Residue Function
- To: mathgroup at smc.vnet.net
- Subject: [mg76060] Residue Function
- From: "Dana DeLouis" <dana.del at gmail.com>
- Date: Tue, 15 May 2007 05:04:28 -0400 (EDT)
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
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