Analytic Continuation of Bernoulli Numbers, a New

arXiv:physics/9705021v231Jul1997AnalyticContinuationofBernoulliNumbers,aNewFormulafortheRiemannZetaFunction,andthePhenomenonofScatteringofZeros1S.C.WoonDepartmentofAppliedMathematicsandTheoreticalPhysicsUniversityofCambridge,SilverStreet,CambridgeCB39EW,UKEmail:[e-mail]ticcontinuationofoperatorsactingintegern-timestocomplexs-times(hep-th/9707206)isappliedtoanoperatorthatgeneratesBernoullinumbersBn(Math.Mag.,70(1),51(1997)).BnandBernoullipolynomialsBn(s)areanalyticcontinuedtoB(s)andBs(z).AnewformulafortheRiemannzetafunctionζ(s)intermsofnestedseriesofζ(n)isderived.Thenewconceptofdynamicsofthezerosofanalyticcontinuedpolynomialsisintroduced,andaninterestingphenonmenonof‘scatterings’ofthezerosofBs(z)isobserved.1PreprintDAMTP-R-97/19on-lineat:BernoulliNumbersBernoullinumbersBnwerediscoveredbyJakobBernoulli(1654-1705)[1].Theyaredefined[2][3]aszez−1=∞Xn=0Bnn!zn,|z|2π,n=1,2,3,...∈Z+(1)Expandingthel.h.s.asaseriesandmatchingthecoefficientsonbothsidesgivesB1=−1/2,Bn=0,oddn6=0,evenn(2)Withthisresult,(1)canberewrittenaszez−1+z2=∞Xn=0B2n2n!z2n(3)Alternatively,BernoullinumberscanbedefinedassatisfyingtherecurrencerelationBn=−1n+1n−1Xk=0n+1kBk,B0=1(4)Bernoullinumbersareinterestingnumbers.Theyappearinconnectionwithawidevarietyofareas,fromEuler-MaclaurinSummationformulainAnalysis[4][5]andtheRiemannzetafunctioninNumberTheory[6][7],toKummer’sregularprimesinspecialcasesofFermat’sLastTheoremandCombinatorics[8].2ATreeforGeneratingBernoulliNumbersItwasshownin[4]howabinaryTreeforgeneratingBernoullinumberscanbeintuitedstep-by-stepandeventuallydiscovered.IntheprocessofcalculatingtheanalyticcontinuationoftheRiemannzetafunctiontothenegativehalfplaneterm-by-term,anemergingpatternwasobserved.ThebigpictureofthestructureoftheTreebecameapparentoncomparingthederivedexpressionswiththeEuler-MaclaurinSummationformula.Inthispaper,westartwiththeTreeandproceedontofindinterestingappli-cations.Whiledoingso,wewillencountersomesurprisingconsequences.TheTreecanbeconstructedusingtwooperators,OLandOR.AteachnodeoftheTreesitsaformalexpressionoftheform±1a!b!....DefineOLandORtoactonlyonformalexpressionsofthisformatthenodesoftheTreeasfollows:OL:±1a!b!...→∓1(a+1)!b!...(5)OR:±1a!b!...→±12!a!b!...(6)Schematically,•OLactingonanodeoftheTreegeneratesabranchdownwardstotheleft(hencethesubscriptLinOL)withanewnodeattheendofthebranch.1+12!S0+14!-12!3!-13!2!S2+12!2!2!=+13!-12!2!S1=OROROLOLOROLOROLOROLOLOROROL=Figure1:ThebinaryTreethatgeneratesBernoullinumbers.•ORactingonthesamenodegeneratesabranchdownwardstotheright.Formafiniteseriesoutofthesumofthetwonon-commutingoperatorsSn=(OL+OR)n+12!=OnL+n−1Xk=0On−1−kLOROkL+···+OnR!+12!(7)Thisisequivalenttosummingtermsonthen-throwofnodesacrosstheTree.BernoullinumbersarethensimplygivenbyBn=n!Sn−1forn≥2(8)eg.,B3=3!S2=3!(OL+OR)2+12!=3!(OL+OR)(OL+OR)+12!=3!(OLOL+OLOR+OROL+OROR)+12!=3!+14!+−12!3!+−13!2!++12!2!2!=0Byobservation,thisSum-across-the-TreerepresentationofSnisexactlyequiv-alenttothefollowingdeterminantknowntogenerateBn,Sn=(−1)n12!1000···013!12!100···014!13!12!10···0.....................1(n−2)!......13!12!101(n−1)!1(n−2)!......13!12!11n!1(n−1)!1(n−2)!......13!12!(9)23AnalyticContinuations3.1AnalyticContinuationofOperatorFirst,weintroducetheideaofanalyticcontinuingtheactionofanoperatorfollowing[9].Weareusedtothinkingofanoperatoractingonce,twice,threetimes,andsoon.However,anoperatoractingintegertimescanbeanalyticcontinuedtoanoperatoractingcomplextimesbymakingthefollowingobservation:AgenericoperatorAactingcomplexs-timescanbeformallyexpandedintoaseriesasAs=w1−w1−As=ws1−h1−1wAis=ws1+∞Xn=1(−1)nn!n−1Yk=0(s−k)#1−1wAn!=ws1+∞Xn=1(−1)nn!n−1Yk=0(s−k)#1+nXm=1−1wmnmAm#!(10)wheres∈C,w∈R,and1istheidentityoperator.TheregionofconvergenceinsandtherateofconvergenceoftheserieswillingeneralbedependentonoperatorA,parameterw,andtheoperandonwhichAacts.3.2AnalyticContinuationoftheTree-GeneratingOperatorJustasin(10),theTree-generatingoperator(OL+OR)acting(s−1)timeson+12!canbesimilarlyexpandedas(OL+OR)s−1+12!=ws1+∞Xn=1(−1)nn!nYk=1(s−k)#1+nXm=1−1wmnm(OL+OR)m#!+12!=ws12+∞Xn=1(−1)nn!nYk=1(s−k)#12+nXm=1−1wmnmBm+1(m+1)!#!(11)whichconvergesforRe(s)(1/w)wheres∈C,w∈R,w0.3.3AnalyticContinuationofBernoulliNumbersBn=n!(OL+OR)n−1+12!=Γ(1+n)(OL+OR)n−1+12!(12)NowthatwecananalyticcontinuetheTree-generatingoperator(OL+OR)with(11),ifwedoso,weturnthesequenceofBernoullinumbersBnintotheiranalyticcontinuation—afunctionB(s)=Γ(1+s)(OL+OR)s−1+12!(13)=wsΓ(1+s)12+∞Xn=1(−1)nn!nYk=1(s−k)#12+nXm=1−1wmnmBm+1(m+1)!#!3whichconvergesforRe(s)(1/w),realw0.Soeffectively,bythemethodofanalyticcontinuationofoperator,wehavenowobtainedthefunctionB(s)astheanalyticcontinuationofBernoullinumbers.02468101214s-0.5-0.2500.250.50.751B(s)Figure2:ThecurveB(s)runsthroughthepointsofallBnexceptB1.AlltheBernoullinumbersBnagreewithB(n),theanalyticcontinuationofBernoullinumbersevaluatedatn,B(n)=Bnforn≥2(14)exceptB(