MA 471-671

ASSIGNMENT 7 SOLUTIONS

Assignment: Page 202: Nos. 1,13,14,15,18

• 1. Without using the computer, predict the structure of the attractor generated by the iterated function system with contraction factor beta and fixed point p_i.

  a) beta=1/3; p_0=(0,0), p_1=(1,0), p_2=(0,1)
  Note: To be consistent with the notation in the book these points can be thought of as column vectors.
  The given data corresponds to the iterated function system
  A_0(x,y)=1/3(x.y)
  A_1(x,y)=1/3(x-1.y) + (1,0)
  A_2(x,y)=1/3(x.y-1) + (0,1)
  which gives rise to a type of Sierpinski triangle which, quite surprisingly, has fractal dimension D=(Log3/Log2). The attractor of this iterated function system (IFS) is
  {(x,y) in KxK | y<= 1-x } where K is the Cantor middle-third set.

  b) beta=1/2; p_0=(0,0), p_1=(1,0)
  The given data corresponds to the iterated function system
  A_0(x,y)=1/2(x.y)
  A_1(x,y)=1/2(x-1.y) + (1,0)
  Note that the unit interval is fixed with respect to this IFS and hence , is its attractor with fractal dimension D=(Log2/Log2)=1.

  c) beta=1/3; p_0=(0,0), p_1=(1,0), p_2=(0,1), p_3=(1,1)
  The given data corresponds to the iterated function system
  A_0(x,y)=1/3(x.y)
  A_1(x,y)=1/3(x-1.y) + (1,0)
  A_2(x,y)=1/3(x.y-1) + (0,1)
  A_3(x,y)=1/3(x-1.y-1) + (1,1)
  The attractor of this IFS is KxK where K is the Cantor middle-third set, and its fractal dimension is D=(Log4/Log3), which is exactly twice that of K.

• 3. Consider the set C obtained from the interval [0,1] by first removing the middle third of the interval and then removing the middle fifths of the two remaining intervals. Now iterate this process, first removing middle thirds, then removing middle fifths. The set C is what remains when this process is repeated infinitely. Is C a fractal? If so, what is its fractal dimension?

  Yes, C is a fractal with fractal dimension D=(Log4)/Log(15/2).

• 13. Can a fractal that is totally disconnected (topological dimension 0) have a fractal dimension larger than 1?

  Yes. Consider the Cantor set that we obtain as a the cartisian product of two Cantor middle-fifths set. Then the fractal dimension would be (by problem 1 and 2)
  

  D = log 4/log5/2 = 2 log 2/log5/2 = 2 * 0.75647... > 1.

• 14. Show that the rational numbers form a subset of the real line that has topological dimension 0. What is the topological dimension of the set of irrationals?

  A set has topological dimension 0 if every point has arbitrarily small neighborhoods whose boundaries do not intersect set. The rationals are such a set since the boundary of a disk with irrational radius fails to intersect the rationals (the sum of a rational and an irrational is irrational).
  It is also true that the irrationals have topological dimension 0. To see this, suppose we had the following lemmas:

  Lemma Let A be a subset of X. Then,

  t-dimA<=t-dimX.

  Lemma For A subset of R^n, t-dimA=n <=> A contains a non-empty open subset of R^n.

  Then by the first lemma, the irrationals have t-dim<=1. But by the second lemma the irrationals can not have t-dim=1 since they contain no open subset. Therefore, the t-dim of the irrationals is zero.
  

• 15. Compute exactly the area of the Koch snowflake.

  Koch's snowflake may be enclosed within a rectangle having length 1 and width 2 sqrt(3)/3, and hence its area is no larger than 2 sqrt(3)/3 square units.
  Let A(0) be the area of the equilateral triangle with sides of length 1. Since the height of this triangle is sqrt(3)/2 units, we have that A(0)=sqrt(3)/4 square units. Now at the k-th step of the Koch process, 3*4^(k-1) equilateral triangles each with sqrt(3)/(4*3^(2k)) are added to the snowflake. The combined area of these triangles is

  3 * 4^(k-1) * sqrt(3)/(4*3^(2k)) = 3 * 4^k/4 * sqrt(3)/(4*9^k) = 3sqrt(3)/16 (4/9)^k .

  Thus, the area of all these triangles is the sum form k=0 to inf of the previos equality, i.e.,

  area = 3sqrt(3)/16 * 4/5 = 3 sqrt(3)/20,

  and so, the area of the Koch snowflake is

  A = sqrt(3)/4 + 3 sqrt(3)/20 = 2 sqrt(3)/5

  square units.
  

• 18. Rework exercise 17, this time replacing each number by a black dot if it is congruent to 1 mod 3 and a white dot otherwise. (That is, points that yield a remainder of 1 upon division by 3 are colored black.) Now descrive the resulting figure. How does it compare with the figure generated in the previous exercise?

  The resulting figure should resemble the Sirpinski triangle as does the figure generated in exercise 17 with slight differences.

Assignment: Page 219: Nos. 4, 7B,C,F, 8A,C, 9, 10, 11.

• 4. What is the formula for the quotient of two complex numbers given in polar representation?

  For the following answer let the symbol @ represent an angle Theta.
  First, let z1=r1e^(@1i) and z2=r2e^(@2i). Then
  

  z1/z2 = r1e^(@1i)/r2e^(@2i) =(r1/r2)e^(@1-@2)i
  assuming r2 doesn't equal zero (ie z2 doesn't equal zero). In other words, the ratio of two complex numbers is a third complex number whose modulus is the ratio of the two moduli and whose argument is the diference of the two arguments.

• 7.For which of the following complex functions does the complex derivative exist?
  

  b) F(x+iy)= x^2 + iy^2.
  This function is not complex differentiable. Let z(t)= (x_0 +t) +iy_0 and w(t)=x_0+i(y_0+t). Then, (all the limits in this section are taken when the real variable t->0)

  lim (F(z(t))-F(z_0))/(z(t)-z_0) =

  lim ((x_0+t)^2+i(y_0)^2-((x_0)^2+i(y_0)^2))/((x_0+t)+iy_0-(x_0+iy_0)) =

  lim ((x_0)^2+2tx_0+t^2+i(y_0)^2-(x_0)^2-i(y_0)^2)/(x_0+t+iy_0-x_0-iy_0) =

  lim t(2x_0+t)/t = 2x_0
  

  but

  lim (F(w(t))-F(z_0))/(w(t)-z_0) =

  lim ((x_0)^2+i(y_0+t)^2-((x_0)^2+i(y_0)^2))/(x_0+i(y_0+t)-(x_0+iy_0)) =

  lim ((x_0)^2+i((y_0)^2+2ty_0+t^2)-(x_0)^2-i(y_0)^2)/(x_0+iy_0+it-x_0-iy_0) =

  lim it(2y_0+t)/it = 2y_0.
  

  c) F(z) = abs(z).

  Since this function is not differentiable in R, then it cannot be differentiable in C.
  

  f)F(z)=2z(i-z).

  Observe that 2z(i-z)=2iz-2z^2, and so F'(z)=21-4z. Indeed, (the limits here are taken as z -> z_0)

  lim (F(z)-F(z_0))/(z-z_0) = lim (2z(i-z)-2z_0(i-z_0))/(z-z_0) =

  lim (2zi-2z_0i-2z^2+2(z_0)^2)/(z-z_0) = lim (2i(z-z_0)-2(z^2-(z_0)^2)/(z-z_0) =

  lim ((2i(z-z_0)-2(z+z_0)(z-z_0))/(z-z_0) = lim (2i-2(z+z_0)) = 2i-4z_0.
  

• 8. Find all fixed points for each of the following complex functions and determine whether they are attracting, repelling, or neutral.
  

  a) Q_2(z) = z^2 + 2.

  The fixed points are

  z^2+2=z => z^2-z+2=0 => z = (1+-sqrt(-7))/7.

  Since Q'_2(z) = 2z,

  abs(Q'_2((1+-sqrt(-7))/2)) = abs (1+-sqrt(-7)) = sqrt(8) = 2sqrt(2)

  which is greater than one, and so the fixed point is repelling.
  

  c) F(z)=iz^2.

  iz^2=z => iz^2-z=0 => (iz-1)z=0 => z=0 or z=1/i=-i.

  Since F'(z)=2iz, we have F'(-i)=2i(-i)=-2i^2=2. In other words, this fixed point is repelling. The origin, on the other hand, is superattracting since F'(0)=0.
  

• 9.Show that z_0=-1+i lies on a cycle of period 2 for Q_i(z)=z^2+i. Is this cycle attracting, repelling, or neutral?

  Since Q_i(i-1) = (i-1)^2+i = i^2+1-2i+i = -i and Q_i(-i)=(-i)^2+i=i-1,these points constitute a 2-cycle for Q_i. Note that Q'_i(z)=2z and

  ((Q_i)^2)'(i-1) = (Q_i)'(i-1)(Q_i)'(-i) = 2(i-1) 2(-i) = 4(1+i).

  Now, since abs(4(1+i))=4 abs(1+i) = 4 sqrt(2)>1, this 2-cycle is repelling.
  

• 10. Show that zo=e^(2PIi/3) lies on a cycle of period 2 for Qo(z)=z^2. Is this cycle attracting, replling, or neutral?

  Qo(e^(2PIi/3))=e^(2PIi/3)^2=e^(4PIi/3) and
  Qo(e^(4PIi/3))=e^(8PIi/3)=e^(2PI)*e^(2PIi/3)=e^(2PIi/3).
  This two cycle is repelling since Qo'(z)=2z and so
  (Qo^2)'(e^(2PIi/3)) = Qo'(e^(2PIi/3))*Qo'(e^(4PIi/3)) = 2e^(2PIi/3)*2e^(4PIi/3) = 4e^(6PIi/3)
  and |4e^(6PIi/3)| = 4 > 1.

• 11. Show that zo=e^(2PIi/7) lies on a cycle of period 3 for Qo(z)=z^2. Is this cycle attracting, replling, or neutral?

  Qo(e^(2PIi/7))=e^(2PIi/7)^2=e^(4PIi/7) and
  Qo(e^(4PIi/7))=e^(8PIi/7) and
  Qo(e^(8PIi/7))=e^(16PIi/7)=e^(2PI)*e^(2PIi/7)=e^(2PIi/7) This two cycle is repelling since Qo'(z)=2z and so
  (Qo^3)'(e^(2PIi/7)) = Qo'(e^(2PIi/7))*Qo'(e^(4PIi/7))*Qo'(e^(8PIi/7)) = 2e^(2PIi/7)*2e^(4PIi/7)*2e^(8PIi/7) = 8e^(14PIi/7)
  and |8e^(14PIi/7)| = 8 > 1.