Finally, for

1 1

< x ¤ 1, Lµ (x) < 1 ’ .

µ µ

So on the range 1 < µ < 2 the behavior of Lµ is as follows: All points

1 1

0 < x < 1 ’ µ steadily increase toward the ¬xed point, 1 ’ µ . All points

1 1

satisfying 1 ’ µ < x < µ steadily decrease toward the ¬xed point. The point

1 1 1

µ satis¬es Lµ ( µ ) = 1 ’ µ and so lands on the non-zero ¬xed point after one

1

application. The points satisfying µ < x < 1 get mapped by Lµ into the

1

interval 0 < x < 1 ’ µ , In other words, they overshoot the mark, but then

steadily increase towards the non-zero ¬xed point. Of course Lµ (1) = 0 which

is always true.

When µ = 2, the points µ and 1 ’ µ coincide and equal 1 with L2 ( 2 ) = 0.

1 1 1

2

1

There is no “steadily decreasing” region, and the ¬xed point, 2 is superattractive

- the iterates zoom into the ¬xed point faster than any geometrical rate.

2 < µ < 3.

2.1.3

1 1 1 1

Here the ¬xed point 1 ’ > while < 2 . The derivative at this ¬xed point

µ 2 µ

is negative:

1

Lµ (1 ’ ) = 2 ’ µ < 0.

µ

1

So the ¬xed point 1 ’ µ is an attractor, but as the iterates converge to the

¬xed points, they oscillate about it, alternating from one side to the other. The

entire interval (0, 1) is in the basin of attraction of the ¬xed point. To see this,

we may argue as follows:

1

The graph of Lµ lies entirely above the line y = x on the interval (0, 1 ’ µ ].

1 1

In particular, it lies above the line y = x on the subinterval [ µ , 1 ’ µ ] and takes

its maximum at 2 . So µ = Lµ ( 2 ) > Lµ (1 ’ µ ) = 1 ’ µ . Hence Lµ maps the

1 1 1 1

4

interval [ µ , 1 ’ µ ] onto the interval [1 ’ µ , µ ]. The map Lµ is decreasing to the

1 1 1

4

1 1

right of 2 , so it is certainly decreasing to the right of 1 ’ µ . Hence it maps

the interval [1 ’ µ , µ ] into an interval whose right hand end point is 1 ’ µ and

1 1

4

whose left hand end point is Lµ ( µ ). We claim that

4

µ 1

Lµ ( ) > .

4 2

This amounts to showing that

µ2 (4 ’ µ) 1

>

16 2

or that

µ2 (4 ’ µ) > 8.

30 CHAPTER 2. BIFURCATIONS

1

0.9

0.8

0.7

0.6

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

1 1

= .4, 1 ’

Figure 2.3: µ = 2.5, = .6

µ µ

31

2.1. THE LOGISTIC FAMILY.

8

Now the critical points of µ2 (4 ’ µ) are 0 and 3 and the second derivative at

8

3 is negative, so it is a local maximum. So we need only check the values of

µ2 (4 ’ µ) at the end points, 2 and 3, of the range of µ we are considering, where

the values are 8 and 9.

The image of [ µ , 1’ µ ] is the same as the image of [ 2 , 1’ µ ] and is [1’ µ , µ ].

1 1 1 1 1

4

µ µ

1 1

The image of this interval is the interval [Lµ ( 4 ), 1 ’ µ ], with 2 < Lµ ( 4 ). If we

1

apply Lµ to this interval, we get an interval to the right of 1 ’ µ with right end

point L2 ( µ ) < Lµ ( 1 ) = µ . The image of the interval [1 ’ µ , L2 ( µ )] must be

1

µ4 µ4

2 4

1µ

strictly contained in the image of the interval [1 ’ µ , 4 ], and hence we conclude

that

µ µ

L3 ( ) > Lµ ( ).

µ

4 4

Continuing in this way we see that under even powers, the image of [ 1 , 1 ’ µ ]

1

2

1

is a sequence of nested intervals whose right hand end point is 1 ’ µ and whose

left hand end points are

1 µ µ

< Lµ ( ) < L3 ( ) < · · · .

µ

2 4 4

1