for loop label/arguement

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Sara Ismail-Sutton
Sara Ismail-Sutton 2021-4-18
评论： Walter Roberson 2021-4-22
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

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回答（1 个）

Walter Roberson
Walter Roberson 2021-4-18
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.
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Sara Ismail-Sutton
Sara Ismail-Sutton 2021-4-21
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
Walter Roberson 2021-4-22
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

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