Math 053-1
Final Exam
Spring 2011
1. For each function f : R → R, find all the periodic points of the function and sketch a
phase portrait of f (x):
(a) f (x) = −x3
(b) f (x) = ex−2 (decimal approximations are sufficient here)
2. Let f : R → R be a continuous function such that f (1) = 2, f (2) = 3, f (3) = 4,
f (4) = 5, and f (5) = 1. Prove that f has a periodic point of prime period 3.
3. Let (X, T ) be a dynamical system. We say that a point p ∈ X is non-wandering (under
T ) if for every open set U containing p, there is a point x ∈ U such that T n (x) ∈ U
for some n > 0. Define N W (T ) to be the set of non-wandering points under T .
(a) Prove that N W (T ) is always a closed set.
(b) Let p be a periodic point for T . Is p non-wandering? Explain.
(c) Suppose p is not periodic, but belongs to the stable set of some attracting periodic
point y. Is p non-wandering? Explain.
(d) Let fr : [0, 1] → [0, 1] be defined by fr (x) = rx(1 − x). Describe N W (fr ) when
0 ≤ r ≤ 3. (There may be different answers depending on the value of r).
(e) Suppose T is an arbitrary chaotic system. Describe N W (T ).
4. Given a continuous function f : [0, 1] → [0, 1], define the double of f to be the function
F : [0, 1] → [0, 1] satisfying the following four properties:
•
•
•
•
F is continuous;
F (2/3) = 0 and F (1) = 1/3;
F is linear on the interval [1/3, 2/3] and also linear on the interval [2/3, 1];
F (x) = 31 f (3x) + 23 if x ∈ [0, 1/3]
(a) Given the following graph of f : [0, 1] → [0, 1], sketch the graph of the double of
f:
1.0
0.8
0.6
0.4
0.2
0.0
0.2
0.4
0.6
0.8
1.0
(b) Suppose F is the double of some unknown continuous function f .
Math 053-1
Final Exam
Spring 2011
i. How many fixed points does F have? Classify the fixed points as attracting,
repelling or neutral (prove your answer).
ii. Suppose p is a periodic point for F . Prove that the prime period of p must
be a power of 2.
iii. Suppose q is a periodic point for f of prime period d. Show q/3 is a periodic
point for F . What is its prime period?
5. For each family of functions given below, sketch a bifurcation diagram. Give all values
of the parameter for which bifurcations occur, and classify each bifurcation.
(a) fr (x) = x2 + rx; r ∈ (−∞, ∞)
(b) fr (x) = x3 + r; r ∈ (−∞, ∞)
6. Let D be the subset of {A, B, C}N consisting of all sequences obeying the following
three rules: first, no two consecutive As are allowed; second, every B must be followed
by an A; third, C cannot be followed by A.
(a) Prove that (D, σ) is chaotic, where σ is the shift map.
(b) Find the topological entropy of (D, σ). A decimal answer here is not sufficient.
(c) Prove that (D, σ) is topologically conjugate to the golden mean shift (the golden
mean shift is the shift on the subset of {0, 1}N consisting of sequences with no
two consecutive zeros).
b→C
b be the rational function f (z) =
7. Let f : C
1
.
z2
(a) Find the fixed point(s) of f and classify them as attracting, repelling, or neutral.
(b) Find the periodic cycle of prime period 2 and classify it as attracting, repelling
or neutral.
(c) Describe the Julia set of f (briefly explain your reasoning).
b →C
b be the rational function g(z) = z 3 − 3z; let I be the interval of real
8. Let g : C
b
numbers [−2, 2], thought of as a subset of C:
(a) Show I is completely invariant under both forward and backward iteration of g.
(b) Show that if |z| > 2, then z ∈ W s (∞).
(c) Explain why g has no attracting periodic point, other than the fixed point at ∞.
(d) “Guess” the topological entropy of ([−2, 2], g). Explain the logic behind your
conjecture. (You don’t necessarily have to guess correctly to receive credit here.)
Math 053-1
Final Exam (Solutions)
Spring 2011
1. (a) 0 is the only fixed point; it is superattracting. There is one periodic cycle of
prime period 2, namely {1, −1 which is repelling. We have W s (0) = (−1, 1) and
W s (∞) = (−∞, −1) ∪ (1, ∞); the phase portrait should show these stable sets,
together with the fact that f interchanges (−∞, 0) and (0, ∞).
(b) This map has two fixed points, namely x1 ≈ .158594 and x2 ≈ 3.14619. x1 is
attracting and W s (x1 ) = (−∞, x2 ). x2 is repelling; W s (∞) = (x2 , ∞). The phase
portrait should show these two fixed points and their stable sets.
2. First, observe that f ([2, 3]) ⊇ [3, 4], so f 2 ([2, 3]) ⊇ [4, 5] and f 3 ([2, 3]) ⊇ [1, 5] ⊇ [2, 3].
Therefore f has a point x ∈ [2, 3] such that f 3 (x) = x. This point may have prime
period 3, but it may be fixed. If it has prime period 3, we are done. If it is fixed, there
may be other fixed points in [2, 3]. Choose a to be the largest of these fixed points
and let b ∈ (a, 3) be such that f (b) < 3 (this can be done by the Intermediate Value
Theorem). Now f ([b, 3]) ⊇ [3, 4], so f 2 ([b, 3]) ⊇ [4, 5] and f 3 ([b, 3]) ⊇ [1, 5] ⊇ [b, 3].
So there is a point in [b, 3], call it y, such that f 3 (y) = y. y cannot be fixed because
y ≥ b > a and a is the largest fixed point of f in [2, 3]. Therefore y has prime period
3 under f .
3. (a) Let W (T ) be the complement of N W (T ); we will show W (T ) is open. Let z ∈
W (T ); by inverting the definition of N W (T ) we see that there is an open set U
containing z such that for all x ∈ U , T n (x) ∈
/ U for all n > 0. Let V be any
open set containing z such that V ⊆ U ; for all points p ∈ V , there is an open set
containing p, namely U , such that for all x ∈ U , T n (x) ∈
/ U for all n > 0. Hence
p ∈ W (T ) by definition, so V ⊆ W (T ). Hence W (T ) is open by definition and
N W (T ) is therefore closed.
(b) Yes; choose x = p in the definition of non-wandering point.
(c) No; here is a heuristic argument why: given such a point p, there is an open set
U containing p but contained in W s (y). Since p 6= y, we can assume U does not
contain y. All points in U eventually leave under U iteration and head towards
y, so p is not non-wandering.
(d) There are two fixed points of fr , namely x = 0 and x = (1 − r)/r. (When r = 1,
there is only one fixed point since these two points coincide and when r = 0, 0
is the only fixed point.) Every other point in [0, 1] belongs to the stable set of
one of these two fixed points, so by part (c) no non-fixed point is non-wandering.
} when r > 0 (and N W (f0 ) = {0} when
Therefore by part (b), N W (fr ) = {0, 1−r
r
r = 0).
(e) If T is chaotic, then periodic points of T are dense in X. Let p ∈ X; given open
set U such that p ∈ U , there is x ∈ U such that x is periodic. Hence x returns
to U and therefore p is non-wandering; since p was arbitrary, all points in X are
non-wandering.
4. (a) Here is the graph of F :
Math 053-1
Final Exam (Solutions)
Spring 2011
1
2
3
1
3
0
1
2
3
3
1
(b) Throughout this solution to part (b), let A = [0, 1/3], B = (1/3, 2/3) and C =
[2/3, 1].
i. Since F (A) ⊆ C and F (C) ⊆ A, F can have no fixed points in A or C.
Restricted to B, F is monotone decreasing so it can have at most one fixed
point in B. Since F (B) ⊇ B, F has at least one fixed point in B. We now
know F has exactly one fixed point in B and therefore exactly one fixed point
in [0, 1].
ii. First, observe F has no periodic points in B other than the fixed point; if
x ∈ B is not fixed, then by the Mean Value Theorem the distance from F n (x)
to the fixed point will increase until x ∈ A ∪ C. Since A ∪ C is invariant under
F , x cannot be periodic. Now, since F (A) ⊆ C and F (C) ⊆ A, no periodic
point in A or C can have odd prime period. Therefore all periodic points
other than the fixed point have even prime period.
iii. We see q/3 ∈ A so by direct calculation, F (q/3) = 31 f (q) + 32 which is in C.
Therefore since F (x) = x − 23 on C, F 2 (q/3) = 13 f (q). Similarly, F 2n (q/3) =
1 n
f (q). If the prime period of q is d, then q/3 will be periodic with prime
3
period 2d.
5. For each family of functions given below, sketch a bifurcation diagram. Give all values
of the parameter for which bifurcations occur, and classify each bifurcation.
(a) This family has two fixed points at x = 0 and x = 1 − r (shown in blue and red
below on the bifurcation diagram). When r ∈ (−1, 1), 0 is attracting and 1 − r
is repelling. When r ∈ (1, 3), 1 − r is attracting and 0 is repelling. When r = 1,
the fixed points coincide since 1 − r = 0; this fixed point is weakly attracting.
At r = 1 we therefore see a transcritical bifurcation. As r decreases past −1 and
increases past 3, we see period-doubling cascades in each direction, the first of
which is indicated below by the green period 2 cycles.
Math 053-1
Final Exam (Solutions)
Spring 2011
3
2
1
-4
-3
-2
-1
1
2
3
4
-1
-2
-3
(b) The bifurcations should occur at a value of r such that the fixed point x satisfies
|fr0 (x)| = 1. Since |fr0 (x)| = 3x2 , we see that x = √13 and therefore since x is
fixed, we can solve for r to obtain r = ± 3√2 3 . Here is the bifurcation diagram
showing the fixed points of fr , graphed against r. The bifurcations at r = ± 3√2 3
are saddle-node bifurcations. The fixed point shown in gold is attracting and the
two outside fixed points, shown in red and blue, are repelling.
1.5
1.0
0.5
-2
-1
1
2
-0.5
-1.0
-1.5
6. (a) First, we show sensitive dependence on initial conditions. Let δ = 1/2 and given
> 0, choose k such that 2−k < . Given x ∈ D, let y ∈ D be such that xj = yj
for all j ∈ [0, k] but xk+2 6= yk+2 . This can be done since whatever symbol xk
is, although xk+1 might be determined, there are always at least two options for
what xk+2 could be. We have d(x, y) < since x and y agree to position k,
but d(σ k+2 (x), σ k+2 (y)) > δ since these shifted sequences disagree on their initial
index. Therefore (D, σ) is sensitive dependent on initial conditions by definition.
Next, we show density of periodic points. Given x ∈ D and > 0, choose k such
that 2−k < . Let y ∈ D be such that xj = yj for all j ∈ [0, k] and yk+1 and yk+2
are chosen so that y0 6= yk+2 . Then for all j > k + 2, set yj = yj−(k+2) . y is a
periodic point in D (of prime period at most k + 2) with d(x, y) < , so periodic
points are dense in D.
Last, we show topological transitivity. Since every nonempty open set in D contains a cylinder, it suffices to show that given two cylinders, some iterate of the
Math 053-1
Final Exam (Solutions)
Spring 2011
first cylinder intersects the second nontrivially. Let E and F be cylinders; suppose E = Cyl(s1 , s2 , ..., sk ; p1 , ..., pk ) and F = Cyl(t1 , t2 , ..., tl ; q1 , ..., ql ). WLOG
assume p1 < p2 < ... < pk and q1 < ... < ql . Define x ∈ D as follows: let y ∈ E
and z ∈ F . Set xj = yj for all j ≤ sk . Then no matter what xsk is, choose xsk +1
and xsk +2 so that xsk +2 = C (this can be done). Then choose xsk +3 and xsk +4 so
that xsk +4 = z0 . Then for j > sk + 4, set xj = zj−(xk +4) . We have x ∈ E but
σ sk +4 (x) ∈ F so (D, σ) is topologically transitive as desired.
(b) The adjacency matrix for this subshift is


0 1 1
D =  1 0 0 ;
0 1 1
by a theorem from class the topological entropy is the logarithm (base 2)
of√ the
√
largest eigenvalue of D. The largest eigenvalue is 12 (1 + 5) so h = log2 1+2 5 .
(c) Let’s call the golden mean shift (Σ, σ). We define a map φ : Σ → D by

 A if xn = 0 and xn+1 = 1
B if xn = 1 and xn+1 = 0
[φ(x)]n =

C if xn = 1 and xn+1 = 1
This map is 1 − 1 and onto. It takes cylinders to cylinders, so it is continuous.
Similarly its inverse is continuous so it is a homeomorphism. One can show by
direct calculation that φ ◦ σ = σ ◦ φ so φ is a topological conjugacy between (D, σ)
and the golden mean shift.
7. (a) The fixed points are z = 1, z = e2πi/3 and z = e4πi/3 ; they are all repelling since
|f 0 (z)| = 2 at all three points.
(b) The periodic cycle of prime period 2 is {0, ∞}; it is attracting since |f 0 (0)mf (∞)| =
0 · 0 = 0.
(c) J(f ) is the unit circle; for points z inside but not on the unit circle, |f 2n (z)| = |z|4n
which tends to zero as n → ∞. For points outside the unit circle, |f 2n (z)| = |z|4n
which tends to ∞ as n → ∞. So any point not on the unit circle is in the stable
set of an attracting periodic cycle, hence is not in J(f ). Thus J(f ) is contained
in the unit circle.
Now, we see that any point e2πip/q where q is a multiple of 3 is periodic under f
(and repelling). Since these points are dense in the unit circle, J(f ) must contain
the unit circle.
8. (a) Here is the graph of g, restricted to I. We see that g(I) = I so I is forward
invariant. For backward invariance, recall that given a rational function of degree
b has d preimages counting multiplicities. From the graph below,
d, every z ∈ C
it is clear that any horizontal line hits the graph in three places between −2 and
Math 053-1
Final Exam (Solutions)
Spring 2011
2, so every z ∈ [−2, 2] (save the critical points) has three preimages in [−2, 2].
Since deg(g) = 3, these z have no other preimages. The critical points ±1 can be
checked directly to see that they have no preimages outside [−2, 2].
2
1
-2
-1
1
2
-1
-2
(b) Let β ∈ (2, |z|). We see that for |z| > 2, |g(z)| = |z 3 −3z| = |z||z 2 −1| > (β 2 −1)|z|.
Similarly, |g n (z)| > (β 2 − 1)n |z| which goes to ∞ as n → ∞ since β 2 − 1 > 1.
(c) If g had an attracting periodic point other than the fixed point at ∞, there would
have to be a critical point of g in the basin of attraction of that critical point. But
both critical points of g (namely, ±1) land on repelling fixed points of g (namely,
±2) under iteration, so this is impossible.
(d) Code points by their itineraries with respect to the partition {[−2, −1), [−1, 1], (1, 2]}
of [−2, 2]; if one does this we see that g restricted to [−2, 2] is (almost) conjugate
to the full 3−shift, which has entropy log2 3.