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154 W. M. Schmidt with t bounded above and below, depending on the interval. Therefore if cβ(P ′′ /Q ′′ ) is the exponent with respect to β, then cβ(P ′′ /Q ′′ ) = c(P/Q) + v/log |Q ′′ | where v is bounded. Thus if we have a sequence of fractions P/Q whose c(P/Q) tends to some finite u ∈ S(α), then the corresponding sequence cβ(P ′′ /Q ′′ ) will also tend to u. Therefore when finite u ∈ S(α), then u ∈ S(β). In a similar way, (6.5), (6.6) imply that ∞ ∈ S(α) implies ∞ ∈ S(β). 7. Algebraic elements in the characteristic p case. From now on we will suppose that char k = p > 0. Further q will denote a positive power of p. Suppose α ∈ k((X −1 ))\k(X) is algebraic over k(X). Following Lasjaunias [8], we will say that α is of Class I if α ∼ α q for some q. Otherwise we will say that α is of Class II. We introduce a subclass of Class I as follows: α is of Class IA if α ≈ α q for some q. Theorem 4. α ≈ α q precisely if the continued fraction of α is of the form (7.1) α = [A0, . . . , An−1, C1, C2, . . .] where for some t ∈ N and some a ∈ k × we have (7.2) Cj+t = aC q j a −1 C q j when j is odd, when j is even. Hence when t is even, the continued fraction is (7.3) [A0, . . . , An−1, C1, . . . , C ←−−−−−−→ t, aCq 1 , a−1C q 2 , . . . , a−1C q ←−−−−−−−−−−−−−−−−→ t , a q+1 C q2 1 , . . . , a−q−1C q2 ←−−−−−−−−−−−−−−−−→ t , aq2 +q+1 q C 3 1 , . . . , . . .], ←−−−−−−−−−−→ and when t is odd, it is (7.4) [A0, . . . , An−1, C1, . . . , C ←−−−−−−→ t, aCq 1 , a−1C q 2 , . . . , aCq ←−−−−−−−−−−−−−−→ t , , . . . , . . .]. ←−−−−−−−−−−→ a q−1 C q2 1 , . . . , aq−1C q2 ←−−−−−−−−−−−−−−−→ t , aq2 −q+1 q C 3 1 Possibly there are no initial terms A0, . . . , An−1. P r o o f. The map α ↦→ αq is an isomorphism of k((X−1 )) into itself. Therefore [B0, B1, . . . , Bm] q = [B q 0 , Bq 1 , . . . , Bq m], and a corresponding relation holds for infinite continued fractions.
Continued fractions and diophantine approximation 155 With α given by (7.1), (7.2), we have αn = [C1, C2, . . .], αn+t = [Ct+1, Ct+2, . . .]. Then αq n = [C q 1 , Cq 2 , . . .], and (7.2) yields αn+t = [aC q 1 , a−1C q 2 , . . .] = aαq n, so that αn+t ≈ αq n. Since α ≈ αn, hence αq ≈ αq n, and since α ≈ αn+t, we obtain α ≈ αq . Conversely, when α ≈ αq , then by Theorem 1 there are n, m, and b ∈ k × such that α = [A0, . . . , An−1, C1, C2, . . .], α q = [B0, . . . , Bm−1, bC1, b −1 C2, . . .]. Then α q = [A q 0 , . . . , Aq n−1 , Cq 1 , Cq 2 , . . .]. If m ≥ n, say m = n+t, we have Cq 1 = Bn, . . . , C q t = Bm−1, C q t+1 = bC1, C q t+2 = b−1 C2, . . . Therefore deg C1 = q deg Ct+1 = q 2 deg C2t+1 = . . . , which is impossible. We may conclude that n > m, say n = m + t. We obtain bC1 = A q m, . . . , aCt = A q n−1 , a−1 Ct+1 = C q 1 , aCt+2 = C q 2 , . . . , where a = b if t is odd, a = b−1 if t is even. Thus indeed (7.2) holds. Since not only α ≈ αn, but in fact α, αn are connected by a fractional linear transformation of determinant (−1) n , and similar for α q ≈ α q n, and in α ≈ αn+t, the implied transformation is of determinant (−1) n+t , and the first half of the proof of Theorem 4 shows that (6.1), (6.2) imply (7.5) α = (Rα q + S)/(T α q + U) with RU − ST = (−1) t a. This can be generalized as follows. Lemma 10. Suppose t ∈ N, and C1, . . . , Ct, B, D in k[X] are given, with BC1, C2, . . . , Ct of positive degree. Set ε = 1 when t is even, ε = 0 when t is odd. Let B, D have Set D | C q−ε i when i is odd, 1 ≤ i ≤ t, B | C q−ε i when i is even, 1 ≤ i ≤ t. Cj+t = (B/D)C q j (D/B)C q j when j ∈ N is odd, when j ∈ N is even. Then each Cj (j = 1, 2, . . .) is a polynomial of positive degree, and α as given by (7.1) satisfies a relation (7.5) with RU − ST = (−1) t BD. P r o o f. The expansion of α is like (7.3) or (7.4), but with B/D in place of a. Setting δ = 1 when t is even, δ = −1 when t is odd, we have, for l = 1, 2, . . . , Cj+lt = q (B/D) l−1 +δq l−2 +...+δ l−1 C ql j (D/B) ql−1 +δq l−2 +...+δ l−1 C ql j when j is odd, when j is even. Since (q l−1 +δq l−2 +. . .+δ l−1 )(q −ε) < q l , we see that Cj+lt is a polynomial divisible by Cj (1 ≤ j ≤ t, l ∈ N), and in fact C1+lt is divisible by BC1. The
Page 1 and 2: ACTA ARITHMETICA XCV.2 (2000) On co
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Page 5 and 6: where (2.9) Continued fractions and
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Page 11 and 12: therefore so that Continued fractio
Page 15: Continued fractions and diophantine
Page 21 and 22: with a, b ∈ k × , so that Contin
Page 25 and 26: Let Now set Continued fractions and

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