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

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#### matteo.basei

• Fractal Phenom
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#### Re: z^z + c

« 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
http://www.matteo-basei.it

"Chaos: when the present determines the future, but the approximate present does not approximately determine the future." Edward Norton Lorenz

#### matteo.basei

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#### Re: z^z + c

« 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.

#### matteo.basei

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#### Re: z^z + c

« 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 »

#### matteo.basei

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#### Re: z^z + c

« 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 $$c = 42$$)

• 3f
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#### Re: z^z + c

« 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??

• 3f
• Posts: 2083

#### Re: z^z + c

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

#### matteo.basei

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#### Re: z^z + c

« 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?

#### matteo.basei

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#### Re: z^z + c

« 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

#### matteo.basei

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

« 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 »

• 3f
• Posts: 2083

#### Re: z^z + c

« 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 $$z \leftarrow e^{c/z}$$.

• Posts: 159

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

« 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.

#### matteo.basei

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

« 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.

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

« Reply #27 on: May 24, 2020, 06:50:42 AM »
This is $$c^z$$, see 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.

#### matteo.basei

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

« Reply #28 on: May 24, 2020, 08:05:25 AM »
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 »

#### matteo.basei

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#### Re: z^z + c

« 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
$\sqrt[z]{c} = c^{1 / z} = \left( e^{c'} \right)^{1 / z} = e^{c' / z}$