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>