POSITIVE FINITENESS OF NUMBER
SYSTEMS
Shigeki Akiyama
Department of Mathematics, Faculty of Science, Niigata University
Ikarashi 2-8050, Niigata 950-2181, Japanakiyama@math.sc.niigata-u.ac.jp

Abstract

We characterize the set of β’s that each polynomial in base β with nonnegative integer coefficients has a finite admissible expression in somenumber systems.

Keywords: Beta expansion, Canonical number system, Pisot number

1.

Introduction

In this note, we study a certain finiteness property of number systems
given by power series in some base β, which are called beta-expansion
and canonical number system.
In relation to symbolic dynamics, an important problem is to determine the set of β’s that each polynomial in base β with non-negative
integer coefficients has a finite expression in the corresponding number
system. However this problem may be pretty difficult in general. We
narrow our scope on the set of such β’s which does not have ‘global’finiteness. Let us explain exactly the problem for beta-expansion (c.f.
[27]).
Let β > 1 be a real number. Each positive x is uniquely expanded
into a beta-expansion:∞Xx=ai β −i(M could be negative)

i=M

under conditions
ai ∈ [0, β) ∩ Z

and∀L ≥ M

0≤x−

L
X
i=M

ai β −i < β −L ,2which is also called greedy expansion. We write this expression asx = xM xM +1 . . . x0 .x1 x2 . . .
following an analogy to the usual decimal expansion. If ai = 0 for sufficiently large i, then the expansion is called finite and the tail 00 . . . canbe omitted as usual. Let Fin(β) be the set of finite beta expansions. It
is obvious that Fin(β) is a subset of Z[1/β]∩[0, ∞) if β were an algebraic
integer1 . Frougny and Solomyak [14] firstly studied the property
Fin(β) = Z[1/β] ∩ [0, ∞)which we call finiteness property (F). If β has the property (F), then β
is a Pisot number, that is, a real algebraic integer greater than one that
all other conjugates of β have modulus less than one.
A polynomial xd −ad−1 xd−1 −· · ·−a0 with ad−1 ≥ ad−2 ≥ · · · ≥ a0 > 0gives a Pisot number β > 1 as a root (c.f. [10]). Then in [14] it is shownthat the property (F) holds for this class of β. The full characterizationof β with (F) among algebraic integers (or among Pisot numbers), is adifficult problem when d ≥ 3 (c.f. [2], [8], [4]).
The expansion of 1 is a digit sequence given by an expression 1 =P∞−i = .c c c . . . such that .0c c . . . is the beta expansion of
1 2 3
2 3
i=1 ci β
1 − c1 /β with c1 = bβc. This expansion play a crucial role to determine
which formal expression could be realized as beta-expansion ([25], [18]).
Especially a formal expression
1=∞X

di β −i = .d1 d2 . . .

i=1

coincides with the expansion of 1 if and only if the digit sequence d1 d2 . . .
is greater than its left shift di di+1 . . . for i > 1 by the natural lexicographical order.
In [14] it is shown that if the expansion of 1 = .c1 c2 . . . has infinite
decreasing digits ( i.e., c1 ≥ c2 ≥ c3 ≥ . . . and ci = ci+1 > 0 fromsome index on), then the set Fin(β) is closed under addition. This isequivalent to the condition:Z+ [β] ⊂ Fin(β)
where Z+ = Z ∩ [0, ∞) and Z+ [β] is the set of polynomials in base βwith coefficients in Z+ . We call this property positive finiteness ((PF)
for short). The author showed in [3] that (PF) implies weak finiteness1 If β is an algebraic integer, thenZ[β] ⊂ Z[1/β].

3Positive finiteness of number systems

which has close connection to Thurston’s tiling generated by Pisot unit
β (c.f. [30], [8]). One motivation to study (PF) comes from this fact.
In [9], Ambrož, Frougny, Masáková and Pelantová gave a characterization of (PF) in terms of ‘transcription’ of minimal forbidden factors.
Our problem in this paper is to characterize β with the property (PF)without (F). By this restriction of the scope, we can give a completecharacterization of such β’s:
Theorem 1. Let β > 1 be a real number with positive finiteness. Theneither β satisfies the finiteness property (F) or β is a Pisot number whose
minimal polynomial is of the form:
xd − (1 + bβc)xd−1 +dXai xd−ii=2

Pdwith ai ≥ 0 (i = 2, . . . , d), ad > 0 and i=2 ai < bβc. In the later case,the expansion of 1 has infinite decreasing digits. Conversely if β > 1 isa root of the polynomial
xd − Bxd−1 +dXai xd−ii=2Pwith ai ≥ 0, ad > 0 and B > 1 + di=2 ai , then this polynomial isirreducible and β is a Pisot number with (PF) without (F). We also
have B = 1 + bβc.
The study of (PF) is reduced to that of (F) by Theorem 1. Unfortunately as a result, nothing new exists in (PF) but the ones alreadyfound in [14].
A parallel problem is solved in another well known number system.
Let α be an algebraic integer 2 of degree d having its absolute norm
|N (α)|. If each element x ∈ Z[α] has an expression:x=X̀ai α i ,ai ∈ A = {0, 1, . . . , |N (α)| − 1}i=0then we say that α gives a canonical number system (CNS for short). If
such expression exists, then it is unique since A forms a complete setof representatives of Z[α]/αZ[α] and the digit string is computed from2 α is used instead of β to distinguish the difference of number systems.

4
the bottom by successive consideration modulo α. If α gives a CNS,
then α must be expanding, that is, all conjugates of α have modulus
greater than one ([22]). Assume that α has the minimal polynomial of
the form x2 + Ax + B. Then α gives a CNS if and only if −1 ≤ A ≤ B
and B ≥ 2 ([19], [20], [15]). When d ≥ 3, the characterization of α’samong expanding algebraic integers is again a difficult question ([6],[28],[7],[11],[12], [5]). It is obvious that CNS is an analogous concept of(F). To pursue this analogy, let us say that α has positive finiteness if
Z+ [α] = A[α], i.e.,XX∀0 ≤ ai ∈ Z ∃bj ∈ {0, 1, . . . , |N (α)| − 1}ai αi =
bj αj .
i≥0j≥0This positive finiteness is in fact weaker than CNS and we can show
Theorem 2. Assume that α has positive finiteness. Then either α gives
a CNS or the minimal polynomial of α is given bydXai xi − C

(1)

i=1

P
with ad = 1, ai ≥ 0 and di=1 ai < C. Conversely if α is a root of theirreducible polynomial (1) with the same condition then α has positive
finiteness but does not give a CNS.
It is not possible to remove irreducibility in the last statement. For
example, x2 + x − 12 = (x − 3)(x + 4) but −4 gives a CNS.
In [26], Pethő introduced a more general concept ‘CNS polynomial’
among expanding polynomials. If the polynomial is irreducible, then theconcept coincides with CNS. It is straightforward to generalize above
Theorem 2 to this framework. In this extended sense, x2 + x − 12 has
positive finiteness.

2.

Proof of Theorem 1.

First we prove the later part of the Theorem 1. Assume that β > 1 isa root of a polynomial:
P (x) = xd −Bxd−1 +

d
X
i=2

ai xd−iwith ai ≥ 0, ad > 0 and B > 1+

d
X

ai .

i=2

By applying Rouché’s Theorem, P (x) and xd − Bxd−1 has the same
number of roots in the open unit disk. Thus β is a Pisot number andP (x) is irreducible. In fact, if P (x) is non trivially decomposed intoPositive finiteness of number systems

5

P1 (x)P2 (x) and P1 (β) = 0, then the constant term of P2 (x) is less than
1 in modulus, and hence it must vanishes. This contradicts ad > 0.
The relation P (β) = 0 formally gives rise to a relation
1 = .B a2 a3 . . . ad
where we put x = −x to simplify the notation. Multiplying β −j (j =1, 2, . . . ) and summing up we have
1 =
+
+
+
=

.B a2 a3 . . . ad
.1 B a2 . . . ad−1 ad.0 1 B . . . ad−2 ad−1 ad....
.(B − 1) (B − 1 − a2 ) (B − 1 − a2 − a3 ) . . . m m m . . .
Pwith m = B − 1 − di=2 ai . As the last sequence is lexicographicallygreater than its left shifts, this gives the expansion of 1 of β with infinite
decreasing digits. By the result of [14], this β has the property (PF).
Now it is clear that B = 1 + bβc. As the expansion of 1 is not finite,
β does not satisfy (F). This is also shown in the following way. Since
P (0) < 0 and P (1) > 0, there is a positive conjugate β 0 ∈ (0, 1). Using
Proposition 1 of [1], β does not satisfy the finiteness property (F).
To prove the first part, we quote two lemmas.
Lemma 3 (Theorem 5 in Handelman [17]). Let β > 1 be an algebraic integer such that other conjugates has modulus less than β andthere are no other positive conjugates. Then β is a Perron-Frobenius
root of a primitive companion matrix.
The proof of this lemma relies on the Perron-Frobenius theorem andthe fact that for any polynomial p(x) without positive roots, (1+x)m p(x)have only positive coefficients for sufficiently large m. (A direct proof ofthis fact will be given in the appendix.) We need anotherLemma 4 (Lemma 2 in [14]). An equality Z+ [β] = Z[β]∩[0, ∞) holds
if and only if β is a Perron-Frobenius root of a primitive companion
matrix.
In the following,√we also use the fact that there are only two Pisot
numbers less than 2. The smallest one, say θ ≈ 1.32372, is a positiveroot of x3 − x − 1 and the next θ1 ≈ 1.38028 is given by x4 − x3 − 1 (c.f.
[24]). C.L. Siegel [29] firstly proved that they are the smallest two Pisot
numbers. In [1], it is shown that θ has property (F). On the other hand,θ1 does not satisfy (PF) since θ1 + 1 has the infinite purely periodic beta
expansion 100.0010000100001 . . . .

6Let us assume that β > 1 has positive finiteness (PF) but does nothave the√ property (F). This implies that β is not an integer and greaterthan 2. Since Z+ ⊂ Fin(β), Proposition 1 of [2] implies that β isa Pisot number. We claim that β has a conjugate β 0 ∈ (0, 1). If not,then by Lemma 3, β is a Perron-Frobenius root of a primitive companion
matrix. Then by Lemma 4, each element of Z[β]∩[0, ∞) has a polynomial
expression in base β with non-negative integer coefficients. Thus (PF)property implies the property (F). This is a contradiction which shows
the claim.
√
By the property (PF), κ = (1 + bβc)/β ∈ Fin(β). Note that β > 2
implies bβc + 1 < β 2 and the beta expansion of κ begins with a0 = 1.
Hence, as κ − 1 < β −1 , we have a beta expansion:
1 + bβc= 1.0a2 a3 . . . a`βPwith a` 6= 0. Set Q(x) = x` − (bβc + 1)x`−1 + `i=2 ai x`−i . Then Q(x)has two sign changes in its coefficients. By Descartes’s law, there exist
at most two positive real roots of Q(x), and therefore they must be β
and β 0 . On the other hand, we see Q(0) = a` > 0. If Q(1) > 0 thenthere are at least two positive root of Q(x) in (0, 1) which is absurd.
P
Thus we have Q(1) < 0 which implies `k=2 ak < bβc. We have alreadyproven under this inequality that Q(x) is irreducible and the expansionof 1 of β has infinite decreasing digits.
A few words should be added to make clear the situation. If bβc + 1
has a finite beta expansion in base β, the above
P procedure yields thesame polynomial Q(x) = x` − (1 + bβc)x`−1 + `k=2 ak x`−k . Since β > 1
is a root of Q(x) and Q(0) > 0, Q(x) has exactly two positive real roots.
If Q(1) < 0, then β has (PF) by the same reasoning. If Q(1) ≥ 0, thenthere is a root η ≥ 1 other than β. Note that this could happen even if β
has property (PF). However in such case, Q(x) must be reducible since β
does not have other positive conjugate if it has property (PF). Especially√
if β satisfies (F), then Q(x) is reducible. For example, β = (1 + 5)/2
satisfies (F) and Q(x) = x3 − 2x2 + 1 = (x2 − x − 1)(x − 1). The
above proof shows, as a consequence, that Q(x) must be irreducible if βsatisfies (PF) without (F).
It is not clear whether the condition Z+ ⊂ Fin(β) implies (PF). We
have difficulty in proving the existence of a positive conjugate β 0 ∈ (0, 1)
under this condition.

7

Positive finiteness of number systems

3.

Proof of Theorem 2.

First we recall that if α has positive finiteness, then α is expanding.
This was proved in CNS case in [22] and the same proof works in positive
finiteness case. (See Lemma 3 and the proof of Theorem 3 in [22]. ) 3
Let us assume that α has positive finiteness but does not give a CNS.
Let P (x) be the minimal polynomial of α. We claim that there exists a
positive conjugate α0 . Suppose not. Then by the remark after Lemma 3,
there is a large integer M that (1+x)M P (x) has only positive coefficients.
P`iThis gives a relation of the form
i=0 ai α = 0 with ai > 0. Thus
each element of Z[α] has an equivalent expression in Z+ [α] which isattained by repeated addition of the above relation. This shows thatZ+ [α] = Z[α] and positive finiteness of α implies that α gives a CNS.
This is a contradiction and the claim is proved. P
Note that α0 > 1.
Let C = |N (α)| and write its expression C = di=0 ai αi with ai ∈ A.
P
Taking modulo α, we see that a0 = 0. Set Q(x) = di=1 ai xi − C. As
Q(0) < 0 and there is only one sign change in the coefficients of Q(x),there exists exactly one positive root of Q(x) which is α0 . Now α0 > 1
Pimplies Q(1) < 0, i.e., di=1 ai < C. Suppose that Q(x) is not irreducible
and Q(x) = P (x)R(x) with deg R ≥ 1. From C = |N (α)|, we deduce|R(0)| = 1 and hence there exists a root η of Q(x) with |η| ≤ 1. Then
¯ d
¯
d
¯X
¯
X
¯
¯
i
0 = |Q(η)| = ¯
ai η − C ¯ ≥ C −
ai
¯
¯
i=1

i=1

gives a contradiction. This shows that Q(x) = P (x) and ad = 1.
Finally we prove the converse.
P Assume that α is a root of the irreducible polynomial Q(x) = di=1 ai xi − C with ad = 1, ai ≥ 0 andPd
i=1 ai < C. Then Q(x) must be expanding since otherwise there exists
a root η with |η| ≤ 1 of Q(x) and we shall meet the same contradiction.
As Q(0) < 0 there exists a positive conjugate α0 . Hence α can not give a
CNS, since −1 can not have finite expansion (c.f. Proposition 6 in [15]).
It remains to show that α has positive finiteness. The idea of this proof
can be traced back to [21].
As α is a root of Q(x), we have an expressionad ad−1 . . . a1 C = 0.
(2)
P
We describe an algorithm from each x = `i=0 di αi with di ∈ Z+ how
to get an equivalent expression in A[β]. Adding κ = bd0 /Cc times the
3 For the later use, it suffices to show an easier fact (Lemma 3 in [22]): ‘each conjugate of α

has modulus not less than one.’

8
relation (2), we have an equivalent expression of x in Z+ [α]:¡¢d` d`−1 . . . d0 + κ × ad ad−1 . . . a1 C = d0`0 d0`0 −1 . . . d00
whose constant term is d00 = d0 − κC ∈ A. Repeat the same process on
d01 to make the coefficients of α1 into A. This process can be continued
in a similar manner. In each step, the sum of digits of the expressionof x is strictly decreasing. Hence we finally get an expression in A[α] infinite steps.

Acknowledgments
The author express his most gratitude to Z.Masáková and E.Pelantová.
This work was motivated by [9] and the discussion with them while mystay in Prague TU. Thanks are also due to J.Thuswaldner with whom
we discussed on positive finiteness of expanding algebraic numbers.

9

Appendix A

Appendix
Handelman showed in [16], as a special case of his wide theory, that for any polynomial p(x) ∈ R[x] having no non-negative roots, there exists a positive integer M
that (1 + x)M p(x) has only positive coefficients (c.f. [23] and [13]). This is a crucial
fact in proving Lemma 3 and Theorem 2. As the statement itself looks elementary,
it may be worthy to note here a direct short proof. To prove this we factorize p(x)into quadratic and linear factors in R[x]. Since a linear factor (x + a) with a > 0
does no harm, we prove that for any x2 + bx + c with b2 < 4c there exists a positive n that (1 + x)n (x2 + bx + c) has positive coefficients. The k-th coefficient of(1 + x)n (x2 + bx + c) is
n
k

!

c

n−k
k−1
+b+
k+1
n−k+1



.

Thus we show that f (k) = c(n − k)(n − k + 1) + b(k + 1)(n − k + 1) + (k − 1)(k + 1) > 0
for k = 0, 1, . . . , n if n is suffciently large. From an expression
f (k) = −1 + b + bn + cn + cn2 + (−c + bn − 2cn)k + (1 − b + c)k2 ,as x2 + bx + c > 0 implies 1 − b + c > 0, the minimum of f (k) is attained whenk = (c − bn + 2cn)/(2 − 2b + 2c). Direct computation showsf (k) ≥

−4 + 8b − 4b2 − 4c + 4bc − c2 + (4b − 4b2 + 4c + 2bc)n + (−b2 + 4c)n2.
4(1 − b + c)

As the coefficient of n2 in the numerator is positive, the assertion is shown.

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