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 show that aalogous coectios with cotiued fractios hold 1 Itroductio ad 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 Householder s iterative method of order p for rootsolvig: x i+1 x i + p (1/f)(p 1) (x i ) (1/f) (p) (x i ), is aother approximatio method Householder s method of order 1 is just Newto s method ad for Householder s method of order oe gets Halley s method 000 Mathematics Suject Classificatio 11A55 Key words ad phrases Cotiued fractios, Householder s iterative methods 1
V PETRIČEVIĆ Let α e a quadratic irratioal ad m N I this paper we study coectios etwee cotiued fractio covergets of α ad Householder s iterative method of order m 1 (with rate of covergece m) for the equatio f(x) (x α)(x α ) 0, where α c d The pricipal questio is: Let the iitial iteratio x 0 p q e the th cotiued fractio coverget of α, is the first iteratio R (m) x 1 also a coverget of α (we will say i that case that R (m) is good approximat)? We will show that 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 (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) (α) as the distace from coverget with m times larger idex: j (m) (13) j (m) s + 1 m( + 1) 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 1 + 1 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
HOUSEHOLDER S APPROX AND CONT FRACTIONS 3 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 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 the period Formulas (14) ad (15) suggest that R () 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 a distace from two times larger idex Dujella also poited out that j( d) is uouded I 005, Dujella ad the author [3 proved that ( d) is uouded, too I 011, the author [11 proved the aalogous results for α 1+ d, d N, d ot a perfect square ad d 1 (mod 4) Sharma [13 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 egis with a 1 He showed that for every such α ad the correspodig Newto s approximat R () p αα q q (p (α+α )q ) it holds R () kl 1 p kl 1 q kl 1, for k 1, ad whe l t ad the period is palidromic the it holds R () kt 1 p kt 1 q kt 1, for k 1
4 V PETRIČEVIĆ Frak ad Sharma [6 discussed geeralizatio of Newto s formula showed that for every α, whose period egis with a 1, it holds They (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 [1, [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 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
HOUSEHOLDER S APPROX AND CONT FRACTIONS 5 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 R (1) 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 Let α e quadratic irratioal whose period egis with a 1 Formula [13, (8) says: For k N it holds (31) (3) (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, whe the period egis with a 1, 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 m+1 too Usig the otatio s kl 1, S R(m) kl 1, t R(1) i 1 ad T R(m) i 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 st αα ST αα s+t α α st αα s+t α α + 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
6 V PETRIČEVIĆ Whe period egis with a 1, we have (see eg [10, 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 egis with a 1 ad whe 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 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
HOUSEHOLDER S APPROX AND CONT FRACTIONS 7 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 i Whe period egis with a 1, it holds 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 [11, 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 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 [ a0, a 1,, a mkl 1, a mkl, i R (m) i 1, [ 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 ) kl 1 R(m) i 1 αα kl 1 + R(m) i 1 α α Lm 31 kl+i 1
8 V PETRIČEVIĆ 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) [ 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) 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 [11, 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 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
HOUSEHOLDER S APPROX AND CONT FRACTIONS 9 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 q3 +1936q5 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 R (5) 0 p8 q 8 514 379 3 160 100 476 403, R(5) 7 p39 q 39 4 993 116 004 999 75 740 560 150, From Theorem 11 we have R(5) 6 p30 q 30 7944493914 119767751, ad also R (5) 8k p 40k+8 q 40k+8 ad R (5) 8k p 40k 10 R (5) 35 487 1 R (5) e q 40k 10 35 501 is ot a coverget of 44, so either R (5) 8k+1 6 51 3 94 44 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 5+ 1 3 [ 9, 5, 6, 1, ad m 3, we have: R m (3) 37p3 7pq +3368q3 81p q 14pq +484q3 q 11 We have R (3) 3 p11 4 4004 659 435 564, ad so o 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 o R (3) 4k+1 p 1k+7 q 1k+7 q 7 q 1k 1 The period is 36 409 3960 (i palidromic Example 37 Let us oserve α 7+ 11 5 [, 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 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 ) (41) +1 d < R (m) d
10 V PETRIČEVIĆ Proof a) From (5) we have (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 [10, 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 R (m) 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 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) ad, 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 (m) d R (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 are covergets of the umer [ a 1, a,, a 1, a The it holds: () d a 0 + a 0p l + q l p l 1
HOUSEHOLDER S APPROX AND CONT FRACTIONS 11 Proof See [10, 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 1 + 1 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 From () it follows: + ( 1)F k ( 1)F k + 1, N ( ( 1)Fk + 1 ) ( ( 1)Fk + 1 ) F k 1 + F k d + F k ( ( 1)Fk + 1 ) + ( 1)Fk 1 + 1 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 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 1 + 1 F l
1 V PETRIČEVIĆ So we get: 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 1 + 1 a 0 + F l 1F l F l F l (46) a 0 + F l 1 F l p l 1 q l 1 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 For each m it holds sup d, lim sup d, { j (m) ( 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}
HOUSEHOLDER S APPROX AND CONT FRACTIONS 13 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 Usig (3) ad (4), 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 1, 30 3, 47 7, 114 17, 161 4, ad let us oserve how other covergets look like ([ M deote matrix form of covergets: [a 0, a 1,, a M ( a 0 ) ( 1 a1 ) ( 1 1 0 1 0 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 1080 7 6 M [ 6, 1 M q 6k+1 q 6k 4 161 1 1 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 47+7 + γ 47 7 γ 47+7 γ 47 7 γ 6+ γ 161+4 + γ 161 4 γ 161+4 From (5) we have γ 161 4 γ 0+3 γ 6 + γ 0 3, γ 0+3 γ 114+17 γ 0 3 + γ 114 17 γ 114+17 γ 114 17 (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± ) 3 161 ± 4, sice 7 1, 47 7 are covergets of, we have 1 pm 1 q m 1 ad 161 4 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 114 17 is a coverget of, we have R (4m+3) 0 p6m+4 q 6m+4
14 V PETRIČEVIĆ 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 l (3) d l (3) / l (3) d l (3) / 3 5 13 1666667 18 36 3040 941176 4 6 1 15 19 71 313157 373684 5 11 165 0 44 193648 6 6 10 1 41 115 195381 7 11 3615 157149 46 796500 0909091 8 1 558900 15 3 157 1 686087 9 1 77 333333 4 66 740880 75 10 14 500 14 5 97 49065 388 11 37 8835 3363636 6 50 9403 193077 1 0 548 1666667 7 113 46055 4185185 13 74698 3461538 8 78 800 785714 14 8 1015 461538 9 171 535517 589655 15 41 915 733333 30 80 41405 666667 16 8 115 175 31 97 90315 31903 17 67 60389 3941176 3 88 8915 75 Tale 1 Upper ouds for l (3), 3 3 [11 the author showed the same iequality for α 1+ d, d N, d 1 (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 of 10 6 ) Other experimets (we tested all m s util 0) give similar upper ouds, ut l (m) is ot a mootoic fuctio i m These experimets suggest that for other Householder s methods similar coectios hold as for Newto s method Eg for the umer of good approximats l (m) it could hold { (m) l } sup : 1 Ackowledgemets The author is grateful to Adrej Dujella for may useful commets ad remarks o the previous versio of this paper
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