HOUSEHOLDER S APPROXIMANTS AND CONTINUED FRACTION EXPANSION OF QUADRATIC IRRATIONALS. Vinko Petričević University of Zagreb, Croatia

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1 HOUSEHOLDER S APPROXIMANTS AND CONTINUED FRACTION EXPANSION OF QUADRATIC IRRATIONALS Viko Petričević Uiversity of Zagre, Croatia Astract There are umerous methods for ratioal approximatio of real umers Cotiued fractio coverget is oe of them ad Newto s iterative method is aother Coectios etwee these two approximatio methods were discussed y several authors Householder s methods are geeralisatio of Newto s method I this paper, we will show that for these methods aalogous coectio with cotiued fractios hold 1 Itroductio ad mai results Let α e a quadratic irratioal, ie α c + d, c, d Q, d > 0 ad d is ot a square of a ratioal umer It is well kow that cotiued fractio expasio of α is periodic, ie has the form α [ a 0, a 1,, a h, a h+1, a h+,, a h+l Here l l(α) deotes the legth of the shortest period i the expasio of α We will oserve quadratic irratioals whose period egis with a 1 We will say that period is palidromic if it holds a 1 a l 1, a a l,, ie the period without the last term is symmetric Cotiued fractio covergets p q [ a 0, a 1,, a give good ratioal approximatios of α Aother approximatio method we otai usig the Householder s iterative method of order p This method is a umerical algorithm for solvig the oliear equatio f(x) 0, where f(x) is a p + 1 times 000 Mathematics Suject Classificatio 11A55 Key words ad phrases Cotiued fractios, Householder s iterative methods 1

2 V PETRIČEVIĆ cotiuously differetiale fuctio ad α is a zero of f ut ot of its derivative Householder s method of order p cosists of a sequece of iteratios x i+1 x i + p (1/f)(p 1) (x i ) (1/f) (p) (x i ) egiig with a iitial guess x 0 Householder s method of order 1 is just Newto s method ad for Householder s method of order oe gets Halley s method I this paper we study coectios etwee cotiued fractio covergets of quadratic irratioal α c + d ad Householder s iterative method of order m 1, m N, m (with rate of covergece m) for the equatio f(x) (x α)(x α ) 0, where α c d Precisely, if the iitial iteratio x 0 p q is the th cotiued fractio coverget of α, the pricipal questio is whether the first iteratio R (m) x 1 also a coverget of α I that case we say that R (m) is good approximat We will show that for quadratic irratioal α whose period of the legth l egis with a 1, there is a good approximat at the ed of the period, ie it holds (11) kl 1 p mkl 1 q mkl 1, for all k N, ad whe period is palidromic ad has eve legth, say l t, there is a good approximat i the half of the period, ie it holds (1) kt 1 p mkt 1 q mkt 1, for all k N I Sectio 3 we show: Theorem 11 To e a good approximat is a periodic property, ie for all k N it holds p s q s kl+ p kml+s q kml+s, ad whe period is palidromic, it is also a palidromic property, ie it holds: p s q s l p ml s q ml s Whe l, from (11) ad (1) it follows that the every approximat is good, ad the it holds R (m) p m(+1) 1 q m(+1) 1 for all 0 So if R (m) is a good approximat, oe might expect that it hits coverget with m-times larger idex However, this is ot always true If R (m) ps q s, we ca defie umers j (m) j (m) (α) as half of the distace from coverget with m times larger idex: (13) j (m) s + 1 m( + 1)

3 HOUSEHOLDER S APPROX AND CONT FRACTIONS 3 We prove that it is uouded y costructig a explicit family of quadratic irratioals, which ivolves the Fioacci umers: Theorem 1 Let F l deote the l-th Fioacci umer Let l > 3, l ±1 (mod 6) The for d l ( F l 3 F l +1) + Fl 3 F l ad M N it holds l( d l ) l ad j (3M 1) 0 ( d l ) j (3M) 0 ( d l ) j (3M+1) 0 ( d l ) l 3 M Coectio etwee Newto s iterative method x i+1 x i f(x i) f (x i ) for solvig oliear equatios f(x) 0 ad cotiued fractios was discussed y several authors So, let us riefly metio what is kow for case m It is well kow that for α d, d N, d ot a perfect square, ad the correspodig Newto s approximat R () eg [1, p 468) 1 ( p q (14) R () kl 1 p kl 1 q kl 1, for k 1 + dq p ) it follows that (see It was proved y Mikusiński [9 (see also Elezović [4) that if l t, the (15) R () kt 1 p kt 1 q kt 1, for k 1 These results imply that if l( d), the all approximats R () are covergets of d I 001, Dujella [ proved the coverse of this result Namely, if all approximats R () are covergets of d, the l( d) Thus, if l( d) >, we kow that some of approximats R () are covergets ad some of them are ot Usig a result of Komatsu [8 from 1999, Dujella also showed that eig a good approximat is a periodic ad a palidromic property, so he defied the umer ( d) as the umer of good approximats i should e coverget whose idex is twice as large whe it is a good approximat However, this is ot always true, ad Dujella defied the umer j( d) as half of the distace from two times larger idex He also poited out that j( d) is uouded I 005, Dujella ad the author [3 proved that ( d) is uouded, too the period Formulas (14) ad (15) suggest that R () I 011, the author [13 proved the aalogous results for α 1+ d, d N, d ot a perfect square ad d 1 (mod 4) Sharma [15 oserved aritrary quadratic surd α c + d, c, d Q, d > 0, d is ot a square of a ratioal umer, whose cotiued fractio period

4 4 V PETRIČEVIĆ of the legth l egis with a 1 correspodig Newto s approximat R () He showed that for every such α ad the it holds R () kl 1 p kl 1 q kl 1, for k 1, p αα q q (p (α+α )q ) ad whe l t ad the period is palidromic the it holds R () kt 1 p kt 1 q kt 1, for k 1 Frak ad Sharma [6 discussed geeralizatio of Newto s formula They showed that for every quadratic irratioal α, whose period egis with a 1, it holds (16) p mkl 1 α(p kl 1 α q kl 1 ) m α (p kl 1 αq kl 1 ) m q mkl 1 (p kl 1 α q kl 1 ) m, for k, m N, (p kl 1 αq kl 1 ) m ad whe l t ad the period is palidromic the it holds (17) p mkt 1 α(p kt 1 α q kt 1 ) m α (p kt 1 αq kt 1 ) m q mkt 1 (p kt 1 α q kt 1 ) m, for k, m N (p kt 1 αq kt 1 ) m Householder s methods Householder s iterative method (see [14, [7, 44) of order p for rootsolvig, cosists of a sequece of iteratios: x i+1 H (p) (x i ) x i + p (1/f)(p 1) (x i ) (1/f) (p) (x i ), (where (1/f) (p) deotes p-th derivatio of 1/f) egiig with a iitial guess x 0 Let f(x) e a p + 1 times cotiuously differetiale fuctio ad α is a zero of f ut ot of its derivative, the, i a eighorhood of α, the covergece has rate p + 1 Aalogous to Newto s method, we will start with fuctio f(x) (x α)(x α ), which satisfies the aove coditios Let us first oserve p-th derivatio of the fuctio 1/f: ( (1/f) (p) 1 ) (p) 1 ( 1 (x α)(x α ) α α x α 1 ) (p) x α ( 1)p p! ( 1 α α (x α) p+1 1 ) (x α ) p+1 So we have (1) H (p) (x) x (x α ) p (x α) p (x α ) p+1 (x α) p+1 (x α)(x α ) α(x α ) p+1 α (x α) p+1 (x α ) p+1 (x α) p+1

5 HOUSEHOLDER S APPROX AND CONT FRACTIONS 5 It is ot hard to show that it holds: () H (p+1) (x) xh (p) (x) αα, for p N H (p) (x) + x α α Formula (16) shows that for aritrary quadratic surd, whose period egis with a 1 ad k N, m, 3,, it holds (3) H (m 1)( p ) kl 1 p mkl 1, q kl 1 q mkl 1 ad whe period is palidromic, ad has eve legth, say l t, from (17) it follows (4) H (m 1)( p ) kt 1 p mkt 1 q kt 1 q mkt 1 Let us recall the defiitio p, ad for m > 1 R (m) H (m 1)( p ), q q ad we say that R (m) is good approximatio, if it is a coverget of α From (3) ad (4) it follows (11) ad (1) From (1) we have (5) ad formula () says (6) R (m+1) α(p α q ) m α (p αq ) m (p α q ) m (p αq ) m, R (1) R (m) αα, for m N, 0, 1, + R (m) α α 3 Good approximats are periodic ad palidromic From ow o, we assume that α is quadratic irratioal whose period of the legth l egis with a 1 From formula [15, (8) we otai: (31) (3) for all k N (a l a 0 )p kl 1 + p kl αα q kl 1, (a l a 0 )q kl 1 + q kl p kl 1 (α + α )q kl 1, Lemma 31 For m, k N ad i 1,,, l, it holds (33) kl+i 1 kl 1 R(m) i 1 αα kl 1 + R(m) i 1 α α Proof For m 1, statemet of the lemma is prove i [5, Thm 1 Suppose that (33) holds for some m N, ad let us show that it holds for

6 6 V PETRIČEVIĆ m+1 too Usig the otatio s kl 1, S R(m) kl 1, t R(1) we have: R (m+1) (6) kl+i 1 kl+i 1 R(m) kl+i 1 αα kl+i 1 + R(m) kl+i 1 α α i 1 st αα ST αα s+t α α st αα s+t α α + ad T R(m) i 1, S+T α α αα ST αα S+T α α α α (st αα )(ST αα ) αα (s+t α α )(S+T α α ) (st αα )(S+T α α )+(ST αα )(s+t α α ) (α+α )(s+t α α )(S+T α α ) (ss αα )(tt αα ) αα (s+s α α )(t+t α α ) (ss αα )(t+t α α )+(tt αα )(s+s α α ) (α+α )(s+s α α )(t+t α α ) ss αα tt αα s+s α α ss αα s+s α α + t+t α α αα tt αα t+t α α α α (6) R (m+1) kl 1 R(m+1) i 1 αα R (m+1) kl 1 + R (m+1) i 1 α α We have (see eg [1, 3) (α a 0 ) [ a l, a l 1,, a, a 1, ad so 1 a 0 a l α [ a l 1,, a, a 1, a l So whe period is palidromic, we have α + α a 0 a l, thus formulas (31) ad (3) i palidromic case ecome (34) (35) a 0 p kl 1 + p kl (α + α )p kl 1 αα q kl 1, a 0 q kl 1 + q kl p kl 1 Lemma 3 For m, k N ad i 1,,, l 1, whe period is palidromic, it holds (36) kl i 1 R(m) kl 1 (R(m) i 1 α α ) + αα i 1 R(m) kl 1 Proof For m 1 we have: kl i 1 p kl i 1 q kl i 1 0 p kl i + p kl i 1 0 q kl i + q kl i 1 [ a 0, a 1,, a kl i 1, a kl i, 0 [ a 0, a 1,, a kl i, a kl i+1,, a kl 1, a 0, 0, a 0, a 1,, a i 1 [ a 0, a 1,, a kl i, a kl i+1,, a kl 1, a 0 p i 1 q i 1 p kl 1(a 0 i 1 ) + p kl q kl 1 (a 0 i 1 ) + q kl (34) (35) kl 1 (R(1) i 1 α α ) + αα i 1 R(1) kl 1

7 HOUSEHOLDER S APPROX AND CONT FRACTIONS 7 Suppose that (36) holds for some m N, ad let us show that it holds for m + 1 too With the same otatio as i the proof of Lemma 31, we have R (m+1) (6) kl i 1 kl i 1 R(m) kl i 1 αα kl i 1 + R(m) kl i 1 α α s(t α α )+αα t s S(T α α )+αα T S αα s(t α α )+αα t s + S(T α α )+αα T S α α (s(t α α )+αα )(S(T α α )+αα ) αα (t s)(t S) (s(t α α )+αα )(T S)+(S(T α α )+αα )(t s) (α+α )(t s)(t S) (ss αα )(tt αα (α+α )(t+t α α ))+αα (s+s α α )(t+t α α ) (tt αα )(s+s α α ) (ss αα )(t+t α α ) ss αα tt αα s+s α α ( tt αα t+t α α ss αα s+s α α t+t α α α α ) + αα (6) R(m+1) kl 1 (R(m+1) R (m+1) i 1 i 1 α α ) + αα R (m+1) kl 1 Let us show that each approximat ca e expressed as the comiatio of coverget with m times larger idex ad carefully selected umers i, which are periodic (so we take them for i 1, l 1) Propositio 33 Let m N For i 1,,, l 1 let The it holds i i 1 q mi 1 p mi 1 p mi i 1 q mi (37) kl+i 1 β(m) i p m(kl+i) + p m(kl+i) 1, for all k 0, i q m(kl+i) + q m(kl+i) 1 ad whe period is palidromic, the (38) kl i 1 p m(kl i) 1 i p m(kl i), for all k 1 q m(kl i) 1 i q m(kl i) i Proof Let us first cosider the cotiued fractio expasio of i [ p mi i 1 0, q mi [13, Lm 3 p mi 1 i 1 q [ 0, a mi, a mi 1,, a 1, a 0 i 1 mi 1 [ 0, a mi, a mi 1,, a 1, a 0 + i 1

8 8 V PETRIČEVIĆ If k 0 we have i p mi + p mi 1 i q mi + q mi 1 [ a0, a 1,, a mi 1, a mi, i [ a 0, a 1,, a mi 1, a mi, 0, a mi, a mi 1,, a 1, a 0 + i 1 ad if k > 0 we have i p m(kl+i) + p m(kl+i) 1 i q m(kl+i) + q m(kl+i) 1 R (m) i 1, [ a0, a 1,, a mkl 1, a mkl, a mkl+1,, a m(kl+i), i [ a 0, a 1,, a mkl 1, a mkl a 0 + p mkl 1 (a mkl a 0 + i 1 i 1 ) + p mkl q mkl 1 (a mkl a 0 + i 1 ) + q mkl (31) (3) p mkl 1 i 1 αα q mkl 1 (3) p mkl 1 + q mkl 1 ( i 1 α α ) Whe period is palidromic we have: p m(kl i) 1 i p m(kl i) q m(kl i) 1 i q m(kl i) kl 1 R(m) i 1 αα kl 1 + R(m) i 1 α α [ a 0, a 1,, a m(kl i) 1, 1 i [ a 0, a 1,, a m(kl i) 1, 0, 0, a mi, a mi 1,, a 1, a 0 i 1 [ a 0, a 1,, a m(kl i) 1, a m(kl i), a m(kl i)+1,, a mkl 1, a 0 i 1 p mkl 1(a 0 i 1 ) + p mkl q mkl 1 (a 0 i 1 ) + q mkl (3) R(m) kl 1 (R(m) i 1 α α ) + αα i 1 R(m) kl 1 (34) (35) Lm 31 kl+i 1 p mkl 1 ( i 1 α α ) + αα q mkl 1 q mkl 1 i 1 p mkl 1 Lm 3 kl i 1 Remark 34 [8, Thm 1 ad [13, Thm are special cases of the last propositio for m ad α d ad α 1+ d, respectively Proof of Thereom 11 The first part for l 1 is (11) ad for 0, 1,, l follows from (37) The secod part follows similarly as i [, Lm 3, ut we have three cases Let R (m) ps q s [ a 0, a 1,, a s If s m( + 1) 1, from (37) we have +1 0, so from (38) we have l p m(l 1) 1 q m(l 1) 1 p ml s q ml s If s > m( + 1) 1, the from (37) we have +1 [ a m(+1)+1, a m(+1)+,, a s [ a m(l 1) 1, a m(l 1),, a ml s

9 HOUSEHOLDER S APPROX AND CONT FRACTIONS 9 From (38) we have l p m(l 1) 1 q m(l 1) 1 +1 p m(l 1) +1 q m(l 1) [ a 0, a 1,, a m(l 1) 1, 1 [ a 0, a 1,, a m(l 1) 1, 0, a m(l 1) 1, a m(l 1),, a ml s [ a 0, a 1,, a ml s 1, 0 p ml s q ml s If s < m( + 1) 1, the from (38) we have From (37) we have l 1 [ a m(+1) 1, a m(+1),, a s+ [ a m(l 1)+1, a m(l 1)+,, a ml s l β(m) l 1 p m(l 1) + p m(l 1) 1 l 1 q p ml s m(l 1) + q m(l 1) 1 q ml s +1 Let us show how Theorem 11 ca e applied The first example shows palidromic situatio, the secod is ot palidromic (ut we accidetally get good approximatio i the half of the period), ad the third shows that good approximats do deped o m Example 35 Let us oserve 44 [ 6, 1, 1, 1,, 1, 1, 1, 1 The period is palidromic ad we have l 8 Let us cosider eg the case m 5 From (5) we have: R (5) p5 +440p3 q +9680pq4 5p 4 q+440p q q5 q 19 From (1) we have R (5) 3 p19 R (5) 11 p59 q 59, R (5) 15 p79 q 79,, R (5) 4k 1 p 0k 1 q 0k 1 Sice, R (5) 0 p8 q ad also R (5) 8k p 40k+8 q 40k+8 ad R (5) 8k p 40k 10 R (5) R (5) e , R(5) 7 p39 q , From Theorem 11 we have R(5) 6 p30 q , q 40k is ot a coverget of 44, so either R (5) 8k is ot a coverget of 44, so either R (5) 8k+ or R(5) 8k 3 will e or R(5) 8k 4 will Example 36 Let us oserve α [ 9, 5, 6, 1, ad m 3 From (5) we have: R m (3) 37p3 7pq +3368q3 We have R (3) 3 p11 q 11 81p q 14pq +484q , ad so R(3) 4k 1 p 1k 1 ot palidromic, ad accidetally we have R (3) 1 p7 case would e p5 q 5 ), ad so R (3) 4k+1 p 1k+7 q 1k+7 q 7 q 1k 1 The period is (i palidromic

10 10 V PETRIČEVIĆ Example 37 Let us oserve α [, 15, 1, 3, 1, 3, 1 For m 3 we have: R (3) 6k 1 p 18k 1 q 18k 1 ; R (3) 1 p7 q 7 ad R (3) 6k+1 p 18k+7 q 18k+7 For m 4 we have: R (4) 6k 1 p 4k 1 q 4k 1 ; R (4) 0 p5 q 5 ad R (4) 6k p 4k+5 q 4k+5 ; R (4) 1 p11 q 11 ad R (4) 6k+1 p 4k+11 q 4k+11 ; R (4) 3 p17 q 17 ad R (4) 6k+3 p 4k+17 q 4k+17 4 Which covergets may appear? From ow o, let us oserve oly quadratic irratioals of the form α d, d N, d ot a perfect square It is well kow that period of such α egis with a 1 ad is palidromic (41) Lemma 41 a) R (m) < d if ad oly if is eve ad m is odd Therefore, R (m) ca e a eve coverget oly if is eve ad m is odd ) Proof a) From (5) we have +1 d < d (4) R (m) d(p q d) m d (p + q d)m (p q d) m O the right side of (4), deomiator is always greater tha 0, ad omiator is less tha 0 if ad oly if is eve (the we have p dq < 0) ad m is odd ) Let us oserve (4) From 1 p 0 < p 1 < p < ad 1 q 0 < q 1 < q < we have < p 0 + dq 0 < p 1 + dq 1 < O the other had, we have 1 > p 0 dq 0 > p 1 dq 1 > (see eg [1, 15), so it holds (41) Let us oserve the defiitio (13) The umer j (m) Lemma 41 a) Usig Theorem 11 we have (43) j (m) j (m) kl+, ad i palidromic case: j(m) j (m) l Propositio 4 For 0 ad m N we have j (m) ( d) m(l/ 1) < Proof Let is a iteger, y p m(+1)+j 1 q m(+1)+j 1 Accordig to (43), it suffices to cosider the case j > 0 ad < l Assume first that l is eve, eg l t We have t 1 pmt 1 q mt 1 l 1 p ml 1 q ml 1 For < t 1, usig (41) we have m( + 1) + j 1 < mt 1, ad j m(t 1) 1 For t 1 ad l 1 we have j 0, ad for ad

11 HOUSEHOLDER S APPROX AND CONT FRACTIONS 11 t 1 < < l 1 we have m(+1)+j 1 < ml 1, or j < ml m(+1) m(t 1), so agai we get j m(l/ 1) 1 Let l is odd, eg l t + 1 If for some, 0 < t holds j m(l/ 1), we would have s : m( + 1) + j 1 ml/ 1 By Theorem 11 it follows l p ml s q ml s, ad ml s ml/ 1 Now it holds d ps q s d p ml s q ml s, thus d R (m) (m) d R l This is ot possile y (41), sice l t For t 1 < < l 1, the proof is the same as i the eve case Propositio 43 Let l N ad a 1,, a l 1 N such that a 1 a l 1, a a l, The umer [ a 0, a 1, a,, a l 1, a 0 is of the form d, d N if ad oly if (44) a 0 ( 1) l 1 p l q l (mod p l 1), where p q [ a 1, a,, a 1, a The it holds: () d a 0 + a 0p l + q l p l 1 Proof See [1, 6 Lemma 44 Let F k deote the k-th Fioacci umer Let N ad k > 1, k 1, (mod 3) For d k () ( ( 1)F k +1) + ( 1)Fk it holds l( d k ()) k ad dk () [ ( 1)F k +1, 1, 1,, 1, 1, ( 1)F }{{} k + 1 k 1 times Proof From (44), it follows: a 0 ( 1) k 1 F k 1 F k ( 1) k 1 F k 1 (F k F k 1 ) ( 1) k 1 ( F k 1) (mod F k ) Now from Cassii s idetity F k F k Fk 1 ( 1)k 1 we have a 0 1 (mod F k ) Whe 3 k, this cogruece is ot solvale, ad if 3 k, the solutio is a 0 F k+1 (mod F k ), ie a 0 F k ( 1)F k ( 1)F k + 1, N From () it follows: ( ( 1)Fk + 1 ) ( ( 1)Fk + 1 ) F k 1 + F k d + F k ( ( 1)Fk + 1 ) + ( 1)Fk 1 + 1

12 1 V PETRIČEVIĆ Remark Periodic cotiued fractios ivolvig Fioacci umers with all a i 1, i 1,, l 1 were kow earlier First example was show i [16 Costructio of such examples, usig Pellia equatios was give i several papers y Molli, see eg [11 ad [10 However, i all such examples, the umers are of the form 1+ d with all a i 1 i symmetric part of the period, ut for the umers of the form d there is at least oe a i 1 I [13, Lemma 5 we costructed, i a similar was as i Lemma 44, all umers of the form 1+ d, d N, d 1 (mod 4), with all a i 1 i symmetric part of the period We have show that all such umers are of the form 1+ d k (), where d k () 4( ) ( F k + 1) + F k 3 + 1, k, N or N whe 3 k Some of those umers was also give i [11, Example 5: D k () 4Fk + (0Fk + 8( 1)k ) + 5, ie it is ot hard to show that it holds D k () d k () It turs out that for the umers i Lemma 44 it holds d k () 1 4 d k ( 1 ), ad whe 3 k, cotiued fractio of d k () have desired form (ad there is o other umer with such period) Proof of Theorem 1 By (13), we have to prove R (3M 1) 0 p Ml q Ml, R (3M) 0 p Ml 1 q Ml 1, R (3M+1) 0 p Ml q Ml We have a 0 F l 3F l +1, ad sice 3 l, F l 3 is odd, thus y Lemma 44 it holds dl [ a 0, 1, 1,, 1, 1, a }{{} 0 l 1 times From Cassii s idetity, sice l is odd (l ±1 (mod 6)), it follows So we get: a 0 F l 3 (F l 1 + F l ) + 1 F l + F l 3 F l F l 1 F l, d a 0 F l 3 F l F l 0 p 0 q 0 a 0, R () 0 p 0 + dq 0 p 0 q 0 a 0 + d a 0 + d a 0 a 0 + F l p l, a 0 a 0 F l 1 q l R (3) 0 p 0(p 0 + 3dq 0) q 0 (3p 0 + dq 0 ) a 0(a 0 + 3d) 3a 0 + d a 0 + a 0(d a 0) 4a 0 + d a 0 F l 1 Fl 3 a 0 + Fl 1 F l + F l a 0 + F l 1F l F l a 0 + F l 1F l F l F l (46) a 0 + F l 1 F l p l 1 q l 1

13 HOUSEHOLDER S APPROX AND CONT FRACTIONS 13 Let us prove the theorem usig iductio o M For provig the iductive step, first oserve that from (6) for m 3 we have: (47) k R() k R(m ) k R () k + d + R (m ) k, k R(3) k R(m 3) k R (3) k + d + R (m 3) k Suppose that for some i {0, l, l 1} it holds p (M 1)l+i q (M 1)l+i We have: p Ml+i q Ml+i [ a 0, 1, 1,, 1, 1, a }{{} 0 + p (M 1)l+i q (M 1)l+i l 1 times p l 1(a 0 + R (m 3) 0 ) + p l (34) p l 1 R (m 3) 0 + dq l 1 q l 1 (a 0 + R (m 3) 0 ) + q (35) l q l 1 R (m 3) 0 + p l 1 (46) R(3) 0 R(m 3) 0 + d R (3) 0 + R (m 3) 0 (47) 0 [ R (m 3) 0 a 0, 1, 1,, 1, 1, a }{{} 0 + R (m 3) 0 l 1 times Corollary 46 Let l( d) l e the legth of the shortest period of the cotiued fractio expasio of d The for each m it holds { j (m) ( d) } +, sup d, lim sup d, { j (m) ( d) } l( m d) 6 5 Numer of good approximats Aalogously as i [, let us defie (m) (α) { : 0 l 1, is a coverget of α} For aritrary m experimetal results suggest that similar properties could hold as for m However, there are some differeces, as the followig example shows Example 51 We have l( ) 6 ad (m) ( { 4, if m (mod 4), ) 6, if m (mod 4) Proof We have [ 6, 1,,,, 1, 1 [ 6, 1,,,, 1, 6, 0 We will deote covergets of regular expasio with p q Usig (3) ad (4),

14 14 V PETRIČEVIĆ p3m 1 q 3m 1 ad 5 p6m 1 q 6m 1 are good approximats, ad y Theorem 11, we oly have to check 0 ad 1 The first few covergets of are sequetially 6 1, 7, ad let us oserve 1, 30 3, 47 7, , how other covergets look like ([ M deote matrix form of covergets: [a 0, a 1,, a M ( a 0 ) ( 1 a1 ) ( a1 ) ( 1 0 p p 1 ) q q 1 ), ad let us write ( 7+ ) 3k γ ( ) ( ) k ( ) p6k+1 p 6k 161 [ 6, 1,,,, 1, 6, 0 k M [ 6, 1 M q 6k+1 q 6k γ 7+ + γ 7 γ 6+ + γ 6, ( ) p6k+3 p 6k+ q 6k+3 q 6k+ ( ) p6k+5 p 6k+4 q 6k+5 q 6k+4 γ 7+ γ 7 γ γ 47 7 γ 47+7 γ 47 7 γ 6+ γ γ γ From (5) we have γ γ 0+3 γ 6 + γ 0 3, γ 0+3 γ γ γ γ γ (p+q ) m +(p q ) m (p +q ) m (p q ) m We see ow that 1 (7+ ) m +(7 ) m (7+ ) m (7 ) m is always a good approximat Namely, from ( 7± ) 47±7 ad ( 7± ) ± 4, sice 7 1, 47 7 are covergets of, we have 1 pm 1 q m 1 ad Fially, let us see whe 0 (6+ ) m +(6 ) m (6+ ) m (6 ) is a coverget m First cosider ( ) 4 6 ± (51) 161 ± 4 ( ) 3 7 ± 3 From (51) we see R (4m) 0 p6m 1 q 6m 1 ad R (4m+1) 0 p6m q 6m, ad sice (6± ) 3 9(114 ± 17 ) ad is a coverget of, we have R (4m+3) 0 p6m+4 q 6m+4 From (6 ± ) 3(7 ± 4 ), ad sice 7 4 is ot a coverget of, either R (4m+) 0 will e a coverget l (m) Let us defie mi{l : there exists d N such that l( d) l ad (m) ( d) } I [3 Dujella ad the author proved that sup { l () : 1 }, ad i [13 the author showed the same iequality for α 1+ d, d N, d 1

15 HOUSEHOLDER S APPROX AND CONT FRACTIONS 15 l (3) d l (3) / l (3) d l (3) / Tale 1 Upper ouds for l (3), 3 3 (mod 4) ad d is ot a perfect square I Tale 1 we show upper ouds for l (3), otaied y experimets, ad correspodig d s (we tested all d s smaller the 10 6 ) Other experimets (we tested all m s util 0) give similar upper ouds, ut (m) ( d) is ot a mootoic fuctio i m Experimetal results lead to the coclusio that for every positive iteger m 3 ad every positive iteger there exist a positive iteger d such that (m) ( d) Moreover, otaied upper ouds for l(m) { (m) l (5) sup suggest that } : 1 for all m I case m families of examples where costructed which show that for every positive iteger there exist a positive iteger d such that () ( d) ad (m) ( d) > l( d)/ To prove the iequality (5) for each m 3 i a similar maer seems early impossile ecause (m) ( d) depeds ot oly o d ut also o m (see Example 51) Ackowledgemets The author is grateful to Adrej Dujella ad the referre for may useful commets ad remarks o the previous versio of this paper Refereces [1 G Chrystal, Algera, Part II, Chelsea, New York, 1964

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