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

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

[Dyuman Joshi]

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?

[food lover]

Feigenbaum delta = 4.669201609102047

script based

and yes i believe so

Answer 1

[Vinayak]

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!

[food lover]

unfortunately there is still an error. Thank you for trying though

[Dyuman Joshi]

@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.

[food lover]

i just inserted his work onto mathlab.