Zeros of the deformed exponential function

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Advances in Mathematics

www.elsevier.com/locate/aim

Zeros of the deformed exponential function

Liuquan Wang^a, Cheng Zhang^b,∗

aSchoolofMathematicsandStatistics,WuhanUniversity,Wuhan430072,Hubei, People’sRepublicofChina

bDepartmentofMathematics,JohnsHopkinsUniversity,Baltimore,MD21218, UnitedStates

a r t i c l e i n f o a b s t r a c t

Articlehistory:

Received7January2018 Receivedinrevisedform4April 2018

Accepted20April2018 Availableonline26May2018 CommunicatedbyGeorgeAndrews

MSC:

primary30C15,11B83,41A60 secondary11M36,34K06

Keywords:

Deformedexponentialfunction Asymptoticexpansion Eisensteinseries Bernoullinumbers

Letf(x)=_∞

n=0 1

n!q^n(n−1)/2xⁿ(0< q <1)bethedeformed exponentialfunction.Itisknownthat thezerosof f(x) are realandforma negativedecreasing sequence(xk) (k≥ 1).

Weinvestigatethecompleteasymptoticexpansionforxkand provethatforanyn≥1,ask→ ∞,

xk=−kq^1−k 1 +

n i=1

Ci(q)k⁻¹⁻ⁱ+o(k⁻¹⁻ⁿ) ,

where Ci(q) are some q series which can be determined recursively.WeshowthateachCi(q)∈Q[A0,A1,A2],where Ai=∞

m=1mⁱσ(m)q^mandσ(m) denotesthesumofpositive divisors of m. When writing Ci as a polynomial inA0,A1

andA2,wefind explicitformulasfor thecoefficients of the linearterms byusing Bernoullinumbers.Moreover,wealso provethat Ci(q)∈ Q[E2,E4,E6],whereE2,E4 andE6 are theclassicalEisensteinseriesofweight2,4and6,respectively.

©2018ElsevierInc.Allrightsreserved.

* Correspondingauthor.

E-mailaddresses:wanglq@whu.edu.cn,mathlqwang@163.com(L. Wang),czhang67@jhu.edu (C. Zhang).

https://doi.org/10.1016/j.aim.2018.05.006 0001-8708/©2018ElsevierInc.Allrightsreserved.

1. Introduction

Considerthefunction

f(x) :=

∞ n=0

xⁿ

n!q^n(n−1)/2 (1.1)

where x,q ∈ C, |q| ≤ 1. The function f(x) is entire and is called the “deformed ex- ponential function” since it reduces to exp(x) when q = 1. It appears naturally and frequentlyinpureandappliedmathematics.Incombinatorics,thefunctionf(x) relates closelyto thegeneratingfunctionforTuttepolynomialsof thecomplete graphKn [24], theenumerationofacyclicdigraphs[18] andinversionsoftrees[15].Italsorelatestothe Whittaker andGoncharovconstants [2] incomplexanalysis,andthepartitionfunction ofone-sitelatticegaswithfugacityxandtwo-particleBoltzmann weightqinstatistical mechanics [19].Moreover,onecanverifythatthisfunctionistheuniquesolutiontothe functional differentialequation

y(x) =y(qx), y(0) = 1, (1.2)

which is aspecialcase ofthe “pantographequation” [6]. Formoredetaileddiscussions onthis function,wemayrefertothenotesfrom AlanSokal’stalks [20].

Surprisingly, many important propertiesof this functionremain open, e.g.,the dis- tribution of its zeros. In 1952, Nassif [17] studied (on Littlewood’s suggestion) the asymptotic behavioursandthezerosoftheentirefunction

∞ n=0

eⁿ²

√2πiz²ⁿ/n!,

which equalsto f(q¹²z²) withq=e^2√2πi. Heused thefactthat√

2 hasaperiodiccon- tinuedfractionexpansion.Later,Littlewood[11,12] consideredgeneralizationstoTaylor serieswhosecoefficientshavesmoothlyvaryingmoduliandargumentsoftheforme^n2απi, whereαisaquadraticirrationality.Seealso[4,10,13,23] forthestudiesonthebehaviours of these functions. To ourknowledge,for generalcomplex numberq satisfying|q| ≤1, the distribution of the zeros of f(x) has not been completely understood up to now.

A theorem of Eremenko cited in[21] considered the case where q lies in any compact set of theopen unit disk D. There are relatively moreworks on themodel casewhere 0< q <1.In1972,Morris et al.[5] usedatheoremof Laguerreto showthatf(x) has infinitelymanyreal zerosandthese zerosareallnegative andsimple.Theyalsoproved thatthere isnootherzerofortheanalyticextension (tothecomplexplane)off(x) by using theso-calledmultipliersequence(amodestgapintheirproofwasfilledbyIserles [8]).Therefore, when0< q <1,thezerosoff(x) formonestrictly decreasingsequence ofnegativenumbers(xk) (k≥1).Weremarkthatinsomepreviousworks(e.g.,[9,22]),

thesubscriptsofthesequencestartwith0ratherthan1.Inthispaper,aswellasin[25], thesubscriptsstartwith1fortheelegance ofnotation.

When0 < q < 1,some conjectures on thezeros xk (k ≥1) have been proposed in [5,8,18]. Forexample,Morris etal.[5] conjecturedthat

k→∞lim xk+1

x_k = 1

q. (1.3)

In1973,Robinson[18] alsoderived(1.2) whencountingthelabeledacyclicdigraphs.He speculatedthat

x_k=−kq^1−k+o(q^1−k). (1.4) These conjectures have been investigated by several authors (see e.g., [7,9,14,22]). In particular,Langley [9] showedthatask→ ∞

xk+1

x_k = 1 q

1 + 1 k

+o(k⁻²). (1.5)

Healso provedthatthere exists apositiveconstantγ, which isindependent ofk, such that

xk=−kq^1−k(γ+o(1)). (1.6) Asaconsequence,(1.3) istrue.Recently,oneoftheauthors[25] refinedLangley’swork andconfirmedtheobservation(1.4).Indeed,heshowedthatask→ ∞,

x_k =−kq^1−k 1 +

∞ m=1

σ(m)q^mk⁻²+o(k⁻²)

. (1.7)

Hereforany positiveintegern,

σ(n) :=

d|n, d>0

d.

LaterDerfeletal.[3] studiedtheasymptoticbehavioursofthezerosofsolutionsof(1.2) withdifferentinitial conditionsinsteadoftherestrictiony(0)= 1.

Ourresearch onthe zerosof thedeformed exponential functionis motivated bythe conjectures introduced by Sokal [21] in his talk at Institut Henri Poincaréin 2009. In this paper, we obtaina complete asymptotic expansionformula for the zerosxk when 0< q <1.To bemorespecific, wewill approximatex_k withremainderterm o(k⁻ⁿ⁻¹) for any n≥1.We alsoestablish the connectionbetween theclassical Eisenstein series and thezerosof the deformedexponential function.To stateour results,we define for i≥0,

A_i=A_i(q) :=

∞ m=1

mⁱσ(m)q^m. (1.8)

Theorem 1.Let0< q <1andn≥1.Thenask→ ∞, x_k=−kq^1−k

1 + n i=1

C_i(q)k⁻¹⁻ⁱ+o(k⁻¹⁻ⁿ)

, (1.9)

where each C_i(q) is a multivariate polynomial in A₀,A₁,...,A_i−1 with rational coeffi- cients.These polynomials canbe determinedrecursively.

Remark 1.For example, C₁ = A₀, C₂ = −A₁, C₃ = −₁₀¹A₀+ ³₅A₁ + ¹₂A₂− ¹³₁₀A²₀. Therecurrencerelationandthebasicstructureofthesepolynomialswillbepresentedin Sections3and4.Moreover,weconjecturethat(1.9) shouldholdforanycomplexnumber qsatisfying0<|q|<1 (andevenforsomeqsatisfying|q|= 1),ifthecomplexzerosare listedaccordingtotheirmultiplicitiesandorderedbyincreasingmodulus.Furthermore, itisinterestingtoconsiderifonecouldobtainsimilarresultsforthezerosoftherescaled Rogers–Ramanujan function[22]

R(x;˜ y, q) = ∞ n=0

xⁿq^n(n−1)/2

(1 +y)(1 +y+y²)· · ·(1 +y+· · ·+yⁿ⁻¹),

whichreduces toa“partialthetafunction”wheny= 0, andthe“deformedexponential function”when y= 1.

Whenn≥4,weobservethattheexpressionofCn(q) intermsofA0,A1,. . . ,A_n−1is not unique. The polynomialgiven by therecurrence relation inTheorem 1is just one candidate.Forexample,wehave

C₄= 1

10A₁−14 15A₂−1

6A₃+23 5A₀A₁

= 1

10A₁−11

10A₂+23

5A₀A₁−6A²₁+ 4A₀A₂. (1.10) Thus we continueto study therelations betweenAi’s. Indeed,the followingidentity is established

A3=A2+ 36A²₁−24A0A2.

DifferentiatingitwithrespecttoqgivesmoresimilaridentitiesonAi’s.Therefore,wefind thatitispossibletoexpressC_nasapolynomialinjustA₀,A₁andA₂.Furthermore,the coefficientsofthelineartermsinthatpolynomialcanbegivenexplicitlyusingBernoulli numbers.LetB_n bethen-thBernoullinumber.ItiswellknownthatB_2m+1= 0 forall m≥1.ThefirstfewvaluesofBi areB0= 1,B1=−¹₂, B2= ¹₆ andB4=−₃₀¹.

Theorem2.Foranyn≥1,Cn canbeexpressedasatrivariatepolynomialinA0,A1and A2 with rational coefficients.Thispolynomial isuniqueandforn≥2,wehave

C2n−1=6B2n

n A0−36B2n

n A1+

1 + 30B2n

n

A2+ higher degree terms, C_2n =−6B_2n

n A₁+ 6B_2n

n −1

A₂+ higher degree terms.

Remark2.Forexample,wefindthat C₅=1

21A₀−2

7A₁+26

21A₂+53

70A²₀+ 22A²₁−36A₀A²₁

−159

35 A0A1−43

2 A0A2+ 2A1A2+737

210A³₀+ 24A²₀A2, C₆=− 1

21A₁−20

21A₂−74

35A₀A₁−1401 35 A²₁−2

5A²₂+705 14 A₀A₂

−101

10 A1A2+1662

5 A0A²₁−321

14 A²₀A1−36 5A³₁

−1132

5 A²₀A₂−864

5 A²₀A²₁+72

5A₀A₁A₂+576 5 A³₀A₂.

TheBernoulli numbersinthelinearterms arefrom Faulhaber’s formulafor thepower sumof the first m positiveintegers(see (2.49)). As showninthe examplesabove, the higherdegreetermslookmuchmorecomplicatedthanthelinearterms,thoughtheycan be explicitlyderivedfrom therecurrencerelation ofCn (see (4.1)),whichis essentially determined by the expressions of the unsigned Stirling numbers of the first kind (see (2.24),(2.46)).

Let

E2=E2(q) := 1−24 ∞ n=1

nqⁿ

1−qⁿ, (1.11)

E4=E4(q) := 1 + 240 ∞ n=1

n³qⁿ

1−qⁿ, (1.12)

E₆=E₆(q) := 1−504 ∞ n=1

n⁵qⁿ

1−qⁿ. (1.13)

It is well knownthatE₂,E₄ and E₆ are classicalEisenstein series on thefull modular group

SL(2,Z) =

a b c d

|ad−bc= 1, a, b, c, d∈Z .

We willshow thatA0,A1 andA2 canbe representedas polynomialsinE2,E4 and E6

with rationalcoefficientsand viceversa.Namely,

Q[A0, A1, A2] =Q[E2, E4, E6].

Since it is well known that E2, E4 and E6 are algebraically independent over C (see e.g.,[16, Lemma117]),itfollowsthatA0,A1andA2 arealsoalgebraicallyindependent over C. So Theorem 2 implies that C_n(q) are in the ring of quasimodular forms on SL(2,Z).

Corollary 1.Letn≥1.ThenCn(q)∈Q[E2,E4,E6].

The paper is organized as follows. In Section 2, we study the series expansion of f(−(k+ak⁻¹)q^1−k) forlarge k,where a=a(k⁻¹) is positiveandbounded forlargek.

Hereonecanthinkof−(k+ak⁻¹)q^1−kasa“prospectiveroot”off.Weobservethatthe first2ktermsofthisseriesdominatetheothers,sothese2ktermsarecarefullyanalyzed in Lemmas 1 and 2. Most of these terms enjoy niceproperties described inLemma 1, while others are more subtle and related to the series expansion of (2.17). Lemma 2 givestheformulaofthecoefficientofeachtermintheseriesexpansionof(2.17).Inthe proof of Lemma 2, the properties of elementary symmetric polynomials and complete homogeneoussymmetricpolynomials playanimportantrole.

InSection3,weproveTheorem1.WefirstdefineCn recursively(see(3.3))byexploit- ingthecoefficientsoftheseriesexpansionof(2.17).Leta=_n−1

i=1 Cik⁻⁽ⁱ⁻¹⁾+λk⁻⁽ⁿ⁻¹⁾. Then weuseLemmas1and2toshowthatifλ=C_n,and inadditionthatλ>0 when n= 1,then

(−1)^k(λ−Cn)f(−(k+ak⁻¹)q^1−k)>0.

This allows usto determine thesigns off(x) attheendpoints of certain intervals. We finish theproof ofTheorem1bytheintermediatevaluetheorem and(1.5).

InSection4,westudyvariousrepresentationsofC_n.InSection4.1,wegivemorede- tailsontherecursiveformulaofCn,andprovethatCn canbe writtenasamultivariate polynomial of A0,A1,. . . ,An−1, in which the coefficient of An−1 is ⁽⁻¹⁾_(n−ⁿ_1)!⁻¹ (Proposi- tion5).InSection4.2,wecontinuetodiscussthestructureofthe multivariatepolynomial representation of Cn by those Ai’s, especiallythe linearterms. In particular, we show inProposition6thatthesumofthecoefficientsofthelineartermsA0,A1,. . . ,An−1 in C_n equalsto (−1)ⁿ⁻¹.Furthermore, wegiveexplicitformulasforthecoefficientsof the linear terms A0 and A1 inPropositions 7and 8. InSection 4.3, we first establish the relations between theclassicalEisenstein series and ourAi’s (Proposition9).Then we show inLemma10thateachA_n canbewrittenas amultivariate polynomialinA₀,A₁ and A2.Combiningthesefacts,wecomplete theproof ofTheorem2.

Remark3.It may be illuminating to consider (1.9) as aformal powerseries. Suppose thatCi(q)=_∞

j=1Cijq^j. Using formalpower series, wedenote Fj(t):=_∞

i=1Cijtⁱ⁺¹. Thenonemayrewrite(1.9) asaformalpowerseries

x_k(q)∼ −kq^1−k 1 +

∞ j=1

F_j(k⁻¹)q^j

. (1.14)

Weobservethattheformalpowerseries(1.14) numericallyagreeswiththeexpansionin q ofthe k-th zerogiven in[21,p. 14]. (Note thatin[21] thesequence (xk) starts with subscript 0and foreach j, Fj(k⁻¹) is arational functionin k.) It was conjecturedby Sokal[21, p. 11] that

Fj(k⁻¹)≥0

forallintegersj,k≥1.Thisconjectureisstill open,evenforfixed k= 1.Inparticular, wewillseethatC_i1= (−1)ⁱ⁺¹ byProposition6,whichimpliesthat

F₁(k⁻¹) = 1 k(k+ 1).

This verifies thecasej = 1.For j ≥2,we find thatthedifficulty to obtainthe closed formofFj(k⁻¹) liesinthecomplexityofthehigherdegreetermsinTheorem2.

2. Preliminaryresults

Throughout this paper, we fixq with 0< q <1. Weuse thenotation “O(k^−m)”to denote the class of functions of k which is bounded by Ck^−m, where C is a constant thatis dependenton the fixed parametersbut independent of k. Let a=a(k⁻¹) be a functioninksatisfyingthata=a₀+O(k⁻¹) wherea₀>0 isaconstant.Themaingoal ofthissectionisto studythevaluesof f(−(k+ak⁻¹)q^1−k) forlargek.

Wefirstobservethat

f(−(k+ak⁻¹)q^1−k) = ∞ n=0

(−1)ⁿu_n, (2.1)

where

u_n =(k+ak⁻¹)ⁿ

n! q^{−n(2k−n−1)/2}.

Inorderto proveTheorem1, wewill selectsomespecialfunctionsasa(seeSection3).

For these a and sufficiently large k, we will see that for thesum in (2.1), the first 2k termsdominatetheothers. Therefore,werewrite(2.1) as

f(−(k+ak⁻¹)q^1−k) =

2k−1 n=0

(−1)ⁿun+ ∞ n=2k

(−1)ⁿun =

k−1

j=0

(−1)^j−1vj+ ∞ n=2k

(−1)ⁿun, (2.2) where wedenote

v_j =u_2k−j−1−u_j (0≤j≤k−1).

Thefollowinglemmashowsthepositivityand“almostmonotonicity”ofthesequencevj. Lemma 1.Thereexistsapositiveinteger K(q)suchthat foranyk≥K(q),

vj >0, 0≤j≤k−1.

Furthermore, there existsa positiveinteger N(q) suchthat forany N ≥N(q) and k≥ q^−3N,

vj< vj+1, 0≤j≤k−N.

Proof. Sincea=a0+O(k⁻¹) anda0>0,wecanfindapositive integerK(q) suchthat a>0 forany k≥K(q).Nowwe assumethatk≥K(q).BytheAM-GMinequality,we have

2k−1−2j

i=1

(j+i)<

⎛

⎜⎜

⎜⎝

2k−1−2j

i=1

(j+i) 2k−1−2j

⎞

⎟⎟

⎟⎠

2k−1−2j

=k^2k−1−2j

<(k+ak⁻¹)^2k−1−2j (0≤j≤k−1). (2.3) This implies

(k+ak⁻¹)^j

j! < (k+ak⁻¹)^2k−1−j

(2k−1−j)! (0≤j≤k−1).

So

uj< u2k−1−j (0≤j≤k−1), whichgivesthefirstinequality.

Note that

vj+1 =q(j+1)(j+2−2k)/2(k+ak⁻¹)^2k−j−2

(2k−2−j)! (1−wj) (2.4)

where

wj= (2k−2−j)!

(j+ 1)!(k+ak⁻¹)^2k−3−2j. Sincewehaveprovedthatvj+1>0,itfollows that

0< w_j<1 BytheAM-GMinequality,

w_j+1 wj

= (k+ak⁻¹)²

(j+ 2)(2k−2−j)>1.

By(2.4),wesee thatvj < vj+1is equivalentto q^j+1−k(1−w_j)> k+ak⁻¹

2k−j−1(1−w_j−1). (2.5) Usingtherelation

w_j−1= (j+ 1)(2k−j−1) (k+ak⁻¹)² wj, weseethat(2.5) isequivalentto

q^−k+j+1− j+ 1 k+ak⁻¹

w_j < q^−k+j+1− k+ak⁻¹

2k−1−j. (2.6) ForsomepositiveintegerN,wedenote

t= 2k−1−j (k+N −1≤t≤2k−1) and

g(t) = 1

t −(2k−t)wk−N

(k+ak⁻¹)² −q^k−t

1−wk−N

k+ak⁻¹

. Directcalculationyields

g(t) =−1

t² + w_k−N (k+ak⁻¹)² +

1−w_k−N k+ak⁻¹

q^k−tlnq (2.7) and

g(t) = 2 t³ −

1−w_k−N k+ak⁻¹

q^k−t(lnq)². (2.8)

Since 0< wj<1,g(t) isdecreasingfort>0.

NotethatwhenNislargeenough(N ≥N1(q)),wehavek≥q^−3N≥max{K(q),N⁶}. Hence

1

k+N−1 =k⁻¹

1−(N−1)k⁻¹+ (N−1)²k⁻²+O(N³k⁻³)

=k⁻¹−(N−1)k⁻²+ (N−1)²k⁻³+O(k^−7/2). (2.9) Similarlywehave

1

(k+N−1)² =k⁻²−2(N−1)k⁻³+O(k^−11/3) (2.10) and

1

(k+N−1)³ =k⁻³+O(k^−23/6). (2.11) Next,

w_k−N =

N−2

t=−N+2

k+t k+ak⁻¹

=

N−2

t=−N+2

1 + t

k 1−a₀

k² +O(k⁻³)

=

N−2

t=−N+2

1 + t

k−a₀ k² −a₀t

k³ +O(k⁻³)

= 1−a0(2N−3)k⁻²+

−N+2≤t1<t2≤N−2

t₁t₂

k² +O(k^−17/6)

= 1−a₀(2N−3)k⁻²−(N−2)(N−1)(2N−3)

6 k⁻²+O(k^−17/6).

Hence

1−wk−N

k+ak⁻¹ =1 k

1−a0k⁻²+O(k⁻³)

·

a₀(2N−3)k⁻²+(N−2)(N−1)(2N−3)

6 k⁻²+O(k^−17/6)

=

a₀(2N−3) + (N−2)(N−1)(2N−3) 6

k⁻³+O(k^−23/6). (2.12) Note thatk≥q^−3N impliesq^−N =O(k^1/3).Nowby(2.11) and(2.12),wededucethat

g(k+N−1) =

2−cNq^1−N(lnq)²

k⁻³+O(k^−7/2) where

cN :=a0(2N−3) +1

6(N−2)(N−1)(2N−3).

Similarly,wehave

g(k+N−1) =

2N−2 +cNq^1−Nlnq

k⁻³+O(k^−7/2).

WhenN issufficientlylarge(N ≥N2(q)≥N1(q)),wewillhave g(k+N−1)<0,

and

g(k+N−1)<0.

Sog(t) andg(t) arealsodecreasingfort≥k+N−1.

Inthesameway,wefindthat g(k+N−1) =

a0(2N−1) + N

6(N−1)(2N−1)−cNq^1−N

k⁻³+O(k^−10/3).

WhenN islargeenough(N≥N(q)≥N2(q)),we have g(k+N−1)<0.

SoifN ≥N(q) andk≥q^−3N,wehave

g(t)≤g(k+N−1)<0, t≥k+N−1.

Therefore, 1

2k−1−j −(j+ 1)w_k−N

(k+ak⁻¹)² −q^−k+j+1

1−w_k−N k+ak⁻¹

<0 (0≤j≤k−N).

Thisimplies

q^−k+j+1− j+ 1 k+ak⁻¹

wk−N< q^−k+j+1− k+ak⁻¹

2k−1−j (0≤j ≤k−N).

Sincewj< wj+1,we have

q^−k+j+1− j+ 1 k+ak⁻¹

wj < q^−k+j+1− k+ak⁻¹

2k−1−j (0≤j≤k−N).

This proves(2.6) andhencethefactthat

vj< vj+1 (0≤j≤k−N). 2

However,thesequencevj maynotbemonotonewhenk−N < j≤k−1.Soweneed moredelicateanalysisonthese vj’s, whichis thecruxoftheproblem.For1≤j≤N,

vk−j=uk+j−1−uk−j =

(k+ak⁻¹)^k+j−1

(k+j−1)! −(k+ak⁻¹)^k−j (k−j)!

q−(k+j−1)(k−j)/2

=

⎛

⎝^j−1

i=1

k+ak⁻¹ k+i −

0 i=1−j

k+i k+ak⁻¹

⎞

⎠(k+ak⁻¹)^k 1

k!q−(k+j−1)(k−j)/2

=

⎛

⎝^j−1

i=0

1 +ak⁻² 1 +ik⁻¹ −

−1 i=1−j

1 +ik⁻¹ 1 +ak⁻²

⎞

⎠1 +ak⁻² k⁻²

·q^j(j−1)/2·(k+ak⁻¹)^k−21

k!q^{−k(k−1)/2}. (2.13)

Wedefine forj≥1,

G(α;x) :=

j−1 i=0

1 +αx²

1 +ix , (2.14)

H(α;x) :=

−1

i=1−j

1 +ix

1 +αx². (2.15)

In particular, when j = 1 we haveG(α;x)= 1+αx² and H(α;x)= 1 since we agree thattheemptyproductequals 1.Itisthen clearfrom(2.13) that

v_k−j =

G(a;k⁻¹)−H(a;k⁻¹)1 +ak⁻²

k⁻² ·q^j(j−1)/2·(k+ak⁻¹)^k−21

k!q^{−k(k−1)/2}. (2.16) Now weanalyzetheseries expansionoftheproduct

G(α;x)−H(α;x) x²

1 +αx²

(2.17) forαbeing apowerseriesofx.

Lemma 2.Let

α(x) :=

∞ i=0

a_ixⁱ. (2.18)

Forn≥1,thecoefficient ofxⁿ⁻¹ intheexpansionof G(α(x);x)−H(α(x);x)

x² (1 +α(x)x²) has theform

μ(an−1+S0(n) +S1(n)ν+S2(n)ν²+...+Sn(n)νⁿ), (2.19) whereμ= 2j−1,ν=j(j−1),andeachSi(n)isapolynomialof a0,a1,...,an−2,which is independentof j and has rational coefficients.In particular,S0(1) =S0(2)= 0 and S1(1)= ¹₆.

TherestofthissectionwillbedevotedtogivingaproofofLemma2.First,wecompute thecoefficientsinthe expansionsG(α;x) and H(α;x) aspowerseries inxregardingα asaparameter.Thatis,

G(α;x) = ∞ N=0

GNx^N, H(α;x) = ∞ N=0

HNx^N,

whereGN and HN arepolynomials ofαwithcoefficientsdependingonj.Forexample, G0=H0= 1,

G₁=H₁=−¹₂j(j−1),

G₂=₂₄¹(j−1)j(j+ 1)(3j−2) +jα, H₂=₂₄¹(j−2)(j−1)j(3j−1) + (1−j)α, G₃=−₄₈¹(j−1)²j²(j+ 1)(j+ 2)−¹₂(j−1)j²α, H3=−₄₈¹(j−3)(j−2)(j−1)²j²+¹₂(j−1)²jα.

TorepresentGN andHN,wedefineforj ≥1 andi≥0 that

q_i(j) :=e_i(1,2,· · ·, j−1), (2.20) Q_i(j) := (−1)ⁱh_i(1,2,· · ·, j−1) (2.21) where

ei(X1, X2,· · ·, Xn) =

1≤n1<n2<···<ni≤n

Xn1Xn2· · ·Xni (2.22) and

h_i(X₁, X₂,· · ·, X_n) =

1≤n1≤n2≤···≤ni≤n

X_n1X_n2· · ·X_ni (2.23)

are elementary symmetric polynomials and complete homogeneous symmetric polyno- mials respectively.Note thathereweagreethat

e₀(X₁, X₂,· · ·, X_n) =h₀(X₁, X₂,· · ·, X_n) = 1

so that q0(j) = Q0(j) = 1. By definition it is clear that qi(j) = Qi(j) = 0 if i ≥ j.

Moreover, weremarkthatqi(j) isjusttheunsignedStirlingnumbersofthefirstkind

qi(j) =c(j, j−i). (2.24)

It isnotdifficultto seethat ∞ k=0

ek(X1, X2,· · ·, Xn)t^k = n i=1

(1 +Xit), (2.25)

∞ k=0

h_k(X₁, X₂,· · ·, X_n)t^k= n i=1

1

1−X_it. (2.26)

These identitiesimplythefollowing well-knownfundamentalrelation: form≥1, m

i=0

(−1)ⁱe_i(X₁,· · ·, X_n)h_m−i(X₁, X₂,· · ·, X_n) = 0. (2.27) Therefore, wehavethefollowing recurrencerelationforQ_k(j):

Q_k(j) =− k i=1

q_i(j)Q_k−i(j), k≥1. (2.28) Recall thegeneralizedbinomialcoefficientfork∈N,z∈C

z k

= z(z−1)· · ·(z−k+ 1)

k! .

Lemma 3.Wehave

GN =

[N/2]

m=0

G(N, m)α^m, N = 0,1,2, ...

where

G(N, m) = j

m

Q_N−2m(j). (2.29)

Moreover,

HN =

[N/2]

m=0

H(N, m)α^m, N = 0,1,2, ...

where

H(N, m) = (−1)^N 1−j

m

qN−2m(j). (2.30)

Proof. Since

j−1

i=0

1 1 +ix =

∞ i=0

Qk(j)x^k, (2.31)

wehave

G(α;x) = j m=0

j m

α^mx^2m

∞ k=0

Qk(j)x^k = ∞ N=0

x^N

[N/2]

m=0

j m

QN−2m(j)α^m. Similarly,

H(α;x) = ∞ m=0

1−j m

α^mx^2m

∞ k=0

qk(j)(−x)^k

= ∞ N=0

x^N

[N/2]

m=0

(−1)^Nq_N−2m(j) 1−j

m

α^m. 2

Forfixedn,both qn(j) andQn(j) arepolynomialsinj.Hencetheycanbenaturally extendedtobe twofunctionsdefinedonthewholereal line.Thefollowing lemmagives arelationbetweenthese twofunctions.

Lemma4.Forn≥0,wehaveQ_n(1−t)= (−1)ⁿq_n(t),t∈R. Proof. Wedenoteforj≥1,

φ_j(x) :=

j−1 i=0

1 1 +ix =

∞ n=0

Q_n(j)xⁿ, (2.32)

ψj(x) :=

j−1 i=1

(1−ix) = ∞ n=0

(−1)ⁿqn(j)xⁿ. (2.33) Inparticular,φ₁(x)=ψ₁(x)= 1.Note that

φj(x) = (1 +jx)φj+1(x) = ∞ n=0

Qn(j+ 1) +jQ_n−1(j+ 1)

xⁿ (2.34)

where wesetQ₋₁(t)= 0.Comparingthecoefficientofxⁿ onbothsides,wededucethat forany n≥0,

Qn(j) =Qn(j+ 1) +jQn−1(j+ 1). (2.35) Similarly,observingthatψ_j+1(x)=ψ_j(x)(1−jx),wededucethatforanyn≥0,

q_n(j+ 1) =q_n(j) +jq_n−1(j) (2.36) where we setq₋₁(t)= 0.Since(2.35) and(2.36) hold forallj ≥1 andboth Q_n(t) and qn(t) arepolynomialsint,weconcludethatforany t∈R

Qn(t) =Qn(t+ 1) +tQn−1(t+ 1), (2.37) q_n(t+ 1) =q_n(t) +tq_n−1(t). (2.38) Now weletQ_n(t)=Q_n(1−t).Then(2.37) implies

Q_n(t+ 1) =Q_n(t)−tQ_n−1(t). (2.39) Comparing(2.39) with(2.38),weseethatthepolynomialsQ_n(t) and(−1)ⁿqn(t) satisfy thesamerecurrencerelation.Next,bydirectcomputation,wefindthat

Q0(j) =q0(j) = 1, Q1(j) =−j(j−1)

2 , q1(j) = j(j−1)

2 . (2.40)

Thus

Q₁(t) =−q₁(t) =−t(t−1) 2 .

Now suppose thatQ_n−1(t)= (−1)ⁿ⁻¹q_n−1(t) for somen≥2.By(2.38) and(2.39) we deduce that

Q_n(t+ 1)−Q_n(t) = (−1)ⁿqn(t+ 1)−(−1)ⁿqn(t).

Summingovert from1toj−1,weobtain

Q_n(j)−Q_n(1) = (−1)ⁿqn(j)−(−1)ⁿqn(1). (2.41) By definition, we have Q_n(1) = (−1)ⁿq_n(1) = 0 for n ≥ 1. Therefore, (2.41) implies that Q_n(j)= (−1)ⁿq_n(j) for any j ≥1.This impliesthat Q_n(t)= (−1)ⁿq_n(t) for any t∈R. 2

SinceG0=H0,G1=H1,weget G(α(x);x)−H(α(x);x)

x² =

∞ N=0

(G_N+2−H_N+2)x^N. Hence

G(α(x);x)−H(α(x);x)

x² (1 +α(x)x²) = ∞ N=0

(G_N+2−H_N+2+ (G_N−H_N)α(x))x^N. (2.42) ToproveLemma2, weneedto computethecoefficientofxⁿ⁻¹ in(2.42).ForN ≥2m, wedefine

Δ(N, m) :=G(N, m)−H(N, m) = j

m

QN−2m(j)−(−1)^N 1−j

m

qN−2m(j).

(2.43) Forexample,

Δ(0,0) = Δ(1,0) = 0,

Δ(2,0) = ¹₆j(j−1)(2j−1), Δ(2,1) = 2j−1,

Δ(3,0) =−₁₂¹(j−1)²j²(2j−1), Δ(3,1) =−¹₂(j−1)j(2j−1), Δ(4,0) = ₂₄₀¹ (−1 +j)j(−1 + 2j)(−4−12j+ 17j²−10j³+ 5j⁴),

Δ(4,1) = ₂₄¹(−1 +j)j(−1 + 2j)(2−3j+ 3j²), Δ(4,2) = 0, Δ(5,0) =−₁₄₄₀¹ (−1 +j)²j²(−1 + 2j)(−12−56j+ 61j²−10j³+ 5j⁴),

Δ(5,1) =−₄₈¹(−1 +j)²j²(−1 + 2j)(6−j+j²), Δ(5,2) = 0.

Proposition 1.Let μ = 2j −1 and ν = j(j −1). Then for N ≥ 2m the polynomial Δ(N,m)can bewritten as

Δ(N, m) =μ(s0+s1ν+· · ·+skν^k), s0, s1, . . . , sk ∈Q, (2.44) wherek≤_2N−3m−1

2

.Moreover, if (N,m)= (2,1),then s₀= 0.

Inordertoprovethisproposition,weneedthefollowinglemmas.

Lemma 5.Any polynomial of j can be written into a polynomial of μ = 2j −1 and ν = j(j−1), in which the degree of μ in each term is at most 1. Moreover, such a representationisunique.Inorderwords,anypolynomialϕ(j)canbeuniquelywrittenas

ϕ(j) =s0,0+s0,1ν+...+s0,n0νⁿ⁰+μ(s1,0+s1,1ν+...+s1,n1νⁿ¹).

Furthermore, ifallthecoefficients ofϕ(j)are rationalnumbers, theneach s_i,l∈Q. Proof. Notethatμ²= 4ν+ 1.Bythebinomialtheorem wehave

jⁿ=

μ+ 1 2

n

= 1 2ⁿ

n k=0

n k

μ^k

= 2⁻ⁿ

⎛

⎝^[n/2]

k=0

n 2k

μ^2k+

[(n−1)/2]

k=0

n 2k+ 1

μ^2k+1

⎞

⎠

= 2⁻ⁿ

[n/2]

k=0

n 2k

(4ν+ 1)^k+ 2⁻ⁿμ

[(n−1)/2]

k=0

n 2k+ 1

(4ν+ 1)^k. (2.45) This provestheassertionsforthepolynomialjⁿ.

Sinceanypolynomialϕ(j) isalinearcombinationsofjⁿ(n= 0,1,· · ·),weknowthat ϕ(j) can be written inthe desired form. If ϕ(j) has rational coefficients, then clearly eachs_i,l∈Q.

Theuniquenessisclear,sinceotherwisewewillhavearelation ϕ1(ν) +μϕ2(ν) = 0

for some nonzero polynomials ϕ1(j) and ϕ2(j), which is impossible because μ² = 4ν+ 1. 2

Remark4.Weseefrom(2.45) thatj²ⁿ=νⁿ+ⁿ₂μνⁿ⁻¹+ lowerdegreetermsandj²ⁿ⁺¹=

1

2μνⁿ+ lowerdegreeterms.

Lemma 6.Fork≥1, q_k(j) = 1

2^kk!j^2k− 2k+ 1

3·2^k(k−1)!j^2k−1+O(j^2k−2), (2.46) Q_k(j) = (−1)^k

2^kk! j^2k+(−1)^k(2k−5)

3·2^k(k−1)!j^2k−1+O(j^2k−2). (2.47) Proof. Wedenote

pm(j) :=

j−1 k=1

k^m. (2.48)

It isknownthat