CHAPTER TWORECURRENCE RELATION

BY Dwee

Background

Computer programming Given a function to find faktorial n! in Pascal

We can make this function in another way, which is defined by recalling itself (recursive function)

function faktorial(n:integer):longint;Var i:integer;Begin faktorial:=1; for i:=1 to n do faktorial:=faktorial*i;End;

Background

Recursive function

This function is more simple but sometimes it use more memory than the function before

Some time we can solve a problem easily by recursive function than using iteration

function faktorial(n:integer):longint;Begin if n=1 then faktorial:=1 else faktorial:=n*faktorial(n-1);End;

Background

After this course, you should be able to Solve a some simple kinds of recurrrence

relation Use forward and backward subtition to find the

solution of recurrence relation Make recurrence relation model from a problem Solve 1st and 2nd order homogenous linear

recurrence relation with constant coeficient Solve 1st and 2nd order non-homogenous linear

recurrence relation with constant coeficient

INTRODUCTION

Let a geometric series: 5, 15, 45, 135, …. If we write the terms as a1, a2, a3, …, an we have

a2/a1=a3/a2=a4/a3=….=an/an-1=3 or we can write the relation between the terms in recurrence relation as

an = 3 an-1, n 0 (*) But recurrence relation (*) doesn’t define a unique

geometric series, because 1,3,9, 27 ,…. also satisfy the relation.

To pinpoint the particular sequence described by an = 3 an-1, n 0, we need to know one of the terms of that sequence

INTRODUCTION

Hence:,an = 3 an-1, n 0, a0=5

uniquely define sequence 5, 15, 45, ….

Recurrence Relation

Recurence Relation : a relation that describe a function a(n), written as an, n 0, where an is depend on the the prior terms an-1, an-2, an-3, …, a1, a0.

Example: Fibonacci Series: 1, 1, 2, 3, 5, 8, ….By recurence relation it is given as

a0= 1, a1=1, an=an-1+an-2 (n2)

Recurrence Relation

Recurrence Relation consists of two parts A set of initial or boundary condition A rule or recurrence part

Example: Fibonacci seriesan=an-1+an-2 (recurrence

part)with a0= 1, a1=1,(n2) (initial condition)

Recurrence Relation

Solution of recurrence relation General Solution Iterative procedure computing

Forward subtitution Backward subtitution

Example:1. Find the general solution of recurrence relation

a. an= 3an-1, a1= 2, n2

b. an = an-1 + n, a0=0, n1

c. an = 2an-1 + 1, a0=0, n1

Recurrence Relation

Example: Find a8 from this recurrence relation using

Forward and Backward subtitutiona0= 1, a1=1, an=an-1+an-2 (n2)

Find a6 from this recurrence relation using forward and backward subtitution

with a1 =1, a2=1, n n3

Modelling Problems to Recurrence Relation

A Bank pay 6% annual interest in savings, compounding the interest monthly. If Bonnie deposits $1000 on the first day on May, how much will this deposit b e worth a year later? Express the problem in the recurrence relation first.

Find a recurrence relation for the number of regions into which the plane is divided by n straight lines if every pair of lines intersect, but no three lines, meet the common point

Find a Recurrence Relation for the number of n digits binary sequences with no consecutive ones.

Modelling Problems to Recurrence Relation

Recurrence Relation with more than one variable Examples: Pascal Identity

C(n,r) = C(n-1, r-1) + C(n-1,r)

with boundary condition C(n,0)=1 and C(n,n)= 1.

Use backward subtition to find C(5,4) by this recurrence relation!

Example: Find Recurrence relation for P(n,r) the number of r-permutation from {x1, x2, …, xn}!

Linear Recurrence Relation with Constant Coefficient

General linear recurrence relation of degree kan + h1(n) an-1+ h2(n)an-2 + …+ hr(n)an-

k=f(n)

where h and f are functions and hr 0.

If f(n) = 0 then the relation is called homogeneous, otherwise the relation is inhomogenous. And f is called inhomogenous part. If h’s are constant functions then the relation is called linear recurrence relation with constant coefficient.

Linear Recurrence Relation with Constant Coefficient

A well defined linear recurrence relation of degree k consist of a recurrence relation part and k initial condition for k consecutive values define one and only one function (solution)

Example : an – 5 an-1 + 6 an-2 = 0 is satisfied by an = C12n + C23n for any coonstant C1 and C2.Let initial condition are a0 = 2 and a1=5, we need

a0 = 2 = C120 + C230 2 = C1 + C2

a1 = 5 = C121 + C231 5 = 2C1 + 3C2

this system has the solution C1=1 and C2 = 1. So the solution of the recurrence relation is an = 2n + 3n

Linear Recurrence Relation with Constant Coefficient

Given a recurrence part of degree k, the strategy is to find a solution with k arbitrary constant C1, C2, …, Ck such that we can satisfy any set of k consecutive initial condition by solving a system of k simultaneous equation (one for each initial condition) in k unknowns (Ci)

Such a solution is called a general solution

Homogenous Linear Recurrence Relation with Constant Coefficient : The Method of Characteristic Root Superposition Principle:

If g1(n) is a solution of an + c1 an-1+ c2an-2 + …+ ckan-k=f1(n) and if g2(n) is a solution of an + c1 an-1+ c2an-2 + …+ ckan-

k=f2(n) then C1g1(n) +C2 g2(n) is a solution of an + c1 an-1+ c2an-2 + …+ ckan-

k=C1 f1(n)+ C2 f2(n) for any constant C1 and C2

Its follow immediately if g1 (n), g2(n), …, gk(n) are solutions of an + c1 an-1+ c2an-2 + …+ ckan-k=0 then so is C1g1 (n)+ C2g2(n)+ …+Ck gk(n)

How do we find different solution gi(n)? One of the choice is to look for solution in form rn for some number r

Homogenous Linear Recurrence Relation with Constant Coefficient : The Method of Characteristic Root Given Recurrence relation:

an + c1 an-1+ c2an-2 + …+ ckan-k=0 Let an = rn , subtituting to the recurrence relation yields

rn + c1rn-1+ c2rn-2 + …+ ckrn-k=0

rn-k(rk + c1rk-1+ c2rk-2 + …+ ck)=0

rk + c1rk-1+ c2rk-2 + …+ ck=0 The last equation is called by characteristic equation.

Its roots are called characteristic roots. Base on the characteristic roots, we have two case of the

solutions Case 1 : All the roots are distinct Case 2 :There are some multiplicity roots

Homogenous Linear Recurrence Relation with Constant Coefficient : The Method of Characteristic Root Case 1: All of characteristic roots are distinct

If r1, r2, …, rk is distinct characteristic root, so by using superposition principle, the general solution of the recurrence relation is an = C1r1

n + C2r2n+…+Ckrk

n

Example: Solve the recurrence relation 1) an + an-1 – 6 an-2 = 0, a0=1, a1=2, n2

2) an = an-1 + an-2, a0=0, a1=1, n2 (Fibonacci series)

3) an=2an-1+an-2-2an-3 , a0 =0, a1=1, a2=1, n3

Homogenous Linear Recurrence Relation with Constant Coefficient : The Method of Characteristic Root If the charactristic roots are complex numbers Recall De Moivre’s Theorem: (cos + i sin )n =

cos n+ i sin (n).If z = x + iy, we can write z = r (cos + i sin ) where r = sqrt(x2+ y2) and (y/x) = tan then zn = (r (cos + i sin ) )n= rn (cos n+ i sin n).

Example: Solve the recurrence relation an = 2an-1-2an-2 , a0=1, a1=2, n2

Homogenous Linear Recurrence Relation with Constant Coefficient : The Method of Characteristic Root Case 2: The characteristic equation has

multiplicity roots Example: Solve the recurrence relation

an = 4an-1-4an-2 , a0 =1, a1=3, n2

If r is a characteristic root of multiplicity m, then it contributes m solution: rn, nrn, n2rn, …, nm-1rn.

Examples:Solve The recurrence relation a0=1, a1=1, a2=2,

an = 4an-1-5an-2+2an-3

Inhomogeneous Recurrence Relation Towers of Hanoi. Consider n circular disks (having different

diameters) with holes in their centers. These disks can be stacked on any of the pegs shown in Fig. 10.11. In the figure, n = 5 and the disks are stacked on peg 1 with no disk resting upon a smaller one. The objective is to transfer the disks one at a time so that we end up with the original stack on peg 3. Each of pegs 1, 2, and 3 may be used as a temporary location for any disk(s), but at no time are we allowed to have a larger disk on top of a smaller one on any peg. What is the minimum number of moves needed to do this for n disks?

Inhomogeneous Linear Recurrence Relation with Constant Coefficient

Recurrence relation: an + c1 an-1+ c2an-2 + …+ ckan-k= f(n) with r consecutive initial conditions

Our strategy is the same as that used in the homogeneous case. Find the general solution to the recurrene part, and use

the initial condition to set up a system of simultaneous equation

By superposition principle: If anh is the solution of an

+ c1 an-1+ c2an-2 + …+ ckan-k= 0 and if anp is the

solution of an + c1 an-1+ c2an-2 + …+ ckan-k= f(n) then an

h + anp is also the solution of the recurence

relation

Inhomogeneous Linear Recurrence Relation with Constant Coefficient

anh is homogeneous part solution and an

p is particular solution

How to find anh ? Using characteristic equation

How to find anp? No general method, but some

techniques are available for certain special case (method of undetermined coefficient).

Example: Solve the recurrence relation: an- 3an-1=5 (7)n, a0=2, n1 Solve the recurrence relation: an+3an-1=4n2-2n, a1=-4, n2 Solve the recurrence relation: an- 4an-1+4an-2= 2n, a0=0

a1=1, n2

Inhomogeneous Linear Recurrence Relation with Constant Coefficient

• Solve the recurrence relation :an- 3an-1=5 (7)n, a0=2, n1• Answer:

Homogenous part: an- 3an-1=0

Characteristic equation: r-3=0, charactristic root :r=3Homogenous part solution: G(n)=c3n

Inhomogenous part: an- 3an-1=5(7)n

We are “guessing” that the particular solution p(n)=A(7n)Subtituting p(n) into RR: A(7n)-3 A(7n-1)=5 (7n)

7n-1 (7A-3A)=7n-1(5.7) 4A = 35 A=35/4

General solution: an = G(n)+p(n) = c 3n + 35/4 (7)n

with a0 = 2, we get c= … so the solution is an = …..

Inhomogeneous Linear Recurrence Relation with Constant Coefficient

• Solve the recurrence relation: an+3an-1=4n2-2n, a1=-4, n2 Answer:

Homogenous part: an+3an-1=0

Homogenous part solution: G(n)=c1(-3)n Inhomogenous part: an+3an-1=4n2-2n

Particular solution is polunomial of degree 2 or higherGuess that p(n) is polynomial of degree 2, p(n) = An2+Bn+C. Subtituting p(n) the RR: An2+Bn+C + 3(A(n-1)2+B(n-1)+C)= 4n2-2n4An2+(4B-6A)n+(3A-3B+4C)=4n2-2nSolving the equation ,we have: A=1, B=1, C=0. The particular solution is p(n) = n2+n

General solution is an=G(n)+p(n)= c1(-3)n + n2+n

Find the solution based on the initial condition

• Solve the recurrence relation an-an-1=n, a0=1, n1 Answer:

Homogenous part: an-an-1= 0

Homogenous part splution part: anh = c(1)n=c

Inhomogenous part: an-an-1=n

Because f(n) =n so the particular solution is polynomial of degree 1 or higher.Find the particular solution using guess p(n) = An +B p(n)= An2 +Bn +C

which one give the solution? Why?

Inhomogeneous Linear Recurrence Relation with Constant Coefficient

How to determine the degree of the polynomial guess?

The degree of anp is the degree of f(n) plus the

multilicity of 1 in the characteristic equation If 1 is not a root of characteristic equation,

the degree of anp is the degree of f

Similarly if f(n) is of the form rn, then the particular solution will be Ankrn, where k is the multiplicity of r in the characteristic equation

Inhomogeneous Linear Recurrence Relation with Constant Coefficient

Solve the recurrence relationan- 4an-1+4an-2= 2n, a0=0 a1=1, n2

Answer: Homogenous part: an- 4an-1+4an-2= 0

Characteristic equation: r2 – 4r + 4 = 0Characteristic root: r = 2 (multiplicity of degree 2)Homogenous part solution: G(n) = ……

Inhomogeneous part: an- 4an-1+4an-2= 2n

Particular solution is p(n)= … The general solution is an = ….

Given a linear nonhomogeneous recurrence relation (with constant coefficients) of the form C0an + C1an-1 + C2an-2 + • • • + Ckan-k = f(n)

Let Gn denote the homogeneous part of the solution an.

EXERCISE. a0=1, an-an-1= n2

a0=1,a1=2, an-5an-1+6an-2 = 2n+1, n>=2 a0=4, an-an-1=2n2-n-1, n>=1 a0=1, a1=0, an-2an-1+an-2=2, n>=2 a0=1, an+3an-1 =2n, n>=1 a0=1, an+2an-1=2n-n2 , n>=1