** Assignment: ** Page 34: Nos. 1 A,C,F; 4 A,B,D; 5. Page 50: Nos. 1 A,b,f,j; 2 B,C; 4 A,C,E.

** Page 34**

**1.**Use graphical analysis to describe the fate of all orbits for each of the following functions. Use different colors for orbits which behave differently.**a)**F(x) = 2x

F is an odd function that fixes {0}, and the rest of the orbits go to +/- inf.

To see the picture for this solution click here.c) F(x) = -2x+1

F fixes {1/3} since-2x+1 = x => 1 = 3x => x = 1/3. Also, | F^n(x)|-> inf for all x except the fixed point.

To see the picture for this solution, click here.**f)**F(x) = x - x^2

This function has a single fixed point at the origin. If 0 < x < 1, then F^n(x) is a positive decreasing sequence of points converging to 0. On the other hand, if x<0, then F^n(x)->-inf as n ->inf. Finallly, note that x=1 is eventually fixed after one iteration.

Expertiments indicate that for 0 < x <1, the orbit of x converges to 0 rather slowly. Similarly, for negative x very close to 0, F^n(x) slowly moves away from the origin. The explanation of this behavior is that the fixed point is neutral.

To see the picture for this solution, click here.**4.**Perform a complete orbit analysis for each of the following functions.**a)**F(x) = 1/2 x - 2

First, let's find a fixed point for F:(1/2)x - 2 = x => -2 = (1/2)x => -4 = x. Thus, F fixes {-4}, and it so happens that F^n(x)-> -4 as n->inf for all x.

To see the picture for this solution, click here.**b)**A(x) = |x|.

Since |x| = x for nonnegative x, we have that F fixes {x : x >=0}. Also, the negative real numbers are eventually fixed since each becomes positive after just one iteration.**d)**F(x) = -x^5.

First, let's compute the fixed points of F:-x^5 = x => 0 = x + x^5 => 0 = x(1 + x^4). Therefore, 0 is a fixed point and the other four fixed points are complex. Now,

F'(x) = -5x^4 => F'(0) = 0, and so 0 is superattracting. Note also that 1 and -1 constitute a repelling 2-cycle for F.

F^2(x) = F(F(x)) = F(-x^5) = x^25 = x => x (x^24 -1) = 0 =>

x = 0, 1 ,-1.Since 0 is the fixed point, 1, -1 is the 2-cycle. It is repelling because

(F^2)'(x) = 25 x^24 => |(F^2)'(+-1)| > 1. This suggests that |F^n(x)| ->inf as n->inf whenever |x|>1. All other orbits escape to infinity.

To see the picture for this solution, click here.**5.**Let F(x) = |xÐ2|. Use graphical analysis to display a variety of orbits of F. Use red to display cycles of period 2, blue for eventually fixed orbits, and green for orbits which are eventually periodic.If x >=2, then |xÐ2| = x => x - 2 = x, which has no solution. On the other hand, if x<2, we have that |x-2| = x => 2 - x = x => 2 = 2x => x=1. Therefore, {1} is de only fixed point of F.

Observe that if x is an odd integer, then so is x-2. Now suppose x is an odd positive integer greater than 2. Then x-2 is an odd positive integer smaller than x, and repeated substraction eventually produces a value of 1 which is fixed by F. In other words, all odd positive integers are eventually fixed. For odd negative integers, we just need to apply the same argument to F(x), since F(x)>=0.

Suppose, now, that x is an even positive integer greater than or equal to 2. Then x-2 is an even positive integer smaller than x. in this case, repeated substractions eventually vanish, but F(0) = |0-2| = 2. And since F(2)=0, we see that the orbit is eventually caught in a cycle of period 2. Similar arguments hold for even negative x. In fact,Periodics points with period 2 of F = {x : 0<=x<=2}, and moreover, every real number is eventually periodic with period 2. To see this, first suppose that x>2. Then there exists an m such that 0

2, and so F(x) is eventually periodic via a similar argument. here.

To see the picture for this solution, click

** Page 50**

**1.**For each of the following functions, find all fixed points and classify them as attracting, repelling, or neutral.**a)**F(x) = x^2-x/2x^2-x/2 = x => x^2 - 3x/2 = 0 => x (x-3/2) = 0 => x=0 or x=3/2. Therefore, the fixed points of F are {0, 3/2}.

F'(x) = 2x-1/2.

F'(0) = -1/2 => 0 is attracting and oscillating.

F'(3/2) = 5/2>1 => 3/2 is repelling.**b)**F(x) = x(1-x)3x(1-x) = x => 2x - 3x^2 = 0 => x(2-3x) = 0 => x=0 or x=2/3. Therefore, the fixed points of F are {0,2/3}.

F'(x) = 3 - 6x.

F'(0) = 3 >1 => 0 is repelling.

F'(2/3) = 3 - 6(2/3) = -1 => 2/3 is neutral.**f)**S(x) = (pi/2) sin x.

The fixed points of S are {0, +/-pi/2}.

S'(x) = (pi/2) cos x.

S'(0) = pi/2>1 => 0 is repelling.

S'(+/-pi/2) = (pi/2) cos(+/-pi/2) = 0 => +/-pi/2 are superattracting.**j)**T(x) = 2x if x <=1/2

2-2x if x>1/2.

Setting each piece of this two part function equal to x yields that the fixed points of T are {0, 2/3}. Also,T'(x) = 2 if x<1/2

-2 if x>1/2since T is pecewise linear, but the derivative of T is not defined at x=2. Therefore, T'(0) = 2 and so 0 is repelling. And since T'(2/3)=-2, 2/3 is also repelling. In fact, no periodic points for T can be attracting.

**2.**For each of the following functions, zero lies on a periodic orbit. Classify this orbit as attracting, repelling, or neutral.**b)**C(x) = (pi/2) cos x.

Since C(0) = pi/2 and C(pi/2) = 0, {0,pi/2} is a 2-cycle.

C'(x) = - (pi/2) sin x.

(C^2)'(0) = C'(0) C'(pi/2) = 0 (- pi/2) = 0, and so this orbit is superattracting.**c)**F(x) = - (1/2)x^3 - (3/2) x^2 + 1.

Since F(0) = 1, F(1) = -1, and F(-1) = 0, we have that {0, +/-1} is a 3-cycle.

F'(x) =-(3/2) x^2 - 3x

(F^3)'(0) = F'(0) F'(-1) = 0 (-9/2) (3/2) = 0, and so this period 3 orbit is superattracting.

**4.**Each of the following functions has a neutral fixed point. Find this fixed point and, using graphical analysis with an accurate graph, determine if it is weakly attracting, weakly repelling, or neither.**a)**F(x) = x + x^2.

Either by inspection of the graph of F, or by solving the equation F(x) = x for x, we see that {0} is a fixed point for F. Since F'(0)=1 (that is, the graph of F is tangent to the diagonal at x=0) this fixed point must be neutral. Moreover, graphical analysis suggest that the fixed point is weakly attracting on the left and weakly repelling on the right. This is simple consequence of the fact that the graph is concave up at the origin (since F''(0)=2>0).

We remark that -1 is eventually fixed after one iteration, and that for all x in the closed interval [-1,0], F^n(x) ->inf as n->inf since 0 is weakly repelling on the right.

To see the picture for this solution, click here.**b)**E(x) = exp (x-1).

The fixed point is at x=1. Locally, in a small neighborhood of 1, this problem is reminiscent of part (a) where the neutral fixed was found to be weakly attracting on the left and weakly repelling on the right. Observe that the graph of E is concave up at the fixed point since E''(1) = 1>0.

To see the picture for this solution, click here.**c)**T(x) = tan x.

We know that {0} is a fixed point for T since tan 0 = 0, and this fixed point is indeed neutral since T'(0) = sec^2 0 = 1. But it is weakly repelling. From the graph, it appears that tan x > x for 0 < x < pi/2 and that tan x < x for -pi/2 < x < 0, and so the origin must be weakly repelling. In fact, it can be shown that all of T's fixed points are repelling.

To see the picture for this solution, click here.**5.**Suppose that F has a neutral fixed point at p with F'(p)=1. Suppose also that F''(p)>0. What can you say about p: is p weakly attracting, weakly repelling , or neither? Use graphical analysis and the concavity of the graph of F near p to support your answer.Since F''(p)>0, the graph of F is concave up at the neutral point p. Graphical analysis clearly shows that p is weakly attracting on the left and weakly repelling on the right.

**6.**Repeat exercise 5, but this time assume that F''(p)<0.When F''(p)<0, the graph of F is concave down at the neutral point p. In this case, p is weakly repelling on the left and weakly attracting on the right.