# Iterated Functions

A function and its first two iterates

This picture is the plot of $g\left(x\right)=3.7x(1-x)$, which is in blue, $g\left(g(x)\right)$, which is in pink and which we create by plugging in $3.7x(1-x)$ in for x. The yellow line is $g\left(g(g(x))\right)$, were we plug in $g\left(g(x)\right)$ for x.

# A More Mathematical Explanation

The iterates of a function is the set of functions $\{f(x),f(f(x)),f(f(f(x))),f(f(f(f(x)))),....\}$, w [...]

The iterates of a function is the set of functions $\{f(x),f(f(x)),f(f(f(x))),f(f(f(f(x)))),....\}$, which we denote by $\{f(x),f^2(x),f^3(x),...f^n(x)\}$.

Now, let's take $f(x)=x^2$. Then we see that

$f^2\left(x\right)=f(f(x))=(x^2)^2=x^4$,

$f^3\left(x\right)=f(f(f(x)))=f(f^2(x))=(x^4)^2=x^8$,

and $f^4\left(x\right)=f(f(f(f(x))))=f(f^3(x))=(x^8)^2=x^{16}$.

We can keep doing this, and write down a general form that $f^n(x)=x^{2^n}$

This set of functions $\{x^2,x^4,x^8,...,x^{2^n}\}$ is the set of iterates of $f(x)=x^2$.

We can also look a the iterates of a single point. To do this, we're going to go back to our function $g(x)=3.7x(1-x)$

Picking an initial value of $x=0.2$, we can look at $g(0.2)=0.8937$, the first iterate of 0.2. We can then look at $g^2(0.2)=g(0.8937)=0.35155$. This is the second iterate of 0.2.

We can keep going on this process, and then plot each of the values to see what the iterates of $x=0.2$ look like. The first 20 iterates are plotted below.

It doesn't appear as though these numbers have any pattern. The iterates are jumping around pretty much at random. So, we say that the system defined by this function is chaotic

Let's pick a similar function to look at some different properties of iterated functions.

Take $k\left(x\right)=3.6x(1-x)$. Now, if we want to find the fixed points of this equation, we solve $k\left(x\right)=x$.

Solving gives us $x=0$ or $x=0.722222$. If we plot the iterates of either of these, we end up with a straight line.

To find the periodic points of period two, we solve the equation $k^2\left(x\right)=x$ which is $12.96 (1 - x) a (1 - 3.6 (1 - x) x)=x$.

This equation has four solutions, $x=0$ and $x=0.722222$ from before and $x=0.408149$ and $x=0.869628$. We see that when we solve for points of period two, we end up getting our fixed points back as well.

The other two points form a cycle of period two. If we start out with $x=0.408149$ and plot its iterates, we can see that they bounce between 0.408149 and 0.869628.

To find the periodic points of period 4, we solve $k^4\left(x\right)=x$. This gives us eight values, including the two fixed points $x=0$ and $x=0.722222$, the points of the two cycle, $x=0.408149$ and $x=0.869628$, and four other points, $x=0.813803$, $x=0.892548$, $x=0.345263$, and $x=0.545499$. These points make up a cycle of period 4, which we can see below as the iterates of 0.813803.

Now, all of these points, including the fixed points, are periodic points. They can be either repelling, attracting, or neutral. We can see that for k(x), 0 is a repelling fixed point, because when we start off very close to 0, the iterates move away.

We can also have attracting periodic points, where iterates of other points will move towards it. Consider $h\left(x\right)=3.4x(1-x)$. This function has a cycle of period two with the points $x=0.451963$ and $x=0.842154$. We can see that the iterates of the point 0.3 eventually go to this cycle.

As you can imagine, a periodic point that is neutral does neither of these things.