ñòð. 74 |

! n| |

ï£° ï£»

nâ†’âˆž

âˆˆNs0

| |62kâˆ’1

ï£« ï£¶

k s

1 i

ï£ âˆ’ x)Ã¿ ï£¸ D f(x):

j

= j (v (9)

!

âˆˆNs i=1 j=0

1 k

âˆˆNs

Ã¿ ;:::;Ã¿

0 0

Ã¿1 +Â·Â·Â·+Ã¿ = k

| |=2k

|Ã¿i |=2; i=1;:::; k

Building on these auxiliary results, we can now state and prove our main theorem. Note that

although Theorem 4 is a deep and nice result on the asymptotic behavior of the multivariate Bernstein

approximants, it does still not yet prove the asymptotic expansion. To do this, a careful analysis of

the coe cient functions in (9) is necessary.

Theorem 5. Let f âˆˆ C 2k (T ) with some k âˆˆ N. Then the sequence of Bernstein approximants

{Bn (f; x)}; deÃ¿ned in (8); possesses an asymptotic expansion of the form

k

c (x)

+ o(nâˆ’k )

Bn (f; x) = f(x) + for n â†’ âˆž: (10)

n

=1

The coe cient functions c (x) can be given explicitly; we have

ï£« ï£¶

| |=2 s

1 i

ï£ âˆ’ x)Ã¿ ï£¸

j

c (x) = j (v (11)

| |âˆ’ ;

!

âˆˆNs i=1 j=0

1

âˆˆNs

=| |âˆ’ Ã¿ ;:::;Ã¿

0 0

Ã¿1 +Â·Â·Â·+Ã¿ =

+16| |62

|Ã¿i |Â¿2; i=1;:::;

with recursively computable numbers ; see (14) below.

i;

Proof. We use the following result, to be found for example in [7].

A sequence {Bn } possesses an asymptotic expansion of the desired form, if and only if for

m = 1; : : : ; k,

mâˆ’1

c

m

lim n Bn âˆ’ f âˆ’ = : cm (12)

n

nâ†’âˆž

=1

exists and is di erent from zero. (Here and below, we set empty sums equal to zero.)

From (12), it is clear that the results due to Lai, as quoted above, are a big step towards the

proof of our Theorem, but as be seen below, there is still something to do.

We Ã¿rst have to make a further analysis of the functions S n . It is clear that if we have points

Ã¿1 ; : : : ; Ã¿ âˆˆ Ns with |Ã¿i |Â¿2; i = 1; : : : ; , and if Â¿ | |=2, then

0

|Ã¿1 + Â· Â· Â· + Ã¿ | Â¿ | |:

322 G. Walz / Journal of Computational and Applied Mathematics 122 (2000) 317â€“328

This means that

ï£« ï£¶

| |âˆ’2 s

n

S (x) n(n âˆ’ 1) Â· Â· Â· (n âˆ’ + 1) i

ï£ âˆ’ x)Ã¿ ï£¸

j

= j (v

n| | n| |

=1 i=1 j=0

Ã¿1 ;:::;Ã¿ âˆˆNs

0

Ã¿1 +Â·Â·Â·+Ã¿ =

|Ã¿i |Â¿2; i=1;:::;

ï£« ï£¶

| |=2 s

n(n âˆ’ 1) Â· Â· Â· (n âˆ’ + 1) i

ï£ âˆ’ x)Ã¿ ï£¸

j

= j (v (13)

n| |

=1 i=1 j=0

1

âˆˆNs

Ã¿ ;:::;Ã¿ 0

Ã¿1 +Â·Â·Â·+Ã¿ =

|Ã¿i |Â¿2; i=1;:::;

with

||

; | | even;

|| 2

=

2 | |âˆ’1

; | | odd:

2

Next, we observe that the expression

n(n âˆ’ 1) Â· Â· Â· (n âˆ’ + 1)

is a polynomial of exact degree in n, say

ni ;

n(n âˆ’ 1) Â· Â· Â· (n âˆ’ + 1) = i;

i=1

with coe cients , which can be computed by the recursion

i;

= 1; = 0; i = 1;

1; 1 i; 1

and

i; +1 := iâˆ’1; âˆ’ ; Â¿1; 16i6 + 1: (14)

i;

In particular,

âˆ’1

=1 and = (âˆ’1) ( âˆ’ 1)! (15)

; 1;

for all .

Together with (13), it follows that

ï£« ï£¶

| |=2 s

n

S (x) i; i

ï£ âˆ’ x)Ã¿ ï£¸ :

j

= j (v (16)

n| | n | |âˆ’i

=1 i=1 i=1 j=0

1

âˆˆNs

Ã¿ ;:::;Ã¿ 0

Ã¿1 +Â·Â·Â·+Ã¿ =

|Ã¿i |Â¿2; i=1;:::;

Rearranging this according to powers of n, we obtain

| |âˆ’1

S n (x) (x)

l;

= (17)

n| | nl

l= | |+1=2

with coe cient functions , which do not depend on n.

l;

G. Walz / Journal of Computational and Applied Mathematics 122 (2000) 317â€“328 323

For later use, we note that for | | even, say | | = 2 , the coe cient of the lowest power of n; ,

;

can be given explicitly: From (16), we deduce that

ï£« ï£¶

s

i

ï£ âˆ’ x)Ã¿ ï£¸

j

(x) = j (v

; ;

i=1 j=0

Ã¿1 ;:::;Ã¿ âˆˆNs

0

Ã¿1 +Â·Â·Â·+Ã¿ =

|Ã¿i |Â¿2; i=1;:::;

ï£« ï£¶

s

i

ï£ âˆ’ x)Ã¿ ï£¸ :

j

= j (v (18)

i=1 j=0

Ã¿1 ;:::;Ã¿ âˆˆNs

0

Ã¿1 +Â·Â·Â·+Ã¿ =2

|Ã¿i |=2; i=1;:::;

From (17), it follows that the sum over all these expressions itself is of the form

1 S n (x) d1; k (x) d2; k (x) dk; k (x)

+ O(nâˆ’(k+1) ):

D f(x) = + + Â·Â·Â· + (19)

!n n n n

|| 2 k

âˆˆNs0

| |62kâˆ’1

We now make the

Claim. For all 6k; the coe cient functions dj; in (19) satisfy

dj; (x) = dj; âˆ’1 (x) j = 1; : : : ; âˆ’ 2

and

d âˆ’1; (x) = d âˆ’1; âˆ’1 (x) + (x) (20)

âˆ’1;

âˆˆNs

0

| |=2 âˆ’2

with from (17).

âˆ’1; 2 âˆ’1

Proof of Claim. The proof is by induction. For k = 1, there is nothing to show, while for k = 2,

the only relation to prove is

d1; 2 (x) = d1; 1 (x) + (x):

1;

âˆˆNs

0

| |=2

But this is true, since d1; 1 = 0.

Now we assume that the claim is true for , and prove it for + 1.

From (17) and (19) and the induction hypothesis,

1 Sn

D f(x)

! n| |

âˆˆNs0

| |62 +1

324 G. Walz / Journal of Computational and Applied Mathematics 122 (2000) 317â€“328

2 âˆ’1 2

d1; (x) d ; (x) (x) (x)

l; l;

âˆ’( +1)

= + Â·Â·Â· + + O(n )+ +

n n nl nl

âˆˆNs

l= l= +1

âˆˆNs 0

0

| |=2 +1

| |=2

d1; +1 (x) d; +1 (x) d +1; +1 (x)

+ O(nâˆ’( +2) )

= + Â·Â·Â· + +

ñòð. 74 |