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

```

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