Correlation dimension computation, how to get it from 3 slopes in a 3d system

[Amiory]

Here I got a 3 dimension system,a IFS(Iterated Function System), while
compute the correlation dimension using ts tool, I got 3 curves in
loglog plot. I know the correlation dimension can be referred from the
slope of the curve and it's user to identify the area of curve that
is usable for estimating the slope, but there will be 3 slope value
from the 3 curves, how to get the one, which is the final correlation
dimension value?
Thank you very much~

[Roger Bagula]

http://en.wikipedia.org/wiki/Hausdorff_dimension

Behaviour under unions and products

If is a finite or countable union, then

This can be verified directly from the definition.

If X and Y are metric spaces, then the Hausdorff dimension of their product satisfies

An example in which the inequality is strict has been constructed by J. M. Marstrand^[2]. It is known that when X and Y are Borel subsets of , the Hausdorff dimension of is bounded from above by the Hausdorff dimension of X plus the upper packing dimension of Y. These facts are discussed in Mattila (1995).

I don't think you
 are going to like this:
two minimum dimensions are:
for projection dimensions:
s(x,y),s(y,z),s(z,x):

smin_1>=Min[s(x,y),s(y,z),s(z,x)]

or the second topological product:

smin_2>=s(x,y)*s(y,z)*s(z,x)

What it amounts to is the minimum dimension
of any of the projections rules the whole.
When determined in three dimensions the result may be larger
( probably is),
but with the projection measure you can only claim the minimum
as I understand it.

Say you have a plane with a dust scattered on it in s(x,y)=0.5.
looking at it edge wise it is a line broken in parts always less than one
in the s(y,z) and s(z,x) directions,
so the dust 0.5 dimension "rules" the set.
The extreme case forms the limit for all the other cases in an Hausdorff type dimension.

The result is the hard logical one:
an intersectional "and" of the Venn  diagrams for dimension.

We would like to say
 s=1+(s(x,y)-1+s(y,z)-1+s(z,x)-1)
but that is probably too big!
We have to think much like the Lyapunov coefficients: L(z,x) the largest
s=1+L(x,y)/L(z,x)+L(y,z)/L(z,x)

If we treat your dimensions s(x,y),s(y,z),s(z,x) like Lyapunov
coefficients, then the "likely" or probable dimension is
s_likely=(s(x,y)+s(y,z)+s(z,x))/Max[s(x,y),s(y,z),s(z,x)]>=Min[s(x,y),s(y,z),s(z,x)]

I hope that helps.

[Roger Bagula]

Roger Bagula wrote:

or the second topological product:

smin_2>=s(x,y)*s(y,z)*s(z,x)

I realized I had made this error right after I posted this:

smin_2>=(s(x,y)*s(y,z)*s(z,x))^(1/3)

That's a geometrical mean.
Another arthematic mean estimate would be:
smin_3> (s(x,y)+s(y,z)+s(z,x))/3
which is certainly less than the
single dimensional max of the Lyapunov type:
s(z,x) =2 as max
(s(x,y)+s(y,z)+2)/2=1+s(x,y)/2+s(y,z)/2

Suppose they are all nearly the same at 1.26:
the Lyapunov gives a false answer of near 3,
but the two average types and the Minimum give about 1.26.