Yes, in fact, this method is definitely able to handle these cases.
There is a subtlety in the implementation which I had not paid
attention to.

So yes, it precisely retrieves the answer to this problem.

Next challenge?

Bill.

On May 2, 2:45 pm, Bill Hart <goodwillh...@googlemail.com> wrote:
> Hmm, very interestingly, the new method *can* be made to output the
> correct answer for this problem.
>
> I'll have to think about whether doing this will screw anything else
> up. I think, in the class of problems it is designed to solve, the
> answer is no, it won't screw others up.
>
> Bill.
>
> On May 2, 2:15 pm, Bill Hart <goodwillh...@googlemail.com> wrote:
>
>
>
>
>
> > It doesn't quite handle this case. But as I'm sure you are aware,
> > neither does the classical algorithm:
>
> > sage: def rel_prec(approx, actual):
> >      return [oo if a==b else -log(abs(a-b)/abs(b), 2) for (a,b) in
> > zip(approx, actual)]
>
> > sage: R.<x> = QQ[]
> > sage: f = R((x^100-10^100)/(x-10))
> > sage: g = x-10
> > sage: ff=f.change_ring(RR)
> > sage: gg=g.change_ring(RR)
>
> > sage: sorted(rel_prec(ff*gg, f*g))
> > [-infinity, -infinity, -infinity, -infinity, -infinity, -infinity, -
> > infinity, -infinity, -infinity, -infinity, -infinity, -infinity, -
> > infinity, -infinity, -infinity, -infinity, -infinity, -infinity, -
> > infinity, -infinity, +Infinity, +Infinity, +Infinity, +Infinity,
> > +Infinity, +Infinity, +Infinity, +Infinity, +Infinity, +Infinity,
> > +Infinity, +Infinity, +Infinity, +Infinity, +Infinity, +Infinity,
> > +Infinity, +Infinity, +Infinity, +Infinity, +Infinity, +Infinity,
> > +Infinity, +Infinity, +Infinity, +Infinity, +Infinity, +Infinity,
> > +Infinity, +Infinity, +Infinity, +Infinity, +Infinity, +Infinity,
> > +Infinity, +Infinity, +Infinity, +Infinity, +Infinity, +Infinity,
> > +Infinity, +Infinity, +Infinity, +Infinity, +Infinity, +Infinity,
> > +Infinity, +Infinity, +Infinity, +Infinity, +Infinity, +Infinity,
> > +Infinity, +Infinity, +Infinity, +Infinity, +Infinity, +Infinity,
> > +Infinity, +Infinity, +Infinity, +Infinity, +Infinity, +Infinity,
> > +Infinity, +Infinity, +Infinity, +Infinity, +Infinity, +Infinity,
> > +Infinity, +Infinity, +Infinity, +Infinity, +Infinity, +Infinity,
> > +Infinity, +Infinity, +Infinity, +Infinity, +Infinity]
>
> > Bill.
>
> > On May 2, 2:58 am, rjf <fate...@gmail.com> wrote:
>
> > > On May 1, 5:42 pm, Bill Hart <goodwillh...@googlemail.com> wrote:
>
> > > ...
>
> > > > So now anything that grows regularly can be multiplied with basically
> > > > zero loss, asymptotically fast. That probably covers most of the
> > > > interesting cases anyhow.
>
> > > The interesting cases are obvious those which are not covered. They
> > > include the
> > > cases where, as one multiples two polynomials, the coefficients in the
> > > result are
> > > zero.  In which case any non-zero result is wrong in ALL places.
>
> > > I suggest testing
> > > P*Q   where P= (x^10-10^10)/(x-10)
> > > and Q=x-10.
>
> > > replace 10 with a larger integer if you wish to make the test more
> > > challenging.
> > > Or compute  (P^2*Q^2)   etc.
>
> > > Will your suggestion get these polynomials exactly right?
>
> > > --
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>
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