Discrete Dynamical System Problem, How to display a value of x that satisfies conditions when inputs to the equation are vectors?
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Basically my original program would input a value of c and spit back a vector x. Most of the time x would diverge to infinity, but there's values of c for which x is bounded. My program runs through these values of c and displays the corresponding values of x (I guess x is a vector?).
Original Program:
n = 100;
c=.2511;
x(1)=0;
for i=1:n;
x(i+1)= (x(i)^2)+c
end
My book states this:
For some values of c, xn stays bounded forever, such as when c = -.5. Find the values of c which make xn stay bounded by looking at several thousand values of c. There are many ways to approach this, try to figure out a good way to do it! (Hint: Narrow your search to only values of c between -3 and 2.) You can assume that xn will become unbounded once abs(xn) gets larger than 2 after 20 iterations, and that xn will stay bounded otherwise. Find the left and right endpoints of the bounded region out to 2 decimal places
My attempt at a solution
c=linspace(-3,2,1000);
x(1)=0;
n=30;
for cindex = 1:1000
for i=1:n;
x(i+1)= (x(i)^2)+c(cindex);
if x(i)>2 && n>20;
disp(x(i))
end
end
end
Please Help me, I know i'm doing something wrong around the if x(i)>2 part
Antworten (1)
Shashank Prasanna
am 7 Mär. 2013
To me it seems like it is bounded for c between (-2,0.2683) Here is the change I made to only print c for bounded solutions:
c=linspace(-3,2,1000);
x(1)=0;
n=30;
for cindex = 1:1000
for i=1:n;
x(i+1)= (x(i)^2)+c(cindex);
end
if x(21)<2;
disp(c(cindex))
end
end
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