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MathGroup Archive 2005

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

  • To: mathgroup at smc.vnet.net
  • Subject: [mg63308] What we from 1^Infinity, Infinity^0, and similar stuff
  • From: ted.ersek at tqci.net
  • Date: Fri, 23 Dec 2005 05:08:32 -0500 (EST)
  • Sender: owner-wri-mathgroup at wolfram.com

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}


Assigning DownValues to Power will not change the way the results above
are done. However, the UpValues below give the results I prefer.  I
thought some users might find this helpful.

In[3]:=
   Unprotect[DirectedInfinity, Indeterminate];
   DirectedInfinity/: Power[1, DirectedInfinity[_]]=1;
   DirectedInfinity/: Power[1, DirectedInfinity[]]=1;
   Indeterminate/: Power[1, Indeterminate]=1;
   DirectedInfinity/: Power[DirectedInfinity[_], 0]=1;
   DirectedInfinity/: Power[DirectedInfinity[], 0]=1;
   Protect[DirectedInfinity, Indeterminate];

In[10]:=
   {1^Infinity, 1^(-Infinity), 1^ComplexInfinity, 1^(0*Infinity),
      Infinity^0, (-Infinity)^0, ComplexInfinity^0}

Out[10]=
    {1,1,1,1,1,1,1}

--------------------------------------------
Note:
FullForm[Infinity] is DirectedInfinity[1]
FullForm[ComplexInfinity] is DirectedInfinity[]


Have a Merry Christmas,

         Ted Ersek



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