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MathGroup Archive 2007

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Re: Simplify and Abs in version 6.0

  • To: mathgroup at smc.vnet.net
  • Subject: [mg78093] Re: [mg78034] Simplify and Abs in version 6.0
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Fri, 22 Jun 2007 06:52:09 -0400 (EDT)
  • References: <200706210958.FAA27636@smc.vnet.net>

It is a good think that mathemaica does not return the answer that  
you suggest:

{-\[ImaginaryI] a, \[ImaginaryI] a}

because it would be completely wrong as in Mathematica {a,b} is an  
ordered pair and not a set. Mathematica has no built in structures  
corresponding to sets and has no way of expressing the answer you  
(seem to) have in mind. You can get the answer with Sqrt[a^2] more  
directly by:

  ComplexExpand[{I*Abs[a], (-I)*Abs[a]}, TargetFunctions -> {Re, Im}]

{I*Sqrt[a^2], (-I)*Sqrt[a^2]}

This, of course, can't be reduced any further without additional  
information about a the sign of a.

Andrzej Kozlowski


On 21 Jun 2007, at 18:58, Michael wrote:

> Hi,
>
> In Mathematica version 6.0, I'm having difficulty trying to coax the
> input
>
> In[1]:=  FullSimplify[{\[ImaginaryI] Abs[a], -\[ImaginaryI] Abs[a]},
> Element[a, Reals]]
>
> to produce the set {-\[ImaginaryI] a, \[ImaginaryI] a}.
> Interestingly, this *does* result for
>
> In[2]:=  FullSimplify[{\[ImaginaryI] Abs[a], -\[ImaginaryI] Abs[a]}, a
>> = 0]
>
> and
>
> In[3]:=  FullSimplify[{Sqrt[-1] Abs[a], -Sqrt[-1] Abs[a]}, a < 0]
>
> Am I missing something here?  I've tried Allan Hayes' suggestion in
> 2003, with
>
> In[4]:=  FullSimplify[{\[ImaginaryI] Abs[a], -\[ImaginaryI] Abs[a]},
> Element[a,Reals],ComplexityFunction -> ((Count[#, _Abs, Infinity]) &)]
> Out[4]=  {\[ImaginaryI] Sqrt[a^2], -\[ImaginaryI] Sqrt[a^2]}
>
> to no avail; of course, I could just add a PowerExand@ to the above
> expression, but this seems like a lot to do, especially when
> Mathematica already has been explicitly told that "a" is a real
> number.
>
> Any tricks or hints would be greatly appreciated!
>
> Regards,
>
> Michael
>
>



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