Re: Dimension of sequences
Fernando Bauer wrote:
>
> Hello All,
>
> A basic question about sequences.
>
> Suppose that a "retangular sequence" is a sequence generated by using
> iteratively
> the function repeat() beginning with an atom. Then, I think we can say that
> the dimension of a retangular sequence is the number of calls to repeat()
> function.
> (atom=dimension 0, vector=dimension 1, matrix=dimension 2, ...).
>
They call it the dimension of a tensor in physics. Well there are two dimensions
because of covariant and contravariant indexes, but Euphoria ignores this.
> Now, let's say we have a sequence like { 1, {1,1} } which is not retangular.
> Then, a question arise:
>
> What is the dimension of a non-retangular sequence ?
>
> a) the maximum depth of the sequence.
> b) an integer number.
> c) a fractal number.
> d) a sequence which depends on the structure.
> e) the dimension concept does not apply.
> f) I don't know.
> g) other.
>
Depends on what you wish to do with the dimension concept.
In keeping with how the dimension of a topological space is defined, I'd say
"the maximum depth" (a/).
> Trying to answer that question, others more basics and related to that arise
> to me (sorry if they are stupid!):
>
> What is the dimension of the circumference ?
> a) 1 , because the area of the circumference is zero. It is a curved 1D
> object.
> b) 2 , because the circumference exists in a bidimensional space.
> c) Both. It has 2 types of dimensions!
>
No, the dimension is clearly one, if only because of the scaling issue. If you
multiply the radius of a circle by c, the length of any arc on this circle is
multiplied by c, which is power(c,1), as well, so the dimension is 1. For a disc,
areas are multiplied by c squared, so discs have dimension 2. That's how
Haüsdorff's dimension is defined, because the relevant measure is the length on
circles and the area on discs.
> Same question for a line not closed as, for example, the form of letter "U".
Same answer: 1. Because there is a length, but not an area.
However, if you do the following iterative process:
1/start with a line segment
2/for each line segment in your objct, replace the middle third segment by the
other two sides of an equiklateral triangle. Try making orientation consistent.
Then the limit object has dimension... ln(4)/ln(3). It isn't either a curve any
longer - even though any finite application of algorithm above results in a curve
- nor a surface yet. Consider that you replace three segments ___ with four _/\_
.
>
> Thanks for your reply,
>
> Regards,
> Fernando Bauer
CChris
|
Not Categorized, Please Help
|
|
