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Re: What is Infinity+Pi*I
*To*: mathgroup at smc.vnet.net
*Subject*: [mg65881] Re: [mg65860] What is Infinity+Pi*I
*From*: "Erickson Paul-CPTP18" <Paul.Erickson at Motorola.com>
*Date*: Thu, 20 Apr 2006 05:15:15 -0400 (EDT)
*Sender*: owner-wri-mathgroup at wolfram.com
*Thread-index*: AcZjmY+EwEH54ILcQpygCGMz0tMFLAAJvq9w
Ted,
OK, this is interesting why:
In[1]:=
Exp[ Infinity + Ï? I ]
Out[1]=
â??
In[2]:=
Limit[ Exp[ x + Ï? I ] , x \[Rule] Infinity ]
Out[2]=
-â??
The reason is, of course,
In[3]:=
Infinity + Ï? I
Out[3]=
â??
Because any finite (even complex finite) value is insignificant against infinity, while any infinite (even though large) maintains the imaginary part. Therefore the limit is evaluated differently than the direct.
In[5]:=
10^10000 + Ï? I//Short
Out[5]//Short=
1000000000000000000000\[LeftSkeleton]9957\[RightSkeleton]\
0000000000000000000000+\[LeftSkeleton]1\[RightSkeleton]
In[6]:=
%[[2]]
Out[6]=
\[ImaginaryI] Ï?
Therefore, we have the interesting result:
In[8]:=
Exp[ Ï? I ] Exp[ Infinity ] â? Exp[ Infinity + Ï? I ]
Out[8]=
True
Whereas:
In[9]:=
Limit[ Exp[ Ï? I ] Exp[ x ] == Exp[ x + Ï? I ], x \[Rule] â??]
Out[9]=
True
So by playing with the order of evaluation to delay Plus:
In[22]:=
SetAttributes[ Plus, HoldAll ]
In[23]:=
Infinity + Ï? I
Out[23]=
Ï? \[ImaginaryI]+â??
In[25]:=
Exp[ Infinity + Ï? I ] //Simplify
From In[25]:=
\!\(\*
RowBox[{\(Simplify::"infd"\), \(\(:\)\(\ \)\), "\<\"Expression \\!\\(-\
\[ExponentialE]\\^\\(\\(\\(
Ï?\\\\ \[ImaginaryI]\\)\\) + â?? - \\(\\(\[ImaginaryI]\\\\ Ï?\\)\\)\\)\\) \
simplified to \\!\\(-â??\\). \\!\\(\\*ButtonBox[\\\"Moreâ?¦\\\", \
ButtonStyle->\\\"RefGuideLinkText\\\", ButtonFrame->None, \
ButtonData:>\\\"Simplify::infd\\\"]\\)\"\>"}]\)
Out[25]=
-â??
Now:
In[27]:=
Exp[ Ï? I ] Exp[ Infinity ] == Exp[ Infinity + Ï? I ] // Simplify
Out[27]=
True
Paul
-----Original Message-----
From: ted.ersek at tqci.net [mailto:ted.ersek at tqci.net]
To: mathgroup at smc.vnet.net
Subject: [mg65881] [mg65860] What is Infinity+Pi*I
I am using Mathematica 4.1, but I suspect all versions do the same in this case.
In[1]:=
Infinity + Pi * I
Out[1]=
Infinity
I think it the input above should return itself.
Am I wrong here? If we do it my way the following would return
-Infinity. (*Negative Infinity*)
In[1]:=
E^( Infinity + Pi * I )
Out[1]=
Infinity
Either way it's an interesting example.
-------
Regards,
Ted Ersek
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