New equation.

f(x)=x^4-2x^3+5x²-8x+4

Find the real and complex roots (there are four roots, if we count a
double).

R.H.

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Le 24/02/2025 à 21:23, Barry Schwarz a écrit :

On Mon, 24 Feb 25 18:52:17 +0000, Richard Hachel <[email protected]>
wrote:

New equation.

f(x)=x^4-2x^3+5x²-8x+4

Find the real and complex roots (there are four roots, if we count a >>double).

R.H.

A quartic always has four roots.

1, 1 ,2i, -2i

Fantastic! Your answer is absolutely correct.

R.H.

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Le 24/02/2025 à 21:23, Barry Schwarz a écrit :

On Mon, 24 Feb 25 18:52:17 +0000, Richard Hachel <[email protected]>

A quartic always has four roots.

Here, I would still put a small caveat.
The fact of saying that an equation of degree n has n roots is perhaps not entirely correct.
I ask myself the question.
If for example we write f(x)=x^3+3x-4, it is indeed an equation of degree
3.
But how many roots, and what are they?
I asked this question to mathematicians, and to artificial intelligence,
and I was given three roots, but they are incorrect, because those who
answer do not seem to understand the real concept of imaginary numbers.
There is in fact only one root.
A very strange root composed of a real root and a complex root. Both
placed on the same point A(1,0) and A(-i,0).

R.H.

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On Mon, 24 Feb 25 18:52:17 +0000, Richard Hachel <[email protected]>
wrote:

New equation.

f(x)=x^4-2x^3+5x�-8x+4

Find the real and complex roots (there are four roots, if we count a
double).

R.H.

A quartic always has four roots.

1, 1 ,2i, -2i

--
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Op 24/02/2025 om 22:11 schreef Richard Hachel:

Le 24/02/2025 à 21:23, Barry Schwarz a écrit :

On Mon, 24 Feb 25 18:52:17 +0000, Richard Hachel <[email protected]>

A quartic always has four roots.

Here, I would still put a small caveat.
The fact of saying that an equation of degree n has n roots is perhaps
not entirely correct.
I ask myself the question.
If for example we write f(x)=x^3+3x-4, it is indeed an equation of
degree 3.
But how many roots, and what are they?
I asked this question to mathematicians, and to artificial intelligence,
and I was given three roots, but they are incorrect, because those who
answer do not seem to understand the real concept of imaginary numbers.
There is in fact only one root.
A very strange root composed of a real root and a complex root. Both
placed on the same point A(1,0) and A(-i,0).

R.H.

https://i.imgur.com/6bUA65Y.png

The conventional definition of complex multiplication is carefully
chosen to preserve geometric interpretations, algebraic properties, and consistency with real numbers. The alternative definition fails to meet
these requirements, making it unsuitable for use in complex analysis and
its applications.

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On Mon, 24 Feb 25 21:11:40 +0000, Richard Hachel <[email protected]>
wrote:

Le 24/02/2025 � 21:23, Barry Schwarz a �crit :

On Mon, 24 Feb 25 18:52:17 +0000, Richard Hachel <[email protected]>

A quartic always has four roots.

Here, I would still put a small caveat.
The fact of saying that an equation of degree n has n roots is perhaps not >entirely correct.
I ask myself the question.
If for example we write f(x)=x^3+3x-4, it is indeed an equation of degree
3.
But how many roots, and what are they?
I asked this question to mathematicians, and to artificial intelligence,
and I was given three roots, but they are incorrect, because those who
answer do not seem to understand the real concept of imaginary numbers.
There is in fact only one root.
A very strange root composed of a real root and a complex root. Both
placed on the same point A(1,0) and A(-i,0).

R.H.

A cubic has three roots.

The roots of your equation are 1, (-1+i*sqrt(15))/2, and
(-1-i*sqrt(15))/2.

I have no idea what you mean by a point A(-i,0).

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Le 25/02/2025 à 09:21, Barry Schwarz a écrit :

On Mon, 24 Feb 25 21:11:40 +0000, Richard Hachel <[email protected]>
wrote:

Le 24/02/2025 à 21:23, Barry Schwarz a écrit :

On Mon, 24 Feb 25 18:52:17 +0000, Richard Hachel <[email protected]>

A quartic always has four roots.

Here, I would still put a small caveat.
The fact of saying that an equation of degree n has n roots is perhaps not >>entirely correct.
I ask myself the question.
If for example we write f(x)=x^3+3x-4, it is indeed an equation of degree >>3.
But how many roots, and what are they?
I asked this question to mathematicians, and to artificial intelligence, >>and I was given three roots, but they are incorrect, because those who >>answer do not seem to understand the real concept of imaginary numbers. >>There is in fact only one root.
A very strange root composed of a real root and a complex root. Both
placed on the same point A(1,0) and A(-i,0).

R.H.

A cubic has three roots.

This is what is generally said, but is it always true?

The roots of your equation are 1, (-1+i*sqrt(15))/2, and
(-1-i*sqrt(15))/2.

That one of the roots is 1, and that it can be represented on a Cartesian coordinate system, is obvious. I then set the point A(1,0).
I then look for the other two roots of the equation, but I realize that I
can't find any others, even complex ones, and that the two complex roots
given are fanciful.
I then start from the principle that the complex roots are the real roots
of the mirror curve, and that the real roots are the complex roots of this other curve, and I find a complex root which is x'=-1.

I therefore obtain the point A(-i,0) which is exactly the same as the
point A(1,0) knowing that i=-1 and -i=+1.

It seems that this curve is its own mirror.

R.H.

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Le 25/02/2025 à 22:36, "Chris M. Thomasson" a écrit :

On 2/25/2025 6:20 AM, Richard Hachel wrote:

Le 25/02/2025 à 09:21, Barry Schwarz a écrit :

On Mon, 24 Feb 25 21:11:40 +0000, Richard Hachel <[email protected]>
wrote:

Le 24/02/2025 à 21:23, Barry Schwarz a écrit :

On Mon, 24 Feb 25 18:52:17 +0000, Richard Hachel <[email protected]> >>>>
A quartic always has four roots.

Here, I would still put a small caveat.
The fact of saying that an equation of degree n has n roots is
perhaps not entirely correct.
I ask myself the question.
If for example we write f(x)=x^3+3x-4, it is indeed an equation of
degree 3.
But how many roots, and what are they?
I asked this question to mathematicians, and to artificial
intelligence, and I was given three roots, but they are incorrect,
because those who answer do not seem to understand the real concept
of imaginary numbers.
There is in fact only one root.
A very strange root composed of a real root and a complex root. Both
placed on the same point A(1,0) and A(-i,0).

R.H.

A cubic has three roots.

This is what is generally said, but is it always true?

The roots of your equation are 1, (-1+i*sqrt(15))/2, and
(-1-i*sqrt(15))/2.

That one of the roots is 1, and that it can be represented on a
Cartesian coordinate system, is obvious. I then set the point A(1,0).
I then look for the other two roots of the equation, but I realize that
I can't find any others, even complex ones, and that the two complex
roots given are fanciful.
I then start from the principle that the complex roots are the real
roots of the mirror curve, and that the real roots are the complex roots
of this other curve, and I find a complex root which is x'=-1.

I therefore obtain the point A(-i,0) which is exactly the same as the
point A(1,0) knowing that i=-1 and -i=+1.

Point (1, 0) = 1+0i
Point (-1, 0) = -1+0i
Point (0, 1) = 0+1i
Point (0, -1) = 0-1i

It seems that this curve is its own mirror.

R.H.

No, no, no, no, no...
I see that you did not understand what I am saying about complex numbers,
and how I would use them in a Cartesian coordinate system.
I use them longitudinally, on the x'Ox axis, but in the opposite
direction.
The complex roots are therefore on the x'Ox axis like the real roots and
are found where the curve g(x) mirror of f(x) passes.


Point (1, 0) = Point (-i,0)
Point (-1, 0) = Point (i,0)
Point (0, 1) = Point (0,1)
Point (0, -1) = Point (0,-1)

Point (5,3) = Point (-5i,3)
Point (-2,-4) = Point (2i,-4)

Imaginary number i is purely ON the x'Ox axe, never elsewhere in
cartesian reference points.

Then there are Argand's representations, where the components of the
complex are perpendicularly dissociated.
But that's something else.


R.H.

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Le 26/02/2025 à 00:03, "Chris M. Thomasson" a écrit :


You don't understand what I'm saying.
But that's okay.
When I use a Cartesian coordinate system, whether in two or three
dimensions, I use two or three real axes.
Ox,Oy,Oz.
So far, so good, everyone understands.
Let's just go back, breathe, blow, to a Cartesian plane, which is very
simple.
I place my "x" on the abscissa, and my "y" on the ordinate.
And finally, I draw my curves...
I draw the curve f(x)=x²+4x+5.
I've been told that this is colossally difficult, and that given the level
of the participants in sci.maths, who are very stupid and barely know how
to draw the line y=2x+1, I shouldn't be talking about curves, and even
less about imaginary numbers.
But I am naturally optimistic, I tell myself that, perhaps, on sci.math
there are intelligent people, more intelligent than the average French
person.
So I will draw my curve, and, surprise! No roots.
So I cannot say that there is a root at A(2,0) and another at B(5,0),
since there is none. There is none.
I repeat (given the stupidity of human beings in general, I have to repeat often), the equation f(x)=x²+4x+5 has no root, and I cannot draw anything
at all on my x'Ox axis.
That is when I realize that, by mirror effect, if I place another mirror
curve that touches the first at the top, my curve will cross my axis at
two points.
Breathe, blow...
This imaginary curve, which is not f(x), I'm going to call it g(x), and
I'm going to give it an equation. g(x)=-x²-4x-3. And there, I'm going to
give it two roots. x'=-3 and x"=-1.
So f(x) has no real roots, but two imaginary roots on its mirror curve,
and g(x) has no imaginary roots, but two real roots.
That said, I cannot grant the real roots of g(x) to f(x), but I can
attribute imaginary mirror roots to it via g(x).
Simply I cannot say that the real roots of f(x) are x'=-3 and x"=-1, I
have to say that they are its imaginary roots of the mirror curve, and to specify it well, it is necessary to write x'=3i and x"=i.
So I can place my points on x'Ox and I place the points A(3i,0) and B(i,0)
on the horizontal axis.
All this remained very simple, and very Cartesian.
At no time did I use the Argand coordinate system (which talks about
totally different things), by giving a perpendicular nature to a+ib,
instead of a simply longitudinal nature in a Cartesian frame.
Imaginary number i in a Cartesian frame, and imaginary number i in an
Argand frame, these are totally different things.
Here, I limit myself to talking about the use of i to find the imaginary
roots of curves in a Cartesian frame.
I repeat: the Argand frame is something completely "different".

R.H.

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Le 26/02/2025 à 01:05, "Chris M. Thomasson" a écrit :

On 2/25/2025 3:58 PM, Richard Hachel wrote:

Le 26/02/2025 à 00:03, "Chris M. Thomasson" a écrit :

You don't understand what I'm saying.
But that's okay.
When I use a Cartesian coordinate system, whether in two or three
dimensions, I use two or three real axes.
Ox,Oy,Oz.
So far, so good, everyone understands.
Let's just go back, breathe, blow, to a Cartesian plane, which is very
simple.
I place my "x" on the abscissa, and my "y" on the ordinate.
And finally, I draw my curves...
I draw the curve f(x)=x²+4x+5.
I've been told that this is colossally difficult, and that given the
level of the participants in sci.maths, who are very stupid and barely
know how to draw the line y=2x+1, I shouldn't be talking about curves,
and even less about imaginary numbers.
But I am naturally optimistic, I tell myself that, perhaps, on sci.math
there are intelligent people, more intelligent than the average French
person.
So I will draw my curve, and, surprise! No roots.
So I cannot say that there is a root at A(2,0) and another at B(5,0),
since there is none. There is none.
I repeat (given the stupidity of human beings in general, I have to
repeat often), the equation f(x)=x²+4x+5 has no root, and I cannot draw
anything at all on my x'Ox axis.
That is when I realize that, by mirror effect, if I place another mirror
curve that touches the first at the top, my curve will cross my axis at
two points.
Breathe, blow...
This imaginary curve, which is not f(x), I'm going to call it g(x), and
I'm going to give it an equation. g(x)=-x²-4x-3. And there, I'm going to
give it two roots. x'=-3 and x"=-1.
So f(x) has no real roots, but two imaginary roots on its mirror curve,
and g(x) has no imaginary roots, but two real roots.
That said, I cannot grant the real roots of g(x) to f(x), but I can
attribute imaginary mirror roots to it via g(x).
Simply I cannot say that the real roots of f(x) are x'=-3 and x"=-1, I
have to say that they are its imaginary roots of the mirror curve, and
to specify it well, it is necessary to write x'=3i and x"=i.
So I can place my points on x'Ox and I place the points A(3i,0) and
B(i,0) on the horizontal axis.
All this remained very simple, and very Cartesian.
At no time did I use the Argand coordinate system (which talks about
totally different things), by giving a perpendicular nature to a+ib,
instead of a simply longitudinal nature in a Cartesian frame.
Imaginary number i in a Cartesian frame, and imaginary number i in an
Argand frame, these are totally different things.
Here, I limit myself to talking about the use of i to find the imaginary
roots of curves in a Cartesian frame.
I repeat: the Argand frame is something completely "different".

I know exactly where to plot say, point 42+21i... Where would you place
in on the plane?

If you read what I just wrote, you can very easily place your point on
your Cartesian coordinate system.
It is impossible in this case (unless we open an Argand coordinate system,
but that is not useful at all here) to place the point anywhere other than
on the x'Ox axis. Thus, your point 42+21i only makes sense in an Argand coordinate system, and not in a Cartesian coordinate system.
But where to put it on x'Ox in the Cartesian coordinate system?
We said that the i-axis is conjunct the x'Ox axis, but inverted. This
means that the abscissa 42+21i is at 42-21=21, and its ordinate at 0 (all ordinates are at y=0).
So we have your point which is at A(21,0).
If your complex was 42-21i, it would be in A(63,0) which you can also
write as A(-63i,0), it is the same thing, written differently.
Be careful, I repeat, here, we are in Cartesian frames,
where i is only a logitudinal elogation carried to a fixed number, and not
in Argand frames which visualize something completely different, and
according to an orthogonal representation.


R.H.

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Op 26/02/2025 om 00:58 schreef Richard Hachel:

Le 26/02/2025 à 00:03, "Chris M. Thomasson" a écrit :

You don't understand what I'm saying.
But that's okay.
When I use a Cartesian coordinate system, whether in two or three
dimensions, I use two or three real axes.
Ox,Oy,Oz.
So far, so good, everyone understands.
Let's just go back, breathe, blow, to a Cartesian plane, which is very simple.
I place my "x" on the abscissa, and my "y" on the ordinate.
And finally, I draw my curves...
I draw the curve f(x)=x²+4x+5.
I've been told that this is colossally difficult, and that given the
level of the participants in sci.maths, who are very stupid and barely
know how to draw the line y=2x+1, I shouldn't be talking about curves,
and even less about imaginary numbers.
But I am naturally optimistic, I tell myself that, perhaps, on sci.math
there are intelligent people, more intelligent than the average French person.
So I will draw my curve, and, surprise! No roots.
So I cannot say that there is a root at A(2,0) and another at B(5,0),
since there is none. There is none.
I repeat (given the stupidity of human beings in general, I have to
repeat often), the equation f(x)=x²+4x+5 has no root, and I cannot draw anything at all on my x'Ox axis.
That is when I realize that, by mirror effect, if I place another mirror curve that touches the first at the top, my curve will cross my axis at
two points.
Breathe, blow...
This imaginary curve, which is not f(x), I'm going to call it g(x), and
I'm going to give it an equation. g(x)=-x²-4x-3. And there, I'm going to give it two roots. x'=-3 and x"=-1.
So f(x) has no real roots, but two imaginary roots on its mirror curve,
and g(x) has no imaginary roots, but two real roots.
That said, I cannot grant the real roots of g(x) to f(x), but I can
attribute imaginary mirror roots to it via g(x).
Simply I cannot say that the real roots of f(x) are x'=-3 and x"=-1, I
have to say that they are its imaginary roots of the mirror curve, and
to specify it well, it is necessary to write x'=3i and x"=i.
So I can place my points on x'Ox and I place the points A(3i,0) and
B(i,0) on the horizontal axis.
All this remained very simple, and very Cartesian.
At no time did I use the Argand coordinate system (which talks about
totally different things), by giving a perpendicular nature to a+ib,
instead of a simply longitudinal nature in a Cartesian frame.
Imaginary number i in a Cartesian frame, and imaginary number i in an
Argand frame, these are totally different things.
Here, I limit myself to talking about the use of i to find the imaginary roots of curves in a Cartesian frame.
I repeat: the Argand frame is something completely "different".

R.H.

Ok so we can visualize this as follows:

https://www.desmos.com/calculator/c3gntlt7kq

the function g1(x) is the reflection of the function g0(x) in your
approach to yield the magenta colored 'roots', while the conventional
approach with complex numbers yields the complex roots colored in yellow.
How about the function g2(x), which has (approximated) roots colored in
green.

https://www.wolframalpha.com/input?i=x%5E2%2B4x-2sqrt%28x%2B3%2F2%29%2Fx+%2B5+%3D0

What roots would your alternative approach yield for that function?

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Le 26/02/2025 à 01:47, sobriquet a écrit :

Op 26/02/2025 om 00:58 schreef Richard Hachel:

Le 26/02/2025 à 00:03, "Chris M. Thomasson" a écrit :

https://www.desmos.com/calculator/c3gntlt7kq

the function g1(x) is the reflection of the function g0(x) in your
approach to yield the magenta colored 'roots', while the conventional approach with complex numbers yields the complex roots colored in yellow.
How about the function g2(x), which has (approximated) roots colored in green.

I don't understand how you mix Cartesian and Argand coordinate systems.
They are not the same thing.
The curve f(x)=x²+4x+5 must be represented on a Cartesian coordinate
system, no relation to a complex Argand coordinate system.
Similarly, the curve g(x)=-x²-4x-3 which is the mirror curve must be represented on a Cartesian coordinate system.
We breathe, we blow.
I will then represent the real roots of the curve g(x), since f(x) does
not have any. So I find x'=-3 and x"=-1. These are the two small majenta
points that you represented. The coordinates are A(-3,0) and B(-1,0).
There is nothing here that is difficult to understand.
Now, I must also place my two small yellow points which are the complex
roots of f(x). Now, I told you that the complex roots of a function are
the real roots of the mirror function, and conversely, the real roots of a function are the complete roots of its mirror function.
You must therefore place your small yellow points ON your small magenta
points.
The way you place them seems to represent an Argand reference frame whose interest here is nil.
The complex roots of f(x) are therefore x'=3i and x"=i.
They are the same as A and B.
But as for g(x), we write them in real form, that is to say A(-3,0) and B(-1,0); for f(x), we must write them in such a way that we know that they
are complex roots and the simplest notation is A(3i,0) and B(i,0).

You can, of course, write in the form A(-2+i,0) and B(-2-i,0) but it is
the same thing (+i=-1 ; -i=+1).

Your yellow and majenta points must however be confused, they are the same roots in fact, but one seen from f(x) and complex; the other seen from
g(x) and real.

Placing the majenta or yellow points elsewhere than on x'Ox" has
absolutely no interest here, except to show that we are mixing a concrete Cartesian frame with an Argand frame which is an abstract frame where i is isolated from the complex, then placed vertically.

R.H.

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Op 26/02/2025 om 02:47 schreef Richard Hachel:

Le 26/02/2025 à 01:47, sobriquet a écrit :

Op 26/02/2025 om 00:58 schreef Richard Hachel:

Le 26/02/2025 à 00:03, "Chris M. Thomasson" a écrit :

https://www.desmos.com/calculator/c3gntlt7kq

the function g1(x) is the reflection of the function g0(x) in your
approach to yield the magenta colored 'roots', while the conventional
approach with complex numbers yields the complex roots colored in yellow.
How about the function g2(x), which has (approximated) roots colored
in green.

I don't understand how you mix Cartesian and Argand coordinate systems.
They are not the same thing.
The curve f(x)=x²+4x+5 must be represented on a Cartesian coordinate
system, no relation to a complex Argand coordinate system.

Not necessarily. You can consider that equation in many number systems.
You can consider it in the integers, in the rational numbers, in the
real numbers, in complex numbers and more (like quaternions).

Similarly, the curve g(x)=-x²-4x-3 which is the mirror curve must be represented on a Cartesian coordinate system.

No, it is just a polynomial equation, but that doesn't dictate which
number system we have to use. We can consider the same polynomial
equation in many number systems.

We breathe, we blow.
I will then represent the real roots of the curve g(x), since f(x) does
not have any. So I find x'=-3 and x"=-1. These are the two small majenta points that you represented. The coordinates are A(-3,0) and B(-1,0).
There is nothing here that is difficult to understand.
Now, I must also place my two small yellow points which are the complex
roots of f(x). Now, I told you that the complex roots of a function are
the real roots of the mirror function, and conversely, the real roots of
a function are the complete roots of its mirror function.
You must therefore place your small yellow points ON your small magenta points.
The way you place them seems to represent an Argand reference frame
whose interest here is nil.

The Argand plane is the plane of complex numbers.

https://en.wikipedia.org/wiki/Complex_plane

The complex roots of f(x) are therefore x'=3i and x"=i.
They are the same as A and B.
But as for g(x), we write them in real form, that is to say A(-3,0) and B(-1,0); for f(x), we must write them in such a way that we know that
they are complex roots and the simplest notation is A(3i,0) and B(i,0).

You can, of course, write in the form A(-2+i,0) and B(-2-i,0) but it is
the same thing (+i=-1 ; -i=+1).

Your yellow and majenta points must however be confused, they are the
same roots in fact, but one seen from f(x) and complex; the other seen
from g(x) and real.

Placing the majenta or yellow points elsewhere than on x'Ox" has
absolutely no interest here, except to show that we are mixing a
concrete Cartesian frame with an Argand frame which is an abstract frame where i is isolated from the complex, then placed vertically.

R.H.

We're talking about some sort of extension of the real numbers here and
the complex numbers are such an extension that ensures that polynomials
always have roots in that number system.

The complex number system has a wide range of applications in science
(like in quantum physics) and your alternative system is not a system at
all, because it fails to yield anything meaningful and consistent.
This is evident when you try out your approach by tweaking the functions
a bit and (for instance) considering rational functions (so allowing for
both negative and positive exponents) or fractional exponents. In that
case the conventional approach will hold up, because it's a *system*
that hangs together in a meaningful and consistent way.

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Le 28/02/2025 à 05:51, Ross Finlayson a écrit :

On 02/27/2025 01:46 AM, efji wrote:

Le 27/02/2025 à 05:19, Ross Finlayson a écrit :

Division in complex numbers is opinionated, not unique.

:)
Hachel has a brother !

So, the natural products and alll their combinations
don't necessarily arrive at "staying in the system".

wow

Furthermore, in things like Fourier-style analysis,
which often enough employ numerical methods a.k.a.
approximations here the small-angle approximation
in their derivations, _always have a non-zero error_.

big time BS :)

Then, something like the "identity dimension", sees
instead of going _out_ in the numbers, where complex
numbers and their iterative products may neatly model
reflections and rotations, instead go _in_ the numbers,
making for the envelope of the linear fractional equation,
Clairaut's and d'Alembert's equations, and otherwise
with regards to _integral_ analysis vis-a-vis the
_differential_ analysis.

Nonsense ala Hachel

These each have things the other can't implement,
yet somehow they're part of one thing.

It's called completions since mathematics is replete.

A BS-philosophical version of Hachel. Let's park them together.

Division in complex numbers most surely is
non-unique, whatever troll you are from
whatever troll rock you crawled out from under.

Oh yes?
Show it to the world, we are listening.
How many results for a/b with a an b complex numbers?

Furthermore, if you don't know usual derivations
of Fourier-style analysis and for example about
that the small-angle approximation is a linearisation
and is an approximation and is after a numerical method,
you do _not_ know.

Then about integral analysis and this sort of
"original analysis" and about the identity line
being the envelope of these very usual integral
equations, it certainly is so.

So, crawl back under your troll rock, troll worm.

Hope you'll not make babies with Hachel :)

FYI: I've been teaching Fourier Transform, complex analysis and a bunch
of other maths for quite a few decades in several universities. Don't
try to learn me anything about your crap :)

I discovered a new equation one time, it's another
expression for factorial, sort of like Stirling's,
upon which some quite usual criteria for convergence die.

Good for you. Any publication ?

--
F.J.

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On 2/27/2025 11:51 PM, Ross Finlayson wrote:

On 02/27/2025 01:46 AM, efji wrote:

Le 27/02/2025 à 05:19, Ross Finlayson a écrit :

Division in complex numbers is opinionated,
not unique.

:)
Hachel has a brother !

A BS-philosophical version of Hachel.
Let's park them together.

Division in complex numbers most surely is
non-unique, [...]

In the complex field,
division is unique, except by 0,
and division by 0 isn't at all.
Because the complex field is a field.

A field has operations '+' '⋅' such that
both are associative and commutative,
have identities 0 1 and inverses -x x⁻¹,
except 0⁻¹, and '⋅' distributes over '+'

Consider inverse(s?) x⁻¹ x⁻¹′
x⁻¹⋅x⋅x⁻¹′ = x⁻¹⋅x⋅x⁻¹′
x⁻¹⋅1 = 1⋅x⁻¹′
x⁻¹ = x⁻¹′

The complex field is a field,
in part, because 𝑖² = -1
from which it must follow that,
for (a+b𝑖)⁻¹ = x+y𝑖 such that
(a+b𝑖)⋅(x+y𝑖) = 1
⎛
⎜ (a+b𝑖)⋅(x+y𝑖) = (ax-by)+(bx+ay)𝑖 = 1
⎜
⎜ ax-by = 1
⎜ bx+ay = 0
⎜
⎜ a²x-aby = a
⎜ b²x+aby = 0
⎜ (a²+b²)x = a
⎜ x = a/(a²+b²)
⎜
⎜ abx-b²y = b
⎜ abx+a²y = 0
⎜ (a²+b²)y = -b
⎜ y = -b/(a²+b²)
⎜
⎜ x+y𝑖 = (a-b𝑖)/(a²+b²)
⎜
⎜ (a+b𝑖)⁻¹ = (a-b𝑖)/(a²+b²)
⎝ uniquely, for a²+b² ≠ 0

It's possible that
you are confusing division with
logarithm or square root or some such.

Furthermore, if you don't know usual derivations
of Fourier-style analysis and for example about
that the small-angle approximation is a linearisation
and is an approximation and is after a numerical method,
you do _not_ know.

Then about integral analysis and this sort of
"original analysis" and about the identity line
being the envelope of these very usual integral
equations, it certainly is so.

Technobabble.
And not in a good way.

[...]

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On 2/28/2025 12:55 PM, Ross Finlayson wrote:

On 02/28/2025 05:50 AM, Jim Burns wrote:

On 2/27/2025 11:51 PM, Ross Finlayson wrote:

Division in complex numbers most surely is
non-unique, [...]

In the complex field,
division is unique, except by 0,
and division by 0 isn't at all.
Because the complex field is a field.

The complex field is a field,
in part, because 𝑖² = -1
from which it must follow that,
for (a+b𝑖)⁻¹ = x+y𝑖 such that
  (a+b𝑖)⋅(x+y𝑖) = 1

⎜ (a+b𝑖)⁻¹ = (a-b𝑖)/(a²+b²)
⎝ uniquely, for a²+b² ≠ 0

It's possible that
you are confusing division with
logarithm or square root or some such.

https://www.youtube.com/watch? v=Uv_6g__03_E&list=PLb7rLSBiE7F5_h5sSsWDQmbNGsmm97Fy5&index=33

What do you say in that video which, in your opinion,
contributes to the discussion?

How about the "yin-yang ad-infinitum" bit
that shows directly a failure of inductive inference,
courtesy a simplest fact of geometry, and graphically.

The complex field does not fail at being a field.

Reliable inductive inferences do not fail at being correct.

A reliable inductive inference
concludes, from a subset being inductive,
that that inductive subset is the whole superset.
It is a _reliable_ inductive inference
only where there is only one inductive subset:
the whole superset.

A failure of inductive inference would be
x and c such that x ∈ {c} ∧ x ≠ c

Are you claiming you have such x and c?

Not.ultimately.untrue, say.

In a finite sequence of claims,
either there is no false claim
or there is a first false claim.

In a finite sequence of claims in which
each claim is true.or.not.first.false,
there is no first.false claim.

In a finite sequence of claims in which
each claim is true.or.not.first.false,
there is no false claim.

--- SoupGate-Win32 v1.05
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On 2/28/2025 6:08 PM, Ross Finlayson wrote:

On 02/28/2025 10:46 AM, Jim Burns wrote:

The complex field does not fail at being a field.

Oh, I read a definition of complex numbers
and point out that division is non-unique.

(c+d𝑖)/(a+b𝑖) = (c+d𝑖)⋅(a+b𝑖)⁻¹

x+y𝑖 such that
(a+b𝑖)⋅(x+y𝑖) = 1
is unique.

(x+y𝑖) = (a+b𝑖)⁻¹ = (a-b𝑖)/(a²+b²)

⎛ (a+b𝑖)⋅(x+y𝑖) = (ax-by)+(bx+ay)𝑖 = 1
⎜
⎜ ax-by = 1
⎜ bx+ay = 0
⎜
⎜ a²x-aby = a
⎜ b²x+aby = 0
⎜ (a²+b²)x = a
⎜ x = a/(a²+b²)
⎜
⎜ abx-b²y = b
⎜ abx+a²y = 0
⎜ (a²+b²)y = -b
⎜ y = -b/(a²+b²)
⎜
⎜ x+y𝑖 = (a-b𝑖)/(a²+b²)
⎜
⎜ (a+b𝑖)⁻¹ = (a-b𝑖)/(a²+b²)
⎝ uniquely, for a²+b² ≠ 0

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On 3/1/2025 5:54 PM, Ross Finlayson wrote:

On 03/01/2025 11:11 AM, Jim Burns wrote:

On 2/28/2025 6:08 PM, Ross Finlayson wrote:

On 02/28/2025 10:46 AM, Jim Burns wrote:

The complex field does not fail at being a field.

Oh, I read a definition of complex numbers
and point out that division is non-unique.

(c+d𝑖)/(a+b𝑖) = (c+d𝑖)⋅(a+b𝑖)⁻¹

x+y𝑖 such that
(a+b𝑖)⋅(x+y𝑖) = 1
is unique.

(x+y𝑖) = (a+b𝑖)⁻¹ = (a-b𝑖)/(a²+b²)

⎛ (a+b𝑖)⋅(x+y𝑖) = (ax-by)+(bx+ay)𝑖 = 1
⎜
⎜ ax-by = 1
⎜ bx+ay = 0
⎜
⎜ a²x-aby = a
⎜ b²x+aby = 0
⎜ (a²+b²)x = a
⎜ x = a/(a²+b²)
⎜
⎜ abx-b²y = b
⎜ abx+a²y = 0
⎜ (a²+b²)y = -b
⎜ y = -b/(a²+b²)
⎜
⎜ x+y𝑖 = (a-b𝑖)/(a²+b²)
⎜
⎜ (a+b𝑖)⁻¹ = (a-b𝑖)/(a²+b²)
⎝ uniquely, for a²+b² ≠ 0

Not the other way around though

What do you mean by that?

Oh, I read a definition of complex numbers
and point out that division is non-unique.

Non.zero complex division is uniquely.valued.

(c+d𝑖)/(a+b𝑖) = x+y𝑖

c+d𝑖 = (a+b𝑖)⋅(x+y𝑖)

c+d𝑖 = (ax-by)+(bx+ay)𝑖

ax-by = c
bx+ay = d

a²x-aby = ac
b²x+aby = bd
(a²+b²)x = ac+bd
x = (ac+bd)/(a²+b²)

abx-b²y = bc
abx+a²y = ad
(a²+b²)y = ad-bc
y = (ad-bc)/(a²+b²)

(c+d𝑖)/(a+b𝑖) = x+y𝑖 =
((ac+bd)+(ad-bc)𝑖)/(a²+b²)

Do you (RF) consider
((ac+bd)+(ad-bc)𝑖)/(a²+b²)
multi.valued?
If so, why?

--- SoupGate-Win32 v1.05
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Le 02/03/2025 à 00:27, Jim Burns a écrit :

On 3/1/2025 5:54 PM, Ross Finlayson wrote:

If so, why?

I think that complex numbers need to be reconsidered from the beginning,
as I did for the special theory of relativity, in order to have something perfectly beautiful, coherent, simple and logical.
Beauty is the splendor of truth.
The later blunders of physicists on the Poincaré-Lorentz transformations (however magnificent) are not beautiful, so they are not true.
Their supposed beauty comes from artifice, not from the real essence of
things, and in the end, SR is rotten with mathematical falsehoods.
It is the same for complex numbers. Professors teach it badly, and most
often without understanding anything of the things they draw or explain.
So all this is not true.
We live in the permanent illusion of understanding things that we do not clearly understand, whereas if we were to remove the conceptual confusion, everything would be clear and obvious.

R.H.

--- SoupGate-Win32 v1.05
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On 3/1/2025 9:11 PM, Ross Finlayson wrote:

On 03/01/2025 03:27 PM, Jim Burns wrote:

On 3/1/2025 5:54 PM, Ross Finlayson wrote:

On 03/01/2025 11:11 AM, Jim Burns wrote:

On 2/28/2025 6:08 PM, Ross Finlayson wrote:

On 02/28/2025 10:46 AM, Jim Burns wrote:

The complex field does not fail at being a field.

Oh, I read a definition of complex numbers
and point out that division is non-unique.

(c+d𝑖)/(a+b𝑖) = (c+d𝑖)⋅(a+b𝑖)⁻¹

x+y𝑖 such that
(a+b𝑖)⋅(x+y𝑖) = 1
is unique.

(x+y𝑖) = (a+b𝑖)⁻¹ = (a-b𝑖)/(a²+b²)

[...]

Not the other way around though

What do you mean by that?

I give up trying to get you to explain your self.

Oh, I read a definition of complex numbers
and point out that division is non-unique.

Non.zero complex division is uniquely.valued.

(c+d𝑖)/(a+b𝑖) = x+y𝑖

[...]

Do you (RF) consider
((ac+bd)+(ad-bc)𝑖)/(a²+b²)
multi.valued?
If so, why?

You started with "non-zero complex division
is uniquely valued", that's called "see rule 1".

I started by stating what I was about to prove,
and then I proved it.

⎛ You have math degree. Right?
⎝ Did you do any proofs while earning it?

----
Do you have a problem with any part of the following>

(c+d𝑖)/(a+b𝑖) = x+y𝑖

c+d𝑖 = (a+b𝑖)⋅(x+y𝑖)

c+d𝑖 = (ax-by)+(bx+ay)𝑖

ax-by = c
bx+ay = d

a²x-aby = ac
b²x+aby = bd
(a²+b²)x = ac+bd
x = (ac+bd)/(a²+b²)

abx-b²y = bc
abx+a²y = ad
(a²+b²)y = ad-bc
y = (ad-bc)/(a²+b²)

(c+d𝑖)/(a+b𝑖) = x+y𝑖 =
((ac+bd)+(ad-bc)𝑖)/(a²+b²)

Do you (RF) consider
((ac+bd)+(ad-bc)𝑖)/(a²+b²)
multi.valued?
If so, why?

Find various different formalisms of complex numbers
from all through the libraries and notice it's
given various ways.

Find a formalism of complex numbers _disagreeing with_
⎛
⎜ a+b𝑖 = c+d𝑖 ⇔ a=c ∧ b=d
⎜
⎜ (a+b𝑖)+(c+d𝑖) = (a+c)+(b+d)𝑖
⎜
⎜ (a+b𝑖)⋅(c+d𝑖) = (ac-bd)+(ad+bc)𝑖
⎝
and show it to me.

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Le 02/03/2025 à 02:49, Richard Hachel a écrit :

Le 02/03/2025 à 00:27, Jim Burns a écrit :

On 3/1/2025 5:54 PM, Ross Finlayson wrote:

If so, why?

I think that complex numbers need to be reconsidered from the beginning, as I did for the special theory of relativity, in order to have something perfectly
beautiful, coherent, simple and logical.
Beauty is the splendor of truth.
The later blunders of physicists on the Poincaré-Lorentz transformations (however magnificent) are not beautiful, so they are not true.
Their supposed beauty comes from artifice, not from the real essence of things,
and in the end, SR is rotten with mathematical falsehoods.
It is the same for complex numbers. Professors teach it badly, and most often without understanding anything of the things they draw or explain.
So all this is not true.
We live in the permanent illusion of understanding things that we do not clearly
understand, whereas if we were to remove the conceptual confusion, everything would be clear and obvious.

R.H.

This is delusional bullshit, Richard.

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Le 02/03/2025 à 13:11, Python a écrit :

Le 02/03/2025 à 02:49, Richard Hachel a écrit :

Le 02/03/2025 à 00:27, Jim Burns a écrit :

On 3/1/2025 5:54 PM, Ross Finlayson wrote:

If so, why?

This is delusional bullshit, Richard.

Bof...

"Monsieur Charles de Gaulle, vos notes en sport ne sont pas mirobolantes,
4/20 au saut en longueur,
7/20 à la course de fond, bien que vous ayez bénéficié de la
mansuétude du jury, 6/20 au lancement du poids, 3/20 en haltérophilie,
et 0/20 à la barre fixe.
Vous ne réussirez jamais rien dans la vie".

R.H.

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Le 02/03/2025 à 16:12, Richard Hachel a écrit :

Le 02/03/2025 à 13:11, Python a écrit :

Le 02/03/2025 à 02:49, Richard Hachel a écrit :

Le 02/03/2025 à 00:27, Jim Burns a écrit :

On 3/1/2025 5:54 PM, Ross Finlayson wrote:

If so, why?

This is delusional bullshit, Richard.

Bof...

"Monsieur Charles de Gaulle, vos notes en sport ne sont pas mirobolantes, 4/20
au saut en longueur,
7/20 à la course de fond, bien que vous ayez bénéficié de la mansuétude du
jury, 6/20 au lancement du poids, 3/20 en haltérophilie, et 0/20 à la barre fixe.
Vous ne réussirez jamais rien dans la vie".

R.H.

De Gaulle didn't pretend that complex numbers was ‘wrong’. He probably learn more than you about them at Saint-Cyr military school.

You suck at everything, he didn't.

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Le 02/03/2025 à 17:32, Richard Hachel a écrit :

Le 02/03/2025 à 06:06, Jim Burns a écrit :
⎜

(a+b𝑖)+(c+d𝑖) = (a+c)+(b+d)𝑖

Yes.

(a+b𝑖)⋅(c+d𝑖) = (ac-bd)+(ad+bc)𝑖

No.

(a+b𝑖)⋅(c+d𝑖) = (ac+bd)+(ad+bc)𝑖

In all thing : (a+b𝑖)⋅(c+d𝑖) = (ac+bd)+(ad+bc)𝑖

This is true in R(j) not in C.

I posted about this yesterday. Did you even read my post?

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Le 02/03/2025 à 06:06, Jim Burns a écrit :
⎜

(a+b𝑖)+(c+d𝑖) = (a+c)+(b+d)𝑖

Yes.

(a+b𝑖)⋅(c+d𝑖) = (ac-bd)+(ad+bc)𝑖

No.

(a+b𝑖)⋅(c+d𝑖) = (ac+bd)+(ad+bc)𝑖

Prove:

(a+b𝑖)⋅(c+d𝑖)= ac+adi+cbi+bidi.

i²=i and not other thing.

(a+b𝑖)⋅(c+d𝑖)= ac+adi+cbi+(-b)(-d)) or ac+adi+cbi+(+b)(+d))

In all thing : (a+b𝑖)⋅(c+d𝑖) = (ac+bd)+(ad+bc)𝑖

R.H.

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Le 02/03/2025 à 18:10, Python a écrit :

Le 02/03/2025 à 17:32, Richard Hachel a écrit :

Le 02/03/2025 à 06:06, Jim Burns a écrit :
 ⎜

(a+b𝑖)+(c+d𝑖) = (a+c)+(b+d)𝑖

 Yes.

(a+b𝑖)⋅(c+d𝑖) = (ac-bd)+(ad+bc)𝑖

 No.

 (a+b𝑖)⋅(c+d𝑖) = (ac+bd)+(ad+bc)𝑖


 In all thing : (a+b𝑖)⋅(c+d𝑖) = (ac+bd)+(ad+bc)𝑖

This is true in R(j) not in C.

I posted about this yesterday. Did you even read my post?

He probably read it and did not understand a single word :)

--
F.J.

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On 3/2/2025 11:32 AM, Richard Hachel wrote:

Le 02/03/2025 à 06:06, Jim Burns a écrit :

On 3/1/2025 9:11 PM, Ross Finlayson wrote:

Find various different formalisms of complex numbers
from all through the libraries and notice it's
given various ways.

Find a formalism of complex numbers _disagreeing with_
⎛
⎜ a+b𝑖 = c+d𝑖 ⇔ a=c ∧ b=d
⎜
⎜ (a+b𝑖)+(c+d𝑖) = (a+c)+(b+d)𝑖
⎜
⎜ (a+b𝑖)⋅(c+d𝑖) = (ac-bd)+(ad+bc)𝑖
⎝
and show it to me.

(a+b𝑖)+(c+d𝑖) = (a+c)+(b+d)𝑖

Yes.

(a+b𝑖)⋅(c+d𝑖) = (ac-bd)+(ad+bc)𝑖

No.

(a+b𝑖)⋅(c+d𝑖) = (ac+bd)+(ad+bc)𝑖

Wrong.
Those are flying rainbow sparkle ponies.
(poneys scintillants arc-en-ciel volants)

Why can't I "vote" for flying rainbow sparkle ponies?

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Le 02/03/2025 à 18:10, Python a écrit :

Le 02/03/2025 à 17:32, Richard Hachel a écrit :

Le 02/03/2025 à 06:06, Jim Burns a écrit :
⎜

(a+b𝑖)+(c+d𝑖) = (a+c)+(b+d)𝑖

Yes.

(a+b𝑖)⋅(c+d𝑖) = (ac-bd)+(ad+bc)𝑖

No.

(a+b𝑖)⋅(c+d𝑖) = (ac+bd)+(ad+bc)𝑖


In all thing : (a+b𝑖)⋅(c+d𝑖) = (ac+bd)+(ad+bc)𝑖

This is true in R(j) not in C.

Non, c'est bon aussi dans C.

C'est (a+b𝑖)⋅(c+d𝑖) = (ac-bd)+𝑖(ad+bc) qui n'est plus bon.

Maintenant, on respire, on souffle.

Quatre cas de figures peuvent se présenter (en fait 3, puisque 2 et 3,
c'est la même chose).

1. (a+b𝑖)⋅(c+d𝑖)
2. (a+b𝑖)⋅(c-d𝑖)
4. (a-b𝑖)⋅(c-d𝑖)

Il ne suffit pas de regarder les petits symboles sur le papier mais il
faut se dire : à quoi correspondent les choses dont je me permets de
parler?

Il faut que les choses aient un sens pratique.

Qu'est ce que je fais si je multiplie, en fait, (a+b𝑖)⋅(c+d𝑖)?

N'est ce pas multiplier entre elles, les deux racines inférieures de
deux polynômes différents?

Question : mais cela me sert à quoi?

Posons une équation quadratique f'(x), dont les racines sont -1 et +3.

Et une autre g(x) dans les racines sont +2 et +9.

Allons multiplier entre elles, les deux racines inférieures de ces deux polynômes différents.

Hourrah, je trouve -2.

Et devant mes sauts de cabris, je fait quoi pour l'Ukraine?

R.H.

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Le 02/03/2025 à 18:54, Richard Hachel a écrit :

Le 02/03/2025 à 18:10, Python a écrit :

Le 02/03/2025 à 17:32, Richard Hachel a écrit :

Le 02/03/2025 à 06:06, Jim Burns a écrit :
⎜

(a+b𝑖)+(c+d𝑖) = (a+c)+(b+d)𝑖

Yes.

(a+b𝑖)⋅(c+d𝑖) = (ac-bd)+(ad+bc)𝑖

No.

(a+b𝑖)⋅(c+d𝑖) = (ac+bd)+(ad+bc)𝑖


In all thing : (a+b𝑖)⋅(c+d𝑖) = (ac+bd)+(ad+bc)𝑖

This is true in R(j) not in C.

Non, c'est bon aussi dans C.

C'est (a+b𝑖)⋅(c+d𝑖) = (ac-bd)+𝑖(ad+bc) qui n'est plus bon.

This is the consequence of the definition of what is called C. Period.

If you have another rule for the product of terms you are not talking
about C but another algebraic structure such as R(j) or R(epsilon).

There is no way to reject a definition as long at it is consistent. And
the definition of C is consistent. Definitions of other structures can be consistent too, see my yesterday post this cannot change what the
definition of C is.

Seriously your stubborn stupidity, ignorance and hypocrisy is boring
Richard.

Il ne suffit pas de regarder les petits symboles sur le papier mais il faut se
dire : à quoi correspondent les choses dont je me permets de parler?

This is exactly what we do and what you don't do.

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Le 02/03/2025 à 18:54, Richard Hachel a écrit :

Il faut que les choses aient un sens pratique.

1/ Speak english!
2/ NO! Not at all. Mathematics are founded on pure abstract concepts,
without any "practical sense". Applications are only byproducts. Many
parts of the contemporary mathematics have no applications at all or
promises of applications (yet!). But you are so far away from the
mathematics developed today, stuck into childish and pathological considerations that have been well known for centuries, and that
millions of people on earth can understand very easily.

PS: I am myself an applied mathematician.

--
F.J.

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Le 02/03/2025 à 19:27, Richard Hachel a écrit :

Le 02/03/2025 à 19:16, efji a écrit :

Le 02/03/2025 à 18:54, Richard Hachel a écrit :

PS: I am myself an applied mathematician.

Like Python?

Poor France.

At least we are not making idiotic blunders like you, asserting that the
graph of f(-x) is the point symetric of the graph of f(x).

Given the amount of fallacies, contradictions and nonsense you've posted, Richard, you'd better not brag and shut up your big mouth.

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Le 02/03/2025 à 19:16, efji a écrit :

Le 02/03/2025 à 18:54, Richard Hachel a écrit :

PS: I am myself an applied mathematician.

Like Python?

Poor France.

R.H.

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Le 02/03/2025 à 19:40, Python a écrit :

At least we are not making idiotic blunders like you, asserting that the graph
of f(-x) is the point symetric of the graph of f(x).

I didn't write this, but artificial intelligence.

Given the amount of fallacies, contradictions and nonsense you've posted, Richard, you'd better not brag and shut up your big mouth.

Shut yours first.

It only says stupid bullshit.

R.H.

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Le 02/03/2025 à 19:45, Richard Hachel a écrit :

Le 02/03/2025 à 19:40, Python a écrit :

At least we are not making idiotic blunders like you, asserting that the graph
of f(-x) is the point symetric of the graph of f(x).

I didn't write this, but artificial intelligence.

But you didn't notice, genius... then you posted and approved it.

You are 1000% out of you domain of competence.

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On 3/2/2025 3:58 PM, Chris M. Thomasson wrote:

On 3/2/2025 9:55 AM, Jim Burns wrote:

On 3/2/2025 11:32 AM, Richard Hachel wrote:

Le 02/03/2025 à 06:06, Jim Burns a écrit :

(a+b𝑖)⋅(c+d𝑖) = (ac-bd)+(ad+bc)𝑖

No.

(a+b𝑖)⋅(c+d𝑖) = (ac+bd)+(ad+bc)𝑖

Wrong.
Those are flying rainbow sparkle ponies.
(poneys scintillants arc-en-ciel volants)

(🐎✨🌈🐦)

Why can't I "vote" for flying rainbow sparkle ponies?

I wonder if
the Flying Spaghetti Monster (FSM) hangs out with
the Flying Rainbow Sparkle Ponies (FRSP)
from time to time? ;^)

Who wouldn't?
https://www.youtube.com/watch?v=pwLgbC5wt54

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Now let's set f(x)=1^x + x

What are the two roots?

R.H.

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