On Wed, Jul 9, 2008 at 9:05 PM, Bill Page <[EMAIL PROTECTED]>
wrote:
> On Wed, Jul 9, 2008 at 8:43 PM, Gabriel Dos Reis
> <[EMAIL PROTECTED]> wrote:
>
> > ...
> > In the field of fraction of polynomial, the fact that 1/(2d) is well
> > defined seals no doubt. It does not matter whether d can take
> > on the value zero. That is just a simple fact. That is one of
> > reasons why simplification of expressions that assume
> > fraction of polynomials prove inadequate some times.
> >
>
> If "d can take on the value zero" doesn't that imply we are thinking
> of '1/(2d)' as a function?
>
> Isn't the following result another example such an error?
>
> (1) -> q:=((x^2-1)/(x-1))
>
> (1) x + 1
> Type: Fraction Polynomial
> Integer
> (2) -> q(1)
>
> (2) 2
> Type: PositiveInteger
>
> Surely the fact that we can write (2) means this simplification is
> unjustified.
Just like Gaby has pointed out a little earlier, in this example q is a
fraction polynomial function, instead of fraction polynomial. The difference
is fundamental. Function has a domain and the corresponding co-domain, which
carry no meaning for the polynomial.
The Axiom (and maybe other CAS) handles it as Fraction Polynomial Integer
because it is the most direct type for it. I would like to consider there is
a missing step that q is "coerced" from a fraction polynomial to a fraction
polynomial function, and then 1 is applied to that function.
yes, x**2/x and x are different functions, as everybody will agree.
>
> > Have a look at this wonderful paper by James Davenport.
> >
> > "Equality in Computer Algebra and Beyond"
> >
>
> Thanks, I will review it again. You have referred to this paper before:
>
>
> http://sourceforge.net/mailarchive/message.php?msg_name=87od8i6aej.fsf_-_%40gauss.cs.tamu.edu
>
> In the abstract from which you quote Davenport writes:
>
> "For example, we refer to Q(x) as "rational functions", even though
> (x^2-1)/(x-1) and x+1 are not equal as functions from R to R, since
> the former is not defined at x=1, even though they are equal as
> elements of Q(x)."
>
> But as I recall he does not deal with this issue in greater
> specificity in the rest of the paper.
>
> For the accessibility challenged here is an alternate (free) link:
>
> http://www.calculemus.net/meetings/siena01/Papers/Davenport.ps
>
> to this paper.
>
> Regards,
> Bill Page.
>
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