Householder s approximants and continued fraction expansion of quadratic irrationals

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1 Householder s approximats ad cotiued fractio expasio of quadratic irratioals Viko Petričević Departmet of Mathematics, Uiversity of Zagre Bijeička cesta 30, 0000 Zagre, Croatia vpetrice@mathhr Astract Let α e a quadratic irratioal It is well kow that the cotiued fractio expasio of α is periodic We oserve Householder s approximat of order m for the equatio x αx α 0 ad x 0 p / : R m αp/q α m α p / α m p / α m p / α m We say that R m is good approximat if R m is a coverget of α Whe period egis with a, there is a good approximat at the ed of the period, ad whe period is palidromic ad has eve legth l, there is a good approximat i the half of the period So whe l, the every approximat is good, ad the it holds R m p m+ q m+ for all 0 Further, we defie the umers j m j m α, y if R m is a good approximat We prove that R m p m+ +j q m+ +j j m is uouded y costructig a explicit family of quadratic irratioals, which ivolves the Fioacci umers Itroductio Cotiued fractios give very good ratioal approximatios of aritrary α R Let p e the th coverget of α The < α p < a + + q a + q Newto s iterative method x k+ x k fx k f x k 000 Mathematics Suject Classificatio: A55 Key words ad phrases: Cotiued fractios, Householder s iterative methods

2 Viko Petričević for solvig oliear equatios fx 0 is aother approximatio method Let α c + d, c, d Q, d > 0 ad d is ot a square of a ratioal umer It is well kow that regular cotiued fractio expasio of α is periodic, ie has the form α [ a 0, a,, a k, a k+, a k+,, a k+l ] Here l lα deotes the legth of the shortest period i the expasio of α Coectios etwee these two approximatio methods were discussed y several authors The pricipal questio is: Let fx x αx α, where α c d ad x 0 p, is x also a coverget of α? It is well kow that for α d, d N, d, ad the correspodig Newto s approximat R p R kl p kl q kl, for k + dq p it holds see eg [C, p 468] It was proved y Mikusiński [M] see also Elezović [E] that if l t, the R kt p kt q kt, for k These results imply that if l d, the all approximats R are covergets of d I 00, Dujella [D] 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 [K] from 999, 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 ad 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 [DP] proved that d is uouded, too I 0, the author [P] proved the aalogous results for α + d, d N, d ad d mod 4 Sharma [S] oserved aritrary quadratic surd α c+ d, c, d Q, d > 0, d is ot a square of a ratioal umer, whose period egis with a He showed that for every such α ad the correspodig Newto s approximat N p αα q p c it holds Nkl p kl q kl, for k,

3 Itroductio 3 ad whe l t ad the period is symmetric except for the last term the it holds N kt p kt q kt, for k Frak ad Sharma [FS] discussed geeralizatio of Newto s formula They showed that for every α, whose period egis with a, it holds 3 p kl kl αp kl α q kl α p kl αq kl p kl α q kl p kl αq kl, for k, N, ad whe l t ad the period is symmetric except for the last term the it holds 4 p kt kt αp kt α q kt α p kt αq kt p kt α q kt p kt αq kt, for k, N I this paper, we discuss coectios etwee cotiued fractio covergets of quadratic surd α c + d whose period egis with a ad aritrary Householder s iterative method see [SG], [H, 3], [H, 44] of order p for rootsolvig: x + H p x x + p /fp x /f p x, where /f p deotes p-th derivatio of /f Householder s method of order is just Newto s method For Householder s method of order oe gets Halley s method, ad Householder s method of order p has rate of covergece p + Aalogous to Newto s method, if we start with fuctio fx x αx α, ad approximat x 0 p, the mai questio is whether x is a coverget of α We show that almost all aalogous results holds as for Newto s method We show that every approximat is good if ut ot oly if l We prove that to e a good approximat is a periodic property ad a palidromic property, whe expasio is palidromic, we costruct a sequece that shows that the distace j m d from coverget with m- times igger idex ca e aritrary large ad suggest that the umer of good approximats i period, m d, could e also ig

4 4 Viko Petričević Householder s methods Let us first oserve m-th derivatio of the fuctio /f: /f m m x αx α α α x α m x α m m! α α x α m+ x α m+ So we have H m x x x α m x α m x α m+ x α m+ x αx α αx α m+ α x α m+ x α m+ x α m+ It is ot hard to show that it holds: H m+ x xh m x αα, for m N H m x + x α α Formula 3 shows that for aritrary quadratic surd, whose period egis with a ad k N, m, 3,, it holds H m pkl p mkl, q kl q mkl ad whe period is eve, say l t ad symmetric except the last term, from 4 it follows 3 H m pkt p mkt q kt q mkt Let us defie R def p, ad for m > R m def H m p We will say that R m is good approximatio, if it is a coverget of α Let us show that good approximatios are periodic ad symmetric if period is palidromic 3 Good approximats are periodic ad symmetric Formula [S, 8] says: For k N it holds 3 3 a l a 0 p kl + p kl q kl d c, a l a 0 q kl + q kl p kl cq kl,

5 3 Good approximats are periodic ad symmetric 5 ad formula says 33 R m+ R R m αα R + R m c, for m N, 0,, Lemma For m, k N ad i,,, l, whe the period egis with a, it holds 34 R m i αα kl+i Rm kl Rm R m kl + Rm i c Proof For m, statemet of the lemma is prove i [F, Thm ] Suppose that 34 holds for some m N, ad let us show that it holds for m + too R m+ kl Rm+ i αα R m+ kl + R m+ i c 33 R kl Rm kl αα R i Rm i αα R kl +Rm kl c R i +Rm i R kl Rm kl αα + R i Rm i αα R kl +Rm kl c R i +Rm i With the otatio s R kl, S Rm kl, t R i c αα c c ad T Rm i, we have: ss αα tt αα αα s + S ct + T c ss αα t + T c + tt αα s + S c cs + S ct + T c st αα ST αα αα s + t cs + T c st αα S + T c + ST αα s + t c cs + t cs + T c, which is y assumptio equal to the R kl+i Rm kl+i αα 33 R kl+i + Rm kl+i c R m+ Whe period egis with a, we have ad see eg [P, 3] kl+i α a 0 [ a, a,, a l, a l ], α a 0 [ a l, a l,, a, a ], a 0 a l α [ a l,, a, a, a l ] So whe period is palidromic, we have c α + α a 0 a l, thus formulas 3 ad 3 i palidromic case ecome a 0 p kl + p kl cp kl + q kl d c, a 0 q kl + q kl p kl

6 6 Viko Petričević Lemma For m, k N ad i,,, l, whe period egis with a ad whe is palidromic, it holds 37 R m kl i Rm kl Rm i c + αα R m i Rm kl Proof For m we have: R kl i p kl i q kl i 0 p kl i + p kl i 0 q kl i + q kl i [ a 0, a,, a kl i, a kl i, 0 ] [ a 0, a,, a kl i, a kl i,, a kl, a 0, 0, a 0, a,, a i ] [ a 0, a,, a kl i, a kl i,, a kl, a 0 p i q i ] p kl a 0 R i + p kl q kl a 0 R i + q kl R kl R i c + αα R i R kl Suppose that 37 holds for some m N, ad let us show that it holds for m + too With the same otatio as i the proof of Lemma, we have R m+ kl Rm+ i c + αα R m+ i R m+ kl 33 ss αα tt αα s+s c t+t c tt αα ss αα t+t c s+s c c + αα ss αα tt αα ct + T c + αα s + S ct + T c tt αα s + S c ss αα t + T c st c + αα ST c + αα αα t st S st c + αα T S + ST c + αα t s ct st S, which is y assumptio equal to the R kl i Rm kl i αα 33 R kl i + Rm kl i c R m+ Theorem Let m N For i,,, l ad β m i whe period egis with a, it holds R m kl+i βm i β m p mi R m i q mi p mi R m i q mi kl i p mkl+i + p mkl+i, for all k 0, i q mkl+i + q mkl+i ad whe period is palidromic except for the last term, the kl i p mkl i β m i p mkl i, for all k q mkl i β m i q mkl i R m

7 3 Good approximats are periodic ad symmetric 7 Proof Let us first cosider the cotiued fractio expasio of β m i β m i [ p mi R m i 0, q ] mi p mi R m i q mi [ 0, a mi, a mi,, a, a 0 + R m i ] If k 0 we have β m i p mi + p mi β m i q mi + q mi [ a 0, a,, a mi, a mi, β m [P, Lm 3] [ 0, a mi, a mi,, a, a 0 R m i ] i ] [ a 0, a,, a mi, a mi, 0, a mi, a mi,, a, a 0 + R m i ] Rm i, ad if k > 0 we have β m i p mkl+i + p mkl+i [ a 0, a,, a mkl, a mkl a 0 + a 0, a,, a mi, a mi, β m q mkl+i + q mkl+i β m i [ a 0, a,, a mkl, a mkl a 0 + R i ] p mkl a mkl a 0 + R m i + p mkl q mkl a mkl a 0 + R m i + q mkl 3 3 p mkl R m i + q mkl d c p mkl + q mkl R m i c Rm kl Rm i + d c Lm R m kl + Rm i c R m Whe period is palidromic we have: [ ] p mkl i β m i p mkl i a q mkl i β m 0, a,, a mkl i, i q mkl i β m i [ ] a 0, a,, a mkl i, 0, 0, a mi, a mi,, a, a 0 R m i [ ] a 0, a,, a mkl i, a mkl i, a mkl i+,, a mkl, a 0 R m i p mkl a 0 R m i + p mkl q mkl a 0 R m i + q mkl Rm kl Rm i c + c d R m i Rm kl kl+i p mkl R m i c d c q mkl q mkl R m i p mkl Lm R m kl i Remark [K, Thm ] ad [P, Thm ] are special cases of last theorem for m ad α d ad α + d respectively Remark Aalogously as i [D, Lm 3], from the last theorem it follows that to e a good approximat is a periodic property, ie for all r N it holds R m p k q k R m rl+ p rml+k q rml+k, i ]

8 8 Viko Petričević ad whe period is palidromic, it is also a palidromic property, ie it holds: R m p k q k R m l p ml k q ml k 4 Which covergets may appear? Let us defie umers P m def R Q m m From 33 it is ot hard to show that it holds 4 P m αq m P αq m p α m Lemma 3 R m < α 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 If is eve, we have p < α, thus p α < 0 Now statemet of lemma follows from 4 Whe is odd, we have p > α, thus from 4 we see that all powers of p α are greater tha 0 Similarly as i [D], if R m p k q k, we ca defie j m j m α, as the distace from coverget with m times larger idex: 4 j m k + m + This is a iteger, y Lemma 3 Usig Remark we have ad i palidromic case: j m α, j m α, kl +, j m α, j m α, l From ow o, let us oserve oly quadratic irratioals of the form α d, d N, d It is well kow that period of such α egis with a ad is palidromic Let us defie umers d m j mi{d : there exists such that j d, j} Tale shows d 3 j, where d 0 6 Tale suggests that for j 3 we get similar results as [D, Tl ] Let us show that for aritrary m we get similar upper ouds for j m as i [D, Prop ] for j

9 4 Which covergets may appear? 9 d 3 j l d k j 3 d, l d 3 j l j 3 d, dj j 3 d, j 3 d, l d Tale : d 3 j for j 73

10 0 Viko Petričević Theorem 43 R m + d < R m d Proof Let us oserve the fuctio gx x + d m + x d m x + d m x d m d d It holds g p k m q k R k d, ad we have g x 4dmx d m d + x m x dm d + x m Thus gx is icreasig whe m is odd or whe x > d, ad decreasig whe x < d ad m is eve Further, g d + e g d e e m d d + e m e e m m, d e m e m so whe e is small eg 0 < e < d it holds 44 g d + e < g d e, whe m is eve, 45 g d + e < g d e, whe m is odd Whe is eve ad m is odd, y Lemma 3 it holds R m + > d Further, p < d g d p + + g d p + + d 44 > g d + p + + d R m + d Whe is odd ad m is eve, y Lemma 3 follows R m < d g d p + + > d Further, g d + p + + d R m + d Let us oserve the case whe m is odd If is eve, y Lemma 3 we have < d < R + m Further, p < d g d p + + [ p, d, we have g d p + + d 45 > g d + p + + d R m + d

11 4 Which covergets may appear? Fially, if is also odd, y Lemma 3 we have R m + < d < R m < d g d p + + d, p ], we have g d p + + d 45 > g d + p + + d R m + d Propositio Whe d, for 0 we have Proof Let R m j m d, < ml/ p m++j q m++j Accordig to Remark, it suffices to cosider the case j > 0 ad < l Assume first that l is eve, eg l s We have R m s p ms q ms R m l p ml q ml For < s, usig 43 we have m++j < ms, ad j ms For s ad l we have j 0, ad for s < < l we have m + + j < ml, or j < ml m + ms, so agai we get j ml/ Let l is odd, eg l s + If for some, 0 < s holds j ml/, we would have k : m + + j ml/ By Remark it follows R m l p ml k q ml k, ad ml k ad ml/ Now it holds d p k q k d p ml k q ml k, thus d R m d R m l This is ot possile y 43, sice l s For s < < l, the proof is the same as i the eve case Theorem 3 Let l N ad a,, a l N such that a a l, a a l, The umer [ a 0, a, a,, a l, a 0 ] is of the form d, d N if ad oly if 46 a 0 l p l q l mod p l, where p q are covergets of the umer [ a, a,, a, a ] The it holds: 47 d a 0 + a 0p l + q l p l Proof See [P, 6] Lemma 4 Let F k deote the k-th Fioacci umer Let N ad k >, k, mod 3 For d k +F k Fk + it holds dk [ F k+,,,,,, F }{{} k + ], k times

12 Viko Petričević ad l d k k Proof From 46, it follows: a 0 k F k F k k F k F k F k k F k mod F k Now from Cassii s idetity F k F k F k k we have a 0 mod F k Whe 3 k, this cogruece is ot solvale, ad if 3 k, the solutio is a 0 F k+ mod F k, ie a 0 F k + From 47 it follows: + F k F k +, N Fk + Fk + F k + F k d + F k Fk + + Fk + Theorem 4 Let F l deote the l-th Fioacci umer Let l > 3, l ± mod 6 The for d l F l 3 F l + + Fl 3 F l + ad M N it holds l d l l ad j 3M d l, 0 j 3M d l, 0 j 3M+ d l, 0 l 3 M Proof By 4, we have to prove R 3M 0 p Ml q Ml, R 3M 0 p Ml q Ml, R 3M+ 0 p Ml q Ml We have a 0 F l 3F l +, ad sice 3 l, F l 3 is odd, thus y Lemma 4 it holds dl [ a 0,,,,,, a }{{} 0 ] l times From Cassii s idetity, sice l is odd l ± mod 6, it follows So we get: a 0 F l 3 F l + F l + F l + F l 3 F l F l F l, d a 0 F l 3 F l + F l R 0 p 0 + dq 0 p 0 q 0 R 0 p 0 q 0 a 0, a 0 + d a 0 a 0 + d a 0 a 0 a 0 + F l F l p l q l,

13 5 Numer of good approximats 3 48 R 3 0 p 0p 0 + 3dq0 q 0 3p 0 + dq0 a 0a 0 + 3d 3a 0 + d a 0 + a 0d a 0 4a 0 + d a 0 F l Fl 3 a 0 + Fl F l + F l a 0 + F l F l F l + a 0 + F l F l F l F l a 0 + F l F l p l q l Let us prove the theorem usig iductio o M For provig the iductive step, first oserve that from 33 for m 3 we have: 49 R m k R k Rm k R k + d + R m k, R m k R3 k Rm 3 k R 3 k + d + R m 3 k Suppose that for some i {0, l, l } it holds p M l+i q M l+i We have: R m 3 0 p Ml+i a 0,,,,,, a q Ml+i }{{} 0 + p M l+i a 0,,,,,, a q M l+i }{{} 0 + R m 3 0 l times l times p l a 0 + R m p l 35 p l R m dq l q l a 0 + R m q 36 l q l R m p l 48 R3 0 R m d R R m R m 0 Corollary For each m it holds sup { j m d, } +, { j m d, } lim sup l m d 6 5 Numer of good approximats Aalogously as i [D], let us defie m α { : 0 l, R m 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 We have l 45 6 ad m { 4, if m mod 4, 45 6, if m mod 4

14 4 Viko Petričević Proof We have 45 [ 6,,,,,, ] Usig ad 3, R m p 3m q 3m ad R m 5 p 6m q 6m are good approximats, ad y Remark, we oly have to check R m 0 ad R m Let us oserve how covergets of 45 look like [ ] M deote matrix form of covergets: [a 0, a,, a ] M a0 0 a 0 a 0 p p p6k+ p 6k [ 6,,,,,,,,,,,,,,,,,,,,, ] q 6k+ q 6k }{{} M k times [ 6,,,,,, 6, 0, 6,,,,,,, 6, 0, 6,,,,,,, 6, 0, 6, ] }{{} M k k k times k k k k k k 45 3k k [ 6, ] M k k 6 Let us write k γ, so we have p6k+3 q 6k+3 p 6k+ q 6k+ p6k+ p 6k q 6k+ q 6k γ γ γ ad p6k+5 p 6k+4 p6k+3 p 6k+ q 6k+5 q 6k+4 q 6k+3 q 6k+ γ γ γ γ γ p6k+ p 6k 5 0 q 6k+ q 6k γ γ γ γ p6k+3 p 6k+ 3 0 q 6k+3 q 6k+ + γ γ γ γ

15 5 Numer of good approximats 5 From 4 we have R m p + 45 m + p 45 m 45 p + 45m p 45 m We see ow that R m m m m is always a good approximat Namely, from 7± 45 47±7 45 m ad 7± ± 4 45, we have R m p m q m Fially, let us see whe R m m m m 6 45 m 45 is a coverget First cosider ± 45 6 ± ± From 5 we see R 4m 0 p 6m q 6m ad R 4m+ 0 p 6m q 6m, ad sice 6± ± 7 45, we have R 4m+3 0 p 6m+4 q 6m+4 From 6 ± ± 4 45, ad sice 7 4 is ot a coverget of 45, either R 4m+ 0 will e a coverget Let us defie l m mi{l : there exists d N such that l d l ad m d } Experimetal results suggests that, as m varies, the umers l m are of similar size, ut they are ot always the same I Tale we show upper ouds for l 3, where 3 8, otaied y experimets, ad correspodig d s we tested all d s smaller of 0 6, ad Tale 3 shows upper ouds for l 4, where 3 36, otaied y experimets, ad correspodig d s we tested all d s smaller tha 0 5 Other experimets we tested all m s util 0 give similar upper ouds, ut for a fixed d, umer of good approximats 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 could hold { m l sup } :

16 6 Viko Petričević l 3 d l 3 / l 3 d l 3 / Tale : Upper ouds for l 3, 3 8

17 5 Numer of good approximats 7 l 4 d l 4 / l 4 d l 4 / Refereces Tale 3: Upper ouds for l 4, 3 36 [C] G Chrystal, Algera, Part II, Chelsea, New York, 964 [D] A Dujella, Newto s formula ad cotiued fractio expasio of d, Experimet Math 0 00, 5 3 [DP] A Dujella ad V Petričević, Square roots with may good approximats, Itegers , #A6 [E] [F] [FS] N Elezović, A ote o cotiued fractios of quadratic irratioals, Math Commu 997, 7 33 E Frak, O cotiued fractio expasios for iomial quadratic surds, Numer Math E Frak ad A Sharma, Cotiued fractio expasios ad iteratios of Newto s formula, J Reie Agew Math [H] A S Householder, Priciples of umerical aalysis, McGraw-Hill, New York, 953 [H] A S Householder, The Numerical Treatmet of a Sigle Noliear Equatio, McGraw-Hill, New York, 970

18 8 Viko Petričević [K] [M] T Komatsu, Cotiued fractios ad Newto s approximats, Math Commu 4 999, J Mikusiński, Sur la méthode d approximatio de Newto, A Polo Math 954, [P] O Perro, Die Lehre vo de Ketterüche I, Dritte ed, BGTeuer Verlagsgesellschaft mh, Stuttgart, 954 [P] V Petričević, Newto s approximats ad cotiued fractio expasio of + d, Math Commu, to appear [SG] P Seah ad X Gourdo, Newto s method ad high order iteratios, preprit, 00 [S] A Sharma, O Newto s method of approximatio, A Polo Math

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

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