Mathematica 9 is now available
Services & Resources / Wolfram Forums
-----
 /
MathGroup Archive
2007
*January
*February
*March
*April
*May
*June
*July
*August
*September
*October
*Archive Index
*Ask about this page
*Print this page
*Give us feedback
*Sign up for the Wolfram Insider

MathGroup Archive 2007

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

Search the Archive

Re: Residue Function

  • To: mathgroup at smc.vnet.net
  • Subject: [mg76370] Re: Residue Function
  • From: CKWong <CKWong.P at gmail.com>
  • Date: Sun, 20 May 2007 02:39:21 -0400 (EDT)
  • References: <f2mcqa$kdg$1@smc.vnet.net>

I'm not sure what's going on here.  As I understand it, residues are
used to evaluate contour integrals because the former is always easier
to calculate than the latter.  Here, every one is going the opposite
direction.
The incompetence of the function Residue is to be deplored but it is
easily remedied.  If one remembers that the function Series  is meant
to provide the Taylor series, it can be easily adapted to get the
Laurent Series.
Take the present example of  finding the residue of Exp[2/z] at z=0,
the proper way to proceed is as follows
Series[ Exp[2x], {x,0,1}] /. x->1/z
There is no need to bring in the pole at infinity, which simply makes
thing worse as you don't seem to know how the residue at infinity is
defined (see my other reply to your earlier posting).
I apologize in advance if I sound over-bearing.  But it's rather
frustrating to witness the mangling of an elegant subject like contour
integrals.



  • Prev by Date: RE: 64-bit again
  • Next by Date: Re: Number of Differing Digits & Another Problem (want to
  • Previous by thread: Re: Residue Function
  • Next by thread: About Condition and HoldAll