Recurrence Relations

How we will proceed

• 1. A definition of recurrence relations.

– A recurrence relationship is a rule by which a sequence is generated.

• 2. Given the initial condition and the rule (i.e. the recurrence relations) what is the sequence?

• 3. Given the rule and a sequence, is the sequence a solution of the recurrence relations?

• Facts about recurrence relations

• Modeling problems with recurrence relations

Definition of a recurrence relation

• 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 of the sequence.

– a_nis the next term in the sequence

• The sequence {a_n} looks like this: a₀, a₁, …a_n-1

– For all integers n with n >= n₀, where n₀ is a nonnegative integer

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

Given the recurrence relation and initial condition, find the sequence

• Let {a_n} be a sequence that satisfies the recurrence relation

_–Rule: a_n = a _n-1 – a _n-2

– Initial conditions: a₀ = 3 and a₁ = 5

– What is the sequence?

Given the rule and a sequence, is the sequence a solution of the recurrence relations?

_•Rule: a_n = 2a _n-1 – a _n-2

• Sequence: a_n = 3n

• Two ways to solve

– Figure out the sequence, and see if it satisfies the rule

the sequence is: a₀= 0, a₁=3, a₂=6, a₃=9, a₄= 12

at n = 4: a₄ ?= 2a₃ – a_{2

12 ?=
2(9) - 6}

– Substitute 3n for n in the rule and simplify
2a _n-1 – a _n-2= 2[3(n-1)] – 3(n-2)
when simplified, the does indeed = 3a

Given the rule and a sequence, is the sequence a solution of the recurrence relations?

_•Rule: a_n = 2a _n-1 – a _n-2

• Sequence: a_n = 2ⁿ

• The sequence is:
a₀= 1, a₁=2, a₂=4, a₃=8, a₄= 16

• Does this fit the rule?

_•At n=4: a₄ ?= 2a₃ – a₂

**16 ?= (2*8) – 4**

• So a_n = 2ⁿ is not a solution

Facts about recurrence relations

• The recurrence relation and initial conditions uniquely determine a sequence

• Any term of the sequence can be found from the initial conditions using the recurrence relation a sufficient number of times

• But, for a certain class of sequences defined by a recurrence relation and initial condition there are better ways to find any term

Overheads

Modeling rabbits

• A young pair of rabbits is placed on an island. After they are two months 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 6 months. After n months. (Assume no rabbits die.)

Towers of Hanoi

• There are three pegs and disks of different sizes.

• The object is to move all the disks from one peg to another

• The Rules

– One disk is moved at a time

– A disk may not be placed on top of a disk of smaller diameter

• Let H_n denote the number of moves needed to solve with n disks.

• Set up a recurrence relation for the sequence {H_n}

Recurrence relation for Towers of Hanoi

• What is the initial condition H₁?

– I.e. how many moves to move one disk?

• We want to develop a rule to tell us how many moves it will take to move n disks in terms of moving n-1 disks

Counting bit strings

• Find a recurrence relation and give initial conditions for the number of bit strings of length n that do not have two consecutive 0s.

• a₁ = num valid bit strings of length 1
{a₁} is 0, 1

• a₂ = num valid bit strings of length 2
{a₂} is 01, 10, 11

• a₃ = num valid bit strings of length 3
{a₃} is 010, 011, 101, 110, 111

• By the sum rule, the total number of bit strings of length n without two consecutive 0 bits is equal to the number ending with 0 plus the number ending with 1

What is the sum rule?

•If a first task can be done in n₁ ways and a second task in n₂ ways, and if these tasks cannot be done at the same time, then there are n₁ + n₂ ways to do either task

Using the sum rule

• Assume n >= 3

• How many strings of length n are there ending with 1?

– It’s all the strings in {a_n-1} with a 1 added

• How many strings of length n are there ending with a 0?

– It’s all the strings in {a_n-2} with a 10 added

• What is {a₃} given {a₂} and {a₁}?

•What is {a₄}?

How many code words are valid?

• A decimal digit code word is valid if it contains an even number of 0 digits

• Find a recurrence relation for a_n, the number of valid code words of length n

_•What is the initial condition (n=1): a₁

– What is {a₁}

– {a₁} = 1,2,3,4,5,6,7,8,9, so a₁= 9

•How can we form valid strings of length n using strings of length n-1?

Forming strings of length n

• Two ways to form a valid string of n digits from a string of n-1 digits

• First way

– Append anything but a 0 to a valid string of n-1 digits

– How many strings does this give us?

_•**9 * a_n-1**

• Second way

– Add a zero to an invalid string of length (n-1)

– How many invalid strings of length (n-1)

• (Total number of strings ) – (number of valid strings)
10^n-1 - a_n-1

^•Add first way and second way
9a_n-1 + (10^n-1 – a_n-1) or 8a_n-1 + 10^n-1

How many cars produced?

• A factory makes custom sports cars at an increasing rate. In the first month only one car is made, in the second month two cars are made, and so on, with n cars made in the nth month.

• Set up a recurrence relation for the number of cars produced in the first n months by this factory.

• How many cars are produced in the first year?

• Find an explicit formula for the number of cars produced in the first n months by this factory.

– An explicit formula is one that used the initial condition rather than the previous term.

• Let Cn = the total number of cars produced in the first n months

• Look on page 76 in Rosen for a table of some useful summation formulae.

Solving recurrence relations

• We will work on linear homogeneous recurrence relations of degree k with constant coefficients.

• Its form is: a_n = c₁a_n-1 + c₂a_n-2 + …+ c_ka_n-k where c₁, c₂, …c_k are real numbers, and c_k != 0

• This is linear since the right hand side is a sum of the multiples of the previous terms

• It is homogeneous since all the terms are multiples of a

• All of the coefficients are constant, rather than depend on n.

• The degree is k because a_n is expressed in terms of the previous k terms of the sequence

The degree of a recurrence relation

• P_n = (11.1)P_n-1 is degree one

• f_n = f_n-1 + f_n-2 is degree two

• what is this? C_n = C_n-5?

• The basic approach to solving these type of problems is to look for solutions of the form a_n = rⁿ

Linear homogeneous recurrence relations of degree 2

_•Let c₁ and c₂ be real numbers. Suppose that
r² – c₁r – c₂ has two distinct roots r₁ and r₂

• Then the sequence {a_n} is a solution of the recurrence relation a_n = c₁a _n-1 + c₂a _n-2 if and only if a_n = "₁r₁ⁿ + "₂r₂ⁿ for n = 0, 1,2,…where "₁and"₂are constants.

• Look at a real problem:

– What is the solution of the recurrence relation:
a_n = a _n-1 + 2a _n-2 with a₀ = 2 and a₁ = 7

Solving linear homogeneous recurrence relations of degree 2

_•First, get the constants C₁ and C₂

• Next, write the characteristic equation

• Then find the roots

• Find "₁and"₂, usually bysolving simultaneous equations and using initial conditions

Recurrence Relations

How we will proceed

• 1. A definition of recurrence relations.

– A recurrence relationship is a rule by which a sequence is generated.

• 2. Given the initial condition and the rule (i.e. the recurrence relations) what is the sequence?

• 3. Given the rule and a sequence, is the sequence a solution of the recurrence relations?

• Facts about recurrence relations

• Modeling problems with recurrence relations

Definition of a recurrence relation

• A recurrence relation for the sequence {an} is an equation that expresses an in terms of one or more of the previous terms of the sequence.

– an is the next term in the sequence

• The sequence {an} looks like this: a0, a1, …an-1

– For all integers n with n >= n0, where n0 is a nonnegative integer

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

Given the recurrence relation and initial condition, find the sequence

• Let {an} be a sequence that satisfies the recurrence relation

– Rule: an = a n-1 – a n-2

– Initial conditions: a0 = 3 and a1 = 5

– What is the sequence?

Given the rule and a sequence, is the sequence a solution of the recurrence relations?

• Rule: an = 2a n-1 – a n-2

• Sequence: an = 3n

• Two ways to solve

– Figure out the sequence, and see if it satisfies the rule

the sequence is: a0= 0, a1=3, a2=6, a3=9, a4 = 12

at n = 4: a4 ?= 2a3 – a2 12 ?= 2(9) - 6

– Substitute 3n for n in the rule and simplify 2a n-1 – a n-2 = 2[3(n-1)] – 3(n-2) when simplified, the does indeed = 3a

Given the rule and a sequence, is the sequence a solution of the recurrence relations?

• Rule: an = 2a n-1 – a n-2

• Sequence: an = 2n

• The sequence is: a0= 1, a1=2, a2=4, a3=8, a4 = 16

• Does this fit the rule?

• At n=4: a4 ?= 2a3 – a2

16 ?= (2*8) – 4

• So an = 2n is not a solution

Facts about recurrence relations

• The recurrence relation and initial conditions uniquely determine a sequence

• Any term of the sequence can be found from the initial conditions using the recurrence relation a sufficient number of times

• But, for a certain class of sequences defined by a recurrence relation and initial condition there are better ways to find any term

Overheads

Modeling rabbits

• A young pair of rabbits is placed on an island. After they are two months 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 6 months. After n months. (Assume no rabbits die.)

Towers of Hanoi

• There are three pegs and disks of different sizes.

• The object is to move all the disks from one peg to another

• The Rules

– One disk is moved at a time

– A disk may not be placed on top of a disk of smaller diameter

• Let Hn denote the number of moves needed to solve with n disks.

• Set up a recurrence relation for the sequence {Hn}

Recurrence relation for Towers of Hanoi

• What is the initial condition H1?

– I.e. how many moves to move one disk?

• We want to develop a rule to tell us how many moves it will take to move n disks in terms of moving n-1 disks

• Find a recurrence relation and give initial conditions for the number of bit strings of length n that do not have two consecutive 0s.

• a1 = num valid bit strings of length 1 {a1} is 0, 1

• a2 = num valid bit strings of length 2 {a2} is 01, 10, 11

• a3 = num valid bit strings of length 3 {a3} is 010, 011, 101, 110, 111

• By the sum rule, the total number of bit strings of length n without two consecutive 0 bits is equal to the number ending with 0 plus the number ending with 1

What is the sum rule?

•If a first task can be done in n1 ways and a second task in n2 ways, and if these tasks cannot be done at the same time, then there are n1 + n2 ways to do either task

• Assume n >= 3

• How many strings of length n are there ending with 1?

– It’s all the strings in {an-1} with a 1 added

• How many strings of length n are there ending with a 0?

– It’s all the strings in {an-2} with a 10 added

• What is {a3} given {a2} and {a1}?

•What is {a4}?

• A decimal digit code word is valid if it contains an even number of 0 digits

• Find a recurrence relation for an, the number of valid code words of length n

• What is the initial condition (n=1): a1

– What is {a1}

– {a1} = 1,2,3,4,5,6,7,8,9, so a1 = 9

•How can we form valid strings of length n using strings of length n-1?

• Two ways to form a valid string of n digits from a string of n-1 digits

• First way

– Append anything but a 0 to a valid string of n-1 digits

– How many strings does this give us?

• 9 * an-1

• Second way

• 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 of the sequence.

– a_nis the next term in the sequence

• The sequence {a_n} looks like this: a₀, a₁, …a_n-1

– For all integers n with n >= n₀, where n₀ is a nonnegative integer

• Let {a_n} be a sequence that satisfies the recurrence relation

_–Rule: a_n = a _n-1 – a _n-2

– Initial conditions: a₀ = 3 and a₁ = 5

_•Rule: a_n = 2a _n-1 – a _n-2

• Sequence: a_n = 3n

the sequence is: a₀= 0, a₁=3, a₂=6, a₃=9, a₄= 12

at n = 4: a₄ ?= 2a₃ – a_{2

12 ?=
2(9) - 6}

– Substitute 3n for n in the rule and simplify
2a _n-1 – a _n-2= 2[3(n-1)] – 3(n-2)
when simplified, the does indeed = 3a

_•Rule: a_n = 2a _n-1 – a _n-2

• Sequence: a_n = 2ⁿ

• The sequence is:
a₀= 1, a₁=2, a₂=4, a₃=8, a₄= 16

_•At n=4: a₄ ?= 2a₃ – a₂

**16 ?= (2*8) – 4**

• So a_n = 2ⁿ is not a solution

• Let H_n denote the number of moves needed to solve with n disks.

• Set up a recurrence relation for the sequence {H_n}

• What is the initial condition H₁?

• a₁ = num valid bit strings of length 1
{a₁} is 0, 1

• a₂ = num valid bit strings of length 2
{a₂} is 01, 10, 11

• a₃ = num valid bit strings of length 3
{a₃} is 010, 011, 101, 110, 111

•If a first task can be done in n₁ ways and a second task in n₂ ways, and if these tasks cannot be done at the same time, then there are n₁ + n₂ ways to do either task

– It’s all the strings in {a_n-1} with a 1 added

– It’s all the strings in {a_n-2} with a 10 added

• What is {a₃} given {a₂} and {a₁}?

•What is {a₄}?

• Find a recurrence relation for a_n, the number of valid code words of length n

_•What is the initial condition (n=1): a₁

– What is {a₁}

– {a₁} = 1,2,3,4,5,6,7,8,9, so a₁= 9

_•**9 * a_n-1**

• (Total number of strings ) – (number of valid strings)
10^n-1 - a_n-1

^•Add first way and second way
9a_n-1 + (10^n-1 – a_n-1) or 8a_n-1 + 10^n-1

• Its form is: a_n = c₁a_n-1 + c₂a_n-2 + …+ c_ka_n-k where c₁, c₂, …c_k are real numbers, and c_k != 0

• The degree is k because a_n is expressed in terms of the previous k terms of the sequence

• P_n = (11.1)P_n-1 is degree one

• f_n = f_n-1 + f_n-2 is degree two

• what is this? C_n = C_n-5?

• The basic approach to solving these type of problems is to look for solutions of the form a_n = rⁿ

_•Let c₁ and c₂ be real numbers. Suppose that
r² – c₁r – c₂ has two distinct roots r₁ and r₂

• Then the sequence {a_n} is a solution of the recurrence relation a_n = c₁a _n-1 + c₂a _n-2 if and only if a_n = "₁r₁ⁿ + "₂r₂ⁿ for n = 0, 1,2,…where "₁and"₂are constants.

– What is the solution of the recurrence relation:
a_n = a _n-1 + 2a _n-2 with a₀ = 2 and a₁ = 7

_•First, get the constants C₁ and C₂

• Find "₁and"₂, usually bysolving simultaneous equations and using initial conditions