Re: Residue Function

*To*: mathgroup at smc.vnet.net*Subject*: [mg76086] Re: Residue Function*From*: Valeri Astanoff <astanoff at gmail.com>*Date*: Wed, 16 May 2007 05:24:17 -0400 (EDT)*References*: <f2bulj$jhe$1@smc.vnet.net>

On 15 mai, 11:31, "Dana DeLouis" <dana.... at gmail.com> wrote: > Hello. I am studying the subject of Residues, and came across the following > 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 arrives > at -2. Anyone familiar with this topic to comment? Thanks in Advance. > > Dana DeLouis Good day, It is correct. It complies with the definition of residue at infinity. You can check with this contour : In[1]:= -(Integrate[Exp[2/z],{z,1,I}]+ Integrate[Exp[2/z],{z,I,-1}]+ Integrate[Exp[2/z],{z,-1,-I}]+ Integrate[Exp[2/z],{z,-I,1}])/(2*Pi*I) Out[1]=-2 V.Astanoff