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MathGroup Archive 2007

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Re: Residue Function

  • To: mathgroup at smc.vnet.net
  • Subject: [mg76111] Re: Residue Function
  • From: dimitris <dimmechan at yahoo.com>
  • Date: Wed, 16 May 2007 05:37:05 -0400 (EDT)
  • References: <f2bulj$jhe$1@smc.vnet.net>

Hi.

As far as I know Complex Analysis you have just enountered a bug.

(*incorect*)
In[43]:=
(Residue[#1, {z, Infinity}] & ) /@ {1/z, Exp[2/z]}

Out[43]=
{-1, -2}

(*correct*)
In[46]:=
(Coefficient[#1, z^(-1)] & )[Normal[{1/z, Exp[2/z]} + O[z,
Infinity]^3]]

Out[46]=
{1,2}

The strange thing that Residue calls the Series function.

In[51]:=
Unprotect[Series];
Series[a___] := Null /; (Print[InputForm[series[a]]]; False)

In[52]:=
Residue[Cos[z]/z, {z, 0}]

>From In[52]:=
series[Cos[z]/z, {z, 0, 1}]

Out[52]=
1

But funnily

In[60]:=
(Residue[#1,{z,Infinity}]&)/@{1/z,Exp[2/z]}

>From In[60]:=
series[z, {z, 0, 1}]

>From In[60]:=
series[E^(2*z), {z, 0, 1}]

Out[60]=
{-1,-2}

Hence, Mathematica correctly makes the transformation 1/z->z.
Nevertheless it returns a incorrect result!

Dimitris

=CF/=C7 Dana DeLouis =DD=E3=F1=E1=F8=E5:
> Hello.  I am studying the subject of Residues, and came across the follow=
ing
> 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 arri=
ves
> at -2.  Anyone familiar with this topic to comment?  Thanks in Advance.
>
> Dana DeLouis



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