Re: Residue Function
- To: mathgroup at smc.vnet.net
- Subject: [mg76097] Re: [mg76060] Residue Function
- From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
- Date: Wed, 16 May 2007 05:29:55 -0400 (EDT)
- References: <200705150904.FAA18673@smc.vnet.net>
On 15 May 2007, at 18:04, Dana DeLouis 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
>
>
>
The residue of f at infinity is usually defined as
Residue[-1/z^2 f[1/z], {z, 0}]
so for f[z]=Exp[2/z] you get exactly -2. In fact it is exactly -(the
coefficient of z^(-1) of the Laurent expansion of f at Infinity).
Andrzej Kozlowski
- References:
- Residue Function
- From: "Dana DeLouis" <dana.del@gmail.com>
- Residue Function