4.6.2 Exponential generating functions The number of r-combinations of multiset S={n1·a1,n2·a2,…, nk·ak} : C(r+k-1,r), generating function: The number of r-permutation of set S={a1,a2,…, ak} :p(n,r), generating function:

C(n,r)=p(n,r)/r! Definition 2: The exponential generating function for the sequence a0,a1,…,an,…of real numbers is the infinite series

Theorem 4.17: Let S be the multiset {n1·a1,n2·a2,…,nk·ak} where n1,n2,…,nk are non-negative integers. Let br be the number of r-permutations of S. Then the exponential generating function g(x) for the sequence b1, b2,…, bk,… is given by g(x)=gn1(x)·g n2(x)·…·gnk(x)，where for i=1,2,…,k, gni(x)=1+x+x2/2!+…+xni/ni! . (1)The coefficient of xr/r! in gn1(x)·g n2(x)·…·gnk(x) is

Example: Let S={1·a1,1·a2,…,1·ak} Example: Let S={1·a1,1·a2,…,1·ak}. Determine the number r-permutations of S. Solution: Let pr be the number r-permutations of S, and

Example: Let S={·a1,·a2,…,·ak}，Determine the number r-permutations of S. Solution: Let pr be the number r-permutations of S, gri(x)=(1+x+x2/2!+…+xr/r!+…)，then g(x)=(1+x+x2/2!+…+xr/r!+…)k=(ex)k=ekx

Example：Let S={2·x1,3·x2}，Determine the number 4-permutations of S. Let pr be the number r-permutations of S, g(x)=(1+x+x2/2!)(1+x+x2/2!+x3/3!) Note: pr is coefficient of xr/r!. Example：Let S={2·x1,3·x2,4·x3}. Determine the number of 4-permutations of S so that each of the 3 types of objects occurs even times. Solution: Let pr be the number r-permutations of S, g(x)=(1+x2/2!)(1+x2/2!)(1+x2/2!+x4/4!)

Example: Let S={·a1,·a2, ·a3}，Determine the number of r-permutations of S so that a3 occurs even times and a2 occurs at least one time. Let pr be the number r-permutations of S, g(x)=(1+x+x2/2!+…+xr/r!+…)(x+x2/2!+…+xr/r! +…) (1+x2/2!+x4/4!+…)=ex(ex-1)(ex+e-x)/2 =(e3x-e2x+ex-1)/2

Example: Let S={·a1,·a2, ·a3}，Determine the number of r-permutations of S so that a3 occurs odd times and a2 occurs at least one time. Let pr be the number r-permutations of S, g(x)=(1+x+x2/2!+…+xr/r!+…)(x+x2/2!+…+xr/r! +…) (x+x3/3!+x5/5!+…) =ex(ex-1)(ex-e-x)/2

4.7 Recurrence Relations P13, P100 Definition: A recurrence relation for the sequence{an} is an equation that expresses an in terms of one or more of the previous terms of the sequence, namely, a0, a1, …, an-1, for all integers n with nn0, where n0 is a nonnegative integer. A sequence is called a solution of a recurrence relation if its terms satisfy the recurrence relation. Initial condition: the information about the beginning of the sequence.

Example(Fibonacci sequence): A young pair rabbits (one of each sex) is placed in enclosure. A pair rabbits dose not breed until they are 2 months old, each pair of rabbits produces another pair each month. Find a recurrence relation for the number of pairs of rabbits in the enclosure after n months, assuming that no rabbits ever die. Solution: Let Fn be the number of pairs of rabbits after n months, (1)Born during month n (2)Present in month n-1 Fn=Fn-2+Fn-1，F1=F2=1

Example (The Tower of Hanoi): There are three pegs and n circular disks of increasing size on one peg, with the largest disk on the bottom. These disks are to be transferred, one at a time, onto another of the pegs, with the provision that at no time is one allowed to place a larger disk on top of a smaller one. The problem is to determine the number of moves necessary for the transfer. Solution: Let h(n) denote the number of moves needed to solve the Tower of Hanoi problem with n disks. h(1)=1 (1)We must first transfer the top n-1 disks to a peg (2)Then we transfer the largest disk to the vacant peg (3)Lastly, we transfer the n-1 disks to the peg which contains the largest disk. h(n)=2h(n-1)+1, h(1)=1

Using Characteristic roots to solve recurrence relations Using Generating functions to solve recurrence relations

4.7.1 Using Characteristic roots to solve recurrence relations Definition: A linear homogeneous recurrence relation of degree k with constant coefficients is a recurrence relation of the form an=h1an-1+h2an-2+…+hkan-k, where hi are constants for all i=1,2,…,k,n≥k, and hk≠0. Definition: A linear nonhomogeneous recurrence relation of degree k with constant coefficients is a recurrence relation of the form an=h1an-1+h2an-2+…+hkan-k+f(n), where hi are constants for all i=1,2,…,k,n≥k, and hk≠0.

Definition: The equation xk-h1xk-1-h2xk-2-…-hk=0 is called the characteristic equation of the recurrence relation an=h1an-1+h2an-2+…+hkan-k. The solutions q1,q2,…,qk of this equation are called the characteristic root of the recurrence relation, where qi(i=1,2,…,k) is complex number. Theorem 4.18: Suppose that the characteristic equation has k distinct roots q1,q2,…,qk. Then the general solution of the recurrence relation is an=c1q1n+c2q2n+…+ckqkn, where c1,c2,…ck are constants.

Example: Solve the recurrence relation an=2an-1+2an-2，(n≥2) subject to the initial values a1=3 and a2=8. characteristic equation : x2-2x-2=0, roots: q1=1+31/2，q2=1-31/2。 the general solution of the recurrence relation is an=c1(1+31/2)n+c2(1-31/2)n, We want to determine c1 and c2 so that the initial values c1(1+31/2)+c2(1-31/2)=3, c1(1+31/2)2+c2(1-31/2)2=8

Theorem 4.19: Suppose that the characteristic equation has t distinct roots q1,q2,…,qt with multiplicities m1,m2,…,mt, respectively, so that mi≥1 for i=1,2,…,t and m1+m2+…+mt=k. Then the general solution of the recurrence relation is where cij are constants for 1≤j≤mi and 1≤i≤t.

Example: Solve the recurrence relation an+an-1-3an-2-5an-3-2an-4=0,n≥4 subject to the initial values a0=1,a1=a2=0, and a3=2. characteristic equation x4+x3-3x2-5x-2=0, roots:-1,-1,-1,2 By Theorem 4.19：the general solution of the recurrence relation is an=c11(-1)n+c12n(-1)n+c13n2(-1)n+c212n We want to determine cij so that the initial values c11+c21=1 -c11-c12-c13+2c21=0 c11+2c12+4c13+4c21=0 -c11-3c12-9c13+8c21=2 c11=7/9,c12=-13/16,c13=1/16,c21=1/8 an=7/9(-1)n-(13/16)n(-1)n+(1/16)n2(-1)n+(1/8)2n

the general solution of the linear nonhomogeneous recurrence relation of degree k with constant coefficients is an=a'n+a n* a'n is the general solution of the linear homogeneous recurrence relation of degree k with constant coefficients an=h1an-1+h2an-2+…+hkan-k a n*is a particular solution of the nonhomogeneous linear recurrence relation with constant coefficients an=h1an-1+h2an-2+…+hkan-k+f(n)

Theorem 4.20: If {a n*} is a particular solution of the nonhomogeneous linear recurrence relation with constant coefficients an=h1an-1+h2an-2+…+hkan-k+f(n), then every solution is of the form {a'n+a n*}, where {a n*} is a general solution of the associated homogeneous recurrence relation an=h1an-1+h2an-2+…+hkan-k. Key:a n*

(1)When f(n) is a polynomial in n of degree t, a n*=P1nt+P2nt-1+…+Ptn+Pt+1 where P1,P2,…,Pt,Pt+1 are constant coefficients (2)When f(n) is a power function with constant coefficient n, if  is not a characteristic root of the associated homogeneous recurrence relation, a n*= Pn ， where P is a constant coefficient. if  is a characteristic root of the associated homogeneous recurrence relation with multiplicities m, a n*= Pnmn ，where P is a constant coefficient. Example: Find all solutions of the recurrence relation an+2an-1=n+1,n1, a0=2

Example: Find all solutions of the recurrence relation h(n)=2h(n-1)+1, n2, h(1)=1 an=an-1+7n,n1, a0=1 If let an*=P1n+P2, P1n+P2-P1(n-1)-P2=7n P1=7n Contradiction let an*=P1n2+P2n

4.7.2 Using Generating functions to solve recurrence relations Example: Solve the recurrence relation an=an-1+9an-2-9an-3,n≥3 subject to the initial values a0=0, a1=1, a2=2

Example: Solve the recurrence relation : an=an-1+9an-2-9an-3,n≥3 subject to the initial values a0=0, a1=1, a2=2 Solution: Let Generating functions of {an} be： f(x)=a0+a1x+a2x2+…+anxn+… , then: -xf(x) =-a0x-a1x2-a2x3…-anxn+1-… -9x2f(x) = -9a0x2-9a1x3-9a2x4-…-9an-2xn-… 9x3f(x) = 9a0x3+9a1x4+…+9an-3xn+… (1-x-9x2+9x3)f(x)=a0+(a1-a0)x+(a2-a1-9a0)x2+ (a3-a2-9a1+9a0)x3+…+(an-an-1-9an-2+9an-3)xn+… a0=0,a1=1, a2=2，and when n≥3,an-an-1-9an-2+9an-3=0， (an=an-1+9an-2-9an-3) thus： (1-x-9x2+9x3)f(x)=x+x2 f(x)=(x+x2)/(1-x-9x2+9x3) =(x+x2)/((1-x)(1+3x)(1-3x)) 1/(1-x)=1+x+x2+…+xn+…； 1/(1+3x)=1-3x+32x2-…+(-1)n3nxn+… 1/(1-3x)=1+3x+32x2+…+3nxn+…；

Example: Find an explicit formula for the Fibonacci numbers, Fn=Fn-2+Fn-1， F1=F2=1。 Solution: Let Generating functions of {Fn} be： f(x)=F0+F1x+F2x2+…+Fnxn+…,then： -xf(x) =-F0x-F1x2-F2x3…-Fnxn+1-… -x2f(x) =-F0x2-F1x3-F2x4-…-Fn-2xn-… (1-x-x2)f(x)=F1x+(F2-F1)x2+(F3-F2-F1)x3+(F4-F3-F2)x4+…+(Fn-Fn-1-Fn-2)xn+… F1=1, F2=1，and when n≥3,Fn-Fn-1-Fn-2=0， (Fn=Fn-1+Fn-2) thus： (1-x-x2)f(x)=x f(x)=x/(1-x-x2) Fn-10.618Fn。 golden section黄金分割。

Exercise P104 18,20,23.Note: By Characteristic roots, solve recurrence relations 23; By Generating functions, solve recurrence relations 18,20. 1.Determine the number of n digit numbers with all digits at least 4, such that 4 and 6 each occur an even number of times, and 5 and 7 each occur at least once, there being no restriction on the digits 8 and 9. 2.a)Find a recurrence relation for the number of ways to climb n stairs if the person climbing the stairs can take one stair or two stairs at a time. b) What are the initial conditions? 3.a) Find a recurrence relation for the number of ternary strings that do not contain two consecutive 0s. b) What are the initial conditions?