Sorry, mistake in my last message Re: roundToZero_F(+inf) = +inf
Oh, of course RoundToZero(+inf) = +inf, since
+inf is a representable number, and
RoundUp(HUGE) = HUGE for the same reason. I guess
I was thinking about numbers strictly between
HUGE and +inf. (Please don't let me distract
you in this case.)
Baker
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Please educate me here. Without actually looking
up what 754 specifies, my intuition would tell
me that roundToZero(+inf) would be HUGE (i.e.
the largest representable number)
and roundUp(HUGE) would be +inf.
(That is, for what it's worth in this context.)
Baker
On 03/01/2012 10:09 AM, Vincent Lefevre wrote:
On 2012-02-29 11:43:45 -0500, Michel Hack wrote:
Vincent Lef?vre wrote, in reply to Dan Zuras:
Coercion: mid_F(X) = if (abs_F(round2Nearest_F(mid(X))) = +inf)
then roundToZero_F(mid(X))
else round2Nearest_F(mid(X))
For X = [0,+inf], this would give:
* mid(X) = +inf
* abs_F(round2Nearest_F(mid(X))) = +inf
* roundToZero_F(mid(X)) = roundToZero_F(+inf) = +inf
while I think you wanted Fmax_F. I don't think you can avoid it,
except artificially, e.g. with nextDown(+inf).
The "roundToZero" applies to the entire operation, in the spirit of 754,
where there is only one rounding after a conceptually-exact computation.
So there is nothing artificial about ending up with Fmax_F.
No, roundToZero does not apply to entire operations (this makes no
sense), but to unrounded results. Here the unrounded result mid(X)
is +inf, because at Level 1:
(inf(X) + sup(X)) / 2 = (0 + (+inf)) / 2 = (+inf) / 2 = +inf
Then the rounded result is roundToZero_F(+inf) = +inf.
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Ralph Baker Kearfott, rbk@xxxxxxxxxxxxx (337) 482-5346 (fax)
(337) 482-5270 (work) (337) 993-1827 (home)
URL: http://interval.louisiana.edu/kearfott.html
Department of Mathematics, University of Louisiana at Lafayette
(Room 217 Maxim D. Doucet Hall, 1403 Johnston Street)
Box 4-1010, Lafayette, LA 70504-1010, USA
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