On conjectures by Csordas, Charalambides and Waleffe

0.41 MB·English

by  Alexander Dyachenko

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<i andtozeroifi =0(itisconvenientsinceallcoefficientsofthepolynomialswedeal
witharenonnegative).
Lemma2. Letα>−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 <n−2(cid:170).
dx n−1 2 n−2
Thedifference(8)−(9)inducestheidentityP(α+1,β)=P(α,β+1)+P(α+1,β+1),sowefinallyhave
n−1 n−1 n−2
(cid:90) 1Pn(α,β−1)·2RedΦn ·wα,β+1dx=η+ζReµ, where η,ζ>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.