Arguments for the beta method not being Kneser's method
#20
I'm very excited to see your take on the manner, Sheldon.

If this turns out to be Kneser, I think that would be really cool. It would imply there are 2 completely different construction methods to Kneser. The theta mapping way, and the infinite composition way. I'm very excited.

I'm also excited to see how you would code this--I'm stuck in a loop in how I've coded it so far. I can't seem to make my code any better; only worse. I'm sure you'll have a much better approach.

Regards, James.


Also in the code I sent you, if you run,

Code:
Abel(0,1)  /*my code is a little iffy, you have to compute a pointwise value before finding Taylor series*/
%22 = 1.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
A = Abel(1+z,1)
%23 = 2.718281828459045235360287471352662497757247093699959574966967627724076630353547594571382178525166427 + 2.967675001354652819296127594565742072515371100795384098700715100444872879616355748110954897095595657*z + 2.357732782678439856207373109446921705172559642201103246412154136509264671254302636421825731027053196*z^2 + 1.972756783820104355828140430188695319933557754505278322338940812272450781077344349039966600447249928*z^3 + 1.521148090181033468925522415936749561332256099614758939957677219761076742954975893121293477491051523*z^4 + 1.132367955367112637558087096821006031223027452473402246898760925916582808906408997101656980327848171*z^5 + 0.8121563103806010230222350911479254900003271460190619053912747122944366657987535989209973326663293676*z^6 + 0.5668056036040309827190929604067804496981862292586028412031220460363419392521542100001466557118432624*z^7 + 0.3861511679501009246106947222279153380577461794798721487665136643264317909051790675972715195270677629*z^8 + 0.2578543148265740343595500532568696208110639088041807087643069410037215153740124745789417971886590220*z^9 + 0.1692037214481119873018313319326908299532593192138533430241298678694714253686213481327033825033120650*z^10 + 0.1093357623807869255131700383434340592586729366813005280713744356159964346365114562733492505352421528*z^11 + 0.06967215724586660529832773486190949838180034194657284765751618014624119610304266010332651433679916640*z^12 + 0.04383424929004270639859632439331405447985109394155763330954570580677886752537726245690955919412589625*z^13 + 0.02725230945725728928653041646527817396175457376567607162605711289589544420029923862597058334312778471*z^14 + 0.01675456714810252605320540728988177648406970282527886136238692483366793046381635208878625743590500442*z^15 + 0.01019336089994092319953180954011473686533749121550453627689868645567636329984469654296962486610763327*z^16 + 0.006142643202590189051063058559475426272016144588458199122627496275555023356411510907892559475747711113*z^17 + 0.0036709175040100352088[+++]
This will give the taylor series about 1, similarly with,


Code:
Abel(z,1)
%24 = 1.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 + 1.091746621076809013884974261863482866515963611560536324138994111751239098282697066968792471034239886*z + 0.2714060762070193171437424955943827931638503717136848165830532882692408669826080900482469467615488910*z^2 + 0.2125525861313469888021478425802562090387068291121784641878304794743661021504702004547199878572094071*z^3 + 0.06977521648514998263674987921378838226309211345190041781551961351764919973313522659458688303371573846*z^4 + 0.04404163417774007948974320281475423239471012247835111677418601284865041866096275647089240474660889988*z^5 + 0.01480629466783694024680606804770874466095025847943622314359234215211918566929634587982547822697524125*z^6 + 0.008561268049536550787110621350376164411249768075478143351453259439601273283709683201952517632614868019*z^7 + 0.002814905681249618769344334929755303131301111679210593232530174245343075739294184023761363767325649788*z^8 + 0.001600984980242967980058067512988028998626599587465129469228613980725879543699254151351823397240636449*z^9 + 0.0005240150126752845344816864863220032432383929844600032134916998405913413929359719890727966534676820259*z^10 + 0.0003017151070992455198885590074112687198266940835895138851311257194250982324280807827139264047511680096*z^11 + 9.024605643550261986785556669340450988716179032759129805040792164935223712423512110510982647458269450 E-5*z^12 + 4.859419988807502511049138781325890892877814919176570632479411035916500915789129780855922960206661452 E-5*z^13 + 8.084794728794402624489481277553499323658745901241299037539670393969783591119684627779672866000859303 E-6*z^14 + 2.704836036866106315173902706242762078649072551241974645493836652948459913867590456573805827341354940 E-6*z^15 - 2.450720613547027454196739715567577929122842285508726144924924557910470116337318420497467650134284414 E-6*z^16 - 6.892515381942296218498173521547907785286934797376940481495402983799448402762944660107195839612213041 E-7*z[+++]



this will give the taylor series about 0. **Note that series precision caps at about 35 for the abel function**

At least, these are the current write ups. They are still a little glitchy, but it's a good base to work from.

You can also run the beta taylor series by writing,


Code:
beta(z,1)
%25 = 0.3047812723715631626923015792598253346867960359900467240879774471150853396719969197860001160926829229 + 0.2581832782858992635060868047102413763939613483259195472236688617584221959696516525587310611841169650*z + 0.09463888612428387194740730932988071510659020889882707705631953397406128888864303628546778237151637377*z^2 + 0.01753438256356050427729870239873798484572480762843257387281852305017368763353319548602777334650882279*z^3 + 0.001759337414570557303275880706685272797080553330810727735490510525753989483713645400676209108170740531*z^4 + 0.0006863979038845097698444711663854233512563664146092614073874910940146297168329600528402038121379649848*z^5 + 0.0004405208937310109260538168072078485540501946830998265386870605812414649715990773648692052914360015455*z^6 + 0.0001283781560448517843991019441521080275738359019385970868544988158381758328687415885543735580411983813*z^7 + 1.127655436026842630431776398940366363564725327178739718557710956416489691941780787467224357901104767 E-5*z^8 - 1.680604395022457192516126485203231558086281982229394781595439105089015023134334729499726466033969271 E-6*z^9 + 9.650906392345079107385085456472781342983407911841549695223564000172509762318796556544110640700563151 E-7*z^10 + 9.390318846181570376336441805537276985437465935052966380388337419322862330907488299598441553527099334 E-7*z^11 + 2.260026187217772870389604410623321795747373698093413868436196203110557225843048456519771194581000847 E-7*z^12 - 1.395526583269469714590317681698606511218377633332715636128915926089879887286949315239326751537315812 E-8*z^13 - 1.638439359347133272252748735159358815820601685906309637165751377872048994785992909128642872108212928 E-8*z^14 + 1.078238513266699145859423752733334882012425749056914213162490031836708930273501784634604646858594594 E-10*z^15 + 2.067291475308126749957840316954606375230489485841092217044166407481420466493201124477973575581880773 E-9*z^16 + 5.23322442162135048345507171457926250010740682929602430387049975932742122583728359988991580006[+++]


which gives the taylor series at 0, again. You can work similarly with beta(A+z) for any point A; also with Abel.  **beta will allow 100 series precision, not sure why this discrepancy is happening**



Also, I know that you were asking me to show the singularities at \( z = j + \pi i \) for \( j\in \mathbb{N} \). And for some reason it wasn't looking like it was diverging. Here's a graph of,


Code:
ploth(X=0,3.14, P = beta(I*X+2,1); [real(P),imag(P)])
%28 = [0.E-307, 3.140000000000000124, -1.3915858263219127444, 1.4403735137825326440]

   

We can clearly see the essential singlarity. And if you run the command:


Code:
ploth(X=0,3.14, P = beta(I*X+1.5,1); [real(P),imag(P)])
%29 = [0.E-307, 3.140000000000000124, 0.E-307, 1.0709593063171574112]



you see that there's no singularity here at beta(Pi*I+1.5,1)

   


And now if you run


Code:
ploth(X=0,1.5, P = beta(X+3.14*I,1); [real(P),imag(P)])
%31 = [0.E-307, 1.5000000000000000000, -189.91240846940786468, 380.2303726792904968]

you can see the pole at \( z = 1+\pi i \) (the other singularities are essential, the only poles which occur are at [\( z=1+(2k+1)\pi i \) for \( k \in \mathbb{Z} \)).

   

This is a better look at how the singularities appear at beta(j+Pi*I,1) for \( j \in \mathbb{N},\,\,j\ge 1 \); and upto the period these are the only singularities.
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Messages In This Thread
RE: Arguments for the beta method not being Kneser's method - by JmsNxn - 09/22/2021, 03:11 AM

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