VL> "Infinity as number" is buggy. For
instance, with the above point of VL> view, if F(x) = (x+1)-x, then F(Inf) = (Inf+1)-Inf
= Inf-Inf = 0.
Under IEEE 754, F(Inf)
= (Inf+1)-Inf = Inf-Inf = NaN, not
0.
- Ian McIntosh
Toronto IBM Lab 8200 Warden D2-445 905-413-3411
----- Forwarded by Ian
McIntosh/Toronto/IBM on 26/02/2009 12:03 PM -----
Vincent Lefevre <vincent@xxxxxxxxxx>
26/02/2009 11:57 AM
Please respond to
Vincent Lefevre <vincent@xxxxxxxxxx>
To
Ian McIntosh/Toronto/IBM@IBMCA
cc
Subject
Re: The current proposal
>> This value is not even properly defined in
many cases,
>> such as when F(x)=x-x or F(x)=(x-1)/(x+1) and A=[0,inf].
>
> If "infinity as number" is true, i.e., if the infinity is
not a
> member of the interval but rather a token for an unbounded real
> number, then it is properly defined:
>
> F(Inf)=Inf-Inf=0
> F(Inf)=(Inf-1)/(Inf+1)=Inf/Inf=1
>
> Nate Hayes
> Sunfish Studio, LLC
"Infinity as number" is buggy. For instance, with the above point
of
view, if F(x) = (x+1)-x, then F(Inf) = (Inf+1)-Inf = Inf-Inf = 0.