# Matrix Transformations and Disk of Convergence in Interpolation Processes

Authors: Chikkanna R. Selvaraj; Suguna Selvaraj
Journal: International Journal of Mathematics and Mathematical Sciences
 (2008)

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## Abstract

LetAρ denote the set of functions analytic in |z|<ρ but not on |z|=ρ(1<ρ<∞). Walsh proved that the difference of the Lagrange polynomial
interpolant of f(z)∈Aρ and the partial sum of the Taylor polynomial
of f converges to zero on a larger set than the domain of definition of f. In
1980, Cavaretta et al. have studied the extension of Lagrange interpolation,
Hermite interpolation, and Hermite-Birkhoff interpolation processes in a similar
manner. In this paper, we apply a certain matrix transformation on the
sequences of operators given in the above-mentioned interpolation processes
to prove the convergence in larger disks.


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## Body

## 1. Introduction

Ifx=(xk) is a complex number sequence and A=[ank] is an infinite matrix, then Ax is the sequence whose nth term is given by(1.1)(Ax)n=∑k=0∞ankxk.A matrix A is called X−Y matrix if Ax is in the set Y whenever x is in X.
For 0≤α<∞,
let(1.2)Gα={x:limsup|xk|1/k≤α}.For various values of α,
this sequence space has been studied extensively by several authors [1–5]. In particular,
Jacob Jr.
[2, page 186] proved the following result.Theorem 1.1.
An infinite matrixA is a Gu−Gv matrix if and only if for each number w such that 0<w<1/v,
there exist numbers B and s such that 0<s<1/u and(1.3)|ank|wn≤Bskfor all n and k.LetAρ denote the collection of functions analytic in
the disk Dρ={z∈𝒞:|z|<ρ} for some 1<ρ<∞ and having a singularity on the circle |z|=ρ.
In Section 2, we state the results proved by
Cavaretta Jr. et
al. [6] on the Lagrange interpolation, Hermite interpolation, and Hermite-Birkhoff interpolation of f(z)∈Aρ in the nth roots of unity, which will be required.
Main results are given in Section 3 and deal with the application of a certain
matrix to the various polynomial interpolants in the above results.
Interestingly, we are able to show that under the matrix transformation the
difference of the interpolant polynomials and the corresponding Taylor
polynomials converges to zero in a larger region.


## 2. Preliminaries

Throughout the paper, we
assume thatf(z)∈Aρ with 1<ρ<∞ and f(z) has the Taylor series expansion(2.1)f(z)=∑k=0∞akzk.For each integer n≥0,
let Ln(z;f) denote the unique Lagrange interpolation
polynomial of degree n which interpolates f(z) in the (n+1)st roots of unity, that is,(2.2)Ln(ω;f)=f(ω),where ω is any (n+1)st root of unity. Setting Sn(z;f)=∑k=0n+1akzk,
the well-known Walsh equiconvergence theorem [7] states that(2.3)limn→∞[Ln(z;f)−Sn(z;f)]=0,∀|z|<ρ2,with the convergence being uniform
and geometric on any closed subset of |z|<ρ2. This theorem has been extended in various ways
by several authors [6, 8–10]. In all that follows, we state some of the
results of [6] which are needed for our main results.Under Lagrange
interpolation, letting(2.4)Sn,j(z;f)=∑k=0nak+j(n+1)zk,j=0,1,…,the authors
[6, Theorem 1, page
156] have proved the following result.Theorem 2.1.
For each positive
integerl,(2.5)limn→∞{Ln(z;f)−∑j=0l−1Sn,j(z;f)}=0,∀|z|<ρl+1and this result is best
possible.In the proof of Theorem2.1, it has been shown by the authors that(2.6)Ln(z;f)−∑j=0l−1Sn,j(z;f)=12πi∫Γf(t)(tn+1−zn+1)(t−z)(tn+1−1)tl(n+1)dt,where Γ is any circle |t|=R with 1<R<ρ.In [6] the authors have studied the Hermite interpolation also in a similar
way. For a fixed positive integer r≥2,
let hr(n+1)−1(z;f) be the unique Hermite polynomial interpolant
to f,f′,…,f(r−1) in the (n+1)st roots of unity, that is,(2.7)hr(n+1)−1(j)(ω;f)=f(j)(ω),j=0,1,…,r−1,where ω is any (n+1)st root of unity.Setting(2.8)Hr(n+1)−1,0(z;f)=∑k=0r(n+1)−1akzk,Hr(n+1)−1,j(z;f)=βj(z)∑k=0nak+(n+1)(r+j−1)⋅zk,j=1,2,…,where(2.9)βj(z)=∑k=0r−1(r+j−1k)(zn+1−1)k,j=1,2,…,in [6, Theorem 3, page 162] the
authors proved the following result.Theorem 2.2.
For each
positive integerl,(2.10)limn→∞{hr(n+1)−1(z;f)−∑j=0l−1Hr(n+1)−1,j(z;f)}=0,∀|z|<ρ1+l/rand this result is best
possible.In the proof of Theorem2.2, it was shown by the authors in [6, page 165] that(2.11)hr(n+1)−1(z;f)−∑j=0l−1Hr(n+1)−1,j(z;f)=12πi∫Γf(t)(t−z)K(t,z)dt,where Γ is any circle |t|=R with 1<R<ρ and(2.12)|K(t,z)|≤M(Rn+1+|z|n+1)(|z|+1)r−1R(r+l)(n+1).Under Hermite-Birkhoff
interpolation, the authors in [6] established several results for different
cases. We consider here only the (0,m) case. Let n and m be integers with n≥m≥0.
Let b2n+1(0,m)(z;f) be the unique Hermite-Birkhoff polynomial of
degree 2n+1 which interpolates f in the (n+1)st roots of unity and whose mth derivative interpolates f(m) in the (n+1)st roots of unity, that is,(2.13)b2n+1(0,m)(ω;f)=f(ω),(b2n+1(0,m)(ω;f))(m)=f(m)(ω),where ω is any (n+1)st root of unity.Setting(2.14)B2n+1,0(0,m)(z;f)=∑k=02n+1akzk,B2n+1,ν(0,m)(z;f)=∑j=0naj+(ν+1)(n+1)⋅zjqj,ν(z),ν=1,2…,where qj,ν(z) is a polynomial of degree (n+1) given by(2.15)qj,ν(z)=zn+1+((ν+1)(n+1)+j)m−(n+1+j)m(n+1+j)m−(j)m(zn+1−1),j=0,1,…,n,with the following conventional
notation(2.16)(j)m={j(j−1)⋯(j−m+1),ifm≤j,0,ifm>j,in [6, Theorem 4, page 170] the
authors proved the following result.Theorem 2.3.
For each
positive integerl,
(2.17)limn→∞{b2n+1(0,m)(z;f)−∑j=0l−1B2n+1,j(0,m)(z;f)}=0,for|z|<ρ1+l/2and this result is best
possible.In the proof, it was shown in [6, Theorem 4, page 171] that(2.18)b2n+1(0,m)(z;f)−∑j=0l−1B2n+1,j(0,m)(z;f)=12πi∫Γf(t)Kn,l(t,z)dt,where Γ is any circle |t|=R with 1<R<ρ and |Kn,l(t,z)| is bounded on the circle |t|=R<|z| by(2.19)|Kn,l(t,z)|≤|z|n+1(|t|n+1+|z|n+1)|t−z||t|(l+1)(n+1)|tn+1−1|+(|z|n+1+1)(|z|n+1+Rn+1)(|z|−R)R(l+2)(n+1).


## 3. Main Results

Our aim in this section is
to apply aGu−Gv matrix to the polynomial sequences of
operators in each of the above three theorems given in Section 2 and prove that
the difference of transformed sequences in each case converges to zero in a
lager disk Dρ^. To simplify, let us denote Ln(z;f) and ∑j=0l−1Sn,j(z;f) in Theorem 2.1 by Ln and Sn,l,
respectively.Theorem 3.1.
Letf(z)∈Aρ and let Γ be any circle |t|=R with 1<R<ρ.
For any ρ^>ρ,
choose u>ρ^/R and 0<v<1.
Let A be a Gu−Gv matrix and define(3.1)λn(z)=∑k=0∞ankLk,σn,l(z)=∑k=0∞ankSk,lfor n=0,1,…. Then, for each l,(3.2)limn→∞[λn(z)−σn,l(z)]=0,∀z∈Dρ^.Proof.
Using the integral representation
given in (2.6), for each l=1,2,…, we have(3.3)λn(z)−σn,l(z)=∑k=0∞ank(Lk−Sk,l)=∑k=0∞ank12πi∫Γf(t)(tk+1−zk+1)(t−z)(tk+1−1)tl(k+1)dt=∑k=0∞ank12πi∫Γf(t)(t−z)tl(k+1)[1−(zt)k+1]tk+1tk+1−1dt.Since |t|=R>1,
we have(3.4)λn(z)−σn,l(z)=∑k=0∞ank12πi∫Γf(t)(t−z)tl(k+1)[1−(zt)k+1]∑q=0∞(1tk+1)qdt=12πi∫Γf(t)(t−z)∑q=0∞1tl+q∑k=0∞ank[1−(zt)k+1](1tl+q)kdt.The interchange of the integral
and the summations is justified by showing that the series(3.5)∑k=0∞ank(1tl+q)k,∑k=0∞ank(zt)k(1tl+q)kconverge absolutely for each q as follows. From (1.3), for any 1<w<1/v we have that |ank|wn≤Bsk,
where s<1/u<R/ρ^<1. Thus, for each q,l, and n,
we have(3.6)∑k=0∞|ank|(1|t|l+q)k≤Bwn∑k=0∞(sRl+q)k=BwnRl+q(Rl+q−s),because s/Rl+q<R/ρ^Rl+q<1 and similarly for |z|<ρ^(3.7)∑k=0∞|ank||zt|k+1(1|t|l+q)k≤Bρ^wnR∑k=0∞(sρ^Rl+q+1)k=Bρ^wnRRl+q+1Rl+q+1−sρ^,because sρ^/Rl+q+1<R/Rl+q+1<1.
Therefore, from (3.6) and (3.7), identities
(3.4) become(3.8)|λn(z)−σn,l(z)|≤B2πwn∫Γ|f(t)||t−z|∑q=0∞1Rl+q[Rl+qRl+q−s+ρ^Rl+qRl+q+1−sρ^]dt=B2πwn∫Γ|f(t)||t−z|∑q=0∞[1Rl+q−s+ρ^Rl+q+1−sρ^]dt.It can be easily proved that the
two series on the right side of the above expression converge by using the
ratio test. Assuming that f(t) is bounded on Γ,w>1  implies that(3.9)limn→∞[λn(z)−σn,l(z)]=0,∀|z|<ρ^.Next, we prove a similar result
of convergence on a larger disk in the case of the Hermite interpolation. To
simplify, let us denote(3.10)hr(n+1)−1(z;f),∑j=0l−1Hr(n+1)−1,j(z;f)in Theorem 2.2 by hn and Hn,l,
respectively.Theorem 3.2.
Letf(z)∈Aρ and let Γ be any circle |t|=R with 1<R<ρ.
For any ρ^>ρ, choose u>ρ^/R and 0<v<1.
Let A be a Gu−Gv matrix and define(3.11)λn(z)=∑k=0∞ankhk,σn,l(z)=∑k=0∞ankHk,lfor n=0,1,…. Then, for each l,(3.12)limn→∞[λn(z)−σn,l(z)]=0,∀z∈Dρ^.Proof.
Using the integral representation
given in (2.11), for each l=1,2,…,
we have(3.13)λn(z)−σn,l(z)=∑k=0∞ank(hk−Hk,l)=∑k=0∞ank12πi∫Γf(t)t−zK(t,z)dt.From (2.12) we obtain
that(3.14)|λn(z)−σn,l(z)|≤∑k=0∞|ank|12π∫Γ|f(t)||t−z|M(Rk+1+|z|k+1)(|z|+1)r−1R(r+l)(k+1)dt≤M(|z|+1)r+12π∫Γ|f(t)||t−z|∑k=0∞|ank|(Rk+1+|z|k+1)R(r+l)(k+1)dt.The interchange of the integral
and the summation is justified by showing that the series(3.15)∑k=0∞|ank|R(r+l−1)(k+1),∑k=0∞|ank|(|z|Rr+l)k+1converge as follows. From (1.3)
we have that for any 1<w<1/v,|ank|wn≤Bsk, where s<1/u<R/ρ^<1.
Thus for any fixed positive integerr≥2 and for each l and n(3.16)∑k=0∞|ank|(1Rr+l−1)k+1≤BwnRr+l∑k=0∞(sRr+l−1)k=Bwn(Rr+l−1−s),because s/Rr+l−1<1/ρ^Rr+l−2<1,
and similarly for |z|<ρ^,(3.17)∑k=0∞|ank|(|z|Rr+l)k+l≤B|z|wnRr+l∑k=0∞(s|z|Rr+l)k=B|z|wn1(Rr+l−s|z|),because s|z|/Rr+l<1/Rr+l−1<1.
Therefore, using (3.16) and
(3.17) in
(3.14) we get that(3.18)|λn(z)−σn,l(z)|≤MB(|z|+1)r+12πwn∫Γ|f(t)||t−z|[1Rr+l−1−s+|z|Rr+l−s|z|]dt.Assuming that f(t) is bounded on Γ,w>1  implies that(3.19)limn→∞[λn(z)−σn,l(z)]=0,for|z|<ρ^.Thus in the Lagrange case, theGu−Gv matrix transformation of the sequence
operators produces new sequences such that the difference between the
transformed sequences of polynomials Ln and Sn,l converges to zero in an arbitrarily large disk Dρ^ by choosing u>ρ^/R.
Similarly, in the Hermite case also, the domain of convergence to zero is
arbitrarily large under Gu−Gv matrix transformation by choosing u>ρ^/R.  But as we see in the following theorem, in the case of
Hermite-Birkhoff interpolation, the domain of convergence to zero of the
difference of the transformed polynomials of Theorem 2.3 is arbitrarily large
only if we choose u>ρ^2/R.To simplify, let us denoteb2n+1(0,m)(z;f) and ∑j=0l−1B2n+1,j(0,m)(z;f) of Theorem 2.3 by bn and Bn,l,
respectively.Theorem 3.3.
Letf(z)∈Aρ and let Γ be any circle |t|=R with 1<R<ρ.
For any ρ^>ρ,
choose u>ρ^2/R and 0<v<1.
Let A be a Gu−Gv matrix and define(3.20)λn(z)=∑k=0∞ankbk,σn,l(z)=∑k=0∞ankBk,lfor n=0,1,…. Then, for each l,(3.21)limn→∞[λn(z)−σn,l(z)]=0,∀z∈Dρ^.Proof.
Using the integral representation
given in (2.18), for each l=1,2,…, we have(3.22)λn(z)−σn,l(z)=∑k=0∞ank12πi∫Γf(t)Kk,l(t,z)dt.From (2.19) we obtain
that(3.23)|λn(z)−σn,l(z)|≤∑k=0∞|ank|12π∫Γ|f(t)||Kk,l(t,z)|dt≤12π∫Γ|f(t)||t−z|∑k=0∞|ank||z|k+1(|t|k+1+|z|k+1)|t|(l+1)(k+1)|tk+1−1|dt+12π∫Γ|f(t)||z|−R∑k=0∞|ank|(|z|k+1+1)(|z|k+1+Rk+1)R(l+2)(k+1)dt.The interchange of the integral
and the summations is justified by showing that the two series in (3.23)
converge as follows. From (1.3) we have that for any 1<w<1/v,|ank|wn≤Bsk,
where s<1/u<R/ρ^2<1.
Since |t|=R and |z|<ρ^,
using the same method used in (3.4) we write the first summation as(3.24)∑k=0∞|ank||z|k+1|t|(l+1)(k+1)(|t|k+1+|z|k+1)(Rk+1−1)≤∑k=0∞Bsk|z|k+1wnR(l+1)(k+1)[1+(|z|R)k+1]∑q=0∞(1Rk+1)q≤Bwn∑q=0∞ρ^Rl+q+1[∑k=0∞(sρ^Rl+q+1)k+∑k=0∞(sρ^Rl+q+1)k(ρ^R)k+1]=Bwn∑q=0∞ρ^Rl+q+1[Rl+q+1(Rl+q+1−sρ^)+ρ^R(Rl+q+2)(Rl+q+2−sρ^2)],because sρ^/Rl+q+1<1/ρ^Rl+q<1 and sρ^2/Rl+q+2<1/Rl+q+1<1 for each l and q.
Now, for the second sum in (3.23) we have(3.25)∑k=0∞|ank|(|z|k+1+1)(|z|k+1+Rk+1)R(l+2)(k+1)≤Bwn∑k=0∞skR(l+2)(k+1)[ρ^2k+2+ρ^k+1+(ρ^R)k+1+Rk+1]=Bwn[ρ^2Rl+2−sρ^2+ρ^Rl+2−sρ^+ρ^Rl+1−sρ^+1Rl+1−s],because for each l,(3.26)sρ^2Rl+2<1Rl+1<1,sρ^Rl+2<1ρ^Rl+1<1,sρ^Rl+1<1ρ^Rl<1,sRl+1<1ρ^2Rl<1.Using (3.24) and
(3.25), the
expression (3.23) becomes(3.27)|λn(z)−σn,l(z)|≤Bwn2π∫Γ|f(t)||t−z|∑q=0∞[ρ^Rl+q+1−sρ^+ρ^2Rl+q+2−sρ^2]dt+B2πwn∫Γ|f(t)||z|−R[ρ^2Rl+2−sρ^2+ρ^Rl+2−sρ^+ρ^Rl+1−sρ^+1Rl+1−s]dt.It can be easily proved that the
two series on the right side of the above expression converge by using the
ratio test. Assuming that f(t) is bounded on Γ,w>1  implies that(3.28)limn→∞[λn(z)−σn,l(z)]=0for each |z|<ρ^.In two of the above three theorems, for anyR and ρ^ satisfying the stated conditions, we have
chosen u>ρ^/R>1 and 0<v<1.
In the case of the last theorem, we have chosen u>ρ^2/R>1 and 0<v<1.
Now, to see if there exists such a Gu−Gv matrix, we give  below an obvious example.Define a matrixA by(3.29)ank=vnpk,p>u.Then for any w such that 0<w<1/v,
we have that(3.30)|ank|wn=(vw)npk<(1p)k,where 1/p<1/u.
Hence by Theorem 1.1, A is a Gu−Gv matrix.



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