09/23/2021, 08:42 AM
Additionally, here are some graphs of beta(z,1). These are called with the code,
Which gives the picture:
This is \( \beta_1(z) \) over \( |\Re(z)|,|\Im(z)| \le 3 \)
And if you run:
you get this picture:
This is \( \beta_1(z) \) over \( -1 \le \Re(z) \le 5 \) and \( |\Im(z)| \le 3 \).
We can also sum Taylor series, if you run
You get the sum of the Taylor series of Abel(z,1) about zero, which converges as it should:
Note, these graphs take about 6 hours to compile o average...
Regards, James
Code:
func(z) = beta(z,1)
%33 = (z)->beta(z,1)
MakeGraph(800,800,-3,3,3,-3,BETA_2PI_I_TEST)Which gives the picture:
This is \( \beta_1(z) \) over \( |\Re(z)|,|\Im(z)| \le 3 \)
And if you run:
Code:
MakeGraph(500,500,-1,3,5,-3, BETA_2PI_I_FURTHEROUT)you get this picture:
This is \( \beta_1(z) \) over \( -1 \le \Re(z) \le 5 \) and \( |\Im(z)| \le 3 \).
We can also sum Taylor series, if you run
Code:
Y= Abel(s,1)
%24 = 1.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 + 1.091746621076809013884974261863482866515963611560536324138994111751239098282697066968792471034239886*s + 0.2714060762070193171437424955943827931638503717136848165830532882692408669826080900482469467615488910*s^2 + 0.2125525861313469888021478425802562090387068291121784641878304794743661021504702004547199878572094071*s^3 + 0.06977521648514998263674987921378838226309211345190041781551961351764919973313522659458688303371573846*s^4 + 0.04404163417774007948974320281475423239471012247835111677418601284865041866096275647089240474660889988*s^5 + 0.01480629466783694024680606804770874466095025847943622314359234215211918566929634587982547822697524125*s^6 + 0.008561268049536550787110621350376164411249768075478143351453259439601273283709683201952517632614868019*s^7 + 0.002814905681249618769344334929755303131301111679210593232530174245343075739294184023761363767325649788*s^8 + 0.001600984980242967980058067512988028998626599587465129469228613980725879543699254151351823397240636449*s^9 + 0.0005240150126752845344816864863220032432383929844600032134916998405913413929359719890727966534676820259*s^10 + 0.0003017151070992455198885590074112687198266940835895138851311257194250982324280807827139264047511680096*s^11 + 9.024605643550261986785556669340450988716179032759129805040792164935223712423512110510982647458269449 E-5*s^12 + 4.859419988807502511049138781325890892877814919176570632479411035916500915789129780855922960206661452 E-5*s^13 + 8.084794728794402624489481277553499323658745901241299037539670393969783591119684627779672866000859297 E-6*s^14 + 2.704836036866106315173902706242762078649072551241974645493836652948459913867590456573805827341354948 E-6*s^15 - 2.450720613547027454196739715567577929122842285508726144924924557910470116337318420497467650134284425 E-6*s^16 - 6.892515381942296218498173521547907785286934797376940481495402983799448402762944660107195839612212864 E-7*s[+++]
\ps 35
func(z) = sum(j=0,34, polcoef(Y,j,s)*z^j)
seriesprecision = 35 significant terms
%25 = (z)->sum(j=0,34,polcoef(Y,j,s)*z^j)
func(0)
%26 = 1.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
MakeGraph(300,300,-1,1,1,-1, TAYLOR_WITH_2PI_I)You get the sum of the Taylor series of Abel(z,1) about zero, which converges as it should:
Note, these graphs take about 6 hours to compile o average...
Regards, James

