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.