Re: Talking Rationally About Surrogate Factoring

jstevh_at_msn.com
Date: 04/18/05 

Date: 18 Apr 2005 14:52:42 -0700

David Kastrup wrote:
> jstevh@msn.com writes:
>
> > Nora Baron wrote:
> >> Rick Decker wrote:
>
> [...]
>
> >> Yes! I have verified this with a bunch of examples. But thanks
> >> for the proof. [Note that in the above, "a" is the larger of the
> >> two factors a and b.]
> >>
> >> What this means is that if you pick integer factors of -T*j^2,
> >> all you get are trivial factors of M.
> >>
> >
> > Wow, why am I not surprised?
> >
> > I think it's more likely that you screwed up in your derivation.
> >
> > My suggestion is that someone use a math package to get the correct
> > answer.
>
> Don't be silly. You claim that the factors of M are non-trivial in
> 50% of the cases.
>
> So it should be a pice of cake for you to provide a _single_ example
> where this is not the case, as that would be all it takes to shoot
> down the proof.
>
> And you can even use small numbers for that: no need for a large
> example here that is only accessible to computers.
>
> > If you used a math package "Nora Baron" then I'm more inclined to
> > take you seriously.
> >
> > Otherwise, I fear that you will find a way to get the result that
> > supports your social position, which is that I'm wrong and my work
> > is not useful.
>
> If you claim a 50% hit rate, then coming up with a _single_ example
> does not seem like too much to ask for you to prove your position.
>

You're not paying attention. Rick Decker claimed to have solved for
b_2 f_1 from the SFT in term of k_1 and k_2, but he screwed up.

Then "Nora Baron" claimed to have solved for b_2 f_1, but now Decker
and "Nora Baron" are claiming that that result proves that for integer
factors of b_2 f_1 you only get trivial factors.

Hey, that may be true, but I'm not taking their word for it, as they're
biased. Neither would like to eat crow by having to acknowledge that
I'm right and they're wrong, so I figure they may find ways to fudge.

So I say, let a computer math package do it.

The equations are as follows so can someone just solve for the answer
with a math package and end this silliness?

b_2 f_1 = (-(Az - 2M^2)+/- sqrt((Az - 2M^2)^2 - 4TM^2))/2

where T = M^2 - j^2, and f_1 is a rational factor of T, and Az is given
by

Az = Ax(Ax +/- sqrt((Ax - 2j^2)^2 + 4Tj^2))/(2Ax - 2M^2)

and Ax is given by

Ax = +/- (k_1 + k_2) + 2j^2

where k_1 k_2 = -Tj^2, and k_1 and k_2 are rationals.

Basically you just work backwards, taking the solution for Ax to solve
for Az and that solution to solve for b_2 f_1.

Can that be done with math software or not?

James Harris

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