MA 471-671

Solutions to Homework Assignment #8

• 1. Describe the filled Julia set for F(z)=z^3.

  Like the quadratic map Q_0, the filled Julia set for F(z)=z^3 is the closed unit disk. Observe that

  F(z)=z^3
  F^2(z)=z^9
  .
  .
  F^n(z)=z^{3^n}
  where the latter may be proved by induction. Now recall that z^k = r^k(Cos(ka) + isin(ka) = r^k e^{ika} where r=|z| and tan(a)=im z/re z. Thus we have F^n(z) = z^{3^n} = r^{3^n} e^{i(3^n)a} Which is analogous to the formula for (Q_0)^n given on page 221 of the text. The rest of the analysis mirrors that of Q_0. For instance, it follows that |z|=1 implies that |F(z)|=1, that is, the unit circle is invariant, since |F(z)| = |z^3| = |z|^3 = 1^3 =1.

• 2. Describe the filled Julia set for F(z) = z^d for d > 4.

  This is a straightforward generalization of Exercise 1. In this case,

  F^n(z) = z^{d^n} = r^{d^n} e^{i(d^n)a} and |F^n(z)| = |r^{d^n} e^{i(d^n)a}| = |r^{d^n}| From this, we see that |F^n(z)| tends to 0 when r is between 0 and 1; |F^n(z)| = 1 if r=1; and |F^n(z)| tends to infinity when r is greater than 1. Therefore, the filled Julia set is again the closed unit disk.

• 3. Consider Q_2(z) = z^2 + 2. Let C = {z : |z|=2}. Give an accurate sketch of (Q_2)^{-1}(C) and (Q_2)^{-2}(C).

  To see a picture of the sets C, (Q_2)^{-1}(C), and (Q_2)^{-2}(C), click here

• 4. Use the techniques of Section 16.2 to conjugate F(z) = z^3 to a ploynomial P(z) via H(z) = z + (1/z). What is P(z)?

  Given the fact that H(F(z)) = P(H(z)), which implies

  z^3 + (1/z^3) = P(z + (1/z)) we want to find out what P(z) is using what was done in Section 16.2. Letting w = z + (1/z) and solving for z gives z_+ = (w + SQRT(w^2 - 4))/2 and z_-=(w-SQRT(w^2-4))/2 But as seen in Section 16.2, we know that z_- = (1/z_+) (and vise versa). This then implies that P(w) = ((w + SQRT(w^2 - 4))/2)^3 + ((w - SQRT(w^2 - 4))/2)^3
  =(1/8)(2w^3 + 6w(w^2 - 4)) After simplifying the above expression we get P(w) = w^3 - 3w.

• 6. The Period-Doubling Bifurcation

  a) Find all complex c-values for whcih Qc(z)=z^2+c has a fixed point zo with Qc'(zo)=-1.
  Solving the system of equations
  {z^2+c=z
  {2z=-1
  gives z=-1/2 and c=-3/4

  b)Show that the points q+(c)=-1/2 + 1/2(Sqrt[-3-4c]) and q-(c)=-1/2 - 1/2(Sqrt(-3-4c) lie on a 2-cycle unless c=-3/4.
  Straightforward calculations yields
  Qc(-1/2 +/- 1/2(Sqrt[-3-4c])= [-1/2 +/- 1/2(Sqrt[-3-4c])]^2 +c
  =1/4 +1/4(-3-4c) -/+ 1/2(Sqrt[-3-4c]) +c
  =1/4 +3/4-c -/+ 1/2(Sqrt[-3-4c]) +c
  =-1/2 -/+ 1/2(Sqrt[-3-4c]
  Note that q+/-(-3/4)=-1/2 -/+ 1/2(Sqrt[-3-4(-3/4)])=-1/2 which agrees with the results of exercise 6a.

  c) Determine whether this cycle is attracting, repelling, or neutral in the two real cases -5/4< c <-3/4 and c >-3/4.
  Using the Chain Rule Along a Cycle given on p. 47 of the text, plus the fact that Qc`(z)=2z, we have
  

  (Qc^2)`(q+)=Qc`(q+)*Qc`(q-)
  =(-1+Sqrt[-3-4c])(-1-Sqrt[-3-4c])
  =1-(-3-4c)
  =4+4c
  =4(1+c)
  and similarly for (Qc^2)`(q-). But
  c >-3/4 => c+1 >1/4 => 4(c+1) >1 which says the 2-cycle is repelling in this range. When -5/4< c <-3/4 we have
  -5/4< c <-3/4 => -1/4< c+1 <1/4 => -1< 4(c+1) <1
  and so the cycle is attracting.

  d) Sketch the locations of the fixed points and q+/1 for Qc is the three real cases -5/4< c <-3/4 , c=-3/4 and c >-3/4. This is the complex period-doubling bifurcation.
  

• 7. Consider the complex function G_lamda(z)=lamda*(z-z^3). Show that the points
  p+/-(lamda)=+/-Sqrt[(lamda +1 )/lamda]
  lie on a cycle of period 2 unless lamda=0 or -1. Discuss the bifurcation that occurs at lamda=-1 by sketching the relative positions of q+/- and the fixed points of G_lamda as lamda increases through -1, assuming only real values.

  It is convenient to write G_lamda(z)=lamda*(z-z^3)=lamda*z(1-z^2).
  G_lamda(+Sqrt[(lamda +1 )/lamda])=
  lamda*Sqrt[(lamda +1 )/lamda]*(1-(Sqrt[(lamda +1 )/lamda])^2)
  =lamda*Sqrt[(lamda +1 )/lamda]*((lamda-(lamda+1))/lamda)
  =lamda*Sqrt[(lamda +1 )/lamda]*(-1/lamda)
  =-Sqrt[(lamda +1 )/lamda]
  and vice versa. Of course, when lamda=0 q+/-(lamda) is undefined and when lamda=-1, q+/-(lamda)=0, but 0 is fixed.
  Computing the fixed points of G_lamda, we have
  lamda*z(1-z^2)=z
  => lamda*(1-z^2)=1 for z not equal to 0
  => 1-z^2=1/lamda
  => 1-1/lamda=z^2
  => (lamda-1)/lamda=z^2
  => +/-Sqrt[(lamda +1 )/lamda]=z
  which we`ll denote as p+ and p-. In the bifurcation diagram for G_lamda, when lamda<1 there are two fixed points and a 2-cycle lying between them. At lamda=-1 the two fixed points persist and have grown in magnitude but the the 2-cycle ends. When lamda>-1 the fixed points persist (growing in magnitude) and the 2-cycle is gone.

  For more information about homework assignments, click here

• 9. The Logistic Functions. The following exercises deal with the family of logistic functions F_L(x) = Lz(1 - z) where both L and z are complex numbers.
  
  • a. Prove that (F_L)'(z) = L(1 - 2z) using complex differentiation.

    By definition,

    (F_L)'(z) = lim(z -> z_0) (F_L(z) - F_L(z_0))/(z - z_0)
    = lim(z -> z_0) (Lz(1 - z) - Lz_0(1 - z_0))/(z - z_0)
    = L lim(z -> z_0) ((z - z^2) - (z_0 - (z_0)^2))/(z - z_0)
    = L lim(z -> z_0) ((z - z_0 ) - (z^2- (z_0)^2))/(z - z_0)
    = L lim(z -> z_0) 1 - (z + z_0) = L(1 - 2z_0) when we let z approach z_0.

  • b. Find all fixed points for F_L.

    In order to find fixed points for F_L(z), we need to solve the equation Lz(1 - z) = z for z. Obviously, 0 is a solution to this equation. Assuming z does not equal zero, we can perform the following calculations:

    Lz(1 - z) = z =>
    L(1 - z) = 1 =>
    1 - z = 1/L =>
    1 - 1/L = z =>
    (L - 1)/L = z
    Therefore, the fixed points of F_L are 0 and (L-1)/L.

  • c. Find all parameter values L for which F_L has an attracting fixed point.

    F_L has an attracting fixed point at the origin when |(F_L)'(0)| is less than 1, and at (L-1)/L when |(F_L)'((L-1)/L)| is less than 1. Given the fact that (F_L)'(z) = L(1 - 2z), we can easily see that (F_L)'(0) = L, which implies that the origin is attracting when L is within the unit disk centered at the origin.
    We also have that

    (F_L)'((L - 1)/L) = L(1 - 2(L - 1)/L)
    = L((L - 2(L - 1))/L)
    =2 - L
    Therefore, the other fixed point at (L-1)/L is attracting when (2-L) is located within the unit disk centered at the origin.