Consider the function \(f(x) = x^2\). Sketch this function and plot the first few points of its orbit \(\left\{ x_0,x_1,x_2,x_3,\ldots\right\}\), i.e. plot the points \(\left(x_0,x_1 = f(x_0)\right)\), \(\left(x_1,x_2 = f(x_1)\right)\), 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 \(\left\{0,0,0,0,\ldots\right\}\). 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 \(\left\{\frac{1}{2},\frac{1}{4},\frac{1}{16},\frac{1}{256},\ldots\right\}\). 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 \(\left\{2,4,16,256,\ldots\right\}\). The iterates of \(x_0 = 2\) are getting larger in magnitude on each iteration and approaching infinity.
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 \(f(\bar{x}) = \bar{x}\). \[\begin{align*} \bar{x}^2 + 3\bar{x} + 1 &= \bar{x}\\ \bar{x}^2 + 2\bar{x} + 1 &= 0\\ (\bar{x}+1)^2 &= 0\end{align*}\] This has one solution, \(\bar{x} = -1\), so \(f(x)\) has just one fixed point at \(\bar{x} = -1\).
Consider the family of functions defined by \(f_c(x) = cx\) where \(c\) is a constant and \(c\neq 0\). 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 \(f_c(\bar{x}) = \bar{x}\), treating \(c\) as a constant. \[\begin{align*} c\bar{x} &= \bar{x}\\ (c-1)\bar{x} &= 0\end{align*}\] For most values of \(c\), this has one solution, \(\bar{x} = 0\), however we can see that when \(c=1\), then all \(\bar{x}\in\mathbb{R}\) are solutions. Thus, the fixed points of \(f_c(x)\) are \(\bar{x} = 0\), \(c\neq 1\) and \(\bar{x}\in\mathbb{R}\), \(c=1\). Notice that when \(c=1\), \(f_1(x) = x\), for which all \(x\in\mathbb{R}\) are clearly fixed points.
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 \(\bar{x}_1\) and \(\bar{x}_2\) such that \(\bar{x}_1 < \bar{x}_2\).
Use a graphical method (i.e. cobweb diagram) to help determine the behaviour of various orbits starting near both \(\bar{x}_1\) and \(\bar{x}_2\). Use your diagram to make an educated guess as to the nature (attractive, repelling, or neither) of each fixed point.
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:
From the graph, we can see that \(f(x)\) intersects the line \(y=x\) twice, and thus has two fixed points.
Cobweb diagram:
From our cobweb diagram we see that the iterates of \(f(x)\) are attracted towards the fixed point \(\bar{x}_1\), 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 \(\bar{x}_2\), so this is likely to be a repelling fixed point.
To find the fixed points of \(f_c(x)\) we need to solve \(f_c(\bar{x}) = \bar{x}\). \[\begin{align*} \bar{x}^2 + c &= \bar{x}\\ \bar{x}^2 - \bar{x} + c &= 0\end{align*}\] Using the quadratic formula, this has solutions \(\bar{x} = \frac{1}{2} \pm \frac{1}{2}\sqrt{1 - 4c}\). This has (real) solutions only when the argument of the square root is greater than (or equal to) zero, i.e. \[\begin{align*} 1 - 4c &\geq 0\\ c &\leq \frac{1}{4}.\end{align*}\] Thus, \(f_c(x)\) has fixed points when \(c\leq \frac{1}{4}\). 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).
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 \(f(\bar{x}) = \bar{x}\). \[\begin{align*} -\bar{x}^3 &= \bar{x}\\ \bar{x} + \bar{x}^3 &= 0\\ \bar{x}(1 + \bar{x}^2) & = 0\\\end{align*}\] This has just one solution, \(\bar{x} = 0\), so \(f(x)\) has one fixed point at \(\bar{x} = 0\).
Now, let’s find any periodic points of period two. We need to solve \(f^{[2]}(\bar{x}) = \bar{x}\). \[\begin{align*} f^{[2]}(x) &= f(f(x))\\ &= -\left(-x^3\right)^3\\ &= x^9\end{align*}\] So we need to solve \(\bar{x}^9 = \bar{x}\). Rearranging, this gives \(\bar{x}(\bar{x}^8-1) = 0\), which has three solutions \(\bar{x} = 0, -1,\) and \(1\). Since \(\bar{x}=0\) is a fixed point, the remaining two points must form a two cycle. Thus, \(\bar{x} = -1\) and \(1\) are the points of period two of \(f(x)\).
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. \[\begin{align*} 1 - \bar{x}^2 &= \bar{x}\\ \bar{x}^2 + \bar{x} - 1 &= 0\\\end{align*}\] Using the quadratic formula, this gives two solutions, \(\bar{x} = \frac{-1}{2}\pm\frac{\sqrt{5}}{2}\), which are the two fixed points of \(f(x)\).
Next, we want to solve for any periodic points of period two. First we find \(f^{[2]}(x)\), \[\begin{align*} f^{[2]}(x) &= 1 - \left(1-x^2\right)^2\\ &= 1 - 1 + 2x^2 - x^4\\ &= 2x^2 - x^4\end{align*}\] and then solve \(f^{[2]}(\bar{x}) = \bar{x}\). \[\begin{align*} 2\bar{x}^2 - \bar{x}^4 &= \bar{x}\\ \bar{x}^4 - 2\bar{x}^2 + \bar{x} &= 0\\ \bar{x}\left(\bar{x}^3 - 2\bar{x} + 1\right) &= 0\end{align*}\] 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 \(f^{[2]}(\bar{x}) = \bar{x}\). This means that \((\bar{x}^2 + \bar{x} - 1)\) must be a factor. Thus we have \[\begin{align*} \bar{x}\left(\bar{x}^2 + \bar{x} - 1\right)\left(\bar{x}-1\right) &= 0\end{align*}\] The two new solutions (which are not fixed points of the original function) are \(\bar{x} = 0\) and \(1\). Thus we must have the two cycle \(\left\{0,1\right\}\).
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 \[\begin{align*} \bar{x} + \cos\left(\bar{x}\right) &= \bar{x}\\ \cos\left(\bar{x}\right) &= 0.\end{align*}\] Considering the unit circle (or a graph of \(\cos(x)\)), we see that this is true when \(\bar{x}\) is an odd multiple of \(90\) degrees, i.e. \(\bar{x} = 90\degree, 270\degree, 450\degree, -90\degree,\ldots\)
We can write this as \(\bar{x}\) is a fixed point of \(f(x)\) when \(\bar{x} = (2k+1)\cdot 90\degree\) for any integer \(k\). Since the set of all integers is an infinite set, this results in an infinite number of fixed points for \(f(x)\).