fibonacci series, while loop
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Remember the Fibonacci series:
a(k+1)=a(k)+a(k−1) ∀k>2, a(0)=1,a(1)=1 (2)
Interestingly the golden ratio can be calculated from the this series:
— Write a script that calculates the golden ratio from the Nth and (N − 1)th Fibonacci numbers.
my code look like this but it didnt work. Can someone help me? Thank you!
a(1) = 1;
a(2) = 1;
n=3
while n <= 16 %the value for this can be changed
a(n) = a(n-1)+a(n-2);
ratio = limit(a(n)/a(n-1),n,inf);
n=n+1;
end
2 个评论
darova
2019-10-6
Where did yu get this information?
John D'Errico
2019-10-6
The limit will indeed approach the golden ratio. This is a well known fact, and pretty easily proven, based on the Binet formula for the Fibonacci sequence.
(sqrt(5) + 1)/2
ans =
1.61803398874989
采纳的回答
John D'Errico
2019-10-6
编辑：John D'Errico
2019-10-6
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; % how far out
a = zeros(1,N); % preallocation
a(1) = 1;
a(2) = 1;
for n = 3:N % a for loop is simpler to write than a while loop
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?
更多回答（1 个）
David Hill
2019-10-6
If you have the symbolic toolbox you can just use the built in fibonacci function.
estGR = fibonacci(n)/fibonacci(n-1);
Otherwise, the code to calculate the nth fibonacci number is not difficult.
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