Miranda Coulter
Tushar Das
MATH 2700
11 May 2010
Solving Fibonacci: Determining the nth Term in the Fibonacci Sequence
Purpose
The purpose of this paper is to use Linear Algebra to determine the nth term in the Fibonacci
sequence without prior knowledge of the previous two terms.
History
In the late twelfth century, the mathematician currently known as Fibonacci was born as
Leonardo of Pisa, or Leonardo Pisano in Italian. The name Fibonacci came from the shortening
of the Latin “filius Bonacci” or son of Bonacci. Known as the greatest European mathematician
of the Middle Ages, he was one of the first to introduce the Hindu-Arabic number system to
number system (the numbering system we use today) to Europe. Fibonacci’s writing of Liber
Abaci, or The Book of Calculations, persuaded many western European scholars and
mathematicians to use the Hindu-Arabic system over the Roman Numeral system, greatly
simplifying the manner in which even basic arithmetic can be approached.
It was in this same book, that the pattern that would later come to be known as the Fibonacci
Sequence was recorded. Although not created by the mathematician, this sequence still bears his
name today.
The Problem
From Fibonacci’s Liber Abaci, Chapter 12,
How Many Pairs of Rabbits Are Created by One Pair in One Year
A certain man had one pair of rabbits together in a certain enclosed place, and one
wishes to know how many are created from the pair in one year when it is the
nature of them in a single month to bear another pair, and in the second month
those born to bear also.
…
Because the above written pair in the first month bore, you will double it; there
will be two pairs in one month.
One of these, namely the first, bears in the second month, and thus there are in the
second month 3 pairs;
of these in one month two are pregnant and in the third month 2 pairs of rabbits
are born, and thus there are 5 pairs in the month;
...
there will be 144 pairs in this [the tenth] month;
to these are added again the 89 pairs that are born in the eleventh month; there
will be 233 pairs in this month.
To these are still added the 144 pairs that are born in the last month; there will be
377 pairs, and this many pairs are produced from the abovewritten [sic] pair in the
mentioned place at the end of the one year. (404-405)
The problem above assumes two things: 1) that the man is in possession of two fully mature
rabbits capable of producing offspring the first month and 2) that there are no deaths throughout
the year. Although Fibonacci clearly outlines the pairs through the first year, how could he
determine the number of pairs years later, once simple addition was no longer practical, if this
pattern continued? For the purposes of this paper, we too will make two assumptions: 1) the man
is in possession of two newborn rabbits incapable of reproducing until after the second month
and 2) that there are no deaths.
The preceding image depicts the question posed by Fibonacci. After the second month, the first
pair of rabbits produces another pair of rabbits capable of producing offspring each month. After
their second month, each new pair of rabbits also produces another pair of rabbits.
n
fn
n
fn
n
fn
n
fn
n
fn
1
1
6
8
11
89
16
987
21
10946
2
1
7
13
12
144
17
1597
22
17711
3
2
8
21
13
233
18
2584
23
28657
4
3
9
34
14
377
19
4181
24
46368
5
5
10
55
15
610
20
6765
25
75025
The above table allows us to quickly see the first 25 values of n and to verify that
𝑓𝑛 = 𝑓𝑛−1 + 𝑓𝑛−2 . In order to solve fn, however, the previous two terms in the sequence must
already be known. Although simple enough, this procedure of adding the two previous terms
together to identify the desired term quickly becomes impractical.
Applying Linear Algebra
To begin, pairs of n values should be written as vectors and the 2x2 matrix A identified.
𝑓
1
[ 1] = [
𝑓2
1
1 𝑓0
1
][ ] = [
𝑓
0 1
1
1 1
] [ ] = 𝐴𝒙
0 0
By raising A to the nth power, the nth and (n+1)th term of the sequence can be found.
[
𝑓𝑛+1
1 1𝑛 1
]=[
] [ ] = 𝐴𝑛 𝑥
𝑓𝑛
1 0 0
Although this equation seems simple enough at first glance, determining An can be both difficult
and time consuming. By utilizing more Linear Algebra techniques, An can be simplified and a
precise definition of fn found.
Diagonalization
To simplify An, a technique known as Diagonalization must be used. A matrix is said to be
diagonalizable when it can written in factored form as PDP-1. With this method An can be
written using the matrices P, P-1, and D (a diagonal matrix). Although several calculations must
first be performed before these three matrices can be identified, expressing An in this form will
ultimately allow for a much simpler expression of fn.
𝐴 = 𝑃𝐷𝑃−1
𝜆
𝐷=[ 0
0
0
]
𝜆1
𝐴2 = 𝑃𝐷𝑃−1 𝑃𝐷𝑃 −1
= 𝑃𝐷𝐼𝐷𝑃 −1
= 𝑃𝐷2 𝑃−1
𝐴𝑛 = 𝑃𝐷𝑛 𝑃 −1
𝐷𝑛 = [
𝜆𝑛0
0
0
]
𝜆1𝑛
To determine the λ values in D, we must now find the eigenvalues and eigenvectors of A.
Eigenvalues and Eigenvectors
We must first determine the eigenvalues of A. An eigenvalue can be defined as a scalar λ such
that the equation 𝐴𝒙 = 𝝀𝒙 has a solution for some nonzero vector x. To determine the
eigenvalue, we set the determinant of (𝐴 − 𝜆𝐼) equal to zero and solve for λ.
0 = det(𝐴 − 𝜆𝐼)
1 1
1
= det ([
] −λ[
1 0
0
= det ([
0
])
1
1 1
λ 0
]−[
])
1 0
0 λ
1−𝜆
=|
1
1
|
−𝜆
= 𝜆2 − 𝜆 − 1
𝜆=
1 ± √5
2
After obtaining the two eigenvalues, calculations for the two eigenvectors, the nonzero vector x
such that 𝐴𝒙 = 𝝀𝒙 for some scalar λ, can now be determined.
𝜆0 =
1−
0
[ ]=
0
[
1 + √5
2
1
1 + √5
2
1 − √5
𝑥+𝑦
𝑥
2
[𝑦 ] =
1 + √5
1 + √5
−
𝑥−
𝑦]
]
[
2
2
1
−𝑦 =
𝑥=
1 − √5
𝑥
2
1 + √5
𝑦
2
𝑥0 = [
1 + √5
2 ]
1
Because there were two eigenvalues, the calculations must be repeated for the second scalar and
a second eigenvector obtained.
1 − √5
2
𝜆1 =
1−
0
[ ]=
0
[
1 − √5
2
1
1 + √5
𝑥+𝑦
𝑥
2
[ ]=
1 − √5 𝑦
1 − √5
−
𝑥
−
𝑦]
[
2 ]
2
1
−𝑦 =
𝑥=
1 + √5
𝑥
2
1 − √5
𝑦
2
𝑥1 = [
1 − √5
2 ]
1
With values for λ, we can now construct P, P-1, and D.
Final Calculations
After applying the diagonalization technique and obtaining both the eigenvalues and
eigenvectors of matrix A, the matrices P, P-1, and D can now be constructed. With these
matrices, a simplified value for 𝐴𝑛 can be obtained and a statement expressing the nth term in the
Fibonacci sequence formulated.
𝐴𝑛 = 𝑃𝐷𝑛 𝑃 −1
1 + √5
2
𝐷=
0
[
0
1 − √5
2 ]
𝑛
0
0
1 − √5
(
)
2
]
𝑛
1 + √5
(
)
2
𝐷 =
𝑛
[
1 + √5 1 − √5
𝑃=[ 2
2 ]
1
1
𝑃−1 =
1
√5
1
[−1
1 − √5
2
1 + √5
]
2
−
With the above matrices we can obtain a statement for the diagonalizable An.
𝑛
1 + √5
(
)
1 + √5 1 − √5
1
2
𝐴𝑛 =
[ 2
2 ]
√5
1
1
0
[
𝑛+1
=
1 + √5
(
)
1
2
√5
𝑛
1 + √5
(
)
2
[
𝑛+1
=
1 + √5
(
)
1
2
√5
0
1
𝑛
1 − √5
(
) [−1
2
]
1 − √5
2
1 + √5
]
2
−
𝑛+1
1 − √5
(
)
2
1 − √5
2
1 + √5
]
2
1
−
𝑛
1 − √5
(
) [−1
2
]
𝑛+1
𝑛
1 − √5
−(
)
2
𝑛
𝑛
1 + √5
1 − √5
(
) −(
)
2
2
𝑛
𝑛+1
1 + √5
1 − √5
(
) −(
)
2
2
[
1 + √5
(
)
2
𝑛+1
1 − √5
−(
)
2
]
By inserting the matrix An into the original equation, a value for fn for any n greater than or equal
to three is finally obtained.
[
𝑓𝑛+1
1 1𝑛 1
]=[
] [ ] = 𝐴𝑛 𝑥
𝑓𝑛
1 0 0
𝑛+1
[
1 + √5
(
)
1
2
𝑓𝑛+1
]=
𝑓𝑛
√5
𝑛+1
1 − √5
−(
)
2
𝑛
𝑛
1 + √5
1 − √5
(
) −(
)
2
2
[
𝑛+1
=
1 + √5
(
)
1
2
√5
𝑛
𝑛
𝑛
1 + √5
1 − √5
(
) −(
)
2
2
𝑛+1
1 + √5
(
)
2
−(
𝑛+1
1 − √5
−(
)
2
𝑛
1 + √5
1 − √5
(
) −(
)
2
2
[
]
1
[ ]
0
𝑛+1
1 − √5
)
2
]
𝑛
1
𝑛
1 + √5
1 − √5
𝑓𝑛 =
[(
) −(
) ]
2
2
√5
The above equation holds true for any n in the Fibonacci sequence when
𝑓
0
[ 0 ] = [ ].
𝑓1
1
The equation can be simply altered, however, to accommodate any two starting values within the
sequence. For example, a precise expression of Fibonacci’s exact riddle would involve a starting
point of
𝑓
1
[ 0] = [ ]
𝑓1
2
and a final equation of
𝑓𝑛 =
1
√5
1+√5
[(
2
𝑛+2
)
1−√5
−(
2
𝑛+2
)
].
Conclusion
The Fibonacci sequence is a relatively simple sequence to solve by simple addition, at least for
the first twenty or so values. With careful application of Linear Algebra techniques, however, a
relatively simple expression for the nth term in the sequence can be found.
References
Fibonacci, Leonardo. "Here Begins Chapter Twelve." Fibonacci's Liber Abaci: a Translation into
Modern English of Leonardo Pisano's Book of Calculation. Trans. L. E. Sigler. New York:
Springer, 2002. 404-05. Print.
Hendel, Russell J. "Approaches to the Formula for the nth Fibonacci Number." The College
Mathematics Journal 25.2 (1994): 139-42. JSTOR. Web. 27 Apr. 2010.
Lay, David C. "Eigenvalues and Eigenvectors." Linear Algebra and Its Applications. Boston:
Pearson/Addison-Wesley, 2006. 301-71. Print.