Fibonacci and Golden Ratio

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Ashley Dunn
Ashley Dunn on 4 Apr 2011
Answered: Guna about 9 hours ago
One of the ways to compute the golden ration
  4 Comments
Jan
Jan on 4 Apr 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
Khan Muhammad Babar on 17 Dec 2020
Is there any way to quantify the Golden mean of Image in MATLAB. Please help.

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Answers (5)

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

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

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Walter Roberson
Walter Roberson on 4 Apr 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 Comments
Walter Roberson
Walter Roberson on 4 Apr 2011
Not completely certain. It worked for the F() values that I tried.
Jack Lê
Jack Lê on 17 Aug 2011
Thanks

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Kishore
Kishore on 8 Jul 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
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 about 9 hours ago
% 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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