Well, what is important to me is you have made a credible effort, and gotten close. In fact, you did compute Fibonacci sequence elements.
First, you should have preallocated the vector a, which would make MATLAB more efficient.
What would i change in the code you show?
N = 16;
a = zeros(1,N);
a(1) = 1;
a(2) = 1;
for n = 3:N
a(n) = a(n-1)+a(n-2);
end
But how about that ratio thing? You cannot use the limit function to compute a limit as you did. Sorry. But the limit is just the ratio of two consecutive numbers from that sequence. So now, you want to see how well those ratios approach the expected limit as N grows larger.
ratio = zeros(1,N-1);
for n = 1:N-1
ratio(n) = a(n)/a(n+1);
end
format long g
ratio'
ans =
1
2
1.5
1.66666666666667
1.6
1.625
1.61538461538462
1.61904761904762
1.61764705882353
1.61818181818182
1.61797752808989
1.61805555555556
1.61802575107296
1.61803713527851
1.61803278688525
What is the expected limit supposed to be? It is approaching that theoretical limit?
Best Answer