Question

Suppose each number Gk+1 is the average of the two previous Gibonacci numbers Gk+1 and Gk and then Gk+2 =1/2(Gk+1+ Gk )

■(Gk+2 ,Gk+1)T =A[Gk+1 ,Gk]T

find the eigenvalus and eignevectors of A
find the limit as n->∞ of the matrix
An=S ∂nS-1
if G0 =0 and G1=1 show that the Gibonacci numbers approach 2/3

Answer 1

U still looking for answer ?

Answer 2

yes please , if you know please adivece me

Answer 3

I think the real formulation is
Vk+1 = A*Vk
where Vk+1=Gk+3 Gk+2
Gk+2 Gk+1
and Vk = Gk+2 Gk+1
Gk+1 Gk

Gk+3 Gk+2= A * Gk+2 Gk+1
Gk+2 Gk+1 Gk+1 Gk
where A = 1/2 1/2
1 0
indeed then A * Gk+1 Gk = 1/2Gk+1+1/2Gk 1/2Gk+1/2Gk-1 =Gk+2 Gk+1
Gk Gk-1 Gk+1 Gk Gk+1 Gk
so we want to understand A
we look for eigenvalues of A, i:e solutions (for lambda) of det(A-lambda.Identity)=0
we find 2 eigen values : 1 and -1/2 and related eigenvectors {1, 1} and {-(1/2), 1}
so A=P*D*P^-1
where D= 1 0
0 -1/2
P= 1 -1/2
1 1
P^-1 = 2/3 1/3
-2/3 2/3
(you can check that P^-1 * P = P*P^-1 = Identity
and that A=P*D*P^-1)
this means Vk+1 = A*Vk=P*D*P^-1*Vk
which means P^-1* Vk+1 = D*P^-1*Vk
P^-1*Vk = Wk
and thus P^-1*Vk+1 = Wk+1 and P^-1*Vk = Wk
and Wk+1 = D*Wk
Wk=(gk+1)/3+(2 gk+2)/3 (gk+0)/3+(2 gk1)/3
(2 gk+1)/3-(2 gk+2)/3 (2 gk+0)/3-(2 gk+1)/3

Wk+1 = D*Wk gives us :
Wk=D^k*W0
D being diagonal, D^k = 1^k 0
0 -1/2^k
when k increases D^k becomes 1 0
0 0
if G0=0 and G1=1 then G2= 1/2(0+1)=1/2
W0 =(g1)/3+(2 g2)/3 (g0)/3+(2 g1)/3
(2 g1)/3-(2 g2)/3 (2 g0)/3-(2 g1)/3
= 2/3 2 /3
1/3 -2 /3

so Wk trds twds vers (2/3 0
0 0)
so(2 gk+0)/3-(2 gk+1)/3 trend twds 0 so gk+1 - gk trds twds 0
and(gk+1)/3+(2 gk+2)/3 trds twds 2/3
as gk=gk+1=gk+2 when k increases
we have (gk)/3+(2 gk)/3 =gk trds twds 2/3