Recurrence Relations

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Atiwong Suchato

Department of Computer Engineering, Chulalongkorn University Discrete Mathematics

 

• Readings:

Recurrence Relations Rosen section 7.1-7.2

Atiwong Suchato

 

• A recurrence relation for the sequence {a n } is an equation that expresses a n in terms of one or more of the previous terms, a 0 ,a 1 ,…,a n-1 .

• A sequence is called a solution of a recurrence relation if its terms satisfy the recurrence relation.

a n = 5a n-1

b n = b n-1 -2 b n-2 +100 c n = c n-3 + c n-4 +log(n)+e n

 

• Example (Rosen p.402):

Determine whether a n =3n, for every nonnegative integer n, is a solution of

a n = 2a n-1 -a n-2 ; n = 2, 3, …

 

The initial conditions specify the terms that precede the first term where the recurrence relation takes effect.

a n = 2a n-1 -a n-2 ; n = 2, 3, …

0 3 6 9 12 ……

1 2 3 4 5 ……

n = 0 1 2 3 4 …… a

_n

=3n

for non-negative int.

a

₀

= 0, a

₁

= 3

a

_n

=n+1

for non-negative int.

a

₀

= 1, a

₁

= 2

defined by the recurrence relation

initial conditions Solution # 1

Solution # 2

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Atiwong Suchato

 

In order to find a unique solution for every non-negative integers to:

b n = b n-2 +b n-4 ; n = 4, 5, …

how many terms of b

_n

needed to be given in the initial conditions?

b 2 b 3 b 4 b 5 b 6

n = 2 3 4 5 6

……

defined by the recurrence relation

initial conditions

b 7

7

b 0 b 1

0 1

Atiwong Suchato

   

To find solutions for doing a task of a size n Find a way to:

Construct the solution at the size n from the solution of the same tasks at smaller sizes.

⌧    

Rules:

Move a disk at a time from one peg to another.

Never place a disk on a smaller disk.

The goal is to have all disk on the 2 nd peg in order of size.

Find H n , the number of moves needed to solve the problem with n disks.

Example:

A man running up a staircase of n stairs. Each step

he takes can cover either 1 or 2 stairs. How

many different ways for him to ascend this

staircase?

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Atiwong Suchato

  

• A linear homogeneous recurrence relation can be solved in a systematic way.

Linear homogeneous recurrence relation of degree k with constant coefficients

a n = c 1 a n-1 + c 2 a n-2 + …+c k a n-k where c 1 , c 2 , …, c k are real numbers

and c k ≠ 0

Atiwong Suchato

  

     

• a n =a n-1 +a 2 n-2

• H n = 2H n-1 +1

• B n = nB n-1

• f n = f n-1 + f n-2

  

 

Recurrence Relation Characteristic Equation

General Solution Unique Solution

solve equation polynomial=0

initial conditions

follow a specific method

solve linear equations for arbitrary variables

  

 

• Recurrence Relation:

• Characteristic equation:

a n - c 1 a n-1 - c 2 a n-2 - … - c k a n-k = 0

r k - c 1 r k-1 - c 2 r k-2 … - c k = 0

Example:

a n = a n-1 +3a n-2

b n = 2b n-1 -b n-2 +5b n-3

d n = d n-2 +d n-5

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Atiwong Suchato

  

 

• Let c 1 , c 2 , …, c k be real numbers. Suppose the characteristic equation:

r k - c 1 r k-1 - c 2 r k-2 … - c k = 0

has k distinct roots r 1 , r 2 , …, r k . Then a sequence {a n } is a solution of the recurrence relation:

a n - c 1 a n-1 - c 2 a n-2 - … - c k a n-k = 0

if and only if:

a n = α 1 r 1 n + α 2 r 2 n + … + α k r k n

for n = 0, 1, 2, …. Where α 1 , α 2 ,…, α k are constants.

Atiwong Suchato

• Example :

What is the solution of the recurrence relation:

a n = a n-1 + 2a n-2 with a 0 = 2 and a 1 = 7?

• Example:

a n = 6a n-1 - 11a n-2 + 6a n-3 with a 0 = 2, a 1 = 5 and a 2 = 15 ?

 

• Suppose the characteristic equation has t distinct roots r 1 , r 2 , …, r t with multiplicities m 1 , m 2 , …, m t .

• Solution:

a n = ( α 1,0 + α 1,1 n + … + α 1,m1-1 n m1-1 )r 1 n +( α 2,0 + α 2,1 n + … + α 2,m2-1 n m2-1 )r 2 n + ………..

+( α t,0 + α t,1 n + … + α t,mt-1 n mt-1 )r t n

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Atiwong Suchato

• Example :

a n = -3a n-1 - 3a n-2 - a n-3 with a 0 = 1, a 1 = -2 and a 2 = -1 ?

Atiwong Suchato

 

   

 

a n = c 1 a n-1 + c 2 a n-2 + … + c k a n-k + F(n)

Associated homogeneous recurrence relation

{a n h } {a n p } {a n } = {a n h } + {a n p }

{a n } = {a n h } + {a n p }

 

   

 

• Key:

1 – Solve for a solution of the associated homogeneous part.

2 – Find a particular solution.

3 – Sum the solutions in 1 and 2

• There is no general method for finding the particular solution for every F(n)

• There are general techniques for some F(n) such as polynomials and powers of constants.

• Example:

Find the solutions of a n = 3a n-1 +2n with a 1 = 3

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Atiwong Suchato

• Example:

Find the solutions of a n = 5a n-1 - 6a n-2 +7 n

Atiwong Suchato

 

F(n) = ( b t n t + b t-1 n t-1 + … + b 1 n + b 0 ) s n where b 0 , b 1 , …, b t and s are real numbers.

When s is not a root of the characteristic equation:

The particular solution is of the form:

( p t n t + p t-1 n t-1 + … + p 1 n + p 0 ) s n When s is a root of multiplicity m:

The particular solution is of the form:

n m ( p t n t + p t-1 n t-1 + … + p 1 n + p 0 ) s n

• Example:

What form does a particular solution of a n = 6a n-1 – 9a n-2 + F(n) have when:

F(n)= 3n, F(n)= n3 n , F(n)= n 2 2 n , F(n)= (n 2 +1)3 n ?

• Example :

Find the solution of ∑

=

= n

k

n k

a

1