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- 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 π·-fractions with Pisot unit base in ππ ((πβπ))
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 πΉπ. 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 π½-fraction, formal power series, Pisot series, π½-expansion, finite field.
INTRODUCTION
The π½-numeration introduced in 1957 by RΓ©nyi [5]
is a new numeration system when we replace the
integer base b with a non-integral base. Let π½ > 1, in the case of a non-integral base, one may write any
π₯ β [0,1] as π₯ = βπβ₯1
π₯π
π½π , where π₯π β {0, β― , [π½]}. The sequence (π₯π)πβ₯1 is called an expansion of π₯ in
π½ base. There is no expansion uniqueness but, among them, the greatest sequence for the lexicographical
order is called the π½-expansion of π₯and it is denoted by ππ½(π₯).
The π½-expansion of π₯ is constructed by the greedy following algorithm. We consider the π½-
transformation
ππ½: [0,1] β [0,1], π₯ β {π½π₯} = π½π₯ β [π½π₯]
and then we define
(π₯π)πβ₯1 = ππ½(π₯) β π₯1π₯2π₯3 β―, where π₯π = [π½ππ½
πβ1
(π₯) ].
In the case π₯ β₯ 1, there exists a unique integer π such that π½πβ1
β€ π₯ < π½π
. So one can write
π₯
π½π = β
π¦π
π½π
πβ₯1 ,
where (π¦π)πβ₯1 is the π½-expansion of
π₯
π½π . Thus, we have
π₯ = β π₯ππ½βπ
β
π=βπ
π π’πβ π‘βππ‘ π₯π = π¦πβπ.
The π½-integer part of π₯ is [π₯]π½ = β π₯ππ½βπ
β
π=βπ and the π½-fractional part of π₯is {π₯}π½ = β π₯ππ½βπ
π>0 .
When {π₯}π½ = 0, we denote by β€π½ the set of all π½-integers.
Obviously, we can present an algorithm of continued fractions similarly to the classical decimal case by
consideration π½ β β (non-integer) and then we get the so called continued π½-fraction, whither the
sequence of partial quotients consists of π½-integers instead of integers.
In [2]
, J. Bernat has showed that the continued π-fraction of π₯ is finite if and only if π₯ β β(π). In [4]
, we
have studied the continued π½-fraction with formal power series over finite fields and we have
characterize elements of π½π((π₯β1
)) having finite π½-fraction when the base π½ is a quadratic Pisot unit.
Throughout this paper, we improve the result given [4]
by studying the case when π½ is only a Pisot unit in
π½π((π₯β1
)). The paper is organized as follows, Section 2, we introduce some basic definitions and results.
In Section 3, we define the continued π½-fraction expansion. In Section 4, we state our main result.
- 2. Kammoun Rania et al. Continued π·-fractions with Pisot unit base in ππ ((πβπ
))
Β© 2017, AJMS. All Rights Reserved. 231
Fields of Formal series π½π((πβπ
))
Let π½π be the field with π elements, π½π[π₯] the ring of polynomials with coefficient in π½π, π½π(π₯) the field
of rational functions, π½π(π₯, π½) the minimal extension of π½πcontaining π₯ and π½ and by π½π[π₯, π½] the
minimal ring containing π₯ and π½. Let π½π((π₯β1
)) be the field of formal power series of the form:
π = β πππ₯π
π
π=ββ
, ππ β π½π,
where π = deg(π) β {
max{π: ππ β 0} for π β 0;
ββ for π = 0.
Define the absolute value |π| = {
πdeg(π )
for π β 0;
0 for π = 0.
As |. | is not Archimedean, it satisfies the strict triangle inequality
|π + π| β€ max(|π|, |π|) and |π + π| = max(|π|, |π|) if |π| β |π|.
Let β π½π((π₯β1
)) , the polynomial part of π is [π] = β πππ₯π
πβ₯0 . We know that the empty sum is always
equal zero. Therefore, the fraction part is [π] β π½π[π₯] and {π} = π β [π] is in the unit disk π·(0,1).
An element π½ β π½π((π₯β1
)) is called a Pisot element if it is an algebraic integer over π½π[π₯], [π½] >
1 and|π½π| < 1 for all conjugates π½. Using the coefficient of minimal polynomial, P. Batman and A.L.
Duquette [1] had characterized the Pisot elements in π½π((π₯β1
)) :
Theorem 2.1.Let π½ β π½π((π₯β1
))be an algebraic integer over π½π[π₯] with the minimal polynomial π(π¦) =
π¦π
β π΄1π¦πβ1
β β― β π΄π , π΄π β π½π[π₯].
Then, π½ is a Pisot elements if and only if |π΄1| > max
2β€iβ€n
|π΄π|.
Let π½ β π½π((π₯β1)) with |π½| > 1. A π½-representation of π is an infinite sequences(ππ)πβ₯1, where ππ β
π½π[π₯] and π = β
ππ
π½π
πβ₯1 . A π½-expansion of π, denoted ππ½(π) = (ππ)πβ₯1 , is a π½-represenation of π such
that:
ππ = [π½ππ½
πβ1(π)] whereππ½: π·(0,1) β π·(0,1) π β π½π β [π½π]. (1)
The π½-expansion can be computed by the following algorithm:
π0 = π and for π β₯ 1 ππ = [π½ππβ1], ππ = π½ππβ1 β ππ.
The π½-expansionππ½(π) is finite if and only if there is π β₯ 0 such that ππ = 0 for all π β₯ π. It is called
ultimately periodic if and only if there is some smallest π β₯ 0 (the pre-period length) and π β₯ 1 (the
period length) for which ππ+π = ππ for all π β₯ π + 1. Using the last notion, let:
πΉππ(π½) = {π β π½π((π₯β1)): ππ½(π)is finite}
and
πππ(π½) = {π β π½π((π₯β1)): ππ½(π)is eventuallly periodic}.
When ππ½(π) = π1π2 β― ππ+1 Β· ππ+2 β― ππ then, we denote by deg(π)π½ = π and ord(π) = π.
For |π| β₯ 1, then there is a unique π β π such that |π½|π
β€ |π| β€ |π½|π+1
. So we have |
π
π½π+1
| < 1 and we
can represent π by shifting ππ½(
π
π½π+1) by π digits to the left. Thus, if ππ½(π) = 0. π1π2 β― , thenππ½(π½π) =
π1. π2π3 β―
Remark 2.1. There is no carry occurring, when we add two polynomials in π½π[π₯] with degree less than
deg(π½). Consequently, if π, π β π½π((π₯β1
)), we get ππ½(π + π) = ππ½(π) + ππ½(π).
In [6], Scheicher has characterized the set πΉππ (π½) when π½ is Pisot.
Theorem 2.2.[6] π½ is a Pisot series if and only if πΉππ(π½) = π½π[π₯, π½β1].
- 3. Kammoun Rania et al. Continued π·-fractions with Pisot unit base in ππ ((πβπ
))
Β© 2017, AJMS. All Rights Reserved. 232
Let π β π½π((π₯β1)), the π½-polynomial part of π is [π]π½ = β πππ½πβπ+1
π+1
π=1 and the π½-fractional part is
{π}π½ = π β [π]π½ = β πππ½πβπ+1
π>π+1 . We define the set of π½-polynomials as follows:
π½π[π₯]π½ = {π β π½π((π₯β1)); {π}π½ = 0}.
Then, clearly π½π[π₯]π½ β π½π[π₯, π½]. Furthermore, we introduce the following set
(π½π[π₯])
β²
= {π β π½π[π₯]π·, deg(π) β€ deg(π½) β 1} = {π β π½π[π₯]π·, ππππ½(π) = 0}.
The set of power series that can be written as a fraction of two π½-polynomials denoted by π½π(π₯)π½. Then,
clearly π½π[π₯]π½ β π½π(π₯, π½). In [3], the authors studied the quantity πΏβ and they define as follows:
πΏβ = min{π β β: β π1, π2 β π½π[π₯]π½; π1π2 β πΉππ(π½) β π½π(π1π2) β π½π[π₯]π½ }.
Theorem 2.3.[3] Let π½be a quadratic Pisot unit series. Then πΏβ = 1.
Continued π·-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 π½ β π½π((π₯β1)) with |π½| > 1.When π½ = π₯, this theory is seems to be
similar to the classical case of continued fractions.
We define the π½-transformation ππ½
β²
by:
ππ½
β²
: π·(0,1) β π·(0,1)
π β
1
π
β [
1
π
]
π½
.
when |π| < 1, we obtain
π =
1
π΄1 +
1
π΄2 +
1
β±
= [0,π΄1, π΄2, β― ]π½
whither (π΄π)πβ₯1 β π½π[π₯]π½and there are defined by π΄π = [
1
πβ²
π½
πβ1
(π)
]
π½
, β π β₯ 1.
For π β π½π((π₯β1)) and π΄0 = [π]π½, we get
π = π΄0 +
1
π΄1 +
1
π΄2 +
1
β±
= [π΄0, π΄1, π΄2, β― ]π½.
The last bracket is called continued π½-fraction expansion of π. The sequence (π΄π)πβ₯0 is called the
sequence of partial π½-quotients of π. We define the ππ‘β
π½-complete quotient of π by π
π =
[π΄0, π΄1, π΄2, β― , π
π]π½. We remark that all (π΄π)πβ₯1are not in π½π.
Main Results
Our main result is an improvement of Theorem 4.1 in [4]
.
Theorem 4.1. Let π½ be a quadratic Pisot unit formal power series over the finite field π½π such that
deg(π½) = π. Let π½ β π½π(π₯, π½) such that the continued π½-fraction of π is given by π =
[π΄0, π΄1, π΄2, β― , π΄π, β― ]. If π β π½π(π₯, π½) then {π΄π/ degπ½(π΄π) > 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 π½ be a unit Pisot series. Then π½π(π₯, π½) = π½π(π₯)π½.
Now, we define two sequences (π
π)πββ and (ππ)πββ in π½π[π₯, π½] by
{
π0 = π0, π1 = π0π1 + 1
π0 = 1 , π1 = π1
and {
π
π = ππππβ1 + ππβ2
ππ = ππππβ1 + ππβ2, βπ β₯ 2
The pair (ππ, ππ) is called reduced π½-fractionary expansion of π for all π β₯ 0.
Proposition 4.1. Let π β π½π(π₯, π½) such that π = [π΄0, π΄1, π΄2, β― , π΄π, β― ]. Then |π β
ππ
ππ
| <
1
|ππ|2.
Proof. Similarly to the classical case.
- 4. Kammoun Rania et al. Continued π·-fractions with Pisot unit base in ππ ((πβπ
))
Β© 2017, AJMS. All Rights Reserved. 233
Proposition 4.2. Let π½ be a quadratic Pisot unit power formal series such that deg(π½) = π and
π1, π2, β― , ππ β π½π[π₯]π½. Then, π½πβ1
π1π2 β― ππ β π½π[π₯]π½.
The proof of the last proposition is an immediate consequence of Thoerem 2.3.
Corollary 4.3. Let π1, π2, β― , π
π β π½π[π₯]π½. Then we have, for all positive integer
π, π½
(πβ1)π
π π1π2 β― ππ β π½π[π₯]π½.
Corollary 4.4.Let (ππ, ππ)πβ₯0 the reduced π½-fractionary expansion of π . Then π½
(πβ1)π
π π
π β π½π[π₯]π½ and
π½
(πβ1)π
π ππ β π½π[π₯]π½.
For π = ππ π½π
+ β― + π0 β π½π[π₯]π½. We denote by πΎ(π) = π degπ½(π) + degπ (ππ ) = 2π +
deg(ππ ).
Lemma 4.5. [4] Let π΄, π΅ β π½π[π₯]π½ with πΎ(π΄) > πΎ(π΅). Then there exists πΆ, π΄1 and π΅1in π½π[π₯]π½, such
that
π΄
π΅
= πΆ +
1
π΄1
π΅1
with πΎ(π΄1) > πΎ(π΅1).
Proof of Theorem 4.1
It is equivalent to prove that there exist π0 β₯ 1, π΄π β (π½π[π₯]π½) β². By Lemma 4.2, we obtain π =
π
π
β
π½π(π₯)π½ such as π, π β π½π[π₯]π½ and (π
π, ππ) the reduced π½-fractionary expansion of π.
By proposition 4.1, |
π
π
β
ππ
ππ
| <
1
|ππ|2 . According to Corollary 4.3 and Lemma 4.5, we have
(π½
(πβ1)
π
(π+1)
(πππ β ππ
π)) inπ½π[π₯]π½ .So, we obtain
1
|π|
< |π½|(πβ1)(π+1)/π
|
πππ β ππ
π
π
| <
|π½|
(πβ1)(π+1)
π
|ππ|
which implies that deg(ππ) β€ deg(π) + (π β 1)(π + 1), where
deg(ππ) = β deg(π΄π)
π
π=1
β€ deg(π) + (π β 1)(π + 1).
Thus β (deg(π΄π) β (π β 1)) β€ deg(π) + (π β 1)
π
π=1 . Finally there exists π0 β₯ 1, such that, for
deg(π΄π) β (π β 1) β€ 0, for all π β₯ π0 and the desired result is reached.
REFERENCES
1. P. Bateman and L. Duquette. The analogue of Pisot- Vijayaraghvan numbers in fields of power
series, Ill. J. Math, 6, (1962), 594-606.
2. J. Bernat. Continued fractions and numeration in the Fibonacci base, Discrete Mathematics, 22,
(2006), 2828-2850.
3. R. Ghorbel, M. Hbaib and S. Zouari. Arithmetics on beta-expansions with Pisot bases over
πΉπ((π₯β1
)), Bull. Belg. Math. Soc. Simon Stevin, 21, (2014), 241-251.
4. M. Hbaib, R. Kammoun. Continued beta-fractions with formal power series over finite fields,
Ramaujan J Math, (2015), DOI 10.1007/s11139-015-9725-5.
5. A. RΓ©nyi. Representations for real numbers and their ergodic properties, Acta Math. Acad. Sci.
Hung, 8, (1957), 477-493.
6. K. Scheicher. Beta-expansions in algebraic function fields over finite fields, finite fields and their
Applications, (2007), 394-410.