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4.7 Recurrence Relations

4.7 Recurrence Relations. P13, P100 Definition: A recurrence relation for the sequence{a n } is an equation that expresses a n in terms of one or more of the previous terms of the sequence, namely, a 0 , a 1 , …, a n-1 , for all integers n with n  n 0 , where n 0 is a nonnegative integer.

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4.7 Recurrence Relations

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  1. 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.

  2. Example(Fibonacci sequence): • 13 世纪初意大利数学家 Fibonacci 研究过著名的兔子繁殖数目问题 • 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

  3. 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

  4. Using Characteristic roots to solve recurrence relations • Using Generating functions to solve recurrence relations

  5. 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.

  6. 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.

  7. 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

  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.

  9. 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+c21=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

  10. 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)

  11. 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*

  12. (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.

  13. Example: Find all solutions of the recurrence relation h(n)=2h(n-1)+1, n2, h(1)=1 • Example: Find all solutions of the recurrence relation • 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

  14. 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

  15. Example: Solve the recurrence relation: • an=an-1+9an-2-9an-3,n≥3 • subject to the initial valuesa0=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+…；

  16. 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。

  17. 期中测验时间： • 11月3日 • 课件 集合，关系，函数，基数,组合数学

  18. ⅠIntroduction to Set Theory • 1. Sets and Subsets • Representation of set: • Listing elements, Set builder notion, Recursive definition • , ,  • P(A) • 2. Operations on Sets • Operations and their Properties • A=?B • AB, and B A • Or Properties • Theorems, examples, and exercises

  19. 3. Relations and Properties of relations • reflexive ,irreflexive • symmetric , asymmetric ,antisymmetric • Transitive • Closures of Relations • r(R),s(R),t(R)=? • Theorems, examples, and exercises • 4. Operations on Relations • Inverse relation, Composition • Theorems, examples, and exercises

  20. 5. Equivalence Relation and Partial order relations • Equivalence Relation • equivalence class • Partial order relations and Hasse Diagrams • Extremal elements of partially ordered sets: • maximal element, minimal element • greatest element, least element • upper bound, lower bound • least upper bound, greatest lower bound • Theorems, examples, and exercises

  21. 6.Everywhere Functions • one to one, onto, one-to-one correspondence • Composite functions and Inverse functions • Cardinality, 0. • Theorems, examples, and exercises

  22. II Combinatorics • 1. Pigeonhole principle • Pigeon and pigeonholes • example，exercise

  23. 2. Permutations and Combinations • Permutations of sets, Combinations of sets • circular permutation • Permutations and Combinations of multisets • Formulae • inclusion-exclusion principle • generating functions • integral solutions of the equation

  24. Applications of Inclusion-Exclusion principle • example,exercise • Applications generating functions and Exponential generating functions • ex=1+x+x2/2!+…+xn/n!+…; • x+x2/2!+…+xn/n!+…=ex-1; • e-x=1-x+x2/2!+…+(-1)nxn/n!+…; • 1+x2/2!+…+x2n/(2n)!+…=(ex+e-x)/2; • x+x3/3!+…+x2n+1/(2n+1)!+…=(ex-e-x)/2; • examples, and exercises • 3. recurrence relation • Using Characteristic roots to solve recurrence relations • Using Generating functions to solve recurrence relations • examples, and exercises

  25. Next: • Graph theory • P115,4.2;P123 4.3;P290 8.1

  26. Exercise P104 18,20,23，Note: By Characteristic roots, solve recurrence relations 23 and By Generating functions, solve recurrence relations 18,20. • 1.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? • 2.a) Find a recurrence relation for the number of ternary strings that do not contain two consecutive 0s. • b) What are the initial conditions?

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