Azzera filtri
Azzera filtri

Use eigendecomposition to compute number in modified Fibonacci sequence

4 visualizzazioni (ultimi 30 giorni)
I constructed the following code to compute x13:
clear all
close all
clc
f(1) = 3;
f(2) = 5;
for i = 3 : 13
f(i) = (f(i-1) + f(i-2))-4;
end
Unfortunately, the desired response is: Use the method of eigendecomposition to compute x13 for this Fibonacci Cat sequence. What is the largest eigenvalue? What is the eigenvector corresponding to this largest eigenvalue?
I was able to use a for-loop to figure out x13, but I do not know how to use the eigendecomposition to compute this value. Thanks!

Risposte (1)

Roger Stafford
Roger Stafford il 11 Ago 2014
First, you can convert the recursion to
y_(k+1) = y_k + y_(k-1)
where y_k = x_k - 4. Then you can write
[y_(k+2);y_(k+1)] = A * [y_(k+1);y_k]
where A = [1,1;1,0]. Then find the eigenvalues and eigenvectors of A.
[V,D] = eig(A);
Hence you would have
[y_13; y_12] = A^11*[y_2;y_1] = (V*D^11*V')*[y_2;y_1]
where D^11 is merely the eleventh power of the diagonal eigenvalues. This expresses the value of y_13 (and therefore x_13) directly in terms of y_2 and y_1 without any iterated steps.
The Wikipedia article at:
http://en.wikipedia.org/wiki/Fibonacci_number
describes this procedure in their section 5 - Matrix Form.
  1 Commento
Nicolas
Nicolas il 13 Ago 2014
Modificato: Nicolas il 13 Ago 2014
Thank you for your response.
I have attempted to utilize the code you provided, but am receiving errors when trying to run it.
This is what I am trying to run:
clear all
close all
clc
y_k = x_k - 4;
A = [1,1;1,0];
y_(k+1) = y_k + y_(k-1);
[y_(k+2);y_(k+1)] = A * [y_(k+1);y_k];
[V,D] = eig(A);
[y_13; y_12] = A^11*[y_2;y_1] = (V*D^11*V')*[y_2;y_1];
I am receiving the error message:
Error: File: hw5exercise3.m Line: 10 Column: 8
Multiple left-hand sides must be separated by commas.
Do you know how I can resolve this problem?
Thanks!

Accedi per commentare.

Categorie

Scopri di più su Linear Algebra in Help Center e File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by