agenda lecture content: recurrence relations solving recurrence relations iteration linear...
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Agenda
• Lecture Content: Recurrence Relations Solving Recurrence Relations
Iteration Linear homogenous recurrence relation of order k with
constant coefficients
• Exercise
Recurrence Relations
Recurrence Relations
Relates the n-th element of a sequence to its predecessors.
Example: an = 9 * an-1
Closely related to recursive algorithms.Suited for analysis of algorithm
Recurrence Relations
Generate a sequence :1. Start with 52. Given any term, add 3 to get the next term 5, 8, 11, 14, …
Sequence {an} = a0, a1, …, an-1, an 1. a0= 5 initial condition2. an = an-1 + 3 , n ≥ 1 recurrence relation
Recurrence Relations for the sequence {an}
is an equation that expresses an in terms of one or more of the previous terms of the sequence, namely, a0, a1, …, an-1 for all integers n with n ≥ n0, where n0 is a nonnegative integer.
A sequence {an} is called a solution of a recurrence relation if its terms an satisfy the recurrence relation.
Motivation
One of the main reason for using recurrence relation is that sometimes it is easier to determine the n-th term of a sequence in terms of its predecessors than it is to find an explicit formula for the n-th term in terms of n.
Example
Let {an} be a sequence that satisfies the recurrence relation an = an-1 – an-2 for n = 2, 3, 4, … and suppose that a0 = 3 and a1 = 5. What are a2 and a3?
a2 = a1 – a0 = 5 – 3 = 2 a3 = a2 – a1 = 2 – 5 = -3 Also a4, a5, …
Exercises
Find the first six terms of the sequence defined by the following recurrence relations:
an = an-1 + an-3
a0 = 1, a1 = 2, a2 = 0
an = n. an-1 + (an-2)2
a0 = -1, a1 = 0
Example
Find the recurrence relation of:- {an} = (0, 1, 2, 3, …)- {an} = (5, 10, 15, 20, …)
Find the first three terms of the sequence defined by the following recurrence relations:- an = 3 * an-1 – an-2 a0 = 3 and a1 = 5- an = an-1 – an-2 + an-3
a0 = 3, a1 = 5 and a2 = 5
Modelling with Recurrence Relations (1)
The number of bacteria in a colony doubles every hour. If a colony begins with five bacteria, how many will be present in n hours?- a0 = 5- a1 = 10- a2 = 20- …
The number of bacteria at the end of n hours: an = 2 * an-1
Modelling with Recurrence Relations (2)
Suppose that a person deposits $ 10,000 in a saving account at a bank yielding 11% per year with interest compounded annually. How much will be in the account after 30 years?
• Pn : the amount in the account after n years• Pn = Pn-1 + 0.11 Pn-1 = 1.11 Pn-1• P0 = 10,000• P1 = 1.11 P0• P2 = 1.11 P1 = 1.11 * 1.11 P0 = (1.11)2 P0• Pn = (1.11)n P0 Pn = (1.11)n P0
Deriving explicit formula from the recurrence relation and its initial condition (Section 7.2)
Converting to a Recursive Algorithm
A recursive function is a function that invokes itself. A recursive algorithm is an algorithm that contains a
recursive function
Input: n, the number of yearsOutput: the amount of money at the end of n years
1. compound_interest (n){2. if (n == 0)3. return 10,0004. return 1.11 * compound_interest (n-1)5. }
Modeling with Recurrence Relation (3)
A young pair of rabbits is placed on an island. A pair of rabbit does not breed until they are 2 months old. After they are 2 months old, each pair of rabbits produces another pair each month. Find a recurrence relation for the number of pairs of rabbit on the island after n months, assuming that no rabbits ever die.
fn : the number of pairs of rabbits after n months
Modeling with Recurrence Relation (3)
Month reproducing pairs young pairs total pairs1 0 1 12 0 1 13 1 1 24 1 2 35 2 3 56 3 5 8
Modeling with Recurrence Relation (3)
fn : the number of pairs of rabbits after n months
f1 = 1 f2 = 1 f3 = 2 fn = the number on the island for the previous
month + the number of newborn pairs fn = fn-1 + fn-2
Fibonacci Numbers
Modelling with Recurrence Relation (4)
Find a recurrence relation and give initial conditions for the number of bit strings of length n that do not have two consecutives 0s.
an : the number of bit strings of length n that do not have two consecutive 0s.
The Tree Diagrams
How many bit strings of length four do not have two consecutive 0s?
bit ke-1 bit ke-2 bit ke-3 bit ke-4
11
1111
00
1100
1100 11 00
11
00 1111 11
0000
11
There are eight bit strings
Modelling with Recurrence Relation (4)
an = an-1 + an-2 for n ≥ 3
a1 = 2 0, 1a2 = 3 (01, 10, 11)a3 = 5 (011, 010, 101, 110, 111)a4 = 8
Solving Recurrence Relations
Solving Recurrence Relations
- Given recurrence relations with sequence: a0 , a1 , …
- Find an explicit formula for the general term:
an
- Two methods:- Iteration- Recurrence relations with constant coefficients
Iteration
Steps:- Use the recurrence relation to write the n-th
term an in terms of certain of its predecessors an-1 , …, a0
- Use the recurrence relation to replace each of an-1 , … by certain of their predecessors.
- Continue until an explicit formula is obtained.
Iteration: Example 1
an = an-1 + 3 n ≥ 1
a0= 5
Solution:
- an-1 = an-2 + 3
- an = an-1 + 3 = (an-2 + 3) + 3 = an-2 + 2.3
= (an-3 + 3) +2.3 = an-3 + 3.3
= an-k + k.3
… k = n
= a0 + n . 3
= 5 + n . 3
Iteration: Example 2(1)
The number of bacteria in a colony doubles every hour. If a colony begins with five bacteria, how many will be present in n hours?
a0 = 5 a1 = 10 a2 = 20 … The number of bacteria at the end of n
hours: an = 2 * an-1
Iteration: Example 2(2)
an = 2 * an-1
= 2 * (2 * an-2 ) = 2 * 2 ( an-2 )
= 2 * 2 * (2 * an-3 ) = 23 ( an-3 )
…
= 2n * a0
= 2n * 5
Iteration: Example 3
The deer population is 1000 at time n = 0 The increase from time n-1 to time n is 10 percent. Find the recurrence relation, initial condition and solve
the recurrence relation at time n
a0 = 1000 an = 1.1 * an-1
an = 1.1 * (1.1 * an-2) = 1.12 * an-2
an = 1.1n * a0 an = 1.1n * 1000
Recurrence Relations with Constant Coefficients
Definition
A linear homogeneous recurrence relation of order k with constant coefficients is a recurrence relation of the form:
an = c1an–1+ c2an–2+ … + ckan–k , ck ≠ 0
Example:
• Sn = 2Sn–1
• fn = fn–1 + fn–2
Not linear homogenous recurrence relations
Example:
• an = 3an–1an–2
• an – an–1 = 2n
• an = 3nan–1
Why?
Solving Recurrence Relations
Example:• Solve the linear homogeneous recurrence
relations with constant coefficients
an = 5an–1 – 6an–2
a0 = 7, a1= 16
Solution:• Often in mathematics, when trying to solve a
more difficult instance of some problem, we begin with an expression that solved a simpler version.
Solving Recurrence Relations
For the first-order recurrence relation, the solution was of the form Vn= tn ; thus for our first attempt at finding a solution of the second-order recurrence relation, we will search for a solution of the form Vn= tn.
If Vn = tn is to solve the recurrence relation, we must have Vn= 5Vn–1–6Vn–2 or tn = 5tn–1 – 6tn–2 or tn – 5tn–1+ 6tn–2 = 0. Dividing by tn–2, we obtain the equivalent equation t2 – 5t1 + 6 = 0. Solving this, we find the solutions t= 2, t= 3. Characteristic
polynomialCharacteristic roots
Solving Recurrence Relations
We thus have two solutions, Sn = 2n and Tn = 3n.
• We can verify that if S and T are solutions of the preceding recurrence relation, then bS + dT, where b and d are any numbers, is also a solution of that relation. In our case, if we define the sequence U by the equation
Un = bSn+ dTn = b2n + d3n,
U is a solution of the given relation.
Solving Recurrence Relations
To satisfy the initial conditions, we must have:
7 = U0 = b20 + d30= b + d,
16 = U1= b21+ d31= 2b + 3d.
Solving these equations for b and d, we obtain b = 5, d = 2.
Therefore, the sequence U defined by Un= 5∙2n + 2∙3n satisfies the recurrence relation and the initial conditions.
We conclude that an= Un= 5∙2n+ 2∙3n, for n = 0, 1, ….
Theorem
Let an = c1an–1+ c2an–2 be a second-order, linear homogeneous recurrence relation with constant coefficients. • If S and T are solutions of the recurrence relation,
then U = bS+ dT is also a solution of the relation.
• If r is a root of t2–c1t–c2= 0, then the sequence rn, n = 0, 1, …, is a solution of the recurrence relation.
• If a is the sequence defined by the recurrence relation, a0 = C0, a1= C1, and r1 and r2 are roots of the preceding equation with r1 ≠ r2, then there exist constants b and d such that an= br1
n + dr2n, n = 0, 1, ….
Example (1)
More Population Growth• Assume that the deer population of Rustic
County is 200 at time n = 0 and 220 at time n = 1 and that the increase from time n–1 to time n is twice the increase from time n–2 to time n–1.
• Write a recurrence relation and an initial condition that define the deer population at time n and then solve the recurrence relation.
Example (1)
Solution:• Let dn denote the deer population at time n.
d0 = 200, d1 = 220
dn – dn-1 = 2(dn-1 – dn-2)
dn = 3dn-1 – 2dn-2
• Solving t2 – 3t + 2 = 0, we have roots 1 and 2. Then the sequence d is of the form dn = b∙1n + c∙2n = b + c2n.
• To meet the initial conditions, we must have 200 = d0= b + c, 220 = d1= b + 2c. Solving for b and c, we find b= 180, and c= 20.
• Thus, dn is given by dn = 180 + 20∙2n.
Example (2)
Find an explicit formula for the Fibonacci sequence:
fn – fn–1 – fn–2= 0, n≥ 3
f1 = 1, f2 = 1
Solution:•We begin by using the quadratic formula to solve t2 –t – 1 = 0. The solutions are t = (1 ±√5)/2. Thus the solution is of the form fn= b((1+√5)/2)n + c((1–√5)/2)n.
–To satisfy the initial conditions, we must have b((1+√5)/2) + c((1–√5)/2)= 1, b((1+√5)/2)2+ c((1–√5)/2)2= 1. Solving these equations for b and d, we obtain b = 1/√5, d = -1/√5.
• •Therefore, fn= 1/√5∙((1+√5)/2)n – 1/√5((1–√5)/2)n.
Theorem
Let an = c1an–1 + c2an–2 be a second-order, linear homogeneous recurrence relation with constant coefficients. • Let a be the sequence satisfying the relation
and a0 = C0, a1= C1.
• If both roots of t2 – c1t – c2 = 0 are equal to r, then there exist constants b and d such that an = brn + dnrn, n = 0, 1, ….
Example
Solve the recurrence relation dn= 4(dn-1 – dn-2) subject to the initial conditions d0= 1 = d1.
– According to the theorem, Sn= rn is a solution, where r is a solution of t2 – 4t + 4 = 0. Thus we obtain the solution Sn = 2n.
– Since 2 is the only solution of the equation, Tn= n2n is also a solution of the recurrence relation.
– Thus the general solution is of the form U = aS+ bT. – We must have U0 = 1 = U1. The last equations become
aS0 + bT0= a+ 0b= 1, aS1+ bT1= 2a+ 2b = 1.
– Solving for a and b, we obtain a = 1, b = -1/2. – Therefore,
the solution is dn= 2n – 1/2n2n = 2n – n2n-1 .
Note
For the general linear homogeneous recurrence relation of order k with constant coefficients c1, c2, …, ck,
• if r is a root of
tk – c1tk-1 – c2tk-2 – … – ck = 0
of multiplicity m, it can be shown that
rn, nrn, … , nm-1rn are solutions of the equation.