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Many further numerical examples are given in the literature [39,41“ 44,50,84].

Appendix A. Stieltjes series and functions

A Stieltjes series is a formal expansion

(’1) j j z j
f(z) = (A.1)

with partial sums
(’1) j j z j :
fn (z) = (A.2)

The coe cients n are the moments of an uniquely given positive measure (t) that has inÿnitely
many di erent values on 06t ¡ ∞ [4, p. 159]:

t n d (t);
= n ∈ N0 : (A.3)
142 H.H.H. Homeier / Journal of Computational and Applied Mathematics 122 (2000) 81“147

Formally, the Stieltjes series can be identiÿed with a Stieltjes integral

d (t)
f(z) = ; |arg(z)| ¡ : (A.4)
1 + zt

If such an integral exists for a function f then the function is called a Stieltjes function. For every
Stieltjes function there exist a unique asymptotical Stieltjes series (A.1), uniformly in every sector
|arg(z)| ¡ ‚ for all ‚ ¡ . For any Stieltjes series, however, several di erent corresponding Stieltjes
functions may exist. To ensure uniqueness, additional criteria are necessary [88, Section 4.3].
In the context of convergence acceleration and summation of divergent series, it is important that
for given z the tails f(z) ’ fn (z) of a Stieltjes series are bounded in absolute value by the next
term of the series,
|f(z) ’ fn (z)|6 n+1 z z¿0: (A.5)
Hence, for Stieltjes series the remainder estimates may be chosen as
!n = (’1)n+1 n+1
n+1 z : (A.6)
fn (z), i.e., to a t˜ variant.
This corresponds to !n =

Appendix B. Derivation of the recursive scheme (148)
(k) (k)
We show that for the divided di erence operator = n [{{x n }}] the identity
(k) (k)
(x n+k+1 + ˜) n+1 ((x)˜ g(x)) ’ (x n + ˜) n ((x)˜ g(x))
((x)˜+1 g(x)) = (B.1)
x n+k+1 ’ x n
holds. The proof is based on the Leibniz formula for divided di erences (see, e.g., [69, p. 50]) that
yields upon use of (x)˜+1 = (x + ˜)(x)˜ and n (x) = x n k; 0 + k; 1
(k+1) (k+1) ( j)
((x)˜+1 g(x)) = ˜ n ((x)˜ g(x)) + n (x) n+j ((x)˜ g(x))
= (x n + ˜) ((x)˜ g(x)) + n+1 ((x)˜ g(x)): (B.2)

Using the recursion relation of the divided di erences, one obtains
(k) (k)
n+1 ((x)˜ g(x)) ’ n ((x)˜ g(x)) (k)
((x)˜+1 g(x)) = (x n + ˜) + n+1 ((x)˜ g(x)): (B.3)
x n+k+1 ’ x n
Simple algebra then yields Eq. (B.1).
Comparison with Eq. (140) shows that using the interpolation conditions gn = g(x n ) = sn =!n and
˜ = k ’ 1 in Eq. (B.1) yields the recursion for the numerators in Eq. (148), while the recursion
for the denominators in Eq. (148) follows for ˜ = k ’ 1 and using the interpolation conditions
gn = g(x n ) = 1=!n . In each case, the initial conditions follow directly from Eq. (140) in combination
with the deÿnition of the divided di erence operator: For k = 0, we use (a)’1 = 1=(a ’ 1) and obtain
n (x n )k’1 gn = (x n )’1 gn = gn =(x n ’ 1).
H.H.H. Homeier / Journal of Computational and Applied Mathematics 122 (2000) 81“147 143

Appendix C. Two lemmata

Lemma C.1. Deÿne
k n+j
—¦ (k)
A= ; (C.1)
(n + j)r+1

—¦ —¦ (k)
(k) j
where is a zero of multiplicity m of (z) = j=0 j z . Then
(’1)m d m (k)
A∼ () (n ’ ∞): (C.2)
r nr+m+1 d xm

Proof. Use

1 1
exp(’at)t r dt;
= a¿0 (C.3)
ar+1 r! 0

to obtain
∞ n ∞
1 —¦
—¦ (k) n+j r (k)
( exp(’t))t r dt:
A= exp(’(n + j)t)t dt = exp(’nt) (C.4)
r! r!
0 0

Taylor expansion of the polynomial yields due to the zero at
(’ )m d m (k) (x)
t m (1 + O(t)):
( exp(’t)) = (C.5)
m! d xm

Invoking Watson™s lemma [6, p. 263 ] completes the proof.

—¦ (k)
Lemma C.2. Assume that assumption (C-3 ) of Theorem 13 holds. Further assume ’ for
n; j j
n ’ ∞. Then; Eq. (280) holds.

Proof. We have
!n+j n+t
j j
∼ exp ∼ exp(j n ) (C.6)
!n n

for large n. Hence,
k k
!n —¦
—¦ (k)
exp( n ))’j = (k)
∼ j( (1= + n) (C.7)
n; j
j=0 j=0

Since the characteristic polynomial (z) has a zero of order at z = 1= according to the
assumptions, Eq. (280) follows using Taylor expansion.
144 H.H.H. Homeier / Journal of Computational and Applied Mathematics 122 (2000) 81“147


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