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MathGroup Archive 2005

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Re: Re: computing residues

  • To: mathgroup at smc.vnet.net
  • Subject: [mg54837] Re: [mg54821] Re: [mg54785] computing residues
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Fri, 4 Mar 2005 05:07:35 -0500 (EST)
  • Sender: owner-wri-mathgroup at wolfram.com

On 3 Mar 2005, at 04:29, Daniel Lichtblau wrote:

> mjumbo wrote:
>
>> Hello,
>>
>> I would like to compute residues of analytical function, but
>> Mathematica's built-in function Residue produces extremely strange
>> results. E.g.
>>
>> Residue[1/Sin[z], {z, w}]
>>
>> gives 0, I expected it produce some conditional expression, or at
>> least to leave this expression unevaluated, but 0 is surely
>> incorrect.
>> What I am doing wrong?
>>
>> MJM
>> __________
>> www.newmail.ru --
>>
>>
>
> You are not doing anything wrong but your expectation is not realistic.
> You would like Residue to encorporate, among other things, a
> (nonexistent) transcendental solver, and an oracle to determine when
> the
> expansion point might be treated as arbitrary. This would also raise
> serious design issues in terms of how to describe the result. To see
> what I mean, consider what output you think might be realistic for the
> following examples.
>
> Residue[(z^2-1)/Sin[z*Pi], {z,z0}]
> Residue[(z^2-1)/Sin[Sin[z*Pi]], {z,z0}]
> Residue[(z^2-1)/Sin[z*t], {z,z0}]
> Residue[1/(z-Tan[z]), {z,z0}]
> Residue[1/(z-t*Tan[z]), {z,z0}]
> Residue[1/Sin[z], {z,z0-Sqrt[z0^2]}]
>
> I can tell you how one might begin to approach this. Use Reduce to find
> zeros of 1/f[z]. Give up when Reduce fails. Otherwise attempt to use
> its
> result as an assumption to Series in order to find information about
> singular points.
>
> Among other issues this assumes that you can in some way recognize what
> is the "variable" on which you will place assumptions. And that Series
> can be used to advantage with appropriate assumptions. For example,
> here
> we find a pole, as desired.
>
> In[19]:== Series[1/Sin[z*Pi], {z,z0,1},
> Assumptions->Element[z0,Integers]]
>
>          Sec[Pi z0]    Sec[Pi z0] Tan[Pi z0]
> Out[19]== ----------- + --------------------- +
>          Pi (z - z0)             2
>
>                                               2
>       Pi Sec[Pi z0]   Pi Sec[Pi z0] Tan[Pi z0]                       2
>>    (------------- + -------------------------) (z - z0) + O[z - z0]
>             6                     4
>
> Here we are not so fortunate.
>
> In[21]:== Series[1/Sin[z], {z,z0,1},
> Assumptions->Element[z0/Pi,Integers]]
>
>                                                        2
> Out[21]== Csc[z0] - Cot[z0] Csc[z0] (z - z0) + O[z - z0]
>
> In summary, I see several things standing in the way of this rather
> vaguely proposed functionality.
>
> (1) Lack of a clear design for what it should do, e.g. how to return
> the
> result.
> (2) Lack of capabilities to discern singular points.
> (3) Lack of an ability to recognize that the expansion point is in some
> way not bound to a specific "value" (that is to say, how do we know to
> treat it as being "possibly a singularity").
> (4) Lack of an ability to recognize what are the assumptions needed to
> place us at such points.
> (5) Lack of a reasonable way to expand series with those assumptions
> that place us at such points.
>
>
> Daniel Lichtblau
> Wolfram Research
>
>
>

It seems to me that all the above arguments are reasonable and
convincing but they still do not justify returning a meanngless 0
whenever Mathematica has no idea of what the answer should be rather
than returning the result unevaluated as is usually done in such cases.
The current approach is quite useless while the "normal" approach has
soem uses, since it allows us to use an "inert" result for later
substitution and evaluation. With the current approach we can construct
(in Maxim's style ;-)) examples  like:

Residue[1/Sin[x], {x, Root[8*#1^3 - 6*#1 - 1 & , 1] -
     Cos[Pi/9]}]

0

FullSimplify[Unevaluated[Residue[1/Sin[x],
     {x, Root[8*#1^3 - 6*#1 - 1 & , 3] - Cos[Pi/9]}]]]

1


It does not appear programistically impossible to change this behaviour
and it should not cause any serious backward compatibility problems. In
fact, even a somewhat more ambitious approach might be possible: use
the method you have indicated to see if you get any useful results you
can and return an If statement just as definite integration with
symbolic limits does. Personally I have doubts whether this is worth
implementing, given that it is not likely to be used very frequently
and a user that needs it can easily do it himself with the help of
Reduce, but simply returning 0 seems to me to be completely useless and
difficult to justify.

Andrzej Kozlowski


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