Finite beta-expansions with negative bases

0.2 MB·

by  Zuzana Krčmáriková

Finite beta-expansions with negative bases
Zuzana Krčmáriková1,∗, Wolfgang Steiner2,†, Tomáš Vávra1,∗
1 Department of Mathematics FNSPE
7
1 Czech Technical University in Prague
0
2 Trojanova 13, 120 00 Praha 2, Czech Republic

n [email protected], [email protected]
a
J
2 IRIF, CNRS UMR 8243, Université Paris Diderot – Paris 7
7
1 Case 7014, 75205 Paris Cedex 13, France

] [email protected]
T
N
.
h
t
a Abstract
m
[ Thefiniteness property is an importantarithmetical property of beta-expansions.

1 We exhibit classes of Pisot numbers β having the negative finiteness property, that is
v the set of finite (−β)-expansions is equal to Z[β 1]. For a class of numbers including
−
9
theTribonaccinumber,wecomputethemaximallengthofthefractionalpartsarising
0
6 in the addition and subtraction of (−β)-integers. We also give conditions excluding
4 the negative finiteness property.
0
.
1
0 1 Introduction
7
1
: Digital expansions in real bases β > 1 were introduced by Rényi [23]. Of particular
v
i interest are bases β satisfying the finiteness property, or Property (F), which means that
X
each element of Z[β 1]∩[0,∞) has a finite (greedy) β-expansion. We know from Frougny
r −
a
and Solomyak [13] that each base with Property (F) is a Pisot number, but the converse is
not true. Partial characterizations are due to [13, 16, 1]. In [2], Akiyama et al. exhibited
an intimate connection to shift radix systems (SRS), following ideas of Hollander [16]. For
results on shift radix systems (with the finiteness property), we refer to the survey [18].
∗Supported by the Czech Science Foundation, grant No. 13-03538S, and by the Grant Agency of the
Czech Technical University in Prague,grant No. SGS14/205/OHK4/3T/14
†Supported by the ANR-FWF project “Fractals and Numeration” (ANR-12-IS01-0002, FWF I1136)
and the ANR project “Dyna3S” (ANR-13-BS02-0003)
1
Numeration systems with negative base −β < −1, or (−β)-expansions, received con-
siderable attention since the paper [17] of Ito and Sadahiro in 2009. They are given by the
(−β)-transformation
T : [ℓ ,ℓ +1) → [ℓ ,ℓ +1), x 7→ −βx−⌊−βx−ℓ ⌋, with ℓ = β ;
β β β β β β β β−+1
−
see Section 2 for details. Certain arithmetic aspects seem to be analogous to those for
positive base systems [12,20], others aredifferent, e.g., both negative and positive numbers
have (−β)-expansions; for β < 1+√5, the only number with finite (−β)-expansion is 0. We
2
say that β > 1 has the negative finiteness property, or Property (−F), if each element of
Z[β 1] has a finite (−β)-expansion. By Dammak and Hbaib [10], we know that β must
−
be a Pisot number, as in the positive case. It was shown in [20] that the Pisot roots of
x2−mx+n, with positive integers m,n, m ≥ n+2, satisfy the Property (−F). This gives
a complete characterization for quadratic numbers, as β does not possess Property (−F)
if β has a negative Galois conjugate, by [20].
First, we give other simple criteria when β doesnot satisfy Property (−F).Surprisingly,
this happens when ℓ has a finite (−β)-expansion, which is somewhat opposite to the
β
positive case, where Property (F) implies that β is a simple Parry number.
Theorem 1. If Tk (ℓ ) = 0 for some k ≥ 1, or if β is the root of a polynomial p(x) ∈ Z[x]
β β
with |p(−1)| = 1,−then β does not possess Property (−F).
The main tool we use is a generalization of shift radix systems. We show that the
(−β)-transformation is conjugated to a certain α-SRS. Then we study properties of this
dynamical system. We obtain a complete characterization for cubic Pisot units.
Theorem 2. Let β > 1 be a cubic Pisot unit with minimal polynomial x3 −ax2 +bx−c.
Then β has Property (−F) if and only if c = 1 and −1 ≤ b < a, |a|+|b| ≥ 2.
Considering Pisot numbers of arbitrary degree, we have the following results.
Theorem 3. Let β > 1 be a root of xd − mxd 1 − ··· − mx − m for some positive
−
integers d,m. Then β has Property (−F) if and only if d ∈ {1,3,5}.
Theorem 4. Let β > 1 be a root of xd − a xd 1 + a xd 2 + ··· + (−1)da ∈ Z[x] with
1 − 2 − d
a ≥ 0 for i = 1,...,d, and a ≥ 2+ d a . Then β has Property (−F).
i 1 i=2 i
These theorems are proved in SecPtion 3. In Section 4, we give a precise bound on the
number of fractional digits arising from addition and subtraction of (−β)-integers in case
β > 1 is a root of x3 −mx2 −mx−m for m ≥ 1. This is based on an extension of shift
radix systems. The corresponding numbers for β-integers have not been calculated yet,
although they can be determined in a similar way.
2
2 (−β)-expansions
For β > 1, any x ∈ [ℓ ,ℓ +1) has an expansion of the form
β β
∞ x
x = i with x = ⌊−βTi 1(x)−ℓ ⌋ for all i ≥ 1.
(−β)i i −β β
−
i=1
X
N
This gives the infinite word d (x) = x x x ··· ∈ A with A = {0,1,...,⌊β⌋}. Since the
β 1 2 3
base is negative, we can repre−sent any x ∈ R without the need of a minus sign. Indeed,
let k ∈ N be minimal such that x ∈ (ℓ ,ℓ +1) and d ( x ) = x x x ···. Then the
( β)k β β β ( β)k 1 2 3
− − −
(−β)-expansion of x is defined as
x ···x x •x x ··· if k ≥ 1,
1 k 1 k k+1 k+2
hxi β = −
− (0•x1x2x3··· if k = 0.
Similarly to positive base numeration systems, the set of (−β)-integers can be defined
using the notion of hxi , by
β
−
Z = {x ∈ R : hxi = x ···x x •0ω} = (−β)kT k(0),
β β 1 k 1 k −β
− − − −
k 0
[≥
where0ω istheinfiniterepetitionofzeros. Theset ofnumbers withfinite(−β)-expansionis
Z
Fin(−β) = {x ∈ R : hxi−β = x1···xk−1xk •xk+1···xk+n0ω} = (−−ββ)n .
n 0
[≥
If hxi = x ···x x •x ···x 0ω with x 6= 0, then fr(x) = n denotes the length
β 1 k 1 k k+1 k+n k+n
of the−fractional pa−rt of x; if x ∈ Z , then fr(x) = 0.
β
−
3 Finiteness
In this section, we discuss the Property (−F) for several classes of Pisot numbers β. Note
that Fin(−β) is a subset of Z[β 1] since β is an algebraic integer, hence Property (−F)
−
means that Fin(−β) = Z[β 1], i.e., Fin(−β) is a ring. We start by showing that bases β
−
satisfying d (ℓ ) = d d ...d 0ω, which can be considered as analogs to simple Parry
β β 1 2 k
−
numbers, do not possess Property (−F). This was conjectured in [19] and supported by
the fact that d (ℓ ) = d d ...d 0ω with d ≥ d + 2 for all 2 ≤ j ≤ k implies that
β β 1 2 k 1 j
d (β−1−d ) =−(d +1)(d +1)···(d +1)1ω. However, the assumption d ≥ d +2 is not
β 1 2 3 k 1 j
−
necessary for showing that Property (−F) does not hold.
We also prove that a base with Property (−F) cannot be the root of a polynomial of
the form a xd +a xd 1 +···+a with | d (−1)ia | = 1.
0 1 − d i=0 i
P
3
Proof of Theorem 1. If Tk (ℓ ) = 0, i.e., d (ℓ ) = d d ...d 0ω, then we have
β β β β 1 2 k
− −
−β d d d
1 2 k
= + +···+
β +1 −β (−β)2 (−β)k
and thus 1 ∈ Z[β 1]. However, we have 1 ∈/ Fin(−β) since T ( 1 ) = 1 , i.e.,
β−+1 − β−+1 β β−+1 β−+1
−
d ( 1 ) = 1ω. Hence β does not possess Property (−F).
β β−+1
−
If p(β) = 0 with |p(−1)| = 1, then write
p(x−1) = xf(x)+p(−1),
with f(x) ∈ Z[x]. Then we have β+11 = |f(β + 1)| ∈ Z[β] and thus −(β−+β1)−j ∈ Z[β−1] for
some j ≥ 0. Now, d β(−β(β+)1−j) = 0j1ω implies that β does not have the Property (−F).
−
The main tool we will be using in the rest of the paper are α-shift radix systems.
An α-SRS is a dynamical system acting on Zd in the following way. For α ∈ R, r =
(r0,r1,...,rd 1) ∈ Rd, and z = (z0,z1,...,zd 1) ∈ Zd, let τr,α be defined as
− −
τr,α(z0,z1,...,zd 1) = (z1,...,zd 1,zd),
− −
where z is the unique integer satisfying
d
0 ≤ r z +r z +···+r z +z +α < 1. (1)
0 0 1 1 d 1 d 1 d
− −
Alternatively, we can say that
τr,α(z0,z1,...,zd 1) = (z1,...,zd 1,−⌊rz+α⌋),
− −
where rz stands for the scalar product.
The usefulness of α-SRS with α = 0 for the study of finiteness of β-expansions was
first shown by Hollander in his thesis [16]. His approach was later formalized in [2] where
the case α = 0 was extensively studied. The symmetric case with α = 1 was then studied
2
in [4]. Finally, general α-SRS were considered by Surer [24].
We say that τr,α has the finiteness property if for each z ∈ Zd there exists k ∈ N such
that τrk,α(z) = 0. The finiteness property of τr,α is closely related to the Property (−F),
thus it is desirable to study the set
D0 = {r ∈ Rd : ∀z ∈ Zd,∃k,τk (z) = 0}.
d,α r,α
The following proposition shows the link between (−β)-expansions and α-SRS.
Proposition 5. Let β > 1 be an algebraic integer with minimal polynomial xd+a xd 1+
1 −
···+a x+a . Set α = β and let (r ,r ,...,r ) ∈ Rd 1 be such that
d 1 d β+1 0 1 d 2 −
− −
xd +(−1)a xd 1 +···+(−1)da = (x+β)(xd 1 +r xd 2 +···+r x+r ),
1 − d − d 2 − 1 0
−
a a
i.e., ri = (−1)d−i d−i +···+ d for i = 0,1,...,d−2.
β βi+1
(cid:18) (cid:19)
Then β has Property (−F) if and only if (r ,r ,...,r ) ∈ D0 .
0 1 d 2 d 1,α
− −
4
Proof. Let r = (r ,r ,...,r ). First we show that for φ : z 7→ rz−⌊rz+α⌋ the following
0 1 d 2
commutation diagram holds,−i.e., the systems (τr,α,Zd−1) and (T β,Z[β]∩[ℓβ,ℓβ +1)) are
−
conjugated.
Zd 1 −−τr−,α→ Zd 1
− −
φ φ
Z[β]∩[ℓ,ℓ +1) −−T−−β→ Z[β]∩[ℓ,ℓ +1)
yβ β yβ β
Since r = (−1)d i 1(βd i 1 + a βd i 2 + ··· + a ) for 0 ≤ i ≤ d − 2, the set {r :
i −− −− 1 −− d i 1 i
0 ≤ i < d} with r = 1 forms a basis of Z[β],−h−ence φ is a bijection. Moreover, we
d 1
have −βr = r +−c with c ∈ Z and r = 0. For z = (z ,z ,...,z ), we have
i i 1 i i 1 0 1 d 2
φ(z) = di=−01rizi−with zd 1 = −⌊rz+α⌋, thu−s −
−
P d 1
−
T β(φ(z)) = −βφ(z)+n = ri 1zi +n′ = φ(z1,...,zd 2,zd 1) = φ(τr,α(z)),
− − − −
i=1
X
wherenandn areintegers; forthethirdequality, wehaveusedthatT (φ(z)) ∈ [ℓ ,ℓ +1).
′ β β β
Therefore, we have r ∈ D0 if and only if for each x ∈ Z[β]∩[ℓ−,ℓ +1) there exists
d 1,α β β
k ≥ 0 such that Tk (x) = 0. S−ince for each x ∈ Z[β 1]∩[ℓ ,ℓ +1) we have Tn (x) ∈ Z[β]
β − β β β
for some n ∈ N, P−roperty (−F) is equivalent to r ∈ D0 . −
d 1,α
−
Thus the problem of finiteness of (−β)-expansions can be interpreted as the problem
of finiteness of the corresponding α-SRS. This problem is often decidable by checking the
finiteness of α-SRS expansions of a certain subset of Zd. A set of witnesses of r ∈ Rd is a
set V ⊂ Zd that satisfies
1. ±e ∈ V where e denotes the standard basis of Rd,
i i
2. if z ∈ V, then τr,0(z),−τr,0(−z) ∈ V.
The following proposition is due to Surer [24] and Brunotte [7].
Proposition 6. Let α ∈ [0,1) and r ∈ Rd. Then r ∈ D0 if and only if there exists a set
d,α
of witnesses that does not contain nonzero periodic elements of τr,α.
Setsofwitnessesforseveralclassesofr ∈ Rd werederivedin[3]. Exploitingtheirexplicit
form, several regions of finiteness can be determined; see in particular [3, Theorems 3.3–
3.5]. An α-SRS analogy of some of those regions was given by Brunotte [7]. Brunotte’s
result, however, is unsuitable for our purposes. The next proposition gives several regions
of finiteness of α-SRS.
Proposition 7. Let r = (r ,r ,...,r ) ∈ Rd and α ∈ [0,1).
0 1 d 1
−
1. If di=−01|ri| ≤ α and ri<0ri > α−1, then r ∈ Dd0,α.
2. If P0 ≤ r ≤ r ≤ ··· ≤Pr ≤ α, then r ∈ D0 .
0 1 d 1 d,α
−
5
3. If di=−01|ri| ≤ α and ri < 0 for exactly one index i = d−k, then r ∈ Dd0,α if and only
if
P
r > α−1. (2)
d jk
−
1 j d/k
≤X≤
Proof. 1. The set V = {−1,0,1}d is closed under τr,0(z) and −τr,0(−z), hence it is
a set of witnesses. For any z ∈ V we have |rz| ≤ α, thus ⌊rz + α⌋ ∈ {0,1}.
Hence any periodic point of τr,α is in {0,−1}d. For z ∈ {0,−1}d we have rz+α ≤
− r +α < 1. Therefore ⌊rz+α⌋ = 0, so the only period is the trivial one.
rj<0 j
2. InPthis case we take as a set of witnesses the elements of {−1,0,1}d with alternating
signs, i.e., z z ≤ 0 for any pair of indices i < j such that z = 0 for each i < k < j.
i j k
For any z ∈ V we have again |rz| ≤ α, thus ⌊rz + α⌋ ∈ {0,1} and τr,α(z) ∈ V.
Therefore, we have τn (z) = (−1,0,...,0) for some n ≥ 0, hence τn+1(z) = 0.
r,α r,α
3. In this case we have V = {−1,0,1}d. As above, all periodic points of τr,α are in
{0,−1}d. If z = (z ,z ,...,z ) is a periodic point with z = −⌊rz + α⌋ = −1,
0 1 d 1 d
−
then we must have z = −1, and consequently z = −1 for all 1 ≤ j ≤ d/k.
d k d jk
− −
Then z = −1 also implies that − r +α ≥ 1, i.e., (2) does not hold. On
d 1 j d/k d jk
≤ ≤ −
the other hand, if (2) holds, then the vector (z ,z ,...,z ) with z = −1 for
0 1 d 1 d jk
P − −
1 ≤ j ≤ d/k, zi = 0 otherwise, is a periodic point of τr,α.
Next we prove Property (−F) when β is a root of a polynomial with alternating coef-
ficients, where the second highest coefficient is dominant.
Proof of Theorem 4. Let β > 1 be a root of p(x) = xd−a xd 1+a xd 2+···+(−1)da ∈
1 − 2 − d
Z[x] with a ≥ 0 for i = 1,...,d, and a ≥ 2+ d a . As d (p(x)x d) ≥ a1 − a1 2 > 0
i 1 i=2 i dx − x2 x−3
for x > 1, the polynomial p(x) has a unique root β > 1, and we have β > a − 1 since
1
P
p(a −1) ≤ −(a −1)d 1+(a −2)(a −1)d 2 < 0. By Proposition 5, Property (−F) holds
1 1 − 1 1 −
if and only if (r ,r ,...,r ) ∈ D0 , with r = a β 1−a β 2+a β 3−···+
0 1 d 2 d 1,α i d i − d i+1 − d i+2 −
(−1)d ia β d+i 1. We hav−e − − − −
− d − −
a −2 a −2 a −2 a −2 1
1 1 1 1
− r ≤ + +···+ ≤ <
i β2 β4 β2 d/2 2 β2 −1 β +1
⌈ ⌉−
rXi<0
6
and
d 1
β +1 − β +1 a +···+a a +···+a a
2 d 3 d d
|r | ≤ + +···+
i
β β β β2 βd 1
i=0 (cid:18) − (cid:19)
X
a +···+a a +2a +···+2a a +2a +···+2a a +2a a
2 d 2 3 d 3 4 d d 1 d d
= + + +···+ − +
β β2 β3 βd 1 βd
−
a −2 2(a −2)−a 2(a −a −2)−a 2(a −a −···−a −2)−a
1 1 2 1 2 3 1 2 d 1 d
≤ + + +···+ −
β β2 β3 βd
1 a −2 a −a −2 a −a −···−a −2
1 1 2 1 2 d 1
≤ 1−2 − − −···− −
β β2 β3 βd
(cid:18) (cid:19)
2 a −2
1
≤ 1− 1− < 1.
β β −1
(cid:18) (cid:19)
Therefore, item 1 of Proposition 7 gives that Property (−F) holds.
NowwecanclassifythecubicPisot unitswithProperty(−F).Thefollowingdescription
of cubic Pisot numbers in terms of the coefficients of the minimal polynomial is due to
Akiyama [1, Lemma 1].
Lemma 8. A number β > 1 with minimal polynomial x3 − ax2 + bx − c is Pisot if and
only if
|b+1| < a+c and b+c2 < sgn(c)(1+ac).
Proof of Theorem 2. Let β > 1 be a cubic Pisot unit with minimal polynomial x3−ax2+
bx − c. If c = −1, then β has a negative conjugate, which contradicts Property (−F)
by [20]. Therefore, we assume in the following that c = 1. Then from Lemma 8 we have
that −a−1 ≤ b < a. By Proposition 5, Property (−F) holds if and only if (r ,r ) ∈ D0 ,
0 1 2,α
with (r ,r ) = (1, b − 1 ) and α = β . We distinguish five cases for the value of b.
0 1 β β β2 β+1
1. b = 0: If a ≥ 2, then we have |r |+|r | = 1 + 1 < α and r +r > 0 > α −1, so
0 1 β β2 0 1
we apply item 3 of Proposition 7. If a = 1, then we have T 1(0) = {0} as β < 1+√5,
−β 2
thus Fin(−β) = {0}. −
2. b = −1: If a ≥ 1, then r +r = − 1 > 1 = α −1. If a ≥ 3, then we also have
0 1 β2 β−+1
|r |+|r | < α and use item 3 of Proposition 7. If a = 2, then r ≈ 0.39, r ≈ −0.55,
0 1 0 1
α ≈ 0.72, {−1,0,1}2 is a set of witnesses, and Property (−F) holds because τr,α acts
on this set in the following way:
(−1,1) 7→ (1,1) 7→ (1,0) 7→ (0,−1) 7→ (−1,−1) 7→ (−1,0) 7→ (0,0),
(0,1) 7→ (1,0), (1,−1) 7→ (−1,−1).
For a = 1, we refer to Theorem 3, which is proved below. If a = 0, then β < 1+√5
2
and thus Fin(−β) = {0}.
7
3. 1 ≤ b ≤ a−2: For b ≥ 2, we have 0 < r < r < α and thus (r ,r ) ∈ D0 by item 2
0 1 0 1 2,α
of Proposition 7. If b = 1, then we can use item 1 of Proposition 7 because r ,r > 0
0 1
and r +r < α.
0 1
4. 1 ≤ b = a−1: We have β = b+ 1 . For b ≥ 3, we have 0 < r < α < r < 1, the
β(β 1) 0 1
set {−1,0,1}2 \{(1,1),(−1,−1)} is−a set of witnesses, and τr,α acts on this set by
(1,0) 7→ (0,−1) 7→ (−1,1) 7→ (1,−1) 7→ (−1,0) 7→ (0,0), (0,1) 7→ (1,−1),
thus Property (−F) holds. If b = 2, then 0 < r < r < α and we can use item 2 of
0 1
Proposition 7. If b = 1, then r ≈ 0.57, r ≈ 0.25, α ≈ 0.64, thus {−1,0,1}2 is a set
0 1
of witnesses, with
(−1,−1) 7→ (−1,1) 7→ (1,0) 7→ (0,−1) 7→ (−1,0) 7→ (0,0),
(0,1) 7→ (1,0), (1,1) 7→ (1,−1) 7→ (−1,0).
5. −a−1 ≤ b ≤ −2: We have −r0−r1+α = −bβ−1 + β12 + β+β1 > 1, thus τr,α(−1,−1) =
(−1,−1), hence (r ,r ) ∈/ D0 .
0 1 2,α
Therefore, β has Property (−F) if and only if −1 ≤ b < a, |a|+|b| ≥ 2.
Finally, we study generalized d-bonacci numbers.
Proof of Theorem 3. Let β > 1 be a root of xd −mxd 1 −···−mx−m with d,m ∈ N.
−
If d = 1 (and m ≥ 2), then β is an integer, and Property (−F) follows from Z = Z;
β
−
see e.g. [20].
If d = 3, then r = (mβ,−mβ −βm2), 0 < r0 < α < −r1 < 1, with α = β+β1, and τr,α satisfies
(0,1) 7→ (1,1) 7→ (1,0) 7→ (0,−1) 7→ (−1,−1) 7→ (−1,0) 7→ (0,0),
with {−1,0,1}2 \{(1,−1),(−1,1)} being a set of witnesses.
If d = 5, then r = (m,−m − m, m + m + m,−m − m − m − m), which gives 0 < r <
β β β2 β β2 β3 β β2 β3 β4 0
α < −r1 < r2 < −r3 < 1 and the τr,α-transitions
(0,1,0,0) 7→ (1,0,0,1) 7→ (0,0,1,0) 7→ (0,1,0,−1) 7→ (1,0,−1,0) 7→ (0,−1,0,0) 7→
(−1,0,0,−1)7→(0,0,−1,−1)7→(0,−1,−1,0)7→(−1,−1,0,0)7→(−1,0,0,0)7→(0,0,0,0),
(0,0,−1,0) 7→ (0,−1,0,1) 7→ (−1,0,1,0) 7→ (0,1,0,−1),
(0,1,1,1) 7→ (1,1,1,1) 7→ (1,1,1,0) 7→ (1,1,0,−1) 7→ (1,0,−1,−1) 7→
(0,−1,−1,−1) 7→ (−1,−1,−1,−1) 7→ (−1,−1,−1,0) 7→ (−1,−1,0,0),
(0,0,0,1)7→(0,0,1,1)7→(0,1,1,0)7→(1,1,0,0)7→(1,0,0,0)7→(0,0,0,−1)7→(0,0,−1,−1),
(−1,−1,0,1) 7→ (−1,0,1,1) 7→ (0,1,1,0).
8
Let V be the set of these states. We have ±e ∈ V, z ∈ V if and only if −z ∈ V and
i
τr,0(z) ∈ V for all z ∈ V, thus V is a set of witnesses. As τr1,1α(z) = (0,0,0,0) for all z ∈ V,
β has Property (−F).
Foroddd ≥ 7,Property(−F)doesnotholdsinceTd 1(m+m+m−1) = m+m+m−1,
−β β2 β3 β4 β2 β3 β4
−
i.e., τd 1(−1,0,0,−1,0,0,...,0) = (−1,0,0,−1,0,0,...,0). For even d ≥ 2, we use the
r,−α
second condition of Theorem 1, or that τr,α(−1,...,−1) = (−1,...,−1).
Therefore, β has Property (−F) if and only if d ∈ {1,3,5}.
4 Addition and subtraction
In this section, we consider the lengths of fractional parts arising in the addition and
subtraction of (−β)-integers; we prove the following theorem.
Theorem 9. Let β > 1 be a root of x3 −mβ2 −mβ −m, m ≥ 1. We have
3 if m = 1 or m is even,
max{fr(x±y) : x,y ∈ Z } = 3m+
β
− (4 if m ≥ 3 is odd.
Throughout the section, let β be as in Theorem 9, r = (r ,r ) = (m,−m − m) and
0 1 β β β2
α = β . Recallthatx,y ∈ Z meansthatTk ( x ) = 0 = Tk ( y ), andfr(x±y) = n
β+1 β β ( β)k β ( β)k
is the minimal n ≥ 0 such tha−t Tk+n( x y ) =−0, w−ith k ≥ 0 suc−h th−at x , y , x y ∈
β ( ±β)k ( β)k ( β)k ( ±β)k
− − − − −
(ℓ ,ℓ +1). To determine fr(x−y), set
β β
s = Tj ( x y )+Tj ( y )−Tj ( x )
j β ( −β)k β ( β)k β ( β)k
− − − − − −
for j ≥ 0. Then we have s = Tj ( x y ) for j ≥ k, and, for all j ≥ 0,
j β ( −β)k
− −
s ∈ −βs +B with B = −A−A+A = {−2m,−2m+1,...,m},
j+1 j
s ∈ [ℓ ,ℓ +1)+[ℓ ,ℓ +1)−[ℓ ,ℓ +1) = (ℓ −1,ℓ +2).
j β β β β β β β β
As s = 0, we have s ∈ Z[β] for j ≥ 0. Therefore, we extend the bijection φ : Z2 →
0 j
Z[β]∩[ℓ ,ℓ +1) to
β β
Φ : Z2 ×{−1,0,1} → Z[β]∩[ℓ −1,ℓ +2), (z,h) 7→ rz−⌊rz+α⌋+h.
β β
Note that Φ(z,0) = φ(z).
Lemma 10. Let z = (z ,z ) ∈ Z2, h ∈ {−1,0,1} and b ∈ B. Then
0 1
−βΦ(z,h)+b = Φ z ,h−⌊rz+α⌋, z −z −h+⌊rz+α⌋ m+ r z +r h−r ⌊rz+α⌋+α +b .
1 1 0 0 1 1 1
(cid:0) (cid:0) (cid:1) (cid:4) (cid:5) (cid:1)
9
Proof. We have
z m
−βΦ(z,h)+b = −z m+z m+ 1 +⌊rz+α⌋β −hβ +b
0 1
β
= r z +r h−⌊rz+α⌋ + z −z −h+⌊rz+α⌋ m+b.
0 1 1 1 0
(cid:0) (cid:1) (cid:0) (cid:1)
Hence, we have sj ∈ Φ(τ˜rj,α(0,0)), where τ˜r,α extends τr,α to a set-valued function by
τ˜r,α : Z2 ×{−1,0,1} → P(Z2 ×{−1,0,1}), (z,h) 7→ (z1,h−⌊rz+α⌋,h′) :
h ∈ {−1,0,1}∩ z −z −h+⌊rz+α⌋ m+ r z +r h−r ⌊rz+α⌋+α +B .
′ 1 0 0 1 (cid:8) 1 1
To give a bound for(cid:0)t(cid:0)he sets τ˜j (0,0), let (cid:1) (cid:4) (cid:5) (cid:1)(cid:9)
r,α
A = {(j,k) : −1 ≤ j < k}, B = {(k,j) : 1 ≤ j ≤ k}, C = {(j,j −k) : 1 ≤ j ≤ k},
k k k
D = {(−j,−k) : 0 ≤ j < k}, E = {(−k,−j) : 2 ≤ j ≤ k},
k k
F = {(−j,k −j) : 2 ≤ j ≤ k +1}.
k
Then {A ,B ,C ,D ,E ,F } forms a partition of Z2 \ {(0,0),(−1,−1)}, with the
k 0 k k k k k k
sets B , C≥ , D , E , F , and E being empty, see Figure 1. If m ≥ 2, then let
0 0 0 0 0 1
S
V = A ∪B ∪C ∪D ∪E ∪F ×{−1,0,1} \{(−1,m,1),(0,m,1)}
k k k k k k
(cid:18)0 k m (cid:19)
≤[≤ (cid:0) (cid:1)
∪ C \{(m+1,0)} ×{1} ∪ D ×{1} ∪ D \{(0,−m−1)} ×{0}
m+1 m+1 m+1
∪(cid:16)D(cid:0) \{(0,−m−1),(cid:1)(−1,−m(cid:17) −(cid:16)1),(−2,−m−(cid:17) 1)(cid:16)}(cid:0)×{−1} (cid:1) (cid:17)
m+1
∪(cid:0) {(0,0),(−1,−1)}∪E \{(−m−1,−m−1)}(cid:1)×{−1,0,1}
m+1
(cid:16) (cid:17)
∪ (cid:0)F \{(−m−2,−1),(−(cid:1)m−1,0)} ×{−1,0} .
m+1
(cid:16) (cid:17)
(cid:0) (cid:1)
If m = 1, then we add the point (−2,0,−1) to this set, i.e.,
V = (0,0),(1,1),(1,0),(0,−1),(−1,−1),(−1,0),(−2,−1) ×{−1,0,1}
∪ (−1,1),(0,1) ×{−1,0} ∪ {(−1,−2)}×{0,1} ∪ (1,−1,1),(0,−2,1),(−2,0,−1) .
(cid:0)(cid:8) (cid:9) (cid:1)
W(cid:0)e(cid:8)call a point z(cid:9)∈ Z2 full i(cid:1)f {(cid:0)z}×{−1,0,1} ⊂ V(cid:1). (cid:8) (cid:9)
The following result is the key lemma of this section.
Lemma 11. Let x,y ∈ [ℓ ,ℓ + 1) such that x − y ∈ [ℓ ,ℓ + 1). Then Tj (x − y) +
β β β β β
Tj (y)−Tj (x) ∈ Φ(V) for all j ≥ 0. −
β β
− −
To prove Lemma 11, we first determine the value of ⌊rz+α⌋ for (z,h) ∈ V.
10