An archived instance of discourse for discussion in undergraduate Complex Dynamics.

# Finding an orbit

mark

Find an orbit of period 5 for the chaotic logistic function $g(x)=4x(1-x)$.

Note: There's a couple of ways to go on this. One way would be to use a computer to solve $g^5(x)=x$. Another way must be doable without a calculator??

SomeCallMeTim

I utilized Mathematica in order to numerically solve for orbits of period 5 for the chaotic logistic function $g(x)=4x(1-x)$.
I first defined the function using the following code:

g[c_][x_] = 4 x - 4 x^2 + c;
g[n_, c_][x_] := Nest[g[c], x, n];

Having defined the function, I next solved for values of x where after 5 iterations, we have returned to the initial value.

per5Points = NSolve[g[5, 0][x] == x, x]

Finally, I extracted and tested one of the nonzero output solutions, taking it through two cycles to be quite certain of the periodic behavior.

p5 = x /. per5Points[[2]]
NestList[g[0], p5, 10]

Which produced the cycle $0.00903565, 0.035816, 0.138133, 0.476209, 0.997736, 0.00903565, 0.035816, 0.138133, 0.476209, 0.997736, 0.00903565$ , which does return to the initial value every 5th iteration.

The full list of solutions generated is:

{x -> 0.}, {x -> 0.00903565}, {x -> 0.010235}, {x -> 0.035816}, {x ->
0.0405211}, {x -> 0.0793732}, {x -> 0.0896183}, {x ->
0.138133}, {x -> 0.155517}, {x -> 0.209972}, {x -> 0.235524}, {x ->
0.292363}, {x -> 0.329169}, {x -> 0.377134}, {x -> 0.411438}, {x ->
0.445821}, {x -> 0.493}, {x -> 0.511038}, {x -> 0.578354}, {x ->
0.615607}, {x -> 0.662237}, {x -> 0.715041}, {x -> 0.716322}, {x ->
0.772761}, {x -> 0.826939}, {x -> 0.928658}, {x -> 0.941514}, {x ->
1.04274}, {x -> 1.04331}, {x -> 1.09213}, {x -> 1.10036}, {x ->
1.12518}

mark

I think this solution by @SomeCallMeTim is awesome! I wonder if anyone else has a different approach? One that could be done by hand??

RedCrayon

Using @SomeCallMeTim's values, we could draw a graph that looks like this

and imagine the cobweb plot superimposed onto it, but it's not obvious that a starting value of $0.00903565$ produces a period $5$ orbit.

Rick

To find a period 5 orbit of g, we can use the binary representation and choose values for c such that

$0_2.\overline{c_1c_2c_3c_4c_5}=0_2\overline{.00001}=\sum_{k=1}^{\infty}\frac{1}{32^k}=\sum_{k=0}^{\infty}\frac{1}{32^{k+1}}=\frac{1}{32}*\sum_{k=0}^{\infty}\frac{1}{32^{k}}=\frac{1}{32} \frac{1}{1-\frac{1}{32}}=\frac{31}{32}\frac{32}{31}=\frac{1}{31}$.

So, the orbit containing $\frac{1}{31}$ is a period 5 orbit of g.

mark

@Rick

Be sure to check your work. For example, if you think that $1/31$ is an orbit of period 5 for $g$, then compute 5 iterates and see where you are.

In reality, I think you've missed a crucial step.

I've added Example 7.9 to our notes on real iteration that might help.