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[87]
(k) (n+1+(ÿ+k’1)= )k
1
Cn ( ; ÿ= ; {{sn }}; {{!n }}) ( n+ÿ)j (n+(ÿ+k)= )k+2
Weniger M transformation
(k) (n+1+ ’(k’1))k
1
Mn ( ; {{sn }}; {{!n }}) (’n’ )j (n+ ’k)k+2
Weniger S transformation
(k) 1
Sn (ÿ; {{sn }}; {{!n }}) 1=(n + ÿ)j (n+ÿ+2k)2
Iterated Aitken process
[2,84]
(k)
An ({{sn }})
(k+1) 2 (k)
( An ({{sn }}))( An )({{sn }})
(k) (k)
=Jn ({{sn }}; {{ sn }}; { n }) Eq. (231) (k) (k)
( An ({{sn }}))( An+1 ({{sn }}))
Overholt process
[64]
Vn(k) ({{sn }})
( sn+k+1 ) [( sn+k )k+1 ]
(k) (k)
=Jn ({{sn }}; {{ sn }}; { n }) Eq. (231) ( sn+k )k+1

a
Refs. [36,38,40].
b
For the deÿnition of the j; n see Eq. (5).
c
Factors independent of n are irrelevant.
H.H.H. Homeier / Journal of Computational and Applied Mathematics 122 (2000) 81“147 97

ˆ (k)
Nn
(k) (k)
Jn ({{sn }}; {{!n }}; { n }) = (k)
ˆ
Dn
with
(0) (1) (k’1)
n ··· n
n
(0) (k)
= 1; = ; k∈N (97)
n n (0) (1) (k’1)
· · · n+1
n+1 n+1

and
˜ (0) ˜ (0)
Dn = 1=!n ; N n = sn =!n ;
˜ (k) ˜ (k’1) (k’1) ˜ (k’1)
Dn = Dn+1 ’ Dn ; k ∈N;
n

˜ (k) ˜ (k’1) (k’1) ˜ (k’1)
N n = N n+1 ’ Nn ; k ∈ N; (98)
n




˜ (k)
Nn
(k) (k)
Jn ({{sn }}; {{!n }}; { n }) = (k)
˜
Dn
with
(0) (1) (k’1)
n+k n+k’1 · · · n+1
(0) (k)
= 1; = ; k ∈ N: (99)
n n (0) (1) (k’1)
··· n
n+k’1 n+k’2
(k)
The quantities n should not be mixed up with the k; n (u) as deÿned in Eq. (43).
ˆ (k) (k)
k
As shown in [46], the coe cients for the algorithm (96) that are deÿned via Dn = n; j =!n+j ,
j=0
satisfy the recursion
(k+1) (k) (k) (k)
= ’ (100)
n; j n; j
n+1; j’1
n
(0) (k)
with starting values n; j = 1. This holds for all j if we deÿne n; j = 0 for j ¡ 0 or j ¿ k. Because
(k) (k)
(k)
n = 0, we have n; k = 0 such that { n; j } is a coe cient set for all k ∈ N0 .
(k)
˜ (k)
Similarly, the coe cients for algorithm (98) that are deÿned via Dn = k ˜n; j =!n+j , satisfy the
j=0
recursion
˜(k+1) = ˜(k) (k) ˜(k)
n+1; j’1 ’ (101)
n; j n n; j
(0) (k)
with starting values ˜n; j = 1. This holds for all j if we deÿne ˜n; j = 0 for j ¡ 0 or j ¿ k. In this
(k) (k)
case, we have ˜n; k = 1 such that { ˜n; j } is a coe cient set for all k ∈ N0 .
Since the J transformation vanishes for {{sn }} = {{c!n }}, c ∈ K according to Eq. (95) for all
(1) (1)
k ∈ N, it is convex. This may also be shown by using induction in k using n; 1 = ’ n; 0 = 1 and the
equation
k+1 k k
(k+1) (k) (k)
(k)
= ’ (102)
n; j n; j
n+1; j
n
j=0 j=0 j=0

that follows from Eq. (100).
98 H.H.H. Homeier / Journal of Computational and Applied Mathematics 122 (2000) 81“147

(k)
Assuming that the limits = limn’∞ exist for all k ∈ N and noting that for k = 0 always
k n
—¦
—¦
= 1 holds, it follows that there exists a limiting transformation J [ ] that can be considered as
0
special variant of the J transformation and with coe cients given explicitly as [46, Eq. (16)]
k’1
—¦ (k) k’j jm
= (’1) ( m) : (103)
j
m=0
j0 +j1 +:::+jk’1 =j;
j0 ∈{0;1};:::; jk’1 ∈{0;1}



As characteristic polynomial we obtain
k k’1
—¦ —¦ (k) j
(k)
(z) = jz = ( jz ’ 1): (104)
j=0 j=0

—¦ —¦
Hence, the J transformation is convex since (k) (1) = 0 due to 0 = 1.
The p J Transformation: This is the special case of the J transformation corresponding to
1
(k)
= (105)
n
(n + ÿ + (p ’ 1)k)2
or to [46, Eq. (18)] 2
±
n+ÿ+2 n+ÿ

 for p = 1;

 p’1 p’1
 k k
(k)
= (106)
n
 k
 n+ÿ+2

 for p = 1

n+ÿ
or to
±
n+ÿ+k ’1 n+ÿ+k +1

 for p = 2;

 p’2 p’2
 k k
(k)
= (107)
n
 k
 n+ÿ+k ’1

 for p = 2;

n+ÿ+k +1
that is,
(k) (k)
p Jn (ÿ; {{sn }}; {{!n }}) = Jn ({{sn }}; {{!n }}; {1=(n + ÿ + (p ’ 1)k)2 }): (108)
—¦ —¦
The limiting transformation p J of the p J transformation exists for all p and corresponds to the J
transformation with k = 1 for all k in N0 . This is exactly the Drummond transformation discussed
in Section 4.2.2, i.e., we have
—¦
(k) (k)
p J n (ÿ; {{sn }}; {{!n }}) = Dn ({{sn }}; {{!n }}): (109)



2
The equation in [46] contains an error.
H.H.H. Homeier / Journal of Computational and Applied Mathematics 122 (2000) 81“147 99


4.2.2. Drummond transformation
This transformation was given by Drummond [19]. It was also discussed by Weniger [84]. It may
be deÿned as
k
[sn =!n ]
(k)
Dn ({{sn }}; {{!n }}) = : (110)
k [1=! ]
n

Using the deÿnition (32) of the forward di erence operator, the coe cients may be taken as
k
(k)
= (’1) j ; (111)
n; j
j
i.e., independent of n. As moduli, one has n = ( <k=2= ) = ˜ (k) . Consequently, the Drummond trans-
k
(k)

formation is given in subnormalized form. As characteristic polynomial we obtain
k
k
(k)
(’1) j z j = (1 ’ z)k :
n (z) = (112)
j
j=0

(k)
Hence, the Drummond transformation is convex since n (1) = 0. Interestingly, the Drummond
transformation is identical to its limiting transformation:
—¦
D (k) ({{sn }}; {{!n }}) = Dn ({{sn }}; {{!n }}):
(k)
(113)
The Drummond transformation may be computed using the recursive scheme

Nn(0) = sn =!n ; (0)
Dn = 1=!n ;
Nn(k) = Nn(k’1) ; (k) (k’1)
Dn = Dn ;
Dn = Nn(k) =Dn :
(k) (k)
(114)

4.2.3. Levin transformation
This transformation was given by Levin [53]. It was also discussed by Weniger [84]. It may be
deÿned as 3
(n + ÿ + k)1’k k
[(n + ÿ)k’1 sn =!n ]
(k)
Ln (ÿ; {{sn }}; {{!n }}) = : (115)
(n + ÿ + k)1’k k [(n + ÿ)k’1 =! ]
n

Using the deÿnition (32) of the forward di erence operator, the coe cients may be taken as
k
(k)
= (’1) j (n + ÿ + j)k’1 =(n + ÿ + k)k’1 : (116)
n; j

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