Family of Formal Power Series with Unbounded Partial Quotients 1 Introduction

International Journal of Algebra, Vol. 4, 2010, no. 28, 1383 - 1392

with Unbounded Partial Quotients

K. Ayadi, M. Hbaib and F. Mahjoub

Département de Mathématiques Faculté des Sciences de Sfax BP 802, 3038 Sfax, Tunisie [email protected] [email protected] [email protected]

Abstract

There is a theory of continued fractions for formal power series in −1 x with coeﬃcients in a ﬁeld Fq. This theory bears a close analogy with classical continued fractions for real numbers with formal power series playing the role of real numbers and the sum of the terms of non-negative degree in x playing the role of the integral part. −1 In this paper we give a family of formal power series in Fq((X )) of algebraical degree qn − 1 with unbounded partial quotients.

Mathematics Subject Classiﬁcation: 11A55, 11R58

Keywords: ﬁnite ﬁelds, formal power series, continued fraction

1 Introduction

Khintchine [2−3] conjectured that if x is a real algebraic number of degree > 2 then x has a continued fraction expansion whose the sequence of partial quotients is unbounded. Note that the answer to this conjecture is far from being aﬀordable. 1384 K. Ayadi, M. Hbaib and F. Mahjoub

More things are known in the case of algebraic formal series over a ﬁnite

field Fq of characteristic p. In 1976, Baum and Sweet have in [1] the first example of an algebraic for- −1 mal series of degree 3 on F2((X) ) whose partial quotients have only a finite number of values as well as examples whose partial quotients have infinite values. This work was continued by Mills and Robbins [5] who gave an example of −1 formal series algebraic over F3((X) ) whose the sequence of partial quotients is unbounded and given explicitly. −1 In [6] David P.Robbins gave a family of cubic formal power series in F2((X) ) with bounded partial quotients. −1 In this work, we give a family of formal power series in Fq((X )) of algebraical degree qn − 1 with unbounded partial quotients. We particu- n larly interested on formal power series verifying f q −1 = P , where P = n n−1 x + an−1x + ...... + a0 ∈ Fq[X] satisfied some conditions.

2 Field of formal series

Let Fq be a finite field with q elements. We denote by Fq[X] the ring of poly- −1 nomials with coefficients in Fq and Fq(X) the field of fractions. Let Fq((X )) be the field of formal power series F −1 −i ∈ Z ∈ F q((X )) = f = i≥n0 fiX : n0 and fi q . −i Let f = i≥n0 fiX . Define the polynomial part of f, noted by [f], as follows: −n0 f0 + f−1X + ... + fn X if n0 ≤ 0 [f]= 0 0 else, the fractional part of f, noted by {f}, as follows: {f} = f − [f] and finally n0 if f =0 γ(f)= +∞ else

−1 Let |.| be the absolute value over Fq((X )), which is not archimedean, such that |.| = q−γ(.). −1 Set Mq = {f ∈ Fq((X )) : |f| < 1} and T the transformation deﬁned by

T : Mq → Mq → 1 − 1 f f [ f ]. Family of formal power series 1385

Then, for any f ∈ Mq, we have

1 f = =[0,a1,a2,...], 1 a1 + 1 a2 + . .. where (ai)i≥0 are polynomials with degree ≥ 1 deﬁned, for any positive integer n, by 1 an = . T n−1(f)

−1 Let f ∈ F2((X )) and a0 =[f], we have

1 f = a0 + =[a0,a1,a2,...]. 1 a1 + 1 a2 + . ..

The later expression is called continued fraction expansion of f. The sequence

(ai)i≥0 is called the sequence of partial quotients of f. If it is bounded, we say that f admits bounded continued fraction expansion.

Deﬁne two sequences of polynomial (pn) and (qn)byp0 = a0, q0 =1,p1 = a0a1 +1,q1 = a1 and for any n ≥ 2,

pn = anpn−1 + pn−2,qn = anqn−1 + qn−2.

We remark that

n−1 pnqn−1 − pn−1qn =(−1) and

pn =[a0,a1,a2....., an], qn pn is called the nth−reduced of f. qn 1386 K. Ayadi, M. Hbaib and F. Mahjoub

Baum and Sweet showed in [1] the following two lemmas in characteristic 2.

Lemma 2.1.

If |qf − p| =2−(deg q+d) ,gcd(p, q)=1, d>0, then for some n,

q = qn, p = pn, and deg an+1 = d.

Lemma 2.2.

−1 Let g1,g2,h1,h2 ∈ F2[X] with g1h2 + g2h1 =0 . Then f ∈ F2((X )) has g1f + h1 bounded partial quotients if and only if has bounded partial quotients. g2f + h2 The generalization of these two lemmas is done in a similar way in characteristic q. In 1976, Baum and Sweet gave in [1] some examples of formal series in −1 F2((X )) whose the sequence of partial quotients is unbounded. They have 1 P 2n−1 proved that the formal series admits unbounded partial quo- P +1 tients which is given explicitly .

3 Results

−1 Let Fq[X] the ring of polynomials with coeﬃcients in Fq and Fq((X )) the ﬁeld of formal power series . Let n integer such that n ≥ 2 and gcd (n, q)=1.

Theorem 3.1. 1 qn−1 P −1 Let P ∈ Fq[X] such that deg P>0, then ∈ Fq((X )) admits P +1 unbounded partial quotients.

Proof :

ni −1 q −1 Let f = 1+P ∈ Fq((X )) (3.1) i≥0

Let m ≥ 0 Family of formal power series 1387

È m−1 n (qn−1)qni (qn−1). qni We have f q −1 = 1+P −1 = lim 1+P −1 i=0 m→+∞ i≥0

nm nm −1 (qn−1) 1−q −1 q −1 = lim (1 + P ) 1−qn = lim 1+P m→+∞ m→+∞ nm =(1+P −1)−1 1 + lim P −q =(1+P −1)−1 m→+∞

1 P = = 1+P −1 P +1 1 qn−1 qn−1 P P −1 So f = , then f = ∈ Fq((X )). P +1 P +1 nm m −1 q − 1 ni Let dm = n − = q . q 1 i=0 d Èm−1 qni Then (1 + P ) m =(1+P ) i=0 m −1 ni = (1 + P )q i=0 m −1 ni qni = P q 1+P −1 i=0 m −1 qni = P dm 1+P −1 i=0 By(3.1), we get

qni P dm f = P dm 1+P −1 i≥0 m −1 qni qni = P dm 1+P −1 1+P −1 i=0 i≥m

qni =(1+P )dm 1+P −1 i≥m nm =(1+P )dm + P dm (P −1)q + K

nm where |K| < |(1 + P )dm + P dm (P −1)q |

nm then |P dm f − (1 + P )dm | = |P −(q −dm)|. (3.2)

If we pose δ = deg P , then (3.2) give

nm P dm f − (1 + P )dm = q−(δdm+δq −2δdm). 1388 K. Ayadi, M. Hbaib and F. Mahjoub

dm dm nm We have gcd (P , (1+P ) ) = 1, moreover, δq −2δdm > 0, then according nm to lemma 2.1, there exists i such that deg(ai+1)=δq − 2δdm.

It is obvious that the degree of (ai+1) is unbounded, so we conclude that the 1 P qn−1 formal series f = admits unbounded partial quotients. P +1

Corollary 3.2. 1 qn−1 ∗ P Let c ∈ F and P ∈ Fq[X] such that deg P>0, then ∈ q P + c −1 Fq((X )) admits unbounded partial quotients.

Before giving the mains theorems, we will introduce these lemmas.

Lemma 3.3.

n n−1 Let P = x + an−1x + ...... + a0 ∈ Fq[X] such that degree of P is a multiple n −1 qn−1 of q − 1, then there is a unique f ∈ Fq((X )) such that f = P .

Proof : Uniqueness of f is obvious. n n−1 n − Let P = x +an−1x + ...... + a0 in which n is a multiple of q 1. −i F −1 qn−1 We seek f = i≥n0 fiX in q((X )) such that f = P so it is suﬃcient to determine the fi step by step. qn−1 n We have f = P then n = −(q − 1)n0 , which gives fn0 =1. n n−1 −(q −1)n0−1 The term of x = x , on the one hand this term equal an−1, on the n − qn−2 other hand equal (q 1)fn0 fn0+1 which gives fn0+1. And so on up to get all the fi.

Lemma 3.4.

Let P ∈ Fq[X] such that gcd (n, q)=1and deg P ≡ 0(mod n). Then the n polynomial P is uniquely expressible as αS − T , where S, T ∈ Fq[X] such that degT < (n − 1)degS and α ∈ Fq\{0} .

Proof : We give the proof of this lemma for q = 2 and n = 3. The generalization is in 3k 3k−1 the same manner. Let P ∈ F2[X] such that P = a3kx + a3k−1x + ... + 2k 2k−1 a2kx + a2k−1x + ... + a0and a3k =1. k i 3 We seek a polynomial S = i=0 biX ∈ F2[X], such that S beginning with Family of formal power series 1389

3k 3k−1 2k 3 3k 3k−1 2k a3kx +a3k−1x +...+a2kx ; i.e S = a3kx +a3k−1x +...+a2kx +H with deg (H) < 2k.

It is obvious that bk =1. 3k−1 The term of x , on the one hand this term equal a3k−1, on the other hand 2 equal bkbn0−1 which gives bn0−1. 3k−2 When we identifying the term in x we get bk−2. Continuos this algorithm, until we identifying the term in x3k−k = x2k we get

b0. Then P = S3 + T and deg (T ) < 2 deg (S).

Theorem 3.5.

n Let P, S, T ∈ Fq[X] such that P = αS −T and T divides P . Then the formal n series g such that gq −1 = P admits unbounded partial quotients.

Proof :

ni −1 q −1 Let f = 1+TP ∈ Fq((X )) (3.3) i≥0

Let m ≥ 0

È m−1 n (qn−1)qni (qn−1). qni We have f q −1 = 1+TP−1 = lim 1+TP−1 i=0 m→+∞ i≥0

nm nm −1 (qn−1) 1−q −1 q −1 = lim (1 + TP ) 1−qn = lim 1+TP m→+∞ m→+∞ nm =(1+TP−1)−1 1 + lim (T −1P )−q =(1+TP−1)−1 m→+∞

1 P = = 1+TP−1 P + T 1 qn−1 qn−1 P P −1 So f = , then f = ∈ Fq((X )). P + T P + T nm m −1 q − 1 ni Let dm = = q . n − q 1 i=0 1390 K. Ayadi, M. Hbaib and F. Mahjoub

−1 d −1 Èm−1 qni Then (1 + T P ) m =(1+T P ) i=0 m −1 ni = (1 + T −1P )q i=0 m −1 ni qni = (T −1P )q 1+TP−1 i=0 m −1 qni =(T −1P )dm 1+TP−1 i=0 By(3.3), we get

qni (T −1P )dm f =(T −1P )dm 1+TP−1 i≥0 m −1 qni qni =(T −1P )dm 1+TP−1 1+TP−1 i=0 i≥m

qni =(1+T −1P )dm 1+TP−1 i≥m nm =(1+T −1P )dm +(T −1P )dm (TP−1)q + K

nm where |K| < |(1 + T −1P )dm +(T −1P )dm (TP−1)q |

nm then |(T −1P )dm f − (1 + T −1P )dm | = |(T −1P )−(q −dm)|. (3.4)

If we pose δ = deg (T −1P ), then (3.4) give

nm (T −1P )dm f − (1 + T −1P )dm =2−(δdm+δq −2δdm).

−1 dm −1 dm nm We have gcd ((T P ) ,(1+T P ) ) = 1, moreover, δq − 2δdm > 0, then nm according to lemma 2.1, there exists i such that deg(ai+1)=δq − 2δdm.

It is obvious that the degree of (ai+1) is unbounded, so we conclude that 1 P qn−1 the formal series f = admits unbounded partial quotients, so P + T 1 1 n qn−1 1 1 n (αS − T ) admits unbounded partial quotients but 1 n α qn−1 S qn−1 α qn−1 S qn−1 is rational, then by lemma 2.2 1 (αSn − T ) qn−1 admits unbounded partial quotients.

Theorem 3.6. ∈ F n − qn−3 Let P , S, T q[X] such that P = αS T ,degT

Proof : We begin the proof as the same manner of the proof of theorem 3.5 until equality (3.4). We multiply (3.4) by T dm , we obtain

nm nm nm P dm f − (T + P )dm = T dm (T −1P )−(q −dm) =2−(βdm+(βq −2βdm−αq ))

where β = deg P and α = deg T We have tow cases: ∗ First case: if h =1 qn−3 nm − − nm Since deg T 0. Moreover, gcd (P dm , (T + P )dm ) = 1 then by lemma 2.1, there exists i such that nm nm deg ai+1 = βq − 2βdm − αq .

It is clear that the degree (ai+1) is unbounded. ∗ Second case: if h =1 There exist P and T ∈ Fq[X] such that P = hP and T = hT with gcd qn−3 (P ,T ) = 1. It is easy to verify that deg T < qn−1 deg P , and then the result is obviously by the ﬁrst case. 1 P qn−1 So in these tow cases, we have f = admits unbounded partial P + T 1 1 n qn−1 quotients, then 1 n (αS − T ) admits unbounded partial quotients α qn−1 S qn−1 1 1 n qn−1 but 1 n is rational, then by lemma 2.2 (αS −T ) admits unbounded α qn−1 S qn−1 partial quotients .

n Example 1. Let f q −1 = αSn −1, then the partial quotient of f is unbounded, by theorem 3.5 n Example 2. Let f q −1 = αSn − x with deg S>1, then the partial quotient of f is unbounded, by theorem 3.5 if x divides S and by theorem 3.6 if x not divides S. 1392 K. Ayadi, M. Hbaib and F. Mahjoub

References

[1]L. E. Baum and H. M. Sweet, Continued fractions of algebraic power series in characteristic 2, Ann. of Math. 103 (1976), 593-610.

[2]A. Khintchine, Continued fractions, Gosudarst. Isdat. Tecn.-Teor. Lit, Mosow-Liningrad, second ed., (1949) (MR 13=274f) in Russian.

[3]A. Khintchine, Kettenbruche, B. G. Teubner, (1956) (MR 18=274f).

[5]W. H. Mills and D. P. Robbins, Continued fractions for certain algebraic power series, J. Number Theory 23, No. 3 (1986), 388-404.

[6]D. P. Robbins, Cubic Laurent Series in Characteristic 2 With bounded Par- tial Quotients, arXiv: math/9903092v1 [math.NT] 16 Mar 1999.

Received: July, 2010