Geeratig Fuctios ad Their Applicatios Agustius Peter Sahaggau MIT Matheatics Departet Class of 2007 18.104 Ter Paper Fall 2006 Abstract. Geeratig fuctios have useful applicatios i ay fields of study. I this paper, the geeratig fuctios will be itroduced ad their applicatios i cobiatorial probles, recurrece equatios, ad physics will be illustrated. 1. Itroductio. Worig with a cotiuous fuctio is soeties uch easier tha worig with a sequece. For exaple, i the aalysis of fuctios, calculus is very useful. However, the discrete ature of sequeces prevets us fro usig calculus o sequeces. A geeratig fuctio is a cotiuous fuctio associated with a give sequece. For this reaso, geeratig fuctios are very useful i aalyzig discrete probles ivolvig sequeces of ubers or sequeces of fuctios. Defiitio 1-1. The geeratig fuctio of a sequece {f } =0 is defied as f(x) = f x, (1-1) =0 for x < R, ad R is the radius of covergece of the series. It is iportat that the series has a ozero radius of covergece, otherwise f(x) would be udefied for all x = 0. Fortuately, for ost sequeces that we would be dealig with, the radius of covergece is positive. However, we ca certaily costruct sequeces for which the series (1-1) is diverget for all x = 0; f = is oe such exaple. Exaple 1-2. As a siple exaple of a geeratig fuctio, cosider a geoetric sequece, f = a. The 1 f(x) = a x =. (1-2) 1 ax =0 This series covergeces absolutely wheever ax < 1. Therefore, the radius of covergece is R = 1/ a. For the rest of the paper, if ot etioed otherwise, x is always chose to be sall eough such that ay series ecoutered i our aalysis coverges absolutely. Now we shall discuss a applicatio of geeratig fuctios to liear recurrece probles. 1
2 18.104 Ter Paper 2. Fro Recursio to Algebra. Geeratig fuctios ca be used to solve a liear recurrece proble. Defiitio 2-1. The proble of liear recurrece is to fid the values of a sequece {u } satisfyig c u + = v, for soe costat v ad ay iteger 0, (2-1) =0 give the iitial values u 0, u 1,..., u 1, where both c 0 ad c are ozero. I order to solve this recurrece proble, we use the followig property of geeratig fuctios. Propositio 2-2. If {u } is a positive iteger, the =0 is a sequece with a geeratig fuctio u(x) ad [ 1 ] 1 u + x = u(x) u j x j. (2-2) x =0 j=0 Proof: We start fro the defiitio of u(x). 1 u(x) = u j x j = u j x j + u j x j. (2-3) j=0 j=0 j= Shiftig the suatio idex i the secod ter to = j, we obtai 1 u(x) = u j x j + u + x j=0 =0 fro which Eq.(2-2) iediately follows. +, (2-4) By ultiplyig both sides of Eq.(2-1) by x ad suig fro = 0 to, we fid v c u + x = v x =0 =0 =0 =. (2-5) 1 x Usig Propositio2-2, each ter o the left had side with fixed ca be expressed i ters of u(x) ad ow costats u 0, u 1,..., u 1. We have ow reduced Eq.(2-5) ito a algebraic equatio for u(x), which ca be easily solved. After fidig u(x), we write dow the Taylor series expasio of u(x) aroud x = 0. Because the Taylor series of a fuctio is uique (if it exists), the coefficiet of x i the Taylor series ust be u. To illustrate this ethod, we shall use it o the followig exaple fro physics. Exaple 2-3. I special relativity, the usual oe diesioal velocity additio for ula v = u + v is odified ito [1, p. 127] u + v v =, (2-6) 1 + uv with v, u, ad v easured i uits of the speed of light c. We will use this velocity additio i the followig proble. Suppose there are ifiitely ay cars labeled by
Geeratig Fuctios ad Their Applicatios 3 itegers 0. The ( + 1)-th car oves to the right relative to the -th car with a relative velocity v (0 < v < 1). I our referece frae, we deote the velocity of the -th car by u. Assuig u 0 = 0, fid u for all 1. Solutio: Notice that u +1 is the additio of u ad v usig the additio forula i Eq.(2-6). u + v u +1 =, for 0. (2-7) 1 + u v This recurrece is ot liear, ad therefore we ay ot apply our previous ethod directly. With a little aipulatio, however, this recurrece ca be trasfored ito a liear recurrece. 1 + u v u v (1 u )(1 v) 1 u +1 = = 1 + u v 1 + v v(1 u ) ( ) 1 1 + v 1 v =. (2-8) 1 u +1 1 v 1 u 1 v Defiig 1 + v v 1 α =, λ =, ad f =, 1 v 1 v 1 u Eq.(2-8) ca be writte as f +1 = αf λ, (2-9) which is a liear recurrece i f. Now ultiply both sides by x ad su fro = 0 to. f +1 x = α f x λ x =0 =0 =0 1( ) λ f(x) f 0 = αf(x), (2-10) x 1 x where we have used Propositio2-2 to siplify the left had side. The iitial coditio of f is give by f 0 = 1/(1 u 0 ) = 1. Solvig for f(x) yields Usig the defiitios of α ad λ, we fid ad hece 1 λx f(x) = 1 αx (1 x)(1 αx) [ ] 1 λ 1 1 f(x) = +. (2-11) 1 αx α 1 1 x 1 αx λ 1 =, (2-12) α 1 2 [ ] 1 1 1 1( ) f(x) = + = α + 1 x = f x. (2-13) 2 1 αx 1 x =0 2 =0 Sice f(x) is the su of two geoetric series, we coclude that the Taylor series aroud 0 has a positive radius of covergece. Therefore, 2f = α +1 by the uiqueess of Taylor series, ad 1 α 1 u = 1 = f α. (2-14) + 1
4 18.104 Ter Paper Sice α > 1, we coclude that 0 < u < 1 for all 1. Physically, this result shows that ay car oves with a speed less tha c (reeber that we are writig u i uits of the speed of light). Special relativity predicts that ay assive object always travels slower tha light [4, p. 119]. 3. Applicatios to Cobiatorial Probles. May cobiatorial probles ca be solved with the aid of geeratig fuctios. I particular, let s cosider the proble of fidig the uber of partitios of a atural uber. Defiitio 3-1. [6, p. 169] A partitio of a atural uber is a way to write as a su of atural ubers, without regard to the orderig of the ubers. Exaple 3-2. 1 + 1 + 3 + 1 is a partitio of 6. With this defiitio, the geeratig fuctio of the uber of partitios of has a siple for. Theore 3-3. [6, p. 169] If p is the uber of partitios of ad p 0 = 1, the 1 p x = 1. (3-1) =0 =1 x Proof: First we eed to establish the covergece of the ifiite product for x = 0. This ifiite product coverges absolutely if the series x coverges absolutely [3, p. 53]. Thus, the right had side coverges absolutely for x < 1. Each factor i the ifiite product ca be expressed as a geoetric series. [ ] 1 = x. (3-2) =1 1 x =1 =0 I this for, we ca see that the coefficiet of the x ter is equal to the uber of ways to choose itegers { 0, = positive itegers} satisfyig =. If we tae the x ter fro the -th factor, the we obtai x. By coparig this with Defiitio 3-1, we coclude that the coefficiet of x is equal to the uber of partitios of. As a chec, let s try expadig the right had side of Eq.(3-1) up to x 4. p x = (1 + x + x2 + x3 + x4 )(1 + x2 + x4 )(1 + x3 )(1 + x4 ) + =0 =1 = (1 + x + x2 + x3 + x4 )(1 + x2 + x3 + 2x4 ) + = 1 + x + 2x2 + 3x3 + 5x4 + O(x5 ), =1 p 1 = 1, p 2 = 2, p 3 = 3, p 4 = 5. (3-3) We ca easily verify that Eq.(3-3) correctly gives the uber of partitios of 1 to 4. Aother iportat cobiatorial proble that ca be easily solved with geeratig fuctios is Catala s proble [6, p. 260].
Geeratig Fuctios ad Their Applicatios 5 Exaple 3-4 (Catala s Proble). Give a product of letters, how ay ways ca we calculate the product by ultiplyig two factors at a tie, eepig the order fixed? As a exaple, for = 3, there are two ways: (a 1 a 2 )a 3 ad a 1 (a 2 a 3 ). Solutio: Deote the solutio for = by K. It is clear that K 2 = 1. For later coveiece, we defie K 1 = 1. Now cosider the = + 1 case. Suppose at the last step of the ultiplicatio, we have bc, where b = a 1 a j, c = a j+1 a +1, ad 1 j. Notice that there are K j ways to ultiply the factors i b, ad there are K j+1 ways to ultiply the factors i c. Thus, for a give j, there are K j K j+1 ways to ultiply a 1 a +1. The total uber of ways ca be obtaied by suig over all possible values of j. K +1 = K j K j+1, 1. (3-4) j=1 Multiply both sides by x +1 ad su fro = 1 to =. K +1 x+1 = K j K j+1 x+1. (3-5) =1 =1 j=1 As usual, we defie the geeratig fuctio for {K } =1. The left had side of Eq.(3-5) is K +1 x+1 = K x =1 =1 Now cosider the expressio for K(x) 2. K(x) = K x. (3-6) =1 K 1 x = K(x) x. (3-7) K(x) 2 = K j x j i K i x. (3-8) j=1 i=1 Let i = j + 1, where 1. For a give, j ca be ay iteger fro 1 to, sice i 1. Thus, we ca rewrite Eq. (3-8) as K(x) 2 = K j K j+1 x j x j+1 = K j K j+1 x+1. (3-9) =1 j=1 =1 j=1 By usig Eqs.(3-7) ad (3-9) i Eq.(3-5), we obtai a quadratic equatio for K(x). 1 1 K(x) 2 K(x) + x = 0, or K(x) = 2 ± 1 4x. (3-10) 2 Notice fro Eq.(3-6) that K(0) = 0, which eas we ust tae the egative sig for the square root. 1 1 4x K(x) =. (3-11) 2
6 18.104 Ter Paper 1 It is clear that the square root has a covergig power series aroud x = 0 for x < 1 ad hece the ifiite series defiig K(x) has a radius of covergece of 4. Use the bioial forula to obtai the power series expasio of 1 4x. ( 1 1 4x = 1 + 2 )(1 2 1)(1 2 2) (1 2 + 1) ( 4x)! =1 = 1 2x + ( 1) 1 1 3 5 (2 3) ( 1) 2 x! =2 1 2 3 (2 2) = 1 2x 2 x!( 1)! =2 ( ) 1 2( 1) = 1 2 x. (3-12) 1 =1 Now we substitute Eq.(3-12) ito Eq.(3-11) to fid ( ) 1 2( 1) K x = x, (3-13) 1 =1 =1 or ( ) 1 2 K +1 =, 0. (3-14) + 1 This proble was solved by Catala i 1838 [6, p. 259-260], ad the Catala ubers are covetioally defied as C = K +1, for 0. There are ay other applicatios of geeratig fuctios i cobiatorial probles that caot be covered here. A wide variety of exaples are discussed i [5, Chapter 3]. 4. Legedre Polyoials. So far, we have oly discussed geeratig fuctios of sequeces of ubers. However, i Sectio1, I etioed that geeratig fuctio ethods ca also be used to aalyze sequeces of fuctios. Oe iterestig exaple is the geeratig fuctio of Legedre polyoials. As we shall see, the geeratig fuctio provides a physical isight, with a deep coectio to electroagetis. There are several ways to defie the Legedre polyoials P (t). For exaple, they ca be defied as solutios to a differetial equatio [2, p. 96]. For our purposes, however, it is ore coveiet to defie P (t) as follows. Defiitio 4-1. The Legedre polyoials {P (t)} =0 are defied i the iterval 1 t 1. [2, p. 100] They satisfy the recurrece relatio ( + 1)P +1 (t) = (2 + 1)tP (t) P 1 (t) for 1, (4-1) with P 0 (t) = 1 ad P 1 (t) = t. Fro this defiitio, it is easy to prove by iductio that P (t) is a polyoial i t of degree. We ow wat to fid the geeratig fuctio of P (t). I order to avoid cofusio i the otatio, we deote the geeratig fuctio of P (t) for fixed t as Q t (x) = P (t)x. (4-2) =0 4,
Geeratig Fuctios ad Their Applicatios 7 Taig the derivative gives [ ] ( + 1)P +1 (t)x+1 = (2 + 1)tP (t) P 1 (t) x [ 2 P (t)x = 2tx P (t)x + x tp +1 (t)x xq t(x) = ( + 1)P (t)x =1 P (t)x. (4-3) Let s ultiply Eq.(4-1) by x +1 ad su fro = 1 to =. =1 =1 =2 =1 =0 The last step follows fro the substitutios = + 1 ad = 1. We ote fro Eqs.(4-2) ad (4-3) that =2 P (t)x =0 while Propositio2-2 iplies t t +1 2 1 2tz s + z s = 0, (4-10) z s = t ± i 1 t 2, z s = 1. (4-11) ( + 1)P (t)x ]. (4-4) = xq (x) P 1 (t)x = xq (x) tx, (4-5) = xq t (x) + Q t(x), (4-6) x P +1 (t)x = Q t (x) P 0 (t) = Q t (x) 1. (4-7) =0 Thus, Eq.(4-4) siplifies to xq t(x) tx = 2tx 2 ( ) Q t(x) + tx Q t (x) 1 ( ) x2 xq t(x) + Q t (x), x t Q t(x) = 1 2tx + x 2 Q t (x). (4-8) By itegratig Eq.(4-8) ad iposig the iitial coditio Q t (0) = P 0 (t) = 1, we obtai Q t (x) = 1 =0 P (t)x =. (4-9) 1 2tx + x 2 To fid the radius of covergece of the power series of Q t (x), we eed to fid the locatio z s (i coplex plae) of the sigularity earest to the origi. Q t (z s ) is sigular if Therefore, Q(z) is aalytic i the regio z < 1, ad its power series coverges absolutely i this regio. I electrostatics, the potetial alog the z axis due to a aziuthally syetric volue charge distributio ρ(r, θ) is give by [2, p. 35] (we set 4πǫ o = 1) π 2 ρ(r, θ ) V (z) = 2π dr r dθ si θ. (4-12) 0 0 z 2 2zr cosθ + r 2
8 18.104 Ter Paper If ρ(r, θ ) is bouded, ρ(r, θ ) = 0 for r > a, ad we are oly iterested i V (z) for z > a, the a 2 π r ρ(r, θ ) V (z) = 2π dr dθ si θ, (4-13) 1 2xcosθ + x 2 q = 2π q = 0 z 0 where x = r /z < 1. Now we ay use Eq.(4-9) with t = cosθ because cosθ 1. Eq.(4-13) ca the be writte as q V (z) =, (4-14) z+1 with a π ( 1) P (t)x =0 dr (r ) +2 dθ 0 0 si θ ρ(r, θ )P (cos θ ) d 3 r r ρ(r, θ )P (cos θ ). (4-15) The ubers q are called the ultipole oets. I particular, q 0 is the oopole oet (or total charge), ad q 1 is the dipole oet [2, p. 146]. Sice the -th oet ter i the potetial falls off as 1/z +1, the first ozero oet q characterizes the behavior of V (z) as z/a. We ca see that the geeratig fuctio of P (t) appears aturally i electroagetis. This techique of expadig the potetial as a series of oets is very useful, ad is called ultipole expasio. A iportat property of P (t) ca be show directly fro the geeratig fuctio by cosiderig Q t ( x). 1 Q t ( x) = P (t)( x) =. (4-16) 1 + 2tx + x 2 =0 Notice that the right had side is also equivalet to Q t (x). =0 =0 = P ( t)x. (4-17) Fro the uiqueess of the power series of Q t ( x), we obtai P ( t) = ( 1) P (t). Therefore P (t) is a odd (eve) polyoial if ad oly if is odd (eve). Aother property ca be obtaied by settig t = 1 i Eq.(4-9). 1 1 P (1)x 1 2x + x 2 1 x =0 =0 = = = x. (4-18) Thus, P (1) = 1 for all. Sice P (t) is odd for odd, we also obtai P ( 1) = ( 1). 5. Useful Tric to Fid a Geeratig Fuctio. I Sectio 2, we saw that we ca easily fid the geeratig fuctio of a sequece if that sequece is defied through a liear recurrece. However, i soe cases, we ay ot have a liear recurrece, such as i the Catala s proble i Sectio3. For soe sequeces without a liear recurrece, it is possible to obtai the geeratig fuctio usig a covolutio property. I fact, we have actually used this property to solve the Catala s proble.
Geeratig Fuctios ad Their Applicatios 9 Defiitio 5-1. A covolutio of two sequeces {f } =0 ad {g } =0 is aother se quece deoted by {(f g) }, with =0 (f g) = f g. (5-1) =0 Theore 5-2. Let {u } =0 ad {v } =0 be two sequeces with geeratig fuc tios u(x) ad v(x) respectively. If w = (u v) ad w(x) is the geeratig fuctio of {w } =0, the w(x) = u(x)v(x). (5-2) The radius of covergece of w(x) is the iiu of the radii of covergece of u(x) ad v(x). Proof: Let r > 0 ad s > 0 be the radii of covergece of u(x) ad v(x). Deote t = i(r, s). Cosider the product i u(x)v(x) = u i x v j x j, (5-3) i=0 j=0 for x < t. Sice both series coverge absolutely, we ay rearrage the ters i the double suatio. Suppose we wat to group the sae powers of x. We ca do this by writig j = i, with 0. For each, i goes fro 0 to, because j is oegative. i u(x)v(x) = u i x v i x i =0 i=0 =0 = w x. (5-4) This is precisely the defiitio of w(x), ad the series o the right had side coverges absolutely for x < t. As we shall see later i this sectio, it is soeties ore coveiet to fid a geeratig fuctio for {a /!} istead of {a }. This is the otivatio to defie the expoetial geeratig fuctio. Defiitio 5-3. The expoetial geeratig fuctio F(x) of a sequece {f } =0 is defied as x F(x) = f. (5-5)! =0 The expoetial geeratig fuctios have the followig property. Lea 5-4. If F(x) is the expoetial geeratig fuctio of a sequece {f } =0, the x f +1 =0! = F (x). (5-6) Proof: Differetiate both sides of the defiitio of F(x) i Eq.(5-5). F 1 1 x x x (x) = f = f = f +1. (5-7)! ( 1)!! =1 =1 =0
10 18.104 Ter Paper We ca state a theore aalogous to Theore5-2 for expoetial geeratig fuctios. Theore 5-5. Let {f } =0 ad {g } =0 be two sequeces with expoetial geer atig fuctios F(x) ad G(x) respectively. If ( ) h = f g, (5-8) =0 ad H(x) is the expoetial geeratig fuctio of {h }, the Solutio: First we eed to fid a recurrece relatio for b. Cosider a set A of ( + 1) distict eleets, A = {a 1, a 2,...,a +1 }. Suppose a 1 is ( cotaied ) i the first subset alog with j other eleets, where 0 j. There are j ways to for this subset, which is the uber of ways to pic j eleets fro {a 2, a 3,, a +1 }. Oce the first subset is fixed, we are left with a set S cotaiig ( j) distict eleets. There are b j ways of partitioig S ito subsets, ad therefore we ay write ( ) ( ) b +1 = b j = b, (5-10) j ( ) ( ) by usig = j ad =. If we defie a sequece {t = 1} =0, the its expoetial geeratig fuctio T(x) is give by =0 H(x) = F(x)G(x). (5-9) The radius of covergece of H(x) is the iiu of the radii of covergece of F(x) ad G(x). Proof: The proof follows the sae steps as the proof of Theore5-2, by substitutig u = f /! ad v = g /!. We shall ow discuss a exaple to illustrate the covolutio ethod i a proble where the expoetial geeratig fuctio is a ore coveiet choice. Exaple 5-6. (Bell ubers) [6, p. 167]. Deote by b the uber of ways to write a set of distict eleets as a uio of disjoit subsets, with b 0 = 1. For = 2, there are two ways: {a 1, a 2 }, ad {a 1 } {a 2 }. Fid a forula for b. j=0 =0 T(x) = x =0! x = e. (5-11) Eq.(5-10) ca be writte as ( ) w = b t, (5-12) =0 with w = b +1. Notice the siilarity betwee Eqs.(5-12) ad (5-8). Applyig Theore5-5 o {b } =0 ad {t }, we obtai =0 x B(x)e B(x)T(x) = W(x) x x = w = b +1, (5-13)!! =0 =0
Geeratig Fuctios ad Their Applicatios 11 where Sice e x has a power series that coverges everywhere, we coclude that B(x) has a ifiite radius of covergece. Let s write the power series expasio of B(x). [ ] 1 x e 1 e x B(x) = e = 1 + e e! =1 [ ] 1 1 1 x 1 1 x = 1 + + = 1 +. (5-17) e! e!! e ( 1)!! =1 =1 =1 =1 =1 x B(x) = b (5-14)! =0 is the expoetial geeratig fuctio of {b } =0. Accordig to Lea5-4, the right had side of Eq.(5-13) ca be writte as B (x). Thus, B (x) = e x B(x). (5-15) We ca itegrate Eq.(5-15) ad use B(0) = b 0 = 1 to fid Therefore, for ay atural uber, x e 1 1 B(x) = e = e ex. (5-16) e 1 1 b =. (5-18) e ( 1)! =1 To chec our aswer, tae = 2. Oe way to fid b 2 is to su the ifiite series i Eq.(5-18). However, there is a sipler way if we otice that B 2 x (x) = ( 1)b, (5-19)! =2 ad thus b 2 = B (0). We ca fid B (x) by differetiatig Eq. (5-16) twice. x x x e 1 e +x 1 B (x) = e e = e, B (x) = (1 + e x x e +x 1 )e. (5-20) Therefore, b 2 = 2 as expected. Icidetally, by usig Eq. (5-18) for = 2, we have prove the followig ifiite series, =1 = 2e. (5-21) ( 1)! Coclusios We have discussed soe basic applicatios of geeratig fuctios, as a ethod to solve a liear recurrece or cobiatorial probles. However, there are certaily ay ore aspects i the subject that are ot discussed here. Readers iterested to lear ore are ivited to read [5] for a very extesive treatet of geeratig fuctios.
12 18.104 Ter Paper Refereces [1] A. P. Frech, Special Relativity, W.W. Norto & Copay, Ic., 1968. [2] J. D. Jacso, Classical Electrodyaics, Third Editio, Wiley, 1998. [3] H. Jeffreys, B. Jeffreys, Methods of Matheatical Physics, Third Editio, Cabridge Uiversity Press, 2000. [4] R. Resic, Itroductio to Special Relativity, Wiley, 1968. [5] H. S. Wilf, Geeratigfuctioology, Secod Editio, Acadeic Press, 1994. [6] R. M. Youg, Excursios i Calculus: A Iterplay of the Cotiuous ad the Discrete, The Matheatical Associatio of Aerica, 1992.