We conclude the week by deriving the celebrated Binet’s formula, an explicit formula for the Fibonacci
numbers in terms of powers of the golden ratio and its reciprical.
4
Lecture 1
The Fibonacci sequence
View this lecture on YouTube
Fibonacci published in the year 1202 his now famous rabbit puzzle:
A man put a malefemale pair of newly born rabbits in a field. Rabbits take a month to
mature before mating. One month after mating, females give birth to one malefemale pair
and then mate again. No rabbits die. How many rabbit pairs are there after one year?
To solve, we construct Table
1.1
. At the start of each month, the number of juvenile pairs, adult
pairs, and total number of pairs are shown.
At the start of January, one pair of juvenile rabbits is
introduced into the population. At the start of February, this pair of rabbits has matured. At the start
of March, this pair has given birth to a new pair of juvenile rabbits. And so on.
month
J
F
M
A
M
J
J
A
S
O
N
D
J
juvenile
1
0
1
1
2
3
5
8
13
21
34
55
89
adult
0
1
1
2
3
5
8
13
21
34
55
89
144
total
1
1
2
3
5
8
13
21
34
55
89
144
233
Table 1.1: Fibonacci’s rabbit population.
We define the Fibonacci numbers
F
n
to be the total number of rabbit pairs at the start of the
n
th
month. The number of rabbits pairs at the start of the 13th month,
F
13
=
233, can be taken as the
solution to Fibonacci’s puzzle.
Further examination of the Fibonacci numbers listed in Table
1.1
, reveals that these numbers satisfy
the recursion relation
F
n
+
1
=
F
n
+
F
n

1
.
(1.1)
This recursion relation gives the next Fibonacci number as the sum of the preceeding two numbers.
To start the recursion, we need to specify
F
1
and
F
2
. In Fibonacci’s rabbit problem, the initial month
starts with only one rabbit pair so that
F
1
=
1. And this initial rabbit pair is newborn and takes one
month to mature before mating so
F
2
=
1.
The first few Fibonacci numbers, read from the table, are given by
1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, . . .
and has become one of the most famous sequences in mathematics.
5
6
LECTURE 1. THE FIBONACCI SEQUENCE
Problems for Lecture 1
1.
The Fibonacci numbers can be extended to zero and negative indices using the relation
F
n
=
F
n
+
2

F
n
+
1
.
Determine
F
0
and find a general formula for
F

n
in terms of
F
n
.
Prove your result using
mathematical induction.
2.
The Lucas numbers are closely related to the Fibonacci numbers and satisfy the same recursion
relation
L
n
+
1
=
L
n
+
L
n

1
, but with starting values
L
1
=
1 and
L
2
=
3. Determine the first 12 Lucas
numbers.
3.
The generalized Fibonacci sequence satisfies
f
n
+
1
=
f
n
+
f
n

1
with starting values
f
1
=
p
and
f
2
=
q
. Using mathematical induction, prove that
f
n
+
2
=
F
n
p
+
F
n
+
1
q
.
(1.2)
4.
Prove that
L
n
=
F
n

1
+
F
n
+
1
.
(1.3)
5.
Prove that
F
n
=
1
5
(
L
n

1
+
L
n
+
1
)
.
6.
The generating function for the Fibonacci sequence is given by the power series
f
(
x
) =
∞
∑
n
=
1
F
n
x
n
.
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 Spring '16
 Computer Science, Fibonacci number, Golden ratio