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

*To*: mathgroup at smc.vnet.net*Subject*: [mg88175] Re: [mg88133] Re: [mg88104] 2*m^z - m^z = ?*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>*Date*: Sun, 27 Apr 2008 04:58:37 -0400 (EDT)*References*: <200804250926.FAA08036@smc.vnet.net> <200804260741.DAA08470@smc.vnet.net>

It is, in fact, possible to write a function "mySimplify" which avoids this particular problem and (I think) gives the same results as Simplify, although it is inevitably a lot slower. Here is the function: SetAttributes[mySimplify, HoldAll] mySimplify[expr_] := Simplify[ReleaseHold[Simplify[Unevaluated[expr] /. x_?NumericQ :> Hold[x]]]] You can check that Table[{2*m^z - m^z, mySimplify[2*m^z - m^z]}, {m, 1, 21}] gives correct answers. It also simplifies other expressions in the expected way: mySimplify[Cos[x]^2 + Sin[x]^2] 1 I am sure mySimplify will not perform well in all sitiations but it maybe a useful alternative to Simplify and perhaps a sophisticated version of it could even be usefully implemented in Mathematica? Andrzej Kozlowski On 26 Apr 2008, at 16:41, Andrzej Kozlowski wrote: > > > > On 25 Apr 2008, at 18:26, Alexey Popkov wrote: > >> Hello, >> What do you think about this: >> >> Table[ >> {2*m^z - m^z, >> FullSimplify[2*m^z - m^z]}, >> {m, 1, 21}] // TableForm >> >> The answer is very interesting (only odd numbers are treated well): >> >> 1 1 >> -2^z + 2^(1 + z) 2^z >> 3^z 3^z >> 2^(1 + 2*z) - 4^z 4^z >> 5^z 5^z >> 2^(1 + z)*3^z - 6^z 2^(1 + z)*3^z - 6^z >> 7^z 7^z >> 2^(1 + 3*z) - 8^z 8^z >> 9^z 9^z >> 2^(1 + z)*5^z - 10^z 2^(1 + z)*5^z - 10^z >> 11^z 11^z >> 2^(1 + 2*z)*3^z - 12^z 2^(1 + 2*z)*3^z - 12^z >> 13^z 13^z >> 2^(1 + z)*7^z - 14^z 2^(1 + z)*7^z - 14^z >> 15^z 15^z >> 2^(1 + 4*z) - 16^z 16^z >> 17^z 17^z >> 2^(1 + z)*9^z - 18^z 2^(1 + z)*9^z - 18^z >> 19^z 19^z >> 2^(1 + 2*z)*5^z - 20^z 2^(1 + 2*z)*5^z - 20^z >> 21^z 21^z >> >> Can anyone explain the reason for this behavior? >> > > > Thera are many such pheonomna in computer algebra. This particular one > I have already once explained on this forum, but as I can't find my > post I will repeat it. > > The reason why this happens is that Mathematica (and all other > Computer ALgebra Systems) performs certain "simplifications", or more > precisely "reductions to canonical form" automatically, even before > applying Simplify or FullSimplify, simply as part of the evaluation > process. One such "simplification" that is done automatically is this: > > 2*6^z > 2^(z + 1)*3^z > > In other words, Mathematica sees that 6 has 2 as a factor, it factors > it out and combines all factor of 2 into a single power. This is a > typical example of "reduction to canonical form". Such reductions make > certian operations much simpler and faster but unfortunately this > particular one does not fit in very well with another reduction that > Mathematica also performs automatically: > > 2^z*3^z > 6^z > > Such "reductions" cannot be undone by Simplify or FullSimplify because > they are performed on evaluation, so even if Simplify "undid" one of > them, the reduction would be immediately performed again. > Unfortunately these two reductions prevent Simplify and FullSimplify > form seeing that > > 2^(z + 1)*3^z-6^z > > can be simplified to 6^z, because Simplify cannot re-write 2^(z + > 1)*3^z as 2*6^z and it also cannot re-write 6^z as 2^z*3^z for reasons > described above. > > In the example you gave the problem can be easily dealt with by > wrapping Evaluate round the first argument of Table. > > Table[Evaluate[{2*m^z - m^z, FullSimplify[2*m^z - m^z]}], {m, 1, 21}] > > However the problem of simplifying > > p=2*6^z-6^z > > and similar expressions is harder to deal with. Note that Mathematica > has no difficulty verifying that > > p == 6^z // Simplify > True > > so it is only transforming p into 6^z by means of FullSimplify that is > difficult. In fact, I don't think it can be done, without using some > tricks. Here is one such trick. We divide p by 3^z, then FullSimplify > and finally again multiply by 3^z: > > FullSimplify[[p/3^z]*3^z > 6^z > > > Andrzej Kozlowski >

**References**:**2*m^z - m^z = ?***From:*Alexey Popkov <popkov@gmail.com>

**Re: 2*m^z - m^z = ?***From:*Andrzej Kozlowski <akoz@mimuw.edu.pl>