigg −1 2√n+1 8n/parenrightbigg2 +O(n−3/2)/parenrightBigg/bracerightBigg ∼exp (−√n/2). Hence, from (5.4.11), rn n∼nn/2exp (−√n/2). (5.4.13) That ﬁnishes the estimation of the three quantities that are needed by Hayman’s theorem. The result, obtained by putting (5.4.9), (5.4.10), and (5.4.13) into (5.4.6) is that an=tn n!∼en 2+√n−1 4 2nn 2√nπ. Finally, if we multiply by n! and use Stirling’s formula, we obtain, for the number of involutions of nletters, tn∼1√ 2nn/2exp/parenleftbigg −n 2+√n−1 4/parenrightbigg . (5.4.14) 188 5 Analytic and asymptotic methods Exercises 1. Use the LIF to show that the (inﬁnite) binomial coeﬃcient sum ξ=/summationdisplay s/parenleftbiggsL+1 s/parenrightbiggA−sL−1 (sL+1 ), forA> 1 and integer L>0, satisﬁesξL−Aξ+1=0 . 2. The Legendre polynomials {Pn(x)}are generated by 1√ 1−2xt+t2=/summationdisplay n≥0Pn(x)tn. Letxbe a ﬁxed complex number that lies outside the real interval [ −1,1], and letτdenote that one of the two roots of the equation τ2−2xτ+1=0 which is>1 in absolute value. Use the method of Darboux to show that, asn→∞ , Pn(x)∼τn+1 /radicalbig nπ(τ2−1). 3. Ifu=u(t) satisﬁesu=tφ(u) andn≥0, show that [un]{φ(u)}n=[tn]/braceleftbiggtu/prime(t) u(t)/bracerightbigg =[tn]1 (1−tφ/prime(u(t))). 4. Deﬁne, for all n≥0,γn=[xn](1 +x+x2)n. (a) Use the result of exercise 3 above to prove that for n≥0, γn=[xn]/braceleftbigg1√ 1−2x−3x2/bracerightbigg . (b) Show that, using the notation of problem 2 above, γn=/parenleftBig√ 3/i/parenrightBign Pn(i/√ 3), and so obtain the asymptotic behavior of the sequence {γn}for largen. 5. Deﬁne, for integer p≥3, Sp(n)=n/summationdisplay k=0/parenleftbiggpn k/parenrightbigg (n≥0). Exercises 189 (a) Exhibit Sp(n)a s[xn] in a certain ordinary power series, which (alas!) itself depends on n. (b) Nevertheless, use the LIF (backwards) to show that /summationdisplay nSp(n)xn(1 +x)−pn−1=1 (1−x)(1−(p−1)x). (c) Deduce from part (b) that the {Sp(n)}satisfy the recurrence /summationdisplay k(−1)k/parenleftbiggpn−(p−1)k k/parenrightbigg Sp(n−k)=(p−1)n+1−1 p−2(n≥0). (d) IfF(u)=/summationtext n≥0Sp(n)un, let x=1 (p−1)−/epsilon1 in part (b) to show that F/parenleftbigg(p−1)p−1 pp/braceleftbigg 1−(p−1)3 2p/epsilon12+···/bracerightbigg/parenrightbigg =p (p−1)(p−2)/epsilon1+O(1) as/epsilon1→0. (e) If g(x)=F/parenleftbigg(p−1)p−1 ppx/parenrightbigg then show that g(x)=1 (p−2)/radicalBigg/parenleftbiggp 2/parenrightbigg1√1−x+O(1). (f) Use Darboux’s method to show that, as n→∞ , Sp(n)∼1 (p−2)/radicalBigg/parenleftbigp 2/parenrightbig nπ/parenleftbiggpp (p−1)p−1/parenrightbiggn . (g) From part (b) show that /summationdisplay n≥0S3(n)/parenleftbigg4u2 27/parenrightbiggn =u u−2 sin(1 3sin−1u)−2u 2u−3 sin (1 3sin−1u). 190 5 Analytic and asymptotic methods 6. Under what additional conditions on a polynomial Pwith nonnegative real coeﬃcients will there exist an Nsuch that for all n>N we have [zn]eP(z)>0? 7. Find the asymptotic behavior (main term) of (1 + /epsilon1n)nif (a)/epsilon1n=na(0<a< 1), (b)/epsilon1n=n−a(0<a< 1), (c)/epsilon1n=n−alogn (1<a< 2). 8. The purpose of this problem is to ﬁnd the asymptotic behavior of the numberanof permutations of nletters whose cycles are all of lengths ≤3, by using Hayman’s method and the Lagrange Inversion Formula. (The use of a symbolic manipulation package on a computer is recommended for this problem, in order to help out with some fairly tedious calculations with power series that will be necessary).) The egf of {an}is f(z) = exp {z+z2 2+z3 3}. (a) Show that fis admissible in the plane. (b) Because rnin this case satisﬁes a cubic equation rather than a quadratic , as in the example in the text, we will use the LIF to ﬁnd the root and its powers with suﬃcient precision. Show that if we write u=1/rn;t=n−1/3;φ(u)=( 1+u+u2)1/3, thenusatisﬁes the equation u=tφ(u), which is in the form (5.1.1). (c) Use the LIF to show that the root rnhas the asymptotic expansion 1 rn=1 n1/3+1 31 n2/3+1 31 n+8 811 n4/3+O(n−5/3). (d) Explain why the number of terms that were retained in part (c) is the minimum number that can be retained and still get the ﬁrst term of the asymptotic expansion of anwith this method. (e) Show that 1 rnn∼n−n 3exp/braceleftbigg1 3n2/3+5 18n1/3/bracerightbigg . (f) Show that b(rn)∼3n. 5.4 Analyticity and asymptotics (III): Hayman’s method 191 (g) Show that f(rn)∼exp/braceleftbigg1 3n+1 6n2/3+5 9n1/3−29 162/bracerightbigg . (h) Combine the results of (d), (e), (f) to show that the number of permutations of nletters that have no cycles of lengths >3i s an∼n2n 3√ 3exp/braceleftbigg −2n 3+1 2n2/3+5 6n1/3−29 162/bracerightbigg . 9. Derive the power series expansion (2.5.16). 10. In this exercise, σ(n,k) is the number of involutions of nletters that have exactly kcycles, and tn=/summationtext kσ(n,k) is the number of involutions of nletters. (a) Show that /summationdisplay n,kσ(n,k) n!xnyk=ey(x+1 2x2). (b) Hence ﬁnd the formula σ(n,k)=n! (n−k)!(2k−n)!2n−k forσ(n,k). (c) Using the results of part (a) and problem 5 of chapter 3, show that the average number of cycles in an involution of nletters is exactly n 2/braceleftbigg 1+tn−1 tn/bracerightbigg . (d) Using (5.4.14), show that the average number of cycles in an in- volution of nletters is =n 2+1 2√n(1 +o(1)) (n→∞ ). Appendix Using Maple∗and Mathematica∗∗ Many branches of mathematics that were formerly thought of as being ﬁt only for humans, are being invaded by computers. First, elementary school students learned how to multiply numbers with many digits and then foundout that little calculators could do it for them. Other kinds of mathematics that are taught in secondary schools that now can be done by computers include expanding and factoring algebraic expressions, solving linear and quadratic equations, plotting graphs of curves and surfaces, doing logarithms and powers, and more. At the university level we ﬁnd now that “computer algebra” programs can diﬀerentiate functions symbolically, do integrals, vector analysis, linear algebra, etc., all symbolically , rather than numerically. Here we want to show how computers can easily handle much of the routine work that is involved in solving problems about generating functions. To emphasize this point, we will show how well computers can do some of the homework problems in this book! Very well indeed, we’re sure you will agree. In this brief Appendix we’ll discuss ﬁrst how computer programs can do extensive manipulations of power series. Next we’ll focus on one such program, Mathematica TM(Version 2.0) , and tell you about its amazing built- inRSolve function. Finally we will look at how MapleTMhandles asymptotics, which can be quite a boon for problems such as those we looked at in the previous chapter. 1. Series manipulation InMathematicaTM, the instruction Series[f,x,x0,m] will display the ﬁrstm+ 1 terms of the power series expansion of faboutx=x0. Thus, to see the ﬁrst 10 terms of the series for sin x/(1 +x), about the origin, you would enter (the MapleTMinstruction that would accomplish the same thing would be series(sin(x)/(1+x),x=0,9) ) Series[Sin[x]/(1+x), {x,0,9}] andMathematicaTMwould respond x−7x3 6+47x5 40−5923x7 5040+426457x9 362880+O (x)10. Perhaps you’d like to check the accuracy of the terms displayed in the series (2.5.10) of Chapter 2, and to see what the next two terms are. If so, then enter ∗Maple is a registered trademark of Waterloo Maple Software. ∗∗Mathematica is a registered trademark of Wolfram Research, Inc. 192 2. The RSolve.m routine 193 Series[(1-Sqrt[1-4x])/(2x), {x,0,9}] and you will see 1+x+2x2+5x3+1 4x4+4 2x5+ 132x6+ 429x7+ 1430x8 + 4862x9+ 16796x10+ 58786x11+O (x)12. If you want to obtain the list of coeﬃcients of the terms of this series, because they are the numbers that the series “generates,” then ask for CoefficientList[%,x] to obtain (the “%” means the result of the computation in the preceding line) {1,1,2,5,14,42,132,429,1430,4862,16796,58786 } and there are the Catalan numbers on display. If you want to see only the coeﬃcient of x7then you would enter Coefficient[%,x,7] instead, and the 429 would appear. A little more work is needed to see sequences that are generated by expo- nential generating functions. Suppose you wanted the ﬁrst 12 Bell numbers. According to theorem 1.6.1 these are the coeﬃcients of xn/n!i n Series[Exp[Exp[x]-1], {x,0,12 }]. If you type exactly that, MathematicaTMwill reply with 1+x+x2+5x3 6+5x4 8+13x5 30+203x6 720+877x7 5040+23x8 224 +1007x9 17280+4639x10 145152+22619x11 1330560+4213597x12 479001600+O (x)13, which isn’t quite what you wanted because, for instance, the coeﬃcient of x8/8! is not readily apparent. One more instruction, such as Table[j! Coefficient[%,x,j], {j,0,12 }] will get the desired display of Bell numbers, {1,1,2,5,15,52,203,877,4140,21147,115975,678570,4213597 }. 2. The RSolve.m routine The RSolve package was written in MathematicaTMby Marko Petkovˇ sek [Pe]. Its purpose is to ﬁnd symbolic solutions to recurrence relations and diﬀerence equations. It can do so by explicitly ﬁnding the ordinary power series or exponential generating function of the unknown sequence. To use it one ﬁrst reads in the package with <<DiscreteMath/RSolve.m 194 Using MapleTMand MathematicaTM One then has a powerful facility for ﬁnding generating function solutions to problems in combinatorial recurrence. Let’s try it on the Fibonacci recurrence, with the call RSolve[ {f[n+2]==f[n+ 1]+f[n],f[0]==0,f[1]==1 },f[n ],n]. It replies, after an order to Simplify[%] , as follows. {{f(n)→/parenleftBig −/parenleftBig 1 2−√ 5 2/parenrightBign +/parenleftBig 1 2+√ 5 2/parenrightBign/parenrightBig If(n≥1,1,0) √ 5}}, which is, of course, the explicit formula for the Fibonacci numbers. If you’re ready for this, let’s change the call above by replacing “ RSolve ”b y“ Gener- atingFunction ,” and adding one more argument, xsay, to tell it the variable to use in the generating function. That means that we enter the request GeneratingFunction[ {f[n+2]==f[n+1]+f[n],f[0]==0,f[1]==1 },f[n],n,x]. And what is the reply? It is {{x 1−x−x2}}, which even in an age of multitudinous computer miracles must leave us in awe. Perhaps you’d rather have the exponential generating function of your numbers. Well then you would change the call to ExponentialGeneratingFunction[ {f[n+2]==f[n+1]+f[n],f[0]==0,f[1]==1 },f[n],n,x] and the computer would inform you that {{−e(1−√ 5)x 2+e(1+√ 5)x 2√ 5}} is the function you seek. Now let’s watch it solve the recurrence (2.2.6) for the number of block fountains of coins that have kcoins in the ﬁrst row. This time the call is GeneratingFunction[f[k]==1+Sum[(k-j) f[j], {j,1,k}]/;k>=1, f[k],k,t], and the response is {{−(−1+t)t 1−3t+t2}} in agreement with (2.2.7). It can even ﬁnd a closed formula for the number of such fountains from the generating function. To get that, ask for Simplify[SeriesTerm[%, {t,0,n}]] and the output will be {{ /parenleftbig 5−√ 5/parenrightbig/parenleftBig 3 2+√ 5 2/parenrightBign 10+/parenleftBig 3 2−√ 5 2/parenrightBign/parenleftbig 5+√ 5/parenrightbig 10 If(n≥0,1,0) −If(n=0,1,0)}}. Exercises 195 As you can see, it did the partial fraction expansion followed by two geometric series manipulations, just as we did to obtain, for instance, (1.3.3). The package can also ﬁnd closed form expressions for the sums of series in which formulas are given for the nth coeﬃcient. A request PowerSum[a n+b, {z,n}] will produce the answer to exercise 1(b) in this book, in the form b 1−z+az (−1+z)2. It can do much harder ones than that, like the gf of the harmonic numbers that we did in Example 5 of chapter 2. That one is the answer to the call PowerSum[Sum[1/j, {j,1,n}],{x,n}], namely −log(1−x) 1−x. The reader who takes the time to experiment with the capabilities of the RSolve.m package will be amply rewarded. 3. Asymptotics in MapleTM InMapleTM, if you type asympt(f,x,n); you will receive nterms of the asymptotic expansion of the function fof the variable x,a sx→∞ . Let’s try Stirling’s formula ﬁrst, by asking for asympt(n!,n,5); The computer’s answer is (we use ‘ Pi’ instead of ‘ π’ etc. because that’s pretty much how it will look on your screen) /parenleftbigg 21/2Pi1/2n1/2+1/1221/2Pi1/2 n1/2+1/28821/2Pi1/2 n3/2−139 5184021/2Pi1/2 n5/2 −571 248832021/2Pi1/2 n7/2+O(1 n9/2)/parenrightbigg /((1/n)nexp(n)). We all know that (1 + 1 /n)n→e, but how fast does it go? The answer given by MapleTMis exp(1)−1/2exp(1) n+11 24exp(1) n2+O(1 n3). In closing, let’s do exercise 8(c) of the previous chapter, which asks for the asymptotic behavior of the nth power of 1 rn=1 n1/3+1 31 n2/3+1 31 n+8 811 n4/3+O(n−5/3). 196 Using MapleTMand MathematicaTM Needless to say, MapleTMis up to the task, and gives n−n 3exp/braceleftbigg1 3n2/3+5 18n1/3/bracerightbigg (1 +O(1)). Exercises On any computer that is available to you, do the following. 1. Exercises 1, 2, 5, 6, 8 of Chapter 1. 2. Check the ﬁrst ﬁve terms of any ﬁve of the series displayed in section 2.5.3. Exercises 1, 2, 4 of chapter 2. 4. Use the “series” command to ﬁnd the ﬁrst 15 values of g(n) of (3.9.1). 5. From (3.8.3), tabulate the number of involutions of nletters, forn≤15. 6. Use the asymptotics capability of Maple TMto ﬁnd the ﬁrst 5 terms of the asymptotic expansions of the following. (a) (1 + 1/√n)n (b)√ n! (c) (1 + 1/n)√n (d) sin (sin 1 /x) Solutions 197 Solutions Answers to problems for chapter 1 1. (a) (xD)(1/(1−x)) =x/(1−x)2 (b) (αxD +β)(1/(1−x)) =αx/(1−x)2+β/(1−x) (c) (xD)2(1/(1−x)) (d) (α(xD)2+βxD +γ)(1/(1−x)) (e)P(xD)(1/(1−x)) (f) 1/(1−3x) (g) 5/(1−7x)−3/(1−4x) 2. (a) (xD)ex=xex (b) (αxD +β)ex=(αx+β)ex (c) (xD)2ex=(x+x2)ex (d) (α(xD)2+βxD +γ)ex (e)P(xD)ex (f)e3x (g) 5e7x−3e4x 3. (a)f(x)+c/(1−x) (b)αf(x)+c/(1−x) (c)xDf(x) (d)P(xD)f(x) (e)f(x)−a0 (f)f(x)−a0−a1x+( 1−a2)x2 (g) (f(x)+f(−x))/2 (h) (f(x)−a0)/x 198 Solutions (i) (f(x)−/summationtexth−1 0ajxj)/xh (j) (f−a0−a1x)/x2+ 3((f−a0)/x)+f (k) (f−a0−a1x)/x2−((f−a0)/x)−f 4. (a)f(x)+cex(b)αf(x)+cex(c)xf/prime(x) (d)P(xD)f(x) (e)f−a0(f)f−a0−a1x+( 1−a2)x2/2 (g) (f(x)+f(−x))/2 (h) f’(x) (i) Dhf(x) (j)f/prime/prime+3f/prime+f (k)f/prime/prime−f/prime−f 5. (a) 2n/n! (b)αn (c) (−1)mifn=2m+ 1 is odd, and 0 else. (d) (an+1−bn+1)/(a−b) (e)/parenleftbigm n/2/parenrightbig 6. (a) We see at once that f/x=3f+2/(1−x), sof=2x/((1−x)(1−3x)). (b)f/x=αf+β/(1−x)s of=βx/((1−x)(1−αx)). (c) Here (f−x)/x2=2f/x−fsof=x/(1−x)2. (d) Sincef/x=f/3+1/(1−x) we havef=3x/((1−x)(3−x)). 8. (a)f/prime=3f+2ex,f(0) = 0 give f=e3x−ex (b)f/prime=αf+βexsof=(β/(1−α))(ex−eαx) (c)f/prime/prime=2f/prime−f,f(0) = 0,f/prime(0) = 1 yield f=xex (d)f/prime=f/3+ex,f(0) = 0 give f=3 2(ex−ex/3) 9.Multiply both sides of the equation f(2n)=f(n)b yx2nand sum over n≥1. Then multiply both sides of f(2n+1 )=f(n)+f(n+1 )b yx2n+1, sum overn≥1, and add to the previous result. Then add f(1)x=xto that result to obtain the functional equation. To ﬁnd the explicit inﬁnite product form of the solution, let’s ﬁrst see how we might guess that answer, and then how we might prove it. Take the functional equation for F, and replace xbyx2throughout, then substitute the result back in the functional equation, to get F(x)=( 1+x+x2)(1 +x2+x4)F(x4). If we now replace xbyx2again, and substitute we’ll get even more factors of Solutions 199 the inﬁnite product. Hence we should suspect that the product is the answer. Toprove that the product is the answer, we have two choices. First, over the ring of formal power series, consider the product as a formal beast which obviously satisﬁes the functional equation for F. Second, analytically, an inﬁnite product/producttext(1+qn) converges if the series/summationtext|qn|does; so, the product converges for |x|small enough, to an analytic function F. 10.For part (a) see section 4.1. (b)p(2) n=/summationtextn j=0Prob(X=j)Prob(X=n−j)=[xn]P(x)2. (c)Pk(x)=P(x)k (d) By part (c) the mean is P/prime k(1)/Pk(1) =/bracketleftbig kP(x)k−1P/prime(x)/P(x)k/bracketrightbig x=1=kµ, and the variance is (logPk(x))/prime+ (logPk(x))/prime/prime/vextendsingle/vextendsingle/vextendsingle/vextendsingle x=1=k(logP(x))/prime+ (logP(x))/prime/prime/vextendsingle/vextendsingle/vextendsingle/vextendsingle x=1, which iskσ2. (e) SinceAk=Bwe havekA/prime/A=B/prime/BorkA/primeB=AB/prime. Equate the coeﬃcients of xnto ﬁnd that nbn=/summationtextn j=1(j(k+1 )−n)ajbn−jfor n≥1, withb0=1 . (f)p∗is the coeﬃcient of x300in (.1x+.2x2+.1x3+.2x4+.2x5+ .2x6)100/(1−x). Numerically, it is about .00000095. (g) The required probability is [xj]1 1−x/parenleftbigx+x2+···+xm m/parenrightbign=1 mn[xj−n](1−xm)n (1−x)n+1. The result follows by expanding the numerator by the binomial theorem, the reciprocal of the denominator by the binomial series, and multiplying. 11.Among these subsets we distinguish those that do contain nand those that don’t. If such a subset contains n, then the rest of that subset is one of the subsets that is counted by f(n−2). If it does not contain nthen the entire subset is one of those that is counted by f(n−1). Thusf(n)= f(n−1)+f(n−2), which together with the starting values f(1) = 2,f(2) = 3 tells us that f(n)=Fn+2, where the F’s are the Fibonacci numbers. 12.As in problem 11, we distinguish those k-subsets that do contain nand those that do not. If such a k-subset does contain nthen the rest of that subset is one of the subsets that is counted by f(n−2,k−1), otherwise the 200 Solutions entirek-subset is one of those that is counted by f(n−1,k). Thusf(n,k)= f(n−1,k)+f(n−2,k−1), fork≥2. If we deﬁne Fk(x)=/summationtext n≥1f(n,k)xn, then after multiplying the recurrence by xnand summing over n≥1 we ﬁnd thatFk(x)=x2Fk−1(x)/(1−x), which together with F1(x)=x/(1−x)2 tells us that Fk(x)=x2k−1/(1−x)k+1. If we expand in the binomial series we ﬁnd that f(n,k)=[xn]Fk(x)=[xn]x2k−1/summationdisplay h≥0/parenleftbiggk+h k/parenrightbigg xh=/parenleftbiggn−k+1 k/parenrightbigg . 13.From problems 11 and 12, it must be that /summationdisplay k/parenleftbiggn−k+1 k/parenrightbigg =Fn+1 (n≥0), which will be proved another way in example 1 of section 4.3. 14.A circular arrangement that does contain nis obtained by taking a linear arrangement of {2,3,...,n −2}, no two consecutive (on the line), adjoining nto it, and laying it out around a necklace. So by exercise 11, there are Fn−1 such arrangements. One that does notcontainnis obtained by taking any linear arrangement of {1,2,...,n −1}and laying it out around a necklace, so there areFn+1of these. Hence there are Fn−1+Fn+1such circular sequences altogether. 15.As in the previous problem, the answer is f(n−3,k−1) +f(n−1,k) wheref(n,k)=/parenleftbign−k+1 k/parenrightbig is the solution to exercise 12. 16.The partial fraction expansion of 1/ (1−x2)2is 1 4(1−x)2+1 4(1−x)+1 4(1 +x)2+1 4(1 +x). Therefore the coeﬃcient of xnin its power series expansion is n+1 4+1 4+(−1)n(n+1 ) 4+(−1)n 4, which is 0 if nis odd and is ( n+2 )/2i fnis even. Otherwise, take the series for 1/(1−u)2and replace ubyx2. Even sneakier would be to use one of the symbolic manipulation programs that are now available on computers. They can produce the general term of such series on demand. 17.Fixj,1≤j≤n, and consider just those permutations σofnletters that haveσ(j)=n. Then no inversions have jas the second member of the pair and exactly n−jinversions have jas the ﬁrst member of the pair. Hence if Solutions 201 we deletenfrom the string of values of σwe obtain a permutation of n−1 letters with n−jfewer inversions. Thus b(n,k)=/summationtext jb(n−1,k−n+j). Multiply by xkand sum on kto obtainBn(x)=( 1 +x+···+xn−1)Bn−1(x). Henceb(n,k) is the coeﬃcient of xkin (1 +x)(1 +x+x2)(1 +x+x2+x3)···(1 +x+x2+x3+···+xn−1). 18. (a) The probability is evidently 1 /k! that the ﬁrst kvalues will decrease, son!/k! of the permutations have this property. (b) The probability that a permutation begins with kdecreasing values followed by an increasing one is 1 /k!−1/(k+ 1)!, if 0 ≤k<n , and is 1/n! whenk=n. The average value of k, weighted with these probabilities, is n−1/summationdisplay k=0k/parenleftbig1 k!−1 (k+ 1)!/parenrightbigg +n n!=n/summationdisplay k=11 k!, and therefore an average permutation of nletters begins with a decreasing sequence whose length is approximately e−1. (c) If we begin with a permutation of n−1 letters that has kruns, then by inserting the letter nin each of the npossible places we manufacture kpermutations of nletters that have kruns andn−k permutations of nletters that have k+ 1 runs. Thus f(n,k)= kf(n−1,k)+(n−k+1 )f(n−1,k−1). 19. (a) It is (1 + x)2(1 +x2)(1 +x5)(1 +x10)2(1 +x20)(1 +x50). (b) The sum represents 38= 6561 integers, each between −99 and 99. Hence these 199 integers are represented an average of 6561 /199 = 32.9..ways, so some integer must be represented at least 33 ways. The required product is (1/x+1+x)2(1/x2+1+x2)(1/x5+1+x5)···(1/x50+1+x50). (c) Ifw1,···,wrare distinct integers, and if Dnis the number of repre- sentations of nas a sumn=w1x1+w2x2+···+wkxkwhere each of thexiis±1, then /summationdisplay nDntn=r/productdisplay i=1(twi+t−wi). 202 Solutions The set of roots is the union of the sets of 2 with roots of −1, for i=1,...,k . Answers to problems for chapter 2 1.The thing to remember is that 1 /(1−u)=1+u+u2+···. (a) We have 1 cosx=1 1−(x2 2−x4 24+···) =1+(x2 2−x4 24+···)+(x2 2−x4 24+···)2+··· =1+x2 2+5x4 24+··· (b) Here we use the binomial theorem with negative exponent. 1 (1 +x)m=( 1+x)−m =/summationdisplay k/parenleftbigg−m k/parenrightbigg xk =1+/parenleftbigg−m 1/parenrightbigg x+/parenleftbigg−m 2/parenrightbigg x2+/parenleftbigg−m 3/parenrightbigg x3+··· =1−mx+m(m+1 ) 2x2−m(m+ 1)(m+2 ) 6x3+··· (c) This is like part (a). We ﬁnd 1 1+(t2+t3+t5+···)=1−(t2+t3+t5+···)+··· =1−t2−t3+t4+t5+···. 2.In each case (except (e)), replace xbyx+bx2+cx3+···, set the result equal tox, and equate the coeﬃcients of like powers of xto 0 to solve for b andc. (a)x+x3 6+··· (b)x−x3 3+··· (c)x−x2+3 2x3+··· (d)x−x3+··· Solutions 203 (e) In this part, note that if yis the inverse function, then log (1 −y)=x, i.e.,y=1−ex=−x−x2/2−x3/6−···. 3.Iff=/summationtext k≥0akxkthen 0=f/prime/prime+f=/summationdisplay k≥0{(k+ 2)(k+1 )ak+2+ak}xk and so ak+2=−ak (k+ 1)(k+2 )(k=0,1,2,...). Ifa0anda1are arbitrarily ﬁxed, then by induction on k a2k=(−1)ka0/(2k)! and a2k+1=(−1)ka1/(2k+ 1)! for allk≥0, and the result follows. 4. (a)x (1−x)2+7 1−x (b)x4 1−x (c)1 1−x2 (d){log (1 1−x)−x−x2 2}/x (e){ex−1−x−x2/2−x3/6−x4/24}/x5 (f)xd dx{x 1−x−x2}=x(1+x2) (1−x−x2)2 (g){(xDxD )+(xD)+1}(ex−1) = (1 +x)2ex−1 5.In the binomial theorem (1 + x)n=/summationtext k/parenleftbign k/parenrightbig xk, letx=1 . 6.We have f(n,k)=/summationdisplay n1+···+nk=nn1n2···nk =[xn](/summationdisplay rrxr)k =[xn]/braceleftbiggx (1−x)2/bracerightbiggk =[xn]xk (1−x)2k. 204 Solutions Hence/summationtext nf(n,k)xn=xk (1−x)2k. Explicitly, since 1 (1−x)2k=/summationdisplay r≥0/parenleftbiggr+2k−1 r/parenrightbigg xr, we ﬁnd that f(n,k)=/parenleftbign+k−1 n−k/parenrightbig . Notice that the answer is 0 when n<k . Explain why. 7.As in problem 6 above, f(n,k,h )=[xn]/braceleftbigg/summationdisplay r≥hxr/bracerightbiggk =[xn]/braceleftbiggxkh (1−x)k/bracerightbigg . 8.1, 1, 1, 1, and 1, respectively. 11.1, 1, 5−1 2, 0, 1, respectively. 13.Letaandbbe relatively prime. Then every divisor dofabis uniquely of the form d=d/primed/prime/primewhered/prime\a,d/prime/prime\b. Hence g(ab)=/summationdisplay d\abf(d)=/summationdisplay d/prime\a d/prime/prime\bf(d/primed/prime/prime) =/summationdisplay d/prime\a d/prime/prime\bf(d/prime)f(d/prime/prime)=/parenleftbig/summationdisplay d/prime\af(d/prime)/parenrightbig/parenleftbig/summationdisplay d/prime/prime\bf(d/prime/prime)/parenrightbig =g(a)g(b). 14.Consider the nfractions 1/n,2/n,...,n/n . If we write them in lowest terms, then each of them will reduce to a fraction h/k, wherek\nandh,k are relatively prime. Further, for a ﬁxed divisor kofn, each of the φ(k) such fractionsh/koccurs in exactly one way, i.e., by reducing exactly one fraction m/n. 15.For Euler’s function, apply M¨ obius inversion to the result of problem 14 above. This gives φ(n)=/summationdisplay d\nµ(n/d)d=n/summationdisplay d\nµ(n/d) n/d=n/summationdisplay d\nµ(d) d. Sinceµis multiplicative, µ(n)/nis a multiplicative function of n, and so, by problem 13, is the last member above. Solutions 205 Forσ(n), suppose a,bare relatively prime. Then every divisor of abis uniquely of the form d/primed/prime/prime, whered/primeandd/prime/primeare divisors of aand ofb, respec- tively, and the result follows. Finally, if a,bare relatively prime, |µ(ab)|=1 iﬀabis squarefree iﬀ aandbare squarefree. 16. (a)ζ(s−1)/ζ(s) (b)ζ(s)ζ(s−1) (c)ζ(s)/ζ(2s) 17. (a)ζ(s−1) (b)ζ(s−α) (c)−ζ/prime(s) (d)ζ(s)ζ(s−q) 18. (a)ζ(s){ζ(s−1)/ζ(s)}=ζ(s−1) (b)ζ(s){1/ζ(s)}=1 (c){1/ζ(s)}{ζ2(s)}=ζ(s) 19.We ﬁnd that F(x)=1 10/parenleftbigg5−√ 5 1−α+x+5+√ 5 1−α−x/parenrightbigg whereα±=( 3±√ 5)/2. Hence there are exactly 5−√ 5 10αk ++5+√ 5 10αk − block fountains whose ﬁrst row contains kcoins. 21. (a)/summationtext nf(n,k,T )xn=(/summationtext t∈Txt)k (b)/summationtext ng(n,k,T )xn=/bracketleftbig yk/bracketrightbig/producttext t∈T(1 +yxt) (c)/summationtext nf(n,k,S,T )xn=/bracketleftbigyk k!/bracketrightbig/producttext t∈T{/summationtext s∈Sysxst s!} 22.Check that the required number is the coeﬃcient of xnin /summationdisplay k≥1(−1)k/parenleftbiggx 1−x/parenrightbiggk =−x, 206 Solutions hencef(n) = 0 for all nexcept that f(1) = −1. 23.On the one hand, x(emx−1) ex−1=/parenleftbiggx ex−1/parenrightbigg (emx−1) =/parenleftbigg/summationdisplay nBn n!xn/parenrightbigg/parenleftbigg/summationdisplay j≥1mjxj j!/parenrightbigg =/summationdisplay n≥0xn n!/braceleftbigg/summationdisplay j≥1/parenleftbiggn j/parenrightbigg Bn−jmj/bracerightbigg . On the other hand, x(emx−1) ex−1=x/parenleftbiggemx−1 ex−1/parenrightbigg =x(1 +ex+e2x+···+e(m−1)x) =xm−1/summationdisplay j=0/summationdisplay r≥0jrxr r! =/summationdisplay r≥0xr+1 r!Sr(m−1), whereSr(m) is the sum of the rth powers of the integers 1 ,...,m .I f w e compare the coeﬃcients of xnwe ﬁnd the explicit formula Sn(m)=1 n+1/summationdisplay r≥1/parenleftbiggn+1 r/parenrightbigg Bn+1−r(m+1 )r, which holds for integers m,n≥1. The ﬁrst few cases are, for n=1,2,3, 1+2+ ···+m=m2 2+m 2 12+22+···+m2=m3 3+m2 2+m 6 13+23+···+m3=m4 4+m3 2+m2 4. 25. (a) If they diﬀer in the jth bit, then their colors diﬀer by jwhich is not 0. If they diﬀer in the jth andkth bits, then their colors diﬀer by j+korj−kmodulo 2n, neither of which can be 0. (c)f(z)=/producttextn k=1(1 +zk). Solutions 207 27. (a)e−x/(1−x). (b) IfD(x)=e−x/(1−x) then (1 −x)D/prime=D−e−x. Match [xn]o n both sides. (c) IfD1(n) is the number with one ﬁxed point then D1(n)=nD(n−1). ThusD(n)−D1(n)=D(n)−nD(n−1) = ( −1)nby part (b). (d) To construct a permutation of nletters that has kﬁxed points, we can choose the kﬁxed points in/parenleftbign k/parenrightbig ways, and the rest of the permutation in D(n−k) ways. Hence Dk(n)=/parenleftbign k/parenrightbig D(n−k). Now multiply by xnyk/n!, sum, and use part (a). 28.We have /summationdisplay d\nµ(n/d)ad(xn/d)=/summationdisplay d\nµ(n/d)/summationdisplay δ\dbd/δ(xnδ/d) in which the coeﬃcient of br(xn/r)i s/summationtext δ\n/rµ(n/(rδ)), which vanishes unless r=nand is 1 in that case. 30. (a)/summationtext n≥1√n/ns=/summationtext n≥11/ns−1/2=ζ(s−1/2). (b) The function is multiplicative and its value at n=pais 0 ifa≥2 and 1 otherwise. Hence by (2.6.6) the generating function is /productdisplay p{1+p−s}=/productdisplay p1−p−2s 1−p−s=ζ(s) ζ(2s). (c) Hereλ(pa)=(−1)afor alla≥0, so by (2.6.6) its generating func- tion Λ(s)i s /productdisplay p{1−p−s+p−2s−··· } =/productdisplay p1 1+p−s=ζ(2s) ζ(s). Finally, equate coeﬃcients of n−son both sides of Λ( s)ζ(s)=ζ(2s). 31. (a) If /summationdisplay n≥1bnxn=/summationdisplay n≥1anxn 1−xn=/summationdisplay nan/summationdisplay m≥1xmn=/summationdisplay rxr/summationdisplay d\rad thusbn=/summationtext d\nad, just as in the theory of Dirichlet series. 208 Solutions (b) Apply part (a) with an=µ(n). (c) It is/summationdisplay n≥1φ(n)xn 1−xn=/summationdisplay n≥1nxn=x (1−x)2 since/summationtext d\nφ(d)=n. 32. (a)f/(1−x) (b)f/(1−x)r (c) By part (b), it is the sequence whose gf is 1 /(1−x)r+1, which, by (2.5.7), is the sequence/parenleftbign+r n/parenrightbig n≥0. (d) It is the coeﬃcient of xnin (/summationtextanxn)/(1−x)r, viz. n/summationdisplay m=0/parenleftbiggm+r−1 m/parenrightbigg an−m (n=0,1,2,...). (e) Thenf(x)/(1−x)r= 1, sof(x)=( 1 −x)randan=/parenleftbigr n/parenrightbig (−1)nfor n≥0. 33. (b) We have Φpa(x)=/productdisplay d\pa(1−xd)µ(pa/d) =1−xpa 1−xpa−1=1+xpa−1+x2pa−1+···+x(p−1)pa−1. (c) Since/producttext m\nΦm(x)=1−xnwe have /productdisplay m\n m>1Φm(1) =n. (∗) Now forn=pkuse induction on k.I fnis not a prime power let pa be the highest power of pthat divides n. Then in ( ∗) above, each divisorpj(1≤j≤a) contributes a factor of p, so all such divisors contributepa. But no higher power of pdividesn,s oΦ n(1) cannot be divisible by p. Sincepwas arbitrary, Φ n(1) must be ±1, and it is easy to rule out −1. Solutions 209 34. (a) This says that each integer r,1≤r≤nis uniquely of the form r=mdwhered\nand gcd(m,n/d ) = 1. But this is clear since we taked=gcd(r,n) andm=r/d. (c) Letx→ωin the result of part (b), and use L’Hospital’s rule. Answers to problems for chapter 3 1.A partition of ninto odd parts looks like n=r1·1+r3·3+r5·5+···. Now substitute the binary expansion of each ri, to get n=( 2a1+2b1+···)·1+( 2a3+2b3+···)·3+( 2a5+2b5+···)·5+···. But now we have a partition of ninto distinct parts, viz., n=2a1+2b1+···+2a3·3+2b3·3+···+2a5·5+2b5·5+···. (What partition corresponds to 39=3+3+7+7+19 ?) The map is uniquely invertible. 2.The deck has a card corresponding to each cyclic permutation of length k,2k,3k,... . The number of these on mkletters is (mk−1)!. The deck enumerator is D(x)=/summationdisplay m≥1(mk−1)! (mk)!xmk=/summationdisplay m≥1xmk mk=1 klog1 1−xk. Hence the hand enumerator, without regard to number of cards in the hand, is H(x) = exp/braceleftbigg1 klog1 1−xk/bracerightbigg =1 (1−xk)1/k =/summationdisplay m≥0/parenleftbigg−1 k m/parenrightbigg (−1)mxmk. The required number is the coeﬃcient of xn/n! here, which is 0 if kdoes not dividen, and is (−1)r/parenleftbigg−1 k r/parenrightbigg n!=n! r!kr(k+ 1)(2k+1 )···((r−1)k+1 ) 210 Solutions ifn/k=ris an integer. 3.It is exp/braceleftbigg/summationtext pxp p!/bracerightbigg . 4. (a) The order is the least common multiple of the cycle lengths. (b) Clearly, ˜g(n,k)=/summationdisplay d\kg(n,d). Hence by M¨ obius inversion (2.6.12) we have g(n,k)=/summationdisplay d\kµ(k/d)˜g(n,d). 5. (a) One ﬁnds by logarithmic diﬀerentiation of the egf (3.8.3) that Tn=Tn−1+(n−1)Tn−2 (n≥2;T0=1 ;T1=1 ). (b) 1, 2, 4, 10, 26, 76 (c) Consider separately those involutions of nletters for which nis a ﬁxed point and those for which nis not ﬁxed. 6.The deck enumerator is D(x)=/summationdisplay n≥4xn n= log1 (1−x)−x−x2/2−x3/3, and so the hand enumerator is H(x)=e−x−x2/2−x3/3 (1−x). 7.A card of weight nis a path or a cycle. If n≥3, there are ( n−1)!/2 ‘cycle cards’ of weight n, and ifn≥2 there are n!/2 ‘path cards.’ Hence D(x)=/parenleftbigg1 (1−x)−log (1 −x)−1−2x−x2/2/parenrightbigg /2, and the hand enumerator H(x)i s sinh/parenleftbigg1 2(1−x)−1 2log (1 −x)−1 2−x−x2 4/parenrightbigg =1x2 2!+4x3 3!+1 5x4 4!+7 2x5 5!+ 435x6 6!+ 3300x7 7!+ 30310x8 8!+···. Solutions 211 Hence the numbers of such graphs on 0 ,1,..., 8 vertices are 0, 0, 1, 4, 15, 72, 435, 3300, 30310. 8.One ﬁnds/bracketleftbign k/bracketrightbig =/bracketleftbign−1 k−1/bracketrightbig +(n−1)/bracketleftbign−1 k/bracketrightbig . This can be proved directly by considering separately those permutations of nletters and kcycles in which nis a ﬁxed point (cycle of length 1) and those in which it is not. 9.Here the deck enumerator is D(x)=/summationdisplay m≥1(2m−1)!x2m (2m)!= log1√ 1−x2. The exponential formula states that the question is answered by sinh D(x), which simpliﬁes to H(x)=x2 2√ 1−x2. The coeﬃcient of xn/n!i sg(n)=0i fnis odd, and g(n)=n! 2n−1/parenleftbiggn−2 n 2−1/parenrightbigg ifn≥2 is even. 10.We ﬁnd that/braceleftbiggn k/bracerightbigg =k/summationdisplay r=1(−1)k−rrn−1 (k−r)!(r−1)!. 12. IfF(n,k) is the number of n-permutations whose cycles have lengths ≤k, thenFhas the egf exp ( x+···+xk/k). Butf(n,k) counts those whose longest cycle has length k, so ifk≥1,f(n,k)=F(n,k)−F(n,k−1), and the required egf is ex+···+xk−1 k−1/parenleftbigg exk k−1/parenrightbigg . 13. (a) It is/summationdisplay i+j+k=ntigjgk i!j!k!=1 (n≥0). (b) If we multiply through by n! to get /summationdisplay i+j+k=nn! i!j!k!tigjgk=n!(n≥0). 212 Solutions then the multinomial coeﬃcient under the summation sign counts the ways of choosing an ordered triple ( R,S,T ) of subsets that par- tition [n],ticounts the involutions of iletters, each of which gets relabeled with the elements of R,gjcounts the 2-regular graphs of jvertices, each of which gets relabeled with the elements of S, etc. Finally, the right side n! countsn-permutations. (c) This elegant solution was found by Mr. Douglas Katzman. Given the triple (τ,G 1,G2), we construct the corresponding permutation σ as follows the cycles of the involution τ, acting onR, become cycles ofσ. For each cycle in the graph G1, locate the smallest numbered vertexvin the cycle. Choose that one of the two possible ways of orienting the cycle which carries vto the larger numbered vertex of its two neighbors. Conversely, in G2select the orientation that carries the smallest numbered vertex of each cycle into the smaller of its two neighbors. 14. (a) From the deﬁning equation eyD(x)=/summationdisplay nφn(y) n!xn, we see that each application of the operator Dymultiplies the left side by another D(x), so the application of some function f(Dy) will multiply it by f(D(x)). If we choose fto be the inverse function D(−1)(Dy), then we will multiply the left side by D(−1)(D(x)), that is, byx. If we multiply the right side of the deﬁning equation by x, we see that it becomes the egf of {nφn−1(y)}, as claimed. (b) In this family, eyD(x)=1 (1−x)y=/summationdisplay n/parenleftbigg−y n/parenrightbigg (−1)nxn =/summationdisplay ny(y+1 )···(y+n−1) n!xn. Thusφn(y) is the ‘rising factorial’ y(y+1 )···(y+n−1). To check the identity, we have ﬁrst that the deck enumerator is D(x)= −log (1 −x). Hence D(−1)(x)=1−e−x.Therefore D(−1)(Dy)φn(y)={1−e−Dy}φn(y). But Taylor’s theorem from diﬀerential calculus is identical with the Solutions 213 assertion that ( eD)f(y)=f(y+ 1) (!!check this!!). Hence (1−e−Dy)φn(y)=φn(y)−φn(y−1) ={y···(y+n−1)}−{ (y−1)···(y+n−2)} =ny(y+1 )···(y+n−2) =nφn−1(y), as required. 15.The result claimed is certainly true if there is only one card in the deck. Then, by the merge, trickle, and ﬂood argument, it is true in general. Part (b) is immediate. For part (c), insert the factors into the product, and outside the product write the reciprocals of all of those factors, to get p(x)=∞/productdisplay k=1exk k∞/productdisplay k=1(1 +xk k)e−xk k =1 (1−x)∞/productdisplay k=1(1 +xk k)e−xk k. Now asx→1−, the inﬁnite product approaches a certain universal constant, viz. C=∞/productdisplay k=1(1 +1 k)e−1 k, hencep(x)∼C/(1−x). The constant is in fact e−γ, whereγis Euler’s constant. 16.Herecnis the coeﬃcient of xn/n!i n (1 +x)x+1=( 1+x)x(1 +x)=( 1+x)/summationdisplay k/parenleftbiggx k/parenrightbigg xk =( 1+x)/summationdisplay kx(x−1)···(x−k+1 ) k!xk =( 1+x)/summationdisplay kxk k!/summationdisplay r(−1)r/bracketleftbiggk r/bracketrightbigg xr. Thus cn n=(n−1)!/braceleftbigg/summationdisplay k(−1)n−k k!(/bracketleftbiggk n−k/bracketrightbigg −/bracketleftbiggk n−k−1/bracketrightbigg )/bracerightbigg . But in the sum the terms all vanish for k≥n, hence the right side is an integer. 214 Solutions 18.The number of cards in the jth deck is 1 for j=1,2 and is 2 for j≥3. Hence by (3.14.6), the hand enumerator is 1 1−x1 (1−x2)/productdisplay j≥31 (1−xj)2=P(x)2 (1−x)(1−x2) whereP(x) is Euler’s generating function (3.16.3) for {p(n)}. 19.In such a tree there is a rooted tree of j vertices attached to one of the edges incident at the root, and a rooted tree of n−1−jvertices attached to the other edge at the root. Further, the full tree is completely determined by this unordered pair of trees, and so the number anof such full trees is equal to the number of unordered pairs of rooted trees, the total number of whose vertices isn−1, i.e., an=1 2/summationdisplay jtjtn−1−j ifn−1 is odd, for then every unordered pair is counted twice by the sum. Ifn−1 is even then we need to consider the number of ways that the two subtrees at the root can be of the same size ( n−1)/2. The number of unordered pairs of not necessarily distinct objects that can be chosen from a set ofadiﬀerent objects if/parenleftbiga+1 2/parenrightbig . Thus in this case the formula above needs an extra term t(n−1)/2/2 added to it, which is equivalent to the result stated. 20.It is 29. 29 cannot be of the form stated, for otherwise we could subtract some multiple of 15 from it to ﬁnd a nonnegative number of the form 6 x+10y. But 29 is not of that form since it is odd, and 14 isn’t either. Next, if nis any integer that is representable then so is n+ 6, so to see that every integer larger than 29 is so representable it is enough to observe that 30 = 6 ·5, 31 = 6 + 10 + 15, 32 = 6 ·2+1 0 ·2, 33 = 6 ·3+1 5 ·1, 34 = 6 ·4+1 0 ·1, and 35 = 10 ·2+1 5 ·1. 21.Iff(n) is that number then /summationdisplay n≥0f(n)xn=1 (1−x)(1−x2)(1−x3)=1 6(1−x)3+1 4(1−x)2 +17 72(1−x)+1 8(1 +x)+1 9(1−ωx)+1 9(1−¯ωx). If we expand each of the fractions on the right we ﬁnd the formula f(n)=1 6/parenleftbiggn+2 2/parenrightbigg +1 4(n+1 )+17 72+(−1)n 8+2 9cos (2nπ 3) which can be rewritten as f(n)=(n+3 )2 12+−7+9 ( −1)n+ 16 cos (2nπ/3) 72. Solutions 215 The second fraction cannot exceed 32 /72<1/2 in absolute value, so f(n) is the unique integer whose distance from ( n+3 )2/12 is less than 1 /2, as required. 22. For a given a1,a2,···, we putn=a1+2a2+···, and we can then construct all possible hands of the desired type by choosing and labeling cards from the given decks as follows. Make an ordered selection of a1cards of size 1 chosen independently from thed1cards of size 1 that are available in deck 1. Then make an ordered selection of a2cards of size 2 from the d2cards of that size that are available in deck 2, etc. The number of ways in which this can be done is da1 1da2 2···. Next, for the a1chosen cards of size 1, choose the 1 label that will appear on each card, which can be done in n!/(n−a1)! ways, but since the order of these cards in the hand is immaterial, this labeling can be done in only n!/(a1!(n−a1)!) ways. Then, for the a2chosen cards of size 2, choose the unordered pairs of labels that will appear on each card. This can be done in /parenleftbiggn−a1 2/parenrightbigg/parenleftbiggn−a1−2 2/parenrightbigg ···/parenleftbiggn−a1−2a2+2 2/parenrightbigg =(n−a1)! (n−a1−2a2)!2!a2 ways (we need only the unordered pairs because the chosen cards have place- holders on them that tell us in what sequence to place the chosen label set on the card). Finally, since the order of the cards of size 2 in the hand is immaterial, there are only (n−a1)! (n−a1−2a2)!2!a2a2! diﬀerent ways to do this. In general, for the ajchosen cards of size j, we can choose the sequence of sets ofjlabels that will appear on each card in exactly (n−a1−2a2−···− (j−1)aj−1)! (n−a1−2a2−···−jaj)!j!ajaj! diﬀerent ways. If we multiply all of these together, for all j≥1, we ﬁnd that the number of hands of the desired speciﬁcation is n!da1 1da2 2··· 1!a12!a2···a1!a2!···. But this is exactly the coeﬃcient of tnxa1 1xa2 2···/n! in the expansion shown in the statement of the problem. 216 Solutions For part (b), in the family of set partitions we have all dj= 1 forj≥1. Use the result of part (a), with x3=x4=···= 1, since we don’t care about classes of size greater than 2, to obtain the joint distribution of classes of sizes 1 and 2 in the form stated. Answers to problems for chapter 4 1.We havepn=( 1−p)n−1pforn≥1, hence {pn}has the opsgf P(x)= px/(1−(1−p)x). The mean is P/prime(1) = 1/p, and from (4.1.3) the variance is σ2= (logP)/prime+ (logP)/prime/prime/vextendsingle/vextendsingle x=1=(1−p) p2. 2. (a) Consider a sequence of ntrials that yields a complete collection for the ﬁrst time at the nth trial. From that sequence we will construct an ordered partition of the set [ n−1] intod−1 classes, as follows: if the ith photo was chosen at the jth trial (1 ≤i≤d, 1≤j≤n−1), then put jinto theith class of the partition. Note thatd−1 of the classes are nonempty. Conversely, from such an ordered partition of [ n−1] we can construct exactly dcollecting sequences, one for each choice of the coupon that wasn’t collectedin the ﬁrst n−1 trials. There are ( d−1)!/braceleftbig n−1 d−1/bracerightbig ordered partitions of [n−1] intod−1 classes, so there are d!/braceleftbign−1 d−1/bracerightbig sequences of trials that obtain a complete collection precisely at the nth trial. There arednunrestricted sequences of ntrials, so the probability of the event described is as shown. (b) By (1.6.5), p(x)=(d−1)!x/summationdisplay n/braceleftbiggn d/bracerightbigg (x d)n=(d−1)!xd (d−x)···(d−(d−1)x). (c)p/prime(1) =d(1 +1 2+···+1 d) (d) From (4.1.3), σ2=d2d/summationdisplay i=11 i2−d/parenleftbig 1+1 2+···+1 d/parenrightbig . (e) About 29 boxes of cereal, with a standard deviation of about 11 boxes. Solutions 217 3.For part (a), the probability p(j,v1,T) has two components. First, with probability d1/(d1+ 1), the walk begins with a step to another vertex of T1. In that case the probability of a ﬁrst return after jsteps is the same as it was inT1, which gives a contribution of d1 d1+1p(j;v1;T1) to the answer. On the other hand, with probability 1 /(1 +d1) the walk begins by using the edge (v1,v2). In that case the required probability will be the probability that the walk takes exactly j−2 steps in the tree T2, ﬁnishing at v2and then crossing back over the edge ( v1,v2) to vertex v1. Fixm≥0, and consider the following event: the sequence of vertices that the walk visits after crossing to v2contains exactly m+ 1 appearances of vertexv2followed by the return to v1. Hence the sequence looks like v2,W1,v2,W2,...,W m,v2, where each of the Wiis a sequence of vertices of T2−v2. The total number of steps in such a walk is j1+···+jm, where the jiare the numbers of steps between consecutive returns to v2. We need the probability that j1+j2+ ···+jm=j−2. But that is /summationdisplay j1+···+jm=j−2p(j1;v2;T2)p(j2;v2;T2)···p(jm;v2;T2)/parenleftbiggd2 d2+1/parenrightbiggm (1 d2+1) =1 d2+1/parenleftbiggd2 d2+1/parenrightbiggm [xj−2]F2(x;v2;T2)m. If we put it all together, we ﬁnd that p(j;v1;T)i s d1 d1+1p(j;v1;T1)+/summationdisplay m≥0/parenleftbigd2 d2+1/parenrightbigm (d1+ 1)(d2+1 )[xj−2]F2(x;v2)m =d1 d1+1p(j;v1;T1)+1 d1+1[xj−2]1 d2+1−d2F2(x;v2). Finally, if we multiply by xjand sum over j, we obtain the result stated. In part (d) one has Pn(x)=x2/(2−Pn−1(x)) forn≥2, withP1= 1. If one assumes Pn(x)=An(x)/Bn(x), thenAn=x2Bn−1andBn=2Bn−1− x2Bn−2. This leads to the result that Pn(x)=x2/parenleftbiggrn−2 ++rn−2 − rn−1 ++rn−1 −/parenrightbigg (n≥2) 218 Solutions wherer±=1±√ 1−x2. 4.The sequence {e≤m}is obviously generated by E(x) 1−x=N(x−1) 1−x. Sincee≥m=N(0)−e≤m−1, it has the gf N(0) 1−x−xE(x) 1−x=N(0)−xN(x−1) 1−x. 5.The board consists of only the diagonal cells of a full n×nboard. To put knonattacking rooks on this board we can choose any kof thencells on the board, sork=/parenleftbign k/parenrightbig . Then (4.2.17) with j= 0 gives /summationdisplay k(n−k)!/parenleftbiggn k/parenrightbigg (−1)k=n!n/summationdisplay k=0(−1)k k! for the answer, in agreement with (4.2.10). 6.We have /summationdisplay m≥0αmxm=/summationdisplay m≥0xm/summationdisplay r≥m(−1)r−mNr =/summationdisplay r≥0(−1)rNr/summationdisplay 0≤m≤r(−1)mxm =/summationdisplay r≥0(−1)rNr/braceleftbigg1+(−1)rxr+1 1+x/bracerightbigg =1 1+x/braceleftbigg e0+/summationdisplay r≥0Nrxr+1/bracerightbigg =e0+xN(x) 1+x=e0+xE(1 +x) 1+x =e0+x{e0+e1(1 +x)+··· } 1+x. Problem 7 is similar. 8.The sum is 1+x+/parenleftbiggn 1/parenrightbigg (x2+x3)+/parenleftbiggn 2/parenrightbigg (x4+x5)+··· =( 1+x)(1 +/parenleftbiggn 1/parenrightbigg x2+/parenleftbiggn 2/parenrightbigg x4+···) =( 1+x)(1 +x2)n. Solutions 219 Iffm(y) denotes the sum in question, then Snake Oil ﬁnds that /summationdisplay mfm(y)xm=( 1+x)(1 +xy+x2)n. See what happens if you try to extend this to the sum that results from replacing ‘ ⌊r/2⌋’b y‘⌊r/3⌋’ in the sum to be found. 9.The ‘objects’ Ω are the λnpossible ways of assigning colors to the vertices ofG. For each edge eof the graph Gthere is a property P(e); a coloring has propertyP(e) if the two endpoints of edge ehave the same color. We seek the number of objects that have exactly 0 properties. Now consider N(⊇S). For a given set Sof edges, this is the number of colorings such that at least all of the edges in Sare badly colored, i.e., have both endpoints the same color. Think of the graph GSwhose vertices are all nof the vertices of the graph G, together with just the edges in S. If all of the edges in Sare badly colored, and if Cis one of the connected components ofGS, then every vertex in Cmust have the same color. So the number of ways of assigning colors to the vertices of Gsuch that the edges of Sare badly colored is N(⊇S)=λκ(S), whereκ(S) is the number of connected components of the graph GS. Hence P(λ;x;G)=/summationdisplay rNr(x−1)r, where Nr=/summationdisplay |S|=rλκ(S). 10.There areknpossible words, and we take these to be our set of objects Ω. A word has property iif the substring woccurs in the word, beginning in position iof the word. Let Sbe a given subset of properties, i.e., a set of places where the substring wis to begin. We seek N(⊇S), which in this case is the number of words of nletters, chosen from an alphabet of kletters, that have the substring wbeginning in all of the positions indicated by S, and maybe elsewhere too. But there are no such words if two of the elements ofSdiﬀer by<m, for then two occurrences of wwould overlap, contrary to the hypothesis that they cannot do so. Hence we suppose that no two elements of Sdiﬀer by<m. Thenrmof the characters in the word are speciﬁed to be occurrences of w, wherer=|S|. That leaves n−rmcharacters to be speciﬁed, and that can be done in N(⊇S)=kn−rmways. Hence Nriskn−rmtimes the number of subsets S ofrelements of [ n−m+ 1] that have no two entries that diﬀer by <m. But how many such subsets are there? 220 Solutions Consider a subset Sofrelements of [ q], no two of whose entries diﬀer by <m. If we delete the elements of Sfrom [q], the remaining q−rintegers are broken intor+1 intervals of consecutive integers whose lengths are t0,t1,...,t r,s a y , where each ti≥m−1 for 1 ≤i≤r−1. The number of ways to choose such integerst0,...,t ris clearly [xq−r]/braceleftbigg1 1−x/bracerightbigg/braceleftbiggxm−1 1−x/bracerightbiggr−1/braceleftbigg1 1−x/bracerightbigg =[xq−r]/braceleftbiggx(m−1)(r−1) (1−x)r+1/bracerightbigg =[xq−r−(m−1)(r−1)]1 (1−x)r+1 =/parenleftbiggq−(m−1)(r−1) r/parenrightbigg . Thus, since q=n−m+1 ,Nr=kn−rm/parenleftbign−mr+r r/parenrightbig , and the number of w-free words is/summationdisplay r(−1)r/parenleftbiggn−mr+r r/parenrightbigg kn−rm. The Snake Oil method tells us that the answer is also the coeﬃcient of xnin 1 1−kx+xm. In turn this suggests that it might have been easier to do this problem by ﬁnding a recurrence relation that is satisﬁed by the answer, instead of by using the sieve method, but we wanted to show you another example of the sieve method in which the N(⊇S)’s do not depend only on the cardinality of the setS. 13.This is an example where the Snake Oil method doesn’t work immedi- ately because the free parameter nappears too often in the summand. As in example 9, the thing to do is to generalize the problem, in this case to the sum/summationdisplay k(−1)k/parenleftbiggn k/parenrightbigg/parenleftbiggn n−m+k/parenrightbigg . The latter responds nicely to Snake Oil, after multiplying by xmetc. Then setm=n. 17.We ﬁnd in part (a) that ∂ ∂x/integraldisplayA −BF(x,y)dy=/integraldisplayA −B∂F ∂xdy =/integraldisplayA −B∂G ∂ydy =G(x,A)−G(x,−B)→0(A,B→∞ ). Solutions 221 18. (a) Since the sum of the d’s is 2n−2, their average is 2 −2/nwhich is less than 2, so at least one of the d’s must be 1. We can suppose w.l.o.g. that d1= 1. Then, in every tree whose degree sequence is ∆=(d1,...,d n), vertex 1 is connected to exactly one other vertex. There is an obvious 1-1 correspondence between the trees of degree sequence ∆ in which vertex 1 is adjacent to vertex j, for some ﬁxed j≥2, and the trees of n−1 vertices 2,3,...,n , in which the vertex degrees are ( d2,...,d j−1,dj−1,dj+1,...,d n). By induction on n, then, the number whose degree sequence is ∆ is n/summationdisplay j=2(n−3)! (d2−1)!···(dj−1−1)!(dj−2)!···(dn−1)! =n/summationdisplay j=2(n−3)!(dj−1) (d2−1)!···(dj−1)!···(dn−1)! = ((2n−3)−(n−1))(n−3)! (d2−1)!···(dn−1)! =(n−2)! (d1−1)!···(dn−1)! as required. (b) By the multinomial theorem (see exercise 20 of chapter 2), Fn(x1,...,x n)=(x1x2···xn)(x1+···+xn)n−2. (d) Let a tree Thave property iif vertexiis an endpoint. If S⊆[n] then the number of trees of nvertices whose set of properties contains S is N(⊇S)=(n−|S|)n−|S|−2(n−|S|)|S|=(n−|S|)n−2, since the ﬁrst factor is the number of trees of n−|S|vertices and the second factor is the number of ways we can attach the |S|endpoints to such a tree. The result now follows from the sieve. (e) In the sieve method, the average number of properties that an object has is always N1/N, which in this case is (n−1)n−2n nn−2=n(1−1 n)n−2∼n e. 19. (a) Evidently we have for all T N(⊆S)=/summationdisplay V⊆SN(=V), 222 Solutions by deﬁnition. Now substitute this for N(⊆S)) under the summa- tion sign in the expression given on the right side of the statement of the problem, interchange the order of summation and verify the resulting identity. (b) It is /summationdisplay n≥0hn(S) n!xn=/productdisplay s∈Sexpds s!xs. Answers to problems for chapter 5 1.Lett=1/A,φ(u)=1+uL, andf(u)=u. Then the equation u=tφ(u) that is treated by the LIF becomes the present equation. The result follows after a small calculation involving the binomial theorem. 3.In the LIF, choose the function f(u) that satisﬁes f/prime(u)=1/φ(u). Then 1 n[un−1]/braceleftbigg f/prime(u)φ(u)n/bracerightbigg =1 u[un−1]φ(u)n−1. On the other hand, if we write z(t)=f(u(t)), then [tn]f(u(t)) = [tn]z(t)=1 n[tn−1]z/prime(t)=1 n[tn−1]tu/prime(t) u(t). For the last equality of the problem, diﬀerentiate u=tφ(u) with respect to t. 4. (a) Putφ(u)=1+u+u2in the result of the previous problem to ﬁnd thatγn=[tn]1 1−t(1+2u), whereu=u(t) satisﬁesu=t(1 +u+u2). By solving the quadratic equation for uand substituting, we ﬁnd the result stated. (b) Letx=i/√ 3 in problem 2. 5. (a) Clearly Sp(n)=[xn]/braceleftbigg (1 +x)pn/(1−x)/bracerightbigg . (b) Takeφ(u)=( 1+u)pandf/prime(u)=1/((1−u)(1 +u)p) in the LIF, Solutions 223 and ﬁnd Sp(n) n+1=1 n+1[un](1 +u)p(n+1) (1−u)(1 +u)p =[tn+1]f(u(t)) =1 n+1[tn]{f/prime(u(t))u/prime(t)} =1 n+1[tn]u/prime(t) (1−u)(1 +u)p =1 n+1[tn]tu/prime(t) (1−u(t))u(t). Sinceu=t(1 +u)p, we ﬁndu/prime=u(1+u) t(1−(p−1)u), and substitution leads to the result stated. (c) Equate coeﬃcients of xnon both sides of the result of part (b). 6.Letxn1,...,xnkbe the powers of xwhose coeﬃcients in P(x) are strictly positive. Then, by Schur’s theorem 3.15.2, what is needed is that gcd(n1,...,n k)=1. 7. (a) It is (1 +na)n∼nnaexp (n1−a−n1−2a/2+···), where the argument of the exponential terminates after the last positive exponent of nis reached. (b) As above without the factor nna. (c) It is ∼1. 8. 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