Sujet : Re: The splendor of true
De : efji (at) *nospam* efi.efji (efji)
Groupes : sci.mathDate : 10. Mar 2025, 10:46:43
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <vqmce3$191rj$2@dont-email.me>
References : 1 2 3 4 5 6
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Le 10/03/2025 à 10:19, Chris M. Thomasson a écrit :
On 3/10/2025 1:33 AM, efji wrote:
Le 10/03/2025 à 03:03, Chris M. Thomasson a écrit :
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Indeed. Sorry for the stupid question, but the following parts of RH's description:
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Z=aa'+bb'+i(ab'+a'b) and not Z=aa'-bb'+i(ab'+a'b).
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Means a = x component, b = y component, right? ;^o
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Well...
In his disturbed mind, (a,b) = a-b on the x axis :)
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But forget the pathetic egotic and just try a+jb=(a,b) un R^2 with the rule (a+ib)*(a'+jb') = aa'+bb'+j(ab'+a'b) which is the rule on the split-complex set, (thus j^2=1).
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Here is what I got for my little test:
https://i.ibb.co/WWwhqh9K/image.png
Oh how fun!
That's what Python found using Wolfram.
That is not surprising :
Mandelbrot set is generated by the iterative sequence:
z_0=(0,0),
z_{n+1} = z_n^2 + c
The set is defined by all the c \in \C such that the sequence (z_n) is bounded.
With the multiplication rule used in the split-complex set, it is probably easy to show that if |c_1|+|c_2|<1, then the sequence is bounded.
BTW, did you know that the name "Mandelbrot set" is quite an usurpation?
Benoit Mandelbrot did not "find" or be the first to give a definition of the Mandelbrot set. It was described by Fatou and Julia before the WW1, without any computer graphics, so they probably did not have a real vision of it, and Mandelbrot was just the first to make a picture of it in the 1980. Adrien Douady, who was a prominent mathematician working on the subject in the 80's (Mandelbrot is not a real mathematician), decided for this name, instead of Fatou-Julia.
-- F.J.