Re: JSH: Resolution now possible

From: Nora Baron (norabaron_at_hotmail.com)
Date: 10/17/04 

Date: 17 Oct 2004 11:50:22 -0700

jstevh@msn.com (James Harris) wrote in message news:<3c65f87.0410170613.507bf41f@posting.google.com>...
> After over two years of arguing on the specifics of my techniques of
> polynomial factorization which adds to even more years of arguing
> before about factorizations I think it's finally clear how to move to
> resolve all the issues:

I am also posting a longer response to your post, but
to forestall your complaints about overly long posts,
in this one I am deleting most of what you say here to focus on
one point. Maybe you can take time out of your busy
schedule to respond just to this.

[snip]

>which leaves posters
>arguing about the details trying to convince that constants are not
>constants, though I can show they are relying on unit factors to make
>their arguments.
>

  Here is your argument about constants.

  First, you define the "constant term" of a function, h(m),
to be h(0).

  That's fine. No problem with that definition. Certainly
h(0) is constant with respect to m. It does not change
when m changes. Values of h(m) are independent of h(0).

  Then you consider a function g1(m) = a1(m)*x + uf,

where a1(m) is a variable function, a(0) = 0, and u and f
are constants with respect to m.

  Then you consider what happens when you divide g1(m)
by another variable function v1(m). You know v1(0) = f.
What you want to show is, v1(m) is not really variable -
it is also a constant function.

  You define:

    h(m) = g1(m)/v1(m) = (a1(m)/v1(m))*x + uf/v1(m).

  Clearly, uf is the constant term of g1(m).

  Therefore uf/v1(m) must be the constant term of h(m).

  But by definition, the constant term of h(m) is h(0):

    h(0) = uf/v1(0) = uf/f = u.

  Therefore, setting constant terms equal, you have

    uf/v1(m) = uf/v1(0) = uf/f = u,

so you conclude that v1(m) = f FOR ALL VALUES OF m.

  Right? Is this your argument?

  Nora B.

Relevant Pages

  • Re: JSH: Resolution now possible
    ... >> polynomial factorization which adds to even more years of arguing ... it's a trivial result from basic algebra that the constant term ... variable of that polynomials. ...
    (sci.math)
  • JSH: What would you do?
    ... I can prove I'm right with basic algebra. ... The people arguing with me get away with arguing about CONSTANTS, ... I've argued with posters like "Nora Baron" about these things before, ... and for your first factor you have a constant term of 7, ...
    (sci.math)
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