Lecture 10
Proof of Li-Yorke Theorem
Relevant section of textbook by Gulick:
Section 1.7, ‘‘Period-3 Points’’, pp.
61-64
We now use the results of the previous two lectures to prove the celebrated Li-Yorke Theorem.
The material presented below closely follows the proof in Gulick’s book (Theorem 1.19), but some
additional explanations are provided here and there.
Theorem: (Li-Yorke) Suppose that f is continuous on the closed interval J, with f (J) ⊂ J. If f
has a period-3 point, then f has points of all other periods, i.e., of period-n, n = 1, 2, · · ·.
Proof: As before, without loss of generality, we assume that f (a) = b, f (b) = c and f (c) = a, with
a < b < c. From Lemma 4 of the previous lecture, f has points of period 1 and 2. It remains to show
that f has points of period n for n > 3.
Here is the main idea of the proof: It will be shown that there is a point p ∈ (b, c) such that
1. f k (p) lies in [b, c] for k = 1, 2, · · · , n − 2.
2. f [ n − 1](p) lies in (a, b).
3. f n (p) = p and lies in (b, c).
This proves that p is a period-n point.
Let
J0 = [b, c] .
Since f (b) = c and f (c) = a, it follows that
[a, c] ⊆ f ([b, c]) .
Since
J0 = [b, c] ⊆ [a, c] ,
90
it follows that
[b, c] ⊆ f ([b, c])
or
J0 ⊆ f (J0 ) .
(1)
From Lemma 2 (previous lecture), it follows that there is a closed interval J1 ⊆ J0 such that
f (J1 ) = J0 .
This implies that
f 2 (J1 ) = f (f (J1 )) = f (J0 ) .
But from (1), J0 ⊆ f (J0 ) which implies that
J0 ⊆ f 2 (J1 ) .
We use Lemma 2 once again to conclude that there exists an interval J2 ⊆ J1 such that
f 2 (J2 ) = J0 .
We repeat this idea one more time so that the reader gets the idea of the iterative process that is
going to be performed. Consider
f 3 (J2 ) = f (f 2 (J2 )) = f (J0 ) .
Once again we use (1) to conclude that
J0 ⊆ f 3 (J2 ) .
Once again, we use Lemma 2 to conclude that there exists an interval J3 ⊆ J2 such that
f 3 (J3 ) = J0 .
The reader should be able to see the pattern. We continue the process to obtain a nested sequence of
closed intervals,
[b, c] = J0 ⊇ J1 ⊇ J2 ⊇ · · · ⊇ Jn−2
(2)
f k (Jk ) = J0 = [b, c] , for k = 1, 2, · · · , n − 2 .
(3)
which obey the properties
Note that the procedure has been stopped at Jn−2 .
91
Aside: The above construction will allow Step 1 in the list presented at the beginning of this proof
to be accomplished. We now have to “move over” to the interval [a, b].
The final step in the above recursive procedure produced subinterval Jn−2 which satisfies
f [n−2] (Jn−2 ) = [b, c] .
Let us now apply the function f to both sides of the above equation:
f [n−1] (Jn−2 ) = f (f [n−2](Jn−2 )) = f ([b, c]) ⊇ [a, c] ⊇ [a, b] .
Once again invoking Lemma 2, we conclude that there is a closed interval Jn−1 ⊆ Jn−2 such that
[a, b] = f [n−1](Jn−1 ) .
(4)
From the above relation and the nested relations in (2), we have
f n (Jn−1 ) = f (f [n−1](Jn−1 )) = f ([a, b]) ⊇ [b, c] ⊇ Jn−2 ⊇ Jn−1 .
(5)
We now use Lemma 3 of the previous lecture to conclude that there exists a point p ∈ Jn−1 , and
therefore in [b, c], that is a fixed point of f n .
The question still remains: Is p a period-n point? After all, a fixed point x̄ of f is also a fixed point
of f n . We now show that this is not possible. It turns out that all doubts about p being a bona fide
period-n point (i.e., a member of a set of distinct points (p1 = p, p2 , · · · , pn )) can be eliminated as
follows.
First of all p ∈ Jn−1 . But from (5), Jn−1 ⊆ Jn−2 which implies that p ∈ Jn−2 . From Eq. (3), it
follows that
f k (p) ∈ [b, c] for k = 0, 1, 2, · · · , n − 2 .
(6)
This means that if we start at p and then apply f to it iteratively, the first n − 2 iterates f k (p) remain
in [b, c]. And then, from Eq. (4), f [n−1](p) lies in [a, b].
We now show that, in fact, f [n−1] (p) lies in [a, b), i.e., b is avoided. If it were true that f [n−1] (p) = b,
then
p = f n (p) = f (f [n−1] (p) = f (b) = c ,
92
which implies that
f (p) = f (c) = a .
But recall from (6) that f (p) ∈ [b, c] which implies that f (p) 6= a. Therefore f [n−1](p) 6= b which
means that f [n−1] lies in [a, b).
In summary, starting at p, the first n − 2 iterates f k (p), 1 ≤ k ≤ n − 2 lie in [b, c]. The (n − 1)st
iterate f [n−1](p) lies in [a, b], which is disjoint with [b, c). Finally, f [n] (p) lies in [b, c]. This shows that
p does have period n. There is no way that it can be part of a m-cycle with m < n.
Finally, the case f (a) = c, f (b) = a and f (c) = b can be proved in a manner similar to the above.
Example: When 3.829 ≤ a ≤ 3.840, the logistic map fa (x) = ax(1 − x) has a three-cycle. The
Li-Yorke Theorem implies that for these a-values, fa (x) has points, hence cycles, of every period. As
Gulick points out, finding such cycles is another story.
Of course, the Li-Yorke Theorem is a remarkable result. But what if we know that a mapping f
has another cycle – for example, a 5-cycle? Can any other cycles exist? Another remarkable result
by the Russian mathematician A.N. Sharkovsky provides a complete answer. We first need to define
what is known as the “Sharkovsky ordering” of the positive integers:
3 ⊣ 5 ⊣ 7 · · · 2 · · · 3 ⊣ 2 · · · 5 ⊣ 2 · · · 7 · · · 22 · · · 3 ⊣ 22 · · · 5 ⊣ 22 · · · 7 ⊣ · · · ⊣ 23 ⊣ 22 ⊣ 1 .
(7)
The statement
m⊣n
means that m appears before, i.e., to the left of, n in the Sharkovsky ordering. We have
17 ⊣ 14 = 2 · 7 ,
and
40 = 23 · 5 ⊣ 64 = 26 .
Are all positive integers generated with this ordering? The answer is “yes” because every positive
integer can be written in the form
2k · (2m + 1)
93
for suitable k ≥ 0 and m ≥ 0.
Theorem (Sharkovsky): Let f be a continuous function defined on the interval J and suppose that
f (J) ⊆ J. If f has a point with period m, then f has a point with period n for all n such that m ⊣ n.
Comments:
• The Li-Yorke Theorem corresponds to the special case of Sharkovsky’s Theorem in which m = 3.
• If f has a period-5 point, then it is guaranteed to points of all periods n > 0 except possibly 3.
In the book by Gulick, an example of a function f which has a period-5 point but no period-3
points is presented (Page 64).
94
Lecture 11
Dynamics of the Logistic Map fa (x) = ax(1 − x) (cont’d)
The Case a = 4: “Chaotic” Behaviour on [0,1]
The case a = 4 represents a rather special dynamical system with very interesting behaviour. For
“almost all” (we’ll explain this term later) initial points x0 ∈ (0, 1), the behaviour of the iterates given
by
xn+1 = 4xn (1 − xn )
n = 0, 1, 2, · · · ,
(8)
demonstrate seemingly random behaviour as shown in the figure below. (The initial condition employed in this computation was x0 = 0.333.)
1
0.9
0.8
0.7
x(n)
0.6
0.5
0.4
0.3
0.2
0.1
0
0
15
30
45
60
75
n
90
105
120
135
150
A plot of the iterates xn for 0 ≤ n ≤ 150 produced by Eq. (8) using the initial condition x0 = 0.333.
The behaviour of these iterates “seems” to be random, but it is not, since the xn were computed
deterministically – no random numbers were involved in the iteration process. Such behaviour is said
to be chaotic. In what follows, we shall provide some basic properties of the map f4 (x) that are
responsible for this chaotic behaviour.
The map f4 (x) = 4x(1 − x) now maps [0, 1] onto itself and is “two-to-one”: Range (f4 ) = Domain
(f4 ) = [0, 1]. Every point x ∈ [0, 1] has two “preimages” y1 , y2 such that f (y1 ) = f (y2 ) = x: See
diagram below. In the case x = 1, we consider the two preimages to be equal, i.e. y1 = y2 = 21 .
95
y
y=x
1
x
y = 4x(1 − x)
y1
y2
1
x
Graph of f4 (x) = 4x(1 − x) showing the two preimages y1 , y2 of a point x ∈ [0, 1].
Recall that the fixed point x̄2 =
3
4
is repulsive.
The fact that the graph of f4 (x) begins at (0, 0), stretches up to
1
2, 1
and comes down to (1, 0)
accounts for some very rich behaviour of iteration sequences xn+1 = 4xn (1 − xn ). It is possible to find
points x close to the repulsive fixed point x̄1 = 0 that, after a finite number of iterations, come close
again to x̄1 – a sample trajectory is sketched below.
y
y=x
x3
x2
x1
x4
1
x0
x
By “sliding” x0 closer to x̄1 = 0, it is possible to bring the iterate xk (it may take more than four
iterations) closer to x̄1 = 0.
In fact, there is a great deal of “regularity” associated with this map, as we now show by examining
higher order periodic orbits of f4 . Consider the map g(x) = f42 (x). The fixed points of g will include
the fixed points of f4 as well as any possible two-cycles (p1 , p2 ). We know that two-cycles exist since
a > 3. The following features extracted from the graph of f4 (x) will allow us to sketch the graph of
f42 (x):
1) f4 (0) = 0 =⇒ f42 (0) = 0; f4 (1) = 0 =⇒ f42 (1) = 0
2) f4
1
2
= 1 and f4 (1) = 0 =⇒ f42
1
2
=0
3) For y1 , y2 the preimages of x = 12 , i.e. f4 (y1 ) = f4 (y2 ) = 21 . Since f4
f42 (y1 ) = f42 (y2 ) = 1.
96
1
2
= 1, it follows that
This information, plus the continuity of f4 (x), leads to the conclusion that the graph of f42 (x) consists
of two copies of f4 (x) that are contracted horizontally in order to fit over the subintervals 0, 21 and
1 2
2 , 1 . The graph is shown on the next page. Note that f4 has four fixed points: the two fixed points
of f4 (x), x̄1 = 0 and x̄2 = 34 , and the two cycle (p1 , p2 ) ∼
= (0.345, 0.904). Thus, there are four periodic
points of period two.
Let us now consider another iteration of f4 (x): the graph of g(x) = f43 (x). Once again, we use
information from the graph of f4 (x):
1) f4 (0) = 0 =⇒ f43 (0) = 0; f4 (1) = 0 =⇒ f42 (1) = 0
2) f4
1
2
= 1 and f4 (1) = 0 =⇒ f43
3) For y1 , y2 the preimages of x =
1
2
1
2,
=0
we found that f42 (y1 ) = f42 (y2 ) = 1. Therefore f43 (y1 ) =
f43 (y2 ) = 0.
4) Each of y1 and y2 has two preimages: f4 (y11 ) = f4 (y12 ) = y1 and f4 (y21 ) = f4 (y22 ) = y2 . From
3),
f43 (y11 ) = f43 (y12 ) = f43 (y21 ) = f43 (y22 ) = 1.
This information leads to the conclusion that the graph of f43 (x) consists of two copies of f42 (x) that
are contracted horizontally to fit over the subintervals 0, 12 and 12 , 1 . The graph of f43 (x) is shown
on the next page. The function f43 (x) has eight fixed points. Therefore there are eight periodic points
of period 3: the two fixed points of f4 (x), x̄1 = 0 and x̄2 = 43 , and two three-cycles (p1 , p2 , p3 ) and
q1 , q2 , q3 ). In the middle two graphs on the previous page, these two three-cycles are shown in relation
to the graph of f4 (x).
The bottom figures show the graphs of f44 (x) and f45 (x). Clearly, a pattern is emerging: The
graph of f4k (x) is obtained by (roughly) shrinking the graph of f4k−1 (x) to produce two copies that are
placed on the subintervals 0, 12 and 12 , 1 . The net result: f4n (x) has 2n fixed points:
f4 (x) = 4x(1 − x) has
2n
periodic points of period n.
This “doubling” of the graph for each application of f4 is a consequence of the two-to-one nature of
f4 on [0, 1].
97
Iterates of logistic map f4(x) = 4x(1 − x)
Logistic map f^2,
a=4
Logistic map f^4,
1
1
0.9
0.9
0.8
0.8
0.7
0.7
0.6
0.6
0.5
0.5
0.4
0.4
0.3
0.3
0.2
0.2
0.1
0.1
0
a=4
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0
0.1
0.2
0.3
f42 (x)
0.4
0.5
0.6
0.7
0.8
0.9
1
0.7
0.8
0.9
1
0.7
0.8
0.9
1
f43 (x)
Logistic map f(x),
a=4
Logistic map f(x),
1
1
0.9
0.9
0.8
0.8
0.7
0.7
0.6
0.6
0.5
0.5
0.4
0.4
0.3
0.3
0.2
0.2
0.1
0.1
0
a=4
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0
0.1
0.2
0.3
0.4
0.5
0.6
The two three-cycles of f4 (x).
Logistic map f^4,
a=4
Logistic map f^5,
1
1
0.9
0.9
0.8
0.8
0.7
0.7
0.6
0.6
0.5
0.5
0.4
0.4
0.3
0.3
0.2
0.2
0.1
0.1
98
0
0
0.1
0.2
0.3
0.4
0.5
f44 (x)
0.6
0.7
0.8
0.9
1
a=4
0
0
0.1
0.2
0.3
0.4
0.5
f45 (x)
0.6
Three ingredients for “chaotic behaviour”: A preview
We now describe three characteristics exhibited by the logistic map for a = 4 which will be responsible
for the chaotic behaviour demonstrated by its iterates.
Ingredient No. 1: Density of repulsive periodic points
Note that all of these periodic points are repulsive. Moreover, as n increases, the graphs of f4n (x)
indicate that these repulsive periodic points are spread out over the entire interval [0, 1]. In fact, some
higher analysis shows that the set of all periodic points of f4 (x) is dense on [0, 1]. We’ll formally
define what “dense” means below. For the moment:
Translation: If you take any point x ∈ [0, 1] and any neighbourhood Nδ (x) of x, where
Nδ (x) = {y | y ∈ (x − δ, x + δ)} ,
(9)
no matter how small δ is, then you will find at least one periodic point of f4 in it.
The existence of so many periodic points so densely packed on the interval suggests a very high degree
of regularity associated with the map f4 (x), since if x0 is a periodic point, there is an n > 0 so that
f n (x0 ) = f 2n (x0 ) = · · · = f kn (x0 ) = x0 . On the other hand, since these fixed points are repulsive,
they treat points near to them as “hot potatoes” under iteration, wishing to throw them away from the
periodic orbits of which they are members. This “regularity” of repulsive fixed points is one ingredient
in the recipe for “chaotic behaviour” of the iteration dynamics xn+1 = 4xn (1 − xn ).
Before moving on, let us formally the idea of a dense subset. For the moment, we’ll restrict
our discussion to the real numbers. Let X ⊆ R – X could be the entire real line, or a subset – for
example, the interval [0, 1]. Now let Y be a proper subset of X, i.e., Y cannot be X itself. We say
that Y is dense in X if, given any x ∈ X, and any δ > 0, we can find a point y ∈ Y such that |x−y| < δ.
Translation: Given any point x ∈ X, you can find points in Y that are arbitrarily close to x. A
standard example is
Y = { set of all rational numbers in X } .
It is well known that the rational numbers form a dense subset of the real numbers. (People often
99
simply say that “the rationals are dense in the reals.”) If we let x ∈ X be an irrational number, then
given any δ > 0, we can always find a rational number y in the interval (x − δ, x + δ). We can consider
δ as the error in approximating the irrational number x by the rational number y. From the density
of the rationals, this implies that an irrational number can be approximated by a real number to any
degree of accuracy.
Ingredient No. 2: Sensitive dependence to initial conditions
The second ingredient, somewhat related to the repulsivity of the fixed points, is the property of
sensitive dependence to initial conditions. This basically means that two distinct points, x and
y, no matter how close they are to each other initially, may, under iteration of f , become separated
significantly. Mathematically “SDIC” is expressed as follows:
A function f : I → I is said to have sensitive dependence to initial conditions (SDIC) at
a point x ∈ I if there exists an ǫ > 0 such that for any δ > 0 there exists a y ∈ Nδ (x) and
an n > 0 such that
|f n (x) − f n (y)| > ǫ .
(10)
|x − y| < δ .
(11)
Recall that y ∈ Nδ (x) means that
Furthermore,
A function f : I → I is said to have sensitive dependence to initial conditions (SDIC) on
the interval I if it has SDIC at all x ∈ I.
We have already seen an example of an SDIC point for a function f (x) – a repulsive fixed point
x̄ = f (x̄) at which |f (x̄)| > 0. For x0 near x̄, iterates xn = f n (x0 ) are mapped away from x̄. This
indicates that the derivative f ′ (x) – provided it exists – plays an important role.
Suppose that f is a C 1 function on an interval I and that there exists a K > 1 such that
|f ′ (x)| ≥ K > 1 ∀ x ∈ I .
100
(12)
Then for any y ∈ I, by the Mean Value Theorem,
|f (x) − f (y)| = |f ′ (c)| |x − y|
(13)
for some c between x and y. From the assumption in (12), it follows that
|f (x) − f (y)| ≥ K|x − y| .
(14)
In other words, f (x) and f (y) are farther apart than x and y were. Now replace x with f (x) and y
with f (y) so that Eq. (14) becomes
|f 2 (x) − f 2 (y)| ≥ K|f (x) − f (y)|
≥ K 2 |x − y|
from Eq. (14) .
(15)
If we keep repeating this procedure, i.e., replacing x with f (x) and y with f (y), we obtain
|f n (x) − f n (y)| ≥ K n |x − y| ,
(16)
for n as long as the points x and y remain in I. Given an ǫ > 0, then for any nonzero initial separation
|x − y| as small as we wish, there exists an N > 0 such that
K N |x − y| > ǫ .
(17)
As such, we have SDIC. Technically, our choice of ǫ > 0 must be such that the iterates f n (x) and
f n (y) remain in the interval I.
101
Lecture 12
Dynamics of the Logistic Map fa (x) = ax(1 − x) (cont’d)
The Case a = 4: “Chaotic” Behaviour on [0, 1]
There are some serious implications of SDIC regarding the numerical “analysis” of chaotic behaviour.
In numerical computations, we rarely, if ever, compute iteration sequences using infinite precision (unless we compute them in exact rational form using a symbolic computation routine such as MAPLE).
As a result, the finite precision approximation p∗ of a periodic point p represents an error ǫ = |p − p∗ |.
Thus, p∗ may be viewed as the “y” in the previous definition of SDIC, and p as x. SDIC implies that
f k (p) and f k (p∗ ) will become separated significantly. If p is a repulsive periodic point, period n, then
there is little chance that f n (p∗ ) is even close to f n (p) = p. In general, small errors in computation
that are introduced by round-off (due to finite precision) may become magnified upon iteration. As
R. Devaney points out in his excellent book, An Introduction to Chaotic Dynamical Systems:
“the results of numerical computation of an orbit, no matter how accurate, may bear no
resemblance whatsoever with the real orbit” (p. 49).
Note: SDIC alone is insufficient for chaotic or irregular behaviour. For example, the linear map
f (x) = 2x demonstrates SDIC: If |x − y| = δ, then |f n (x) − f n (y)| = 2n δ → ∞ as n → ∞. However,
the behaviour of f n (x) = 2n x is rather straightforward. A nonlinear map that brings larger valued
points back is necessary to produce more complicated behaviour. The logistic map f4 (x) does this.
The SDIC property of chaotic dynamical systems is illustrated on the next page. The results of
two numerical experiments are presented, in which the “exact” (at least to a few hundred digits of
accuracy) orbits of two points x0 and y0 , initially close, are followed. The graphs indicate that no
matter how close x0 and y0 may be, their iterates xn and yn eventually travel far apart – almost
to opposite ends of the interval [0, 1] – only to come arbitrarily close again at another time. The
process of separation and near-merging proceed in a seemingly random manner. Note, however, that
this dynamical process is not a random one but rather a deterministic process: A point xn+1 is
determined uniquely by its predecessor xn .
Very shortly, we shall examine SDIC a little more closely, actually proving its existence for some
simple maps.
102
Logistic map f4(x) = 4x(1 − x):
Sensitive Dependence to Initial Conditions
1
0.8
0.8
0.6
0.6
0.4
0.4
0.2
0.2
0
20
40
60
80
100
0
20
40
60
80
100
-0.2
-0.2
-0.4
-0.4
-0.6
-0.6
-0.8
-0.8
-1
Page 1
Page 1
The above graphs summarize the results of two numerical experiments that illustrate the SDIC property of the logistic map f4 (x). In each experiment, two initial x0 and y0 , with an initial separation of
e0 = y0 − x0 , were chosen. The iteration sequences xn = 4xn−1 (1 − xn−1 ) and yn = 4yn−1 (1 − yn−1 )
were then computed to n = 100, using 1000 digits of floating-point precision. In the above graphs
are plotted the quantities en = yn − xn vs. n. In both cases, the separation en assumes values that
are both arbitrarily small as well as arbitrarily close to 1 or −1. We may interpret y0 as an initial
approximation to x0 with error e0 .
Left graph: x0 = 0.1 and y0 = 0.1001, so that e0 = 0.0001. The iterates xn and yn remain close to
about n = 5.
103 e0 = 0.0000001. The iterates xn and yn remain
Right graph: x0 = 0.1 and y0 = 0.1000001, so that
close to about n = 15.
Ingredient No. 3: Transitivity
The final ingredient of chaotic behaviour, which is responsible for the great separation and nearmerging of iterates mentioned above, is transitivity, which is a kind of mixing property of the
mapping f (x). Mathematically, transitivity may be defined as follows:
f : I → I is transitive on I if, for any two points x, y ∈ I and any two neighbourhoods
N (x) and N (y) of these points, there exists a point p ∈ N (x) and an n > 0 such that
f n (p) ∈ N (y).
Transitivity implies that there are no special intervals in I ⊂ R or regions in I ⊂ Rn that can be
completely avoided by orbits of points from other parts of I. For example, the map f2 (x) = 2x(1 − x)
cannot be transitive on [0, 1], since Range (f2 ) = 0, 12 . Thus, no points are mapped into the interval
1
2
2
2 , 1 . Even the map f (x) = x cannot be transitive on [0, 1] since, if xn ∈ (0, 1), xn+1 = xn < xn . In
other words a point xn ∈ (0, 1) can never be mapped to a value y > xn .
Summary: Ingredients for a chaotic dynamical system
In summary, the ingredients for a chaotic dynamical system xn+1 = f (xn ), where f : I → I are:
1. Regularity: The set of all periodic points of f is dense on I,
2. Sensitive Dependence to Initial Conditions,
3. Transitivity of f .
Here are some other examples of chaotic mappings on [0, 1]:
y
1
1. The “Tent Map”

 2x,
0 ≤ x ≤ 12 ,
T (x) =
 2 − 2x, 1 < x ≤ 1 ,
2
a “straightened” version of the logistic map f4 (x).
0
104
1
x
y
1
2. Another “unimodal” map”
f (x) = sin πx,
0≤x≤1
0
1
x
1
x
y
1
3. The “Baker map” on [0, 1]
B(x) = 2x mod 1, 0 ≤ x ≤ 1 .
0
In general, it may be rather complicated to show that a function f (x) defined a chaotic dynamical
system on an interval I, using the three ingredients listed earlier. However, if the action of f on points
x ∈ I can be put into one-to-one correspondence with another map g : I → I that is known to be
chaotic, then f is chaotic. Historically, the logistic map f4 was shown to be chaotic by means of a
1 − 1 correspondence it has with the “tent map” in Example 1. The tent map, being piecewise linear,
is easier to analyze.
A return to “Sensitive Dependence on Initial Conditions” (SDIC)
Let us now get a little more experience with SDIC. First, recall the definition from the previous lecture:
A function f : I → I is said to have sensitive dependence to initial conditions (SDIC) at
a point x ∈ I if there exists an ǫ > 0 such that for any δ > 0 there exists a y ∈ Nδ (x) – or
“there exists a y, |x − y| < δ” – and an n > 0 such that
|f n (x) − f n (y)| > ǫ .
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|x − y| < δ .
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Recall that y ∈ Nδ (x) means that
and
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A function f : I → I is said to have sensitive dependence to initial conditions (SDIC) on
the interval I if it has SDIC at all x ∈ I.
Translation: No matter how close one is forced to start near x – the role of δ > 0 – one can find a y
and n such that x and y are far apart – the role of ǫ > 0 – after n iterations.
y
1
Example 1: Consider the tent map defined earlier:

 2x,
0 ≤ x ≤ 21 ,
T (x) =
 2 − 2x, 1 < x ≤ 1 .
2
0
1
x
Problem 1(a): Show that T has SDIC at x = 0.
Proof: Notice that T n (0) = 0 for all n ≥ 1. Also notice that for y > 0, y close to 0,
T n (y) = 2n y .
We are free to choose any ǫ > 0. Well, any “reasonable” ǫ > 0. We couldn’t choose ǫ = 5, because the
1
length of the interval [0, 1] on which T is operating has only length one! Let’s choose ǫ = . For any
3
δ > 0, we can find a y > 0, with |x − y| < δ which, in this case, means
0<y<δ
and an n > 0 such that
T n (y) = 2n y >
1
.
3
From the above, we see that n satisfies
n > log2
1
3y
.
Problem 1(b): Show that T has SDIC on [0, 1], i.e., T has SDIC at all x ∈ [0, 1].
Proof: We shall need the following result, which was simply stated in class but not proved. A short
proof is given after this problem is solved.
Theorem: Let x be a dyadic point in (0,1], i.e.,
x=
a
, k ≥ 1,
2k
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1 ≤ a ≤ 2k .
Then
T n (x) = 0 for n > k , i.e., n = k + 1, k + 2, · · · .
In other words, any dyadic point in [0, 1] is eventually mapped to 0 after a finite number of iterations.
(Trivially, in the case a = 2k , T (1) = 0.)
We have already proved the desired result for x = 0 in Problem 1(a). We must now consider two cases
involving x ∈ (0, 1].
Case 1: The point x ∈ (0, 1] is a dyadic point. In this case, pick a y ∈ (0, 1) near x which is not a
dyadic point, i.e., for any δ > 0, find a y such that |x − y| < δ and
y 6=
b
2l
l ≥ 1,
1 ≤ b ≤ 2l .
If we apply the operator T to x and y iteratively, then after k + 1 iterations, x is mapped to 0 but
T n (y) 6= 0. (We should actually prove this, but we won’t – the proof actually follows from the proof
of the “dyadic result” which is presented after this proof.) At this point, we may simply “reset” the
clock, pretending that we are starting with the point z = T n (y) in relation to the point T n (x) = 0.
1
1
If z > = ǫ, we are done, i.e., we have shown SDIC. But if 0 < z ≤ , then we simply repeat the
3
3
argument we used in Problem 1(a) to show that after a sufficient number, n > k, of iterations,
|T n (x) − T n (y)| = |T n (y)| >
1
.
3
Case 2: x is not a dyadic point. In this case, choose y to be a dyadic point satisfying |x − y| < δ.
We know that there exists an m > 0 such that T m (y) = 0. But T m (x) cannot be 0. We now use the
same argument as in Case 1: The roles of x and y are now reversed. If the point z = T m (x) > 13 , then
1
we are done, i.e., we have shown SDIC. But if 0 < z ≤ , we once again repeat the argument used in
3
Problem 1(a) to show that after a sufficient number, n > k, of iterations,
|T n (x) − T n (y)| = |T n (x)| >
1
.
3
The proof is complete.
Let us now return to prove the following result.
Theorem: If x is a dyadic point, i.e.,
x=
a
, k ≥ 1,
2k
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1 ≤ a ≤ 2k ,
then T k (x) = 0.
Proof: Applying T to any point y ∈ [0, 1] yields one of the following two results,
1. 2y if y ∈ [0, 1/2]
2. 2 − 2y if y ∈ (1/2, 1].
Actually, if y = 1/2, both formulas can be used, and they yield the value 1.
Let’s apply T one more time to compute T 2 (y). There are four possibilities - we won’t worry where
y has to be in order to choose each of them:
1. 2(2y) = 4y
2. 2(2 − 2y) = 4 − 4y
3. 2 − 2(2y) = 2 − 4y
4. 2 − 2(2 − 2y) = 2 − 4 + 4y = −2 + 4y.
There seems to be a pattern emerging here. First of all, the final formula is linear in y. This is a
consequence of the fact that the two “pieces” of T are linear in their argument. Secondly, the multiple
of y is a power of 2. On the basis of this observation, we conjecture that T n (y) has the form
T n (y) = A ± 2n y ,
where A is an integer – in fact, an even integer, which will depend on the position of y. Its actual
value is irrelevant, as we’ll see.
Let’s go back to the above four possible formulas. Given a y ∈ [0, 1], we can actually determine which
of them can be applied to y: The one or ones which yield a number T (y) which lies in [0, 1].
For example, if y = 2/5, the first mapping, 4y, yields 8/5 > 1, so it can’t be used. The second
mapping 4 − 4y, yields 4 − 8/5 = 12/5 > 1, so it can’t be used. The third mapping 2 − 4y yields
2 − 8/5 = 2/5, so it can be used. The fourth mapping −2 + 4y yields −2 + 8/5 = −2/5, so it can’t be
used. Net result, T 2 (2/5) = 2/5, which can be verified from the original definition of T :
T (2/5) = 4/5 , T (4/5) = 2 − 8/5 = 2/5 .
(Note that we discovered, by accident, a two-cycle of T .)
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What about the special point y = 1/2? The first formula yields 2, which is unacceptable. The
second one yields 4 − 2 = 2 which is also unacceptable. The third yields 2 − 2 = 0 which is fine. The
fourth yields −2 + 2 = 0, which is also fine. In summary, T 2 (1/2) = 0, which we already knew.
a
, the result will have the form,
2k
a
a
T k k = −A ± 2k k = −A ± a .
2
2
If we now apply T k times to the dyadic point
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The only value(s) of A that are possible – corresponding to the way in which we chose the various
compositions of T – are those which yield a value that lies in the interval [0, 1]. Since A and a are
integers, there are only two possibilities: 0 and 1. If −A + a = 0, we are done, since T k (x) = 0, in
agreement with the original statement of the theorem. If −A + a = 1, it follows that T k (x) = 1 which
implies that T k+1 (x) = 0, once again in agreement with the theorem. The proof of the theorem is
complete.
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