A New Approach to Newton-Type Polynomial Interpolation

Research ArticleA New Approach to Newton-Type PolynomialInterpolation with Parameters

Le Zou 123 Liangtu Song23 Xiaofeng Wang 1 Thomas Weise4 Yanping Chen1

and Chen Zhang1

1School of Artificial Intelligence and Big Data Hefei University Hefei 230601 Anhui China2Hefei Institutes of Physical Science Chinese Academy of Sciences Hefei 230031 China3University of Science and Technology of China Hefei 230027 China4Institute of Applied Optimization School of Artificial Intelligence and Big Data Hefei University Hefei 230601 Anhui China

Correspondence should be addressed to Xiaofeng Wang xfwanghfuueducn

Received 9 June 2020 Revised 28 September 2020 Accepted 10 October 2020 Published 17 November 2020

Academic Editor Massimiliano Ferrara

Copyright copy 2020 Le Zou et al is is an open access article distributed under the Creative Commons Attribution License whichpermits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

Newtonrsquos interpolation is a classical polynomial interpolation approach and plays a significant role in numerical analysis andimage processinge interpolation function of most classical approaches is unique to the given data In this paper univariate andbivariate parameterized Newton-type polynomial interpolation methods are introduced In order to express the divided dif-ferences tables neatly the multiplicity of the points can be adjusted by introducing new parameters Our new polynomialinterpolation can be constructed only based on divided differences with one or multiple parameters which satisfy the interpolationconditions We discuss the interpolation algorithm theorem dual interpolation and information matrix algorithm Since theproposed novel interpolation functions are parametric they are not unique to the interpolation data erefore its value in theinterpolant region can be adjusted under unaltered interpolant data through the parameter values Our parameterized Newton-type polynomial interpolating functions have a simple and explicit mathematical representation and the proposed algorithms aresimple and easy to calculate Various numerical examples are given to demonstrate the efficiency of our method

1 Introduction

e interpolation problem has been the subject of manyclassic studies in approximation theory [1ndash5] In the lastseveral years many researchers have been focusing on thissubject and have obtained interesting results e existingapproaches can be divided into polynomial and rationalinterpolation methods both of which are applicable tonumerical approximation [1 2] image interpolation [3ndash6]and arc structuring and surface modeling [7ndash10] Newtonrsquospolynomial interpolation and ielersquos interpolating con-tinued fractions can be incorporated to generate variousinterpolation schemes based on rectangular grids As onecan see the (partial) inverse differences and the (partial)divided differences play a great role on the study of thepolynomial and rational interpolation Many scholars haveconstructed a variety of blending rational interpolation

Vonza [11] developed rational interpolation-based associ-ated continued fractions Varsamis and Karampetakis [12]presented recursive algorithms for the Newton polynomialinterpolation of a given bivariate function Tan et al con-structed Newtonndashiele [13] and ielendashNewton blendingrational interpolation [1] However blending rationalinterpolants strongly depend on the existence of so-calledblending differences By using the method of divide andconquer Tan and Tang [14] proposed composite interpo-lation over triangular subgrids By dividing the original set ofsupport points into some blocks Tan and Zhao proposed theblock-based iele-like blending rational interpolation [15]and the block-based Newton-type blending rational inter-polation [16] Tang and Liang [17] developed the bivariateblending ielendashWerner osculatory rational interpolationBased on iele interpolating continued fraction polyno-mial interpolation and barycentric rational interpolation

HindawiMathematical Problems in EngineeringVolume 2020 Article ID 9020541 15 pageshttpsdoiorg10115520209020541

many types of blending rational interpolation were studiedIn recent years Dyn and Floater [18] studied multivariatepolynomial interpolation based on lower subsets Cuyt andSalazar Celis [19] developed a generalized multivariate datafitting model to solve a variety of scientific computingproblems such as graphics filtering metamodeling com-putational finance and more which captured linear as wellas nonlinear phenomena Interpolation was also applied tographic image morphing and image processing [1 3ndash620ndash23] He et al [3 20] studied ielendashNewton rationalinterpolation in the polar coordinates applied it to imagesuper-resolution and obtained better performance Karimand Saaban [21] proposed a novel rational bicubic ballfunction with one parameter by using tensor product ap-proach and used it for image upscaling Zhan et al [22]developed a nonlocal and local image interpolation modelbased on nonlocal bounded variation regularization andlocal total variation and obtained better performance Heet al [24] proposed an image inpainting algorithm by usingcontinued fractions rational interpolation In order to obtainbetter repaired results ielersquos rational interpolation wascombined with Newtonndashiele rational interpolation torepair damaged images In [25] the Newton interpolationmethod was adopted in capacitance gradient and the cali-bration of cantilever stiffness Based on polynomial ap-proximation of local surface Rahman et al [5] presented anovel local patch descriptor which is invariant to the changesin viewpoint scale orientation and illumination

e mathematical description and the constructionmethod of the curve and surfaces are important problems incomputer-aided geometric design [7ndash9 26 27] ere aremany ways to handle this problem [1 7ndash9 28 29] includingBezier and NURBS-based approaches as well as the poly-nomial spline methodese methods are used for instancefor shaping the outer hull of aircraft and a ship [28]However the disadvantage of polynomial interpolation liesin its global property that is it is difficult to control the localconstraint of a given interpolation point under the conditionthat the values of the interpolant points are fixed [29 30] Toconstruct the polynomial spline the derivative values of theinterpolating data are required additionally to the functionvalues Unfortunately in many practical fields it is difficultto obtain these derivative values

In recent years many scholars have studied parametricspline interpolation Zhang et al [29 30] proposed newtypes of weighted blending spline interpolation By selectingappropriate parameters and different coefficients the valueof the spline interpolation function can be modified at anypoint in the interpolant interval under the condition that thevalues of the interpolant points are fixed so that the geo-metric surfaces can be adjusted e drawback is that thecomputation is complicated e bivariate rational inter-polation with parameters based only on the function valueshas been studied in [29] Based on function values andpartial derivatives of the function being interpolated Duanet al [31] proposed a new bivariate rational interpolationwhich had a simple and explicit rational mathematical

representation with parameters Based on function valuesDuan et al [32] constructed a rational cubic spline withparametersis spline is C1-continuous in the interpolatinginterval By selecting suitable parameters it can also becomeC2-continuous Reddy et al [33] presented univariate andbivariate rational cubic fractal interpolation functions withshape parameters which can be used for surface modelingwhen the partial derivatives of the given function are ir-regular In [34] the authors presented a weighted bivariaterational bicubic spline interpolation based on functionvalues which is C1-continuous for any positive parameterse shape of the interpolating surface can be modified bychanging the parameters under the condition that the valuesof the interpolating nodes are fixed which is convenient inengineering and useful in computer-aided geometric design[28ndash34] Tang and Zou [35] studied and presented generalinterpolant frames with many classical interpolation for-mulas Univariate and bivariate iele-like rational inter-polation continued fractions with parameters were studiedin [36] ere the interpolation function can deal withinterpolation problems where inverse differences do notexist rough the selection of proper parameter values thevalue of the proposed interpolation function can be changedat any point in the interpolant region under the conditionthat the values of the interpolating nodes are fixed Howeverhow to select appropriate parameters is a hard problem It isalso difficult to define such an interpolant function Poly-nomials have many advantages such as a simple structureeasy differentiation integration and having derivatives ofany order e polynomial function is a good tool for ap-proximating smooth function In computer-aided geometricdesign there is still a great demand for complicated modelsand the integration of design and fabrication but there arestill two unresolved problems [29 30]

(1) How can one construct a suitable polynomial in-terpolation function with an explicit mathematicalrepresentation simple calculation which is alsoconvenient to use and to study theoretically

(2) For given data how can one remodel the curves orsurfaces shape to meet the actual requirements in thecomputer-aided geometric design

Given n + 1 points (x0 x1 xn isin R) and n + 1 values(y0 y1 yn isin R) in the traditional interpolationmethodthe classical interpolating polynomial Pn(x) of degree less orequal to n is unique to the given interpolation points so itcannot be an answer to the above questions At presentNewtonrsquos polynomial interpolation is in the center of re-search on polynomial interpolation methods While it hasshown good interpolation performance its disadvantage isthat it is unique to the given interpolation points [28] Toovercome the above shortcomings we propose a novelNewton interpolation polynomial only based on (partial)divided differences by introducing one or multiple pa-rameters which can be seen as a new approach to inter-polation method of Newton polynomials Similar to theinterpolation functions in [28ndash32] the proposed method

2 Mathematical Problems in Engineering

allows for modifying the curve or surface shape under thecondition that the values of the interpolant points are fixedAt the same time the proposed method has many advan-tages such as a simple and explicit mathematical repre-sentation and better performance e interpolant functiononly depends on the values of the function being interpo-lated and the (partial) divided differences so the compu-tation is simple

e rest of the paper is organized as follows First wedevelop one univariate parameterized Newton-type poly-nomial interpolation and give the interpolation algorithmsand theorems in Section 2 en we present one kind ofbivariate parameterized Newton-type polynomial interpo-lation and give the interpolation algorithms theorems andthe dual interpolation polynomial in Section 3 In Section 4to illustrate the effectiveness of the proposed algorithmsnumerical examples are given

2 Parameterized Univariate Newton-TypePolynomial Interpolation

We now develop the univariate Newton-type polynomialinterpolation problems in one variable Suppose functiony f(x) and we are given the support points set(x0 y0) (x1 y1) (xn yn)1113864 1113865 on interval [a b] We obtainthe following classic Newton polynomial representation [1]

Pn(x) f x0( 1113857 + f x0 x11113858 1113859 x minus x0( 1113857 + middot middot middot

+ f x0 x1 xn1113858 1113859 x minus x0( 1113857 x minus x1( 1113857 x minus xnminus1( 1113857

(1)

where f(x0) f[x0 x1] f[x0 x1 xn] represent thedivided differences of f(x)

If the interpolation polynomial function exists for thegiven n + 1 interpolation data points then classical inter-polating polynomial by Pn(x) of degree less or equal to n isunique Zhao and Tan [16] developed the block-basedNewton-type blending rational interpolation For a given setof interpolant data they divided the interpolation into manysubsets (blocks) based on different block point data manytypes of polynomial or rational interpolation can be con-structed e block-based divided difference is calculatedbased on the block interpolation points first en theblock-based Newton-type blending rational interpolationover all blocks is constructed is requires a lot of calcu-lation which is inconvenient for interpolation applicationHence we construct a novel Newton interpolation poly-nomial with a single parameter based on one virtual pointe point is adjusted strategically and the univariateNewton-type polynomial interpolation with one parameteris constructed

By introducing a new parameter λ an arbitrary point ofthe original points (xk yk)(k 0 1 n) is regarded as avirtual double point while the multiplicity of the otherpoints remains the same

Let

y0i yi i 0 1 n (2)

When (j 1 k + 1) for (i j j + 1 n)

yji

yjminus1i minus y

jminus1jminus1

xi minus xjminus1 (3)

For (i k + 1 k + 2 n)

zk+1i

yk+1i minus λ

xi minus xk

(4)

When (j k + 2 k + 3 n) for (i j j + 1 n)

zji

zjminus1i minus z

jminus1jminus1


Based on equations (2)ndash(5) the novel divided differencesare given as shown in Table 1

Compared with the classical Newton polynomial in-terpolation it is easy to see that the method in this paper issimple and convenient and is the same as that of the classicalNewton polynomial interpolation

e Newton-type polynomial interpolation with a pa-rameter λ can be constructed

P(0)n (x) c0 + c1 x minus x0( 1113857 + c2 x minus x0( 1113857 x minus x1( 1113857 + middot middot middot

+ ck x minus x0( 1113857 x minus x1( 1113857 middot middot middot x minus xkminus1( 1113857

+ c0k+1 x minus x0( 1113857 x minus x1( 1113857 middot middot middot x minus xkminus1( 1113857 x minus xk( 1113857

+ ck+1 x minus x0( 1113857 x minus x1( 1113857 middot middot middot x minus xkminus1( 1113857 x minus xk( 11138572

+ ck+2 x minus x0( 1113857 x minus x1( 1113857 middot middot middot x minus xkminus1( 1113857

middot x minus xk( 11138572

x minus xk+1( 1113857

+ middot middot middot + cn x minus x0( 1113857 x minus x1( 1113857 middot middot middot x minus xkminus1( 1113857


x minus xk+1( 1113857 middot middot middot x minus xnminus1( 1113857

(6)

where

ci y

ii i 0 1 k

zii i k + 1 k + 2 n

⎧⎨

⎩ c0k+1 λ (7)

Theorem 1 Given the interpolation data ( (x0 y0) (x11113864

y1) (xn yn))

P(0)n xi( 1113857 yi i 0 1 n (8)

Proof Suppose (0le ile k)Equation (6) is the classical Newton polynomial inter-

polation so P(0)n (xi) yi (i 0 1 k) For i k + 1

Mathematical Problems in Engineering 3

P(0)n xi( 1113857 c0 + c1 xi minus x0( 1113857 + c2 xi minus x0( 1113857 xi minus x1( 1113857 + middot middot middot + ck xi minus x0( 1113857 xi minus x1( 1113857 middot middot middot xi minus xkminus1( 1113857

+ c0k+1 xi minus x0( 1113857 xi minus x1( 1113857 middot middot middot xi minus xkminus1( 1113857 xi minus xk( 1113857 + ck+1 xi minus x0( 1113857 xi minus x1( 1113857 middot middot middot xi minus xkminus1( 1113857 xi minus xk( 1113857

2

c0 + c1 xi minus x0( 1113857 + c2 xi minus x0( 1113857 xi minus x1( 1113857 + middot middot middot + ck xi minus x0( 1113857 xi minus x1( 1113857 middot middot middot xi minus xkminus1( 1113857

+ λ xi minus x0( 1113857 xi minus x1( 1113857 middot middot middot xi minus xkminus1( 1113857 xi minus xk( 1113857 +y

k+1i minus λ

xi minus xk

xi minus x0( 1113857 xi minus x1( 1113857 middot middot middot xi minus xkminus1( 1113857 xi minus xk( 11138572


+ yk+1i xi minus x0( 1113857 xi minus x1( 1113857 middot middot middot xi minus xkminus1( 1113857 xi minus xk( 1113857

middot middot middot y0k+1 yk+1

If n ge i ge k + 2



2+

middot ck+2 xi minus x0( 1113857 xi minus x1( 1113857 middot middot middot xi minus xkminus1( 1113857 xi minus xk( 11138572

xi minus xk+1( 1113857 + middot middot middot +

middot ci xi minus x0( 1113857 xi minus x1( 1113857 middot middot middot xi minus xkminus1( 1113857 xi minus xk( 11138572

xi minus xkminus1( 1113857 middot middot middot xi minus ximinus1( 1113857



2+



middotz

iminus1i minus z

iminus1iminus1

xi minus ximinus1xi minus x0( 1113857 xi minus x1( 1113857 middot middot middot xi minus xkminus1( 1113857 xi minus xk( 1113857

2xi minus xk+1( 1113857 middot middot middot xi minus ximinus1( 1113857



2+



middot ziminus1i xi minus x0( 1113857 xi minus x1( 1113857 middot middot middot xi minus xkminus1( 1113857 xi minus xk( 1113857


Table 1 Divided differences table

Nodes 0 order divideddifferences

1 order divideddifferences

k order divideddifferences

k + 1 order divideddifferences


n + 1 order divideddifferences

x0 y00

x1 y01 y1

1⋮ ⋮ ⋮ ⋱xk y0

k y1k middot middot middot yk

k

xk y0k y1

k middot middot middot ykk λ

xk+1 y0k+1 y1

k+1 middot middot middot ykk+1 yk+1

k+1 zk+1k+1

⋮ ⋮ ⋮ middot middot middot ⋮ ⋮ ⋮ ⋱xn y0

n y1n middot middot middot y

kn

yk+1n zk+1

n middot middot middot znn


middot middot middot c0 + c1 xi minus x0( 1113857 + c2 xi minus x0( 1113857 xi minus x1( 1113857 + middot middot middot + ck xi minus x0( 1113857 xi minus x1( 1113857 middot middot middot xi minus xkminus1( 1113857

+ c0k+1 xi minus x0( 1113857 xi minus x1( 1113857 middot middot middot xi minus xkminus1( 1113857 xi minus xk( 1113857 + z

k+1i xi minus x0( 1113857 xi minus x1( 1113857 middot middot middot xi minus xkminus1( 1113857 xi minus xk( 1113857

2



k+1i minus λ

xi minus xk





(9)

So we have

P(0)n xi( 1113857 yi i 0 1 n (10)

erefore the results followSimilarly we can also consider the Newton-type poly-

nomial interpolation with two parameters with the proposedalgorithm It also can be divided into two cases One case isthat an arbitrary point from the original points is treated as avirtual treble point another case is that any two points in theoriginal points are regarded as two virtual double pointseNewton interpolation polynomial with more parameters canbe constructed similarly

3 Bivariate Parameterized Newton-TypePolynomial Interpolation

Now we generalize some applications of the previous resultsto the multivariate Newton-type polynomial interpolation

Suppose (1113937mn sub D sub R2) is a set of two dimensionalpoints on the rectangular regionD f(x y) is a bivariate realfunction defined on the rectangular region D and let

f xi yj1113872 1113873 fij i 0 1 m j 0 1 n (11)

e bivariate Newton polynomial interpolation is asfollows

N(x y) a00 + a01 y minus y0( 1113857 + a02 y minus y0( 1113857 y minus y1( 1113857 + middot middot middot + a0m y minus y0( 1113857 middot middot middot y minus ymminus1( 1113857

+ a10 + a11 y minus y0( 1113857 + a12 y minus y0( 1113857 y minus y1( 1113857 + middot middot middot + a1m y minus y0( 1113857 middot middot middot y minus ymminus1( 11138571113872 1113873 x minus x0( 1113857

+ middot middot middot + an0 + an1 y minus y0( 1113857 + an2 y minus y0( 1113857 y minus y1( 1113857 + middot middot middot + anm y minus y0( 1113857 middot middot middot y minus ymminus1( 11138571113872 1113873 x minus x0( 1113857 middot middot middot x minus xnminus1( 1113857

(12)

and aij (i 0 1 m j 0 1 n) represent the bi-variate partial divided differences

Theorem 2 (see [2]) Suppose (1113937mn sub D sub R2) is a set oftwo dimensional points on rectangular region D and f(x y)

is a real function defined on rectangular region D then

N xi yj1113872 1113873 fij forall xiyj1113872 1113873 isin1113945mn

i 0 1 m j 0 1 n

(13)

31 Bivariate Newton-Type Polynomial Interpolation with aSingle Parameter By introducing a new parameter λ we useany point of the original points (xk yl fkl)(k 0 1

m l 0 1 n) as a virtual double point while the mul-tiplicity of the other points remains 1 e Newton-typepolynomial interpolation based on parameter λ can beconstructed using Algorithm 1

Algorithm 1Step 1 initialization

f(00)ij f xi yj1113872 1113873 i 0 1 m j 0 1 n

(14)

Step 2 for (j 0 1 n p 1 2 m i p p+

1 m)

f(p0)ij

f(pminus10)ij minus f

(pminus10)pminus1j

xi minus xpminus1 (15)

Step 3 For(i 0 1 k minus 1 k + 1 m q

1 2 n j q q + 1 n)

f(iq)

ij f

(iqminus1)ij minus f

(iqminus1)iqminus1

yj minus yqminus1 (16)


Step 4 by introducing parameter λ into the formulaf

(k0)kj (j 0 1 n) and using formulas (2)ndash(5) the

final results are

ak0 ak1 akl a0kl+1 akl+1 akl+2 akl+3 akn1113872 1113873

T

(17)

Step 5 using the data in formulas (16) and (17) theNewton-type polynomial interpolation with a singleparameter with regard to y can be constructed

Ai(y)

f(i0)i0 + f

(i1)i1 y minus y0( 1113857 + f

(i2)i2 y minus y0( 1113857 y minus y1( 1113857 + middot middot middot + f

(in)in y minus y0( 1113857 y minus y1( 1113857 middot middot middot y minus ynminus1( 1113857

i 0 1 middot middot middot k minus 1 k + 1 middot middot middot m

ak0 + ak1 y minus y0( 1113857 + middot middot middot + akl y minus y0( 1113857 middot middot middot y minus ylminus1( 1113857 + a0kl+1 y minus y0( 1113857 middot middot middot y minus ylminus1( 1113857 y minus yl( 1113857

+akl+1 y minus y0( 1113857 middot middot middot y minus ylminus1( 1113857 y minus yl( 11138572

+ akl+2 y minus y0( 1113857 middot middot middot y minus ylminus1( 1113857 y minus yl( 11138572

y minus yl+1( 1113857 + middot middot middot +

akn y minus y0( 1113857 middot middot middot y minus ylminus1( 1113857 y minus yl( 11138572

y minus yl+1( 1113857 middot middot middot y minus ynminus1( 1113857 i k

⎧⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎩

(18)

Step 6 let

N0mn(x y) A0(y) + A1(y) x minus x0( 1113857

+ A2(y) x minus x0( 1113857 x minus x1( 1113857 + middot middot middot

+ Am(y) x minus x0( 1113857 x minus x1( 1113857 middot middot middot x minus xmminus1( 1113857

(19)

en N0mn(x y) is a bivariate Newton-type polynomial

interpolation with a single parameter

Theorem 3 Given the interpolation points f(00)ij f(xi

yj)(i 0 1 m j 0 1 n) the bivariate Newton-type polynomial interpolation with a single parameterN0

mn(x y) satisfies

N0mn xi yj1113872 1113873 fij forall xiyj1113872 1113873 isin Πmn

i 0 1 m j 0 1 n(20)

Proof For an arbitrary point (xi yj fij) if (i 0 1

k minus 1 k + 1 m) it holds that

Ai yj1113872 1113873 f(i0)i0 + f

(i1)i1 yj minus y01113872 1113873 + f

(i2)i2 yj minus y01113872 1113873 yj minus y11113872 1113873

+ middot middot middot + f(ij)ij yj minus y01113872 1113873 yj minus y11113872 1113873

middot middot middot yj minus yjminus11113872 1113873 middot middot middot f(i0)ij

(21)

If (i k)

Ai(y) ak0 + ak1 y minus y0( 1113857 + middot middot middot + akl y minus y0( 1113857 middot middot middot y minus ylminus1( 1113857 + a0kl+1 y minus y0( 1113857 middot middot middot y minus ylminus1( 1113857 y minus yl( 1113857




middot akn y minus y0( 1113857 middot middot middot y minus ylminus1( 1113857 y minus yl( 11138572

y minus yl+1( 1113857 middot middot middot y minus ynminus1( 1113857

(22)

Regardless of (0le jlt l l j nge jgt l) from eorem 1we can derive

Ai yj1113872 1113873 ak0 + ak1 yj minus y01113872 1113873 + middot middot middot + akl yj minus y01113872 1113873 middot middot middot yj minus ylminus11113872 1113873 + a0kl+1 yj minus y01113872 1113873 middot middot middot yj minus ylminus11113872 1113873 yj minus yl1113872 1113873

+ akl+1 yj minus y01113872 1113873 middot middot middot yj minus ylminus11113872 1113873 yj minus yl1113872 11138732


yj minus yl+11113872 1113873

+ middot middot middot + akj yj minus y01113872 1113873 middot middot middot yj minus ylminus11113872 1113873 yj minus yl1113872 11138732

yj minus yl+11113872 1113873 middot middot middot yj minus yjminus11113872 1113873 middot middot middot f(i0)ij

(23)

From eorem 4 (forall(xiyj) isin Πmn)

N0mn xi yj1113872 1113873 A0 yt( 1113857 + A1 yj1113872 1113873 xi minus x0( 1113857 + A2 yj1113872 1113873 xi minus x0( 1113857 xi minus x1( 1113857 + middot middot middot + As yj1113872 1113873 xi minus x0( 1113857 xi minus x1( 1113857 middot middot middot xi minus ximinus1( 1113857

f(00)0j + f

(10)1j xi minus x0( 1113857 + f

(20)2j xi minus x0( 1113857 xi minus x1( 1113857 + middot middot middot + f

(i0)ij xi minus x0( 1113857 xi minus x1( 1113857 middot middot middot xi minus ximinus1( 1113857 middot middot middot f

(00)ij fij

(24)


en we have proved eorem 3Similarity we can also consider the Newton-type

polynomial interpolation with two parameters ere arethree cases one case is that a point was viewed as a virtualtreble point another case is that two points were viewed asvirtual double points in the same column the third case istaking two points as virtual double points in different col-umns e bivariate Newton-type polynomial interpolationwith three or more parameters can be derived in a similarfashion

32 Dual Bivariate Newton-Type Polynomial InterpolationwithaSingle Parameter FromAlgorithm 1 according to thepartial divided difference of x first and then concerning thepartial divided difference of y the novel bivariate Newton-type polynomial interpolation was constructed In fact wecan also construct it by first regarding the partial divideddifference of y and then the partial divided difference of xBy introducing a new parameter θ any point in the originalinterpolant points (xk yl fkl)(k 0 1 m l

0 1 n) is regarded as a virtual double point en theNewton-type polynomial interpolation with parameter θ canbe obtained using Algorithm 2


f(00)ij f xi yj1113872 1113873 i 0 1 m j 0 1 n (25)

Step 2 if (i 0 1 m q 1 2 n j q q+

1 n) let

f(0q)

ij f

(0qminus1)

ij minus f(0qminus1)

iqminus1


Step 3 for (j 0 1 l minus 1 l + 1 n p 1 2

m i p p + 1 m)

f(pj)

ij f

(pminus1j)ij minus f

(pminus1j)pminus1j


Step 4 by introducing a parameter θ into the f(0l)il (i

0 1 m) and calculating them using formulas(2)ndash(5) the final results are

a0l a1l akl a0k+1l ak+1l ak+2l aml (28)

Step 5 by using the data in formulas (27) and (28) theunivariate Newton-type polynomial interpolation witha single parameter in regard to x can be constructed asfollows

Aj(x)

f(0j)0j + f

(1j)1j x minus x0( 1113857 + f

(2j)2j x minus x0( 1113857 x minus x1( 1113857 + middot middot middot + f

(mj)mj x minus x0( 1113857 x minus x1( 1113857 middot middot middot x minus xmminus1( 1113857

j 0 1 l minus 1 l + 1 n

a0l + a1l x minus x0( 1113857 + middot middot middot + akl x minus x0( 1113857 middot middot middot x minus xlminus1( 1113857 + a0k+1l x minus x0( 1113857 middot middot middot x minus xlminus1( 1113857 x minus xl( 1113857

+ak+1l x minus x0( 1113857 middot middot middot x minus xlminus1( 1113857 x minus xl( 11138572

+ ak+2l x minus x0( 1113857 middot middot middot x minus xlminus1( 1113857 x minus xl( 11138572

x minus xl+1( 1113857 + middot middot middot +

aml x minus x0( 1113857 middot middot middot x minus xlminus1( 1113857 x minus xl( 11138572

x minus xl+1( 1113857 middot middot middot x minus xmminus1( 1113857 j l

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

(29)

Step 6 let

N2mn(x y) A0(x) + A1(x) y minus y0( 1113857

+ A2(x) y minus y0( 1113857 y minus y1( 1113857 + middot middot middot

+ Am(x) y minus y0( 1113857 y minus y1( 1113857 middot middot middot y minus ymminus1( 1113857

(30)

en N2mn(x y) is a dual bivariate Newton-type

polynomial interpolation with a single parameterWe call N2

mn(x y) as the dual interpolation ofN0

mn(x y) As a matter of fact the bivariate Newton-typepolynomial interpolation with a single parameter can alsobe calculated by the information matrix method is al-gorithm borrows elementary transformation thought oflinear algebra so it is easy to implement We can calculatethe partial divided differences in Algorithms 1 and 2 in amanner of the information matrix method Here we onlygive the information matrix method of the dual bivariateNewton-type polynomial interpolation with a single

parameter Algorithm 2 can also be described by Algo-rithm 3 based on information matrix as follows

Algorithm 3Step 1 initialization Let

f(00)ij f xi yj1113872 1113873 (i 0 1 m j 0 1 n)

(31)

One can define the following initial informationmatrix

M1

f(00)00 f

(00)10 middot middot middot f

(00)m0

f(00)01 f


(00)m1

⋮ ⋮ middot middot middot ⋮

f(00)0n f

(00)1n middot middot middot f

(00)mn

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(32)

Step 2 for (i 0 1 middot middot middot middot middot middot m q 1 2 middot middot middot n j q

q + 1 middot middot middot n) let


f(0q)ij

f(0qminus1)ij minus f

(0qminus1)iqminus1


e above recursive process aims to change the initialinformation matrix M1 into the matrix M2 by usingrow transformation

M2

f(00)00 f


(00)m0

f(01)01 f


(01)m1


f(0n)0n f

(0n)1n middot middot middot f

(0n)mn

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(34)

Step 3 for (j 0 1 middot middot middot l minus 1 l + 1 middot middot middot n p 1 2 middot middot middot

m i p p + 1 middot middot middot m) let

f(pj)ij

f(pminus1j)ij minus f

(pminus1j)pminus1j


e above recursive process aims to change the(0 1 l minus 1 l + 1 m) columns in the initial in-formation matrix M2 into the matrix M3 by usingcolumn transformation

M3

f(00)00 f


(m0)m0

f(01)01 f


(m1)m1


f(0lminus1)0 lminus1 f

(1lminus1)1 lminus1 middot middot middot f

(mlminus1)m lminus1

f(0l)0 l f

(0l)1 l middot middot middot f

(0l)m l

f(0l+1)0 l+1 f

(1l+1)1 l+1 middot middot middot f

(ml+1)m l+1


f(0n)0 n f

(1n)1 n middot middot middot f

(mn)m n

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(36)

By introducing a parameter θ into thef(0l)

il (i 0 1 m) in the information matrix M3and calculating them using formulas (2)ndash(5) the finalresults are



Aj(x)

f(0j)0j + f

(1j)1j x minus x0( 1113857 + f



j 0 1 middot middot middot l minus 1 l + 1 middot middot middot n







⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

(38)

Step 5 let




(39)

en N2mn(x y) is the dual bivariate Newton-type

polynomial interpolation with a single parameter

Theorem 4 Given the interpolation data(xiyjfij)(i 01 mj 01 n) the dual bivariateNewton-type polynomial interpolation with a single parameterN2

mn(xy) satisfies

N2mn xi yj1113872 1113873 fij

forall xiyj1113872 1113873 isin Πmn i 0 1 m j 0 1 n

(40)

Proof For an arbitrary point (xi yj fij) if (j 0 1

l minus 1 l + 1 n ) we have

Aj xi( 1113857 f(i0)i0 + f

(i1)i1 xi minus x0( 1113857 + f

(i2)i2 xi minus x0( 1113857 xi minus x1( 1113857

+ middot middot middot + f(ij)ij xi minus x0( 1113857 xi minus x1( 1113857 middot middot middot xi minus ximinus1( 1113857

middot middot middot f(0j)ij

(41)


If j l

Aj(x) a0l + a1l x minus x0( 1113857 + middot middot middot + akl x minus x0( 1113857 middot middot middot x minus xlminus1( 1113857 + a0k+1l x minus x0( 1113857 middot middot middot x minus xlminus1( 1113857 x minus xl( 1113857



x minus xl+1( 1113857

+ middot middot middot + aml x minus x0( 1113857 middot middot middot x minus xlminus1( 1113857 x minus xl( 11138572

x minus xl+1( 1113857 middot middot middot x minus xmminus1( 1113857

(42)

Regardless of (0le ilt k i k nge igt k) fromeorem 1we have

Aj xi( 1113857 a0l + a1l xi minus x0( 1113857 + middot middot middot + akl xi minus x0( 1113857 middot middot middot xi minus xlminus1( 1113857 + a0k+1l xi minus x0( 1113857 middot middot middot xi minus xlminus1( 1113857 xi minus xl( 1113857

+ ak+1l xi minus x0( 1113857 middot middot middot xi minus xlminus1( 1113857 xi minus xl( 11138572


xi minus xl+1( 1113857

+ middot middot middot + ail xi minus x0( 1113857 middot middot middot xi minus xlminus1( 1113857 xi minus xl( 11138572

xi minus xl+1( 1113857 middot middot middot xi minus ximinus1( 1113857 middot middot middot f(0j)

ij

(43)

So we have

N2mn xi yj1113872 1113873 A0 xi( 1113857 + A1 xi( 1113857 yj minus y01113872 1113873 + A2 xi( 1113857 yj minus y01113872 1113873 yj minus y11113872 1113873 + middot middot middot + At xi( 1113857 yj minus y01113872 1113873 yj minus y11113872 1113873 middot middot middot yj minus yjminus11113872 1113873

f(00)i0 + f

(01)i1 yj minus y01113872 1113873 + f

(02)i2 yj minus y01113872 1113873 yj minus y11113872 1113873 + middot middot middot + f

(0j)ij yj minus y01113872 1113873 yj minus y11113872 1113873 middot middot middot yj minus yjminus11113872 1113873 middot middot middot f

(00)st

fst

(44)

en we can obtain the resultIn addition it can also construct other dual bivariate

parameterized Newton-type polynomial interpolationssimilar to the proposed algorithm in Section 32 Further-more according to the previous discussion we can alsoconstruct parameterized Newton-type polynomial interpo-lation in both x and y directions e formulas of every newparameterized Newton-type polynomial interpolation de-pend on the selected parameters We can construct differentparameterized Newton-type polynomial interpolationfunctions according to the application needs At the sametime the value of the interpolant function can be changed atany point in the interpolant region under the condition thatthe values of the interpolant points are fixed by selectingappropriate parameters

As a matter of fact one can also construct a one-pa-rameter family of interpolating polynomials

Pλ(x) ϕn(x) + Pn(x) (45)

with Pn(x) being the classical interpolating polynomial andΦn(x) 1113937

nk (x minus xk)

e proposed parameterized Newton-like interpolationpolynomial is another approach to the one-parameter familyof interpolating polynomials In this paper the proposedalgorithms are only based on (partial) divided differenceswhich seems a little complicated In fact the calculation isthe same as that of the classical Newton interpolationpolynomial We can also discuss the univariate case withmore parameters and bivariate case with one or more pa-rameters similarly e algorithm given in this paper de-scribes a new approach method for Newton interpolationpolynomial

e parameterized Newton-type polynomial interpola-tion defined in this paper is a local interpolation in eachsubinterval or subrectangle which only depends on thefunction values ere are some parameters in each inter-polating region and when the parameters vary the inter-polant function will be changed under the condition that thevalues of the interpolant points are fixed erefore theinterpolation curves and surfaces will change with differentparameters us the shape of the interpolant curves orsurface can be adjusted for the given interpolation pointsHowever the question of how to select the optimal


parameters of the proposed interpolant algorithms is still achallenging problem which we will consider in our futurework

4 Numerical Examples

In this section we give some examples to demonstrate howthe proposed algorithms are implemented and how theparameters can be adjusted to change the curve or surfaceshape

Example 1 Let the interpolation data be as given in Table 2From the interpolation data in the Table 2 we get the

Newton interpolation polynomial as

p(x) 12

x(x minus 1) (46)Take (2 1) as a virtual double point and the multiplicity

of the other points remains 1 e Newton-type polynomialinterpolation with one parameter λ derived via Algorithm 1is

p(0)

(x) 12

x(x minus 1) + λx(x minus 1)(x minus 2) minus λx(x minus 1)(x minus 2)2

(47)

It is easy to verify that

p xi( 1113857 p(0)

xi( 1113857 fi (i 0 1 2) (48)

If the interpolant points are given the shape of inter-polating curves is fixed is is because of the uniqueness ofthe interpolation function e parameterized Newton-typepolynomial interpolation methods defined in this paper areparametric By changing the parameters the interpolationfunction can be adjusted while observing the condition thatthe values of the interpolant points are fixed e localityconstraint in the interpolation curves is to make the functionvalue at a specific point (x y) equal to a given real number inactual design [29 30] e central point in the interpolatingcurve is more important than other points for shape control[29 30] so we consider the central point value controlproblem

In fact if we take M 15 based on the parameterizedNewton-type polynomial interpolation we can get λ minus2and the function value of the central point is p(0)(15) 15Figure 1(b) shows the plot of the interpolation p(0)(x) If wetake M 0 it is easy to prove λ (23) and Figure 1(c)shows the plot of the interpolation p(0)(x) In this case thefunction value of central point is p(0)(15) 0 If we takeM minus1 it is easy to prove λ (229) and Figure 1(d) showsthe plot of the interpolation p(0)(x) the function value ofcentral point is p(0)(15) minus1

By choosing different values of parameter λ thefunction p(0)(x) invariably satisfies the given interpolationdata e value of the interpolation function p(0)(x) can bechanged in the interpolant region by choosing appropriateparameter λ

Example 2 Let the interpolation data be as given in Table 3We obtain the bivariate Newton polynomial interpola-

tion by the method in [2] as

P(x y) 1 + y + x(1 + y) (49)

Following Algorithm 1 in this paper we increase themultiplicity of node (0 0 1) to 2 the multiplicity of theother points remains 1 and then we construct the Hermiteinterpolation with first-order derivative in point (0 0 1) byintroducing parameter λ We obtain the Newton-typepolynomial interpolation with parameter λ with regard toy

P(1)

(x y) 1 + λy +(1 minus λ)y2

+ x(1 + y) (50)

Following Algorithm 2 we derive the Newton-typepolynomial interpolation with parameter β with regard to x

P(2)

(x y) 1 + βx +(1 minus β)x2

+ y(1 + x) (51)

Following Algorithms 1 and 2 we derive the Newton-type polynomial interpolation with parameters δ η withregard to x and y

P(3)

(x y) 1 + δx +(1 minus δ)x2

+ ηy +(1 minus η)y2

+ xy

(52)

We can also obtain the bivariate Newton polynomialinterpolation by adding times with parameter c

P(4)

(x y) 1 + x + y + xy + cxy(x minus 1)(y minus 1) (53)

It is easy to verify

P xi yj1113872 1113873 P(1)

xi yj1113872 1113873 P(2)

xi yj1113872 1113873 P(3)

xi yj1113872 1113873

P(4)

xi yj1113872 1113873 fij (i j 0 1)

(54)

It is easy to verify that the function P(1)(x y) invariablysatisfies the given interpolation data with the different valuesof parameter λ Figure 2 shows the plot of the classicalbivariate Newton polynomial interpolation P(x y)

Similar to interpolant curves we consider the local pointcontrol problem of the interpolation surface that is how tomake the function value of the interpolation at a point (x y)

equal to a given real number S in the actual design Wecannot adjust the shape of the classical Newton interpolationpolynomial P(1)(x y) We can use the algorithm in thisarticle to solve this problem Compared with other pointsthe center point is the key area of the interpolation patche function value of the central point is P(05 05) 225If we set S minus1 based on the parameterized Newton-type

Table 2 Interpolation data

i 1 2 3 4xi 0 1 2 3fi 0 0 1 3


0 05 1 15 2 25 3x

0

05

1

15

2

25

3

(a)

0 05 1 15 2 25 3x

ndash2

ndash1

0

1

2

3

(b)

0 05 1 15 2 25 3x

0

05

1

15

2

25

3

(c)

0 05 1 15 2 25 3x

ndash1

0

1

2

3

4

5

(d)

Figure 1 (a) e plot of p(x) (b) e plot of p(0)(x) with λ minus2 (c) e plot of p(0)(x) with λ (23) (d) e plot of p(0)(x) withλ (229)


(xi yj) x0 0 x1 1

y0 0 1 2y1 1 2 4

435

325

215

11

05

0 002

0406

081

y + x (y + 1) + 1

xy

Figure 2 e plot of P(x y)


4

3

2

1

0

ndash1

x (y + 1) ndash 12 y + 13 y2 + 1

002

0406

081

x

1

05

0y

(a)

4

3

2

1

8 y + x (y + 1) ndash 7 y2 + 1

002

0406

081

x

1

05

0y

(b)

Figure 3 e plot of P(1)(x y) (a) e plot of P(1)(x y) with λ minus12 (b) e plot of P(1)(x y) with λ 8

y (x + 1) ndash 12 x + 13 x2 + 1

43210

ndash1

002

0406

081

x

1

05

0y

(a)

8 x + y (x + 1) ndash 7 x2 + 1

4

3

2

1

002

0406

081

x

1

05

0y

(b)

Figure 4 e plot of P(2)(x y) (a) e plot of P(2)(x y) with β minus12 (b) e plot of P(2)(x y) with β 8

x y ndash 6 y ndash 5 x + 6 x2 + 7 y2 + 1

4

3

2

1

0

ndash1

0 02 04 06 08 1

x

1

05

0y

(a)


4

3

2

1

0

ndash1

0 02 04 06 08 1

x

1

05

0y

(b)

Figure 5 Continued


polynomial interpolation we can get λ minus12 and the shapeof the surface will ldquodroprdquo In this case the function value ofthe central point is P(1)(05 05) minus1 If we set S 4 it iseasy to prove λ 8 and the shape of the surface ldquorisesrdquo Inthis case the function value of central point isP(1)(05 05) 4 Figures 3(a) and 3(b) show the plot of theparameterized Newton-type polynomial interpolationP(1)(x y) with λ minus12 and λ 8 respectively Similarly wecan set S minus1 and S 4 for P(i)(05 05) minus1 (i 2 3 4)

and P(i)(05 05) 4 (i 2 3 4) we can discuss polyno-mial interpolation P(2)(x y) P(3)(x y) and P(4)(x y)Figures 4(a) and 4(b) show the plot of the parameterizedNewton-type polynomial interpolation P(2)(x y) with β

minus12 and β 8 respectively Figures 5(a)ndash5(d) show the plotof the parameterized Newton-type polynomial interpolationP(3)(x y) with δ minus5 η minus6 and δ minus9 η minus2 and δ

8 η 1 and δ 3 η 6 respectively In fact we can chooseother combinations of parameters δ η that satisfy certain

conditions so that P(3)(05 05) minus1 and P(3)(05 05) 4Figures 6(a) and 6(b) show the plot of the parameterizedNewton-type polynomial interpolation P(4)(x y) with c

28 and c minus52 respectivelyFor the given interpolation points we get four Newton

polynomial interpolations with one or two parameters epolynomial P(4)(x y) has a higher degree than the proposedmethod in this paper In terms of the central point valuecontrol problem the proposed method in this paper has alarge adjustment space as shown in Figures 3ndash5 Meanwhilewe can adjust the shape of surface in one or in two directionsor add parameters to adjust in both directions according toactual needs e interpolation function can be adjustedunder the condition that the values of the interpolant pointsare fixed is is achieved by adjusting the parameters of ourNewton-type polynomial interpolation methods We canalso modify the new Newton-type polynomial interpolationby parameter λ so our methods give a new choice for the

x ndash 8 y + x y ndash 7 y2 + 1

4

3

2

11

05

0y

002

0406

081

x

(c)

6 x + 3 y + x y ndash 5 x2 ndash 2 y2 + 1

1

05

0y

002

0406

081

x

435

325

215

1

(d)

Figure 5 e plot of P(3)(x y) (a) e plot of P(3)(x y) with δ minus5 η minus6 (b) e plot of P(3)(x y) with δ minus9 η minus2 (c) e plot ofP(3)(x y) with δ 8 η 1 (d) e plot of P(3)(x y) with δ 3 η 6

y + x (y + 1) ndash 52 x y (x ndash 1) (y ndash 1) + 1

4

3

2

ndash1

0

1

1

05

0y

002

0406

081

x

(a)

y + x (y + 1) + 28 x y (x ndash 1) (y ndash 1) + 1

1

05

0y

002

0406

081

x

435

325

215

1

(b)

Figure 6 e plot of P(4)(x y) (a) e plot of P(4)(x y) with c minus52 (b) e plot of P(4)(x y) with c 28


application and a new method for studying the polynomialinterpolation theory According to the other algorithms inthis paper we can also construct many other novel Newton-type polynomial interpolations with more parameterssimilarly

5 Conclusions and Future Work

We presented a new approach to interpolation method ofNewton polynomials We constructed several kinds of uni-variate and bivariate parameterized Newton-type polynomialinterpolation We discussed the interpolation theorem thealgorithms the dual bivariate parameterized Newton-typepolynomial interpolation and algorithm based on informa-tionmatrix Ourmethods are easy to be used and theoreticallysound e value of the Newton-type polynomial interpolantfunction can be adjusted in the interpolant region by choosingappropriate parameter values while observing the conditionthat the values of the interpolant points are fixed Accordingto the actual geometric design needs the shape of the in-terpolation curves or surfaces can be adjusted which isconvenient in engineering and useful in computer-aidedgeometric design [28 30] e proposed method has ad-vantages such as a simple and explicit mathematical repre-sentation and ease of computation But how to select theoptimal parameter for application is a difficult problem Ourfuture work will concentrate on the following aspects

(1) How to select the optimal parameter values andsuitably adjust the shape of the interpolant curves orsurface according to the actual design requirements

(2) Investigating how the presented parameterizedNewton-type polynomial interpolation method canbe applied in other pixel-level image processingsteps such as super-resolution reconstruction imageupscaling image rotation removal of salt and peppernoise image metamorphosis and image inpainting

(3) How to construct multivariate composite interpo-lation and blending rational interpolation overlacunary triangular pyramid-type and other ir-regularity grids based on the parameterized Newton-type polynomial interpolation iele continuedfraction interpolation and other classical interpo-lation methods e proposed method can also begeneralized to Hermite interpolation and osculatoryrational interpolation

In summary the parameterized Newton-type polyno-mial interpolation algorithm has many advantages such as asimple and explicit mathematical representation and ease ofcomputation the value of the Newton-type polynomialinterpolant function can be adjusted in the interpolant re-gion by choosing appropriate parameter values according tothe actual geometric design needs the shape of the inter-polation curves or surfaces can be adjusted

In addition with the help of the Samlson generalizedinverse it is easy to generalize the parameterized Newton-type polynomial interpolation method to vector-valuedcases or matrix-valued cases [1 15 16]

Data Availability

No data were used to support this study

Conflicts of Interest

e authors declare that there are no conflicts of interest

Acknowledgments

is study was supported by the National Natural ScienceFoundation of China (nos 61672204 61673359 and61806068) the Natural Science Foundation of AnhuiProvince (nos 1908085MF184 and 1908085QF285) in partby the Key Scientific Research Foundation of EducationDepartment of Anhui Province (nos KJ2018A0555 andKJ2019A0833) and in part by the Key Technologies RampDProgram of Anhui Province (no 1804a09020058)

References

[1] R H Wang and G Q Zhu Approximation of Rational In-terpolation and its Application Science Publishers BeijingChina 2004 in Chinese

[2] J Stoer and R Bulirsch Introduction to Numerical AnalysisSpringer-Verlag New York NY USA 2nd edition 2002

[3] L He J Tan Z Su X Luo and C Xie ldquoSuper-resolution bypolar Newton-ieles rational kernel in centralized sparsityparadigmrdquo Signal Processing Image Communication vol 31pp 86ndash99 2015

[4] T Bai J Tan M Hu and Y Wang ldquoA novel algorithm forremoval of salt and pepper noise using Continued Fractionsinterpolationrdquo Signal Processing vol 102 pp 247ndash255 2014

[5] M Rahman M J Khan M A Asghar et al ldquoImage localfeatures description through polynomial approximationrdquoIEEE Access vol 7 pp 183692ndash183705 2019

[6] D N H anh V B S Prasath L M Hieu and S DvoenkoldquoAn adaptive method for image restoration based on high-order total variation and inverse gradientrdquo Signal Image andVideo Processing vol 14 no 6 pp 1189ndash1197 2020

[7] G Farin Curves and Surfaces for Computer Aided GeometricDesign A Practical Guide Academic Press New York NYUSA 1988

[8] C de BoorA Practical Guide to Splines Springer-Verlag NewYork NY USA Revised edition 2001

[9] W-B Zhong X-C Luo W-L Chang et al ldquoToolpath in-terpolation and smoothing for computer numerical controlmachining of freeform surfaces a reviewrdquo InternationalJournal of Automation and Computing vol 17 no 1 pp 1ndash162020

[10] S Vats B B Sagar K Singh A Ahmadian and B A PanseraldquoPerformance evaluation of an independent time optimizedinfrastructure for big data analytics that maintains symme-tryrdquo Symmetry vol 12 no 8 p 1274 2020

[11] S M Vonza ldquoNewton-Tile-Type interpolational formula inthe form of two-dimensional continued fraction with non-equivalent variablesrdquo Nathfmatіcal Nfthpds and Phys-icomfchanical Fields vol 47 pp 67ndash72 2004

[12] D N Varsamis and N P Karampetakis ldquoOn the Newtonbivariate polynomial interpolation with applicationsrdquo Mul-tidimensional Systems and Signal Processing vol 25 no 1pp 179ndash209 2014


[13] J Tan and Y Fang ldquoNewton-ielersquos rational interpolantsrdquoNumerical Algorithms vol 24 no 12 pp 141ndash157 2000

[14] J Tan and S Tang ldquoComposite schemes for multivariateblending rational interpolationrdquo Journal of Computationaland Applied Mathematics vol 144 no 1-2 pp 263ndash275 2002

[15] Q-J Zhao and J Tan ldquoBlock-based iele-like blendingrational interpolationrdquo Journal of Computational and AppliedMathematics vol 195 no 1-2 pp 312ndash325 2006

[16] Q J Zhao and J Q Tan ldquoBlock-based Newton-type blendingrational interpolationrdquo Journal of Computational Mathe-matics vol 24 pp 515ndash526 2006

[17] S Tang and Y Liang ldquoBivariate blending iele-Wernerrsquososculatory rational interpolationrdquo Numerical Mathematics AJournal of Chinese Universities (English Series) vol 3pp 271ndash288 2007

[18] N Dyn and M S Floater ldquoMultivariate polynomial inter-polation on lower setsrdquo Journal of Approximation eoryvol 177 pp 34ndash42 2014

[19] A Cuyt and O Salazar Celis ldquoMultivariate data fitting witherror controlrdquo BIT Numerical Mathematics vol 59 no 1pp 35ndash55 2019

[20] L He J Tan Y Xing et al ldquoSuper-resolution reconstructionbased on Continued Fractions interpolation kernel in thepolar coordinatesrdquo Journal of Electronic Imaging vol 27Article ID 043035 2018

[21] S A Karim and A Saaban ldquoShape preserving interpolationusing rational cubic ball function and its application in imageinterpolationrdquo Mathematical Problems in Engineeringvol 2017 Article ID 7459218 13 pages 2017

[22] Y Zhan S J Li and M Li ldquoLocal and nonlocal regularizationto image interpolationrdquo Mathematical Problems in Engi-neering vol 2014 Article ID 230348 9 pages 2014

[23] R Behl A J Alsolami B A Pansera W M Al-HamdanM Salimi and M Ferrara ldquoA new optimal family ofschroderrsquos method for multiple zerosrdquo Mathematics vol 7no 11 p 1076 2019

[24] L He Y Xing K Xia et al ldquoAn adaptive image inpaintingmethod based on Continued Fractions interpolationrdquo Dis-crete Dynamics in Nature and Society vol 2018 Article ID9801361 17 pages 2018

[25] Y Zheng M Zhao P Sun et al ldquoOptimization of electrostaticforce system based on Newton interpolation methodrdquo Journalof Sensors vol 2018 Article ID 7801597 8 pages 2018

[26] H L Minh T T H Hanh and D N H anh ldquoMonotonefinite-difference schemes with second order approximationbased on regularization approach for the dirichlet boundaryproblem of the gamma equationrdquo IEEE Access vol 8pp 45119ndash45132 2020

[27] A Pourmoslemi M Ferrara B A Pansera and M SalimildquoProbabilistic norms on the homeomorphisms of a grouprdquoSoft Computing vol 24 no 10 pp 7021ndash7028 2020

[28] Q Duan LWang and E H Twizell ldquoA new bivariate rationalinterpolation based on function valuesrdquo Information Sciencesvol 166 no 1-4 pp 181ndash191 2004

[29] Y Zhang Q Duan F Bao and C Zhang ldquoA rational in-terpolation surface model and visualization constraintrdquo Sci-entia Sinica Mathematica vol 44 no 7 pp 729ndash740 2014

[30] Y F Zhang F X Bao M Zhang et al ldquoA weighted bivariateblending rational interpolation function and visualizationcontrolrdquo Journal Computational Analysis and Applicationsvol 14 no 7 pp 1303ndash1320 2012

[31] Q Duan Y Zhang and E H Twizell ldquoA bivariate rationalinterpolation and the propertiesrdquo Applied Mathematics andComputation vol 179 no 1 pp 190ndash199 2006

[32] Q Duan L Wang and E H Twizell ldquoA new C2 rationalinterpolation based on function values and constrainedcontrol of the interpolant curvesrdquo Applied Mathematics andComputation vol 161 no 1 pp 311ndash322 2005

[33] K M Reddy and A K B Chand ldquoConstrained univariate andbivariate rational fractal interpolationrdquo International Journalfor Computational Methods in Engineering Science and Me-chanics vol 20 no 5 pp 404ndash422 2019

[34] W-X Huang and G-J Wang ldquoA weighted bivariate blendingrational interpolation based on function valuesrdquo AppliedMathematics and Computation vol 217 no 9 pp 4644ndash46532011

[35] L Zou and S Tang ldquoA new approach to general interpolationformulae for bivariate interpolationrdquo Abstract and AppliedAnalysis vol 2014 Article ID 421635 11 pages 2014

[36] L Zou L Song X Wang et al ldquoBivariate iele-Like rationalinterpolation Continued Fractions with parameters based onvirtual pointsrdquo Mathematics vol 8 no 71 2020


many types of blending rational interpolation were studiedIn recent years Dyn and Floater [18] studied multivariatepolynomial interpolation based on lower subsets Cuyt andSalazar Celis [19] developed a generalized multivariate datafitting model to solve a variety of scientific computingproblems such as graphics filtering metamodeling com-putational finance and more which captured linear as wellas nonlinear phenomena Interpolation was also applied tographic image morphing and image processing [1 3ndash620ndash23] He et al [3 20] studied ielendashNewton rationalinterpolation in the polar coordinates applied it to imagesuper-resolution and obtained better performance Karimand Saaban [21] proposed a novel rational bicubic ballfunction with one parameter by using tensor product ap-proach and used it for image upscaling Zhan et al [22]developed a nonlocal and local image interpolation modelbased on nonlocal bounded variation regularization andlocal total variation and obtained better performance Heet al [24] proposed an image inpainting algorithm by usingcontinued fractions rational interpolation In order to obtainbetter repaired results ielersquos rational interpolation wascombined with Newtonndashiele rational interpolation torepair damaged images In [25] the Newton interpolationmethod was adopted in capacitance gradient and the cali-bration of cantilever stiffness Based on polynomial ap-proximation of local surface Rahman et al [5] presented anovel local patch descriptor which is invariant to the changesin viewpoint scale orientation and illumination

e mathematical description and the constructionmethod of the curve and surfaces are important problems incomputer-aided geometric design [7ndash9 26 27] ere aremany ways to handle this problem [1 7ndash9 28 29] includingBezier and NURBS-based approaches as well as the poly-nomial spline methodese methods are used for instancefor shaping the outer hull of aircraft and a ship [28]However the disadvantage of polynomial interpolation liesin its global property that is it is difficult to control the localconstraint of a given interpolation point under the conditionthat the values of the interpolant points are fixed [29 30] Toconstruct the polynomial spline the derivative values of theinterpolating data are required additionally to the functionvalues Unfortunately in many practical fields it is difficultto obtain these derivative values

In recent years many scholars have studied parametricspline interpolation Zhang et al [29 30] proposed newtypes of weighted blending spline interpolation By selectingappropriate parameters and different coefficients the valueof the spline interpolation function can be modified at anypoint in the interpolant interval under the condition that thevalues of the interpolant points are fixed so that the geo-metric surfaces can be adjusted e drawback is that thecomputation is complicated e bivariate rational inter-polation with parameters based only on the function valueshas been studied in [29] Based on function values andpartial derivatives of the function being interpolated Duanet al [31] proposed a new bivariate rational interpolationwhich had a simple and explicit rational mathematical

representation with parameters Based on function valuesDuan et al [32] constructed a rational cubic spline withparametersis spline is C1-continuous in the interpolatinginterval By selecting suitable parameters it can also becomeC2-continuous Reddy et al [33] presented univariate andbivariate rational cubic fractal interpolation functions withshape parameters which can be used for surface modelingwhen the partial derivatives of the given function are ir-regular In [34] the authors presented a weighted bivariaterational bicubic spline interpolation based on functionvalues which is C1-continuous for any positive parameterse shape of the interpolating surface can be modified bychanging the parameters under the condition that the valuesof the interpolating nodes are fixed which is convenient inengineering and useful in computer-aided geometric design[28ndash34] Tang and Zou [35] studied and presented generalinterpolant frames with many classical interpolation for-mulas Univariate and bivariate iele-like rational inter-polation continued fractions with parameters were studiedin [36] ere the interpolation function can deal withinterpolation problems where inverse differences do notexist rough the selection of proper parameter values thevalue of the proposed interpolation function can be changedat any point in the interpolant region under the conditionthat the values of the interpolating nodes are fixed Howeverhow to select appropriate parameters is a hard problem It isalso difficult to define such an interpolant function Poly-nomials have many advantages such as a simple structureeasy differentiation integration and having derivatives ofany order e polynomial function is a good tool for ap-proximating smooth function In computer-aided geometricdesign there is still a great demand for complicated modelsand the integration of design and fabrication but there arestill two unresolved problems [29 30]

(1) How can one construct a suitable polynomial in-terpolation function with an explicit mathematicalrepresentation simple calculation which is alsoconvenient to use and to study theoretically

(2) For given data how can one remodel the curves orsurfaces shape to meet the actual requirements in thecomputer-aided geometric design

Given n + 1 points (x0 x1 xn isin R) and n + 1 values(y0 y1 yn isin R) in the traditional interpolationmethodthe classical interpolating polynomial Pn(x) of degree less orequal to n is unique to the given interpolation points so itcannot be an answer to the above questions At presentNewtonrsquos polynomial interpolation is in the center of re-search on polynomial interpolation methods While it hasshown good interpolation performance its disadvantage isthat it is unique to the given interpolation points [28] Toovercome the above shortcomings we propose a novelNewton interpolation polynomial only based on (partial)divided differences by introducing one or multiple pa-rameters which can be seen as a new approach to inter-polation method of Newton polynomials Similar to theinterpolation functions in [28ndash32] the proposed method








(1)




Let

y0i yi i 0 1 n (2)

When (j 1 k + 1) for (i j j + 1 n)

yji

yjminus1i minus y

jminus1jminus1


For (i k + 1 k + 2 n)

zk+1i

yk+1i minus λ

xi minus xk

(4)


zji

zjminus1i minus z

jminus1jminus1











x minus xk+1( 1113857




(6)

where

ci y

ii i 0 1 k

zii i k + 1 k + 2 n

⎧⎨

⎩ c0k+1 λ (7)


y1) (xn yn))

P(0)n xi( 1113857 yi i 0 1 n (8)






2



k+1i minus λ

xi minus xk





If n ge i ge k + 2



2+







2+



middotz

iminus1i minus z

iminus1iminus1





2+












x0 y00

x1 y01 y1

1⋮ ⋮ ⋮ ⋱xk y0


k

xk y0k y1


xk+1 y0k+1 y1


k+1 zk+1k+1



kn

yk+1n zk+1






2



k+1i minus λ

xi minus xk





(9)

So we have

P(0)n xi( 1113857 yi i 0 1 n (10)






f xi yj1113872 1113873 fij i 0 1 m j 0 1 n (11)





(12)





i 0 1 m j 0 1 n

(13)




f(00)ij f xi yj1113872 1113873 i 0 1 m j 0 1 n

(14)


1 m)

f(p0)ij


(pminus10)pminus1j



1 2 n j q q + 1 n)

f(iq)

ij f


(iqminus1)iqminus1





final results are


T

(17)


Ai(y)

f(i0)i0 + f

(i1)i1 y minus y0( 1113857 + f










⎧⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎩

(18)

Step 6 let




(19)





mn(x y) satisfies


i 0 1 m j 0 1 n(20)



Ai yj1113872 1113873 f(i0)i0 + f

(i1)i1 yj minus y01113872 1113873 + f




(21)

If (i k)







(22)





yj minus yl+11113872 1113873



(23)



f(00)0j + f

(10)1j xi minus x0( 1113857 + f



(00)ij fij

(24)







f(00)ij f xi yj1113872 1113873 i 0 1 m j 0 1 n (25)

Step 2 if (i 0 1 m q 1 2 n j q q+

1 n) let

f(0q)

ij f

(0qminus1)


iqminus1



m i p p + 1 m)

f(pj)

ij f


(pminus1j)pminus1j






Aj(x)

f(0j)0j + f

(1j)1j x minus x0( 1113857 + f



j 0 1 l minus 1 l + 1 n







⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

(29)

Step 6 let




(30)







f(00)ij f xi yj1113872 1113873 (i 0 1 m j 0 1 n)

(31)


M1

f(00)00 f


(00)m0

f(00)01 f


(00)m1


f(00)0n f


(00)mn

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(32)




f(0q)ij


(0qminus1)iqminus1



M2

f(00)00 f


(00)m0

f(01)01 f


(01)m1


f(0n)0n f


(0n)mn

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(34)



f(pj)ij


(pminus1j)pminus1j



M3

f(00)00 f


(m0)m0

f(01)01 f


(m1)m1




(mlminus1)m lminus1

f(0l)0 l f


(0l)m l

f(0l+1)0 l+1 f


(ml+1)m l+1


f(0n)0 n f


(mn)m n

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(36)





Aj(x)

f(0j)0j + f

(1j)1j x minus x0( 1113857 + f










⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

(38)

Step 5 let




(39)




mn(xy) satisfies

N2mn xi yj1113872 1113873 fij


(40)



Aj xi( 1113857 f(i0)i0 + f

(i1)i1 xi minus x0( 1113857 + f




(41)


If j l




x minus xl+1( 1113857



(42)








ij

(43)

So we have


f(00)i0 + f

(01)i1 yj minus y01113872 1113873 + f



(00)st

fst

(44)




Pλ(x) ϕn(x) + Pn(x) (45)


nk (x minus xk)









p(x) 12



p(0)

(x) 12


(47)


p xi( 1113857 p(0)

xi( 1113857 fi (i 0 1 2) (48)






P(x y) 1 + y + x(1 + y) (49)


P(1)


+ x(1 + y) (50)


P(2)


+ y(1 + x) (51)


P(3)



+ xy

(52)


P(4)



P xi yj1113872 1113873 P(1)

xi yj1113872 1113873 P(2)

xi yj1113872 1113873 P(3)

xi yj1113872 1113873

P(4)

xi yj1113872 1113873 fij (i j 0 1)

(54)





i 1 2 3 4xi 0 1 2 3fi 0 0 1 3


0 05 1 15 2 25 3x

0

05

1

15

2

25

3

(a)

0 05 1 15 2 25 3x

ndash2

ndash1

0

1

2

3

(b)

0 05 1 15 2 25 3x

0

05

1

15

2

25

3

(c)

0 05 1 15 2 25 3x

ndash1

0

1

2

3

4

5

(d)



(xi yj) x0 0 x1 1

y0 0 1 2y1 1 2 4

435

325

215

11

05

0 002

0406

081

y + x (y + 1) + 1

xy



4

3

2

1

0

ndash1

x (y + 1) ndash 12 y + 13 y2 + 1

002

0406

081

x

1

05

0y

(a)

4

3

2

1

8 y + x (y + 1) ndash 7 y2 + 1

002

0406

081

x

1

05

0y

(b)


y (x + 1) ndash 12 x + 13 x2 + 1

43210

ndash1

002

0406

081

x

1

05

0y

(a)

8 x + y (x + 1) ndash 7 x2 + 1

4

3

2

1

002

0406

081

x

1

05

0y

(b)



4

3

2

1

0

ndash1

0 02 04 06 08 1

x

1

05

0y

(a)


4

3

2

1

0

ndash1

0 02 04 06 08 1

x

1

05

0y

(b)

Figure 5 Continued










4

3

2

11

05

0y

002

0406

081

x

(c)


1

05

0y

002

0406

081

x

435

325

215

1

(d)



4

3

2

ndash1

0

1

1

05

0y

002

0406

081

x

(a)


1

05

0y

002

0406

081

x

435

325

215

1

(b)











Data Availability




Acknowledgments


References













































(1)




Let

y0i yi i 0 1 n (2)

When (j 1 k + 1) for (i j j + 1 n)

yji

yjminus1i minus y

jminus1jminus1


For (i k + 1 k + 2 n)

zk+1i

yk+1i minus λ

xi minus xk

(4)


zji

zjminus1i minus z

jminus1jminus1











x minus xk+1( 1113857




(6)

where

ci y

ii i 0 1 k

zii i k + 1 k + 2 n

⎧⎨

⎩ c0k+1 λ (7)


y1) (xn yn))

P(0)n xi( 1113857 yi i 0 1 n (8)






2



k+1i minus λ

xi minus xk





If n ge i ge k + 2



2+







2+



middotz

iminus1i minus z

iminus1iminus1





2+












x0 y00

x1 y01 y1

1⋮ ⋮ ⋮ ⋱xk y0


k

xk y0k y1


xk+1 y0k+1 y1


k+1 zk+1k+1



kn

yk+1n zk+1






2



k+1i minus λ

xi minus xk





(9)

So we have

P(0)n xi( 1113857 yi i 0 1 n (10)






f xi yj1113872 1113873 fij i 0 1 m j 0 1 n (11)





(12)





i 0 1 m j 0 1 n

(13)




f(00)ij f xi yj1113872 1113873 i 0 1 m j 0 1 n

(14)


1 m)

f(p0)ij


(pminus10)pminus1j



1 2 n j q q + 1 n)

f(iq)

ij f


(iqminus1)iqminus1





final results are


T

(17)


Ai(y)

f(i0)i0 + f

(i1)i1 y minus y0( 1113857 + f










⎧⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎩

(18)

Step 6 let




(19)





mn(x y) satisfies


i 0 1 m j 0 1 n(20)



Ai yj1113872 1113873 f(i0)i0 + f

(i1)i1 yj minus y01113872 1113873 + f




(21)

If (i k)







(22)





yj minus yl+11113872 1113873



(23)



f(00)0j + f

(10)1j xi minus x0( 1113857 + f



(00)ij fij

(24)







f(00)ij f xi yj1113872 1113873 i 0 1 m j 0 1 n (25)

Step 2 if (i 0 1 m q 1 2 n j q q+

1 n) let

f(0q)

ij f

(0qminus1)


iqminus1



m i p p + 1 m)

f(pj)

ij f


(pminus1j)pminus1j






Aj(x)

f(0j)0j + f

(1j)1j x minus x0( 1113857 + f



j 0 1 l minus 1 l + 1 n







⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

(29)

Step 6 let




(30)







f(00)ij f xi yj1113872 1113873 (i 0 1 m j 0 1 n)

(31)


M1

f(00)00 f


(00)m0

f(00)01 f


(00)m1


f(00)0n f


(00)mn

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(32)




f(0q)ij


(0qminus1)iqminus1



M2

f(00)00 f


(00)m0

f(01)01 f


(01)m1


f(0n)0n f


(0n)mn

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(34)



f(pj)ij


(pminus1j)pminus1j



M3

f(00)00 f


(m0)m0

f(01)01 f


(m1)m1




(mlminus1)m lminus1

f(0l)0 l f


(0l)m l

f(0l+1)0 l+1 f


(ml+1)m l+1


f(0n)0 n f


(mn)m n

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(36)





Aj(x)

f(0j)0j + f

(1j)1j x minus x0( 1113857 + f










⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

(38)

Step 5 let




(39)




mn(xy) satisfies

N2mn xi yj1113872 1113873 fij


(40)



Aj xi( 1113857 f(i0)i0 + f

(i1)i1 xi minus x0( 1113857 + f




(41)


If j l




x minus xl+1( 1113857



(42)








ij

(43)

So we have


f(00)i0 + f

(01)i1 yj minus y01113872 1113873 + f



(00)st

fst

(44)




Pλ(x) ϕn(x) + Pn(x) (45)


nk (x minus xk)









p(x) 12



p(0)

(x) 12


(47)


p xi( 1113857 p(0)

xi( 1113857 fi (i 0 1 2) (48)






P(x y) 1 + y + x(1 + y) (49)


P(1)


+ x(1 + y) (50)


P(2)


+ y(1 + x) (51)


P(3)



+ xy

(52)


P(4)



P xi yj1113872 1113873 P(1)

xi yj1113872 1113873 P(2)

xi yj1113872 1113873 P(3)

xi yj1113872 1113873

P(4)

xi yj1113872 1113873 fij (i j 0 1)

(54)





i 1 2 3 4xi 0 1 2 3fi 0 0 1 3


0 05 1 15 2 25 3x

0

05

1

15

2

25

3

(a)

0 05 1 15 2 25 3x

ndash2

ndash1

0

1

2

3

(b)

0 05 1 15 2 25 3x

0

05

1

15

2

25

3

(c)

0 05 1 15 2 25 3x

ndash1

0

1

2

3

4

5

(d)



(xi yj) x0 0 x1 1

y0 0 1 2y1 1 2 4

435

325

215

11

05

0 002

0406

081

y + x (y + 1) + 1

xy



4

3

2

1

0

ndash1

x (y + 1) ndash 12 y + 13 y2 + 1

002

0406

081

x

1

05

0y

(a)

4

3

2

1

8 y + x (y + 1) ndash 7 y2 + 1

002

0406

081

x

1

05

0y

(b)


y (x + 1) ndash 12 x + 13 x2 + 1

43210

ndash1

002

0406

081

x

1

05

0y

(a)

8 x + y (x + 1) ndash 7 x2 + 1

4

3

2

1

002

0406

081

x

1

05

0y

(b)



4

3

2

1

0

ndash1

0 02 04 06 08 1

x

1

05

0y

(a)


4

3

2

1

0

ndash1

0 02 04 06 08 1

x

1

05

0y

(b)

Figure 5 Continued










4

3

2

11

05

0y

002

0406

081

x

(c)


1

05

0y

002

0406

081

x

435

325

215

1

(d)



4

3

2

ndash1

0

1

1

05

0y

002

0406

081

x

(a)


1

05

0y

002

0406

081

x

435

325

215

1

(b)











Data Availability




Acknowledgments


References









































2



k+1i minus λ

xi minus xk





If n ge i ge k + 2



2+







2+



middotz

iminus1i minus z

iminus1iminus1





2+












x0 y00

x1 y01 y1

1⋮ ⋮ ⋮ ⋱xk y0


k

xk y0k y1


xk+1 y0k+1 y1


k+1 zk+1k+1



kn

yk+1n zk+1






2



k+1i minus λ

xi minus xk





(9)

So we have

P(0)n xi( 1113857 yi i 0 1 n (10)






f xi yj1113872 1113873 fij i 0 1 m j 0 1 n (11)





(12)





i 0 1 m j 0 1 n

(13)




f(00)ij f xi yj1113872 1113873 i 0 1 m j 0 1 n

(14)


1 m)

f(p0)ij


(pminus10)pminus1j



1 2 n j q q + 1 n)

f(iq)

ij f


(iqminus1)iqminus1





final results are


T

(17)


Ai(y)

f(i0)i0 + f

(i1)i1 y minus y0( 1113857 + f










⎧⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎩

(18)

Step 6 let




(19)





mn(x y) satisfies


i 0 1 m j 0 1 n(20)



Ai yj1113872 1113873 f(i0)i0 + f

(i1)i1 yj minus y01113872 1113873 + f




(21)

If (i k)







(22)





yj minus yl+11113872 1113873



(23)



f(00)0j + f

(10)1j xi minus x0( 1113857 + f



(00)ij fij

(24)







f(00)ij f xi yj1113872 1113873 i 0 1 m j 0 1 n (25)

Step 2 if (i 0 1 m q 1 2 n j q q+

1 n) let

f(0q)

ij f

(0qminus1)


iqminus1



m i p p + 1 m)

f(pj)

ij f


(pminus1j)pminus1j






Aj(x)

f(0j)0j + f

(1j)1j x minus x0( 1113857 + f



j 0 1 l minus 1 l + 1 n







⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

(29)

Step 6 let




(30)







f(00)ij f xi yj1113872 1113873 (i 0 1 m j 0 1 n)

(31)


M1

f(00)00 f


(00)m0

f(00)01 f


(00)m1


f(00)0n f


(00)mn

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(32)




f(0q)ij


(0qminus1)iqminus1



M2

f(00)00 f


(00)m0

f(01)01 f


(01)m1


f(0n)0n f


(0n)mn

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(34)



f(pj)ij


(pminus1j)pminus1j



M3

f(00)00 f


(m0)m0

f(01)01 f


(m1)m1




(mlminus1)m lminus1

f(0l)0 l f


(0l)m l

f(0l+1)0 l+1 f


(ml+1)m l+1


f(0n)0 n f


(mn)m n

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(36)





Aj(x)

f(0j)0j + f

(1j)1j x minus x0( 1113857 + f










⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

(38)

Step 5 let




(39)




mn(xy) satisfies

N2mn xi yj1113872 1113873 fij


(40)



Aj xi( 1113857 f(i0)i0 + f

(i1)i1 xi minus x0( 1113857 + f




(41)


If j l




x minus xl+1( 1113857



(42)








ij

(43)

So we have


f(00)i0 + f

(01)i1 yj minus y01113872 1113873 + f



(00)st

fst

(44)




Pλ(x) ϕn(x) + Pn(x) (45)


nk (x minus xk)









p(x) 12



p(0)

(x) 12


(47)


p xi( 1113857 p(0)

xi( 1113857 fi (i 0 1 2) (48)






P(x y) 1 + y + x(1 + y) (49)


P(1)


+ x(1 + y) (50)


P(2)


+ y(1 + x) (51)


P(3)



+ xy

(52)


P(4)



P xi yj1113872 1113873 P(1)

xi yj1113872 1113873 P(2)

xi yj1113872 1113873 P(3)

xi yj1113872 1113873

P(4)

xi yj1113872 1113873 fij (i j 0 1)

(54)





i 1 2 3 4xi 0 1 2 3fi 0 0 1 3


0 05 1 15 2 25 3x

0

05

1

15

2

25

3

(a)

0 05 1 15 2 25 3x

ndash2

ndash1

0

1

2

3

(b)

0 05 1 15 2 25 3x

0

05

1

15

2

25

3

(c)

0 05 1 15 2 25 3x

ndash1

0

1

2

3

4

5

(d)



(xi yj) x0 0 x1 1

y0 0 1 2y1 1 2 4

435

325

215

11

05

0 002

0406

081

y + x (y + 1) + 1

xy



4

3

2

1

0

ndash1

x (y + 1) ndash 12 y + 13 y2 + 1

002

0406

081

x

1

05

0y

(a)

4

3

2

1

8 y + x (y + 1) ndash 7 y2 + 1

002

0406

081

x

1

05

0y

(b)


y (x + 1) ndash 12 x + 13 x2 + 1

43210

ndash1

002

0406

081

x

1

05

0y

(a)

8 x + y (x + 1) ndash 7 x2 + 1

4

3

2

1

002

0406

081

x

1

05

0y

(b)



4

3

2

1

0

ndash1

0 02 04 06 08 1

x

1

05

0y

(a)


4

3

2

1

0

ndash1

0 02 04 06 08 1

x

1

05

0y

(b)

Figure 5 Continued










4

3

2

11

05

0y

002

0406

081

x

(c)


1

05

0y

002

0406

081

x

435

325

215

1

(d)



4

3

2

ndash1

0

1

1

05

0y

002

0406

081

x

(a)


1

05

0y

002

0406

081

x

435

325

215

1

(b)











Data Availability




Acknowledgments


References










































2



k+1i minus λ

xi minus xk





(9)

So we have

P(0)n xi( 1113857 yi i 0 1 n (10)






f xi yj1113872 1113873 fij i 0 1 m j 0 1 n (11)





(12)





i 0 1 m j 0 1 n

(13)




f(00)ij f xi yj1113872 1113873 i 0 1 m j 0 1 n

(14)


1 m)

f(p0)ij


(pminus10)pminus1j



1 2 n j q q + 1 n)

f(iq)

ij f


(iqminus1)iqminus1





final results are


T

(17)


Ai(y)

f(i0)i0 + f

(i1)i1 y minus y0( 1113857 + f










⎧⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎩

(18)

Step 6 let




(19)





mn(x y) satisfies


i 0 1 m j 0 1 n(20)



Ai yj1113872 1113873 f(i0)i0 + f

(i1)i1 yj minus y01113872 1113873 + f




(21)

If (i k)







(22)





yj minus yl+11113872 1113873



(23)



f(00)0j + f

(10)1j xi minus x0( 1113857 + f



(00)ij fij

(24)







f(00)ij f xi yj1113872 1113873 i 0 1 m j 0 1 n (25)

Step 2 if (i 0 1 m q 1 2 n j q q+

1 n) let

f(0q)

ij f

(0qminus1)


iqminus1



m i p p + 1 m)

f(pj)

ij f


(pminus1j)pminus1j






Aj(x)

f(0j)0j + f

(1j)1j x minus x0( 1113857 + f



j 0 1 l minus 1 l + 1 n







⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

(29)

Step 6 let




(30)







f(00)ij f xi yj1113872 1113873 (i 0 1 m j 0 1 n)

(31)


M1

f(00)00 f


(00)m0

f(00)01 f


(00)m1


f(00)0n f


(00)mn

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(32)




f(0q)ij


(0qminus1)iqminus1



M2

f(00)00 f


(00)m0

f(01)01 f


(01)m1


f(0n)0n f


(0n)mn

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(34)



f(pj)ij


(pminus1j)pminus1j



M3

f(00)00 f


(m0)m0

f(01)01 f


(m1)m1




(mlminus1)m lminus1

f(0l)0 l f


(0l)m l

f(0l+1)0 l+1 f


(ml+1)m l+1


f(0n)0 n f


(mn)m n

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(36)





Aj(x)

f(0j)0j + f

(1j)1j x minus x0( 1113857 + f










⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

(38)

Step 5 let




(39)




mn(xy) satisfies

N2mn xi yj1113872 1113873 fij


(40)



Aj xi( 1113857 f(i0)i0 + f

(i1)i1 xi minus x0( 1113857 + f




(41)


If j l




x minus xl+1( 1113857



(42)








ij

(43)

So we have


f(00)i0 + f

(01)i1 yj minus y01113872 1113873 + f



(00)st

fst

(44)




Pλ(x) ϕn(x) + Pn(x) (45)


nk (x minus xk)









p(x) 12



p(0)

(x) 12


(47)


p xi( 1113857 p(0)

xi( 1113857 fi (i 0 1 2) (48)






P(x y) 1 + y + x(1 + y) (49)


P(1)


+ x(1 + y) (50)


P(2)


+ y(1 + x) (51)


P(3)



+ xy

(52)


P(4)



P xi yj1113872 1113873 P(1)

xi yj1113872 1113873 P(2)

xi yj1113872 1113873 P(3)

xi yj1113872 1113873

P(4)

xi yj1113872 1113873 fij (i j 0 1)

(54)





i 1 2 3 4xi 0 1 2 3fi 0 0 1 3


0 05 1 15 2 25 3x

0

05

1

15

2

25

3

(a)

0 05 1 15 2 25 3x

ndash2

ndash1

0

1

2

3

(b)

0 05 1 15 2 25 3x

0

05

1

15

2

25

3

(c)

0 05 1 15 2 25 3x

ndash1

0

1

2

3

4

5

(d)



(xi yj) x0 0 x1 1

y0 0 1 2y1 1 2 4

435

325

215

11

05

0 002

0406

081

y + x (y + 1) + 1

xy



4

3

2

1

0

ndash1

x (y + 1) ndash 12 y + 13 y2 + 1

002

0406

081

x

1

05

0y

(a)

4

3

2

1

8 y + x (y + 1) ndash 7 y2 + 1

002

0406

081

x

1

05

0y

(b)


y (x + 1) ndash 12 x + 13 x2 + 1

43210

ndash1

002

0406

081

x

1

05

0y

(a)

8 x + y (x + 1) ndash 7 x2 + 1

4

3

2

1

002

0406

081

x

1

05

0y

(b)



4

3

2

1

0

ndash1

0 02 04 06 08 1

x

1

05

0y

(a)


4

3

2

1

0

ndash1

0 02 04 06 08 1

x

1

05

0y

(b)

Figure 5 Continued










4

3

2

11

05

0y

002

0406

081

x

(c)


1

05

0y

002

0406

081

x

435

325

215

1

(d)



4

3

2

ndash1

0

1

1

05

0y

002

0406

081

x

(a)


1

05

0y

002

0406

081

x

435

325

215

1

(b)











Data Availability




Acknowledgments


References









































final results are


T

(17)


Ai(y)

f(i0)i0 + f

(i1)i1 y minus y0( 1113857 + f










⎧⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎩

(18)

Step 6 let




(19)





mn(x y) satisfies


i 0 1 m j 0 1 n(20)



Ai yj1113872 1113873 f(i0)i0 + f

(i1)i1 yj minus y01113872 1113873 + f




(21)

If (i k)







(22)





yj minus yl+11113872 1113873



(23)



f(00)0j + f

(10)1j xi minus x0( 1113857 + f



(00)ij fij

(24)







f(00)ij f xi yj1113872 1113873 i 0 1 m j 0 1 n (25)

Step 2 if (i 0 1 m q 1 2 n j q q+

1 n) let

f(0q)

ij f

(0qminus1)


iqminus1



m i p p + 1 m)

f(pj)

ij f


(pminus1j)pminus1j






Aj(x)

f(0j)0j + f

(1j)1j x minus x0( 1113857 + f



j 0 1 l minus 1 l + 1 n







⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

(29)

Step 6 let




(30)







f(00)ij f xi yj1113872 1113873 (i 0 1 m j 0 1 n)

(31)


M1

f(00)00 f


(00)m0

f(00)01 f


(00)m1


f(00)0n f


(00)mn

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(32)




f(0q)ij


(0qminus1)iqminus1



M2

f(00)00 f


(00)m0

f(01)01 f


(01)m1


f(0n)0n f


(0n)mn

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(34)



f(pj)ij


(pminus1j)pminus1j



M3

f(00)00 f


(m0)m0

f(01)01 f


(m1)m1




(mlminus1)m lminus1

f(0l)0 l f


(0l)m l

f(0l+1)0 l+1 f


(ml+1)m l+1


f(0n)0 n f


(mn)m n

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(36)





Aj(x)

f(0j)0j + f

(1j)1j x minus x0( 1113857 + f










⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

(38)

Step 5 let




(39)




mn(xy) satisfies

N2mn xi yj1113872 1113873 fij


(40)



Aj xi( 1113857 f(i0)i0 + f

(i1)i1 xi minus x0( 1113857 + f




(41)


If j l




x minus xl+1( 1113857



(42)








ij

(43)

So we have


f(00)i0 + f

(01)i1 yj minus y01113872 1113873 + f



(00)st

fst

(44)




Pλ(x) ϕn(x) + Pn(x) (45)


nk (x minus xk)









p(x) 12



p(0)

(x) 12


(47)


p xi( 1113857 p(0)

xi( 1113857 fi (i 0 1 2) (48)






P(x y) 1 + y + x(1 + y) (49)


P(1)


+ x(1 + y) (50)


P(2)


+ y(1 + x) (51)


P(3)



+ xy

(52)


P(4)



P xi yj1113872 1113873 P(1)

xi yj1113872 1113873 P(2)

xi yj1113872 1113873 P(3)

xi yj1113872 1113873

P(4)

xi yj1113872 1113873 fij (i j 0 1)

(54)





i 1 2 3 4xi 0 1 2 3fi 0 0 1 3


0 05 1 15 2 25 3x

0

05

1

15

2

25

3

(a)

0 05 1 15 2 25 3x

ndash2

ndash1

0

1

2

3

(b)

0 05 1 15 2 25 3x

0

05

1

15

2

25

3

(c)

0 05 1 15 2 25 3x

ndash1

0

1

2

3

4

5

(d)



(xi yj) x0 0 x1 1

y0 0 1 2y1 1 2 4

435

325

215

11

05

0 002

0406

081

y + x (y + 1) + 1

xy



4

3

2

1

0

ndash1

x (y + 1) ndash 12 y + 13 y2 + 1

002

0406

081

x

1

05

0y

(a)

4

3

2

1

8 y + x (y + 1) ndash 7 y2 + 1

002

0406

081

x

1

05

0y

(b)


y (x + 1) ndash 12 x + 13 x2 + 1

43210

ndash1

002

0406

081

x

1

05

0y

(a)

8 x + y (x + 1) ndash 7 x2 + 1

4

3

2

1

002

0406

081

x

1

05

0y

(b)



4

3

2

1

0

ndash1

0 02 04 06 08 1

x

1

05

0y

(a)


4

3

2

1

0

ndash1

0 02 04 06 08 1

x

1

05

0y

(b)

Figure 5 Continued










4

3

2

11

05

0y

002

0406

081

x

(c)


1

05

0y

002

0406

081

x

435

325

215

1

(d)



4

3

2

ndash1

0

1

1

05

0y

002

0406

081

x

(a)


1

05

0y

002

0406

081

x

435

325

215

1

(b)











Data Availability




Acknowledgments


References












































f(00)ij f xi yj1113872 1113873 i 0 1 m j 0 1 n (25)

Step 2 if (i 0 1 m q 1 2 n j q q+

1 n) let

f(0q)

ij f

(0qminus1)


iqminus1



m i p p + 1 m)

f(pj)

ij f


(pminus1j)pminus1j






Aj(x)

f(0j)0j + f

(1j)1j x minus x0( 1113857 + f



j 0 1 l minus 1 l + 1 n







⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

(29)

Step 6 let




(30)







f(00)ij f xi yj1113872 1113873 (i 0 1 m j 0 1 n)

(31)


M1

f(00)00 f


(00)m0

f(00)01 f


(00)m1


f(00)0n f


(00)mn

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(32)




f(0q)ij


(0qminus1)iqminus1



M2

f(00)00 f


(00)m0

f(01)01 f


(01)m1


f(0n)0n f


(0n)mn

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(34)



f(pj)ij


(pminus1j)pminus1j



M3

f(00)00 f


(m0)m0

f(01)01 f


(m1)m1




(mlminus1)m lminus1

f(0l)0 l f


(0l)m l

f(0l+1)0 l+1 f


(ml+1)m l+1


f(0n)0 n f


(mn)m n

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(36)





Aj(x)

f(0j)0j + f

(1j)1j x minus x0( 1113857 + f










⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

(38)

Step 5 let




(39)




mn(xy) satisfies

N2mn xi yj1113872 1113873 fij


(40)



Aj xi( 1113857 f(i0)i0 + f

(i1)i1 xi minus x0( 1113857 + f




(41)


If j l




x minus xl+1( 1113857



(42)








ij

(43)

So we have


f(00)i0 + f

(01)i1 yj minus y01113872 1113873 + f



(00)st

fst

(44)




Pλ(x) ϕn(x) + Pn(x) (45)


nk (x minus xk)









p(x) 12



p(0)

(x) 12


(47)


p xi( 1113857 p(0)

xi( 1113857 fi (i 0 1 2) (48)






P(x y) 1 + y + x(1 + y) (49)


P(1)


+ x(1 + y) (50)


P(2)


+ y(1 + x) (51)


P(3)



+ xy

(52)


P(4)



P xi yj1113872 1113873 P(1)

xi yj1113872 1113873 P(2)

xi yj1113872 1113873 P(3)

xi yj1113872 1113873

P(4)

xi yj1113872 1113873 fij (i j 0 1)

(54)





i 1 2 3 4xi 0 1 2 3fi 0 0 1 3


0 05 1 15 2 25 3x

0

05

1

15

2

25

3

(a)

0 05 1 15 2 25 3x

ndash2

ndash1

0

1

2

3

(b)

0 05 1 15 2 25 3x

0

05

1

15

2

25

3

(c)

0 05 1 15 2 25 3x

ndash1

0

1

2

3

4

5

(d)



(xi yj) x0 0 x1 1

y0 0 1 2y1 1 2 4

435

325

215

11

05

0 002

0406

081

y + x (y + 1) + 1

xy



4

3

2

1

0

ndash1

x (y + 1) ndash 12 y + 13 y2 + 1

002

0406

081

x

1

05

0y

(a)

4

3

2

1

8 y + x (y + 1) ndash 7 y2 + 1

002

0406

081

x

1

05

0y

(b)


y (x + 1) ndash 12 x + 13 x2 + 1

43210

ndash1

002

0406

081

x

1

05

0y

(a)

8 x + y (x + 1) ndash 7 x2 + 1

4

3

2

1

002

0406

081

x

1

05

0y

(b)



4

3

2

1

0

ndash1

0 02 04 06 08 1

x

1

05

0y

(a)


4

3

2

1

0

ndash1

0 02 04 06 08 1

x

1

05

0y

(b)

Figure 5 Continued










4

3

2

11

05

0y

002

0406

081

x

(c)


1

05

0y

002

0406

081

x

435

325

215

1

(d)



4

3

2

ndash1

0

1

1

05

0y

002

0406

081

x

(a)


1

05

0y

002

0406

081

x

435

325

215

1

(b)











Data Availability




Acknowledgments


References







































f(0q)ij


(0qminus1)iqminus1



M2

f(00)00 f


(00)m0

f(01)01 f


(01)m1


f(0n)0n f


(0n)mn

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(34)



f(pj)ij


(pminus1j)pminus1j



M3

f(00)00 f


(m0)m0

f(01)01 f


(m1)m1




(mlminus1)m lminus1

f(0l)0 l f


(0l)m l

f(0l+1)0 l+1 f


(ml+1)m l+1


f(0n)0 n f


(mn)m n

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(36)





Aj(x)

f(0j)0j + f

(1j)1j x minus x0( 1113857 + f










⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

(38)

Step 5 let




(39)




mn(xy) satisfies

N2mn xi yj1113872 1113873 fij


(40)



Aj xi( 1113857 f(i0)i0 + f

(i1)i1 xi minus x0( 1113857 + f




(41)


If j l




x minus xl+1( 1113857



(42)








ij

(43)

So we have


f(00)i0 + f

(01)i1 yj minus y01113872 1113873 + f



(00)st

fst

(44)




Pλ(x) ϕn(x) + Pn(x) (45)


nk (x minus xk)









p(x) 12



p(0)

(x) 12


(47)


p xi( 1113857 p(0)

xi( 1113857 fi (i 0 1 2) (48)






P(x y) 1 + y + x(1 + y) (49)


P(1)


+ x(1 + y) (50)


P(2)


+ y(1 + x) (51)


P(3)



+ xy

(52)


P(4)



P xi yj1113872 1113873 P(1)

xi yj1113872 1113873 P(2)

xi yj1113872 1113873 P(3)

xi yj1113872 1113873

P(4)

xi yj1113872 1113873 fij (i j 0 1)

(54)





i 1 2 3 4xi 0 1 2 3fi 0 0 1 3


0 05 1 15 2 25 3x

0

05

1

15

2

25

3

(a)

0 05 1 15 2 25 3x

ndash2

ndash1

0

1

2

3

(b)

0 05 1 15 2 25 3x

0

05

1

15

2

25

3

(c)

0 05 1 15 2 25 3x

ndash1

0

1

2

3

4

5

(d)



(xi yj) x0 0 x1 1

y0 0 1 2y1 1 2 4

435

325

215

11

05

0 002

0406

081

y + x (y + 1) + 1

xy



4

3

2

1

0

ndash1

x (y + 1) ndash 12 y + 13 y2 + 1

002

0406

081

x

1

05

0y

(a)

4

3

2

1

8 y + x (y + 1) ndash 7 y2 + 1

002

0406

081

x

1

05

0y

(b)


y (x + 1) ndash 12 x + 13 x2 + 1

43210

ndash1

002

0406

081

x

1

05

0y

(a)

8 x + y (x + 1) ndash 7 x2 + 1

4

3

2

1

002

0406

081

x

1

05

0y

(b)



4

3

2

1

0

ndash1

0 02 04 06 08 1

x

1

05

0y

(a)


4

3

2

1

0

ndash1

0 02 04 06 08 1

x

1

05

0y

(b)

Figure 5 Continued










4

3

2

11

05

0y

002

0406

081

x

(c)


1

05

0y

002

0406

081

x

435

325

215

1

(d)



4

3

2

ndash1

0

1

1

05

0y

002

0406

081

x

(a)


1

05

0y

002

0406

081

x

435

325

215

1

(b)











Data Availability




Acknowledgments


References







































If j l




x minus xl+1( 1113857



(42)








ij

(43)

So we have


f(00)i0 + f

(01)i1 yj minus y01113872 1113873 + f



(00)st

fst

(44)




Pλ(x) ϕn(x) + Pn(x) (45)


nk (x minus xk)









p(x) 12



p(0)

(x) 12


(47)


p xi( 1113857 p(0)

xi( 1113857 fi (i 0 1 2) (48)






P(x y) 1 + y + x(1 + y) (49)


P(1)


+ x(1 + y) (50)


P(2)


+ y(1 + x) (51)


P(3)



+ xy

(52)


P(4)



P xi yj1113872 1113873 P(1)

xi yj1113872 1113873 P(2)

xi yj1113872 1113873 P(3)

xi yj1113872 1113873

P(4)

xi yj1113872 1113873 fij (i j 0 1)

(54)





i 1 2 3 4xi 0 1 2 3fi 0 0 1 3


0 05 1 15 2 25 3x

0

05

1

15

2

25

3

(a)

0 05 1 15 2 25 3x

ndash2

ndash1

0

1

2

3

(b)

0 05 1 15 2 25 3x

0

05

1

15

2

25

3

(c)

0 05 1 15 2 25 3x

ndash1

0

1

2

3

4

5

(d)



(xi yj) x0 0 x1 1

y0 0 1 2y1 1 2 4

435

325

215

11

05

0 002

0406

081

y + x (y + 1) + 1

xy



4

3

2

1

0

ndash1

x (y + 1) ndash 12 y + 13 y2 + 1

002

0406

081

x

1

05

0y

(a)

4

3

2

1

8 y + x (y + 1) ndash 7 y2 + 1

002

0406

081

x

1

05

0y

(b)


y (x + 1) ndash 12 x + 13 x2 + 1

43210

ndash1

002

0406

081

x

1

05

0y

(a)

8 x + y (x + 1) ndash 7 x2 + 1

4

3

2

1

002

0406

081

x

1

05

0y

(b)



4

3

2

1

0

ndash1

0 02 04 06 08 1

x

1

05

0y

(a)


4

3

2

1

0

ndash1

0 02 04 06 08 1

x

1

05

0y

(b)

Figure 5 Continued










4

3

2

11

05

0y

002

0406

081

x

(c)


1

05

0y

002

0406

081

x

435

325

215

1

(d)



4

3

2

ndash1

0

1

1

05

0y

002

0406

081

x

(a)


1

05

0y

002

0406

081

x

435

325

215

1

(b)











Data Availability




Acknowledgments


References












































p(x) 12



p(0)

(x) 12


(47)


p xi( 1113857 p(0)

xi( 1113857 fi (i 0 1 2) (48)






P(x y) 1 + y + x(1 + y) (49)


P(1)


+ x(1 + y) (50)


P(2)


+ y(1 + x) (51)


P(3)



+ xy

(52)


P(4)



P xi yj1113872 1113873 P(1)

xi yj1113872 1113873 P(2)

xi yj1113872 1113873 P(3)

xi yj1113872 1113873

P(4)

xi yj1113872 1113873 fij (i j 0 1)

(54)





i 1 2 3 4xi 0 1 2 3fi 0 0 1 3


0 05 1 15 2 25 3x

0

05

1

15

2

25

3

(a)

0 05 1 15 2 25 3x

ndash2

ndash1

0

1

2

3

(b)

0 05 1 15 2 25 3x

0

05

1

15

2

25

3

(c)

0 05 1 15 2 25 3x

ndash1

0

1

2

3

4

5

(d)



(xi yj) x0 0 x1 1

y0 0 1 2y1 1 2 4

435

325

215

11

05

0 002

0406

081

y + x (y + 1) + 1

xy



4

3

2

1

0

ndash1

x (y + 1) ndash 12 y + 13 y2 + 1

002

0406

081

x

1

05

0y

(a)

4

3

2

1

8 y + x (y + 1) ndash 7 y2 + 1

002

0406

081

x

1

05

0y

(b)


y (x + 1) ndash 12 x + 13 x2 + 1

43210

ndash1

002

0406

081

x

1

05

0y

(a)

8 x + y (x + 1) ndash 7 x2 + 1

4

3

2

1

002

0406

081

x

1

05

0y

(b)



4

3

2

1

0

ndash1

0 02 04 06 08 1

x

1

05

0y

(a)


4

3

2

1

0

ndash1

0 02 04 06 08 1

x

1

05

0y

(b)

Figure 5 Continued










4

3

2

11

05

0y

002

0406

081

x

(c)


1

05

0y

002

0406

081

x

435

325

215

1

(d)



4

3

2

ndash1

0

1

1

05

0y

002

0406

081

x

(a)


1

05

0y

002

0406

081

x

435

325

215

1

(b)











Data Availability




Acknowledgments


References







































0 05 1 15 2 25 3x

0

05

1

15

2

25

3

(a)

0 05 1 15 2 25 3x

ndash2

ndash1

0

1

2

3

(b)

0 05 1 15 2 25 3x

0

05

1

15

2

25

3

(c)

0 05 1 15 2 25 3x

ndash1

0

1

2

3

4

5

(d)



(xi yj) x0 0 x1 1

y0 0 1 2y1 1 2 4

435

325

215

11

05

0 002

0406

081

y + x (y + 1) + 1

xy



4

3

2

1

0

ndash1

x (y + 1) ndash 12 y + 13 y2 + 1

002

0406

081

x

1

05

0y

(a)

4

3

2

1

8 y + x (y + 1) ndash 7 y2 + 1

002

0406

081

x

1

05

0y

(b)


y (x + 1) ndash 12 x + 13 x2 + 1

43210

ndash1

002

0406

081

x

1

05

0y

(a)

8 x + y (x + 1) ndash 7 x2 + 1

4

3

2

1

002

0406

081

x

1

05

0y

(b)



4

3

2

1

0

ndash1

0 02 04 06 08 1

x

1

05

0y

(a)


4

3

2

1

0

ndash1

0 02 04 06 08 1

x

1

05

0y

(b)

Figure 5 Continued










4

3

2

11

05

0y

002

0406

081

x

(c)


1

05

0y

002

0406

081

x

435

325

215

1

(d)



4

3

2

ndash1

0

1

1

05

0y

002

0406

081

x

(a)


1

05

0y

002

0406

081

x

435

325

215

1

(b)











Data Availability




Acknowledgments


References







































4

3

2

1

0

ndash1

x (y + 1) ndash 12 y + 13 y2 + 1

002

0406

081

x

1

05

0y

(a)

4

3

2

1

8 y + x (y + 1) ndash 7 y2 + 1

002

0406

081

x

1

05

0y

(b)


y (x + 1) ndash 12 x + 13 x2 + 1

43210

ndash1

002

0406

081

x

1

05

0y

(a)

8 x + y (x + 1) ndash 7 x2 + 1

4

3

2

1

002

0406

081

x

1

05

0y

(b)



4

3

2

1

0

ndash1

0 02 04 06 08 1

x

1

05

0y

(a)


4

3

2

1

0

ndash1

0 02 04 06 08 1

x

1

05

0y

(b)

Figure 5 Continued










4

3

2

11

05

0y

002

0406

081

x

(c)


1

05

0y

002

0406

081

x

435

325

215

1

(d)



4

3

2

ndash1

0

1

1

05

0y

002

0406

081

x

(a)


1

05

0y

002

0406

081

x

435

325

215

1

(b)











Data Availability




Acknowledgments


References















































4

3

2

11

05

0y

002

0406

081

x

(c)


1

05

0y

002

0406

081

x

435

325

215

1

(d)



4

3

2

ndash1

0

1

1

05

0y

002

0406

081

x

(a)


1

05

0y

002

0406

081

x

435

325

215

1

(b)











Data Availability




Acknowledgments


References















































Data Availability




Acknowledgments


References
































































Documents

A New Approach to Newton-Type Polynomial Interpolation