Re: Residue Function

*To*: mathgroup at smc.vnet.net*Subject*: [mg76111] Re: Residue Function*From*: dimitris <dimmechan at yahoo.com>*Date*: Wed, 16 May 2007 05:37:05 -0400 (EDT)*References*: <f2bulj$jhe$1@smc.vnet.net>

Hi. As far as I know Complex Analysis you have just enountered a bug. (*incorect*) In[43]:= (Residue[#1, {z, Infinity}] & ) /@ {1/z, Exp[2/z]} Out[43]= {-1, -2} (*correct*) In[46]:= (Coefficient[#1, z^(-1)] & )[Normal[{1/z, Exp[2/z]} + O[z, Infinity]^3]] Out[46]= {1,2} The strange thing that Residue calls the Series function. In[51]:= Unprotect[Series]; Series[a___] := Null /; (Print[InputForm[series[a]]]; False) In[52]:= Residue[Cos[z]/z, {z, 0}] >From In[52]:= series[Cos[z]/z, {z, 0, 1}] Out[52]= 1 But funnily In[60]:= (Residue[#1,{z,Infinity}]&)/@{1/z,Exp[2/z]} >From In[60]:= series[z, {z, 0, 1}] >From In[60]:= series[E^(2*z), {z, 0, 1}] Out[60]= {-1,-2} Hence, Mathematica correctly makes the transformation 1/z->z. Nevertheless it returns a incorrect result! Dimitris =CF/=C7 Dana DeLouis =DD=E3=F1=E1=F8=E5: > Hello. I am studying the subject of Residues, and came across the follow= ing > example in an article. Basically, it says the Residue of the following > equation is 2. However, both Mathematica 5.2 & 6.0 (windows) gives it as > -2. > > equ = Exp[2/z]; > > Residue[equ, {z, Infinity}] > > -2 > > >From what little I know, the residue comes from the ^-1 term in the > series... > > Normal[Series[equ, {z, Infinity, 3}]] > > 1 + 4/(3*z^3) + 2/z^2 + 2/z > > Coefficient[%, z^(-1)] > > 2 > > The above shows 2, as in the article. I don't see where Mathematica arri= ves > at -2. Anyone familiar with this topic to comment? Thanks in Advance. > > Dana DeLouis