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

  1. Consider the function f(x)=x2. Sketch this function and plot the first few points of its orbit {x0,x1,x2,x3,}, i.e. plot the points (x0,x1=f(x0)), (x1,x2=f(x1)), etc..., for the starting values x0=0,12, and 2. Describe what is happening to the orbit of f(x) for each of these starting values.

    Solution:

    x0=0:

    The orbit of x0=0 under f(x)=x2 is {0,0,0,0,}. The point x0=0 is a fixed point of f(x).

    x0=12:

    The orbit of x0=12 under f(x)=x2 is {12,14,116,1256,}. The iterates of x0=12 are getting smaller in magnitude on each iteration and approaching zero.

    x0=2:

    The orbit of x0=2 under f(x)=x2 is {2,4,16,256,}. The iterates of x0=2 are getting larger in magnitude on each iteration and approaching infinity.

  2. Let f(x)=x2+3x+1. Find all of the fixed points of f(x).

    Solution: To find the fixed points we need to solve f(x¯)=x¯. x¯2+3x¯+1=x¯x¯2+2x¯+1=0(x¯+1)2=0 This has one solution, x¯=1, so f(x) has just one fixed point at x¯=1.

  3. Consider the family of functions defined by fc(x)=cx where c is a constant and c0. Determine all of the fixed points of fc(x).

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

    Solution: To find the fixed points of fc(x) we solve fc(x¯)=x¯, treating c as a constant. cx¯=x¯(c1)x¯=0 For most values of c, this has one solution, x¯=0, however we can see that when c=1, then all x¯R are solutions. Thus, the fixed points of fc(x) are x¯=0, c1 and x¯R, c=1. Notice that when c=1, f1(x)=x, for which all xR are clearly fixed points.

    1. Consider the function f(x)=x212. 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 x¯1 and x¯2 such that x¯1<x¯2.

    2. Use a graphical method (i.e. cobweb diagram) to help determine the behaviour of various orbits starting near both x¯1 and 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 fc(x)=x2+c where c is a constant. For what values of c do fixed points of fc(x) exist? Some sketches of the graphs of fc(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 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 x¯2, so this is likely to be a repelling fixed point.

    3. To find the fixed points of fc(x) we need to solve fc(x¯)=x¯. x¯2+c=x¯x¯2x¯+c=0 Using the quadratic formula, this has solutions x¯=12±1214c. This has (real) solutions only when the argument of the square root is greater than (or equal to) zero, i.e. 14c0c14. Thus, fc(x) has fixed points when c14. Notice that when c=14, fc(x) has one fixed point (only one solution to the quadratic formula) and when c<14, fc(x) has two fixed points (two solutions to the quadratic formula).

  4. Let f(x)=x3. Find all fixed points and periodic points of period two of f(x).

    Solution: First, let’s find any fixed points by solving f(x¯)=x¯. x¯3=x¯x¯+x¯3=0x¯(1+x¯2)=0 This has just one solution, x¯=0, so f(x) has one fixed point at x¯=0.

    Now, let’s find any periodic points of period two. We need to solve f[2](x¯)=x¯. f[2](x)=f(f(x))=(x3)3=x9 So we need to solve x¯9=x¯. Rearranging, this gives x¯(x¯81)=0, which has three solutions x¯=0,1, and 1. Since x¯=0 is a fixed point, the remaining two points must form a two cycle. Thus, x¯=1 and 1 are the points of period two of f(x).

  5. CHALLENGE Let f(x)=1x2. Find all fixed points and periodic points of period two of f(x).

    Solution: First, let’s find the fixed points. 1x¯2=x¯x¯2+x¯1=0 Using the quadratic formula, this gives two solutions, x¯=12±52, 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), f[2](x)=1(1x2)2=11+2x2x4=2x2x4 and then solve f[2](x¯)=x¯. 2x¯2x¯4=x¯x¯42x¯2+x¯=0x¯(x¯32x¯+1)=0 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](x¯)=x¯. This means that (x¯2+x¯1) must be a factor. Thus we have x¯(x¯2+x¯1)(x¯1)=0 The two new solutions (which are not fixed points of the original function) are x¯=0 and 1. Thus we must have the two cycle {0,1}.

  6. 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 x¯+cos(x¯)=x¯cos(x¯)=0. Considering the unit circle (or a graph of cos(x)), we see that this is true when x¯ is an odd multiple of 90 degrees, i.e. x¯=90°,270°,450°,90°,

    We can write this as x¯ is a fixed point of f(x) when x¯=(2k+1)90° 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).