MathGroup Archive 2007

[Date Index] [Thread Index] [Author Index]

Search the Archive

Residue Function


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




  • Prev by Date: Re: Compatibility woes
  • Next by Date: About Condition and HoldAll
  • Previous by thread: Re: Hole/Disk function
  • Next by thread: Re: Residue Function