Re: Simplify and Abs in version 6.0

*To*: mathgroup at smc.vnet.net*Subject*: [mg78077] Re: Simplify and Abs in version 6.0*From*: dh <dh at metrohm.ch>*Date*: Fri, 22 Jun 2007 06:43:57 -0400 (EDT)*References*: <f5djsj$t7i$1@smc.vnet.net>

Hi Michael, FullSimplify works on (ordered) lists not (unordered) sets. Therefore, mathematica is right. If you want something else, you can still do this e.g. by replacement rules. hope this helps,Daniel 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 > >