This document discusses continued β-fractions in the field of Laurent series over a finite field Fq. Specifically: 1) It introduces the concept of β-expansions and continued β-fractions where the base β is a unit Pisot series in Fq((x-1)). 2) It summarizes previous work characterizing elements of Fq((x-1)) having finite β-fractions when β is a quadratic Pisot unit. 3) The main result of the paper is to improve upon this by studying the case when β is a Pisot unit (not necessarily quadratic) in Fq((x-1)).

1. *Corresponding Author: Rania Kammoun, Email: raniakammoun32@gmail.com
RESEARCH ARTICLE
Available Online at www.ajms.in
Asian Journal of Mathematical Sciences 2017; 1(6):230-233
Continued 𝜷-fractions with Pisot unit base in 𝑭𝒒 ((𝒙−𝟏))
Rania Kammoun*
* University of Sfax, Faculty of Sciences, Department of Mathematics, Algebra Laboratory, Geometry and Spectral
Theory (AGTS) LR11ES53, BP 802, 3038 Sfax, Tunisia.
Receivedon:15/11/2017,Revisedon:01/12/2017,Acceptedon:29/12/2017
ABSTRACT
In this paper, we are interested in introducing a new theory of continued fractions based on the beta-
expansion theory in the field of Laurent series over a finite field 𝐹𝑞. We will characterize all elements
having finite continued beta-fraction where the base is a unit Pisot quadratic series.
Classification Mathematic Subject: 11R06, 37B50.
Key words: Continued 𝛽-fraction, formal power series, Pisot series, 𝛽-expansion, finite field.
INTRODUCTION
The 𝛽-numeration introduced in 1957 by Rényi [5]
is a new numeration system when we replace the
integer base b with a non-integral base. Let 𝛽 > 1, in the case of a non-integral base, one may write any
𝑥 ∈ [0,1] as 𝑥 = ∑𝑘≥1
𝑥𝑘
𝛽𝑘 , where 𝑥𝑘 ∈ {0, ⋯ , [𝛽]}. The sequence (𝑥𝑘)𝑘≥1 is called an expansion of 𝑥 in
𝛽 base. There is no expansion uniqueness but, among them, the greatest sequence for the lexicographical
order is called the 𝛽-expansion of 𝑥and it is denoted by 𝑑𝛽(𝑥).
The 𝛽-expansion of 𝑥 is constructed by the greedy following algorithm. We consider the 𝛽-
transformation
𝑇𝛽: [0,1] → [0,1], 𝑥 → {𝛽𝑥} = 𝛽𝑥 − [𝛽𝑥]
and then we define
(𝑥𝑘)𝑘≥1 = 𝑑𝛽(𝑥) ≔ 𝑥1𝑥2𝑥3 ⋯, where 𝑥𝑘 = [𝛽𝑇𝛽
𝑘−1
(𝑥) ].
In the case 𝑥 ≥ 1, there exists a unique integer 𝑖 such that 𝛽𝑖−1
≤ 𝑥 < 𝛽𝑖
. So one can write
𝑥
𝛽𝑖 = ∑
𝑦𝑘
𝛽𝑘
𝑘≥1 ,
where (𝑦𝑘)𝑘≥1 is the 𝛽-expansion of
𝑥
𝛽𝑖 . Thus, we have
𝑥 = ∑ 𝑥𝑘𝛽−𝑘
∞
𝑘=−𝑛
𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡 𝑥𝑘 = 𝑦𝑘−𝑛.
The 𝛽-integer part of 𝑥 is [𝑥]𝛽 = ∑ 𝑥𝑘𝛽−𝑘
∞
𝑘=−𝑛 and the 𝛽-fractional part of 𝑥is {𝑥}𝛽 = ∑ 𝑥𝑘𝛽−𝑘
𝑘>0 .
When {𝑥}𝛽 = 0, we denote by ℤ𝛽 the set of all 𝛽-integers.
Obviously, we can present an algorithm of continued fractions similarly to the classical decimal case by
consideration 𝛽 ∈ ℝ (non-integer) and then we get the so called continued 𝛽-fraction, whither the
sequence of partial quotients consists of 𝛽-integers instead of integers.
In [2]
, J. Bernat has showed that the continued 𝜙-fraction of 𝑥 is finite if and only if 𝑥 ∈ ℚ(𝜙). In [4]
, we
have studied the continued 𝛽-fraction with formal power series over finite fields and we have
characterize elements of 𝔽𝑞((𝑥−1
)) having finite 𝛽-fraction when the base 𝛽 is a quadratic Pisot unit.
Throughout this paper, we improve the result given [4]
by studying the case when 𝛽 is only a Pisot unit in
𝔽𝑞((𝑥−1
)). The paper is organized as follows, Section 2, we introduce some basic definitions and results.
In Section 3, we define the continued 𝛽-fraction expansion. In Section 4, we state our main result.

2. Kammoun Rania et al. Continued 𝜷-fractions with Pisot unit base in 𝑭𝒒 ((𝒙−𝟏
))
© 2017, AJMS. All Rights Reserved. 231
Fields of Formal series 𝔽𝒒((𝒙−𝟏
))
Let 𝔽𝑞 be the field with 𝑞 elements, 𝔽𝑞[𝑥] the ring of polynomials with coefficient in 𝔽𝑞, 𝔽𝑞(𝑥) the field
of rational functions, 𝔽𝑞(𝑥, 𝛽) the minimal extension of 𝔽𝑞containing 𝑥 and 𝛽 and by 𝔽𝑞[𝑥, 𝛽] the
minimal ring containing 𝑥 and 𝛽. Let 𝔽𝑞((𝑥−1
)) be the field of formal power series of the form:
𝑓 = ∑ 𝑓𝑘𝑥𝑘
𝑙
𝑘=−∞
, 𝑓𝑘 ∈ 𝔽𝑞,
where 𝑙 = deg(𝑓) ≔ {
max{𝑘: 𝑓𝑘 ≠ 0} for 𝑓 ≠ 0;
−∞ for 𝑓 = 0.
Define the absolute value |𝑓| = {
𝑞deg(𝑓 )
for 𝑓 ≠ 0;
0 for 𝑓 = 0.
As |. | is not Archimedean, it satisfies the strict triangle inequality
|𝑓 + 𝑔| ≤ max(|𝑓|, |𝑔|) and |𝑓 + 𝑔| = max(|𝑓|, |𝑔|) if |𝑓| ≠ |𝑔|.
Let ∈ 𝔽𝑞((𝑥−1
)) , the polynomial part of 𝑓 is [𝑓] = ∑ 𝑓𝑘𝑥𝑘
𝑘≥0 . We know that the empty sum is always
equal zero. Therefore, the fraction part is [𝑓] ∈ 𝔽𝑞[𝑥] and {𝑓} = 𝑓 − [𝑓] is in the unit disk 𝐷(0,1).
An element 𝛽 ∈ 𝔽𝑞((𝑥−1
)) is called a Pisot element if it is an algebraic integer over 𝔽𝑞[𝑥], [𝛽] >
1 and|𝛽𝑖| < 1 for all conjugates 𝛽. Using the coefficient of minimal polynomial, P. Batman and A.L.
Duquette [1] had characterized the Pisot elements in 𝔽𝑞((𝑥−1
)) :
Theorem 2.1.Let 𝛽 ∈ 𝔽𝑞((𝑥−1
))be an algebraic integer over 𝔽𝑞[𝑥] with the minimal polynomial 𝑃(𝑦) =
𝑦𝑛
− 𝐴1𝑦𝑛−1
− ⋯ − 𝐴𝑛 , 𝐴𝑖 ∈ 𝔽𝑞[𝑥].
Then, 𝛽 is a Pisot elements if and only if |𝐴1| > max
2≤i≤n
|𝐴𝑖|.
Let 𝛽 ∈ 𝔽𝑞((𝑥−1)) with |𝛽| > 1. A 𝛽-representation of 𝑓 is an infinite sequences(𝑑𝑖)𝑖≥1, where 𝑑𝑖 ∈
𝔽𝑞[𝑥] and 𝑓 = ∑
𝑑𝑖
𝛽𝑖
𝑖≥1 . A 𝛽-expansion of 𝑓, denoted 𝑑𝛽(𝑓) = (𝑑𝑖)𝑖≥1 , is a 𝛽-represenation of 𝑓 such
that:
𝑑𝑖 = [𝛽𝑇𝛽
𝑖−1(𝑓)] where𝑇𝛽: 𝐷(0,1) → 𝐷(0,1) 𝑓 → 𝛽𝑓 − [𝛽𝑓]. (1)
The 𝛽-expansion can be computed by the following algorithm:
𝑟0 = 𝑓 and for 𝑖 ≥ 1 𝑑𝑖 = [𝛽𝑟𝑖−1], 𝑟𝑖 = 𝛽𝑟𝑖−1 − 𝑑𝑖.
The 𝛽-expansion𝑑𝛽(𝑓) is finite if and only if there is 𝑘 ≥ 0 such that 𝑑𝑖 = 0 for all 𝑖 ≥ 𝑘. It is called
ultimately periodic if and only if there is some smallest 𝑝 ≥ 0 (the pre-period length) and 𝑠 ≥ 1 (the
period length) for which 𝑑𝑖+𝑠 = 𝑑𝑖 for all 𝑖 ≥ 𝑝 + 1. Using the last notion, let:
𝐹𝑖𝑛(𝛽) = {𝑓 ∈ 𝔽𝑞((𝑥−1)): 𝑑𝛽(𝑓)is finite}
and
𝑃𝑒𝑟(𝛽) = {𝑓 ∈ 𝔽𝑞((𝑥−1)): 𝑑𝛽(𝑓)is eventuallly periodic}.
When 𝑑𝛽(𝑓) = 𝑑1𝑑2 ⋯ 𝑑𝑙+1 · 𝑑𝑙+2 ⋯ 𝑑𝑚 then, we denote by deg(𝑓)𝛽 = 𝑙 and ord(𝑓) = 𝑚.
For |𝑓| ≥ 1, then there is a unique 𝑘 ∈ 𝑁 such that |𝛽|𝑘
≤ |𝑓| ≤ |𝛽|𝑘+1
. So we have |
𝑓
𝛽𝑘+1
| < 1 and we
can represent 𝑓 by shifting 𝑑𝛽(
𝑓
𝛽𝑘+1) by 𝑘 digits to the left. Thus, if 𝑑𝛽(𝑓) = 0. 𝑑1𝑑2 ⋯ , then𝑑𝛽(𝛽𝑓) =
𝑑1. 𝑑2𝑑3 ⋯
Remark 2.1. There is no carry occurring, when we add two polynomials in 𝔽𝑞[𝑥] with degree less than
deg(𝛽). Consequently, if 𝑓, 𝑔 ∈ 𝔽𝑞((𝑥−1
)), we get 𝑑𝛽(𝑓 + 𝑔) = 𝑑𝛽(𝑓) + 𝑑𝛽(𝑔).
In [6], Scheicher has characterized the set 𝐹𝑖𝑛 (𝛽) when 𝛽 is Pisot.
Theorem 2.2.[6] 𝛽 is a Pisot series if and only if 𝐹𝑖𝑛(𝛽) = 𝔽𝑞[𝑥, 𝛽−1].

3. Kammoun Rania et al. Continued 𝜷-fractions with Pisot unit base in 𝑭𝒒 ((𝒙−𝟏
))
© 2017, AJMS. All Rights Reserved. 232
Let 𝑓 ∈ 𝔽𝑞((𝑥−1)), the 𝛽-polynomial part of 𝑓 is [𝑓]𝛽 = ∑ 𝑑𝑖𝛽𝑙−𝑖+1
𝑙+1
𝑖=1 and the 𝛽-fractional part is
{𝑓}𝛽 = 𝑓 − [𝑓]𝛽 = ∑ 𝑑𝑖𝛽𝑙−𝑖+1
𝑖>𝑙+1 . We define the set of 𝛽-polynomials as follows:
𝔽𝑞[𝑥]𝛽 = {𝑓 ∈ 𝔽𝑞((𝑥−1)); {𝑓}𝛽 = 0}.
Then, clearly 𝔽𝑞[𝑥]𝛽 ⊆ 𝔽𝑞[𝑥, 𝛽]. Furthermore, we introduce the following set
(𝔽𝑞[𝑥])
′
= {𝑃 ∈ 𝔽𝑞[𝑥]𝜷, deg(𝑃) ≤ deg(𝛽) − 1} = {𝑃 ∈ 𝔽𝑞[𝑥]𝜷, 𝑑𝑒𝑔𝛽(𝑃) = 0}.
The set of power series that can be written as a fraction of two 𝛽-polynomials denoted by 𝔽𝑞(𝑥)𝛽. Then,
clearly 𝔽𝑞[𝑥]𝛽 ⊆ 𝔽𝑞(𝑥, 𝛽). In [3], the authors studied the quantity 𝐿⊙ and they define as follows:
𝐿⊙ = min{𝑛 ∈ ℕ: ∀ 𝑃1, 𝑃2 ∈ 𝔽𝑞[𝑥]𝛽; 𝑃1𝑃2 ∈ 𝐹𝑖𝑛(𝛽) ⇒ 𝛽𝑛(𝑃1𝑃2) ∈ 𝔽𝑞[𝑥]𝛽 }.
Theorem 2.3.[3] Let 𝛽be a quadratic Pisot unit series. Then 𝐿⊙ = 1.
Continued 𝜷-fraction algorithm
We begin by introduce a generalization of the algorithm of the expansion in continued fraction in the
field of formal power series in base 𝛽 ∈ 𝔽𝑞((𝑥−1)) with |𝛽| > 1.When 𝛽 = 𝑥, this theory is seems to be
similar to the classical case of continued fractions.
We define the 𝛽-transformation 𝑇𝛽
′
by:
𝑇𝛽
′
: 𝐷(0,1) → 𝐷(0,1)
𝑓 →
1
𝑓
− [
1
𝑓
]
𝛽
.
when |𝑓| < 1, we obtain
𝑓 =
1
𝐴1 +
1
𝐴2 +
1
⋱
= [0,𝐴1, 𝐴2, ⋯ ]𝛽
whither (𝐴𝑘)𝑘≥1 ∈ 𝔽𝑞[𝑥]𝛽and there are defined by 𝐴𝑘 = [
1
𝑇′
𝛽
𝑘−1
(𝑓)
]
𝛽
, ∀ 𝑘 ≥ 1.
For 𝑓 ∈ 𝔽𝑞((𝑥−1)) and 𝐴0 = [𝑓]𝛽, we get
𝑓 = 𝐴0 +
1
𝐴1 +
1
𝐴2 +
1
⋱
= [𝐴0, 𝐴1, 𝐴2, ⋯ ]𝛽.
The last bracket is called continued 𝛽-fraction expansion of 𝑓. The sequence (𝐴𝑘)𝑘≥0 is called the
sequence of partial 𝛽-quotients of 𝑓. We define the 𝑛𝑡ℎ
𝛽-complete quotient of 𝑓 by 𝑓
𝑛 =
[𝐴0, 𝐴1, 𝐴2, ⋯ , 𝑓
𝑛]𝛽. We remark that all (𝐴𝑘)𝑘≥1are not in 𝔽𝑞.
Main Results
Our main result is an improvement of Theorem 4.1 in [4]
.
Theorem 4.1. Let 𝛽 be a quadratic Pisot unit formal power series over the finite field 𝔽𝑞 such that
deg(𝛽) = 𝑚. Let 𝛽 ∈ 𝔽𝑞(𝑥, 𝛽) such that the continued 𝛽-fraction of 𝑓 is given by 𝑓 =
[𝐴0, 𝐴1, 𝐴2, ⋯ , 𝐴𝑛, ⋯ ]. If 𝑓 ∈ 𝔽𝑞(𝑥, 𝛽) then {𝐴𝑖/ deg𝛽(𝐴𝑖) > 0} is finite.
So as to prove the above Theorem, first we need to recall some results given in [4]
and we use the
following Lemmas and Propositions.
Lemma 4.2. [4] Let 𝛽 be a unit Pisot series. Then 𝔽𝑞(𝑥, 𝛽) = 𝔽𝑞(𝑥)𝛽.
Now, we define two sequences (𝑃
𝑛)𝑛∈ℕ and (𝑄𝑛)𝑛∈ℕ in 𝔽𝑞[𝑥, 𝛽] by
{
𝑃0 = 𝑎0, 𝑃1 = 𝑎0𝑎1 + 1
𝑄0 = 1 , 𝑄1 = 𝑎1
and {
𝑃
𝑛 = 𝑎𝑛𝑃𝑛−1 + 𝑃𝑛−2
𝑄𝑛 = 𝑎𝑛𝑄𝑛−1 + 𝑄𝑛−2, ∀𝑛 ≥ 2
The pair (𝑃𝑛, 𝑄𝑛) is called reduced 𝛽-fractionary expansion of 𝑓 for all 𝑛 ≥ 0.
Proposition 4.1. Let 𝑓 ∈ 𝔽𝑞(𝑥, 𝛽) such that 𝑓 = [𝐴0, 𝐴1, 𝐴2, ⋯ , 𝐴𝑛, ⋯ ]. Then |𝑓 −
𝑃𝑛
𝑄𝑛
| <
1
|𝑄𝑛|2.
Proof. Similarly to the classical case.

4. Kammoun Rania et al. Continued 𝜷-fractions with Pisot unit base in 𝑭𝒒 ((𝒙−𝟏
))
© 2017, AJMS. All Rights Reserved. 233
Proposition 4.2. Let 𝛽 be a quadratic Pisot unit power formal series such that deg(𝛽) = 𝑚 and
𝑃1, 𝑃2, ⋯ , 𝑃𝑚 ∈ 𝔽𝑞[𝑥]𝛽. Then, 𝛽𝑚−1
𝑃1𝑃2 ⋯ 𝑃𝑚 ∈ 𝔽𝑞[𝑥]𝛽.
The proof of the last proposition is an immediate consequence of Thoerem 2.3.
Corollary 4.3. Let 𝑃1, 𝑃2, ⋯ , 𝑃
𝑚 ∈ 𝔽𝑞[𝑥]𝛽. Then we have, for all positive integer
𝑛, 𝛽
(𝑚−1)𝑛
𝑚 𝑃1𝑃2 ⋯ 𝑃𝑚 ∈ 𝔽𝑞[𝑥]𝛽.
Corollary 4.4.Let (𝑃𝑛, 𝑄𝑛)𝑛≥0 the reduced 𝛽-fractionary expansion of 𝑓 . Then 𝛽
(𝑚−1)𝑛
𝑚 𝑃
𝑛 ∈ 𝔽𝑞[𝑥]𝛽 and
𝛽
(𝑚−1)𝑛
𝑚 𝑄𝑛 ∈ 𝔽𝑞[𝑥]𝛽.
For 𝑃 = 𝑎𝑠𝛽𝑠
+ ⋯ + 𝑎0 ∈ 𝔽𝑞[𝑥]𝛽. We denote by 𝛾(𝑃) = 𝑚 deg𝛽(𝑃) + deg𝑠(𝑎𝑠) = 2𝑚 +
deg(𝑎𝑠).
Lemma 4.5. [4] Let 𝐴, 𝐵 ∈ 𝔽𝑞[𝑥]𝛽 with 𝛾(𝐴) > 𝛾(𝐵). Then there exists 𝐶, 𝐴1 and 𝐵1in 𝔽𝑞[𝑥]𝛽, such
that
𝐴
𝐵
= 𝐶 +
1
𝐴1
𝐵1
with 𝛾(𝐴1) > 𝛾(𝐵1).
Proof of Theorem 4.1
It is equivalent to prove that there exist 𝑛0 ≥ 1, 𝐴𝑛 ∈ (𝔽𝑞[𝑥]𝛽) ′. By Lemma 4.2, we obtain 𝑓 =
𝑃
𝑄
∈
𝔽𝑞(𝑥)𝛽 such as 𝑃, 𝑄 ∈ 𝔽𝑞[𝑥]𝛽 and (𝑃
𝑛, 𝑄𝑛) the reduced 𝛽-fractionary expansion of 𝑓.
By proposition 4.1, |
𝑃
𝑄
−
𝑃𝑛
𝑄𝑛
| <
1
|𝑄𝑛|2 . According to Corollary 4.3 and Lemma 4.5, we have
(𝛽
(𝑚−1)
𝑚
(𝑛+1)
(𝑃𝑄𝑛 − 𝑄𝑃
𝑛)) in𝔽𝑞[𝑥]𝛽 .So, we obtain
1
|𝑄|
< |𝛽|(𝑚−1)(𝑛+1)/𝑚
|
𝑃𝑄𝑛 − 𝑄𝑃
𝑛
𝑄
| <
|𝛽|
(𝑚−1)(𝑛+1)
𝑚
|𝑄𝑛|
which implies that deg(𝑄𝑛) ≤ deg(𝑄) + (𝑚 − 1)(𝑛 + 1), where
deg(𝑄𝑛) = ∑ deg(𝐴𝑖)
𝑛
𝑖=1
≤ deg(𝑄) + (𝑚 − 1)(𝑛 + 1).
Thus ∑ (deg(𝐴𝑖) − (𝑚 − 1)) ≤ deg(𝑄) + (𝑚 − 1)
𝑛
𝑖=1 . Finally there exists 𝑛0 ≥ 1, such that, for
deg(𝐴𝑖) − (𝑚 − 1) ≤ 0, for all 𝑖 ≥ 𝑛0 and the desired result is reached.
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