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

**Follow-Ups**:**Re: What we from 1^Infinity, Infinity^0, and similar stuff***From:*Murray Eisenberg <murray@math.umass.edu>