Download Fibonacci Sequence Example

Section 4.8: Fibonacci Numbers and Applications to Difference Equations
In this section, we see how the theoretical concepts of a basis of a vector space can be used to solve
an actual down-to-earth math problem. In his book Liber Abaci (1202), Leonardo of Pisa, better
known as Fibonacci, introduced the sequence of numbers: 1, 1, 2, 3, 5, 8, 13, 21, 34, ... to model the
growth of rabbit populations. These numbers are known as the Fibonacci numbers and are formally
defined by the recursion relation:
Fn+2 = Fn+1 + Fn , for n ≥ 0, with F0 = F1 = 1
One can ask whether there is a non-recursive formula for the nth Fibonacci number Fn . We have
√
1 Proposition 1. Fn = √ φn+1 − (1 − φ)n+1 , where φ = 12 (1 + 5) is the golden ratio.
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There are many proofs of this formula, but linear algebra gives a particularly simple proof.
Proof. Let V be the set of all sequences (a0 , a1 , a2 , . . . ) with ai ∈ R and let H be the set of all
sequences (a0 , a1 , a2 , . . . , ) satisfying the equation:
xn+2 = xn+1 + xn , for n ≥ 0.
(*)
The Fibonacci sequence F = (F0 , F1 , F2 , . . . ) defines one element F ∈ H. We now prove the formula
for Fn by making a number of observations.
1. V is a vector space.
2. H is a vector subspace of V .
Proof: Given a sequence α = (αi ) ∈ H, we have αn+2 = αn+1 + αn . Multiplying by c, we
have cαn+2 = cαn+1 + cαn . Thus the sequence cα = (cαi ) is in H and H satisfies closure of
scalar multiplication. Additive closure follows similarly. Lastly, the zero element 0 = (0, 0, 0, . . . )
satisfies the equation (*), hence 0 ∈ H. Thus H is a subspace of V .
3. dim H = 2.
Proof: Given any two real numbers a0 , a1 ∈ R, there is a unique element h ∈ H with h =
(a0 , a1 , a2 , . . . ) and the rest of the terms ai are determined by the recurrence equation in (*).
Let v1 ∈ H be the term that begins v1 = (1, 0, . . . ), and let v2 ∈ H be the element that begins
v2 = (0, 1, . . . ). Then h = a0 v1 + a1 v2 as they both have the same first two elements. Hence v1 , v2
span V . They are also linearly independent as the only solution to av1 + bv2 = 0 is a = b = 0.
Thus {v1 , v2 } is basis for H and dim H = 2.
√
4. Let φ = 12 (1 + 5) and define w1 = (1, φ, φ2 , . . . ), w2 = (1, 1 − φ, (1 − φ)2 , . . . ). Then w1 , w2 ∈ H.
Proof: The element R = (1, r, r2 , . . . , ) is in H if and only if rn+2 = rn+1 +rn , for n ≥ 0. Dividing
by rn , this equation is equivalent to r2 − r − 1 = 0, which has the two solutions r = φ, (1 − φ).
Hence w1 , w2 ∈ H.
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5. The set S = {w1 , w2 } is a basis for H.
Proof: If we can show that S is linearly independent, then S is a basis, as any linearly independent
set of H is a subset of a basis. Since dim H = 2 and S has size 2, S must already be a basis for
H. To show S is linearly independent, we consider the equation cw1 + dw2 = 0. Looking at the
first two coordinates, we obtain
c(1, φ) + d(1, 1 − φ) = (0, 0)
Rearranging, we have
c
A
!
d
=
!
0
0
, with A =
!
1
1
φ
1−φ
!
0
−1
Since det A 6= 0, the only solution to this equation is A
.
0
!
0
. Hence S is linearly
0
=
independent.
6. F = φw1 + (1 − φ)w2 .
Proof: Since {w1 , w2 } are a basis for H, we know F = cw1 + dw2 for some real numbers c, d ∈ R.
Since elements of H are determined by their first two coordinates, we need to solve the equation
(1, 1) = c(1, φ) + d(1, 1 − φ).
In other words,
c+d=1
cφ + d(1 − φ) = 1
or
Since det A 6= 0, A is invertible and
!
!
c
1
−1 1
=A
= √
− 5
d
1
A
c
!
d
=
1
1
!
, with A =
! !
(1 − φ) −1
1
−φ
1
1
=
!
1
1
φ
(1 − φ)
√
φ/ 5
!
√
−(1 − φ)/ 5
1 7. Fn = √ φn+1 − (1 − φ)n+1 .
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1
Proof: Fn is the (n + 1)st coordinate of F . Since F = √ [φw1 − (1 − φ)w2 ], looking at the
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(n + 1)st coordinates, we have:
1
Fn = √ [φ(φn ) − (1 − φ)(1 − φ)n ]
5
1 n+1
=√ φ
− (1 − φ)n+1 .
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