Directory UMM :Data Elmu:jurnal:J-a:Journal of Computational And Applied Mathematics:Vol101.Issue1-2.1999:
Journal of Computational and Applied Mathematics 101 (1999) 167–175
A note on the relation between two convergence acceleration
methods for ordinary continued fractions
Paul Levrie a; b; ∗ , Adhemar Bultheel b
b
a
Department IWT, Karel de Grote - Hogeschool, Salesianenlaan 30, B-2660 Hoboken, Belgium
Department of Computer Science, K.U. Leuven, Celestijnenlaan 200A, B-3001 Heverlee, Belgium
For Haakon Waadeland’s seventieth birthday
Received 12 December 1997; received in revised form 15 September 1998
Abstract
In this note we relate two methods of convergence acceleration for ordinary continued fractions, the rst one is due
to Lorentzen and Waadeland (Jacobsen and Waadeland, 1980, 1998), the second one to Waadeland (1986, 1987, 1987,
c 1999 Elsevier Science B.V. All rights reserved.
1987, 1988).
AMS classication: 40A15; 65B99; 65Q05
Keywords: Continued fractions; Convergence acceleration
1. Introduction and notations
In this note we discuss the relationship between the method of Lorentzen and Waadeland [1,
2] and the method of Waadeland [7–11] for accelerating the convergence of ordinary continued
fractions. We consider the continued fraction
K
an
bn
=
a1
a2
ak
+
+ ··· +
+ ···;
b1
b2
bk
ak ; bk ∈ C; ak 6= 0
and we assume that this continued fraction converges in C.
If we denote by Sn the linear fractional transformation
Sn (w) =
∗
a1
a2
an
+
+ ··· +
;
b1
b2
bn + w
Corresponding author. E-mail: [email protected].
c 1999 Elsevier Science B.V. All rights reserved.
0377-0427/99/$ – see front matter
PII: S 0 3 7 7 - 0 4 2 7 ( 9 8 ) 0 0 2 1 7 - 9
(1.1)
168
P. Levrie, A. Bultheel / Journal of Computational and Applied Mathematics 101 (1999) 167–175
then it is easy to prove that
Sn (w) =
An + wAn−1
;
Bn + wBn−1
where An and Bn are dened by
A−1 = 1; A0 = 0;
B−1 = 0; B0 = 1;
An = bn An−1 + an An−2 ;
Bn = bn Bn−1 + an Bn−2 ; n ¿ 1
An and Bn are called the nth numerator and nth denominator of (1.1), respectively.
The nth approximant fn (in the classical sense) of the continued fraction (1.1) is dened by
fn = Sn (0) = An =Bn .
The nth tail f(n) is given by
f(n) =
an+1
an+2
+
+ ··· :
bn+1
bn+2
The value f = f(0) of the continued fraction (1.1) is dened by
f = lim fn = lim Sn (0):
n→∞
n→∞
(1.2)
The classical way to evaluate continued fraction (1.1) is by calculating Sn (0) for a suciently
large n. Over a number of years various methods were developed to accelerate the convergence of
limit process (1.2). In most of these methods a new sequence fn∗ is produced that converges faster
to f than the sequence of classical approximants Sn (0):
f − fn∗
= 0:
n → ∞ f − fn
lim
(1.3)
In most cases Sn (0) is replaced by Sn (w(n) ) where the sequence {w(n) } is suitably chosen. The
sequence Sn (w(n) ) is called a modication of continued fraction (1.1). In view of relationship
f = Sn (f(n) );
the most obvious choice for w(n) is
w(n) = lim f(k) = w
k→∞
if this limit exists (see [5, 6]). Sn (w) will converge faster than Sn (0). Other choices lead to even
faster convergence. Assume
f(n) = w + n ;
then it is a matter of estimating n by some approximation n′ and under certain conditions Sn (w + n′ )
will converge faster than Sn (w). This is the approach taken in [1, 2]. We have discussed this method
and shown equivalence with a method by Thron and Waadeland in [4].
In this note we consider two other methods which are based on the following idea.
P. Levrie, A. Bultheel / Journal of Computational and Applied Mathematics 101 (1999) 167–175
169
Let us assume that we know another continued fraction K(a˜n ; b˜ n ) with a˜n ∼ an ; b˜ n ∼ bn and
(n)
˜
f ∼ f(n) for n → ∞. (pn ∼ qn for n → ∞ if pn =qn → 1 as n → ∞). Then we can write
f(n) = f˜(n) + n :
So the problem is to nd estimates of n .
Lorentzen and Waadeland [1, 2] show that for limit-periodic continued fractions n is the minimal
solution of a nonlinear recurrence relation. We show that with a linearization of the recurrence
relation, we nd an approximant ˜n for which Sn ( f˜(n) + ˜n ) converges faster than Sn ( f˜(n) ).
Another method by Waadeland [7–11] uses the following approach. Continued fraction (1.1) is
interpreted as a function of the variables a1 ; b1 ; a2 ; b2 ; : : : . This function is then expanded into a
Taylor series and the partial sums of the series are used as approximations to f(n) . If we only use
the linear terms the resulting modication is called the Taylor-modication of order 1.
We show that ˜n corresponds to this Taylor-modication of order 1. Furthermore, it is shown that
a similar linearization can also lead to Taylor-modications of order 2.
This is discussed in the next paragraph. In the third paragraph we look at an application of this
result.
2. Comparison of the two methods
2.1. The method of Lorentzen and Waadeland [1; 2]
First of all we look in some more detail at the method developed by Lorentzen and Waadeland
[1, 2].
We want to compute the value of the convergent continued fraction K(an =bn ), with tail values
f(n) . Let us assume that we know a continued fraction K(a˜n = b˜ n ) and all its tail values f˜(n) , with
an ∼ a˜n ;
bn ∼ b˜ n as n → ∞:
Let us assume that this implies: f(n) ∼ f˜(n) as n → ∞. Then we can write
f(n) = f˜(n) + n
with lim
n→∞
n
= 0:
˜
f(n)
Using this together with the recurrence relation satised by the tails
f(n−1) =
an
;
bn + f(n)
we deduce that yn = n is a solution of the following nonlinear recurrence relation:
f˜(n−1) yn + (bn + f˜(n) + yn )yn−1 = n
with n = an − f˜(n−1) (bn + f˜(n) ).
In what follows, we assume that the continued fraction we start with is limit-periodic:
lim an = a ∈ C0 ;
n→∞
lim bn = b ∈ C
n→∞
(2.1)
170
P. Levrie, A. Bultheel / Journal of Computational and Applied Mathematics 101 (1999) 167–175
and that the roots w1 and w2 of the auxiliary equation x2 + bx − a = 0 satisfy |w1 |¡| w2 |. In this
case the continued fraction converges, limn → ∞ n = 0, and if we choose
f˜(n) = w1
then it is possible to prove that (1.3) holds with fn∗ = Sn ( f˜(n) ).
We shall continue, however, to write things down in the more general setting.
Let us assume that
lim
n→∞
n
=t ∈C
n−1
(with |t| 6 1). Then we also have: ([1, 2])
n
lim
=t
n → ∞ n−1
and, using this in (2.1):
n
∼ f˜(n−1) t + b + f˜(n)
n−1
as n → ∞:
We can rewrite this in the following way:
n ∼
n+1
b + f˜(n+1) + t f˜(n)
as n → ∞:
Hence, if we dene
ˆn =
n+1
(n+1)
˜
b+f
+ t f˜(n)
then ˆn is an approximation to n , and in [1, 2] it is shown that
f − Sn ( f˜(n) + ˆn )
= 0:
n→∞
f − Sn ( f˜(n) )
lim
This is the method of Lorentzen and Waadeland discussed in [1, 2]. Note that in [3] we use another
choice for w(n) based on:
n ∼ n =
n+1
˜
bn+1 + f(n+1) + t f˜(n)
as n → ∞:
Let us look at a variation of this method. Instead of using recurrence relation (2.1) we replace it
by a linear one
f˜(n−1) yn + (bn + f˜(n) )yn−1 = n
(2.2)
leaving out the nonlinear part yn yn−1 which will be small compared to the other terms. It is easy
to prove that this new recurrence relation has a minimal solution (see [3]) yn = ˜n :
lim
n→∞
˜n
=0
zn
(2.3)
P. Levrie, A. Bultheel / Journal of Computational and Applied Mathematics 101 (1999) 167–175
171
for every solution zn of the homogeneous equation associated with (2.2). Furthermore it also satises
lim
n→∞
˜n
= t:
˜n−1
(To prove this we can use the same method as in [3].)
From (2.1) we get
f˜(n−1) ˜n = ˜n−1 + (bn + f˜(n) )
n = ˜n−1
= lim
=1
n → ∞ n =n−1
n→∞ f
˜(n−1) n =n−1 + (bn + f˜(n) + n )
lim
and, hence,
lim
n→∞
n
=1
˜n
and
lim
n→∞
ˆn
= 1:
˜n
Note that we could equally well have used
f˜(n−1) yn + (b + f˜(n) )yn−1 = n
(2.4)
instead of Eq. (2.2).
Using the methods from [1, 2], we can now prove the following result:
Theorem. Under the assumptions given above; we have
f − Sn ( f˜(n) + ˜n )
= 0:
n→∞
f − Sn ( f˜(n) )
lim
2.2. The method of Waadeland [7–11]
Since ˜n is minimal solution of recurrence relation (2.2), it is possible to compute it using backward
recursion: if we denote by ˜n(N ) the solution of (2.2) with
˜N(N+1) = 0;
then we have
lim ˜n(N ) = ˜n :
N →∞
(This follows from
˜n(N ) = ˜n −
˜N +1
zn ;
zN +1
where zn is as in (2.3).)
We now calculate an explicit expression for ˜n . Using (2.2) and starting from ˜N(N+1) = 0, we get
˜N(N ) =
N +1
bN +1 + f˜ (N +1)
172
P. Levrie, A. Bultheel / Journal of Computational and Applied Mathematics 101 (1999) 167–175
and
)
˜N(N−1
=
N
f˜ (N −1) N +1
:
−
bN + f˜ (N ) (bN + f˜ (N ) )(bN +1 + f˜ (N +1) )
Continuing in this way, we nd
˜n(N ) =
N
−n
X
i=0
i
Y
f˜(n)
f˜(m+n)
−
f˜(i+n) (bi+n+1 + f˜(i+n+1) ) m=1
bm+n + f˜(m+n)
!
i+n+1 :
We now let N → ∞,
˜(n)
˜n = f
∞
X
i=0
i
Y
1
f˜(i+n) (bi+n+1 + f˜(i+n+1) ) m = 1
f˜(m+n)
−
bm+n + f˜(m+n)
!
i+n+1 :
For n = 0 this gives us
˜(0)
˜0 = f
∞
X
i=0
i
Y
1
f˜(m)
−
f˜(i) (bi+1 + f˜(i+1) ) m = 1
bm + f˜(m)
!
i+1 :
The modication Sn ( f˜(n) + ˜n ) of the continued fraction K(an =bn ) is exactly the Taylor-modication
of the rst-order dened and discussed in [7–11] for the special case bn = 1 for all n. However, the
method by which it is obtained is very dierent from ours.
The idea behind Taylor-modications is the following: A convergent continued fraction K(an =1)
can be seen as a function of an innite number of variables
F(a1 ; a2 ; : : : ; an ; : : :) = K
an
:
1
If we assume that all its tail values f(n) are nite, the derivative of F with respect to one of its
variables is given by
Y
@F
f(m)
f n−1
−
=
@an an m = 1
1 + f(m)
!
(see [8]). We now expand the function F in a Taylor series around the point (a1 ; a2 ; : : : ; an ; : : :) =
(a˜1 ; a˜2 ; : : : ; a˜n ; : : :):
∞
X
@F
F(a1 ; a2 ; : : : ; an ; : : :) = F(a˜1 ; a˜2 ; : : : ; a˜n ; : : :) +
@a
i=1
∞
i−1
X
1 Y
˜(0)
⇒ F(a1 ; a2 ; : : : ; an ; : : :) = f˜(0) + f
i=1
a˜i
m=1
· (ai − a˜i ) + higher order terms;
i aj = a˜j
f˜(m)
−
1 + f˜(m)
!
i + h: o: terms
and similarly
∞
X
1
˜(n)
F(an+1 ; an+2 ; : : :) = f˜(n) + f
i=1
a˜i+n
i−1
Y
m=1
f˜(m+n)
−
1 + f˜(m+n)
!
i+n + h: o: terms:
173
P. Levrie, A. Bultheel / Journal of Computational and Applied Mathematics 101 (1999) 167–175
If we replace the tail f(n) of the original continued fraction by the sum of the rst two terms in
this expression, we call the result the Taylor-modication of order 1.
Using the same method it is possible to recuperate the Taylor-modication of second order (see
[11]). If we put
f(n) = f˜(n) + ˜n + n
in the recurrence relation for the tails, then we get a recurrence relation for the sequence n :
( f˜(n−1) + ˜n−1 )n + (bn + f˜(n) + ˜n + n )n−1 = − ˜n ˜n−1 :
We leave out the products of ’s
˜ and ’s, assuming these terms will be small compared to the others,
and we dene ˜ n as the minimal solution of the recurrence relation:
f˜(n−1) yn + (bn + f˜(n) )yn−1 = − ˜n−1 ˜n :
(2.5)
Note that the left-hand side of this equation is the same as that of Eq. (2.3), and so it is easy to
see that
˜(n)
˜ n = − f
∞
X
i=0
1
i
Y
f˜(i+n) (bi+n+1 + f˜(i+n+1) ) m = 1
f˜(m+n)
−
bm+n + f˜(m+n)
!
˜n+i ˜n+i+1 :
We have checked that Sn ( f˜(n) + ˜n + ˜ n ) coincides with the second-order Taylor-modication in the
special case
bn = 1:
3. An application
We look at the special case
an = a + C · T n ;
bn = 1
with a ∈ C\] − ∞; − 41 ]; |T | ¡ 1
which has been studied in [11]. Note that for the continued fraction K(a=1) we have for all n:
f˜(n) =
with
the root of x2 + x − a with smallest modulus. The other root is −(1 + ). (We use the
notations from [11].)
Using the results from the previous paragraph, it is now very easy to nd the Taylor-modications
of the rst and the second order for K(an =1). We only have to calculate ˜n and ˜ n . Now, ˜n is the
minimal solution of (2.2):
yn + (1 + )yn−1 = CT n :
This recurrence relation clearly has a solution of the form
yn = A1 · T n
(3.1)
174
P. Levrie, A. Bultheel / Journal of Computational and Applied Mathematics 101 (1999) 167–175
and this will be the minimal solution, since the homogeneous equation associated with (3.1) has the
solution
(−1)n
1+
n
for which we have
A1 · T n
n = 0:
n → ∞ (−1)n ((1 + )= )
lim
(Note that |(1 + )= |¿1.)
We can determine the coecient A1 using (3.1),
A1 T n + (1 + )A1 T n−1 = CT n ⇒ A1 =
1+
CT
:
+ T
Hence,
˜n =
CT n+1
:
1+ + T
It follows from the theorem in the previous paragraph that Sn ( + ˜n ) converges faster than Sn ( ).
We calculate ˜ n using the same method. yn = ˜ n satises (2.5)
yn + (1 + )yn−1 = − ˜n ˜n−1 ⇔
yn + (1 + )yn−1 = −
C 2 T 2n+1
:
(1 + + T )2
(3.2)
This equation has a minimal solution of the form
yn = A2 T 2n :
Substituting this into Eq. (3.2) we nd
A2 = −
(1 +
C 2T 3
+ T )2 (1 +
+ T 2)
⇒ ˜ n = −
(1 +
C 2 T 2n+3
+ T )2 (1 +
+ T 2)
:
It is easy to prove (using the method from [3]) that Sn ( + ˜n + ˜ n ) converges faster than Sn ( + ˜n ).
The two modications discussed above may be found in [11]. We note, however, that we can
continue in the same way to get better and better appproximations to f˜(n) , with much less eort
than in [11]. For instance, if we assume that
f(n) =
+ ˜n + ˜ n + n ;
then the corresponding approximation yn = ˜n to n can be shown to satisfy
yn + (1 + )yn−1 = − ˜n−1 ˜ n − ˜n ˜ n−1
(using the same technique used to obtain (2.5)). This is equivalent with
yn + (1 + )yn−1 =
(1 +
C 3 (1 + T )T 3n+2
+ T )3 (1 + + T 2 )
P. Levrie, A. Bultheel / Journal of Computational and Applied Mathematics 101 (1999) 167–175
175
and solving this equation for the minimal solution we nd
˜n =
(1 +
C 3 (1 + T )T 3n+5
+ T )3 (1 + + T 2 )(1 +
+ T 3)
:
Sn ( + ˜n + ˜ n + ˜n ) converges faster than Sn ( + ˜n + ˜ n ).
References
[1] L. Jacobsen, H. Waadeland, An asymptotic property for tails of limit periodic continued fractions, Rocky Moutain
J. Math. 20(1) (1990) 151–163.
[2] L. Jacobsen, H. Waadeland, Convergence acceleration for limit periodic continued fractions under asymptotic side
conditions, Numer. Math. 53 (1988) 285–298.
[3] P. Levrie, Improving a method for computing non-dominant solutions of certain second-order recurrence relations of
Poincare-type, Numer. Math. 56 (1989) 501–512.
[4] P. Levrie, A. Bultheel, A note on two convergence acceleration methods for ordinary continued fractions, J. Comput.
Appl. Math. 24 (1988) 403– 409.
[5] L. Lorentzen, Computation of limit periodic continued fractions. A survey, Numer. Algorithms 10 (1995) 69–111.
[6] L. Lorentzen, H. Waadeland, Continued Fractions with Applications, North-Holland, Amsterdam, 1992.
[7] H. Waadeland, A note on partial derivatives of continued fractions, Analytic Theory of Continued Fractions, vol. 2,
Lecture Notes Math., vol. 1199, Springer, Berlin, 1986, pp. 294 –299.
[8] H. Waadeland, Local properties of continued fractions, Rational Approximation and Applications in Mathematics
and Physics, Lecture Notes Math., vol. 1237, Springer, Berlin, 1987, pp. 239–250.
[9] H. Waadeland, Derivatives of continued fractions with applications to hypergeometric functions, J. Comput. Appl.
Math. 19 (1987) 161–169.
[10] H. Waadeland, Linear approximations to continued fractions K(zn =1), J. Comput. Appl. Math. 20 (1987) 403– 415.
[11] H. Waadeland, Some recent results in the analytic theory of continued fractions, in: A. Cuyt (Ed.), Nonlinear
Numerical Methods and Rational Approximation, Reidel Publ., Dordrecht, 1988, pp. 299–333.
A note on the relation between two convergence acceleration
methods for ordinary continued fractions
Paul Levrie a; b; ∗ , Adhemar Bultheel b
b
a
Department IWT, Karel de Grote - Hogeschool, Salesianenlaan 30, B-2660 Hoboken, Belgium
Department of Computer Science, K.U. Leuven, Celestijnenlaan 200A, B-3001 Heverlee, Belgium
For Haakon Waadeland’s seventieth birthday
Received 12 December 1997; received in revised form 15 September 1998
Abstract
In this note we relate two methods of convergence acceleration for ordinary continued fractions, the rst one is due
to Lorentzen and Waadeland (Jacobsen and Waadeland, 1980, 1998), the second one to Waadeland (1986, 1987, 1987,
c 1999 Elsevier Science B.V. All rights reserved.
1987, 1988).
AMS classication: 40A15; 65B99; 65Q05
Keywords: Continued fractions; Convergence acceleration
1. Introduction and notations
In this note we discuss the relationship between the method of Lorentzen and Waadeland [1,
2] and the method of Waadeland [7–11] for accelerating the convergence of ordinary continued
fractions. We consider the continued fraction
K
an
bn
=
a1
a2
ak
+
+ ··· +
+ ···;
b1
b2
bk
ak ; bk ∈ C; ak 6= 0
and we assume that this continued fraction converges in C.
If we denote by Sn the linear fractional transformation
Sn (w) =
∗
a1
a2
an
+
+ ··· +
;
b1
b2
bn + w
Corresponding author. E-mail: [email protected].
c 1999 Elsevier Science B.V. All rights reserved.
0377-0427/99/$ – see front matter
PII: S 0 3 7 7 - 0 4 2 7 ( 9 8 ) 0 0 2 1 7 - 9
(1.1)
168
P. Levrie, A. Bultheel / Journal of Computational and Applied Mathematics 101 (1999) 167–175
then it is easy to prove that
Sn (w) =
An + wAn−1
;
Bn + wBn−1
where An and Bn are dened by
A−1 = 1; A0 = 0;
B−1 = 0; B0 = 1;
An = bn An−1 + an An−2 ;
Bn = bn Bn−1 + an Bn−2 ; n ¿ 1
An and Bn are called the nth numerator and nth denominator of (1.1), respectively.
The nth approximant fn (in the classical sense) of the continued fraction (1.1) is dened by
fn = Sn (0) = An =Bn .
The nth tail f(n) is given by
f(n) =
an+1
an+2
+
+ ··· :
bn+1
bn+2
The value f = f(0) of the continued fraction (1.1) is dened by
f = lim fn = lim Sn (0):
n→∞
n→∞
(1.2)
The classical way to evaluate continued fraction (1.1) is by calculating Sn (0) for a suciently
large n. Over a number of years various methods were developed to accelerate the convergence of
limit process (1.2). In most of these methods a new sequence fn∗ is produced that converges faster
to f than the sequence of classical approximants Sn (0):
f − fn∗
= 0:
n → ∞ f − fn
lim
(1.3)
In most cases Sn (0) is replaced by Sn (w(n) ) where the sequence {w(n) } is suitably chosen. The
sequence Sn (w(n) ) is called a modication of continued fraction (1.1). In view of relationship
f = Sn (f(n) );
the most obvious choice for w(n) is
w(n) = lim f(k) = w
k→∞
if this limit exists (see [5, 6]). Sn (w) will converge faster than Sn (0). Other choices lead to even
faster convergence. Assume
f(n) = w + n ;
then it is a matter of estimating n by some approximation n′ and under certain conditions Sn (w + n′ )
will converge faster than Sn (w). This is the approach taken in [1, 2]. We have discussed this method
and shown equivalence with a method by Thron and Waadeland in [4].
In this note we consider two other methods which are based on the following idea.
P. Levrie, A. Bultheel / Journal of Computational and Applied Mathematics 101 (1999) 167–175
169
Let us assume that we know another continued fraction K(a˜n ; b˜ n ) with a˜n ∼ an ; b˜ n ∼ bn and
(n)
˜
f ∼ f(n) for n → ∞. (pn ∼ qn for n → ∞ if pn =qn → 1 as n → ∞). Then we can write
f(n) = f˜(n) + n :
So the problem is to nd estimates of n .
Lorentzen and Waadeland [1, 2] show that for limit-periodic continued fractions n is the minimal
solution of a nonlinear recurrence relation. We show that with a linearization of the recurrence
relation, we nd an approximant ˜n for which Sn ( f˜(n) + ˜n ) converges faster than Sn ( f˜(n) ).
Another method by Waadeland [7–11] uses the following approach. Continued fraction (1.1) is
interpreted as a function of the variables a1 ; b1 ; a2 ; b2 ; : : : . This function is then expanded into a
Taylor series and the partial sums of the series are used as approximations to f(n) . If we only use
the linear terms the resulting modication is called the Taylor-modication of order 1.
We show that ˜n corresponds to this Taylor-modication of order 1. Furthermore, it is shown that
a similar linearization can also lead to Taylor-modications of order 2.
This is discussed in the next paragraph. In the third paragraph we look at an application of this
result.
2. Comparison of the two methods
2.1. The method of Lorentzen and Waadeland [1; 2]
First of all we look in some more detail at the method developed by Lorentzen and Waadeland
[1, 2].
We want to compute the value of the convergent continued fraction K(an =bn ), with tail values
f(n) . Let us assume that we know a continued fraction K(a˜n = b˜ n ) and all its tail values f˜(n) , with
an ∼ a˜n ;
bn ∼ b˜ n as n → ∞:
Let us assume that this implies: f(n) ∼ f˜(n) as n → ∞. Then we can write
f(n) = f˜(n) + n
with lim
n→∞
n
= 0:
˜
f(n)
Using this together with the recurrence relation satised by the tails
f(n−1) =
an
;
bn + f(n)
we deduce that yn = n is a solution of the following nonlinear recurrence relation:
f˜(n−1) yn + (bn + f˜(n) + yn )yn−1 = n
with n = an − f˜(n−1) (bn + f˜(n) ).
In what follows, we assume that the continued fraction we start with is limit-periodic:
lim an = a ∈ C0 ;
n→∞
lim bn = b ∈ C
n→∞
(2.1)
170
P. Levrie, A. Bultheel / Journal of Computational and Applied Mathematics 101 (1999) 167–175
and that the roots w1 and w2 of the auxiliary equation x2 + bx − a = 0 satisfy |w1 |¡| w2 |. In this
case the continued fraction converges, limn → ∞ n = 0, and if we choose
f˜(n) = w1
then it is possible to prove that (1.3) holds with fn∗ = Sn ( f˜(n) ).
We shall continue, however, to write things down in the more general setting.
Let us assume that
lim
n→∞
n
=t ∈C
n−1
(with |t| 6 1). Then we also have: ([1, 2])
n
lim
=t
n → ∞ n−1
and, using this in (2.1):
n
∼ f˜(n−1) t + b + f˜(n)
n−1
as n → ∞:
We can rewrite this in the following way:
n ∼
n+1
b + f˜(n+1) + t f˜(n)
as n → ∞:
Hence, if we dene
ˆn =
n+1
(n+1)
˜
b+f
+ t f˜(n)
then ˆn is an approximation to n , and in [1, 2] it is shown that
f − Sn ( f˜(n) + ˆn )
= 0:
n→∞
f − Sn ( f˜(n) )
lim
This is the method of Lorentzen and Waadeland discussed in [1, 2]. Note that in [3] we use another
choice for w(n) based on:
n ∼ n =
n+1
˜
bn+1 + f(n+1) + t f˜(n)
as n → ∞:
Let us look at a variation of this method. Instead of using recurrence relation (2.1) we replace it
by a linear one
f˜(n−1) yn + (bn + f˜(n) )yn−1 = n
(2.2)
leaving out the nonlinear part yn yn−1 which will be small compared to the other terms. It is easy
to prove that this new recurrence relation has a minimal solution (see [3]) yn = ˜n :
lim
n→∞
˜n
=0
zn
(2.3)
P. Levrie, A. Bultheel / Journal of Computational and Applied Mathematics 101 (1999) 167–175
171
for every solution zn of the homogeneous equation associated with (2.2). Furthermore it also satises
lim
n→∞
˜n
= t:
˜n−1
(To prove this we can use the same method as in [3].)
From (2.1) we get
f˜(n−1) ˜n = ˜n−1 + (bn + f˜(n) )
n = ˜n−1
= lim
=1
n → ∞ n =n−1
n→∞ f
˜(n−1) n =n−1 + (bn + f˜(n) + n )
lim
and, hence,
lim
n→∞
n
=1
˜n
and
lim
n→∞
ˆn
= 1:
˜n
Note that we could equally well have used
f˜(n−1) yn + (b + f˜(n) )yn−1 = n
(2.4)
instead of Eq. (2.2).
Using the methods from [1, 2], we can now prove the following result:
Theorem. Under the assumptions given above; we have
f − Sn ( f˜(n) + ˜n )
= 0:
n→∞
f − Sn ( f˜(n) )
lim
2.2. The method of Waadeland [7–11]
Since ˜n is minimal solution of recurrence relation (2.2), it is possible to compute it using backward
recursion: if we denote by ˜n(N ) the solution of (2.2) with
˜N(N+1) = 0;
then we have
lim ˜n(N ) = ˜n :
N →∞
(This follows from
˜n(N ) = ˜n −
˜N +1
zn ;
zN +1
where zn is as in (2.3).)
We now calculate an explicit expression for ˜n . Using (2.2) and starting from ˜N(N+1) = 0, we get
˜N(N ) =
N +1
bN +1 + f˜ (N +1)
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P. Levrie, A. Bultheel / Journal of Computational and Applied Mathematics 101 (1999) 167–175
and
)
˜N(N−1
=
N
f˜ (N −1) N +1
:
−
bN + f˜ (N ) (bN + f˜ (N ) )(bN +1 + f˜ (N +1) )
Continuing in this way, we nd
˜n(N ) =
N
−n
X
i=0
i
Y
f˜(n)
f˜(m+n)
−
f˜(i+n) (bi+n+1 + f˜(i+n+1) ) m=1
bm+n + f˜(m+n)
!
i+n+1 :
We now let N → ∞,
˜(n)
˜n = f
∞
X
i=0
i
Y
1
f˜(i+n) (bi+n+1 + f˜(i+n+1) ) m = 1
f˜(m+n)
−
bm+n + f˜(m+n)
!
i+n+1 :
For n = 0 this gives us
˜(0)
˜0 = f
∞
X
i=0
i
Y
1
f˜(m)
−
f˜(i) (bi+1 + f˜(i+1) ) m = 1
bm + f˜(m)
!
i+1 :
The modication Sn ( f˜(n) + ˜n ) of the continued fraction K(an =bn ) is exactly the Taylor-modication
of the rst-order dened and discussed in [7–11] for the special case bn = 1 for all n. However, the
method by which it is obtained is very dierent from ours.
The idea behind Taylor-modications is the following: A convergent continued fraction K(an =1)
can be seen as a function of an innite number of variables
F(a1 ; a2 ; : : : ; an ; : : :) = K
an
:
1
If we assume that all its tail values f(n) are nite, the derivative of F with respect to one of its
variables is given by
Y
@F
f(m)
f n−1
−
=
@an an m = 1
1 + f(m)
!
(see [8]). We now expand the function F in a Taylor series around the point (a1 ; a2 ; : : : ; an ; : : :) =
(a˜1 ; a˜2 ; : : : ; a˜n ; : : :):
∞
X
@F
F(a1 ; a2 ; : : : ; an ; : : :) = F(a˜1 ; a˜2 ; : : : ; a˜n ; : : :) +
@a
i=1
∞
i−1
X
1 Y
˜(0)
⇒ F(a1 ; a2 ; : : : ; an ; : : :) = f˜(0) + f
i=1
a˜i
m=1
· (ai − a˜i ) + higher order terms;
i aj = a˜j
f˜(m)
−
1 + f˜(m)
!
i + h: o: terms
and similarly
∞
X
1
˜(n)
F(an+1 ; an+2 ; : : :) = f˜(n) + f
i=1
a˜i+n
i−1
Y
m=1
f˜(m+n)
−
1 + f˜(m+n)
!
i+n + h: o: terms:
173
P. Levrie, A. Bultheel / Journal of Computational and Applied Mathematics 101 (1999) 167–175
If we replace the tail f(n) of the original continued fraction by the sum of the rst two terms in
this expression, we call the result the Taylor-modication of order 1.
Using the same method it is possible to recuperate the Taylor-modication of second order (see
[11]). If we put
f(n) = f˜(n) + ˜n + n
in the recurrence relation for the tails, then we get a recurrence relation for the sequence n :
( f˜(n−1) + ˜n−1 )n + (bn + f˜(n) + ˜n + n )n−1 = − ˜n ˜n−1 :
We leave out the products of ’s
˜ and ’s, assuming these terms will be small compared to the others,
and we dene ˜ n as the minimal solution of the recurrence relation:
f˜(n−1) yn + (bn + f˜(n) )yn−1 = − ˜n−1 ˜n :
(2.5)
Note that the left-hand side of this equation is the same as that of Eq. (2.3), and so it is easy to
see that
˜(n)
˜ n = − f
∞
X
i=0
1
i
Y
f˜(i+n) (bi+n+1 + f˜(i+n+1) ) m = 1
f˜(m+n)
−
bm+n + f˜(m+n)
!
˜n+i ˜n+i+1 :
We have checked that Sn ( f˜(n) + ˜n + ˜ n ) coincides with the second-order Taylor-modication in the
special case
bn = 1:
3. An application
We look at the special case
an = a + C · T n ;
bn = 1
with a ∈ C\] − ∞; − 41 ]; |T | ¡ 1
which has been studied in [11]. Note that for the continued fraction K(a=1) we have for all n:
f˜(n) =
with
the root of x2 + x − a with smallest modulus. The other root is −(1 + ). (We use the
notations from [11].)
Using the results from the previous paragraph, it is now very easy to nd the Taylor-modications
of the rst and the second order for K(an =1). We only have to calculate ˜n and ˜ n . Now, ˜n is the
minimal solution of (2.2):
yn + (1 + )yn−1 = CT n :
This recurrence relation clearly has a solution of the form
yn = A1 · T n
(3.1)
174
P. Levrie, A. Bultheel / Journal of Computational and Applied Mathematics 101 (1999) 167–175
and this will be the minimal solution, since the homogeneous equation associated with (3.1) has the
solution
(−1)n
1+
n
for which we have
A1 · T n
n = 0:
n → ∞ (−1)n ((1 + )= )
lim
(Note that |(1 + )= |¿1.)
We can determine the coecient A1 using (3.1),
A1 T n + (1 + )A1 T n−1 = CT n ⇒ A1 =
1+
CT
:
+ T
Hence,
˜n =
CT n+1
:
1+ + T
It follows from the theorem in the previous paragraph that Sn ( + ˜n ) converges faster than Sn ( ).
We calculate ˜ n using the same method. yn = ˜ n satises (2.5)
yn + (1 + )yn−1 = − ˜n ˜n−1 ⇔
yn + (1 + )yn−1 = −
C 2 T 2n+1
:
(1 + + T )2
(3.2)
This equation has a minimal solution of the form
yn = A2 T 2n :
Substituting this into Eq. (3.2) we nd
A2 = −
(1 +
C 2T 3
+ T )2 (1 +
+ T 2)
⇒ ˜ n = −
(1 +
C 2 T 2n+3
+ T )2 (1 +
+ T 2)
:
It is easy to prove (using the method from [3]) that Sn ( + ˜n + ˜ n ) converges faster than Sn ( + ˜n ).
The two modications discussed above may be found in [11]. We note, however, that we can
continue in the same way to get better and better appproximations to f˜(n) , with much less eort
than in [11]. For instance, if we assume that
f(n) =
+ ˜n + ˜ n + n ;
then the corresponding approximation yn = ˜n to n can be shown to satisfy
yn + (1 + )yn−1 = − ˜n−1 ˜ n − ˜n ˜ n−1
(using the same technique used to obtain (2.5)). This is equivalent with
yn + (1 + )yn−1 =
(1 +
C 3 (1 + T )T 3n+2
+ T )3 (1 + + T 2 )
P. Levrie, A. Bultheel / Journal of Computational and Applied Mathematics 101 (1999) 167–175
175
and solving this equation for the minimal solution we nd
˜n =
(1 +
C 3 (1 + T )T 3n+5
+ T )3 (1 + + T 2 )(1 +
+ T 3)
:
Sn ( + ˜n + ˜ n + ˜n ) converges faster than Sn ( + ˜n + ˜ n ).
References
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