GENERATING FUNCTIONS AND RECURRENCE RELATIONS Generating Functions Recurrence Relations Suppose a0; a1; a2;:::; an;:::;is an infinite sequence. A recurrence recurrence relation is a set of equations an = fn(an−1; an−2;:::; an−k ): (1) The whole sequence is determined by (6) and the values of a0; a1;:::; ak−1. Generating Functions Linear Recurrence Fibonacci Sequence an = an−1 + an−2 n ≥ 2: a0 = a1 = 1. Generating Functions n bn = jBnj = jfx 2 fa; b; cg : aa does not occur in xgj: b1 = 3 : a b c b2 = 8 : ab ac ba bb bc ca cb cc bn = 2bn−1 + 2bn−2 n ≥ 2: Generating Functions bn = 2bn−1 + 2bn−2 n ≥ 2: Let (b) (c) (a) Bn = Bn [ Bn [ Bn (α) where Bn = fx 2 Bn : x1 = αg for α = a; b; c. (b) (c) (b) Now jBn j = jBn j = jBn−1j. The map f : Bn ! Bn−1, f (bx2x3 ::: xn) = x2x3 ::: xn is a bijection. (a) Bn = fx 2 Bn : x1 = a and x2 = b or cg. The map (a) (b) (c) g : Bn ! Bn−1 [ Bn−1, g(ax2x3 ::: xn) = x2x3 ::: xn is a bijection. (a) Hence, jBn j = 2jBn−2j. Generating Functions Towers of Hanoi Peg 1 Peg 2 Peg 3 Hn is the minimum number of moves needed to shift n rings from Peg 1 to Peg 2. One is not allowed to place a larger ring on top of a smaller ring. Generating Functions xxx H n-1 moves 1 move H n-1 moves Generating Functions We see that H1 = 1 and Hn = 2Hn−1 + 1 for n ≥ 2. So, H H 1 n − n−1 = : 2n 2n−1 2n Summing these equations give H H 1 1 1 1 1 n − 1 = + + ··· + = − : 2n 2 2n 2n−1 4 2 2n So n Hn = 2 − 1: Generating Functions A has n dollars. Everyday A buys one of a Bun (1 dollar), an Ice-Cream (2 dollars) or a Pastry (2 dollars). How many ways are there (sequences) for A to spend his money? Ex. BBPIIPBI represents “Day 1, buy Bun. Day 2, buy Bun etc.”. un = number of ways = un;B + un;I + un;P where un;B is the number of ways where A buys a Bun on day 1 etc. un;B = un−1; un;I = un;P = un−2. So un = un−1 + 2un−2; and u0 = u1 = 1: Generating Functions If a0; a1;:::; an is a sequence of real numbers then its (ordinary) generating function a(x) is given by 2 n a(x) = a0 + a1x + a2x + ··· anx + ··· and we write n an = [x ]a(x): For more on this subject see Generatingfunctionology by the late Herbert S. Wilf. The book is available from https://www.math.upenn.edu// wilf/DownldGF.html Generating Functions an = 1 1 a(x) = = 1 + x + x2 + ··· + xn + ··· 1 − x an = n + 1. 1 a(x) = = 1 + 2x + 3x2 + ··· + (n + 1)xn + ··· (1 − x)2 an = n. x a(x) = = x + 2x2 + 3x3 + ··· + nxn + ··· (1 − x)2 Generating Functions Generalised binomial theorem: α an = n 1 X α a(x) = (1 + x)α = xn: n n=0 where α α(α − 1)(α − 2) ··· (α − n + 1) = : n n! m+n−1 an = n 1 1 1 X −m X m + n − 1 a(x) = = (−x)n = x n: (1 − x)m n n n=0 n=0 Generating Functions General view. Given a recurrence relation for the sequence (an), we (a) Deduce from it, an equation satisfied by the generating P n function a(x) = n anx . (b) Solve this equation to get an explicit expression for the generating function. n (c) Extract the coefficient an of x from a(x), by expanding a(x) as a power series. Generating Functions Solution of linear recurrences an − 6an−1 + 9an−2 = 0 n ≥ 2: a0 = 1; a1 = 9. 1 X n (an − 6an−1 + 9an−2)x = 0: (2) n=2 Generating Functions 1 X n anx = a(x) − a0 − a1x n=2 = a(x) − 1 − 9x: 1 1 X n X n−1 6an−1x = 6x an−1x n=2 n=2 = 6x(a(x) − a0) = 6x(a(x) − 1): 1 1 X n 2 X n−2 9an−2x = 9x an−2x n=2 n=2 = 9x2a(x): Generating Functions a(x) − 1 − 9x − 6x(a(x) − 1) + 9x2a(x) = 0 or a(x)(1 − 6x + 9x2) − (1 + 3x) = 0: 1 + 3x 1 + 3x a(x) = = 1 − 6x + 9x2 (1 − 3x)2 1 1 X X = (n + 1)3nxn + 3x (n + 1)3nxn n=0 n=0 1 1 X X = (n + 1)3nxn + n3nxn n=0 n=0 1 X = (2n + 1)3nxn: n=0 n an = (2n + 1)3 : Generating Functions Fibonacci sequence: 1 X n (an − an−1 − an−2)x = 0: n=2 1 1 1 X n X n X n anx − an−1x − an−2x = 0: n=2 n=2 n=2 2 (a(x) − a0 − a1x) − (x(a(x) − a0)) − x a(x) = 0: 1 a(x) = : 1 − x − x2 Generating Functions 1 a(x) = − (ξ1 − x)(ξ2 − x) 1 1 1 = − ξ1 − ξ2 ξ1 − x ξ2 − x ! 1 ξ−1 ξ−1 = 1 − 2 ξ1 − ξ2 1 − x/ξ1 1 − x/ξ2 where p p 5 + 1 5 − 1 ξ = − and ξ = 1 2 2 2 are the 2 roots of x2 + x − 1 = 0: Generating Functions Therefore, −1 1 −1 1 ξ1 X −n n ξ2 X −n n a(x) = ξ1 x − ξ2 x ξ1 − ξ2 ξ1 − ξ2 n=0 n=0 1 −n−1 −n−1 X ξ − ξ = 1 2 xn ξ1 − ξ2 n=0 and so −n−1 −n−1 ξ1 − ξ2 an = ξ1 − ξ2 0 p !n+1 p !n+11 1 5 + 1 1 − 5 = p @ − A : 5 2 2 Generating Functions Inhomogeneous problem 2 an − 3an−1 = n n ≥ 1: a0 = 1. 1 1 X n X 2 n (an − 3an−1)x = n x n=1 n=1 1 1 1 X X X n2xn = n(n − 1)xn + nxn n=1 n=2 n=1 2x2 x = + (1 − x)3 (1 − x)2 x + x2 = (1 − x)3 1 X n (an − 3an−1)x = a(x) − 1 − 3xa(x) n=1 = a(x)(1 − 3x) − 1: Generating Functions x + x2 1 a(x) = + (1 − x)3(1 − 3x) 1 − 3x A B C D + 1 = + + + 1 − x (1 − x)2 (1 − x)3 1 − 3x where x + x2 =∼ A(1 − x)2(1 − 3x) + B(1 − x)(1 − 3x) + C(1 − 3x) + D(1 − x)3: Then A = −1=2; B = 0; C = −1; D = 3=2: Generating Functions So −1=2 1 5=2 a(x) = − + 1 − x (1 − x)3 1 − 3x 1 1 1 1 X X n + 2 5 X = − xn − xn + 3nxn 2 2 2 n=0 n=0 n=0 So 1 n + 2 5 a = − − + 3n n 2 2 2 3 3n n2 5 = − − − + 3n: 2 2 2 2 Generating Functions General case of linear recurrence an + c1an−1 + ··· + ck an−k = un; n ≥ k: u0; u1;:::; uk−1 are given. X n (an + c1an−1 + ··· + ck an−k − un) x = 0 It follows that for some polynomial r(x), u(x) + r(x) a(x) = q(x) where k 2 k Y q(x) = 1 + c1x + c2x + ··· + ck x = (1 − αi x) i=1 and α1; α2; : : : ; αk are the roots of p(x) = 0 where k k k−1 p(x) = x q(1=x) = x + c1x + ··· + c0. Generating Functions Products of generating functions 1 1 X n X n a(x) = anx ; b(x)) = bnx : n=0 n=0 2 a(x)b(x) = (a0 + a1x + a2x + ··· ) × 2 (b0 + b1x + b2x + ··· ) = a0b0 + (a0b1 + a1b0)x + 2 (a0b2 + a1b1 + a2b0)x + ··· 1 X n = cnx n=0 where n X cn = ak bn−k : k=0 Generating Functions Derangements n X n n! = d : k n−k k=0 n Explanation: k dn−k is the number of permutations with n exactly k cycles of length 1. Choose k elements ( k ways) for which π(i) = i and then choose a derangement of the remaining n − k elements. So n X 1 d − 1 = n k k! (n − k)! k=0 1 1 n ! X X X 1 d − xn = n k xn: (3) k! (n − k)! n=0 n=0 k=0 Generating Functions Let 1 X dm d(x) = xm: m! m=0 From (3) we have 1 = ex d(x) 1 − x e−x d(x) = 1 − x 1 n X X (−1)k = xn: k! n=0 k=0 So n k dn X (−1) = : n! k! k=0 Generating Functions Triangulation of n-gon Let an = number of triangulations of Pn+1 n X = ak an−k n ≥ 2 (4) k=0 a0 = 0, a1 = a2 = 1. k +1 n+1 1 Generating Functions Explanation of (4): ak an−k counts the number of triangulations in which edge 1; n + 1 is contained in triangle1 ; k + 1; n + 1. There are ak ways of triangulating1 ; 2;:::; k + 1; 1 and for each such there are an−k ways of triangulating k + 1; k + 2;:::; n + 1; k + 1. Generating Functions 1 1 n ! X n X X n x + anx = x + ak an−k x : n=2 n=2 k=0 But, 1 X n x + anx = a(x) n=2 since a0 = 0; a1 = 1. 1 n ! 1 n ! X X n X X n ak an−k x = ak an−k x n=2 k=0 n=0 k=0 = a(x)2: Generating Functions So a(x) = x + a(x)2 and hence p p 1 + 1 − 4x 1 − 1 − 4x a(x) = or : 2 2 But a(0) = 0 and so p 1 − 1 − 4x a(x) = 2 1 ! 1 1 X (−1)n−1 2n − 2 = − 1 + (−4x)n 2 2 n22n−1 n − 1 n=1 1 X 12n − 2 = xn: n n − 1 n=1 So 12n − 2 a = : n n n − 1 Generating Functions Exponential Generating Functions Given a sequence an; n ≥ 0, its exponential generating function (e.g.f.) ae(x) is given by 1 X an a (x) = xn e n! n=0 x an = 1; n ≥ 0 implies ae(x) = e : 1 a = n!; n ≥ 0 implies a (x) = n e 1 − x Generating Functions Products of Exponential Generating Functions Let ae(x); be(x) be the e.g.f.’s respectively for (an); (bn) respectively.