Solving Second-Order Recursive Relations

Lecture 42

Section 8.3

Mon, Apr 16, 2007

Second-Order Linear Homogeneous Recurrence Relations with Constant Coefficients

Second-order – Each term is defined in terms of the previous two terms.

Linear – The terms of the sequence appear to the first power in the relation.

Homogeneous – There is no constant term. Constant coefficients – The coefficients of

the terms are constants.

The Form of Such Recurrence Relations

Such a recurrence relation is of the form

an = Aan – 1 + Ban – 2

for all integers n 2, with initial terms a0 and a1 given.

The Fibonacci sequence is a very simple example.

The Characteristic Equation

The characteristic equation of

an = Aan – 1 + Ban – 2

is

t2 – At – B = 0. For the Fibonacci sequence, the

characteristic equation is t2 – t – 1 = 0.

The Characteristic Equation

Let r be a root of the characteristic equation.

Then the sequence an = rn satisfies the recurrence relation.

Solving Such Recurrence Relations – Case I

Theorem: If the characteristic equation has roots r and s which are distinct real numbers, then the recursive sequence is given by

an = Crn + Dsn,

where C and D are constants, determined by the values of a0 and a1.

Solving Such Recurrence Relations – Case I

Proof: Since rn and sn satisfy the recurrence

relation, it follows easily that Crn + Dsn also satisfies it.

By using the initial conditions, we can solve for C and D.

The resulting formula satisfied the recurrence relation with the initial conditions.

Example

Solve the recurrence relationa0 = 2,

a1 = 3,

an = an – 1 + 2an – 2, for all n 2.

The first few terms are 2, 3, 7, 13, 27, 53, 107, 213, …

Example

The roots of the characteristic equation are r = 2 and s = -1.

So the general form is an = C(2n) + D(-1)n.

Solve a0 = C + D = 2 and a1 = 2C – D = 3. We get C = 5/3 and D = 1/3. The sequence is

.3

)1(25)1(

3

12

3

5 nnnn

na

Example

Find a non-recursive formula for the Fibonacci numbers.

Find a non-recursive formula for the Lucas numbers:L1 = 1

L2 = 3

Ln = Ln – 1 + Ln – 2, for all n 2.

Solving Such Recurrence Relations – Case II

Theorem: If the characteristic equation has double root r, then the recursive sequence is given by

an = Crn + Dnrn = (C + Dn)rn,

where C and D are constants determined by the values of a0 and a1.

Example

Solve the recurrence relationa0 = 0,

a1 = 4,

an = an – 1 – (¼)an – 2, for all n 2.

The first few terms are 0, 4, 4, 3, 2, …

Example

The root of the characteristic equation is the double root r = ½.

So the general form is an = (C + Dn)(½)n.

Solve a0 = C = 0 and a1 = (C + D)(½) = 4. We get C = 0 and D = 8. The sequence is

.22

8

2

18

3

nn

n

n

nnna

Example

Find a nonrecursive formula for the recursive sequencea0 = 1

a1 = 10

an = 2an – 1 – an – 2, for all n 2.

Case III – Complex Roots

The sequences become more interesting when the roots of the characteristic equation are complex numbers.

Write the first few terms of the sequencea0 = 0

a1 = 1

an = 2an – 1 – 5an – 2, for all n 2.