Accelerations of generalized Fibonacci sequences

0.16 MB·English

by  Marco Abrate

ACCELERATIONS OF GENERALIZED FIBONACCI SEQUENCES
MARCO ABRATE,STEFANO BARBERO,UMBERTO CERRUTI, ANDNADIRMURRU
3
1
0 Abstract. In this paper we study how to accelerate the convergence of the ra-
2 tios (xn) of generalized Fibonacci sequences. In particular, we provide recurrent

n formulas in order to generate subsequences (xgn) for every linear recurrent se-
a quence (gn) of order 2. Using these formulas we prove that some approximation
J methods,assecant,Newton,HalleyandHouseholdermethods,cangeneratesub-

5 sequencesof(xn). Moreover,interestingpropertiesonFibonaccinumbersariseas
1 anapplication. Finally,weapplyalltheresultstotheconvergentsofaparticular
continued fraction which represents quadraticirrationalities.

]
T
N
.
h
t
a 1. Introduction
m
[ The problem of finding sequences of approximations for a solution of a certain

equation f(t) = 0 is really intriguing. There are various and famous methods to
1
v generate such sequences in literature, as Newton, secant and Halley methods. But
7 another interesting problem is how to increase the rate of convergence for the dif-
7
ferent methods. One of the answer consists in selecting some subsequences of the
4
starting approximations sequences which accelerate the convergence process. In the
3
. particular case when f(t) corresponds to a second degree polynomial, we may think
1
about it as a characteristic polynomial of a generalized Fibonacci sequence. The ra-
0
3 tio of its consecutive terms generates a new sequence which converges to the root of
1 larger modulusof f(t), when it exists and it is real. Many authors have given meth-
:
v ods to accelerate the rate of convergence of this sequence. McCabe and Phillips [5]
i
X have studied how some methods of approximations, like Newton and secant meth-
ods, provide an acceleration of such sequence for suitable initial values. Gill and
r
a Miller [2] have found similar results for the Newton method, only for the ratios of
consecutive Fibonacci numbers. Many other works have been developed about this
argument. In[4]and[6],e.g., theresultsofMcCabeandPhillips[5]havebeengener-
alized for ratios of non–consecutive elements of generalized Fibonacci sequences. In
[7]theseresultshavebeenextendedtootherapproximationmethods,liketheHalley
method. The aim of this paper is to give new interesting points of view to prove
in an easier way some known results about accelerations of generalized Fibonacci
sequences, providing recurrent formulas in order to generate these subsequences.
The accelerations related to approximation methods become particular cases of our
work. In the first section we will expose our approach, showing in the second sec-
tion the relationship with known approximation methods as Newton, secant, Halley
and Householder methods. Moreover our results, in addition to a straightforward
proof of some identities about Fibonacci numbers, can be applied to convergents of
a particular continued fraction representing quadratic irrationalities. We will prove
in the last section that particular subsequences of these convergents correspond to
1
2 MARCOABRATE,STEFANOBARBERO,UMBERTOCERRUTI,ANDNADIRMURRU
Newton,Halley andsecantapproximationsfortherootoflargermodulusofasecond
degree polynomial, when it exists and it is real.
2. Accelerations of linear recurrent sequences of the second order
Definition 2.1. We define, over an integral domain with unit R, the set S(R) of
sequences a = (a )+∞, and the set W(R) of linear recurrent sequences. Further-
n n=0
more, we indicate with a = (a )+∞ = W(α,β,p,q) the linear recurrent sequence of
n n=0
order 2 with characteristic polynomial t2 −pt+ q and initial conditions α and β,
i.e.,
a = α
0
a = β
 1
an = pan−1−qan−2 ∀n≥ 2 .
If we consider the sequence U = W(0,1,p,q), it is well–known that the sequence

x = (x )+∞ defined by
n n=2
U
n
x = , ∀n≥ 2 (2.1)
n
Un−1
converges to the root of larger modulus (when it is real and it exists) of the char-
acteristic polynomial of U. Thus, every subsequence (x )+∞ (for some sequence
gn n=0
(g )+∞) can be considered as a convergence acceleration of the starting sequence
n n=0
x. Cerruti and Vaccarino [1] have deeply studied the applications of the companion
matrix of linear recurrent sequences. Here we consider some of these results which
are useful to find the explicit formulas for the acceleration of the sequences U and
x. We define
0 1
M = ,
−q p
(cid:18) (cid:19)
as the companion matrix of a linear recurrent sequence with characteristic polyno-
mial t2 −pt+q. We recall that some authors may use the transpose of M as the
companion matrix.
Lemma 2.2. The sequences U = W(0,1,p,q) and T = W(1,0,p,q) satisfy the
following relations
T = −qU
n+1 n , ∀n ≥0. (2.2)
(Un+1 = Tn+pUn
T U T U
Mn = n n = n n , (2.3)
T U −qU T +pU
n+1 n+1 n n n
(cid:18) (cid:19) (cid:18) (cid:19)
where M is the companion matrix of U and T.
Proof. Using the Binet formula and initial conditions, we easily check that
αn −βn αβn −βαn
U = , T = , ∀n ≥ 0,
n α−β n α−β
where α and β are the zeros of the characteristic polynomial t2 − pt + q. Thus,
rememberingthat αβ = q, we immediately obtain (2.2) . In order to prove (2.3), we
observe that the characteristic polynomial of U and T annihilates M, i.e., we have
M0 = 1 0 , M = 0 1 , Mn = pMn−1−qMn−2, ∀n ≥ 2.
0 1 −q p
(cid:18) (cid:19) (cid:18) (cid:19)
ACCELERATIONS OF GENERALIZED FIBONACCI SEQUENCES 3
So, using (2.2), we clearly find that the entries of Mn in the first column are the
terms T , T of T, and in the second column are the terms U , U of U. (cid:3)
n n+1 n n+1
Dealing with definition(2.1)of x, theratio of two consecutive terms of any sequence
a ∈ W(R) has a closed formula, as we show in the following
Theorem 2.3. Let us consider any sequence a = W(a ,a ,p,q) and the sequence
0 1
a
y = (y )+∞, with y = n , then
n n=2 n an−1
a a x −a q
y = n = 1 n 0 ∀n≥ 2 , (2.4)
n an−1 a0xn+a1−a0p
where x is the n–th term of the sequence x defined by (2.1).
n
Proof. By direct calculation we have
a a U +a T a U +a T
n 1 n 0 n 1 n 0 n
y = = =
n an−1 a1Un−1+a0Tn−1 a1Un−1+a0(pUn−1−qUn−2−pUn−1)
U
n
a −a q
= a1Un −a0qUn−1 = 1Un−1 0 =
a1Un−1+a0Un −a0pUn−1 Un
a +a −a p
0 1 0
Un−1
a x −a q
1 n 0
= .
a x +a −a p
0 n 1 0
(cid:3)
Now, as a very important consequence of the previous theorem, we can evaluate the
terms of the sequence x shifted by m positions
Corollary 2.4. If x is the sequence defined by (2.1), then for all m ≥ 2−n
x x −q
m+1 n
x = .
n+m x +x −p
n m+1
Proof. Let us consider a = W(U ,U ,p,q), that is a = U for all n ≥ 0. So
m m+1 n n+m
the sequence y, introduced in the previous theorem, becomes y = (y = x )+∞,
n n+m n=2
and, from (2.4), we obtain
U x −U q x x −q
m+1 n m m+1 n
x = = . (2.5)
n+m U x +U −U p x +x −p
m n m+1 m n m+1
(cid:3)
ThebasissequencesU andT playacentralroleinthefollowingresults. Theacceler-
ationsforsuchsequencesyieldexplicitformulasfortheaccelerationoftheconvergent
sequence x = (x )+∞. Moreover, some interesting properties on Fibonacci numbers
n n=2
arise as an application. We recall without proof an important theorem about the
generalized Fibonacci sequences.
Theorem 2.5 ([1], Th 3.1 and Cor. 3.3). Denote by V = W(2,p,p,q) the Lucas
sequence with parameters p and q, then
W (0,1,p,q) = W (0,1,p,q)·W (0,1,V ,qm),
mn m n m
4 MARCOABRATE,STEFANOBARBERO,UMBERTOCERRUTI,ANDNADIRMURRU
W (1,0,p,q) = W (1,0,V ,qm)+W (1,0,p,q)W (0,1,V ,qm).
mn n m m n m
Theorem 2.6. Given the sequences U = W(0,1,p,q), T = W(1,0,p,q), and x as
defined in equation (2.1), if we choose a sequence g = W(i,j,s,t), for some i, j ≥ 2,
then
U = a U(s) T(−t) +b T(s) U(−t) +(a b +a b )U(s) U(−t)
gn 2 gn−1 gn−2 2 gn−1 gn−2 1 2 2 4 gn−1 gn−2
T = T(s) T(−t) +a U(s) T(−t) +b T(s) U(−t) +(a b +a b )U(s) U(−t) ,
gn gn−1 gn−2 1 gn−1 gn−2 1 gn−1 gn−2 1 1 2 3 gn−1 gn−2
for every n ≥ 2, where U(s) =W(0,1,V ,qs), T(s) = W(1,0,V ,qs) and
s s
Ms = a1 a2 , M−t = b1 b2 .
a a b b
3 4 3 4
(cid:18) (cid:19) (cid:18) (cid:19)
Moreover, if we choose x and x , for some i, j ≥ 2, as the initial steps of the
i j
acceleration of x, then
q2a x(s) +q2b x(−t) −q(a b +a b )x(s) x(−t)
x = 2 gn−1 2 gn−2 1 2 2 4 gn−1 gn−2 , ∀n≥ 2 , (2.6)
gn q2−qa x(s) −qb x(−t) +(a b +a b )x(s) x(−t)
1 gn−1 1 gn−2 1 1 2 3 gn−1 gn−2
U(s)
where x(s) = −q .
T(s)
Proof. We know as a consequence of Theorem 2.5 that
(s) (s) (s)
T +a U a U
Msgn−1 = gn−1 1 gn−1 2 gn−1 ,
(s) (s) (s)
a3Ugn−1 Tgn−1 +a4Ugn−1!
(−t) (−t) (−t)
M−tgn−2 = Tgn−2 +b1Ugn−2 b2Ugn−2 .
(−t) (−t) (−t)
b3Ugn−2 Tgn−2 +b4Ugn−2!
So from
Tgn Ugn = Mgn = Msgn−1−tgn−2 = Msgn−1M−tgn−2 ,
T U
(cid:18) gn+1 gn+1(cid:19)
weeasily findtheexplicitexpressionsof U andT . Finally, tocomplete theproof,
gn gn
(s) (−t)
we only need to divide both U and T by the product T T and consider
gn gn gn−1 gn−2
their resulting ratio multiplied by −q
(s) (−t) (s) (−t)
U U U U
a gn−1 +b gn−2 +(a b +a b ) gn−1 gn−2
−qUgn = −q 2Tg(ns−)1 2Tg(n−−t)2 1 2 2 4 Tg(ns−)1 Tg(n−−t)2 ,
T (s) (−t) (s) (−t)
gn 1+a Ugn−1 +b Ugn−2 +(a b +a b )Ugn−1Ugn−2
1 (s) 1 (−t) 1 1 2 3 (s) (−t)
T T T T
gn−1 gn−2 gn−1 gn−2
rearranging the terms we obtain the equality (2.6). (cid:3)
Corollary 2.7. When g = W(2,3,1,−1), i.e., gn−3 = Fn, ∀n ≥ 3, where F =
W(0,1,1,−1) is the Fibonacci sequence, then for all n ≥ 5
U = U T +T U +pU U =
Fn Fn−1 Fn−2 Fn−1 Fn−2 Fn−1 Fn−2
= −q UFn−1UFn−2−1+UFn−1−1UFn−2 +pUFn−1UFn−2 ,
x(cid:0) = qxFn−1 +qxFn−2 −pxFn−1x(cid:1)Fn−2 .
Fn q−x x
Fn−1 Fn−2
ACCELERATIONS OF GENERALIZED FIBONACCI SEQUENCES 5
Proof. The identities immediately follow from Theorem 2.6, observing that in this
case we have s = 1, t = −1, consequently Ms = M = M−t, U(s) = U = U(−t) and
T(s) = T = T(−t). (cid:3)
Remark 2.8. When U is the Fibonacci sequence F, the previous Corollary provides
a direct proof of the interesting formula
FFn = FFn−1FFn−2−1+FFn−1−1FFn−2 +FFn−1FFn−2 ,
for all n ≥ 5, avoiding many algebraic manipulations based on the basic properties
of Fibonacci numbers. This formula is also true for n = 3 and n = 4, as a little
calculation can show.
The acceleration shown in the previous corollary is the same found by McCabe
and Phillips in [5] with the secant method applied to the sequence (x )+∞ shifted
n n=2
by one step.
If we consider
T2−qU2 2T U +pU2
M2n = MnMn = n n n n n ,
−q 2T U +pU2 T2−qU2+p 2T U +pU2
n n n n n n n n
(cid:18) (cid:19)
we find U = 2T U +pU2,(cid:0)T = T2−qU(cid:1) 2 , and (cid:0) (cid:1)
2n n n n 2n n n
U 2qx −px2
x = −q 2n = n n . (2.7)
2n T q−x2
2n n
If we start from x , a repeated use of equation (2.7) yields to the subsequence
2
x2,x4,x8,...,x2n,..., corresponding to the subsequence obtained in [5], using the
Newton method on the sequence (x )+∞ shifted by one step.
n n=2
Proposition 2.9. Using the notation of Theorem 2.6, for g = W(h,h + k,2,1),
i.e., g = kn+h ∀n≥ 0, then for all n ≥ 2
n
1
U = qU2 U +2T U T +pU2 T −U T2
gn qgn−2 gn−1 gn−2 gn−1 gn−1 gn−2 gn−1 gn−2 gn−2 gn−1
(cid:16) (cid:17)
1
T = T2 T +pT2 U −qT U2 +2qT U U
gn qgn−2 gn−1 gn−2 gn−1 gn−2 gn−2 gn−1 gn−1 gn−1 gn−2
(cid:16) (cid:17)
and
x2 x +2qx −px2 −qx
x = gn−1 gn−2 gn−1 gn−1 gn−2 .
gn q−px −x2 +2x x
gn−2 gn−1 gn−1 gn−2
Proof. We need to evaluate
Mgn = M2gn−1M−gn−2.
Considering that
M−n = 1 p −1 n = 1 Tn+pUn −Un ,
qn q 0 qn qUn Tn
(cid:18) (cid:19) (cid:18) (cid:19)
we have
1 T2 −qU2 2T U +pU2 T +pU −U
Mgn = gn−1 gn−1 gn−1 gn−1 gn−1 gn−2 gn−2 gn−2 ,
qgn−2(cid:18) ... ... (cid:19)(cid:18) qUgn−2 Tgn−2 (cid:19)
6 MARCOABRATE,STEFANOBARBERO,UMBERTOCERRUTI,ANDNADIRMURRU
and
1
U = qU2 U +2T U T +pU2 T −U T2 ,
gn qgn−2 gn−1 gn−2 gn−1 gn−1 gn−2 gn−1 gn−2 gn−2 gn−1
(cid:16) (cid:17)
1
T = T2 T +pT2 U −qT U2 +2qT U U .
gn qgn−2 gn−1 gn−2 gn−1 gn−2 gn−2 gn−1 gn−1 gn−1 gn−2
(cid:16) (cid:17)
Finally, dividing both U and T by T2 T , their ratio becomes
gn gn gn−1 gn−2
2 2
U U U U U
q gn−1 gn−2 +2 gn−1 +p gn−1 − gn−2
U T T T T T
gn = (cid:18) gn−1(cid:19) gn−2 gn−1 (cid:18) gn−1(cid:19) gn−2
Tgn 1+pUgn−2 −q Ugn−1 2+2qUgn−1Ugn−2
T T T T
gn−2 (cid:18) gn−1(cid:19) gn−1 gn−2
U
from which, rememberingthat x = −q gn, with simplecalculations weobtain the
gn T
thesis. gn (cid:3)
Remark 2.10. Applying the last Proposition to the Fibonacci numbers, we get an-
other example of a beautiful identity, easily proved without hard calculations. In
fact, when we consider g = kn, we have for all n ≥ 2
n
Fkn =(−1)k(n−2)(−Fk2(n−1)Fk(n−2)+2Fk(n−1)Fk(n−1)−1Fk(n−2)−1
+Fk2(n−1)Fk(n−2)−1−Fk(n−2)Fk2(n−1)−1),
which in particular , for k = 1, becomes
Fn = (−1)n −Fn2−1Fn−2+2Fn−1Fn−2Fn−3+Fn2−1Fn−3−Fn3−2 .
(cid:0) 3. Approximation methods (cid:1)
Intheprevioussectionwehavealreadyseenhowsomeaccelerations areconnected
tosomeapproximationmethods,likeNewtonandsecantmethods. Inthissection,as
particular cases of our work, we exactly retrieve the accelerations provided by these
methods and studied in the papers [2] and [5]. Furthermore, usingour approach, we
canshowsimilarresultsfordifferentapproximationmethods,liketheHalleymethod.
Finally, we will examine the acceleration provided by Householder method, which
is a generalization of Newton and Halley methods. First of all, we recall that the
secantmethod,appliedtotheequationf(t)= 0,providesrationalapproximationsof
a root, starting from two approximations x and x , through the recurrencerelation
0 1
f(xn−1)(xn−1−xn−2)
xn = xn−1− f(xn−1)−f(xn−2) , ∀n≥ 2,
which becomes, when f(t)= at2−bt−c
axn−1xn−2−c
x = .
n axn−1+axn−2−b
Theorem 3.1. Let us consider the Fibonacci sequence F and the sequences U =
W(0,1,p,q), T = W(1,0,p,q), and x = (x )+∞, defined by equation (2.1). The
n n=2
subsequence (x )+∞ corresponds to the approximations of the root of larger
Fn+2+1 n=0
modulus of t2−pt+q provided by secant method.
ACCELERATIONS OF GENERALIZED FIBONACCI SEQUENCES 7
Proof. We take the sequence g defined by g = F + 1 for every n ≥ 0. Since
n n+2
gn−1 = Fn+2, we have the recurrent formula gn = gn−1+gn−2−1. Let M be the
companion matrix of the sequence U. From the matrix product
Mgn = Mgn−1Mgn−2M−1
we immediately observe that
1 p
U = U U − T T , T = U T +U T + T T .
gn gn−1 gn−2 q gn−1 gn−2 gn gn−2 gn−1 gn−1 gn−2 q gn−1 gn−2
Thus
U U 1
gn−1 gn−2 −
U T T q
gn = gn−1 gn−2 ,
T U U p
gn gn−2 + gn−1 +
T T q
gn−2 gn−1
and
U x x −q
x = −q gn = gn−1 gn−2 ,
gn T x +x −p
gn gn−2 gn−1
clearly proving the thesis. (cid:3)
The Newton method provides rational approximations for a solution of the equa-
tion f(t)= 0. When f(t)= at2−bt−c, given an initial approximation y , we recall
0
that the Newton approximations sequence satisfies the recurrence relation
yn = yn−1− ayn2−2a1y−n−b1yn−−1b−c = 2aayyn2n−−11+−cb, ∀n≥ 1 . (3.1)
On the other hand, the Halley method generates approximations for a solution of
t2−pt+q = 0 through the following recurrence relation
yn = yn−1+ (yn23−y1n2−−1p−yn3−p1y+n−q1)(+pp−22−ynq−1), ∀n≥ 1 , (3.2)
for a given initial step y . In the next theorem we will show how particular sub-
0
sequences of the sequence x, defined by equation (2.1), generate the Newton and
Halley approximations to the root of larger modulus of t2−pt+q .
Theorem 3.2. Given the sequences U = W(0,1,p,q), T = W(1,0,p,q), and x,
defined by equation (2.1), then the recurrence relations
x2 −q
n
x2n−1 = 2x −p, (3.3)
n
(x2n−1−pxn−1+q)(p−2xn−1)
x3n−2 = xn−1+ 3x2n−1−3pxn−1+p2−q , (3.4)
respectively generates the subsequences of x producing Newton and Halley approxi-
mations for the root of larger modulus of t2−pt+q .
Proof. The proof of relation (3.3) is a direct consequence of Corollary 2.4
x2 −q
n
x2n−1 = xn+n−1 = 2x −p .
n
8 MARCOABRATE,STEFANOBARBERO,UMBERTOCERRUTI,ANDNADIRMURRU
Starting from x , a repeated use of this equation yields the subsequence
2
(x2,x3,x5,...,x2n+1,...) ,
which consists of the Newton approximations for the root of larger modulus of
t2 −pt+q . Relation (3.3) and Corollary 2.4 also provide the proof of recurrence
(3.4) for Halley approximations
x2n−1xn−q
x3n−2 = xn+2n−2 = xn+x2n−1−p =
x2 −q
n x −q
= 2xn −p n = x3n−3qxn+pq =
x2 −q 3x2 −3px +p2−q
x + n −p n n
n 2x −p
n
(x2n−1−pxn−1+q)(p−2xn−1)
= xn−1+ 3x2n−1−3pxn−1+p2−q .
(cid:3)
Tocomplete ouroverview onapproximation methodsandgeneralize theresultsof
previous theorem, we consider the Householder method [3], which provides approxi-
mations fora solution of anonlinear equation f(t)= 0. TheHouseholder recurrence
relation is
(1/f)(d−1)(y )
n
y = y +d , ∀n≥ 0 (3.5)
n+1 n (1/f)(d)(y )
n
for a given initial approximation y , where f(d) is the d–th derivative of f. The
0
Newton and Halley method clearly are particular cases of this method, for d = 1
and d = 2 respectively. Indeed, when f(t) = t2−pt+q from (3.5), when d = 1 or
d= 2, it is easy to retrieve the formulas (3.1) and (3.2).
Theorem 3.3. With the same hypotheses of Theorem 3.2 , if y = x for some
n k
k ≥ 2, then yn+1 = x(d+1)k−d, where yn+1 is obtained from relation (3.5) when
f(t)= t2−pt+q.
Proof. Let α , α be the roots of t2−pt+q, we can write
1 2
1 (d) (t−α )−1−(t−α )−1 (d) (−1)dd! (t−α )d+1−(t−α )d+1
1 2 2 1
= = .
t2−pt+q α −α (α −α )(t−α )d+1(t−α )d+1
! 1 2 ! 1 (cid:0)2 1 2 (cid:1)
Thus, relation (3.5) becomes
(y −α )(y −α ) (y −α )d−(y −α )d
n 1 n 2 n 2 n 1
y = y − ,
n+1 n (y −α )d+1−(y −α )d+1
n 2 (cid:0) n 1 (cid:1)
and with some algebraic manipulations we obtain
α (y −α )d+1−α (y −α )d+1
1 n 2 2 n 1
y = .
n+1 (y −α )d+1−(y −α )d+1
n 2 n 1
ACCELERATIONS OF GENERALIZED FIBONACCI SEQUENCES 9
U αk −αk
If we set y = x = k = 1 2 , then
n k Uk−1 α1k−1−α2k−1
αk −α αk−1 d+1 α αk−1−αk d+1
α 1 2 1 −α 1 2 2
1 αk−1−αk−1! 2 αk−1−αk−1!
1 2 1 2
y = =
n+1 αk −α αk−1 d+1 α αk−1−αk d+1
1 2 1 − 1 2 2
αk−1−αk−1! αk−1−αk−1!
1 2 1 2
α(d+1)k−d(α −α )d+1−α(d+1)k−d(α −α )d+1
= 1 1 2 2 1 2 =
α(d+1)k−d−1(α −α )d+1−α(d+1)k−d−1(α −α )d+1
1 1 2 2 1 2
U(d+1)k−d
= = x(d+1)k−d .
U(d+1)k−d−1
(cid:3)
4. Applications to continued fractions
In order to show some interesting applications of the exposed results, we study
the acceleration of a sequence of rationals which come from the sequence of con-
vergents of a certain continued fraction. The continued fraction, which we will
introduce, provides a periodic representation of period 2 for every quadratic irra-
tionality (except for the square roots). Surely its convergents do not generate the
best approximations for these irrationalities (because we use rational partial quo-
tients, instead of integers), but between them, using the acceleration method of the
previous section, we can find at the same time the approximations derived from the
the secant method, the Newton method and the Halley method.
We remember that a continued fraction is a representation of a real number α
through a sequence of integers as follows:
1
α = a + ,
0
1
a +
1
1
a +
2 a +···
3
where the integers a ,a ,... can be evaluated with the recurrence relations
0 1
a = [α ]
k k
1 k = 0,1,2,... ,
α = if α is not an integer
 k+1 α −a k
k k
for α0 = α(see, e.g., [8]). A continued fraction can be expressed in a compact way
using the notation [a ,a ,a ,a ,...]. The finite continued fraction
0 1 2 3
p
n
[a ,...,a ] = C = , n = 0,1,2,... ,
0 n n
q
n
is a rational number and is called the n–th convergent of [a ,a ,a ,a ,...] and the
0 1 2 3
a ’s are called partial quotients. Furthermore, the sequences (p )+∞ and (q )+∞
i n n=0 n n=0
10 MARCOABRATE,STEFANOBARBERO,UMBERTOCERRUTI,ANDNADIRMURRU
are recusively defined by the equations
pn =anpn−1+pn−2 , ∀n≥ 2 , (4.1)
(qn = anqn−1+qn−2
and initial conditions p = a , p = a a +1, q = 1, q = a . In the following we
0 0 1 0 1 0 1 1
will use rational numbers as partial quotients, instead of the usual integers.
Remark 4.1. In general a continued fraction can have complex numbers as partial
quotients. In this case given a real number there are no algorithms which provide
the partial quotients. However, such a continued fraction can converge to a real
number. In the following, we will study continued fractions of period 2 with rational
partial quotients which are convergents to the root of larger modulus of a quadratic
equation. For a deep study of the convergence of continued fractions, i.e., for an
analytic theory of these objects, see [9].
a a
0 1
Remark 4.2. A continued fraction with rational partial quotients , ,... has
b b
(cid:20) 0 1 (cid:21)
an equivalent form as
a b
0 1
+ ,
b b b
0 1 2
a +
1
b b
2 3
a +
2
a +...
3
but the representation
a 1
0
+
b a 1
0 1
+
b a 1
1 2
+
b2 a3+...
b
3
is more suitable for the study of the convergents, as we will see soon.
a a a
0 1 2
Whenwestudythecontinuedfraction , , ,... ,theequations(4.1)become
b b b
(cid:20) 0 1 2 (cid:21)
a
n
pn = pn−1+pn−2
b
 n , ∀n ≥ 2 , (4.2)
a
 n
qn = qn−1+qn−2
b
n

 a0 a0a1 a1
with initial conditions p = , p = + 1, q = 1, q = , providing two
0 1 0 1
b b b b
0 0 1 1
sequences of rationals . In the next Proposition we study how to determine the
convergents through the ratio of two recurrent sequences of integers.
a a a
Proposition 4.3. Given the continued fraction 0, 1, 2,... , let (p )+∞ and
b b b n n=0
(q )+∞ be the sequences which provide the sequenc(cid:20)e o0f co1nve2rgen(cid:21)ts (C )+∞ defined
n n=0 n n=0