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Fractal Math, Chaos Theory & Research => Mandelbrot & Julia Set => Topic started by: tonymasiello on May 10, 2017, 06:10:41 PM




Title: number of arms in spiral/snowflake structures
Post by: tonymasiello on May 10, 2017, 06:10:41 PM
Hi,

I am rather new to the forum, so I do not know all of the lingo yet.

I have been playing with Kalles Fraktaler recently and have enjoyed going on random explorations and looking for interesting patterns.

One thing that has fascinated me are the complex spiral/snowflake type structures. I have noticed that structures will most often first appear with a prime number of arms and then as one zooms in, more complex arm structures will arise, often by increasing factors of two. I have been surprised that these structures will often first appear with five or seven arms and then multiply. The case of seven arm structures are quite interesting as the arms often consist of smaller seven and fourteen arm spirals. I have also seen structures that periodically increase by a factor of two, so a series of arm numbers such as 7, 14, 28, 56, 112 will arise. I've also seen odd separations, where a 28 arm structure will evolve into an 8 arm structure with four arms consolidating siz arms each, and the remaining four arms continuing.

What is the largest prime number of spiral arms that have been found in a Mandelbrot? I am curious if there are 11, 13 and 17 armed structures. Also, do complexities ever arise in multiples larger than two... For instance a 3 to 9 structure or a 5 to 15, etc...

Thanks!


Title: Re: number of arms in spiral/snowflake structures
Post by: Chillheimer on May 10, 2017, 10:41:27 PM
welcome to ff!

you can find ALL positive integer numbers of arms in the mandelbrot set.
the number of arms rises to infinity the deeper you dive into the valleys.
this visualizes the number of arms:
https://en.wikipedia.org/wiki/Mandelbrot_set#/media/File:Mandelbrot_Set_%E2%80%93_Periodicities_coloured.png

I don't see what you mean that structures will most often appear as prime numbers. you find ALL natural number of arms, so it totally depends on where you chose to zoom. if you chose to zoom to a period 6 bulb you will get 6 arms.
I have not yet encountered a special connection between prime numbers and the mandelbrot-set but wouldn't be surprised if there is one.


the doubling you encounter is called period doubling. it is just what happens in the mset. as soon as you zoom towards a minibrot each pattern will double.
I created a video some time ago that might help explaining the phenomenon in relatively easy words.
https://www.youtube.com/watch?v=Ojhgwq6t28Y

if you don't use z->z²+c but z->z³+c as base formula you will indeed find the structures don't double but triple. you can try this in kalles fraktaler by going to action/iterations and set "power" to the exponent you want.


cheers,
Chillheimer


Title: Re: number of arms in spiral/snowflake structures
Post by: tonymasiello on May 11, 2017, 02:10:41 AM
Thanks Chillheimer for the reply and the video link. I just watched the video, it is great.

So, if you follow the doubling backwards, the number of arms should always originate with a prime number. The highest starting spiral I've seen is seven arms. So for spirals with less than 200 arms, the possible combinations are:

128/64/32/16/8/4/2
192/96/48/24/12/6/3
160/80/40/20/10/5
112/56/28/14/7

So my question is two fold... Has anyone seen prime spirals higher than seven? Also do the spirals ever multiply by more than two? Is there ever a case where they triple?


Title: Re: number of arms in spiral/snowflake structures
Post by: Kalles Fraktaler on May 11, 2017, 12:16:44 PM
Do you mean 11 arms like this?


Title: Re: number of arms in spiral/snowflake structures
Post by: Chillheimer on May 11, 2017, 01:57:30 PM
So, if you follow the doubling backwards, the number of arms should always originate with a prime number. The highest starting spiral I've seen is seven arms.
nope. as said, you can start with ANY number of arms. you just need to zoom deeper into the valleys. see the attached image, I haven't counted the number but it's a base period from which doubling will start if you zoom further into that area. next doubling will be 2 times the number of arms in this picture.

so if you look close at the wikipedia-image link in my first post, you will see a period 3, 5, 7 and then a period 9 bulb with 9 arms.
This 9-bulb is the base. any doubling starts from 9. 18, 36. no 4.5 or 3.
or on the right side you see a period 6 bulb as base, no 3 below it, doubling starts with 2*6.

So my question is two fold... Has anyone seen prime spirals higher than seven? Also do the spirals ever multiply by more than two? Is there ever a case where they triple?
as mentioned in my first answer in the power-2 mandelbrot set no, there will always be multiplication by 2. but in the power-x mandelbrot-set you have the multiplication * x, triple, quadruple, whatever z^x you choose as x.


Title: Re: number of arms in spiral/snowflake structures
Post by: Adam Majewski on May 11, 2017, 03:41:08 PM
https://commons.wikimedia.org/wiki/File:Conformal_Warping_Around_The_Cardioid_In_The_Mandelbrot_Set.gif
https://commons.wikimedia.org/wiki/File:Zoom_around_principal_Misiurewicz_point_for_periods_from_2_to_1024.gif
http://math.bu.edu/people/bob/papers/farey.ps


Title: Re: number of arms in spiral/snowflake structures
Post by: tonymasiello on May 12, 2017, 12:41:50 PM
Yes! That's exactly what I've been looking for! Thanks!

Do you mean 11 arms like this?


Title: Re: number of arms in spiral/snowflake structures
Post by: tonymasiello on May 12, 2017, 01:10:50 PM
so if you look close at the wikipedia-image link in my first post, you will see a period 3, 5, 7 and then a period 9 bulb with 9 arms.
This 9-bulb is the base. any doubling starts from 9. 18, 36. no 4.5 or 3.
or on the right side you see a period 6 bulb as base, no 3 below it, doubling starts with 2*6.
as mentioned in my first answer in the power-2 mandelbrot set no, there will always be multiplication by 2. but in the power-x mandelbrot-set you have the multiplication * x, triple, quadruple, whatever z^x you choose as x.

Yes, I see it now... Thank you for the explanation!!!