Weighted (0;0,2)-interpolation on the unit circle

Adv. Inequal. Appl. 2018, 2018:1 https://doi.org/10.28919/aia/3402 ISSN: 2050-7461

WEIGHTED (0;0,2)-INTERPOLATION ON THE UNIT CIRCLE

SWARNIMA BAHADUR AND SARIYA BANO∗

Department of Mathematics and Astronomy, University of Lucknow, Lucknow 226007, India

Copyright c2018 Bahadur and Bano. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract. In this paper, we study the explicit representation of P´al-type weighted(0; 0,2)-interpolation on the

unit circle on two pairwise disjoint sets of nodes obtained by projecting vertically the zeros of 1−x2 Pn(x)

and 1−x2

P00_n(x)respectively onto the unit circle, where Pn(x)stands fornthLegendre polynomial.

Keywords:Legendre polynomial; weighted interpolation; P´al-type interpolation; existence; explicit forms.

2010 AMS Subject Classification:41A05, 30E10.

1. Introduction

The problem of Lacunary interpolation was initiated by P. Tur ´an [12] on the zeros of 1−x2P_n0₋₁(x), whereP_n−1(x)is the Legendre polynomial of degree(n−1). The problem of (0,2) interpola-tion on the roots of unity was first studied by O. Ki ˇs[9]. Later on several mathematicians have

studied such kind of interpolatory polynomials on the different set of nodes viz. real line, unit

circle etc. L.G.P´al [11] considered two sets of nodes with one additional condition,where the

function values are prescribed on one set and the derivative on other one and the functional

value at the additional point and he obtained a unique polynomial of degree 2n−1. In 1990,

∗_{Corresponding author}

E-mail address: [email protected]

Received June 12, 2017

of ∏n(x)and ∏

0

n(x) and obtained the fundamental polynomials. After that J.Szabad ´os and

A.K.Varma [8] considered convergence of(0,2)interpolation process on the zeros of∏n(x)and

obtained existence,explicit representation and convergence theorem . In 2010, M.L ´en ´ard [10]

considered the P´al-type interpolation

on

line. H.P. Dikshit [2] considered the existence of P´al -type interpolation on the non-uniformly

distributed nodes on the unit circle. In 2011, K. K. Mathur and author1 [3] established con-vergence of weighted (0,2) interpolation on unit circle. In 2012, she [4] considered (0; 0,2)

interpolation on projected nodes of 1−x2Pn(x)andP

0

n(x)on the unit circle.Later on she [5]

considered the(0,2; 0)interpolation on the same set of nodes. After that she (with M. Shukla)

[6] considered weighted P´al-type (0; 1) interpolation on the two disjoint set of nodes ,which

are obtained by projecting vertically the zeros of 1−x2P_n(α,β)(x)andP(α,β)

0

n (x)respectively

onto the unit circle, where Pn(α,β)(x) stands for the Jacobi polynomial. Recently, authors [7]

considered weighted(0,2)interpolation on the nodes obtained by projecting vertically the

ze-ros of 1−x2P_n0(x)onto the unit circle obtained the existence and established a convergence theorem for the same.

These have motivated us to consider(0; 0,2)interpolation on two pairwise disjoint set of nodes

onto the unit circle, in which the Lagrange data is prescribed on the first set of nodes where

as Lacunary data on the other one .We obtained regularity and explicit forms of interpolatory

polynomials.

In this paper, we consider two pairwise disjoint set of the nodes Z_n and G_n, where Lagrange

data onZnand weighted (0,2) data onGn are prescribed , whereZnandGnare given as,

(1) Z_n=

    

   

z₀=1, z_2n+1=−1, z_k=cosθ_k+i sinθ_k,

z_n+k=z_k, k=1(1)n

    

   

(2) G_n=

    

   

t₀=1, t_2n−3=−1, t_k=cosϕ_k+i sinϕ_k,

t_n+k=t_k, k=1(1)n−2

    

In section 2,we give some Preliminaries, in section 3, we describe the problem and regularity,

in section 4, we present the explicit forms of P´al-type weighted(0; 0,2)-interpolation onto the

unit circle.

2. Preliminaries

In this section, we shall give some well-known results, which we shall use.

The differential equation satisfied byPn(x)is

(3) 1−x2P_n00(x)−2xP_n0(x) +n(n+1)P_n(x) =0

(4) W(z) =

2n

∏

k=1

(z−z_k) =K_nP_n

1+z2 2z

zn

(5) R(z) = z2−1 W(z)

(6) H(z) =

2n−4

∏

k=1

(z−t_k) =K_n∗P_n00

1+z2 2z

zn−2

(7) N(z) = z2−1 H(z)

We shall require the fundamental polynomial of Lagrange interpolation based on the zeros

ofR(z),H(z)andN(z)are respectively given as:

(8) L_k(z) = R(z)

R0(z_k) (z−z_k),k=0(1)2n+1

(9) l_k(z) = H(z)

H0(t_k) (z−t_k), k=1(1)2n−4

(10) l1k(z) =

N(z)

(11) J_k(z) = Z z

0

t l_k(t)dt

(12) J_n(z) =

Z z

0

H(t)dt

which satisfies,

(13) Jn(−z) =− Jn(z)

We will also use the following results

(14) W 0(z_k) = 1

2Kn z

2 k −1

zn_k−2P_n0(x_k)

(15) W 00(z_k) = K_n(n−1) z2_k −1−1z_kn−3P_n0(x_k)

(16) W 0(t_k) = 1

2Kn

n(n+3) t_k2−1+4

t_k2+1 t

n−1 k Pn(x

∗

k),

(17) W 00(t_k) = 1

2Kn

n(n−1)(n−1) t_k2 −1−1

t_k2 +1 t

n−2 k Pn(x

∗

k),

(18) R0(zk) = z2k−1

W0(zk)

(19) R00(z_k) =4z_kW0(z_k) + z2_k −1W 00(z_k)

(20) R0(tk) = t2k−1

W0(tk) +2tkW(tk)

(21) R00(t_k) =4t_kW0(t_k) + t_k2 −1W 00(t_k) +2W(t_k)

(22) H0(t_k) = K

∗

n

2 t

2 k −1

t_kn−4P_n000(x∗_k),

(23) H 00(t_k) =K_n∗ (n−5) t_k2 −1−5 t_kn−5P_n000(x∗_k),

(24) N0(z_k) = z2_k−1H0(z_k) +2z_kH(z_k)

(26) N0(tk) = t2k−1

H0(tk)

(27) N 00(t_k) =4t_kH 0(t_k) + t_k2 −1H00(t_k)

3. The Problem and Regularity

Let {z_k}2n₀ +1 and {t_k}2n₀ −3 be two disjoint set of nodes obtained by projecting vertically the zeros of 1−x2Pn(x)and 1−x2

P00_n(x)onto the unit circle respectively, where Pn(x)stands

for nth Legendre polynomial. Here we are interested to determine the polynamial Rn(z) of

degree≤ 6n−5 satisfying the conditions:

(28)

    

   

R_n(z_k) = αk , k=0(1)2n+1

R_n(t_k) = β_k , k=1(1)2n−4

[p(z)R_n(z)]00_z=t

k =γk , k=0(1)2n−3

where, p(z) is weight function such that[p(z)W(z)N(z)]00_z=t

k =0,and αk ,βk and γk are arbitrary complex constants.

Theorem 3.1:Rn(z)is regular onZnandGn.

Proof :It is sufficient, if we show, the unique solution of (28) is Rn(z)≡0,when all dataαk=

β_k=γ_k=0.

In this case, we have

Rn(z) =W(z)N(z)q(z),

where,q(z)is a polynomial of degree≤2n−3.

Obviously,R_n(z_k) =0 fork=0(1)2n+1,

and alsoRn(tk) =0 fork=1(1)2n−4, from

[p(z)R_n(z)]00_z=t

k =0

we get,

q0(t_k) =0

Therefore , we have

whereJ_n(z)is given in (12).

Now for z=1 &−1, we get

a=b=0

Therefore, Rn(z)≡0

Hence the theorem follows.

4. Explicit Representation Of Interpolatory Polynomials

We shall writeR_n(z)satisfying (28) as:

(29) Rn(z) = 2n+1

∑

k=0

αkAk(z) +

2n−4

∑

k=1

β_kBk(z) +

2n−3

∑

k=0

γ_kCk(z)

where A_k(z),B_k(z)andC_k(z)are unique polynomial ,each of degree at most 6n−5 satisfying

the conditions :

Fork=0(1)2n+1

(30)

    

   

A_k z_j = δjk, j=0(1)2n+1

A_k t_j = 0, j=1(1)2n−4

[p(z)A_k(z)]00_z=t

j =0, , j=0(1)2n−3

Fork=1(1)2n−4

(31)

    

   

B_k z_j =0, j=0(1)2n+1

B_k tj

=δjk j=1(1)2n−4

[p(z)B_k(z)]00_z=tj = 0 , j=0(1)2n−3

Fork=0(1)2n−3

(32)

    

   

C_k z_j =0, j=0(1)2n+1

C_k tj

= 0, j=1(1)2n−4

Theorem 4.1: Fork=0(1)2n−3 , we have

(33) C_k(z) = 1

2t_k R(t_k) p(t_k)H 0(t_k) R(z) H(z) Z z

0

t l_k(t) dt

Proof:Let,

(34) C_k(z) = c_k R(z) H(z) J_k(z)

where J_k(z)is given in (11).

Obviously,C_k z_j=0,for each j=0(1)2n+1.

and C_k t_j=0,for each j=1(1)2n−4.

One can see that [p(z)C_k(z)]00_z=t

j =0 , for j6=k and for j=k, we get

(35) c_k= 1

2t_kR(t_k) p(t_k)H0(t_k)

using (35) in (34) we get (33).

Theorem 4.2: Fork=1(1)2n−4, we have (36)

B_k(z) = l 2

1k (z)W(z)

W(t_k) −

W(z)N(z)

H0(t_k)R(t_k)

Z z

0

[l_1k0(t)−l_1k 0(t_k)l_1k(t)]

(t−t_k) dt+bk Z z

0

t l_k(t)

where,

(37) b_k= 1 t_k

"

p00(t_k)

2p(t_k) +

W 00(t_k)

2W(t_k) +

p0(t_k)W 0(t_k) p(t_k)W(t_k) +4lk

0_(t

k)

(

p0(t_k) p(t_k) +

W 0(t_k) W(t_k)

)#

Proof: Let

(38) B_k(z) = l 2

1k (z)W(z)

W(t_k) +

W(z)N(z)

H 0(t_k) R(t_k){Sk(z) +bkJk(z)}

where J_k(z)is given in (11).

Obviously,

B_k zj

B_k t_j = δjk , for j=1(1)2n−4

Further from [ p(z)B_k(z)]00_z=t

j =0, for j6=k, we get,

(39) S_k(z) =−

Z z

0

l_1k0 (t)−l_1k0 (t_k)l_1k(t) (t−t_k) dt

again for j=k, we get (37).

using (37) and (39) in (38) ,we get (36).

Theorem 4.3: Fork=0(1)2n+1, we have (40)

A_k(z) = Lk(z)N(z)H(z) N(z_k)H(z_k) −

N(z)W(z)

H2(z_k)R0(z_k)N(z_k) Z z

0

t2−1[H

0_{(t) H(z}

k)−H 0(zk) H(t)] (t−z_k) dt

Proof:Let ,

(41) A_k(z) = Lk(z)N(z)H(z) N(z_k)H(z_k) +

N(z)W(z)

H 2(z_k) R0(z_k) N(z_k)Mk(z)

Obviously, A_k zj

= δjk, for j=0(1)2n+1,

and A_k t_j = 0, for j=1(1)2n−4.

one can check that,[ p(z)A_k(z)]00_z=t

j =0, for j=k. Further from [ p(z)A_k(z)]00_z=t

j =0,

for

,

(42) M_k(z) =− Z z

0

t2−1

h

H0(t)H(z_k)−H0(z_k)H(t)i

(t−z_k) dt

Using (42) in (41) we get (40).

Conflict of Interests

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