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Re: What we from 1^Infinity, Infinity^0, and similar stuff


Same result in Mathematica 5.2.

ted.ersek at tqci.net wrote:
> I am using Mathematica 4.1, and version 5 may work different in this case.
> 
> It seems I can compute (1^z) where (z) has any numeric value and
> Mathematica returns the Integer 1. I can also compute (z^0) where (z) is
> any non zero value and Mathematica returns the Integer 1. Hence I think
> the following should return {1,1,1,1,1,1,1}. Can someone explain why that
> would be wrong?
> 
> In[1]:= Off[Power::indet, Infinity::indet];
> 
>         {1^Infinity, 1^(-Infinity), 1^ComplexInfinity,
>           1^(0*Infinity), Infinity^0, (-Infinity)^0, ComplexInfinity^0}
> 
> Out[2]= {Indeterminate, Indeterminate, Indeterminate, Indeterminate,
>            Indeterminate, Indeterminate, Indeterminate}
> 

-- 
Murray Eisenberg                     murray at math.umass.edu
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University of Massachusetts                413 545-2859 (W)
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Amherst, MA 01003-9305


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