A nice contributor pointed out that the imaginary universe based on i,
which is an interesting idea to find roots to equations that do not have
any, that is to say, roughly, to find the roots of the symmetric curve pointed at $(0,y) in the Hachel system.
Although physicists use incorrect complex products, since for me, the
real part of a complex product is (aa'+bb'), and not (aa'-bb'), they nevertheless manage to find pretty figures.
So I wondered, what would happen if, instead of working with their
equations, we worked with mine.
Into what strange world would we fall, if, instead of using Z=aa'- bb'+i(ab'+a'b), we used the much more logical and natural equation Z=aa'+bb'+i(ab'+a'b).
How would the "Mandelbrot" or the "Julia" obtained be less pretty?
How would the "Mandelbrot" or the "Julia" obtained be less pretty?
You have not the slightest idea of the connection between fractals and complex numbers, but you talk from you psychiatric hospital, copying
fancy names you just discovered yesterday. Pathetic.
F.J.
Into what strange world would we fall, if, instead of using Z=aa'-bb'+i(ab'+a'b), we used the much more logical and natural equation Z=aa'+bb'+i(ab'+a'b).
A nice contributor pointed out that the imaginary universe based on i, which is
an interesting idea to find roots to equations that do not have any, that is to
say, roughly, to find the roots of the symmetric curve pointed at $(0,y) in the
Hachel system.
Although physicists use incorrect complex products, since for me, the real part
of a complex product is (aa'+bb'), and not (aa'-bb'), they nevertheless manage to
find pretty figures.
So I wondered, what would happen if, instead of working with their equations, we
worked with mine.
Into what strange world would we fall, if, instead of using Z=aa'-bb'+i(ab'+a'b), we used the much more logical and natural equation Z=aa'+bb'+i(ab'+a'b).
A nice contributor pointed out that the imaginary universe based on i,
which is an interesting idea to find roots to equations that do not have
any,
that is to say, roughly, to find the roots of the symmetric curve
pointed at $(0,y) in the Hachel system.
Although physicists use incorrect complex products, since for me, the
real part of a complex product is (aa'+bb'), and not (aa'-bb'), they nevertheless manage to find pretty figures.
So I wondered, what would happen if, instead of working with their
equations, we worked with mine.
Into what strange world would we fall, if, instead of using Z=aa'- bb'+i(ab'+a'b), we used the much more logical and natural equation Z=aa'+bb'+i(ab'+a'b).
How would the "Mandelbrot" or the "Julia" obtained be less pretty?
Isn't beauty the splendor of truth?
R.H. (suivi sci.math)
Le 09/03/2025 à 00:54, Richard Hachel a écrit :
...
Into what strange world would we fall, if, instead of using
Z=aa'-bb'+i(ab'+a'b), we used the much more logical and natural equation
Z=aa'+bb'+i(ab'+a'b).
What makes you claim that ( aa' + bb', ab' + a'b ) is "more logical and natural"
than ( aa' - bb', ab' + a'b) ?
On 3/8/2025 3:54 PM, Richard Hachel wrote:
So, give me your versions of the functions using your new stuff, mainly (addition and multiplication), and I will give it a go when I get some
free time to burn. Giving me a length would also help me. Otherwise I
will just use the traditional version. I am interested in what it might actually look like via a plot...
Op 09/03/2025 om 00:54 schreef Richard Hachel:
A nice contributor pointed out that the imaginary universe based on i,
which is an interesting idea to find roots to equations that do not
have any,
This is perhaps a misunderstanding on your part. Complex numbers will
also crop up for polynomials that have real roots where you still need
to deal with complex numbers in order to obtain these real roots.
On 3/8/2025 3:54 PM, Richard Hachel wrote:
A nice contributor pointed out that the imaginary universe based on i,
which is an interesting idea to find roots to equations that do not
have any, that is to say, roughly, to find the roots of the symmetric
curve pointed at $(0,y) in the Hachel system.
Although physicists use incorrect complex products, since for me, the
real part of a complex product is (aa'+bb'), and not (aa'-bb'), they
nevertheless manage to find pretty figures.
So I wondered, what would happen if, instead of working with their
equations, we worked with mine.
Into what strange world would we fall, if, instead of using Z=aa'-
bb'+i(ab'+a'b), we used the much more logical and natural equation
Z=aa'+bb'+i(ab'+a'b).
How would the "Mandelbrot" or the "Julia" obtained be less pretty?
Isn't beauty the splendor of truth?
Give me your version of the addition and multiplication functions of
your new version, and I should be able to try it out.
Here are mine, using my JavaScript code from:
(note: the vectors are 2-ary here)
https://fractallife247.com/ct_main.js
For addition:
__________________________
function ct_vec2_add(v0, v1) {
return [v0[0] + v1[0], v0[1] + v1[1]];
}
__________________________
For Multiplication:
__________________________
function ct_vec2_complex_mul(v0, v1) {
var x = v0[0] * v1[0] - v0[1] * v1[1];
var y = v0[0] * v1[1] + v0[1] * v1[0];
return [x, y];
}
__________________________
Also, a length of a vector might be useful to me even though I can infer
it from your versions of the two functions above:
__________________________
function ct_vec2_length(v0) {
return Math.sqrt(v0[0] * v0[0] + v0[1] * v0[1]);
}
function ct_vec2_complex_abs(v0) {
return ct_vec2_length(v0);
}
__________________________
So, give me your versions of the functions using your new stuff, mainly (addition and multiplication), and I will give it a go when I get some
free time to burn. Giving me a length would also help me. Otherwise I
will just use the traditional version. I am interested in what it might actually look like via a plot...
Am 09.03.2025 um 01:23 schrieb sobriquet:
Op 09/03/2025 om 00:54 schreef Richard Hachel:
A nice contributor pointed out that the imaginary universe based on
i, which is an interesting idea to find roots to equations that do
not have any,
This is perhaps a misunderstanding on your part. Complex numbers will
also crop up for polynomials that have real roots where you still need
to deal with complex numbers in order to obtain these real roots.
Actually, from a historical point of view, that was the first time a
number - now called i - with the property i*i = -1 was used in
calculations. So at least i (defined this way) *is* useful. (*sigh*)
https://www.sciencefocus.com/science/a-brief-introduction-to-imaginary- numbers
Seems that Hachel is a complete idiot.
Python already gave the answer. Please read. This silly rule transforms
the Mandelbrot set into a square :)
But the dumb man [Hachel] ]never never read the answers. He is lost in his sick mind.
Python already gave the answer. Please read. This silly rule transforms
the Mandelbrot set into a square :)
But the dumb man [Hachel] never never read the answers. He is lost in his sick mind.
Le 09/03/2025 à 00:54, Richard Hachel a écrit :
Into what [...] world would we fall, if, instead of using Z = aa'-bb' + i(ab'+a'b), we used the [...] equation Z = aa'+bb' + i(ab'+a'b).
Le 09/03/2025 à 00:54, Richard Hachel a écrit :
Into what [...] world would we fall, if, instead of using Z = aa'-bb' + i(ab'+a'b), we used the [...] equation Z = aa'+bb' + i(ab'+a'b).
You are a fractal of misunderstandings and confusions. Even your
confusions are confused...
Am 09.03.2025 um 01:28 schrieb Python:
Le 09/03/2025 à 00:54, Richard Hachel a écrit :
Into what [...] world would we fall, if, instead of using Z = aa'-bb'
+ i(ab'+a'b), we used the [...] equation Z = aa'+bb' + i(ab'+a'b).
In the world of split-complex numbers which differs from the world of
complex numbers. *sigh*
Le 09/03/2025 à 01:28, Python a écrit :
Le 09/03/2025 à 00:54, Richard Hachel a écrit :
...
Into what strange world would we fall, if, instead of using
Z=aa'-bb'+i(ab'+a'b), we used the much more logical and natural equation >>> Z=aa'+bb'+i(ab'+a'b).
What makes you claim that ( aa' + bb', ab' + a'b ) is "more logical and natural"
than ( aa' - bb', ab' + a'b) ?
Several reasons.
First, the simplicity of writing, all in positive terms.
The concordance with statistical science.
The ease of natural understanding (Plougastel high school).
The perpetual monitoring of derivations using i=-1 as a simple numerical concordance check.
The simplicity of the explanation and the generalization of i^x=-1
The visualization of the associated curves symmetrical in $(0,y).
Roots often much simpler to calculate and write...
So basically, it's simpler and more beautiful.
Now what does that change for fractals, for Gauss's representation, for experimentally validated equations, I don't know.
Indeed. Sorry for the stupid question, but the following parts of RH's description:
Z=aa'+bb'+i(ab'+a'b) and not Z=aa'-bb'+i(ab'+a'b).
Means a = x component, b = y component, right? ;^o
On 3/10/2025 1:33 AM, efji wrote:
Le 10/03/2025 à 03:03, Chris M. Thomasson a écrit :
Indeed. Sorry for the stupid question, but the following parts of
RH's description:
Z=aa'+bb'+i(ab'+a'b) and not Z=aa'-bb'+i(ab'+a'b).
Means a = x component, b = y component, right? ;^o
Well...
In his disturbed mind, (a,b) = a-b on the x axis :)
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).
Humm... I am not all that familiar with the split complex numbers.
Something like this?
glm::vec2
ct_complex_split_mul(
const glm::vec2& z1,
const glm::vec2& z2
) {
return {
z1.x * z2.x + z1.y * z2.y,
z1.x * z2.y + z1.y * z2.x
};
}
Where:
glm::vec2 s0 = { 0, 1 };
std::cout << "s0 * s0 = " << ct_complex_split_mul(s0, s0) << "\n";
outputs:
s0 * s0 = (1, 0)
?
On 3/10/2025 1:33 AM, efji wrote:
Le 10/03/2025 à 03:03, Chris M. Thomasson a écrit :
Indeed. Sorry for the stupid question, but the following parts of
RH's description:
Z=aa'+bb'+i(ab'+a'b) and not Z=aa'-bb'+i(ab'+a'b).
Means a = x component, b = y component, right? ;^o
Well...
In his disturbed mind, (a,b) = a-b on the x axis :)
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).
Here is what I got for my little test:
https://i.ibb.co/WWwhqh9K/image.png
Oh how fun!
Well...
In his disturbed mind, (a,b) = a-b on the x axis :)
On 3/9/2025 1:33 AM, efji wrote:
Le 09/03/2025 à 05:35, Chris M. Thomasson a écrit :
On 3/8/2025 3:54 PM, Richard Hachel wrote:
A nice contributor pointed out that the imaginary universe based on
i, which is an interesting idea to find roots to equations that do
not have any, that is to say, roughly, to find the roots of the
symmetric curve pointed at $(0,y) in the Hachel system.
Although physicists use incorrect complex products, since for me, the
real part of a complex product is (aa'+bb'), and not (aa'-bb'), they
nevertheless manage to find pretty figures.
So I wondered, what would happen if, instead of working with their
equations, we worked with mine.
Into what strange world would we fall, if, instead of using Z=aa'-
bb'+i(ab'+a'b), we used the much more logical and natural equation
Z=aa'+bb'+i(ab'+a'b).
How would the "Mandelbrot" or the "Julia" obtained be less pretty?
Isn't beauty the splendor of truth?
Give me your version of the addition and multiplication functions of
your new version, and I should be able to try it out.
Here are mine, using my JavaScript code from:
(note: the vectors are 2-ary here)
https://fractallife247.com/ct_main.js
For addition:
__________________________
function ct_vec2_add(v0, v1) {
return [v0[0] + v1[0], v0[1] + v1[1]];
}
__________________________
For Multiplication:
__________________________
function ct_vec2_complex_mul(v0, v1) {
var x = v0[0] * v1[0] - v0[1] * v1[1];
var y = v0[0] * v1[1] + v0[1] * v1[0];
return [x, y];
}
__________________________
Also, a length of a vector might be useful to me even though I can
infer it from your versions of the two functions above:
__________________________
function ct_vec2_length(v0) {
return Math.sqrt(v0[0] * v0[0] + v0[1] * v0[1]);
}
function ct_vec2_complex_abs(v0) {
return ct_vec2_length(v0);
}
__________________________
So, give me your versions of the functions using your new stuff,
mainly (addition and multiplication), and I will give it a go when I
get some free time to burn. Giving me a length would also help me.
Otherwise I will just use the traditional version. I am interested in
what it might actually look like via a plot...
Python already gave the answer. Please read. This silly rule transforms
the Mandelbrot set into a square :)
But the dumb man never never read the answers. He is lost in his sick mind. >>
Indeed. Sorry for the stupid question, but the following parts of RH's description:
Z=aa'+bb'+i(ab'+a'b) and not Z=aa'-bb'+i(ab'+a'b).
Means a = x component, b = y component, right? ;^o
Le 10/03/2025 à 09:33, efji a écrit :
Well...
In his disturbed mind, (a,b) = a-b on the x axis :)
Yes.
It's what i said.
Le 10/03/2025 à 14:37, Richard Hachel a écrit :
Le 10/03/2025 à 09:33, efji a écrit :
Well...
In his disturbed mind, (a,b) = a-b on the x axis :)
Yes.
It's what i said.
Then you get a 1d fractal...
Le 10/03/2025 à 14:52, efji a écrit :
Le 10/03/2025 à 14:37, Richard Hachel a écrit :
Le 10/03/2025 à 09:33, efji a écrit :
Well...
In his disturbed mind, (a,b) = a-b on the x axis :)
Yes.
It's what i said.
Then you get a 1d fractal...
No.
It is true that my method is much simpler and more elegant to find the complex
roots of functions, and that it is much more detailed. I explain in concrete terms
what we do as rotation and with what, and I explain what i is, and its universality.
Therefore, in the case of orthonormal Cartesian frames, there is no longer any
need to try to place abstract roots in 3D that have nothing to do with the subject. A simple x'Ox axis is enough without even using the y'Oy dimension. It is
enough to cross an inverse axis i'Oi (from right to left) to place my complex coordinates there.
It is very simple.
We can do complex additions there if we want.
Z=(a+a')+i(b+b').
Now, if we talk about products of complexes, it is not the same thing.
We must place a second axis in the horizontal plane x'2Ox2, similar to x'Ox, with its own counter-axis i'Oi, without needing to touch y'Oy which remains as is.
We will therefore be able, on this horizontal plane, to practice the multiplication of complex numbers.
This has nothing to do with Gaussian geometry, which is "something else".
Le 10/03/2025 à 15:24, Richard Hachel a écrit :
It is true that my method is much simpler and more elegant to find
the complex roots of functions, and that it is much more detailed.
On 3/10/2025 12:45 PM, Moebius wrote:
Am 10.03.2025 um 18:15 schrieb Python:
Le 10/03/2025 à 15:24, Richard Hachel a écrit :
Delusion:
It is true that my method is much simpler and more elegant to find
the complex roots of functions, and that it is much more detailed.
Holy shit! What a moron.
A picture of Hachel/Lengrand?
https://as2.ftcdn.net/v2/jpg/03/62/42/65/1000_F_362426568_9Zmxa9SgYiUcGdIrg9LLxtBfgoJ9IebH.jpg
Wow. What is he looking at? ;^)
Anyway I need to thank everybody for making me give split complex
numbers a go. Thanks.
Am 10.03.2025 um 18:15 schrieb Python:
Le 10/03/2025 à 15:24, Richard Hachel a écrit :
Delusion:
It is true that my method is much simpler and more elegant to find
the complex roots of functions, and that it is much more detailed.
Holy shit! What a moron.
A picture of Hachel/Lengrand?
https://as2.ftcdn.net/v2/jpg/03/62/42/65/1000_F_362426568_9Zmxa9SgYiUcGdIrg9LLxtBfgoJ9IebH.jpg
Python wrote:
f(x) = x^2
i^4 = -1
Absolument.
f(i^4) = (i^4)^2 = i^8 = -1
Oui.
i^8=-1.
f(-1) = (-1)^4 = 1
Oui.
[...]
Hence 1 = -1 or a = b does not imply that f(a) = f(b) anymore. Such is
his "system". Sigh.
On 3/8/2025 3:54 PM, Richard Hachel wrote:
The sad part is that everyone but Lengrand/Hachel learnt something here.
Le 10/03/2025 à 22:21, "Chris M. Thomasson" a écrit :
On 3/8/2025 3:54 PM, Richard Hachel wrote:
<http://nemoweb.net/jntp?gpYr5eUNsnuIlHzO4tsg1kWypUg@jntp/Data.Media:1>
What is this?
R.H.
Le 11/03/2025 à 00:19, Python a écrit :
Le 11/03/2025 à 00:09, Richard Hachel a écrit :
Le 10/03/2025 à 22:21, "Chris M. Thomasson" a écrit :
On 3/8/2025 3:54 PM, Richard Hachel wrote:
<http://nemoweb.net/jntp?gpYr5eUNsnuIlHzO4tsg1kWypUg@jntp/Data.Media:1>
What is this?
R.H.
In C (complex numbers as defined in math i.e. R[X]/(X^2 + 1)) consider
the sequence :
(z_n and c are complex numbers)
z_0 = 0 + 0i
z_(n+1) = (z_n)^2 + c
if (z_(n)) does not go to infinity then c is a member of Mandelbrot's set
(it is also the set of c for which another set (Julia's set J_c) is
connexe)
You can explore it there on line (zoom on the border, it is quite
fascinating) :
https://mandel.gart.nz/
There the set itself is the black part, the colors at the border
represents how fast the sequence diverges.
Note that the small "copies" of the whole set you would notice look not
quite right at high zoom, this is because of limited (fixed) precision
of floating point numbers in JavaScript.
I implemented this on a SUN Workstation in C on SunView Desktop
Environment (no X11) for UNIX SunOS, as a student, back in 1992 using variable (arbitrary) precision floating point numbers (long floats). Computing was slow back then but small copies of the set looked exactly
as the whole set even at high zoom.
Le 11/03/2025 à 00:09, Richard Hachel a écrit :
Le 10/03/2025 à 22:21, "Chris M. Thomasson" a écrit :
On 3/8/2025 3:54 PM, Richard Hachel wrote:
<http://nemoweb.net/jntp?gpYr5eUNsnuIlHzO4tsg1kWypUg@jntp/Data.Media:1>
What is this?
R.H.
In C (complex numbers as defined in math i.e. R[X]/(X^2 + 1)) consider the sequence :
(z_n and c are complex numbers)
z_0 = 0 + 0i
z_(n+1) = (z_n)^2 + c
if (z_(n)) does not go to infinity then c is a member of Mandelbrot's set
(it is also the set of c for which another set (Julia's set J_c) is connexe)
You can explore it there on line (zoom on the border, it is quite fascinating) :
https://mandel.gart.nz/
There the set itself is the black part, the colors at the border represents how
fast the sequence diverges.
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