Enumerative & Asymptotic Combinatorics

Size: px

Start display at page:

Download "Enumerative & Asymptotic Combinatorics"

Dominick Rich
6 years ago
Views:

1 C50 Eumerative & Asymptotic Combiatorics Notes 4 Sprig 2003 Much of the eumerative combiatorics of sets ad fuctios ca be geeralised i a maer which, at first sight, seems a bit umotivated I this chapter, we develop a small amout of this large body of theory Motivatio We ca loo at -aalogues i several ways: The -aalogues are, typically, formulae which ted to the classical oes as 1 Most basic is the fact that a 1 lim 1 1 = a for ay real umber a (this is immediate from l Hôpital s rule) There is a formal similarity betwee statemets about subsets of a set ad subspaces of a vector space, with cardiality replaced by dimesio For example, the iclusio-exclusio rule for sets becomes U V + U V = U + V dim(u +V ) + dim(u V ) = dim(u) + dim(v ) for vector spaces Now, if the uderlyig field has elemets, the the umber of 1-dimesioal subspaces of a -dimesioal vector space is ( 1)/( 1), which is exactly the -aalogue of 1

2 The aalogy ca be iterpreted at a much higher level, i the laguage of braided categories I will ot pursue this here You ca read more i various papers of Shah Majid, for example Braided Groups, J Pure Appl Algebra 86 (1993), ; Free braided differetial calculus, braided biomial theorem ad the braided expoetial map, J Math Phys 34 (1993), I coectio with the secod iterpretatio, ote the theorem of Galois: Theorem 1 The cardiality of ay fiite field is a prime power Moreover, for ay prime power, there is a uiue field with elemets, up to isomorphism To commemorate Galois, fiite fields are called Galois fields, ad the field with elemets is deoted by GF() [ Defiitio The Gaussia coefficiet, or -biomial coefficiet,, where ad are atural umbers ad a real umber differet from 1, is defied by Propositio 2 [ (a) lim 1 = ( 1)( 1 1) ( +1 1) ( 1)( 1 1) ( 1) [ = ( ) (b) If is a prime power, the the umber of -dimesioal [ subspaces of a - dimesioal vector space over GF() is eual to Proof The first assertio is almost immediate from lim 1 ( 1)/( 1) = For the secod, ote that the umber of choices of liearly idepedet vectors i GF() is ( 1)( ) ( 1 ), sice the ith vector must be chose outside the spa of its predecessors Ay such choice is the basis of a uiue -dimesioal subspace Puttig =, we see that the umber of bases of a -dimesioal space is ( 1)( ) ( 1 ) Dividig ad cacellig powers of gives the result 2

3 The -biomial theorem The -biomial coefficiets satisfy a aalogue of the recurrece relatio for biomial coefficiets [ [ [ [ [ 1 1 Propositio 3 = = 1, = + for 0 < < 0 1 Proof This comes straight from the defiitio Suppose that 0 < < The [ [ ( 1 )[ 1 1 = ( = )[ [ = 1 The array of Gaussia coefficiets has the same symmetry as that of biomial coefficiets From this we ca deduce aother recurrece relatio Propositio 4 (a) For 0, [ [ = (b) For 0 < <, [ [ [ 1 1 = + 1 Proof (a) is immediate from the defiitio For (b), [ [ = [ [ 1 1 = + 1 [ [ 1 1 = + 1 We come ow to the -aalogue of the biomial theorem, which states the followig 3

4 Theorem 5 For a positive iteger, a real umber 1, ad a idetermiate z, we have (1 + i 1 z) = =0 ( 1)/2 z [ Proof The proof is by iductio o ; startig the iductio at = 1 is trivial Suppose that the result is true for 1 For the iductive step, we must compute ( 1 [ ) 1 (1 ( 1)/2 z + 1 z ) =0 The coefficiet of z i this expressio is [ [ 1 1 ( 1)/2 + ( 1)( 2)/ ( [ 1 = ( 1)/2 [ ) = ( 1)/2 [ by Propositio 4(b) Elemetary symmetric fuctios I this sectio we touch briefly o the theory of elemetary symmetric fuctios Let x 1,,x be idetermiates For 1, the th elemetary symmetric fuctio e (x 1,,x ) is the sum of all moomials which ca be formed by multiplyig together distict idetermiates Thus, e has ( ) terms, ad ( ) e (1,1,,1) = For example, if = 3, the elemetary symmetric fuctios are e 1 = x 1 + x 2 + x 3, e 2 = x 1 x 2 + x 2 x 3 + x 3 x 1, e 3 = x 1 x 2 x 3 We adopt the covetio that e 0 = 1 Newto observed that the coefficiets of a polyomial of degree are the elemetary symmetric fuctios of its roots, with appropriate sigs: 4

5 Propositio 6 (z x i ) = =0 ( 1) e (x 1,,x )z Cosider the geeratig fuctio for the e : E(z) = e (x 1,,x )z =0 A slight rewritig of Newto s Theorem shows that E(z) = (1 + x i z) Hece the biomial theorem ad its -aalogue give the followig specialisatios: Propositio 7 so (a) If x 1 = = x = 1, the E(z) = (1 + z) = e (1,1,,1) = =0 ( ) z, ( ) (b) If x i = i 1 for i = 1,,, the so E(z) = (1 + i 1 z) = Partitios ad permutatios =0 ( 1)/2 z [, e (1,,, 1 ) = ( 1)/2 [ The umber of permutatios of a -set is! The liear aalogue of this is the umber of liear isomorphisms from a -dimesioal vector space to itself; this is eual to the umber of choices of basis for the -dimesioal space, which is ( 1)( ) ( 1 ) These liear maps form a group, the geeral liear group GL(,) Usig the -biomial theorem, we ca trasform this multiplicative formula ito a additive formula: 5

6 Propositio 8 GL(,) = ( 1) ( 1)/2 i=0( 1) (+1)/2[ Proof We have GL(,) = ( 1) ( 1)/2 (1 i ), ad the right-had side is obtaied by substitutig z = i the -biomial theorem The total umber of matrices is 2, so the probability that a radom matrix is ivertible is As, we have p () = (1 i ) p () p() = (1 i ) i 1 Accordig to Euler s Petagoal Numbers Theorem, we have p() = ( 1) (3 1)/2 = Z So, for example, p(2) = is the limitig probability that a large radom matrix over GF(2) is ivertible What is the -aalogue of the Stirlig umber S(,), the umber of partitios of a -set ito parts? This is a philosophical, ot a mathematical uestio; I argue that the -aalogue is the Gaussia coefficiet [ The umber of surjective maps from a -set to a -set is!s(,), sice the preimages of the poits i the -set form a partitio of the -set whose parts ca be mapped to the -set i ay order The -aalogue is the umber of surjective liear maps from a -space V to a -space W Such a map is determied by its erel U, a ( )- dimesioal subspace of V, ad a liear isomorphism from V /U to W So the aalogue of S(,) is the umber of choices of U, which is [ = [ 6

7 Irreducible polyomials Though it is ot really a -aalogue of a classical result, the followig theorem comes up i various places Recall that a polyomial of degree is moic if the coefficiet of x is eual to 1 Theorem 9 The umber f () of moic irreducible polyomials of degree over GF() satisfies f () = Proof We give two proofs, oe depedig o some algebra, ad the other a rather ice exercise i maipulatig formal power series First proof: We use the fact that the roots of a irreducible polyomial of degree over GF() lie i the uiue field GF( ) of degree over GF() Moreover, GF( ) GF( ) if ad oly if ; ad every elemet of GF( ) geerates some subfield over GF(), which has the form GF( ) for some dividig Now each of the elemets of GF( satisfies a uiue miimal polyomial of degree for some ; ad every irreducible polyomial arises i this way, ad has distict roots So the result holds Secod proof: All the algebra we use i this proof is that each moic polyomial of degree ca be factorised uiuely ito moic irreducible factors If the umber of moic irreducibles of degree is m, the we obtai all moic polyomials of degree by the followig procedure: Express = a, where a are o-egative itegers; Choose a moic irreducibles of degree from the set of all m such, with repetitios allowed ad order ot importat; Multiply the chose polyomials together Altogether there are moic polyomials x +c 1 x 1 + +c of degree, sice there are choices for each of the coefficiets Hece ( ) m + a 1, (1) = where the sum is over all seueces a 1,a 2, of atural umbers which satisfy a = 7 a

8 Multiplyig by x ad summig over, we get 1 1 x = x 0 = a 1,a 2, = 1 a 0 1 ( m + a 1 a ( m + a 1 a = (1 x ) m 1 ) x a ) (x ) a Here the maipulatios are similar to those for the sum of cycle idices i Chapter 2; we use the fact that the umber of choices of a thigs from a set of m, with repetitio allowed ad order uimportat, is ( m+a 1) a, ad i the fourth lie we ivoe the Biomial Theorem with egative expoet Taig logarithms of both sides, we obtai x 1 = log(1 x) = m log(1 x ) 1 x = m r 1 r 1 r The coefficiet of x i the last expressio is the sum, over all divisors of, of m /r = m / This must be eual to the coefficiet o the left, which is / We coclude that = m, (2) as reuired Note how the very complicated recurrece relatio (1) for the umbers m chages ito the much simpler recurrece relatio (2) after taig logarithms! We will see how to solve such a recurrece i the sectio o Möbius iversio 8

Math 155 (Lecture 3)

Math 155 (Lecture 3) Math 55 (Lecture 3) September 8, I this lecture, we ll cosider the aswer to oe of the most basic coutig problems i combiatorics Questio How may ways are there to choose a -elemet subset of the set {,,,