7.2 Solving Recurrence Relations

Definition 1 (p. 460)- LHRR-K

Def: A linear homogeneous recurrence relations of degree k with constant coefficients (referred to as “LHRR-K”) is a recurrence relation of the form

an = c1an-1 + c2 an-2 + … +ckan-k where c1, c2, …ck are real numbers, and ck≠0.

------------------------------Note: These can be explicitly solved in a

systematic way.

First, are the following examples or non-examples?

• a n = 3 a n-1

• a n = a n-1 + a n-2 2

• f n = f n-1 + f n-2

• H n = 2H n-1 +1

• B n = n B n-1

• a n = a n-1 +3 a n-2 , a0 =1, a1 =2

Solving LHRR-K:

Step 1: Find a characteristic equation: an=rn is a solution of an = c1an-1

+ c2 an-2 + … +ckan-k iff

rn = c1 rn-1 + c2 rn-2 + … +ck rn-k

Divide by rn-k: Then rk= c1 rk-1 + c2 rk-2 + … +ck

So: rk - c1 rk-1 - c2 rk-2 - … - ck =0

For degree 2: the characteristic equation is r2-c1r –c2=0 (roots are used to find explicit formula)

Basic Solution: an=α1r1n+ α2r2

n where r1 and r2 are roots of the characteristic equation

Thm. 1 (for 2nd degree equations)

Theorem 1: Let c1, c2 be elements of the real numbers.

Suppose r2-c1r –c2=0 has two distinct roots r1 and r2,

Then the sequence {a n} is a solution of the recurrence relation an = c1an-1 + c2 an-2

iff an=α1r1n+ α2r2

n for n=0, 1, 2… where α1 and α2 are constants.----------------------------

First an example… Proof: later…

Ex. 1. Let an=7an-1 – 10 an-2 for n≥2; a0=2, a1=1

Find the characteristic equation …get r=2, 5

Let an= α1r1

n+ α2r2n … solve

system and get α1 = 3 and α2 = -1

So, basic solution is: an= 3*2n-5n

Prove this is a solution (as we did in sec 7.1):

Sketch of Proof of Thm. 1: (p. 462)

Step 1: Show that an=α1 r1n+ α2 r2

n is a solution of

an =c1an-1+c2an-2

c1an-1+c2an-2

=c1(α1 r1n-1+ α2 r2

n-1 )+c2 (α1 r1n-2+ α2 r2

n-2) why?

= α1r1n-2(c1 r1 + c2 ) + α2r2

n-2(c1 r2 + c2 ) algebra

=α1r1n-2r1

2 + α2r2n-2r2

2

reason: r1 r2 are roots of r2-c1r-c2=0

so r12=c1r1 +c2 and

r2 2= c1r2 +c2

=α1r1n + α2r2

n

=an

Pf- step 2Step 2: Show that there exist constants α1 α2 such that

an=α1 r1n+ α2 r2

n satisfies the initial conditions a0=C0 and a1=C1.

a 0 = C 0 = …

a 1 = C 1 = …

Next solve system of 2 equations and 2 variables and get…

α1 = (C1-C0r2)/(r1-r2) α2 = (C0r1 – C1) / (r1-r2)

Ex: 2. an =6an-1 -8an-2 for n≥2; a0=4 and a1=10.

Find characteristic equation Find solution

Ex 2 - Prove it is a solution

•

Ex: 3. Fibonacci numbers:

fn =fn-1 +fn-2 for n≥2; f0=0 and f1=1.

Find characteristic equationr2 = r+1 find r1 and r2

Thm. 1 says fn= α1r1n+ α2r2

n

Solve for α1, α2

Solution:

Note: Thm. 1 is only for r1≠r2

Thm. 2 is for r1 = r2

Theorem 2: Let c1, c2 be elements of the real numbers.

Suppose r2-c1r –c2=0 has only one root r0 ,

Then the sequence {a n} is a solution of the recurrence relation an = c1an-1 + c2 an-2 iff an=α1r0

n+ α2 n r0n

for n=0, 1, 2… where α1 and α2 are constants.

Ex: 4. (Recall this ex from section 5.1)

an =2an-1 -an-2 for n≥2; a0=0 and a1=3

Find characteristic equation Find solution Prove it is a solution

Ex: 5. an = - 6an-1 -9an-2 for n≥2; a0=3 and a1= - 3

Find characteristic equation Find solution

Prove it is a solution

Ex: 6. an =8an-1 -16an-2 for n≥2; a0=2 and a1=20.

Find characteristic equation Find solution Prove it is a solution

Summarize k degree solution

•

Ex: (#12 in book) an =2an-1 +an-2 -2an-3 for n≥3; a0=3 ,a1=6, a2=0

Find characteristic equation r3 – 2 r2- r +2=0Use synthetic division to get (r-1)(r+1)(r-2)=0 Find solution Prove it is a solution

Ex: (#15 bk) an =2an-1 +5an-2 -6an-3 for n≥3; a0=7 ,a1= - 4, a2=8

Find characteristic equation Use synthetic division to get Find solution Prove it is a solution

Ex: 7. an =5an-2 -4an-4 ; a0=3, a1=2, a2=6 and a3=8

Find characteristic equation Find solution

Prove it is a solution

Ex: (#8 in book—modeling number of lobsters caught)

Find characteristic equation Find solution

Prove it is a solution