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Re: The actual math, advanced polynomial factorization
Posted:
Sep 27, 2004 8:08 PM


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 polynomial defined over x] Correction in superseded post: this is 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



