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
a)F(x) = 2x
F is an odd function that fixes {0}, and the rest of the orbits go to +/- inf.
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c) F(x) = -2x+1
F fixes {1/3} since
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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.
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a) F(x) = 1/2 x - 2
Thus, F fixes {-4}, and it so happens that F^n(x)-> -4 as n->inf for all x.
b) A(x) = |x|.
d) F(x) = -x^5.
Therefore, 0 is a fixed point and the other four fixed
points are complex. Now,
and so 0 is superattracting. Note also that 1 and
-1 constitute a repelling 2-cycle for F.
Since 0 is the fixed point, 1, -1 is the
2-cycle. It is repelling because
This suggests that |F^n(x)| ->inf as n->inf whenever
|x|>1. All other orbits escape to infinity.
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.
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
First, let's find a fixed point for F:
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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.
First, let's compute the fixed points of F:
x = 0, 1 ,-1.
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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,
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Page 50
a) F(x) = x^2-x/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)
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,
since 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.
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.
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.
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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.
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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.
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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.
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.