GENERATING FUNCTIONS AND RANDOM WALKS

Transcription

1 GENERATING FUNCTIONS AND RANDOM WALKS SIMON RUBINSTEIN-SALZEDO 1 A illustrative example Before we start studyig geeratig fuctios properly, let us look a example of their use Cosider the umbers a, defied by the recurrece relatio a +1 = 2a + 1, a 0 = 0 We would like a geeral formula for a We might try workig out a few terms to see if we ca detect a patter: a 0 = 0, a 1 = 1, a 2 = 3, a 3 = 7, a 4 = 15, a 5 = 31 If we re lucky, we otice that it looks like a = 2 1 The, we ca try to prove this formula with iductio But that oly worked because we got lucky ad happeed to otice ad guess the patter How could we figure out a geeral formula without ay guesswork? Let s solve the above example, without ay guesswork What we will do is to set up a object that carries all the iformatio about the recurrece, i a easy-to-use format To do this, let us defie a object A(x) = a x = a 0 + a 1 x + a 2 x 2 + a 3 x 3 + We wat to use the recurrece relatio i order to lear more about A(x) So, let us take the relatio a +1 = 2a + 1, multiply by x to get a +1 x = 2a x + x, ad the sum from = 0 to : a +1 x = 2 a x + x Some of the terms i this sum are familiar objects: the first term o the right side is 2A(x), ad the secod term is a geometric series, which sums to 1 1 x 1 But what about the left side? Date: I case you re worried about covergece, do t be Whe we use geeratig fuctios, we like to trouble ourselves as little as possible with issues of covergece This will oly be a issue if we ever try to plug somethig i for x Sometimes we will do that, ad the we will eed to be careful, but for ow it will ot be ecessary 1

2 2 SIMON RUBINSTEIN-SALZEDO We ca express that i terms of A(x) as well, as A(x) a 0 Sice a x 0 = 0, this meas that the left side is A(x) Thus, we have x A(x) = 2A(x) + 1 x 1 x We ca ow solve for A(x) to get A(x) = x (1 x)(1 2x) Now, we ve figured out what A(x) is How do we recover the a s i a useful form? The usual thig to do at this poit is to use partial fractios, ie, to try to write A(x) = c 1 x + d 1 2x, for some costats c ad d There are may ways to fid c ad d, but oe of the easiest is to take the above equatio ad clear deomiators, to get (1) x = c(1 2x) + d(1 x) ad plug i ice values for x like 1 ad 1, so that we get c = 1 ad d = 1, so that 2 A(x) = 1 1 2x 1 1 x 2 Fially, we have expressed A(x) as a sum of two terms, each of which is the sum of a geometric series So, we usum them to get A(x) = 2 x x = (2 1)x = a x Equatig coefficiets of the x terms, 3 we fid that a = Geeratig fuctios The power series A(x) = a x that we used above is called the geeratig fuctio for the sequece (a ) As we saw above, they are really useful for solvig problems ivolvig recurreces For example, by slightly modifyig the above approach, we ca prove Biet s formula for Fiboacci umbers: (2) F = 1 5 (( 1 + ) ( ) ) 5 2 (It s a little bit aoyig, sice we have to do partial fractios with terms that have 5 s i them, but a slightly uglier computatio is the mai extra difficulty we ecouter I would t wat to do it o the board with people watchig me) 2 I hear you complaiig that I promised ot to plug thigs i for x sice we cavalierly disregarded covergece issues earlier, ad the I did so ayway If this bothers you, feel free to solve for c ad d by equatig costat ad liear terms i equatio (1) 3 We ca do this, sice two such power series are equal if ad oly if all their coefficiets are equal

3 GENERATING FUNCTIONS AND RANDOM WALKS 3 However, some recurreces have a more complicated form tha the liear recurreces defiig a above or the Fiboacci umbers May of these come from sequeces of combiatorial iterest Example The th Catala umber C is the umber of paths from (0, 0) to (2, 0) cosistig of steps of the form (1, 1) ad steps of the form (1, 1) that ever pass below the x-axis 4 Such paths are called Dyck paths Here are all the Dyck paths with = 3: The first few Catala umbers are C 0 = 1, C 1 = 1, C 2 = 2, C 3 = 5, C 4 = 14, C 5 = 42 The Catala umbers have a ice recurrece relatio: Theorem 1 The Catala umbers satisfy the recurrece relatio C +1 = C 0 C + C 1 C C 1 C 1 + C C 0 = C i C i Based o this recurrece, we ca compute the geeratig fuctio C(x) = C x for the Catala umbers But before we do that, we should ivestigate what happes whe we add ad multiply geeratig fuctios: Theorem 2 Let A(x) = a x ad B(x) = b x be two geeratig fuctios The A(x) + B(x) = (a + b )x, A(x)B(x) = ( ) a k b k x Let us ow work out the geeratig fuctio for the Catala umbers, based o the recurrece relatio: doig the usual trick of multiplyig by x ad summig over gives ( ) C +1 x = C k C k x 4 This is very far from the oly way of defiig the Catala umbers Richard Staley, i his book Eumerative Combiatorics, Volume 2, gives 66 iterpretatios of the Catala umbers i his (i)famous Exercise 619 He ow lists several hudreds more o his website, at ec/cataddpdf, ad also i his ew book, simply titled Catala Numbers k=0 k=0 i=0

4 4 SIMON RUBINSTEIN-SALZEDO The left side is C(x) 1, ad the right side is C(x) 2 Hece we have x C(x) 1 = C(x) 2 x Solvig for C(x) usig the quadratic formula gives C(x) = 1 ± 1 4x 2x Ah, but we re ot quite doe, because we do t kow whether it s the positive or egative root But we ca figure this out, i oe of two ways: (1) The hoest way: We ca do this by computig some terms ad seeig which oe is right Usig the positive root ad applyig the biomial theorem, we would get C(x) = 1 1 x x 2x2 +, so we would have C 1 = 1, C 0 = 1, ad so forth But this is wrog If we istead try the egative root, we would get C(x) = 1 + x + 2x 2 + 5x 3 +, which is right Hece we have C(x) = 1 1 4x 2x (2) Cheatig: If we plug x = 0 ito C(x), we should get C(0) = C 0 = C 0 = 1 So, let s try it If we try it with the positive root, we get C(0) = 2, which is disastrous 0 O the other had, if we try it with the egative root, we get C(0) = 0 Much better! 0 Geeratig fuctios are very powerful whe tacklig combiatorial problems Ideed, there is a eormous amout to be said about geeratig fuctios ad combiatorics, but I wo t say it here, because I d like to move o to radom walks ad probability (See the list of refereces at the ed) 3 Radom walks George Pólya was the first perso to study radom walks He became iterested i them whe he repeatedly ra ito the same couple while takig his afteroo walks, despite choosig differet routes Embarrassed that this coicidece might be costrued as soopig, he wodered whether this would happe by chace, ad the subject of radom walks was bor Let us set up the problem A radom walker starts at the origi i R d At every secod, the walker moves oe uit i a directio parallel to oe of the coordiate axes (So, the walker takes a step of the form (0, 0,, 0, ±1, 0,, 0), each with equal probability) Sice there are 2d possible steps, each oe occurs with probability 1 2d The problem is to determie the probability that the walker evetually returs to the origi It will ot be easy to treat all dimesios at oce, so we shall begi with a 1-dimesioal radom walk Here, the walker moves oe step to the right with probability 1/2, ad oe step to the left with probability 1/2 Let us defie p as the probability that the walker is

5 GENERATING FUNCTIONS AND RANDOM WALKS 5 at the origi after time, ad let f be the probability that the walker has retured to the origi for the first time at time Note that, if is odd, the p = f = 0 There are two ways of proceedig First method: Let s work out what f 2 is I order for the walker to have retured to the origi for the first time at time 2, the walker must have stayed etirely positive or etirely egative before that Let us assume that the walker stayed etirely i the positives I order for this to happe, the walker must have made oe positive step at the begiig, the doe a Dyck path of legth 2 2, the take oe egative step at the ed The umber of ways for this to happe is C 1, ad there are 2 2 possible paths i total Hece, the probability that this happes is C But sice the walker could have stayed etirely egative as well, we have to multiply this umber by 2 So, the probability that the walker returs for the first time at time 2 is f 2 = C What is the probability that the walker ever returs to the origi? Well, i ay walk that evetually returs to the origi, there must be some first retur, so the probability of a retur is simply the sum of the f s: f 2 I terms of the Catala umbers, this is C 1 2 = 2 1 C From the geeratig fuctio for the Catala umbers, we ca work this out as a umber, sice it s almost what we get whe we plug i x = 1 i C(x): 4 C C(1/4) = 4 Hece the desired probability is 1 C(1/4) = 1 Thus we have proved: 2 Theorem 3 I a oe-dimesioal radom walk, the walker returs to the origi with probability 1 Secod method: This method ivolves meshig the f s ad the p s better To do this, we fid a recurrece relatio ivolvig the f s ad p s: if we re at the origi after 2 steps, the we must have had a first retur to the origi at 2k steps, for some k, ad the made a path that goes back to the origi i 2 2k steps Hece, we have p 2 = f 0 p 2 + f 2 p f 2 2 p 2 + f 2 p 0, whe 1 (Whe = 0, this relatio does t work, because there is o first retur) Let us set F (x) = f 2x ad P (x) = p 2x We ca fid a relatio betwee F ad P by usig the recurrece: multiply by x ad sum over 1 to get ( ) p 2 x = f 2k p 2 2k x The left side is P (x) p 0 = P (x) 1, ad the right side is F (x)p (x) f 0 p 0 = F (x)p (x) (sice f 0 = 0) Thus we have P (x) 1 = F (x)p (x), k=0

6 6 SIMON RUBINSTEIN-SALZEDO or F (x) = 1 1 P (x) Now, we compute P (x) To compute p 2, we ote that, i order to be back at the origi after 2 steps, we must have take positive steps ad egative steps, ad these ca be arraged i ay order Each possible sequece of positive ad egative steps occurs with probability 1/4, so p 2 = 1 ( ) 2 4 Hece P (x) = ( ) 2 (x ) 4 By the biomial theorem ad some maipulatios, we fid that this is equal to 1 1 x Hece F (x) = 1 1 x, so F (1) = 1, oce agai So, we have ow show that with probability 1, the walker returs home But how log does it take, o average? Let us determie how to work out the average value of a radom variable It is easiest to uderstad what this meas by first doig a example: the average value of a die roll The average value of a die roll is 7 = 1 ( ) There is a probability of 1/6 of 2 6 each result, so we multiply the result by the probability ad sum over all the possibilities Hece, i this case, the average time it takes to get back to the origi for the first time is 2f 2 There is a stadard trick to evaluate such sums: take a derivative! If A(x) = a x, the A (x) = a x 1 I particular, if we have a radom variable that takes o oly oegative iteger values, with a beig the probability of gettig (so that i particular a = 1), the the expected or mea value of the radom variable is A (1) I our case, F (x) = f 2 x 1 Hece, the average time to the first retur home is 2F (1), with the 2 comig because we multiplied f 2 by x rather tha x 2 We ca compute this directly, sice F (x) = 1 1 x; we get F (x) = 1 2, so it looks like 2F (1) is ifiite This meas that, eve though the 1 x walker returs i fiite time with probability 1, the average amout of time it takes is ifiite!

7 GENERATING FUNCTIONS AND RANDOM WALKS 7 4 Higher-dimesioal radom walks Now that we ve take care of the oe-dimesioal radom walk, let s move o to higher dimesios There are surprises waitig for us here! Let s do the 2-dimesioal case Here, there are four possible steps: (1, 0), ( 1, 0), (0, 1), ad (0, 1) We defie f (2) 2 ad p (2) 2 to be the probability of a first retur or retur (respectively) to the origi at time 2 We will compute f (2) 2 ad p (2) 2 by followig Method 2 i the oe-dimesioal case We start by computig p (2) 2 I order to retur to the origi after 2 steps, we must take k steps right, k steps left, k steps up, ad k steps dow, for some k with 0 k The probability of takig k steps right, k steps left, k steps up, ad k steps dow is 1 (2)! 4 2 k! 2 ( k)! = 1 (2)!! ! 2 k! 2 ( k)! = 1 ( )( ) k Thus p (2) 2 = k= ( 2 )( k ) 2 = ( ) 2 2, where the last equality follows from problem 9 Now, the recurrece relatig the f s ad the p s still holds, so if we let F (2) be the geeratig fuctio for the f s, ad P (2) the geeratig fuctio for the p s, we have oce agai F (2) (x) = 1 1 P (2) (x) This time it wo t be so easy to evaluate P (2) (x) exactly, so we ll have to resort to a approximatio But what we care about is whether we will ecessarily retur That is, is F (2) (1) = 1? This will happe if ad oly if P (2) (1) is ifiite So, let us try to work that out We do that by applyig Stirlig s formula (problem 11), which says that! 2π e We ca use this to approximate ( ) 2 = (2)! as well: we get! ( 2 ) 2 22 π Hece, Now, p (2) (2 2 ) 2 π = 1 π P (2) 2 (1) = p (2) 2 which diverges Hece P (2) 2 (1) also diverges, so F (2) (2) (1) = 1, so agai, the walker always returs to home 1 π,

8 8 SIMON RUBINSTEIN-SALZEDO We might ow be ready to cojecture that the walker always returs home But there is a importat rule i cojecture-makig: wheever you re about to make a cojecture, try oe more case 5 So, let s do the 3-dimesioal case The setup i three dimesios is very similar We ow compute p (3) 2, ad we fid it to be p (3) 2 = j,k (2)! j! 2 k! 2 ( j k)! 2 = ( 2 ) j,k ( ) 2 1! 3 j!k!( j k)! This is hard to evaluate exactly, but we ca boud it: the term i paretheses is maximized whe j, k, j k are all aroud /3, so we ca replace all the terms with what we get whe we plug i /3 for j ad k (Ad it s o particular harm to assume that is divisible by 3, but we wo t be that careful, sice we re just goig to use Stirlig s formula ayway) Whe we do this, we fid that the largest value M of! j!k!( j k)! is roughly c We ow boud p (3) 2 : p (3) 2 1 ( = 1 ( ) 2 M 2 2 = ( 2 1 π M c π 3/2 ) j,k 3 3 ) M ( ) M! 3 j!k!( j k)!, for some costat c But the sum f (3) 2 c π 3/2 coverges, ad hece so does the smaller sum p(3) 2 Thus is strictly less tha 1, ie the walker does ot always retur home! So, surprisigly, we see differet behavior i three dimesios from oe or two dimesios I fact, the two-dimesioal case is critical: i two dimesios or fewer, the walker returs home, whereas i more tha two dimesios, the walker does ot ecessarily retur The reaso for puttig the cutoff at 2, rather tha at 3, is that if we were to come up with a sesible defiitio for a radom walk i d dimesios, where d is ot ecessarily a iteger, the walker would always retur if d 2, but ot if d > 2 (Ideed, oe ca come up with sesible defiitios of this) The reaso for this is that the sum i questio that we have to evaluate is quite close to 1 d/2 5 Okay, so I just made that up But I still thik it s a good rule!

9 GENERATING FUNCTIONS AND RANDOM WALKS 9 5 Refereces The best place to get started with geeratig fuctios is udoubtedly Herbert Wilf s geeratigfuctioology, which ca be dowloaded for free from his website 6 Lookig back at the book after may years (I read it whe I was i high school), I ca see that, without eve tryig, I came up with several of the same examples as he did, presumably because that book so strogly shaped how I thik about geeratig fuctios A good place to lear about the aalytic theory of geeratig fuctios (pluggig stuff ito them), especially its applicatios to umber theory, is the first chapter of Doald Newma s Aalytic Number Theory The material o radom walks ca be foud i may places, icludig Charles Gristead ad Laurie Sell s Itroductio to Probability 6 Problems (1) Prove Biet s formula (equatio (2)) for the Fiboacci umbers usig geeratig fuctios (2) Fid a closed formula for the Catala umbers (3) Prove Theorem 1 (4) Use geeratig fuctios to solve the recurrece a +1 = 2a + (5) Let p() deote the umber of partitios of, ie, the umber of ways of writig as the sum of positive itegers, discoutig the order of the terms (For example, p(5) = 7: the 7 ways are 5, 4+1, 3+2, 3+1+1, 2+2+1, , ) (a) Show that p x 1 = 1 x (b) Show that the umber of partitios of ito all odd parts is the same as the umber of partitios of ito distict parts, usig geeratig fuctios (For example, whe = 5, there are 3 of each The partitios ito odd parts are 5, 3+1+1, ; the partitios ito distict parts are 5, 4+1, 3+2) (c) Do part (b) by givig a bijectio Observe that this is substatially harder! (6) Is it possible to relabel the sides of two dice with positive iteger labels so that the probability of gettig ay sum from 2 to 12 is the same as that of ordiary dice? (7) Is it possible to put weights o the sides of two ordiary dice so that the probability of gettig each sum from 2 to 12 is 1? 11 (8) (Putam 2003) For a set S of oegative itegers, let r S () deote the umber of ordered pairs (s 1, s 2 ) such that s 1, s 2 S, s 1 s 2, ad s 1 + s 2 = Is it possible to partitio the oegative itegers ito two sets A ad B (ie, so that A B = N, ad A B = ) i such a way that r A () = r B () for all? (9) Show that k=0 ( ) 2 = k 6 ( ) 2

10 10 SIMON RUBINSTEIN-SALZEDO (10) Evaluate the sum 2 2 (11) (a) Show that e <! < +1 e (b) Harder: Show that! C e, for some costat C (c) Eve harder: Show that! 2π e This is Stirlig s approximatio (12) I a oe-dimesioal radom walk, what is the expected umber of visits to the origi by time 2? (13) Show that, i a oe- or two-dimesioal radom walk, we evetually lad o each lattice poit with probability 1 (14) Let d be a positive iteger Suppose we have two walkers o the iteger lattice Z d, both startig from the origi, both doig idepedet radom walks (a) For which values of d is it the case that two walkers will lad o the same vertex at the same time (after the iitial meetig) with probability 1? (b) For which values of d is it the case that the two walkers will both lad o the same poit other tha the origi, possibly at differet times, with probability 1? Euler Circle, Palo Alto, CA address: simo@eulercirclecom