On conjectures by Csordas, Charalambides and ∗ Waleffe Alexander Dyachenko Galina van Bevern 12th June 2015 InthepresentnoteweobtainnewresultsontwoconjecturesbyCsordasetal. regardingthe interlacing property of zeros of special polynomials. These polynomials came from the Jacobi 5 1 taumethodsfortheSturm-Liouvilleeigenvalueproblem. Theircoefficientsarethesuccessive 0 (α,β) evenderivativesoftheJacobipolynomialsP evaluatedatthepointone.Thefirstconjecture 2 n n statesthatthepolynomialsconstructedfromPn(α,β)andPn(α−,1β)areinterlacingwhen−1<α<1and u −1<β.WeproveitinarangeofparameterswiderthanthatgivenearlierbyCharalambidesand J Waleffe.Wealsoshowthatwithinnarrowerboundsanotherconjectureholds.Itassertsthatthe 2 (α,β) (α,β) polynomialsconstructedfromP andP arealsointerlacing. 1 n n−2 ] Keywords: Jacobipolynomials·Interlacingzeros·Taumethods·Hurwitzstability·Hermite-Biehlertheorem A MathematicsSubjectClassification(2010): 33C45·26C10·30C15 C . h t 1 Introduction a m [ Thisstudyisdevotedtopropertiesofzerosofpolynomialswhichoriginatefrommathematicalphysics.The orthogonalpolynomialsprovedtobeaveryhelpfultoolforthediscretizationoflineardifferentialoperators.The 2 v mainfeatureofthetaumethodsistheadoptionofapolynomialbasiswhichdoesnotautomaticallysatisfythe 4 boundaryconditions.Thisinducesaproblemattheboundary(i.e.thepoints±1intheJacobicase).Thefamily 9 3 ofpolynomialsstudiedhereisconnectedthiswaytotheeigenproblemu(cid:48)(cid:48)(x)=λu(x)ontheintervalx∈(−1,1) 3 withvarioushomogeneousboundaryconditions(forthedetailssee[4,3]).Weplaceourmainemphasisonthe 0 . analyticpropertiesoftheconsideredfamilyitself,leavingasidethecorrespondingpropertiesoftheoriginal 1 0 differentialoperators.Moreinformationonthetaumethodscanbefoundine.g.[1,§10.4.2]. 5 TheJacobipolynomials(seetheirdefinitionandbasicpropertiesine.g.[7,Ch.IV]) 1 : v (cid:195) (cid:33) (cid:34) (cid:35) Xi Pn(α,β)(x)= n+nα 2F1 α−+n,1 n+α+β+1;1−2x , n=1,2,... r a appearregularlyinapplicationsasclassicalorthogonalpolynomials. Theyaremoregeneralthanthoseof Chebyshev, Legendre and Gegenbauer. The Jacobi polynomials are orthogonal with respect to the meas- urewα,β(x)=(1−x)α(1+x)βontheinterval(−1,1)wheneverboththeparametersαandβaregreaterthan−1: (cid:90) 1 −1Pn(α,β)(x)Pk(α,β)(x)wα,β(x)dx=0 if k(cid:54)=n. TheusualnormalizationsupposesthatP(α,β)(1)=(cid:161)n+α(cid:162)= (α+1)n,whereweappliedtheso-calledPochhammer n n n! symbolortherisingfactorialdefinedas (α+1) :=(α+1)·(α+2)···(α+n). n ∗ ThisworkwasfinanciallysupportedbytheEuropeanResearchCouncilundertheEuropeanUnion’sSeventhFrameworkProgramme (FP7/2007–2013)/ERCgrantagreementno.259173. 1 2 OnconjecturesbyCsordas,CharalambidesandWaleffe Inthisnotationwehave P(α,β)(x)= (α+1)n (cid:88)∞ (−n)k(n+α+β+1)k (cid:181)1−x(cid:182)k, n=1,2,.... (1) n n! k!(α+1) 2 k=0 k Definition(see[2,p.17]). Wesaythatthezerosofthepolynomialsg(x)andh(x)interlace(orinterlacestrictly)if thefollowingconditionsholdsimultaneously: • allzerosofg(x)andh(x)aresimple,realanddistinct(i.e.thepolynomialsarecoprime), • betweeneachtwoconsecutivezerosofg(x)thereisexactlyonezeroofthepolynomialh(x),and • betweeneachtwoconsecutivezerosofh(x)thereisexactlyonezeroofthepolynomialg(x). Wesaythatthezerosofthepolynomialsg(x)andh(x)interlacenon-strictlyiftheirzerosarerealandbecome strictlyinterlacingafterdividingbothpolynomialsbythegreatestcommondivisorgcd(g,h).Roughlyspeaking, thezerosoftwopolynomialsinterlacenon-strictlyiftheycanmeetbutneverpassthrougheachotherwhen changingcontinuouslyfromastrictlyinterlacingstate. Definition. Apair(cid:161)g(x),h(x)(cid:162)iscalledrealifforanyrealnumbersA,B thecombinationAg(x)+Bh(x)hasonly realzeros.Thisisequivalenttothenon-strictinterlacingpropertyofg(x)andh(x),whichisshownine.g.[2, ChapterI]. Remark. Thephrases“g(x)andh(x)interlace”,“g(x)andh(x)possesstheinterlacingproperty”,“g(x)interlaces h(x)”,“g(x)andh(x)haveinterlacingzeros”and“thezerosofg(x)andh(x)areinterlacing”weusesynonymously. Itiswell-knownthattheorthogonalpolynomialsonthereallinehaverealinterlacingzeros(duetothe so-calledthree-termrecurrence;seee.g.[7,pp.42–47,Sections3.2–3.3]).Thatis,inparticular,thezerosof (α,β) (α,β) P andP interlaceforallnaturaln.Inthepresentnotewestudyzerosofpolynomialsthatdonotsatisfy n n−1 thethree-termrecurrence.Morespecifically,weconsider (cid:175) φ(α,β)(µ):=[n(cid:88)/2] d2k P(α,β)(x)(cid:175)(cid:175) ·µk= (α+1)n [n(cid:88)/2](−n)2k(n+α+β+1)2k (cid:179)µ(cid:180)k, n=1,2,..., (2) n k=0 dx2k n (cid:175)(cid:175)x=1 n! k=0 (α+1)2k 4 wherethenotation[a]standsfortheintegerpartofthenumbera. TheoremCCW(Csordas,CharalambidesandWaleffe[4]). Foreverypositiveintegern(cid:202)2thepolynomialφ(α,β)(µ), n −1<α<1,−1<β,hasonlyrealnegativezeros. Theproofofthistheoremgivenin[4]restsontheHermite-Biehlertheory(seeTheoremHBherein). Remark(toTheoremCCW). Infact,theauthorshaveshownthatφ(α,β)(µ)interlacesφ(α+1,β+1)(µ).Asaresult, n n−1 thetheoremremainsvalidwhen1(cid:201)α<2and0<β.Notethattheinterlacingpropertyhereisstrict,soitimplies thesimplicityofthezeros. Furthermore,thepolynomialφ(α,β)(µ)hasonlysimplenegativezerosfor−1<α<0 n and−2<β,aswell.ThisfollowsasastraightforwardconsequenceofTheoremsCW*andLemma2ofthepresent study. BasedonTheoremCCWtheauthorsof[4]conjecturedthatthesepolynomialsalsohavethefollowingproperty. ConjectureA([4,p.3559]). For−1<α<1,−1<βandn(cid:202)4thezerosofthepolynomialsφ(α,β)andφ(α,β)interlace. n n−1 Inparticular,thisassertionwouldimplythatthespectraofpolynomialapproximationstothecorresponding differentialoperatorarenegativeandsimple(see[3]). In[3],ConjectureAwasprovedfor−1<α,β<0and 0<α,β<1:seeTheoremCWbelow.Infact,theupperboundonβisredundant.TheoremCW*withashorter proofstatesthattheconjectureholdstruefor −1<α<0, −1<β or 0(cid:201)α<1, 0<β or 1(cid:201)α<2, 1<β. (3) Additionally,westudyanotherassertionaboutthesamepolynomials. ConjectureB([4,p.3559]). Forφ(α,β)(µ)asinTheoremCCWandforalln(cid:202)5,thezerosofthepolynomialsφ(α,β)(µ) n n andφ(α,β)(µ)interlace. n−2 Originally, thisconjecturewasstatedfor−1<α<1andβ>−1. However, numericalcalculationsshow AlexanderDyachenkoandGalinavanBevern 3 thatitfailsforsomevaluessatisfying−1<β<0<α<1. Our(partial)solutiontoConjectureBisgivenin Theorem10: itholdstruefor−1<α<0<β or 0<α<1<β. Weapproachbyextendingtheideaof[3]to anotherpairofauxiliarypolynomials.Certainly,thereexistsarelationbetweenConjectureAandConjectureB asdiscussedinSection5. Vieta’sformulaeimplythatthesumofallzerosofφ(α,β)(µ)tendsto−1/2forevennandto−1/6foroddn n asn→∞.Thus,theassertionofConjectureBgivesthatthezeropointsofφ(α,β)(µ)convergemonotonically n inn outsideofanyfixedintervalcontainingtheorigin. Iftheassertionsofbothconjectureshold,thenthe fractionφ(α,β)(µ)/φ(α,β)(µ)mapstheupperhalfofthecomplexplaneintoitselfandconvergestoafunction 2n−1 2n meromorphicoutsideofanydiskcentredattheorigin.Thissituationresembleshowthequotientsoforthogonal polynomialsofthefirstandsecondkindsbehave. Section2ofthepresentpaperintroducesconnectionsbetweenpolynomialsφ(α,β)(µ)withdifferentn,α n andβ. Theseconnectionsallowustoextendandclarifytheresult[3]inSection3(seeTheoremCW*). We showthatConjectureAholdstrueundertheconditions(3). Section4containstheproofofConjectureBfor −1<α<0<βand0<α<1<β(seeTheorem10).Inthelastsectionweshowthatthestudiedconjecturesare actuallyrelated. 2 Basicrelationsbetweenthepolynomialsφ(α,β) forvariousαandβ n BeingconnectedwiththeJacobipolynomials,thefamily(cid:179)φ(α,β)(cid:180) ,wheren=2,3,...,inheritssomeoftheir n n properties. The formulae induced by the corresponding relations for the Jacobi case include (we omit the argumentµofφ(α,β)forbrevity’ssake): n (2n+α+β)φ(α,β−1)=(n+α+β)φ(α,β)+(n+α)φ(α,β), (4) n n n−1 (2n+α+β)φ(α−1,β)=(n+α+β)φ(α,β)−(n+β)φ(α,β), (5) n n n−1 (n+α+β)φ(α,β)=(=4=)+=(=5=) (n+β)φ(α,β−1)+(n+α)φ(α−1,β), (6) n n n φ(α,β)=(=4=)−=(=5=) φ(α,β−1)−φ(α−1,β). (7) n−1 n n Thelattertwoidentitiescontainlabelsofequationsabovetheequalitysign:weusetheconventionthatthis explainshowtheequalitiescanbeobtained(uptosomepropercoefficients).Forexample,2n+α+βtimes(6) isthesumofn+βtimes(4)andn+αtimes(5),whereas2n+α+βtimes(7)isthedifference(4)−(5). The identities(4)and(5)canbecheckedbyapplyingtheformulae(see,e.g.,[6,p.737]) (2n+α+β)P(α,β−1)(x)=(n+α+β)P(α,β)(x)+(n+α)P(α,β)(x), (8) n n n−1 (2n+α+β)P(α−1,β)(x)=(n+α+β)P(α,β)(x)−(n+β)P(α,β)(x), (9) n n n−1 respectively,totheleft-handsideofthedefinition(2).Therelationsinvolvingderivatives(notsurprisingly) differfromthosefortheJacobipolynomials: (n+α)φ(α,β+1)=nφ(α,β)−2µ(cid:179)φ(α,β)(cid:180)(cid:48), (10) n−1 n n (n+α+β+1)φ(α,β+1)=(=1=0=)a=n=d=(=4=) (n+α+β+1)φ(α,β)+2µ(cid:179)φ(α,β)(cid:180)(cid:48), (11) n n n (n+α)φ(α−1,β+1)==(5=),=t=h=en==(1=1=)−=(=10==) αφ(α,β)+2µ(cid:179)φ(α,β)(cid:180)(cid:48), (12) n n n n+β 2µ(cid:179)φ(α,β−1)(cid:180)(cid:48)=(=1=1=)a=n=d=(=6=) (n+α)φ(α−1,β)−αφ(α,β), (13) n+α+β n n n 2µ(cid:179)φ(α,β−1)(cid:180)(cid:48)=(=1=0=)a=n=d=(=7=) nφ(α−1,β)−αφ(α,β). (14) n n n−1 Soweseethatthepresentedidentitiesarenotindependentinthesensethattheycanbeobtainedascom- binations of others with various α and β. We derive the formula (10) with the help of the right-hand side 4 OnconjecturesbyCsordas,CharalambidesandWaleffe of(2). nφ(α,β)−2µ(cid:179)φ(α,β)(cid:180)(cid:48)= (α+1)n [n(cid:88)/2](−n)2k(n+α+β+1)2k (cid:179)µ(cid:180)k(n−2k) n n n! (α+1) 4 k=0 2k =(n+α)(α+1)n−1[n(cid:88)/2](−n+2k)(−n)(−n+1)···(−n+2k−1)(n+α+β+1)2k (cid:179)µ(cid:180)k (n−1)! (−n)(α+1) 4 k=0 2k =(n+α)φ(α,β+1). n−1 Certainly,wecancombineformulaefurtherobtaininge.g. nφ(α,β)−2µ(cid:179)φ(α,β)(cid:180)(cid:48)=(=1=0=) (n+α)φ(α,β+1)=(=1=1=) (n+α)φ(α,β)+ n+α 2µ(cid:179)φ(α,β)(cid:180)(cid:48), (15) n n n−1 n−1 n+α+β n−1 nφ(α,β)−(n+α)φ(α,β)==(1=1=)−=(=1=0=) 2n+α+β2µ(cid:179)φ(α,β−1)(cid:180)(cid:48)==ddµ=(=4=) 2µ(cid:179)φ(α,β)(cid:180)(cid:48)+ n+α 2µ(cid:179)φ(α,β)(cid:180)(cid:48). (16) n n−1 n+α+β n n n+α+β n−1 Thekeyrolefurtherplaysthefollowingcombinationof(10)and(11),whichisvalidforanarbitraryreal A, n+α+βφ(α,β)+Aφ(α,β)= (1+A)n+α+βφ(α,β−1)+2µ1−A (cid:179)φ(α,β−1)(cid:180)(cid:48). (17) n+α n n−1 n+α n n+α n Thenextidentitystandsapartandcanbecheckedexplicitlywiththehelpof(2) φ(α,β)(µ)−φ(α,β)(0)= 1(n+α+β+1) ·µφ(α+2,β+2)(µ). n n 4 2 n−2 ItreflectsthestandardformulaforthederivativeoftheJacobipolynomial(e.g.[6,p.737]): (cid:179)Pn(α,β)(x)(cid:180)(m)=2−m(n+α+β+1)mPn(α−+mm,β+m)(x). (18) Remark. Notethattheequalities(4)–(18)areofformalnature,andthereforetheirvalidityrequiresnoortho- gonalityfromtheJacobipolynomials.Thatis,theseequalitiesholdstrueifallcoefficientsaredefined,notonly forα,β>−1. Remark. ApolynomialwithonlyrealzerosinterlacesitsderivativebyRolle’stheorem.1 Consequently,theyboth interlaceanyrealcombinationofthem.2 Soifinoneoftheformulae(10)–(12)thefirsttermontheright-hand sidehasonlyrealzeros,thenallthreeinvolvedpolynomialsarepairwiseinterlacing.Forexample,ifφ(α,β)has onlyrealzeros,thenφ(α,β+1),φ(α,β)and2µ(cid:179)φ(α,β)(cid:180)(cid:48)arepairwiseinterlacingwhichisprovidedby(10).n n−1 n n Remark. The identities (6) and (7) show that the interlacing property of φ(α,β)(µ) and φ(α,β)(µ) can also be n n−1 expressed as the interlacing property of φ(α,β−1)(µ) and φ(α−1,β)(µ). Analogously, from the relations (13) and(14)itcanbeseenthatthisisalsoequivanlenttotheinternlacingpropertyof(cid:179)φ(α,β−1)(µ)(cid:180)(cid:48)andφ(α−1,β)(µ). n n Lemma 1. If −1 < α < 1, −1 < β or if 1 (cid:201) α < 2, 0 < β, then the polynomials (cid:179)φ(α,β)(µ)(cid:180)(cid:48), (cid:179)φ(α,β)(µ)(cid:180)(cid:48) and n n−1 (cid:179)φ(α,β−1)(µ)(cid:180)(cid:48)arepairwiseinterlacing. n Proof. By Theorem CCW the polynomials φ(α,β)(µ), φ(α,β)(µ) have only (simple) negative zeros. Then the n n−1 relation (10) shows that the polynomial φ(α,β+1)(µ) with negative zeros interlaces (strictly) both φ(α,β)(µ) n−1 n andµ(cid:179)φ(α,β)(µ)(cid:180)(cid:48).Moreover,thesignofφ(α,β+1)(µ)attheoriginandattherightmostzeroofφ(α,β)(µ)isthe n n−1 n same,andtherefore(cid:179)φ(α,β)(µ),µφ(α,β+1)(µ)(cid:180)isarealcoprimepair.Asimilarconsiderationoftherelation(11) n n−1 givesthatthepair(cid:179)µφ(α,β)(µ),φ(α,β+1)(µ)(cid:180)isalsorealandcoprime. n−1 n−1 Inparticular,eachintervalbetweentwoconsequentzerosofφ(α,β+1)(µ)containsexactlyonezeroofφ(α,β)(µ) n−1 n 1Non-strictlywheneverthepolynomialhasamultiplezero. 2Seethedefinitionofarealpaironthepage2. AlexanderDyachenkoandGalinavanBevern 5 aswellasofφ(α,β)(µ).Takingthesignsofthelasttwopolynomialsattheendsoftheseintervalsintoaccount n−1 shows that the difference in the left-hand side of (16), and hence (cid:179)φ(α,β−1)(µ)(cid:180)(cid:48), changes its sign between n consecutivezerosofφ(α,β+1)(µ).Sincedeg(cid:179)φ(α,β−1)(cid:180)(cid:48)(cid:201)degφ(α,β+1),thepolynomialµ(cid:179)φ(α,β−1)(µ)(cid:180)(cid:48)necessarily n−1 n n−1 n interlacesφ(α,β+1)(µ).Thentheright-handsideof(16)showsthat n−1 (cid:179)φ(α,β)(µ)(cid:180)(cid:48)+ n+α (cid:179)φ(α,β)(µ)(cid:180)(cid:48) (19) n n+α+β n−1 andφ(α,β+1)(µ)haveinterlacingzeros.Atthesametime,bydifferentiatingtheequality(16)weobtainthat n−1 n(cid:179)φ(α,β)(µ)(cid:180)(cid:48)−(n+α)(cid:179)φ(α,β)(µ)(cid:180)(cid:48) (20) n n−1 isproportionalto(cid:179)µ(cid:179)φ(α,β−1)(µ)(cid:180)(cid:48)(cid:180)(cid:48)and,hence,interlacesµ(cid:179)φ(α,β−1)(µ)(cid:180)(cid:48).Putinotherwords,thepolynomi- n n als(19)and(20)areinterlacing. Withappropriatefactors, theirsumgives(cid:179)φ(α,β)(µ)(cid:180)(cid:48) andtheirdifference n gives(cid:179)φ(α,β)(µ)(cid:180)(cid:48).Thisyieldsthelemma. n−1 3 ResultsonConjectureA TheoremHB(Hermite-Biehler,forrealpolynomials). Arealpolynomial f(z):=p(z2)+zq(z2)isstable3ifandonly ifp(0)·q(0)>0andallzerosofp(z)andzq(z)arenonpositiveandstrictlyinterlacing. Thiswell-knownfactplaysacrucialrolein[4]forprovingTheoremCCW.HerethestatementoftheHermite- Biehlertheoremisexpressedcloselytotheonegivenin[5,p.228].Thereplacementofq(z)withzq(z)provides the desired order of zeros. That is, the zero of p(z) and q(z) closest to the origin belongs to the former polynomial.Themoregeneralstatementcanbefoundine.g.[2,p.21].ThepresentstudyalsousesTheoremHB asamaintool. SomeboundsontheparametersαandβarenecessaryevenforTheoremCCW(i.e.aresatisfiedifallzeros of φ(α,β)(µ) are negative for every n (cid:202) 2). The restriction α > −1 corresponds to positivity of the coeffi- n cients (and to negativity of all zeros). The parameter α is bounded from above by 3.37228.... Indeed, if the polynomial φ(α,β)(µ)=:(cid:80)m b µk, m =[n/2], has only negative4 zeros, then by Rolle’s theorem, the n k=0 k zerosofp(µ):=µmφ(α,β)(µ−1)andp(cid:48)(µ)areinterlacing. So,TheoremHBthenimpliesthatthepolynomial n p(µ2)+µp(cid:48)(µ2)isstable. Therefore,whenb >0theHurwitzcriterion(seee.g.[2,ChapterI])givesusthe 0 easy-to-checkconditionsb >0and 1 (cid:175) (cid:175) (cid:175)mb (m−1)b (m−2)b (cid:175) (cid:175)(cid:175)(cid:175) b 0 b 1 b 2(cid:175)(cid:175)(cid:175)(cid:202)0, thatis b12 (cid:202) 2m >2, (21) (cid:175) 0 1 2 (cid:175) b b m−1 (cid:175)(cid:175) 0 mb0 (m−1)b1(cid:175)(cid:175) 0 2 whicharenecessarilytruewhenthepolynomialp(µ)hasonlyrealzeros.Inourcasewehave b12 = n(n−1)(α+3)2(n+α+β+1)2 −n−→−−∞→ (α+3)(α+4), b b (n−2)(n−3)(α+1) (n+α+β+3) (α+1)(α+2) 0 2 2 2 sothecondition(21)failstobetrue(alongwiththeassertionofTheoremCCW)foreveryn bigenoughand (cid:112) (cid:112) every β when α > 1+ 33 = 3.37228... or α < 1− 33 = −2.37228.... Computer experiments show that the 2 2 polynomialsφ(α,β)withpositivecoefficientscanhavezerosoutsidetherealaxisfora(largeenough)negativeβ. n So,somelowerconstraintontheparameterβisalsorequired. 3Thatis,Hurwitzstable: f(z)=0 =⇒ Rez<0. 4Herewesupposethatφ(α,β)(µ)hasonlysimplezeros;thecaseofmultiplezerosfollowsbycontinuity. n 6 OnconjecturesbyCsordas,CharalambidesandWaleffe Definition. Denotetheithzeroofapolynomialpwithrespecttothedistancefromtheoriginbyzr (p).Putzr (p) i i equalto−∞ifdegp−1andn+α+β>0,n=4,5,.... Thezerosofthepolynomialsφ(α,β) andφ(α,β) arenegative n n−1 andinterlacenon-strictly(strictly)ifandonlyifthepolynomialφ(α,β−1) hasonlyrealzeros(onlysimplerealzeros, n respectively).Moreover,ifφ(α,β−1)hasonlyrealzeros,thenzr (cid:179)φ(α,β)(cid:180)(cid:201)zr (cid:179)φ(α,β)(cid:180). n 1 n−1 1 n Proof. Therelation(17)with A=1and A=−(n+α+β)/n impliesthateachcommonzeroofthepolynomi- alsφ(α,β)andφ(α,β)isamultiplezeroofφ(α,β−1).Theconverseresultisgivenby(10)and(11). n−1 n n Supposethatφ(α,β−1)hasonlyrealzeros.Thecoefficientsofthispolynomialarepositiveundertheassump- n tionsofthelemma,andhenceallofitszerosarenegative.ByRolle’stheorem,thepair(cid:179)φ(α,β−1),µ(cid:179)φ(α,β−1)(cid:180)(cid:48)(cid:180)is n n real.Therefore,thepolynomialsφ(α,β)andφ(α,β)haveonlyrealzerosbytheformulae(10)and(11),respectively. n−1 n Thezerosarenegativeautomaticallysincethecoefficientsofpolynomialsarepositive.Moreover,wehavethat thefirstzeroofφ(α,β) isclosertotheoriginthanthatofφ(α,β). Therelation(17)holdsforallreal A,which n n−1 yieldsthatthepolynomialsφ(α,β)andφ(α,β)formarealpairandthushave(non-strictly)interlacingzeros. n−1 n Letφ(α,β)andφ(α,β)havenegativeinterlacingzeros.Thenanyoftheirrealcombinationsonlyhasrealzeros. n−1 n Thisistrueforφ(α,β−1)accordingtotheidentity(4). n Corollary3. For−1<α<1,β>0or1(cid:201)α<2,β>1thezerosofthepolynomialsφ(α,β) andφ(α,β), n=4,5,..., n n−1 interlace.(Furthermore,thepolynomialsφ(α,β)andµφ(α,β)interlace.) n n−1 Proof. ThiscorollaryisprovidedbyTheoremCCW(seealsotheremarkonit)andLemma2. TheoremCW(Theorem3.10,Charalambidesetal.[3]). ConjectureAholdstruefor−1<α,β<0andfor0<α,β<1. ThistheoremreliesonTheorem3.8andTheorem3.9ofthesameworkandonTheoremCCW.Infact,the original proof (which we extend in the next section to treat Conjecture B) does not need any upper bound onβ. ItbecomesmoreevidentonrecallingthattheregionofpositiveβiscoveredbyCorollary3(i.e. isa straightforwardconsequenceofLemma2andTheoremCCW). TheoremCW*. ConjectureAholdstruefor−1<α<0,−1<β or 0(cid:201)α<1,0<β or 1(cid:201)α<2,1<β. Proof. Corollary3suitsthecaseofpositiveα.FortheregionwithnegativeαitisenoughtoproveonlyTheorem4 (whichisstatedbelow)andTheorem3.9isnotneeded.Indeed,accordingtoTheorem4andTheoremHBwe havethatallzerosofφ(α,β−1)aresimpleandrealforalln,soLemma2isapplicable. n Theorem4(EquivalenttoTheorem3.8,Charalambidesetal.[3]). If−1<α<0,−1<βandn=4,5,...,thenthe zerosofthepolynomialΦ (1;µ),where n Φ (x;µ):= (cid:88)n dk P(α,β−1)(x)µk, n k=0dxk n lieintheopenlefthalfofthecomplexplane. Remark. Itisworthnotingthatthistheoremcannotbeextendedtothefullrange−1<β<0<α<1.According tocomputerexperiments, ConjectureAseemstoholdinthisrange, whileTheorem4fails, e.g., forn=12 whenβ=−0.8andα(cid:39)0.97842,orwhenβ=−0.9andα(cid:39)0.97140. ThereasonisthatprovingConjectureA onlyrequiresnegativesimplezerosofthepolynomialφ(α,β−1)(µ),Theorem4neverthelessassertsadditional n propertiesofφ(α+1,β)(µ)asgivenbytheHermite-Biehlertheorem. n−1 AlexanderDyachenkoandGalinavanBevern 7 Proof. Thisproofisakinto[3,Theorem3.8]butusesotherrelationsfortheJacobipolynomials.Thepolyno- mialΦ (x;µ)satisfiesthedifferentialequation(hereweconsiderµasaparameter) n Φ (x;µ)=P(α,β−1)(x)+µdΦn(x;µ). n n dx dΦ dΦ dΦ LetΦn:=Φn(x;µ)forbrevityandlet n denoteacomplexconjugateof n.Multiplyingby nwα,β+1, dx dx dx wherewα,β+1:=wα,β+1(x)=(1−x)α(1+x)β+1,andintegrationovertheinterval(−1,1)givesus (cid:90)−11ΦnddΦxnwα,β+1dx=(cid:90)−11Pn(α,β−1)ddΦxnwα,β+1dx+µ(cid:90)−11(cid:175)(cid:175)(cid:175)(cid:175)ddΦxn(cid:175)(cid:175)(cid:175)(cid:175)2wα,β+1dx. SelectµsothatΦ (1;µ)=0.Toestimatetherealpartofµweaddthelastequationtoitscomplexconjugate n andobtain (cid:90)−11d(cid:161)|dΦxn|2(cid:162)wα,β+1dx=(cid:90)−11Pn(α,β−1)·2ReddΦxn ·wα,β+1dx+2Reµ(cid:90)−11(cid:175)(cid:175)(cid:175)(cid:175)ddΦxn(cid:175)(cid:175)(cid:175)(cid:175)2wα,β+1dx. (22) Sincewα,β+1increaseson(−1,1)andlimx→−1+Φnwα,β+1=limx→1−Φnwα,β+1=0,theleft-handsidesatisfies (cid:90) 1 d(cid:161)|Φ |2(cid:162) (cid:90) 1 −1 dxn wα,β+1dx=− −1|Φn|2wα(cid:48),β+1dx<0. (α,β−1) Applyingtherelation(8)tothepolynomialP threetimesgivesus n P(α,β−1)= n+α+β P(α,β)+ n+α P(α,β)= (n+α+β)2 P(α,β+1)+(n+α+β)(n+α)P(α,β+1) n 2n+α+β n 2n+α+β n−1 (2n+α+β) n (2n+α+β) n−1 2 2 +(n+α)(n+α+β)P(α,β+1)+ (n+α−1)2 P(α,β+1), (2n+α+β−1) n−1 (2n+α+β−1) n−2 2 2 thatis, P(α,β−1)= (n+α+β)2 P(α,β+1)+ 2(n+α)(n+α+β) P(α,β+1)+ (n+α−1)2 P(α,β+1). n (2n+α+β) n (2n+α+β−1)(2n+α+β+1) n−1 (2n+α+β−1) n−2 2 2 BythedefinitionofΦ andtheformula(18), n RedΦn =2−1(n+α+β)P(α+1,β)+Reµ·2−2(n+α+β) P(α+2,β+1)+(cid:169)polynomialsofdegree 0. −1 dx Nowthetermsoftherelation(22)areestimated,andityields0>Reµ. 4 ResultsonConjectureB WejusthaveshownthattheresultontheConjectureAin[3]canbeobtainedinashorterwayifweconsiderpoly- nomialswithshiftedparametervalues(weusedφ(α,β−1))insteadofrealcombinationsofthepolynomialsφ(α,β) n n andφ(α,β).Atthesametime,toverifytheConjectureBwecancombineboththeseideas. n−1 Foranyfixedn>3considertheintermediarypolynomial (cid:195) (cid:33) f(x;µ):= (cid:88)n µk A dk P(α,β)(x)+µ dk P(α,β)(x) . k=0 dxk n dxk n−1 8 OnconjecturesbyCsordas,CharalambidesandWaleffe Lemma5. Thepolynomial f(1;µ)isHurwitz-stableprovidedthat−1<α<1andβ,A>0. Proof. Fromthedefinitionof f(x,µ)thedifferentialequation µ d f(x;µ)+AP(α,β)(x)+µP(α,β)(x)=f(x;µ) dx n n−1 follows.Multiplicationby f(x;µ)wα−1,β(x)givesus f ddxf wα−1,β+APn(α,β)(x)µf wα−1,β+Pn(α−,1β)(x)fwα−1,β= µ1|f|2wα−1,β, whereweput f :=f(x;µ)andwα−1,β:=wα−1,β(x)=(1−x)α−1(1+x)βforbrevity.Addingtothisequalityits complexconjugateandintegratingyields (cid:90)−11d(cid:161)d|fx|2(cid:162)wα−1,βdx+A(cid:90)−11Pn(α,β)(x)(cid:195)µf +µf (cid:33)wα−1,βdx+(cid:90)−11Pn(α−,1β)(x)(cid:179)f +f(cid:180)wα−1,βdx (cid:181)1 1(cid:182)(cid:90) 1 = µ+µ |f|2wα−1,βdx. (23) −1 Takeµsothat f(1;µ)=0.Thenthepolynomial f(x;µ)canberepresentedas n−1 f(x;µ)=(1−x)(cid:88)c P(α,β)(x) k k k=0 withsomecomplexconstantsck dependingonµ(generallyspeaking). Observethatcn−1<0: denotingthe leadingcoefficientinx bylc,weobtain cn−1= lc(cid:179)l−c(cid:161)xfP((xα;,βµ))((cid:162)x)(cid:180) =−Al·cl(cid:179)cP(cid:179)P(αn(,αβ,)β()x()x(cid:180))(cid:180) =(=1=) −A(n+nα!+2nβ·+(n1+)nα·(+nβ−)n1−)!12n−1 =−A(22nn+(nα++αβ+−β1))2 <0. n−1 n−1 Thenwehave (cid:90) 1 (cid:90) 1 n−1 −1Pn(α,β)(x)fwα−1,βdx= −1Pn(α,β)(x)k(cid:88)=0ckPk(α,β)(x)wα,βdx=0, (24) (cid:90) 1Pn(α−,1β)(x)fwα−1,βdx=cn−1(cid:90) 1(cid:179)Pn(α−,1β)(x)(cid:180)2wα,βdx<0 −1 −1 asaconsequenceoforthogonalityoftheJacobipolynomials.Notethatif−1<α<1andβ>0,then wα(cid:48)−1,β=(cid:161)β(1−x)−(α−1)(1+x)(cid:162)·wα−2,β−1>0 and x→lim−1+|f|2wα−1,β=xl→im1−|f|2wα−1,β=0. Therefore,integratingbypartsyields (cid:90) 1 d(cid:161)|f|2(cid:162) (cid:90) 1 −1 dx wα−1,βdx=− −1(cid:161)|f|2(cid:162)wα(cid:48)−1,βdx<0. (25) Bytheformulae(24)–(25),theleft-handsideoftheequation(23)isnegative,thusnecessarilyReµ<0. Consideralsoanotherintermediarypolynomialforafixedn>3,namely (cid:195) (cid:33) g(x;µ):= (cid:88)n µk µ dk P(α,β)(x)+A dk P(α,β)(x) . k=0 dxk n dxk n−1 Lemma6. Thepolynomialg(1;µ)isHurwitz-stableprovidedthat−1<α<0andβ,A>0. AlexanderDyachenkoandGalinavanBevern 9 Proof. ThisproofisanalogoustotheproofsofTheorem4andLemma5.Fromthedefinitionofg(x,µ)wehave µdg(x;µ)+µP(α,β)(x)+AP(α,β)(x)=g(x;µ) dx n n−1 whichgivesus(weputg :=g(x;µ)andg(cid:48):= dg(x;µ) forbrevity’ssake) dx µg(cid:48)g(cid:48)wα,β+µPn(α,β)(x)g(cid:48)wα,β+APn(α−,1β)(x)g(cid:48)wα,β=gg(cid:48)wα,β (cid:48) aftermultiplicationbyg wα,β.Addingtotheequationitscomplexconjugateandintegratingyields (µ+µ)(cid:90) 1|g(cid:48)|2wα,βdx+(cid:90) 1Pn(α,β)(x)(cid:179)µg(cid:48)+µg(cid:48)(cid:180)wα,βdx+A(cid:90) 1Pn(α−,1β)(x)(cid:179)g(cid:48)+g(cid:48)(cid:180)wα,βdx −1 −1 −1 (cid:90) 1(cid:179) (cid:180) = gg(cid:48)+gg(cid:48) wα,βdx. (26) −1 Observethatg isapolynomialofdegreeninxanditsleadingcoefficientisgivenbyP(α,β)(x)·µ.Consequently, n (cid:179) (α,β) (cid:180) substitutingtheexplicitexpressionforlc P (x) fromtheformula(1)gives n lc(g(cid:48))=nlc(cid:179)P(α,β)(x)(cid:180)µ= (2n+α+β−1)2 ·(n+α+β)n−1µ= (2n+α+β−1)2lc(cid:179)P(α,β)(x)(cid:180)µ. n 2(n+α+β) (n−1)!2n−1 2(n+α+β) n−1 Thisallowsustocalculatethethirdsummandontheleft-handsideof(26): (cid:90)−11Pn(α−,1β)(x)(cid:179)g(cid:48)+g(cid:48)(cid:180)wα,βdx=(cid:90)−11Pn(α−,1β)(x)(2n2(+nα++αβ+−β)1)2lc(cid:179)Pn(α−,1β)(x)(cid:180)(cid:161)µ+µ(cid:162)xn−1wα,βdx =(cid:161)µ+µ(cid:162)(2n2(+nα++αβ+−β)1)2(cid:90)−11(cid:179)Pn(α−,1β)(x)(cid:180)2wα,βdx. (27) Additionally,wehave (cid:90) 1 Pn(α,β)(x)g(cid:48)wα,βdx=0. (28) −1 Takeµsothatg(1;µ)=0.Then wα(cid:48),β=(cid:161)β(1−x)−α(1+x)(cid:162)·wα−1,β−1>0 and x→lim−1+|g|2wα,β=xl→im1−|g|2wα,β=0 since−1<α<0andβ>0.Integratingbypartsweobtain (cid:90) 1(cid:179) (cid:180) (cid:90) 1 gg(cid:48)+gg(cid:48) wα,βdx=− |g|2wα(cid:48),βdx<0. (29) −1 −1 Nowletusbringtogethertherelations(26)–(29): (µ+µ)(cid:90)−11(cid:181)|g(cid:48)|2+A(2n2(+nα++αβ+−β)1)2(cid:179)Pn(α−,1β)(x)(cid:180)2(cid:182)wα,βdx=−(cid:90)−11|g|2wα(cid:48),βdx, hence 2Reµ=µ+µ<0. Thatis,anyzeroµofthepolynomialg(1;µ)residesinthelefthalfofthecomplexplane. Corollary7. Foranypositive A,thezerosofthepolynomials 2Aφ(α,β)(µ)+(cid:161)n+α+β(cid:162)µφ(α+1,β+1)(µ) and A(cid:161)n+α+β+1(cid:162)φ(α+1,β+1)(µ)+2φ(α,β)(µ) (30) n n−2 n−1 n−1 areinterlacingprovidedthat−1<α<1andβ>0.Ifinaddition−1<α<0,thezerosofthepolynomials (cid:161)n+α+β+1(cid:162)µφ(α+1,β+1)(µ)+2Aφ(α,β)(µ) and 2φ(α,β)(µ)+A(cid:161)n+α+β(cid:162)φ(α+1,β+1)(µ) (31) n−1 n−1 n n−2 arealsointerlacing. 10 OnconjecturesbyCsordas,CharalambidesandWaleffe Proof. To get the assertion we apply the relation (18) to the even and odd parts of the polynomials f(x;µ) andg(x;µ).Theevenpartof f(x;µ)is f(x;µ)+f(x;−µ) =A[n(cid:88)/2]µ2k d2k P(α,β)(x)+µ2[n(cid:88)/2]µ2k d2k+1 P(α,β)(x) 2 k=0 dx2k n k=0 dx2k+1 n−1 =A[n(cid:88)/2]µ2k d2k P(α,β)(x)+n+α+βµ2[n/(cid:88)2]−1µ2k d2k P(α+1,β+1)(x). k=0 dx2k n 2 k=0 dx2k n−2 Analogously,fortheoddpartwehave f(x;µ)−f(x;−µ) =A[(n−(cid:88)1)/2]µ2k+1 d2k+1 P(α,β)(x)+[(n−(cid:88)1)/2]µ2k+1 d2k P(α,β)(x) 2 k=0 dx2k+1 n k=0 dx2k n−1 =An+α+β+1[(n−(cid:88)1)/2]µ2k+1 d2k P(α+1,β+1)(x)+[(n−(cid:88)1)/2]µ2k+1 d2k P(α,β)(x). 2 k=0 dx2k n−1 k=0 dx2k n−1 Thesamemanipulationswithg(x;µ)give g(x;µ)+g(x;−µ) =[(n−(cid:88)1)/2]µ2k+2 d2k+1 P(α,β)(x)+A[(n−(cid:88)1)/2]µ2k d2k P(α,β)(x) 2 k=0 dx2k+1 n k=0 dx2k n−1 = n+α+β+1µ2[(n−(cid:88)1)/2]µ2k d2k P(α+1,β+1)(x)+A[(n−(cid:88)1)/2]µ2k d2k P(α,β)(x) and 2 k=0 dx2k n−1 k=0 dx2k n−1 g(x;µ)−g(x;−µ) =[n(cid:88)/2]µ2k+1 d2k P(α,β)(x)+A[(n−(cid:88)1)/2]µ2k+1 d2k+1 P(α,β)(x) 2 k=0 dx2k n k=0 dx2k+1 n−1 =µ(cid:195)[n(cid:88)/2]µ2k d2k P(α,β)(x)+An+α+β[n/(cid:88)2]−1µ2k d2k P(α+1,β+1)(x)(cid:33). k=0 dx2k n 2 k=0 dx2k n−2 The polynomials f(1;µ) and g(1;µ) are stable by Lemma 5 and Lemma 6, respectively. Thus, the pairs of polynomialsmentionedin(30)and(31)haveinterlacingzerosbyTheoremHB. Lemma8(seee.g.[8,Lemma3.4]or[3,Lemma3.5]). Letrealpolynomialsp(x)andq(x)suchthatp(0),q(0)>0 (cid:161) (cid:162) haveonlynegativezeros.Then p(x),xq(x) isarealpairifandonlyifthecombinations r (x):=r (x;A,B):=Ap(x)+Bxq(x) and r (x):=r (x;C,D):=Cp(x)+Dq(x) (32) 1 1 2 2 arenonzerooutsidethereallineforall A,B,C,D>0. Recall that p(x) and xq(x) is a real pair whenever they interlace (non-strictly if p(x) and xq(x) have a commonzero).Forp andq asinthislemmawethushavedegq(cid:201)degp(cid:201)1+degq automatically. Proof. Withoutlossofgenerality,assumeintheproofthatthepolynomialsp andq havenocommonzeros:if not,thezerosarerealandwecanfactorthemoutofr andr .Thepresenceofcommonzerospreventsp andq 1 2 frombeingstrictlyinterlacing. (cid:161) (cid:162) Let p(x),xq(x) be a real pair. The polynomials p and xq interlace exactly when between each pair of consecutivezeroszri(p),zri−1(p)thepolynomialq hasexactlyonechangeofsign,i =2,...,degp.Thatis,the interlacingpropertyofthesepolynomialsisequivalentto R(z):= q(z) =γ+(cid:88)n Ai implying zR(z)= zq(z) =γz+(cid:88)n A +(cid:88)n zri(p)Ai , p(z) z−zr (p) p(z) i z−zr (p) i=1 i i=1 i=1 i whereγ(cid:202)0andtheresidues A = q(zri(p)) arepositiveforalli.Thestraightforwardcheckshowsthat i p(cid:48)(zri(p)) signImR(z)=−signIm(zR(z))=−signImz foranyz∈(cid:67)suchthatp(z)(cid:54)=0.Consequently,thecombinations(32)havezerosonlyontherealline.

On conjectures by Csordas, Charalambides and Waleffe

Similar On conjectures by Csordas, Charalambides and Waleffe