Fw: A question Level 1 <---> level 2 mappings; arithmetic versus applications
(On representing semi-bounded intervals in mid-rad)
Ian McIntsoh wrote:
> For the particular case of [0, +oo], one option would be mid=maxfinite
> and rad=mid, but maybe a better one would be mid=maxfinite/2 (rounded up)
> and rad=mid.
(and similar proposals for [a, +oo] with a>0)
The problem with this approach is that containment is lost. This would
become painfully obvious if the result is converted to a type with a
larger exponent range and the computation allowed to proceed...
Chenyi Hu wrote:
> When one endpoint of an interval [a,b] is unbounded, would left-rad
> and right-rad be helpful?
Yes, this form (preferred by mechanical drafting people) would work,
if you stretch the meaning of "mid", by picking "mid" to be near the
endpoint with the smallest magnitude, and using a small exact value
for the corresponding side-radius. The other endpoint (say sup) is
then (inf+lrad)+(sup-(inf+lrad)) which can recover sup exactly by
choosing lrad to be such that no power-of-base is crossed by the
central addition. For example, if inf and sup are both positive, we
can use mid=inf and rad=sup-inf. Even if rad equals sup (because inf
was so small), sup can be recovered because in that case sup+inf = sup
(in terms of FP arithmetic, not Real arithmetic).
A mid-rad computation can produce a result that cannot be converted
exactly to inf-sup, but as Dan Z pointed out repeatedly, it can be
converted (as part of the mid-rad finalization) to a representation
that *is* convertible exactly to inf-sup, possibly at the cost of a
slight widening.
Michel.
---Sent: 2010-07-01 00:10:17 UTC