Generating Functions

Transcription

1 Geeratig Fuctios Geeratig Fuctios are a powerful tool i combiatorics Wilf describes them, quite appropriately, as a clotheslie o which we hag up a sequece of umber for display What does he mea by that? Let s cosider the well ow recurrece for Fiboacci umbers: F + F + F, where F deotes the th Fiboacci umber How do we fid a ice formula for F? Noe of the approaches that we had leart earlier (Domai/Rage Trasformatios, Master method wor i this settig But GFs come to our rescue Formally speaig, a geeratig fuctio for a sequece of umbers (a 0 is the power series 0 a x, or as Wilf puts it the clotheslie The most fudametal geeratig fuctio: the sequece,,,,, has the power series 0 x /( x (for the momet, we do ot focus o the restrictio x < From Recurreces to Closed Form Let s try the geeratigfuctioologist approach to get a closed form for the Fiboacci umbers Defie F (x : 0 F x x + F 2 x 2 + F 3 x 3 +, sice F 0 0 ad F The multiplyig both sides of the recurrece F + F + F by x ad summig for all we get F x After solvig for F (x we obtai F + x F (x + F 2 x + F 3 x 2 + F (x + F 0 x + F x 2 + F 2 x 3 + F (x x x F (x F (x + xf (x x x x 2 x ( xφ + ( xφ, where φ ± ( ± 5/2 Usig the partial fractio expasio o the RHS we further get ( F (x (φ + φ xφ + xφ 5 0(φ + φ x, where i the last equality we use the expasio of the fudametal geometric fuctio By equatig the coefficiets of x o both sides we obtai F 5 (φ + φ From this formula we also get the asymptotics Solvig recurreces is but oe applicatio of GFs There are may other applicatios: to prove idetities; fid a ew recurrece formula from the gf; fidig a asymptotic estimate where exact formula may ot be possible; prove uimodality, covexity ad other such properties We explore some of these applicatios here

2 Let s try to get a closed form for a more familiar sequece: let f(, be the umber of subsets of size of a elemet set We ow f(, ( but evertheless let s derive a geeratig fuctio for f(, We have the followig recurrece: f(, f(, + f(, A iterpretatio of this recurrece is as follows: fix a elemet, say, i the set []; the all the - sized subsets of the set [] ca be partitioed ito two classes: those that cotai, of which there are f(, sets, ad those that do ot cotai, of which there are f(, sets Let s defie B (x : 0 f(, x The multiplyig the recurrece above by x ad summig both from 0 we get B (x xb (x + B (x ( + xb (x ( + x B 0 (x Note B 0 (x, as there is exactly oe subset of the empty set, amely itself Thus we are iterested i the coefficiet of x i ( + x To get hold of it, we do the followig stadard tric: coeff of x is equal to the th derivative of ( + x /! evaluated at x 0; also, evidet by Taylor expasio at x 0 The term so obtaied is ( ( + /! (, as expected Let s try to aswer aother very similar questio: How may partitios are there of a -elemet set? Or, i other words, how may equivalece relatios ca we have o a elemet set? We first try to aswer a simpler questio: I how may ways ca we partitio a -elemet set ito boxes such that o box remais empty, ad every elemet goes ito some box? Let { be this umber (called the Stirlig umber of secod id So, for istace, {, sice there is oly oe way to partitio the set [] ito a sigle, amely itself; with some effort we ca verify that { 2 2 Give, the defiitio of {, the aswer to our first questio is { 0 Ca we get a recurrece, similar to f(,? Let s proceed with the iterpretatio we had give earlier: cosider the elemet ; the partitios of [] ito boxes are of two types: partitios of the first type are those i which is the oly member i its box, or shares a box with other elemets I the first case, the remaiig boxes are filled with the elemets of the set [ ], which ca be doe i { { way i the secod case, there are ways to distribute elemets ito boxes; but for each such partitio of [ ] ito boxes, we ca put ito oe of the boxes to get a partitio of [] ito boxes Thus we have the recurrece: { + It is atural to defie { { 0, if >, ad 0 0, for all How do we defie the GF for {? We had overlooed this matter earlier, or were rather lucy i our choice, sice there are at least two ways of defiig the GF: { A (x : 0 x ad B (x : 0 x Which of the two shall we choose? If we pic the first oe, as we did for the biomial coefficiets, the we see that to do the sum { x we eed to do a differetiatio, which would mae it more complicated (we eed to solve a differetial equatio If, however, we choose the secod oe the we do t face this problem Thus taig the GF-approach, we multiply the recurrece by x ad sum o both sides to get B (x x + x xb (x + xb (x 0 0 Thus B (x x x B x (x ( x( 2x ( x We are iterested i owig the coefficiet of x o the RHS, which is the same as the coefficiet of x i ( x( 2x ( x 2

3 Ufortuately, our Taylor s formula approach does t wor here (differetiatig the fractios, oly gives more fractios What we eed is to cosider the partial fractio expasio of ( x( 2x ( x a x Multiplyig both sides by ( x, ad pluggig x / we get Thus the coefficiet of x i is our expressio for a a ( (!(! a x a ( + (x + (x 2 + ( (!(! (!(! ( That gives us a expressio for the umber of partitios of [] ito boxes What about our origial questio of the total umber of partitios of [] As we had metioed earlier, this umber, called the th Bell umber, is B : 0 But we observe oe ice fact about the formula i ( Recall that we had defied { 0, for > The formula i (, automatically ecodes this defiitio sice (!(!! 0 ( ( But the RHS ca be derived from (x 0 ( ( x as follows: apply times the operator xd/dx to (x, divide by!, ad plug x to get the term o the RHS However, if >, the (x will always divide (xd/dx (x Thus ( vaishes for > We ca cleverly use this fact i gettig a closed form for B Let m > the the observatio we ust made implies B m 0 m 0 0 ( m!(! 0! m ( m (! 0 m! 0 (! Sice our choice of m was a arbitrary umber greater tha, we ca let m, which gives us B e 0! (2 Though the above equatio has its usefuless ad simplicity, it does ot led itself to computatio as we have a idefiite sum o the RHS (it is good for estimatig Ca we derive a recurrece for B? We ext see how GFs ca lead to recurreces, thus brigig us a complete circle If we proceed alog familiar lies to derive a GF for B we get 0 B x (x! 3!( x

4 which does t led itself to further maipulatio However, there is o hard ad fast rule o choosig the clotheslie Give the similarity betwee B ad the series for e x, we should istead try 0 B x! e e (x e x! (e x e! e ex The geeratig fuctio ust derived is called a expoetial geeratig fuctio; the earlier oes are called ordiary geeratig fuctios Cotiuig further, we ca use the geeratig fuctio to derive a recurrece for B as follows Taig logarithm o both sides of 0 x B! x ee!! we obtai l 0 B x! ex Differetiatig both sides ad multiplyig by x we get B x (! xex 0 Comparig the coefficiets of x o both sides we get a formula that leds itself to computatio 2 Formal Power Series ( B B, 0 x B! x ex B! Formal power series ca be imagied as polyomials with ifiite terms More precisely, they are defied as A : 0 a x We will ofte use the otatio {a to deote the formal power series, ad switch betwee the fuctioal form A ad the coefficiet form {a I this sectio we treat them purely as algebraic obects It is ot hard to show that the set of formal power series is a rig, deoted by R[[x]], where R is the uderlyig rig (eg, itegers from which the coefficiets are chose; additio ad subtractio are doe coefficiet wise, ad multiplicatio is defied aturally as a x b x ( a b x (3 The reciprocal of a formal power series A is aother series B if AB ; the coefficiets of B come from the ratioal field correspodig to R ad are uiquely defied The followig lemma characterizes the existece of iverses Lemma A formal power series a x has a reciprocal iff a 0 0 A distictio is made from power series, where the otio of covergece plays a importat role 4

5 Proof Let {b be the iverse of {a The we must have b 0 a 0, which implies a 0 0 Moreover, from the multiplicatio rule (3 it follows that a 0 b a b Coversely, if a 0 0 the we ca use the relatio above to solve for b s QED Aalogous to fuctioal calculus, we defie oe more otio: the derivative of a formal power series A a x is the series A : a x The derivative is defied to be as such; we do ot defie it by a limitig process as is usually doe Nevertheless, we carry over the stadard rules of differetiatig sums, products, quotiets, ad compositios The followig two results will be of cosequece: If A 0 the A is a costat If A A the A ce x Let s see why The equality states that a a Thus iductively, a a 0 /!, for all, which implies that A a 0 e x Based upo the formalizatio above, we ow proceed to give the details of rules for maipulatig formal power series to obtai geeratig fuctios for various coefficiet sequeces By the symbol A o {a we mea that A is the ordiary power series geeratig fuctio for the sequece {a, ie, A a x Similarly defie A e {a to mea that A is the expoetial power series geeratig fuctio for {a, ie, A a x /! Ordiary Power Series GF Expoetial Power Series GF Let f o {a Let f e {a (f a 0 /x o {a + Df e {a + I geeral, f 0 ax o x {a+ D f e {a + Eg, F +2 F + + F, we get (f x/x 2 f/x + f We get f f + f; solutio f (eφ + x e φ x 5 2 xdf o {a, D is derivative operator This rule remais the same: What geerates { 2 a? It is (xd 2 f I geeral, (xd f o { a Thus, A(x R[x] the we have A(xDf o {A(a Q: Ogf for { ( 2 + /!? We ow e x o {/! A(xDf e {A(a Applyig the rule gives us ((xd 2 + e x is the desired ogf 3 Covolutio: If f o {a, g o {b the fg o { a b f e {a, g e {b the fg e { ( a b What is the result if g /( x? f/( x o { 0 a If g e x the fe x e { Also, f o { i +i 2+ +i a i a i2 a i f e { i +i 2+ +i!i! i!a i a i2 a i Apply to get the sum of squares of first umbers What about H? To demostrate the differece betwee the two types of gf, we see some applicatios ( b What is the egf for 0 ( 2 (? Example : Give pairs of parethesis, let f( be the umber of ways to arrage them i a legal maer, ie, whe scaig a strig of parethesis from left to right the umber of left parethesis always exceed the umber of right parethesis (similar, to the role of parethesis i programmig laguages; clearly, f(0, the empty strig So, for istace, for 3, we have the followig five legal strigs (((, (((, (((, (((, ((( Ca we get a closed form for f(? Every legal strig cotais withi it smaller legal strigs I particular, there is a first legal strig, ie, there is a smallest umber, let s call it the miimal legal idex, such that the first 2 characters form a legal strig; i the example above, the umber is, 2, 3, 3, resp We call a strig with pairs of parethesis primitive if the miimal legal idex is ; i the example above, the third ad fourth strigs are primitive Let f(, be the umber of legal strigs cotaiig a first legal strig of 5

6 legth 2 Sice every strig has a uique miimal legal idex, it follows that f( f(, Ca we get a recursio for f(,? Let g( be the umber of primitive strigs of legth 2 The for every strig i f(,, the first 2 characters form a primitive strig, ad the remaig 2( form a legal strig (ot ecessarily primitive Thus f(, g(f( But what is g(? Every primitive strig i g( is of the form ( legal strig of legth 2(, ie, the first left parethesis has to be matched with the last right parethesis (it caot be matched aywhere i betwee, as that would give us a cotradictio Moreover, give a legal strig of legth 2( we ca costruct a primitive strig i g( This biective correspodece implies g( f( Hece we have the recursio f( f( f( This equatios suggests that we use the covolutio rule for ogfs Let F o {f( The we have that xf o {f( Multiplyig the recurrece by x ad summig for, by the covolutio rule we get xf 2 o the RHS ad F o the left, ie, F xf 2 Solvig for F i this quadratic equatio we get F ± 4x 2x Which sig to choose? If we choose the positive sig the lettig x 0, we get that F (0 f(0, which is ot correct If we choose the egative sig the lettig x 0 we get F (0 as desired Hece F (x 4x 2x, ad hece we ca show that f( ( 2 /( + These umbers are called the Catala umbers 2 A Combiatorial Proof of Catala Numbers: To derive this proof, we first cosider a differet coutig problem Cosider a grid A mootoe path from (0, 0 to (, is a path that either goes right or up by oe uit The total umber of paths from the origi to (, are ( 2 Let C be the umber of good mootoe paths, ie, mootoe paths that do ot cross the diagoal y x (but may touch it It is ot hard to see that the umber of legal paretheses strig is the same as the umber of good mootoe paths If we ca cout the umber of bad mootoe paths, the we ca subtract it from the total umber of mootoe paths to get C A bad mootoe path must cross the diagoal Cosider the first istace whe this happes The coordiates of that poit must be of the form (, + The rest of the path must cotai ( right-moves ad ( up-moves Now reflect the path startig from (, +, ie, wheever we go up i the origial path we go right, ad vice versa The reflected path will thus tae ( up-moves ad ( right moves ad ed up at the poit ( +, + + (, + That is all mootoe paths to (, + correspod to bad mootoe paths to (, Sice the umber of such paths are ( 2, we get that C ( ( 2 2 ( Example 2: A deragemet of letters is a permutatio without a fixed poit We wat to compute D, the umber of deragemets of letters Let P (, be the umber of permutatios that have exactly fixed poits; P (, 0 D( The! 0 P (, But each permutatios i P (, is set of fixed poits, ad the remaiig letters form a deragemet Thus P (, ( D, ad hece! ( D 0 The equatio above fits the patter of covolutio of egf Thus let D(x e {D The dividig the equatio above by!, multiplyig it by x, ad summig for 0, we get x ex D(x, 2 Catala was a mathematicia from Belgium, famous for his coecture i 844 that the oly two cosecutive powers of atural umbers are 8 ad 9 This was proved i 2003 by 6

7 or D(x e x /( x Thus by the covolutio rule for ogfs, we get D! + 2! 3! + + (! Later we will see aother approach to derive the same result 3 The Aalytic Aspects of Power Series As we have see above, the algebraic, or formal, aspects of power series are useful whe we are tryig to get a precise cout or closed formula for the coutig fuctio, which is the th term of the series Ofte, however, this might ot be possible, ad eve whe it is, it might be desirable to get a asymptotic uderstadig of the growth of the coefficiets of the series So, for istace, we ow from above that the Catala umbers C ( 2 /( + Usig Stirlig s approximatio! 2π(/e, we get that C ( 4π(2/e 2 2π(/e π 3 This ca be iterpreted as a special case of a class of geeratig fuctios for which the asymptotics is of the form E(θ(, where E( is a expoetial factor i ad θ( is some sub-expoetial factor Ca we derive the above asymptotics directly? The aalytic aspects of the GF C(z ( 4z/2z help us i this regard The crucial aspect here is the study of the sigularities of the GF, i this case C(z Ituitively, sigularities are poits i the complex plae where either the fuctio or oe of its derivatives is ot cotiuous If we cosider C(z the as z /4, C(z /2, however, its first derivative teds to ifiity If ρ is the sigularity with the smallest absolute value the we will see that the expoetial factor E( ρ ; that explais 4 i C The sub-expoetial factor θ( is govered by the ature of the sigularity whether it is simple or ot, algebraic, etc These two remars may be cosidered as the two fudametal priciples of coefficiet asymptotics To explai the first priciple, we eed to start with the otio of radius of covergece, R, of a power series f(z 0 a z Ituitively, R for a series f(z is the radius of the smallest disc cetered at origi, such that for all poits i this disc the fuctio f has the series expasio with coefficiets a More precisely, we have the followig theorem: Theorem 2 (Cauchy-Hadamard The series a z coverges for all values of z, st z < R ad diverges for all z, st z > R, where R : (4 lim sup a / To uderstad this defiitio, we eed to uderstad the otio of lim sup A umber B is called the limit superior of a sequece {x if B is fiite ad (a for every ɛ > 0 all except fiitely may x satisfy x < B + ɛ, ad iifiitely may x satisfy x > B ɛ; 2 B ad for every M > 0 there is a such that x > M; 3 B ad for every umber x there are oly fiitely may such that x > x Q:What is the limit supremum of ( ( + /? Remars: As the example shows, there may be ifiitely may exceptios to x > B ɛ Also, limit superior always exists With this defiitio i had let s try to prove the theorem above Proof Let s assume R < The we wat to show that the series coverges for all z, z < R For ay such z, there exists a ɛ > 0 such that for sufficietly large, say larger tha N, a ( R + ɛ 7

8 Thus the series 0 a x ca be show to coverge absolutely, as log as ɛ is chose such that z (/R+ɛ <, or, z < R/( + Rɛ But such a ɛ exists sice z < R Similarly, whe z > R we ca show that there are ifiitely may, such that a > (/R ɛ, ad hece the series caot coverge absolutely as log as z (/R ɛ >, or z > R/( Rɛ; agai we ca chose a ɛ satisfyig this costrait as z > R QED The defiitio of R i (4 is ot very hady because we eed to ow a first However, the radius of covergece has a ice relatio with the sigularity with the smallest absolute value: Theorem 3 A fuctio f aalytic at the origi with a fiite radius of covergece R has a sigularity o the boudary of the disc cetered at origi with radius R Proof If ot the we ca icrease the radius of covergece by some ɛ > 0 The fuctio is aalytic o a ɛ-aulus of the boudary, ad by Cauchy s formula for coefficiets we ow a < M/(R + ɛ Thus the series coverges o a larger radius QED Remar: (Prigsheim s Theorem: If all the coefficiets a are o-egative the z R is a sigularity for f; eg, /(z has as its sigularity Example: f(z z/(e z R 2πi ot 0 as by L Hospital s rule the limit as z 0 is oe However, the sigularities are 2πi, 4πi So the radius of covergece is 2π, the absolute value of the smallest sigularity, ad hece for sufficietly large a (/(2π + ɛ, ad for ifiitely may, a (/(2π ɛ 30 Poles Theorem 2 suggests that if we ca icrease the radius of covergece the we would expect a tighter estimate Oe way to obtai this is to fid aother fuctio g that has the same sigularities as f o the circle z R The the fuctio f g is aalytic, ie has a coverget power series expasio, i some larger dis R > R Therefore, Theorem 2 it follows that, the coefficiets of f ad g differ by at most (/R + ɛ for sufficietly large This suggests if we ca get hold of the coefficiets of g, the the coefficiets of f differ by at most the error term metioed above This approach wors specially well for meromorphic fuctios, ie, a fuctio that is aalytic everywhere except at fiitely may poits i C, called its poles I a eighborhood of a pole α such a fuctio f, we ca express f(z r a (z α, ie, as a Lauret expasio aroud α; r is called the order of α The ratioal fuctio formed by the egative terms is called the pricipal part of the f aroud α, deoted by P (f, α The fuctio f P (f, α is clearly aalytic at α By a similar process, we ca costruct a fuctio that is aalytic at all the poles α 0,, α s of f of a give radius R by deletig the correspodig primitive parts P (f, α i, i 0,, s Thus by our earlier remar the coefficiets of i P (f, α i ad f are asymptotically quite close Let s see a applicatio of this result Ordered Bell umbers, B(, are the umber of ways to partitio a set [] where the orderigs of the partitio matter We ow that { are the umber of ways to partitio ito equivalece classes If the orderig of the classes matter the we have! { ways of partitioig [] Thus B(! ( ( 8

9 Agai, usig the observatio that the secod sum o the RHS vaishes for ay >, we ca choose m > ad rewrite B( lim m m 0 0 lim m m m ( ( ( + ( What is a geeratig fuctio for B(? Multiplyig by z /! ad summig over 0, we obtai the followig egf for B( B(z z 2 +! (z 2 + e z! 2 + ( 2 e z /2 2 e z The poles of this fuctio are l 2 + 2iπ The pricipal part correspodig to the pole earest to zero is /2(z l 2 The coefficiet of z is /2(l 2 + The fuctio B(z + 2(z l 2 is has a pole at l 2 + 2πi, ad hece ( B(! 2(l O (l π 2 4 Uimodality of Sequeces A sequece {a is said to be uimodal if there exists a idex such that a 0 a a a a + a Biomial coefficiets (, 0,,, are a typical example of a uimodal sequece, where the maximum is obtaied at /2 I this sectio, we see how geeratig fuctios ca be used to decided uimodality of a sequece A stroger otio tha uimodality is log-cocavity: A sequece {a is said to be log-cocave, if for all log a log a + log a +, 2 or equivaletly a 2 a a + Ulie uimodality, log-cocavity is a local coditio Why is it a stroger otio? Because, if a sequece is ot uimodal the it is ot log-cocave as well, sice there must be a idex such that a > a < a +, which implies a 2 < a a + Thus, if we ca detect log-cocavity the we ca show that the sequece is uimodal If all the roots of the polyomial 0 a x are o-positive, the its coefficiet sequece is a log-cocave sequece Biomial coefficiets: ( 0 x ( + x has as a root of multiplicity Sterlig umber of secod id: A (x : { 0 x Defie A (x : { 0 x ; ote the differece from the geeratig fuctio we had earlier Usig the recurrece relatio for { we ca show that e x A (x x(e x A (x Assume iductively that the roots of A (x are all o-positive; the so are the roots of e x A (x Now, we wat to accout for roots of A (x From Rolle s theorem we ow that there are 2 o-positive roots of (e x A (x ; to these add zero as a root, to get roots To fid the last missig root, we observe that as x, e x A (x goes to zero; so the derivative (e x A (x must have oe more root smaller tha all the roots of e x A (x Thus we have accouted all the real o-positive roots of A (x 9