A Theorem of Gopengauz Type with Added Interpolatory Conditions T. Kilgore & J. Prestin Abstract
The theorem of Gopengauz guarantees the existence of a polynomial which well approximates a function f 2 C q [?1; 1], while at the same time its kth derivative (k q) well approximates the kth derivative of the function, and moreover the polynomial and its derivatives respectively interpolate the function and its derivatives at 1. With more generality, we shall prescribe that the polynomial interpolate the function at up to q + 1 points near 1 and up to q + 1 points near ?1. The points may coalesce, in which case one also interpolates at the coalescent point a number of derivatives one less than the multiplicity of coalescence. Aside from intrinsic theoretical interest, our results are clearly applicable in describing more precisely the error incurred in certain linear processes of simultaneous approximation, such as interpolation with added nodes near 1. The original theorem of Gopengauz will be shown to follow as a special case.
AMS Subject Classi cation: Primary 41, Secondary 65
Introduction: Let q be a natural number. Then for n 2q +1 we consider sets of points Sn := fs ;n . . . ; sq;n; t ;n ; . . . ; tq;ng; in which ?1 s ;n . . . sq;n < ?1 + n2 and 1 t ;n . . . tq;n > 1 ? n2 : Now that we have indicated clearly the dependence of the elements of Sn upon n, the second subscript, n, is clearly redundant and will be suppressed. Associated with each set Sn is also the polynomial Qn de ned by 1
0
0
0
1
0
Qn(x) := (x ? s ) . . . (x ? sq )(x ? tq ) . . . (x ? t ) 0
0
which has zeroes which depend on n but is always of degree 2q + 1.
We recall that the notation !(f ; ) denotes the modulus of continuity of the function f , de ned by
!(f ; ) := supjx?yj jf (x) ? f (y)j; and we are ready to state our results.
Theorem 1: Let f 2 C q [?1; 1], with q 0. Then for each n 2q + 1, there
exists a polynomial Pn of degree at most n such that for k = 0; . . . ; q and for ?1 x 1
p
!
q?k 1 ? x 1 q; + ! f jf (x)j K n n with K independent of n and f and also satisfying k
( )
(x) ? Pnk
2
( )
( )
2
p
1?x + 1 n n 2
2
!
(1)
jf (x) ? Pn(x)j
q 0v 0 11 u j ( x ? t )( x ? s ) j u ` m K min @t jQn (x)j q @ AA : (2) i; m6 j nq i;j j(x ? ti )(x ? sj )j ! f ; `6 min n ( )
=
=
A related result is: Theorem 2: Let f 2 C q [?1; 1] and let Pn satisfy (1) for some constant K . If Pn also interpolates f on Sn , then Pn also satis es (2) with a constant CK , in which C is an absolute constant. From Theorem 1 and Theorem 2 and their proofs we have a fortiori the following: Corollary: Let Sn;j1 ;j2 be a set containing points s ; . . . ; sj1 in [?1; ?1 + n2 ] and points t ; . . . ; tj2 in [1 ? n2 ; 1], the count in both cases including any multiplicity. If j ; j q and if f 2 C q [?1; 1], then there is a polynomial Pn which interpolates f on Sn;j1 ;j2 and satis es (1) with a constant K independent of n and f . Furthermore, if Pn is any polynomial interpolating f on Sn;j1 ;j2 and satisfying (1) with any constant K (whether independent of the other quantities or not), then for 0 x 1 1
0
1
0
1
2
jf (x) ? Pn(x)j
q 11 0v p !q?j2 0 j x ? t j u ` 1 ) j j ( x ? t ) . . . ( x ? t 1 ? x CK min @u j 2 t AA : !@f q ; min `6 i nj2 ij2 jx ? ti j n +n n 2
0
( )
2
0
2
=
A similar inequality involving the points s ; . . . ; sj1 is valid for ?1 x 0. An important special case occurs when in the set Sn we have sj = ?1 and tj = 1 for j = 0; . . . ; q. Then as a consequence of Theorem 1 and Theorem 2 we get (2), and then because of the fact that f k (1) ? Pnk (1) = 0 for k = 0; . . . ; q we again get (2) for each of the derivatives (with q replaced, of course, with q ? k). Therefore, Theorem 1 and Theorem 2 together imply the original theorem of Gopengauz [3], which states Theorem: Let f 2 C q [?1; 1]. Then for each n 4q + 5 (we need only n 2q + 1 here), there exists a polynomial Pn of degree at most n such that for k = 0; . . . ; q and for ?1 x 1 0
( )
jf
k
( )
(x) ? Pnk
( )
(x)j K
p
1?x n
2
!q?k
( )
q
! f ;
p
1?x n
( )
2
!
(3)
with K independent of n and f . We turn now to the proofs of our two theorems. Because Theorem 2 will be used in the proof of Theorem 1, we will prove Theorem 2 rst and then Theorem 1.
Proof of Theorem 2: The proof of Theorem 2 divides itself naturally into two cases: that q = 0 is one of the cases, and that q > 0 is the other. The Case q = 0: We begin with (1) for some given constant K and with
jf (t ) ? Pn(t )j = 0: 0
0
We must show the stronger inequality
0 q 1 j ( x ? t )( x ? s ) j A: jf (x) ? Pn(x)j C 0 K !@f ; n 0
(4)
0
If x = t or if x = s , then (4) follows no matter what the constant is. More than this, in Balazs and Kilgore ([1]) it is shown that if t = 1 and s = ?1 we get 0
0
0
0
jf (x) ? Pn (x)j (7K + 7) ! f ; 3
p
!
1?x : n 2
(5)
Applying (5) on the interval [s ; t ], of length not less than 2(1 ? n2 ), we have there immediately the estimate 1
0
0
! 0 qj(x ? t )(x ? s )j 1 A: jf (x) ? Pn(x)j (7K + 7) n n? 1 !@f ; n 2
0
0
2
Therefore, it remains to show (4) for ?1 x < s and for t < x 1, in both p of which cases we have 1 ? x < n . Since these two cases are similar, we will assume with no loss of generality that t < x 1. Taking into account that f (t ) = Pn(t ), we may estimate 0
0
1
2
0
0
0
jf (x) ? Pn (x)j jf (x) ? f (t )j + jPn(x) ? Pn(t )j; 0
and
(6)
0
1 0 q j ( x ? t )( x ? s ) j A: jf (x) ? f (t )j !(f ; jx ? t j) !@f ; n 0
0
0
0
(7)
Now we notice that for arbitrary t 2 [?1; 1]
jPn (t) ? Pn(t )j jPn (t) ? f (t)j + jf (t) ? f (t )j; 0
0
whence also for arbitrary t 2 [?1; 1]
p
!
(8) jPn(t) ? Pn(t )j K ! f ; 1n? t + n1 + !(f ; jt ? t j): Now one needs to obtain an estimate for jPn (x) ? Pn (t )j of the form 2
0
0
2
0
1 0 q j ( x ? t )( x ? s ) j A; jPn (x) ? Pn(t )j const !@f ; n 0
0
0
(9)
in which t < x 1. If one wishes to estimate the constant, an economical but tedious method for obtaining this estimate is laid out in Balazs and Kilgore [1]. One begins by letting t = cos , t = cos , x = cos x , and Pn(x) ? Pn(t ) = Tn(), an even trigonometric polynomial. Then, using an identity of Balazs and Kilgore (cf. [1] or [2]) for the derivative of a trigonometric polynomial we can write 0
0
0
0
0
1
nj Xn (?1)j Tn0 () = 81n Tn( + j ) @ sin j A (sin j ) n sin j 4
2
+1
2
1
0
=1
1
2
2
2
Xn sin nj A 1 1 @ 8n jTn ( + j )j j (sin j ) n sin j 4
2
2
1
=1
2
4
2
2
(10)
in which
j := 2j4?n 1 :
Inserting the estimate (8) into (10) and using the fact that !() ( + 1)!(), one obtains an estimate for jTn0 ()j valid on the interval [x; ], and then one obtains (9) by integration from x to . An alternative is to use Brudnyi's inequality, in which the constant has also been estimated in Balazs and Kilgore [2]. We omit further details here. This completes the argument for the case that q = 0. The Case q > 0: It remains to show for q 1 that if the polynomials Pn satisfy (1) with constant K and also interpolate f on the set Sn , then for i; j = 0; . . . ; q and for ` 6= i; m 6= j we can obtain 0
0
jf (x) ? Pn(x)j
q 1 0 1 jQn (x)j A !@f q ; j(x ? t`)(x ? sm)j A : (11) j(x ? ti )(x ? sj )j n p We must begin by considering the relationship between 1 ? x + n and a typical factor of jQn j, of the form j(x ? ti)(x ? sj )j, for arbitrary i and j , on various subsets of the interval [?1; 1]. On the subinterval [?1 + n2 ; 1 ? n2 ] 0v u u 00 ? q @ C Kn t
( )
2
1
we have
p
p
1?x + n 1 1 ? x + 1 j(x ? ti )(x ? sj )j 2 q n (1 ? n2 ) ? x 1
In turn,
1
2
2
p
p
2
q 1 ? x + n = 1 ? x + n 1 ? x + n2 1 ? x ? n2 + n4 (1 ? n2 ) ? x 1
2
1
2
2
1
2
2
2
2
p
2
2
1
1
:
! 21
1
(12)
If we assume that 1 ? x nc , where c is suciently large (for example p c 2), then (12) gives a decreasing function of 1 ? x , and therefore we get for these x 2
2
p
0
11
2 c2 + c + 1 ? x + 2 2 2A n n n n @ q pc + 1 : = c2 c ?2 (1 ? n2 ) ? x n2 ? n2 + n4 1
p
1
2
2
2
2
2
1
1
2
Now if 1 ? x nc we must consider two subcases. p (i) If sj x ti and 1 ? x nc , then j(x ? ti )(x ? sj )j nc . 2
2
5
(13)
(ii) If ti x 1 or if ?1 x sj , then without loss of generality let ti x 1. Then jx ? ti j n2 , and jx ? sj j 2. Thus in this case p j(x ? ti )(x ? sj )j 12 n . We are ready now to obtain estimates for the constant C 00 in (11). Choosing for example c = 2, we recall that !() ([] + 1)!() and apply the p estimate (13) to (1) to obtain for 1 ? x n that 1
2
2
2
jf (x) ? Pn(x)j
v p !q ! u u j Q ( x ) j 1 ? x 1 3 n ? q q t p Kn j(x ? ti )(x ? sj )j ! f ; n + n 2 2
( )
2
v
!
q
0
1
u q u jQn(x)j 3 ? q @f q ; j(x ? t`)(x ? sm)j A : (14) ! 3 p Kn t j(x ? ti )(x ? sj )j n 2 p On the other hand, if 1 ? x n , then it follows from our previous analysis that for ` 6= i; m 6= j we can estimate Qn(x) q? (x ? ti )(x ? sj )(x ? t` )(x ? sm) 2 ; and hence using Rolle's theorem we obtain ( )
2
2
1
q
p !q ?
jf (x) ? Pn(x)j j(x ? ti )(x ? sj )j q q j(x ? t` )(x ? sm )j jQn (x)j
j(f (x) ? Pn(x))(x ? ti )(x ? sj )j 2 n jQn (x)j p q? q q (q! 2) (15) nq? jf (z ) ? Pn (z )j; p for some intermediate point z lying between x and 1. Since 1 ? z p 1 ? x n , we may now further estimate 1
1
( )
( )
1
2
2
2
2 q? p1 ? x 1 ! ! f q ; p n?x2 + n2 2 q? p 2 K n jf q (z ) ? Pnq (z )j n ?x + n +n pj x?t x?sm j n n2 ` 2 q ! f q ; n q 3K n ; j(x ? t`)(x ? sm)j 1
( )
1
( )
2
1
1
( )
2
( )
(
1
)(
1
)
in which we have also used the fact that ? !() 2? !() for 0 < < . p Therefore, for 1 ? x n we obtain 1
2
2
6
1
jf (x) ? Pn(x)j q v 0 1 j ( x ? t )( x ? s ) j u ` m j Q ( x ) j 3 2q K u n t @ q A: q! nq j(x ? ti )(x ? sj )j ! f ; n
(16)
( )
The estimate (14) is therefore seen to be dominant on the entire interval [?1; 1], with the much stronger inequality (16) holding near the ends. The estimate for C 00 in (14) can be improved for large values of q by allowing the parameter c to vary with q. Indeed, we can obtain a constant C 00 which is independent of q as well as of the other quantities involved and is 1 thus absolute. To do this, we can choose c = (q!) q . Then similarly as in (15) and (16) we can estimate
v u u 1 jf (x) ? Pn(x)j nq t j(x ?jQt n)((xx)?j s )j jf q (z ) ? Pnq (z )j i j ( )
( )
v p ! u u K j Q ( x ) j 1 ? x 1 n q t nq j(x ? t )(x ? s )j ! f ; n + n ; (17) i j 2
( )
2
p
1
valid for 1 ? x qn q . Now, from Stirling's formula we know that 2
( !)
1
e? q (q!) q : 1
p
Hence, if we now let 1 ? x eqn , we will get an estimate which is also 1 p valid for 1 ? x qn q . Thus, choosing now c = qe in (13) we obtain (for q > 4) 0 p 1q ! 2q 1 ? x + 2 q + 3 e n @q A 1+ ee ; q ? 2 e (1 ? n2 ) ? x and from this we obtain 2
2
( !)
1
2
1
2
2
2
2
v p ! e Ku u e j Q 1?x 1 n (x)j q t jf (x) ? Pn(x)j nq j(x ? t )(x ? s )j ! f ; n + n (18) i j 2
( )
2
p
for 1 ? x qe , and, combining (18) with (17) we see that the estimate (17) in fact holds on the entire interval [?1; 1]. 2
7
Now, just as was done in deriving (14) we can obtain from (18) the estimate
jf (x) ? Pn(x)j v
0
u 3 ee K u t jQn (x)j @ q q n j(x ? ti )(x ? sj )j ! f ; ( )
p
q
1
j(x ? t`)(x ? sm) A
(19)
n
p
for 1 ? x n . If on the other hand 1 ? x n we have this estimate already, as a consequence of (16). Therefore, (19) holds on the entire interval [?1; 1], and our proof is completed in the case q > 0, with C 00 minf pq+1q ; 3eeg. For the constant C we must now take the maximum of C 0 and C 00, and the proof of Theorem 2 is concluded. 2
2
2
2
3
(
2)
Proof of Theorem 1: We will construct a polynomial Pn which satis es both (1) and (2), with a constant independent of both n and f . According to a result of Trigub [4], a polynomial pn exists which satis es (1) with a constant K independent of n and f . Beginning with this polynomial pn , we will construct polynomials pn; ; . . . ; pn;q . For each j = 0; . . . ; q the polynomial pn;j will be seen still to satisfy (1) (with a new constant Kq depending on q and K ) and to satisfy 0
(f ? pn;j )(tk ) = (f ? pn;j )(tk ) = 0 for k = 0; . . . ; j: Then we will de ne
(20)
Pn := pn;q :
(21)
It is clear from Theorem 2 that the polynomial Pn will then satisfy both (1) and (2). We de ne rst a polynomial pn; ; by 0 0
pn; ; (x) = (f (s ) ? pn (s )) 0 0
0
0
1 ? x 2
+ (f (t ) ? pn (t )) 0
0
1 + x 2
:
Then both (1) and (20) follow for pn + pn; ; in place of pn. Thus we can let pn; := pn + pn; ; and, de ning Pn by (21), we have (1) with a new constant K 2K . If q = 0 we are now nished. We now construct the polynomial pn; in the eventuality that q > 0. Let us assume inductively that we have constructed a polynomial pn;j? for 0 0
0
0 0
0
0
1
8
which (20) holds and for which (1) holds with a constant of Kj? . We now show how to construct the polynomial pn;j . To this end, we let m denote the greatest even integer such that mq < n, and we let Tm(x) := cos(m arccos x), the Chebyshev polynomial of degree m. The properties of Tm which we need here are that Tm(1) = 1 ; kTmk 1 (by k k we mean here the supremum norm on [?1; 1]); and Tm0 (1) = m ; Tm0 (?1) = ?m . Using these properties, we de ne for each j = 0; . . . ; q Tm;j (x) := Tm( 2x s? ?sj t? tj ); j j noting that Tm;j (sj ) = Tm;j (tj ) = 1, while 0 (t ) = ?T 0 (s ) = 2m Tm;j j m;j j tj ? s j and that there are constants and depending upon q but otherwise independent of m and n such that if ?1 x ?1 + n2 and 1 ? n2 y 1 we have minfTm;j (x); Tm;j (y)g maxfTm;j (x); Tm;j (y)g : (22) 1
2
2
2
1
1
Now, let us assume that we have de ned a polynomial pn;j? such that (20) holds for pn;j? , and (1) holds with a constant Kq? . We now construct for each k = 0; . . . ; j a polynomial pn;j;k as follows: 1
1
1
! jY ? t ? x j pn;j;j (x) := f(f (sj ) ? pn;j? (sj )) t ? s g 11 ?? TTm;`((sx)) (23) j j m;` j ! ` jY ? +f(f (tj ) ? pn;j? (tj )) sj ? x g 1 ? Tm;`(x) : (24) sj ? tj ` 1 ? Tm;`(tj ) The polynomial pn;j;j enjoys the property that 1
1
=1
1
1
=1
pn;j;j (sk ) = pn;j;j (tk ) = 0 for k = 0; . . . ; j ? 1; while pn;j;j also interpolates f ? pn;j? exactly at sj and tj . Since the order of the points s ; . . . ; sj and t ; . . . ; tj was actually not relevant in this construction, we can by relabelling the points in a suitable manner similarly construct polynomials pn;j;k respectively interpolating f ? pn;j? on the points sk and tk for each k = 0; . . . ; j ? 1. Having done this, we can de ne 1
0
0
1
pn;j := pn;j? + 1
9
j X
k
=0
pn;j;k ;
(25)
and clearly (20) is satis ed at the index j . To see that (1) is also satis ed, it suces to show that (26) kpn;j;k k Mj Kj n?q !(f q ; n1 ); in which Mj and Kj are constants depending upon j , with K 2K . And to see this we may without loss of generality choose k = j . Rewriting (23) as ( )
2
0
pn;j;j (x) =
!
jY ? (s ? s )(1 ? T (x) (f (sj ) ? pn;j? (sj )) tj ? x j ` m;` f (s ? s ) . . . (s ? s ) t ? s g 1 ? Tm;`(sj ) j j j? j j ` 1
1
0
1
=1
!
jY ? +f (f (tj ) ? pn;j? (tj )) sj ? x g (tj ? t`)(1 ? Tm;`(x)) ; (tj ? t ) . . . (tj ? tj? ) sj ? tj 1 ? Tm;`(tj ) ` 1
1
0
1
=1
we easily get (26) by using Rolle's theorem, taking also (22) into consideration. From (26) we now clearly get (1) for the polynomial pn;j constructed in (25) with a new constant Kj , and we can repeat the induction until we reach j = q. As previously stated, we now let Pn := pn;q . At this point, it remains only to show that (2) follows from the fact that Pn interpolates f on Sn . This follows from Theorem 2, and then the proof of Theorem 1 is completed, with a new value of K equal to CKq , where C is the constant obtained in Theorem 2.
References [1] K. BALA ZS and T. KILGORE, On some constants in simultaneous approximation, International Journal of Mathematics and Mathematical Sciences, to appear. [2] K. BALA ZS and T. KILGORE, Some identities and inequalities for derivatives, submitted. [3] I. GOPENGAUZ, A theorem of A. F. Timan on the approximation of functions by polynomials on a nite segment, Mat. Zametki 1 (1967), 163-172 (in Russian). 10
[4] R. TRIGUB, Approximation of functions by polynomials with integral coecients (in Russian), Isv. Akad. Nauk. SSSR, Ser. Matem. 26 (1962), 261-280.
T. Kilgore Mathematics Auburn University Auburn, AL 36849 U.S.A.
[email protected] J. Prestin FB Mathematik Universitat Rostock Universitatsplatz 1 18051 Rostock Germany
[email protected]
11
