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Journal of Number Theory 131 (2011) 15531574
Contents lists available atScienceDirect
Journal of Number Theory
www.elsevier.com/locate/jnt
Asymptotic normality of additive functions on polynomialsequences in canonical number systems
Manfred G. Madritsch a,, Attila Petho ba Department of Analysis and Computational Number Theory, Graz University of Technology, A-8010 Graz, Austriab Department of Computer Science, University of Debrecen, Number Theory Research Group,
Hungarian Academy of Sciences and University of Debrecen, P.O. Box 12, H-4010 Debrecen, Hungary
a r t i c l e i n f o a b s t r a c t
Article history:
Received 19 October 2010Revised 3 February 2011Accepted 8 February 2011Available online 15 April 2011
Communicated by David Goss
MSC:
11K1611A6360F05
Keywords:
Additive functionsCanonical number systemsExponential sums
The objective of this paper is the study of functions which onlyact on the digits of an expansion. In particular, we are interestedin the asymptotic distribution of the values of these functions. Thepresented result is an extension and generalization of a result ofBassily and Ktai to number systems defined in a quotient ring ofthe ring of polynomials over the integers.
2011 Elsevier Inc. All rights reserved.
1. Introduction
In this paper we investigate the asymptotic behavior ofq-additive functions. But before we start weneed an idea of additive functions and the number systems they are living in. Note that a function fis said to be q-additiveif it acts only on the q-adic digits, i.e., f(0)=0 and
f(n)=
h=0f
ah(n)qh
for n=
h=0ah(n)q
h,
whereah(n)N:= {0, . . . , q 1} are the digitsof the q-adic expansion ofn.
* Corresponding author.E-mail addresses:[email protected](M.G. Madritsch), [email protected](A. Petho).
0022-314X/$ see front matter 2011 Elsevier Inc. All rights reserved.doi:10.1016/j.jnt.2011.02.015
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One of the first results dealing with the asymptotic behavior of such a q-additive function is thefollowing, which is due to Bassily and Ktai [3].
Theorem.Let f be a q-additive function such that f(aqh )
=O(1)as h
and a
N. Furthermore let
mh,q:=1
q
aN
f
aqh
, 2h,q:=1
q
aN
f2
aqh m2h,q,
and
Mq(x):=N
h=0mh,q, D
2q (x)=
Nh=0
2h,q
with N= [logqx]. Assume that Dq(x)/(logx)1/3
as x and let P be a polynomial with integercoefficients, degree d and positive leading term. Then, as x ,
1
x#
n
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The simplest version of an additive function is the sum-of-digits function sb defined by
sb(g) := h0ah(g).
The result by Bassily and Ktai was first generalized to number systems in the Gaussian integersby Gittenberger and Thuswaldner [7] who gained the following
Theorem.Let b Z[i]and (b,N)be a canonical number system in Z[i]. Let f be a b-additive function suchthat f(abh) =O(1)as h and aN. Furthermore let
mh,b:=1
N(b)
aN
f
abh
, 2h,b:=1
N(b)
aN
f2
abh m2h,b,
and
Mb(x):=L
h=0mh,b, D
2b (x)=
Lh=0
2h,b
withNthe norm of an element overQ and L= [logN(b)x].Assume that Db(x)/(logx)
1/3 as x and let P be a polynomial of degree d with coefficientsin Z[i]. Then, as N ,
1
#zZ[i]
N(z)
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Therefore we define R(T1, . . . , Tr) to be the set
R(T1, . . . , Tr):= R:(i)
Ti , 1 i r. (1.1)Now we use Lemma 1.1 to bound the area R(T1, . . . , Tn)such that we reach all elements of a certainlength. Thus for a fixed Twe set Ti for 1 i nsuch that
log Ti=log Tlog |b(i)|nlog |N(b)| . (1.2)
Furthermore we will write for short R(T):=R(T1, . . . , Tr) with Ti as in (1.2).Finally one can extend the definition of a number system also for negative powers of b. Then for
K such that
=
h=ahb
h withahN
we call
:=
h=0ahb
h and {} :=h1
ahbh
the integer partand fractional partof, respectively.With all these tools we now can state the generalization of the theorem of Bassily and Ktai to
arbitrary number fields.
Theorem 1.2. (See [15].) Let (b,N) be a number system in R and f be a b -additive function such that
f(abh) =O(1)as h and aN. Furthermore let
mh,b:=1
N(b) aNf
abh
, 2h,b:=
1
N(b) aNf2
abh
m2h,b,
and
Mb(x):=L
h=0mh,q, D
2b(x)=
Lh=0
2h,q
with L= [logN(b)x].Assume that there exists an >0such that Db(x)/(logx)
as x and let PK[X]be a poly-nomial of degree d. Then, as T
let Ti be as in(1.2),
1
#R(T)#
zR(T)
f(P(z)) Mb(Td)Db(T
d)
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2. Definitions and result
The objective of this paper is generalizations of number systems to quotient rings of the ring ofpolynomials over the integers. Our aim is to extend Theorem 1.2 to such rings. To formulate our
results we have to introduce the relevant notions. In particular we use the following definition inorder to describe number systems in quotient rings of the ring of polynomials over the integers.
Definition 2.1. Let p Z[X] be monic of degree n and let N be a subset ofZ. The pair (p,N) iscalled a number system if for every gZ[X] \ {0}there exist unique N and ahN, h=0, . . . , ;a=0 such that
g
h=0ah(g)X
h (mod p). (2.1)
In this case ah are called the digits and = (a) the length of the representation.This concept was introduced in [17] and was studied among others in [1,2,11,12]. It was proved
in [2], that Nmust be a complete residue system modulo p(0) including 0 and the zeroes of p arelying outside or on the unit circle. However, following the argument of the proof of Theorem 6.1of [17], which deal with the case p square free, one can prove that none of the zeroes of p are lyingon the unit circle.
If p is irreducible then we may replace Xby one of the roots of p. Then we are in the caseofZ[X]/(p)= Z[] being an integral domain in an algebraic number field (cf. Section 1). Then wemay also denote the number system by the pair (,N) instead of (p,N). For example, let q 2 bea positive integer, then (p,N) with p
=X
q gives a number system in Z, which corresponds to
the number systems (q,N). Furthermore for n a positive integer and p= X2 + 2n X+(n2 + 1)we getnumber systems in Z[i].
Now we want to come back to these more general number systems and consider additive functionswithin them.
Definition 2.2.Let (p,N) be a number system. A function f is called additiveif f(0)=0 and
f(g)
h=0f
ah(g)Xh
(mod p) forg
h=0ah(g)X
h (mod p).
Since we have defined the analogues of number systems and additive functions to the definitionsfor number fields above, we now need to extend the length estimation of Lemma 1.1 in order tosuccessfully state the result. But before we start we need a little linear algebra. We fix a numbersystem (p,N) and factor p by
p:=t
i=1p
mii
with pi Z[X] irreducible and degpi=ni . Furthermore we denote by ik the roots of pi for i=1, . . . ,tand k
=1, . . . ,ni .
Then we define by
R := Z[X]/(p)=t
i=1Ri withRi=Z[X]/
p
mii
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for i=1, . . . ,t the Z-module under consideration and in the same manner by
K
:=Q
[X
]/(p)
=
t
i=1
Ki withKi=
Q
[X
]/pmii
for i=1, . . . ,tthe corresponding vector space. Finally we denote by Kthe completion ofKaccordingto the usual Euclidean distance.
ObviouslyRis a free Z-module of rank n. Let :RRbe a linear mapping and{z1, . . . ,zn}beany basis ofR. Then
(zj )=n
i=1ai jzi (j=1, . . . ,n)
with ai jZ. The matrix M()=(ai j ) is called the matrix of with respect to the basis{z1, . . . ,zn}.For an element rR we define by r:RR the mapping of multiplication by r; that is r(z)= rzfor every zR. Then we define the norm N(r) and the trace Tr(r) of an element r R as thedeterminant and the trace of M(r), respectively, i.e.,
N(r):=detM(r), Tr(r):=TrM(r).Note that these are unique despite of the used basis {z1, . . . ,zn}. We can canonically extend thesenotions to K and Kby everywhere replacing Zby Qand R, respectively.
In the following we will need parameters which help us bounding the length of the expansion ofan element gR. Therefore let g Z[X]be a polynomial, then we put
B i jk(g) :=dj1gdXj1
X=ik
(i=1, . . . ,t; j=1, . . . , mi; k=1, . . . ,ni ).
In connection with these values we define the house function H as
H(g)
:=
tmax
i=
1
mimaxj=
1
nimaxk=
1 B i jk(g).We want to investigate the elements with bounded maximum length of expansion. To this end
we need a proposition which estimates the length of expansion in connection with properties of thenumber itself. The proof of this proposition will be presented in the following section.
Proposition 2.1.Assume that(p,N)is a number system. Let N=max{|a|: aN}and we set
M(g):=max
log |B i jk(g)|log |ik|
: i=1, . . . ,t; j=1, . . . , mi; k=1, . . . ,ni
.
If g Z[X]is of degree at most n 1, then for any>0there exists L=L()such that if(g) >L then
(g) M(g) C. (2.2)
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Our theorem shows that the distribution properties of patterns in the sequence of digits dependneither on the polynomial p nor on the quotient ring R. They are intimate properties of the back-ward division algorithm defined in [17].
We will show the main theorem in several steps.
(1) In the following section we will show properties of number systems which we need on the onehand to estimate the length of expansion and on the other hand to provide us with an Urysohnfunction, that helps us counting the occurrences of a fixed pattern of digits in the expansion.
(2) Equipped with these tools we will estimate the exponential sums occurring in the proof in Sec-tion 4. Therefore we need to split the module R up into its components and consider each ofthem separately. We also show that we may neglect the nilpotent elements.
(3) Now we take a closer look at the Urysohn function, which will count the occurrences of our pat-tern in the expansion, and estimate the number of hits of the border of this function in Section 5.In particular, we count the number of hits of the area, where the function value lies between 0and 1 as this area corresponds to the error term.
(4) In Section 6 we will show that any chosen patterns of digit and position occurs uniformly in theexpansions. This will be our central tool in the proof of Theorem 2.2.
(5) Finally we draw all the thinks together. The main idea here is to use the growth rate of the devi-ation together with the FrchetShohat Theorem to cut of the head and the tail of the expansion.Then an application of the central Proposition 5.1 and juxtaposition of the moments will provethe result.
3. Number system properties
In this section we want to show two properties we need in the sequel. The first deals with theabove mentioned estimation of the length of an expansion (Proposition 2.1). We will need this resultin order to justify our choice of Tas in (2.5). Secondly we construct the Urysohn function for indi-
cating the elements starting with a certain digit. The main idea is to embed the elements ofRin Rnand to use the properties of matrix number systems in this field.
We start with
Proof of Proposition 2.1. In the proof we combine ideas from [2] and [12].We may assume g=0. As{p,N} is a number system there are =(g) and ah N for h=
0, . . . , ; a=0 such that
g
h=0ahX
h (mod p),
i.e.,
g=
h=0ahX
h + rp
with a polynomial r Z[X]. For j 1 this implies
dj1g
dXj
1=
h=j1
h!
(h j+ 1)!ahX
hj+1
+
j1
s=0
j 1
s
dsr
dXs
dj1sp
dXj
1
s. (3.1)
Consider a zero ik of p, which has multiplicity mi . As we noticed in the Introduction, the argumentof the proof of Theorem 6.1 of [17] allows us to prove that|ik|>1 for all i=1, . . . ,t;k=1, . . . ,ni .Inserting ik into (3.1) we obtain
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B i jk(g) =dj1gdXj1
X=ik
=
h=j1
h!(h j+ 1)!ah
hj+1ik
for i=1, . . . ,tand j=1, . . . ,mi . This implies by taking absolute valueB i jk(g) N
h=j1
h!(h j+ 1)! |ik|
hj+1
N !
( j+ 1)!j+1
ik
h=j1
h(h 1) (h j+ 1)( 1) ( j+ 1) |ik|
h
Nj1|ik|
|ik
| 1
,
which verifies the lower bound for , because|ik|> 1.Now we turn to prove the upper bound. Denote by V= Vp the following mapping: for gZ[X]of
degree at most n 1 choose an aNsuch that g(0) a (mod p(0)). Such an a exists by Theorem 6.1of [17]. Putting q= g(0)a
p(0) , let V(g)= gqpa
X . Obviously V(g) Z[X] and has degree at most n 1,
thus Vcan be iterated. Moreover we have
gu
h=0ahX
h +Xu+1 Vu+1(g) (mod p) (3.2)
with ahN for h=0, . . . , u.Choose u the largest integer satisfying|B i jk(g)| u
j |ik |uj+1|ik |1 for all i=1, . . . ,t, j=1, . . . ,mi and
k=1, . . . ,ni . Then u (1 + /2)M(A). Proceeding like in the previous case we get
B i jk(g)=dj1gdXj1
X=ik
=u
h=j
h!(h j+ 1)!ah
hjik
+j
s=0
j 1s
(u+ 1)!
(u+ 1 s)! u+1sik
djs1 Vu+1(g)dXjs1
X=ik
.
By its definition Vu+1(g) has integer coefficients. Divide the last equation by (u+1)!(u+1j)!
u+1jik
andconsider the obtained equations for i=1, . . . , tandk=1, . . . ,ni and for fixed i andk for j=1, . . . ,misuccessively, then using the choice ofu and that|ik|>1 we conclude that
djs1 Vu+1(g)dXjs1
X=ik
< c,
wherecis a constant depending only on Nas well as the size and the multiplicities of the zeroes of p.These can be considered asn inequalities for the n unknown coefficients of Vu+1(g). Furthermore the
determinant of the coefficient matrix is not zero (cf. [2]). Thus the solutions are bounded. As theyare integers there are only finitely many possibilities for Vu+1(g). As{p,N} is a number system,Vu+1(g) has a representation, which length is bounded by a constant, say c1, which depends onlyon Nas well as the size and the multiplicities of the zeroes of p. Thus (g) u+c1 (1+ )M(g)and the proposition is proved.
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Now we turn our attention back to the counting of the numbers and in particular to the con-struction of the Urysohn function. In order to properly count the elements we need the fundamentaldomain, which is defined as the set of all numbers whose integer part is zero. Since this is not soeasy to define in this context we want to consider its embedding in Rn . The main idea is to use the
corresponding matrix of the polynomial p and to use properties of matrix number systems. This ideaessentially goes back to Grchenig and Haas [8]. The following definitions are standard in that areaand we mainly follow Gittenberger and Thuswaldner [7] and Madritsch [15].
We note that if (p,N) is a number system then X is an integral power base of K, i.e.,{1,X, . . . ,Xn1} is an R-basis for K. Thus we define the embedding by
:KRn,a1+ a2X+ + anXn1 (a1, . . . , an).
Now let p= bn1Xn1 + + b1X+b0 . Then we define the corresponding matrix B by
B=
0 0 b01 0 0
...
0 1 . . .
......
.... . .
. . . . . .
......
.... . . 1 0
...
0 0 0 1 bn1
. (3.3)
One easily checks that (X
p)
=B
(p). Since B is invertible we can extend the definition of
by setting for an integer h
Xh p :=B h (p). (3.4)
By this we define the (embedded) fundamental domain by
F :=
zRnz=
h1
Bhah, ah (N)
.
Following Grchenig and Haas [8] we get that
(F+g1) (F+g2)=0
for every g1,g2 Zn with g1= g2 , where denotes the n-dimensional Lebesgue measure. Thus(B, (N)) is a matrix number system and a so-called just touching covering system. Therefore we areallowed to apply the results of the paper by Mller et al. [16].
We now follow the lines of Madritsch [15] where the ideas of Gittenberger and Thuswaldner [7]were combined with the results of Ktai and Krnyei [10] and Mller et al. [16].
Our main interest is the fundamental domain consisting of all numbers whose first digit equalsa
N, i.e.,
Fa=B1F+ (a).
Imitating the proof of Lemma 3.1 of [7] we get the following.
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Lemma 3.1.For all aNand all vN there exist a1
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4. Estimation of the Weyl sum
Before we continue with the estimation of the number of points inside the fundamental domainand those hitting the border, we want to estimate the exponential sums, which will occur in the
following sections. In particular we want to prove the following.
Proposition 4.1.Let T 0and Ti jk as in(2.5). Let L be the maximum length of the b-adic expansion of zR(T)and let C1 and C2 be sufficiently large constants. Furthermore let l 1, . . . , lh be positions and h1, . . . , hhbe corresponding n-dimensional vectors. If
C1 log L l1< l2 1 or not. The latterreduces to an estimation of the sum in an algebraic number field. Whereas for the case of mi >1we have to deal with nilpotent elements. Therefore we divide Ri into the radical and the nilpotentelements. Thus we define Ri as
Ri:=Z[X]/(pi ) and Ri (T):=g Ri :
B i jk(g) Ti jk (4.3)and the set Ni to be the nilpotent elements, i.e.,
Ni
:= {g
Ri : g
0 mod pi
} and Ni(T)
:= g
Ri(T): g
0 mod pi. (4.4)
But before we start with the proof we need to show that the estimation is good compared withthe trivial one. Thus we will show the following.
Lemma 4.2.Let Ti jk for i=1, . . . ,t, j=1, . . . ,mi , k=1, . . . ,ni be positive reals. Then
#R(T)=t
i=1#Ri(T),
#Ri(T)=ci ni
k=1 Ti1k
mi
+OTmini10 ,#Ni(T)=ci
nik=1
Ti1k
mi1+OT(mi1)ni10 ,
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where
T0=t
maxi=1
max
1, (Ti11 Ti,mi ,ni )
1mini
and the constants ci will be defined in Lemma4.3.
Proof. The first assertion follows immediately from the definition ofR(T). Since the Ri are indepen-dent we fix an i and focus on Ri (T). Obviously we have that
Ri=Z[X]/p
mii
= Z[X]/(pi )mi . (4.5)Thus we concentrate on Z[X]/(piZ[X]) which is an order in a number field Ki of degree ni over Q.For Ki let () (1 r1) be the real conjugates and (m) and (m+r2) (r1+ 1 m r1+ r2) bethe pairs of complex conjugates of. Note that r1
+2r2
=ni . We will apply the following lemma.
Lemma 4.3.(See [15, Lemma 3.3].) Let Tk(1 k r1+ r2)be positive integers and set Tr1+r2+k=Tr1+k for1 k r2 . Then
#
a Z[X]/(pi ):a(k) Tk=ci T1 Tni+OTni10 ,
where T0=max(1, (T1 Tni )1/ni )and ci is a constant depending on Z[X]/(pi ).
Furthermore since (4.5) holds, we get that there exists a Z linear mapping Mi such that
Mi (Ti11, . . . , Ti,mi ,ni )= (Ti11, . . . ,Ti,mi ,ni )
and
#Ri (T)=mi
j=1#
a Z[X]/(pi ):a(k)Ti jk, 1 k ni.
As the value of Ti jk is independent of j, an application of Lemma 4.3 yields
#Ri (T)=ci
nik=1
Ti1k
mi+OTmini1i0 ,
whereci depends on ci and Mi and
Ti0=max1, (Ti11 Ti,mi ,ni )
1mi ni
.
For the estimate involving Ni (T) we note that
Ni= {gRi: g0 mod pi}=Z[X]/(pi )mi1
and the result follows in the same way as for Ri (T).
With help of all these tools we can show Proposition 4.1.
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Proof of Proposition 4.1. The first step consists in splitting the sum over R(T) up into those overRi (T). Therefore let i:RRi be the canonical projections. Then :=(1, . . . , t) is an isomor-phism by the Chinese Remainder Theorem. Furthermore let i be the embedding defined by
i:KiRni mi ,a1+ a2X+ + ami niXnimi1 (a1, . . . , ami ni )
for i=1, . . . ,t. Finally we define the matrix Mto be such that
M (z):= 1 1(z) , . . . , t t(z).We note that for Pi:= iP and lZ we have
M P(z)Xl= 1P1(z1)Xl, . . . , tP(zt)Xl.Thus
zR(T)
e
hr=1
hr,
P(z)Xlr1
= ti=1
ziRi (T)
e
hr=1
hri , i
Pi(zi )X
lr1
where (hr1, . . . , hrt) := hrM1 .Now we will consider each sum over Ri (T)separately. Therefore we fix until the end of the proofa 1 i tand distinguish two cases according to whether m i=1 or not. Case 1. mi=1: In this case we set =i1 and observe that K= Ki= Q(i1) and Ri= Z[].
Furthermore let OKbe the maximum order aka the ring of integers of K, then clearly Z[i1] OK.We denote by ik= (k) the conjugates of . Now we will proceed as in the proof of Proposition 6.1of Madritsch [15].
Therefore we need some parameters of the field Kand for short we set n=ni during this case.Then we order the conjugates by denoting with (k) for 1 k r1 the real conjugates, whereas (k)
and (k+r2) denote the pairs of complex conjugates, where n=r1+2r2. Let Tr be the trace of anelement of K over Q, then we define
(z):= Tr(z), Tr(z) , . . . , Trn1z= i(z), (4.6)where =V VT and Vis the Vandermonde matrix
V=
1 1 1 (2) ((n))n1...
......
n1 ((2))n1 ((n))n1
.
We set (hr1, . . . ,hrn ) := hri 1 and note that
hri , i
Pi(zi )X
lr1= hTri 1Pi(zi )lr1=Tr
nk=1
hrk klr2 Pi(zi)
.
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Thus we may rewrite the sum under consideration as follows
zRi (T)
e h
r=1hri , iPi(zi )X
lr1= zRi (T) eTr h
r=1
n
k=1
hrkklr2 Pi(zi ).
Now we need an approximation lemma which essentially goes back to Siegel [19]. Therefore let be the different of K over Qand be the absolute value of the discriminant of K. Then we have thefollowing.
Lemma 4.4. Let N1, . . . , Nr1+r2 be real numbers and let N= n
N1 Nr1 (Nr1+1 Nr1+r2 )2 be their geo-metric mean. Suppose that N>
1n, then, corresponding to any K , there exist qOKand a1 such
that
q(k)(k) a(k)
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Case 1.1.H(q) (log T)2 : We apply Lemma 4.5 and get
ziRi (T)
e h
r=1
n
k=1
hrk Pi (zi)
lr+
1 Tn(log T)0 .
Case 1.2. 2 H(q) < (log T)2 : In the last two cases we need Minkowskis lattice theory (cf. [9]).Let 1 be the first successive minimum of the Z-lattice 1 . Then we get
h
r=1
(k)r
((k))lr+1
a(k)q(k)
1q(k)2 1
1q(k) 1q(k)2 1
1
2q(k) (log T)2 ,
which implies
(log T)2
hr=1
(k)r
((k))lr+1
h
r=1 |(k)r ||(k)|l1+1 .
Since|(k)|> 1 and (4.2) we have
(k)l1+1 (k)(log T)2 n(log T)2+1 ,which yields
l1(2+ 1)log |(k)| loglog T
contradicting the lower bound of l1 for sufficiently large C1 in (4.1). Case 1.3. 0
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lh+ 1 nd log|N()|T c(loglog|N()|T)
contradicting the upper bound for sufficiently large C2.
Case 2.m i > 1: Now we have to go one step further and to take a closer look at Ri . In particular
we divide every element ziRi into its radical and its nilpotent part. We fix an element zR andset zi:= i (z).
On the one hand, since Ri=RiNi we have for ziRi the unique representation
zi=zi1+zi2 (4.8)
with zi1Ri and zi2Ni . This motivates the definition of the linear map i j such that i j (z):=zi jfor i=1, . . . ,tand j=1, 2.
On the other hand, since Ri=(Z[X]/(pi ))mi we have for every ziRi the unique representation
zi=mi
j=1ai jp
j1i
=mi
j=1
nik=1
ai jkXk1pj1i
with a i jZ[X]and ai jkZ, respectively. We clearly have
i1(z)=ni
k=1ai1kX
k1 and i2(z)=mi
j=2
nik=1
ai jkXk1pj1i .
Thus we define for ziRi the embeddings i1 and i2 by
i1
i1(zi)=(ai11, . . . , ai1ni ) and i2i2(zi )=(ai21, . . . ,ai,mi ,ni ).
Then there exists an invertible matrix Mi such that
Mi
i i(z)= i1 i1(z), i2 i2(z).
Now we can divide the sum up as follows.
ziRi (T)
e
hr=1
hri , i
Pi (zi)X
lr1
=
zi1Ri (T)
zi2Ni (T)
e
hr=1
hri , i
Pi (zi1+zi2)Xlr1
= zi1Ri (T)
e
h
r=
1hri1, i1Pi1(zi1)Xlr1
zi2N
i (T)
e
h
r=
1hri2, i2Pi2(zi2)Xlr1
,
where we have set Pi j= i jP for j=1, 2.Since for the first sum we have that mi=1 we may follow Case 1 above and use Lemma 4.3 for
trivially estimating the second one to prove the proposition for this case.
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5. Treatment of the border
In Section 3 above we have constructed the Urysohn function we need in order to properly countthe number of elements within the fundamental domain. In this construction we also used an axe-
parallel tube in order to cover the border of the fundamental domain. The number of hits of this tubegives rise to the error term which we will consider in this section.
We fix a positive integer v , which will be chosen later, and a real vector T. Furthermore for l 0we define Fl to be the number of hits of the border of the Urysohn function which is
Fl:=#
zR(T) Bl1P(z)
aNPv,amod B
1Zn
. (5.1)
As indicated above we are interested in an estimation of Fl .
Proposition 5.1.Let
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where is the n-dimensional Lebesgue measure and S is the number of elements in R(T). ByLemma 4.2 we have that
S=t
i=1 ci ni
k=1 Ti1k
mi
+OTn10 . (5.4)Now we apply Lemma 5.2 to get
D S
Bl1
P(z)
zR(T) 2
H+ 1+
0
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Proof. We recall our Urysohn function ua (defined in (3.5)) and set for Rn
t()=ub1Bl11 ubh B
lh1,where B is the matrix defined in (3.3).
Now we want to apply the Fourier transformation, which we developed in Lemma 3.2. Thereforewe set
M := M= (h1, . . . , hh) hrZn, forr=1, . . . , h.An application of Lemma 3.2 yields
t()=
MMTMe
hr=1
hrBlr1
, (6.2)
where TM=h
r=1 cmr1,...,mrn . Combining this with the definition of Fl in (5.1) we get
zR(T)
t
P(z) Fl1+ +Flh . (6.3)
Plugging (6.2) into (6.3) together with an application of Lemma 3.2 for the coefficients yields
= c1 ct|det(B)|hTn +
0 =MM
TMe
hr=1
hr, B
lr1
P(z)+O
hr=1
Flr
.
Now an application of Proposition 4.1 to treat the exponential sums, of Proposition 5.1 for the bor-der Fl with vloglog Tand the observation that
MM |
TM
| 2h
|det B
|2hv
(log T)t0/2,
where we used the definition of in (3.6), proves the proposition.
7. Proof of Theorem 2.2
For this proof we mainly follow the proof of the theorem of Bassily and Ktai [3]. In the samemanner we cut of the head and tail of the expansion and show the theorem for a truncated versionof the additive function. In particular we set C:=max(C1, C2), A:= [Clog L] and B:=LA , whereL, C1 and C2 are defined in the statement of Proposition 6.1. Furthermore we define the truncated
function fto be
f
P(z)= B
j=Af
aj
P(z)
bj
.
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By the definition of A and f(abj )1 with aNwe get that f(P(z))= f(P(z))+O(log L). In thesame manner we define the truncated mean and standard deviation
M(T):=B
j=A mj and D
2
(T):=B
j=A 2
j.
At this point we need that the deviation D tends sufficiently fast do infinity. In particular, wecould refine the statement, if we shrink the part, which is cut of. Since M(T) M(T)=O(log L)andD2(T) D 2(T)=O(log L)we get that it suffices to show that
1
#R(T)#
zR(T)
f(P(z)) M(Td)D (Td)
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