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Recurrence Relations (RRs)

Recurrence Relations (RRs). A “ Recurrence Relation ” for a sequence {a n } is an equation that expresses a n in terms of one or more of the previous terms in the sequence (i.e., a 0 ,a 1 ,a 2 ,…,a n-1 ) for all n ≥n 0 . Examples: a 0 =1, a 1 =3, a 2 =4; for n ≥3, a n = a n-1 +a n-2 -a n-3

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Recurrence Relations (RRs)

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  1. Recurrence Relations (RRs) • A “Recurrence Relation” for a sequence {an} is an equation that expresses an in terms of one or more of the previous terms in the sequence (i.e., a0,a1,a2,…,an-1) for all n≥n0. • Examples: a0=1, a1=3, a2=4;for n ≥3, an= an-1+an-2-an-3 1,3,4,6,7,9,10,12,13,15,16,18,19,21,22,… a0=2, a1=4, a2=3;for n ≥3, an= an-1+an-2-an-3 2,4,3,5,4,6,5,7,6,8,7,9,8,10,9,11,… a0=4, a1=3, a2=3;for n ≥3, an= an-1+an-2-an-3 4,3,3,2,2,1,1,0,0,-1,-1,-2,-2,-3,-3,-4,-4,… UCI ICS/Math 6A, Summer 2007

  2. Solutions to Recurrence Relations • A sequence {an} is called a “solution” of the recurrence relation if its terms satisfy the recurrence relation. Examples: For n ≥3, an= an-1+an-2-an-3; Initial conditions: a0=1, a1=3, a2=4.Solution: 1,3,4,6,7,9,10,12,13,15,16,18,19,21,22,… For n ≥3, an= an-1+an-2-an-3; Initial conditions: a0=2, a1=4, a2=3. Solution: 2,4,3,5,4,6,5,7,6,8,7,9,8,10,9,11,… For n ≥3, an= an-1+an-2-an-3; Initial conditions: a0=4, a1=3, a2=3. Solution: 4,3,3,2,2,1,1,0,0,-1,-1,-2,-2,-3,-3,-4,-4,… Every recurrence relationship has many solutions each determined uniquely by its own initial conditions. We say a function f:N→R is a “solution” to a recurrence relation if the sequence {f(n)} is a solution to it.Example: f(n)=5 ∙2n is a solution to the recurrence relation an=2∙an-1. UCI ICS/Math 6A, Summer 2007

  3. Solving Recurrence Relations • If an=4an-1 and a0=3, find a function f such that f(n)=an. an=4an-1=42an-2=43an-3=...=4na0=34n. • If an=an-1+n and a0=4, find a function f such that f(n)=an. an=an-1+n=an-2+n+(n-1) =an-2+n+(n-1)+(n-2)=...=a0+n+(n-1)+(n-2)+...+1=4+n(n+1)/2=(n2+n+8)/2. • If an=2nan-1 and a0=5, find a function f such that f(n)=an. an=2nan-1=22n(n-1)an-2=23n(n-1)an-3=...=2nn!a0=5 2nn!. • If an=2an-1+1 and a0=0, find a function f such that f(n)=an. an=2an-1+1=22an-2+2+1=23an-3+4+2+1=24an-4+8+4+2+1=...=2n-1+2n-2+2n-3+...+8+4+2+1=2n-1. UCI ICS/Math 6A, Summer 2007

  4. Modeling with Recurrence Relations • An initial deposit of P0 dollars deposited at 7% annual interest. Pn=(1+0.07)Pn-1 is the value after n years. Pn=(1+0.07)nP0 • Rabbits on an island. Each pair producing a new pair every month. Fibonacci Numbers: f0=0, f1=1; for n ≥2, fn= fn-1+fn-2. • Tower of Hanoi Disks of decreasing diameter on 1 of 3 pegs. Move disks to another peg, always maintaining decreasing disk diameters on each peg. Hn = number of moves to transfer n disks from 1 peg to another. H1=1; for n ≥2,Hn= Hn-1+1+Hn-1 =2Hn-1+1. 1,3,7,15,31,63,127,255,511,1023,2047,4095,8191,16383,32767,65535, Hn= 2n-1. UCI ICS/Math 6A, Summer 2007

  5. RRs for Counting Bit Strings • How many bit strings of length n do not contain 2 consecutive 0’s? b1=2 ({0,1}), b2=3 ({01,10,11}) For n ≥2: 01counted or 1counted bn = bn-2 + bn-1 . • Recognize this? • How many bit strings of length n contain 2 consecutive 0’s? b0=b1=0 • For n ≥2: 00any or 01counted or 1counted bn = 2n-2 + bn-2 + bn-1 . • How many bit strings of length n contain 3 consecutive 0’s? b0=b1=b2=0 • For n ≥3: 000any or 001counted or 01counted or 1counted bn = 2n-3 + bn-3 + bn-2 + bn-1 . UCI ICS/Math 6A, Summer 2007

  6. Counting Code Words • How many strings of n digits have an even number of 0’s? • a1=9 • For n ≥2: Either the first digit isn’t 0 and the rest has an even number of 0’s (there are 9an-1 of these) or the first digit is 0 and the rest does not have an even number of digits (there are 10n-1-an-1 of these) an = 9an-1+(10n-1-an-1)=8an-1+10n-1 UCI ICS/Math 6A, Summer 2007

  7. Catalan Numbers • Cn = number of ways to parenthesize the product of n+1 numbers.C1=1: (1x2); C2=2: (1x(2x3)), ((1x2)x3) • C3=5: (1x(2x(3x4))), (1x((2x3)x4)), ((1x2)x(3x4)), ((1x(2x3))x4), (((1x2)x3)x4) • C0=C1=1 • Cn=C0Cn-1+C1Cn-2+C2Cn-3+…+Cn-3C2+Cn-2C1+Cn-1C0(Covered in Sec.7.4, Ex.41) Cn = Number of extended binary trees with n internal nodes. 2 5 14 UCI ICS/Math 6A, Summer 2007

  8. Solving Recurrence Relations • We’ve seen several examples of Recurrence Relations an= a∙an-1 bn= n∙bn-1 fn =fn-2 +fn-1 bn=2n-2+bn-2+bn-1 bn=2n-3+bn-3+bn-2+bn-1 . bn= bn-1+bn-2-bn-3 Hn=2Hn-1+1 • In each case, many sequences satisfy the relationship and one also needs to set/have initial conditions get a unique solution. • In some cases, we also have nice (“closed form”) function, f, for the sequence, giving the nth term in a formula that doesn’t depend on earlier ones, making it so that the sequence is just {f(n)}. • (For example, an=an and Hn=2n-1.) • We look at some classes of recurrence relations where we can do this. UCI ICS/Math 6A, Summer 2007

  9. Recurrence Relations • When c1, c2, …, ck are constants and ck≠0, an= c1an-1+c2an-2+…+ckan-k+F(n) • is said to be a linear recurrence relation of degree k with constant coefficients . • Linear = each aj appears only to the 1st power, no aj2, aj3, . . . • degree k = k previous terms, ck≠0. • constant coefficients = cj are constants, not functions cj(n). • Additionally, if F(n)=0, the relation is called homogeneous. • Homogeneouslinear recurrence relation of degree k with constant coefficients:an= c1an-1+c2an-2+…+ckan-k where ck≠0 • We like these because we can solve them explicitly.  UCI ICS/Math 6A, Summer 2007

  10. Examples of Recurrence Relations • HLRRwCC = Homogeneous linear recurrence relation with constant coefficients • an= a∙an-1 HLRRwCC of degree 1 • an= (an-1)2Not Linear • an= n∙an-1 Coefficients are not constant • fn =fn-2 +fn-1 HLRRwCC of degree 2 • bn=2n-2+bn-2+bn-1 . Not Homogeneous • bn=2n-3+bn-3+bn-2+bn-1 Not Homogeneous • an= an-1+an-2-an-3 HLRRwCC of degree 3 • Hn=2Hn-1+1 Not Homogeneous • Remember: A recurrence relation has many solutions. Only when the initial conditions are specified is the solution unique. For degree k, you need k contiguous initial conditions. UCI ICS/Math 6A, Summer 2007

  11. Solving Linear (degree 1) HLRRwCC • The relation is an= c∙an-1 • All solutions are of the form an= d∙cnThe function is fd(n)= d∙cn • With the initial condition a0 specified, the unique solution is an= a0cn • We saw this in computing compound interest An initial deposit of P0 dollars deposited at 7% annual interest.Pn=(1+0.07)Pn-1 is the value after n years. Pn=(1+0.07)nP0 UCI ICS/Math 6A, Summer 2007

  12. Fibonacci Solved • When we had the HLRRwCC of degree 1, f(n)=c∙f(n-1), we had such good luck with fd(n)=d∙cn , let’s try F(n)=rn on the Fibonacci numbers where we have the HLRRwCC of degree 2: F(n)=F(n-1)+F(n-2). • If we have F(n)=rn and F(n)=F(n-1)+F(n-2), then r2=r+1.There are actually 2 solutions to this: r1=(1+√5)/2 and r2=(1-√5)/2 • Letting Fi(n)= ri n (for i=1,2), we have Fi(n-1)+Fi(n-2) = ri n-1+ri n-2 = ri n-2(ri +1) = ri n-2ri 2 =ri n = Fi(n)which is just what we want except we also want F(0)=0 and F(1)=1. • Combining our 2 solutions (see lemma below) as F(n)=d1∙F1(n)+d2∙F2(n), we get 0=d1+d2 and 1=d1∙r1+d2∙r2 ,and so 1=d1∙(r1-r2 )=d1∙(√5) • Giving the Fibonacci numbers as UCI ICS/Math 6A, Summer 2007

  13. Solutions to HLRRwCC • We like HLRRwCC because we can solve them explicity.That is, we can find functions f:N→R so that thesequence {f(n)} solves/satisfies the recurrence relation.Here’s part of the reason why. • Lemma: If functions f and g are solutions to the HLRRwCC an= c1an-1+c2an-2+c3an-3+…+ckan-k then f+g is also a solution as is d∙f for any constant d. • Definitions: The characteristic equation of this HLRRwCC is rk= c1rk-1+c2rk-2+c3rk-3+…+ck-1r+ckThe solutions (roots) of this equation are called thecharacteristic roots of the recurrence relation UCI ICS/Math 6A, Summer 2007

  14. (Fibonacci) Degree 2 HLRRwCC • 1) Write out the characteristic equation. (For Fibonacci: r2=r+1) • 2) Find the characteristic roots (r1 and r2). • 3) Any/Every function of the form f(n)=d1∙r1 n+d2∙r2 n(d1 and d2 constants) satisfies/solves the recurrence relation. • 4a) If the characteristic roots are distinct, we can pick d1 and d2 to produce the required initial values. • 4b) If there is only 1 characteristic root (r1=r2=r1), then any/every function of the form f(n)=(d1+d2∙n)∙rn (d1 and d2 constants) satisfies/solves the recurrence relation. • Example: a0=1, a1=4; for n ≥2, an= 4an-1-4an-2 1,4,12,32,80,192,448, Characteristic Equation: r2=4r-4  r2-4r+4=0  r=2 (twice) f(n)=(d1+d2∙n)∙2n & 1=d1 & 4=2d1+2d2d1=d2=1; f(n)=(1+n)∙2nand checking (“just for fun”) f(6)=(1+6)∙26=7∙64=448 UCI ICS/Math 6A, Summer 2007

  15. All Solutions For Any HLRRwCC • Thrm: If rk= c1rk-1+c2rk-2+…+ck-1r+ck (ck≠0) has k distinct roots, r1,r2,…,rk, then every solution of the recursion relation an=c1an-1+c2an-2+…+ckan-k has the form an= d1r1n+d2r2n+…+dkrkn for some d1, d2, …, dk and every such sequence solves/satisfies the recursion relation. If there are t distinct roots, each with multiplicity mi, the sequences {an} solving the recursion relation are given by UCI ICS/Math 6A, Summer 2007

  16. Degree 3 HLRRwCC Examples • a0=2, a1=5, a2=15; for n ≥3, an= 6∙an-1-11∙an-2- 6∙an-3 • r3= 6∙r2-11∙r-6 r3-6∙r2+11∙r+6=0  (r-1)(r-2)(r-3)=0 • General solution: an= x∙1n+y∙2n+z∙3n . • Initial values: 2=x+y+z; 5=x+2y+3z; 15=x+4y+9z  x=1; y=-1; z=2 • Specific solution: an= 1-2n+2∙3n . • a0=1, a1=-2, a2=-1; for n ≥3, an= -3∙an-1-3∙an-2-an-3 • r3= -3∙r2-3∙r-1 r3+3∙r2+3∙r+1=0  (r+1)3=0  r=-1 with multiplicity 3 • General solution: an=(x+y∙n+z∙n2)∙(-1)n . • Initial values: 1=x; 2=x+y+z; -1=x+2y+4z  x=1; y=3; z=-2 • Specific solution: an= (1+3n-2n2) (-1)n. UCI ICS/Math 6A, Summer 2007

  17. Inhomogeneous Recurrence Relations • Thrm: If f:N→Ris any solution to the recurrence relation an= c1an-1+c2an-2+…+ckan-k+F(n) (1)and g:N→Ris a solution to the corresponding homogeneous RR an= c1an-1+c2an-2+…+ckan-k (2) • then f-g his also a solution to (1). • Moreover, if h:N→Ris a solution to (1) then, f-h is a solution to (2). • Thrm: With (1) and (2) as above, if F(n) has the form F(n)=(btnt+bt-1nt-1+ . . .+b1n+b0) snthen there is a particular solution to (1) of the form nm(ptnt+pt-1nt-1+ . . .+p1n+p0) snwhere m is the multiplicity of s as a characteristic root of (2)and m=0 if s is not such a root. UCI ICS/Math 6A, Summer 2007

  18. Inhomogeneous RR Example • To Solve: an= 3∙an-1+2∙n; a1=3 • an= 3∙an-1 has as its general solution an= c∙3n • F(n)=2n. • We take s=1 and look for a solution of the form f(n)=bn+d an= 3∙an-1+2∙n  bn+d=3(b(n-1)+d)+2n  2n(b+1)+(2d-3b)=0  Pick b=-1, d=3b/2=-3/2 f(n)=-n-3/2 is one solution and/but its initial value is -5/2 (we get the sequence -5/2, -7/2, -9/2, -11/2, . . .) • General solution: an = c∙3n-n-3/2 • Initial condition: 3 = a1= c∙3-1-3/2 = 3c-5/2  c=11/6 • Specific solution: an = -(2n+3-11∙3n-1)/2 UCI ICS/Math 6A, Summer 2007

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