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For simplicity; the argument f of the divided di erences in the following formulas will be dropped.
(i) If Ã¿:=bk+1 âˆˆ C \ {b1 ; : : : ; bk } corresponding to uk+1 (z) = 1=(z âˆ’ Ã¿) then
j0
k k
c1; j (Ã¿ âˆ’ 1)âˆ’ c2; j (Ã¿ âˆ’ 2) (Ã¿ âˆ’ 1 )(Ã¿ âˆ’ 2)
k+1 k k âˆ’jâˆ’1
c1; j = + (c1; âˆ’ c2; )Ã¿ ; j = 1; : : : ; j0 ;
2âˆ’ 1 âˆ’
2 1 =j+1

(57)
k k
c1; j (Ã¿ âˆ’ 1 )( 2 âˆ’ bj ) âˆ’ c2; j (Ã¿ âˆ’ 2 )( 1 âˆ’ bj )
k+1
c1; j =
( 2 âˆ’ 1 )(Ã¿ âˆ’ bj )
jr+1 k k
c1; âˆ’ c2;
(Ã¿ âˆ’ 1 )(Ã¿ âˆ’ 2)
âˆ’ jr Â¡ j6jr+1 ; r = 0; : : : ; q âˆ’ 1; (58)
âˆ’ (Ã¿ âˆ’ bj ) âˆ’j+1
2 1 =j+1
ï£« ï£¶
j0 qâˆ’1 jr+1 k k
c1;âˆ’ c2;
(Ã¿ âˆ’ 1 )(Ã¿ âˆ’ 2) ï£­ ï£¸:
k+1 k k âˆ’1
c1; k+1 = (c1; âˆ’ c2; )Ã¿ + (59)
2âˆ’ (Ã¿ âˆ’ b ) âˆ’jr
1 r=0
=1 =jr +1

(ii) If Ã¿:=bk+1 = Ã¿i âˆˆ C corresponding with uk+1 (z) = 1=(z âˆ’ Ã¿i )ni +1 then
j0
k k
c1; j (Ã¿ âˆ’ 1)âˆ’ c2; j (Ã¿ âˆ’ 2) (Ã¿ âˆ’ 1 )(Ã¿ âˆ’ 2)
k+1 k k âˆ’jâˆ’1
c1; j = + (c1; âˆ’ c2; )Ã¿ ; j = 1; : : : ; j0 ;
2âˆ’ 1 âˆ’
2 1 =j+1

(60)
k k
c1; j (Ã¿ âˆ’ 1 )( 2 âˆ’ bj ) âˆ’ c2; j (Ã¿ âˆ’ 2 )( 1 âˆ’ bj )
k+1
c1; j =
( 2 âˆ’ 1 )(Ã¿ âˆ’ bj )
jr+1 k k
c1; âˆ’ c2;
(Ã¿ âˆ’ 1 )(Ã¿ âˆ’ 2)
+ ; jr Â¡ j6jr+1 ; r = 0; : : : ; i âˆ’ 2; i; i + 1; : : : ; q âˆ’ 1;
âˆ’ (Ã¿ âˆ’ bj ) âˆ’j+1
2 1 =j+1

(61)
k k
c1; jiâˆ’1 +1 (Ã¿ âˆ’ 1)âˆ’ c2; jiâˆ’1 +1 (Ã¿ âˆ’ 2)
k+1
c1; jiâˆ’1 +1 =
2âˆ’ 1
ï£® ï£¹
j0 qâˆ’1 jr+1 k k
c1; âˆ’ c2; ï£º
(Ã¿ âˆ’ 1 )(Ã¿ âˆ’ 2) ï£¯ k k âˆ’1
+ (c1; âˆ’ c2; )Ã¿ + ï£»; (62)
ï£°
âˆ’ (Ã¿ âˆ’ b ) âˆ’jr
2 1 r=0
=1 =jr +1
r=iâˆ’1
G. Muhlbach / Journal of Computational and Applied Mathematics 122 (2000) 203â€“222 217

k k
c1; j (Ã¿ âˆ’ 1) âˆ’ c2; j (Ã¿ âˆ’ 2) (Ã¿ âˆ’ 1 )(Ã¿ âˆ’ 2)
k+1 k k
c1; j = + (c1; jâˆ’1 âˆ’ c2; jâˆ’1 ); j = jiâˆ’1 + 2; : : : ; ji ;
2âˆ’ 1 âˆ’
2 1
(63)

(Ã¿ âˆ’ 1 )(Ã¿ âˆ’ 2)
k+1 k k
c1; k+1 = (c1; ji âˆ’ c2; ji ): (64)
âˆ’
2 1

(iii) If Ã¿:=bk+1 = âˆž corresponding with uk+1 (z) = z n0 then
qâˆ’1 k
k k k
c1; 1 âˆ’ c2; 1 âˆ’ =0 (c1; j +1 âˆ’ c2; j +1 )
2 1
k+1
c1; 1 = ; (65)
âˆ’
2 1

k k k k
c1; j âˆ’ c2; j âˆ’ (c1; jâˆ’1 âˆ’ c2; jâˆ’1 )
2 1
k+1
c1; j = ; j = 2; : : : ; j0 ; (66)
2âˆ’ 1

k k k k k k
c1; j âˆ’ c2; j âˆ’ (c1; j âˆ’ c2; j )bj âˆ’ (c1; j+1 âˆ’ c2; j+1 )
2 1
k+1
c1; j = ; j0 Â¡ j Â¡ k; j = j1 ; j2 ; : : : ; jq ; (67)
2âˆ’ 1

k k k k
c1; j âˆ’ c2; j âˆ’ (c1; j âˆ’ c2; j )bj
2 1
k+1
c1; j = ; j = ji ; i = 1; : : : ; q; (68)
2âˆ’ 1

k k
c1; j0 âˆ’ c2; j0
k+1
c1; k+1 =âˆ’ : (69)
2âˆ’ 1

Proof. According to (36) if Ã¿:=bk+1 âˆˆ C we have
k+1
k+1 k+1
p1 = c1; j uj
j=1
ï£« ï£¶ ï£« ï£¶
k k
Ã¿âˆ’ 2 Ã¿âˆ’ Ã¿âˆ’ 1 Ã¿âˆ’
1 2
=ï£­ c1; j uj ï£¸ 1 + âˆ’ï£­ c2; j uj ï£¸ 1 +
k k
zâˆ’Ã¿ 2âˆ’ zâˆ’Ã¿ 2âˆ’
1 1
j=1 j=1
k
1 k k
= (c1; j (Ã¿ âˆ’ 1) âˆ’ c2; j (Ã¿ âˆ’ 2 ))uj
2âˆ’ 1 j=1

k
(Ã¿ âˆ’ 1 )(Ã¿ âˆ’ 2) 1
k k
+ (c1; j âˆ’ c2; j )uj : (70)
âˆ’ zâˆ’Ã¿
2 1 j=1

If Ã¿ = âˆž according to (37) we have
k k
k+1 k k
c1; j uj (z âˆ’ 2) âˆ’ c2; j uj (z âˆ’ 1)
j=1 j=1
k+1 k+1
p1 = c1; j uj =
1âˆ’ 2
j=1
k k
k k k k
(c1; j âˆ’ c2; j 1 )uj âˆ’ (c1; j âˆ’ c2; j )uj z
2
j=1 j=1
= : (71)
2âˆ’ 1
218 G. Muhlbach / Journal of Computational and Applied Mathematics 122 (2000) 203â€“222

Now by partial fraction decomposition
âˆ’1
1 1
Ã¿âˆ’ âˆ’1
z = z +Ã¿ ; (72)
zâˆ’Ã¿ zâˆ’Ã¿
=0

1 1 âˆ’1 1 1 1
= + ; (73)
(z âˆ’ b) +1 z âˆ’ Ã¿ (Ã¿ âˆ’ b) (z âˆ’ b) +1âˆ’ (Ã¿ âˆ’ b) +1 z âˆ’ Ã¿
+1
=0

1 zâˆ’b+b 1 1
z= = +b : (74)
(z âˆ’ b) +1 (z âˆ’ b) +1 (z âˆ’ b) (z âˆ’ b) +1
Eq. (73) is readily veriÃ¿ed by multiplying both sides by (z âˆ’ b) +1 (z âˆ’ Ã¿) and making use of the
Ã¿nite geometric series. Eq. (72) follows from
1 (z âˆ’ Ã¿ + Ã¿)
Ã¿ âˆ’ (z âˆ’ Ã¿) âˆ’1
z = =
zâˆ’Ã¿ zâˆ’Ã¿ =0
âˆ’1
Ã¿ âˆ’1
âˆ’ âˆ’ âˆ’1
= + Ã¿ z (âˆ’Ã¿) :
zâˆ’Ã¿ =1 =0

âˆ’1
Here the second sum can be extended over = 0; : : : ; âˆ’ 1 since the binomial coe cients
vanish for the extra summands. By interchanging the two summations we obtain
âˆ’1
1 Ã¿ âˆ’1
z Ã¿âˆ’ âˆ’1 âˆ’1
z = + (âˆ’1) (âˆ’1) :
zâˆ’Ã¿ zâˆ’Ã¿ =1
=0

The second sum equals
âˆ’1
(âˆ’1) âˆ’ + (âˆ’1) +1 +1
(âˆ’1) = (âˆ’1)
=0

= f( ) ( ) = 0, where
since the sum in the last equation is the forward di erence 1 f(0)

(x âˆ’ 1)(x âˆ’ 2) Â· Â· Â· (x âˆ’ )
xâˆ’1
f(x) = =
!
is a polynomial of degree and 6 âˆ’ 1 Â¡ :
Let now := j (bj ) denote the multiplicity of bj in (b1 ; : : : ; bjâˆ’1 ). Consider case (i): Ã¿ = bk+1 âˆˆ
C; Ã¿ âˆˆ {b1 ; : : : ; bk }: Then from (72) and (73) (for simplicity we drop the argument z)
jâˆ’2
1
Ã¿ ujâˆ’1âˆ’ + Ã¿jâˆ’1 uk+1 ;
uj = 16j6j0 ; (75)
zâˆ’Ã¿ =0

1 âˆ’1 1
uj = ujâˆ’ + uk+1 ; j0 Â¡ j6k: (76)
zâˆ’Ã¿ (Ã¿ âˆ’ bj ) (Ã¿ âˆ’ bj ) +1
+1
=0
G. Muhlbach / Journal of Computational and Applied Mathematics 122 (2000) 203â€“222 219

In case (ii): Ã¿ = bk+1 = Ã¿i âˆˆ C, from (72) and (73)
jâˆ’2
1
Ã¿ ujâˆ’1âˆ’ + Ã¿jâˆ’1 ujiâˆ’1 +1 ;
uj = 16j6j0 (77)
zâˆ’Ã¿ =0

1 âˆ’1 1
uj = ujâˆ’ + uj +1 ; j0 Â¡ j6jiâˆ’1 or ji Â¡ j6k; (78)
zâˆ’Ã¿ (Ã¿ âˆ’ bj ) (Ã¿ âˆ’ bj ) +1 iâˆ’1
+1
=0

1
uj = uj+1 ; jiâˆ’1 Â¡ j Â¡ ji ; (79)
zâˆ’Ã¿

1
uj = uk+1 ; j = ji : (80)
zâˆ’Ã¿
In case (iii): Ã¿ = bk+1 = âˆž we have

uj z = uj+1 ; j = 1; : : : ; j0 âˆ’ 1; (81)

uj z = z n0 = uk+1 ; j = j0 ; (82)

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