18.06 Final Solution
Hold on Tuesday, 19 May 2009 at 9am in Walker Gym.
Total: 100 points.
Problem 1:
A sequence of numbers
f
0
,f
1
,f
2
,...
is deﬁned by the recurrence
f
k
+2
= 3
f
k
+1

f
k
,
with starting values
f
0
= 1,
f
1
= 1. (Thus, the ﬁrst few terms in the sequence are
1
,
1
,
2
,
5
,
13
,
34
,
89
,...
.)
(a) Deﬁning
u
k
=
±
f
k
+1
f
k
²
, reexpress the above recurrence as
u
k
+1
=
A
u
k
, and
give the matrix
A
.
(b) Find the eigenvalues of
A
, and use these to predict what the ratio
f
k
+1
/f
k
of
successive terms in the sequence will approach for large
k
.
(c) The sequence above starts with
f
0
=
f
1
= 1, and

f
k

grows rapidly with
k
.
Keep
f
0
= 1, but give a
diﬀerent
value of
f
1
that will make the sequence (with
the
same recurrence
f
k
+2
= 3
f
k
+1

f
k
) approach
zero
(
f
k
→
0) for large
k
.
Solution
(18 points = 6+6+6)
(a) We have
±
f
k
+2
f
k
+1
²
=
±
3

1
1
0
²±
f
k
+1
f
k
²
⇒
A
=
±
3

1
1
0
²
.
(b) Eigenvalues of
A
are roots of det(
A

λI
) =
λ
2

3
λ
+ 1 = 0. They are
λ
1
=
3 +
√
5
2
and
λ
2
=
3

√
5
2
. Note that
λ
1
> λ
2
, so the ratio
f
k
+1
/f
k
will
approach
λ
1
=
3 +
√
5
2
for large
k
.
(c) Let
v
1
,
v
2
be the eigenvectors with eigenvalues
λ
1
and
λ
2
, respectively. So,
we can write
u
0
=
c
1
v
1
+
c
2
v
2
and then
u
k
=
c
1
λ
k
1
v
1
+
c
2
λ
k
2
v
2
. If we need
f
k
→
0,
1