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Topic: The actual math, advanced polynomial factorization
Replies: 14   Last Post: Sep 29, 2004 9:43 AM

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Andrzej Kolowski

Posts: 353
Registered: 12/6/04
Re: The actual math, advanced polynomial factorization
Posted: Sep 27, 2004 8:08 PM
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jstevh@msn.com (James Harris) wrote in message news:<3c65f87.0409270231.71cdcf01@posting.google.com>...
> David Kastrup <dak@gnu.org> wrote in message news:<x51xgp8b3o.fsf@lola.goethe.zz>...
> > jstevh@msn.com (James Harris) writes:
> >
> > > David Kastrup <dak@gnu.org> wrote in message news:<x53c16xb8r.fsf@lola.goethe.zz>...
> > >> jstevh@msn.com (James Harris) writes:
> > >>
> > >> > The following factorization is useful:
> > >>
> > >> For what?
> > >>
> > >> > f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) =
> > >> >
> > >> > (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)
> > >> >
> > >> > Note that x, m, f, and u are all independent variables.
> > >>
> > >> No, they aren't independent. They are related by the equation
> > >>
> > >> f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) =
> > >> (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)
> > >>
> > >
> > > No. They are not. Notice that you can give any value you wish for x,
> > > m, f and u without checking each variable against the other because
> > > they are not constrained against each other.
> > >
> > >
> > > But notice that you can't just give a value for a_1, a_2 and a_3
> > > because they ARE constrained by the values of x, m, f and u.
>
>
> That's the mathematics, but notice what I face, which is what I've
> faced from sci.math posters for YEARS, in the reply by David Kastrup.
>
> >
> > That's nothing to "notice", that is what you may want to define.
> > Prima facie, the above equation just expresses a relation between all
> > of the variables and does not differentiate anything among them. In
> > the context of defining a particular task associated with the
> > equation, you can assign meaning to various variables. You always
> > muddy up things wildly. In particular, for the purpose you want, you
> > _choose_ the following characterizations:
> >
> > a) x is a symbolic variable used for defining a polynomial. It does
> > not assume any value at all.
> > b) m, f and u are _declared_ to be free variables: you try making a
> > statement that will hold for all choices of them.
> > c) a_1, a_2 and a_3 are then (though not necessarily uniquely)
> > determined by that choice and [are in the coefficient space for the
> > polynomial defined over x] Correction in superseded post: this is
> > of course not the case when the polynomial can't be factored into
> > linear terms in the given coefficient space, like when factoring
> > the real polynomial x^2+1.
> >
> > In order to make something like this make sense, you have to declare
> > _how_ m, f and u are supposed to be chosen: what possible values they
> > may assume. In addition, you have to declare the ring or other
> > algebraic structure in which a_1, a_2 and a_3 are supposed to be.
> > Only then can you start making meaningful statements.
> >
> > And in fact, you have repeatedly changed your mind about those basic
> > premises you have not bothered to fix in APF, and are lying about
> > this when defending it.
>
>
> f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) =
>
> (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)
>
> and f, m, x and u are independent variables, while a_1, a_2, and a_3
> are dependent variables.
>
> Such a simple thing, but notice all the verbiage from David Kastrup.
>

Yes, f, m, x, and u are independent and a_1, a_2, and a_3 are not.
But the statement you are discussing, i.e., divisibility of
a_1, a_2 or a_3 by f in the algebraic integers is not really a
statement about f. It is a statement about a_1, a_2, and a_3,
conditional on f. That is, the statement itself is about
*dependent* variables. So any vague principle you have in
your head about independent variables does not apply. This
is a bogus argument based on a nonmathematical "principle"
that you think applies in this case. It does not.

Andrzej



> It's not about what's mathematically correct with these people.
>
> I say they are YOU, and they are the modern math world.
>
> You've learned that math is just some way to b.s. people too
> intimidated to check you, while you've accepted things that are
> convenient but false, like the rather naive and wrong ideas that base
> what you call Galois Theory.
>
> Evariste Galois was right, but you people twisted his research.
>
>
> James Harris




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