Proof. Direct veri¬cation.

3.2. CONJUGACY OF T AND L4 61

Let us understand the importance of this result. The general quadratic

transformation f depends on three parameters a, b and d. But if we are inter-

ested in the qualitative behavior of the iterates of f , it su¬ces to examine the

one parameter family Cc . Any quadratic transformation (with non-vanishing

leading term) has the same behavior (in terms of its iterates) as one of the Qc .

The family of possible behaviors under iteration is one dimensional, depending

on a single parameter c. We may say that the family Qc (or for that matter

the family Lµ ) is universal with respect to quadratic maps as far as iteration is

concerned.

Conjugacy of T and L4

3.2

Let T : [0, 1] ’ [0, 1] be the map de¬ned by

1 1

T (x) = 2x, 0 ¤ x ¤ T (x) = ’2x + 2, ¤ x ¤ 1.

,

2 2

So the graph of T looks like a tent, hence its name. It consists of the straight

1

line segment of slope 2 joining x = 0, y = 0 to x = 2 , y = 1 followed by the

segment of slope ’2 joining x = 1 , y = 1 to x = 1, y = 0.

2

Of course, here L4 is our old friend, L4 (x) = 4x(1 ’ x). We wish to show

that

L4 —¦ h = h —¦ T

where

πx

h(x) = sin2 .

2

In other words, we claim that the diagram of section 1 commutes when f =

T, g = L4 and h is as above. The function sin θ increases monotonically from 0

to 1 as θ increases from 0 to π/2. So, setting

πx

θ= ,

2

we see that h(x) increases monotonically from 0 to 1 as x increases from 0 to 1.

It therefore is a one to one continuous map of [0, 1] onto itself, and thus has a

continuous inverse. It is di¬erentiable everywhere with h(x) > 0 for 0 < x < 1.

But h (0) = h (1) = 0. So h’1 is not di¬erentiable at the end points, but is

di¬erentiable for 0 < x < 1.

To verify our claim, we substitute

= 4 sin2 θ(1 ’ sin2 θ)

L4 (h(x))

= 4 sin2 θ cos2 θ

= sin2 2θ

= sin2 πx.

1

So for 0 ¤ x ¤ we have veri¬ed that

2

L4 (h(x)) = h(2x) = h(T (x)).

62 CHAPTER 3. CONJUGACY

1

0.9

0.8

0.7

0.6

T(x)

0.5

0.4

0.3

0.2

0.1

0

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

x

Figure 3.1: The tent map.

3.2. CONJUGACY OF T AND L4 63

1

0.9

0.8

0.7

0.6

h(x)

0.5

0.4

0.3

0.2

0.1

0

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

x

Figure 3.2: h(x) = sin2 πx

.

2

64 CHAPTER 3. CONJUGACY

1

< x ¤ 1 we have

For 2