Join for FREE | Take the Tour Lost Password?
Shop deviantART for the
holidays and save BIG!
Click here! :holly:
[x]

deviantART

 

Which spots belong to the M set? EDIT

Tue Jan 6, 2009, 10:18 AM
Along with this journal there are four illustrations uploaded under scraps:

MandelJuliaRelation [link]
c = +1 [link]
c = -1 [link]
c = 0 + i [link]

They will be linked again in the text below.

The Mandelbrot (M) set is defined as those parameters “c” for which the variable “z” forever stays within a radius of 2 under the dynamical process z -> z^2 + c when z = 0 is the starting number. The full definition is given in article 5 in my Chaotic series [link] . Please check it out! It’s a short article ;) The generation is graphically illustrated in the lower part in my scrap JuliaAndMandel-loop [link] . This can be written as:
0 -> 0^2 + c = c -> c^2 + c -> (c^2 + c)^2 + c, etc.
However, instead of being tedious, let’s take some hardcore examples:


1) Does c = 1 belongs to the Mandelbrot set?

Let’s test:
0 -> 0^2 + 1 = 1 -> 1^2 + 1 = 2 -> 2^2 +1 = 5 -> 5^2 +1 = 26, etc.
Obviously c = 1 does NOT belongs to the M set as the variable “z” very rapidly starts to increase towards infinity. This means that the Julia set for c = 1 is a Cantor dust, see [link] . In fact, ones the variable “z” has passed a radius of 2 (it may be after for example 3, 33, or 3 millions of iterations) it starts to increase very rapidly with an increasing acceleration towards infinity :sprint:


2) Does c = 0 belongs to the Mandelbrot set?

Well:
0 -> 0^2 + 0 = 0 -> 0^2 + 0 = 0 etc.
Yeah! And the Associated Julia set is the border of the closed unit disc which was the subject in my journal “The circle is closed” [link] . The parameter c = 0 is situated in the “center” of the cardeoid-shaped body (see [link] ). Every Julia set that is generated with a parameter “c” situated inside the cardeoid, is a Julia set where one of its two fixpoints is attractive (attracts orbits from inside its closed domain).


3) Does c = -1 belongs to the Mandelbrot set?

Let’s test again:
0 -> 0^2 + (-1) = -1 -> (-1)^2 + (-1) = 1 – 1 = 0 -> 0^2 + (-1) = -1, etc. Here the orbit of z = 0 is immediately caught into the 2-periodic cycle 0 -> -1 -> 0 -> -1, etc. In fact c = -1 is the very center of the head of the Mandelbrot set. Every c-value from the head gives rise to filled-in Julia sets with a 2-periodic attractive cycle. This is shown in the illustration [link] . The above cycle further more is super attractive. However that will not be dealt with here ;) Note that every start value of “z” that are in a domain enclosed by the fractal border (the Julia set) will be caught into this periodic cycle. The most obvious example is z = +1, which has the orbit: +1 -> 1^2 + (-1) = 0 -> 0^2 + (-1) = -1 -> (-1)^2 + (-1) = 0, etc. That is:
+1 -> 0 -> -1 -> 0 -> -1, etc. After one iteration the orbit is caught by the attractive cycle (0, -1)


4) Does c = -2 belongs to the Mandelbrot set?

Let’s test again:
0 -> 0^2 + (-2) = -2 -> (-2)^2 + (-2) = 4-2 = 2 -> 2^2 + (-2) = 4 – 2 = 2, etc.
That is, for c = -2, z = 0 has the orbit 0 -> -2 -> 2 -> 2 etc. In fact c = - 2 is the very tip of the spike of the entire Mandelbrot set. The resulting Julia set for this c-value is a straight line from z = -2 + 0i to z = +2 + 0i. This one is dealt with in [link] promoted in the journal [link] .


5) Does c = i belongs to the Mandelbrot set?

This is our first example on a non-real number of “c” :hmm: However we have learned a little bit how to handle complex numbers in [link] :) So let’s make a try ;)
0 -> 0^2 + i -> i -> i^2 + i = -1 + i -> (-1 + i)^2 + i = (1 – i – i + i^2) + i = (1 – 2i –1) + i = -i -> (-i)^2 + i = -1 + i, etc. So simply speaking we have the orbit: 0 -> i -> -1 + i -> -i -> -1 + i -> -i, etc. After two iterations we have been caught in the 2-periodic (repelling) cycle -1 + i -> -i. That is, c = i no doubt belongs to the M set :) Like c = -2, c = i is the very tip of a branch. In this case on the right branch of the “Y” at the northern part of the M set. Moreover it is the tip of a very slow right turning spiral. I will return to this spot in my next journal. The resulting Julia set, shown in my illustration [link] , is infinite thin, but connected.


6) Does c = - i belongs to the Mandelbrot set?

We will guess “yeah” since this spot is mirrored around the real axis. So let’s check out the orbit:
0 -> 0^2 + (-i) = -i -> (-i^2) + (-i) = -1 – i -> (-1 – i)^2 + (-i) = (1 + i + i + (-i)^2) + (-i) = 1 + 2i –1) + (-i) = i -> i^2 + (-i) = -1 – i, etc. That’s the orbit; : 0 -> -i -> -1 - i -> i -> -1 - i -> i, etc.


The above was rather obvious examples where we only had to deal with integers. However in the vast majority of spots in a calculation of a Julia or Mandelbrot motive, long floating points numbers are to be used. Moreover the length of these floating point numbers will increase the deeper the motives are zoomed in.

But this we leave to our dear computers ;) In the next journal I well mentioned something about external rays to the border of the M set. Then we will come back to our earlier angle-doubling and shift operation :)

  • Mood: Wow!
  • Listening to: the silence
  • Reading: Comments on DA
  • Watching: DA
  • Playing: I'm serious
  • Eating: something
  • Drinking: Often

Devious Comments

love 0 0 joy 0 0 wow 2 2 mad 0 0 sad 0 0 fear 0 0 neutral 0 0
:iconscootnoodles:
Dang Ingvar, from now on you are Professor Ingvar to me!!! :p

--
:peace:


People take different roads seeking fulfillment and happiness.
Just because they’re not on your road doesn’t mean they’ve gotten lost. -Dalai Lama
:iconfractalmonster:
"Wannabe mathematician" is the word ;)
..and you have not checked out my four illustrations yet ;)

--
Regards,
Ingvar
[link]
:iconpugluvr:
:faint:
Haha, and I just barely started solving equations with 3 variables in Alg. II XD
But I still got it ;)

--
Salazar: I've sent my right hand to dispose of you.
Leon: Your right hand comes off?
:O_o:

"MY COOKIES EXPLODED!" -Crystal
:iconscootnoodles:
Forgive me friend, I have now taken the time to view them and read the associated information. Some being familiar to me but only this view gave them some relevance to my fractalization attempts. You are teaching me even when I don't think I am ready to learn anything! (dang you're good!)

--
:peace:


People take different roads seeking fulfillment and happiness.
Just because they’re not on your road doesn’t mean they’ve gotten lost. -Dalai Lama
:iconfractalmonster:
Pardon Scott. The daily duties, I know :hmm: I just noticed there where no visitors on my new illustrations when I wrote that. Glad you appreciate my efforts :) And the articles are meant as a library to visit and revisit, so do not stress ;)

--
Regards,
Ingvar
[link]
:iconfractalmonster:
Sounds more difficult than the algoritm for the M set :hmm:
Hope you don't became overloaded ;)

--
Regards,
Ingvar
[link]
:iconscootnoodles:
No stress! That is what fractals do best, release stress in an illustrated way! :)

--
:peace:


People take different roads seeking fulfillment and happiness.
Just because they’re not on your road doesn’t mean they’ve gotten lost. -Dalai Lama
:iconfractalmonster:
Yeah, fractals are good therapy, both looking at and create :)

--
Regards,
Ingvar
[link]
:iconscootnoodles:
Indeed they are! Thanks for being a great source of Fractal Info!!! ;)

--
:peace:


People take different roads seeking fulfillment and happiness.
Just because they’re not on your road doesn’t mean they’ve gotten lost. -Dalai Lama
:iconfractalmonster:
As long as at least a few people try to grasp it, it's worth the effort and a great pleasure :) Moreover I've noticed it's a god way to establish the my own fractal knowledge :dance:

--
Regards,
Ingvar
[link]

Site Map