Re: Residue Function
- To: mathgroup at smc.vnet.net
- Subject: [mg76168] Re: Residue Function
- From: dimitris <dimmechan at yahoo.com>
- Date: Thu, 17 May 2007 05:57:40 -0400 (EDT)
- References: <f2bulj$jhe$1@smc.vnet.net>
I thinl I am only one that stated that the residue should be 2.
Searching more I figure that 2 or -2 is just a matter of convention.
It is just how someone defines the residue at infinity.
The book I ussually consult (Ablowitz & Fokas) takes the residues
at infinity to be the same as in zero. In general the residue at
inifnity is
equal to the sum of residues in the rest of complex plane.
Reading another book today I saw that you can also take the residue at
infinity
in such a way so that the sum of all the residues (including infinity)
to be 0.
In the first case Exp[2/z] has residue at infinity 2.
In the latter case the function has residue at inifnity -2.
Now I learn something!
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