ON A WEIGHTED INTERPOLATION OF FUNCTIONS WITH CIRCULAR MAJORANT

ON A WEIGHTED INTERPOLATION OF FUNCTIONS WITH CIRCULAR MAJORANT Received: 31 July, 2008 Accepted: 06 February, 2009 Communicated by: SIMON J SMITH Department of Mathematics and Statistics La Trobe University, PO Box 199, Bendigo Victoria 3552, Australia EMail: ssmith@latrobeeduau QI Rahman 2000 AMS Sub Class: 41A05, 41A10 Key words: Abstract: Interpolation, Lagrange interpolation, Weighted interpolation, Circular majorant, Projection norm, Lebesgue constant, Chebyshev polynomial Denote by L n the projection operator obtained by applying the Lagrange interpolation method, weighted by (1 x 2 ) 1/2, at the zeros of the Chebyshev polynomial of the second kind of degree n + 1 The norm L n = max Lnf, f 1 where denotes the supremum norm on 1, 1, is known to be asymptotically the same as the minimum possible norm over all choices of interpolation nodes for unweighted Lagrange interpolation Because the projection forces the interpolating function to vanish at ±1, it is appropriate to consider a modified projection norm L n ψ = Lnf, where ψ C 1, 1 is a given max f(x) ψ(x) function (a curved majorant) that satisfies 0 ψ(x) 1 and ψ(±1) = 0 In this paper the asymptotic behaviour of the modified projection norm is studied in the case when ψ(x) is the circular majorant w(x) = (1 x 2 ) 1/2 In particular, it is shown that asymptotically L n w is smaller than L n by the quantity 2π 1 (1 log 2) Page 1 of 20

1 Introduction 3 2 Some Lemmas 7 3 Proof of the Theorem 17 Page 2 of 20

1 Introduction Suppose n 1 is an integer, and for any s, let θ s = θ s,n = (s + 1)π/(n + 2) For i = 0, 1,, n, put x i = cos θ i The x i are the zeros of the Chebyshev polynomial of the second kind of degree n + 1, defined by U n+1 (x) = sin(n + 2)θ/ sin θ where x = cos θ and 0 θ π Also let w be the weight function w(x) = 1 x 2, and denote the set of all polynomials of degree n or less by P n In the paper 5, JC Mason and GH Elliott introduced the interpolating projection L n of C 1, 1 on {wp n : p n P n } that is defined by (11) (L n f)(x) = w(x) n l i (x) f(x i) w(x i ), where l i (x) is the fundamental Lagrange polynomial n x x k U n+1 (x) (12) l i (x) = = x i x k U n+1(x i )(x x i ) k=0 k i Mason and Elliott studied the projection norm L n = max L n f, f 1 where denotes the uniform norm g = max 1 x 1 g(x), and obtained results that led to the conjecture (13) L n = 2 π log n + 2 (log 4π ) π + γ + o(1) as n, Page 3 of 20 where γ = 0577 is Euler s constant Smith 8 This result (13) was proved later by

As pointed out by Mason and Elliott, the projection norm for the much-studied Lagrange interpolation method based on the zeros of the Chebyshev polynomial of the first kind T n+1 (x) = cos(n + 1)θ, where x = cos θ and 0 θ π, is 2 π log n + 2 π (log 8π ) + γ + o(1) (See Luttmann and Rivlin 4 for a short proof of this result based on a conjecture that was later established by Ehlich and Zeller 3) Therefore the norm of the weighted interpolation method (11) is smaller by a quantity asymptotic to 2π 1 log 2 In addition, (13) means that L n, which is based on a simple node system, has (to within o(1) terms) the same norm as the Lagrange method of minimal norm over all possible choices of nodes and the optimal nodes for Lagrange interpolation are not known explicitly (See Brutman 2, Section 3 for further discussion and references on the optimal choice of nodes for Lagrange interpolation) Now, an immediate consequence of (11) is that for all f, (L n f)(±1) = 0 Thus L n is particularly appropriate for approximations of those f for which f(±1) = 0 This leads naturally to a study of the norm (14) L n ψ = max f(x) ψ(x) L nf, where ψ C 1, 1 is a given function (a curved majorant) that satisfies 0 ψ(x) 1 and ψ(±1) = 0 Evidently L n ψ L n In this paper we will look at the particular case when ψ(x) is the circular majorant w(x) = 1 x 2 Note that studies of this nature were initiated by P Turán in the early 1970s, in the context of obtaining Markov and Bernstein type estimates for p if p P n satisfies p(x) w(x) for x 1, 1 see Rahman 6 for a key early paper in this area Our principal result is the following theorem, the proof of which will be developed in Sections 2 and 3 Page 4 of 20

Theorem 11 The modified projection norm L n w, defined by (14) with w(x) = 1 x2, satisfies (15) L n w = 2 π log n + 2 (log 8π ) π + γ 1 + o(1) as n Observe that (15) shows L n w is smaller than L n by an amount that is asymptotic to 2π 1 (1 log 2) Before proving the theorem, we make a few remarks about the method to be used By (11), ( ) L n w = max 1 x 1 w(x) n l i (x) Since the x i are arranged symmetrically about 0, then w(x) n l i(x) is even, and so by (12), L n w = max 0 θ π/2 F n (θ), where (16) F n (θ) = sin(n + 2)θ n + 2 n cos θ cos θ i Figure 1 shows the graph of a typical F n (θ) if n is even, and it suggests that the local maximum values of F n (θ) are monotonic increasing as θ moves from left to right, so that the maximum of F n (θ) occurs close to π/2 For n odd, similar graphs suggest that the maximum occurs precisely at π/2 These observations help to motivate the strategy used in Sections 2 and 3 to prove the theorem the approach is akin to that used by Brutman 1 in his investigation of the Lebesgue function for Lagrange interpolation based on the zeros of Chebyshev polynomials of the first kind Page 5 of 20

2 1 0 0 π/4 π/2 Figure 1: Plot of F 12 (θ) for 0 θ π/2 Page 6 of 20

2 Some Lemmas This section contains several lemmas that will be needed to prove the theorem The first such lemma provides alternative representations of the function F n (θ) that was defined in (16) Lemma 21 If j is an integer with 0 j n + 1, and θ j 1 θ θ j, then ( j 1 ) j sin(n + 2)θ n F n (θ) = ( 1) n + 2 cos θ i cos θ + (21) cos θ cos θ i=j i j 1 = ( 1) j 2 sin(n + 2)θ (22) sin(n + 1)θ + n + 2 cos θ i cos θ Proof The result (21) follows immediately from (16) For (22), note that the Lagrange interpolation polynomial for U n (x) based on the zeros of U n+1 (x) is simply U n (x) itself, so, with l i (x) defined by (12), U n (x) = n l i (x)u n (x i ) = U n+1(x) n + 2 n 1 x 2 i x x i (This formula appears in Rivlin 7, p 23, Exercise 132) Therefore sin(n + 1)θ = sin(n + 2)θ n + 2 n cos θ cos θ i If this expression is used to rewrite the second sum in (21), the result (22) is obtained Page 7 of 20

We now show that on the interval 0, π/2, the values of F n (θ) at the midpoints between consecutive θ-nodes are increasing this result is established in the next two lemmas Lemma 22 If j is an integer with 0 j n, then where n,j := (n + 2) ( F n (θ j+1/2 ) F n (θ j 1/2 ) ) = 2 sin θ j sin θ 1/2 n,j, j (23) n,j := (j n 1) + cot θ (2j+2i 3)/4 cot θ (2j 2i 1)/4 Proof By (22), + cot θ j 1/4 cot θ 1/2 + 1 2 csc θ j 1/4 csc θ j+1/4 n,j = 2(n + 2) sin θ j sin θ 1/2 j sin 2 j 1 θ i + 2 cos θ i cos θ j+1/2 cos θ i cos θ j 1/2 From the trigonometric identity Page 8 of 20 (24) sin 2 A cos A cos B = 1 ( ) ( ) B A B + A cot 2 sin B + cot cos A cos B, 2 2

it follows that j Therefore sin 2 j 1 θ i cos θ i cos θ j+1/2 = 1 2 sin θ j+1/2 2j+2 i j+1 cos θ i cos θ j 1/2 cot θ (2i 3)/4 1 2 sin θ j 1/2 2j i j cos θ j (j + 1) cos θ j+1/2 + j cos θ j 1/2 cot θ (2i 3)/4 2j = cos θ j sin θ 1/2 cot θ (2i 3)/4 + (2j + 2) sin θ j sin θ 1/2 + 1 2 sin θ j+1/2 ( cot θj 1/4 + cot θ j+1/4 ) 2j (25) n,j = (4j 2n) sin θ j sin θ 1/2 + 2 cos θ j sin θ 1/2 cot θ (2i 3)/4 Next consider j sin θ j + cos θ j 2j = cot θ (2i 3)/4 + sin θ j+1/2 ( cot θj 1/4 + cot θ j+1/4 ) j ( ) sin θj + cos θ j cot θ(2i 3)/4 + cot θ (4j 2i 1)/4 Page 9 of 20

(26) Also = sin θ j j 1 + cos θ j sin θ (2i 3)/4 sin θ (4j 2i 1)/4 j = sin θ j cot θ (4j 2i 1)/4 cot θ (2i 3)/4 j = sin θ j cot θ (2j+2i 3)/4 cot θ (2j 2i 1)/4 sin θ j+1/2 ( cot θj 1/4 + cot θ j+1/4 ) (27) = 2 sin θ j cos θ j sin θ j+1/2 sin θ j 1/4 sin θ j+1/4 2 cos θ j+1/4 sin θ j+1/4 + sin θ 1/2 = sin θ j sin θ j 1/4 sin θ j+1/4 2 cos θ j+1/4 = sin θ j sin θ 1/2 + csc θ j 1/4 csc θ j+1/4 sin θ j 1/4 sin θ 1/2 = sin θ j sin θ 1/2 2 + 2 cot θj 1/4 cot θ 1/2 + csc θ j 1/4 csc θ j+1/4 The lemma is now established by substituting (26) and (27) into (25) Lemma 23 If j is an integer with 0 j (n 1)/2, then F n (θ j+1/2 ) > F n (θ j 1/2 ) Proof By Lemma 22 we need to show that n,j > 0, where n,j is defined by (23) Now, if 0 < a < π/4 and 0 < b < a, then cot(a + b) cot(a b) > cot 2 a Page 10 of 20

Also, x csc 2 x is decreasing on (0, π/4), so csc 2 x > π/(2x) if 0 < x < π/4 Thus and so j + j cot θ (2j+2i 3)/4 cot θ (2j 2i 1)/4 > j + j cot 2 θ (j 1)/2 = j csc 2 θ (j 1)/2 > (n + 2)j j + 1, n,j > 1 + cot θ j 1/4 cot θ 1/2 + 1 2 csc θ j 1/4 csc θ j+1/4 n + 2 j + 1 > csc 2 θ (4j 3)/8 + 1 2 csc θ j 1/4 csc θ j+1/4 n + 2 j + 1 Because θ (4j 3)/8 < π/4, the first term in this expression can be estimated using csc 2 x > π/(2x), while the second term can be estimated using csc x > 1/x Therefore n + 2 n,j > j + 5/4 n + 2 (n + 2) 2 + j + 1 2π 2 (j + 3/4)(j + 5/4) n + 2 > 1 (j + 1)(j + 5/4) 4 + n + 2 2π 2 This latter quantity is positive if n 3 Since 0 j (n 1)/2, the only unresolved cases are when j = 0 and n = 1, 2, and it is a trivial calculation using (23) to show that n,j > 0 in these cases as well Page 11 of 20 We next show that in any interval between successive θ-nodes, F n (θ) achieves its maximum in the right half of the interval

Lemma 24 If j is an integer with 0 j (n + 1)/2, and 0 < t < 1/2, then (28) F n (θ j 1/2+t ) F n (θ j 1/2 t ) Proof If j = (n + 1)/2, then θ j 1/2 = π/2, so equality holds in (28) because F n (θ) is symmetric about π/2 Thus we can assume j n/2 For convenience, write a = j 1/2 t, b = j 1/2 + t Since sin(n + 2)θ a = sin(n + 2)θ b = ( 1) j cos tπ, it follows from (21) that F n (θ b ) F n (θ a ) has the same sign as G n,j (t) := n i=j (cos θ b cos θ i )(cos θ a cos θ i ) j 1 If j = 0 this is clearly positive, and otherwise G n,j (t) > 2j 1 i=j (cos θ b cos θ i )(cos θ a cos θ i ) j 1 (cos θ i cos θ b )(cos θ i cos θ a ) (cos θ i cos θ b )(cos θ i cos θ a ) j 1 sin 2 θ 2j i 1 = (cos θ b cos θ 2j i 1 )(cos θ a cos θ 2j i 1 ) (cos θ i cos θ b )(cos θ i cos θ a ) Page 12 of 20

We will show that each term in this sum is positive Because sin θ 2j i 1 > sin θ i, this will be true if for 0 i j 1, sin θ 2j i 1 (cos θ i cos θ b )(cos θ i cos θ a ) sin θ i (cos θ b cos θ 2j i 1 )(cos θ a cos θ 2j i 1 ) > 0 By rewriting each difference of cosine terms as a product of sine terms, it follows that we require sin θ 2j i 1 sin θ (j+i 1/2+t)/2 sin θ (j+i 1/2 t)/2 sin θ i sin θ (3j i 3/2+t)/2 sin θ (3j i 3/2 t)/2 > 0 To establish this inequality, note that sin θ 2j i 1 sin θ (j+i 1/2+t)/2 sin θ (j+i 1/2 t)/2 sin θ i sin θ (3j i 3/2+t)/2 sin θ (3j i 3/2 t)/2 = 1 cos θt 1 (sin θ 2j i 1 sin θ i ) sin θ 2j i 1 cos θ j+i+1/2 + sin θ i cos θ 3j i 1/2 2 = cos θ t 1 sin θ j i 3/2 cos θ j 1/2 1 sin θj 2i 5/2 + sin θ 3j 2i 3/2 4 = cos θ j 1/2 cos θ t 1 sin θ j i 3/2 1 2 sin θ 2j 2i 2 = cos θ j 1/2 sin θ j i 3/2 cos θt 1 cos θ j i 3/2 > 0, Page 13 of 20 and so the lemma is proved The final major step in the proof of the theorem is to show that in each interval between successive θ-nodes, the maximum value of F n (θ) is achieved essentially at the midpoint of the interval

Lemma 25 If n, j are integers with n 2 and 0 j (n + 1)/2, then (29) max θ j 1 θ θ j F n (θ) = F n (θ j 1/2 ) + O ( (log n) 1), where the O ((log n) 1 ) term is independent of j Proof By Lemma 24, it is sufficient to show that G n,j,t := F n (θ j 1/2+t ) F n (θ j 1/2 ) is bounded above by an O ((log n) 1 ) term that is independent of j and t for 0 t 1/2 Now, by (22) we have (210) G n,j,t = 2 j 1 n + 2 cos tπ cos θ i cos θ j 1/2+t + 2 sin (n + 1)tπ 2(n + 2) sin cos θ i cos θ j 1/2 ( (2j + 1)π 2(n + 2) (n + 1)tπ 2(n + 2) Since cos tπ 1 4t 2 if 0 t 1/2, then each summation term can be estimated by cos tπ 4t 2 cos θ i cos θ j 1/2+t cos θ i cos θ j 1/2 cos θ i cos θ j 1/2+t From (2x)/π sin x x for 0 x π/2, it follows that cos θ i cos θ j 1/2+t = 2 sin θ (j+i 1/2+t)/2 sin θ (j i 5/2+t)/2 8(i + 1) 2 π 2 (j i)(j + i + 2), ) Page 14 of 20

and so j 1 cos tπ cos θ i cos θ j 1/2+t cos θ i cos θ j 1/2 32t2 π 2 = 32t2 π 2 j 1 (i + 1) 2 (j i)(j + i + 2) j 1 2 + j + 1 2 2j+1 1 k (211) 16t2 (j + 1) (log(j + 1) 1), π2 where the final inequality follows from Also, (212) sin 2j+1 1 k 1 + log(j + 1) ( ) (n + 1)tπ (2j + 1)π 2(n + 2) sin (n + 1)tπ sin tπ 2(n + 2) 2(n + 2) 2 tπ2 (j + 1) 2(n + 2) sin (2j + 1)π 2(n + 2) We now return to the characterization (210) of G n,j,t By (212), G n,0,t π 2 /(2(n + 2)) For j 1, it follows from (211) and (212) that (213) G n,j,t 2π2 t(j + 1) 1 16t π 6 j + 1 log(j + 1), n + 2 π4 32(n + 2) log(j + 1) Page 15 of 20

where the latter inequality follows by maximizing the quadratic in t On the interval 1 j (n + 1)/2, the maximum of (j + 1)/ log(j + 1) occurs at an endpoint, so { } j + 1 2 (214) log(j + 1) max log 2, n + 3 2 log((n + 3)/2) The result (29) then follows from (213) and (214) Page 16 of 20

3 Proof of the Theorem Since L n w = max 0 θ π/2 F n (θ), it follows from Lemmas 23 and 25 that ( F π ) n 2 + O ((log n) 1 ) if n is odd, L n w = ( ) F π(n+1) n + O ((log n) 1 ) 2(n+2) if n is even To obtain the asymptotic result (15) for L n w we use a method that was introduced by Luttmann and Rivlin 4, Theorem 3, and used also by Mason and Elliott 5, Section 9 If n is odd, then by (22) with n = 2m 1, (31) ( π ) F n = 2 = m 1 2 2m + 1 m 2 2m + 1 cos θ i (k 1/2)π csc sin 2m + 1 (k 1/2)π, 2m + 1 where the second equality follows by reversing the order of summation Now, π 2m + 1 m (k 1/2)π csc 2m + 1 π m = 2m + 1 (k 1/2)π csc 2m + 1 + 2m + 1 (k 1/2)π m 1 k 1/2 The asymptotic behaviour as m of each of the sums in this expression is given Page 17 of 20

by lim m π 2m + 1 m (k 1/2)π csc 2m + 1 = 2m + 1 (k 1/2)π π/2 0 csc x 1 dx x and Also, m sin m 2m 1 k 1/2 = 2 (k 1/2)π 2m + 1 = csc m 1 k 1 k = log 4 π = log(4m) + γ + o(1) π mπ 4m + 2 sin2 4m + 2 = 2m + 1 + O(1) π Substituting these asymptotic results into (31) yields the desired result (15) if n is odd On the other hand, if n = 2m is even, then by (22) and (24), (32) F n ( π(n + 1) 2(n + 2) ) π = sin 4m + 4 + 1 m + 1 = 1 m + 1 m 1 ( 2m+2 1 2 cos π 4m + 4 cos θ i cos (2m+1)π 4m+4 ) m 1 (2i 1)π cot 8m + 8 cos θ i + O(m 1 ) Page 18 of 20

The sum of the cotangent terms can be estimated by a similar argument to that above, using to obtain Also, π/2 2m+2 1 cot 2m + 2 m 1 1 cos θ i = m + 1 0 (cot x x 1 ) dx = log 2 π, (2i 1)π 8m + 8 = 2 π 1 2(m + 1) (cos = 2 π + O(m 1 ) ( log 16m ) π + γ + o(1) mπ 4m + 4 csc π 4m + 4 ) 2 If these asymptotic results are substituted into (32), the result (15) is obtained if n is even, and so the proof of Theorem 11 is completed Page 19 of 20

References 1 L BRUTMAN, On the Lebesgue function for polynomial interpolation, SIAM J Numer Anal, 15 (1978), 694 704 2 L BRUTMAN, Lebesgue functions for polynomial interpolation a survey, Ann Numer Math, 4 (1997), 111 127 3 H EHLICH AND K ZELLER, Auswertung der Normen von Interpolationsoperatoren, Math Ann, 164 (1966), 105 112 4 FW LUTTMANN AND TJ RIVLIN, Some numerical experiments in the theory of polynomial interpolation, IBM J Res Develop, 9 (1965), 187 191 5 JC MASON AND GH ELLIOTT, Constrained near-minimax approximation by weighted expansion and interpolation using Chebyshev polynomials of the second, third, and fourth kinds, Numer Algorithms, 9 (1995), 39 54 6 QI RAHMAN, On a problem of Turán about polynomials with curved majorants, Trans Amer Math Soc, 163 (1972), 447 455 7 TJ RIVLIN, Chebyshev Polynomials From Approximation Theory to Algebra and Number Theory, 2nd ed, John Wiley and Sons, New York, 1990 8 SJ SMITH, On the projection norm for a weighted interpolation using Chebyshev polynomials of the second kind, Math Pannon, 16 (2005), 95 103 Page 20 of 20