MT5821 Advanced Combinatorics

Transcription

1 MT5821 Advaced Combiatorics 3 Formal power series Geeratig fuctios are the most powerful tool available to combiatorial eumerators. This week we are goig to look at some of the thigs they ca do. 3.1 Commutative rigs with idetity I studyig formal power series, we eed to specify what kid of coefficiets we should allow. We will see that we eed to be able to add, subtract ad multiply coefficiets; we eed to have zero ad oe amog our coefficiets. Usually the itegers, or the ratioal umbers, will work fie. But there are advatages to a more geeral approach. A favourite object of some group theorists, the so-called Nottigham group, is defied by power series over a fiite field. A commutative rig with idetity is a algebraic structure i which additio, subtractio, ad multiplicatio are possible, ad there are elemets called 0 ad 1, with the followig familiar properties: additio ad multiplicatio are commutative ad associative; the distributive law holds, so we ca expad brackets; addig 0, or multiplyig by 1, do t chage aythig; subtractio is the iverse of additio; 0 1. Examples icude the itegers Z (this is i may ways the prototype); ay field (for example, the ratioals Q, real umbers R, complex umbers C, or itegers modulo a prime p, F p. Let R be a commutative rig with idetity. A elemet u R is a uit if there exists v R such that uv = 1. The uits form a abelia group uder the operatio of multiplicatio. Note that 0 is ot a uit (why?). 1

2 3.2 Formal power series A formal power series is, iformally, a expressio of the form a(x) = a 0 + a 1 x + a 2 x 2 + a 3 x 3 +, like a polyomial but cotiuig for ever. However, if you start askig what x is, or what the ifiite sum meas, you will quickly ru ito trouble. So we proceed i a more careful mathematical maer. Oe spi-off is that we get a defiitio of a polyomial at o extra cost. Defiitio A formal power series is a ifiite sequece (a 0,a 1,a 2,...) of elemets take from a commutative rig with idetity R. A polyomial is a formal power series (a 0,a 1,a 2 ) for which there is some atural umber such that a i = 0 for i > ; the smallest such is the degree of the polyomial. We always thik of a formal power series as represeted i the form a 0 + a 1 x + a 2 x 2 + a 3 x 3 + = a x. At the momet, this is just a coveiet way of writig it, but we will see that it is ot ucoected with the otio of power series i aalysis. We use laguage based o this; we ofte call a the coefficiet of x, ad refer to a 0 as the costat term of the power series. There are two typical occurreces of formal power series i combiatorial eumeratio. Suppose we have umbers b 0,b 1,..., for which b solves some coutig problem o a set of size (for example, the umber of partitios of the set). The we defie the ordiary geeratig fuctio or o.g.f. of the umbers is the formal power series (b 0,b 1,b 2,b 3,...) = b x ; the expoetial geeratig fuctio or e.g.f. of the umbers is the formal power series b x (b 0 /0!,b 1 /1!,b 2 /2!,b 3 /3!,...) =!. 2

3 The reaso for the ame expoetial geeratig fuctio is the fact from aalysis that the Taylor series of the expoetial fuctio is e x x =!, the expoetial geeratig fuctio of the very simple series 1,1,1,... of umbers. Icidetally, we will write the expoetial fuctio as exp(x) rather tha e x. 3.3 Operatios o formal power series There are may operatios o formal power series; this is what gives them their flexibility ad wide applicability. Here we meet some of these. I every case, although our iformal way of writig power series appears to ivolve a ifiite sum, the defiitios oly ever ask us to add or multiply a fiite umber of elemets. Additio or said otherwise, We defie the sum of two formal power series term by term: (a 0,a 1,...) + (b 0,b 1,...) = (a 0 + b 0,a 1 + b 1,...), a x + b x = (a + b )x. From ow o we will just use the secod form; but we ca always go back to the first form if required. Multiplicatio We defie the product of two formal power series by the covolutio formula ( ) a x )( b x = c x, where c = a k b k. k=0 Note that this is just what the iformal represetatio of power series would lead you to expect, though it appears somewhat umotivated if power series are writte as ifiite sequeces. 3

4 Propositio 3.1 Let R be a commutative rig with idetity. The the set R[[x]] of all formal power series over R is a commutative rig with idetity, ad the subset R[x] of all polyomials over R is also a commutative rig with idetity. We will ot stop to prove this, which just ivolves somewhat tedious verificatio of axioms. But there is much more that formal power series ca do! Example: Summig a geometric series. Verify that (1 ax) a x = 1, so it makes sese to write a x = 1 1 ax = (1 ax) 1. Ifiite sums Let a (0),a (1),a (2),... be formal power series over R. Suppose that the followig coditio holds: for ay 0, there exists m 0 such that the first coefficiets of a (i) are all zero if i > m. The the ifiite sum a (0) + a (1) + a (2) + = a (i) i 0 makes sese, sice if we look i positio (the coefficiet of x ), there are oly fiitely may o-zero coefficiets to add. (We adopt the covetio that addig zero ifiitely ofte has o effect.) Various ice properties of ifiite sums, such as the ifiite distributive law, ow hold. Note that our iterpretatio of the formal power series (a 0,a 1,...) as the ifiite sum a x is justified by this defiitio, sice a x is itself a very simple formal power series (with a i positio ad zero elsewhere), ad these series satisfy our coditio for a ifiite sum to exist. 4

5 Ifiite products Just as addig 0 ifiitely ofte is supposed to leave thigs uchaged, similarly multiplyig by 1 ifiitely ofte does ot chage thigs. As a result, if a (0),a (1),a (2),... satisfy our coditio for ifiite sums to exist, the the ifiite product (1 + a (0) )(1 + a (1) ) = (1 + a () ) is defied. For we get a term i x i the product by selectig terms where the expoets of x sum to, ad choosig 1 from all the remaiig factors; ad by assumptio, oly fiitely may terms ca be obtaied i this way. This is less familiar tha ifiite sums, so let us have a small example. Cosider (1 + x m ). m 1 If we write this as a power series a x, what is the coefficiet a? We obtai a term i x by writig = m 1 +m m r, where m 1,m 2,...,m r are distict positive itegers, ad choosig x m i from the bracket 1 + x m i for i = 1,...,r ad 1 from all the other brackets. So Thus, The coefficiet a is the umber of expressios for as a sum of distict positive itegers. (1 + x m ) = 1 + x + x 2 + 2x 3 + 2x 4 + 3x 5 + 4x 6 +, m 1 where the coefficiet 4 of x 6 comes from the expressios 6 = = = Differetiatio We ca differetiate formal power series; o calculus ivolved, except that we steal from calculus the idea that the derivative of x is x 1. So d dx a x = a x 1. 1 We usually write the differetial operator as D or D x istead of d/dx. 5

6 Now the usual rules for differetiatig a sum or product of two fuctios hold. Ideed, Leibiz rule for arbitrary derivatives of a product hold: ( ) D ( f g) = D k f D k g, k as is easily proved by iductio. k=0 Substitutio Whe ca we substitute a power series g ito a power series f? If f = a x, we would like to have f (g) = a g. So we eed the powers g to satisfy the coditio for ifiite sums to exist, that is, o-zero terms are pushed further to the right i higher powers. This will hold if the costat term of g is equal to 0, sice the the first terms of g will all be zero. So the rule is: Substitutio f (g) is defied if the costat term of g is equal to zero. Now everythig you would expect to hold does hold, icludig the Chai Rule: D( f (g)) = (D f )(g) Dg. More geerally, Faà di Bruo s formula describes the th derivative of f (g), for ay two formal power series f ad g. I wo t descibe it here. (To my kowledge, Faà di Bruo is the oly mathematicia who was also a Catholic sait; amog his studets were C. Segre ad G. Peao.) 3.4 Elemetary properties Multiplicative iverses Propositio 3.2 A formal power series a x is a uit i the rig R[[x]] if ad oly if its costat term a 0 is a uit i the rig R. 6

7 Proof Suppose that ( ) a x )( b x = 1. The, cosiderig the costat term, we see that a 0 b 0 = 1, so that a 0 is a uit. Coversely, suppose that a 0 is a uit, ad we wat to fid a formal power series b x satisfyig the above displayed equatio. We see that a 0 b 0 = 1, so b 0 is the iverse of a 0, which is uique (Why?) For > 0, we wat k=0 a k b k = 0. Movig all terms except a 0 b to the right ad multiplyig by b 0 (the iverse of a 0 ), we see that ( b = b 0 a k b k ). k=1 Thus b is determied by the kow coefficiets a i ad the coefficiets b 0,b 1,...,b 1. These ca be foud recursively, obtaiig the iverse of the give power series. Example: Fiboacci umbers the recurrece relatio The Fiboacci umbers F 0,F 1,F 2,... satisfy F = F 1 + F 2 for 2. There is some argumet about what the first two terms are: three commo covetios are F 0 = 0, F 1 = 1; F 0 = F 1 = 1; F 0 = 1, F 1 = 2. Clearly, oce the first two terms are chose, the rest of the sequece is determied; ad the sequeces resultig from these three covetios differ oly by a shift. All three covetios have somethig to recommed them. The secod ad third solve coutig problems: the secod gives the umber of compositios of as a sum of oes ad twos, while the third gives the umber of biary sequeces with o two cosecutive oes. [Exercise: Prove these!] The first covetio has o 7

8 such simple coutig iterpretatio, but has the ice properties that gcd(f m,f ) = F gcd(m,), ad that F is prime oly if is prime; also that F 12 = 12 2 (the oly time this happes apart from F 1 = 1 2 ). Fiboacci s famous rabbits seem to follow the secod covetio, but Fiboacci actually used the third covetio, as this page from his book Liber Abaci shows: the calculatio o the right of the picture below (from Wikipedia) gives F 12 = 377. Here, I will adopt the middle covetio. Let φ(x) = F x, ad calculate the coefficiet of x i (1 x x 2 )φ(x) = (1 x x 2 )(1 + x + 2x 2 + ). The costat term is clearly 1, ad the coefficiet of x is 0. For 2, the coefficiet of x is F F 1 F 2 = 0. So (1 x x 2 )φ(x) = 1, ad φ(x) = (1 x x 2 ) 1. From this we ca fid a formula for the Fiboacci umbers. Write 1 x x 2 = (1 αx)(1 βx), 8

9 where α ad β are roots of the quadratic y 2 y 1 = 0. The write (1 x x 2 ) 1 as partial fractios: 1 1 x x 2 = A 1 αx + B 1 βx = A α x + B β x, by solvig two liear equatios for A ad B. The we ca read off the formula: F = Aα + Bβ. We see that, for ay sequece satisfyig the Fiboacci recurrece for 2, whatever the first two terms are, the geeratig fuctio will have the form (a + bx)(1 x x 2 ) 1. So we will obtai a formula of the same shape, with the same α ad β, but with a differet A ad B. Example: Coected permutatios The purpose of this example is to covice you that eve power series with zero radius of covergece ca cout somethig iterestig. A permutatio π of the set {1,...,} is said to be coected if there does ot exist a umber k with 0 < k < such that π permutes the umbers 1,...k amog themselves (ad so also the umbers k + 1,..., amog themselves). Let c be the umber of coected permutatios. Ca we calculate c? Give ay permutatio π, let k be the smallest positive umber for which π permutes amog themselves the umbers 1,...,k. Thus π is coected if ad oly if k =. Now π is obtaied by stitchig together a coected permutatio of {1,...,k} ad a arbitrary permutatio of {k + 1,...,}. Thus,! = c k ( k)!. k=1 Put A(x) =!x (a power series which coverges oly for x = 0) ad C(x) = 1 c x. What is the coefficiet of x i C(x)A(x)? It is clearly 1 for = 0. If 1 > 0, this coefficiet is! k=1 So C(x)A(x) = 1, ad C(x) = A(x) 1. c k ( k)! = 0. 9

10 Substitutio iverses Recall that, to substitute g ito f, we require that g has costat term 0. Iverses for substitutio should the satisfy f (g(x)) = x = g( f (x)). For this we require that both f ad g have zero costat term. Arguig i a similar way to the last result, we eed the coefficiet of x to be a uit i R, ad this coditio is also sufficiet. Lagrage iversio, which we may or may ot get to, gives a formula for the iverse of a power series satisfyig these two coditios (costat term zero ad coefficiet of x a uit). I particular, the set of formal power series x+ (costat term 0, coefficiet of x equal to 1) forms a group uder the operatio of compositio. This group is sometimes referred to as the Nottigham group. 3.5 Coectio with aalysis We ve see that we ca maipulate formal power series without payig ay regard to whether or ot they coverge we ca eve look at formal power series over rigs where covergece would make o sese. But the good ews is that, over the real or complex umbers, if our series are coverget for some o-zero values of x, the we ca use all the tools of aalysis o them. The most importat case of this is the followig priciple: Ay idetity betwee real or complex power series, ivolvig additio, multiplicatio (possibly ifiite sums ad products) ad substitutio, is a idetity i the rig of formal power series. This is because of the uiqueess of the Taylor series for a aalytic fuctio. We will see may examples later. Aother, less commo but occasioally useful, is the use of Cauchy s formula to extract the coefficiets of a power series. If f (z) = a z is aalytic i a disc, the a = 1 f (z) dz, 2πi z+1 where the cotour cotais the origi ad is cotaied i the disc of covergece. 10

11 ( ) a The biomial series We already defied the biomial coefficiet for ay k real or complex umber a ad ay o-egative iteger k, by the rule ( ) a a(a 1) (a k + 1) =. k k! Now the Biomial Theorem gives the Taylor series of the aalytically defied fuctio (1 + x) a (which is aalytic i the ope uit disc): Theorem 3.3 (Biomial Theorem for arbitrary expoet) ( ) a (1 + x) a = x k. k 0 k This is a theorem of aalysis, sice the fuctio (a + x) a has o combiatorial defiitio. Ideed, we ca take it as a defiitio of the fuctio (1 + x) a. Nevertheless, it is very useful to us. For example, it ca be proved aalytically that (1 + x) a (1 + x) b = (1 + x) a+b is valid i the uit disc; so it is a idetity for formal power series, by our geeral priciple. This meas that the Vadermode covolutio ( ) ( )( ) a + b a b = k k k=0 holds for ay real or complex values of a ad b. The expoetial ad logarithmic series These are two further importat series: exp(x) = x!, log(1 + x) = ( 1) x. 1 The first is coverget everywhere, the secod iside the uit disc. They satisfy all the familiar idetities, by our geeral priciple: for example, log(1 + x) a = a log(1 + x), exp(log(1 + x)) = 1 + x, log(1 + (exp(x) 1) = x. 11

12 Ofte it is coveiet to use a slightly differet form of the logarithmic series: x log(1 x) = 1. Note that we ca oly substitute a power series ito aother if its costat term is zero. So the slightly covoluted form of the third idetity is because exp(x) has costat term 1, so we must subtract 1 before we substitute it; ad i the first idetity, we have to observe that (1 + x) a = 1 + g(x), where g is a formal power series with costat term 0. The logarithm has the property that it coverts products ito sums; i fact, it coverts ifiite products ito ifiite sums. I the ext sectio, I will use this property, for series with o-zero radius of covergece, ad so I ca appeal to our aalytic priciple. But i fact it holds geerally, ad so is a combiatorial property, which really requires a combiatorial proof! 3.6 Further examples We ed this chapter with a couple of examples to show geeratig fuctios at work: to show the existece of fiite fields, ad to fid a explicit formula for the umber of bracketigs of a o-associative product. The first ivolves takig logs, the secod the biomial theorem with o-iteger expoet. Irreducible polyomials Before we pluge i to this topic, we eed a elemetary combiatorial result. We kow that the umber of ways of choosig k objects from a set of size, if we are ot( allowed ) to repeat a object i the selectio ad we do t care about the order, is. What if we are allowed repetitios? k Propositio 3.4 The umber of selectios ( of k objects) from a set of, with repetitios allowed ad order ot importat, is. + k 1 k The proof is i the appedix. Galois showed that there exists a fiite field of ay give prime power order. The way to costruct a fiite field of order q, if you kow oe of order q, is to fid a irreducible polyomial of degree over the smaller field, ad adjoi a root of it. (You ca take q to be prime here, sice the the itegers mod q will do for 12

13 a startig field.) So we eed to prove that there is a irreducible polyomial of degree over a field of order q, for ay ad q. How do we do this? By coutig the irreducible polyomials! We eed oly cosider moic polyomials. We regard q as fixed, ad the startig field F of order q also as fixed. Let a be the umber of irreducible polyomials of degree over F. The total umber of moic polyomials of degree is q. We have a uique factorisatio theorem to say that each is uiquely a product of irreducibles. Suppose that there are m 1 factors of degree 1, m 2 of degree 2, ad so o, with m 1 + 2m 2 + =. The umber of ways of choosig m( i irreducibles ) from a ai + m i 1 set of size a i (with replacemet, ad order uimportat) is. So m 1 +2m 2 + = i 1 ( ) ai + m i 1 = q. This is a recurrece relatio, sice there is a term a (correspodig to irreducible polyomials) ad all the other terms ivolve as with smaller idex. But it looks like a ightmare to solve! But let s pluge i qx = q x = x = m 1,m 2,... m i m 1 +2m 2 + = i 1 i 1 ( ai + m i 1 m i ( ) ai + m i 1 m i ) x im i. I the last lie, istead of summig over, ad the for each summig over m 1,m 2,... satisfyig m 1 + 2m 2 + =, we simply sum over all m 1,m 2,... idepedetly; ad we split the x over the terms i the sum as show. Now I claim that the last sum is equal to i 1 m 0 ( ai + m 1 m ) x im. For this expressio is a product of sums; we expad the brackets ad get a sum of terms, oe from each bracket (but to make thigs work, choosig 1 from all but fiitely may of the brackets). Note that our ifitie product coditio is satisfied. 13 m i

14 Now a simple calculatio shows that ( ) ( ) a + m 1 a = ( 1) m ; m m so fially we obtai or (1 qx) 1 = (1 x i ) a i. i 1 At this poit, we ca take logs, ad fid log(1 qx) = a i log(1 x i ), i 1 (qx) 1 Equatig coefficiets of powers of x gives or x = a i ik i 1 k 1 k. q = a i /i, i q = ia i. i This is a much simpler recurrece! It is easy to see directly that we must have a i > 0, so that irreducible polyomials exist. Moreover, Möbius iversio, which we meet later, ca be used to solve it explicitly for a. Fially, it is possible to show that there are just eough irreducible polyomials to build oe fiite field of order q ; so the uiqueess follows as well. Just to see how much simpler the ew recurrece is, here is the origial recurrece for q = 2, = 5: a 5 + a 4 a 1 + a 3 a 2 + a 3 ( a ) ( a ) a 1 + a 2 ( a ) ( ) a = 2 5 ; 5 if you kow a 1 = 2, a 2 = 1, a 3 = 2, a 4 = 3 ad have a bit of perseverace, you ca fid that a 5 = 6. The ew recurrece is simply 5a 5 + a 1 =

15 Bracketigs of a formula Suppose that is a biary operatio which is ot ecessarily associative, so that the value of a product of terms might deped o the way the terms are bracketed. How may differet bracketigs are there? For example, for = 4, there are five bracketigs: ((a b) c) d,(a (b c)) d,(a b) (c d),a ((b c) d),a (b (c d)). Let C be the umber of bracketigs. Propositio 3.5 C = 1 ( ) I do ot kow ay simple direct way to prove this formula. We will detour via geeratig fuctios. Let F(x) = C x. 1 Sice ay bracketig of terms (for > 1) is obtaied by bracketig the first k ad the last k ad fially combiig the results, we have C = 1 C k C k for > 1. k=1 We see o the right the covolutio product, so that F(x) 2 is almost the same as F(x): they agree for all powers of x greater tha the first. But F(x) has a term x, whereas F(x) 2 begis with x 2. So F(x) = x + F(x) 2. Now this is a quadratic equatio F 2 F + x = 0, which we ca solve: F(x) = 1 2 (1 ± 1 4x). Which sig to choose? Sice the costat term is zero, we must choose the egative sig: F(x) = 1 2 (1 1 4x). Now we ca obtai C by extractig the coefficiet of x, usig the biomial theorem with expoet 1 2 : C = 1 ( ) 1/2 ( 4) 2 15

16 = (2 2)! = = 1 ( 1)!! ( ). (2 3) 2 ( 4) /! The umbers C are the Catala umbers, about which we shall have a lot more to say. 3.7 Appedix: selectio with repetitio Propositio 3.6 The umber of selectios ( of k objects) from a set of, with repetitios allowed ad order ot importat, is. + k 1 k Proof To specify such a selectio, we eed to say, for each of the objects o 1,...,o, the umber of times it is chose. These umbers x 1,...,x are oegative itegers ad sum to k. So we( have to prove ) that ( the umber ) of choices of + k 1 + k 1 o-egative itegers with sum k is =. k 1 To do this, take a row of + k 1 boxes i a row, ad put barriers ito 1 of the boxes. The take x 1 to be the umber of boxes before the first barrier, x 2 the umber betwee the first ad secod barriers,..., ad x the umber of boxes after the last barrier. Clearly the required coditios hold. Coversely, if x 1,...,x are o-egative with sum k, the put the first barrier after x 1 empty boxes, the secod after aother x 2 empty boxes, ad so o. So the ( required) umber is the umber of choices of positios for the barriers, + k 1 that is,. 1 For example, if = 4 ad k = 5, the the picture correspods to x 1 = 3, x 2 = 0, x 3 = 1, x 4 = 1, i other words, to the selectio [o 1,o 1,o 1,o 3,o 4 ]. 16

17 Exercises 3.1. Now is the time to have a serious attempt at the last questio o Sheet I the equatio exp(x +y) = exp(x) exp(y), calculate the terms of total degree i the two variables, ad deduce the Biomial Theorem for positive iteger expoets We saw that F = Aα + Bβ. Calculate α ad β, ad A ad B for each of the three covetios i the text Show that ay sequece satisfyig the Fiboacci recurrece F = F 1 +F 2 for 2 is give by the formula i the precedig questio, with the same values of α ad β but possibly differet A ad B Let A(x) be the geeratig fuctio for a sequece (a ). Show that the geeratig fuctio for the sequece of partial sums is A(x)/(1 x) Let F(x) = (1 x ) 1. 1 (a) If F(x) = a x, show that a is the umber of partitios of, expressios for as a sum of positive itegers, where the order is ot sigificat. (b) Let F(x) 1 = b x. Give a descriptio of the coefficiets b. (c) Work out the first few umbers b. What do you fid? 3.7. Fid the iverse (uder substitutio) of the power series x + x 2. If your aswer is the series f (x), calculate the first few terms of f (x) + f (x) 2 to check your result I a electio, there are two cadidates, A ad B. The umber of votes cast is 2, ad each cadidate receives votes. Show that the umber of ways i which the votes ca be couted so that at every itermediate stage, A has more votes tha B, is the Catala umber C. [Hit: Let the umber be f (). A leads by just oe vote after the first vote is couted. Suppose that this ext happes after 2i + 1 votes have bee couted. Show that there are f (i) ways to cout the votes before this poit, ad f ( i) ways after this poit.] Now show that the umber of ways i which the votes ca be couted so that at every itermediate stage, A has at least as may votes as B, is C +1. [Hit: Give 17

18 A oe extra vote at the start ad B oe extra vote at the ed; the the coditios of the precedig paragraph are satisfied.] 3.9. A clow stads o the edge of a swimmig pool, holdig a bag cotaiig red ad blue balls. He draws the balls out oe at a time ad discards them. If he draws a blue ball, he takes oe step back; if a red ball, oe step forward. (All steps have the same size.) Show that the probability that the clow remais dry is 1/( + 1) [*] A simple versio of the QUICKSORT algorithm to sort a list L ito the correct order ca be specified recursively as follows: Let a be the first elemet of the list, ad partitio the remaider ito L (the elemets less tha a) ad L + (the elemets greater tha a). Sort L ad L +. Retur (sort(l ),a,sort(l + )). Suppose that q is the average umber of comparisos required to sort a list L of legth (i radom order). Prove that q = k 1 + q k ) = 1 + k=1(q 2 1 q k. k=0 [Hit: a is equally likely to be the smallest, secod,..., largest elemet of L.] Let Q(x) = q x. Multiply the recurrece relatio by x ad sum: q x = ( 1)x + 2 Deduce that xq (x) = 2x2 (1 x) 3 + 2x 1 x Q(x). Solve this differetial equatio to obtai Deduce that Q(x) = q = 2( + 1) i=1 2(x + log(1 x)) (1 x) 2. ( 1 q i )x. i=0 ( ) 1 4 = 2log + O(). i 18