Tent map

Consider continuous tent map fc(x)
    f(x) = { cx ,
c(1 - x) ,   		0 ≤ x ≤ 1/2
1/2 < x ≤ 1
for c = 2. Similar to sawtooth map in the binary number system multiplying x by 2 corresponds to the left shift by one bit site. If before the shift the upper bit b1 = 1 (i.e. x ≥ 1/2) then (using 2 = 1.111...2 = 1.(1)2 ) we get
    f(x) = 2 - 2x = 1.111... - 1.b2 b3 b4 ... = 0.u2 u3 u4 ... .
where uk = 1 - bk is inversion of the bit bk. Thus after the left shift the upper bit is truncated again but if it is 1 then all the rest bits are inverted.
It is evident that symbolic sequence σ = (s1, s2, s3 ...) is obtained from the truncated bits.

Conversion of symbolic sequences to coordinate X

To get rules for conversion of symbolic sequences to coordinate it is useful to recode bits of x by means of substitution 0 → 1 and 1 → -1. Then multiplication of bits by -1 inverts them. Therefore after the fist left shift it is enough to multiply all the bits by the first one p1 and so on. Then the tent map dynamics is
0 . p1   p2   p3   p4 ...
s1= p1 . (p1 p2 )   (p1 p3 )   (p1 p4 ) ...
s2= p1 p2 . (p1 p2 p1 p3 )   (p1 p2 p1 p4 ) ...   or taking into account that pk2 = 1
. (p2 p3 )   (p2 p4 ) ...
s3= p2 p3 . (p3 p4 )   (p3 p5 ) ...
...
sn= pn-1 pn   . (pn pn+1 )   (pn pn+2 ) ...
For the n-th term of the corresponding symbolic sequence we get
    sn = pn-1 pn .
Multiplying this formula by pn-1 we get
    pn = sn pn-1 ,   p1 = s1 .
By means of this formula we get x recursively from symbolic sequence. In the standard notation these rules are
    bn = { bn-1
1 - bn-1   		if   sn = 0
if   sn = 1
For example symbolic sequence σ = {111...} = (1) corresponds to the unstable fixed point x1 = 0.(10) = 102 /112 = 2/3. For period-2 orbit σ = (01) we get x1 = 0.(0110) = 01102 /11112 = 2/5 and x2 = 0.(1100) = 11002 /11112 = 4/5.
Now you can repeat all the "chaos story" told for the sawtooth map before. Later it will be repeated for the quadratic map for c = -2.

Cantor set and the Tent map dynamics
Fig.1 shows, how the classic Cantor set emerges in dynamics of the tent map fc(x) for c = 3 . Really, as since xo = 0 is an unstable fixed point, therefore iterations for x outside the [0, 1] interval diverge to infinity. f(x) maps open interval (1/3,2/3) beyond [0,1], thus we can throw away these points too. As since intervals [0,1/3] and [2/3,1] are mapped linearly onto [0,1], therefore we can continue this process ad infinitum. In every iteration we cut the central one third of an interval. The limit Cantor set is nowhere dense as since it has holes in any small interval.
The lengh of all removed intervals is
    1/3 + 2/9 + 4/27 + ... = 1/3 ∑n=0,∞ 2n / 3n = 1/3 [1 / (1-2/3)] = 1 ,
therefore Lebesque measure (length) of the classic Cantor set is zero. Its fractal dimension is log2 / log3 . Cantor set with zero measure will appear again for any c > 2.
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updated 12 July 06