Fibonacci and Golden Ratio

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Ashley Dunn
Ashley Dunn 2011-4-4
回答： Guna 2024-4-16
One of the ways to compute the golden ration
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Jan
Jan 2011-4-4
@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
Khan Muhammad Babar 2020-12-17
Is there any way to quantify the Golden mean of Image in MATLAB. Please help.

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回答（5 个）

Clemens
Clemens 2011-8-17
Actually the Golden Ratio is exactly:
( 1 + sqrt(5) ) / 2
so no need for iteration. Proof is easy through z-transform.
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Walter Roberson
Walter Roberson 2011-8-17
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
Jan 2011-8-17
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.

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Walter Roberson
Walter Roberson 2011-4-4
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.
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Sean de Wolski
Sean de Wolski 2011-4-4
Soya sausages? That's like one term in the Taylor-series expansion of sausages.
Walter Roberson
Walter Roberson 2011-8-17
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....)

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Walter Roberson
Walter Roberson 2011-4-4
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.
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Walter Roberson
Walter Roberson 2011-4-4
Not completely certain. It worked for the F() values that I tried.

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Kishore
Kishore 2023-7-8
Ran in:
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
gr = 1
gr = 2
gr = 1.5000
gr = 1.6667
gr = 1.6000
gr = 1.6250
gr = 1.6154
gr = 1.6190
gr = 1.6176
gr = 1.6182
gr = 1.6180
gr = 1.6181
gr = 1.6180
gr = 1.6180
gr = 1.6180
gr = 1.6180
gr = 1.6180
gr = 1.6180
gr = 1.6180
disp(fib)
Columns 1 through 16 0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 Columns 17 through 21 987 1597 2584 4181 6765
Guna
Guna 2024-4-16
Ran in:
% 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:');
Approximations of the golden ratio using Fibonacci sequence:
disp(golden_ratio_approximations);
1.0000 2.0000 1.5000 1.6667 1.6000 1.6250 1.6154 1.6190 1.6176 1.6182 1.6180 1.6181 1.6180 1.6180 1.6180 1.6180 1.6180 1.6180

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