| > Could someone tell me what the Peano Function is? I know that
> it is a function whose value passes through every point of the unit
> square, but what is the function itself?
I'll try to reconstruct my memories of it.
We're looking for a transformation which is applied on
-----------
to give some other line, eg:
---+ +---
| |
+---+
and is then applied iteratively.
For the above example, we can say:
5X = (3^D)X, where X is the surface covered by the limit line, and
D is the fractal dimension of the surface. (This is, roughly,
just the quantification of the notion of self-similarity.)
5 is the number, N, of line segments, and (1/3) is the ratio,(1/R), of
the subline length to the original length.
So we want to have:
N >= R^2
in order to cover the whole plane.
So, something like
*
/ \
/ \
/ \
/ \
where the angle at the top is right
will be large enough, but no way will it cover a square, or
(aha!)
+---+
| |
---+---+---
| |
+---+
This is going to cover a square of which the start and end-points
are opposite diagonals. (d'you see?: I can't draw fractals
very well in ASCII chars).
Moreover, the graph is Eulerian (only 2 vertices of odd order) so
that we can define a function of time which navigates the shape.
The infinite iteration of this function is going to be a function
of the form your looking for.
Is there anyone knows a good book to learn about this kind of thing?
It's all fascinating, but I'm pretty ignorant about the details.
Andrew.
|
| � Is there anyone knows a good book to learn about this kind of thing?
� It's all fascinating, but I'm pretty ignorant about the details.
There's Falconer's book, "The Geometry of Fractal Sets", Cambridge,
as well as Mandelbrot's classic, "The Fractal Geometry of Nature",
Freeman. The former is an inexpensive paperback and good on the
finer mathematical details. There are many others - Barnsley's book,
"Fractals Everywhere", Academic has a lot of detail on these.
If you can get to a decent library with the Advances in Mathematics serial,
an attractive paper is
F. M. Dekking
"Recurrent Sets"
Advances in Mathematics, Vol 44, 1982
It contains many nice drawings.
These curves and fractals are nearly all based on finitely generated
groups of linear automorphisms of the affine plane. I've recently been
playing with some which are based on conformal automorphisms of the
Riemann sphere - fractional linear transformations of the form
(a z + b) / (c z + d). These preserve circles instead of parallel
lines and generate amazingly delicate wrought-iron figures, with much
resemblance to some Julia sets. When I get a chance, I'll post some
results of folks are at all interested. They look beautiful on a
good PostScript printer.
- Jim
|
| There is an easy way to define a Peano function. It is
in Rudin's book on real analysis, in one of the problems
(with an attribution to the source where Rudin found it).
I can't include that since the book is at home.
This process gives a way to define a continuous function
on I = [0,1] (the unit interval) which maps the Cantor
set onto I^k.
The Cantor set is a compact, uncountable perfect set obtained
by the following process. Let C[0] = I. Given C[n], define
C[n+1] by removing the (open) "middle thirds" of the segments
of C[n]. So for example, C[1] = [0,1/3] union [2/3,1]. Then
C[2] is [0,1/9] union [2/9,1/3] union [2/3,7/9] union [8/9,1].
Then C[0], C[1], ... is a nested sequence of closed subsets
of the unit interval. The Cantor set C is the intersection of
all of the C[n]. (It has measure zero if you are into that
kind of stuff.)
A number like 3/10 = 0.3 has a finite decimal representation,
I just gave it, but it also has an infinite decimal representation,
3/10 = 0.3 = 0.299999.... Likewise, each element of [0,1] has a
unique infinite ternary (base three) representation of the form
0.abcd.... where the "digits" are either 0, 1, or 2. You can
verify that the Cantor set C is the set of all x in [0,1] such
that the infinite ternary representation of x consists of all
0's and 2's, i.e., no 1's. Example, 1/3 = 0.0222222....
The Peano function constructed will take any x in I = [0,1]
into I^k, and will be continuous. If furthermore x is in the
Cantor set C, then the function will take that infinite stream
of 0's and 2's in its ternary representation and split it apart
into k infinite streams of 0's and 1's which will be the binary
representation of the k coordinates of f(x). (The 0's stay 0 and
the 2's become 1's). This is an extremely clever idea and I wish
I had the reference to its author.
So define g:R -> [0,1] so that it is continuous, periodic with
period 2, is 0 on [0,1/3], and is 1 on [2/3,1]. For concreteness,
let g(y) be 0 on [0,1/3], linear from 0 to 1 on [1/3,2/3], 1 on
[2/3,1], linear from 1 down to 0 on [1,2], and periodic with
period 2. g as defined is continuous.
Now consider g(3^n x) for some x in the Cantor set and some nonnegative
integer n. Take x in its ternary representation. Then 3^n x is
the same stream of base three digits, but with the "decimal" point
moved n places to the right. All of the digits are 0's and 2's,
so the integral part of this number (to the left of the "decimal"
point) is even. Since g has period 2, we can therefore ignore
this integral part, and g(3^n x) will be g of the part of the
representation to the right of the shifted "decimal" point. If the
first base three digit there is a 0, then g of that number will
be 0. If the first base three digit there is a 2, then g of that
number will be 1.
So g(3^n x) for x in the Cantor set is 0 or 1 depending on whether
the n+1st digit (indexing those starting at one) of the infinite
ternary representation of x is a 0 or 2. Also g(3^n x) is continuous.
Now consider the function f(x) = sum(n=0 to oo) g(3^n x)/2^(n+1).
This infinite series is uniformly convergent and so its limit, f,
is also continuous. Each term is between 0 and 1/2^(n+1) so f takes
on values between 0 and 1/2 + 1/4 + ... = 1. Finally, each number
in I is the value of f(x) for some x in the Cantor set ... take the
binary representation of the number in I, change the 1's to 2's and
that gives the ternary representation of x in the Cantor set such that
f(x) is that number. So f:R -> I, and f[C] is *all* of I.
You can cover I^k for some finite k in the same manner. Let
fi for i = 0, ..., k-1 be given by
fi(x) = sum(n=0 to oo) g(3^(kn+i) x)/2^(n+1)
Let f(x) = (f0(x),...,fk-1(x)). Then just as above each fi is
continuous, fi:R->I, the image of fi on C is all of I, f:R->I^k,
and the image of f on C is all of I^k. (watch it, that last one
does not follow from merely the fact that each fi covers I, but
from the fact that each k-tuple of infinite binary sequences comes
from an x in the Cantor set). Restrict the domain of f from R to
I (or even to C) and you have a continuous function from I (or C)
which covers I^k.
I don't know what the function will look like, and I wonder if it
will also be a function which is nowhere differentiabe, which is
another nice example. But I haven't really considered that problem.
Dan
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| re .3
In Rudin's book Principles of Mathematical Analysis
it is on page 168, problem 14 in the problems at the
end of chapter 7. He states
(This simple example of a so-called
"space-filling curve" is due to I. J.
Schoenberg, Bull. A.M.S., vol. 44,
1938, pp. 519.)
Furthermore in Munkres's Topology a first course (one
of my favorites) in Chap. 7-2 exercise 5, page 274 it
states
(a) Let X be a Hausdorff space. Show that if there
is a continuous surjective mapping f:I->X, then X
is compact, connected, locally connected, and
metrizable.
(b) The converse also holds; it is a famous theorem
of point set topology called the Hahn-Mazurkiewicz
theorem (see [H-Y], p. 129). Assuming this theorem,
show there is a continuous surjective map f:I->I^omega.
A Hausdorff space that is the continuous image of the
closed unit interval is often called a Peano space.
Of course he gives no hint as to a proof of (b) [don't you
just love when they do that? (-: ]. The reference [H-Y] is
to Hocking and Young, Topology, Addison-Wesley.
Dan
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