Re: 2*m^z - m^z = ?
- To: mathgroup at smc.vnet.net
- Subject: [mg88174] Re: 2*m^z - m^z = ?
- From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
- Date: Sun, 27 Apr 2008 04:58:27 -0400 (EDT)
- References: <200804250926.FAA08036@smc.vnet.net> <firstname.lastname@example.org> <email@example.com>
I never wrote that all computer algebra systems have exactly the same particular problem. I only wrote that they all use canonical forms of expressions and such problems occur (see any book on symbolic algebra). There are likely to be such problems in other algebra system although of course not exactly the same one. This particular problem could have been avoided by a different choice of canonical forms. It probably was caused by an oversight but may be bow difficult to fix because quite likely many other simplifications depend on these particular reductions. You seem to take everything in a very literal sense... Demanding "comments" from Wolfram Research on this forum is not likely to lead to a success. Participation in the MathGroup by Wolfram staff is purely voluntary and unofficial and they respond to posts only if they happen to read them and like to respond to them. Writing that "comments are expected" and such stuff, is, in my opinion, likely to make them less inclined to do so. If you need technical support or believe you have found a bug, write to Wolfram's Technical Support, not to the MathGroup. Andrzej Kozlowski On 27 Apr 2008, at 00:12, Alexey Popkov wrote: > On 26 =C1=D0=D2, 11:41, Andrzej Kozlowski <a... at mimuw.edu.pl> wrote: >> The reason why this happens is that Mathematica (and all other >> Computer ALgebra Systems) > > I do not agree about "all other Computer ALgebra Systems". Try inter > in free Maxima 0.7.5 the following: > 2*6^z-6^z; > and press Enter. > The result is: > 6^z > > It means that it is specifical Mathematica problem... And comments > from Wolfram Research are expected... > > In Mathematica: > In:= 2*6^z-6^z > Out= 2^(1+z) 3^z-6^z > And even > In:= FullSimplify[2*6^z-6^z] > Out= 2^(1+z) 3^z-6^z
- 2*m^z - m^z = ?
- From: Alexey Popkov <firstname.lastname@example.org>
- 2*m^z - m^z = ?