Theorem :-  is strongly chaotic on Z
Proof. First we will to show that the set of periodic points of  is dense in Z. to that
end, let ...z3 z 2 z1.z0 z1z2 ... be an arbitrary element of Z, and let n be an arbitrary
positive integer. If x is the doubly-repeating two-sided sequence z n ...z3 z 2 z1.z0 z1z2 ...zn
, then it follows that x is periodic (with period 2n 1 ). Moreover, zk  xk for k  n , so
that x  z  1 2n 1 . Thus the periodic points are dense in Z. to show that  has
sensitive dependence on initial conditions, let z be in Z,   1 2,   0, and n so large
that 1 2n 1   . If x is chosen with xk  zk for all k and such that k  n but xn 1  zn 1 ,
then x  z  1 2n 1   . However,
 n 1 ( x)  ...xn .xn 1... and  n 1 ( z )  ...zn .zn 1...
So that  n 1 ( x)   n 1 ( x)  1   . Therefore  has sensitive dependence on initial
conditions. Finally, we will show that  is transitive. To see this, let the forward
portion of the two-sided sequence Z * have the form
0 1 000 001 010 011 100 101 110 111 00000 00001 ...
(where for each positive odd integer n, all possible n-tuples appear in order), and let
the backward portion of Z * have the form
... 0100 0011 0010 0001 0000 11 10 01 00
(where for each positive even integer n, all possible n-tuples appear in backward
order). Then it is possible to show that the orbit of Z * is dense in Z . with this we
have completed the proof that  is strongly chaotic on Z .
Definition :- Let V be a subset of R2 . And let map, F : V  R 2 have coordinate maps
with continuous partial derivatives. Also assume that F has an attractor AF , and that
v0 is in AF . Finally, assume that 1 (v0 )  1  2 (v0 ), . Then the Lyapunov dimension of
AF at v0 , deno dim ted dim L AF (v0 ). is given by ,
dim L AF (v0 )  1 
ln 1 (v0 )
ln 2 (v0 )
In the event that 1 (v0 ) and 2 (v0 ) are independent of v0 (except possibly for isolated
points v0 ), we write 1 and 2 for 1 (v0 ) and 2 (v0 ) , respectively. In that case we
define the Lyapunov dimension of AF by the formula
dim L AF  1 
ln 1
ln 2
By the definition of Lyapunov dimension and with the help a computer, one Next we
will find the Lyapunov dimension of a map without the aid of a computer.
Example 1 :- Let m be the horseshoe map Find the Lyapunov dimension of the
attractor AM .
Solution:- If v and w are in the attractor AM and are very near to each other , then M
shrinks the distance between v and w vertically by a factor a  1 3 , and expands the
distance horizontally by a factor b  3 . therefore
 a 0

DM (v)  
 0 3
For each v in AM , the eigenvalues of DM (v) are a and 3. now fix v0 in AM . Then the
iterates of v0 are also in AM , so by (3) and result of section 3.1, the absolute values of
the eigenvalues d n1 (v0 ) and dn 2 (v0 ) of Dn M (v0 ) are given by,
d n1 (v0 )  3n and d n 2 (v0 )  a n
Notice that dn1 (v0 )  dn 2 (v0 ) because 3 1  a by hypothesis. Therefore the Lypunov
numbers 1 (v0 ) and 2 (v0 ) are given by
1 (v0 )  dn1 (v0 )1 n  3 and 2 (v0 )  dn2 (v0 )1 n  a
Since these numbers are independent of v0 , we find that
1  3 and 2  a
Finally , by (5) the Lyapunov dimension of the attractor AM is given by
dim L AM  1 
ln 3
ln a