in 7-dimensional simplex On the polynomial sharp upper estimate conjecture Journal of Number Theory

(1)

Contents lists available atScienceDirect

Journal of Number Theory

www.elsevier.com/locate/jnt

On the polynomial sharp upper estimate conjecture in 7-dimensional simplex

^✩

Stephen S.-T. Yau^a,∗, Beihui Yuan^b, Huaiqing Zuo^c

a DepartmentofMathematicalSciences,TsinghuaUniversity,Beijing100084, PR China

b DepartmentofMathematics,CornellUniversity,Ithaca,NY,14853,USA

cYauMathematicalSciencesCenter,TsinghuaUniversity,Beijing100084, PR China

a r t i c l e i n f o a bs t r a c t

Article history:

Received4May2015

Receivedinrevisedform14May 2015

Accepted31August2015 Availableonline13October2015 CommunicatedbyDavidGoss DedicatedtoProfessorJohnHenry Coatesontheoccasionofhis70th birthday

MSC:

primary11P21 secondary11Y99 Keywords:

Sharpestimate Integralpoints Simplex

Because ofitsimportanceinnumbertheoryandsingularity theory, the problem of ﬁnding a polynomial sharp upper estimate of the number of positive integral points in an n-dimensional(n≥3)polyhedronhas receivedattentionby a lot of mathematicians. The ﬁrst named author proposed the Number Theoretic Conjecture for the upper estimate.

The previous results on the Number Theoretic Conjecture in low dimension cases (n < 7) are proved by using the sharpGLYconjecturewhichistrueonlyforlowdimensional case.Thustheproofcannotbegeneralizedtohighdimension.

In this paper, we oﬀer a uniform approach to prove the Number Theoretic Conjecture for all dimensions by simply usingtheinductionmethodandtheYau–Zhang[19]estimates (see Lemmas 2.3–2.5). As a result, the Number Theoretic Conjecture isprovenfor n = 7.Animportant estimatefor all dimensions is also obtained (Propositions 3.1 and 3.2) whichwillbeusefultoprovethegeneralcaseoftheNumber Theoretic Conjecture. Asan application, wegive a sharper

✩ This work waspartially supported by NSFC (grant nos. 11401335, 11531007), TsinghuaUniversity InitiativeScientiﬁcResearchProgramandTsinghuaUniversitystart-upfund.

* Correspondingauthor.

E-mailaddresses:[email protected](S.S.-T. Yau),[email protected](B. Yuan), [email protected](H. Zuo).

(2)

estimateoftheDickman–DeBruijnfunctionψ(x,y) for5≤ y <19,comparedwiththeresultobtainedbyEnnola.

1. Introduction

LetT(a₁,a₂,. . . ,a_n) beann-dimensionalsimplexdescribedby x1

a₁ +x2

a₂ +· · ·+xn

a_n ≤1, x1, x2, . . . , xn≥0 (1) where a₁ ≥ a₂ ≥ · · · ≥ a_n ≥ 1 are real numbers. Let P_n = P(a₁,a₂,. . . ,a_n) and Qn = Q(a1,a2,. . . ,an) be the number of positive integer solutions and nonnegative integersolutionsof(1), respectively.Wecanseethereisarelation

Q(a₁, . . . , a_n) =P(a₁(1 +a), . . . , a_n(1 +a)) wherea= _a¹

1 +· · ·+_a¹

n.

The estimate of Pn and Qn can be appliedin number theory. A smooth number is anumberwith only small primefactors. Smooth numbersplayimportant roles infac- toring and primality testing [12]. Given an integery, the number m = p^l₁¹p^l₂². . . p^l_nⁿ is called y-smooth ifall its primefactors p_i ≤y for i= 1,. . . ,n. Number theorists want to know thenumberof y-smooth integerslessthan or equalto x, whichis denotedby ψ(x,y), called theDickman–DeBruijn function.Oneof thecentraltopics incomputa- tionalnumbertheoryistheestimateofψ(x,y) (see[5]).Itturnsoutthatthecomputation of ψ(x,y) is equivalent to compute the number of integral points in an k-dimensional tetrahedron (a₁,a₂,· · ·,a_k) with real vertices (a₁,0,· · ·,0),· · ·,(0,· · ·,0,a_k). Let p₁ < p₂<· · ·< p_k denote theprimes uptoy. Itisclearthatp^l₁¹p^l₂²· · ·p^l_k^k ≤xwhich is alsoequivalenttocounting thenumberof(l₁,l₂,· · ·,l_k)∈Zⁿ_≥0 suchthat

l₁ a1

+ l₂ a2

+· · ·+ l_k

ak ≤1, where a_i= logx logpi

.

Therefore, ψ(x,y) is precisely the number Qk of (integer) lattice points inside the n-dimensionaltetrahedron(1)withai= _log^log_p^x

i,n=k,and1≤i≤k.In[4],Ennolagave bothlowerand upperboundsfortheψ(x,y):

(logx)^k k!

k i=1

logpi

< ψ(x, y)≤

(logx+ k i=1

logpi)^k k!

k i=1

logpi

(2)

whichyieldsthefollowingresult.

(3)

Theorem 1.1.(SeeEnnola [4].) Uniformlyfor2≤y≤

logxlog₂x,we havethat ψ(x, y) = 1

k!

p≤y

(logx

logp)[1 +O( y² logxlogy)].

Numbers P_n andQ_n also haveapplications in geometry and singularity theory. Let f : (Cⁿ,0)→(C,0) be aholomorphic functionwithisolatedcritical pointat theorigin and V = {z ∈ Cⁿ : f(z) = 0}. The geometric genus pg is deﬁned to be dim Γ(V − {0},Ωⁿ⁻¹)/L²(V − {0},Ωⁿ⁻¹), where Ωⁿ⁻¹ is the sheaf of germs of holomorphic (n− 1)-forms on V − {0}. If f(z₁,. . . ,z_n) is weighted homogeneous of type (w₁,. . . ,w_n), wherew₁,. . . ,w_n areﬁxedpositiverationalnumbers,i.e.,f canbeexpressedasalinear combination ofmonomialszⁱ₁¹. . . z_nⁱⁿ forwhichi1/w1+· · ·+in/wn= 1,thenMerleand Teissier[9]showedthatpg isexactlythenumberP(w1,. . . ,wn).

TherearealotofpapersonﬁndingtheexactformulaforP_norQ_n,incasea₁,. . . ,a_n are integers. For example, Mordell [11] gave an exact formula for Q₃ with a₁,a₂ and a3 relatively prime.Pommersheim [13] extendedthis result to arbitraryintegersa1,a2

and a3 using toricvarietytechniques anda resultof Ehrhart[3]. Theexact formulais complicated, it involves generalized Dedekind sum. It is hard to ﬁgure out how large the sumis fromtheexactformula. Therefore,sometimeswewantto getasharp upper estimateofPn intermsofapolynomialina1,. . . ,an.Suchapolynomialupperestimate havemanyimportant applications.Forexample, itcanbeused inthefollowing Durfee Conjecture:

Conjecture. (See Durfee [2].) Let (V,0) be an isolated hypersurface singularity deﬁned by aholomorphicfunctionf : (Cⁿ,0)→(C,0).Let

μ= dimC{z1, . . . , zn}/(fz1, . . . , fzn)

be theMilnor numberof thesingularity. Thenn!pg ≤μ wherepg isthegeometricgenus of (V,0).

Iff isweightedhomogeneousoftype(w1,. . . ,wn),MilnorandOrlik[10]provedthat μ= (w1−1). . .(wn−1).ThereforeDurfeeconjecture isaspecialcaseofthefollowing theorem,whichwasprovedbyYauand Zhang[19]:

Theorem1.2 (GLYroughestimate).Leta1,. . . ,an bepositiverealnumbersgreaterthan or equalto1andn≥3.Then

n!P(a1, . . . , an)<(a1−1)(a2−1). . .(an−1). (3) Theestimateintheabovetheoremisnice.However,itisnotsharpenoughtoprovide asolutionofthefollowingproblem:

(4)

Problem. (See [21,20].) Let f : (Cⁿ,0) → (C,0) be a holomorphic function with an isolated critical point at the origin. Find an intrinsic characterization for f to be a homogeneouspolynomial.

In1971,Saito [14]gaveanintrinsic characterizationforf to beaweightedhomoge- neouspolynomial

Theorem 1.3 (Saito). Let f : (Cⁿ,0) →(C,0) be a holomorphic function with isolated criticalpointattheorigin.Thenf isaweightedhomogeneouspolynomialafterabiholo- morphicchange of coordinatesifand onlyif μ=τ,where

μ= dimC{z1, . . . , zn}/(fz1, . . . , fzn) and

τ= dimC{z1, . . . , zn}/(f, fz1, . . . , fzn).

To ﬁnda necessary and suﬃcient condition forf to be a homogeneouspolynomial, Yaumadethefollowingconjecturein1995:

Conjecture (YauGeometricConjecture).Letf : (Cⁿ,0)→(C,0)beaweightedhomoge- neouspolynomialwithanisolatedsingularityattheorigin.Letμ,pgandν betheMilnor number,geometricgenusandmultiplicityofsingularityV ={z:f(z)= 0},respectively.

Then

μ−h(ν)≥n!pg (4)

whereh(ν)= (ν−1)ⁿ−ν(ν−1). . .(ν−n+ 1).Theequality holdsif andonlyif f isa homogeneouspolynomialafterabiholomorphic change of coordinates.

The Yau Geometric Conjecture together with Theorem 1.3 will give an intrinsic characterization for a holomorphic function f to be a homogeneous polynomial after a biholomorphic change of coordinates. In order to prove Yau Geometric Conjecture, Lin,Yau[6]andGranvillehaveformulatedGLYRoughEstimateandthefollowingGLY SharpConjecture:

Conjecture (GLY SharpEstimate).Letn≥3.If a1≥a2≥ · · · ≥an ≥n−1.Then n!P_n≤f_n :=Aⁿ₀ +s(n, n−1)

n Aⁿ₁+

n−2

l=1

s(n, n−1−l) _n−1

l

Aⁿ_l⁻¹ (5)

wheres(n,k) istheStirling numberof theﬁrstkind deﬁnedby thefollowinggenerating function:

(5)

x(x−1). . .(x−n+ 1) =

n

k=0

s(n, k)x^k

and Aⁿ_k isdeﬁned as

Aⁿ_k = ( n

i=1

ai)(

1≤i1≤i2≤···≤ik≤n

1 ai1ai2. . . aik

)

fork= 1,2,. . . ,n−1.Equalityin (5)holdsif andonly ifa1=· · ·=an areintegers.

The above GLY Sharp Estimate is true for n = 4,5,6 (cf. [22,1]) and there is a counter-examplefor n= 7[16]. In[16], Wangand Yaualsomodify GLY Conjectureas follows:

Conjecture (ModiﬁedGLYConjecture).Thereexistsanintegery(n)whichdependsonly on nsuchthat thesharp estimate (5)holdswhen a₁≥a₂≥ · · · ≥a_n ≥y(n).

In order to overcome the diﬃcultythat GLY Sharp Estimate is only true when an

is largerthan y(n), Yauproposedanewsharpupper estimatewhichismotivedforthe YauGeometricConjecture:

Conjecture (YauNumberTheoretic Conjecture).Let P_n =P_n(a₁, a₂, . . . , a_n) = #{(x₁, . . . , x_n)∈Zⁿ+: x1

a1

+x2

a2

+· · ·+xn

an ≤1}, where n≥3,a₁≥a₂≥ · · · ≥a_n>1are realnumbers.If P_n>0,then

n!Pn≤(a1−1)(a2−1). . .(an−1)−(an−1)ⁿ+an(an−1). . .(an−(n−1)) (6) and equalityholds ifand onlyif a1=a2=· · ·=an areintegers.

There is an intimate relation between the Yau Geometric Conjecture and the Yau Number Theoretic Conjecture. Let f : (Cⁿ,0) → (C,0) be a weighted homogeneous polynomialwithanisolatedsingularityat theorigin,thenthemultiplicity ν off at the origin isgiven by inf{n∈Z+ :n≥inf{w1,. . . ,wn}}, where wi is theweightof xi. In general, the weight w_i is a rational number. Incase the minimal weight is an integer, the Yau Geometric Conjecture and Yau Number Theoretic Conjecture are the same.

However,ingeneral,thesetwoconjecturesdonotimplyeachother.

TheYauNumberTheoreticConjecturehasalreadybeenveriﬁedforn= 3 byXuand Yau[17,18]andforn= 4,5 byLin,Luo,YauandZuo[7,8].Liang,YauandZuo[5]gave thefollowing resultforn= 6.Inthispaper,wewill provetheconjectureforn= 7:

(6)

Theorem 1.4 (Main Theorem). Let P7 = P7(a1,a2,. . . ,a7) = #{(x1,. . . ,x7) ∈ Z⁷+ :

x1

a1 +^x_a²

2+· · ·+^x_a⁷

7 ≤1},wherea1≥a2≥ · · · ≥a7>1arereal numbers.IfP7>0,then 7!P₇≤(a₁−1)(a₂−1). . .(a₇−1)−(a₇−1)⁷+a₇(a₇−1). . .(a₇−6) (7) andequalityholdsif andonlyif a1=a2=· · ·=a7 areintegers.

Let

g_n(a₁, . . . , a_n) := (a₁−1). . .(a_n−1)−(a_n−1)ⁿ+a_n(a_n−1). . .(a_n−(n−1)) betherighthandof(6).Incasen= 7,

g7(a1, . . . , an) := (a1−1). . .(a7−1)−(a7−1)⁷+a7(a7−1). . .(a7−6).

In[7],we cansee that,whenn= 5, thenumberof subcasesincreases from4 (when n= 4)to 11. Theauthors of [7]([5]resp.) simplifythose 11 (21 resp.)subcasesinto 5 (6 resp.)majorclasses.Theydividethewholerangeintofiveintervalsandclassifythose subcases by which interval the last variable an is in. The benefit of this classification is that the number of classes will increase only by 1 as the dimension increase by 1.

However,theproofsof n≤6 reliedontheGLY SharpEstimate,whichis onlytruefor n≤6.Thereforetheproofcannotbegeneralizedto higherdimension.Inthispaper,we avoidentirelytheGLYSharp Estimate,andwewill proveourmain theorem purelyby induction.Thisisasigniﬁcantimprovementsinceitsuggests awaytoprovethegeneral case.

Asanapplication,we willalsoprovethat

Theorem 1.5 (Estimate of ψ(x,y)). Letψ(x,y) be the functionas before. We have the followingupperestimate for5≤y <17:

(I)when 5≤y <7andx>5,wehave ψ(x, y)≤ 1

6{ 1

log 2 log 3 log 5(logx+ log 15)(logx+ log 10)(logx+ log 6)

− 1

log³5[(logx+ log 6)³

−(logx+ log 6 + log 5)(logx+ log 6)(logx+ log 6−log 5)]}; (II)when7≤y <11andx>7,wehave

ψ(x, y)≤ 1

24{ 1

log 2 log 3 log 5 log 7(logx+ log 105)(logx+ log 70)

·(logx+ log 42)(logx+ log 30)

− 1

log⁴7[(logx+ log 30)⁴

(7)

−(logx+ log 7 + log 30)(logx+ log 30)

·(logx+ log 30−log 7)(logx+ log 30−2 log 7)]}; (III)when 11≤y <13andx>11,wehave

ψ(x, y)≤ 1

120{ 1

log 2 log 3 log 5 log 7 log 11(logx+ log 1155)(logx+ log 770)

·(logx+ log 462)(logx+ log 330)(logx+ log 210)

− 1

log⁵11[(logx+ log 210)⁵

−(logx+ log 11 + log 210)(logx+ log 210)

·(logx+ log 210−log 11)(logx+ log 210−2 log 11)

·(logx+ log 210−3 log 11)]}. (IV) when13≤y <17 andx>13,we have

ψ(x, y)≤ 1

720{ 1

log 2 log 3 log 5 log 7 log 11 log 13(logx+ log 15015)(logx+ log 10010)

·(logx+ log 6006)(logx+ log 4290)(logx+ log 2730)

·(logx+ log 2310)− 1

log⁶13[(logx+ log 2310)⁶

−(logx+ log 13 + log 2310)(logx+ log 2310)(logx+ log 2310−log 13)

·(logx+ log 2310−2 log 13)(logx+ log 2310−3 log 13) (logx+ log 2310−4 log 13)]}.

(V) when 17≤y <19 andx>17,we have ψ(x, y)≤ 1

5040{ 1

log 2 log 3 log 5 log 7 log 11 log 13 log 17(logx+ log 255255)

(logx+ log 170170)(logx+ log 102102)(logx+ log 72930)(logx+ log 46410)

·(logx+ log 39270)(logx+ log 30030)− 1

log⁷17[(logx+ log 30030)⁷

−(logx+ log 17 + log 30030)(logx+ log 30030)(logx+ log 30030−log 17)

·(logx+ log 30030−2 log 17)(logx+ log 30030−3 log 17)

·(logx+ log 30030−4 log 17)(logx+ log 30030−5 log 17)]}.

Remark.Forcomparison,welisttheEnnola’supperbounds(see(2))for5≤y <19 as follows:

(8)

(1):5≤y <7 andx>5,

ψ(x, y)≤ (logx+ log30)³ 6log2log3log5 (2):7≤y <11 and x>7,

ψ(x, y)≤ (logx+ log210)⁴ 24log2log3log5log7 (3):11≤y <13 andx>11,

ψ(x, y)≤ (logx+ log2310)⁵ 120log2log3log5log7log11 (4):13≤y <17 andx>13,

ψ(x, y)≤ (logx+ log30030)⁶ 720log2log3log5log7log11log13 (5):17≤y <19 andx>17,

ψ(x, y)≤ (logx+ log255255)⁷

5040log2log3log5log7log11log13log17

Itiseasyto seethatourupperboundof ψ(x,y) issubstantiallybetterthantheone obtainedbyEnnola.Forexample,in17≤y <19 andx>17 case,thoughthecoeﬃcient of(logx)⁷inourestimate issameas Ennola’s,butourcoeﬃcientof(logx)⁶ is

1

5040(log255255 + log170170 + log102102 + log72930 + log46410 + log39270 + log30030 log2log3log5log7log11log13log17

− 14

log⁶17)≈0.00599482679 whichissmallerthanEnnola’s

1 5040

7log255255

log2log3log5log7log11log13log17≈0.0620120969.

Weuse thesymbolic computation software,Sage, to deal withtremendous involved computation.Besides,wehavefoundaquickwaytojudgethepositivityofapolynomial ina restricteddomain. Wealso simplify the process of computationby making use of somecharacteristics ofthosepolynomials.

(9)

2. Somelemmas

Wewillfrequentlyusethefollowingtwolemmastodecidethepositivityofpolynomials insomerestricteddomains.

Lemma 2.1.(See[16]Lemma3.1.) Letf(β) beapolynomial deﬁnedby

f(β) =

n

i=0

ciβⁱ (8)

where β∈(0,1).If forany k= 0,1,. . . ,n

k

i=0

ci≥0 (9)

then f(β)≥0forβ∈(0,1).

Lemma 2.1iseasytouse.However,theconditionofLemma 2.1maynotbe satisﬁed insomesituation.Inthatcase,we shallmakeuseofthefollowinglemma.

Lemma 2.2 (Sturm’s Theorem). Starting from a given polynomial X = f(x), let the polynomials X₁,X₂,. . . ,X_r be determinedbyEuclidean algorithm asfollows:

X1 = f(x) , X = Q₁X₁−X₂, X1 = Q2X2−X3, . . . . . . . . . .

Xr−1 = QrXr (10)

where degXk >degXk+1 fork = 1,. . . ,r−1.For every real number a whichis not a root of f(x)letw(a)be thenumberof variationsin signinthenumber sequence

X(a), X1(a), . . . , Xr(a)

in whichall zeros are omitted. If b and c are any numbers (b < c) for which f(x) does notvanish,thenthenumberof thevariousrootsintheintervalb≤x≤c(multipleroots to becountedonly once) isequalto

w(b)−w(c).

(10)

Proof. See[15]. 2

TheconditionofLemma 2.2isnecessaryandsuﬃcient, soitcanbe appliedtojudge thepositivityofanysuchpolynomialsinsomeintervals.ThecomputationinLemma 2.2 ismorecomplicated than thatinLemma 2.1. Therefore,we preferLemma 2.1 when it works.

Lemma2.3.(See[19]Proposition 3.1.) Givenanypositiverealnumberβ where0< β <

1,leta>1beanynumbersuchthatβ=a− a ,wherea denotesthegreatestpositive integer lessthanorequaltoa.If n≥3,then

a−1>(n+ 1)

a−1

k=0

(k+β)ⁿ

aⁿ . (11)

Lemma2.4.(See[19]Lemma3.3.) Leta_j−1,a_j,. . . ,a_n+1berealnumbersandβ=a_n+1− an+1 .Assumethat a_j−1>1andaj ≥aj+1≥ · · · ≥an≥an+1>1.If _a^a_n+1ⁿ β ≥1,and

n+1

i=j

(ai−1)>(n+ 1)

an+1−1 k=0

[(k+β)^j⁻¹ a^j_n+1⁻¹

n

i=j

( ai

an+1

(k+β)−1)] (12) then

n+1

i=j−1

(a_i−1)>(n+ 1)

an+1−1 k=0

[(k+β)^j−2 a^j−2_n+1

n i=j−1

( a_i an+1

(k+β)−1)]. (13)

Lemma2.5.(See[19]Lemma3.4.) Leta_j−1,aj,. . . ,an+1berealnumbersandβ=an+1− an+1 .Assumethat aj−1>1andaj ≥aj+1≥ · · · ≥an≥an+1>1.If _a^aⁿ

n+1β <1,and

n+1

i=j

(ai−1)>(n+ 1)

an+1−1 k=1

[(k+β)^j−1 a^j_n+1⁻¹

n

i=j

( ai

a_n+1(k+β)−1)] (14) then

n+1

i=j−1

(ai−1)>(n+ 1)

an+1−1 k=1

[(k+β)^j⁻² a^j_n+1⁻²

n i=j−1

( a_i an+1

(k+β)−1)]. (15)

3. Proofofthemaintheorem

We will prove the main theorem by induction. Notice that P_n can be obtained by recursion:letkbethepossibleintegersuchthat1≤k≤ a_n ,wherea_n isthebiggest integerlessthanor equaltoan.Foreachk,wehavean(n−1)-dimensionalsimplex

(11)

x1

a₁ +x2

a₂ +· · ·+ k

a_n ≤1, x1, x2, . . . , xn−1≥0. (16) LetP_n−1^(k) bethenumberof positiveintegersolutionof (16).Clearly,

P_n=

an k=1

P_n^(k)₋₁. (17)

Therefore,sincewealreadyknowthattheYauNumberTheoreticConjectureistruefor n = 6 [5], we wantto prove thatg7(a1,. . . ,a7) isgreater than or equal to the sumof g₆’s, theupperestimateof 6-dimensionallayersinT(a₁,a₂,. . . ,a₇).

Let m be number of 6-dimensional layers in 7-dimensional simplex, i.e. P₆^(m) > 0 and P₆^(m+1) = 0, where P₆^(k) = #{(x1,. . . ,x6) ∈Z⁶+ : ^x_a¹

1 +· · ·+^x_a⁶

6 + _a^k

7 ≤1}, where 1≤k≤m, a₁≥a₂≥ · · · ≥a₆≥1 arerealnumbers.Let

Δ_m:=g₇(a₁, . . . , a₇)−7

m

k=1

g₆(a₁(1− k a7

), . . . , a₆(1− k a7

))

= (a₁−1). . .(a₇−1)−(a₇−1)⁷+a₇(a₇−1). . .(a₇−6)

−7[

m

k=1

(a₁(1− k a7

)−1). . .(a₆(1− k a7

)−1)−(a₆(1− k a7

)−1)⁶ +a6(1− k

a7

)(a6(1− k a7

)−1). . .(a6(1− k a7

)−5)]

be the diﬀerence between g₇(a₁,. . . ,a₇) and the sum of g₆’s. We should prove that Δm≥0 undertheconditionofmain theorem.

Since P₆^(m)= #{(x1,. . . ,x6)∈Z⁶+: ^x_a¹

1 +· · ·+^x_a⁶

6 +_a^m

7 ≤1}, letα= 1−_a^m₇ ∈(0,1), Ai=aiα,fori= 1,. . . ,6,thenwehave

x₁ A1

+ x₂ A2

+· · ·+ x₆

A6 ≤1 (18)

and

g₆(m) :=

m

k=1

g₆(m−k+kα

mα A₁, . . . ,m−k+kα mα A₆) Δm(A1, . . . , A6, α) =g7(A₁

α , . . . ,A₆ α , m

1−α)−7g6(m).

Let B_6,k be e_k(A₁,. . . ,A₆), the elementary symmetric polynomial, for k = 0,. . . ,6, thatis,B_6,k=A₁· · ·A₆

1≤i1<···≤ik≤6 1

Ai1...A_ik.Forexample,B_6,0=A₁· · ·A₆,B_6,5= A1+· · ·+A6 andB6,6= 1.Then

(12)

g₆(m) =

m

k=1

(m−k+kα

mα A₁−1)· · ·(m−k+kα

mα A₆−1)−

m

k=1

(m−k+kα

mα A₆−1)⁶ +

m

k=1

m−k+kα

mα A₆(m−k+kα

mα A₆−1)· · ·(m−k+kα

mα A₆−5)

=

m

k=1

[(m−k+kα

mα )⁶B6,0−(m−k+kα

mα )⁵B6,1+ (m−k+kα mα )⁴B6,2

−(m−k+kα

mα )³B6,3+ (m−k+kα

mα )²B6,4−(m−k+kα

mα )B6,5+B6,6] +

m

k=1

[−9(m−k+kα

mα )⁵A⁵₆+ 70(m−k+kα

mα )⁴A⁴₆−205(m−k+kα mα )³A³₆ + 259(m−k+kα

mα )²A²₆−114(m−k+kα

mα )A6−1].

Tomakeg₆(m) apolynomialofm,wemusttransformthefunctiontoavoidtheappear- anceofminthesumsymbol.Let

Sq:=

m

k=1

(m−k+kα

mα )^q, forq= 1, . . . ,6.

WewillusetheﬁrstsixSq inthelatercomputation:

S1= 1 mα[1

2m(m+ 1)α+1

2m(m−1)]

S2= ( 1 mα)²[1

6m(m+ 1)(2m+ 1)(α−1)²+m²(m+ 1)(α−1) +m³] S3= ( 1

mα)³[1

4m²(m+ 1)²(α−1)³+1

2m²(m+ 1)(2m+ 1)(α−1)² +3

2m³(m+ 1)(α−1) +m⁴] S₄= ( 1

mα)⁴[ 1

30m(m+ 1)(2m+ 1)(3m²+ 3m−1)(α−1)⁴+m³(m+ 1)²(α−1)³ +m³(m+ 1)(2m+ 1)(α−1)²+ 2m⁴(m+ 1)(α−1) +m⁵]

S5= ( 1 mα)⁵[ 1

12m²(m+ 1)²(2m²+ 2m−1)(α−1)⁵ +1

6m²(m+ 1)(2m+ 1)(3m²+ 3m−1)(α−1)⁴+5

2m⁴(m+ 1)²(α−1)³ +5

3m⁴(m+ 1)(2m+ 1)(α−1)²+5

2m⁵(m+ 1)(α−1) +m⁶] S6= ( 1

mα)⁶[ 1

42m(m+ 1)(2m+ 1)(3m⁴+ 6m³−3m+ 1)α⁶ +1

2m³(m+ 1)²(2m²+ 2m−1)(α−1)⁵

(13)

+1

2m³(m+ 1)(2m+ 1)(3m²+ 3m−1)(α−1)⁴+ 5m⁵(m+ 1)²(α−1)³ +5

2m⁵(m+ 1)(2m+ 1)(α−1)²+ 3m⁶(m+ 1)(α−1) +m⁷].

In large part of this paper, we determine the positivity of the polynomial in some restricteddomainbyusingtheinitialvalueofallpartialderivatives.Tomakethispoint clear, weintroducethefollowing lemmas:

Lemma 3.1.Letf(m) beapolynomial ofm,whose degree iss.If (1) _∂m^∂^s^fs >0,

(2) _∂m^∂^k^fk|m=m0>0fork= 0,. . . ,s−1.

Then f(m)>0form≥m0. Proof. Itistrivial. 2

Lemma 3.2. Consider αand m asparameters and letΔ_m(A₁,A₂,. . . ,A₆,α) be apoly- nomial of A1,. . . ,A6.If

(1) Δm(A⁽⁰⁾₁ ,. . . ,A⁽⁰⁾₆ ,α)≥0, (2) ^∂Δ_∂A^m

i ≥0, _∂A^∂²^Δ^m

i∂A6 ≥0and ^∂_∂A⁵^Δ5^m

6 ≥0forall1≤i≤5,A₁≥A⁽⁰⁾₁ ,. . . ,A₆≥A⁽⁰⁾₆ , (3) ^∂_∂A^k^Δk^m

6 |_A₁_=A⁽⁰⁾

1 ,...,A6=A⁽⁰⁾₆ ≥0forall1≤k≤4.

Then Δ_m(A₁,A₂,. . . ,A₆,α)≥0forA₁≥A⁽⁰⁾₁ ,. . . ,A₆≥A⁽⁰⁾₆ .

Proof. Supposef(A1,. . . ,A6) is apolynomialof A1,. . . ,A6. Toprove f ≥0 forA1 ≥ A⁽⁰⁾₁ ,. . . ,A6≥A⁽⁰⁾₆ , weonlyneedtoshow

(1) f(A⁽⁰⁾₁ ,. . . ,A⁽⁰⁾₆ )≥0 and (2) _∂A^∂f

i ≥0,forall1≤i≤6,A₁≥A⁽⁰⁾₁ ,. . . ,A₆≥A⁽⁰⁾₆ .

In particular, we can apply this method to show Δ_m ≥ 0 and _∂A^∂^k^Δ^m

i1...∂A_ik ≥ 0, where 1≤i₁≤ · · · ≤i_k≤6.Inorder toshow _∂A^∂^k^Δ^m

i1...∂A_ik ≥0,weonlyneedtoshow (1) _∂A^∂^k^Δ^m

i1...∂A_ik|_A₁_=A⁽⁰⁾

1 ,...,A6=A⁽⁰⁾₆ ≥0 and (2) _∂A^∂

j(_∂A^∂^k^Δ^m

i1...∂A_ik)≥0,forall1≤j ≤6,A1≥A⁽⁰⁾₁ ,. . . ,A6≥A⁽⁰⁾₆ .

(14)

Noticethatfork≥2,^∂_∂A^k^Δk^m

6 onlycontainsonevariableA6,i.e., _∂A^∂^k+1^Δ^m

i∂^kA6 = 0 for1≤i≤5.

Therefore,giventhethreeconditionsinthepropositionstatement,byinductionwecan provethatΔm(A1,A2,. . . ,A6,α)≥0 forA1≥A⁽⁰⁾₁ ,. . . ,A6≥A⁽⁰⁾₆ . 2

Sowecanusetheinitial valueofall partialderivativesto determinethesign ofΔ_m byapplyingLemma 3.2. Thefollowingproposition givesresultsabout thesignofsome partial derivatives of Δm in general n-dimensional case. This proposition can save us somelaborofcomputing.

Proposition3.1. Let

gn(a1, . . . , an) := (a1−1). . .(an−1)−(an−1)ⁿ+an(an−1). . .(an−(n−1)) be the polynomial upper estimate of Pn(a1,. . . ,an) in the YauNumber Theoretic Con- jecture. And let m be the number of (n−1)-dimensional layers in the n-dimensional simplex,i.e.,P_n−1(m)>0andP_n−1(m+ 1)= 0.Letα= 1−_a^m_n ∈(0,1),A_i=a_iα,for i= 1,. . . ,n−1and

g_n−1(m) :=

m

k=1

g_n−1(m−k+kα

mα A1, . . . ,m−k+kα mα A_n−1) Δ_m(A₁, . . . , A_n−1, α) =g_n(A₁

α , . . . ,A_n−1 α , m

1−α)−ng_n−1(m) then

∂Δm

∂Ai

>0 and

∂²Δm

∂Ai∂An−1

>0

foralli= 1,. . . ,n−2,A1≥ · · · ≥An−1≥₁^mα₋_α,α∈(0,1).

Proof. Notice that A₁,. . . ,A_n−2 are symmetric in the polynomial. Therefore we only needto prove ^∂Δ_∂A^m

1 >0 and _∂A^∂²^Δ^m

1∂An−1 >0 forA₁ ≥ · · · ≥A_n−1 ≥ _1−α^mα, α∈(0,1).Let k=an −k,β =an− an ,

∂Δ_m

∂A1

= 1 α{

n i=2

(a_i−1)−n

an−1 k=an−m

[(k+β an

)

n−1

i=2

(a_i an

(k+β)−1)]}

and

(15)

∂²Δm

∂A1∂An−1

= 1 α²{

n−2 i=2

(ai−1)(an−1)−n

an−1 k=an−m

[(k+β an

)²

n−2 i=2

(ai

an

(k+β)−1)]}. Ourgoalistoshow that

n

i=2

(ai−1)−n

an−1 k=an−m

[(k+β an

)

n−1 i=2

(ai

an

(k+β)−1)]>0 (19) and

n−2 i=2

(ai−1)(an−1)−n

an−1 k=an−m

[(k+β an

)²

n−2 i=2

(ai

an

(k+β)−1)]>0. (20) Wearegoingto considertwo cases.

Case 1:an−1< m< an

In this case, m=a_n . Since P_n(m) >0, _a¹

1 +· · ·+_a¹

n−1 +_a^m

n ≤1 must holds. Thus

1 an−1 +_a^m

n ≤1 anditisequivalentto _a¹

n−1 +^aⁿ_a^−β

n ≤1,so ^aⁿ⁻¹_a

n β ≥1.ByLemma 2.3, a_n−1> n

an−1 k=0

(k+β)ⁿ⁻¹ aⁿ⁻¹n

. (21)

Sincewehavea₁≥a₂≥ · · · ≥a_n−2≥a_n−1≥a_n >1,ifwerepeatedlyapplyLemma 2.4 to (21), thenaftern−2 timeswewill have

n

i=2

(ai−1)−n

an−1 k=0

[(k+β a_n )

n−1 i=2

(ai

a_n(k+β)−1)]>0.

Notice that we also have a1 ≥ a2 ≥ · · · ≥ an−2 ≥ an > 1, so if we repeatedly apply Lemma 2.4to (21)onlyn−3 timeswewillhave

n−2 i=2

(ai−1)(an−1)−n

an−1 k=0

[(k+β an

)²

n−2 i=2

(ai

an

(k+β)−1)]>0.

Notice thatfork ≥0,

(k+β a_n )

n−1 i=2

(ai

a_n(k+β)−1)≥0 (22)

(k+β an

)²

n−2

i=2

(ai

an

(k+β)−1)≥0. (23)

Thus wehave(19)and(20).