Right now (testing with 5.21.0 and the current CVS), we have
(%i3) limit(2/x-1/x,x,0);
(%o3) infinity
but
(%i6) limit(2/x,x,0)-limit(1/x,x,0);
(%o6) 0
That just does not smell right.
Shouldn't we have
inf + inf = inf (NOT 2 inf),
inf - inf = und (NOT 0),
minf + minf = minf (NOT 2 minf),
minf - minf = und,
inf - minf = inf,
minf - inf = minf,
infinity + inf = infinity + minf = infinity + infinity = und,
infinity - inf = infinity - minf = infinity - infinity = und?
Multiplication can also be a problem; e.g., we have
(%i12) infinity/infinity;
(%o12) 1
but
(%i13) limit(infinity/inf,x,0);
(%o13) infinity
but in both cases, I think und would be the only valid answer.
On the other hand, I realize that some simplifications may (for better or
for worse) rely on the current behavior.
Also, I noticed that the example for und (type ? und) is one that actually
yields infinity.
Viktor
-----Original Message-----
From: maxima-bounces at math.utexas.edu [mailto:maxima-bounces at math.utexas.edu]
On Behalf Of Richard Fateman
Sent: Tuesday, September 07, 2010 5:29 PM
To: Jean Pellegri; maxima List
Subject: Re: [Maxima] inf - inf = 0 ??
On 9/7/2010 1:49 PM, Jean Pellegri wrote:
Salut
Maxima says :
(%i) inf - inf;
(%o) 0
This is obviously incorrect
not if they are the SAME inf. e.g. limit(x-x,x,inf).
More seriously, what would you like to see? There are various proposals,
e.g.
here's one.
each use of inf is labeled. then inf[1]-inf[2] is different from
inf[1]-inf[1].
With interval arithmetic, you have similar problems. e.g. if -1<x<1 is
represented by interval(-1,1),
what is interval(-1,1) - interval(-1,1)? maxima says zero, but it should
be interval(-2,2).
unless they are the SAME interval.
RJF