Re: MidRad and reproducibility
On 2009-09-30 11:54:52 +0200, Arnold Neumaier wrote:
> >>>2. Something about the midpoint of an interval. If you evaluate
> >>>yy = f(xx), what's the relation between mid(yy) and f(mid(xx))?
> >>>Do we have correct rounding or some error bound?
> >>Please make a specific suggestions of the relation that you'd want
> >>to see.
> >
> >Something like mid(yy) = o(f(mid(xx))) with correct rounding or some
> >error bound (that's the simplest solution, and for applications where
> >this would be a bad idea, infsup may be a better choice anyway).
>
> I do not understand the formula. Please explain.
What I mean is that the midpoint of the output interval can be
obtained by evaluating f on the midpoint of the input interval
and rounding its value.
> >>Forcing the two equal forces overestimation in the width of the
> >>product by a factor of up to 1.5. Would that be compatible with an
> >>accurate mode?
> >
> >I don't understand what you mean.
>
> [0,2r]*[0,2r]=[0,4r^2] has a width of 4r^2 but the centered product
> <r,r>*<r,r>=<r,3r^2> has a width of 6r^2, so the width is
> overestimated by a factor of 1.5, even for tiny r.
But that's entirely your fault: you're not using midrad with tiny
intervals. Indeed, if you consider <r,r>, the error is huge!
> In this case, the multiprecision part costs O(N log N) and the
> bounding part costs O(1) both for midrad and for triplex.
> This means that for high precision the bounding part is negligible,
> no matter how it is done.
I agree.
> In my opinion, this favors triplex, since that hasn't the
> disatvantages of midrad for large intervals.
But midrad is still (slightly) better for tiny intervals. Now,
triplex would still be much better than InfSup here.
--
Vincent Lefèvre <vincent@xxxxxxxxxx> - Web: <http://www.vinc17.net/>
100% accessible validated (X)HTML - Blog: <http://www.vinc17.net/blog/>
Work: CR INRIA - computer arithmetic / Arénaire project (LIP, ENS-Lyon)