Logistic map bifurcation

Question

Ok I am trying to do this on matlab, but I need to understand how to find the bifurcation values for logistic map by hand first.

So here is the logistic map: $$ x_{i+1} = f(x_i) \qquad \text{where} \qquad f(x) = rx(1 - x) $$

So far, I have found the $x$ values by letting $$ rx(1 - x) = x $$ and then $$ x = 1 - \frac{1}{r}. $$ And, this is period one.

Now what? Please help me understand this.

Answer 1

Here's a Maple version of what you want: restart; f := rsolve({s(1) = .3, s(n+1) = as(n)(1-s(n))}, s, makeproc); g := proc (n, i) options operator, arrow; eval(f(n), a = i) end proc; t := time(); L := [seq(seq([i, g(k, (1/320)*i)], i = 800 .. 1280), k = 200 .. 300)]; time()-t; plot(L, x = 800 .. 1280, y = 0 .. 1, style = point, symbol = solidcircle, colour = black, symbolsize = 5)

I tried to buy a home version of Matlab, but no dice, so its Maple, sorry.

Answer 2

I need to understand how to find the bifurcation values for logistic map by hand first.

If you want to understand how to compute periodic points then after period 1 is period 2 ( period doubling ).

Period 2 points can be found by solving equation ( z is on vertical axis) :

$ f(f(z)) = z$

period 3 :

$f(f(f(z))) = z $

and so on.

The next thing is that above image shows only attracting periodic points, not repelling. After doubling are also other cascades.

There are many more things in real iterations, some you can find here

HTH