Re: Re: computing residues

*To*: mathgroup at smc.vnet.net*Subject*: [mg54859] Re: [mg54821] Re: [mg54785] computing residues*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>*Date*: Fri, 4 Mar 2005 05:07:53 -0500 (EST)*References*: <200503010658.BAA25262@smc.vnet.net> <200503030329.WAA21091@smc.vnet.net> <4a6a68e0a91addc250bf47ab9ab03e74@mimuw.edu.pl>*Sender*: owner-wri-mathgroup at wolfram.com

I have to admit Mathematica is smarter than I had thought and in fact: Residue[1/Sin[x],{x,Root[8*#1^3-6*#1-1&,3]-Cos[Pi/9]}] 1 I made a mistake by using Root[8*#1^3-6*#1-1&,1] instead of Root[8*#1^3-6*#1-1&,3] in the first part of my example below. In fact Residue deals with this case impressively well. This certainly seems to reduce the strength of my argument, though I still would prefer to get an unevaluated input in the non-numerical case. Andrzej On 3 Mar 2005, at 08:47, Andrzej Kozlowski wrote: > > 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 -- óçåë ñâîáîäíûõ êîììóíèêàöèé. >>> >>> >> >> 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 >

**References**:**computing residues***From:*mjumbo <mjumbo@nm.ru>

**Re: computing residues***From:*Daniel Lichtblau <danl@wolfram.com>

**Re: FindRoot/NDSolve problem in VS5**

**Front End Programming question**

**Re: Re: computing residues**

**Re: computing residues**