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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