Non-Homogeneous Recurrence Relations

Vasileios Hatzivassiloglou

University of Texas at Dallas

Review Problems

• Solve the recurrence relation an = 4an-1 – 4an-2 with initial conditions a0 = 3 and a1 = 8.

• Which recurrence relation has characteristic roots 2, 2, and -1?

• Find the general form of the solution of the recurrence relation an = 8an-2 – 16an-4.

Allowing multiple roots

• Theorem 4: If the above characteristic equation has t distinct roots r1, r2, ..., rt with multiplicities m1, m2, ..., mt (such that mi≥1 and m1+m2+...+mt = k) then {an} is a solution of the recurrence if and only if

• for all n≥0, where bi,j are constantsnt

mmttt

nmm

nmmn

rnbnbb

rnbnbb

rnbnbba

t

t)(

)(

)(

11,1,0,

21

1,21,20,2

11

1,11,10,1

2

2

1

1

Example

• Suppose the roots are 2,2,2,3,3,5. What is the form of the solution {an}?

• an = (b1,0 + b1,1n + b1,2n2)2n + (b2,0 + b2,1n)3n + b3,05n

Linear non-homogeneous recurrence relations

• Still constant coefficients• Non-homogeneous:

– We now have one or more additional terms which depend on n but not on previous values of an

• Examples:– an= 2an-1 + 1, an=an-1 + n, an=an-2 + n2 + 1

• General form:– an = c1an-1 + c2an-2 + ... + ckan-k + F(n)

Associated homogeneous recurrence relation

• If we ignore F(n) in the previous form, we obtain the homogeneous recurrence relation associated with the non-homogeneous one we are trying to solve

• Theorem 5: If {an(p)} is a particular solution for

a non-homogeneous recurrence relation, then all solutions are of the form {an

(p)}+{an(h)},

where {an(h)} is a solution of the associated

homogeneous recurrence relation

Proving Theorem 5

• Suppose {an(p)} is such a particular solution and {bn} is

another solution• an

(p) = c1an-1(p) + c2an-2

(p) + ... + ckan-k(p) + F(n)

• bn = c1bn-1 + c2bn-2 + ... + ckbn-k + F(n)• By subtracting the first equation from the second,• bn – an

(p) = c1(bn-1 – an-1(p)) + c2(bn-2 – an-2

(p)) + ... + ck(bn-k – an-k

(p))

• bn – an(p) is a solution of the associated homogeneous

recurrence relation

Finding solutions

• Finding a particular solution is the tricky part

• There are general solutions for certain classes of functions F(n) but not for every possible F(n)

• Sometimes, we have to guess on possible forms based on F(n)

Example

• an = 2an-1 + 2n, a1 = 6• The homogeneous solution is

– an(h) = b2n where b is some constant

• For a particular solution, guess• an

(p) = cn + d = 2an-1(p) + 2n = 2(c(n-1) +

d) + 2n = (2c + 2)n – 2c + 2d• (c + 2)n + (d – 2c) = 0• c + 2 = 0 ⇔ c = -2• d – 2c = 0 ⇔ d + 4 = 0 ⇔ d = -4

an(p) = cn + d

Example continued

• We determined that any solution is of the form

• an = -2n – 4 + b2n

• With a1 = 6, we have

• a1 = -2 – 4 + b21 = -6 + 2b = 6 ⇔• 2b = 12 ⇔ b = 6• an = -2n – 4 + 6·2n

Polynomial and exponential F(n)

• Theorem 6: If the function F(n) is of the form F(n) = (btnt + bt-1nt-1 + ... + b0)sn, then– If s is not a root of the characteristic equation

of the associated homogeneous recurrence relation, there is a solution of the form (ptnt + pt-1nt-1 + ... + p0)sn

– If s is such a root with multiplicity m, then there is a solution of the form nm(ptnt + pt-1nt-1 + ... + p0)sn

Sum of integers

• an = 1 + 2 + ... + n• Recurrence relation:• Associated homogeneous RR:• Solving the homogeneous RR:• F(n) = n = n·1n so t=1, b1=1, b0 = 0, s=1• The homogeneous RR’s characteristic equation

is• so s is a characteristic root with multiplicity m=1

an = an-1 + nan

(h) = an-1(h)

r - 1 = 0

an(h) = b·1n = b

Sum of integers continued• A particular solution is of the form• an = n1(p1n + p0)1n = p1n2 + p0n

• an = an-1 + n ⇔• p1n2 + p0n = p1(n-1)2 + p0(n-1) + n = p1(n2 – 2n + 1)

+ p0(n-1) + n = p1n2 + (-2p1 + p0 + 1)n + (p1-p0) ⇔

• (-2p1 + 1)n + (p1-p0) = 0

• For this to be true, -2p1 + 1 = 0 ⇔ p1 = 1/2• and p0 = p1 = 1/2

Sum of integers continued

• Therefore the particular solution is

• and all solutions are of the form

• From the initial condition a1=1, we obtain

• a1 = 1·(1+1)/2 + b = 1 + b, so b=0

2)1(

2

2

212

21

02

1

nnnnnnnpnpan

bnnan

2)1(

Reading

• Section 7.2 (Subsection on Non-homogeneous recurrence relations)

Review Problems

• Solve the recurrence relation an = 2an-1 + 3·2n with a0 = 2.

• Can you apply Theorem 6 for finding particular solutions to the recurrence relation an = 5an-1 – 6an-2 + 2n + 3n with a0=3 and a1=5?

• Solve the previous recurrence with the extra knowledge that it has a particular solution of the form qn2n + p1n + p0 where q, p1, and p0 are constants.