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App lied Mathem atics E-Notes, 22(2022), xx-x x c

ISSN 1607-2510

Available free at mirror sites of http://www.ma th.nthu.edu.tw/amen/

Extension Of Pál Type Hermite-Fejér Interpolation Onto The

Unit Circle

Swarnima Bahadury, Varunz, Vishnu Narayan Mishrax

Received 20 July 2021

Abstract

The paper is devoted to the study of a Pál type (0; 1) interpolation problem on the unit circle

considering two disjoint sets of nodes. The nodal points are obtained by projecting vertically the zeros of

the Jacobi polynomial P(;)

n(x)and its derivative P(;)0

n(x), together with 1onto the unit circle. The

Lagrange data are prescribed on the …rst set of nodes, the Hermite data are prescribed on the second

one and generalized Hermite-Fejér boundary conditions are prescribed at 1. An explicit representation

of the interpolatory polynomial is given and the convergence is studied for analytic functions on the unit

disk. The results are of interest to approximation theory.

1. Introduction

Interpolation problems on the unit circle have been an area of constant investigation during the past few

years. A considerable amount of literature got accumulated on Lacunary, Birkho¤ or Pál-type interpolation

on the unit circle.

Throughout this paper, we denote the Jacobi polynomial of degree n by P(;)

n(x). Pál [14] proved that

there does not exist a unique polynomial of degree 2n2, when values of the function are prescribed on

the set of nodes with npoints and those of their derivatives on another set of (n1) points. To obtain a

unique solution, he imposed an extra condition and provided the explicit representation of the interpolatory

polynomial. Since then, researchers look forward to more general Pál-type interpolation problems. Lénárd

[13] considered a (0;2) type Pál interpolation problem and obtained regularity and explicit representation

for the same.

In 1960, Kiˆs [10] was the initiator of interpolation processes on the unit circle. He considered the (0;2) and

(0;1; :::::::::; r 2; r)interpolation for an integer r2on the nth roots of unity. Brück [5] studied Lagrange

interpolation of a function considering nodes z

kn =T(wkn), where wkn = exp 2ik

2n+ 1; n 0; k = 1(1)2n

and T=z

1z ;0< < 1is a Mobius transformation of the unit disk into itself.

In 2003, Dikshit [8] considered the Pál-type interpolation on non-uniformly distributed nodes on the unit

circle. Bruin [6] considered Pál-type interpolation problem and studied the e¤ect of interchanging the value

nodes and the derivative nodes on the problem’s regularity. Bahadur and Shukla [1] considered weighted

(0; 1) Pál-type interpolation problem on the vertically projected zeros of (1 x2)P(;)

n(x)and P(;)0

n(x)

onto the unit circle. Explicit representation and convergence was studied for analytic functions on the unit

disk. Many researchers ([2,3,4,11,12]) worked in similar direction.

In the present paper, we extended Pál-type Hermite-Fejér interpolation onto the unit circle by prescribing

Lagrange data on nodes obtained by vertically projecting zeros of P(;)

n(x)as well as Hermite data on nodes

Mathematics Subject Classi…cations: 41A10, 97N 50, 41A05, 30E10.

yDepartment of Mathematics and Astronomy, University of Lucknow, Lucknow, Uttar P radesh, India

zCorresponding author. Department of Mathem atics and Astronomy, University of Lucknow, Lucknow, Uttar Pradesh

xCorresponding author. Department of Mathematics, Indira Gandhi National Tribal University, Lalpur, Amarkantak, Anup-

pur, M adhya Pradesh 484 887, India

PREPRINT

2Extension of Pál Type Hermite-Fejér Interpolation onto the Unit Circle

obtained by vertically projecting zeros of P(;)0

n(x)onto the unit circle. The novelty of this paper is that

we took generalized Hermite-Fejér boundary conditions at 1. To obtain the explicit representations of

the interpolatory polynomial is our …rst aim, since the problem is regular. We also obtained the order of

convergence of such interpolatory polynomial.

The paper has been organized in the following manner. Section 2is assigned to preliminaries. The

interpolation problem and explicit representation of the interpolatory polynomial are de…ned in Section

3. Sections 4and 5are devoted to …nding estimates and establishing a convergence theorem respectively.

Conclusions have been covered in Section 6.

2. Preliminaries

This section includes the following results, which we shall use. The di¤erential equation satis…ed by P(;)

n(x)

is

(1 x2)P(;)00

n(x)+[(++ 2)x]P(;)0

n(x) + n(n+++ 1)P(;)

n(x) = 0:

Using the Szeg½

o transformation x=1 + z2

2z, we have

(z21)4P(;)00

n(x)+4z(z21)f(++ 2)z2+ 1g(z21) 2z3()P(;)0

n(x)

16z6n(n+++ 1)P(;)

n(x)=0:

Let Z2nand T2n2be two distinct sets of nodes such that

Z2n=fzk=xk+iyk= cos k+isin k;zn+k=zk;k= 1;2; :::; n ;xk; yk2Rg

and

T2n2=ftk=x

k+iy

k= cos k+isin k;tn+k=tk;k= 1;2; :::; (n1) ; x

k; y

k2Rg;

which are obtained by projecting vertically the zeros of P(;)

n(x)and P(;)0

n(x)respectively on the unit

circle.

The nodal polynomials W(z)and W1(z)de…ned on Z2nand T2n2are given by (2.1) and (2.2) respectively.

W(z) =

2n

Y

k=1

(zzk) = KnP(;)

n 1 + z2

2z!zn(2.1)

and

W1(z) =

2n2

Y

k=1

(ztk) = K

nP(;)0

n 1 + z2

2z!zn1;(2.2)

where

Kn= 22nn!(++n+ 1)

(++ 2n+ 1)

and

K

n= 22n1(n1)! (++n+ 1)

(++ 2n+ 1):

The fundamental polynomials of Lagrange interpolation on the zeros of W(z)and W1(z)are respectively

given by (2.3) and (2.4).

lk(z) = W(z)

(zzk)W0(zk); k = 1;2; :::; 2n(2.3)

and

l

k(z) = W1(z)

(ztk)W0

1(tk); k = 1;2; :::; (2n2) :(2.4)

PREPRINT

Bahadur et al. 3

We can write z=x+iy, where x; y 2R. If jzj= 1, then

z21= 2p1x2(2.5)

and

jzzkj=p2r1xxkp1x2q1x2

k:(2.6)

To evaluate the estimates of the fundamental polynomials formed in the next Section 3, we will use the

following (refer to pg.164–166 of [16]).

For 1x1, we have

(1 x2)1=2jP(;)

n(x)j=O(n1);(2.7)

P(;)

n(x)=O(n);(2.8)

P(;)0

n(x)=O(n+2);(2.9)

P(;)00

n(x)=O(n+4):(2.10)

Considering set of nodes Z2n, where xk=cosk; k = 1;2; :::; n are the zeros of P(;)

n(x). Then

(1 x2

k)1 k

n!2

;(2.11)

P(;)

n(xk)k1

2n;(2.12)

P(;)0

n(xk)k3

2n+2;(2.13)

P(;)00

n(xk)k5

2n+4:(2.14)

Let f(z)be continuous for jzj 1, analytic for jzj<1and f(r)2Lip ; = 1 + , > 0. Then, there exists

a polynomial Fn(z)of degree 4n+ 2r1satisfying Jackson’s inequality (see [9]):

jf(z)Fn(z)j C !r+1(f; n1); r 0(2.15)

and also an inequality by O. Kiˆs [10]

F(m)

n(z)C nm!r+1(f ; n1); m 2Z+;(2.16)

where !r(f; n1) = O(nr+1)denotes the rth modulus of continuity of f(z)as well as Cis a constant

independent of nand z.

3. The Problem & Explicit Representation of Interpolatory Polynomial

Here, we are interested in determining the convergence of interpolatory polynomial Rn(z)of degree

4n+ 2r1on the set of nodes Z2nand T2n2with Hermite-Fejér boundary conditions at 1satisfying the

conditions. 8

>

<

>

:

Rn(zk) = kfor k= 1;2; :::; 2n;

R0

n(tk) = kfor k= 1;2; :::; (2n2);

R(m)

n(1) = 0 for m= 0;1; :::; r;

(3.1)

where kand kare complex constants and r < 1.

PREPRINT

4Extension of Pál Type Hermite-Fejér Interpolation onto the Unit Circle

Rn(z)can be written in the form given below

Rn(z) =

2n

X

k=1

kAk(z) +

2n2

X

k=1

kBk(z):(3.2)

Here, Ak(z)and Bk(z)are the …rst and second fundamental polynomials each of degree 4n+ 2r1

satisfying (3.3) and (3.4) respectively. For k= 1;2; :::; 2n;

8

>

<

>

:

Ak(zj) = kj for j= 1;2; :::; 2n;

A0

k(tj) = 0 for j= 1;2; :::; 2n2;

A(m)

k(1) = 0 for m= 0;1; :::; r;

(3.3)

and for k= 1;2; :::2n2;8

>

<

>

:

Bk(zj) = 0 for j= 1;2; :::; 2n;

B0

k(tj) = kj for j= 1;2; :::; 2n2;

B(m)

k(1) = 0 for m= 0;1; :::; r:

(3.4)

Explicit expressions of the polynomials Bk(z)and Ak(z)are given in Theorems 1and 2respectively.

Remark 1. The equations (3.5) and (3.6) have been developed while deriving out explicit representation

of the interpolatory polynomial. Readers can get the motivation to form such expression from the idea to

maintain the degree of the polynomial as well as simultaneously satisfy the conditions required to form the

fundamental polynomial.

Jk(z) = Zz

0

zn+1(z21)rl

k(z)dz (3.5)

and

J1j(z) = Zz

0

znj(z21)rW1(z)dz ;j= 0;1;(3.6)

where J1j(1) = (1)j+1J1j(1).

Theorem 1. For k= 1;2; :::; (2n2), second fundamental polynomial is given by (3.7)

Bk(z) = znW(z)bkJk(z) + b0kJ10(z) + b1kJ11 (z);(3.7)

where

bk=1

W(tk)tk(t2

k1)r;(3.8)

b1k=bk(Jk(1) + Jk(1))

2J11(1) (3.9)

and

b0k=bk(Jk(1) Jk(1))

2J10(1) :(3.10)

Proof. Consider (3.7), where Bk(z)is atmost of the degree (4n+ 2r1) satisfying the conditions given in

(3.4). At z=zj,j= 1;2; :::; n

Bk(zj) = zn

jW(zj)bkJk(zj) + b0kJ10(zj) + b1kJ11 (zj):

PREPRINT

Bahadur et al. 5

Since zj’s are the zeros of the polynomial W(z), so Bk(zj) = 0. Di¤erentiating Bk(z)with respect to z gives

us

B0

k(z) = nzn1W(z) + znW0(z)bkJk(z) + b0kJ10 (z) + b1kJ11(z)

+znW(z)bkJ0

k(z) + b0kJ0

10(z) + b1kJ0

11(z)

=hnzn1nKnP(;)

n1 + z2

2zzno+KnznnP(;)0

n1 + z2

2zzn+P(;)

n1 + z2

2znzn1oi

bkJk(z) + b0kJ10(z) + b1kJ11 (z)+znW(z)bkJ0

k(z) + b0kJ0

10(z) + b1kJ0

11(z)

and

B0

k(z) = hKnP(;)0

n1 + z2

2zibkJk(z) + b0kJ10(z) + b1kJ11 (z)

+znW(z)bkJ0

k(z) + b0kJ0

10(z) + b1kJ0

11(z):

Since tj’s are the zeros of the polynomial W1(z), we see that B0

k(tj) = tn

jW(tj)bkJ0

k(tj).

Using (2.4) and (3.5), we have

Bk0(tj) = tn

jW(tj)bktn+1

j(t2

j1)rl

k(tj) = tjW(tj)bk(t2

j1)rkj :

Using condition B0

k(tj) = kj given in (3.4), at j=k, we get (3.8). One can verify the results for j6=k.

Also, from B(m)

k(1) = 0 for m= 0;1; :::; r, we get (3.9) and (3.10). Hence, Theorem 1follows.

Theorem 2. For k= 1;2; :::; 2n, …rst fundamental polynomial is given by (3.11)

Ak(z) = (z21)r+1 lk(z)W1(z)

(z2

k1)r+1W1(zk)+znW(z)Sk(z) + a0kJ10 (z) + a1kJ11(z);(3.11)

where

Sk(z) = Zz

0

zn(z21)r

W0(zk)(z2

k1)r+1W1(zk)(z21)W0

1(z) + ckW1(z)

(zzk)dz; (3.12)

a0k=Sk(1) Sk(1)

2J10(1) ;(3.13)

a1k=(Sk(1) + Sk(1))

2J11(1) (3.14)

and

ck=(1 zk2)W0

1(zk)

W1(zk):(3.15)

Proof. Consider (3.11), where Ak(z)is atmost of the degree (4n+ 2r1) satisfying the conditions given in

(3.3). At z=zj,j= 1;2; :::; 2n, we have

Ak(zj) = (z2

j1)r+1 lk(zj)W1(zj)

(z2

k1)r+1W1(zk)+zn

jW(zj)Sk(zj) + a0kJ10(zj) + a1kJ11 (zj):

Since zj’s are the zeros of the polynomial W(z), we see that

Ak(zj) = (z2

j1)r+1 lk(zj)W1(zj)

(z2

k1)r+1W1(zk)=(z2

j1)r+1 kj W1(zj)

(z2

k1)r+1W1(zk)=kj :

PREPRINT

6Extension of Pál Type Hermite-Fejér Interpolation onto the Unit Circle

Di¤erentiating Ak(z)with respect to z;

A0

k(z) = 1

(z2

k1)r+1W1(zk)hW1(z)2z(r+ 1)(z21)rlk(z)+(z21)r+1 l0

k(z)

+(z21)r+1lk(z)W0

1(z)i+nnzn1W(z) + znW0(z)oSk(z) + a0kJ10 (z) + a1kJ11(z)

+znW(z)S0

k(z) + a0kJ0

10(z) + a1kJ0

11(z):

At z=tj, we have

A0

k(tj) = 1

(z2

k1)r+1W1(zk)hW1(tj)2tj(r+ 1)(t2

j1)rlk(tj)+(t2

j1)r+1l0

k(tj)

+(t2

j1)r+1lk(tj)W0

1(tj)i+nntn1

jW(tj) + tn

jW0(tj)oSk(tj) + a0kJ10(tj) + a1kJ11 (tj)

+tn

jW(tj)S0

k(tj) + a0kJ0

10(tj) + a1kJ0

11(tj):

Since tj’s are the zeroes of W1(z), we see that

A0

k(tj) = (t2

j1)r+1lk(tj)W0

1(tj)

(z2

k1)r+1W1(zk)+nntn1

jW(tj) + tn

jW0(tj)oSk(tj) + a0kJ10(tj) + a1kJ11 (tj)

+tn

jW(tj)S0

k(tj):

From the second condition given in (3.3), we have

0 = (t2

j1)r+1lk(tj)W0

1(tj)

(z2

k1)r+1W1(zk)+nntn1

jW(tj) + tn

jW0(tj)oSk(tj) + a0kJ10(tj) + a1kJ11 (tj)

+tn

jW(tj)S0

k(tj)

and

tn

jW(tj)S0

k(tj) = (t2

j1)r+1lk(tj)W0

1(tj)

(z2

k1)r+1W1(zk)

+nntn1

jW(tj)tn

jW0(tj)oSk(tj) + a0kJ10(tj) + a1kJ11 (tj)

=(t2

j1)r+1lk(tj)W0

1(tj)

(z2

k1)r+1W1(zk)

+(ntn1

jKnP(;)

n 1 + t2

j

2tj!tn

jtn

j(KnP(;)0

n 1 + t2

j

2tj!tn

j

+nKnP(;)

n 1 + t2

j

2tj!tn1

j)) Sk(tj) + a0kJ10(tj) + a1kJ11 (tj)!:

After a little computation, we get

S0

k(tj) = (t2

j1)r+1lk(tj)W0

1(tj)

tn

jW(tj)(z2

k1)r+1W1(zk):

We can write above equation as

S0

k(z) = zn(z21)r

W0(zk)(z2

k1)r+1W1(zk)(z21)W0

1(z) + ckW1(z)

(zzk):(3.16)

PREPRINT

Bahadur et al. 7

Integrating (3.16) provides us with a polynomial Sk(z)of degree (3n+ 2r1) given by (3.12).

To establish the validity of Sk(z), we must have [(z21)W0

1(z) + ckW1(z)]jz=zk= 0, which in turn gives

(3.15). Similarly, the constants a0kand a1kcan be found out by satisfying the condition

A(m)

k(1) = 0 for m= 0;1; :::; r:

Hence, Theorem 2follows.

4. Estimates of Fundamental Polynomials

We need to calculate estimates in order to obtain the rate of convergence of interpolatory polynomials.

Lemma 1. Let Ak(z)be given by (3.11). Then for jzj 1;

2n

X

k=1 jAk(z)j=O(1 x2)r=2nr+1 log n;(4.1)

where 1< r

2.

Lemma 2. Let Bk(z)be given by (3.7). Then for jzj 1;

2n2

X

k=1 jBk(z)j=O(1 x2)r=2nrlog n;(4.2)

where 1< r1

2.

Proof of Lemma 1.From (3.11) we have

2n

X

k=1 jAk(z)j

2n

X

k=1 (z21)r+1lk(z)W1(z)

(z2

k1)r+1W1(zk)

| {z }

I1

+

2n

X

k=1 znW(z)Sk(z)

| {z }

I2

+

2n

X

k=1 znW(z)(a0kJ10(z) + a1kJ11 (z))

| {z }

I3

:

We can write as

2n

X

k=1 jAk(z)j I1+I2+I3:(4.3)

Using (2.3) we have

I1=

2n

X

k=1 (z21)r+1W(z)W1(z)

(z2

k1)r+1(zzk)W0(zk)W1(zk)

=

2n

X

k=1

(z21)r+1nKnP(;)

n1+z2

2zznonK

nP(;)0

n1+z2

2zzn1o

(z2

k1)r+1(zzk)nKnP(;)

n1+z2

2zzno0

jz=zknK

nP(;)0

n1+z2

k

2zkzn1

ko:

PREPRINT

8Extension of Pál Type Hermite-Fejér Interpolation onto the Unit Circle

Since P(;)

n(xk)=0and jzkj= 1, we get

I1= 2

2n

X

k=1 (z21)r+1P(;)

n(x)P(;)0

n(x)

j(z2

k1)r+2j jzzkjP(;)0

n(xk)

2:

Using (2.5) and (2.6), we get

I1= 2

2n

X

k=1

2r+1(1 x2)r+1

2P(;)

n(x)P(;)0

n(x)

2r+2(1 x2

k)r+2

2p2q1xxkp1x2p1x2

kP(;)0

n(xk)

2

=1

p2

2n

X

k=1

(1 x2)r+1

2P(;)

n(x)P(;)0

n(x)q1xxk+p1x2p1x2

k

(1 x2

k)r+2

2p(1 xxk)2(1 x2)(1 x2

k)P(;)0

n(xk)

2

=1

p2

2n

X

k=1

(1 x2)r+1

2P(;)

n(x)P(;)0

n(x)q1xxk+p(1 xxk)2(xxk)2

(1 x2

k)r+2

2jxxkjP(;)0

n(xk)

2

2n

X

k=1

(1 x2)r+1

2P(;)

n(x)P(;)0

n(x)p1xxk

(1 x2

k)r+2

2jxxkjP(;)0

n(xk)

2:

For jxxkj 1

21x2

k, we have

I1

2n

X

k=1

2(1 x2)r+1

2P(;)

n(x)P(;)0

n(x)p1xxk

(1 x2

k)r+2

2(1 x2

k)P(;)0

n(xk)

2

2n

X

k=1

2p2(1 x2)r

2np(1 x2)P(;)

n(x)oP(; )0

n(x)

(1 x2

k)r+4

2P(;)0

n(xk)

2:

Using (2.7), (2.9), (2.11) and (2.13), we get

I1=O (1 x2)r=2nr+1

2n

X

k=1

1

kr2+1 !:

From r2+ 1 1, we get

I1=O(1 x2)r=2nr+1 log nn1< r

2o:(4.4)

The reader can verify that estimate remains the same in the case where jxxkj<1

21x2

k. Following

similar scheme as above gives

I2=O(1 x2)r=2nrlog nn1< r

2o;(4.5)

and

I3=O(1 x2)r=2nrlog nn1< r

2o:(4.6)

PREPRINT

Bahadur et al. 9

Combining (4.4), (4.5) and (4.6) gives Lemma 1.

Proof of Lemma 2.Consider (3.7), we have

jBk(z)j=znW(z)bkJk(z) + b0kJ10(z) + b1kJ11 (z);(4.7)

2n2

X

k=1 jBk(z)j

2n2

X

k=1 znW(z)bkJk(z)

| {z }

M1

+

2n2

X

k=1 znW(z)((b0kJ10(z) + b1kJ11 (z))

| {z }

M2

(4.8)

and 2n2

X

k=1 jBk(z)j M1+M2:(4.9)

Using (3.8) and (3.5), we have

M1=

2n2

X

k=1 znW(z)(1

W(tk)tk(t2

k1)r)Zz

0

zn+1(z21)rl

k(z)dz:

Using (2.4), we get

M1=

2n2

X

k=1 znW(z)(1

W(tk)tk(t2

k1)r)Zz

0

zn+1(z21)r(W1(z)

(ztk)W0

1(tk))dz

2n2

X

k=1

znW(z)

W(tk)tk(t2

k1)rmax

jzj=1 Zz

0

zn+1(z21)r(W1(z)

(ztk)W0

1(tk))dz

2n2

X

k=1

znW(z)(z21)rW1(z)

W(tk)tk(t2

k1)r(ztk)W0

1(tk)Zz

0

zn+1 dz:

Using (2.1) and (2.2), we get

M1

2n2

X

k=1

znnKnP(;)

n1+z2

2zzno(z21)rnK

nP(;)0

n1+z2

2zzn1o

nKnP(;)

n1+t2

k

2tktn

kotk(t2

k1)r(ztk)nK

nP(;)0

n1+z2

2zzn1o0

jz=tk

jzn+2 j

n+ 2

Since P(;)0

n(x

k)=0and jtkj= 1, we get

M12

n+ 2

2n2

X

k=1 P(;)

n(x)(z21)rP(;)0

n(x)

P(;)

n(x

k)j(t2

k1)r+1j jztkjP(;)00

n(x

k)

:

Owing to (2.5) and (2.6), we have

M11

(n+ 2)

2n2

X

k=1

(1 x2)r=2P(;)

n(x)P(;)0

n(x)p1xx

k

(1 x2

k)r+1

2P(;)

n(x

k)jxx

kjP(;)00

n(x

k)

:

For jxx

kj 1

21x2

k, we have

M12

(n+ 2)

2n2

X

k=1

(1 x2)r=2P(;)

n(x)P(;)0

n(x)p1xx

k

(1 x2

k)r+1

2P(;)

n(x

k)j1x2

kjP(;)00

n(x

k)

(4.10)

2p2

(n+ 2)

2n2

X

k=1

(1 x2)r=2P(;)

n(x)P(;)0

n(x)

(1 x2

k)r+3

2P(;)

n(x

k)P(;)00

n(x

k)

:

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10 Extension of Pál Type Hermite-Fejér Interpolation onto the Unit Circle

Using (2.8), (2.9), (2.11), (2.12) and (2.14), we get

M1=O (1 x2)r=2nr

2n

X

k=1

1

kr2!:

From r21, we get

M1=O(1 x2)r=2nrlog n 1< r1

2:(4.11)

The estimate remains the same in the case, where jxx

kj<1

21x2

k. Similarly, we have

M2=O(1 x2)r=2nrlog n 1< r1

2:(4.12)

Combining (4.11) and (4.12) give us desired Lemma 2.

5. Convergence

Theorem 1. Let f(z)be continous for jzj 1and analytic for jzj<1and f(r) Lip ; = 1 + , > 0.

Let the arbitrary numbers k’s is such that

jkj=On !r+1(f; n1); k = 1;2; :::; (2n2):(5.1)

Then sequence fRn(z)gis de…ned by

Rn(z) =

2n

X

k=1

f(zk)Ak(z) +

2n2

X

k=1

kBk(z);(5.2)

satis…es the following relation for jzj 1

jRn(z)f(z)j=O(1 x2)r=2nr+1!r+1 (f; n1) log n;(5.3)

where !r+1(f ; n1)be the (r+ 1)th modulus of continuity of f(z).

Proof. Since Rn(z)be the uniquely determined polynomial of degree 4n+ 2r1and the polynomial

Fn(z)satisfying equation (2.15) can be expressed as

Fn(z) =

2n

X

k=1

Fn(zk)Ak(z) +

2n2

X

k=1

F0

n(zk)Bk(z);(5.4)

we can write

jRn(z)f(z)j jRn(z)Fn(z)j+jFn(z)f(z)j:(5.5)

Using (5.2) and (5.4), we have

jRn(z)f(z)j

2n

X

k=1 jf(zk)Fn(zk)jjAk(z)j+

2n2

X

k=1 jkF0

n(zk)jjBk(z)j+jFn(z)f(z)j

2n

X

k=1 jf(zk)Fn(zk)jjAk(z)j

| {z }

N1

+

2n2

X

k=1 jkjjBk(z)j

| {z }

N2

+

2n2

X

k=1 jF0

n(zk)jjBk(z)j

| {z }

N3

+jFn(z)f(z)j

| {z }

N4

PREPRINT

Bahadur et al. 11

and

jRn(z)f(z)j N1+N2+N3+N4;(5.6)

where

N1=

2n

X

k=1 jf(zk)Fn(zk)jjAk(z)j:

From (2.15) and (4.1), we have

N1=O!r+1(f ; n1)(1 x2)r=2nr+1 log n(5.7)

and

N2=

2n2

X

k=1 jkjjBk(z)j:

From (5.1) and (4.2), we have

N2=On !r+1(f; n1))(1 x2)r=2nrlog n(5.8)

and

N3=

2n2

X

k=1 jF0

n(zk)jjBk(z)j:

From (2.16) and (4.2), we have

N3=On !r+1(f; n1)(1 x2)r=2nrlog n(5.9)

and

N4=jFn(z)f(z)j:

From (2.15), we have

N4=O!r+1(f ; n1):(5.10)

Using (5.7)–(5.10) in (5.6), we get

jRn(z)f(z)j=O(1 x2)r=2nr+1!r+1 (f; n1) log n:

Hence, Theorem 1follows.

6. Conclusion

This research article poses a completely new problem by introducing the generalized Hermite-Fejér boundary

conditions at the points 1. Since these additional nodes gradually increase the degree of the interpolatory

polynomial. So, the order of convergence must also depend on that increment which can be seen in (5.3)

as we require the (r+ 1)th modulus of continuity for the convergence purpose. Since the present problem

is posed considering generalized Hermite-Fejér boundary conditions only at 1, a subtle open problem is to

consider the generalized Hermite-Fejér boundary conditions at 1as well as on all the nodal points, where

Lagrange and Hermite data are prescribed (i.e 1[Z2n[T2n2). This will provide a much broader aspect

of convergence and comparisons to the present problem.

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12 Extension of Pál Type Hermite-Fejér Interpolation onto the Unit Circle

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