Fractal Geometry of Nature

[Turtle]

I felt bad I missed that show last week. We had just moved and had left our

TV. At least we have one now. I will have to try and watch it off of pbs.org

website.

 

...

Somebody earlier in this post mentioned Julia Sets and they simply where

the constant C in equation z <- z^2 + C is the coordinate in the iteration.

 

maddog

 

When you get a chance to watch the program, check me on a bit I recall hearing in it concerning Julia sets. To whit, that the Mandelbrot set contains all Julia sets. :shrug: I may have to ask Rac for the book back if he still has it, as I don't recall that association from it.

[Qfwfq]

When you get a chance to watch the program, check me on a bit I recall hearing in it concerning Julia sets. To whit, that the Mandelbrot set contains all Julia sets.

Dubious statement.

 

It may be a sound-alike of "The set of all points for which J is connected is known as the Mandelbrot set." in this page on Julis sets, but the sentence seems to refer only to the case of [imath]R(z)=z^2+c[/imath] meaning all values of c which give a connected Julia set for this form.

[Turtle]

Originally Posted by Turtle

When you get a chance to watch the program, check me on a bit I recall hearing in it concerning Julia sets. To whit, that the Mandelbrot set contains all Julia sets.

Dubious statement.

 

It may be a sound-alike of "The set of all points for which J is connected is known as the Mandelbrot set." in this page on Julis sets, but the sentence seems to refer only to the case of [imath]R(z)=z^2+c[/imath] meaning all values of c which give a connected Julia set for this form.

 

:hyper: You didn't watch the show did you? At any rate, the show is the basis for my dubiosity and perhaps I missheard, or perhaps they misstated, or perhaps mathworld isn't the final word.

[Qfwfq]

You didn't watch the show did you?

No, I didn't, but the sentence I quoted seems to match up with what I can find about the topic although some sources can be confusing.

 

The fact is that the word contains in the show's sentence would need to be given a precise meaning. What Mathworld et al. say more exactly is presumeably what the show meant but said more vaguely.

 

But the king is not gallopping on the horse's back, it's the horse that's gallopping, with the king on its back! :hihi:

[Turtle]

..But the king is not gallopping on the horse's back, it's the horse that's gallopping, with the king on its back! :doh:

 

:) Roger that. The King & his Horse go galloping everywhere together. :)

 

Here's the program link and I'll schedule some time this week to watch it again. NOVA | Hunting the Hidden Dimension | PBS Meantime Rac is going to try & root out Mandelbrot's Fractal Geometry of Nature for me and I hope to find an address for Benoit as I'd like to ask him about Bucky Fuller's work. :hihi:

[maddog]

Dubious statement.

 

It may be a sound-alike of "The set of all points for which J is connected is known as the Mandelbrot set." in this page on Julis sets, but the sentence seems to refer only to the case of [imath]R(z)=z^2+c[/imath] meaning all values of c which give a connected Julia set for this form.

 

As I understand it, I think you both have it right. The Julia Set is about a coordinate

point. So the Set of all Julia Sets would span the space and therefore would

be the Mandlebrot Set. Both are Iteration Functions Systems (IFS).

 

ps: I do have to start practicing that math/imath analogue to LaTeX. ;-)

 

maddog

[freeztar]

When you get a chance to watch the program, check me on a bit I recall hearing in it concerning Julia sets. To whit, that the Mandelbrot set contains all Julia sets. :shrug: I may have to ask Rac for the book back if he still has it, as I don't recall that association from it.

 

"In 1980, he created an equation of his own*, one that combined all of the julia sets into a single image. When Mandelbrot iterated his equation, he got his own set of numbers. Graphed on a computer it was a kind of roadmap of all the julia sets and quickly became famous as the emblem of fractal geometry, the Mandelbrot set."

 

*[math]f(z)=z^2+c[/math]

 

-From the second part of the episode on Nova's website

 

Off to watch part 3. :)

[Turtle]

"In 1980, he created an equation of his own*, one that combined all of the julia sets into a single image. When Mandelbrot iterated his equation, he got his own set of numbers. Graphed on a computer it was a kind of roadmap of all the julia sets and quickly became famous as the emblem of fractal geometry, the Mandelbrot set."

 

*[math]f(z)=z^2+c[/math]

 

-From the second part of the episode on Nova's website

 

Off to watch part 3. :)

 

You rock! Again. :hihi: I checked with Racoon & he thinks he still has Mandelbrot's book The Fractal Geometry of Nature somewhere and will look for it, and I wrote a communique de la electronique to Mr. Mandelbrot at Yale today inquiring if he had any communication with Fuller or familiarity/influence with/from Fuller's Synergetics:Explorations in the Geometry of Thinking. We shall see what we shall see.

[freeztar]

I still have issue with the quote though. :)

 

How could *all* Julia sets be represented in one image?

 

From the section just before they stated such, they identified a julia set as iterative functions (ie feedback loop of one equation). Is it not reasonable to assume that with infinite function equations one could also have infinite julia sets? How could Mandelbrot's 'set' incorporate 'infinite sets'? :hihi:

[Turtle]

I still have issue with the quote though. ;)

 

How could *all* Julia sets be represented in one image?

 

From the section just before they stated such, they identified a julia set as iterative functions (ie feedback loop of one equation). Is it not reasonable to assume that with infinite function equations one could also have infinite julia sets? How could Mandelbrot's 'set' incorporate 'infinite sets'? :hihi:

 

I'll leave that one to Q or Maddog, other than to observe that only certain types of IFS's produce the Julia sets. Change the equation and get a different type of set. Earlier in the thread I give a link to some free software that originally came with Jame's Gleik's book Chaos, and it lets you mess about with a number of different types of fractals and their innards. I'll brb with it to post again. It's a peach I tell ya...a peach! :)

 

>> Chaos Downloads from Rudy Rucker's Wesbsite

[Qfwfq]

I still have issue with the quote though. :(

 

How could *all* Julia sets be represented in one image?

 

From the section just before they stated such, they identified a julia set as iterative functions (ie feedback loop of one equation). Is it not reasonable to assume that with infinite function equations one could also have infinite julia sets? How could Mandelbrot's 'set' incorporate 'infinite sets'? :confused:

Indeed the program should have at least said "all quadratic Julia sets". :hyper:

[freeztar]

Indeed the program should have at least said "all quadratic Julia sets". :confused:

 

Which begs the question...how many quadratic julia sets exist?

[Turtle]

I received a reply from Mr. Mandelbrot's office; he never met Mr. Fuller. :confused:

[Qfwfq]

Which begs the question...how many quadratic julia sets exist?

Plenty!

 

For every complex value there is a Julia set. I don't know if the map is injective but cardinality-wise my guess would be that it's [imath]2^{\aleph_0}[/imath] anyway.

 

The actual point though was that not all Julia sets are the quadratic ones, whereas the Mandelbrot set concerns only these.

[maddog]

Plenty!

 

For every complex value there is a Julia set. I don't know if the map is injective but cardinality-wise my guess would be that it's [imath]2^{\aleph_0}[/imath] anyway.

It is at least as you say injective, though I don't believe it is bijective.

Any one Julia Set so formed is likely to be in more than one algebraic map.

So 1-1 correspondence I think is out.

 

The actual point though was that not all Julia sets are the quadratic ones, whereas the Mandelbrot set concerns only these.

I am not sure of the exact definition of the Mandlebrot set being of only

quadratic equations. I had plotted many such maps base on nothing more

than the assumption that any Analytic Function over a piecewise defined

domain can form a Mandlebrot set.

 

For example [imath]f(z) = cos(z) + c[/imath] or [imath]f(z) = z^4 + c[/imath], etc.

 

maddog

[Qfwfq]

OK when saying "the" Mandelbrot set we usually mean "that" one i. e. the one given by the form [imath]z^2-c[/imath] so the relation is with the quadratic Julia sets. As for the map, it's definitely surjective over the quadatic Julia sets, by definition, so if the map is injective then it's also bijective. :)

[maddog]

OK when saying "the" Mandelbrot set we usually mean "that" one i. e. the one given by the form [imath]z^2-c[/imath] so the relation is with the quadratic Julia sets. As for the map, it's definitely surjective over the quadatic Julia sets, by definition, so if the map is injective then it's also bijective. :)

Like you, I don't know if Mandelbrot ever considered other analytic functions as being claim to being the set named after him or just referring to only quadratic function. As for the rest, I defer to your more concise definition. My Analysis course (injective + surjective = bijective) is a bit rusty.

 

I will admit I am always thinking in the largest of terms by using the largest container.

 

I conjectured a long time ago that an Iterative Function System f(z) can be formed from any analytic function being defined over a piecewise continuous domain.

 

maddog