Discrete Structure Chapter 6-Recurrence Relation

8/10/2019 Discrete Structure Chapter 6-Recurrence Relation

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RECURRENCE

1. Sequence2. Recursively defined sequence3. Finding an explicit formula for

recurrence relation


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Learning OutcomesYou should be able to solve

first-order and second-order linearhomogeneous recurrence relation withconstant coefficients


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Preamble What is recurrence and how does it

relate to a sequence?


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Sequences A sequenceis an ordered list of objects (or events). Like a set,

it contains members(also called terms) Sequences can be finiteor infinite.

2,4,6,8,

for i1 ai= 2i(explicit formula)infinite sequence with infinite distinct values

-1,1,-1,1, for i 1 bi= (-1)i

infinite sequence with finite distinct values

For 1


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Ways to define sequence Write the first few terms:

3,5,7,

Use explicit formula for its nth term

an= 2n for n1

Use recursion How to define a sequence using a

recursion?


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Recursively defined sequencesRecursion can be used to defined a sequence.

This requires:

A recurrence relation: a formula that relates each term ak

to some previous terms ak-1

, ak-2

,

ak = ak-1+ 2ak-2

The initial conditions: the values of the first few terms a0, a1,

Example: For all integers k 2, find the terms b2, b3and b4:bk= bk-1+ bk-2 (recurrence relation)

b0= 1 and b1= 3 (initial conditions)

Solution:b2= b1 + b0 = 3 + 1 = 4

b3= b2+ b1 = 4 + 3 = 7b4= b3+ b2 = 7 + 4 = 11


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Explicit formula and

recurrence relation Show that the sequence 1,-1!,2!,-3!,4!,,(-1)nn!,

for n0, satisfies the recurrence relationsk= (-k)sk-1 for all integers k1.

The general term of the sequence: sn=(-1)nn!substitute k and k-1 for n to getsk=(-1)

kk! sk-1=(-1)k-1(k-1)!

Substitute sk-1 into recurrence relation:

(-k)sk-1= (-k)(-1)k-1

(k-1)!= (-1)k(-1)k-1(k-1)!= (-1)(-1)k-1 k(k-1)!= (-1)k k! = sk


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Examples of recursively

sequence Famous recurrences

Arithmetic sequences: ak= ak-1+ d

e.g. 1,4,7,10,13,geometric sequences: ak= ark-1e.g. 1,3,9,27,Factorial: f(n) = n . f(n-1)

Fibonacci numbers: fk= fk-1+fk-21,1,2,3,5,8,Tower of Hanoi problem


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Tower of Hanoi


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Application of recurrence Analysis of algorithm containing recursive function such as

factorial function.

Algorithm f(n)

/input: A nonnegative integer/output: The value of n!If n = 0 return 1else return f(n-1)*n

No. of operations (multiplication) determines the efficiency ofalgo.

Recurrence relation is used to express the no. of operation inthe algorithm.


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Solving Recurrence relation by

Iteration It is helpful to know an explicit formula for a

sequence. An explicit formula is called a solutionto the

recurrence relation Most basic method is iteration- start from the initial condition- calculate successive terms until a patterncan be seen- guess an explicit formula


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Some examplesLet a0,a1,a2, be the sequence defined recursively as follows: For

all integers k1,(1) ak= ak-1+2(2) a0= 1

Use iteration to guess an explicit formula for the sequence.a0=1a1=a0+2a2=a1+2=(1+2)+2 = 1+2.2a3=a2+2=(1+2.2)+2 = 1+3.2

a4=a3+2=(1+3.2)+2 = 1+4.3.Guess: an=1+n.2=1+2nThe above sequence is an arithmetic sequence.


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Geometric Sequence

Let r be a fixed nonzero constant, and suppose a sequencea0,a1,a2, is defined as follows:

ak= rak-1for all integers k 1a0= a

Use iteration to guess an explicit formula for the sequencea0=aa1=ra0=raa2=ra1=r(ra)=r

2aa3=ra2=r(r

2a)=r3a

Guess: an=rna = arn for all integers n0The above sequence is geometric sequence and r is a common

ratio.


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Explicit formula for tower of Hanoi

mn= 2n1. (exponential order)

To move 1 disk takes 1 second

m64= 2641 = 1.844674 * 1019 seconds

= 5.84542 * 1011years

= 584.5 billion years.


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Second-Order Linear Homogeneous withconstant coefficients

A second-order linear homogeneousrecur. relation with c.c. is a recur.

relation of the formak= Aak-1+ Bak-2for all integers k some fixed integer,

where A and B are fixed real numberswith B 0.


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Terminology

ak= Aak-1+ Bak-2 Second order: ak contains the two previous

terms Linear: ak-1 and ak-2appear in separate terms

and to the first power

Homogeneous: total degree of each term is

the same (no constant term) Constant coefficients: A and B are fixed real

numbers


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Examples

Second-Order Linear Homogeneous with constantcoefficients

ak= 3a

k-1+ 2a

k-2 - yes.

bk= bk-1+ bk-2+ bk-3 - no

dk= (dk-1)2+ dk-1dk-2 - no; not linear

ek= 2ek-2 - yes; A = 0, B = 2.

fk= 2fk-1+ 1 - no; not homogeneous


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Using the characteristic equation tofind sequences

Example: Consider the following recurrence relationak = ak-1+2ak-2for all k >= 2.Find sequences that satisfy the relation.Solution: For the given relation, A=1 and B=2.Relation is satisfied by the sequence 1,t,t2,t3, iff t satisfies the

characteristic equationt2AtB = 0 ort2t2 = 0(t2)(t + 1) = 0.

t = 2 or t = -1.Sequences: 1,2,22,23, and

1,-1,(-1)2,(-1)3, or 1,-1,1,-1, ,(-1)n,


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Linear combination ofsequences

Lemma

If r0,r1,r2,,rn,.. and s0,s1,s2,,sn, are sequences thatsatisfy the same second-order linear homogeneous

recurrence relation with c.c., and if C and D are anynumbers, then the sequence a0,a1,a2,defined by the formula

an= C.rn + D.snfor all integer n>=0

also satisfies the same recurrence relation. C and D can be calculated using initial conditions.


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Two possible solutions

For the characteristic equation

t2AtB = 0

there are two possible solutions:

- Distinct-roots case

- Single-root case


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Steps for finding explicitformula

1. Form the characteristic equation.

2. Solve the equationlet r and s be theroots. ( r s)

3. Set up an explicit formula:

ak= C.rk+ D.sk

4. Find C and D using initial conditions.


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Quadratic equation

ax2+ bx + c = 0

x = [-b +- (b24.a.c)]/(2.a)


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Single-Root Case

Two sequences that satisfy the relation

ak= A.ak-1+B.ak-2

where r is the root of t

2

- A.t - B = 0.Explicit formula for the new sequence

an= C.rn+ D.nrn


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Example

A sequence b0, b1, b2, satisfies the rec. relation

bk= 4bk-14bk-2for k>=2 with initial conditions b0=1and b

1

=3.

Find explicit formula for the sequence.

Solution: A=4 and B=-4

Charac eq: t24t +4 =0

(t-2)2

=0. t=2.Seq: 1,2,22, , 2n,..

0,2,2.22,3.23,,n.2n,

Explicit formula: bn= C.rn+ D.nrn

= C.2n+ D.n2n


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Example

bn= C.2n+ D.n2n

b0= 1 = C.20+ D.0.20= C

b1= 3 = C.2 + D.2 3 = 1.2 + 2D

Hence D = .

Therefore bn= 2n+ (1/2).n.2n

= 2n (1+ n/2) for integer n>=0. Sequence: 1,3,8,20,


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Question

?????


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Summary

How to express the sequence

Find explicit formula for first-order

recurrence relation

Find explicit formula for second-orderrecurrence relation (distinct and single

root)


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THANK YOU

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Discrete Structure Chapter 6-Recurrence Relation