index exceeds matrix dimensions

P=0.4;
T=0.5;
a=2;
for i=1:max_it
     if x(i)>=0 && x(i)<P   %(Error Here)
         x(i+1)=x(i)/P;
     end
    if x(i)>=P && x(i)<=T
        x(i+1)=(x(i)-T)/(P-T);
    end
    if x(i)>=T && x(i)<=1
        x(i+1)=a*((x(i)-T)/1-T)*(1-(x(i)-T)/(1-T));
    end
%     if x(i)>=1-P && x(i)<1
%         x(i+1)=(1-x(i))/P;
%     end    
    G(i) =x(i);
end

Answer 1

[Walter Roberson]

We do not know what you initialized x as.

You probably did not pre-allocate x; you are probably counting on it growing as you go. However notice that you only ever grow x conditionally: you stop growing x if you encounter an element that is negative or greater than 1.

Answer 2

[deepak kumar]

P=0.4;
T=0.5;
a=2;
max_it=100;
x(1)=0.7;
Value=1;
O=zeros(1,max_it);
for i=1:max_it
     if x(i)>=0 && x(i)<P %(Error Here)
         x(i+1)=x(i)/P;
     end
    if x(i)>=P && x(i)<=T
        x(i+1)=(x(i)-T)/(P-T);
    end
    if x(i)>=T && x(i)<=1
        x(i+1)=a*((x(i)-T)/1-T)*(1-(x(i)-T)/(1-T));
    end
%     if x(i)>=1-P && x(i)<1
%         x(i+1)=(1-x(i))/P;
%     end    
    G(i) =x(i)*Value;
end


    case 7
%Sine map
for i=1:max_it
     x(i+1) = sin(pi*x(i));
     G(i)=(x(i))*Value;
 end

[Walter Roberson]

The second x you generate is negative, and as I pointed out before, you stop growing x as soon as you find a negative value.