Hello. Feigenbaum delta from the logistic map. Been stuck for hours and i dont know how to solve it. Please tell me my mistake and if its completly wrong please tell me

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THE QUESTION
Compute the Feigenbaum delta from the logistic map. The logistic map is given by
,
and the Feigenbaum delta is defined as
where
and where is the value of μ for which is in the orbit of the period-N cycle with .
Here is a resonable outline:
Loop 1 Start at period- with , and increment n with each iteration
Compute initial guess for using , and .
Loop 2 Iterate Newton's method, either a fixed number of times or until convergence
Initialize logistic map
Loop 3 Iterate the logistic map times
Compute x and
Loop 3 (end)
One step of Newton's method
Loop 2 (end)
Save and compute
Loop 1 (end)
Grading will be done on the converged values of up to . Set .
clc
clear all;
start_time = clock;
a0 = 2; a1 = 1+ sqrt(5); d=4;
mu(1)=a0;
mu(2)=a1;
for k = 3:15
a = a1 + (al-a0)/d;
for i = 1:2
res = 0.5; der = 0;
for j = 2:2^(k-1)+1
der = res*(1-res)+ a*(1-2*res)*der;
end
a= a-(res-0.5)/der;
end
%d = (a1-a0)/(a-a1); % approxima
fins a 4.66919841237705
d = (vpa(a1)-vpa(a0))/(vpa(a)-vpa(a1));
% approxima a 4.69201587522386
que, s clarament
millor
fprintf('Approximaci n mero % us %.15f/n', k,double(d));
a0 = a1,
a1 = a;
mu(k)=a;
end
end_time= clock;
total_time= end_time-start_time
  2 个评论
Dyuman Joshi
Dyuman Joshi 2024-1-22
What is the expected output from the question?
Is it a script-based problem or a function-based problem?
Also, some pointers for the code -
> You don't need to use vpa() here, so remove it.
> tic-toc will be a better fit here instead of clock
However, is it necessary to calculate the time taken by the code in the question?

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回答(1 个)

Vinayak
Vinayak 2024-1-19
Hi,
After analysing your code and the required map, I found a few issues. It is my understanding that you are referring to variable ‘a1’ instead of ‘al’ inside the for loop. I have modified the calculation of “d” according to the equation shared. Here is the updated code:
clc
clear all;
start_time = clock;
a0 = 2; a1 = 1 + sqrt(5); d = 4;
mu = zeros(1, 15); % Initialize mu array
mu(1) = a0;
mu(2) = a1;
for k = 3:15
a = a1 + (a1 - a0) / d;
for i = 1:2
res = 0.5; der = 0;
for j = 2:2^(k - 1) + 1
der = res * (1 - res) + a * (1 - 2 * res) * der;
end
a = a - (res - 0.5) / der;
end
% Corrected indexing for mu array
d = (mu(k-1) - mu(k-2)) / (mu(k) - mu(k-1));
fprintf('Approximation number %d is %.15f\n', k, double(d));
a0 = a1;
a1 = a;
mu(k) = a;
end
end_time = clock;
total_time = etime(end_time, start_time);
fprintf('Total time: %f seconds\n', total_time);
The above updated code correctly computes the Feigenbaum delta from the logistic map.
Hope this helps!
  3 个评论
Dyuman Joshi
Dyuman Joshi 2024-1-22
@food lover, There is no "num_doublings" parameter in the answer above, nor is there a for loop with n as the loop index.
So the error does not stem from the code given by @Vinayak.
Unless we know what code you have provided as the solution and what exactly is the question asking, it is difficult to provide any suggestions.

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