Grade 11/12 Math Circles
Dynamical Systems and Fractals Part 1 - Solutions

Exercise Solutions

Exercise 1

Find all of the fixed points of the function $f(x)=x^2-2$.

Exercise 1 Solution

Set $f(\overline{x})=\overline{x}$ and solve. $$\begin{array}{rl}\to {\overline{x}}^2-2& =\overline{x}\\ {\overline{x}}^2-\overline{x}-2& =0\\ (\overline{x}+1)(\overline{x}-2)& =0\end{array}$$ This has two solutions: ${\overline{x}}_1=-1$ and ${\overline{x}}_2=2$, both of which must be fixed points of $f(x)=x^2-2$.

Let’s check!
$f(-1)=(-1{)}^2-2=1-2=-1$
$f(2)=2^2-2=4-2=2$

Exercise 2

Given that $f(x)=\frac{1}{x}$, find the periodic points of period two of $f(x)$.

Hint: You may want to find the fixed points of $f(x)$ first.

Exercise 2 Solution

First, find the fixed points of $f(x)$ by solving $f(\overline{x})=\overline{x}$. $$\begin{array}{rl}\frac{1}{\overline{x}}& =\overline{x}\\ {\overline{x}}^2=1\end{array}$$ This has solutions ${\overline{x}}_1=-1$ and ${\overline{x}}_2=1$. so these must be our fixed points.

Now we want to solve for the fixed points of $f^{[2]}(x)$. $$\begin{array}{rl}{f}^{[2]}(x)& =f(f(x))\\ & =\frac{1}{1/x}\\ & =x\end{array}$$ Setting $f^{[2]}(\overline{x})=\overline{x}$ gives $\overline{x}=\overline{x}$. This is true for all values of $\overline{x}$. Does this mean that all values of $\overline{x}$ are periodic points of period two of $f(x)$? Almost!

Since the points ${\overline{x}}_1=-1$ and ${\overline{x}}_2=1$ are fixed points of $f(x)$, they cannot also be periodic points of period two. We also need to be careful and consider the domain of $f(x)$, which excludes the point $x=0$ (since $f(x)=\frac{1}{x}$ is undefined when $x=0$). This means that $x=0$ cannot be a periodic point of $f(x)$. What we are left with is that all $x\in \mathbb{R}$ except for $x=-1,1$, and $0$ are periodic points of period two of $f(x)$. We could write the set of periodic points of period two of $f(x)$ as $\{x\in \mathbb{R}|x\ne -1,1,0\}$.

Problem Set Solutions

1. Consider the function $f(x)=x^2$. Sketch this function and plot the first few points of its orbit $\{x_0,x_1,x_2,x_3,\dots \}$, i.e. plot the points $(x_0,x_1=f(x_0))$, $(x_1,x_2=f(x_1))$, etc..., for the starting values $x_0=0,\frac{1}{2}$, and $2$. Describe what is happening to the orbit of $f(x)$ for each of these starting values.

   Solution:

   $x_0=0$:

   The orbit of $x_0=0$ under $f(x)=x^2$ is $\{0,0,0,0,\dots \}$. The point $x_0=0$ is a fixed point of $f(x)$.

   

   $x_0=\frac{1}{2}$:

   The orbit of $x_0=\frac{1}{2}$ under $f(x)=x^2$ is $\{\frac{1}{2},\frac{1}{4},\frac{1}{16},\frac{1}{256},\dots \}$. The iterates of $x_0=\frac{1}{2}$ are getting smaller in magnitude on each iteration and approaching zero.

   

   $x_0=2$:

   The orbit of $x_0=2$ under $f(x)=x^2$ is $\{2,4,16,256,\dots \}$. The iterates of $x_0=2$ are getting larger in magnitude on each iteration and approaching infinity.

   

2. Let $f(x)=x^2+3x+1$. Find all of the fixed points of $f(x)$.

   Solution: To find the fixed points we need to solve $f(\overline{x})=\overline{x}$. $$\begin{array}{rl}{\overline{x}}^2+3\overline{x}+1& =\overline{x}\\ {\overline{x}}^2+2\overline{x}+1& =0\\ (\overline{x}+1{)}^2& =0\end{array}$$ This has one solution, $\overline{x}=-1$, so $f(x)$ has just one fixed point at $\overline{x}=-1$.

3. Consider the family of functions defined by $f_c(x)=cx$ where $c$ is a constant and $c\ne 0$. Determine all of the fixed points of $f_c(x)$.

   Hint: You may end up with different fixed points depending on the value of $c$.

   Solution: To find the fixed points of $f_c(x)$ we solve $f_c(\overline{x})=\overline{x}$, treating $c$ as a constant. $$\begin{array}{rl}c\overline{x}& =\overline{x}\\ (c-1)\overline{x}& =0\end{array}$$ For most values of $c$, this has one solution, $\overline{x}=0$, however we can see that when $c=1$, then all $\overline{x}\in \mathbb{R}$ are solutions. Thus, the fixed points of $f_c(x)$ are $\overline{x}=0$, $c\ne 1$ and $\overline{x}\in \mathbb{R}$, $c=1$. Notice that when $c=1$, $f_1(x)=x$, for which all $x\in \mathbb{R}$ are clearly fixed points.

4. 1. Consider the function $f(x)=x^2-\frac{1}{2}$. Sketch $f(x)$ and $y=x$ on the same set of axes and show graphically that $f(x)$ has two fixed points. Label these fixed points on your sketch as ${\overline{x}}_1$ and ${\overline{x}}_2$ such that ${\overline{x}}_1<{\overline{x}}_2$.

   2. Use a graphical method (i.e. cobweb diagram) to help determine the behaviour of various orbits starting near both ${\overline{x}}_1$ and ${\overline{x}}_2$. Use your diagram to make an educated guess as to the nature (attractive, repelling, or neither) of each fixed point.

   3. Now consider the family of functions $f_c(x)=x^2+c$ where $c$ is a constant. For what values of $c$ do fixed points of $f_c(x)$ exist? Some sketches of the graphs of $f_c(x)$ for various values of $c$ may help, but they are not necessary.

   Solution:

   1. From the graph, we can see that $f(x)$ intersects the line $y=x$ twice, and thus has two fixed points.

      

   2. Cobweb diagram:

      

      From our cobweb diagram we see that the iterates of $f(x)$ are attracted towards the fixed point ${\overline{x}}_1$, so we can guess that this is an attractive fixed point. On the other hand, the iterates of $f(x)$ move away from the fixed point ${\overline{x}}_2$, so this is likely to be a repelling fixed point.

   3. To find the fixed points of $f_c(x)$ we need to solve $f_c(\overline{x})=\overline{x}$. $$\begin{array}{rl}{\overline{x}}^2+c& =\overline{x}\\ {\overline{x}}^2-\overline{x}+c& =0\end{array}$$ Using the quadratic formula, this has solutions $\overline{x}=\frac{1}{2}\pm \frac{1}{2}\sqrt{1-4c}$. This has (real) solutions only when the argument of the square root is greater than (or equal to) zero, i.e. $$\begin{array}{rl}1-4c& \ge 0\\ c& \le \frac{1}{4}.\end{array}$$ Thus, $f_c(x)$ has fixed points when $c\le \frac{1}{4}$. Notice that when $c=\frac{1}{4}$, $f_c(x)$ has one fixed point (only one solution to the quadratic formula) and when $c<\frac{1}{4}$, $f_c(x)$ has two fixed points (two solutions to the quadratic formula).

5. Let $f(x)=-x^3$. Find all fixed points and periodic points of period two of $f(x)$.

   Solution: First, let’s find any fixed points by solving $f(\overline{x})=\overline{x}$. $$\begin{array}{rl}-{\overline{x}}^3& =\overline{x}\\ \overline{x}+{\overline{x}}^3& =0\\ \overline{x}(1+{\overline{x}}^2)& =0\end{array}$$ This has just one solution, $\overline{x}=0$, so $f(x)$ has one fixed point at $\overline{x}=0$.

   Now, let’s find any periodic points of period two. We need to solve $f^{[2]}(\overline{x})=\overline{x}$. $$\begin{array}{rl}{f}^{[2]}(x)& =f(f(x))\\ & =-{(-x^3)}^3\\ & =x^9\end{array}$$ So we need to solve ${\overline{x}}^9=\overline{x}$. Rearranging, this gives $\overline{x}({\overline{x}}^8-1)=0$, which has three solutions $\overline{x}=0,-1,$ and $1$. Since $\overline{x}=0$ is a fixed point, the remaining two points must form a two cycle. Thus, $\overline{x}=-1$ and $1$ are the points of period two of $f(x)$.

6. CHALLENGE Let $f(x)=1-x^2$. Find all fixed points and periodic points of period two of $f(x)$.

   Solution: First, let’s find the fixed points. $$\begin{array}{rl}1-{\overline{x}}^2& =\overline{x}\\ {\overline{x}}^2+\overline{x}-1& =0\end{array}$$ Using the quadratic formula, this gives two solutions, $\overline{x}=\frac{-1}{2}\pm \frac{\sqrt{5}}{2}$, which are the two fixed points of $f(x)$.

   Next, we want to solve for any periodic points of period two. First we find $f^{[2]}(x)$, $$\begin{array}{rl}{f}^{[2]}(x)& =1-{(1-x^2)}^2\\ & =1-1+2x^2-x^4\\ & =2x^2-x^4\end{array}$$ and then solve $f^{[2]}(\overline{x})=\overline{x}$. $$\begin{array}{rl}2{\overline{x}}^2-{\overline{x}}^4& =\overline{x}\\ {\overline{x}}^4-2{\overline{x}}^2+\overline{x}& =0\\ \overline{x}({\overline{x}}^3-2\overline{x}+1)& =0\end{array}$$ Factoring the cubic part of this expression could be difficult, but luckily we know that the fixed points of $f(x)$ are also solutions to $f^{[2]}(\overline{x})=\overline{x}$. This means that $({\overline{x}}^2+\overline{x}-1)$ must be a factor. Thus we have $$\begin{array}{rl}\overline{x}({\overline{x}}^2+\overline{x}-1)(\overline{x}-1)& =0\end{array}$$ The two new solutions (which are not fixed points of the original function) are $\overline{x}=0$ and $1$. Thus we must have the two cycle $\{0,1\}$.

7. CHALLENGE Consider the function $f(x)=x+\cos (x)$. Show that $f(x)$ has an infinite number of fixed points.

   Solution: To find the fixed points of $f(x)$ we must solve $$\begin{array}{rl}\overline{x}+\cos \left(\overline{x}\right)& =\overline{x}\\ \cos \left(\overline{x}\right)& =0.\end{array}$$ Considering the unit circle (or a graph of $\cos (x)$), we see that this is true when $\overline{x}$ is an odd multiple of $90$ degrees, i.e. $\overline{x}=90\mathrm{°},270\mathrm{°},450\mathrm{°},-90\mathrm{°},\dots$

   We can write this as $\overline{x}$ is a fixed point of $f(x)$ when $\overline{x}=(2k+1)\cdot 90\mathrm{°}$ for any integer $k$. Since the set of all integers is an infinite set, this results in an infinite number of fixed points for $f(x)$.