On Mon, Nov 2, 2009 at 12:43 AM, DIETER ENSSLEN <[email protected]> wrote: > 1+x:^o.0j1 does indeed produce a 0 > > as opposed to 1+ 1x1^(1p1*0j1) which produces 0j1.22465e_16 which is a > 'practical' zero but not a say 1.1e_330 type of zero > > and both I trust represents e^(pi* i) +1 = 0 > > I would guess there are many other situations where 0, pi, and e and such do > not happen as neatly as in your example. In a CAS the answer to the above is > 0 no matter how you write it.
Then, 1+ 1x1^(1p1*0j1) would not an example of a subset of J which is a CAS. -- Raul ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
