Base -1 Daniel Fellow Posts: 248 Threads: 80 Joined: Aug 2007 06/19/2022, 11:34 PM See fractals at tetration of -1  Mandelbrot of exponential map  at -1 Julia set of exponential map at -1 Daniel JmsNxn Ultimate Fellow Posts: 1,176 Threads: 123 Joined: Dec 2010 06/20/2022, 12:40 AM (This post was last modified: 06/20/2022, 12:40 AM by JmsNxn.) Hey, Daniel--could you elaborate further on how you are constructing these graphs/the mathematical theory behind this? I know you are using the fixed point formula $$(-1)^{-1} = -1$$ but could you elaborate further? Which branch of the exponential are you using particularly. I assume this is the Schroder iteration (your Bell matrix approach). But which branch of $$(-1)^z$$ are you choosing. Which is to mean: $$(-1)^z = f_k(z) = e^{\pi i(2k+1) z}$$ for some $$k \in \mathbb{Z}$$. And each has a repelling fixed point at $$z=-1$$ with multiplier $$(2k+1)\pi i$$. I assume that you are doing the entire iteration about these fixed points (every entire function about a repelling fixed point admits an entire iteration). Just curious because this looks really interesting. I'm just interested to know more about the backstory of how these graphs are made!   Please, elaborate! Regards, James. Daniel Fellow Posts: 248 Threads: 80 Joined: Aug 2007 06/20/2022, 01:01 AM (06/20/2022, 12:40 AM)JmsNxn Wrote: Hey, Daniel--could you elaborate further on how you are constructing these graphs/the mathematical theory behind this? I know you are using the fixed point formula $$(-1)^{-1} = -1$$ but could you elaborate further? Which branch of the exponential are you using particularly. I assume this is the Schroder iteration (your Bell matrix approach). But which branch of $$(-1)^z$$ are you choosing. Which is to mean: $$(-1)^z = f_k(z) = e^{\pi i(2k+1) z}$$ for some $$k \in \mathbb{Z}$$. And each has a repelling fixed point at $$z=-1$$ with multiplier $$(2k+1)\pi i$$. I assume that you are doing the entire iteration about these fixed points (every entire function about a repelling fixed point admits an entire iteration). Just curious because this looks really interesting. I'm just interested to know more about the backstory of how these graphs are made!   Please, elaborate! Regards, James. These fractals were made thirty years ago with FractInt, a versatile fractal generator with a programming language. As you can see in the code, the algorithms are simple that generated the fractals.   Tetration (exponential map) Mandelbrot set Code:TetrationM (XAXIS) {;   z = pixel:    z = pixel ^ z     |z| <= 100000   }  Tetration (exponential map) Julia set Code:TetraJ (XAXIS) {;   z = pixel:    z = P1 ^ z     |z| <= 100000   } Daniel JmsNxn Ultimate Fellow Posts: 1,176 Threads: 123 Joined: Dec 2010 06/20/2022, 02:37 AM Daniel, please explain better. I get that that makes sense to you. Please elaborate further. At the risk of sounding stupid. Explain more. Elaborate. Can you elaborate further from the Fractint reference? I didn't get much from this that I could use to answer my original question. Any help would be greatly appreciated. Are these just $$f(z) = e^{\pi i z}$$ and $$F(s)$$ is the iterate? Please, elaborate futher. Daniel Fellow Posts: 248 Threads: 80 Joined: Aug 2007 06/20/2022, 03:21 AM (This post was last modified: 06/20/2022, 03:24 AM by Daniel.) (06/20/2022, 01:01 AM)Daniel Wrote: (06/20/2022, 12:40 AM)JmsNxn Wrote: Hey, Daniel--could you elaborate further on how you are constructing these graphs/the mathematical theory behind this? I know you are using the fixed point formula $$(-1)^{-1} = -1$$ but could you elaborate further? Which branch of the exponential are you using particularly. I assume this is the Schroder iteration (your Bell matrix approach). But which branch of $$(-1)^z$$ are you choosing. Which is to mean: $$(-1)^z = f_k(z) = e^{\pi i(2k+1) z}$$ for some $$k \in \mathbb{Z}$$. And each has a repelling fixed point at $$z=-1$$ with multiplier $$(2k+1)\pi i$$. I assume that you are doing the entire iteration about these fixed points (every entire function about a repelling fixed point admits an entire iteration). Just curious because this looks really interesting. I'm just interested to know more about the backstory of how these graphs are made!   Please, elaborate! Regards, James. These fractals were made thirty years ago with FractInt, a versatile fractal generator with a programming language. As you can see in the code, the algorithms are simple that generated the fractals.   Tetration (exponential map) Mandelbrot set Code:TetrationM (XAXIS) {;   z = pixel:    z = pixel ^ z     |z| <= 100000   }  Tetration (exponential map) Julia set Code:TetraJ (XAXIS) {;   z = pixel:    z = P1 ^ z     |z| <= 100000   } (06/20/2022, 02:37 AM)JmsNxn Wrote: Daniel, please explain better. I get that that makes sense to you. Please elaborate further. At the risk of sounding stupid. Explain more. Elaborate. Can you elaborate further from the Fractint reference? I didn't get much from this that I could use to answer my original question. Any help would be greatly appreciated. Are these just $$f(z) = e^{\pi i z}$$ and $$F(s)$$ is the iterate? Please, elaborate futher. Tetration (exponential map) Mandelbrot set - the Mandelbrot set with the quadratic equation replaced by a=pixel and $a^z$  Code:TetrationM (XAXIS) {; \\ x axis symmetry   z = pixel:          \\ initialize setting z to the value of the current pixel converted to a complex number    z = pixel ^ z      \\ iterate once the original complex value of the pixel     |z| <= 100000     \\ iterate until |z| becomes larger than 100,000. The number of iterations is the escape value                       \\ represented by the color of the pixel.   }  Tetration (exponential map) Julia set Code:TetraJ (XAXIS) {;   \\ x axis symmetry   z = pixel:        \\ initialize setting z to the value of the current pixel converted to a complex number    z = P1 ^ z       \\ P1 is a user defined complex variable set at runtime.                     \\ iterate once P1 to the power of z     |z| <= 100000   \\ iterate until |z| becomes larger than 100,000. The number of iterations is the escape value                     \\ represented by the color of the pixel.   } Daniel Catullus Fellow Posts: 213 Threads: 47 Joined: Jun 2022 07/06/2022, 09:37 AM (This post was last modified: 08/14/2022, 10:55 PM by Catullus.) $\dpi{110} -1\uparrow\uparrow1=-1$. $\dpi{110} -1$ is a fixed point of $\dpi{110} -1\uparrow x$. A fixed point is also a 1-cycle, so the issue with iterated exponentials at n-cycles happens with $\dpi{110} -1\uparrow\uparrow1$. What if $\dpi{110} -1\uparrow\uparrow x=-1\forall x$? Please remember to stay hydrated. ฅ(ﾐ⚈ ﻌ ⚈ﾐ)ฅ Sincerely: Catullus /ᐠ_ ꞈ _ᐟ\ « Next Oldest | Next Newest »

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