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Story So Far

In the first three weeks we have defined a dynamical system $(S,f)$. It is a set of states $S$ together with an iterator function/ rule of evolution $f:S\rightarrow S$. We take an initial state/ seed point $x_o\in S$ and examine the orbit of $x_0$:

$\displaystyle \text{orb}(x_0)=\{x_0,x_1,x_2,x_3,\dots\}$,

where the states $x_1,x_2,\dots$ are produced iteratively by the iterator function:

$x_1=f(x_0)$ and $x_n=f(x_{n-1})$.

We developed the Logistic Model of Population Growth, and this comprises an important example of a dynamical system which we will examine in more depth a little later on.

We studied fixed points. These are states $x_f\in S$ such that if an orbit of a point hits’ $x_f$ then the orbit will remain fixed at $x_f$. Thus fixed points are points with the property that

$f(x_f)=x_f$.

So the fixed points of a function $f:S\rightarrow S$ are points such that the output of the function equals the input.  Note that when we graph functions, the $x$-axis comprises the inputs and the $y$-axis the outputs, and if we are looking for fixed points/ points such that output equals input we need to look for points such that $y=x$. This means that if we graph $latex$y=f(x)\$ then the fixed points of $f(x)$ are the points on the graph of $f(x)$ that intersect $y=x$.

Similarly periodic points are states/ points $x\in S$ such that if an orbit of a point hits’ $x$ then the orbit will keep returning to $x$ after, say $N$ iterations of $f$; that is $f^N(x)=x$:

$x_0,x_1,x_2,\dots,x,f(x),\dots,f^{N-1}(x),x,f(x),\dots,f^{N-1}(x),x,f(x),\dots.$

We also noted that a period-2 point would also be period-6 for example:

$\{\alpha,\beta,\alpha,\beta,\alpha,\beta,\alpha,\beta,\alpha,\beta,\dots\}$

Here $\alpha$ is period-6 but the lowest period is two. We call this the prime period of $\alpha$. Finally we proved some little theorems about periodicity.

Finding periodic points, say period-2 points means finding points $x\in S$ such that if we apply the iterator function twice, then we get back to $x$:

$f(f(x))=f^2(x)=x$.

Solving this equation is not necessarily that easy but we proved that if $f:S\rightarrow S$, then the fixed-point factor-theorem applies: $f(x)-x$ divides into $f^2(x)-x$ and this helps immensely.

Also we expect that Theorems 1 & 2 hold and we proved these.

Exercises

For future weeks I will try and organise these exercises a little better…

Test

The test will take place on February 20. Everything up to but not including section 3.4 in the typeset notes is examinable: we should have this covered this week or next.