Elementary definitions
In this section , we recall some important definitions which need them :
Definition
We say a map
is a linear transformation if:
1.
2.
A linear transformation
may be represented by a matrix
That is,
Definition
Let
be a metric space and
the first iterate of
be a map ,let
for
More generally , if n is any positive integer , and
for f , then
That is,
then
is the (n+1)st iterate of
is the n-th iterate of
.
the second iterate of
Definition
We call the sequence
)=
Examples:
of iterates of
the orbit of
1.
,4,16,256,…}
2.
2,3,8,63,…}
3.
1
,5,26,…}
4.
Definition
let
be in the domain of
.Then
is a fixed point of
if
Graphically , a point p in the domain of f is a fixed point of f if and only if
the graph of f touches (or crosses) the line y=x at (p , p) (Figure (a)).
y
y
Y=x
Y=x
f
f ( x)  sin( x)
x


Fixed point
Figure a
Example:
Let
, to find the fixed points:
Figure b
x
Definition
Let
b a fixed point of
The point
interval
and in
The point
interval
and in
is an attracting fixed point of
containing
provided that there is an
such that if
is in the domain of
, then
is a repelling fixed point of
provided that there is an
containing
is in the domain of
but
such that if
then
Theorem Suppose that is differentiable at a fixed point:
1. If
then
is attracting.
2. If
then
is repelling.
3. If
then
can be attracting, repelling or neither.
Proof: To begin our proof of (a), we notice that since
f / ( p )  1 , the
definition of derivative implies that there is a positive constant A<1 and
an open interval J   p   , p    such that if x is in J and x  p, then
f ( x)  f ( p )
A
x p
Therefore f ( x)  f ( p)  A x  p , for all x in J . For each such x, this means
that
f ( x)  p  f ( x)  f ( p)  A x  p
(1)
So that f(x) is in J because 0<A<1. thus f(x) is at least as close to p as x is.
Let x be fixed in J. If f [ n] ( x)  p for some n, then f [ n] ( x)  p as n
increases without bound, so we will assume henceforth that f [ n] ( x)  p
for all n. Next we will use that Law of Induction to prove that
f [ n ] ( x )  p  An x  p
for all n  1
(2) By (1), the inequality
holds for n=1. Next, we assume that (2) holds for a given n>1. Then
f [ n ] ( x) is in J since 0  An  A  1 . Therefore by (1) with f [ n ] ( x) substituted
for
x,
and
then
by
(2),
we
find
that
f [ n 1] ( x)  p  f ( f [ n ] ( x))  p  A f [ n ] ( x)  p  A( An x  p )
So that f [ n 1] ( x)  p  An 1 x  p . By the Law of Induction we deduce that
(2) holds for all integers n  1 . Since An  0 as n increases without
bound, it follows that f [ n] ( x)  p for every x in J. thus (a) is proved.
Example
Let
, to find the type of the fixed points:
0, thus 0 is an attracting fixed point
,thus 1 is repelling