ñòð. 47 |

. .

. .

Ljâˆ’1 ; u1 : : : : : : : : : Ljâˆ’1 ; uN

1

u1 ::: ::: ::: uN

â€˜j = (31)

det V Lj+1 ; u1 : : : : : : : : : Lj+1 ; uN

. .

. .

. .

LN ; u1 : : : : : : : : : LN ; uN

the adjoint Vadj of V equals

Vadj = (det V )C (32)

where C has entries cj; t deÃ¿ned by (29). Hence

V âˆ’1 = C : (33)

It is remarkable [6] that in case of q simple Ã¿nite poles and a pole at inÃ¿nity of multiplicity

n0 ; q + n0 = N; and N simple nodes the inverse of V can be factorized as

D1 0

V âˆ’1 = V D2 ; (34)

0 H (s)

where D1 ; D2 are diagonal matrices of dimensions q and N , respectively, and where H (s) is a

triangular Hankel matrix of the form

ï£« ï£¶

s1 s2 s3 : : : snq

ï£¬s ï£·

s3 : : : Â·

ï£¬2 ï£·

ï£¬ ï£·

ï£¬s ::: Â· ï£·

H (s) = ï£¬ 3 ï£·:

ï£¬Â· Â· ï£·

ï£¬ ï£·

ï£Â· ï£¸

snq

2.2. The Nevilleâ€“Aitken interpolation formula

In [7] a Neville-Algorithm is given which computes the whole triangular Ã¿eld

u1 ; : : : ; uk

k=1; :::; N

(pik f)i=1; :::; N âˆ’k+1 ; pik f = p f âˆˆ Uk (35)

ai ; : : : ; ai+kâˆ’1

of interpolants recursively where pik f agrees with the function f on {ai ; : : : ; ai+kâˆ’1 }. In [7] this

algorithm was derived from the general Nevilleâ€“Aitken algorithm [11,3] via explicit formulas for

the Cauchyâ€“Vandermonde determinants [17]. In [5] we were going the other way around and have

given a di erent proof of the Nevilleâ€“Aitken algorithm which is purely algebraic. In this survey we

will derive the algorithm by a simple direct argument.

G. Muhlbach / Journal of Computational and Applied Mathematics 122 (2000) 203â€“222 209

Proposition 3. Let k âˆˆ N;

âˆˆ Ck+1 ;

Ak+1 = (a1 ; : : : ; ak ; ak+1 ) = ( 1 ; : : : ; 1; 2; : : : ; pâˆ’1 ; ; p; : : : ; p)

m1 mp

with 1 ; : : : ; p pairwise distinct and m1 + Â· Â· Â· + mp = k + 1; Ak = (a1 ; : : : ; ak ); Ak = (a2 ; : : : ; ak+1 )

and a1 = ak+1 . Let Uk+1 = (u1 ; : : : ; uk+1 ) be a CV-system associated with the pole system Bk+1 =

(b1 ; : : : ; bk+1 ). Suppose Ak+1 âˆ© Bk+1 = âˆ…. Let p1 âˆˆ Uk interpolate f at Ak and p2 âˆˆ Uk interpolate

f at Ak and let p3 âˆˆ Uk+1 interpolate f at Ak+1 .

(i) If bk+1 âˆˆ C then

p1 (z)(z âˆ’ ak+1 )(bk+1 âˆ’ a1 ) âˆ’ p2 (z)(z âˆ’ a1 )(bk+1 âˆ’ ak+1 )

p3 (z) = : (36)

(ak+1 âˆ’ a1 )(z âˆ’ bk+1 )

(ii) If bk+1 = âˆž then

p1 (z)(z âˆ’ ak+1 ) âˆ’ p2 (z)(z âˆ’ a1 )

p3 (z) = : (37)

a1 âˆ’ ak+1

Proof. (i) Call the right-hand side of (36) p3 . It belongs to Uk+1 in view of

Ëœ

z âˆ’ ak+1 bk+1 âˆ’ ak+1 z âˆ’ a1 bk+1 âˆ’ a1

=1+ and =1+

z âˆ’ bk+1 z âˆ’ bk+1 z âˆ’ bk+1 z âˆ’ bk+1

by the partial fraction decomposition theorem. Obviously, p3 interpolates f at Ak+1 if all nodes are

Ëœ

simple since the weights add to one and each of the unknown values p1 (ak+1 ); p2 (a1 ) has factor 0.

It is a consequence of Leibnizâ€™ rule that this holds true also in case of multiple nodes. In fact,

d

p3 |z= i

Ëœ

dz

bk+1 âˆ’ a1 d z âˆ’ ak+1 bk+1 âˆ’ ak+1 d z âˆ’ a1

= p1 (z) âˆ’ p2 (z)

ak+1 âˆ’ a1 d z z âˆ’ bk+1 z= i ak+1 âˆ’ a1 dz z âˆ’ bk+1 z= i

âˆ’

bk+1 âˆ’ a1 d d z âˆ’ ak+1

= p1 (z)|z= |z=

ak+1 âˆ’ a1 dz dz z âˆ’ bk+1

i i

=0

âˆ’

bk+1 âˆ’ ak+1 d d z âˆ’ a1

âˆ’ p2 (z)|z= |z=

ak+1 âˆ’ a1 dz dz z âˆ’ bk+1

i i

=0

âˆ’

bk+1 âˆ’ a1 d d z âˆ’ ak+1

= f(z)|z= |z=

ak+1 âˆ’ a1 dz dz z âˆ’ bk+1

i i

=0

âˆ’

bk+1 âˆ’ ak+1 d d z âˆ’ a1

âˆ’ f(z)|z= |z=

ak+1 âˆ’ a1 dz dz z âˆ’ bk+1

i i

=0

d bk+1 âˆ’ a1 z âˆ’ ak+1 bk+1 âˆ’ ak+1 z âˆ’ a1

= f(z) âˆ’ f(z)

dz ak+1 âˆ’ a1 z âˆ’ bk+1 ak+1 âˆ’ a1 z âˆ’ bk+1 z= i

d

= f(z)|z= i

dz

since the weights add to one. This is evident for all i = 1; : : : ; p and = 0; : : : ; mp âˆ’ 1 except for

i = 1 and = m1 âˆ’ 1 or i = p and = mp âˆ’ 1. If i = 1 and = m1 âˆ’ 1 then the unknown derivative

210 G. Muhlbach / Journal of Computational and Applied Mathematics 122 (2000) 203â€“222

(d=d z)m1 âˆ’1 p2 (z)|z= 1 has the factor (z âˆ’ a1 )=(z âˆ’ bk+1 )|z= 1 which vanishes. Similarly, for i = p and

= mp âˆ’ 1 the unknown derivative (d=d z)mp âˆ’1 p1 (z)|z= p has the factor (z âˆ’ ak+1 )=(z âˆ’ bk+1 )|z= p

which vanishes.

(ii) Obviously, the right-hand side of (37) belongs to Uk+1 and satisÃ¿es the interpolation conditions

as is shown by the same argument used in the proof of (i).

Remarks.

1. Letting bk+1 â†’ âˆž in (36) gives an alternative proof of (37). Observe that (37) is the classical

Nevilleâ€“Aitken recursion for interpolation by polynomials. It has to be used anytime a pole âˆž

is inserted.

2. p3 (z) is a convex combination of p1 (z) and p2 (z) if a1 ; ak+1 ; z; bk+1 are real and a1 6z6ak+1 Â¡ bk+1

or bk+1 Â¡ a1 6z6ak+1 .

3. Proposition 2 constitutes an algorithm of arithmetical complexity O(N 2 ) to compute the triangular

Ã¿eld (35) recursively from initializations piâ€˜ f âˆˆ Uâ€˜ (â€˜Â¿1) which are generalized Taylor inter-

polants: piâ€˜ f agrees with f at (ai ; : : : ; ai+â€˜âˆ’1 ) where all nodes are identical ai = Â· Â· Â· = ai+â€˜âˆ’1 =: a:

From (26) and (27) immediately

lâˆ’1

d

piâ€˜ (f) = f(a)!1 (38)

dz

=0

with

1 â€˜âˆ’ âˆ’1 (d=d z) Bâ€˜ (a) +

!1 = (z âˆ’ a) âˆˆ Uâ€˜ (39)

Bâ€˜ (z) =0 !

are derived.

2.3. Newtonâ€™s formula and the interpolation error

Given a complex function f which is deÃ¿ned and su ciently often di erentiable at the multiple

nodes of system (8) then (9) constitutes a system of linear equations for the coe cients cj of the

generalized polynomial

N

p = pf =: cj uj âˆˆ UN ; (40)

j=1

where the coe cient

u1 ; : : : ; uN f

N

cj = c1; j (f) =

a1 ; : : : ; aN j

will be referred to as the jth divided di erence of f with respect to the systems UN and AN . In

[6,9] for consistently ordered poles which are assumed to be simple if Ã¿nite and for simple nodes

algorithms for solving system (9) recursively are derived whose arithmetical complexity is O(N 2 ).

In this section we are going to derive Newtonâ€™s formula for the interpolant obtained in [14] and

a procedure to compute the divided di erences

u1 ; : : : ; uk+1 f

ai ; : : : ; ai+k k + 1

G. Muhlbach / Journal of Computational and Applied Mathematics 122 (2000) 203â€“222 211

recursively [14], see also [10,16]. For the latter a new short proof is given. This way we will

establish an algorithm to compute the interpolant (40) in the general case of multiple nodes and

multiple poles in Newtonâ€™s form recursively whose arithmetical complexity again is O(N 2 ). Later,

in Proposition 6 of Section 4, we will derive an algorithm solving the linear system (9) recursively

in the general case of multiple poles and multiple nodes avoiding the additional partial fraction

decomposition.

Proposition 4. If p1 f = : k c1; j (f)uj âˆˆ Uk interpolates f at Ak = (a1 ; : : : ; ak ) and p1 f =

k+1

k k

j=1

: k+1 c1; j (f)uj âˆˆ Uk+1 interpolates f at Ak+1 = (a1 ; : : : ; ak ; ak+1 ); then

k+1

j=1

k+1 k k+1 k

p1 f = p1 f + c1; k+1 (f)r1 uk+1 ; (41)

where

Ak (z) Bk (bk+1 )

k k

r1 uk+1 (z) = uk+1 (z) âˆ’ p1 uk+1 (z) = (42)

Bk+1 (z) Ak (bk+1 )

with Ak the node polynomial associated with Ak ; Ak (bk+1 ):= k âˆ— (bk+1 âˆ’ aj ) and with Bk ; Bk+1

j=1

the pole polynomials associated with the pole systems Bk and Bk+1 ; respectively. Furthermore;

u ; : : : ; uk+1 f

c1; k+1 (f) = 1

k+1

a1 ; : : : ; ak+1 k + 1

with ï£±

u1 ; : : : ; uk f u1 ; : : : ; uk f

ï£´

ï£´ âˆ’

ï£´

ï£´ a2 ; : : : ; ak+1 k a1 ; : : : ; ak k

ï£´

ï£´

ï£´ if a1 = ak+1 ;

ï£²

u1 ; : : : ; uk+1 f Akâˆ’1 (bk )

Bk (bk+1 )

ak+1 âˆ’a1

Â· Â·

= (43)

(ak+1 âˆ’bk+1 )âˆ—

a1 ; : : : ; ak+1 k + 1 Ak (bk+1 ) Bkâˆ’1 (bk )

ï£´

ï£´ ï£´

ï£´

ï£´ u1 ; : : : ; ukâˆ’1 ; f u1 ; : : : ; ukâˆ’1 ; uk

ï£´ det V

ï£´ det V

ï£³

a; : : : ; a; a a; : : : ; a; a

k âˆ—

if in the second case all nodes are identical a1 = Â· Â· Â· = ak+1 = a. Here Akâˆ’1 (bk+1 ):= j=2 (bk+1 âˆ’ aj ).

For any z âˆˆ C \ Bk

Ak (z)

k k

r1 f(z) = f(z) âˆ’ p1 f(z) = [a1 ; : : : ; ak ; z](Bk f) (44)

Bk (z)

with

k

[a1 ; : : : ; ak ; z](Bk f) = [a1 ; : : : ; ai ]Bk [ai ; : : : ; ak ; z]f + [a1 ; : : : ; ak ; z]Bk f(z) (45)

i=1

denoting the ordinary divided di erence where [a1 ; : : : ; ak ; z]Bk = 0 i at least one pole is at inÃ¿nity.

Moreover; if bk+1 âˆˆ C is chosen arbitrarily;

Ak (z)Bk (bk+1 )

u ; : : : ; uk+1 f

r1 f(z) = 1

k

: (46)

a1 ; : : : ; ak ; z k + 1 Bk+1 (z)Ak (bk+1 )

k+1

Proof. Consider linear system (9) for the coe cients of p1 f:

ï£« ï£¶ ï£« ï£¶

k+1

c1; 1 (f) L1 ; f

u1 ; : : : ; uk+1 . .

ï£¬ ï£·ï£¬ ï£·

. .

V ï£¸=ï£

ï£ ï£¸

. .

L1 ; : : : ; Lk+1 k+1

Lk+1 ; f

c1; k+1 (f)

212 G. Muhlbach / Journal of Computational and Applied Mathematics 122 (2000) 203â€“222

bordered by the equation

k+1

ñòð. 47 |