Subject: Strange thing in Maxima: reduction of fraction
From: Jaime E. Villate
Date: Thu, 07 Dec 2006 17:05:53 +0000
On Thu, 2006-12-07 at 11:42 -0500, Raymond Toy wrote:
> >> (%i2) (q - p) / abs(q - p)^2;
> >>
> >> 1
> >> (%o2) -----
> >> q - p
> >>
> >> Obviously, the output of (%o1) is correct, because I've not
> made any
> >> assumptions about p & q in advance. However, why did Maxima
> produce the
> >> output of (%o2) when I swapped the order of p & q as in the
> above?
> >>
> >> Thanks a lot!
>
> Leo> Looks like a bug.
>
> Is it? Maxima doesn't always simplify everything that you type in.
> Isn't %o2 correct, independent of any assumptions on p & q (except
> that I think maxima is implicitly assuming p and q are real)?
Sure, if (q - p) is real, then, abs(q - p)^2 = (q - p)(q - p)
but that makes me wonder about the following:
(%i3) a/a;
(%o3) 1
this seems to me true only if a is not zero. %i3 is not the same as
(%i4) limit(a/a,a,0);
(%o4) 1
which I believe is correct.
Jaime