(Question) Are the mandelbrot sets generated different in appearance to the "actual" set?

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Offline abbaddon1001

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« on: May 15, 2018, 01:37:56 AM »
Hey guys,

I looked up somewhere that the mandelbrot sets we see on computer are an approximation to the true mandelbrot set. How far off could the difference between the generated and the "true" appearance of the set be?

Offline pauldelbrot

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« Reply #1 on: May 15, 2018, 01:48:52 AM »
Not very. I seem to recall there is a theorem that, unless the precision is too low (enough to cause detectable artifacting), the color of a pixel is the correct color for some point inside of that pixel.

Offline gerrit

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« Reply #2 on: May 15, 2018, 01:57:46 AM »
Not very. I seem to recall there is a theorem that, unless the precision is too low (enough to cause detectable artifacting), the color of a pixel is the correct color for some point inside of that pixel.
That's "backward stability". AFAIK it is not proven but a conjecture supported numerically. See https://fractalforums.org/fractal-mathematics-and-new-theories/28/perturbation-theory/487/msg2365#msg2365

Offline claude

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« Reply #3 on: May 15, 2018, 04:48:27 AM »
It can be quite far off.

The usual assumption when sampling a digital image (or audio, or whatever else) is that the source image has been bandlimited before sampling (with a low pass filter), to avoid aliasing (whereby high frequencies get folded into lower frequencies, in a way that is impossible to correct after sampling).  This is not possible for the Mandelbrot set, so images are always going to suffer from aliasing.  Supersampling can help a lot, so use as much as possible if you want improved images.

For example, find a minibrot near -2+0i at very deep zooms, chances are you'll end up with a Moire mess of stars, instead of concentric rings of rays (because the rays will be regularly spaced finer than the pixel spacing, leading to interference).

Offline gerrit

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« Reply #4 on: May 15, 2018, 05:27:16 AM »
For example, find a minibrot near -2+0i at very deep zooms, chances are you'll end up with a Moire mess of stars, instead of concentric rings of rays (because the rays will be regularly spaced finer than the pixel spacing, leading to interference).
Yes, though each pixel has the correct value, things go wrong at display time.
Below an illustration of such a location, Bad with just computing each pixel, good with using 1) oversampling by factor 8 2) jitter (random displacements of the center of the pixel withing the square of the actual pixel) 3) Gaussian blur 4) downsampling.
However I think we can say that if all these things are taken care of the result is accurate.

Offline gerrit

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« Reply #5 on: May 15, 2018, 05:49:11 AM »
I looked up somewhere that the mandelbrot sets we see on computer are an approximation to the true mandelbrot set. How far off could the difference between the generated and the "true" appearance of the set be?
I recall reading that the M-set is "weakly computable" or something like that, meaning that if you specify a required accuracy, you can write a program that computes it to that accuracy (disregarding display issues). It is however strictly uncomputable, as there is no algorithm that can tell you whether a given point c is in the set or not. It's easy to understand: if you iterate up to N and it escapes you know it's not in the set. If it doesn't escape it may escape after iteration N, and you know nothing. You could set \( N = \infty \) but then your program may not terminate.

However if you have two very close points c and d and let's say c escapes but d does not, then you know the boundary is on the line between c and d. This way you can narrow down where it is without ever being able to say exactly where it is. This works because of special properties of the M-set. AFAIK the argument does not apply to for example \( z \leftarrow e^z +c \), where the outside points are some Cantor sets made out of lines.

Offline abbaddon1001

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« Reply #6 on: May 15, 2018, 09:58:00 AM »
Really interesting replies, I was wondering if the same can be said for any other types of fractals, including 3d. Are there fractals that can be computed with near perfect accuracy?

Offline claude

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« Reply #7 on: May 15, 2018, 11:08:39 AM »
I think fractals that are based on iterated function systems with rational coefficients will be easier to calculate more exactly:

- Cantor middle-thirds dust
- Sierpinski triangle (right-angled variant)
- Sierpinski carpet

And with field extensions (subsets of algebraic numbers) you could do the equliateral Sierpinski triangle too.

But in general, computers have trouble with arbitrary real numbers, as most of them aren't computable.

Offline Ebanflo

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« Reply #8 on: June 26, 2018, 01:29:28 AM »
The Mandelbrot set, the Cantor set, Weierstrass curve, Apollonian gasket, Sierpinski triangle, and every other fractal that I know of all exist inside Euclidean space. This means that they have uncountably many points to them (even the Cantor set), whereas a monitor has only finitely many pixels, and a computer has finitely many bits of memory.

Thus, its impossible for a computer to render or display any fractal perfectly. It seems like the discussion is focused on some other notion of "error," but I thought it was important that this was pointed out.

Offline gerrit

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« Reply #9 on: June 26, 2018, 05:19:58 AM »
Thus, its impossible for a computer to render or display any fractal perfectly. It seems like the discussion is focused on some other notion of "error," but I thought it was important that this was pointed out.
I'm sure we are all aware that our monitors have only a finite amount of pixels.

I think the idea is you render at resolutions 100X100, then 1000X1000, then 10000X10000 etc. and when things stop looking different we converged to what it "really should look like".

Offline Kalter Rauch

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« Reply #10 on: July 09, 2018, 01:59:51 AM »
I read a long time ago
that the actual size of the Mandelbrot Set would span the orbit of Saturn! ;D

Offline Fraktalist

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« Reply #11 on: July 09, 2018, 09:22:50 AM »
Really interesting replies, I was wondering if the same can be said for any other types of fractals, including 3d. Are there fractals that can be computed with near perfect accuracy?
I don't think so - one of the core features of mathematical fractals is that they are scale invariant to infinity. so as long as you don't have a computer with infinite power or infinite time to calculate it as well as a display with infinite resolution, there is no way of compute it with perfect accuracy.
On the other way, for strictly self similar fractals the cantor set, this is only a theoretical 'problem', because the pattern will stay exactly similar, no matter if it's at iteration 10 or 10e100000, so there's no point going deeper and deeper...

Thus, its impossible for a computer to render or display any fractal perfectly. It seems like the discussion is focused on some other notion of "error," but I thought it was important that this was pointed out.
Even if it was possible for a computer to render it perfectly - you would never be able to see it as it really is, because your eyes too have a finite resolution. You don't even see the world around you as it is, because your vision is limited in so many ways (small bandwidth of wavelength, intensity, resolution).

I read a long time ago
that the actual size of the Mandelbrot Set would span the orbit of Saturn! ;D
That's nothing!
Word on the street is the Mset even would span the circumference of your mother!!  :o

Offline Adam Majewski

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« Reply #12 on: July 09, 2018, 03:28:16 PM »

Below an illustration of such a location, Bad with just computing each pixel, good with using 1) oversampling by factor 8 2) jitter (random displacements of the center of the pixel withing the square of the actual pixel) 3) Gaussian blur 4) downsampling.

I like these images. Can you post precise location ? I would like to put it to the wikibooks page about image noisehttps://en.wikibooks.org/wiki/Fractals/Image_noise

TIA

Offline gerrit

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« Reply #13 on: July 09, 2018, 04:37:58 PM »
I like these images. Can you post precise location ? I would like to put it to the wikibooks page about image noisehttps://en.wikibooks.org/wiki/Fractals/Image_noise

TIA
I don't know what the precise location was, but most likely a mini near \( i \) at around 1e2000 magnification.

Offline Kalter Rauch

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« Reply #14 on: July 09, 2018, 07:39:23 PM »


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