for loop label/arguement

Need to iterate over the bifurcation map n_trans time (n_trans specified above), this is the relevant bit of the code for my question.. which is pretty basic.. what is wrong with the below? I get the error message

"Error in solution: Line: 25 Column: 20

Invalid array indexing."

(no longer line 25 ofc...)

many thanks

    
    x(1)=0.5;
     for j=2:(n_trans-1)
         x(j+1)=mu*x(j)(1-x(j));
        
     end
    

Answer 1

[Walter Roberson]

That looks suspiciously like an Octave message rather than a MATLAB message...

x(j+1)=mu*x(j)(1-x(j));
             ^^

MATLAB has no implied multiplication, not anywhere -- not even inside the binary language of the moisture vaporators.

[Sara Ismail-Sutton]

sorry, i'm really confused with your answer, thanks ..

[Steven Lord]

You need to explicitly include * between two terms if you want to multiply.

x = 5;
y = 3*x % works
y = 3x % does not work
z = xy % does not work

[Sara Ismail-Sutton]

okay sure, my bad that makes sense.. but that's not what is effecting the error message here is it?

[Walter Roberson]

Yes it is. It is seeing variable(index)(index) and saying that you cannot do two () index operations in a row.

[Sara Ismail-Sutton]

okay many thanks, now I am getting the error

" Unable to perform assignment because the left and right sides have a different number of elements.

Error in solution (line 25)

x(j)=mu*x(j-1)*(1-x(j-1));"

[Walter Roberson]

Your mu appears to be non-scalar.

[Sara Ismail-Sutton]

how so?

[Walter Roberson]

We do not have your complete source code. But if j is non-scalar then x(j-1) would be non-scalar and x(j-1)*(1-x(j-1)) would be an error unless j-1 is a square matrix, in which case you would get a result the same size as j-1 . But that would be fine for storing into x(j) . So the problem is not the j, and must be the mu. For example mu might be empty.

[Sara Ismail-Sutton]

ok, the task is bifurcation diagram for logistic map, the assignment provides a template, just the for loops that I have wrote.

So, from what you say, I believe mu is a no-scalar , overall, but this concerns the second loop, so here it will be fixed and a scalar?

Thanks

mu_min=2.4; mu_max=4; %range of mu values
n_mu=500; %number of mu pixels
n_x=400; %number of x pixels
mu_edges=linspace(mu_min,mu_max,n_mu+1); %edges of mu pixels
mu=(mu_edges(1:n_mu)+mu_edges(2:n_mu+1))/2; %values of mu on which to perform computation
x_edges=linspace(0,1,n_x+1); %edges of x pixels

n_trans=200000; %transient iterations
n_data=100000;  %number of x values per mu value

x_data=zeros(n_data,n_mu); %x-data used to construct figure

x_0=0.5; %initial condition

% WRITE THE COMPUTATIONAL ENGINE OF THE CODE.
% USE THE ALREADY DEFINED PARAMETERS AND VARIABLES: n_mu, mu, x_0, n_trans, n_data.
% YOUR FINAL RESULT WILL BE THE VARIABLE x_data, and this variable will be assessed.



for i=mu_min:mu_max
    
    x(1)=0.5;
     for j=2:n_trans
         x(j)=mu*x(j-1)*(1-x(j-1));
        
     end
    
 
   x(1)= x(n_trans)
    for k=2:n_data
   
        x(k+1)=mu*x(k)*(1-x(k));
       
        
            
       
    end
end








%%%%% bin data and plot image %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
x_histogram=zeros(n_x,n_mu); %binned values of x
for i=1:n_mu
x_histogram(:,i)=histcounts(x_data(:,i),x_edges);
x_histogram(:,i)=255*x_histogram(:,i)/max(x_histogram(:,i));
end
colormap(flipud(gray(256))); brighten(-0.8); cmap=colormap;
im=image([mu_edges(1) mu_edges(end)], [x_edges(1) x_edges(end)], x_histogram);
set(gca,'YDir','normal');
xlabel('$\mu$','Interpreter','latex','FontSize',14);
ylabel('$x\;\;$','Interpreter','latex','FontSize',14);
title('Logistic Map Bifurcation Diagram','Interpreter','latex','FontSize',16)

[Walter Roberson]

for i=mu_min:mu_max

Should be

for i=1:n_mu

and

           x(j)=mu*x(j-1)*(1-x(j-1));

should refer to mu(i)

Possibly some output should stored indexed at i