Advanced Counting Techniques. 7.1 Recurrence Relations

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1 Chapter 7 Advanced Counting Techniques 71 Recurrence Relations We have seen that a recursive definition of a sequence specifies one or more initial terms and a rule for determining subsequent terms from those that precede them A rule similar to this is called a recurrence relation Definition 1 A recurrence relation for the sequence is an equation that express terms of one or more of the previous terms of the sequence A sequence called a solution of a recurrence if it satisfies the recurrence relation Example 1 Let be a sequence that satisfies the recurrence relation for n=2,3,4,, and suppose that and What are and? Solution: Clearly =5-3=2, =2-5=-3 and so on Modeling with Recurrence Relations Example 3 Suppose that a person deposits $10,000 in a savings account at a bank yielding 11% per year with interest compound annually How much will be in the account after 30 years? Solution: Let Pn: the amount in the account after n years Then P0=10,000; P1=(111)P0, P2=111P1,, Pn=111P(n-1) That is can prove the validity of this formula We So n=30, P0=$10,000 by the formula gives us =$228,92297 Example 4 Rabbits and Fibonacci Numbers A young pair of rabbits is placed on an island A pair of rabbits does not breed until they are 2 month old After they are 2 month old, each pair of rabbits produces another pair each month Find a recurrence relation for the number of pairs of rabbits on the island after n month, assuming that no rabbits ever die 1

2 Solution: Let fn be the number of pairs of rabbits after n months We will show that fn, for n=1, 2, 3,, are the terms of the Fibonacci sequence At the end of first month, the number of pairs of rabbits f1=1 At the end of second month, the number is still f2=1, because no breeding From the third month on, after n months, the number of pairs fn is then the number of the pairs of the previous month f(n-1) plus the number of newborn pairs f(n-2), because each new born pair comes from a pair at least 2 months old So for n 3 This is of course the Fibonacci sequence Example 5 The Tower of Hanoi There is 3 pegs mounted on board together with disks of different sizes Initially these disks are placed on the first peg in order of size, with the largest on the bottom The disks are allowed to be moved one at a time from one peg to another as long as a disk is never placed on top of a smaller disk The goal is to have all the disks on the second peg in order of size, with the largest on the bottom See Figure2 Let Hn be the number of moves needed to solve the problem with n disks Set up a recurrence relation for the sequence Solution: We can transfer the top n-1 disks, by the rules, to peg 3 using H(n-1) moves See Figure3 We did not move the largest disk Now we move the largest disk to the second peg, this one move We can transfer n-1 disks on peg 3 to peg 2 using additional H(n-1) moves, placing them on top of the largest disk, which stays fixed on the bottom of peg 2 This shows that The initial condition is H1=1 Moving one disk needs one step For the rest 2

3 72 Solving Linear Recurrence Relations We study one important class of recurrence relations that express terms of a sequence as a linear combination of previous terms, which can be solved explicitly in a systematic way Definition 1 A linear homogenous recurrence relation of degree k with constant coefficients is a recurrence relation of the form an= c1an-1+ c2an-2+ + ckan-k, where c1, c2,, ck R and ck 0 Keywords: linear, homogeneous, constants- c1, c2,, ck ; degree - k Example 1 The recurrence relation Pn=(111)Pn-1 is linear, homogeneous recrel of degree one The rec rel fn=fn-1+ fn-2 is linear, homogeneous, recrel of degree 2 The rec rel an=an-5 is linear, homogeneous, recrel of degree 5 Example 2 The recurrence relation is homogeneous, degree of two, but not linear The rec rel Hn=2Hn-1+1 is linear, degree of one, but not homogeneous The rec rel Bn=nBn-1 is homogeneous, linear, degree of one, but the coefficient is not constant Solving Linear Homogeneous Recurrence Relations with Constant Coefficients We look solutions of the form for constant r is a solution of the rec rel an= c1an-1+ c2an-2+ + ckan-k The last equation is called the characteristic equation of the rec rel and the solutions of the equation is called the characteristic roots of the rec rel We treat only linear recrel of degree two Theorem 1 3

4 Let c1, c2 R Suppose that r*r- c1r- c2=0 has two real roots r1 r2 Then the sequence {an} is a solution of the recrel an= c1an-1+ c2an-2 for n=0,1, 2,, where α1 and α2 are constants Example 3 What is the solution of the recurrence relation an=an-1+2an-2 with a0=2 and a1=7? Solution: We use Theorem 1 The characteristic equation of recrel is r*r r-2=0 So the roots are r=2 and r=-1 and so the solution of the recrel is for some constant α1 and α2 From the initial condition we know that a0=2=α1 + α2, a1=7=2α1 +(-1)α2 Thus α1 = 3 and α2 =-1 Example 4 Find an explicit formula for the Fibonacci numbers Solution: Fibonacci numbers: fn= fn-1+ fn-2 and the initial conditions: f0=0 and f1=1 The treatment is similar to the Example 3 Theorem 2 Let c1, c2 R and c2 0 Suppose that r*r- c1r- c2=0 has only one root r0 Then the sequence {an} is a solution of the recrel an= c1an-1+ c2an-2 for n=0,1, 2,, where α1 and α2 are constants Example 5 What is the solution of the recurrence relation an=6an-1-9an-2 with initial conditions a0=1 and a1=6? Solution: The root to r*r-6r+9=0 is r=3 So the solution to this rec rel is for some constant α1 and α2 The rest is similar to the Example 3 4

5 Linear Non-homogeneous Recurrence Relations with Constant Coefficients A linear non-homogeneous recurrence relation with constant coefficients is a rec rel of the form an= c1an-1+ c2an-2+ + ckan-k +F(n), (E*) where c1, c2,, ck R and F(n) is a function only depends on n and not identically zero Here an= c1an-1+ c2an-2+ + ckan-k (E**) is the associated homogeneous recurrence relation Theorem 5 If is a particular solution of the recrel (E*), then every solution is the from, where is the solution of the associated rec rel (E**) Example 10 Find all solutions of the recurrence relation an=3an-1+2n What is the solution with a1=3? Solution: The associated rec rel is an=3an-1 and so the solutions are, where α1 is a constant Since F(n)=2n, so the trial solution is a linear function in n, ie, pn=nc+d So the equation an=3an-1+2n becomes nc+d=3((n-1)c+d)+2n Solving the equation we get c=-1 and d=-3/2 So the particular solution is now pn=-n-3/2 By Theorem 5, all solutions are of the form an= pn+ =-n-3/2+, where α1 is a constant With a1=3 from the general solution we get 3=-1-3/2+, so α1 =11/6 Thus an=-n-3/2+(11/6) Example 11 Find all solutions of the recurrence relation Solution: The particular solution is pn=c By the rec rel we find C=49/20 If we find the solution of associated rec rel, then the problem is solved 5