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Solving Recurrence Relations Using Characteristic Roots

Learn to solve recurrence relations through examples using characteristic roots and generating functions, including linear homogeneous and nonhomogeneous equations. Explore the Tower of Hanoi and Fibonacci sequence problems with step-by-step solutions.

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Solving Recurrence Relations Using Characteristic Roots

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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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