CONTINUED FRACTIONS (From real numbers to formal numbers) by A. Lasjaunias 1. Introduction The study of continued fractions is an ancient part of elementary Number Theory. It was studied by Leonhard Euler in the 18-th century. Actually, a remarkable paper from him was translated from Latin language into English and published thirty years ago [1]. The subject has been treated very deeply by Oskar Perron at the beginning of the 20 th century, in a famous book which has been edited several times [2]. It can also be found in several books on Number Theory, among them a famous one is “An Introduction to the Theory of Numbers” due to Hardy and Wright which was reedited many times until recently [3]. In the previous cited works, continued fractions were considered for real numbers. In the 1930’s, L. Carlitz initiated formal arithmetic, replacing natural integers by polynomials over a finite field. Since then, arithmetic in function fields was developed and from the middle of last century, both number fields and function fields were studied simultaneously (see for in- stance Hasse’s treatise on the topic [4]). Our main focus will be here contin- ued fractions in function fields, and particularly function fields over a finite field. The study of certain continued fractions in this last setting begins in the 1970’s with the works of Baum and Sweet [5, 6], later developed by Mills and Robbins [7]. Continued fractions are the tool to study rational approximation of numbers or functions. For general information concerning Diophantine ap- proximation and continued fractions in power series fields, and also many arXiv:1711.11276v1 [math.NT] 30 Nov 2017 references, one can see [8],[9] and [10, Chap. 9]. 2. Fields of formal power series Let K be field and T a formal indeterminate. We have the following analogy : 1 1 K∗ ± ∩ ∩ Z K[T ] ←→ ∩ ∩ Q K(T ) ←→ ∩ ∩ R K((1/T )) We replace the number expanded in base b n x = X anb R ∈ n≤k by the formal power series n α = X anT K((1/T )). ∈ n≤k Hence a non-zero element of K((1/T )) can be written as i α = X uiT where i Z, ui K and ui0 = 0. ∈ ∈ 6 i≤i0 An ultrametric absolute value is defined over this field by α = T i0 where | | | | T is a fixed real number greater than 1 (not specified). Note that the field | | K((1/T )) is the completed of the field K(T ) for this absolute value, just as R is the completed of Q for the usual absolute value. The integer (polynomial) part of α is u T i K[T ] and it will be P0≤i≤i0 i ∈ denoted [α] as in the classical case. We have x + y max( x , y ) for x,y K((T −1) (ultrametric absolute | | ≤ | | | | ∈ value). If P,Q K(T ) then P = T deg(P ) and P/Q = T deg(P )−deg(Q). ∈ | | | | | | | | Note that, with this metric, T −n tends to zero when the integer n tends to infinity. In the sequel, Fq is the finite field of characteristic p, with q elements (q = ps, with s 1 integer). The field K will be either F or Q. We simply ≥ q denote F(q) the field of power series in 1/T over Fq. The importance of the fields F(q) is due to the finiteness of the base field and to the existence of the t Frobenius isomorphism. Indeed, let p be the characteristic of Fq and r = p where t 0 is an integer, then the mapping x xr is a field-isomorphism ≥ → of F(q). 2 In the case of real numbers, a number is rational if and only if the sequence of digits is ultimately periodic. It is interesting to notice that the same is true in F(q) : i α = X uiT Fq(T ) (ui)i≤0 is ultimately periodic. ∈ ⇐⇒ i≤i0 A very simple rational element is obtained by the following expansion : T/(T 1)=1+ T −1 + + T −n + in F (T ) or Q(T ). − · · · · · · p From this example, we can observe that the property stated above fails in Q((T −1)). Indeed we have by formal differentiation : (T/(T 1))′ = 1/(T 1)2 = T −2 2T −3 + nT −n−1 − − − − − ···− −··· Exercise 1: What is the power series expansion of the rational function (T 2/(T 2 3T + 2))5 in F(5) ? − Exercise 2: Show that there is α F(2) such that α3 =1+1/T . Give the ∈ first digits of α and write an algorithm to obtain as many digits as wanted. 3. Continuants and finite continued fractions Let W = w1, w2, . , wn be a finite sequence of elements belonging to a ring A (or simply formal variables). We set W = n for the length of the | | word W . We define the following operators for the word W . ′ ′ W = w2, w3, . , wn or W = if W = 1 ′′ ′′ ∅ | | W = w1, w2, . , wn−1 or W = if W = 1 ∗ ∅ | | W = wn, wn−1, . , w1. We consider the finite continued fraction associated to W to be 1 [W ] = [w1, w2, . , wn]= w1 + . 1 w2 + . 1 .. + wn This continued fraction is a quotient of multivariate polynomials, usually called continuants, built from the variables w1, w2, . , wn. [w1]= w1 [w1, w2] = (w1w2 + 1)/w2 3 and [w1, w2, w3] = (w1w2w3 + w1 + w3)/(w2w3 + 1). The continuant built on W will be denoted by W . We now recall h i the definition of this sequence of multivariate polynomials. Set = 1. If the sequence W has only one element, then we have W = h∅i h i W . Hence, with the above notations, the continuants can be computed, recursively on the length W , by the following formula | | W = w W ′ + (W ′)′ for W 2 (1) h i 1h i h i | |≥ or equivalently W = W ′′ w + (W ′′)′′ for W 2. (1 bis) h i h i n h i | |≥ − Thus, with these notations, for any finite word W , the finite continued fraction [W ] satisfies W [W ]= h i . (2) W ′ h i It is easy to check that the polynomial W is, in a certain sense, symmetric h i in the variables w , w , . , w : hence we have W ∗ = W . This symmetry 1 2 n h i h i implies the classical formula W [W ∗]= h i . (2 bis) W ′′ − h i In the sequel, for n 1, we consider the finite continued fraction ≥ z = [a , a , , a ], where the a ’s are positive natural integers, in the real 1 2 · · · n i case or polynomial in T of positive degree, in the formal case. Consequently, it is clear that z Q in the first case or z K(T ) in the second case. For ∈ ∈ 1 k n, we denote [a , a , , a ] by z (hence z = z). Hence, we can ≤ ≤ k k+1 · · · n k 1 write z = [a , a , , a ] = [a , , a , z ] for 1 k n 1. 1 2 · · · n 1 · · · k k+1 ≤ ≤ − The ai are called the partial quotients of the expansion, while the zi are called the complete quotients or the tails of the expansion. For 1 k n, ≤ ≤ we also define the following elements x = a , , a and y = a , , a . k h 1 · · · ki k h 2 · · · ki 4 Due to the recursive definition of the continuants, we see that both sequences (xk)1≤k≤n and (yk)1≤k≤n satisfy the same recurrence relation K = a K + K for 1 k n k+1 k+1 k k−1 ≤ ≤ with different initial conditions in both cases : (x0,x1) = (1, a1) or (y0,y1) = (0, 1). For 1 k n, according to each case, real or formal, x and y are positive ≤ ≤ k k integers or polynomial of positive degree and the rational xk/yk is called a convergent of z. Moreover, we clearly have z = xn/yn. Since z = [a , , a , z ], by the property (1 bis) of the continuants, we 1 · · · k k+1 − get a , , a , z a , , a z + a , , a z = h 1 · · · k k+1i = h 1 · · · ki k+1 h 1 · · · k−1i. a , , a , z a , , a z + a , , a h 2 · · · k k+1i h 2 · · · ki k+1 h 2 · · · k−1i This implies the following important formula : x x z + x n = k k+1 k−1 for 1 k n 1. (3) yn ykzk+1 + yk−1 ≤ ≤ − If α is an element of Q with α 1 or an element of K(T ) with ≥ α T then α can be expanded as a finite continued fraction in a unique | | ≥ | | way. That is to say, with the above choice for the ai’s in each case, we have α = [a , a , , a ]. 1 2 · · · n The only restriction to this unicity of the expansion may happen in the real case if the last partial quotient is equal to 1. As for example : [5] = [4, 1] or [5, 2] = [5, 1, 1]. In both cases, the partial quotient ai is the integral part of the com- plete quotients zi. There are obtained by induction. So a1 = [α], the integer part of α. Since α = [a , z ], we get z = 1/(α a ) and a is the integer 1 2 2 − 1 2 part of z2, etc...This can be resumed by the formulas α = a + u , 0 u < 1 (or u < 1) 1 1 ≤ 1 | 1| 1/u = z = a + u , 0 u < 1 (or u < 1) 1 2 2 2 ≤ 2 | 2| 1/u = z = a + u , 0 u < 1 (or u < 1) 2 3 3 3 ≤ 3 | 3| · · · 5 This is known as the continued fraction algorithm.