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10/22/2022, 12:59 PM
(This post was last modified: 10/22/2022, 05:01 PM by Daniel.)
See

Flow for background.

I'm testing my tetration software by creating a pentation Fantou set. The two following fractals are are generated by Mathematica and the second by FractInt. The tetration equation that is iterated to get to pentation is \[^z(\sqrt{2})\]

Note that the fractals are essentially the same.

Tetration compared to iterated exponentiation. The x-axis is x while the y-axis is \[^z(\sqrt{2})\].

Daniel

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10/22/2022, 02:11 PM
(This post was last modified: 10/22/2022, 02:31 PM by bo198214.)
(10/22/2022, 12:59 PM)Daniel Wrote: I'm testing my tetration software by creating a pentation Fantou set. The two following fractals are are generated by Mathematica and the second by FractInt. The tetration equation that is iterated to get to pentation is \[^z(\sqrt{2})\]

Wait, wait, wait - if we call the fractal of \(z^{z^{z^{\dots}}}\) tetration fractal (though that naming seems not right to me already - I would call it "exponentiation fractal" because the repeated operation is exponentiation), then I would assume we call the fractal of \({}^{{}^{{}^{\dots}z}z}z\) the pentation fractal and not the fractal of \({}^{{}^{{}^{z}\sqrt{2}}\sqrt{2}}\dots\), which would follow from your formula.

Or more precisely: If \(c\) is the pixel value as a complex number, then the "tetration" fractal would be obtained by the iteration of \(f_c(z)=c^z\), so the "pentation" fractal would be obtained by the iteration of the function \(f_c(z)={}^zc\) while your fractal is obtained by iteration of which function?

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(10/22/2022, 02:11 PM)bo198214 Wrote: (10/22/2022, 12:59 PM)Daniel Wrote: I'm testing my tetration software by creating a pentation Fantou set. The two following fractals are are generated by Mathematica and the second by FractInt. The tetration equation that is iterated to get to pentation is \[^z(\sqrt{2})\]

Wait, wait, wait - if we call the fractal of \(z^{z^{z^{\dots}}}\) tetration fractal (though that naming seems not right to me already - I would call it "exponentiation fractal" because the repeated operation is exponentiation), then I would assume we call the fractal of \({}^{{}^{{}^{\dots}z}z}z\) the pentation fractal and not the fractal of \({}^{{}^{{}^{z}\sqrt{2}}\sqrt{2}}\dots\), which would follow from your formula.

Or more precisely: If \(c\) is the pixel value as a complex number, then the "tetration" fractal would be obtained by the iteration of \(f_c(z)=c^z\), so the "pentation" fractal would be obtained by the iteration of the function \(f_c(z)={}^zc\) while your fractal is obtained by iteration of which function?

I'm iterating \[^z(\sqrt{2})\].

Daniel

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(10/22/2022, 02:56 PM)Daniel Wrote: (10/22/2022, 02:11 PM)bo198214 Wrote: Or more precisely: If \(c\) is the pixel value as a complex number, then the "tetration" fractal would be obtained by the iteration of \(f_c(z)=c^z\), so the "pentation" fractal would be obtained by the iteration of the function \(f_c(z)={}^zc\) while your fractal is obtained by iteration of which function?

I'm iterating \[^z(\sqrt{2})\].

You mean, you iterate \(f(z)={^z}(\sqrt{2})\) with initial value \(z_0=c\).

But a pentation fractal (in your terminology) would be to iterate \(f(z)={^z}c\) with initial value \(z_0=c\) or even \(z_0=0\) or \(z_0=1\).

Do you agree?

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10/22/2022, 11:06 PM
(This post was last modified: 10/22/2022, 11:30 PM by Daniel.)
(10/22/2022, 09:54 PM)bo198214 Wrote: (10/22/2022, 02:56 PM)Daniel Wrote: (10/22/2022, 02:11 PM)bo198214 Wrote: Or more precisely: If \(c\) is the pixel value as a complex number, then the "tetration" fractal would be obtained by the iteration of \(f_c(z)=c^z\), so the "pentation" fractal would be obtained by the iteration of the function \(f_c(z)={}^zc\) while your fractal is obtained by iteration of which function?

I'm iterating \[^z(\sqrt{2})\].

You mean, you iterate \(f(z)={^z}(\sqrt{2})\) with initial value \(z_0=c\).

But a pentation fractal (in your terminology) would be to iterate \(f(z)={^z}c\) with initial value \(z_0=c\) or even \(z_0=0\) or \(z_0=1\).

Do you agree?

\(z_0=1\) because \({^0}(\sqrt{2})=1\).

What I have done here could be done with a number of different methods. The fractals could provide insight as to with methods are equivalent.

\({\sqrt{2}\uparrow^4 n} = \underbrace{^{^{^{^{^1{\sqrt{2}}}\cdot}\cdot}\sqrt{2}}\sqrt{2}}_{n \sqrt{2}}\)

Daniel

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10/23/2022, 07:32 AM
(This post was last modified: 10/23/2022, 07:39 AM by bo198214.)
(10/22/2022, 11:06 PM)Daniel Wrote: (10/22/2022, 09:54 PM)bo198214 Wrote: You mean, you iterate \(f(z)={^z}(\sqrt{2})\) with initial value \(z_0=c\).

But a pentation fractal (in your terminology) would be to iterate \(f(z)={^z}c\) with initial value \(z_0=c\) or even \(z_0=0\) or \(z_0=1\).

Do you agree?

\(z_0=1\) because \({^0}(\sqrt{2})=1\).

Daniel, a fractal is made up of pixels, each representing a complex value \(c\), for each pixel the color of the pixel is associated with the escape behaviour of the iteration of a function \(f_c\). If that function does not depend on \(c\) and the initial value for the iteration also does not depend on \(c\) - like when you suggest that \(f_c(z)={^z}(\sqrt{2})\) and \(z_0=1\) - then the fractal will have the same color for each pixel. Clearly this is not the case in the pictures given by you, hence the function and initial value can not be as you gave them.

So what is really used in the fractal?