c^z, z^z + c, sin(z) + c and so on...

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Offline matteo.basei

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« Reply #15 on: May 18, 2020, 01:03:34 PM »
What kind of in coloring are you using? If I increase the iterations on my image I lost the black part of yours (Reply #13).

That's correct, I don't use colors for non-chaotic regions, it's always the same method I used in other images.
In this case it is only a compromise.
I prepared a sequence to show it (100, 150, 200 and 250 iterations).
Matteo Basei
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"Chaos: when the present determines the future, but the approximate present does not approximately determine the future." Edward Norton Lorenz

Offline matteo.basei

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« Reply #16 on: May 18, 2020, 01:41:30 PM »
Sorry my ignorance but (recall the real part of Γ function) means ...

The Euler \( \Gamma \) function, the extension of the factorial function to complex numbers.
This is a 3D animation in wireframe: https://www.youtube.com/watch?v=WH6BcjHR4ig
and a image which shows the similarity with reply #8.

Offline matteo.basei

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« Reply #17 on: May 22, 2020, 07:39:50 AM »
Same shape from \( \sin(z) + c \).
« Last Edit: May 22, 2020, 05:05:24 PM by matteo.basei »

Offline matteo.basei

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« Reply #18 on: May 22, 2020, 07:50:47 AM »
Another interesting function that show the same shape: \( f(z) = \sqrt[z]{c} \).

"Question mark"
\( c = 9.2 + i \, 8.9 \)

(the answer could be in \( c = 42 \))

Offline gerrit

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« Reply #19 on: May 22, 2020, 08:42:49 AM »
Another interesting function that show the same shape: \( f(z) = \sqrt[z]{c} \).

"Question mark"
\( c = 9.2 + i \, 8.9 \)

(the answer could be in \( c = 42 \))
Found a nice one at \( c = 8.878338613 - i \, 9.315404091 \), using "M-set" (2nd image) as parameter guide. I use critical orbit from \( \infty \) so I take \( z_0 = c^0 = 1 \). Not sure how to start from the singular orbit at z=0, what should z_1 be??

Offline gerrit

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« Reply #20 on: May 22, 2020, 09:14:55 AM »
Found another question mark at \( c = 11.61401443 - i \ 7.4007905351 \), fun formula.

Offline matteo.basei

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« Reply #21 on: May 22, 2020, 01:03:24 PM »
Found a nice one at \( c = 8.878338613 - i \, 9.315404091 \)

Beautiful.
If I may say, I think you write a lot of digits more than necessary
(for example \( c = 8.878338613 - i \, 9.315404091 \) gives almost the same image of \( c = 8.88 - i \, 9.32 \)).



Not sure how to start from the singular orbit at z=0, what should z_1 be??

Sorry, I'm not sure I understand. Do you mean M-set with \( z_0 = 0 \)?
For M-set do you mean fixed \( z_0 \) and \( c = \) "point to evaluate", it's correct?

Offline matteo.basei

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« Reply #22 on: May 22, 2020, 01:04:44 PM »
Found another question mark at \( c = 11.61401443 - i \ 7.4007905351 \), fun formula.

Changing \( c \) you can find a lot of question (marks), but only one anwser :)

Offline matteo.basei

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« Reply #23 on: May 22, 2020, 01:53:08 PM »
Again \( f(z) = \sqrt[z]{c} \).

I tried a new simpler formula for lightness, just \( 1 - \frac{1}{\epsilon + 1} \) without pow or log.
It gives a value in \( \left[ 0 , 1 \right) \) and it seems good (I will try with other \( f(z) \)).

In this image saturation is set to 0, so I not calculated hue (that anyway was not so interesting with this function).
To (almost) complete the spiral I had to do 500 iterations.

PS: I changed the title of thread, is this a bad practices or is ok?
« Last Edit: May 23, 2020, 02:39:27 PM by matteo.basei »

Offline gerrit

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« Reply #24 on: May 22, 2020, 07:13:41 PM »
Beautiful.
If I may say, I think you write a lot of digits more than necessary
(for example \( c = 8.878338613 - i \, 9.315404091 \) gives almost the same image of \( c = 8.88 - i \, 9.32 \)).
Sorry, I'm not sure I understand. Do you mean M-set with \( z_0 = 0 \)?
For M-set do you mean fixed \( z_0 \) and \( c = \) "point to evaluate", it's correct?
Yeah, I just paste the numbers from UF. Yes by M-st I mean parameter space starting at a critical point.
I think this fractal is equivalent to \( z \leftarrow e^{c/z} \).

Offline C0ryMcG

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« Reply #25 on: May 24, 2020, 12:39:37 AM »
Something I found interesting from that last formula... I couldn't really get images as nice as yours (it's sensitive to escape radii and coloring algorithms and would take too much fiddling, I think) but I noticed how BIG the julia's were, compared to other formulae.
One thing I like to do is apply circle inversion to julia sets to see the details in a different way, and this reminded me of an already-inverted julia image.

So here's what I got when I inverted these. For every C value, inverting would result in these repeated patterns to infinity, at different angles depending on C.

Offline matteo.basei

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« Reply #26 on: May 24, 2020, 04:07:57 AM »
So here's what I got when I inverted these. For every C value, inverting would result in these repeated patterns to infinity, at different angles depending on C.

This is \( c^z \), see reply #10.

Offline C0ryMcG

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« Reply #27 on: May 24, 2020, 06:50:42 AM »
This is \( c^z \), see reply #10.
I've seen reply #10.
Are you saying that c^z is the same as c^1/z after circle inversion? I would have expected that of c^-z maybe...

Looking into this, I think you guess incorrectly. It does have very similar behavior, but not the same. Here's an image with C set to -0.27-1.52i like the file name in the second image of post number 10 suggests, and it's not the same shape.

Offline matteo.basei

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« Reply #28 on: May 24, 2020, 08:05:25 AM »
I've seen reply #10.
Are you saying that c^z is the same as c^1/z after circle inversion? I would have expected that of c^-z maybe...

Looking into this, I think you guess incorrectly. It does have very similar behavior, but not the same. Here's an image with C set to -0.27-1.52i like the file name in the second image of post number 10 suggests, and it's not the same shape.

\( -z \) is just a reflection, instead
\[ \frac{1}{z} = \frac{z^*}{z \, z^*} = \frac{z^*}{\left| z \right|^2} \]
is a reflection plus an inversion.

See www.matteo-basei.it/argand (I'm sorry it's in italian, but maybe Google can translate).
« Last Edit: May 25, 2020, 07:06:11 AM by matteo.basei »

Offline matteo.basei

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« Reply #29 on: May 25, 2020, 06:53:58 AM »
Yeah, I just paste the numbers from UF.

I suppose UF is a program to make fractal.



I think this fractal is equivalent to \( z \leftarrow e^{c/z} \).

It's just a parameter substitution.
Let \( c' = \ln c \), so \( c = e^{c'} \) and
\[ \sqrt[z]{c} = c^{1 / z} = \left( e^{c'} \right)^{1 / z} = e^{c' / z} \]