How to find a particular solution for a non-homogeneous recurrence relation
Consider a non-homogeneous linear recurrence relation of order k with constant coefficients:
c0 an + c1 an−1 + . . . + ck an−k = f (n).
(1)
Let
a(p)
n be a particular solution of the recurrence relation (1)
and let
a(h)
n be the general solution of the corrsponding homogeneous recurrence relation
c0 an + c1 an−1 + . . . + ck an−k = 0.
(2)
(h)
Note that an involves k parameters. Then
(p)
an = a(h)
n + an
(3)
is the general solution of the non-homogeneous recurrence relation (1). (Prove this!) Note that
an also involves k parameters. And conversely: the general solution of the non-homogeneous
(p)
recurrence relation (1) is of the form (3) for any fixed particular solution an . (Prove this, too!)
(p)
How do we find a particular solution an ?
(p)
1. Find an appropriate trial set-up for an in the table below.
(p)
2. Adjust your set-up using the additional guidelines. Your an will involve a number of
parameters.
(p)
3. Find the values for these parameters by observing that an should be a solution of the
non-homogeneous recurrence relation (1).
(p)
Trial set-up for an
f (n)
b0 nt + b1 nt−1 + . . . + bt−1 n + bt
rn
nt r n
(p)
Trial set-up for an
B0 nt + B1 nt−1 + . . . + Bt−1 n + Bt
Arn
rn (B0 nt + B1 nt−1 + . . . + Bt−1 n + Bt )
Additional guidelines:
(p)
• If f (n) = f1 (n) + f2 (n), then an should be the sum of particular solutions corresponding
to f1 (n) and f2 (n).
• If f (n) (or a term in f (n)) is a solution of the corresponding homogeneous recurrence
(p)
(p)
relation, then multiply the trial an by ns for the smallest integer s such that ns an (and
every summand thereof) is NOT a solution of the homogeneous recurrence relation.
Example 1. Consider the second-order non-homogeneous linear recurrence relation with
constant coefficients an = aan−1 + ban−2 + f (n) with characteristic roots r1 = 2 and r2 = 5.
Depending on f (n), we seek a particular solution in the following form:
f (n)
3
5n2 − 3n + 2
(6n2 + 1) · 3n
5 · 2n
5 · 2n + 6 · 3n
5 · 2n + 6 · 5n
(p)
Set-up for an
A
An2 + Bn + C
(An2 + Bn + C) · 3n
A2n · n
An2n + B · 3n
An2n + Bn5n
Example 2. Consider the second-order non-homogeneous linear recurrence relation with
constant coefficients an = aan−1 + ban−2 + f (n) with a double characteristic root r1 = r2 = −2.
Depending on f (n), we seek a particular solution in the following form:
f (n)
3
5n2 − 3n + 2
6 · 3n
5 · (−2)n
5n · (−2)n
5n2 · (−2)n
(p)
Set-up for an
A
An2 + Bn + C
A · 3n
A(−2)n · n2
(An + B)(−2)n · n2
(An2 + Bn + C)(−2)n · n2
2