MA 471-671

Solutions to Homework Assignment #9

1 Prove that Q_c has a periodic point of prime period 2 at each root of the equation z^2+z+c+1=0.

Notice first that Q^2_c(z)=(z^2+c)^2 + c = z^4+2z^2c+c^2+c. Hence, we need to solve Q^2_c(z)=z, that is, find the root of the polynomial

z^4+2cz^2-z+c^2+c=0.

Since a fixed point for Q_c are also roots of this polynomial, we can factorize via long division and obtain

z^4+2cz^2-z+c^2+c=(z^2-z+c)(z^2+z+c+1)=0.

Hence, Q_c must have periodic points of prime period 2 at each root of the polynomial z^2+z+c+1=0.

2 Show that Q_c has an attracting 2-cycle inside the circle of radius 1/4 centered at -1.

>From the last exercise, we see that the 2-cycle is given by the roots

z+=1/2(-1 +sqrt[1-4(c+1)]) and z-=1/2(-1-sqrt[1-4(c+1)]).

Moreover

| DQ^2_c(z±) | = | DQ_c(z+) |*| DQ_c(z-) | = 2|z+|*2|z-| = 4|z+*z-|

since z+ is the conjugate of z-, we get

| DQ^2_c(z±) | = 4 | z+|^2 = 4|1+c|.

Then, the set of c-values for which Q_c has an attracting 2-cycle is given by the inequality |1+c| < 1/4.

3 Prove that the Mandelbrot set is symmetric about the real axis.

We need to exhibit a homeomorphism H that conjugates Q_c with Q_c, where c denotes the conjugate value of c. Then, the orbit of the critical point under Q_c must have the same fate as the orbit under Q_c. Define H as the conjugation map (so H(z)=z in our freaky notation). Clearly, H is a homeomorphism (is injective, surjective and continuous. Since H is its own inverse, clearly has a continuous inverse). Moreover, the conjugation equation holds as

H°Q_c(z)=z^2+c, and Q_c°H(z)=(z)^2 +c.

The Logistic Functions. The following exercises deal with the logistic family F_a(z)=az(1-z) where both a and z are complex numbers.

4 Prove that, for a =/= 0, |F_a(z)| > |z| provided |z| > 1/|a| + 1. Use this to give an analog of the escape criterion for the logistic family.

Assume |z| > 1/|a| + 1. Then |z|-1 > 1/|a| and by the triangle inequality, |z-1| >|z|-1 > 1/|a|. Multiplying both sides by |z| (provided z=/=0), we obtain

|z|*|z-1| > |z|/|a|, that is |az|*|1-z| > |z|.

Hence, |F_a(z)| > |z| if |z| > 1/|a| + 1, a=/=0. Notice that 1/|a| + 1 approaches 1 as |a|-> infinity. Hence our criteria works for

|z| > max { 1, 1/|a| + 1},

i.e. at least outside the unit circle.

5 Show that if |a| >2+sqrt[2], the orbit of the critical point of F_a escapes to infinity.

The critical point is obtained as the solution of

DF_a(z)=0 <==> a-2az=0, <==> z=1/2

for all a-values. Since the c.pt. lies inside the unit disk, we cannot apply the coterie given in ex #4.. So we must require that at least, F_z(1/2) lie outside the unit disk. That is,

|F_a(1/2)| > 1/|a| + 1 implies |F^2_a(1/2)| > |F_a(1/2)|

and so the orbit of c.pt. will escape to infinity. Thus

|F_a(1/2)| =|a|/2=1/|a| + 1 <==> |a^2|-4|a|-4=0.

Computing the roots of this quadratic polynomial, we obtain

a+=2+2sqrt[2] and a-=2-2sqrt[2]< 0,

so we discard the last solution. Then we must have |a| > 2+2sqrt[2] to ensure that |F_a(1/2)| > 1/|a| + 1, so it implies

|F^2_a(1/2)| > |F_a(1/2)|.

Hence the orbit of 1/2 must escape under F_a.

6 Show that, if a=/=0, the logistic function f_a is conjugates to the complex quadratic function Q_c(z)=z^2+c, where c=a/2-a^2/4. Let c=V(a) be this correspondence between a and c. Why does this result fail if a=0?

Remember that x^2+c is conjugated to ax(1-x) via H(x)=ux+v, with u,v real numbers. Then we can use the same idea: we will conjugate Q_c with F_a via H(z)=uz+v, with u,v complex numbers and with V(a)=a/2-a^2/4=c (exercise #8 explains why V is defined like this). Using the conjugate equation, we obtain

H°Q_c(z) = u(z^2+V(a))+v = uz^2 + uV(a) + v, and
F_a°H(z) = a(uz+v)(1-uz-v) = -auz^2 + auz(2v-1)+v(v-1).

Comparing terms, we obtain the next equations

quadratic terms	-au^2=u	equation (1)
linear terms	-au(2v-1) = 0	equation (2)
constant terms	-v(v-1)=v+u(a/2+a-^2/4)	equation (3)

From (1), u=0 or u=-1/a. But if u=0, H(z) reduces to a constant function!, hence we must only consider u=-1/a.

From (2), we obtain v=1/2 (now, equation (3) determines two more values for v, compute them and define the maps H(z) associated to those values. Are those maps also conjugacies?).

Hence H(z)=-z/a + 1/2 and is clearly a homeomorphism. Checking the conjugacy equation, we get

H°Q_c(z) = -1/a (z^2 +a/2-a^2/4)+1/2 = -z^2/a +a/4, and
F_a°H(z) = a(-z/a+1/2)(a+z?a-1/2) = -z^2/a + a/4.

Now, if a=0, the correspondence V(a) send zero into zero. But while Q_0 reduces to z^2, F_0 reduces to the constant zero map, hence with no dynamics to conjugate with z^2.

7 Let L_1 be the set of a-values for which F_a has an attracting fixed point. Compute L_1 and sketch its image in the complex plane.

F_a(z)=z <==> az-az^2=z <==> z(az+1-a) = 0. Hence we have two fixed points:

z_1=0 and z_2=(a-1)/a.

To determine when we have an attracting fixed point, we need to compute | DF_a(z) | = |a|*|1-2z| for z_1 and z_2.

For z_1, we obtain |DF_a(0)| = |a|. Hence, the origin is attracting <==> |a|<1. Hence L_1 is the unit disk centered at the origin.

For z_2, we obtain |DF_a(z_2)| = |2-a|. Hence, the second fixed point is attracting for any a-value that lies inside the disk centered at 2 and with radius 1. This disk is L_1 for the second fixed point.

8 Show that the image of L_1 under the correspondence V in exercise 6 is the main cardioid in the Mandelbrot set.

Recall V(a) = a/2 a^2/4. Hence, if we parametrize the unit circle by exp(it), with 0 <= t <= 2pi, we have

V(exp(it)) = exp(it)/2 - exp(2it)/4

which is the equation of the main cardioid as given in page 252. Hence V takes L_1(0) into the main cardioid of the Mandelbrot set.

Now, parametrize the boundary of L_1 for the second fixed point by exp(it)-2, again 0 <= t <= 2pi. Then

V(exp(it)-2) = exp(it)/2 +1 - exp(2it)/ 4 - exp(it) -1 = - exp(it)/2 -exp(2it)/4

That is, V takes L_1(z_2) into the main cardioid of the Mandelbrot set, but runs through the curve with the clockwise direction.

9 Identify geometrically the set L_2 of a-values that are mapped to the period 2 bulb in the Mandelbrot set by the correspondence V. How many regions are in this set?

Since V is a quadratic map, its inverse should be a square root map, hence we must expect to regions in the a-plane sent to the c-plane by V. To compute the inverse of V (denoted by U+ and U-) we need to solve

c = a/2 - a^2/4

for a in terms of c. Then we obtain a quadratic a^2 -2a + 4c = 0 and two roots that are the inverse maps for V

U+(c) = 1 + sqrt[1-4c] and U-(c) = 1- sqrt[1-4c].

Hence, the preimages under V of the 2-bulb in the Mandelbrot set are given by two circles in the a-plane, with centers at U+(-1)=1+sqrt[5] and U-(-1)=1-sqrt[5].

Higher Degree Polynomials. The following eight exercises deal with polynomials of the form P_(d,c)(z)=z^d+c.

11 Prove that P'_(d,c)(z)=dz^(d-1). Conclude that, for each integer d>1, P_(d,c) has a single critical point at 0.

Clearly DP_(d,c)= d*z^(d-1) when differentiating with respect to z. Now

DP_(d,c)(z) = 0 <==> dz^(d-1) = 0 <==> d=0 or z=0.

Hence, for d>1, DP_(d,c) is a polynomial of degree d-1, and z=0 is the unique critical point of P_(d,c).

12 Prove the following escape criterion for P_(d,c). Show that if |z|>=|c| and |z^(d-1)|>2, then |P^n_(d,c)(z)| -> infinity as n-> infinity.

This is analogous to case d=2.

Assume |z| >= c and |z|^(d-1) > 2. We want to show

|P_(d,c)(z)| > |z|.

To do so, notice that since |z|^(d-1) > 2, then |z|^d > 2|z| (provided z=/=0). Then

|z|^d - |z| > |z| ==> |z|^d - |c| > = |z|^d - |z| > |z|.

By the triangle inequality,

|z| < |z|^d - |c| = < |z^d+c| = | P_(d,c)(z) | .

Hence, if |z| >= |c| and |z|^(d-1) > 2, then P^n_(d,c) diverges as n --> infinity.

13 Show that if |c|>2^(1/(d-1)), the orbit of the critical point of P_(d,c) escapes to infinity.

By exercise #11, c.pt. is z=0. Then P_(d,c)(0) = c. In order to use the escape criterion given in ex. #12, we need to apply it to the second image of the critical point. That is, we require

|c|^(d-1) > 2,

which is the hypothesis given after taking the (d-1)-root. Hence

|P^2_(d,c)(0)| > |P_(d,c)(0)| = |c|,

and so the orbit of zero escapes.

--Covey_of_Quail_508_000--