MA 471-671

ASSIGNMENT 1 SOLUTIONS

Assignment: Page 26: Nos. 1,3,5,6,11,12,13

• 1. Let F(x)=x^2. Compute the first five points on the oribit of 1/2.
  F^1(1/2)=F(1/2)=1/4
  F^2(1/2)=F(1/4)=1/16
  F^3(1/2)=F(1/16)=1/256
  F^4(1/2)=F(1/256)=1/65536
  F^5(1/2)=F(1/65536)=1/4294967296
  

• 3. Let F(x)=x^2-2. Computer F^2(x) and F^3(x).
  The second iterate of F is
  F^2(x) = F(F(x)) = F(x^2-2) = [x^2-2]^2 - 2 = x^4 - 4x^2 + 2
  The third iterate of F is
  F^3(x) = F(F^2(x)) = F(x^4-4x^2+2) =[x^4-4x^2+2]^2 - 2=x^8 - 8x^6 + 20x^4 - 16x^2 + 2

• 5. Let F(x)=x^2. Compute F^2(x), F^3(x) and F^4(x). What is the formula for F^n(x)?
  F^2(x) = F(F(x)) = F(x^2) = [x^2] ^2 = x^4
  F^3(x) = F(F^2(x)) = F(x^4) = [x^4] ^2 = x^8
  F^4(x) = F(F^3(x)) = F(x^8) = [x^8] ^2 = x^16
  In general,
  F^n(x) = [x]^(2^n)
  

• 6. Let A(x) = |x| . Compute A^2(x) and A^3(x).
  
  • Recall, by definition:
    |x|={ x if x >= 0
    or, {-x if x <= 0
    
  Therefore, since A(x) >= 0 for all x
  A^2(x) = A(A(x)) = |A(x)| = A(x)
  Similarly,
  A^3(x) = A(A^2(x)) = A(A(x)) = |A(x)| = A(x)
  In fact,
  A^n(x) = A(x) for n>=1.
  

  
  

• 7. Find all real fixed points.
  a) F(x)=3x+2
  3x+2=x -> x=-1
  

  b) F(x)=X^2-2
  x^2-2=x -> x^2-x-2=0 -> x=2,-1
  

  c) F(x)=x^2+1
  x^2+1=x -> x^2-x+1=0 -> no real fixed points

  d) F(x)=x^3-3x
  x^3-3x=x -> x^3-4x=0 -> x(x^2-4)=0 -> x=0,+2,-2
  

  e) F(x)=|x|
  |x|=x -> |x|-x=0 -> all x>=0
  

  f)F(x)=x^5
  x^5=x -> x^5-x=0 -> x(x^4-1)=0 -> x=0,+1,-1
  

• 11. For each of the following seeds, discuss the behavior of the resulting orbit under D.
  To recall the definition and graph of the doubling function click here.
  a) x=0.3
  D(0.3)=0.6 , D(0.6)=0.2 , D(0.2)=0.4 , D(0.4)=0.8 and D(0.8)=0.6
  {ie. 0.3 -> 0.6 -> 0.2 -> 0.4 -> 0.8 -> 0.6}
  Therefore, the orbit of 0.3 is eventually periodic with preperiod 1.
  b) x=0.7
  0.7 -> 0.4
  Therefore, it follows by part a) that 0.7 is eventually periodic with preperiod 1.
  c) x=1/8
  1/8 -> 2/8 -> 4/8 -> 0 which is fixed by D
  Therefore, 1/8 is eventually fixed with preperiod 3.
  d) x=1/16
  1/16 -> 2/16 -> 4/16 -> 8/16 -> 0
  Therefore, 1/16 is eventually fixed with preperiod 4.
  e) x=1/7
  1/7 -> 2/7 -> 4/7-> 1/7
  Therefore, 1/7 is periodic with period 3 (ie. lies on a three cycle)
  f) x=1/14
  1/14 -> 2/14 =1/7
  Therefore, by part e), 1/14 is eventually periodic(period 3) with preperiod 1.
  g) x=1/11
  1/11 -> 2/11 -> 4/11-> 8/11 -> 5/11 -> 10/11-> 9/11 -> 7/11 -> 3/11-> 6/11 -> 1/11
  Therefore, 1/11 is periodic with period 10.
  h) x=3/22
  3/22 -> 6/22 = 3/11
  Therefore, by part g), 3/22 is eventually periodic(period 10) with preperiod 1.
  
  
• 12. Give an explicit formula for D^2(x),D^3(x) and D^n(x).
  Recall:
  D(x)=
  { 2x for 0 <= x < 1/2
  { 2x-1 for 1/2 <= x < 1
  

  D^2(x)=
  { 4x for 0 <= x < 1/4
  { 4x-1 for 1/4 <= x < 1/2
  { 4x-2 for 1/2 <= x < 3/4
  { 4x-3 for 3/4 <= x < 1
  

  D^3(x)=
  { 8x for 0 <= x < 1/8
  { 8x-1 for 1/8 <= x < 1/4
  { 8x-2 for 1/4 <= x < 3/8
  { 8x-3 for 3/8 <= x < 1/2
  { 8x-4 for 1/2 <= x < 5/8
  { 8x-5 for 5/8 <= x < 3/4
  { 8x-6 for 3/4 <= x < 7/8
  { 8x-7 for 7/8 <= x < 1
  

  D^n(x)=
  { 2^n*x for 0 <= x < 1/(2^n)
  { 2^n*x-1 for 1/(2^n <= x < 2/(2^n)
  { 2^n*x-2 for 2/(2^n) <= x < 3/(2^n)
  { .
  { .
  { .
  { 2^n*x-(2^n -1) for (2^n -1)/2^n <= x < 1
  or
  2^n*x -K for K/2^n <= x < (K+1)/2^n
  

• 13. Graphs of D^2(x),D^3(x) and D^n(x).
  

  The graph of D^2 consists of 4 straight lines, each with slope 4. The graph of D^3 consists of 8 straight lines, each with slope 8. The graph of D^n consists of 2^n straight lines, each with slope 2^n.