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Re: Residue Function

• To: mathgroup at smc.vnet.net
• Subject: [mg76271] Re: Residue Function
• From: "Dana DeLouis" <dana.del at gmail.com>
• Date: Fri, 18 May 2007 06:32:24 -0400 (EDT)

Thank you everyone for your help.  Here's what I have so far.
The original question was to find the Residue of Exp[2/z].
The article showed using Mathematic returning 2.  However, it returns
unevaluated for me.
Residue[Exp[2/z], {z, 0}]

Series doesn't help here either.

Series[Exp[2/z], {z, 0, 5}]

In fact, most of the examples in books I am studying are not evaluated using
the function "Residue" in Mathematica.
It appears to not work well with Exp, and Trig Functions examples.
I believe you have to use Infinity in the above example, but I am not sure
yet as to the proper way to do it.

I have received feedback that the Residue function did in fact work in older
versions of Mathematica.  However, beginning in version 5, some
functionality was removed as it was causing errors elsewhere.  At least I
know now why most of my examples are not working.
I was also given this version, but it's a little over my head at this point.

(1/(2*Pi*I))*Integrate[Exp[(2/r)*Exp[(-I)*t]]*((-I)*r*Exp[I*t]), 
     {t, 0, 2*Pi}, Assumptions -> r > 0]

-2

Valeri also posted an interesting equation that returned -2 using version
5.2.
However, in Mathematica 6.0, it returns 0.    Hmmm! 
I received an email stating that this "appears to be a bug" with version 6.0
at this time.

Another major concern mentioned was that this returned 0 also.  
FullSimplify[-(Integrate[Exp[2/z], {z, 1, I}] + 
          Integrate[Exp[2/z], {z, I, -1}] + Integrate[Exp[2/z], 
            {z, -1, -I}] + Integrate[Exp[2/z], {z, -I, 1}])/(2*Pi*I)]
0

But this returned -2.

Chop[-(NIntegrate[Exp[2/z], {z, 1, I}] + NIntegrate[Exp[2/z], 
            {z, I, -1}] + NIntegrate[Exp[2/z], {z, -1, -I}] + 
          NIntegrate[Exp[2/z], {z, -I, 1}])/(2*Pi*I)]

-2

It appears to be an Integrate bug somewhere with 6.0, but I'm not an expert
here to comment on it.

Thank you everyone for your time.  Interesting subject "Residues."

-- 
Dana DeLouis
Windows XP & Mathematica 6  (&  5.2 Help Files)


"Dana DeLouis" <dana.del at gmail.com> wrote in message
news:f2bulj$jhe$1 at smc.vnet.net...
> 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
> 
> 
>

• Follow-Ups:
  • Re: Re: Residue Function
    • From: Carl Woll <carlw@wolfram.com>