5.3 Second Order Recurrence Relations
In the previous section we saw how to solve first order linear recurrence relations. This
is when an is given by a linear formula of an-1, i.e.
an = pnan-1 + sn
where pn and sn are given sequences. In this section and the next we look at second order
linear recurrence relations when an is given by a linear formula of an-1 and an-2, i.e
an = pnan-1 + qnan-2 + sn
where pn, qn and sn are given sequences. For simplicity we concentrate on the constant
coefficient case when pn and qn don't vary with n, i.e.
(1)
an = pan-1 + qan-2 + sn
where p and q are just numbers. In this section we look at the situation where the
recurrence relation is homogeneous which is when rn = 0 for all n, i.e.
(2)
an = pan-1 + qan-2
In the next section we look at the inhomogeneous case (1).
We illustrate the method of solution of equations of the form (2) with the following
example.
Example 1. Consider the equation
(3)
an = an-1 + 2an-2
along with the initial conditions
(4)
a0 = 2
and
a1 = 3
(3) is the special case of (2) when p = 1 and q = 2.
To solve equations of the form (2) we start by looking for solutions which have the
special form
(5)
an = rn
where r is a number to be determined. To determine r we substitute into (2). We
illustrate this with (3). If an is given by (3) then
(6)
an-1 = rn-1
5.3 - 1
and
(7)
an-2 = rn-2
Substituting (5), (6) and (7) into (3) gives
rn = rn-1 + 2rn-2
Divide this by rn-2 giving
r2 = r + 2
or
r2 - r - 2 = 0
This is called the characteristic equation. It is a quadratic equation. The roots are the
values of r in the solutions an = rn. To solve we either factor or use the quadratic formula.
In this case we can factor.
(r – 2)(r + 1) = 0
This equation has two solutions
r1 = 2
and
r2 = - 1
Recall the r is a number such that (5) is a solution to (3). This gives the following two
solutions to (3)
an = 2n
and
an = (- 1)n
Neither of these solutions satisfy the initial conditions (4). In order to get a solution
which satisfies (4) we need to take a superposition of these two solutions, i.e. multiply
them by constants and add. We can do this because of the following
Superposition Principle. If an and bn are two solutions of the equation (2) then so are
an + bn
Aan
Aan + Bbn
for any constants A and B.
Proof. By hypothesis we have
5.3 - 2
(8)
an = pan-1 + qan-2
and
(9)
bn = pbn-1 + qbn-2
If we add the equations (8) and (9) we get
(an + bn) = p(an-1 + bn-1) + q(an-2 + bn-2)
which shows that an + bn is a solution. If we multiply equation (8) by A we get
(Aan) = p(Aan-1) + q(Aan-2)
which shows that Aan is a solution. If we multiply equation (8) by A and equation (9) by
B and add we get
(Aan + Bbn) = p(Aan-1 + Bbn-1) + q(Aan-2 + Bbn-2)
which shows that Aan + Bbn is a solution. //
It follows from the superposition principle that an = A 2n + B (- 1)n is a solution to (3) for
any constants A and B. Now we choose the constants A and B to satisfy the initial
conditions (4). Plugging in n = 0 we get
2 = a0 = A 20 + B (- 1)0 = A + B
Plugging in n = 1 we get
3 = a1 = A 21 + B (- 1)1 = 2A - B
This is a system of two equations and two unknowns. We multiply them by numbers to
get the coefficient of one of the unknowns the same and then add or subtract. In this case
we can just add the equations
A + B = 2
2A - B = 3
3A
= 5
5
 A = 3
5
1
 B = 2–A = 2-3 = 3
5
1
So an = 3 2n + 3 (- 1)n. If we were interested in the behavior for large n then an = O(2n).
If the roots of the characteristic equation are equal then an = nrn is also a solution where r
is a root of the characteristic equation. Then the general solution is an = Arn + Bnrn.
5.3 - 3
Example 2. Consider the equation
(10)
an = 4an-1 - 4an-2
along with the initial conditions
(11)
a0 = 2
and
a1 = 5
Try an = rn. Substituting into (10) gives
rn = 4rn-1 - 4rn-2  r2 = 4r – 4  r2 - 4r + 4 = 0  (r – 2)2 = 0 
r1 = r2 = 2
So an = 2n is a solution. As indicated, when the roots of the characteristic equation are
equal, then an = nrn is a solution. So in this case an = n2n is a solution. We can check this
by plugging into (10). When we do this we get
?
n2n = 4(n – 1)2n-1 - 4(n – 2)2n-2
Dividing by 2n-2 gives
?
?
4n = 8(n – 1) - 4(n – 2)  4n = 8n – 8 - 4n + 8
which is true.
We get the general solution by taking a superposition of an = 2n and an = n2n. So
an = A 2n + Bn2n is a solution to (10) for any constants A and B. Now we choose the
constants A and B to satisfy the initial conditions (4). Plugging in n = 0 we get
2 = a0 = A 20 + B (0) 20 = A
Plugging in n = 1 we get
5 = a1 = A 21 + B (1) 21 = 2A + 2B = (2)(2) + 2B.
1
1
So B = 2 and an = 2 2n + 2 n2n = (n + 4)2n-1. If we were interested in the behavior for
large n then an = O(n2n).
Example 3 (Fibonacci sequence). Recall the Fibonacci sequence fn is defined by the
recurrence relation
(12)
fn = fn-1 + fn-2
along with the initial conditions
5.3 - 4
(13)
f0 = 1
and
f1 = 1
To solve we look for solutions of the form an = rn where r determined by substituting int
(12). Doing this gives
rn = rn-1 + rn-2
Divide this by rn-2 giving
r2 = r + 1
or
r2 - r - 1 = 0
This doesn't look easy to factor so we use the quadratic formula.
r =
1 1+4
1 5
=
2
2
This equation has two solutions
r1 =
1+ 5
1 + 2.24

= 1.62
2
2
r2 =
1- 5
1 - 2.24

= - 0.62
2
2
This gives the following two solutions to (3)
1 + 5n
fn =  2   1.62n


and
1 - 5n
fn =  2   (- 0.62)n


By the superposition principle the general solution is
fn
1 + 5n
1 - 5n


= A
+B 2 
 2 


is a solution to (3) for any constants A and B. Now we choose the constants A and B to
satisfy the initial conditions (13). Plugging in n = 0 we get
1 = f0 = A + B
So B = 1 – A. Plugging in n = 1 we get
5.3 - 5
1 + 5
 1 - 5
1 = f1 = A  2  + B  2 




Using B = 1 – A gives
1 =
5A +
1- 5
2
or
A =
1+ 5
5+ 5
5 + 2.24
=

= 0.724
10
10
2 5
B = 1 - A = 1 -
5+ 5
5- 5
5 - 2.24
=

= 0.276
10
10
10
So
fn =
n
n
5 + 5 1 + 5 5 - 5 1 - 5

 +

  0.724 (1.62)n + 0.276 (- 0.62)n
10  2 
10  2 
Note that (- 0.62)n  0 as n  , so fn  0.724 (1.62)n for large n. In particular,
 1 + 5 n 
fn = O  2  .

 
Example 4. Let Sn be the number of n bit strings that don't contain two consecutive 1's.
Find a formula for Sn.
We discussed this in section 5.1 where we saw that Sn satisfied the recurrence relation
and initial conditions
if n  3
Sn-1 + Sn-2
if n = 1
Sn = 2
3
if n = 2
This is the same recurrence relation as the Fibonacci sequence fn except the initial
conditions are different. In fact S1 = f2 and S2 = f3. It follows that Sn = fn+1 for all n. So
5 + 5 1 + 5
5 - 5 1 - 5

 +


10  2 
10  2 
n
n
5 + 3 5 1 + 5 5 - 3 5 1 - 5

 +


=
10  2 
10  2 
n
 1.17 (1.62) - 0.17 (- 0.62)n
n+1
n+1
Sn =
5.3 - 6