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