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(k) (n+1+(Ã¿+kâˆ’1)= )k

1

Cn ( ; Ã¿= ; {{sn }}; {{!n }}) ( n+Ã¿)j (n+(Ã¿+k)= )k+2

Weniger M transformation

(k) (n+1+ âˆ’(kâˆ’1))k

1

Mn ( ; {{sn }}; {{!n }}) (âˆ’nâˆ’ )j (n+ âˆ’k)k+2

Weniger S transformation

(k) 1

Sn (Ã¿; {{sn }}; {{!n }}) 1=(n + Ã¿)j (n+Ã¿+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 deÃ¿nition 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) 81â€“147 97

Ë† (k)

Nn

(k) (k)

Jn ({{sn }}; {{!n }}; { n }) = (k)

Ë†

Dn

with

(0) (1) (kâˆ’1)

n Â·Â·Â· n

n

(0) (k)

= 1; = ; kâˆˆN (97)

n n (0) (1) (kâˆ’1)

Â· Â· Â· n+1

n+1 n+1

and

Ëœ (0) Ëœ (0)

Dn = 1=!n ; N n = sn =!n ;

Ëœ (k) Ëœ (kâˆ’1) (kâˆ’1) Ëœ (kâˆ’1)

Dn = Dn+1 âˆ’ Dn ; k âˆˆN;

n

Ëœ (k) Ëœ (kâˆ’1) (kâˆ’1) Ëœ (kâˆ’1)

N n = N n+1 âˆ’ Nn ; k âˆˆ N; (98)

n

Ëœ (k)

Nn

(k) (k)

Jn ({{sn }}; {{!n }}; { n }) = (k)

Ëœ

Dn

with

(0) (1) (kâˆ’1)

n+k n+kâˆ’1 Â· Â· Â· n+1

(0) (k)

= 1; = ; k âˆˆ N: (99)

n n (0) (1) (kâˆ’1)

Â·Â·Â· n

n+kâˆ’1 n+kâˆ’2

(k)

The quantities n should not be mixed up with the k; n (u) as deÃ¿ned in Eq. (43).

Ë† (k) (k)

k

As shown in [46], the coe cients for the algorithm (96) that are deÃ¿ned via Dn = n; j =!n+j ,

j=0

satisfy the recursion

(k+1) (k) (k) (k)

= âˆ’ (100)

n; j n; j

n+1; jâˆ’1

n

(0) (k)

with starting values n; j = 1. This holds for all j if we deÃ¿ne n; j = 0 for j Â¡ 0 or j Â¿ k. Because

(k) (k)

(k)

n = 0, we have n; k = 0 such that { n; j } is a coe cient set for all k âˆˆ N0 .

(k)

Ëœ (k)

Similarly, the coe cients for algorithm (98) that are deÃ¿ned via Dn = k Ëœn; j =!n+j , satisfy the

j=0

recursion

Ëœ(k+1) = Ëœ(k) (k) Ëœ(k)

n+1; jâˆ’1 âˆ’ (101)

n; j n n; j

(0) (k)

with starting values Ëœn; j = 1. This holds for all j if we deÃ¿ne Ëœn; j = 0 for j Â¡ 0 or j Â¿ k. In this

(k) (k)

case, we have Ëœn; k = 1 such that { Ëœn; j } is a coe cient set for all k âˆˆ N0 .

Since the J transformation vanishes for {{sn }} = {{c!n }}, c âˆˆ K according to Eq. (95) for all

(1) (1)

k âˆˆ N, it is convex. This may also be shown by using induction in k using n; 1 = âˆ’ n; 0 = 1 and the

equation

k+1 k k

(k+1) (k) (k)

(k)

= âˆ’ (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) 81â€“147

(k)

Assuming that the limits = limnâ†’âˆž exist for all k âˆˆ N and noting that for k = 0 always

k n

â—¦

â—¦

= 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)]

kâˆ’1

â—¦ (k) kâˆ’j jm

= (âˆ’1) ( m) : (103)

j

m=0

j0 +j1 +:::+jkâˆ’1 =j;

j0 âˆˆ{0;1};:::; jkâˆ’1 âˆˆ{0;1}

As characteristic polynomial we obtain

k kâˆ’1

â—¦ â—¦ (k) j

(k)

(z) = jz = ( jz âˆ’ 1): (104)

j=0 j=0

â—¦ â—¦

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 + Ã¿ + (p âˆ’ 1)k)2

or to [46, Eq. (18)] 2

ï£±

n+Ã¿+2 n+Ã¿

ï£´

ï£´ for p = 1;

ï£´

ï£´ pâˆ’1 pâˆ’1

ï£² k k

(k)

= (106)

n

ï£´ k

ï£´ n+Ã¿+2

ï£´

ï£´ for p = 1

ï£³

n+Ã¿

or to

ï£±

n+Ã¿+k âˆ’1 n+Ã¿+k +1

ï£´

ï£´ for p = 2;

ï£´

ï£´ pâˆ’2 pâˆ’2

ï£² k k

(k)

= (107)

n

ï£´ k

ï£´ n+Ã¿+k âˆ’1

ï£´

ï£´ for p = 2;

ï£³

n+Ã¿+k +1

that is,

(k) (k)

p Jn (Ã¿; {{sn }}; {{!n }}) = Jn ({{sn }}; {{!n }}; {1=(n + Ã¿ + (p âˆ’ 1)k)2 }): (108)

â—¦ â—¦

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

â—¦

(k) (k)

p J n (Ã¿; {{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) 81â€“147 99

4.2.2. Drummond transformation

This transformation was given by Drummond [19]. It was also discussed by Weniger [84]. It may

be deÃ¿ned as

k

[sn =!n ]

(k)

Dn ({{sn }}; {{!n }}) = : (110)

k [1=! ]

n

Using the deÃ¿nition (32) of the forward di erence operator, the coe cients may be taken as

k

(k)

= (âˆ’1) j ; (111)

n; j

j

i.e., independent of n. As moduli, one has n = ( <k=2= ) = Ëœ (k) . Consequently, the Drummond trans-

k

(k)

formation is given in subnormalized form. As characteristic polynomial we obtain

k

k

(k)

(âˆ’1) j z j = (1 âˆ’ 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:

â—¦

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(kâˆ’1) ; (k) (kâˆ’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

deÃ¿ned as 3

(n + Ã¿ + k)1âˆ’k k

[(n + Ã¿)kâˆ’1 sn =!n ]

(k)

Ln (Ã¿; {{sn }}; {{!n }}) = : (115)

(n + Ã¿ + k)1âˆ’k k [(n + Ã¿)kâˆ’1 =! ]

n

Using the deÃ¿nition (32) of the forward di erence operator, the coe cients may be taken as

k

(k)

= (âˆ’1) j (n + Ã¿ + j)kâˆ’1 =(n + Ã¿ + k)kâˆ’1 : (116)

n; j

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