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1. *Corresponding Author: I. Hassairi, Email: imen.hassairi@yahoo.fr
RESEARCH ARTICLE
Available Online at
Unbounded transcendent formal power series in Fq((X-1
))
1
I. Hassairi* and 2
R.Ghorbel
1,2
Faculty of Sciences of Sfax, Department of Mathematics, Sfax 3000, Tunisia.
Received on: 15/02/2017, Revised on: 27/02/2017, Accepted on: 06/03/2017
ABSTRACT
Let Fq be a finite field and Fq((X-1
)) the field of formal power series with coefficients in Fq. In this work,
we construct a family of transcendental continued fractions in Fq((X-1
)) with unbounded partial quotients
from algebraic elements.
Key Words: continued fraction, formal power series, transcendence, finite fields.
AMS Subject classification: 11A55, 11J81.
INTRODUCTION
In [9, 3], Maillet and Baker studied the real number x = [a0, a1,..] where (ai)i ≥0 is the sequence of partial
quotients of x such that an = an+1 = … = an+ (n) 1, for infinitely positive integers n where λ(n) is a sequence
of integers verifying some increasing properties. The authors showed that x is transcendent and their
proof is based on the theorem of Davenport and Roth [5] and taking into account the assumption on (ai)i
≥0 which is bounded.
Later, Adamczewski and Bugeaud [1] improved new transcendence criteria for continued fractions by
using the Schmidt subspace Theorem given in [13] where the author showed that if an irrational number
is very well approximated by quadratic numbers then it is quadratic or transcendental.
Unfortunately, in the field of formal series, we have not similar theorems to these of Roth and Schmidt.
In 1976, Baum and Sweet [4] proved that the unique solution in F2((X-1
)) of
the cubic equation
X α3
+ α + X = 0
has a continued fraction expansion with partial quotients of bounded degree. They observed that no real
algebraic number of degree ≥ 3 has yet been shown to have bounded or unbounded partial quotients.
In 1986, Mills and Robbins [10] provided an example of algebraic formal series over F2((X-1
)) whose
sequence of partial quotients is unbounded.
In 2004, Mkaouar [12] gave a similar result to the Baker one [3] concerning the transcendence of formal
series over a finite field.
In 2006, Hbaib, Mkaouar and Tounsi [6] proved a result which allows the construction of a family of
transcendent continued fractions over Fq((X-1
)) from an algebraic formal series of degree more than 2.
Recently, Ammous, Driss and Hbaib [2] improved a new transcendence criterion depending only on the
length of specific blocks appearing in the sequence of partial quotients.
The goal of this work is to generalize the result in [2]. We will construct unbounded continued fractions
and transcendent in Fq((X-1
)) from algebraic formal power series: exactly, this paper aims to approximate
transcendental series by a family of algebraic series.
This article is organized as follows: In the next section, we set up the problem and we give some useful
definitions and known results in the field of formal power series and the continued fraction expansions
over this field. In section 3, we treat the objective of this paper and we establish a new general result on a
transcendence criterion. Later, we give an example to illustrate the importance of our result.
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2. I. Hassairi et al./ Unbounded transcendent formal power series in Fq((X
-1
))
32
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Field of formal series Fq((X-1
))
Let Fq((X -1
)) be a field with q > 1 elements of characteristic p > 0. We denote by Fq[X] the ring of
polynomials with coefficient in Fq and Fq(X) the field of rational functions. Let Fq((X -1
)) be the field of
formal series, i.e., for any f 2 Fq((X 1
)), f is written as follows,
where bn € Fq and n0 € Z. A formal serie f = has a unique decomposition as f =│f│+ {f}
with the polynomial part│f│2 € Fq[X] and the fractional part verifying {│f│}< 1. Here we define the
non-Archimedean absolute value as follows:
(2.1)
Thus, │f + g│ ≤ max(│f│; │g│) and │f + g│= max(│f│, │g│) if (│f│≠│g│). Let us recall that the
continued fraction expansion of a formal series is written as
f = a0 +
1
= [a0; a1; a2; :::];
a1 +
1
1
a2 + .
..
where a0 = [f], for any i ≥1, ai = [fi] € Fq[X] with deg(ai) ≥1, (ai)I ≥ 0 is the sequence of partial quotients
of f and fi = . We denote by fn = [an; an+1; :::] the n-th complete quotient of f.
Remark -1
(i) If (deg(ai))i≥0 is bounded then f have a bounded continued fraction expansion.
(ii) The expansion is finite if and only if f € Fq(X).
(iii) The sequence of partial quotients of f is ultimately periodic if and only if f is quadratic over Fq(X).
Now, we define two sequences of polynomials (Pn)n≥0 and (Qn)n≥0 by
P0 = a0; Q0 = 1; P1 = a0a1 + 1; Q1 = a1
Pn = anPn-1 + Pn-2; Qn = anQn-1 + Qn-2; for n ≥ 2
We can check that
PnQn-1 – Pn-1Qn = (-1)n-1; for n ≥ 1;
= [a0; a1; :::; an]; for n _ 0:
is called the nth convergent of f and it satisfies
lim Pn
= f = [a0; a1; :::; an; :::]:
n!1 Qn
According to the non-archimedean absolute value, we find the following important equality
│f- │= │ │=│ │-1
=│ │-1
│ │-2
Let f be an algebraic formal series of minimal polynomial P (Y ) = AmY m
+ Am-1Y m-1
+… + A0 where Ai
€ Fq[X]. We define H(f) = max0 ≤ i ≤ m│Ai│and (f) = Am.
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3. I. Hassairi et al./ Unbounded transcendent formal power series in Fq((X
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))
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MAIN RESULTS
Theorem- 3
Let f 2 Fq((X 1
)) such that f = [a1; a2; ::; an; ::] = [K1; K2; :::; Kn; ::] where (Ki)i≥1 is a sequence of finite
blocks of polynomials in Fq[X] such that
•
Ki = Bi; Bi
q
Ai; Bi
q2
Ai:::Bi
qλi-1
, for any i ≥1 with (Bi)i ≥1; (Ai)i ≥1 are two bounded
sequences in € Fq[X].
•
( λi)i ≥ 1 is an increasing sequence of positive integers.
If = +∞ ,
then f is transcendent.
To prove Theorem (3) we will need the following lemmas.
Lemma :2 [6] Assume that f is an algebraic formal series of degree d such that f = [a1; a2; :::; at; h] where
for any 1 ≤ i ≤t, ai € Fq[X], h € Fq((X-1
)). If │f│≥1 and │h│> 1 then h is algebraic of degree d and
H(h)≤ H(f)│ │d-2
Lemma: 3 [2] Let P(Y) = AmY m + Am-1Y m-1
+ … + A0 be a reduced polynomial in Fq((X󲐀1)) with
Ai € Fq[X]. Then P admits a unique root f with │f│ > 1 and [f] = [ - ].
Moreover, P is irreducible.
Lemma: 4. [14] Let f = [a0; a1; …] and g = [b0; b1; …] be two formal series having the same
n + 1 first terms of partial quotients. Then
│f-g│≤
Lemma: 5. [6] Let f and g be two formal series of degree d and m respectively. If g reduced and f
≠ g,
then
│f-g│≥
Lemma :6. The equation Afq+1
– ABfq
- 1 = 0 has a unique solution in Fq((X-1
)) with [f] ≠ 0 and its
continued fractions expansion is given by
f = [B, Bq
A, …, Bq n-1
,….] with deg(B) ≥ 1.
Proof: The existence and uniqueness of f is due to Lemma 3.3. Let f0 f, f1 = … fn = be the
complete quotients of f. If f = [a0; a1;::;::] then by induction we prove that for each n € N, fn verifies the
following equation:
An Bn +Dn = 0
Where An = ,
Bn =
and Dn = An
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))
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Finally, by Lemma 3.3 we obtain that an = [ ] = Bq n-1 .
Proof of Theorem 1: Assume that f is algebraic of degree d > 2 (it is clear that f is not quadratic). Let gi
be the formal power series which verifies the following equation:
Ai Yq+1
- AiBiYq
-1= 0 (1)
Applying Lemma 6 we get that gi is algebraic with algebraic degree (q+1), H(gi) =
│AiBi│, σ(gi) =
│Ai│and the continued fraction expansion of gi is[Bi, Ai, …, ,….] .
We denote by hi the (λ1 + … + λi-1 +1) complete quotient of f, i.e., hi =f λ1 + … + λ i-1 +1 =[Bi, Ai, …,
,….]
On the one hand, since hi and gi have the same first λi partial quotients so from Lemma 4, we have
│hi – gi │≤ = (2)
where is the (λi)-reduced of gi and (ai)i≥1 is the sequence of partial quotients of f.
On the other hand, according to Lemma 5 we obtain
│hi – gi │≥ (3)
By Lemma 2, we get
H(hi) ≤ H(f)
Then equation (3) becomes,
│hi – gi │≥
≥ (4)
From the inequalities (2) and (4) we obtain
≤ (5)
Taking the q-logarithm in both sides, we get
2
2deg +( +1− 2)deg⁡( ) (6)
Since │ai│≥ │Bi│, for any we obtain,
2
).
Hence,
−1−1+−1 −1 +1 1≤ ≤ −1
deg +( −2) 1+ 2+…+ −1deg +( +1− 2) 1+ 2+…+ −1deg .
The fact that (Ai) and (Bi) are bounded, for each i ≥1, then there exists M > 0 such that,
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))
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Finally, we have
< (d-2) M, which contradicts the hypothesis.
As application, we give the following:
Application 1: Let f € F2((X-1
)) such that f = [a1, a2, …an, …] = [K1,K2, …,Kn, ..] where
for any i ≥1. Let us set Ai = Bi = X if i is even, Ai = Bi = X
+ 1 otherwise and λi = 2i! for each i ≥1. Then f is transcendent since
Application 2: Let f € F2((X-1
)) such that f = [a1, a2, …an, …] = [K1,K2, …,Kn, ..] where
for any i ≥1. Let us set Bi = X+ where represents
the class of i in Fq and λi = qi!
for each i≥ 1. Then f is transcendent since
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