The goal is to completely do without Argand geometry (which is strictly useless
in Cartesian planes). Here, we have a curve f(x), which gives us a real root x=1,
since f(x)=(x-1)²(x²+4), but also two complex roots, which are no longer systematically x'=2i and x"=-2i.
This is very strange and worthy of interest.
Thus, f'(x)=x²+4 does not have the same complex roots as f(x)=(x-1)²(x²+4).
This clearly confuses all mathematicians, and even makes them laugh.
And yet it's true.
Just as real roots are unalterable (I can multiply by as many roots as I want, I
would always have my first roots continued in the equation), this is no longer
true for complex roots. This is very strange.
<http://nemoweb.net/jntp?MgzF4TKP_46j3u1MBdnuxy16Yhg@jntp/Data.Media:1>
<https://www.nemoweb.net/?DataID=MgzF4TKP_46j3u1MBdnuxy16Yhg@jntp>
x'=2i
Yes, but...
x"=-0.38829i and not x"=-2i
R.H.
Le 20/03/2025 à 17:45, Richard Hachel a écrit :
Thus, f'(x)=x²+4 does not have the same complex roots as f(x)=(x-1)²(x²+4).
in addition of ignoring the signification of "root", "associativity", "equality", "inclusion", "absorbing element", he proves now that he
doesn't know the notion of "derivative" and the basic notations of
maths. His ignorance and conceit are infinite...
Thus, f'(x)=x²+4 does not have the same complex roots as f(x)=(x-1)²(x²+4).
Le 20/03/2025 à 18:20, efji a écrit :
Le 20/03/2025 à 17:45, Richard Hachel a écrit :
Thus, f'(x)=x²+4 does not have the same complex roots as
f(x)=(x-1)²(x²+4).
in addition of ignoring the signification of "root", "associativity",
"equality", "inclusion", "absorbing element", he proves now that he
doesn't know the notion of "derivative" and the basic notations of
maths. His ignorance and conceit are infinite...
He is using confusing notations on purpose. Like all cranks his point is
to introduce confusions, WM does the same.
Le 20/03/2025 à 18:20, efji a écrit :
Le 20/03/2025 à 17:45, Richard Hachel a écrit :
He is using confusing notations on purpose.
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