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. 49
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For simplicity; the argument f of the divided di erences in the following formulas will be dropped.
(i) If ÿ:=bk+1 ∈ C \ {b1 ; : : : ; bk } corresponding to uk+1 (z) = 1=(z ’ ÿ) then
j0
k k
c1; j (ÿ ’ 1)’ c2; j (ÿ ’ 2) (ÿ ’ 1 )(ÿ ’ 2)
k+1 k k ’j’1
c1; j = + (c1; ’ c2; )ÿ ; j = 1; : : : ; j0 ;
2’ 1 ’
2 1 =j+1

(57)
k k
c1; j (ÿ ’ 1 )( 2 ’ bj ) ’ c2; j (ÿ ’ 2 )( 1 ’ bj )
k+1
c1; j =
( 2 ’ 1 )(ÿ ’ bj )
jr+1 k k
c1; ’ c2;
(ÿ ’ 1 )(ÿ ’ 2)
’ jr ¡ j6jr+1 ; r = 0; : : : ; q ’ 1; (58)
’ (ÿ ’ bj ) ’j+1
2 1 =j+1
« 
j0 q’1 jr+1 k k
c1;’ c2;
(ÿ ’ 1 )(ÿ ’ 2)  :
k+1 k k ’1
c1; k+1 = (c1; ’ c2; )ÿ + (59)
2’ (ÿ ’ b ) ’jr
1 r=0
=1 =jr +1

(ii) If ÿ:=bk+1 = ÿi ∈ C corresponding with uk+1 (z) = 1=(z ’ ÿi )ni +1 then
j0
k k
c1; j (ÿ ’ 1)’ c2; j (ÿ ’ 2) (ÿ ’ 1 )(ÿ ’ 2)
k+1 k k ’j’1
c1; j = + (c1; ’ c2; )ÿ ; j = 1; : : : ; j0 ;
2’ 1 ’
2 1 =j+1

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

(61)
k k
c1; ji’1 +1 (ÿ ’ 1)’ c2; ji’1 +1 (ÿ ’ 2)
k+1
c1; ji’1 +1 =
2’ 1
® 
j0 q’1 jr+1 k k
c1; ’ c2; 
(ÿ ’ 1 )(ÿ ’ 2)  k k ’1
+ (c1; ’ c2; )ÿ + »; (62)
°
’ (ÿ ’ b ) ’jr
2 1 r=0
=1 =jr +1
r=i’1
G. Muhlbach / Journal of Computational and Applied Mathematics 122 (2000) 203“222 217

k k
c1; j (ÿ ’ 1) ’ c2; j (ÿ ’ 2) (ÿ ’ 1 )(ÿ ’ 2)
k+1 k k
c1; j = + (c1; j’1 ’ c2; j’1 ); j = ji’1 + 2; : : : ; ji ;
2’ 1 ’
2 1
(63)

(ÿ ’ 1 )(ÿ ’ 2)
k+1 k k
c1; k+1 = (c1; ji ’ c2; ji ): (64)

2 1

(iii) If ÿ:=bk+1 = ∞ corresponding with uk+1 (z) = z n0 then
q’1 k
k k k
c1; 1 ’ c2; 1 ’ =0 (c1; j +1 ’ c2; j +1 )
2 1
k+1
c1; 1 = ; (65)

2 1

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

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

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

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

Proof. According to (36) if ÿ:=bk+1 ∈ C we have
k+1
k+1 k+1
p1 = c1; j uj
j=1
«  « 
k k
ÿ’ 2 ÿ’ ÿ’ 1 ÿ’
1 2
= c1; j uj  1 + ’ c2; j uj  1 +
k k
z’ÿ 2’ z’ÿ 2’
1 1
j=1 j=1
k
1 k k
= (c1; j (ÿ ’ 1) ’ c2; j (ÿ ’ 2 ))uj
2’ 1 j=1

k
(ÿ ’ 1 )(ÿ ’ 2) 1
k k
+ (c1; j ’ c2; j )uj : (70)
’ z’ÿ
2 1 j=1

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


Now by partial fraction decomposition
’1
1 1
ÿ’ ’1
z = z +ÿ ; (72)
z’ÿ z’ÿ
=0


1 1 ’1 1 1 1
= + ; (73)
(z ’ b) +1 z ’ ÿ (ÿ ’ b) (z ’ b) +1’ (ÿ ’ b) +1 z ’ ÿ
+1
=0


1 z’b+b 1 1
z= = +b : (74)
(z ’ b) +1 (z ’ b) +1 (z ’ b) (z ’ b) +1
Eq. (73) is readily veriÿed by multiplying both sides by (z ’ b) +1 (z ’ ÿ) and making use of the
ÿnite geometric series. Eq. (72) follows from
1 (z ’ ÿ + ÿ)
ÿ ’ (z ’ ÿ) ’1
z = =
z’ÿ z’ÿ =0
’1
ÿ ’1
’ ’ ’1
= + ÿ z (’ÿ) :
z’ÿ =1 =0

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

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

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

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


1 ’1 1
uj = uj’ + uk+1 ; j0 ¡ j6k: (76)
z’ÿ (ÿ ’ bj ) (ÿ ’ bj ) +1
+1
=0
G. Muhlbach / Journal of Computational and Applied Mathematics 122 (2000) 203“222 219


In case (ii): ÿ = bk+1 = ÿi ∈ C, from (72) and (73)
j’2
1
ÿ uj’1’ + ÿj’1 uji’1 +1 ;
uj = 16j6j0 (77)
z’ÿ =0



1 ’1 1
uj = uj’ + uj +1 ; j0 ¡ j6ji’1 or ji ¡ j6k; (78)
z’ÿ (ÿ ’ bj ) (ÿ ’ bj ) +1 i’1
+1
=0



1
uj = uj+1 ; ji’1 ¡ j ¡ ji ; (79)
z’ÿ

1
uj = uk+1 ; j = ji : (80)
z’ÿ
In case (iii): ÿ = bk+1 = ∞ we have

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

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

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. 49
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