THIS is cool!!

[SelfSim]

One of the advantages of writing your own programs is being able to generate Multibrot sets which are iterations of the function f(z) = zⁿ + c where z and c are complex numbers and n > 2.
For n=2 is the familiar Mandelbrot set.

For n =3 is a symmetrical Multibrot set.

I'm willing to admit that I've never managed to completely wrap my mind around the significance (if any) of the black parts. Aren't they the iteration sets where the outputs of the equations 'escape' to infinity? What is the physical significance of that, (I mean: given the eagerness to adopt the obvious self similarity-at-all-scales characteristic as having physical significance)?
(Let's face it: there's a lot of black bits in this fractal, which then more or less begs the question of physical significance, I think?)

 

[FrumiousBandersnatch]

I'm willing to admit that I've never managed to completely wrap my mind around the significance (if any) of the black parts. Aren't they the iteration sets where the outputs of the equations 'escape' to infinity? What is the physical significance of that, (I mean: given the eagerness to adopt the obvious self similarity-at-all-scales characteristic as having physical significance)?
(Let's face it: there's a lot of black bits in this fractal, which then more or less begs the question of physical significance, I think?)

The black bits are where the function has absolute bounds, i.e. doesn't escape to infinity.

 

[SelfSim]

The black bits are where the function has absolute bounds, i.e. doesn't escape to infinity.

Hmm .. I'm not sure I see how that might answer my original question, though(?)
(This is probably just my problem here .. not yours )

Can you perhaps illustrate with an example, (inputs then outputs), using a couple of data points, where the outputs of the function are assigned blackness?

 

[sjastro]

Hmm .. I'm not sure I see how that might answer my original question, though(?)
(This is probably just my problem here .. not yours )

Can you perhaps illustrate with an example, (inputs then outputs), using a couple of data points, where the outputs of the function are assigned blackness?

First of all some basics.
The complex number z is of the form z = x + yi where x and y are real numbers and i = √-1 is an imaginary number.
The modulus of z or |z| = √(x² + y²)
The equation x² + y² = |z|² is simply the equation of a circle of radius |z|.

One can also have functions of complex variables of the form f(z).
One particular function is f(z) = z² +c where z and c are complex numbers.
Lets suppose we input z = 0 and let c vary then the output is f(0) = c.
f(0) now becomes our new variable which is inputted back into the function which is now of the form f(f(0)).
This is known as an iteration which can be repeated by now making f(f(0)) the new input and the function takes the form f(f(f(0)).
The process can be repeated to any number of iterations.
At each iteration the modulus of |f(0)|, |f(f(0))|, |f(f(f(0))|…… can be calculated.

The condition for a Mandelbrot set is that if we perform an infinite number of iterations the modulus for each term |f(0)|, |f(f(0))|, |f(f(f(0))|…… is always less than or equal to 2 which depends on c.
If the condition is met then c is an element in the Mandelbrot set.

It’s obviously not possible to iterate the function an infinite number of times in a computer program.
I set the limit to 400 iterations and am fairly confident if the modulus of the 400th iteration is still less than or equal to 2 it will remain that way.

Here are some examples.
For c = 0 + 1i, |f(0)| = 1, |f(f(0))| = 1.4142136, |f(f(f(0))| = 1, f(f(f(f(0)))) = 1.4142136.
The modulus alternates between 1 and 1.4142136 hence the point c is assigned a black value when plotted.
For c = 0.5 + 0.5i the successive moduli for each iterative term is 0.70710678, 1.11803399, 1.52069063, 1.70591801, 3.54935426, 13.3049962, 177.531916, 31517.1012, 9.9332767e8, 9.8669986e17, 9.73576613e35, 9.47851422e71, 8.98422318e143, overflow error.
After the 4th iteration the condition is no longer met and c is plotted with a light gray value.

The yellow values in the plot of the Mandelbrot set occurs where the value exceeds 2 but in the interval of 20-50 iterations.

 

[SelfSim]

First of all some basics.
The complex number z is of the form z = x + yi where x and y are real numbers and i = √-1 is an imaginary number.
The modulus of z or |z| = √(x² + y²)
The equation x² + y² = |z|² is simply the equation of a circle of radius |z|.

One can also have functions of complex variables of the form f(z).
One particular function is f(z) = z² +c where z and c are complex numbers.
Lets suppose we input z = 0 and let c vary then the output is f(0) = c.
f(0) now becomes our new variable which is inputted back into the function which is now of the form f(f(0)).
This is known as an iteration which can be repeated by now making f(f(0)) the new input and the function takes the form f(f(f(0)).
The process can be repeated to any number of iterations.
At each iteration the modulus of |f(0)|, |f(f(0))|, |f(f(f(0))|…… can be calculated.

The condition for a Mandelbrot set is that if we perform an infinite number of iterations the modulus for each term |f(0)|, |f(f(0))|, |f(f(f(0))|…… is always less than or equal to 2 which depends on c.
If the condition is met then c is an element in the Mandelbrot set.

It’s obviously not possible to iterate the function an infinite number of times in a computer program.
I set the limit to 400 iterations and am fairly confident if the modulus of the 400th iteration is still less than or equal to 2 it will remain that way.

Here are some examples.
For c = 0 + 1i, |f(0)| = 1, |f(f(0))| = 1.4142136, |f(f(f(0))| = 1, f(f(f(f(0)))) = 1.4142136.
The modulus alternates between 1 and 1.4142136 hence the point c is assigned a black value when plotted.
For c = 0.5 + 0.5i the successive moduli for each iterative term is 0.70710678, 1.11803399, 1.52069063, 1.70591801, 3.54935426, 13.3049962, 177.531916, 31517.1012, 9.9332767e8, 9.8669986e17, 9.73576613e35, 9.47851422e71, 8.98422318e143, overflow error.
After the 4th iteration the condition is no longer met and c is plotted with a light gray value.

The yellow values in the plot of the Mandelbrot set occurs where the value exceeds 2 but in the interval of 20-50 iterations.

Thanks so much for that ..
I think I had it completely round the wrong way then .. The black values actually denote those sets which fully meet the Mandelbrot criteria and those which don't, are assigned non black values with their colours being graded on how fast they move away from that modulus of 2.

Never truly got that .. much clearer now.
Thanks kindly!

Also thanks to @FrumiousBandersnatch .. (what he meant is a lot clearer now).
Cheers

 

[sjastro]

The computer code for generating the Mandelbrot set is usually based on an escape algorithm as described in a previous post; the number of iterations for the modulus of the function |f(z)| where f(z) = z² + c to exceed a circle of radius 2 is used to plot c.

There is another method for generating the Mandelbrot set where the period of the iteration is used.
The period of the iteration is where input value x and output values are the same according to the equation fₙ(x) = x where n is the nth iteration.
By using the period n instead of the escape radius for plotting c a much more detailed Mandelbrot set is obtained.

The left hand Mandelbrot set as shown previously was obtained using the escape algorithm, the right hand the period method.
The black plotted points indicate the shortest period.

 

[sjastro]

Whereas the Mandelbrot set is based on the iteration of the function f(z) = z² + c and took 3 minutes to generate the set on my computer there are interesting sets are which are somewhat more complicated to write computer code for and take much longer to generate the corresponding set.

One such function is f(z) =z⁻² + c.
As with real numbers dividing a complex number z by zero presents problems and programming challenges.
The set for f(z) =z⁻² + c is;

This took 6 hours to generate the set.
Considerably more detail can be extracted by applying some late 19th century maths to dynamic systems.
The computer code is more complicated and took 17 hours to generate the set.

 

[jamesbond007]

I want to be in the Mandelbrot set.

Outside who knows what you'll get haha.

 

[Ophiolite]

I'm not familiar with it.
I decided to write a program in Basic not only for the fun of it but to also familiarize myself with the bounded properties of the Mandelbrot set.
For example the number of iterations used to determine whether a point in the complex plane remains bounded or not varies from point to point.
By colour coding each point against the number of iterations a psychedelic version of the Mandelbrot set can be generated.
This is centered around (-0.761574, -0.0847596i) "zoomed" to around 10000X.

looking at your illustration, I think I had that for lunch once, in a side street in Macau, shortly after Yom Kippur.

 

[sjastro]

My last post had a few bugs in the computer code for the equation f(z) =z⁻² + c.
The last image in the post was not quite right.
I rewrote the program to give the following image which is a far more accurate representation.
The red region is where the iteration remains finite.

 

[FrumiousBandersnatch]

My last post had a few bugs in the computer code for the equation f(z) =z⁻² + c.
The last image in the post was not quite right.
I rewrote the program to give the following image which is a far more accurate representation.
The red region is where the iteration remains finite.

Fractal Pacman...

 

[Brightmoon]

Those are cool!

 

[chilehed]

I've learned more about fractals in this one thread than in my entire prior life. Thanks for the response!

 

[sjastro]

Cosmologists come up with physical models of the universe.
Mathematicians do the same except mathematical universes are either difficult to visualize or seem to be physically impossible.
Mathematicians are far less constrained than cosmologists.
One such mathematical universe is an infinitely large universe which rotates.
An object rotates relative to its surrounding space but when the entire universe rotates...........
The constraints for generating fractals in this thread are based on the escape or periodic method as described previously.
I decided on a different method by generating the fractal based on the condition abs(|f(z)|- |c|) < 10⁻⁸ where abs is the absolute function and f(z) =z⁻² + c.

It's totally different from the previous versions and is remarkably symmetrical.

The image reminds me of the Victorian artist Louis Wain whose cat pictures showed Wain's descent into schizophrenia with time.

The second image from the top showing the cat's fear and the artificial background indicate the early onset of schizophrenia.
Wain's images evolved into "fractals".

 

[Strathos]

Cosmologists come up with physical models of the universe.
Mathematicians do the same except mathematical universes are either difficult to visualize or seem to be physically impossible.
Mathematicians are far less constrained than cosmologists.
One such mathematical universe is an infinitely large universe which rotates.
An object rotates relative to its surrounding space but when the entire universe rotates...........

When it comes to fractals the constraints for generating fractals are based on escape or periodic method as described previously.
I decided on a different method by generating the fractal based on the condition abs(|f(z)|- |c|) < 10⁻⁸ where abs is the absolute function and f(z) =z⁻² + c.

It's totally different from the previous versions and is remarkably symmetrical.

The image reminds me of the Victorian artist Louis Wain whose cat pictures showed Wain's descent into schizophrenia with time.

The second image from the top showing the cat's fear and the artificial background indicate the early onset of schizophrenia.
Wain's images evolved into "fractals".

Looks like the cat is having what they call a 'bad trip'