EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 12, No. 2, 2019, 418-431 ISSN 1307-5543 – www.ejpam.com Published by New York Business Global Convergence of β-Modified Jacobi-Perron algorithm over the field of formal power series Amara Chandoul1, Fahad Aljuaydi2,∗ 1 Departamento de Matemática, Universidade de Braśılia, Campus Universitário Darcy Ribeiro Braśılia - DF 70910-900, Brazil 2 Department of Mathematics, College of Sciences and Humanities, Prince Sattam bin Abdulaziz University, Al-Kharj, Saudi Arabia Abstract. The aim of this paper is to study multidimentional β-continued fraction algorithm over the field of formal power series. In the case of the Modified Jacobi-Perron algorithm, we prove that it converges. 2010 Mathematics Subject Classifications: 40A15 Key Words and Phrases: β-Continued fractions, Modidied Jacobi-Perron algorithm, conver- gence, formal power series. 1. Introduction In [4], we studied multidimensional continued fraction algorithm over the field of formal power series. In the case of the Brun algorithm by using its homogenous version, we prove that it converges. In this paper, we study multidimentional β-continued fraction in the case of the Modified Jacobi Perron algorithm (MJPA), we prove that it converges. 2. The field of formal power series In order to state our results, we need to introduce some basic notion of the field of formal power series. Let Fq be a field with q elements of characteristic p, Fq[X] the set of polynomials of coefficients in Fq and Fq(X) its field of fractions. The set Fq((X−1)) is the field of formal power series over Fq Fq((X−1)) = {f = +∞∑ j=s fjX −j : fj ∈ Fq, s ∈ Z}. ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v12i2.3391 Email addresses: amarachandoul@yahoo.fr (A. Chandoul), f.m.427@hotmail.com (F. Aljuaydi) http://www.ejpam.com 418 c© 2019 EJPAM All rights reserved. A. Chandoul, F. Aljuaydi / Eur. J. Pure Appl. Math, 12 (2) (2019), 418-431 419 Let f = +∞∑ j=s fjX −j ∈ Fq((X−1)), where fs 6= 0. We denote its polynomial part by [f ] and by {f} its fractional part. We remark that f = [f ] + {f}. We define a non-archimedean absolute value on Fq((X−1)) by | f |= e−s and | 0 |= 0. It is clear that, for any P ∈ Fq[X], | P |= edegP and, for any Q ∈ Fq[X], such that Q 6= 0, | P Q |= edegP − degQ. Let β0 ∈ Fq((X−1)) \ {0}, then, we define L = {ω ∈ Fq((X−1)), |ϕ| < |β0|}, which is a compact abelian group with the addition and the metric d(ϕ, ω) = |ϕ − ω| : ∀ϕ, ω ∈ Fq((X−1)). Now, for 1 ≤ j ≤ n, we put L(n) j = { (ϕ1, · · · , ϕn) ∈ Ln, { |ϕj | > |ϕi| for 1 ≤ i < j, |ϕj | ≥ |ϕi| for j < i ≤ n } , and Lnj = L× L× · · · × L, n times then L(n) j ⊂ Lnj and Ln = ⋃ 1≤i≤n L(n) i . 3. β-Continued fraction in Fq((X−1)) Let β = (βi)i∈Z with βi ∈ Fq((X−1)) \ {0}, such that deg(βi)i∈Z is a strictly increasing sequence of integers. β is called base sequence. Let S = {(di)−∞| β0 | for all z ∈ H(β). We can define the β-continued fraction by the β-transformation Tβ on D(0, |β0|), which is given by the following mapping Tβ : D(0, |β0|) → D(0, |β0|) ω 7→  { β20 ω } β if f 6= 0 0 else For any base sequence β, the so-called β-continued fraction is introduced in [7]. A β-continued fraction is an expression of the form ω = a0 + β20 a1 + β20 · · ·+ β20 an + · · · = [a0; a1, · · · ]β, where a0 ∈ I(β) and ai ∈ H(β) for i ≥ 1. It is easy to prove that deg ai > deg β0 for all i ≥ 1. Remark 2. If β = (Xi)i∈N, then the transformation Tβ describe the regular continued fraction over the field of formal power series and has been introduced by Artin [8]. A. Chandoul, F. Aljuaydi / Eur. J. Pure Appl. Math, 12 (2) (2019), 418-431 421 4. Multidimensional continued fractions Let B ⊂ E and T : B −→ B be a map. The pair (B, T ) is called a fibred system if there exists a finite or countable partition {B(P ) : P ∈ I} of B, where I ⊂ Fq[X]n, such that the restriction of T to any B(P ) is an injective map. As E is a normed space, we assume that a system defines an algorithm of multidimensional continued fractions if for all P = (P1, . . . , Pn) ∈ I, there exists an (n+ 1)× (n+ 1) invertible matrix α(P ) = (Ci,j) with entries in Fq[X] such that if y = Tf where f ∈ B(P ), then yi = Ci0 + ∑n j=1Cijfj C00 + ∑n j=1C0jfj for all 1 ≤ i ≤ n. The map T is called a multidimensional continued fraction algorithm. For all 1 ≤ i ≤ n, if f ∈ B(P (1)), then T if ∈ B(P (i)). The sequence P (1), P (2), . . . , P (n), . . . is called the expansion of f by the algorithm T. Let β(P ) = (Bi,j) be the inverse matrix of α(P ), we set β(P (1), · · · , P (s)) = β(P (1)) · · ·β(P (s)) = ((B (s) ij )), where 0 ≤ i, j ≤ n, then y = T sf if, and only if, fi = B (s) i0 + n∑ g=1 B (s) ig yg B (s) 00 + n∑ g=1 B (s) 0g yg . The algorithm T is said convergent, if for all f ∈ B, lim s→+∞ ( B (s) 10 B (s) 00 , . . . , B (s) n0 B (s) 00 ) = f. (4.1) The vectors ( B (s) 10 B (s) 00 , . . . , B (s) n0 B (s) 00 ) are the convergents of f. 5. Definitions In this section, we define a map Tβ which is arisen from β-MJPA. Let β = (βi)i∈Z be a base sequense. The map Tβ : Ln → Ln by Tβ(ϕ1, . . . , ϕn) = ( ϕ2 ϕj , . . . , ϕj−1 ϕj , { β20 ϕj } β , { ϕj+1 ϕj } β , . . . , { ϕn ϕj } β ) = ( ϕ2 ϕj , . . . , ϕj−1 ϕj , 1 ϕj − an+1, ϕj+1 ϕj − aj+1, . . . , ϕn ϕj − an ) A. Chandoul, F. Aljuaydi / Eur. J. Pure Appl. Math, 12 (2) (2019), 418-431 422 where an+1 [ β20 ϕj ] β and ai = [ ϕi ϕj ] β : i ≥ j + 1, for (ϕ1, . . . , ϕn) ∈ Lnj , (ϕ1, . . . , ϕn) 6= (0, . . . , 0) and Tβ(0, . . . , 0) = (0, . . . , 0). For s ≥ 1, we put (ϕ (s) 1 , . . . , ϕ (s) n ) = T sβ(ϕ1, . . . , ϕn) and a (s) i = ai(ϕ (s−1) 1 , . . . , ϕ (s−1) n ) =0, · · · , 0, [ β20 ϕ (s−1) j ] β , [ ϕ (s−1) j+1 ϕ (s−1) j ] β , . . . , [ ϕ (s−1) n ϕ (s−1) j ] β  = (0, · · · , 0, an+1, aj+1, · · · , an) for 1 ≤ i ≤ n+ 1, that is T sβ(ϕ1, . . . , ϕn) = T (ϕ (s−1) 1 , . . . , ϕ (s−1) n ) = ϕ(s−1) 2 ϕ (s−1) j , . . . , ϕ (s−1) j−1 ϕ (s−1) j , { β20 ϕ (s−1) j } β , { ϕ (s−1) j+1 ϕ (s−1) j } β , . . . , { ϕ (s−1) n ϕ (s−1) j } β  = ( ϕ (s−1) 2 ϕ (s−1) j , . . . , ϕ (s−1) j−1 ϕ (s−1) j , 1 ϕ (s−1) j − a(s)n+1, ϕ (s−1) j+1 ϕ (s−1) j − a(s)j+1, . . . , ϕ (s−1) n ϕ (s−1) j − a(s)n ) for (ϕ (s−1) 1 , . . . , ϕ (s−1) n ) ∈ Lnj . Also we put κ(s) := j such that degϕ (s−1) j > ϕ (s−1) i for 1 ≤ i < j and degϕ (s−1) j ≥ ϕ(s−1) i for j < i ≤ n. 5.1. The matrix Let (ϕ1, . . . , ϕn) ∈ Lnj , (ϕ1, . . . , ϕn) 6= (0, . . . , 0). We define the (n+1)×(n+1) matrix M = (mi1i2), mi1i2 ∈ Fq((X−1)), associated to (ϕ1, . . . , ϕn) in the following way : (i) 1 ≤ i2 ≤ n, i2 6= j mi1i2 = δi1i2 for 1 ≤ i1 ≤ n+ 1, (1) (ii) i2 = j mi1i2 = { 1 for i1 = n+ 1 0 for 1 ≤ i1 ≤ n (2) (iii) i2 = n+ 1, 1 ≤ i1 ≤ n+ 1 mi1i2 = ai1 , (3) A. Chandoul, F. Aljuaydi / Eur. J. Pure Appl. Math, 12 (2) (2019), 418-431 423 that is, M = M(ϕ1, . . . , ϕn) =  1 0 . . . 0 0 0 . . . 0 0 0 1 . . . 0 0 0 . . . 0 0 ... ... . . . ... ... ... . . . ... ... 0 0 . . . 1 0 0 . . . 0 0 0 0 . . . 0 0 0 . . . 0 1 0 0 . . . 0 0 1 . . . 0 aj+1 ... ... . . . ... ... ... . . . ... ... 0 0 . . . 0 0 0 . . . 1 an 0 0 . . . 0 1 0 . . . 0 an+1  . (4) For (ϕ1, . . . , ϕn) = (0, . . . , 0), we define M the (n+ 1)× (n+ 1) unit matrix In+1. We put M (0) = In+1, M (s) = M(ϕ (s−1) 1 , . . . , ϕ(s−1) n ) for s ≥ 1, where (ϕ (0) 1 , . . . , ϕ (0) n ) = (ϕ1, . . . , ϕn). Since, we consider the columns of the matrixM (1), . . . ,M (s), we denote M (1) . . .M (s) =  A (s) 11 . . . . . . A (s) 1n B (s) 1 ... ... ... A (s) κ(s)1 . . . . . . A (s) κ(s)n B (s) j ... ... ... A (s) n1 . . . . . . A (s) nn B (s) n A (s) 01 . . . . . . A (s) 0n B (s) 0  . and M (0) =  B (−d) 1 . . . B (−1) 1 B (0) 1 ... ... ... B (−n) n . . . A (s) nn B (0) n A (−n) 0 . . . B (−1) 0 B (0) 0  . Using definition of B (s) 0 , it is clear that degB (s) 0 = s∑ i=1 deg a (i) n+1 which we use often. B (s) 0 will be the denominator of the s−th convergent and B (s) i , 1 ≤ i ≤ n, will be numerator. A. Chandoul, F. Aljuaydi / Eur. J. Pure Appl. Math, 12 (2) (2019), 418-431 424 Evidently, M (1) . . .M (s) =  A (s−1) 11 . . . . . . A (s−1) 1n B (s−1) 1 ... ... ... A (s−1) κ(s)1 . . . . . . A (s−1) κ(s)n B (s−1) κ(s) ... ... ... A (s−1) n1 . . . . . . A (s−1) nn B (s−1) n A (s−1) 01 . . . . . . A (s−1) 0n B (s−1) 0   1 0 . . . 0 0 0 . . . 0 0 0 1 . . . 0 0 0 . . . 0 0 ... ... . . . ... ... ... . . . ... ... 0 0 . . . 1 0 0 . . . 0 0 0 0 . . . 0 0 0 . . . 0 1 0 0 . . . 0 0 1 . . . 0 aj+1 ... ... . . . ... ... ... . . . ... ... 0 0 . . . 0 0 0 . . . 1 an 0 0 . . . 0 1 0 . . . 0 an+1  =  A (s−1) 11 . . . A (s−1) 1,κ(s)−1 B (s−1) 1 A (s−1) 1,κ(s)+1 . . . A (s−1) 1n B (s) 1 ... ... ... ... ... ... ... ... A (s−1) κ(s)1 . . . A (s−1) κ(s),κ(s)−1 B (s−1) κ(s) A (s−1) κ(s),κ(s)+1 . . . A (s−1) κ(s),n B (s) κ(s) ... ... ... ... ... ... ... ... A (s−1) n1 . . . A (s−1) n,κ(s)−1 B (s−1) n A (s−1) n,κ(s)+1 . . . A (s−1) nn B (s) n A (s−1) 01 . . . A (s−1) 0,κ(s)−1 B (s−1) 0 A (s−1) 0,κ(s)+1 . . . A (s−1) 0n B (s) 0  . (5) where B (s) i = A (s−1) iκ(s) + n∑ k=κ(s)+1 a (s) k A (s−1) ik + asn+1B s−1 i , 0 ≤ i ≤ n. Since detM (1) . . .M (s) = ±1, which follows from (4), we see that B (s) 0 , . . . , B (s) n−1 and B (s) n have no non-trivial common factor. By a simple calculation for (ϕ1, . . . , ϕn) ∈ Lnκ(s) , we see that (i) i2 6= κ(s), n+ 1 A (s) i1i2 = A (s−1) i1i2 for 1 ≤ i1 ≤ n, (6) (ii) i2 = κ(s) A (s) i1i2 = B (s−1) i1 for 0 ≤ i1 ≤ n, (7) (iii) i2 = n+ 1 A (s) i1i2 = B (s) i1 = B (s) i = A (s−1) iκ(s) + n∑ k=κ(s)+1 a (s) k A (s−1) ik + asn+1B s−1 i for 0 ≤ i ≤ n. (8) From (5), we find that B (s) i increases as s increases and degB (s) i1 > degA (s) i1,κ(s) > degA (s) i1,i2 A. Chandoul, F. Aljuaydi / Eur. J. Pure Appl. Math, 12 (2) (2019), 418-431 425 if i2 6= κ(s), n+ 1 for 0 ≤ i1 ≤ n. We put M (1) . . .M (s)  ϕ (s) 1 ... ϕ (s) n 1  =  A (s) 11 ϕ (s) 1 + . . .+A (s) 1nϕ (s) n +B (s) 1 ... A (s) n1ϕ (s) 1 + . . .+A (s) nnϕ (s) n +B (s) n A (s) 01 ϕ (s) 1 + . . .+A (s) 0nϕ (s) n +B (s) 0  and obtain following theorem. Theorem 1. For any (ϕ1, . . . , ϕn) ∈ Ln, we have ϕi = A (s) i1 ϕ (s) 1 + . . .+A (s) in ϕ (s) n +B (s) i A (s) 01 ϕ (s) 1 + . . .+A (s) 0nϕ (s) n +B (s) 0 , for 1 ≤ i ≤ n, whenever T s ′ β (ϕ1, . . . , ϕn) 6= (0, . . . , 0), for any 0 ≤ s′ ≤ s. Proof. We prove the theorem using the method of mathematical induction. For n = 1, we have from the definition, for (ϕ1, . . . , ϕn) ∈ L(n) j , Tβ(ϕ1, . . . , ϕn) = (ϕ (1) 1 , . . . , ϕ (1) n ) = ( ϕ2 ϕj , . . . , ϕj−1 ϕj , 1 ϕj − a(1)n+1, ϕj+1 ϕj − a(1)j+1, . . . , ϕn ϕj − a(1)n ) Then ϕi =  1.ϕ (1) i 1.ϕ (1) j + a (1) n+1 for 1 ≤ i < j 1 1.ϕ (1) j + a (1) n+1 for i = j 1.ϕ (1) i + a (1) i 1.ϕ (1) j + a (1) n+1 for j < i ≤ n (9) On the other hand, for (ϕ1, . . . , ϕn) ∈ L(n) j , A (1) i1 ϕ (1) 1 + . . .+A (1) in ϕ (1) n +B (1) i A (1) 01 ϕ (1) 1 + . . .+A (1) 0nϕ (1) n +B (1) 0 =  1.ϕ (1) i 1.ϕ (1) j + a (1) n+1 for 1 ≤ i < j 1 1.ϕ (1) j + a (1) n+1 for i = j 1.ϕ (1) i + a (1) i 1.ϕ (1) j + a (1) n+1 for j < i ≤ n (10) From (9) and (10), the assertion of theorem holds for s = 1. Now, we assume that the assertion of the theorem holds by s, and we will show that the assertion holds for s + 1. Note that κ(s+ 1) is chosen by (ϕ (s) 1 , . . . , ϕ (s) n ) ∈ L(n) κ(s+1), A. Chandoul, F. Aljuaydi / Eur. J. Pure Appl. Math, 12 (2) (2019), 418-431 426 A (s+1) i1 ϕ (s+1) 1 + . . .+A (s+1) in ϕ (s+1) n +B (s+1) i A (s+1) 01 ϕ (s+1) 1 + . . .+A (s+1) 0n ϕ (s+1) n +B (s+1) 0 = κ(s+1)∑ k=1 A (s+1) ik ϕ (s) k ϕ (s) κ(s+1) +A (s+1) i,κ(s+1)( 1 ϕ (s) κ(s+1) − a(s)n+1) + n∑ k=κ(s+1)+1 A (s+1) ik ( ϕ (s) k ϕ (s) κ(s+1) − a(s)n+1) +B (s+1) i κ(s+1)∑ k=1 A (s+1) 0k ϕ (s) k ϕ (s) κ(s+1) +A (s+1) 0,κ(s+1)( 1 ϕ (s) κ(s+1) − a(s)n+1) + n∑ k=κ(s+1)+1 A (s+1) 0k ( ϕ (s) k ϕ (s) κ(s+1) − a(s)n+1) +B (s+1) 0 = κ(s+1)∑ k=1 A (s+1) ik ϕ (s) k ϕ (s) κ(s+1) +B (s) i ( 1 ϕ (s) κ(s+1) − a(s)n+1) + n∑ k=κ(s+1)+1 A (s) ik ( ϕ (s) k ϕ (s) κ(s+1) − a(s)n+1) +B (s+1) i κ(s+1)∑ k=1 A (s+1) 0k ϕ (s) k ϕ (s) κ(s+1) +B (s) 0 ( 1 ϕ (s) κ(s+1) − a(s)n+1) + n∑ k=κ(s+1)+1 A (s) 0k ( ϕ (s) k ϕ (s) κ(s+1) − a(s)n+1) +B (s+1) 0 From (8), A (s+1) i1 ϕ (s+1) 1 + . . .+A (s+1) in ϕ (s+1) n +B (s+1) i A (s+1) 01 ϕ (s+1) 1 + . . .+A (s+1) 0n ϕ (s+1) n +B (s+1) 0 = κ(s+1)∑ k=1 A (s) ik ϕ (s) k ϕ (s) κ(s+1) +B (s) i . 1 ϕ (s) κ(s+1) + n∑ k=κ(s+1)+1 A (s) ik ϕ (s) k ϕ (s) κ(s+1) +A (s) i,κ(s+1) κ(s+1)∑ k=1 A (s) 0k ϕ (s) k ϕ (s) κ(s+1) +B (s) 0 . 1 ϕ (s) κ(s+1) + n∑ k=κ(s+1)+1 A (s) 0k ϕ (s) k ϕ (s) κ(s+1) +A (s) 0,κ(s+1) = A (s) i1 ϕ (s) 1 + . . .+A (s) in ϕ (s) n +B (s) i A (s) 01 ϕ (s) 1 + . . .+A (s) 0nϕ (s) n +B (s) 0 = ϕi. Thus the assertion holds for s+ 1, completing the proof. � The vector V (s) 0 = ( B (s) 1 B (s) 0 , . . . , B (s) n B (s) 0 ) is called the s-th convergent of ϕ = (ϕ1, . . . , ϕn) by the β-MJPA and M (1) . . .M (s) the matrices expansion by this algorithm. Morover the expansion by the β-MJPA is said to be finite or infinite if T sβ(ϕ1, . . . , ϕn) = (0, . . . , 0) for some s ≥ 0 or T sβ(ϕ1, . . . , ϕn) 6= (0, . . . , 0) for any s ≥ 0, respectively. A. Chandoul, F. Aljuaydi / Eur. J. Pure Appl. Math, 12 (2) (2019), 418-431 427 6. Convergence of A-Modified Jacobi-Perron algorithm over the field of formal power series Now, we give the main result. Theorem 2. Let ϕ = (ϕ1, . . . , ϕn) ∈ Ln and V (s) 0 = V (s) 0 (ϕ) for all s ≥ 1, then the sequence (V (s) 0 )s≥1 converges to ϕ. In order to prove this theorem we need the following lemma Lemma 2. For any sequence M (1), · · · ,M (s+1), · · · of the form (5) |B(s) 0 | ∣∣∣∣∣B(s+1) i B (s+1) 0 − B (s) i B (s) 0 ∣∣∣∣∣ ≤ e−1 holds for any s ≥ 1. Proof. We prove this result by using the mathematical induction on s. Note that κ(s) = min 1≤i≤n+1 {i : m (s) i,n+1 6= 0} where m (s) i,n+1 is the (i, n+ 1) component of M (s). Then if 1 ≤ κ(1) < κ(2), ∣∣∣∣∣B(2) i B (2) 0 − B (1) i B (1) 0 ∣∣∣∣∣ = ∣∣∣∣∣ a (2) i a (1) n+1a (2) n+1 ∣∣∣∣∣ for 1 ≤ i ≤ n. Since deg a (s) n+1 ≥ 1 and deg a (s) n+1 ≥ deg a (s) i , 1 ≤ i ≤ n, for s ≥ 1, we have |B(1) 0 | ∣∣∣∣∣B(2) i B (2) 0 − B (1) i B (1) 0 ∣∣∣∣∣ ≤ e−1 (11) Moreover, if κ(1) = κ(2), we have ∣∣∣∣∣B(2) i B (2) 0 − B (1) i B (1) 0 ∣∣∣∣∣ =  ∣∣∣∣∣ a (2) i (1 + a (1) n+1a (2) n+1)a (1) n+1 ∣∣∣∣∣ for 1 ≤ i ≤ κ(1) ∣∣∣∣∣ a (1) n+1a (2) i − a (1) i (1 + a (1) n+1a (2) n+1)a (1) n+1 ∣∣∣∣∣ for κ(1) ≤ i ≤ n (12) and if κ(2) < κ(1) ≤ n, we have ∣∣∣∣∣B(2) i B (2) 0 − B (1) i B (1) 0 ∣∣∣∣∣ =  ∣∣∣∣∣∣ a (2) i (a (1) n+1a (2) n+1 + a (2) κ(1) ∣∣∣∣∣∣ for 1 ≤ i ≤ κ(1)∣∣∣∣∣∣ a (1) n+1a (2) i − a (1) i a (2) κ(1) (a (2) κ(1) + a (1) n+1a (2) n+1)a (1) n+1 ∣∣∣∣∣∣ for κ(1) ≤ i ≤ n (13) A. Chandoul, F. Aljuaydi / Eur. J. Pure Appl. Math, 12 (2) (2019), 418-431 428 Then similarly, we have ∣∣∣B(1) 0 ∣∣∣ ∣∣∣∣∣B(2) i B (2) 0 − B (1) i B (1) 0 ∣∣∣∣∣ ≤ e−1 (14) Now we suppose the assertion of (Lemma 2) holds by s− 1. For s ≥ 2 ∣∣∣∣∣B(s+1) i B (s+1) 0 − B (s) i B (s) 0 ∣∣∣∣∣ = ∣∣∣∣∣∣∣∣∣∣∣ A (s) iκ(s+1) + n∑ k=κ(s+1)+1 a (s+1) k A (s) ik + a (s+1) n+1 B (s) i A (s) 0κ(s+1) + n∑ k=κ(s+1)+1 a (s+1) k A (s) 0k + a (s+1) n+1 B (s) 0 − B (s) i B (s) 0 ∣∣∣∣∣∣∣∣∣∣∣ = ∣∣∣∣∣∣∣∣∣∣∣ A (s) iκ(s+1)B (s) 0 −A (s) 0κ(s+1)B (s) i + n∑ k=κ(s+1)+1 a (s+1) k (A (s) ik B (s) 0 −A (s) 0kB (s) i ) (A (s) 0κ(s+1) + n∑ k=κ(s+1)+1 a (s+1) k A (s) 0k + a (s+1) n+1 B (s) 0 )B (s) 0 ∣∣∣∣∣∣∣∣∣∣∣ Using the fact that deg a (s+1) k A (s) 0k < a (s) n+1B (s) 0 ∣∣∣∣∣B(s+1) i B (s+1) 0 − B (s) i B (s) 0 ∣∣∣∣∣ = ∣∣∣∣∣∣ n∑ k=κ(s+1) a (s+1) k (A (s) ik B (s) 0 −A (s) 0kB (s) i ) ∣∣∣∣∣∣∣∣∣a(s+1) n+1 (B (s) 0 )2 ∣∣∣ = 1∣∣∣a(s+1) n+1 B (s) 0 ∣∣∣ ∣∣∣∣∣∣ n∑ k=κ(s+1) a (s+1) k A (s) 0k ( A (s) ik A (s) 0k − B (s) i B (s) 0 )∣∣∣∣∣∣ By (6) and (7), we replace A (s) ik by B (lk) i , for some lk, lk < s. ∣∣∣B(s) 0 ∣∣∣ ∣∣∣∣∣B(s+1) i B (s+1) 0 − B (s) i B (s) 0 ∣∣∣∣∣ = 1∣∣∣a(s+1) n+1 ∣∣∣ ∣∣∣∣∣∣ n∑ k=κ(s+1) a (s+1) k B (lk) 0 ( B (lk) i B (lk) 0 − B (s) i B (s) 0 )∣∣∣∣∣∣ ≤ 1∣∣∣a(s+1) n+1 ∣∣∣ ∣∣∣∣∣∣ n∑ k=κ(s+1) a (s+1) k B (lk) 0 s−1∑ k=1 ( B (l) i B (l) 0 − B (l+1) i B (l+1) 0 )∣∣∣∣∣∣ ≤ 1∣∣∣a(s+1) n+1 ∣∣∣ max 1≤k≤s−1 max k≤l≤s−1 ∣∣∣B(k) 0 ∣∣∣ ∣∣∣∣∣B(l) i B (l) 0 − B (l+1) i B (l+1) 0 ∣∣∣∣∣ ≤ e−2 ≤ e−1 A. Chandoul, F. Aljuaydi / Eur. J. Pure Appl. Math, 12 (2) (2019), 418-431 429 Then, from the assumption of the induction, ∣∣∣B(s) 0 ∣∣∣ ∣∣∣∣∣B(s+1) i B (s+1) 0 − B (s) i B (s) 0 ∣∣∣∣∣ ≤ e−1 completing the proof. � Proof. (of theorem 2). We see∣∣∣∣∣ϕi − B (s) i B (s) 0 ∣∣∣∣∣ = ∣∣∣∣∣A(s) i1 ϕ (s) 1 + . . .+A (s) in ϕ (s) n +B (s) i A (s) 01 ϕ (s) 1 + . . .+A (s) 0nϕ (s) n +B (s) 0 − B (s) i B (s) 0 ∣∣∣∣∣ = ∣∣∣∣∣∣∣∣∣∣ n∑ k=1 (A (s) ik B (s) 0 −A (s) 0kB (s) i )ϕ (s) k (A (s) 01 ϕ (s) 1 + . . .+A (s) 0nϕ (s) n +B (s) 0 )B (s) 0 ∣∣∣∣∣∣∣∣∣∣ = ∣∣∣∣∣∣∣∣∣∣∣ n∑ k=1 ( A (s) ik A (s) 0k − B (s) i B (s) 0 ) A (s) 0k ϕ (s) k A (s) 01 ϕ (s) 1 + . . .+A (s) 0nϕ (s) n +B (s) 0 ∣∣∣∣∣∣∣∣∣∣∣ For each k, 1 ≤ k ≤ n, there exists an increasing sequence lk, lk < s, such that j(lk) = k, one gets A (s) ik = B (lk) i . Then we have A. Chandoul, F. Aljuaydi / Eur. J. Pure Appl. Math, 12 (2) (2019), 418-431 430 ∣∣∣∣∣ϕi − B (s) i B (s) 0 ∣∣∣∣∣ = ∣∣∣∣∣∣∣∣∣∣∣ n∑ k=1 ( A (s) ik A (s) 0k − B (s) i B (s) 0 ) A (s) 0k ϕ (s) k A (s) 01 ϕ (s) 1 + . . .+A (s) 0nϕ (s) n +B (s) 0 ∣∣∣∣∣∣∣∣∣∣∣ = ∣∣∣∣∣∣∣∣∣∣∣ n∑ k=1 ( A (lk) ik A (lk) 0k − B (s) i B (s) 0 ) B (lk) 0 ϕ (s) k A (s) 01 ϕ (s) 1 + . . .+A (s) 0nϕ (s) n +B (s) 0 ∣∣∣∣∣∣∣∣∣∣∣ = n∑ k=1 ∣∣∣B(lk) 0 ∣∣∣ ∣∣∣∣∣ ( A (lk) ik A (lk) 0k − B (s) i B (s) 0 ) ϕ (s) k ∣∣∣∣∣∣∣∣A(s) 01 ϕ (s) 1 + . . .+A (s) 0nϕ (s) n +B (s) 0 ∣∣∣ ≤ max 1≤l≤s−1 ∣∣∣B(lk) 0 ∣∣∣ ∣∣∣∣∣ ( A (lk) ik A (lk) 0k − B (s) i B (s) 0 ) ϕ (s) k ∣∣∣∣∣∣∣∣A(s) 01 ϕ (s) 1 + . . .+A (s) 0nϕ (s) n +B (s) 0 ∣∣∣ Using (Lemma 2), we get∣∣∣∣∣ϕi − B (s) i B (s) 0 ∣∣∣∣∣ ≤ e−1 ∣∣∣ϕ(s) k ∣∣∣∣∣∣A(s) 01 ϕ (s) 1 + . . .+A (s) 0nϕ (s) n +B (s) 0 ∣∣∣ Since degB (s) 0 = s∑ k=1 deg a (k) n+1 ≥ s, then, for any ε > 0, there exists s0 ≥ 1 such that∣∣∣∣∣ϕi − B (s) i B (s) 0 ∣∣∣∣∣ ≤ e−1∣∣∣B(s) 0 ∣∣∣ < ε, for any s ≥ s0 This implies lim s−→∞ B (s) i B (s) 0 = ϕi for 1 ≤ i ≤ n which proves the theorem. � REFERENCES 431 References [1] A. 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