Extension Of Pál Type Hermite-Fejér Interpolation Onto The Unit Circle

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^(α, β)(x) and its derivative P_n^(α, β)' (x), together with ±1 onto the unit circle. The Lagrange data are prescribed on the first 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.

Full text

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 2n2, when values of the function are prescribed on
the set of nodes with npoints and those of their derivatives on another set of (n1) 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 r2on the nth roots of unity. Brück [5] studied Lagrange
interpolation of a function considering nodes z
kn =T(wkn), where wkn = exp 2ik
2n+ 1; n 0; k = 1(1)2n
and T=z
1z ;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
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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
(z21)4P(;)00
n(x)+4z(z21)f(++ 2)z2+ 1g(z21) 2z3()P(;)0
n(x)
16z6n(n+++ 1)P(;)
n(x)=0:
Let Z2nand T2n2be 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
T2n2=ftk=x
k+iy
k= cos k+isin k;tn+k=tk;k= 1;2; :::; (n1) ; 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 T2n2are given by (2.1) and (2.2) respectively.
W(z) =
2n
Y
k=1
(zzk) = KnP(;)
n 1 + z2
2z!zn(2.1)
and
W1(z) =
2n2
Y
k=1
(ztk) = K
nP(;)0
n 1 + z2
2z!zn1;(2.2)
where
Kn= 22nn!(++n+ 1)
(++ 2n+ 1)
and
K
n= 22n1(n1)! (++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)
(zzk)W0(zk); k = 1;2; :::; 2n(2.3)
and
l
k(z) = W1(z)
(ztk)W0
1(tk); k = 1;2; :::; (2n2) :(2.4)
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Bahadur et al. 3
We can write z=x+iy, where x; y 2R. If jzj= 1, then
z21= 2p1x2(2.5)
and
jzzkj=p2r1xxkp1x2q1x2
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 1x1, we have
(1 x2)1=2jP(;)
n(x)j=O(n1);(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=cosk; k = 1;2; :::; n are the zeros of P(;)
n(x). Then
(1 x2
k)1 k
n!2
;(2.11)
P(;)
n(xk)k1
2n;(2.12)
P(;)0
n(xk)k3
2n+2;(2.13)
P(;)00
n(xk)k5
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+ 2r1satisfying Jackson’s inequality (see [9]):
jf(z)Fn(z)j  C !r+1(f; n1); r 0(2.15)
and also an inequality by O. Kiˆs [10]
F(m)
n(z)C nm!r+1(f ; n1); m 2Z+;(2.16)
where !r(f; n1) = O(nr+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+ 2r1on the set of nodes Z2nand T2n2with Hermite-Fejér boundary conditions at 1satisfying the
conditions. 8
>
<
>
:
Rn(zk) = kfor k= 1;2; :::; 2n;
R0
n(tk) = kfor k= 1;2; :::; (2n2);
R(m)
n(1) = 0 for m= 0;1; :::; r;
(3.1)
where kand kare complex constants and r < 1.
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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) +
2n2
X
k=1
kBk(z):(3.2)
Here, Ak(z)and Bk(z)are the …rst and second fundamental polynomials each of degree 4n+ 2r1
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; :::; 2n2;
A(m)
k(1) = 0 for m= 0;1; :::; r;
(3.3)
and for k= 1;2; :::2n2;8
>
<
>
:
Bk(zj) = 0 for j= 1;2; :::; 2n;
B0
k(tj) = kj for j= 1;2; :::; 2n2;
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(z21)rl
k(z)dz (3.5)
and
J1j(z) = Zz
0
znj(z21)rW1(z)dz ;j= 0;1;(3.6)
where J1j(1) = (1)j+1J1j(1).
Theorem 1. For k= 1;2; :::; (2n2), second fundamental polynomial is given by (3.7)
Bk(z) = znW(z)bkJk(z) + b0kJ10(z) + b1kJ11 (z);(3.7)
where
bk=1
W(tk)tk(t2
k1)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+ 2r1) satisfying the conditions given in
(3.4). At z=zj,j= 1;2; :::; n
Bk(zj) = zn
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) = nzn1W(z) + znW0(z)bkJk(z) + b0kJ10 (z) + b1kJ11(z)
+znW(z)bkJ0
k(z) + b0kJ0
10(z) + b1kJ0
11(z)
=hnzn1nKnP(;)
n1 + z2
2zzno+KnznnP(;)0
n1 + z2
2zzn+P(;)
n1 + z2
2znzn1oi
bkJk(z) + b0kJ10(z) + b1kJ11 (z)+znW(z)bkJ0
k(z) + b0kJ0
10(z) + b1kJ0
11(z)
and
B0
k(z) = hKnP(;)0
n1 + z2
2zibkJk(z) + b0kJ10(z) + b1kJ11 (z)
+znW(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) = tn
jW(tj)bkJ0
k(tj).
Using (2.4) and (3.5), we have
Bk0(tj) = tn
jW(tj)bktn+1
j(t2
j1)rl
k(tj) = tjW(tj)bk(t2
j1)rkj :
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) = (z21)r+1 lk(z)W1(z)
(z2
k1)r+1W1(zk)+znW(z)Sk(z) + a0kJ10 (z) + a1kJ11(z);(3.11)
where
Sk(z) = Zz
0
zn(z21)r
W0(zk)(z2
k1)r+1W1(zk)(z21)W0
1(z) + ckW1(z)
(zzk)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+ 2r1) satisfying the conditions given in
(3.3). At z=zj,j= 1;2; :::; 2n, we have
Ak(zj) = (z2
j1)r+1 lk(zj)W1(zj)
(z2
k1)r+1W1(zk)+zn
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
j1)r+1 lk(zj)W1(zj)
(z2
k1)r+1W1(zk)=(z2
j1)r+1 kj W1(zj)
(z2
k1)r+1W1(zk)=kj :
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6Extension of Pál Type Hermite-Fejér Interpolation onto the Unit Circle
Di¤erentiating Ak(z)with respect to z;
A0
k(z) = 1
(z2
k1)r+1W1(zk)hW1(z)2z(r+ 1)(z21)rlk(z)+(z21)r+1 l0
k(z)
+(z21)r+1lk(z)W0
1(z)i+nnzn1W(z) + znW0(z)oSk(z) + a0kJ10 (z) + a1kJ11(z)
+znW(z)S0
k(z) + a0kJ0
10(z) + a1kJ0
11(z):
At z=tj, we have
A0
k(tj) = 1
(z2
k1)r+1W1(zk)hW1(tj)2tj(r+ 1)(t2
j1)rlk(tj)+(t2
j1)r+1l0
k(tj)
+(t2
j1)r+1lk(tj)W0
1(tj)i+nntn1
jW(tj) + tn
jW0(tj)oSk(tj) + a0kJ10(tj) + a1kJ11 (tj)
+tn
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
j1)r+1lk(tj)W0
1(tj)
(z2
k1)r+1W1(zk)+nntn1
jW(tj) + tn
jW0(tj)oSk(tj) + a0kJ10(tj) + a1kJ11 (tj)
+tn
jW(tj)S0
k(tj):
From the second condition given in (3.3), we have
0 = (t2
j1)r+1lk(tj)W0
1(tj)
(z2
k1)r+1W1(zk)+nntn1
jW(tj) + tn
jW0(tj)oSk(tj) + a0kJ10(tj) + a1kJ11 (tj)
+tn
jW(tj)S0
k(tj)
and
tn
jW(tj)S0
k(tj) = (t2
j1)r+1lk(tj)W0
1(tj)
(z2
k1)r+1W1(zk)
+nntn1
jW(tj)tn
jW0(tj)oSk(tj) + a0kJ10(tj) + a1kJ11 (tj)
=(t2
j1)r+1lk(tj)W0
1(tj)
(z2
k1)r+1W1(zk)
+(ntn1
jKnP(;)
n 1 + t2
j
2tj!tn
jtn
j(KnP(;)0
n 1 + t2
j
2tj!tn
j
+nKnP(;)
n 1 + t2
j
2tj!tn1
j)) Sk(tj) + a0kJ10(tj) + a1kJ11 (tj)!:
After a little computation, we get
S0
k(tj) = (t2
j1)r+1lk(tj)W0
1(tj)
tn
jW(tj)(z2
k1)r+1W1(zk):
We can write above equation as
S0
k(z) = zn(z21)r
W0(zk)(z2
k1)r+1W1(zk)(z21)W0
1(z) + ckW1(z)
(zzk):(3.16)
PREPRINT

Bahadur et al. 7
Integrating (3.16) provides us with a polynomial Sk(z)of degree (3n+ 2r1) given by (3.12).
To establish the validity of Sk(z), we must have [(z21)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;
2n2
X
k=1 jBk(z)j=O(1 x2)r=2nrlog n;(4.2)
where 1<  r1
2.
Proof of Lemma 1.From (3.11) we have
2n
X
k=1 jAk(z)j 
2n
X
k=1 (z21)r+1lk(z)W1(z)
(z2
k1)r+1W1(zk)
| {z }
I1
+
2n
X
k=1 znW(z)Sk(z)
| {z }
I2
+
2n
X
k=1 znW(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 (z21)r+1W(z)W1(z)
(z2
k1)r+1(zzk)W0(zk)W1(zk)
=
2n
X
k=1 
(z21)r+1nKnP(;)
n1+z2
2zznonK
nP(;)0
n1+z2
2zzn1o
(z2
k1)r+1(zzk)nKnP(;)
n1+z2
2zzno0
jz=zknK
nP(;)0
n1+z2
k
2zkzn1
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 (z21)r+1P(;)
n(x)P(;)0
n(x)
j(z2
k1)r+2j jzzkjP(;)0
n(xk)
2:
Using (2.5) and (2.6), we get
I1= 2
2n
X
k=1
2r+1(1 x2)r+1
2P(;)
n(x)P(;)0
n(x)
2r+2(1 x2
k)r+2
2p2q1xxkp1x2p1x2
kP(;)0
n(xk)
2
=1
p2
2n
X
k=1
(1 x2)r+1
2P(;)
n(x)P(;)0
n(x)q1xxk+p1x2p1x2
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
2P(;)
n(x)P(;)0
n(x)q1xxk+p(1 xxk)2(xxk)2
(1 x2
k)r+2
2jxxkjP(;)0
n(xk)
2

2n
X
k=1
(1 x2)r+1
2P(;)
n(x)P(;)0
n(x)p1xxk
(1 x2
k)r+2
2jxxkjP(;)0
n(xk)
2:
For jxxkj  1
21x2
k, we have
I1
2n
X
k=1
2(1 x2)r+1
2P(;)
n(x)P(;)0
n(x)p1xxk
(1 x2
k)r+2
2(1 x2
k)P(;)0
n(xk)
2

2n
X
k=1
2p2(1 x2)r
2np(1 x2)P(;)
n(x)oP(; )0
n(x)
(1 x2
k)r+4
2P(;)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
kr2+1 !:
From r2+ 1 1, we get
I1=O(1 x2)r=2nr+1 log nn1<  r
2o:(4.4)
The reader can verify that estimate remains the same in the case where jxxkj<1
21x2
k. Following
similar scheme as above gives
I2=O(1 x2)r=2nrlog nn1<  r
2o;(4.5)
and
I3=O(1 x2)r=2nrlog nn1<  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=znW(z)bkJk(z) + b0kJ10(z) + b1kJ11 (z);(4.7)
2n2
X
k=1 jBk(z)j 
2n2
X
k=1 znW(z)bkJk(z)
| {z }
M1
+
2n2
X
k=1 znW(z)((b0kJ10(z) + b1kJ11 (z))
| {z }
M2
(4.8)
and 2n2
X
k=1 jBk(z)j  M1+M2:(4.9)
Using (3.8) and (3.5), we have
M1=
2n2
X
k=1 znW(z)(1
W(tk)tk(t2
k1)r)Zz
0
zn+1(z21)rl
k(z)dz:
Using (2.4), we get
M1=
2n2
X
k=1 znW(z)(1
W(tk)tk(t2
k1)r)Zz
0
zn+1(z21)r(W1(z)
(ztk)W0
1(tk))dz

2n2
X
k=1 
znW(z)
W(tk)tk(t2
k1)rmax
jzj=1 Zz
0
zn+1(z21)r(W1(z)
(ztk)W0
1(tk))dz

2n2
X
k=1 
znW(z)(z21)rW1(z)
W(tk)tk(t2
k1)r(ztk)W0
1(tk)Zz
0
zn+1 dz:
Using (2.1) and (2.2), we get
M1
2n2
X
k=1 
znnKnP(;)
n1+z2
2zzno(z21)rnK
nP(;)0
n1+z2
2zzn1o
nKnP(;)
n1+t2
k
2tktn
kotk(t2
k1)r(ztk)nK
nP(;)0
n1+z2
2zzn1o0
jz=tk
jzn+2 j
n+ 2
Since P(;)0
n(x
k)=0and jtkj= 1, we get
M12
n+ 2
2n2
X
k=1 P(;)
n(x)(z21)rP(;)0
n(x)
P(;)
n(x
k)j(t2
k1)r+1j jztkjP(;)00
n(x
k)
:
Owing to (2.5) and (2.6), we have
M11
(n+ 2)
2n2
X
k=1
(1 x2)r=2P(;)
n(x)P(;)0
n(x)p1xx
k
(1 x2
k)r+1
2P(;)
n(x
k)jxx
kjP(;)00
n(x
k)
:
For jxx
kj  1
21x2
k, we have
M12
(n+ 2)
2n2
X
k=1
(1 x2)r=2P(;)
n(x)P(;)0
n(x)p1xx
k
(1 x2
k)r+1
2P(;)
n(x
k)j1x2
kjP(;)00
n(x
k)
(4.10)
2p2
(n+ 2)
2n2
X
k=1
(1 x2)r=2P(;)
n(x)P(;)0
n(x)
(1 x2
k)r+3
2P(;)
n(x
k)P(;)00
n(x
k)
:
PREPRINT

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
kr2!:
From r21, we get
M1=O(1 x2)r=2nrlog n 1<  r1
2:(4.11)
The estimate remains the same in the case, where jxx
kj<1
21x2
k. Similarly, we have
M2=O(1 x2)r=2nrlog n 1<  r1
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
jkj=On !r+1(f; n1); k = 1;2; :::; (2n2):(5.1)
Then sequence fRn(z)gis de…ned by
Rn(z) =
2n
X
k=1
f(zk)Ak(z) +
2n2
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; n1) log n;(5.3)
where !r+1(f ; n1)be the (r+ 1)th modulus of continuity of f(z).
Proof. Since Rn(z)be the uniquely determined polynomial of degree 4n+ 2r1and the polynomial
Fn(z)satisfying equation (2.15) can be expressed as
Fn(z) =
2n
X
k=1
Fn(zk)Ak(z) +
2n2
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+
2n2
X
k=1 jkF0
n(zk)jjBk(z)j+jFn(z)f(z)j

2n
X
k=1 jf(zk)Fn(zk)jjAk(z)j
| {z }
N1
+
2n2
X
k=1 jkjjBk(z)j
| {z }
N2
+
2n2
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 ; n1)(1 x2)r=2nr+1 log n(5.7)
and
N2=
2n2
X
k=1 jkjjBk(z)j:
From (5.1) and (4.2), we have
N2=On !r+1(f; n1))(1 x2)r=2nrlog n(5.8)
and
N3=
2n2
X
k=1 jF0
n(zk)jjBk(z)j:
From (2.16) and (4.2), we have
N3=On !r+1(f; n1)(1 x2)r=2nrlog n(5.9)
and
N4=jFn(z)f(z)j:
From (2.15), we have
N4=O!r+1(f ; n1):(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; n1) 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[T2n2). This will provide a much broader aspect
of convergence and comparisons to the present problem.
PREPRINT

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