Newton s formula and continued fraction expansion of d

Transcription

1 Newton s formula and continued fraction expansion of d ANDREJ DUJELLA Abstract It is known that if the period s(d) of the continued fraction expansion of d satisfies s(d), then all Newton s approximants R n ( pn + dqn p n ) are convergents of d, and moreover we have R n pn+ + for all n 0. Motivated with this fact we define two numbers j j(d, n) and b b(d) by R n pn++j ++j if R n is a convergent of d; b {n : 0 n s and R n is a convergent of d}. The question is how large the quantities j and b can be. We prove that j is unbounded and give some examples which support a conjecture that b is unbounded too. We also discuss the magnitude of j and b compared with d and s(d). Introduction Let d be a positive integer which is not a perfect square. The simple continued fraction expansion of d has the form d [a0 ; a, a,..., a s, a 0 ]. Here s s(d) denotes the length of the shortest period in the expansion of d. Moreover, the sequence a,..., a s is symmetrical, i.e. a i a s i for i,..., s. This expansion can be obtained using the following algorithm: a n a0 +b n c n a 0 d, b a 0, c d a 0,, b n a n c n b n, c n d b n c n for n Mathematics Subject Classification: A55 Key words and phrases: continued fractions, Newton s formula ()

2 ANDREJ DUJELLA (see [Sierpiński 987, p. 39]). Let pn be the nth convergent of d. Then (a n+ + )q n < d p n < a n+ q n () (see [Schmidt 980, p. 3]). Furthermore, if there is a rational number p q with q such that d p q < q, (3) then p q equals one of the convergents of d. Another method for the approximation of d is by Newton s formula x k+ (x k + d x k ). (4) In this paper we will discuss connections between these two methods. More precisely, if pn is a convergent of d, the questions is whether R n ( pn + d p n ) is also a convergent of d. This question was discussed by several authors. It was proved by Mikusiński [954] (see also [Clemens at al. 995; Elezović 997; Sharma 959]) that R ks p ks q ks, and if s t then R kt p kt q kt for all positive integers k. These results imply that if s(d) or, then all approximants R n are convergents of d. Moreover, under these assumptions we have R n p n+ + (5) for all n 0.

3 NEWTON S FORMULA AND CONTINUED FRACTIONS 3 Which convergents may appear? Lemma R n d q ( n pn ) d p n Proof. (R n d) ( p n d) + ( d d) ( p n d) p n ( p n d) p n dqn p n ( p n d) Theorem If R n p k q k, then k is odd. Proof. Since p l q l > d if and only if l is odd, and by Lemma we have R n > d, we conclude that k is odd. Assume that R n is a convergent of d. Then by Theorem we have R n p n++j ++j for an integer j j(d, n). We have already seen that if s(d) then j(d, n) 0. In [Elezović 997; Komatsu 999; Mikusiński 954] some examples can be found with j ±. We would like to investigate the problem how large j can be. The following result of Komatsu [999] shows that all periods of the continued fraction expansions of d have the same behavior concerning the questions in which we are interested, i.e. we may concentrate our attention on R i for 0 i s. Lemma (Komatsu 999) For n 0,,..., s/ there exist α n such that R ks+n R ks n α np ks+n + p ks+n α n q ks+n + q ks+n for all k 0, and p ks n α n p ks n q ks n α n q ks n for all k. The following lemma reduces further our problem to the half-periods.

4 4 ANDREJ DUJELLA Lemma 3 Let 0 n s/. If R n p n++j ++j, then R s n p (s n )+ j q (s n )+ j. Proof. If ( ) pn++j ++j p n+j +j ( ) ( ) ( ) an++j an+3 pn+ q n+ 0 0 p n+ + ( ) ( ) d c pn+ +, (6) f e p n+ + then ( ) ( ps n j q s n j e f p s n 3 j q s n 3 j c d By the assumption and formula (6), we have Now Lemma and formula (7) imply R s n p s n 3 d c q s n 4 s 3 d c q s n 4 R n p n++j p n+ + d c p n+ ++j + + d c q. n+ ) ( ) ps n 3 q s n 3. (7) p s n 4 q s n 4 p s n 3 j q s n 3 j p (s n )+ j q (s n )+ j. Lemma 4 Proof. R n+ < R n The statement of the lemma is equivalent to ( ) n (d + p n p n+ ) > 0. (8) If n is even, then pn < d and p n+ + > d. Furthermore, since p n+ + d < d p n qn, we have pn + p n+ + < d. Therefore p n pn+ [( pn < + p )/ n+ ] < d + + and inequality (8) is satisfied. If n is odd, the proof is completely analogous.

5 NEWTON S FORMULA AND CONTINUED FRACTIONS 5 Proposition If d is a square-free positive integer such that s(d) >, then for all n 0. j(d, n) s(d) 3 Proof. According to Lemma 3 it suffices to consider the case j > 0. Let R n p n++j ++j. By Lemma there is no loss of generality in assuming that n < s. Assume first that s is even, say s t. Then R t p s q s and R s p s q s. If n < t, then Lemma 4 clearly implies that n + + j s and j s 3. Since s is even, we have j s 4. For n t or n s we obtain j 0. If t < n < s, then n + + j s and j s 3 n s 3. Thus we have again j s 4. Assume now that s is odd, say s t +. Instead of applying Newton s method for x 0 p t q t, we will apply regula falsi method for x 0 p t q t x pt q t. It was proved by Frank [96] that with this choice of x 0 and x we have R t,t x 0 x + d x 0 + x p s q s. If t < n < s, then from R s p s q s we obtain that j s 3 as above. Thus, assume that n t. Since the number x 0x +d x 0 +x lies between the numbers x 0 and x, we conclude that R t,t d < R t d. Hence, by Lemma 4, we have n + + j s and j s 3. and The following lemma shows that the estimate from Proposition is sharp. Lemma 5 Let t and m 5 be integers such that m ± (mod 6) and let d F m [(F m t F m 4 ) + 4]/4. Then d [ F m (F m t F m 4 ); t,,,...,,, t, F }{{} m (F m t F m 4 )]. (9) Therefore, s(d) m. Furthermore, R 0 p m q m j(d, km ) m 3 for k. m 3 and hence j(d, 0) m 3 m 3, j(d, km) and

6 6 ANDREJ DUJELLA Proof. Since m ± (mod 6), the number F m F m 4 is an integer. It is clear that a 0 d F m (F m t F m 4 ). Then we have d + a a 0 d + a 0 a0 d a0 d a 0 Fm Fm t. t F m 4 F m Let Then d a0 + a + α. d a0 + F m F m 3 α Fm and Since d a 0 + F m a 0 + F m a 0 we have a 0 + F m F m α > F m 3 F m. (0) < a 0 + F m a 0 a 0 + F m F m F m α < F m F m + F m F m 3 F m F m F m 3 + F m F m F m F m. () From inequalities (0) and () we conclude that [0;,,...,, y] () α }{{} m 3 and a a 3 a m. Furthermore, from () we have α yf m 3 + F m 4 yf m + F m 3

7 NEWTON S FORMULA AND CONTINUED FRACTIONS 7 and y α F m 4 F m 3 F m α F m 3 d + a0 d + a0 Fm + F m 3 a 0 + F m 3 d F m + F m 3 a 0 + F m 3 d F m + F m 3 a 0 F m 3 d F m ( d a 0 ) d + a0 F m [F m + F m 3 ( d + a 0 )] [ + F m 3F m (t )]. (3) Let z y (t ). From (3) we obtain z F m + F m F m 3 ( d + a 0 ) d a0 + F m F m 3 > a 0F m F m 3 + F m F m a 0 a 0 +. We have a m y t and a m a 0 +. But now from [Perron 954, Satz 3.3] it follows that a m a 0 and s(d) m. Let us consider now the approximant R 0 From (9) we have (a 0 + d a 0 ) a 0 + d a 0 F m [(F m t F m 4 ) + ] (F m t F m 4 ) p m q m a 0 + a 0 + a + F a m F m F m tf m F m 4 R 0, d F m F m (F m t F m 4 ). F m (t )F m + F m 3 and j(d, 0) m 3 as we claimed. Now Lemmas and 3 imply that j(d, km) m 3 and j(d, km ) m 3 for k. Corollary sup{ j(d, n) } + lim sup { j(d, n) } s(d)

8 8 ANDREJ DUJELLA It remains the question how large can be j compared with d. In [Cohn 977] it was proved that s(d) < 7 π d log d + O( d). However, under the extended Riemann Hypothesis for Q( d) one would expect that s(d) O( d log log d) (see [Williams 98; Patterson and Williams 985]) and therefore j(d, n) O( d log log d). Let d(j) min{d : there exist n such that j(d, n) j}. In Table we list values of d(j) for j 48 such that d(j) > d(j ) for j < j. We also give corresponding values n and k such that R n p k q k p n++j ++j. We don t have enough data to support any conjecture about the rate of growth of d(j). In particular, it remains open whether lim sup{ j(d,n) d } > 0. 3 Number of good approximants Proposition If a n+ > d +, then Rn is a convergent of d. Proof. From () and Lemma we have R n d < p n qna 3. n+ Let R n u v, where (u, v). Then certainly v p n, and d u v < 8p nqn 4p n a n+ < v ( ) d + d + a n+ qn < v, which proves the proposition. Theorem R n is a convergent of d for all n 0 if and only if s(d). Proof. As we mentioned in the introduction, the result of Mikusiński [954] imply that if s(d), then all R n are convergents of d. Assume now that R n is a convergent of d for all n 0. Then we must have R n p n+ p n+ for all n 0. Indeed, this is a consequence of the fact that R s p s q s, Corollary and Lemma 4. Therefore, R 0 p q and R ks p ks+ q ks+ (4)

9 NEWTON S FORMULA AND CONTINUED FRACTIONS 9 d(j) s(d) n k j(d, n) log d(j)/ log j(d, n) d(j)/j(d, n) Table : d(j) for j 4

10 0 ANDREJ DUJELLA for all n 0. Let d [a 0 ; a,..., a s, a 0 ] and d a 0 + t. Then where R ks αp ks+ + p ks+ αq ks+ + q ks+, (5) a 0 a t α (a a + )t a 0 (see [Komatsu 999, Corollary ]). From (4) and (5) it follows that α 0 and therefore t a 0 a. It is well known (see e.g. [Sierpiński 987, p. 3]) that if d a 0 + t, where t is a divisor of a 0, then s(d). If R n is a convergent of d, then we will say that R n is a good approximant. Let b(d) {n : 0 n s and R n is a convergent of d}. Theorem shows that if s(d) > then s(d) b(d) >. Komatsu [999] proved that if d (x + ) + 4 then b(d) 3, s(d) 5 (see also [Elezović 997]) and if d (x + 3) 4 then b(d) 4, s(d) 6. Example If d 6x 4 6x 3 x +6x 4, where x, then s(d) 8 and b(d) 6. Using algorithm () it is straightforward to check that d [(x + )(x ); x,,, x x,,, x, (x + )(x )]. Hence, s(d) 8. Now the direct computation shows that R 0 p 3 x(4x 3) q 3 x + R p 5 (x )(8x4 8x + ) q 5 x(x ) R 3 p 7 (x )(6x 4 6x + ) q 7 x(x + )(4x 3) R 5 p 9 (x )(8x8 56x x 4 3x + ) q 9 4x(x )(8x 4 8x + ) R 6 p q x(4x 3)(64x 6 96x x 3) (x + )(8x 3 6x )(8x 3 6x + ) R 7 p 5 (8x4 8x + )(56x 8 5x x 4 64x + ) q 5 x(x + )(x )(4x 3)(6x 4 6x. + )

11 NEWTON S FORMULA AND CONTINUED FRACTIONS Hence, b(d) 6. In the same manner we can check that for d 6x x 3 + 5x + 3x +, x, we have also s(d) 8 and b(d) 6. Let s b min{s : there exist d such that s(d) s and b(d) b}. We know that s, s, s 3 5, s 4 6 and s 6 8. In Table we list upper bounds for s b obtained by experiments. b s b s b /b b s b s b /b Table : upper bounds for s b Questions: Is it true that inf{s b /b : b 3} 4 3? What can be said about sup{s b /b : b }? Example Let d 5[(0x + ) + 4]. Then d [50x + 5; x, 9,, x, 4,, 4x,,,,, x,,, 5x +, 4x,,, x,,,,, x,,, 4x, 5x +,,, x,,,,, 4x,, 4, x,, 9, x, 00x + 0].

12 ANDREJ DUJELLA Hence, s(d) 43. Furthermore, b(d) 5. Indeed, it may be verified that R n p k q k for (n, k) {(0, 3), (3, ), (6, 5), (, 3), (4, 7), (5, 35), (8, 4), (3, 43), (6, 49), (7, 57), (30, 6), (35, 69), (38, 73), (4, 8), (4, 85)}. We expect that Example may be generalized to yield positive integers d with b(d) arbitrary large. In this connection, we have the following conjecture. Conjecture Let d F m[(f m x ± F m 3 ) + 4], where m ± (mod 6). Then b(d) 3F m. We have checked Conjecture for m 5. We have also the more precise form of Conjecture. Namely, we have noted that if d Fm[(F m x + F m 3 ) +4], where x is sufficiently large, then in the sequence a, a,..., a s the numbers x, x, 4x and 4x appear F n F n 3 3, F n 3 +, L n 3 + and F n 3 times, respectively, and the number a 0 appears once. If this conjecture on the sequence a, a,..., a s is true, then at d least 3F n elements in that sequence are greater then +, and Proposition implies b(d) 3F n. We have also noted similar phenomena for d Fm[(F m x F m 3 ) + 4]. As in the case of j(d, n), we are also interested in the question how large can be b(d) compared with d. Let d b min{d : b(d) b}. In Table 3 we listed values of d b for b 0 such that d b > d b for b < b. Consider the expression log d b log b. Conjecture implies that { log db } sup log b : b 4 and Table 3 suggests that this bound might be less than 4. It would be interesting to find exact value for sup{ log d b log b : b }.

13 NEWTON S FORMULA AND CONTINUED FRACTIONS 3 d b s(d b ) b log d b / log b Table 3: d b for b 0

14 4 ANDREJ DUJELLA REFERENCES [Clemens et al. 995] L. E. Clemens, K. D. Merrill and D. W. Roeder, Continued fractions and series, J. Number Theory 54 (995), [Cohn 977] J. H. E. Cohn, The length of the period of the simple continued fraction of d /, Pacific J. Math. 7 (977), 3. [Elezović 997] N. Elezović, A note on continued fractions of quadratic irrationals, Math. Commun. (997), [Frank 96] E. Frank, On continued fraction expansions for binomial quadratic surds, Numer. Math. 4 (96), [Komatsu 999] T. Komatsu, Continued fractions and Newton s approximants, Math. Commun. 4 (999), [Mikusiński 954] J. Mikusiński, Sur la méthode d approximation de Newton, Ann. Polon. Math. (954), [Patterson and Williams 985] C. D. Patterson and H. C. Williams, Some periodic continued fractions with long periods, Math. Comp. 44 (985), [Perron 954] O. Perron, Die Lehre von den Kettenbrüchen, Teubner, Stuttgart, 954. [Schmidt 980] W. M. Schmidt, Diophantine Approximation, Lecture Notes in Math. 785, Springer, Berlin, 980. [Sharma 959] A. Sharma, On Newton s method of approximation, Ann. Polon. Math. 6 (959), [Sierpiński 987] W. Sierpiński, Elementary Theory of Numbers, PWN, Warszawa; North-Holland, Amsterdam, 987. [Williams 98] H. C. Williams, A numerical inverstigation into the length of the period of the continued fraction expansion of d, Math. Comp. 36 (98), Department of Mathematics, University of Zagreb, Bijenička cesta 30, 0000 Zagreb, Croatia address: duje@math.hr