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Re: 2*m^z - m^z = ?

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  • Subject: [mg88174] Re: 2*m^z - m^z = ?
  • From: Andrzej Kozlowski <akoz at>
  • Date: Sun, 27 Apr 2008 04:58:27 -0400 (EDT)
  • References: <> <fuumc1$88s$> <>

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> 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[1]:= 2*6^z-6^z
> Out[1]= 2^(1+z) 3^z-6^z
> And even
> In[2]:= FullSimplify[2*6^z-6^z]
> Out[2]= 2^(1+z) 3^z-6^z

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