Fibonacci and Golden Ratio
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One of the ways to compute the golden ration
4 comentarios
Jan
el 4 de Abr. de 2011
@Ashley: Don't give up: As soon as you edit the question and add any details - and a question! - you will get meaningful answers.
Khan Muhammad Babar
el 17 de Dic. de 2020
Is there any way to quantify the Golden mean of Image in MATLAB. Please help.
Respuestas (5)
Clemens
el 17 de Ag. de 2011
Actually the Golden Ratio is exactly:
( 1 + sqrt(5) ) / 2
so no need for iteration. Proof is easy through z-transform.
2 comentarios
Walter Roberson
el 17 de Ag. de 2011
But that gets back to my original answer, "The Golden Ratio is an irrational number, and thus an infinite number. It is not possible to compute its decimal expansion in a finite amount of time."
Jan
el 17 de Ag. de 2011
Fortunately the universe is finite. Therefore I do not believe, that an infinite number will match into it. While there is a minimal Planck length and a minimal Plank time, I propose a Planck eps for irrational numbers. According to Rupert Sheldrake, I claim that PI has as many numbers as has been calculated already. And after reading http://scientopia.org/blogs/goodmath/2010/12/08/really-is-wrong/ I'm not sure at all anymore about this fuzzy digits stuff.
Walter Roberson
el 4 de Abr. de 2011
The Golden Ratio is an irrational number, and thus an infinite number. It is not possible to compute its decimal expansion in a finite amount of time.
8 comentarios
Sean de Wolski
el 4 de Abr. de 2011
Soya sausages? That's like one term in the Taylor-series expansion of sausages.
Walter Roberson
el 17 de Ag. de 2011
Jan, Soya Beans used for the production of soya products are the dried fruit of the soya plant, and thus were not covered by the Veggi-Toolbox in R2011a (which, I understand, is still withheld from production due to legal battles over whether Tomatoes are fruits or vegetables....)
Walter Roberson
el 4 de Abr. de 2011
Let F(t) be Fibonacci number #t. Then
y = 100; %initial guess
x = (F(t+2) * y + F(t+1)) / (F(t+1) * y + F(t));
while x ~= y;
y = x;
x = (F(t+2) * y + F(t+1)) / (F(t+1) * y + F(t));
end
When the loop finishes (no more than a few centuries later, I'm sure), x and y will be the Golden ratio.
3 comentarios
Walter Roberson
el 4 de Abr. de 2011
Not completely certain. It worked for the F() values that I tried.
Kishore
el 8 de Jul. de 2023
fib=[0 1];
i=3;
while(i<=21)
fib(i)=fib(i-1)+fib(i-2);
gr=fib(i)/fib(i-1)
i=i+1;
end
disp(fib)
0 comentarios
Guna
el 16 de Abr. de 2024
% Function to calculate Fibonacci sequence up to a certain number of terms
function fib_sequence = fibonacci(n)
fib_sequence = zeros(1, n);
fib_sequence(1) = 0;
fib_sequence(2) = 1;
for i = 3:n
fib_sequence(i) = fib_sequence(i-1) + fib_sequence(i-2);
end
end
% Calculate the golden ratio using Fibonacci sequence
n = 20; % Number of Fibonacci terms to generate
fib_seq = fibonacci(n);
% Calculate the ratio of consecutive Fibonacci numbers
golden_ratio_approximations = fib_seq(3:end) ./ fib_seq(2:end-1);
% Display the approximations of the golden ratio
disp('Approximations of the golden ratio using Fibonacci sequence:');
disp(golden_ratio_approximations);
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