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