How to introduce anonymous function in this code?

Hi! So here is a code which I made in class to make a bifurcation diagram but it takes a lot of time to execute. As I've seen how anonymous function speed up the computation considerably, I was wondering if there is a way to implement the said handler in this code:

f=inline('x*(1+r*(1-x))');
for r=0.5:0.01:3
  x=0.4;
  for i=1:100
      x=f(r,x);
  end
  for i=1:100
      x=f(r,x);
      plot(r,x,'*')
      hold on
  end
end

My goal is to achieve the same graph as attached.

Answer 1

[John D'Errico]

f = @(r,x) x.*(1+r.*(1-x));

To be honest, I'm not sure what your goal is here. It looks like you are trying to solve an equation using a fixed point iteration, but not done terribly efficiently.

A quick check on what SHOULD be happening. You are effectively trying to solve the equation

x*(1+r*(1-x)) - x = 0

Which reduces to

-r*x^2 + r*x = 0

So the value of r is irrelevant to the solution, as long as r is not zero. This problem has two unique roots, at x=0 and x=1.

Solve will tell us that.

syms x r
x = solve(x.*(1+r.*(1-x)) == x)
x =
 0
 1

So this problem has two solutions, regardless of r. But what happens when we apply fixed point iteration to it, as has been done here?

I'll let the iterative loop go a LONG way out, as this will be quite fast now.

f = @(r,x) x.*(1+r.*(1-x));
r = 0.5:0.01:3;
x = 0.4;
for i=1:10000
  x = f(r,x);
end

plot(r,x)
grid on

So it converges to a nice, stable result for r less than 2. Above that point, things go crazy. This is a common behavior of fixed point iteration methods. You need to watch the derivatives of f. But that is a complete lecture on fixed point iteration.

[Shuk Tombstone]

Actually, if you run my code you will see that the graph you should get is a line that kind of "branch off" as it goes along the horizontal axis.