Re: Re: Residue Function

*To*: mathgroup at smc.vnet.net*Subject*: [mg76324] Re: [mg76271] Re: Residue Function*From*: Carl Woll <carlw at wolfram.com>*Date*: Sat, 19 May 2007 04:52:28 -0400 (EDT)*References*: <200705181032.GAA13845@smc.vnet.net>

Dana DeLouis wrote: >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. > > One possibility is to use NResidue: Needs["NumericalCalculus`"] In[3]:= NResidue[Exp[2/z], {z, 0}, Radius -> 1] // Chop Out[3]= 2. As the help for Residue mentions, Residue will usually only work Mathematica is able to find a power series at that point. Since Exp[2/z] has a Laurent expansion but not a Taylor series expansion at z=0, Series doesn't work. Since Exp[2/z] does have a "Taylor" series expansion at the point z=Infinity, another possibility is to find this series and pick off the residue: In[4]:= s = Series[Exp[2/z], {z, Infinity, 3}] // Normal Out[4]= 1+2/z+2/z^2+4/(3 z^3) In[5]:= Residue[s, {z, 0}] Out[5]= 2 Carl Woll Wolfram Research >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." > > >

**References**:**Re: Residue Function***From:*"Dana DeLouis" <dana.del@gmail.com>

**Re: Close Evaluation Cell in Version 6.0**

**Re: All permutations of a sequence**

**Re: Residue Function**

**Re: Residue Function**