An Introduction to Dynamical Systems - 2nd Edition

10.1. ’Il'ansition Graphs 429

is allowable provided that there is an edge in the transition graph from sj to s,-+1
for all 0 3 j .

An allowable in nite sequence of symbols s = sosl... is called periodic
provided that there exists an n Z 1 such that s,,.,.;, = sk for all k Z 0, so
s = (sosl ...s,,_1)°°. An allowable word so . . . s,, is called periodic provided that
s,, = so, so (sosl . . . s,,_1)°° is an allowable periodic sequence.

An allowable periodic word w = sosl . . . so is called reducible, provided that
n = mp with p > 1 and sosl ...s,,_1s,, = (so. ..s,.,,_1)"s,,. An allowable period.ic
word s = sosl . . . s,, that is not reducible is said to be irreducible and to have least
period n. An irreducible periodic word w = sosl ...s,, is called n-periodic.

An allowable periodic sequence s = (sos, is said to have least period n
and be n-periodic provided that s = (so . . . s,,_1)°° and the word so. . .s,,_1so is
irreducible.

De nition 10.11. Assume that a function f is continuous on a partition of closed

intervals {I, We use the subscript on the interval, I,, as a symbol for the

given interval, so s is a choice from S = { 1, . . . , N Given a string of such symbols,

so. ..s,, with each sj E S, we seek to nd a point :1: such that fj(a:) is a point in

the interval I,’ for 0 3 j 3 n. In such a string of intervals we allow a given interval

to be used more than once. We let

I,,,___,,_ = {:0 = fj(a:) e 1,, for 0 5;" 5 rt} = Q f-1'(1,,).

051511

Most of the following theorem follows directly from Theorem 10.2. For part

(c) we assume that we start with a partition for f , which allows us to conclude the

point has least period n for f. .

Theorem 10.3. Let f be a function from IR to IR and it is continuous on a partition
of closed intervaL9 {I,-}§V=1. Let w = so . . . s,, be an allowable word for the transition
graph for the partition.

a. The set I,,,...s.. 75 ill, and there is a point 2:" such that fj(:c“) is in I,’ for
0 $1 S n-

b. If the word so. ..s,, is periodic with so = so, then the point 2:" of part (a)
can be chosen so that f"(a:‘) = 2:‘. Thus, the period of 1:‘ must divide n.

c. Assume that w = so . ..s,, is n-periodic and irreducible (i.e., n is the least
period of the string). Assume that the end points of Isa are not in I,,,___,,_, or
0(Iso) O I,,,,____,,_ = (ll. Then, the point ac" has least period n for f.

Remark 10.12. We usually verify the assumption in part (c) that the end points
of I,,,, are not in I,,°___,,_ by using the fact that they are periodic with a period
different than n.

Proof. Let J, = I,,. Let K; for 0 3 i 3 n be the intervals constructed in the
proof of Theorem 10.2. Parts (a) and (b) follow directly from that earlier theorem.

(c) We assumed that 6(I,,,,) I,,,___,,_ = 0, so f"(::") = :i:’ G int(I,,,,) = int(I,,_).
Since f"(int(Ko)) = int(I,,,), ac" E int(Ko). Then, fJ(a:') G f’(int(Ko)) =
int(K,-) C int(I,,,) must be in the open intervals corresponding to each I_,,. Since

430 10. Itineraries

we started with a partition, the open intervals int(I,) are disjoint. If f"(a:') = :z:"

for 0 < k < n, then the word w = so . ..s,, would have period k < n which is

impossible since it is irreducible. This shows that 1:‘ cannot have period less than

n and must have least period n. Cl

Remark 10.13. When we list the string of symbols for a periodic orbit, we usually
write an in nite sequence of symbols, rather than the nite one used in the state-
ment of the theorem. In the preceding theorem, a string such as 8gS]_8g83 = RRLR
corresponds to a period-3 point. The orbit continues to pass through this same
string of intervals for higher iterates. Therefore, we can write an in nite sequence
of symbols for a periodic point, which we call the itinerary of the point. For the
preceding example, we write

30$, = (RRL)°° = RRLRRLRRLRRL - -- ,

because fii (aso) is i11 I,’ for all j 2 0. For a word w, in order for an in nite sequence
w°° to be allowable, it is necessary for the rst symbol in the word w to be able
to follow the last symbol in the word w. Thus, for the string w = RRL, for w°°
to be an allowable sequence, it is necessary that the rst symbol R can follow the
last symbol L. In Section 10.3, an in nite sequence of symbols (an itinerary) is
assigned to nonperiodic points, as well as to periodic points. Therefore, we start
using this notation at this time.

We can apply the previous theorem to a function with a period-3 point to
obtain the following result by T. Li and J . Yorke, [Li,75].

Proposition 10.4 and Yorke). Assume that f is a continuous function from
lR to itself. Assumeassume that either (a) f has a period-3 point or (b) there is a
point mo such that either (i) f3(:|:o) 3 a:o < f(:co) < f2(:z:o) or (ii) f3(a:o) Z aso >
f(:1:o) > f2(zo). Then, for each n G N, f has a period-n point.

Proof. Case (b) includes case (a). Case (b.ii) is similar to (b.i) by reversing the
orientation on the line, so we assume that condition (b.i) holds: f (2:2) 3 so <
f (mo) = $1 < _f(a:1) = 1:2. See Figure 6 for the graph of a function with period
three. Take the partition of closed intervals IL = [:z:o,:z:|] and IR = [21,12]. By
following the iterates of the end points, it follows that

f(IL)D[171,£1.‘g]= IR and f(IR) 3 |:z:o,:c2] = IRUIL.

The transition graph for IL = [zo, 2:1] and IR = [$1, 1:2] is given in Figure 5. Table
1 indicates which sequence of symbols corresponds to a periodic point of the period
indicated.

- Symbol Sequence
@8

(RL)°°
(RRL)°°

£50310»-3 (Rn—lL)oo

Table 1. Sequence of symbols for periods for a function with a period-3 point

10.1. '1lansition Graphs 431

Note that the points on the boundaries of the intervals all have period three, so the

point given by Theorem l0.3(b) must be in the open interva.ls and it so must have

the period indicated. Since all periods appear, the fimction must have points of all

periods. El

f (=1)

:1:

Figure 6. Plot of a period-3 orbit using the stair step plot

10.1.1. Sharkovskii Theorem. We rst de ne the special ordering of the pos-
itive integers necessary to state the Sharkovskii theorem about the set of periods
that can occur for a continuous function on IR.

We want to see which periods are forced by other periods. For a function f, let

9(f) = {k : f has a. point of period k

Proposition 10.4 shows that if 3 is in .9’(f) for a continuous function on IR, then
9(f) is the set of all positive integers.

De nition 10.14. We introduce a.n ordering of the positive integers, which are
the possible periods, called the Sharkovskii ordering. When we write m D n, we
want the existence of a period-m point to force the existence of a period-n point.
So 3 l> n for any positive integer n. The correct ordering is as follows. Let j be
the set of all odd integers greater than one. These odd integers f are listed in the
Sharkovskii ordering, with the ordering opposite the one they have in terms of how
they appear on the line,

3>5>7r>91>111>---.
Next, all the integers which are equal to 2 times an integer in / are added to the
list, then 22 times an integer in j are added to the list, and then increasing powers
of two times an integer in / are added to the list:

3>5r>7>--->2~3>2-51>2-7>---
>2'=-s>2'=-5|>2'°-7»---|>2*+1-s>2'=+‘-5>2'=+'-7>---.

432 10. Itineraries

Finally, all the powers of 2 are added in decreasing powers:

3D5l>7[>--~l>2k-3l>2k-5>2k-7!>---
-~-1>2'°+‘>2'°>2"'1|>--->22l>2l>1.

We have now listed all the positive integers.

Theorem 10.5 (Sharkovskii). a. Let f be a continuous function from IR to it-
self. Assume that f has a period-n point and n|> lc. Then, f has a period-k
point. Thus, ifn is in .9”(f) and nl> k, then k is in 9(f).

-- b. For any n G N, there ea:i.sts a continuous f : IR —> 1R such that 9(f) = {k :

n>k '_

Key steps in the proof involve the clever use of Theorem 10.3. We do not prove
the general statement of part (a), but present some of the main ideas through the
consideration of special cases. See [Rob99] gives a proof using Stephan cycles
de ned below. P. D. Straf n [Str78], B.-S. Du [Du,02], and K. Burns and B. Has-
selblatt [Burll] have published articles proving the Sharkovskii theorem that are
accessible to anyone reading this book. We do give constructions to show that part
(b) of the theorem is true and partial results toward the proof of part (a) in Section
10.8 at the end of the chapter.

Many of the proofs of this theorem (e.g., the one by Block, Guckenheimer,
Misiurewicz, and Young [Blo80]) use the fact that, if there is a point of odd period
n, then there is possibly another period-n point for which the order of its iterates in
the real line is especially simple. This latter periodic orbit is called a Stefan cycle.
Piecewise linear maps of the type given in Example 10.16 show the existence of a
map with exactly the periods forced by an odd integer in the Sharkovskii order.

De nition 10.15. A Stefan cycle is a. period-n orbit with n odd, for which the
order of the iterates zcj = f1 ($0) on the line is one of the following:

:r,,_1<---<a:4 <z2<a:0<a:1<:r3<---<a:,,_2,

the exact reverse order

I,-,._2 <"‘<.123<IL'1 <IQ<$g <IE4 <"'<In_1,

or one of these two types of orbits starting at a different point on the orbit.

A periodic orbit of period 3 is always a Stefan cycle.

Example 10.16. Assume that f from R to itself is continuous and has a period-7
Stefan cycle :05 = f1(:r:o) for 0 3 j 3 6, with order on the line ass < 2:4 < $2 <
:00 < 2:1 < 2:3 < 2:5. The associated partition of closed intervals for Stefan cycle
is I1 = [$0.11]. I2 = [$2,110], Ia = [Imva], I4 = [=v4.=v2]. Is = [ra.-1=sl» and
I6 = [$6, a:4]. See Figure 7. By considering the image of the end points of each I11

10.1. Transition Graphs 433

$5

Is

$3
Ia

Ii
I1

$0 —
I2

I2

I4
$4

Is

$6
$6 I6 I4 I4 11212 $011 I113 $315 I5

Figure 7. Example 10.16: Graph of an f with period-7 Stephan cycle

we get that

f(I1)3 l$2,$1l = I1 U12,
f(I2) 3 l-"?1.Ial_ = Ia,
f(11) D [$4.12] = I4.
f(I-1) 3 l1'Fa.I5l = Is.
f(Is) 3 [$6.114] = 16,

D [$5,230] = I1 U I3 U I5.

The graph of a piecewise linear map f with the preceding properties and its transi-
tion graph are shown in Figures 7 and 8. The Table 2 gives sequences of symbols for
the various periods that are possible. The end points of I1 have period 7 so cannot
have the symbol sequence given in the chart, so the point must have the period-n
indicated. In particular, it shows that .9”(f) must contain all positive integers ex-
cept possibly 3 and 5. In fact, if the function f is linear between the points :i:, as is
the case for the function in the gure, then .97’(f) = {Ic : 7l>k } = {lc 2 k 96 3,5
Other functions with a period-7 Stephan cycle could have other periods.

Example 10.17 (Doubling the periods). In this example, we show how to take
a function f from [0,1] to itself and make a new function g de ned on [0,1] such
that

(10-1) 9(9) = {1}U29’(f)

(i.e., g has a xed point, together with twice all the periods of f).

434 10. Itineraries

/3

@

/ca®: \@5@/

Figure 8. Example 10.16: Tra.nsition graph

n Sequence Period .

of Symbol Exists?

1°° W Yes

(56)°° Yes

None No

(3456)°° Yes

None No

(123456)°'° Yes

l503UIJ>~0Ol\)>—~ (1"_523456)°° Yes

Thble 2. Strings of symbols giving various periods for a function with a.
period-7 Stefan=‘cycle, Example 10.16.

De ne g by

§+§;(a¢) if05:c$§,

<10-2) ya) = 12+ r<1)1(§ — w) iii s I s %»

I-§ n§5z5i

See Fig1n"e 9. Note that the graph of §f (3:c) reproduces the graph of f, but
one-third as large. For the function g this graph is raised by the constant

The interval [1/3,2/3] maps over itself by g, so there is a xed point for g in
[1/3,2/3]. No other interval maps to [1/3, 2/3], so there a.re no other periodic points
in this interval.

The two intervals [0,1/3] and [2/3,1] are interchanged by g,

N%DC%M wd WMD=NW,

so all periodic points in these two intervals must have periods that are a multiple
of 2. On [0,1/3],

g’(t) = 9 (§ + §1o:»>) = §/(Bx).

Exercises 10.1 435

1 1 ------------------------------- -1- ------------- --1

f (I) 9(1) 5 ,

--------------

.............. --.|----.....- .-.-.,........_......
I

0 sI1 0 "I ' "l ll!

(-1) 12 1

“ (b) °

Figure 9. (a) Graph of a map f(:i:) and (b) graph of g(m) that has double the
periods of [(2)

which simply reproduces a copy of f, but on a smaller scale. Thus, the periods of
g2 on [0,1/3] a.re the same as the periods of f. This is also true for g2 on [2/3, 1].
Therefore, the periods of g for points starting in either [0,1/3] or [2/3,1] are just
twice the periods of f . This completes the discussion about the periods of g.

For any n E N, we now know how to construct acontinuous function f : R —> R
such that .92'(f) = {k : n > k}. Using a function with only xed points and
a piecewise linear function with a period-n Stephan cycle, we can get examples
for any odd n. Repeatedly applying the doubling of periods construction of the
preceding example, we can get functions with n = 2" m for m odd. Exercise 5 in
this section shows that there is a function whose periods contain all powers of two
2" but no other periods.

For the proof of Theorem 10.5(a), the existence of a Stefan cycle of odd period
implies the existence of all periods forces by the Sharkovskii ordering. We do not
give a complete proof but prove some fin-ther results in Section 10.8 at the end of
the chapter that illustrate the techniques used in the proof.

Exercises 10.1

1. Assume there is a period-9 Stefan cycle for a continuous map f on R.
a. Determine the transition graph. In particular, show that the interval J3 =
[p3,p6| maps over J, for all odd t.
b. Show that f must have all periods except, possibly, 3, 5, and 7.

2. Let f be a continuous function de ned on the interval [1,6] with f (1) = 5,

f(2) = 6, f(3) = 4, [(4) = 1, [(5) = 2, and f(6) = 3. Assume that the

mction is linear between these integers.
a. Sketch the graph of f.
b. Label the intervals between the integers and give the transition graph.

10. Itineraries

c. For which n is there a. period-n orbit? Determine a symbol sequence in
terms of the intervals that shows each period that exists.

Let f be a continuous function de ned on the interval [1,4], with f(1) = 4,
f(2) = 3, f(3) = 1, and _f(4) = 2. Assmne that the function is linear between
these integers.

a. Sketch the graph of f.
b. Label the intervals between the integers and give the transition graph.
c. For which n is there a period-n orbit? Determine a. symbol sequence in

temis of the intervals that shows each period that exists.

Let p(z) = (:02 — 1)(:c2 — 4) = .124 — 52:2 +4. Let N(a:) = N,,(:c) be the Newton

map associated with p given by equation (9.1). Notice that N (sv) goes to ioo

at a1 = —\/E, G2 = 0, and a3 = The general shape of the graph of N(:r)

is the same as that given in Figure 22, but this example is more symmetrical

about 0.

a. Sketch the graph of N (12).

b. Show that 2: = —2, -1, 1, and 2 are xed point sinlcs of N.

c. Let I1 be the closed interval in (a1,a.2) that is mapped onto [a.1,a3] by N;

let I2 be the closed interval in (a2,a;;) that is mapped onto [a1,a3] by N.

Give the transition graph for I1 and I2.
d. Show that the Newton map N has points of all periods. (Note that these

are points for which the Newton iteration does not converge to a zero of

the polynomial.)

e. Show that the basin of attraction of the xed point 2 for N contains the

interval oo(A).similar argument should show that the basin of

attractionlof the xed point -2 contains the interval (—oo, —\/ ).)

Let fo be the function that is identically equal to 1/3 on [0,1]. As in Example
10.17, let f1 be the function de ned in terms of fo by equation (10.2). (The
periods of fl and fq are related by equation (10.1).) By induction, let f,,
be the fu_nction de ned in terms of f,,_1 by equation (10.2). De ne foo by
f°°(:c) = lim,,_.m f,,(z).

a. Using equation (10.2) and induction, prove that each of the functions fn
has periods of points exactly equal to {1, 2, . . . , 2"}.

b. Explain why foo is continuous. In particular, why is f,,,,(0) = 1 and
foo continuous at :1: = 0? Hint: Draw the graphs of fg, _f|, fa, and f3.
Show by induction that for k Z n, f;,(:c) = f,,(a:) for 1/3" 3 :1: 5 1 and

fg($)Z-§(1+%+"'+y,i1_r) f0I‘0s£C§1/3".

c. Prove that the periods of foo are exactly {2* : 0 5 t < oo}. Further,
explain why foo has exactly one orbit of each period 2".

Assume that f is continuous and takes the closed interval [a, b] into itself. Prove
that f has a xed point in [a, b].

Suppose that I1, I2, I3, and I4 are disjoint closed intervals and f(sv) is a contin-

UOUS map SUCI1 that = I2, = I3, = 14,3-Dd. 3 I1 UI2UI3-

Show that f has a period-3 point.

10.2. Tbpological 'Il'ansitivity 437

8. (Burns and Hasselblatt) Assume that a. map f from lll to itself has a period-9
orbit, :z:_.; = f-" (:i:) with 2:9 = zcq, whose order on the line is

!lI5<I4<£l72<$7<IQ<$1 <$3<$g<£L‘5.

(Note this orbit is not a Stefan cycle.) Denote six subintervals by

I1 = [$0,911], I2 = [I1,$1l» Is = l$4,$1l,
14 = l$4,I1l, Is = [10,15]. Is = l=Pa»I4l-

(Note that the interiors are not disjoint, but 816 F1 Ij = (D for 1 3 j 5 5.)
a. Show that these interval force at least the covering indicated by the tran-
sition graph in Figure 8. (There are many other coverings as well.)
b. Using Theorem 10.3(c) and strings that start with the symbol 6, show
that f has all periods forced by period 7.

10.2. Topological Transitivity

We next consider maps that have orbits that come close to every point in a whole
interval. In this section we use binary expa.nsions of a number to nd such an orbit,
while in the next section, we use symbolic dynamics. To make this concept precise,
we de ne what we mean for a subset to be dense in the set which contains it. Since
we are often considering a set of points which is taken to itself, we next de ne an
invariant set. Then, if a map has an orbit that is dense in an invariant set, we call
the map topologically transitive. We use binary expansions of numbers to show
that the doubling map is topologically transitive on the unit interval.

Binary expansions '

Any number :|: in [0, 1] can be represented as a binary ea.-pension

Ill: .F’11a8s
kl.

where each aj is either 0 or 1. Just as with the decimal expansion, some numbers
have two expansions. For example,

H8=MN "3.'—' OM82“n- _ "3.'—' r-I "Iv-' I-‘IQ IO»-I
J: £_Z
"Kw-' —-r-I LiNi-I

Notice that we used the formula for a geometric series. If —1 < r < 1, then

1. |—‘ D15 i
IP13

In general, a number whose expansion ends in an in nite string of 1’s is the same

as the number whose expansion changes the previous 0 to a 1 and replaces all the

1's with 0’s,

3 y- 8 3 n- 1
_ aj 1 _ _ Gj

_E2i+_Z 2.1 _E2:i+2'\'
]=1 ]=-n+1 ]=1

438 10. Itineraries

If we start with a repeating binary expansion, we can nd the rational number
that it represents. For example,

£+£..+.£+_1_+l+£+...=(£+l)<]_+!'.+i+...)
22*2°2‘2°2° 24 as’

_s 1 _a s_s

"z(t)-z';";~
@|-I

If :1: = 1, then :1: = If 2: is in [0,1), then we ca.n nd its binary

expansion as follows: Multiply :1: by 2 and let a1 be the integer part of 2:: and f1
the fractional part, 22: = a1 + fl with a1 = 0 or 1 and 0 3 fl < 1; multiply f1 by
2 and let a2 be the integer part of 2f1_ and f2 the fractional part, 2f1 = 0.2 + f2;

continue by induction to nd a_,- and f,-. Consider this process for the number 4/7:

a-=-=1+- m=1md =— ~
\lD—I ~1|-I

2-—|u—b-I- =—@Nl\1 =0+— ag=0andf2=—

RIQ ~Il\3

|§~1 03=08.I1df3=-
2-—-=-== o + ,
m=lmd =—
\INl\1 ~|-|-is

a-=-=1+-
~i.:=- ~1o~1 ~z-was

and the process is beginning to repeat, a1+3,- = 1 and a2+;;,- = a;;,- = 0. Indeed,
this binary expansion does sum to 4/7 as desired:

5l—' .|_o¥+§0 5+?F-I +...=E(1+_+8_D-I2+...)=E|—<' i1_F-I é)

k>Q-l—II <00 ~1_-am’-'

Ternary expansions

In the same way, any number a: in [0, 1] can be represented as a ternary expan-
sion

X a‘

_ _-7
3: - 31 '

:1=1
where each aj is either a 0, a 1, or a 2. A number whose ternary expansion ends
in an in nite string of 2’s is the same as the number whose expansion adds one to
the previous entry and replaces all the 2’s with 0’s; that is,

an X 2 a,.+l

s"+ E131! 3" '
j =n+

Dense sets

De nition 10.18. For a set A C S C IR", A is dense in S provided that
arbitrarily close to each point p in S, there is a point in A. More precisely, for
each p in S and each c > 0, there is a point a in A such that ||a— pll < ¢~

10.2. Topological Tl-ansitivity 439

For sets in the line, this can be expressed by saying that for each p in S and each
e > 0, the interval (p — e.p + c) contains a point of A,

(p-e,p+c) A79@.

Example 10.19. Let Q be the set of rational numbers, and Q‘ = R \ Q be the
set of irrational numbers. Then, both Q [0, 1] and Q‘ |"| [0, 1] are dense in [0, 1].
This says that, arbitrarily close to any number, there is a rational number and an
irrational number.

To see that the rational numbers are dense in [0, 1], take the decimal expansion

of an arbitrary number in [0, 1], a: = di/10.‘i with each dj a choice from the

set {0, 1, . . ., 9 For any nite n, the truncated expansion 12,, = E;'=1d1'/105 is

a rational number within 10'" of az. This shows that the rational numbers are
dense.

The irrational numbers can be shown to be dense, by showing that any number
as can be approximated by \/2(1’/q), where p and q are integers.

Example 10.20. Consider the set S = 2 0 3 p 5 210° is an integer This

set is within a distance of 2"“ of each point in [0,1], but this distance is not
arbitrarily small. In particular, S does not intersect the open interval (0, 2'1°°).
Therefore, S is not dense in [0,1].

Invariant sets and topological transitivity

Since we eventually are going to consider an orbit that is dense in a set which
is taken to itself by the map, we start by de ning an invariant set. In the material
through Chapter 11, the ambient space 'is line IR, but we give the de nition for a
general space X which can be IR" in later chapters.

De nition 10.21. Let f be a function from a set X to itself. A subset A of X
is called positively invariant provided that, if :1: is in A then f(:|:) is in A (i.e.,
f (A) C A). A subset A of X is called invariant, provided that (i) it is positively
invariant and (ii) for every point b in A, there is some a in A with f(a) = b (i.e.,

f(A) = A)-

De nition 10.22. If f is a map from X to itself, where X has a distance is
de ned (e.g., a subset of some JR"), then f is topologically transitive on an invariant
subset A C X, or just tmnsitive, provided that there is a point 2:’ in A such that
the orbit O’T(a:‘) is dense in A.

Doubling map

We show that the doubling map D(a:) = 2 :1: (mod 1) is topologically transitive

on [0, 1] by using binary expansions. Consider a number 2: = “i/21', where

each aj is either a 0 or a 1. Using this expansion,

D( %~22 )=a1+ZJ2.aj2—s_, (mod1)= iii

:P._’l8 :l"_l3 PN’

440 _ 10. Itineraries

The nu‘ iterate simply shifts the expansion by n places:

S3 2
_as') =g“%':_
Kl.

1
D(a:)

Z

° 0.5 .1

Figure 10. Graph of the doubling map

If a point 0: has an ‘expansion that repeats every n places, a,-.,.,, = aj for all j,
then D"(:v) = :1: and a: is a period-n point.

The rst n places in the expansion of a number also determine which subinterval

of length 2"" contains the point 1:; if as = “i/21' and a:,, = Z;”=,°1/21', then

2,, § a: 3 run + 2—n .

We can now show that D has a point with a dense orbit in the interval [0, 1|.

Theorem 10.6. The doubling map D is topologically transitive on [0, 1]. In other
words, there is a point 2:‘ in [0,1] such that the orbit of :i:‘ by the doubling map
D is dense in [0,1] (i.e., the orbit of 0:" comes arbitrarily close to every point in
[0,1]).

Proof. We describe the point 1:‘ by giving its binary expansion. Let bf = 0 and
b; = 1; this lists both possible bits. Then we list all the strings of bits of length
two, b§b§ = 00, b§b§ = 10, b;b§ = 01, and b§b§0 = 11. At the third stage, we use
bf, through b§,, to list all the strings of bits of length three: 000, 100, 010, 110, 001,
101, 011, and 111. Continuing by induction, at the j"‘ stage, we list all the strings
of bits of length j . Let 2:’ be the point with this particular binary expansion: that
is, :i:‘ = E;-‘=1 bi/21'.

We claim that the orbit of as" comes arbitrarily close to every point in [0, 1]-

Take any a point a: = "1/21, where we have given its binary expansion. For

any k, the rst Ic bits (1102 . . . at appear as a string of bits for a:"; that is, a_,~ = b§+m

Exercises 10.2 441

for some xed m and 1 3 j 3 k. Then, the binary expansion of D"‘(:c') agrees

with a: in these rst k bits, and ~—

ID1n(I-u)_$|= oo b3'- "l'"%— a"- ' S Z°° 1
Z

j=k+l :i=k+1

=F1 1 =211 -

Thus, the orbit of :0‘ comes within a distance less than 2_;¢ of sc. Since 2_k.IS

arbitrarily small, we are done. El

Although we do not prove that the following example has a dense orbit until
Chapter 11, we refer to it several times, so we state the result at this time.

Example 10.23. Let a be an irrational number, e.g., a = ‘/5/2. Let R0, be the
map given by

R,,(a:) = zv + a (mod 1).

Although we do not make this formal, we often identify the interval modulo one
with the circle, and speak of this map Ra as an irrational rotation on the circle by
a percentage of a full turn. Then,

R§(:i:) = at + na (mod 1).

In Example 11.2, we show that orbit 0;“ (0) is dense in [0, 1], so Ra, is topologically
transitive on [0, 1].

Exercises 10.2

1. Which of the following sets are dense in [0, 1]?
a. S1 is the set of all real numbers in [0, 1] of the form 22;, where p and n are
arbitrary positive integers.
b. S2 is the set of all real numbers in [0, 1] except those of the form 22;, where
p and n are arbitrary positive integers (i.e., S2 = [0,1] \ S1).
c. The set of numbers zv in [0, 1| which have decimal expansions that use the
digits 0, 2, 4, 6, and 8, but not 1, 3, 5, 7, or 9; that is,

HM8S3= :v= —"Q'Q. u. .:d,-is0,2,4,6,or8 .
Kb.

2. Express the number 5/24 in terms of a ternary expansion.
3. Using the ternary expansion of numbers, show that the tripling map

f (sv) = 3:: (mod 1) is topologically transitive.
4. In terms of its binary expansion, describe an irrational number that does not

have a dense orbit for the doubling map.

442 V 10. Itineraries

5. Let S be the set of all irrational numbers that have a dense orbit for the
doubling map. Show that S is dense in [0,1]. Hint: Consider numbers of the
form 227:, %3’l + 51;; :i:‘, where 1:‘ is the point with a dense orbit for the doubling
map.

10.3. Sequences of Symbols

In Section 10.1, we showed the existence of periodic points by means of specifying
their itineraries, the sequence of intervals through which they pass. In this sec-
tion, we extend these ideas to nonperiodic itineraries and orbits by using in nite
sequences of symbols. In particular, we show that the tent and logistic maps are
topologically transitive on [0, 1|. We start by giving a de nition of the shift space.

Shift Space

De nition 10.24. We consider a symbol space S with N‘symbols. For this de ni-
tion, we take the symbols as S = {0, . . . , N— 1}. When there are just two symbols,
we often use S = { L, R }, as we do when considering the tent map in this section.

The full shift space on N -symbols, denoted by 21,, is the set of all in nite
sequences of symbols s = 508182 . . . , where each s,- is an element of S (i.e., sj = 0,
. . . , or N — 1). The N indicates that there are N symbols; the plus sign indicates
that the symbols are de ned only for j Z 0. The term “full” refers to the fact that
any symbol is allowed for each s,-. If the context is clear, we just call Z}, the shift
space.

The shift map 0 from E‘,f,, to itself is de ned by dropping the rst symbol so
in the sequence and shifting all the other symbols to the left one place,

a(s0s1...s,,...) =s1s3...s,,...,

so that, 0(5) = s’, where sf, = s,-+1.

De nition 10.25. For two symbols s and t in S, with 0 3 s,t 3 N — 1, let

6 ,t = 0 ifs = t ‘

(S ) {I ifs gét.

For two elements s and t in E1, (two sequences of symbols), we de ne the distance

between them as 0°

d(s, t) __ 2 _5(3$j,.i_;'l.

J=0

The distance function cl is called a metric on 27,}.

Proposition 10.7. The metric d on EX. just de ned has the following properties
that justify calling it a distance:

i. d(s,t) Z 0.
ii. d(s,t) = O if and only ifs = t.
iii. d(s,t) = d(t,s).
iv. d(r, s) + d(s, t) Z d(r, t).

10.3. Sequences of Symbols 443

The proof of this proposition and the proofs of the following two results are
given in Section 10.8.

The next result shows that two sequences s and t are close in terms of the
metric d provided they agree on the rst nite number of entries.

Proposition 10.8. Let Z} be the shift space on N symbols, with distance d as
previously de ned. Given t xed in E}. Then,

{s€2§:s,~=t,- for03j3Ic} = {s€E,f,:d(s,t)33"‘2‘1

Proposition 10.9. The shift map 0 is continuous on E}

Theorem 10.10. The shift map o on EX, has the following two properties:
a. The periodic points are dense.
b. There exists s‘ in EX, such that 0;‘ (s') is dense in E}, so the shift map

0 is topologically transitive on Ex.

Proof. (a) Given any t in EX, and n 2 1, let w = t0...t,,_, and

S=W°°=in...in_1t()...in_1tQ...tn_1-'- .

Since o"(w°°) = w°°, it is aperiodic point for o. The distance d(s,t) 3 3"‘+12'1.
Since n is arbitrary, the periodic points are dense.

(b) The construction of a point s‘ with a dense orbit is similar to that given in
Theorem 10.6 for the doubling map. Let s‘ be the symbol sequence that includes
all the symbol strings of length 1; then, all the symbol strings of length 2; and then,
by induction, all the symbol strings of length n for each n. For N = 2,

s" = 0100 O1 1011000 00} 010 011100101110 111

Given any t in 2:, and any k > 0, there is an iterate m such that the

rst k + 1 symbols of a"‘(s") are t0...tk. By Proposition 10.8, the distance

d(t,o"'(S')) S 3"‘2'1. Since k is arbitrary, this can be made arbitrarily small.

This shows that the orbit of s" is dense. El

De nition 10.26. Throughout the rest of this chapter and the next, we indicate
the length of an interval J by /\(J) where /\ is the lower case Greek letter lambda.

Tent map

Consider the tent map with partition of closed intervals

IL = [0,0.5] and IR = [0.5, I].

Each of these intervals in the partition has length 1/2, /\(IL) = /\(IR) = 1/2. Like the
doubling map, both intervals IL and IR are mapped all the way across the union
[0, 1] = IL U IR, T([0,0.5]) = [0, 1] = T([0.5,1]). See Figure 11. The difference
from the doubling map is that the tent map is decreasing on the second half of the
interval, [0.5,1]. Therefore, rather than using a binary expansion of numbers, we
specify a point by the sequence of intervals through which it passes, using the labels
L and R on the two intervals in the partition as the symbols. These symbols play
the role of 0 and 1 in the binary expansion for the doubling map.

444 10. Itineraries

1
T(:1:)

SB

0 IL 0.5 IR 1

Figure 11. Intervals for the tent map _T

For a point :r, let

__ L if T1'(1) e 1,, = [o,o.s],
3’ ' {R if T1'(¢) e In = [o.s,1],

and let

h(t) =s=s0sl...s,,...

be the in nite sequence of symbols of L's or R’s associated with :c. This map h is
called the itinerary map,’ and h(:c) gives the itinerary of the point :1: in terms of the
partition. The only arbitrary choice of symbols occurs when T“ (rr) = 0.5: to de ne
h we set sk = R, but sh = L is just as reasonable. However, if T*'(a:) = 0.5, then
Tk+‘(z)=1andT‘(a:)=0foriZ /c+2, so sk.,.1 =Rand si =LforiZk:+2.
Thus, for any nite string of symbols w, wLRL°° and wRRL°° could both be
associated to the same point. (This ambiguity is related to the nonuniqueness of
the decimal or binary expansions of a number.)

If h.(:n) = s = {3]'}j20, then T-"'(T(a:)) = Tj+1(a:), is i11 the interval I_.,,.+,, so
the symbol sequence associated with T(:c) is the shift of the sequence s by 0,
0'(s) = o(h(:|:)), or

l"~(T(==)) = ¢'(l1(r))-

Thus, h. satis es the conjugacy equation for T and 0.

We next explain how we can start with a symbol sequence and associate a point
in the interval. As we have done previously, let

Tl

1.,,___,,,_ = {==T1'(¢) e 1,, for 0 5;" g n } = Q T-1'(I,,.)

i=0

be the interval associated with this nite string of these symbols. Since T-l (T(a:)) =
T1+1(a:) E I,j+,, z E I,o____,,_ if and only if :1: E I,,,, and T(a;) G I,,____,,_, so

Isg...s,, = I59 n T_1(Ia1...sn)~

10.3. Sequences of Symbols 445

The subinterval [O,1/4] C IL is mapped onto I1, = [U,1/2], so

|:0,1/4] = IL Fl T_l(IL) = ILL.

Similarly, [1/4.1/2] C I1, is mapped onto IR = [1/2,1], so

[1/4.1/2] = IL Q T_1(IR) = ILR-

Notice that, T restricted to I1, = [0,1/2] is increasing, so the order of I1, and IR is
reproduced with ILL to the left of ILR. In the same way, [1/2. 3/4] C IR is mapped
onto IR = [1/2, 1] and [3/4, 1] C IR is mapped onto IL = [0,1/2], so

[1/2,3/4]=IR T‘1(I;1)=IRR and [3/4,1]=1RnT-1(1L)=IRL.

Notice that, T restricted to In = [1/2,1] is decreasing, so the order of IL and IR
is reversed with IRR to the left of IRL. Each of the intervals, with two symbols
speci ed, has length 24, )\(I,°,,) = 2-2. See Figure 12.

I-I2::

I-I P"

ILL ILR IRR IRL

Figure 12. Intervals for strings of two symbols

To determine the order of intervals for which three symbols are speci ed, place
the intervals with two symbols along the second axis of the plot. See Figure 13.
The tent map T restricted to IL = [0,0.5] is increasing, so the order of ILL, ILR,
IRR, and I51, is reproduced with the order of intervals for three symbols whose rst
symbols is L:

ILLL = IL Fl T_l(ILL)» ILLR = IL Fl T'l(ILR),
ILRR = IL Fl T_l(IRR): ILRL = IL Fl T_1(IRL)-

On the other hand, T restricted to IR = [0.5, 1] is decreasing, so the order of the
intervals for three symbols whose rst symbols is R is reversed:

IRRL = IL Fl T-l(IRL), Inna = IL Fl T_1(I1=uz),
IRLR = IL F7 T_l(ILR)a IRLL = IL Fl T‘1((ILL)-

446 10. Itineraries

IRL

IRR

ILR

I| _-I__I “"’I|I I| I| I -.___|___

ILL '-—-l-—_——-|

-‘"“'"“|'-f-— ____1_I___ _-_-_'-__.'-_ .'I _ _ |__ _l_ _ - _'"1" '- ______I

ILLL ILRR IRRL_ IRLR

ILLR ILRL Inna IRLL

Figure 13. Intervals for strings of three symbols

Each of the intervals with three symbols speci ed has length 2‘3, ,\(I,o_.,,,,) = 2‘3.
The eight intervals together cover all of [0, 1], or

low ll = U I80-1182'
80,81-82

Continuing this process, each of the intervals obtained by specifying n symbols
has length 2"", )\(I,,,,___";__,) = 2‘". The 2" such intervals together cover all of [0, 1],

I lovll = U I80---0»-1'

8° ,.. .,8y|_ 1

The order of these intervals could be determined.
Finally, we want to see that an in nite sequence of symbols determines a unique

point. Just as a point can be speci ed by giving an in nite decimal expansion, a
in nite sequence of symbols determines a point by the sequence of interval through
which the point passes under iteration by the map. By the preceding construction,
the intervals I,,,,____.,,, are nested; that is,

I00 3 Is°s| D ‘ ' ' 3 Iso....s,,_1 D Ie0...a,.-

Since these intervals are closed, the intersection is nonempty. (See Appendix A.2
for more discussion of the intersection of closed bounded sets.) In particular, if
I,°____.,,_ has a left-end point of an and right-end point of b,,, then

30$“!s"'San—lSan<b'nsbn—1S"'Sb1£b0-

The sequence of left-end points a,, is increasing and bounded above by bo (or any
of the b,,), so it must converge to a point aw; the sequence of right b,, is decreasing
and is bounded below by ao, so it must converge to a point boo, and aw S bw
For each n and j Z n, an 5 a._,- 5 bj 3 b,,, so an 5 aw 3 boo § b,,. The distance
between the two end points of I,,,___,,, is 2“""1, boo — am 3 b,-, — a,, = 2"‘"1 I01‘
all n, so am = bog and the intersection is a single point. Thus, we have shown that

10.3. Sequences of Symbols 447

the in nite sequence of symbols determines a unique point, which we denote by
Ic(s). This map k is essentially the inverse of the itinerary map h, and goes from
the space of sequences, E; , to the points in the interval.

The next theorem summarizes these results.

Theorem 10.11. Let h be the itinerary map for the tent map with partition
{I;,, IR}. Let s in E; be any symbol sequence ofL’s and R’s.

a. Then there is a unique point 2,, = k(s) in [0,1] such that T7 (2,) is in I,’
for all j, or

{$5} = F] T_j(Is,) = H Iso...s,,_;
i=0 3=o

Moreover, the map k from E3’ to [0, 1] is continuous but not one to one,
k(wLRL°°) = k:(wRRL°°) for any nite string w.

b. The map It is a semiconjugacy between the shift map cr and the tent map T;

that is

Tok(s) = k:oa(s). \

c. The maps k and h are essentially inverses of each other, k o h(:c0) = 1:0. On
the other hand, h o k(s) _= s, except that, for s = wLRL°° where w is a word of
length j, T-"'(Is:(s)) = k(a-"(s)) = k(LRL°°) = 0.5 and h o k(s) = wRRL°°.

Proof. (a) The proof that k(s) is a single point is given in the discussion preceding
the statement of the theorem. We prove the continuity using 6’s and e’s. Given
any c > 0, there is an 1n such that 2'(""Hl < e. Let 6 = 1/(2.3'"). If d(s,t) 3 6,
then sj = tj for 0 3 j 3 m. This implies that both k(s) and h(t) are in the same
interval I_,°...s,,_. The length ).(I_.,,,...,,_) =, 2“(""Hl, so

|k(s) - /¢(¢)| 5 2-("Ml < 6.

This proves the continuity of k.

($1) Let 1',, = k(s). The p(oiE1t)§F7 in the interval I5“, for all j 2 0,

so Tzs) isin I,,,,,___ and Tks =Icos.

(c) The maps are inverses by construction, except as noted. El

Remark 10.27. This map In gives us another way to show the existence of periodic
points for the map T. It also allow us show that the tent map is topologically
transitive.

We have previously considered the intervals I,°___s,,_, for nite strings of sym-
bols, for which there is a whole interval of points. The theorem says that, for an
in nite sequence of symbols, there is a unique point which has this itinerary.

Note that an in nite binary expansion, which is used in the discussion of the
doubling map, speci es a point by giving its location to greater and greater preci-
sion. By contrast, an in nite sequence of symbols from the itinerary speci es the
point by giving the rough location for more and more iterates. (For the doubling
map, the binary expansion and the itinerary are essentially linked together.)

Theorem 10.12. Let Ic be the map from symbols in 2;’ to points in [0,1] for the
tent map T given in the preceding theorem. Let h be the itinerary map.

448 10. Itineraries

a. If s = (so . ..s,,_1)°° is a symbol sequence that is periodic with period n
(i.e., a period-n point for the shift map 0), then k(s) is a period-n point for T. If
5 is a symbol sequence which is eventually periodic with period n, then k(s) is an
eventually period-n point for T.

b. If mo is a period-n point for T, then h(a:0) is a period-n point for 0. If 2:0
is an eventually period-n point for T, then h(a:g) is an eventually period-n point for

U.

c. The periodic points for T are dense in [0,1].

d. The tent map T is topologically transitive on the interval [0,1]. In other
words, there is a point :i:‘ in [0, 1] such that the orbit of :2‘ by the tent map T is
dense in [0, 1].

Proof. (a-b) If the map lc were one to one, then parts (a) and (b) would follow
from Proposition 9.17. However, the only way in which k.is not one to one and h
is not uniquely de ned involves points that go through 0.5 and these orbits are not
periodic. Therefore, the parts (a) and (b) really follow from the earlier result. We
leave the details to the reader.

(c) Let 1:0 be an arbitrary point in [0,1]. Given any c > 0, there is a n such
that 2‘" < e. We need to show that there is a periodic point within 2'" of 1:0.
The point 1:0 G I_.,,,___,,,_, for some nite string w = so . . .s,,_1. Let w°° be the
in nite sequence that repeats the string w over and over. Let p = lc(w°°) be the
point for the periodic sequence w°°. Both p a.nd T"(p) have the same symbol
sequence, so p = T"(p) is periodic. Both 1:0 and p are in I“, with ).(I.,,.) = 2"‘,
so the periodic point p satis es |:ro — p| 3 2'" < e.

(d) The proof is very similar to that for the doubling map. Let s" be the
symbol sequence constructed in the proof of Theorem 10.10(b). Let 2:‘ = lc(s')
be the point with this symbol sequence. For any interval I,,,___,__,, this string of
symbols to . . . t,,_1 appears somewhere in the sequence s". Let 1n be the iterate, so
that a"‘(s) has to . . .t,,_1 as the rst n symbols. Then,

T”'(I') 5 Ila---in-1'

Since the lengths of these intervals are arbitrarily small, )\(I,,,___¢,__,) = 2"", this

proves that the orbit of as" is dense in the interval [0, 1]. El

Logistic map

Consider the logistic function for a = 4, G(y) = g4(y). Let I1, = [0, 0.5] and
IR = [0.5, 1] be the same intervals as those for the tent map. Then G takes each
of these intervals onto [0,1], since G(0.5) = 1. See Figure 14. The results for the
logistic map G(y) are the same as those for the tent map. Let IL and IR be the
same intervals as those for the tent map. See Figure 14. Given a symbol sequence
s, de ne the intervals

71

= {y = (Po) e 1., for 0 52' s n 1 = Fl G"(I.,>-
j=0

10.3. Sequences of Symbols 449

1
G (1)

I

0 IL 0.5 In 1

Figure 14. Intervals for the logistic map G(y) = g4 (y)

We showed in Proposition 9.16 that the map

y= is = e) = —.@_ rra: 1 — cos rra:

is a conjugacy from T(a:) to G(y). Because h is a conjugacy between T and G,
with h([0,0.5]) = [0,0.5] and h([0.5,1]) = [O.5,1], we have

n—1 n.—1

1§f,___,n_, = Q G-1'(1,,)'= Q h.OT_j on-1(1,,)

J'=° .'i=°

= n ( r-1'(I,,))

j=0

= h (IZ0...s,,_,) 7

where IZ;__,,"_ 1 is the interval for the tent map. There are bounds on the derivative
of the conjugacy equation,

h'(:B) = — sin(1r:r) and

3N-l=l

|h'(I)| 5 -
IO

Let :20 and 2:1 be the end points of the interval IZ;___,,“_,, so |a:1—:1:0| = 2'", and let
y0 = h(:r0) and y1 = h(:r1) be the corresponding end points of the interval I§i___,,'__,.
By the mean value theorem, there is a point :22 between :00 and :21 such that

2/1 - yo = M11)“ ll/($0)
= h'(:1:2) (a:1 — 2:0).

450 10. Itineraries

Therefore, the length of the interval is

*(I§-s.n) =|v1-yd

= lh'(I2)|l==1 — Iol
S gl$1-Id

= 1r2_"_l.

Since the length of these intervals is going to zero as n goes to in nity, their inter-
section is a single point, which we de ne as k(s),

F1 <r1'<r..>-{i<s>}-

.‘i=° ' I

We summarize the results in the next theorem.

Theorem 10.13. Let ha be the itinerary map for the logistic map G. Let s in E;
be any symbol sequence of L 's and R ’s.

a. The intervals Ig,____,,n_, have length at most 1r2‘"‘1, )\(Ig____,'__l) 3 1r2""1.

b. There is a unique point 2:, = kG(s) in [0,1] such that G5(a:,,) is in I,’ for
all j,

IX

{$8} = n G_'1(Ie,-) = n I?0...s,,_;
_7'=O 'n=O

Moreover, the map k from E3‘ to [0, 1] is continuous but not one to one,
kG(wLRL°°) = kG(wRR.L°°) for any nite string w.

c. The map kc isia semiconjugacy between the shift map o and the logistic

map G; that is, GokG@)=kGoa@)

d. The maps kc and ha are essentially inverses of each other, kGohG(.'c) = a:.
On the other hand, ha okG(s) _= s, except that, for s = wLRL°° where w is a word
oflengthj, G1(kG(s)) = kG(oJ (s)) = kG(LRL°°) = 0.5 and hGokG(s) = wRRL°°.

e. If s is a symbol sequence, which is periodic with period n (i.e., o period-n
point for the shift map 0), then kG(s) is a period-n point for G. If s is a symbol
sequence which is eventually periodic, with period n, then Ic(s) is an eventually
period-n point for G.

f. If y0 is a period-n point for G, then hG(y0) is a period-n point for a. If y0
is an eventually period-n point for G, then hG(y0) is an eventually period-n point
for o.

g. The periodic points for G are dense in [0,1].

h. The logistic map G is topologically transitive on [0,1]. In other words, there
is o point y‘ = lcG(s') in [0, 1] such that the orbit of y‘ by G is dense in [0,1].

10.4. Sensitive Dependence 451

Exercises 10.3

1. Consider the intervals for the quadratic map G. Let 2:9 be the point corre-
sponding to the symbol string LRRLLRRL. Is 2:0 less than or greater than
1/2'? Is G5(:r:0) less than or greater than 1/2?

2. Consider the tripling map f(a:) = 3a: (mod 1). Use the left, center, and right
intervals given by I1, = [0,1/3], Ic = [1/3,2/3], and IR = [2/3,1]. Give all the
intervals for symbol strings of length loss than or equal to three. What is the
order of these intervals on the line?

3. Consider the “saw-toothed map" de ned by

S(:r)= 3:: ifO§:t§%,

2~3:1: if§5a.-5%,

3.1:-2 ir§g1g1.

Use the three symbols L, C, and R, with corresponding intervals I1, = [0,1/3],

Ia = ll/8,2/8]. and In = l%.1l-

a. Give the order in the line of the nine intervals that correspond to strings of
2 symbols (e.g., ICR). (What you are asked to determine is which interval
is the furthest to the left, which interval is the next to the fLn'thest to the
left, and straight through to the interval which is fin'thest to the right.)

b. Consider the interval ICRL. It is a subset of which of the intervals involving
two symbols, I,,,,,'? This interval L0,, is divided into three subintervals
when giving intervals for strings of three symbols. Is I¢R1, the furthest left
of these three intervals, the center interval, or the furthest right of these
three intervals.

4. Give all the sequences in E; that are period-3 points for a. Which of these
sequences are on the same orbit for 0'?

10.4. Sensitive Dependence on Initial Conditions

In this section, we de ne sensitive dependence on initial conditions: the property of
two orbits which start close to each other and move apart. This property is central
to the notion of “chaotic dynamics". We start by considering the doubling map.

Example 10.28. Let D(a:) = 2:: (mod 1) be the doubling map on [0,1]. This
map demonstrates the property of sensitive dependence on initial conditions. Let
so = 1/3 and yo = $0 + 0.0001. Table 3 shows the rst 14 iterates of these points
and the distances between their iterates. Notice that the rst 12 iterates get farther
and farther apart, and by the 12"‘ iterate, they are more than a distance 1/3 apart.
After this nimiber of iterates, they can get closer or farther apart on subsequent
iterates, and their dynamic behavior seems independent of each other. Also, see
Figures 15. When looking at the gures, remember that 2:0 = 1/3 has period 2, so
the iteration keeps repeating; the orbit for yo = 0:0 + 0.0001 wanders off.

452 10. Itineraries

3 $11 Un lyn " Inl

0.3333 0.3334 0.0001

0.6667 0.6669 0.0002

0.3333 0.3337 0.0004

0.6667 0.6675 0.0008

0.3333 0.3349 0.0016

0.6667 0.6699 0.0032

0.3333 0.3397 0.0064

0.6667 0.6795 0.0128

@\1O)U!|bODI\J—1© 0.3333 0.3589 0.0256

9 0.6667 0.7179 0.0512

10 0.3333 0.4357 0.1024

ll 0.6667 0.8715 - 0.2048

12 0.3333 0.7429 ' 0.4096

13 0.6667 0.4859 0.1808

14 0.3333 0.9717 0.6384

'Ihble 3. Sensitive dependence for the doubling map

1!!’1 3 ‘("3 1
A—" $11; yn

0.8

II!!!’ 0.6
0.4

0.2

0- Tl.

i°246810 132 14

Figure 15. Iteration of doubling map showing sensitive dependence on initial
oonditions: :0 = 1/3 and yo = 1:0 +0.0001. (a) Graphical iteration. (b) iterate
1,, and yn as functions of n.

We now proceed to give the formal de nition. The second de nition concerning
sensitive dependence when restricted to an invariant set is used in our discussion
of attractors, especially in higher dimensions.

De nition 10.29. A map f on a space X has sensitive dependence on initial
conditions at points of A C X provided that there exists 1' > 0 such that for every
1:0 in A and for any 6 > 0, there is a yo within 6 of 2:0, (i.e., |y0 — a:o| < 6) and an
iterate It > 0 such that

|fk(!/0) — fk($o)| 2 T-

If the set A is invariant by f, then we say that the map has sensitive dependence
on initial conditions when restricted to A provided that the new point yo C8-fl be

10.4. Sensitive Dependence 453

chosen within the set A; that is, there is an 1' > 0 such that, for any point 2:0 in A
and any 6 > 0, there is a yo in A with ll/0 — a:Q| < 6 and an iterate It Z 0 such that

|f"(yo) — f"( =o)| 2 T-

We will show that D, T, and G each have sensitive dependence on initial
conditions on [0, 1].

The next de nition of expansive diifers from sensitive dependence in that all
pairs of distinct points move apart.

De nition 10.30. A map f is expansive on an invariant subset A in X, provided
that there exists 1' > 0 such that, for any yo and 20 in A with yo 96 1:0, there is
some iterate k Z 0 with

lfkfyol — fk(I0)| Z 1‘-

From the foregoing de nitions, any expansive map on A has sensitive depen-
dence on initial conditions when restricted to A, but the converse is not necessarily
true.

Remark 10.31. A map with sensitive dependence on initial conditions has the
“butter y effect” mentioned in the historical introduction. A small change in initial
conditions can be ampli ed to create large differences. E. Lorenz introduced this
latter term because he interpreted it to mean that a butter y apping its wings in
one part of the world could affect the weather in another part of the world a month
later.

Theorem 10.14. The doubling map D is expansive with expansive constant 1' = 1/3
and so has sensitive dependence on initial conditions at points in the whole domain

[0111 '

Proof. The doubling map is given by D(a:) = 21: — k, where It = 0 for 0 5 :2 < 1/2.
k=1for1/2§a:< 1,andk=2fora:=1.

Let :00 < yo be two points in |0,1], rcj = D1(:|:0), and yj = D5(yo). Let It,-
a.nd Is} be determined by 1:,-+1 = 22:, — Icj and y_,-+1 = 2y, — lf :c_,- < yj are

both in [0,1/2) or both in [1/2,1), then ltj = kg and

1/1+1 - f'=1+1 = 2%" " 1%" - (21% - l°1)= 2041 - $1)-

By induction, yj — :r:_,- = 2-"(yo — $0) for 0 5 j 5 n until either (i) y,, — 22,, Z 1/3 or
(ii) 0 < yn —a:,, < 1/3 and 2,, and yn are not in the same subinterval so kg = It; +1.
In case (i), we are done. In case (ii),

3:n+l - :1/n+1 = 2311 “ kn _ (2911 _ kn -1)

= 1 " 2(1/n _ 3n)

> 1 - 2 (ii - 1-

Since the distance cannot double at each step and stay less than 1/3 forever, even-

tually the distance must get greater than 1/3. El

Theorem 10.15. The tent map T and logistic map G both have sensitive depen-
dence on initial conditions when restricted to the interval [0,1].

454 10. Itineraries

Proof. A similar proof shows that either map has sensitive dependence on initial
conditions, with constant 1' = 1/2. We write the proof for the logistic map, which
in nonlinear.

Let 2:0 be a point in [0,1]. Let hG(a:0) = s be the symbol sequence for 1:0.

Case (i): Assume that s does not end in repeated L’s, hG(:c0) gé s0 . . . s,,L°°,
so $0 does not eventually go to the xed point at 0. There is an arbitrarily large n
for which an = R. Consider the sequence of symbols t, with tj = s,- for 0 5 j < n
and tj = L for j Z n. Let y0 = kG(t) be the point with this symbol sequence. Both
a:0 and y0 are in Ig____,n_1, so |a:0—y0| 5 /\(I§f,,__,,,__,) S "/2"+1 by Theorem 10.13(a).

We can malze this arbitrarily small by taking n large with s,, = R. Taking the nth
iterate, hG(G"(y0)) = L°° so G"‘(y0) = 0, while G"(:r0) is in IR, so G"(a:0) 2 0.5;
therefore, |G"(a:0) — G"(y0)| Z 0.5. This shows that there is a point arbitrarily
close to 1:0 and an iterate such that the distance between G"(:v0) and G"(y0) is
greater than 1/2; that is, G has sensitive dependence on initial conditions.

Case (ii): s = h(:z0) = s0 . . . s,,,_1L°°. Let t = s0. ..s,,_1R°° for large n. 2 m.

Then, |a:0-y0| 5 /\(I_,,,___,,__,) 5 "/2"+1 and G"(:r0) = lc(L°°) = 0, while G"(y0) =

k(R°°) 2 1/2. Again, |G"(:z0) — G"(y0)| Z 0.5. E]

The proof that the logistic and tent maps have sensitive dependence use?sym-
bolic dynamics and did not use a boimd on the derivative directly, as is done for
the doubling map. However, a common feature of the doubling map and the tent
map is that they have absolute value of the derivative bigger than one (i.e., they
stretch intervals). In order to have the interval [0, 1] forward invariant by a map
which stretches, the map must either fold like the tent map or cut like the doubling
map. When we consider chaotic attractors in Section 11.2, we return to this idea
of stretching and folding or stretching and cutting.

Exercises 10.4

1. Let f (I) Q 3 a; (mod 1) be the tripling map. Let $0 < y0 be two distinct points
in [0,1], :|:_,- = fj(:z:0), and y_.,- = fj(y0). Consider the subintervals I0 = [0,1/3),
I1 = [1/3,2/3), and I2 = [2/3,1).
a. If a:,- < y_.,- and both are in the same subinterval I0, then show that
1/1+1 — I.-i+1 = 30/1 - $1")-
b. If :v_,- < yj are in different adjacent intervals with yj —a:,- 5 1/4, then show
that f($jl — ff!/jl=1— 30/1' _ $1‘) Z 1/4-
c. Show that f is expansive with expansive constant 1/4.
d. Find a pair of points whose distance is not tripled by the map.

2. Consider the “saw-toothed map" S(:z:) de ned in Exercise 10.3.3. Show that S
has sensitive dependence on initial conditions.

3. Let p be a xed point for f such that |f'(p)| > l. Prove that f has sensitive
dependence on initial conditions at p.

4. Let p be a period-n point for f such that |(f")'(p)| < 1. Prove that f does not
have sensitive dependence on initial conditions at p.

10.5. Cantor Sets 455

5. Write out the details showing that the logistic map G has sensitive dependence
on initial conditions.

6. Use a computer program to investigate the sensitive dependence for the qua-
dratic map G(:r) = 4a:(1 — zr). How many iterates does it take to get separation
by 0.1 and 0.3 for the two different initial conditions a:0 and :20 + 6, for the
choices 2:0 = 0.1 and 0.48 and 6 = 0.01 and 0.001? (Thus, there are four pairs
of points.)

10.5. Cantor Sets

S0 far we have used symbolic dynamics for maps that have an invariant interval,
with no points leaving the interval. In this section, we consider the case in which
some points leave the interval. The result is a very “thin” invariant set made up
of in nitely many points, but containing no intervals. This set is called a Cantor
set, named after a mathematician who worked around 1900 and introduced many
of the ideas of rigorous set theory into mathematics.

Previously, we have used the tent map with slope 2 quite extensively. In this
section, we consider tent maps with larger slopes to construct the Cantor set.

For 1- > 0, let T, be the tent map of slope 1- given by

T ra: if 2: 5 é,

r($)_ r(1—a;) if:cZ

See Figure 16.

Ts(-'1?)
1

0 IL % Ill
g IR 1

Figure 16. Graph of the tent map T3

456 10. Itineraries

Lemma 10.16. Let r > 2. If TZ(:r0) is bounded for allj 2 0, then TZ(:1:0) G [0, 1]
for all j Z 0.

Proof. If :00 < 0, then T1'1($Q) = rjzo, which goes to minus in nity as j goes to
in nity.

If :20 > 1, then T,(a:0) < 0, and T3-+'(:vo) = Tg-(T3(x0)) = 3-"T3(:vo) goes to

minus in nity as j goes to in nity, by the previous case. El

We consider all the points that stay in the interval [0, 1] for the rst n iterates,

K,,={a::T§(a:)€[0,l]for05jgn}. \

Note that K0 = [0, 1]. P1-orn now on we consider the map T3 of slope 3, but most of
what we do is valid for a.ny T, with 'r > 2. The Sl.l_blI1‘l5€1'V8.lS de ned in the argument
would have to be modi ed for other values of 1'.

If 1/3 < mo < 2/3, then T3(a:o) > 1, so (1/3, 2/3) is not in K1. However,
T3 ([0, 1/3]) = [0,1] and T3 ([2/3, 1]) = [0, 1], so K1 is the union of two intervals

IL = [0,1/3] and Ia = [2/3,1] ,

each of length 1/3. The total length of K1 is 2 (1/3) = 2/3, )\(K1) = 2/3. Again, we
use A to denote the total length of a union of intervals.

KI, K i

_L '7__ __ __K3

Figure 17. The sets K0, ...K3 for the Cantor set

A point 2: in K2 has T3(a:), and 1"§(:r) in [0, 1], so T3(z) is in K1. Also, K2 C K1,
so it has a portion in IL and a portion in IR. Thus, the set K2 is the set of the
points in I1, UIR that map to K1, or

Kg = (IL T3'1(K1)) U (IR T3'1(K1)).

Since T3 is monotonic with an expansion by a. factor of 3 on each interval, IL and
IR, the sets I1, O T3‘l(K1) and IR ('1 T3_1(K1) are each made up of two intervals of
length (1/(5)2:

IL nT3_l(K1) = [011/9] U [2/9,1/3] 1
IR T3_1(K1) = [2/3,7/9] U [B/9,1] -
The part in I1, is obtained by shrinking K1 by a factor of 1/3, and the part in IR is
obtained by shrinking K1 by a. factor of 1/3 a.nd ipping it over. The total set
K2 = [0.1/9] U [2/9,1/a] U [2/3.7/9] U [8/9,1]

is the union of 4 = 22 intervals, each of length 1/9 = (1/3)2; the total length of the
intervals in K2 is (2/3)2, ,\(K2) = (2/3)2.

10.5. Cantor Sets 457

Repeating by induction, we see that each of

I1, T3_1(K,,_1) and

IR n T3_1(Kn—1)

is the union of 2"“ intervals, each of length (1/3) (1/3)n_1 = (1/3)", and

K,, = (IL nT;‘(1<,,_,)) u (11, n T3“(K,,_1))

is the union of 2"“ + 2"‘! = 2" intervals, ealch of length (1/3)"; the total length
of the intervals in K,, is (2/3)", )\(K,,) = (2/3) .

Let

K= nKn={z:T§(z)6[0,1] for05j<oo
n20

The set K is called the middle-third Cantor set, since the middle third of each
interval is removed at each stage. The set K is contained in each of the sets
K,,, so it must have length less than (2/3)" for any n. Therefore, the “length” or
“Lebesgue measure” of K is zero, )\(K) = 0. (See Section 11.4 for a more complete
discussion of Lebesgue measure.)

Just as we did for the tent map T2 in Section 10.3, we can label each of the
intervals in K,, by means of the dynamics of T3 rather than the order on the real
line. For a nite string s0s1 . . . s,,_1, where each s_.,- equals R or L, let

I,,,,,,l___,,__, = {z 2T;-f'(a) G1,’. for0 53' $n— 1}.

In particular, ILL = [0,1/9], ILR = [2/9,1/3], I113 = [2/3,7/9], and Im, = [3/9,1].
See Figure 18. In this case, these intervals are disjoint for different strings of the
the same length, and K,, is the union of all the I,,,_,,___,,,__, over all the choices of
the strings of symbols,

Kn = U{I,o,,___,__, :s_,- G {L,R} for 0 §_j <

We explore some of the properties of this set K. At each stage, the set Kn has
2"“ end points, which we label as E". Thus,

0- I0 n-I IO

E2 ={0| _9 1_91_31 -31 9~1 ‘I 0-4
G<1O‘| %—’

Let E be the union of all these sets; that is,

E = U E,,,

n20

which we call the set of all the end points of the Cantor set K. In the construction,
the part removed from Km to form K,,,+1 comes from the interior of Km, so the
points in En are never removed, so En and E are contained in K. Since each of the
sets E,, is nite, the union E is a countable set of points (i.e., there is a function
from the positive integers onto the set E).

We want to see that there are more points in the Cantor set than just the end
points. One way to see this is to show that the Cantor set is uncountable, while we
have shown that the set of end points is countable. An in nite set S is uncountable

458 10. Itineraries

In --- QII— --- -1.‘- ---r ---

III II I I I III I II II
--|-_- --'--
.-1. .114..- nnnlnncu

I I I I I I III I I III I I

IL —_QQ--u- --"P-__-. ---r--- --_-- --§-- -1---.

ILL ILR Inn IRL

Figure 18. intervals for strings of two symbols for T3

if it is not countable (i.e., there is no function from the positive integers onto the set
S). A standard result in analysis is that the unit interval [0, 1] is uncountable. See
[Ma.r93] or [Wad00]_., In the next theorem, we show that there is a function from
the Cantor set onto the, unit interval [0, 1], so the Cantor set is also uncountable.
Thus, there are many more points in the entire Cantor set than just the end points.

Theorem 10.17. a. The points in the set K are exactly those which can be given
in a ternary expansion using only 0's and 2 ’s, and such an expansion for points in
K is unique. The set K is uncountable.

b. The set E is countable. The points in E have ternary expansions, which
end in either repeated 0 ’s (the e:vpan.sion is nite), or repeated 2 's.

c. The set

K \ E 75 (ll,

and points in this set have ternary expansions with only 0's and 2 ’s, which do not
end in repeated 0 ’s or in repeated 2's.

Proof. (a) Using the ternary expansion, a point a; in [0, 1] can be written as

:n= —8‘*3.
H[.\8'1H

with each a_,- equal to 0, 1, or 2. Notice that

Z§2=aZ2 §F=1 §'12_‘—1 i=§2"a3=1'
J'21 1'20 3

so we can write the number 1 with a.n expansion of the preceding form.

10.5. Cantor Sets 459

The sets K1 and K do not contain the open interval (1/3,2/3), whose points
have a1 = 1 and 1/3 as part of their expansion. All the points in [2/3, 1] start with
a1 = 2, while all those in [0, 1/3) start with a1 = 0. The point 1/3 is in K1 and K,
but it can be written using only 0's and 2’s in its expansion:

:i22 1'20 3

Thus, all the points in K1 and K can be represented with an expansion, with
a1 either 0 or 2. The left end points of the intervals in K1 all have expansions
which end in repeated 0's, while the right end points have expansions which end in
repeated 2’s.

At the second stage,

K2 = K1 \ ((1/9,2/9) U (7/9,3/9)) -

The interval (1/9, 2/9), which contains points that start with the expansion 0/3 + 1/31,
is removed, and the interval (7/9,3/9), which contains points that start with the
expansion 2/3 + 1/32, is removed: both of these intervals have points with a 1 in their
expansions. The points in

[0, 1/9] u [2/3, 7/9]

have O2 = 0, while those in

[2/9,1/s] U [8/9,1]

have G2 = 2. Again, the right end points can all be expressed using repeated 2’s:

5=Z and

3'23 3;

57 = 52 + 2 52-
3'23

Thus, all the points in K2 and K can be represented with an expansion with a2
either 0 or 2. The left end points of the intervals in K2 all have expansions that end
in repeated 0’s, while the right end points have expansions which end in repeated
2's.

At the n‘h stage, all the points which need to have 1/3" in their expansion are
removed. The points at the right end points of Kn all have expansions that end in
repeated 2’s, and the left end points of Kn all have expansions that end in repeated
0's. Thus, any point in K can be given by a ternary expansion that uses only 0’s
and 2's and no l’s. By making these choices, for a point in K, the ternary expansion
is unique.

Now, form the map F from K to [0, 1], de ned by

_ =°°w_*1!Kb.
‘3-.5?
- 5.3.21;/‘H.

This map realizes all possible binary expansions, so F is onto [0,1]. Since [0, 1] is
uncountable, it follows that K is uncountable.

460 10. Itineraries

(b) We have already argued that the set of end points is countable. The proof
of part (a) explains why they have the type of ternary expansions claimed in the
theorem.

(c) Since E is countable, there are uncountably many points in K \ E, and

K \ E 75 @. El

Example 10.32. To nd a point in K \ E, we take a ternary expansion that does
not end in either repeated 0’s or in repeated 2’s. For example, if aj is 0 and for
odd j and aj is 2 for even j, then

:t—'322+32‘ + —-92+922 +

=-29 @+-+971 +~)=- -92(—1 -l¢5 )

2 9 ,_,<o»-

‘O W uh

Therefore, 1/4 is in K but not in E.

Example 10.33. Let us check to see whether the point 9/13 is in K. To nd its
ternary expansion, we multiply by 3 and take the integer part, obtaining,

a3=Z=2+% m=Z
13 13 13 a,=o,

3-i=o+-3, a3 ==0v
13 13

9

3' E13L ==O'+ _1-31

a3=§=2+i, a=a
13 13 13

and the process is beginning to repeat. Therefore, a1+3,- = 2 and (1g+3_-,' = a3, = 0.

Since the expansion uses only 0's and 2's, the point 9/13 is in K. Since the expansion

does not end in repeated 0’s or just repeated 2's, it is not in E.

The two intervals, IL = [0,1/3] and In = [2/3,1], can be used to de ne a map
kT from E; to K,

{kT(5)} = n I90---8""

Since I,,,___,,_ I¢,,___¢,, = 0 if so . . .s.,, 79 to . . .t,,, kT is one to one. We get the next
theorem using this mction and symbolic dynamics as before.

Theorem 10.18. Consider the tent map T3. Use this to de ne the middle-third
Cantor set K.

a. The Cantor set K is an invariant set for T3.
b. lcT : E; —> K is a one-to-one conjugacy from 0 on E; to T3 on K.
c. The periodic points for T3 are dense in K.
d. The map T3 has sensitive dependence on initial conditions when restricted
to K.
e. The map T3 is topologically transitive on K.

170.5. Cantor Sets 461

Length of preimages of intervals

When we turn to the logistic map, we need a comparison of the length of
an interval and its image by a nonlinear map. This lemma is also used to give
an estimate on the length of intervals de ned by a symbol sequence. When the
derivative is always greater than one, this will allow us to conclude that there is a
unique point that realizes the symbol sequence.

Lemma 10.19. Assume that B > 0 and f is a continuously di erentiable map from
a closed interval [a, b] into IR such that | f’ (a:)| Z ,6 for all a < :2 < b. Assume that
J1 is a closed subinterval of [a, b] that is mapped onto the interval J2 = _f(J1).
Then the lengths of J1 and J2 are related by

1

/\(-T1) S B X(J2)-

If f(:c) is linear on [(1, b] with |f’(a:)| = ,6, then /\(J1) = ,1;/\(J2).

Proof. Let 2:1 and 2:2 be the end points of J1. By the mean value theorem, U

f(=v1) - f(==2) = f'(Ia)(I1 — $2),
where $3 is a point between :01 and 2:2. So

/\(J2) = |f(=v1) - f(Iz)| = |f'(Ia)i - I111 — Izl
2 /3|¢1- 4112i = I3/\(-T1),

>~(J1) s §A<J2)-

Logistic family

The results for the tent map T3 go over to the logistic family g,,(a:) = aa:(1 — 2:)
for a > 4. Our proof uses the fact that the absolute value of the derivative is
always greater than one on the Cantor set, which is true for a > 2 + \/5 is 4.24.
Our argument about the Lebesgue measure of the Cantor set being zero uses a >
2 + 2\/2 z 4.83. Both conditions are true for a = 5, so we take g(m) = g5(a:) to
simplify the discussion.

The points that g maps to 1 can be found by solving g(m) = l:

5:1: — 52:2 = 1,

0 = 51’ - 5:: + 1,
xi = €5 1 -,/1—0fa?20 = -21 5: —~1/05-.

We let Ii = [0,:n_] and Ii, = [a:+,1]. The derivative ofg is g'(z) = 5 — 101:, and

y’(-'¢-)=5—5+\/5=\/5>2.
g'(:|:.,.)=5—5—\/5=—\/5< -2.

Since the absolute value of the derivative of g is even larger in IQL U I9R than at the
points xi, |g’(:r)| Z \/5 for all I in IiUI-11,.

462 10. Itineraries
_
95 (iv)

1

\

U’ IL £1I_ 3+ Inx 1

Figure 19. Graph of the logistic map g5

Since g(I§°) = [0, 1] for so = L or R, /\(I§°) < '\(l°'1])/\/5 = 1/A < 1/2 by
Lemma 10.19. Again, each of the intervals HM, maps onto the interval I , by g.
By Lemma 10.19, the length of any I308, satis es

1A<1:,.,>g »\(I£,) s

Continuing by induction, each of the intervals I§0____," is mapped onto the interval
I§,____,,__. By Lemma 10.19, the lengths of the intervals I§°___,,'_ and I§,___,,n satisfy

A<I2...._..> s i<Is,.....> 5 = ( fii.

where the second inequality follows from the induction hypothesis. Let K3, be the
union of the 2" possible I§o___,n_, and Ag = nzo K2,. We can de ne a map
kg : 2;’ —> Ag by

{l¢"(5)} = n I§.,....-»..-

The total length /\(Kf,+1) 3 (2/,/§)n+1 goes to zero as n goes to in nity, so
Ag has Lebesgue measure zero. The set Ag has many of the properties of the
middle-third Cantor set as summarized in the following theorem.

Theorem 10.20. Consider the logistic map g = g,,,' with a Z 2 + 2\/2, and its
invariant set Ag = nn20g""([O,1]).

a. The set Ag is uncountable and has Lebesgue measure equal lo zero.
b. kg : E; —> Ag is a one-to-one conjugacy from a on )3; to g on Ag.
c. The periodic points for g are dense in Ag.

Exercises 10.5 463

d. The map g has sensitive dependence on initial conditions when restricted to
Ag-

e. The map g is topologically transitive on Ag.

In Section 10.8 at the end of the chapter, we show that the sets K for T3 and
Ag for g5 (:0) = 5a:(1 — 1:) have the properties which characterize sets that are called
Cantor sets.

Exercises 10.5

1. A sequence of intervals J ;, is called nested, provided that J1. D J;,+1 for lc Z 1.
a. Give a nested sequence of closed intervals J k such that

Q J,, = 0.

1:21

b. Give a nested sequence of bounded intervals J k such that

n Jk = ll.
#31

2. Let

5:1: :1: § 0.5,

T13) ' T511”) {5 (1 - 1) 1: 3 0.5.

a. Sketch the graph of T.
b. What are the intervals that make up the set of points :1: such that 1: and

T(a:) are both in [0, 1]?
c. Describe the set of points a: such that 1:, T(a:), and T2(:c) are all in [0, 1];

that is, describe the set

K;={:1::Tj(:|:) e [0,1] for05j g 2}.

It is made up of how many intervals of what length each? What is its total
length?
d. Without giving the intervals exactly, how many intervals a.re there, and
what is length of each interval in the set

Kn = {e 1 Ti(a:) e [0,1] for all 0 5;" 5 1.}?

e. Let _
K = {:2 : TJ(:t') G [0,1] for allj Z 0}.

Explain which numbers in [0,1] belong to K in terms of the numbers

expansion base 5, or z = "1/5:".

f. Give a number in K that is not an end point of one of the intervals in the

nite process de ning K.

g. Are the numbers 23/25 and 25/31 in the set K?

3. Consider the logistic map g5(a:) = 5a: (1 — 2:), with intervals I§§___,,", as de ned
in Section 10.5. Show that If,’ and Ifn have different lengths.

4. Consider the function F(:z:) = 60:3 — 5:: on [-1, 1]. The points where F(a:) = 1

are 1 and “1/2 :1: \/5/s. The points where F(:2) = —1 are -1 and 1/2 zh \/5/5.

464 10. Itineraries

a. Sketch the graph of F.
b. Describe the set of points :1: such that both :1: and F(:|:) are in [-1, 1], or

K1 = {:c:Fj(a:)€ [-1.1] forogjgl}

= l—1.11nF“‘(l—1,11>.

It is made up of how many intervals? What is the maximum length of
these intervals?
c. Describe the set of points

K2 = {¢=F1‘(=;) e [-1,1] for0 5;; 5 2}

2

= Q F"(l—1-1l)-

j=0

It is made up of how many intervals? What bound can you put on the
length of each of the intervals in K2? Hint: Use a lower bound on the
absolute value of the derivative on K1 (i.e., IF’ 2 /\ for some A > 1
and all :1: in K1).
d. Tell what bound you can put on the length of one of the intervals in

K,,={a::Fj(a:)€[—1,1]for0§j§n}

= Q F"'(l~1.1l)-
i=0

e. Explain why F has an invariant set that is like a Cantor set.

5. Consider the cotangent function f(rs) = cot(a:) on [0,21r]. Explain why it has
an invariant set tl1at'-‘is like a Cantor set made up of points that stay in [0, 21r]
for all iterates.

6. Let

5:1: + 4 for :1: $ -0.4,
f(a:)= -51: for —0.45:|:50.4,

5a:——4 forO.45:c.

a. Sketch the graph off. Notice that f(—1) = -1, f(—0.4) = 2, f(0.4) = —2,
and f(1) = 1.

b. Consider the sets Kn = {rc : fj(:1:) 6 [—-1,1] for 0 53' 5 n} How many
intervals do K1, K2, and Kn contain? What is the length of each of these
intervals in these sets? What is the total length of Kn?

10.6. Piecewise Expanding Maps and Subshifts

In the preceding three sections, we used symbolic dynamics (itineraries) to nd a
dense orbit in an interval or invariant set and to show that a map has sensitive
dependence on initial conditions. All the cases considered so far have allowed all
possible transitions between intervals. On the other hand, in Section 10.1, we used
transition graphs where only certain transitions were allowed between symbols.
In this section, we combine transition graphs with symbolic dynamics to show
that certain other maps are topologically transitive on an interval. The space

10.6. Expanding Maps 465

of sequences of symbols where there are restrictions on the transitions are called
subshifts of nite type.

In the section on transition graphs for the Sharkovskii theorem, we got the
existence of periodic orbits, but not uniqueness. In this section, we use the as-
sumption that the map has a derivative with absolute value greater than one to get
uniqueness of the point with as given symbol sequence.

Example 10.34. We de ne what is called the shed map by

_ :c+c if0§:z:5c,
f($)_{(1—:1:) ifc§a:§1,

with 5 > 1 and 0 < c < 1/2. See Figure 20. The absolute value of the derivative
of this function is greater than one on two adjacent intervals, with a point of
nondiiferentiability at the point where the intervals meet. To insure that the map
is continuous and takes on the same value 1 at a: = c, we need

Ii c + c = 1 and
(1 — c) ,6 = 1.

For these equations to hold, we need .
1

c=—1+—/3‘

1=<1-cm=(1-

l3+1=(l3+1—1)l3.
0= 2— —l, so

p=Li@~1.61s>1.

For these choices, f is continuous on [0, 1] and differentiable on the two open
intervals (0, c) and (c, 1) with |f’(:|:)| =,B> 1.

Notice that

f(l0.¢l) = l¢-1] and
f(l¢»1l)= [0.1] = [0-Cl U l¢- ll»

so the image of the left interval covers the right interval, and the image of the right
interval covers both the left and the right intervals. Thus, the transition graph
9’ allows either R or L to follow R, but only allows R to follow L, Any sequence
s = sosl .. . that follows these rules is called an allowable symbol sequence. Let E;
be the set of all allowable symbol sequences.

The length of I1, is c and the length of IR is 1 — c, so the length of each of these
intervals is less than or equal to 1 ~ c. Let s = {$1-}§";0 be an allowable sequence
based on the transition graph. The interval I_.,,,___,,_ maps to I_.,,___,,_, so applying
Lemma 10.19 and induction on 11,

)\(Is°...s,,) S _l A(Is1...s.,) S £3-2 /\(Isg...s,,) 5 ' ' ' S _n /\(Is,,) S _n (1 - C)-

466 _ 10. Itineraries

1 _ i.

f(I)

C

Q Q__ HH

Figure 20. Shed map

Thus, for an allowable symbol sequence s = sosl ..., the intersection of these
intervals I_,,°____,,_ for all positive n has length zero and contains a single point Ic(s):

H = {1=<s>}¢ @-

1120

Any orbit of the shed map can be determined by a sequence of symbols that
is allowable in terms of its transition graph because the image of each interval in
the partition is the union of intervals in the partition. Because the union of the
intervals in the partition is not positively invariant for maps like the tent map with
slope greater than two, we express the following and subsequent de nitions in a
way that does not require this condition.

Remember that a function f is continuous on a pa.rtition of closed bounded
intervals {Ii},-'1, provided that (i) int(I,) int(IJ-) = 0 for 1'96 j and (ii) f is
continuous on U12, I,-. (We do not assume that _f is continuous everywhere.) If
the function is one to one on I,-, then f “(I1-) L; is a single interval whenever it
is nonempty.

De nition 10.35. Assume that f is a real-valued function de ned on a domain
Q C_ IR. A Markov partition for f is a partition of closed bounded intervals
{I1,'. . . ,IN } such that (i) f is continuous on the partition, (ii) f is one to one
on each I,-, and (iii) if the image of the interior of one of the Ij by f intersects the
interior of any I,-, then the image f (I,-) goes all the way across L; that is,

if f(mr(1,-)) n mc(I,-) 7* lb, then f(1.) 3 1,-.

Condition (iii) is called the Markov property for the partition.

If a function is continuous on a partition and one to one on each interval, then
it satis es the Markov property if and only if, for each end point e of one of the
intervals, f(e) is not in any of the open intervals int(I,-).

10.6. Expanding Maps 467

A Markov partition for f determines a transition graph as de ned in Section
10.1 with an edge from i to j if and only if f(I,-) 3 I,-. In this case, Theorem
10.22(c) proves that all the points that stay in the union of the intervals of the
partition a.re coded by allowable sequences.

We have seen several examples of Markov partitions. (1) The tent map T2
and the logistic map G = g4 each has {[0,0.5], [0.5,1]} as a Markov parti-
tion. (2) The tent map T3 has Markov partition {[0,1/3], [2/3,1] However,

{ [0,2/5] , [3/5,1] } and { [0, 1/2] , [1/2, 1] } are two other Markov partitions for T3.
(3) The shed map given in Example 10.34 has {[0, c], [c, 1] } as a Markov partition.

De nition 10.36. Given a Markov partition {Ij for f, we say that f is

expanding on the Markov partition with expanding factor ,6 > 1 provided that f

is differentiable on each open interval int(I,-) and | f’ (a:)| Z ,6 for all :1: in int(IJ-).

If a function is expanding on a Markov partition, then we can show that each
allowable sequence of intervals has a unique point with that itinerary. See Theorem
10.22(c).

The shed map and functions in the homework for this section satisfy the
stronger requirements of the following de nition that imply they are are expanding
on their Markov partition.

De nition 10.37. A map f from 1R to itself is called a continuous piecewise ex-
panding on an interval [a, b] provided that it is continuous on [a, b] and there are a

nite set of points

0.=P|)<P1<"'<PN=b

and aconstant [3 > 1 such that |f’(:z:)] 2 for p_,-_1< 1: < pj and forj =1,...,N.
The number ,6 is called the expanding factor or stretching factor. In all of our
examples, the derivative is a constant on each of the intervals (p,-_|,p,-), so =
min{ |f'(1:_,-)| }, where a:_,- is any point in (p,-_1,pj).

Given the points {p0,p1, . . . , p,v }, we denote the induced closed intervals by
Ij = [pa-_1,pj] {p0,p1,...,pN }, and int(I)) = (p,-_1,p,-) the corresponding open
intervals on which f is differentiable. We call either the points {p0, . . . , pN } or the
intervals {I,- the partition for the expanding map.

Note that, for a continuous piecewise expanding map f, the partition {I_,- }_§-"=1
automatically satis es conditions (i) and (ii) of a Markov partition. Therefore, we
only need to check the images f(p,-) of the points in the partition.

The fact that f is continuous means that the limits of f from the two sides
of the p_,- give the same value, and there are not any jumps at the p_,- points. The
tent maps T2 and T3 are examples of continuous piecewise expanding maps. The
doubling map is piecewise expanding but not continuous. Much of what we say
in this sections applies when f is allowed to have discontinuous jumps at the p,-‘s:
We say a few words about this at the end of the section, but do not state a formal
theorem that applies to these functions. The logistic map g5 is not piecewise
expanding because g§(0.5) = 0, but it is expanding on its natural Markov partition,

{l0.rr-l- l=v+.1l}-

468 10. Itineraries

Example 10.38. Let f (IE) be de ned by

+ for 0 5 2: 3 1-,
f($) = wo:hiln=- — Oil|=~waHs H for i $ 1: 5 1.

Then. 1(0) = 2/3. P(O) = r(%) = 4/3=. /3(0) = r(4/an = 2%». and mo) = i/31"

for some integer i that does not have 3 as a divisor. All of the points f7(0) would
have to be end points of any Markov partition. Since this set of points does not
repeat and so is in nite, there cannot be a nite set of dividing points for a Markov
pa.rtition (i.e., f does not have a Markov partition of all of [0, 1]). This continuous
piecewise expanding map does not have a Markov partition of [0, 1] even though
the slopes are rational.

De nition 10.39. Let Q be a transition graph'.with N vertices S. A sequence s
in 2}} is called allowable provided that there is an edge in 9’ from s_.,- to s,-+1 for
all j Z 0. Let E; be the subset of EX, of all allowable sequences. We refer to E;
as the shift space. We denote by Ug the restriction of a to 2;. This shift map
Ug takes E; to itself, so it is is called a subshifl; it is called a subshift of nite
type because the rules for allowable sequences can be expressed using a. transition
graph with a nite number of edges.

We next de ne the properties of the transition graph that are needed to imply
topological transitivity.

De nition 10.40. A transition graph fl is called reducible provided that there is
some pair of vertices i and j such that there is no ( nite) path in ff from the i"‘
vertex to the jth vertex. (If i = j, then it would mean that there is no path in the
graph that leaves the vertex and returns again to the vertex.) A transition graph 9
is called irreducible provided that it is not reducible (i.e., for every pair of vertices
i and j, including the case of i = j, there is some path in g h'om the i‘h vertex to
the j"h vertex).

The transition graph for the shed map in Example 10.34 is irreducible.

Example 10.41. The transition graph given in Figure 21(a) is reducible: there is
no transition from vertices 3 or 4 back to vertices 1 or 2. A piecewise expanding
map with this transition graph is the function given in Figure 21(b):

f(-1') = 2-2:: for0§:z:§1,
4a:—4 for1§a:§2,
8-21: for2§a:$3,
—4+2a: for3§rc§4.

A function with a reducible transition graph could not possibly be topologically
transitive on the union of all the intervals.

Theorem 10.21. Assume that if is an irreducible tmnsition graph with N Z 2
vertices. Let E; be the set of all allowable sequences for the transition graph if
and Ug be the restriction of the shift map o to 2;.

a. Then cry is topologically transitive on 2;.

“<@>\“<@r W<>@ @1» .NIIII10.6. Expanding Maps 469

Figure 21. Reducible transition graph for Example 10.41

b. If in addition if has some vertex that has more than one edge going out,
then Ug has sensitive dependence on initial conditions on 2;.

Proof. (a) We need to de ne a symbol sequence s‘ which realizes all allowable
string of all lengths. Let w,-j be a nite string of symbols such that iw,,-j is an
allowable string, starting at i and ending at j. If j can follow i, then wij can be taken
to be the empty string. Now, let s‘ be the in nite sequence described as follows.
Write down 1, followed by ww, followed by 2, continuing until wi,_1;,, followed by
lc. This gives all the strings of length one. Next, write down all allowable strings
with two symbols, with a transition between them that makes the total sequence
allowable. Continue, by induction, writing down all the allowable strings with n
symbols, with a transition between them, to make the total sequence allowable. In
this way s‘ has all possible nite allowable strings. By the proof as before, s‘ has
a dense orbit in 2g.

(b) There is a periodic orbit that goes through the vertex with more than one

edge going out. This periodic orbit can be used in a way similar to the xed point

0 in the proof of sensitive dependence for the tent map. We leave the details to the

reader. El

The next theorem summarizes the result for a function that is expanding on
a Markov partition, which includes the case of a continuous piecewise expanding
map. This result contains earlier results about the tent map, logistic map, and tent
maps with invariant Cantor sets. It does not apply to maps such as the doubling
map that are not continuous.

Theorem 10.22. Assume that f is expanding on a Markov partition {I_,-
with N Z 2 and ea.-panding factor > 1. Note that f must be continuous on
X = U;-vzl I,-. Let L = ma.x{ )r(I_,-) : 1 53' 5 N} be the maarimum of the length of
the intervals. Let ff be the transition graph induced by the partition, and 2; and
cg be the associated subshift of nite type.

8.

A,= f-"(x)={¢ex=f1'(e)ex foralljZ0}

1120

470 10. Itineraries

is a nonempty invariant set for f . Note that if f (X) = X is invariant, then
AI = X.

b. Any nite allowable string so . . . s,, of symbols corresponds to a nonempty
closed interval I_,,,___,,,_ of length, at most, L,B‘".

c. An in nite allowable sequence s in 2; corresponds to a unique point 2:0 =
Ic(s) in IR such that f-’(a:Q) is in the interval 1,1. for all j 2 0, {k(s)} =

nnzo Is°...s., -

Moreover, the map 'k from 2; to A; C X C IR is continuous and onto A1-.
It is a semiconjugacy between the shift map Ug on 2; and f on Af.

d. If s is an allowable period-n sequence, then k:(s) is a period-n point for f. If

s is an allowable eventually.period-n sequence, then k(s) is an eventually period-n

point for f. If k(s) is neither periodic nor eventually periodic, then s is neither

periodic nor eventually periodic. .

e. Assume that the transition graph 9 is irreducible and has some vertea: that
has more than one edge going out. (This latter assumption is automatic if the
union of the intervals X is invariant.) Then the map f has sensitive dependence
on initial conditions when restricted to A f.

f. If the transition graph 9 is irreducible, then f is topologically transitive on
A ;.

Idea of Proof. Parts (a), (b), (c), and (d) follow as before. The analysis of
Example 10.34 actually proves parts (b) and (c). Note that the points in A ; are
exactly those points that stay in X for all iterates and so have an in nite itinerary
de ned. For a map likethe tent map T, with r > 2, Af is a Cantor set.

Parts (e) and (f) follow from Theorem 10.21 since k is a semiconjugacy from

2; onto AI: :0‘ = k(s‘) has a dense orbit in AI. El

Remark 10.42. The point :22‘ passes through all intervals I_,,,___,,n. For this dis-
cussion, assume that all transitions are allowable. For some string of 100 iterates,
fj (a:") is in interval I1. Later, it passes through the string of intervals I1, I2, ...,
IN a million times. Thus, the future behavior of the orbit is not at all predictable
on the basis of the past and current location. This feature justi es calling this orbit
chaotic.

Remark 10.43. Sometimes, we want to consider discontinuous expanding maps,
such as the doubling map. In this case, all allowable sequences correspond to
orbits except possibly, those sequences that would give the end points on one of
the subintervals. For the doubling map, if we take 1/2 in [1/2, 1] = I1, then 0 =

D (1/2) = D-7 (1/2) is in Io for all j Z 1. Thus an orbit that goes through the point

1/2 ends in a symbol sequence 10°°. On the other hand, the symbol sequence 01°”
should correspond to a point p with p 5 1/2 and D1'(1/2) = 1 for all j Z 1. Since
such an orbit does not exist, there is no point that has a symbol sequence ending
in 01°°. (If we used D(1/2) = 1 and D(1) = 1, then we do have a point for the
symbol sequence 01°°, but none with the symbol sequence 10°°.) The point is that
although symbolic dynamics could be used to analyze the doubling map or other

10. 6. Expanding Maps 471

discontinuous maps which are piecewise expanding, the corresponding shift space
is not of nite type and the preceding theorem does not apply directly.

10.6.1. Counting Periodic Points for Subshifts of Finite Type. In Section
10.1, we used transition graphs to determine periodic points. In this section, we in-
troduce the transition matrix which makes it easier to count the number of periodic
points for a subshift of nite type.

De nition 10.44. Assume that we start with a transition graph if with N vertices,
labeled 1, ..., N. The associated transition matria: for the subshift Eg is the
N x N matrix T = (t,-J-) of 0's and 1’s, in which an entry t,-j = 1 whenever there
is a transition from vertex i to vertex j, and t,-_,~ = 0 whenever this transition is
not allowed. Let ET be the set of sequences of symbols s = {s;,}@0, where each
transition s;,sk+1 is allowed, t,,,_,,+, = 1. By the de nitions, ET = Eg. Let a-1~ be
the restriction of the shift map to ET. The space ET with the map my is ca.lled
a subshzft of nite type for the matrix T. Another term for ET is is a topological
Markov chain

If we start with the matrix, and not with the transition graph, then we need
to specify the properties it must have to call it a transition matrix. A transition
matrix T = (t,-J-) must satisfy the following properties: (i) It is a square N x N
matrix, where each entry t;_,- is either 0 or 1. (ii) For i xed between 1 and N, the
sum on the row 21- ti; Z 1. (This means that it is possible to go to some other
symbol from the symbol Notice, that if zit” = 0, then there is no way to get
back to the symbol j.

A nite string of symbols w = sk . . .s,,+,,, is called an allowable word or an
allowable string provided that t,,._.,J.+, = 1 for j = lc ...k +m — 1.

The trace of T is the sum of the entiies down the diagonal, tr(T) = tn + - - - +
t NN. For a transition matrix, the trace gives the number of symbols that can follow
themselves (i.e., the symbols s for which s°° is an allowable sequence of symbols).
These sequences correspond to xed points of the shift map. Therefore, the trace
of T gives the number of xed points of the shift map 0-r. The next theorem gives
the corresponding result for higher powers.

Theorem 10.23. Let T be a transition mat1-ix, and let ET be the corresponding
shift space, with shift map cr-1-. Then, the number of red points of all‘; on ET,
N(Ic), equals the trace of Tl‘, tr(T'°).

This theorem follows from the next lemma about the number of words of length
Ic + 1 that start at symbol i and end at symbol j.

Lemma 10.24. Assume that the ij-entry of T'° is p, (T'°)iJ. = p. Then, there are
p allowable strings of symbols of length lc + 1, starting at i and ending at j e.,
strings of the farm is1s2...sk_1j).

Proof. We prove the result by induction on k. Let N (lc; i, j) be the number of
strings of symbols of length lc + 1 starting at i and ending at j .

The result is true for Ic = 1, since t,-j = (T),-j equals 0 or 1, depending on
whether the transition from i to j is allowed or not (i.e., whether the string ij is
allowed or not), (T),-j = N(1;i,j).

472 10. Itineraries

Assume that the result is true for Ic — 1, for all choices of i and j. Then, by
matrix multiplication,

(Tklij = (Tk—1T)ij = 2(Tk_1);mt"1.1' = 2 (Tk-Ila-n"

By the induction hypothesis, we see that (T'°‘1),m is the number of allowable
strings, starting at i and ending at m, which is the same as the number of allowable
strings of length lc + 1 that start at i and end in j and are equal to m in the
next-to-last position. By adding up all of these for which it is possible to make a
transition from m toj (i.e., for which tmj = 1), we get N(k:;i,j), or

(T'°),j = 2 N(k —1;i,m) = N(lc;i,j).

t,,,,=1 I

This completes the induction step, the proof of the lemma, and the proof of the

theorem. El

Example 10.45. Consider the transition matrix

T = (10 11).

Then, the sequence of N (k) starts as = 3.

1v(1) = tr(T) =1 and N(2) = tr

We leave it to the exercises to verify that

:- tr(T") = tr(Tk“2) + tr(T"") or

'-N(k) = N(lc—-2) +N(k -1).

Thus, N(3) = 1+3 = 4, N(4) = 3+4 = 7, and N(5) = 4+7 = 11. The

recurrence relationship N (lc) = N (k —- 2) + N(lc - 1) for a sequence of integers is
called the Fibonacci recurrence relation. The usual sequence starts with N (1) = 1
and N(2) = 1, but our sequence starts with N(1) = 1 and N(2) = 3.

Since #Fix(o-'f.) = N(lc), #Per(k,a-1-) equals #Fix(al‘r) minus the points
of lower period, and the number of orbits of period-k satis es #0Ib(k,UT) =

, Table 4 gives the number of period-It points and orbits up to period

SGVBII. .

There are certain types of transition matrices that we want to distinguish.

De nition 10.46. A transition matrix T is a permutation matria: provided that
Zj tij = 1 for all i and Z,-t,-j = 1 for all j (i.e., ea/ch symbol i goes to a unique
symbol j and there is a unique symbol i which goes to each other symbol j).

A transition matrix T is called reducible provided that there is a pair (i, j)

with 1 5 11,1‘ s n such that (T'=).,- = 0 for all k 3 1. (This means that there

is no way to get from the symbol i to the symbol j.) A transition matrix T is
called irreducible provided that it is not reducible; that is, for each pair (i, j) with
1 3 i, j § n, there is an integer k that depends on the pair such that (T")ij > 0
(i.e., there is some allowable sequence from the symbol i to the symbol j). Notice

Exercises 10.6 473

# Fix(a.'§.) Lower Period # Per(k, 0'1-) #Orb(k, o-r)
1
1 |-lCnr—*¢.0|-Ir—*O 1 1
3 2
4 3 ihto n-Ir-1
7 4
11 10
18 12
P\IO>U\|r>0Ot\Jr—~ 29 28

Table 4. Number of Periodic Points for 01- in Example 10.45

that, a transition matrix is irreducible if and only if its associated transition graph
is irreducible.

Theorem 10.25. For a transition main}: T, the following two conditions are equiv-
alent:

i. The matria: T is irreducible.
ii. The shift map 01- has a dense forward orbit in 2?; and Z, ti, Z 1 for each j.

Proof. First, we assume condition (i) and prove that condition (ii) follows. For a
xed j, since it is possible to get from any symbol to j, we must have Z, t,-j Z 1.
The proof that 0'-r has a dense forward orbit in 2-} is essentially the same as that
of 10.21(f) so we do not repeat it. Thus, we have shown that (i) implies (ii).

Now assume that condition (ii) is satis ed. Take an arbitrary pair (i, j). Let
a be a sequence with ao = i. Since it is -possible to reach j from another symbol,
there is a sequence b with bm = j , for some m Z 1. The orbit of s‘ comes close
to both a and b. Thus, there is some kl 2 0 such that the rst symbol of oh‘ (s)
is i, so sf“ = i. Because, the orbit comes near to b, there is a kg > kl such that

the rst symbol of 0"” (s) is j, so .9}: = j . Thus, 8;, ...sf,2 is an allowable string

that starts with i a.nd ends with j. This shows that (T'°"'“),_,- 96 0. Since (i, j) is

arbitrary, it follows that T is irreducible and condition (i) is satis ed. Cl

Exercises 10.6

1. Consider the continuous piecewise expanding map

§+ 2 1: if 0 g 1 g §,
f(I)= 2-—3a: if§5:c5 §,

2¢_§ if§§:cgl.

a. Draw a graph of f (i.e., draw f(ac) versus 2:, not the transition graph).
b. What is a Markov partition for f? What is the expanding factor for f on

[0,1]-'2
c. What is the transition graph for the Markov partition?

474 10. Itineraries

d. Why does f have a point with a dense orbit in [0,1]?
e. How many periodic points does f have of each period, considering the

periods up to and including 6?

2. Consider the discontinuous map

§ + 22: if 0 3 a: < 51;,
g(:r)= 3:i:—1 if§$a:§%,

§—2a: if§ 51:31.

a. Draw the graph of g (i.e., draw g(m) versus rs, not the transition graph).
b. What is a Markov partition for g? What is the expanding factor for g on

[9, 1]?
c. What is the transition graph Q for the»l\/Iarkov partition?
d. Let Eg be the set of allowable symbol sequences in terms of the transition

graph. Let h be the itinerary function for g. \ Since the function g is
discontinuous, it is possible that an allowable symbol sequence s in {lg
does not correspond to an orbit of g (i.e., s is not equal to h(:z:) for any :1:
in [0,1]). What is the set of allowable symbol sequences s in {lg for which
s 96 h(:c) for all a: in [0,1]?

3. Consider the continuous piecewise expanding map

f($)_ \/5:1: for0$a:5_1/,/5,

2—\/§a:for1/~/§5a:§1.

a. Find a Markov partition for the function on the interval [0, 1]. Hint: Take
enough iterates of the points 0. 1/\/5. and 1 until they start repeating.

b. Given the transition graph for the Markov partition.

4. Show that the function

f(x)={§+§:v, for0$a:§§,
§(1—:v) forg zc l

does not have a Markov partition.

5. Consider the transition matrix
110

T=O1l.
101

How many periodic points does my have for each period, considering the periods
up to and including 4?

6. Let

T ~ (10 11)

and
Th: at bk)
<¢k d/=

a. For a vector (:cQ,y0), let (:r|,,yk) = (z0,yo)T". Talcing y_1 = 1:0, Show
that :1:;,+1 = yk and yk+1 = yk + y;,_1. (Thus, the yi, satisfy the Fibonacci
recurrence relation.)

10. 7. Applications 475

b. Use the fact that (1,0)T" = (a;,,b;,) and (0,1)T" = (c|,,d;,) to show that
ak = a.;,_1+a;,_g and dk = d;,_1+d,,_2 (i.e., they also satisfy the Fibonacci
recurrence relation).

c. Show that tr(T") = tr(T'°_2) + tr(T"‘1).

7. Using symbolic dynamics, show that the shed map of Example 10.34 has sen-
sitive dependence on initial conditions. Do not quote the theorem, but give an
explicit argument.

10.7. Applications

10.7.1. Newton Map with Nonconvergent Orbits. In this subsection, we re-
turn to considering the Newton map Nf for nding roots of polynomials. We show
by the methods of Section 10.6, that this map can have many nonperiodic orbits
that do not converge to the xed points of Nf (i.e., do not converge to a zero of f).
The invariant set given by the subshift of nite type has zero Lebesgue measure
(“length”), so still most orbits converge to one of the xed points. These Newton
maps are not continuous at all points, since they have vertical asymptotes. How-
ever, we are still able to use symbolic dynamics to show the existence of points that
do not converge to xed points or periodic orbits.

Theorem 10.26. Let f be a polynomial of degree d Z 4 with all real roots that
are distinct. Then, the associated Newton map N_,-(rs) has many orbits that do not
converge to xed points and are not eventually periodic. In fact, the orbits can be
speci ed by symbolic dynamics of subshift of nite type on 2d — 4 symbols.

Proof. We have assumed that the roots a.re all real and distinct, so

f($) = A (1 ~ =Fo)(1"- I1) ' ' ' (1' - 114-1),

with 1:0 < 1:1 < < a;d_1. For simplicity, assume that the coei cient of zvd, A = 1.
By Rolle’s theorem, between successive zeroes of f(:12), there is a zero of f' (:i:): so
there are

Ilij < 0,3‘ < Ij+1 “flth f'((Zj) =0 fOl.' 0 s (1-2.

Because f’(m) has degree d — 1, the aj are exactly the zeros of f’ Similarly, by
Rolle’s theorem applied to f'(:i:), between successive zeroes of f'(a:) there is a zero
of f”(:|:): so there are

(1j_1< bj < 0]" With f”(b_7') =0 f0l.' 1 s d—2.

Again, the zeros of f”(a:) are exactly bj for 1 j 5 d — 2.

Let N (:i:) = N;(a:) be the Newton map associated with f. The map N(:12) and
N’(1:) are

N(a:) -_ a: _ ff(,<('I=)) and

,,,“, = 1 _ /'(=>* - re>r"<¢) Z /(¢>r"<:=>_

f’(w)’ f’(I)’

The Newton map has xed points at the as; and vertical asymptotes at the values

a_,- for 0 3 j 3 d — 2. Since only f'(:|:) changes sign as :1: crosses one of the aj, the

map N goes to oo on one side of Gj and to —oo on the other side. Since we a.re

476 10. Itineraries

assuming the leading coef cient A = 1, N (a:) goes to oo for on the right side of
a,,_¢. Thus, N(2) goes to —oo on the left side of ad_2. This pattern has to repeat
at each of the a_,-.

The tangent line for N (:i:) is horizontal at the points a:,- for 0 3 j 3 d — 1, and
at bj, for 1 3 j 3 d — 2. In particular, the xed points of N are superattracting,
as we noted in the preceding chapter. Thus, there are two critical points 11:; and
bj between the vertical asymptotes at a,-_1 and a_,- for 1 3 j 3 d — 2. To make
the argument simpler, we assume that f (ar) and f”(:c) do not have any common
roots, so :|:_,- 96 b_,-. The order of these critical points :v_,- and bj can vary. In Figure
22, we sketch the graph of N (2) a. polynomial of degree four for which b1 < ml
and 2:2 < bg.

'10 91 I12

........................................... .. a2

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ 02 --- ---------------------n q --aI n--. £2+

---------------------------- -- 1,, = C;

______ _____ _ _ _ _ _ _ _ _ _ _ _ _ _ __ I2 = C; ---E .......-.-.--..~-.-.-- .---E-.-.. gg
;
---- -- |l |I| I1||| ||I|II| I| :1-_c,+
5
---- -- -- I>1=c;I-1 I| II I| I| II I| II I|
0.0 I
I|III| I |||

----r --_4-_-,.- - - - - 1- |.-:.- :3-la
, _||: J4
--.5 ...................... “gt...-..} . - . . . - .. an
J1
-"+ +- +

3/2 Z2 G2 22 yi

Figure 22. Intervals for the Newton map of polynomial of degree four. a.
Plot from below ao to above ag. b. Enlargement of plot between a1 and 02.

For each of the intervals [a,~,aJ-+1], we de ne two subintervals to use for the
symbolic dynamics. We start by letting yf, for 1 3 j 3 d — 2, be chosen so that

N(v,~+) = "0,

N(yJ-_) = ad_2, and

a_,-_1 <3/J7 < yr < a,-.

If we were to use symbolic dynamics based on these d — 2 intervals [y_,-',y;L] f0I‘
1 3 j 3 d — 2, the result is the full shift on d — 2 symbols. Many of the orbits are
not xed. Since these intervals contain the superattracting xed points raj and we
want orbits which stay away from the xed points, we remove open intervals about
:v_,- which lie in their basin of attractions. By this process, we get 2d — 4 closed
intervals without any xed points for N. The resulting 2d — 4 intervals induce a
subshift of nite type, which is not a permutation for lc Z 4.

10. 7. Applications 477

We are assuming that f (az) and f” (1) do not have any common roots, so
1173' 75 bj. Let

c; = min{a:,~, bj} and

C; = Hl3.X{£lIj, bj},

so c; < cf. Let zjl and 237 be chosen such that

N(z_;*) = c;',
N(zJ-') = cg, and
yJ-‘<2;-"<cJ7 §$j§C;-<27-_<y;-.

See Figure 22. For 1 3j 3 d — 2, let

J2,-_1 = [yj',zJ?L] and
-721 =11,-‘.1/,*l,

so my in not in J2,~_1 nor in J2,-. By the choices,

N(J2j_1)=[C;,Gd_2lDlZJT,(.'td-213 J25 U - - - U-T44-4 and
N(J2,-) = [a.0,cJT] D [ag,z_;*] D J1 U"‘UJ2j_1.

For d = 5, the transition matrix is

I-OI—IO—*O I-*©l—*OI—‘ r—IO—~O|-I n—IO.O>—IO|-I O|—IOI—IO|-I

This irreducible transition matrix de nes a transitive subshift of nite type. By the

construction, the intervals Jj do not contain any of the xed points of N. Using

these 2d — 4 intervals, we get symbolic dynamics of orbits whose itineraries pass

through these intervals. Each point that corresponds to one of these sequences of

intervals has an orbit that does not converge to a zero or f(:i:). In fact, for the

case considered, in which the zeros are distinct, it can be shown that each symbol

sequence corresponds to exactly one point, but we do not give those details. For

d = 3, all we get is one period-2 orbit, so we assume d Z 4. El

Remark 10.47. For more details about the complicated dynamics of Newton
maps, see [Sa.a84] or [Hur84].

10.7.2. Complicated Dynamics for Population Growth Models. I.n Section
9.7.2, we discussed the xed points for several population growth models. In this
section, we indicate that these models can exhibit more complicated dynamics. The
reader could also see the book by R. May [May75] or the more recent book [Bra01]
by Brauer and Castillo—Chavez.

The population growth model given by Ricker is

R,,_;,(a:) = a av e"“’

478 10. Itineraries

for both a and b positive. The most interesting range of parameters is for a > 1. For
a m 22.2 and b = 1, the map exhibits a period-3 orbit for :20 z 0.05. See Figure 23.
In fact, by Proposition 10.4(b), all we need is that I-i’§2_2,1(0.05) 5 0.05 to imply that
R22_2_1(:|:) must have periodic points of all periods. The dynamics is not as simple as
that discussed in Section 9.7.2; in particular, the population does not always tend
to a constant level. All of these periodic points could be unstable and the total set
could be a Cantor set, so it does not follow that a randomly chosen point exhibits
complicated dynamics. However, Figure 23 is the plot of a single orbit starting at
:00 = 0.1, and it seems to ll much of the interval from 0 up to 8.1, indicating that
there is an invariant set with complicated dynamics. This occurs even though the
absolute value of the derivative is not always greater than one: in this sense, this
example is more like the logistic map G than the piecewise expanding maps. The
next chapter, where we discuss chaotic -attractors, describes much more carefully
the types of complicated dynamics necessary to call an invariant set chaotic.

R22.2,1(-Tl

8.

6/

4-

A2

I

0 24 6 s

Figure 23. Period-3 orbit for a, z 22.2, b = 1, and 1:9 z 0.05.

Exercises 10.7

1. Let p(:|:) = (:1: — a:o)(:|: — :c1)(a: —~ :v2)(a: — :c3)(z — $4) be a polynomial of degree
ve, with all real roots that are distinct and with

:l2g<;'C1<;'C2<I3<J35.

Describe the symbolic dynamics that results from picking intervals in a manner
analogous to those used in the proof of Theorem 10.26 for four real roots.
2. Consider the polynomial p(:z:) = 0:2 + 1, which has no real zeros.

a. Determine the Newton map N,,(a:) and plot its graph.
b. Let D(:1:) = 2a: (mod 1) be the doubling map and C(:c) = cot(1ra:) be the

cotangent function, which takes the interval (0,1) onto all of IR. Show that
C(D(a:)) = N,(C(:c)), so C is a conjugacy of D and Np. Hint: Use the