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(n)
1 (z) =âˆ’ ; n âˆˆ N0 ; (4.19b)
[ n+2 z âˆ’ n+1 ] âˆ’ n+1 [ n+3 z âˆ’ n+2 ]
n+3

(n)
Nk+1 (z)
(n) (n+3)
k+1 (z) = (z) âˆ’ (n) ; k; n âˆˆ N0 ; (4.19c)
k
Dk+1 (z)
(n) (n+2) (n+2) (n)
(z)][ n+3k+2 z âˆ’ n+3k+1 + 2 k (z)]
Nk+1 (z) = [ + (z)]{[ n+3k+3 +
n+3k+3 k k
(n+1) (n) (n+2)
(z)]2 âˆ’ [ n+3k+1 +
+[ n+3k+2 + k (z)][ n+3k+3 + (z)]}; (4.19d)
k k

(n) (n+2) (n)
(z)][ n+3k+2 z âˆ’ n+3k+1 + 2
Dk+1 (z) = [ + k (z)]
n+3k+3 k
(n) (n+1)
2
âˆ’[ + k (z)][ n+3k+3 z âˆ’ n+3k+2 + (z)]: (4.19e)
n+3k+1 k
(n+2) (n+2)
(z) and 2 k (z) are deÃ¿ned by (2:15).
Here, k
A comparison of (4.6) and (4.18) yields
(n) (n)
k (z) = Gk + O(z); z â†’ 0: (4.20)
(n) (n)
Consequently, the z-independent part Gk of k (z) is the prediction for the Ã¿rst coe cient n+3k+1
(n)
not used for the computation of Jk .
If we set z = 0 in the recursive scheme (4:19) and use (4.20), we obtain the following recursive
(n)
scheme for the predictions Gk :
(n)
G0 = 0; n âˆˆ N0 ; (4.21a)
2
n+3 {[ n+2 ] âˆ’2 n+1 n+3 }
(n)
G1 =âˆ’ ; n âˆˆ N0 ; (4.21b)
n+1 n+2

(n)
Fk+1
(n) (n+3)
Gk+1 = Gk âˆ’ (n) ; k; n âˆˆ N0 ; (4.21c)
Hk+1
346 E.J. Weniger / Journal of Computational and Applied Mathematics 122 (2000) 329â€“356

(n) (n+2) (n+1) (n) (n+2)
âˆ’ Gk ]2 âˆ’ 2[
Fk+1 = [ âˆ’ Gk ]{[ âˆ’ Gk ][ âˆ’ Gk ]}; (4.21d)
n+3k+3 n+3k+2 n+3k+1 n+3k+3

(n) (n) (n+1)
Hk+1 = [ âˆ’ Gk ][ âˆ’ Gk ]: (4.21e)
n+3k+1 n+3k+2
(n) (n) (n) (n)
The z-independent parts Ck of Rk (z) and Gk of k (z), respectively, are connected. A com-
parison of (4.13), (4.15), (4.18), and (4.20) yields
(n) (n)
Gk = Ck + n+3k+1 : (4.22)
As in the case of Aitkenâ€™s iterated 2 process or Wynnâ€™s epsilon algorithm, a new approximation
to the limit will be computed after the computation of each new partial sum. Thus, if the index m
of the last partial sum fm (z) is a multiple of 3, m = 3 , we use as approximation to the limit f(z)
the transformation
{f0 (z); f1 (z); : : : ; f3 (z)} â†’ J (0) ; (4.23)
if we have m = 3 + 1, we use the transformation
â†’ J(1) ;
{f1 (z); f2 (z); : : : ; f3 +1 (z)} (4.24)
and if we have m = 3 + 2, we use the transformation
â†’ J(2) ;
{f2 (z); f3 (z); : : : ; f3 +2 (z)} (4.25)
These three relationships can be combined into a single equation, yielding [95, Eq. (10:4-7)]
(mâˆ’3<m=3=)
{fmâˆ’3<m=3= (z); fmâˆ’3<m=3=+1 (z); : : : ; fm (z)} â†’ J<m=3= ; m âˆˆ N0 : (4.26)

5. Applications

In this article, two principally di erent kinds of results were derived. The Ã¿rst group of
results â€” the accuracy-through-order relationships (2.13), (3.9), and (4.13) and the corresponding
(n)
recursive schemes (2:14), (3.9), and (4:14) â€” deÃ¿nes the transformation error terms z n+2k+1 Rk (z),
z n+2k+1 r2k (z), and z n+3k+1 Rk (z). These quantities describe how the rational approximants A(n) , 2k ,
(n) (n) (n)
k
(n)
and Jk di er from the function f(z) which is to be approximated. Obviously, the transformation
error terms must vanish if the transformation process converges.
The second group of results â€” (2.19), (3.14), and (4.18) and the corresponding recursive schemes
(n) (n) (n)
(2:20), (3:15), and (4:19) â€” deÃ¿nes the terms z n+2k+1 k (z), z n+2k+1 â€™2k (z), and z n+3k+1 k (z). These
(n) (n) (n)
quantities describe how the rational approximants Ak , 2k , and Jk di er from the partial sums
fn+2k (z) and fn+3k (z), respectively, from which they were constructed. Hence, the Ã¿rst group of
results essentially describes what is still missing in the transformation process, whereas the second
group describes what was gained by constructing rational expressions from the partial sums.
The recursive schemes (2:14), (3.9), and (4:14) of the Ã¿rst group use as input data the remainder
terms
âˆž
fn (z) âˆ’ f(z)
=âˆ’ n+ +1 z : (5.1)
z n+1 =0

In most practically relevant convergence acceleration and summation problems, only a Ã¿nite number
of series coe cients are known. Consequently, the remainder terms (5.1) are usually not known
E.J. Weniger / Journal of Computational and Applied Mathematics 122 (2000) 329â€“356 347

explicitly, which means that the immediate practical usefulness of the Ã¿rst group of results is quite
limited. Nevertheless, these results are of interest because they can be used to study the convergence
As an example, let us consider the following series expansion for the logarithm
âˆž
(âˆ’z)m
ln(1 + z)
= 2 F1 (1; 1; 2; âˆ’z) = ; (5.2)
z m+1
m=0

which converges for all z âˆˆ C with |z| Â¡ 1. The logarithm possesses the integral representation
1
ln(1 + z) dt
= ; (5.3)
z 1 + zt
0

which shows that ln(1+z)=z is a Stieltjes function and that the hypergeometric series on the right-hand
side of (5.2) is the corresponding Stieltjes series (a detailed treatment of Stieltjes functions and
Stieltjes series can for example be found in Section 5 of Baker and Graves-Morris [8]). Consequently,
ln(1 + z)=z possesses the following representation as a partial sum plus an explicit remainder which
is given by a Stieltjes integral (compare for example Eq. (13:1-5) of Weniger [95]):
n
(âˆ’z)m 1
t n+1 dt
ln(1 + z)
+ (âˆ’z)n+1
= ; n âˆˆ N0 : (5.4)
z m+1 1 + zt
0
m=0

For |z| Â¡ 1, the denominator of the remainder integral on the right-hand side can be expanded.
Interchanging summation and integration then yields
âˆž
1
t n+1 dt (âˆ’1)n+m+1 z m
n+1
(âˆ’1) = : (5.5)
1 + zt m=0 n + m + 2
0

Next, we use for 06n66 the negative of these remainder integrals as input data in the recursive
schemes (2:14), (3.9), and (4:14), and do a Taylor expansion of the resulting expressions. Thus, we
obtain according to (2.13), (3.9), and (4.13)
421z 7 796321z 8 810757427z 9
ln(1 + z)
A(0) + O(z 10 );
= + âˆ’ + (5.6a)
3
z 16537500 8682187500 4051687500000
z7 31z 8 113z 9
ln(1 + z)
(0)
+ O(z 10 );
= + âˆ’ + (5.6b)
6
z 9800 77175 120050
z7 19z 8 z9
ln(1 + z)
(0)
+ O(z 10 ):
J2 = + âˆ’ + (5.6c)
z 37800 198450 4725
All calculations were done symbolically, using the exact rational arithmetics of Maple. Consequently,
the results in (5.6) are exact and free of rounding errors.
The leading coe cients of the Taylor expansions of the transformation error terms for A(0) and
3
(0) (0)
J2 are evidently smaller than the corresponding coe cients for 6 . This observation provides
considerable evidence that Aitkenâ€™s iterated 2 process and Brezinskiâ€™s iterated theta algorithm are
in the case of the series (5.2) for ln(1 + z)=z more e ective than Wynnâ€™s epsilon algorithm which
according to (3.2) produces PadÃƒ approximants.
e
This conclusion is also conÃ¿rmed by the following numerical example in Table 1, in which the
convergence of the series (5.2) for ln(1+z)=z is accelerated for z =0:95. The numerical values of the
348 E.J. Weniger / Journal of Computational and Applied Mathematics 122 (2000) 329â€“356

Table 1
âˆž
(âˆ’z)m =
Convergence of the transformation error terms. Transformation of ln(1 + z)=z = m=0
(m + 1) for z = 0:95
âˆž (âˆ’1)n+m z m (nâˆ’2<n=2=) (nâˆ’2<n=2=) (nâˆ’3<n=3=)
z n+1 R<n=2= z n+1 r2<n=2= z n+1 R<n=3=
n (z) (z) (z)
m=0 n+m+2
Eq. (2.13) Eq. (3.9) Eq. (4.13)

0:312654 Â· 100
0 0 0 0
âˆ’0:197206 Â· 100
1 0 0 0
0:143292 Â· 100 0:620539 Â· 10âˆ’2 0:620539 Â· 10âˆ’2
2 0
âˆ’0:112324 Â· 100 âˆ’0:230919 Â· 10âˆ’2 âˆ’0:230919 Â· 10âˆ’2 0:113587 Â· 10âˆ’2
3
0:922904 Â· 10âˆ’1 0:109322 Â· 10âˆ’3 0:156975 Â· 10âˆ’3 âˆ’0:367230 Â· 10âˆ’3
4
âˆ’0:782908 Â· 10âˆ’1 âˆ’0:333267 Â· 10âˆ’4 âˆ’0:466090 Â· 10âˆ’4 0:148577 Â· 10âˆ’3
5
0:679646 Â· 10âˆ’1 0:131240 Â· 10âˆ’5 0:413753 Â· 10âˆ’5 0:137543 Â· 10âˆ’5
6
âˆ’0:600373 Â· 10âˆ’1 âˆ’0:371684 Â· 10âˆ’6 âˆ’0:108095 Â· 10âˆ’5 âˆ’0:392983 Â· 10âˆ’6
7
0:537619 Â· 10âˆ’1 0:111500 Â· 10âˆ’7 0:110743 Â· 10âˆ’6 0:131377 Â· 10âˆ’6
8
âˆ’0:486717 Â· 10âˆ’1 âˆ’0:311899 Â· 10âˆ’8 âˆ’0:266535 Â· 10âˆ’7 0:412451 Â· 10âˆ’9
9
0:444604 Â· 10âˆ’1 0:689220 Â· 10âˆ’10 0:298638 Â· 10âˆ’8 âˆ’0:139178 Â· 10âˆ’9
10
âˆ’0:409189 Â· 10âˆ’1 âˆ’0:199134 Â· 10âˆ’10 âˆ’0:678908 Â· 10âˆ’9 0:475476 Â· 10âˆ’10
11
0:378992 Â· 10âˆ’1 0:282138 Â· 10âˆ’12 0:808737 Â· 10âˆ’10 âˆ’0:316716 Â· 10âˆ’12
12

remainder terms (5.5) were used as input data in the recursive schemes (2:14), (3.9), and (4.13) to
compute numerically the transformation error terms in (2.13), (3.9), and (4.13). The transformation
error terms, which are listed in columns 3â€“5, were chosen in agreement with (2.27), (3.21), and
(4.26), respectively.
The zeros, which are found in columns 3â€“5 of Table 1, occur because Aitkenâ€™s iterated 2 process
and Wynnâ€™s epsilon algorithm can only compute a rational approximant if at least three consecutive
partial sums are available, and because the iteration of Brezinskiâ€™s theta algorithm requires at least
four partial sums.
The result in Table 1 show once more that Aitkenâ€™s iterated 2 process and Brezinskiâ€™s iterated
theta algorithm are in the case of series (5.2) for ln(1 + z)=z apparently more e ective than Wynnâ€™s
epsilon algorithm.
The second group of results of this article â€” (2.19), (3.14), and (4.18) and the corresponding
recursive schemes (2:20), (3:15), and (4:19) â€” can for example be used to demonstrate how rational
approximants work if a divergent power series is to be summed.
Let us therefore assume that the partial sums, which occur in (2.19), (3.14), and (4.18), diverge
if the index becomes large. Then, a summation to a Ã¿nite generalized limit f(z) can only be
(n) (n)
accomplished if z n+2k+1 k (z) and z n+2k+1 â€™2k (z) in (2.19) and (3.14), respectively, converge to the
(n)
negative of fn+2k (z), and if z n+3k+1 k (z) in (4.18) converges to the negative of fn+3k (z).
Table 2 shows that this is indeed the case. We again consider the inÃ¿nite series (5.2) for ln(1+z)=z,
but this time we choose z = 5:0, which is clearly outside the circle of convergence. We use the
numerical values of the partial sums n (âˆ’z)m =(m+1) with 06n610 as input data in the recursive
m=0
schemes (2:20); (3:15), and (4:19) to compute the transformation terms in (2.19), (3.14), and (4.18).
The transformation terms, which are listed in columns 3â€“5 of Table 2, were chosen in agreement
with (2.27), (3.21), and (4.26), respectively. All calculations were done using the oating point
arithmetics of Maple.
E.J. Weniger / Journal of Computational and Applied Mathematics 122 (2000) 329â€“356 349

Table 2
Convergence of transformation terms to the partial sums. Transformation of ln(1 + z)=z =
âˆž
(âˆ’z)m =(m + 1) for z = 5:0
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