Re: y=f(x)=(x²)²+2x²+3

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Sujet : Re: y=f(x)=(x²)²+2x²+3
De : r.hachel (at) *nospam* tiscali.fr (Richard Hachel)
Groupes : sci.math
Date : 06. Feb 2025, 23:15:30
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Le 06/02/2025 à 21:52, sobriquet a écrit :
 Ok, but that's a bit like people saying that 3 + 5 = 7 and then claiming that usually mathematicians say that 3 + 5 = 8, but they have different concepts that are more correct.
 Unless you're able to demonstrate that your alternative concepts are superior than the conventional way of defining these concepts, it
seems a bit silly.
The concept of a complex number didn't fall from the sky.. it was developed over many centuries by multiple generations of mathematicians. So it seems unlikely that someone can come along and claim their way to conceive of a complex number is superior or more correct.
You may be right.
But I also know that mathematicians and physicists can also be wrong.
Descartes' works are full of errors, Berkeley and Newton did not agree on the calculation of an infinitely small increment (I think Berkeley was right), Lorentz wandered for years on relativistic transformations before Poincaré gave it to him.
If you observe carefully, you will realize that basically, the notion of complex numbers is very quickly presented, then very quickly skipped. You are given an i²=-1, in order to square an awkward discriminant, and, if the principle is correct (we multiply by 1, then we pose that i²=-1 by convention and this allows us to have a positive root).
But no one has ever explained where this being comes from, which, in its being, is an entity whose square is equal to -1. It is a convention.
Personally, I wish that we put this being under the microscope once and for all, he fuuses imaginary.
It would seem that i is a special entity, which, like 1, can be used in such a way that, whatever its exponent, it remains identical to itself in its abstract being. -i is always equal to -i.
We can do as with 1^x give all the possible and imaginable powers to x, always, 1^x=1.
It seems that it is the same thing with i=-1.
i²=-1.
i^(-1/2)=-1
i^4=-1
i°=-1
etc...etc...etc...
So we have to define i, because then, in the calculations, big sign errors can appear. The biggest one can be that, sometimes, depending on the concept i²=-1 or i²=1 if we simplify too quickly.
Let's take z1=16+9i and z2=14+3i and make a product.
We have: aa'+i(ab'+ba')+(ib')(ib)
If we make the product as is, we see that if i=-1 then (-9)(-3)=+27,
but if we square it, we have i²=-1 and i²bb'=-27
The error is then colossal:
Hachel finds Z=251+174i.
Mathematicians find Z=197+174i
We need to think about this and check what is correct and why it is correct.
R.H.

Date Sujet#  Auteur
6 Feb 25 * y=f(x)=(x²)²+2x²+321Richard Hachel
6 Feb 25 `* Re: y=f(x)=(x²)²+2x²+320sobriquet
6 Feb 25  `* Re: y=f(x)=(x²)²+2x²+319Richard Hachel
6 Feb 25   `* Re: y=f(x)=(x²)²+2x²+318sobriquet
6 Feb 25    +* Re: y=f(x)=(x²)²+2x²+34Chris M. Thomasson
6 Feb 25    i+- Re: y=f(x)=(x²)²+2x²+31FromTheRafters
7 Feb 25    i`* Re: y=f(x)=(x²)²+2x²+32Richard Hachel
7 Feb 25    i `- Re: y=f(x)=(x²)²+2x²+31FromTheRafters
6 Feb 25    `* Re: y=f(x)=(x²)²+2x²+313Richard Hachel
6 Feb 25     +- Re: y=f(x)=(x²)²+2x²+31Python
7 Feb 25     `* Re: y=f(x)=(x²)²+2x²+311sobriquet
7 Feb 25      `* Re: y=f(x)=(x²)²+2x²+310Richard Hachel
7 Feb 25       +* Re: y=f(x)=(x²)²+2x²+38Alan Mackenzie
7 Feb 25       i+* Re: y=f(x)=(x²)²+2x²+34Richard Hachel
7 Feb 25       ii`* Re: y=f(x)=(x²)²+2x²+33Python
7 Feb 25       ii +- Re: y=f(x)=(x²)²+2x²+31Richard Hachel
22 Feb 25       ii `- Re: y=f(x)=(x²)²+2x²+31Moebius
7 Feb 25       i+- Re: y=f(x)=(x²)²+2x²+31Richard Hachel
22 Feb 25       i`* Re: y=f(x)=(x²)²+2x²+32Moebius
22 Feb 25       i `- Re: y=f(x)=(x²)²+2x²+31Alan Mackenzie
7 Feb 25       `- Re: y=f(x)=(x²)²+2x²+31sobriquet

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