Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2013 Article ID 642818 10 pageshttpdxdoiorg1011552013642818

Research ArticleA New Algorithm to Approximate Bivariate MatrixFunction via Newton-Thiele Type Formula

Rongrong Cui12 and Chuanqing Gu1

1 Department of Mathematics Shanghai University Shanghai 200444 China2 School of Mathematical Science Yancheng Teachers University Yancheng 224000 China

Correspondence should be addressed to Chuanqing Gu cqgustaffshueducn

Received 22 October 2012 Revised 6 December 2012 Accepted 10 December 2012

Academic Editor Juan Manuel Pena

Copyright copy 2013 R Cui and C Gu This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

A new method for computing the approximation of bivariate matrix function is introduced It uses the construction of bivariateNewton-Thiele type matrix rational interpolants on a rectangular gridThe rational interpolant is of the formmotivated by Tan andFang (2000) which is combined by Newton interpolant and branched continued fractions with scalar denominator The matrixquotients are based on the generalized inverse for a matrix which is introduced by C Gu the author of this paper and it is effectivein continued fraction interpolation The algorithm and some other important conclusions such as divisibility and characterizationare given In the end two examples are also given to show the effectiveness of the algorithm The numerical results of the secondexample show that the algorithm of this paper is better than the method of Thieletype matrix-valued rational interpolant in Gu(1997)

1 Introduction

Matrix-valued rational interpolation and approximation the-ory have further practical application in many fields suchas in automatic control theory computer science and ele-mentary particle physics [1] Kuchminska et al proved ananalog of the van Vlerk theorem and constructed an inter-polation formular of the Newton-Thiele type in [2] Tan andFang in [3] put emphasis on the study of Newton-Thielebivariate rational interpolants Graves-Morris provided apractical Thiele-fraction method for rational interpolationof vectors based on the Samelson inverse in [4] Gu et algeneralized the definition of Samelson inverse to the case ofmatrices and applied it to deal with the problems of rationalinterpolation of matrices [5ndash9] Bose and Basu gave theexistence nonuniqueness and recursive computation of thetwo-dimensional matrix Pade approximants in [10]

In this paper we introduce a bivariate matrix-valuedrational interpolant with scalar denominator in Newton-Thiele form motivated by [3] The construction of theinterpolant is combined by the classic methods Newtoninterpolant and branched continued fractions As we knowbranched continued fraction has been studied by Cuyt and

Verdonk and Siemaszko [11 12] and many other authors Theinterpolant of this paper is based on using the generalizedinverse ofmatrices A new algorithm to approximate bivariatematrix function is given and some examples are also put toshow the effectiveness of the algorithm which is much betterthan the method mentioned in [9]

First we will give the definition of the so-called gen-eralized inverse of a matrix Let 119862

119906times119907 consists of all 119906 times 119907

matrices with their elements in the complex plan 119862 and let119860 = (119886

119894119895) 119861 = (119887

119894119895) and 119860 119861 isin 119862

119906times119907

Definition 1 (see [8]) The scalar product of matrices119860 and 119861

is defined by

119860 sdot 119861 =

119906

sum

119894=1

119907

sum

119895=1

119886119894119895119887119894119895

= tr (119860119861lowast

) (1)

where 119861lowast denotes the transpose of 119861 and the Euclidean norm

of 119860 is given by

119860 = (

119906

sum

119894=1

119907

sum

119895=1

10038161003816100381610038161003816119886119894119895

10038161003816100381610038161003816

2

)

12

(2)

brought to you by COREView metadata citation and similar papers at coreacuk

provided by Crossref

2 Journal of Applied Mathematics

It follows from Definition 1 and (2) that

119860 sdot 119860 =

119906

sum

119894=1

119907

sum

119895=1

10038161003816100381610038161003816119886119894119895

10038161003816100381610038161003816

2

= 1198602

= tr (119860119860lowast

) (3)

where 119860 denotes the complex conjugate matrix of 119860 and 119860lowast

denotes the complex conjugate transpose matrix of119860 On thebasis of (2) and (3) the generalized inverse of the matrix119860 isdefined as

119860minus1

119903=

1

119860=

119860

1198602 119860 = 0 119860 isin 119862

119906times119907

(4)

in particular for 119860 isin 119877119906times119907

119860minus1

119903= 1119860 = 119860119860

2

119860 = 0

Bymeans of generalized inverse119860minus1119903formatrices wewant

to define bivariate Newton-Thiele rational interpolants andgive the algorithm and some properties on a rectangular grid

2 Newton-Thiele Interpolation Formula

Let Λ119898119899

= (119909119897 119910119896) 119897 = 0 1 119898 119896 = 0 1 119899 (119909

119897 119910119896) isin

R2A matrix-valued set

Π119898119899

= 119884119897119896

119884119897119896

= 119884 (119909119897 119910119896) isin 119862119906times119907

(119909119897 119910119896) isin Λ

119898119899 (5)

We need to find a bivariate matrix rational function

119877119898119899

(119909 119910) =119873 (119909 119910)

119863 (119909 119910) (6)

whose numerator 119873(119909 119910) is a complex or real polynomialmatrix and denominator 119863(119909 119910) is a real polynomial and

119877119898119899

(119909119897 119910119896) =

119873 (119909119897 119910119896)

119863 (119909119897 119910119896)

= 119884119897119896 (119909

119897 119910119896) isin Λ

119898119899 (7)

First some notations need to be given as follows

12059300

(119909119897 119910119896) = 119884119897119896 forall (119909

119897 119910119896) isin Λ

119898119899

12059310

(119909119894 119909119895 119910119896) =

12059300

(119909119895 119910119896) minus 12059300

(119909119894 119910119896)

119909119895minus 119909119894

1205931198970

(119909119901 119909

119902 119909119894 119909119895 119910119896)

= (120593119897minus10

(119909119901 119909

119902 119909119895 119910119896)

minus 120593119897minus10

(119909119901 119909

119902 119909119894 119910119896)) times (119909

119895minus 119909119894)minus1

1205931198971

(119909119901 119909

119902 119909119894 119909119895 119910119903 119910119904)

= (119910119904minus 119910119903) times (120593

1198970(119909119901 119909

119902 119909119894 119909119895 119910119904)

minus1205931198970

(119909119901 119909

119902 119909119894 119909119895 119910119903))minus1

120593119897119896

(119909119901 119909

119902 119909119894 119909119895 119910119886 119910

119887 119910119903 119910119904)

= (119910119904minus 119910119903)

times (120593119897119896minus1

(119909119901 119909

119902 119909119894 119909119895 119910119886 119910

119887 119910119904)

minus120593119897119896minus1

(119909119901 119909

119902 119909119894 119909119895 119910119886 119910

119887 119910119903))minus1

(8)

where the first subscript 119897 of 120593 means the number of nodes119909119901 119909

119902 119909119894 119909119895minus 1 and the second subscript 119896 of 120593

means the number of nodes 119910119886 119910

119887 119910119903 119910119904minus 1

For simplicity we let the nodes 119909119901 119909

119902 119909119894 119909119895be

replaced by 1199090 119909

119897 and the nodes 119910

119886 119910

119887 119910119903 119910119904be

replaced by 1199100 119910

119896 then we can get the following defini-

tion

Definition 2 By means of generalized inverse (4) we definebivariate Newton-Thiele type matrix blending differences asfollows

12059300

(119909119897 119910119896) = 119884119897119896

12059310

(1199090 1199091 119910119896) =

12059300

(1199091 119910119896) minus 12059300

(1199090 119910119896)

1199091minus 1199090

1205931198970

(1199090 119909

119897 119910119896)

= (120593119897minus10

(1199090 119909

119897minus2 119909119897 119910119896)

minus120593119897minus10

(1199090 119909

119897minus2 119909119897minus1

119910119896)) times (119909

119897minus 119909119897minus1

)minus1

1205931198971

(1199090 119909

119897 1199100 1199101)

=(1199101minus 1199100)

1205931198970

(1199090 119909

119897 1199101) minus 1205931198970

(1199090 119909

119897 1199100)

1205930119896

(119909119897 1199100 119910

119896)

= (119910119896minus 119910119896minus1

)

times (1205930119896minus1

(119909119897 1199100 119910

119896minus2 119910119896)

minus1205930119896minus1

(119909119897 1199100 119910

119896minus2 119910119896minus1

))minus1

120593119897119896

(1199090 1199091 119909

119897 1199100 119910

119896)

= (119910119896minus 119910119896minus1

)

times (120593119897119896minus1

(1199090 119909

119897 1199100 119910

119896minus2 119910119896)

minus120593119897119896minus1

(1199090 119909

119897 1199100 119910

119896minus2 119910119896minus1

))minus1

(9)

Journal of Applied Mathematics 3

We assume that for all 119897 119909119897

= 119909119897minus1

and

120593119897119896minus1

(1199090 119909

119897 1199100 119910

119896minus2 119910119896)

= 120593119897119896minus1

(1199090 119909

119897 1199100 119910

119896minus2 119910119896minus1

) forall119897 119896

(10)

From (9) matrix-valued Newton-Thiele type continuedfractions for two-variable function can be constructed asfollows

119877119898119899

(119909 119910)

= 1198800(119910) + 119880

1(119910) (119909 minus 119909

0) + sdot sdot sdot + 119880

119898(119910) (119909 minus 119909

0)

times (119909 minus 1199091) sdot sdot sdot (119909 minus 119909

119898minus1)

(11)

where for 119897 = 0 1 119898

119880119897(119910) = 120593

1198970(1199090 119909

119897 1199100) +

119910 minus 1199100

1205931198971

(1199090 119909

119897 1199100 1199101)

+ sdot sdot sdot +119910 minus 119910119899minus1

|

| 120593119897119899

(1199090 119909

119897 1199100 119910

119899)

(12)

Remark 3 Because of the construction119863(119909 119910) in 119877119898119899

(119909 119910)

mentioned as (6) is actually119863(119910) and each119880119897(119910) depends on

119899

We can also construct the antithetical form of bivariatematrix continued fraction in light of Definition 4

Definition 4 By means of generalized inverse (4) we definebivariate antithetical Newton-Thiele type matrix blendingdifferences

12059500

(119909119897 119910119896) = 119884119897119896

12059501

(119909119897 1199100 1199101) =

(12059500

(119909119897 1199101) minus 12059500

(119909119897 1199100))

(1199101minus 1199100)

1205950119896

(119909119897 1199100 119910

119896)

= (1205950119896minus1

(119909119897 1199100 119910

119896minus2 119910119896)

minus1205950119896minus1

(119909119897 1199100 119910

119896minus2 119910119896minus1

))

times (119910119896minus 119910119896minus1

)minus1

1205951119896

(1199090 1199091 1199100 119910

119896)

=(1199091minus 1199090)

1205950119896

(1199091 1199100 119910

119896) minus 1205950119896

(1199090 1199100 119910

119896)

1205951198970

(1199090 119909

119897 119910119896)

= (119909119897minus 119909119897minus1

)

times (120595119897minus10

(1199090 119909

119897minus2 119909119897 119910119896)

minus120595119897minus10

(1199090 119909

119897minus1 119910119896))minus1

120595119897119896

(1199090 1199091 119909

119897 1199100 119910

119896)

= (119909119897minus 119909119897minus1

)

times (120595119897minus1119896

(1199090 119909

119897minus2 119909119897 1199100 119910

119896)

minus120595119897minus1119896

(1199090 119909

119897minus1 1199100 119910

119896))minus1

(13)

From (13) we can get the antithetical formula for two-variable function

119898119899

(119909 119910) = 1198810(119909) + 119881

1(119909) (119910 minus 119910

0)

+ sdot sdot sdot + 119881119899(119909) (119910 minus 119910

0) (119910 minus 119910

1)

sdot sdot sdot (119910 minus 119910119899minus1

)

(14)

where for 119896 = 0 1 119899

119881119896(119909) = 120595

0119896(1199090 1199100 119910

119896)

+119909 minus 1199090

1205951119896

(1199090 1199091 1199100 119910

119896)

+ sdot sdot sdot +119909 minus 119909119898minus1

|

| 120595119898119896

(1199090 119909

119898 1199100 119910

119896)

(15)

Now we will give two theorems about 119877119898119899

(119909 119910) as in (11)and (12) and

119898119899(119909 119910) as in (14) and (15) First of all some

notations need to be definedIn (12) for 119897 = 0 1 119898 let

119880(119904)

119897(119910) = 120593

119897119904(1199090 1199091 119909

119897 1199100 119910

119904)

+119910 minus 119910119904

119880(119904+1)

119897(119910)

119904 = 0 1 119899 minus 1(16)

where

119880(119899)

119897(119910) = 120593

119897119899(1199090 1199091 119909

119897 1199100 119910

119899)

119880(0)

119897(119910) = 119880

119897(119910)

(17)

Similarly in (15) for 119896 = 0 1 119899 let

119881(119904)

119896(119909) = 120595

119904119896(1199090 1199091 119909

119904 1199100 119910

119896)

+119909 minus 119909119904

119881(119904+1)

119896(119909)

119904 = 0 1 119898 minus 1(18)

where

119881(119898)

119896(119909) = 120595

119898119896(1199090 1199091 119909

119898 1199100 119910

119896)

119881(0)

119896(119909) = 119881

119896(119909)

(19)

Theorem 5 Let

(i) 120593119897119896(1199090 119909

119897 1199100 119910

119896) 119897 = 0 1 119898 119896 = 0

1 119899 exist and be nonzero (except for 1205931198970(1199090

119909119897 1199100))

4 Journal of Applied Mathematics

(ii) 119880(119904)

119897(119910) = 120593

119897119904(1199090 1199091 119909

119897 1199100 119910

119904) + ((119910 minus

119910119904)119880(119904+1)

119897(119910)) satisfy119880

(119904+1)

119897(119910119904) = 0 119904 = 0 1 119899minus1

then 119877119898119899

(119909 119910) as in (11) and (12) exists such that119877119898119899

(119909119897 119910119896) = 119884119897119896 (119909119897 119910119896) isin Λm119899

Proof For simpleness let

120593119897119896

(1199090 119909

119897 1199100 119910

119896) = 120593119897119896

(20)

if the conditions hold thus for 119897 = 0 1 119898 119896 = 0 1 119899

119880119897(119910119896) = 119880

(0)

119897(119910119896)

= 1205931198970

+119910119896minus 1199100

1205931198971

+ sdot sdot sdot +119910119896minus 119910119896minus1

|

| 120593119897119899

+119910119896minus 119910119896|

| 119880(119896+1)

119897(119910119896)

(21)

(119910119896minus119910119896)119880(119896+1)

119897(119910119896) = 0 since119880

(119896+1)

119897(119910119896) = 0 then we get

119880119897(119910119896)

= 1205931198970

(1199090 119909

119897 1199100) +

119910119896minus 1199100

1205931198971

(1199090 119909

119897 1199100 1199101)+ sdot sdot sdot +

119910119896minus 119910119896minus2

|

| 120593119897119896minus1

(1199090 119909

119897 1199100 119910

119896minus1)

+119910119896minus 119910119896minus1

|

| 120593119897119896

(1199090 119909

119897 1199100 119910

119896)

= 1205931198970

(1199090 119909

119897 1199100) +

119910119896minus 1199100

1205931198971

(1199090 119909

119897 1199100 1199101)+ sdot sdot sdot

+119910119896minus 119910119896minus2

|

| 120593119897119896minus1

(1199090 119909

119897 1199100 119910

119896minus1)

+119910119896minus 119910119896minus1

|

| (119910119896minus 119910119896minus1

) times (120593119897119896minus1

(1199090 119909

119897 1199100 119910

119896minus2 119910119896) minus 120593119897119896minus1

(1199090 119909

119897 1199100 119910

119896minus1))minus1

= 1205931198970

(1199090 119909

119897 1199100) +

119910119896minus 1199100

1205931198971

(1199090 119909

119897 1199100 1199101)+ sdot sdot sdot +

119910119896minus 119910119896minus2

|

| 120593119897119896minus1

(1199090 119909

119897 1199100 119910

119896minus2 119910119896)

= sdot sdot sdot = 1205931198970

(1199090 119909

119897 1199100) +

119910119896minus 1199100

1205931198971

(1199090 119909

119897 1199100 119910119896)

= 1205931198970

(1199090 119909

119897 119910119896)

(22)

We can use (9) (11) and (22) to find that119877119898119899

(119909119897 119910119896)

= 1198800(119910119896) + 1198801(119910119896) (119909119897minus 1199090)

+ sdot sdot sdot + 119880119897(119910119896) (119909119897minus 1199090) (119909119897minus 1199091) sdot sdot sdot (119909

119897minus 119909119897minus1

)

= 119884 (119909119897 119910119896)

(23)

We can similarly prove the following result

Theorem 6 Let(i) 120595119897119896(1199090 119909

119897 1199100 119910

119896) 119897 = 0 1 119898 119896 = 0

1 119899 exist and be nonzero (except for 1205950119896

(1199090

1199100 119910

119896))

(ii) 119881(119904)

119896(119909) = 120595

119904119896(1199090 1199091 119909

119904 1199100 119910

119896) + ((119909 minus

119909119904)119881(119904+1)

119896(119909)) satisfy119881(119904+1)

119896(119909119904) = 0 119904 = 0 1 119898minus1

then 119898119899

(119909 119910) as in (14) and (15) exists such that119898119899

(119909119897 119910119896) = 119884119897119896 (119909119897 119910119896) isin Λ

119898119899

Now we give the Newton-Thiele Matrix InterpolationAlgorithm (NTMIA)

Algorithm 7 (NTMIA) Input (119909119894 119910119895) 119884119894119895 (119894 = 0 1

119898 119895 = 0 1 119899)

Output 119898(119909 119910)

Step 1 For all (119909119894 119910119895) isin Λ

119898119899 let 120593(119909

119894 119910119895) = 119884119894119895

Step 2 For 119895 = 0 1 119899 119901 = 1 2 119898 119894 = 119901 119901 +

1 119898 compute

1205931199010

(1199090 119909

119901minus1 119909119894 119910119895)

=

120593119901minus10

(1199090 119909

119901minus2 119909119894 119910119895)minus120593119901minus10

(1199090 119909

119901minus1 119910119895)

119909119894minus119909119901minus1

(24)

Journal of Applied Mathematics 5

Step 3 For 119894 = 0 1 119898 119902 = 1 2 119899 119895 = 119902 119902 + 1 119899compute

120593119894119902

(1199090 119909

119894 1199100 119910

119902minus1 119910119895)

= (119910119895minus 119910119902minus1

) times (120593119894119902minus1

(1199090 119909

119894 1199100 119910

119902minus2 119910119895)

minus120593119894119902minus1

(1199090 119909

119894 1199100 119910

119902minus1))minus1

= ((119910119895minus 119910119902minus1

) (120593119894119902minus1

(1199090 119909

119894 1199100 119910

119902minus2 119910119895)

minus120593119894119902minus1

(1199090 119909

119894 1199100 119910

119902minus1)))

times (10038171003817100381710038171003817120593 (1199090 119909

119894 1199100 119910

119902minus2 119910119895)

minus120593 (1199090 119909

119894 1199100 119910

119902minus1)10038171003817100381710038171003817

2

)

minus1

(25)

Step 4 For 119894 = 0 1 119898 119895 = 0 1 119899 compute 119886119894119895

=

120593119894119895(1199090 119909

119894 1199100 119910

119895)

Step 5 For 119894 = 0 1 119898 119895 = 1 119899 judge if 119886119894119895

= 0 if yesgo to Step 6 otherwise exist and show ldquothe algorithm is notvalidrdquo

Step 6 For 119894 = 0 1 119898 119895 = 119899 minus 2 119899 minus 3 0

119876119894119899

= 1 119875119894119899

= 119886119894119899

119876119894119899minus1

=1003817100381710038171003817119886119894119899

1003817100381710038171003817

2

119875119894119899minus1

= 119886119894119899minus1

119876119894119899minus1

+ (119910 minus 119910119899minus1

) 119875119894119899

119876119894119895

=10038171003817100381710038171003817119886119894119895+1

10038171003817100381710038171003817

2

119876119894119895+1

+ 2 (119910 minus 119910119895+1

) tr (119886119894119895+1

119875119894119895+2

lowast

)

+ (119910 minus 119910119895+1

)2

119876119894119895+2

119875119894119895

= 119886119894119895119876119894119895

+ (119910 minus 119910119895) 119875119894119895+1

(26)

Step 7 For 119894 = 0 1 119898 119895 = 119899 119899 minus 1 1 let 119861119894119895(119910) =

119875119894119895(119910)119876

119894119895(119910) if 119861

119894119895(119910119895minus1

) = 0 then go to Step 8 otherwiseexist and show ldquothe algorithm is not validrdquo

Step 8 For 119894 = 0 1 119898 let 119861119894(119910) = 119861

1198940(119910) = 119875

1198940(119910)

1198761198940(119910)

Step 9 For 119896 = 1 119898 let 0(119909 119910) = 119861

0(119910) then compute

119896(119909 119910) =

119896minus1(119909 119910) + sdot sdot sdot + (119909 minus 119909

0) sdot sdot sdot (119909 minus 119909

119896minus1)119861119896(119910)

3 Some Properties

Lemma 8 (see [5]) Define 1198800(119910) = 120593

00(1199090 1199100) + (119910 minus

1199100)12059301

(1199090 1199100 1199101) + sdot sdot sdot + (119910minus119910

119899minus1)1205930119899

(1199090 1199100 119910

119899) if we

use generalized inverse by a tail-to-head rationalization thena polynomial matrix 119864

0(119910) and a real polynomial 119865

0(119910) exist

such that(i) 1198800(119910) = 119864

0(119910)1198650(119910) 1198650(119910) ge 0

(ii) 1198650(119910) | 119864

0(119910)2ldquo|rdquo means the sign of divisibility

Definition 9 119877(119909 119910) = 119873(119909 119910)119863(119909 119910) is defined to be oftype [119897119905] if deg119873

119894119895 le 119897 for 1 le 119894 le 119906 1 le 119895 le 119907 deg119873

119894119895 = 119897

for some (119894 119895) and deg119863 = 119905 where119873(119909 119910) = (119873119894119895(119909 119910)) isin

119862119906times119907

Lemma 10 (see [5]) Let1198800(119910) be defined as in Lemma 8 and

1198800(119910) = 119864

0(119910)1198650(119910) if 119899 is even119880

0(119910) is of type [119899119899] if 119899 is

odd 1198800(119910) is of type [119899119899 minus 1]

Theorem 11 Let 120593119897119896

isin 119862119906times119907 (119909

119897 119910119896) isin Λ

119898119899 and 119909 119910 isin

119877 Define 119880119897(119910) 119900119891 119877

119898119899(119909 119910) as in (7) by a tail-to-head

rationalization using generalized inverse and suppose everyintermediate denominator be nonzero in the operation thena square polynomial matrix 119873(119909 119910) and a real polynomial119863(119909 119910) exist such that

(i) 119877119898119899

(119909 119910) = 119873(119909 119910)119863(119909 119910)

(ii) 119863(119909 119910) ge 0

(iii) 119863(119909 119910) | 119873(119909 119910)2

Proof Consider the following algorithm for the constructionof 119873(119909 119910) 119863(119909 119910) and 119877

119898119899(119909 119910)

Initialization let 1198630(119909 119910) = 1 and

1198730(119909 119910) = 119880

0(119910) = 120593

00+

119910 minus 1199100

12059301

+ sdot sdot sdot +119910 minus 119910119899minus1

|

| 1205930119899

(27)

By Lemma 8 a polynomial matrix 1198640(119910) and a real polyno-

mial 1198650(119910) exist such that

(a1) 1198730(119909 119910) = 119864

0(119910)1198650(119910)

(a2) 1198650(119910) ge 0

(a3) 1198650(119910) | 119864

0(119910)2

Recursion for 119895 = 1 2 119898 let

119878(119895)

(119909 119910) = 119878(119895minus1)

(119909 119910) + 119880119895(119910) (119909 minus 119909

0) sdot sdot sdot (119909 minus 119909

119895minus1)

(28)

with 119880119895(119910) = 119864

119895(119910)119865119895(119910) 119865119895(119910) | 119864

119895(119910)2 having the

representation

(b1) 119878(119895)

(119909 119910) = 119873119895(119909 119910)119863

119895(119909 119910)

(b2) 119863119895(119909 119910) ge 0

(b3) 119863119895(119909 119910) | 119873

119895(119909 119910)

2

where 119863119895(119909 119910) is a real polynomial

6 Journal of Applied Mathematics

Then we can get

119878(119895+1)

(119909 119910) = 119878(119895)

(119909 119910) + 119880119895+1

(119909 minus 1199090) sdot sdot sdot (119909 minus 119909

119895)

=119873119895(119909 119910)

119863119895(119909 119910)

+119864119895+1

(119910)

119865119895+1

(119910)(119909 minus 119909

0) sdot sdot sdot (119909 minus 119909

119895)

= (119873119895(119909 119910) 119865

119895+1(119910) + 119863

119895(119909 119910) 119864

119895+1(119910)

times (119909 minus 1199090) sdot sdot sdot (119909 minus 119909

119895))

times (119863119895(119909 119910) 119865

119895+1(119910))minus1

=119873119895+1

(119909 119910)

119863119895+1

(119909 119910)

(29)

where

119873119895+1

(119909 119910) = 119873119895(119909 119910) 119865

119895+1(119910)

+ 119863119895(119909 119910) 119864

119895+1(119910) (119909 minus 119909

0)

sdot sdot sdot (119909 minus 119909119895)

119863119895+1

(119909 119910) = 119863119895(119909 119910) 119865

119895+1(119910)

(30)

Obviously

10038171003817100381710038171003817119873119895+1

10038171003817100381710038171003817

2

=10038171003817100381710038171003817119873119895(119909 119910)

10038171003817100381710038171003817

2

1198652

(119895+1)(119910) + 119863

2

119895(119909 119910)

times10038171003817100381710038171003817119864(119895+1)

(119910)10038171003817100381710038171003817

2

(119909 minus 1199090)2

sdot sdot sdot (119909 minus 119909119895)2

+ 119873119895(119909 119910) sdot 119864

119895+1(119910) 119865119895+1

(119910)119863119895(119909 119910)

times (119909 minus 1199090) sdot sdot sdot (119909 minus 119909

119895)

+ 119873119895(119909 119910) sdot 119864

119895+1(119910) 119865119895+1

(119910)119863119895(119909 119910)

times (119909 minus 1199090) sdot sdot sdot (119909 minus 119909

119895)

(31)

since

119863119895(119909 119910) |

10038171003817100381710038171003817119873119895(119909 119910)

10038171003817100381710038171003817

2

119865119895+1

(119910) |10038171003817100381710038171003817119864119895+1

(119910)10038171003817100381710038171003817

2

119863119895+1

(119909 119910) |10038171003817100381710038171003817119873119895+1

(119909 119910)10038171003817100381710038171003817

2

(32)

Termination 119878(119898)(119909 119910) = 119877119898119899

(119909 119910)

Theorem 12 (Characterization) Let 119877119898119899

(119909 119910) = 119873(119909 119910)

119863(119909 119910) be expressed as

119877 = 119877119898119899

(119909 119910) = 1198800(119910) + 119880

1(119910) (119909 minus 119909

0)

+ sdot sdot sdot + 119880119898

(119910) (119909 minus 1199090) sdot sdot sdot (119909 minus 119909

119898minus1)

(33)

where 119880119897(119910) = 119864

119894119865119894 as in (12) we assume 119865

119894 119865119895have no

common factor 119894 119895 = 0 1 119898 119894 = 119895 then

(i) if n is even 119877 is of type [119898119899 + 119898 + 119899119898119899 + 119899](ii) if n is odd 119877 is of type [119898119899 + 119899119898119899 + 119899 minus 119898 minus 1]

Proof The proof is recursive Let 119899 be evenFor 119898 = 0 119877

0119899(119909 119910) = 119880

0(119910) = 119864

01198650 by Lemma 10 we

find that it is of type [119899119899]For 119898 = 1 because deg119864

0 = deg119865

0 = 119899 deg119864

1 =

deg1198651 = 119899

1198771119899

(119909 119910) =1198731

1198631

= 1198800(119910) + 119880

1(119910) (119909 minus 119909

0)

=1198640

1198650

+1198641

1198651

(119909 minus 1199090)

=11986401198651+ 11986411198650(119909 minus 119909

0)

11986501198651

(34)

is of type [2119899 + 12119899]For 119898 = 2 from the formula above we know that

deg1198731 = 2119899 + 1 deg119863

1 = 2119899 deg119864

2 = 119899 and deg119865

2 =

119899 one has1198772119899

(119909 119910) =1198732

1198632

=1198731

1198631

+1198642

1198652

(119909 minus 1199090) (119909 minus 119909

1)

=11987311198652+ 11986421198631(119909 minus 119909

0) (119909 minus 119909

1)

11986311198652

(35)

It is easy to find that 1198772119899

(119909 119910) is of type [3119899 + 23119899]For 119898 = 119896 let 119877

119896119899(119909 119910) = 119873

119896119863119896be of type [(119896 + 1)119899 +

119896(119896 + 1)119899]For 119898 = 119896 + 1 we consider that

119877119896+1119899

(119909 119910) =119873119896

119863119896

+119864119896+1

119865119896+1

(119909 minus 1199090) sdot sdot sdot (119909 minus 119909

119896)

=119873119896119865119896+1

+ 119864119896+1

119863119896(119909 minus 119909

0) sdot sdot sdot (119909 minus 119909

119896)

119863119896119865119896

=119873119896+1

119863119896+1

(36)

wheredeg 119873

119896+1 = (119896 + 1) (119899 + 1) + 119899

= (119896 + 2) 119899 + (119896 + 1)

deg 119863119896+1

= (119896 + 1) 119899 + 119899 = (119896 + 2) 119899

(37)

Therefore when 119899 is even 119877 is of type [119898119899 + 119898 + 119899119898119899 + 119899]Similarly if 119899 is odd for 119898 = 0 119877

0119899(119909 119910) = 119880

0(119910) =

11986401198650 by Lemma 10 we find that it is of type [119899119899 minus 1]

For 119898 = 1 because deg1198640 = deg119864

1 = 119899 deg119865

0 =

deg1198651 = 119899 minus 1

1198771119899

(119909 119910) =1198731

1198631

= 1198800(119910) + 119880

1(119910) (119909 minus 119909

0)

=1198640

1198650

+1198641

1198651

(119909 minus 1199090)

=11986401198651+ 11986411198650(119909 minus 119909

0)

11986501198651

(38)

is of type [21198992119899 minus 2]

Journal of Applied Mathematics 7

For119898 = 2 from the formula above we know that deg1198731

= 2119899 deg1198631 = 2119899 minus 2 deg119864

2 = 119899 and deg119865

2 = 119899 minus 1

one has

1198772119899

(119909 119910) =1198732

1198632

=1198731

1198631

+1198642

1198652

(119909 minus 1199090) (119909 minus 119909

1)

=11987311198652+ 11986421198631(119909 minus 119909

0) (119909 minus 119909

1)

11986311198652

(39)

We get that 1198772119899

(119909 119910) is of type [31198993119899 minus 3]For 119898 = 119896 let

119877119896119899

(119909 119910) =119873119896

119863119896

(40)

be of type [(119896 + 1)119899(119896 + 1)(119899 minus 1)]For 119898 = 119896 + 1 we consider that

119877119896+1119899

(119909 119910) =119873119896

119863119896

+119864119896+1

119865119896+1

(119909 minus 1199090) sdot sdot sdot (119909 minus 119909

119896)

=119873119896119865119896+1

+ 119864119896+1

119863119896(119909 minus 119909

0) sdot sdot sdot (119909 minus 119909

119896)

119863119896119865119896

=119873119896+1

119863119896+1

(41)

where

deg 119873119896+1

= (119896 + 1) (119899 minus 1) + 119896 + 1 = (119896 + 2) 119899

deg 119863119896+1

= (119896 + 1) (119899 minus 1) + (119899 minus 1) = (119896 + 2) (119899 minus 1)

(42)

Therefore when 119899 is odd119877 is of type [(119898+1)119899(119898+1)(119899minus

1)]

Definition 13 A matrix-valued Newton-Thiele type ratio-nal fraction 119877

119898119899(119909 119910) = 119873(119909 119910)119863(119909 119910) is defined to

be a bivariate generalized inverse and rational interpolant(BGIRINT) on the rectangular grid Λ

119898119899if

(i) 119877119898119899

(119909119897 119910119896) = 119884119897119896 (119909119897 119910119896) isin Λ

119898119899

(ii) 119863(119909119897 119910119896) = 0 (119909

119897 119910119896) isin Λ

119898119899

(iii) (a) if 119899 is even

deg 119873 (119909 119910) = 119898119899+ 119898+119899 deg 119863 (119909 119910) = 119898119899+119899

(43)

(b) if n is odd

deg 119873 (119909 119910) = 119898119899+119899 deg 119863 (119909 119910) = 119898119899+119899minus119898minus1

(44)

(iv) 119863(119909 119910) | 119873(119909 119910)2

(v) 119863(119909 119910) is real and 119863(119909 119910) ge 0

Now let us turn to the error estimate of the BGIRINTFirstly we give Lemma 14

Lemma 14 Let

119877lowast

119898119899(119909 119910) = 119880

lowast

0(119910) + 119880

lowast

1(119910) (119909 minus 119909

0)

+ sdot sdot sdot + 119880lowast

119898(119910) (119909minus119909

0) (119909minus119909

1) sdot sdot sdot (119909minus119909

119898minus1)

(45)

where for 119897 = 0 1 119898

119880lowast

119897(119910) = 120593

1198970(1199090 119909

119897 1199100) +

119910 minus 1199100

1205931198971

(1199090 119909

119897 1199100 1199101)

+ sdot sdot sdot +119910 minus 119910119899minus1

|

| 120593119897119899

(1199090 119909

119897 1199100 119910

119899)

+119910 minus 119910119899|

| 120593119897119899+1

(1199090 119909

119897 y0 119910

119899 119910)

(46)

then 119877lowast

119898119899(119909 119910) satisfies

119877lowast

119898119899(119909119894 119910) = 119891 (119909

119894 119910) 119894 = 0 1 119898

119877lowast

119898119899(119909 119910119895) = 119877119898119899

(119909 119910119895) 119895 = 0 1 119899

(47)

Remark 15 We delete the proof of the above lemma since itcan be easily generalized from [3]

Now the error estimate Theorem 16 will be put which isalso motivated by [3] We let

119877119898119899

(119909 119910) =119873 (119909 119910)

119863 (119909 119910) 119877

lowast

119898119899(119909 119910) =

119873lowast

(119909 119910)

119863lowast (119909 119910) (48)

Theorem 16 Let a matrix 119891(119909 119910) bemax(119898 + 1 119899 + 1) timescontinuously differentiable on a set119863 = 119886 le 119909 le 119887 119888 le 119910 le 119889

containing the points Λ119898119899

= (119909119897 119910119896) 119897 = 0 1 119898 119896 =

0 1 119899 (119909119897 119910119896) isin R2 then for any point (119909 119910) isin 119863 there

exists a point (120585 120578) isin 119863 such that

119891 (119909 119910) minus 119877119898119899

(119909 119910)

=1

119863 (119909 119910)119863lowast (119909 119910)

times 120596119898+1

(119909)

(119898 + 1)

120597119898+1

120597119909119898+11198641(120585 119910) +

120603119899+1

(119910)

(119899 + 1)

120597119899+1

120597119910119899+11198642(119909 120578)

(49)

where

120596119898+1

(119909) = (119909 minus 1199090) (119909 minus 119909

1) sdot sdot sdot (119909 minus 119909

119898)

120603119899+1

(119910) = (119910 minus 1199100) (119910 minus 119910

1) sdot sdot sdot (119910 minus 119910

119899)

1198641(119909 119910) = 119863 (119909 119910)119863

lowast

(119909 119910) [119891 (119909 119910) minus 119877lowast

119898119899(119909 119910)]

1198642(119909 119910) = 119863 (119909 119910)119863

lowast

(119909 119910) [119877lowast

119898119899(119909 119910) minus 119877

119898119899(119909 119910)]

(50)

8 Journal of Applied Mathematics

Proof Let

1198641(119909 119910) = 119863 (119909 119910)119863

lowast

(119909 119910) [119891 (119909 119910) minus 119877lowast

119898119899(119909 119910)]

1198642(119909 119910) = 119863 (119909 119910)119863

lowast

(119909 119910) [119877lowast

119898119899(119909 119910) minus 119877

119898119899(119909 119910)]

119864 (119909 119910) = 1198641(119909 119910) + 119864

2(119909 119910)

(51)

From Lemma 14 we know

1198641(119909119894 119910) = 0 119894 = 0 1 119898 (52)

which results in

1198641(119909 119910) =

120596119898+1

(119909)

(119898 + 1)

120597119898+1

120597119909119898+11198641(119909 119910)

100381610038161003816100381610038161003816100381610038161003816119909=120585

(53)

where 120596119898+1

(119909) = (119909 minus 1199090)(119909 minus 119909

1) sdot sdot sdot (119909 minus 119909

119898) and 120585 is a

number contained in the interval (119886 119887) which may dependon 119910 Similarly from

1198642(119909 119910119895) = 0 119895 = 0 1 119899 (54)

we can get that

1198642(119909 119910) =

120603119899+1

(119910)

(119899 + 1)

120597119899+1

120597119910119899+11198642(119909 119910)

100381610038161003816100381610038161003816100381610038161003816119910=120578

(55)

where120603119899+1

(119910) = (119910minus1199100)(119910minus119910

1) sdot sdot sdot (119910minus119910

119899) and 120578 is a number

contained in the interval (119888 119889) which may depend on 119909Therefore

119864 (119909 119910) = 1198641(119909 119910) + 119864

2(119909 119910)

=120596119898+1

(119909)

(119898 + 1)

120597119898+1

120597119909119898+11198641(119909 119910)

100381610038161003816100381610038161003816100381610038161003816119909=120585

+120603119899+1

(119910)

(119899 + 1)

120597119899+1

120597119910119899+11198642(119909 119910)

100381610038161003816100381610038161003816100381610038161003816119910=120578

(56)

The proof is thus completed

4 Numerical Examples

Example 17 Let (119909119894 119910119895) and 119884

119894119895 119894 119895 = 0 1 2 be given in

Table 1

11987722

(119909 119910) = 1198800(119910) + 119880

1(119910) 119909 + 119880

2(119910) 119909 (119909 minus 1) (57)

Using Algorithm 7 we can get

1198800(119910) = (

1 0

0 0) +

119910

( 0 00 1

)+

119910 minus 1 |

| (0 0

minus25 minus15)

1198801(119910) = (

0 minus1

0 0) +

119910

(0 12

minus12 0)

+119910 minus 1 |

| ( 0 11 0

)

1198802(119910) = (

0 1

1

20) +

119910

(0 minus35

15 0)

+119910 minus 1 |

| (0 minus1514

minus514 minus57)

(58)

Based on generalized inverse (4) we obtain

11987722

(119909 119910)

= (11988611

11988612

11988621

11988622

) times (701199106

minus 3681199105

+ 8921199104

minus12401199103

+10401199102

minus512119910+128)minus1

=119873 (119909 119910)

119863 (119909 119910)

(59)

where

11988611

= 701199106

minus 3681199105

+ 8921199104

minus 12401199103

+ 10401199102

minus 512119910 + 128

11988612

= 1781199105

119909 minus 8581199104

119909 + 16881199103

119909

minus 17761199091199102

+ 1024119909119910 minus 256119909 minus 381199105

1199092

+ 2621199104

1199092

minus 6401199103

1199092

minus 5121199092

119910

minus 51199106

1199092

+ 51199106

119909 + 8001199092

1199102

+ 1281199092

11988621

= 1081199105

minus 2001199104

+ 2001199103

minus 1121199102

+ 32119910

minus 281199106

+ 601199106

119909 minus 3441199105

119909 + 7361199104

119909 minus 7841199103

119909

+ 4081199091199102

minus 32119909119910 minus 64119909 minus 241199105

1199092

+ 161199104

1199092

minus 961199092

119910 + 101199106

1199092

+ 401199092

1199102

+ 641199092

11988622

= 681199105

minus 1401199104

+ 1601199103

minus 961199102

+ 32119910 minus 141199106

minus 501199106

1199092

+ 2701199105

1199092

minus 6401199104

1199092

+ 8201199103

1199092

minus 5601199092

1199102

+ 1601199092

119910 + 501199106

119909 minus 2701199105

119909

+ 6401199104

119909 minus 8201199103

119909 + 5601199091199102

minus 160119909119910

(60)

FromDefinition 13 we know that 11987722

(119909 119910) is a BGIRINTsince 119877

22(119909119894 119910119895) = 119884119894119895and deg 119873(119909 119910) = 8 deg 119863(119909 119910) =

6 and 119863(119909 119910) | 119873(119909 119910)2

In paper [9] a bivariateThieletypematrix-valued rationalinterpolant 119877

119879(119909 119910) = 119866

00(119910) + (119909 minus 119909

0)11986610

(119910) + sdot sdot sdot + (119909 minus

119909119898minus1

)1198661198990

(119910) (page 73) where for 119897 = 0 1 119899

1198661198970

(119910) = 1198611198970

(1199090 119909

119897 1199100)

+119910 minus 1199100

1198611198971

(1199090 119909

119897 1199100 1199101)

+ sdot sdot sdot +119910 minus 119910119898minus1

|

| 119861119897119898

(1199090 119909

119897 1199100 119910

119898)

(61)

Here 119861119897119896(1199090 119909

119897 1199100 119910

119896) 119897 = 0 1 119899 119896 = 0 1 119898

is defined different from that of this paper (see [9]) We

Journal of Applied Mathematics 9

Table 1

119910119895119909119894

1199090= 0 119909

1= 1 119909

2= 2

1199100= 0 (

1 0

0 0) (

1 minus1

0 0) (

1 0

1 0)

1199101= 1 (

1 0

0 1) (

1 0

minus1 1) (

1 minus1

0 1)

1199102= 2 (

1 0

minus1 0) (

1 1

minus1 0) (

1 1

0 minus1)

Table 2

119884(13 13) 119877NT(13 13) 119877NT minus 119884119865

(minus0856888 05155013 3669296

130000 366929 2600) (

minus0856888 05155010 3669297

130000 366929 2600) 109119890 minus 006

119877119879(13 13) 119877

119879minus 119884119865

(minus098999 014111 547394

129999 547394 2999) 2613697

119884(15 15) 119877NT(15 15) 119877NT minus 119884119865

(minus0989999 01411200 4481689

150000 4481689 300) (

minus0989992 01411203 4481688

150000 4481688 300) 915119890 minus 007

119877119879(15 15) 119877

119879minus 119884119865

(minus099829372 minus005837 547394

149999 547394 31999) 143144

now give another numerical example to compare the twoalgorithms which shows that the method of this paper isbetter than the one of [9]

Example 18 Let

119884 (119909 119910) = (cos (119909 + 119910) sin (119909 + 119910) 119890

119910

119909 119890119910

119909 + 119910) (62)

and we suppose 1199090 1199091 1199092 1199093 1199094 1199095 = 119910

0 1199101 1199102 1199103 1199104 1199105

= 10 12 14 16 18 20 The numerical results 119877NT(119909 119910)

in Step 9 of Algorithm 7 and 119877119879(119909 119910) in [9] are given in

Table 2

Remark 19 InTable 2 from the F-normof119877NTminus119884 and119877119879minus119884

we can see that the error using Newton-Thiele type formulais much less than the one usingThieletype formula

5 Conclusion

In this paper the bivariate generalized inverse Newton-Thieletype matrix interpolation to approximate a matrix function119891(119909 119910) is given and we also give a recursive algorithmaccompanied with some other important conclusions such asdivisibility characterization and some numerical examplesFrom the second example it is easy to find that the approxi-mant using Newton-Thiele type formula is much better thanthe one usingThieletype formula

Acknowledgments

The work is supported by Shanghai Natural Science Foun-dation (10ZR1410900) and by Key Disciplines of ShanghaiMunicipality (S30104)

References

[1] G A Baker and P R Graves-Morris Pade Approximation PartII Addison-Wesley Publishing Company Reading Mass USA1981

[2] I Kh Kuchminska OM Sus and SMVozna ldquoApproximationproperties of two-dimensional continued fractionsrdquo UkrainianMathematical Journal vol 55 pp 36ndash54 2003

[3] J Tan and Y Fang ldquoNewton-Thielersquos rational interpolantsrdquoNumerical Algorithms vol 24 no 1-2 pp 141ndash157 2000

[4] P R Graves-Morris ldquoVector valued rational interpolants IrdquoNumerische Mathematik vol 42 no 3 pp 331ndash348 1983

[5] G Chuan-qing and C Zhibing ldquoMatrix valued rational inter-polants and its error formulardquo Mathematica Numerica Sinicavol 17 pp 73ndash77 1995

[6] G Chuan-qing ldquoMatrix Pade type approximat and directionalmatrix Pade approximat in the inner product spacerdquo Journal ofComputational and Applied Mathematics vol 164-165 pp 365ndash385 2004

[7] C Gu ldquoThiele-type and Lagrange-type generalized inverserational interpolation for rectangular complexmatricesrdquo LinearAlgebra and Its Applications vol 295 no 1ndash3 pp 7ndash30 1999

[8] C Gu ldquoA practical two-dimensional Thiele-type matrix Padeapproximationrdquo IEEE Transactions on Automatic Control vol48 no 12 pp 2259ndash2263 2003

10 Journal of Applied Mathematics

[9] C Gu ldquoBivariate Thiele-type matrix-valued rational inter-polantsrdquo Journal of Computational and Applied Mathematicsvol 80 no 1 pp 71ndash82 1997

[10] N K Bose and S Basu ldquoTwo-dimensional matrix Pade approx-imants existence nonuniqueness and recursive computationrdquoIEEE Transactions on Automatic Control vol 25 no 3 pp 509ndash514 1980

[11] A Cuyt and B Verdonk ldquoMultivariate reciprocal differencesfor branched Thiele continued fraction expansionsrdquo Journal ofComputational and Applied Mathematics vol 21 no 2 pp 145ndash160 1988

[12] W Siemaszko ldquoThiele-tyoe branched continued fractions fortwo-variable functionsrdquo Journal of Computational and AppliedMathematics vol 9 no 2 pp 137ndash153 1983

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Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Mathematical PhysicsAdvances in

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Operations ResearchAdvances in

Journal of

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Algebra

Discrete Dynamics in Nature and Society

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Journal of Applied Mathematics 3

We assume that for all 119897 119909119897

= 119909119897minus1

and

120593119897119896minus1

(1199090 119909

119897 1199100 119910

119896minus2 119910119896)

= 120593119897119896minus1

(1199090 119909

119897 1199100 119910

119896minus2 119910119896minus1

) forall119897 119896

(10)

From (9) matrix-valued Newton-Thiele type continuedfractions for two-variable function can be constructed asfollows

119877119898119899

(119909 119910)

= 1198800(119910) + 119880

1(119910) (119909 minus 119909

0) + sdot sdot sdot + 119880

119898(119910) (119909 minus 119909

0)

times (119909 minus 1199091) sdot sdot sdot (119909 minus 119909

119898minus1)

(11)

where for 119897 = 0 1 119898

119880119897(119910) = 120593

1198970(1199090 119909

119897 1199100) +

119910 minus 1199100

1205931198971

(1199090 119909

119897 1199100 1199101)

+ sdot sdot sdot +119910 minus 119910119899minus1

|

| 120593119897119899

(1199090 119909

119897 1199100 119910

119899)

(12)

Remark 3 Because of the construction119863(119909 119910) in 119877119898119899

(119909 119910)

mentioned as (6) is actually119863(119910) and each119880119897(119910) depends on

119899

We can also construct the antithetical form of bivariatematrix continued fraction in light of Definition 4

Definition 4 By means of generalized inverse (4) we definebivariate antithetical Newton-Thiele type matrix blendingdifferences

12059500

(119909119897 119910119896) = 119884119897119896

12059501

(119909119897 1199100 1199101) =

(12059500

(119909119897 1199101) minus 12059500

(119909119897 1199100))

(1199101minus 1199100)

1205950119896

(119909119897 1199100 119910

119896)

= (1205950119896minus1

(119909119897 1199100 119910

119896minus2 119910119896)

minus1205950119896minus1

(119909119897 1199100 119910

119896minus2 119910119896minus1

))

times (119910119896minus 119910119896minus1

)minus1

1205951119896

(1199090 1199091 1199100 119910

119896)

=(1199091minus 1199090)

1205950119896

(1199091 1199100 119910

119896) minus 1205950119896

(1199090 1199100 119910

119896)

1205951198970

(1199090 119909

119897 119910119896)

= (119909119897minus 119909119897minus1

)

times (120595119897minus10

(1199090 119909

119897minus2 119909119897 119910119896)

minus120595119897minus10

(1199090 119909

119897minus1 119910119896))minus1

120595119897119896

(1199090 1199091 119909

119897 1199100 119910

119896)

= (119909119897minus 119909119897minus1

)

times (120595119897minus1119896

(1199090 119909

119897minus2 119909119897 1199100 119910

119896)

minus120595119897minus1119896

(1199090 119909

119897minus1 1199100 119910

119896))minus1

(13)

From (13) we can get the antithetical formula for two-variable function

119898119899

(119909 119910) = 1198810(119909) + 119881

1(119909) (119910 minus 119910

0)

+ sdot sdot sdot + 119881119899(119909) (119910 minus 119910

0) (119910 minus 119910

1)

sdot sdot sdot (119910 minus 119910119899minus1

)

(14)

where for 119896 = 0 1 119899

119881119896(119909) = 120595

0119896(1199090 1199100 119910

119896)

+119909 minus 1199090

1205951119896

(1199090 1199091 1199100 119910

119896)

+ sdot sdot sdot +119909 minus 119909119898minus1

|

| 120595119898119896

(1199090 119909

119898 1199100 119910

119896)

(15)

Now we will give two theorems about 119877119898119899

(119909 119910) as in (11)and (12) and

119898119899(119909 119910) as in (14) and (15) First of all some

notations need to be definedIn (12) for 119897 = 0 1 119898 let

119880(119904)

119897(119910) = 120593

119897119904(1199090 1199091 119909

119897 1199100 119910

119904)

+119910 minus 119910119904

119880(119904+1)

119897(119910)

119904 = 0 1 119899 minus 1(16)

where

119880(119899)

119897(119910) = 120593

119897119899(1199090 1199091 119909

119897 1199100 119910

119899)

119880(0)

119897(119910) = 119880

119897(119910)

(17)

Similarly in (15) for 119896 = 0 1 119899 let

119881(119904)

119896(119909) = 120595

119904119896(1199090 1199091 119909

119904 1199100 119910

119896)

+119909 minus 119909119904

119881(119904+1)

119896(119909)

119904 = 0 1 119898 minus 1(18)

where

119881(119898)

119896(119909) = 120595

119898119896(1199090 1199091 119909

119898 1199100 119910

119896)

119881(0)

119896(119909) = 119881

119896(119909)

(19)

Theorem 5 Let

(i) 120593119897119896(1199090 119909

119897 1199100 119910

119896) 119897 = 0 1 119898 119896 = 0

1 119899 exist and be nonzero (except for 1205931198970(1199090

119909119897 1199100))

4 Journal of Applied Mathematics

(ii) 119880(119904)

119897(119910) = 120593

119897119904(1199090 1199091 119909

119897 1199100 119910

119904) + ((119910 minus

119910119904)119880(119904+1)

119897(119910)) satisfy119880

(119904+1)

119897(119910119904) = 0 119904 = 0 1 119899minus1

then 119877119898119899

(119909 119910) as in (11) and (12) exists such that119877119898119899

(119909119897 119910119896) = 119884119897119896 (119909119897 119910119896) isin Λm119899

Proof For simpleness let

120593119897119896

(1199090 119909

119897 1199100 119910

119896) = 120593119897119896

(20)

if the conditions hold thus for 119897 = 0 1 119898 119896 = 0 1 119899

119880119897(119910119896) = 119880

(0)

119897(119910119896)

= 1205931198970

+119910119896minus 1199100

1205931198971

+ sdot sdot sdot +119910119896minus 119910119896minus1

|

| 120593119897119899

+119910119896minus 119910119896|

| 119880(119896+1)

119897(119910119896)

(21)

(119910119896minus119910119896)119880(119896+1)

119897(119910119896) = 0 since119880

(119896+1)

119897(119910119896) = 0 then we get

119880119897(119910119896)

= 1205931198970

(1199090 119909

119897 1199100) +

119910119896minus 1199100

1205931198971

(1199090 119909

119897 1199100 1199101)+ sdot sdot sdot +

119910119896minus 119910119896minus2

|

| 120593119897119896minus1

(1199090 119909

119897 1199100 119910

119896minus1)

+119910119896minus 119910119896minus1

|

| 120593119897119896

(1199090 119909

119897 1199100 119910

119896)

= 1205931198970

(1199090 119909

119897 1199100) +

119910119896minus 1199100

1205931198971

(1199090 119909

119897 1199100 1199101)+ sdot sdot sdot

+119910119896minus 119910119896minus2

|

| 120593119897119896minus1

(1199090 119909

119897 1199100 119910

119896minus1)

+119910119896minus 119910119896minus1

|

| (119910119896minus 119910119896minus1

) times (120593119897119896minus1

(1199090 119909

119897 1199100 119910

119896minus2 119910119896) minus 120593119897119896minus1

(1199090 119909

119897 1199100 119910

119896minus1))minus1

= 1205931198970

(1199090 119909

119897 1199100) +

119910119896minus 1199100

1205931198971

(1199090 119909

119897 1199100 1199101)+ sdot sdot sdot +

119910119896minus 119910119896minus2

|

| 120593119897119896minus1

(1199090 119909

119897 1199100 119910

119896minus2 119910119896)

= sdot sdot sdot = 1205931198970

(1199090 119909

119897 1199100) +

119910119896minus 1199100

1205931198971

(1199090 119909

119897 1199100 119910119896)

= 1205931198970

(1199090 119909

119897 119910119896)

(22)

We can use (9) (11) and (22) to find that119877119898119899

(119909119897 119910119896)

= 1198800(119910119896) + 1198801(119910119896) (119909119897minus 1199090)

+ sdot sdot sdot + 119880119897(119910119896) (119909119897minus 1199090) (119909119897minus 1199091) sdot sdot sdot (119909

119897minus 119909119897minus1

)

= 119884 (119909119897 119910119896)

(23)

We can similarly prove the following result

Theorem 6 Let(i) 120595119897119896(1199090 119909

119897 1199100 119910

119896) 119897 = 0 1 119898 119896 = 0

1 119899 exist and be nonzero (except for 1205950119896

(1199090

1199100 119910

119896))

(ii) 119881(119904)

119896(119909) = 120595

119904119896(1199090 1199091 119909

119904 1199100 119910

119896) + ((119909 minus

119909119904)119881(119904+1)

119896(119909)) satisfy119881(119904+1)

119896(119909119904) = 0 119904 = 0 1 119898minus1

then 119898119899

(119909 119910) as in (14) and (15) exists such that119898119899

(119909119897 119910119896) = 119884119897119896 (119909119897 119910119896) isin Λ

119898119899

Now we give the Newton-Thiele Matrix InterpolationAlgorithm (NTMIA)

Algorithm 7 (NTMIA) Input (119909119894 119910119895) 119884119894119895 (119894 = 0 1

119898 119895 = 0 1 119899)

Output 119898(119909 119910)

Step 1 For all (119909119894 119910119895) isin Λ

119898119899 let 120593(119909

119894 119910119895) = 119884119894119895

Step 2 For 119895 = 0 1 119899 119901 = 1 2 119898 119894 = 119901 119901 +

1 119898 compute

1205931199010

(1199090 119909

119901minus1 119909119894 119910119895)

=

120593119901minus10

(1199090 119909

119901minus2 119909119894 119910119895)minus120593119901minus10

(1199090 119909

119901minus1 119910119895)

119909119894minus119909119901minus1

(24)

Journal of Applied Mathematics 5

Step 3 For 119894 = 0 1 119898 119902 = 1 2 119899 119895 = 119902 119902 + 1 119899compute

120593119894119902

(1199090 119909

119894 1199100 119910

119902minus1 119910119895)

= (119910119895minus 119910119902minus1

) times (120593119894119902minus1

(1199090 119909

119894 1199100 119910

119902minus2 119910119895)

minus120593119894119902minus1

(1199090 119909

119894 1199100 119910

119902minus1))minus1

= ((119910119895minus 119910119902minus1

) (120593119894119902minus1

(1199090 119909

119894 1199100 119910

119902minus2 119910119895)

minus120593119894119902minus1

(1199090 119909

119894 1199100 119910

119902minus1)))

times (10038171003817100381710038171003817120593 (1199090 119909

119894 1199100 119910

119902minus2 119910119895)

minus120593 (1199090 119909

119894 1199100 119910

119902minus1)10038171003817100381710038171003817

2

)

minus1

(25)

Step 4 For 119894 = 0 1 119898 119895 = 0 1 119899 compute 119886119894119895

=

120593119894119895(1199090 119909

119894 1199100 119910

119895)

Step 5 For 119894 = 0 1 119898 119895 = 1 119899 judge if 119886119894119895

= 0 if yesgo to Step 6 otherwise exist and show ldquothe algorithm is notvalidrdquo

Step 6 For 119894 = 0 1 119898 119895 = 119899 minus 2 119899 minus 3 0

119876119894119899

= 1 119875119894119899

= 119886119894119899

119876119894119899minus1

=1003817100381710038171003817119886119894119899

1003817100381710038171003817

2

119875119894119899minus1

= 119886119894119899minus1

119876119894119899minus1

+ (119910 minus 119910119899minus1

) 119875119894119899

119876119894119895

=10038171003817100381710038171003817119886119894119895+1

10038171003817100381710038171003817

2

119876119894119895+1

+ 2 (119910 minus 119910119895+1

) tr (119886119894119895+1

119875119894119895+2

lowast

)

+ (119910 minus 119910119895+1

)2

119876119894119895+2

119875119894119895

= 119886119894119895119876119894119895

+ (119910 minus 119910119895) 119875119894119895+1

(26)

Step 7 For 119894 = 0 1 119898 119895 = 119899 119899 minus 1 1 let 119861119894119895(119910) =

119875119894119895(119910)119876

119894119895(119910) if 119861

119894119895(119910119895minus1

) = 0 then go to Step 8 otherwiseexist and show ldquothe algorithm is not validrdquo

Step 8 For 119894 = 0 1 119898 let 119861119894(119910) = 119861

1198940(119910) = 119875

1198940(119910)

1198761198940(119910)

Step 9 For 119896 = 1 119898 let 0(119909 119910) = 119861

0(119910) then compute

119896(119909 119910) =

119896minus1(119909 119910) + sdot sdot sdot + (119909 minus 119909

0) sdot sdot sdot (119909 minus 119909

119896minus1)119861119896(119910)

3 Some Properties

Lemma 8 (see [5]) Define 1198800(119910) = 120593

00(1199090 1199100) + (119910 minus

1199100)12059301

(1199090 1199100 1199101) + sdot sdot sdot + (119910minus119910

119899minus1)1205930119899

(1199090 1199100 119910

119899) if we

use generalized inverse by a tail-to-head rationalization thena polynomial matrix 119864

0(119910) and a real polynomial 119865

0(119910) exist

such that(i) 1198800(119910) = 119864

0(119910)1198650(119910) 1198650(119910) ge 0

(ii) 1198650(119910) | 119864

0(119910)2ldquo|rdquo means the sign of divisibility

Definition 9 119877(119909 119910) = 119873(119909 119910)119863(119909 119910) is defined to be oftype [119897119905] if deg119873

119894119895 le 119897 for 1 le 119894 le 119906 1 le 119895 le 119907 deg119873

119894119895 = 119897

for some (119894 119895) and deg119863 = 119905 where119873(119909 119910) = (119873119894119895(119909 119910)) isin

119862119906times119907

Lemma 10 (see [5]) Let1198800(119910) be defined as in Lemma 8 and

1198800(119910) = 119864

0(119910)1198650(119910) if 119899 is even119880

0(119910) is of type [119899119899] if 119899 is

odd 1198800(119910) is of type [119899119899 minus 1]

Theorem 11 Let 120593119897119896

isin 119862119906times119907 (119909

119897 119910119896) isin Λ

119898119899 and 119909 119910 isin

119877 Define 119880119897(119910) 119900119891 119877

119898119899(119909 119910) as in (7) by a tail-to-head

rationalization using generalized inverse and suppose everyintermediate denominator be nonzero in the operation thena square polynomial matrix 119873(119909 119910) and a real polynomial119863(119909 119910) exist such that

(i) 119877119898119899

(119909 119910) = 119873(119909 119910)119863(119909 119910)

(ii) 119863(119909 119910) ge 0

(iii) 119863(119909 119910) | 119873(119909 119910)2

Proof Consider the following algorithm for the constructionof 119873(119909 119910) 119863(119909 119910) and 119877

119898119899(119909 119910)

Initialization let 1198630(119909 119910) = 1 and

1198730(119909 119910) = 119880

0(119910) = 120593

00+

119910 minus 1199100

12059301

+ sdot sdot sdot +119910 minus 119910119899minus1

|

| 1205930119899

(27)

By Lemma 8 a polynomial matrix 1198640(119910) and a real polyno-

mial 1198650(119910) exist such that

(a1) 1198730(119909 119910) = 119864

0(119910)1198650(119910)

(a2) 1198650(119910) ge 0

(a3) 1198650(119910) | 119864

0(119910)2

Recursion for 119895 = 1 2 119898 let

119878(119895)

(119909 119910) = 119878(119895minus1)

(119909 119910) + 119880119895(119910) (119909 minus 119909

0) sdot sdot sdot (119909 minus 119909

119895minus1)

(28)

with 119880119895(119910) = 119864

119895(119910)119865119895(119910) 119865119895(119910) | 119864

119895(119910)2 having the

representation

(b1) 119878(119895)

(119909 119910) = 119873119895(119909 119910)119863

119895(119909 119910)

(b2) 119863119895(119909 119910) ge 0

(b3) 119863119895(119909 119910) | 119873

119895(119909 119910)

2

where 119863119895(119909 119910) is a real polynomial

6 Journal of Applied Mathematics

Then we can get

119878(119895+1)

(119909 119910) = 119878(119895)

(119909 119910) + 119880119895+1

(119909 minus 1199090) sdot sdot sdot (119909 minus 119909

119895)

=119873119895(119909 119910)

119863119895(119909 119910)

+119864119895+1

(119910)

119865119895+1

(119910)(119909 minus 119909

0) sdot sdot sdot (119909 minus 119909

119895)

= (119873119895(119909 119910) 119865

119895+1(119910) + 119863

119895(119909 119910) 119864

119895+1(119910)

times (119909 minus 1199090) sdot sdot sdot (119909 minus 119909

119895))

times (119863119895(119909 119910) 119865

119895+1(119910))minus1

=119873119895+1

(119909 119910)

119863119895+1

(119909 119910)

(29)

where

119873119895+1

(119909 119910) = 119873119895(119909 119910) 119865

119895+1(119910)

+ 119863119895(119909 119910) 119864

119895+1(119910) (119909 minus 119909

0)

sdot sdot sdot (119909 minus 119909119895)

119863119895+1

(119909 119910) = 119863119895(119909 119910) 119865

119895+1(119910)

(30)

Obviously

10038171003817100381710038171003817119873119895+1

10038171003817100381710038171003817

2

=10038171003817100381710038171003817119873119895(119909 119910)

10038171003817100381710038171003817

2

1198652

(119895+1)(119910) + 119863

2

119895(119909 119910)

times10038171003817100381710038171003817119864(119895+1)

(119910)10038171003817100381710038171003817

2

(119909 minus 1199090)2

sdot sdot sdot (119909 minus 119909119895)2

+ 119873119895(119909 119910) sdot 119864

119895+1(119910) 119865119895+1

(119910)119863119895(119909 119910)

times (119909 minus 1199090) sdot sdot sdot (119909 minus 119909

119895)

+ 119873119895(119909 119910) sdot 119864

119895+1(119910) 119865119895+1

(119910)119863119895(119909 119910)

times (119909 minus 1199090) sdot sdot sdot (119909 minus 119909

119895)

(31)

since

119863119895(119909 119910) |

10038171003817100381710038171003817119873119895(119909 119910)

10038171003817100381710038171003817

2

119865119895+1

(119910) |10038171003817100381710038171003817119864119895+1

(119910)10038171003817100381710038171003817

2

119863119895+1

(119909 119910) |10038171003817100381710038171003817119873119895+1

(119909 119910)10038171003817100381710038171003817

2

(32)

Termination 119878(119898)(119909 119910) = 119877119898119899

(119909 119910)

Theorem 12 (Characterization) Let 119877119898119899

(119909 119910) = 119873(119909 119910)

119863(119909 119910) be expressed as

119877 = 119877119898119899

(119909 119910) = 1198800(119910) + 119880

1(119910) (119909 minus 119909

0)

+ sdot sdot sdot + 119880119898

(119910) (119909 minus 1199090) sdot sdot sdot (119909 minus 119909

119898minus1)

(33)

where 119880119897(119910) = 119864

119894119865119894 as in (12) we assume 119865

119894 119865119895have no

common factor 119894 119895 = 0 1 119898 119894 = 119895 then

(i) if n is even 119877 is of type [119898119899 + 119898 + 119899119898119899 + 119899](ii) if n is odd 119877 is of type [119898119899 + 119899119898119899 + 119899 minus 119898 minus 1]

Proof The proof is recursive Let 119899 be evenFor 119898 = 0 119877

0119899(119909 119910) = 119880

0(119910) = 119864

01198650 by Lemma 10 we

find that it is of type [119899119899]For 119898 = 1 because deg119864

0 = deg119865

0 = 119899 deg119864

1 =

deg1198651 = 119899

1198771119899

(119909 119910) =1198731

1198631

= 1198800(119910) + 119880

1(119910) (119909 minus 119909

0)

=1198640

1198650

+1198641

1198651

(119909 minus 1199090)

=11986401198651+ 11986411198650(119909 minus 119909

0)

11986501198651

(34)

is of type [2119899 + 12119899]For 119898 = 2 from the formula above we know that

deg1198731 = 2119899 + 1 deg119863

1 = 2119899 deg119864

2 = 119899 and deg119865

2 =

119899 one has1198772119899

(119909 119910) =1198732

1198632

=1198731

1198631

+1198642

1198652

(119909 minus 1199090) (119909 minus 119909

1)

=11987311198652+ 11986421198631(119909 minus 119909

0) (119909 minus 119909

1)

11986311198652

(35)

It is easy to find that 1198772119899

(119909 119910) is of type [3119899 + 23119899]For 119898 = 119896 let 119877

119896119899(119909 119910) = 119873

119896119863119896be of type [(119896 + 1)119899 +

119896(119896 + 1)119899]For 119898 = 119896 + 1 we consider that

119877119896+1119899

(119909 119910) =119873119896

119863119896

+119864119896+1

119865119896+1

(119909 minus 1199090) sdot sdot sdot (119909 minus 119909

119896)

=119873119896119865119896+1

+ 119864119896+1

119863119896(119909 minus 119909

0) sdot sdot sdot (119909 minus 119909

119896)

119863119896119865119896

=119873119896+1

119863119896+1

(36)

wheredeg 119873

119896+1 = (119896 + 1) (119899 + 1) + 119899

= (119896 + 2) 119899 + (119896 + 1)

deg 119863119896+1

= (119896 + 1) 119899 + 119899 = (119896 + 2) 119899

(37)

Therefore when 119899 is even 119877 is of type [119898119899 + 119898 + 119899119898119899 + 119899]Similarly if 119899 is odd for 119898 = 0 119877

0119899(119909 119910) = 119880

0(119910) =

11986401198650 by Lemma 10 we find that it is of type [119899119899 minus 1]

For 119898 = 1 because deg1198640 = deg119864

1 = 119899 deg119865

0 =

deg1198651 = 119899 minus 1

1198771119899

(119909 119910) =1198731

1198631

= 1198800(119910) + 119880

1(119910) (119909 minus 119909

0)

=1198640

1198650

+1198641

1198651

(119909 minus 1199090)

=11986401198651+ 11986411198650(119909 minus 119909

0)

11986501198651

(38)

is of type [21198992119899 minus 2]

Journal of Applied Mathematics 7

For119898 = 2 from the formula above we know that deg1198731

= 2119899 deg1198631 = 2119899 minus 2 deg119864

2 = 119899 and deg119865

2 = 119899 minus 1

one has

1198772119899

(119909 119910) =1198732

1198632

=1198731

1198631

+1198642

1198652

(119909 minus 1199090) (119909 minus 119909

1)

=11987311198652+ 11986421198631(119909 minus 119909

0) (119909 minus 119909

1)

11986311198652

(39)

We get that 1198772119899

(119909 119910) is of type [31198993119899 minus 3]For 119898 = 119896 let

119877119896119899

(119909 119910) =119873119896

119863119896

(40)

be of type [(119896 + 1)119899(119896 + 1)(119899 minus 1)]For 119898 = 119896 + 1 we consider that

119877119896+1119899

(119909 119910) =119873119896

119863119896

+119864119896+1

119865119896+1

(119909 minus 1199090) sdot sdot sdot (119909 minus 119909

119896)

=119873119896119865119896+1

+ 119864119896+1

119863119896(119909 minus 119909

0) sdot sdot sdot (119909 minus 119909

119896)

119863119896119865119896

=119873119896+1

119863119896+1

(41)

where

deg 119873119896+1

= (119896 + 1) (119899 minus 1) + 119896 + 1 = (119896 + 2) 119899

deg 119863119896+1

= (119896 + 1) (119899 minus 1) + (119899 minus 1) = (119896 + 2) (119899 minus 1)

(42)

Therefore when 119899 is odd119877 is of type [(119898+1)119899(119898+1)(119899minus

1)]

Definition 13 A matrix-valued Newton-Thiele type ratio-nal fraction 119877

119898119899(119909 119910) = 119873(119909 119910)119863(119909 119910) is defined to

be a bivariate generalized inverse and rational interpolant(BGIRINT) on the rectangular grid Λ

119898119899if

(i) 119877119898119899

(119909119897 119910119896) = 119884119897119896 (119909119897 119910119896) isin Λ

119898119899

(ii) 119863(119909119897 119910119896) = 0 (119909

119897 119910119896) isin Λ

119898119899

(iii) (a) if 119899 is even

deg 119873 (119909 119910) = 119898119899+ 119898+119899 deg 119863 (119909 119910) = 119898119899+119899

(43)

(b) if n is odd

deg 119873 (119909 119910) = 119898119899+119899 deg 119863 (119909 119910) = 119898119899+119899minus119898minus1

(44)

(iv) 119863(119909 119910) | 119873(119909 119910)2

(v) 119863(119909 119910) is real and 119863(119909 119910) ge 0

Now let us turn to the error estimate of the BGIRINTFirstly we give Lemma 14

Lemma 14 Let

119877lowast

119898119899(119909 119910) = 119880

lowast

0(119910) + 119880

lowast

1(119910) (119909 minus 119909

0)

+ sdot sdot sdot + 119880lowast

119898(119910) (119909minus119909

0) (119909minus119909

1) sdot sdot sdot (119909minus119909

119898minus1)

(45)

where for 119897 = 0 1 119898

119880lowast

119897(119910) = 120593

1198970(1199090 119909

119897 1199100) +

119910 minus 1199100

1205931198971

(1199090 119909

119897 1199100 1199101)

+ sdot sdot sdot +119910 minus 119910119899minus1

|

| 120593119897119899

(1199090 119909

119897 1199100 119910

119899)

+119910 minus 119910119899|

| 120593119897119899+1

(1199090 119909

119897 y0 119910

119899 119910)

(46)

then 119877lowast

119898119899(119909 119910) satisfies

119877lowast

119898119899(119909119894 119910) = 119891 (119909

119894 119910) 119894 = 0 1 119898

119877lowast

119898119899(119909 119910119895) = 119877119898119899

(119909 119910119895) 119895 = 0 1 119899

(47)

Remark 15 We delete the proof of the above lemma since itcan be easily generalized from [3]

Now the error estimate Theorem 16 will be put which isalso motivated by [3] We let

119877119898119899

(119909 119910) =119873 (119909 119910)

119863 (119909 119910) 119877

lowast

119898119899(119909 119910) =

119873lowast

(119909 119910)

119863lowast (119909 119910) (48)

Theorem 16 Let a matrix 119891(119909 119910) bemax(119898 + 1 119899 + 1) timescontinuously differentiable on a set119863 = 119886 le 119909 le 119887 119888 le 119910 le 119889

containing the points Λ119898119899

= (119909119897 119910119896) 119897 = 0 1 119898 119896 =

0 1 119899 (119909119897 119910119896) isin R2 then for any point (119909 119910) isin 119863 there

exists a point (120585 120578) isin 119863 such that

119891 (119909 119910) minus 119877119898119899

(119909 119910)

=1

119863 (119909 119910)119863lowast (119909 119910)

times 120596119898+1

(119909)

(119898 + 1)

120597119898+1

120597119909119898+11198641(120585 119910) +

120603119899+1

(119910)

(119899 + 1)

120597119899+1

120597119910119899+11198642(119909 120578)

(49)

where

120596119898+1

(119909) = (119909 minus 1199090) (119909 minus 119909

1) sdot sdot sdot (119909 minus 119909

119898)

120603119899+1

(119910) = (119910 minus 1199100) (119910 minus 119910

1) sdot sdot sdot (119910 minus 119910

119899)

1198641(119909 119910) = 119863 (119909 119910)119863

lowast

(119909 119910) [119891 (119909 119910) minus 119877lowast

119898119899(119909 119910)]

1198642(119909 119910) = 119863 (119909 119910)119863

lowast

(119909 119910) [119877lowast

119898119899(119909 119910) minus 119877

119898119899(119909 119910)]

(50)

8 Journal of Applied Mathematics

Proof Let

1198641(119909 119910) = 119863 (119909 119910)119863

lowast

(119909 119910) [119891 (119909 119910) minus 119877lowast

119898119899(119909 119910)]

1198642(119909 119910) = 119863 (119909 119910)119863

lowast

(119909 119910) [119877lowast

119898119899(119909 119910) minus 119877

119898119899(119909 119910)]

119864 (119909 119910) = 1198641(119909 119910) + 119864

2(119909 119910)

(51)

From Lemma 14 we know

1198641(119909119894 119910) = 0 119894 = 0 1 119898 (52)

which results in

1198641(119909 119910) =

120596119898+1

(119909)

(119898 + 1)

120597119898+1

120597119909119898+11198641(119909 119910)

100381610038161003816100381610038161003816100381610038161003816119909=120585

(53)

where 120596119898+1

(119909) = (119909 minus 1199090)(119909 minus 119909

1) sdot sdot sdot (119909 minus 119909

119898) and 120585 is a

number contained in the interval (119886 119887) which may dependon 119910 Similarly from

1198642(119909 119910119895) = 0 119895 = 0 1 119899 (54)

we can get that

1198642(119909 119910) =

120603119899+1

(119910)

(119899 + 1)

120597119899+1

120597119910119899+11198642(119909 119910)

100381610038161003816100381610038161003816100381610038161003816119910=120578

(55)

where120603119899+1

(119910) = (119910minus1199100)(119910minus119910

1) sdot sdot sdot (119910minus119910

119899) and 120578 is a number

contained in the interval (119888 119889) which may depend on 119909Therefore

119864 (119909 119910) = 1198641(119909 119910) + 119864

2(119909 119910)

=120596119898+1

(119909)

(119898 + 1)

120597119898+1

120597119909119898+11198641(119909 119910)

100381610038161003816100381610038161003816100381610038161003816119909=120585

+120603119899+1

(119910)

(119899 + 1)

120597119899+1

120597119910119899+11198642(119909 119910)

100381610038161003816100381610038161003816100381610038161003816119910=120578

(56)

The proof is thus completed

4 Numerical Examples

Example 17 Let (119909119894 119910119895) and 119884

119894119895 119894 119895 = 0 1 2 be given in

Table 1

11987722

(119909 119910) = 1198800(119910) + 119880

1(119910) 119909 + 119880

2(119910) 119909 (119909 minus 1) (57)

Using Algorithm 7 we can get

1198800(119910) = (

1 0

0 0) +

119910

( 0 00 1

)+

119910 minus 1 |

| (0 0

minus25 minus15)

1198801(119910) = (

0 minus1

0 0) +

119910

(0 12

minus12 0)

+119910 minus 1 |

| ( 0 11 0

)

1198802(119910) = (

0 1

1

20) +

119910

(0 minus35

15 0)

+119910 minus 1 |

| (0 minus1514

minus514 minus57)

(58)

Based on generalized inverse (4) we obtain

11987722

(119909 119910)

= (11988611

11988612

11988621

11988622

) times (701199106

minus 3681199105

+ 8921199104

minus12401199103

+10401199102

minus512119910+128)minus1

=119873 (119909 119910)

119863 (119909 119910)

(59)

where

11988611

= 701199106

minus 3681199105

+ 8921199104

minus 12401199103

+ 10401199102

minus 512119910 + 128

11988612

= 1781199105

119909 minus 8581199104

119909 + 16881199103

119909

minus 17761199091199102

+ 1024119909119910 minus 256119909 minus 381199105

1199092

+ 2621199104

1199092

minus 6401199103

1199092

minus 5121199092

119910

minus 51199106

1199092

+ 51199106

119909 + 8001199092

1199102

+ 1281199092

11988621

= 1081199105

minus 2001199104

+ 2001199103

minus 1121199102

+ 32119910

minus 281199106

+ 601199106

119909 minus 3441199105

119909 + 7361199104

119909 minus 7841199103

119909

+ 4081199091199102

minus 32119909119910 minus 64119909 minus 241199105

1199092

+ 161199104

1199092

minus 961199092

119910 + 101199106

1199092

+ 401199092

1199102

+ 641199092

11988622

= 681199105

minus 1401199104

+ 1601199103

minus 961199102

+ 32119910 minus 141199106

minus 501199106

1199092

+ 2701199105

1199092

minus 6401199104

1199092

+ 8201199103

1199092

minus 5601199092

1199102

+ 1601199092

119910 + 501199106

119909 minus 2701199105

119909

+ 6401199104

119909 minus 8201199103

119909 + 5601199091199102

minus 160119909119910

(60)

FromDefinition 13 we know that 11987722

(119909 119910) is a BGIRINTsince 119877

22(119909119894 119910119895) = 119884119894119895and deg 119873(119909 119910) = 8 deg 119863(119909 119910) =

6 and 119863(119909 119910) | 119873(119909 119910)2

In paper [9] a bivariateThieletypematrix-valued rationalinterpolant 119877

119879(119909 119910) = 119866

00(119910) + (119909 minus 119909

0)11986610

(119910) + sdot sdot sdot + (119909 minus

119909119898minus1

)1198661198990

(119910) (page 73) where for 119897 = 0 1 119899

1198661198970

(119910) = 1198611198970

(1199090 119909

119897 1199100)

+119910 minus 1199100

1198611198971

(1199090 119909

119897 1199100 1199101)

+ sdot sdot sdot +119910 minus 119910119898minus1

|

| 119861119897119898

(1199090 119909

119897 1199100 119910

119898)

(61)

Here 119861119897119896(1199090 119909

119897 1199100 119910

119896) 119897 = 0 1 119899 119896 = 0 1 119898

is defined different from that of this paper (see [9]) We

Journal of Applied Mathematics 9

Table 1

119910119895119909119894

1199090= 0 119909

1= 1 119909

2= 2

1199100= 0 (

1 0

0 0) (

1 minus1

0 0) (

1 0

1 0)

1199101= 1 (

1 0

0 1) (

1 0

minus1 1) (

1 minus1

0 1)

1199102= 2 (

1 0

minus1 0) (

1 1

minus1 0) (

1 1

0 minus1)

Table 2

119884(13 13) 119877NT(13 13) 119877NT minus 119884119865

(minus0856888 05155013 3669296

130000 366929 2600) (

minus0856888 05155010 3669297

130000 366929 2600) 109119890 minus 006

119877119879(13 13) 119877

119879minus 119884119865

(minus098999 014111 547394

129999 547394 2999) 2613697

119884(15 15) 119877NT(15 15) 119877NT minus 119884119865

(minus0989999 01411200 4481689

150000 4481689 300) (

minus0989992 01411203 4481688

150000 4481688 300) 915119890 minus 007

119877119879(15 15) 119877

119879minus 119884119865

(minus099829372 minus005837 547394

149999 547394 31999) 143144

now give another numerical example to compare the twoalgorithms which shows that the method of this paper isbetter than the one of [9]

Example 18 Let

119884 (119909 119910) = (cos (119909 + 119910) sin (119909 + 119910) 119890

119910

119909 119890119910

119909 + 119910) (62)

and we suppose 1199090 1199091 1199092 1199093 1199094 1199095 = 119910

0 1199101 1199102 1199103 1199104 1199105

= 10 12 14 16 18 20 The numerical results 119877NT(119909 119910)

in Step 9 of Algorithm 7 and 119877119879(119909 119910) in [9] are given in

Table 2

Remark 19 InTable 2 from the F-normof119877NTminus119884 and119877119879minus119884

we can see that the error using Newton-Thiele type formulais much less than the one usingThieletype formula

5 Conclusion

In this paper the bivariate generalized inverse Newton-Thieletype matrix interpolation to approximate a matrix function119891(119909 119910) is given and we also give a recursive algorithmaccompanied with some other important conclusions such asdivisibility characterization and some numerical examplesFrom the second example it is easy to find that the approxi-mant using Newton-Thiele type formula is much better thanthe one usingThieletype formula

Acknowledgments

The work is supported by Shanghai Natural Science Foun-dation (10ZR1410900) and by Key Disciplines of ShanghaiMunicipality (S30104)

References

[1] G A Baker and P R Graves-Morris Pade Approximation PartII Addison-Wesley Publishing Company Reading Mass USA1981

[2] I Kh Kuchminska OM Sus and SMVozna ldquoApproximationproperties of two-dimensional continued fractionsrdquo UkrainianMathematical Journal vol 55 pp 36ndash54 2003

[3] J Tan and Y Fang ldquoNewton-Thielersquos rational interpolantsrdquoNumerical Algorithms vol 24 no 1-2 pp 141ndash157 2000

[4] P R Graves-Morris ldquoVector valued rational interpolants IrdquoNumerische Mathematik vol 42 no 3 pp 331ndash348 1983

[5] G Chuan-qing and C Zhibing ldquoMatrix valued rational inter-polants and its error formulardquo Mathematica Numerica Sinicavol 17 pp 73ndash77 1995

[6] G Chuan-qing ldquoMatrix Pade type approximat and directionalmatrix Pade approximat in the inner product spacerdquo Journal ofComputational and Applied Mathematics vol 164-165 pp 365ndash385 2004

[7] C Gu ldquoThiele-type and Lagrange-type generalized inverserational interpolation for rectangular complexmatricesrdquo LinearAlgebra and Its Applications vol 295 no 1ndash3 pp 7ndash30 1999

[8] C Gu ldquoA practical two-dimensional Thiele-type matrix Padeapproximationrdquo IEEE Transactions on Automatic Control vol48 no 12 pp 2259ndash2263 2003

10 Journal of Applied Mathematics

[9] C Gu ldquoBivariate Thiele-type matrix-valued rational inter-polantsrdquo Journal of Computational and Applied Mathematicsvol 80 no 1 pp 71ndash82 1997

[10] N K Bose and S Basu ldquoTwo-dimensional matrix Pade approx-imants existence nonuniqueness and recursive computationrdquoIEEE Transactions on Automatic Control vol 25 no 3 pp 509ndash514 1980

[11] A Cuyt and B Verdonk ldquoMultivariate reciprocal differencesfor branched Thiele continued fraction expansionsrdquo Journal ofComputational and Applied Mathematics vol 21 no 2 pp 145ndash160 1988

[12] W Siemaszko ldquoThiele-tyoe branched continued fractions fortwo-variable functionsrdquo Journal of Computational and AppliedMathematics vol 9 no 2 pp 137ndash153 1983

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Stochastic AnalysisInternational Journal of

4 Journal of Applied Mathematics

(ii) 119880(119904)

119897(119910) = 120593

119897119904(1199090 1199091 119909

119897 1199100 119910

119904) + ((119910 minus

119910119904)119880(119904+1)

119897(119910)) satisfy119880

(119904+1)

119897(119910119904) = 0 119904 = 0 1 119899minus1

then 119877119898119899

(119909 119910) as in (11) and (12) exists such that119877119898119899

(119909119897 119910119896) = 119884119897119896 (119909119897 119910119896) isin Λm119899

Proof For simpleness let

120593119897119896

(1199090 119909

119897 1199100 119910

119896) = 120593119897119896

(20)

if the conditions hold thus for 119897 = 0 1 119898 119896 = 0 1 119899

119880119897(119910119896) = 119880

(0)

119897(119910119896)

= 1205931198970

+119910119896minus 1199100

1205931198971

+ sdot sdot sdot +119910119896minus 119910119896minus1

|

| 120593119897119899

+119910119896minus 119910119896|

| 119880(119896+1)

119897(119910119896)

(21)

(119910119896minus119910119896)119880(119896+1)

119897(119910119896) = 0 since119880

(119896+1)

119897(119910119896) = 0 then we get

119880119897(119910119896)

= 1205931198970

(1199090 119909

119897 1199100) +

119910119896minus 1199100

1205931198971

(1199090 119909

119897 1199100 1199101)+ sdot sdot sdot +

119910119896minus 119910119896minus2

|

| 120593119897119896minus1

(1199090 119909

119897 1199100 119910

119896minus1)

+119910119896minus 119910119896minus1

|

| 120593119897119896

(1199090 119909

119897 1199100 119910

119896)

= 1205931198970

(1199090 119909

119897 1199100) +

119910119896minus 1199100

1205931198971

(1199090 119909

119897 1199100 1199101)+ sdot sdot sdot

+119910119896minus 119910119896minus2

|

| 120593119897119896minus1

(1199090 119909

119897 1199100 119910

119896minus1)

+119910119896minus 119910119896minus1

|

| (119910119896minus 119910119896minus1

) times (120593119897119896minus1

(1199090 119909

119897 1199100 119910

119896minus2 119910119896) minus 120593119897119896minus1

(1199090 119909

119897 1199100 119910

119896minus1))minus1

= 1205931198970

(1199090 119909

119897 1199100) +

119910119896minus 1199100

1205931198971

(1199090 119909

119897 1199100 1199101)+ sdot sdot sdot +

119910119896minus 119910119896minus2

|

| 120593119897119896minus1

(1199090 119909

119897 1199100 119910

119896minus2 119910119896)

= sdot sdot sdot = 1205931198970

(1199090 119909

119897 1199100) +

119910119896minus 1199100

1205931198971

(1199090 119909

119897 1199100 119910119896)

= 1205931198970

(1199090 119909

119897 119910119896)

(22)

We can use (9) (11) and (22) to find that119877119898119899

(119909119897 119910119896)

= 1198800(119910119896) + 1198801(119910119896) (119909119897minus 1199090)

+ sdot sdot sdot + 119880119897(119910119896) (119909119897minus 1199090) (119909119897minus 1199091) sdot sdot sdot (119909

119897minus 119909119897minus1

)

= 119884 (119909119897 119910119896)

(23)

We can similarly prove the following result

Theorem 6 Let(i) 120595119897119896(1199090 119909

119897 1199100 119910

119896) 119897 = 0 1 119898 119896 = 0

1 119899 exist and be nonzero (except for 1205950119896

(1199090

1199100 119910

119896))

(ii) 119881(119904)

119896(119909) = 120595

119904119896(1199090 1199091 119909

119904 1199100 119910

119896) + ((119909 minus

119909119904)119881(119904+1)

119896(119909)) satisfy119881(119904+1)

119896(119909119904) = 0 119904 = 0 1 119898minus1

then 119898119899

(119909 119910) as in (14) and (15) exists such that119898119899

(119909119897 119910119896) = 119884119897119896 (119909119897 119910119896) isin Λ

119898119899

Now we give the Newton-Thiele Matrix InterpolationAlgorithm (NTMIA)

Algorithm 7 (NTMIA) Input (119909119894 119910119895) 119884119894119895 (119894 = 0 1

119898 119895 = 0 1 119899)

Output 119898(119909 119910)

Step 1 For all (119909119894 119910119895) isin Λ

119898119899 let 120593(119909

119894 119910119895) = 119884119894119895

Step 2 For 119895 = 0 1 119899 119901 = 1 2 119898 119894 = 119901 119901 +

1 119898 compute

1205931199010

(1199090 119909

119901minus1 119909119894 119910119895)

=

120593119901minus10

(1199090 119909

119901minus2 119909119894 119910119895)minus120593119901minus10

(1199090 119909

119901minus1 119910119895)

119909119894minus119909119901minus1

(24)

Journal of Applied Mathematics 5

Step 3 For 119894 = 0 1 119898 119902 = 1 2 119899 119895 = 119902 119902 + 1 119899compute

120593119894119902

(1199090 119909

119894 1199100 119910

119902minus1 119910119895)

= (119910119895minus 119910119902minus1

) times (120593119894119902minus1

(1199090 119909

119894 1199100 119910

119902minus2 119910119895)

minus120593119894119902minus1

(1199090 119909

119894 1199100 119910

119902minus1))minus1

= ((119910119895minus 119910119902minus1

) (120593119894119902minus1

(1199090 119909

119894 1199100 119910

119902minus2 119910119895)

minus120593119894119902minus1

(1199090 119909

119894 1199100 119910

119902minus1)))

times (10038171003817100381710038171003817120593 (1199090 119909

119894 1199100 119910

119902minus2 119910119895)

minus120593 (1199090 119909

119894 1199100 119910

119902minus1)10038171003817100381710038171003817

2

)

minus1

(25)

Step 4 For 119894 = 0 1 119898 119895 = 0 1 119899 compute 119886119894119895

=

120593119894119895(1199090 119909

119894 1199100 119910

119895)

Step 5 For 119894 = 0 1 119898 119895 = 1 119899 judge if 119886119894119895

= 0 if yesgo to Step 6 otherwise exist and show ldquothe algorithm is notvalidrdquo

Step 6 For 119894 = 0 1 119898 119895 = 119899 minus 2 119899 minus 3 0

119876119894119899

= 1 119875119894119899

= 119886119894119899

119876119894119899minus1

=1003817100381710038171003817119886119894119899

1003817100381710038171003817

2

119875119894119899minus1

= 119886119894119899minus1

119876119894119899minus1

+ (119910 minus 119910119899minus1

) 119875119894119899

119876119894119895

=10038171003817100381710038171003817119886119894119895+1

10038171003817100381710038171003817

2

119876119894119895+1

+ 2 (119910 minus 119910119895+1

) tr (119886119894119895+1

119875119894119895+2

lowast

)

+ (119910 minus 119910119895+1

)2

119876119894119895+2

119875119894119895

= 119886119894119895119876119894119895

+ (119910 minus 119910119895) 119875119894119895+1

(26)

Step 7 For 119894 = 0 1 119898 119895 = 119899 119899 minus 1 1 let 119861119894119895(119910) =

119875119894119895(119910)119876

119894119895(119910) if 119861

119894119895(119910119895minus1

) = 0 then go to Step 8 otherwiseexist and show ldquothe algorithm is not validrdquo

Step 8 For 119894 = 0 1 119898 let 119861119894(119910) = 119861

1198940(119910) = 119875

1198940(119910)

1198761198940(119910)

Step 9 For 119896 = 1 119898 let 0(119909 119910) = 119861

0(119910) then compute

119896(119909 119910) =

119896minus1(119909 119910) + sdot sdot sdot + (119909 minus 119909

0) sdot sdot sdot (119909 minus 119909

119896minus1)119861119896(119910)

3 Some Properties

Lemma 8 (see [5]) Define 1198800(119910) = 120593

00(1199090 1199100) + (119910 minus

1199100)12059301

(1199090 1199100 1199101) + sdot sdot sdot + (119910minus119910

119899minus1)1205930119899

(1199090 1199100 119910

119899) if we

use generalized inverse by a tail-to-head rationalization thena polynomial matrix 119864

0(119910) and a real polynomial 119865

0(119910) exist

such that(i) 1198800(119910) = 119864

0(119910)1198650(119910) 1198650(119910) ge 0

(ii) 1198650(119910) | 119864

0(119910)2ldquo|rdquo means the sign of divisibility

Definition 9 119877(119909 119910) = 119873(119909 119910)119863(119909 119910) is defined to be oftype [119897119905] if deg119873

119894119895 le 119897 for 1 le 119894 le 119906 1 le 119895 le 119907 deg119873

119894119895 = 119897

for some (119894 119895) and deg119863 = 119905 where119873(119909 119910) = (119873119894119895(119909 119910)) isin

119862119906times119907

Lemma 10 (see [5]) Let1198800(119910) be defined as in Lemma 8 and

1198800(119910) = 119864

0(119910)1198650(119910) if 119899 is even119880

0(119910) is of type [119899119899] if 119899 is

odd 1198800(119910) is of type [119899119899 minus 1]

Theorem 11 Let 120593119897119896

isin 119862119906times119907 (119909

119897 119910119896) isin Λ

119898119899 and 119909 119910 isin

119877 Define 119880119897(119910) 119900119891 119877

119898119899(119909 119910) as in (7) by a tail-to-head

rationalization using generalized inverse and suppose everyintermediate denominator be nonzero in the operation thena square polynomial matrix 119873(119909 119910) and a real polynomial119863(119909 119910) exist such that

(i) 119877119898119899

(119909 119910) = 119873(119909 119910)119863(119909 119910)

(ii) 119863(119909 119910) ge 0

(iii) 119863(119909 119910) | 119873(119909 119910)2

Proof Consider the following algorithm for the constructionof 119873(119909 119910) 119863(119909 119910) and 119877

119898119899(119909 119910)

Initialization let 1198630(119909 119910) = 1 and

1198730(119909 119910) = 119880

0(119910) = 120593

00+

119910 minus 1199100

12059301

+ sdot sdot sdot +119910 minus 119910119899minus1

|

| 1205930119899

(27)

By Lemma 8 a polynomial matrix 1198640(119910) and a real polyno-

mial 1198650(119910) exist such that

(a1) 1198730(119909 119910) = 119864

0(119910)1198650(119910)

(a2) 1198650(119910) ge 0

(a3) 1198650(119910) | 119864

0(119910)2

Recursion for 119895 = 1 2 119898 let

119878(119895)

(119909 119910) = 119878(119895minus1)

(119909 119910) + 119880119895(119910) (119909 minus 119909

0) sdot sdot sdot (119909 minus 119909

119895minus1)

(28)

with 119880119895(119910) = 119864

119895(119910)119865119895(119910) 119865119895(119910) | 119864

119895(119910)2 having the

representation

(b1) 119878(119895)

(119909 119910) = 119873119895(119909 119910)119863

119895(119909 119910)

(b2) 119863119895(119909 119910) ge 0

(b3) 119863119895(119909 119910) | 119873

119895(119909 119910)

2

where 119863119895(119909 119910) is a real polynomial

6 Journal of Applied Mathematics

Then we can get

119878(119895+1)

(119909 119910) = 119878(119895)

(119909 119910) + 119880119895+1

(119909 minus 1199090) sdot sdot sdot (119909 minus 119909

119895)

=119873119895(119909 119910)

119863119895(119909 119910)

+119864119895+1

(119910)

119865119895+1

(119910)(119909 minus 119909

0) sdot sdot sdot (119909 minus 119909

119895)

= (119873119895(119909 119910) 119865

119895+1(119910) + 119863

119895(119909 119910) 119864

119895+1(119910)

times (119909 minus 1199090) sdot sdot sdot (119909 minus 119909

119895))

times (119863119895(119909 119910) 119865

119895+1(119910))minus1

=119873119895+1

(119909 119910)

119863119895+1

(119909 119910)

(29)

where

119873119895+1

(119909 119910) = 119873119895(119909 119910) 119865

119895+1(119910)

+ 119863119895(119909 119910) 119864

119895+1(119910) (119909 minus 119909

0)

sdot sdot sdot (119909 minus 119909119895)

119863119895+1

(119909 119910) = 119863119895(119909 119910) 119865

119895+1(119910)

(30)

Obviously

10038171003817100381710038171003817119873119895+1

10038171003817100381710038171003817

2

=10038171003817100381710038171003817119873119895(119909 119910)

10038171003817100381710038171003817

2

1198652

(119895+1)(119910) + 119863

2

119895(119909 119910)

times10038171003817100381710038171003817119864(119895+1)

(119910)10038171003817100381710038171003817

2

(119909 minus 1199090)2

sdot sdot sdot (119909 minus 119909119895)2

+ 119873119895(119909 119910) sdot 119864

119895+1(119910) 119865119895+1

(119910)119863119895(119909 119910)

times (119909 minus 1199090) sdot sdot sdot (119909 minus 119909

119895)

+ 119873119895(119909 119910) sdot 119864

119895+1(119910) 119865119895+1

(119910)119863119895(119909 119910)

times (119909 minus 1199090) sdot sdot sdot (119909 minus 119909

119895)

(31)

since

119863119895(119909 119910) |

10038171003817100381710038171003817119873119895(119909 119910)

10038171003817100381710038171003817

2

119865119895+1

(119910) |10038171003817100381710038171003817119864119895+1

(119910)10038171003817100381710038171003817

2

119863119895+1

(119909 119910) |10038171003817100381710038171003817119873119895+1

(119909 119910)10038171003817100381710038171003817

2

(32)

Termination 119878(119898)(119909 119910) = 119877119898119899

(119909 119910)

Theorem 12 (Characterization) Let 119877119898119899

(119909 119910) = 119873(119909 119910)

119863(119909 119910) be expressed as

119877 = 119877119898119899

(119909 119910) = 1198800(119910) + 119880

1(119910) (119909 minus 119909

0)

+ sdot sdot sdot + 119880119898

(119910) (119909 minus 1199090) sdot sdot sdot (119909 minus 119909

119898minus1)

(33)

where 119880119897(119910) = 119864

119894119865119894 as in (12) we assume 119865

119894 119865119895have no

common factor 119894 119895 = 0 1 119898 119894 = 119895 then

(i) if n is even 119877 is of type [119898119899 + 119898 + 119899119898119899 + 119899](ii) if n is odd 119877 is of type [119898119899 + 119899119898119899 + 119899 minus 119898 minus 1]

Proof The proof is recursive Let 119899 be evenFor 119898 = 0 119877

0119899(119909 119910) = 119880

0(119910) = 119864

01198650 by Lemma 10 we

find that it is of type [119899119899]For 119898 = 1 because deg119864

0 = deg119865

0 = 119899 deg119864

1 =

deg1198651 = 119899

1198771119899

(119909 119910) =1198731

1198631

= 1198800(119910) + 119880

1(119910) (119909 minus 119909

0)

=1198640

1198650

+1198641

1198651

(119909 minus 1199090)

=11986401198651+ 11986411198650(119909 minus 119909

0)

11986501198651

(34)

is of type [2119899 + 12119899]For 119898 = 2 from the formula above we know that

deg1198731 = 2119899 + 1 deg119863

1 = 2119899 deg119864

2 = 119899 and deg119865

2 =

119899 one has1198772119899

(119909 119910) =1198732

1198632

=1198731

1198631

+1198642

1198652

(119909 minus 1199090) (119909 minus 119909

1)

=11987311198652+ 11986421198631(119909 minus 119909

0) (119909 minus 119909

1)

11986311198652

(35)

It is easy to find that 1198772119899

(119909 119910) is of type [3119899 + 23119899]For 119898 = 119896 let 119877

119896119899(119909 119910) = 119873

119896119863119896be of type [(119896 + 1)119899 +

119896(119896 + 1)119899]For 119898 = 119896 + 1 we consider that

119877119896+1119899

(119909 119910) =119873119896

119863119896

+119864119896+1

119865119896+1

(119909 minus 1199090) sdot sdot sdot (119909 minus 119909

119896)

=119873119896119865119896+1

+ 119864119896+1

119863119896(119909 minus 119909

0) sdot sdot sdot (119909 minus 119909

119896)

119863119896119865119896

=119873119896+1

119863119896+1

(36)

wheredeg 119873

119896+1 = (119896 + 1) (119899 + 1) + 119899

= (119896 + 2) 119899 + (119896 + 1)

deg 119863119896+1

= (119896 + 1) 119899 + 119899 = (119896 + 2) 119899

(37)

Therefore when 119899 is even 119877 is of type [119898119899 + 119898 + 119899119898119899 + 119899]Similarly if 119899 is odd for 119898 = 0 119877

0119899(119909 119910) = 119880

0(119910) =

11986401198650 by Lemma 10 we find that it is of type [119899119899 minus 1]

For 119898 = 1 because deg1198640 = deg119864

1 = 119899 deg119865

0 =

deg1198651 = 119899 minus 1

1198771119899

(119909 119910) =1198731

1198631

= 1198800(119910) + 119880

1(119910) (119909 minus 119909

0)

=1198640

1198650

+1198641

1198651

(119909 minus 1199090)

=11986401198651+ 11986411198650(119909 minus 119909

0)

11986501198651

(38)

is of type [21198992119899 minus 2]

Journal of Applied Mathematics 7

For119898 = 2 from the formula above we know that deg1198731

= 2119899 deg1198631 = 2119899 minus 2 deg119864

2 = 119899 and deg119865

2 = 119899 minus 1

one has

1198772119899

(119909 119910) =1198732

1198632

=1198731

1198631

+1198642

1198652

(119909 minus 1199090) (119909 minus 119909

1)

=11987311198652+ 11986421198631(119909 minus 119909

0) (119909 minus 119909

1)

11986311198652

(39)

We get that 1198772119899

(119909 119910) is of type [31198993119899 minus 3]For 119898 = 119896 let

119877119896119899

(119909 119910) =119873119896

119863119896

(40)

be of type [(119896 + 1)119899(119896 + 1)(119899 minus 1)]For 119898 = 119896 + 1 we consider that

119877119896+1119899

(119909 119910) =119873119896

119863119896

+119864119896+1

119865119896+1

(119909 minus 1199090) sdot sdot sdot (119909 minus 119909

119896)

=119873119896119865119896+1

+ 119864119896+1

119863119896(119909 minus 119909

0) sdot sdot sdot (119909 minus 119909

119896)

119863119896119865119896

=119873119896+1

119863119896+1

(41)

where

deg 119873119896+1

= (119896 + 1) (119899 minus 1) + 119896 + 1 = (119896 + 2) 119899

deg 119863119896+1

= (119896 + 1) (119899 minus 1) + (119899 minus 1) = (119896 + 2) (119899 minus 1)

(42)

Therefore when 119899 is odd119877 is of type [(119898+1)119899(119898+1)(119899minus

1)]

Definition 13 A matrix-valued Newton-Thiele type ratio-nal fraction 119877

119898119899(119909 119910) = 119873(119909 119910)119863(119909 119910) is defined to

be a bivariate generalized inverse and rational interpolant(BGIRINT) on the rectangular grid Λ

119898119899if

(i) 119877119898119899

(119909119897 119910119896) = 119884119897119896 (119909119897 119910119896) isin Λ

119898119899

(ii) 119863(119909119897 119910119896) = 0 (119909

119897 119910119896) isin Λ

119898119899

(iii) (a) if 119899 is even

deg 119873 (119909 119910) = 119898119899+ 119898+119899 deg 119863 (119909 119910) = 119898119899+119899

(43)

(b) if n is odd

deg 119873 (119909 119910) = 119898119899+119899 deg 119863 (119909 119910) = 119898119899+119899minus119898minus1

(44)

(iv) 119863(119909 119910) | 119873(119909 119910)2

(v) 119863(119909 119910) is real and 119863(119909 119910) ge 0

Now let us turn to the error estimate of the BGIRINTFirstly we give Lemma 14

Lemma 14 Let

119877lowast

119898119899(119909 119910) = 119880

lowast

0(119910) + 119880

lowast

1(119910) (119909 minus 119909

0)

+ sdot sdot sdot + 119880lowast

119898(119910) (119909minus119909

0) (119909minus119909

1) sdot sdot sdot (119909minus119909

119898minus1)

(45)

where for 119897 = 0 1 119898

119880lowast

119897(119910) = 120593

1198970(1199090 119909

119897 1199100) +

119910 minus 1199100

1205931198971

(1199090 119909

119897 1199100 1199101)

+ sdot sdot sdot +119910 minus 119910119899minus1

|

| 120593119897119899

(1199090 119909

119897 1199100 119910

119899)

+119910 minus 119910119899|

| 120593119897119899+1

(1199090 119909

119897 y0 119910

119899 119910)

(46)

then 119877lowast

119898119899(119909 119910) satisfies

119877lowast

119898119899(119909119894 119910) = 119891 (119909

119894 119910) 119894 = 0 1 119898

119877lowast

119898119899(119909 119910119895) = 119877119898119899

(119909 119910119895) 119895 = 0 1 119899

(47)

Remark 15 We delete the proof of the above lemma since itcan be easily generalized from [3]

Now the error estimate Theorem 16 will be put which isalso motivated by [3] We let

119877119898119899

(119909 119910) =119873 (119909 119910)

119863 (119909 119910) 119877

lowast

119898119899(119909 119910) =

119873lowast

(119909 119910)

119863lowast (119909 119910) (48)

Theorem 16 Let a matrix 119891(119909 119910) bemax(119898 + 1 119899 + 1) timescontinuously differentiable on a set119863 = 119886 le 119909 le 119887 119888 le 119910 le 119889

containing the points Λ119898119899

= (119909119897 119910119896) 119897 = 0 1 119898 119896 =

0 1 119899 (119909119897 119910119896) isin R2 then for any point (119909 119910) isin 119863 there

exists a point (120585 120578) isin 119863 such that

119891 (119909 119910) minus 119877119898119899

(119909 119910)

=1

119863 (119909 119910)119863lowast (119909 119910)

times 120596119898+1

(119909)

(119898 + 1)

120597119898+1

120597119909119898+11198641(120585 119910) +

120603119899+1

(119910)

(119899 + 1)

120597119899+1

120597119910119899+11198642(119909 120578)

(49)

where

120596119898+1

(119909) = (119909 minus 1199090) (119909 minus 119909

1) sdot sdot sdot (119909 minus 119909

119898)

120603119899+1

(119910) = (119910 minus 1199100) (119910 minus 119910

1) sdot sdot sdot (119910 minus 119910

119899)

1198641(119909 119910) = 119863 (119909 119910)119863

lowast

(119909 119910) [119891 (119909 119910) minus 119877lowast

119898119899(119909 119910)]

1198642(119909 119910) = 119863 (119909 119910)119863

lowast

(119909 119910) [119877lowast

119898119899(119909 119910) minus 119877

119898119899(119909 119910)]

(50)

8 Journal of Applied Mathematics

Proof Let

1198641(119909 119910) = 119863 (119909 119910)119863

lowast

(119909 119910) [119891 (119909 119910) minus 119877lowast

119898119899(119909 119910)]

1198642(119909 119910) = 119863 (119909 119910)119863

lowast

(119909 119910) [119877lowast

119898119899(119909 119910) minus 119877

119898119899(119909 119910)]

119864 (119909 119910) = 1198641(119909 119910) + 119864

2(119909 119910)

(51)

From Lemma 14 we know

1198641(119909119894 119910) = 0 119894 = 0 1 119898 (52)

which results in

1198641(119909 119910) =

120596119898+1

(119909)

(119898 + 1)

120597119898+1

120597119909119898+11198641(119909 119910)

100381610038161003816100381610038161003816100381610038161003816119909=120585

(53)

where 120596119898+1

(119909) = (119909 minus 1199090)(119909 minus 119909

1) sdot sdot sdot (119909 minus 119909

119898) and 120585 is a

number contained in the interval (119886 119887) which may dependon 119910 Similarly from

1198642(119909 119910119895) = 0 119895 = 0 1 119899 (54)

we can get that

1198642(119909 119910) =

120603119899+1

(119910)

(119899 + 1)

120597119899+1

120597119910119899+11198642(119909 119910)

100381610038161003816100381610038161003816100381610038161003816119910=120578

(55)

where120603119899+1

(119910) = (119910minus1199100)(119910minus119910

1) sdot sdot sdot (119910minus119910

119899) and 120578 is a number

contained in the interval (119888 119889) which may depend on 119909Therefore

119864 (119909 119910) = 1198641(119909 119910) + 119864

2(119909 119910)

=120596119898+1

(119909)

(119898 + 1)

120597119898+1

120597119909119898+11198641(119909 119910)

100381610038161003816100381610038161003816100381610038161003816119909=120585

+120603119899+1

(119910)

(119899 + 1)

120597119899+1

120597119910119899+11198642(119909 119910)

100381610038161003816100381610038161003816100381610038161003816119910=120578

(56)

The proof is thus completed

4 Numerical Examples

Example 17 Let (119909119894 119910119895) and 119884

119894119895 119894 119895 = 0 1 2 be given in

Table 1

11987722

(119909 119910) = 1198800(119910) + 119880

1(119910) 119909 + 119880

2(119910) 119909 (119909 minus 1) (57)

Using Algorithm 7 we can get

1198800(119910) = (

1 0

0 0) +

119910

( 0 00 1

)+

119910 minus 1 |

| (0 0

minus25 minus15)

1198801(119910) = (

0 minus1

0 0) +

119910

(0 12

minus12 0)

+119910 minus 1 |

| ( 0 11 0

)

1198802(119910) = (

0 1

1

20) +

119910

(0 minus35

15 0)

+119910 minus 1 |

| (0 minus1514

minus514 minus57)

(58)

Based on generalized inverse (4) we obtain

11987722

(119909 119910)

= (11988611

11988612

11988621

11988622

) times (701199106

minus 3681199105

+ 8921199104

minus12401199103

+10401199102

minus512119910+128)minus1

=119873 (119909 119910)

119863 (119909 119910)

(59)

where

11988611

= 701199106

minus 3681199105

+ 8921199104

minus 12401199103

+ 10401199102

minus 512119910 + 128

11988612

= 1781199105

119909 minus 8581199104

119909 + 16881199103

119909

minus 17761199091199102

+ 1024119909119910 minus 256119909 minus 381199105

1199092

+ 2621199104

1199092

minus 6401199103

1199092

minus 5121199092

119910

minus 51199106

1199092

+ 51199106

119909 + 8001199092

1199102

+ 1281199092

11988621

= 1081199105

minus 2001199104

+ 2001199103

minus 1121199102

+ 32119910

minus 281199106

+ 601199106

119909 minus 3441199105

119909 + 7361199104

119909 minus 7841199103

119909

+ 4081199091199102

minus 32119909119910 minus 64119909 minus 241199105

1199092

+ 161199104

1199092

minus 961199092

119910 + 101199106

1199092

+ 401199092

1199102

+ 641199092

11988622

= 681199105

minus 1401199104

+ 1601199103

minus 961199102

+ 32119910 minus 141199106

minus 501199106

1199092

+ 2701199105

1199092

minus 6401199104

1199092

+ 8201199103

1199092

minus 5601199092

1199102

+ 1601199092

119910 + 501199106

119909 minus 2701199105

119909

+ 6401199104

119909 minus 8201199103

119909 + 5601199091199102

minus 160119909119910

(60)

FromDefinition 13 we know that 11987722

(119909 119910) is a BGIRINTsince 119877

22(119909119894 119910119895) = 119884119894119895and deg 119873(119909 119910) = 8 deg 119863(119909 119910) =

6 and 119863(119909 119910) | 119873(119909 119910)2

In paper [9] a bivariateThieletypematrix-valued rationalinterpolant 119877

119879(119909 119910) = 119866

00(119910) + (119909 minus 119909

0)11986610

(119910) + sdot sdot sdot + (119909 minus

119909119898minus1

)1198661198990

(119910) (page 73) where for 119897 = 0 1 119899

1198661198970

(119910) = 1198611198970

(1199090 119909

119897 1199100)

+119910 minus 1199100

1198611198971

(1199090 119909

119897 1199100 1199101)

+ sdot sdot sdot +119910 minus 119910119898minus1

|

| 119861119897119898

(1199090 119909

119897 1199100 119910

119898)

(61)

Here 119861119897119896(1199090 119909

119897 1199100 119910

119896) 119897 = 0 1 119899 119896 = 0 1 119898

is defined different from that of this paper (see [9]) We

Journal of Applied Mathematics 9

Table 1

119910119895119909119894

1199090= 0 119909

1= 1 119909

2= 2

1199100= 0 (

1 0

0 0) (

1 minus1

0 0) (

1 0

1 0)

1199101= 1 (

1 0

0 1) (

1 0

minus1 1) (

1 minus1

0 1)

1199102= 2 (

1 0

minus1 0) (

1 1

minus1 0) (

1 1

0 minus1)

Table 2

119884(13 13) 119877NT(13 13) 119877NT minus 119884119865

(minus0856888 05155013 3669296

130000 366929 2600) (

minus0856888 05155010 3669297

130000 366929 2600) 109119890 minus 006

119877119879(13 13) 119877

119879minus 119884119865

(minus098999 014111 547394

129999 547394 2999) 2613697

119884(15 15) 119877NT(15 15) 119877NT minus 119884119865

(minus0989999 01411200 4481689

150000 4481689 300) (

minus0989992 01411203 4481688

150000 4481688 300) 915119890 minus 007

119877119879(15 15) 119877

119879minus 119884119865

(minus099829372 minus005837 547394

149999 547394 31999) 143144

now give another numerical example to compare the twoalgorithms which shows that the method of this paper isbetter than the one of [9]

Example 18 Let

119884 (119909 119910) = (cos (119909 + 119910) sin (119909 + 119910) 119890

119910

119909 119890119910

119909 + 119910) (62)

and we suppose 1199090 1199091 1199092 1199093 1199094 1199095 = 119910

0 1199101 1199102 1199103 1199104 1199105

= 10 12 14 16 18 20 The numerical results 119877NT(119909 119910)

in Step 9 of Algorithm 7 and 119877119879(119909 119910) in [9] are given in

Table 2

Remark 19 InTable 2 from the F-normof119877NTminus119884 and119877119879minus119884

we can see that the error using Newton-Thiele type formulais much less than the one usingThieletype formula

5 Conclusion

In this paper the bivariate generalized inverse Newton-Thieletype matrix interpolation to approximate a matrix function119891(119909 119910) is given and we also give a recursive algorithmaccompanied with some other important conclusions such asdivisibility characterization and some numerical examplesFrom the second example it is easy to find that the approxi-mant using Newton-Thiele type formula is much better thanthe one usingThieletype formula

Acknowledgments

The work is supported by Shanghai Natural Science Foun-dation (10ZR1410900) and by Key Disciplines of ShanghaiMunicipality (S30104)

References

[1] G A Baker and P R Graves-Morris Pade Approximation PartII Addison-Wesley Publishing Company Reading Mass USA1981

[2] I Kh Kuchminska OM Sus and SMVozna ldquoApproximationproperties of two-dimensional continued fractionsrdquo UkrainianMathematical Journal vol 55 pp 36ndash54 2003

[3] J Tan and Y Fang ldquoNewton-Thielersquos rational interpolantsrdquoNumerical Algorithms vol 24 no 1-2 pp 141ndash157 2000

[4] P R Graves-Morris ldquoVector valued rational interpolants IrdquoNumerische Mathematik vol 42 no 3 pp 331ndash348 1983

[5] G Chuan-qing and C Zhibing ldquoMatrix valued rational inter-polants and its error formulardquo Mathematica Numerica Sinicavol 17 pp 73ndash77 1995

[6] G Chuan-qing ldquoMatrix Pade type approximat and directionalmatrix Pade approximat in the inner product spacerdquo Journal ofComputational and Applied Mathematics vol 164-165 pp 365ndash385 2004

[7] C Gu ldquoThiele-type and Lagrange-type generalized inverserational interpolation for rectangular complexmatricesrdquo LinearAlgebra and Its Applications vol 295 no 1ndash3 pp 7ndash30 1999

[8] C Gu ldquoA practical two-dimensional Thiele-type matrix Padeapproximationrdquo IEEE Transactions on Automatic Control vol48 no 12 pp 2259ndash2263 2003

10 Journal of Applied Mathematics

[9] C Gu ldquoBivariate Thiele-type matrix-valued rational inter-polantsrdquo Journal of Computational and Applied Mathematicsvol 80 no 1 pp 71ndash82 1997

[10] N K Bose and S Basu ldquoTwo-dimensional matrix Pade approx-imants existence nonuniqueness and recursive computationrdquoIEEE Transactions on Automatic Control vol 25 no 3 pp 509ndash514 1980

[11] A Cuyt and B Verdonk ldquoMultivariate reciprocal differencesfor branched Thiele continued fraction expansionsrdquo Journal ofComputational and Applied Mathematics vol 21 no 2 pp 145ndash160 1988

[12] W Siemaszko ldquoThiele-tyoe branched continued fractions fortwo-variable functionsrdquo Journal of Computational and AppliedMathematics vol 9 no 2 pp 137ndash153 1983

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Differential EquationsInternational Journal of

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Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Operations ResearchAdvances in

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

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Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Decision SciencesAdvances in

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Journal of Applied Mathematics 5

Step 3 For 119894 = 0 1 119898 119902 = 1 2 119899 119895 = 119902 119902 + 1 119899compute

120593119894119902

(1199090 119909

119894 1199100 119910

119902minus1 119910119895)

= (119910119895minus 119910119902minus1

) times (120593119894119902minus1

(1199090 119909

119894 1199100 119910

119902minus2 119910119895)

minus120593119894119902minus1

(1199090 119909

119894 1199100 119910

119902minus1))minus1

= ((119910119895minus 119910119902minus1

) (120593119894119902minus1

(1199090 119909

119894 1199100 119910

119902minus2 119910119895)

minus120593119894119902minus1

(1199090 119909

119894 1199100 119910

119902minus1)))

times (10038171003817100381710038171003817120593 (1199090 119909

119894 1199100 119910

119902minus2 119910119895)

minus120593 (1199090 119909

119894 1199100 119910

119902minus1)10038171003817100381710038171003817

2

)

minus1

(25)

Step 4 For 119894 = 0 1 119898 119895 = 0 1 119899 compute 119886119894119895

=

120593119894119895(1199090 119909

119894 1199100 119910

119895)

Step 5 For 119894 = 0 1 119898 119895 = 1 119899 judge if 119886119894119895

= 0 if yesgo to Step 6 otherwise exist and show ldquothe algorithm is notvalidrdquo

Step 6 For 119894 = 0 1 119898 119895 = 119899 minus 2 119899 minus 3 0

119876119894119899

= 1 119875119894119899

= 119886119894119899

119876119894119899minus1

=1003817100381710038171003817119886119894119899

1003817100381710038171003817

2

119875119894119899minus1

= 119886119894119899minus1

119876119894119899minus1

+ (119910 minus 119910119899minus1

) 119875119894119899

119876119894119895

=10038171003817100381710038171003817119886119894119895+1

10038171003817100381710038171003817

2

119876119894119895+1

+ 2 (119910 minus 119910119895+1

) tr (119886119894119895+1

119875119894119895+2

lowast

)

+ (119910 minus 119910119895+1

)2

119876119894119895+2

119875119894119895

= 119886119894119895119876119894119895

+ (119910 minus 119910119895) 119875119894119895+1

(26)

Step 7 For 119894 = 0 1 119898 119895 = 119899 119899 minus 1 1 let 119861119894119895(119910) =

119875119894119895(119910)119876

119894119895(119910) if 119861

119894119895(119910119895minus1

) = 0 then go to Step 8 otherwiseexist and show ldquothe algorithm is not validrdquo

Step 8 For 119894 = 0 1 119898 let 119861119894(119910) = 119861

1198940(119910) = 119875

1198940(119910)

1198761198940(119910)

Step 9 For 119896 = 1 119898 let 0(119909 119910) = 119861

0(119910) then compute

119896(119909 119910) =

119896minus1(119909 119910) + sdot sdot sdot + (119909 minus 119909

0) sdot sdot sdot (119909 minus 119909

119896minus1)119861119896(119910)

3 Some Properties

Lemma 8 (see [5]) Define 1198800(119910) = 120593

00(1199090 1199100) + (119910 minus

1199100)12059301

(1199090 1199100 1199101) + sdot sdot sdot + (119910minus119910

119899minus1)1205930119899

(1199090 1199100 119910

119899) if we

use generalized inverse by a tail-to-head rationalization thena polynomial matrix 119864

0(119910) and a real polynomial 119865

0(119910) exist

such that(i) 1198800(119910) = 119864

0(119910)1198650(119910) 1198650(119910) ge 0

(ii) 1198650(119910) | 119864

0(119910)2ldquo|rdquo means the sign of divisibility

Definition 9 119877(119909 119910) = 119873(119909 119910)119863(119909 119910) is defined to be oftype [119897119905] if deg119873

119894119895 le 119897 for 1 le 119894 le 119906 1 le 119895 le 119907 deg119873

119894119895 = 119897

for some (119894 119895) and deg119863 = 119905 where119873(119909 119910) = (119873119894119895(119909 119910)) isin

119862119906times119907

Lemma 10 (see [5]) Let1198800(119910) be defined as in Lemma 8 and

1198800(119910) = 119864

0(119910)1198650(119910) if 119899 is even119880

0(119910) is of type [119899119899] if 119899 is

odd 1198800(119910) is of type [119899119899 minus 1]

Theorem 11 Let 120593119897119896

isin 119862119906times119907 (119909

119897 119910119896) isin Λ

119898119899 and 119909 119910 isin

119877 Define 119880119897(119910) 119900119891 119877

119898119899(119909 119910) as in (7) by a tail-to-head

rationalization using generalized inverse and suppose everyintermediate denominator be nonzero in the operation thena square polynomial matrix 119873(119909 119910) and a real polynomial119863(119909 119910) exist such that

(i) 119877119898119899

(119909 119910) = 119873(119909 119910)119863(119909 119910)

(ii) 119863(119909 119910) ge 0

(iii) 119863(119909 119910) | 119873(119909 119910)2

Proof Consider the following algorithm for the constructionof 119873(119909 119910) 119863(119909 119910) and 119877

119898119899(119909 119910)

Initialization let 1198630(119909 119910) = 1 and

1198730(119909 119910) = 119880

0(119910) = 120593

00+

119910 minus 1199100

12059301

+ sdot sdot sdot +119910 minus 119910119899minus1

|

| 1205930119899

(27)

By Lemma 8 a polynomial matrix 1198640(119910) and a real polyno-

mial 1198650(119910) exist such that

(a1) 1198730(119909 119910) = 119864

0(119910)1198650(119910)

(a2) 1198650(119910) ge 0

(a3) 1198650(119910) | 119864

0(119910)2

Recursion for 119895 = 1 2 119898 let

119878(119895)

(119909 119910) = 119878(119895minus1)

(119909 119910) + 119880119895(119910) (119909 minus 119909

0) sdot sdot sdot (119909 minus 119909

119895minus1)

(28)

with 119880119895(119910) = 119864

119895(119910)119865119895(119910) 119865119895(119910) | 119864

119895(119910)2 having the

representation

(b1) 119878(119895)

(119909 119910) = 119873119895(119909 119910)119863

119895(119909 119910)

(b2) 119863119895(119909 119910) ge 0

(b3) 119863119895(119909 119910) | 119873

119895(119909 119910)

2

where 119863119895(119909 119910) is a real polynomial

6 Journal of Applied Mathematics

Then we can get

119878(119895+1)

(119909 119910) = 119878(119895)

(119909 119910) + 119880119895+1

(119909 minus 1199090) sdot sdot sdot (119909 minus 119909

119895)

=119873119895(119909 119910)

119863119895(119909 119910)

+119864119895+1

(119910)

119865119895+1

(119910)(119909 minus 119909

0) sdot sdot sdot (119909 minus 119909

119895)

= (119873119895(119909 119910) 119865

119895+1(119910) + 119863

119895(119909 119910) 119864

119895+1(119910)

times (119909 minus 1199090) sdot sdot sdot (119909 minus 119909

119895))

times (119863119895(119909 119910) 119865

119895+1(119910))minus1

=119873119895+1

(119909 119910)

119863119895+1

(119909 119910)

(29)

where

119873119895+1

(119909 119910) = 119873119895(119909 119910) 119865

119895+1(119910)

+ 119863119895(119909 119910) 119864

119895+1(119910) (119909 minus 119909

0)

sdot sdot sdot (119909 minus 119909119895)

119863119895+1

(119909 119910) = 119863119895(119909 119910) 119865

119895+1(119910)

(30)

Obviously

10038171003817100381710038171003817119873119895+1

10038171003817100381710038171003817

2

=10038171003817100381710038171003817119873119895(119909 119910)

10038171003817100381710038171003817

2

1198652

(119895+1)(119910) + 119863

2

119895(119909 119910)

times10038171003817100381710038171003817119864(119895+1)

(119910)10038171003817100381710038171003817

2

(119909 minus 1199090)2

sdot sdot sdot (119909 minus 119909119895)2

+ 119873119895(119909 119910) sdot 119864

119895+1(119910) 119865119895+1

(119910)119863119895(119909 119910)

times (119909 minus 1199090) sdot sdot sdot (119909 minus 119909

119895)

+ 119873119895(119909 119910) sdot 119864

119895+1(119910) 119865119895+1

(119910)119863119895(119909 119910)

times (119909 minus 1199090) sdot sdot sdot (119909 minus 119909

119895)

(31)

since

119863119895(119909 119910) |

10038171003817100381710038171003817119873119895(119909 119910)

10038171003817100381710038171003817

2

119865119895+1

(119910) |10038171003817100381710038171003817119864119895+1

(119910)10038171003817100381710038171003817

2

119863119895+1

(119909 119910) |10038171003817100381710038171003817119873119895+1

(119909 119910)10038171003817100381710038171003817

2

(32)

Termination 119878(119898)(119909 119910) = 119877119898119899

(119909 119910)

Theorem 12 (Characterization) Let 119877119898119899

(119909 119910) = 119873(119909 119910)

119863(119909 119910) be expressed as

119877 = 119877119898119899

(119909 119910) = 1198800(119910) + 119880

1(119910) (119909 minus 119909

0)

+ sdot sdot sdot + 119880119898

(119910) (119909 minus 1199090) sdot sdot sdot (119909 minus 119909

119898minus1)

(33)

where 119880119897(119910) = 119864

119894119865119894 as in (12) we assume 119865

119894 119865119895have no

common factor 119894 119895 = 0 1 119898 119894 = 119895 then

(i) if n is even 119877 is of type [119898119899 + 119898 + 119899119898119899 + 119899](ii) if n is odd 119877 is of type [119898119899 + 119899119898119899 + 119899 minus 119898 minus 1]

Proof The proof is recursive Let 119899 be evenFor 119898 = 0 119877

0119899(119909 119910) = 119880

0(119910) = 119864

01198650 by Lemma 10 we

find that it is of type [119899119899]For 119898 = 1 because deg119864

0 = deg119865

0 = 119899 deg119864

1 =

deg1198651 = 119899

1198771119899

(119909 119910) =1198731

1198631

= 1198800(119910) + 119880

1(119910) (119909 minus 119909

0)

=1198640

1198650

+1198641

1198651

(119909 minus 1199090)

=11986401198651+ 11986411198650(119909 minus 119909

0)

11986501198651

(34)

is of type [2119899 + 12119899]For 119898 = 2 from the formula above we know that

deg1198731 = 2119899 + 1 deg119863

1 = 2119899 deg119864

2 = 119899 and deg119865

2 =

119899 one has1198772119899

(119909 119910) =1198732

1198632

=1198731

1198631

+1198642

1198652

(119909 minus 1199090) (119909 minus 119909

1)

=11987311198652+ 11986421198631(119909 minus 119909

0) (119909 minus 119909

1)

11986311198652

(35)

It is easy to find that 1198772119899

(119909 119910) is of type [3119899 + 23119899]For 119898 = 119896 let 119877

119896119899(119909 119910) = 119873

119896119863119896be of type [(119896 + 1)119899 +

119896(119896 + 1)119899]For 119898 = 119896 + 1 we consider that

119877119896+1119899

(119909 119910) =119873119896

119863119896

+119864119896+1

119865119896+1

(119909 minus 1199090) sdot sdot sdot (119909 minus 119909

119896)

=119873119896119865119896+1

+ 119864119896+1

119863119896(119909 minus 119909

0) sdot sdot sdot (119909 minus 119909

119896)

119863119896119865119896

=119873119896+1

119863119896+1

(36)

wheredeg 119873

119896+1 = (119896 + 1) (119899 + 1) + 119899

= (119896 + 2) 119899 + (119896 + 1)

deg 119863119896+1

= (119896 + 1) 119899 + 119899 = (119896 + 2) 119899

(37)

Therefore when 119899 is even 119877 is of type [119898119899 + 119898 + 119899119898119899 + 119899]Similarly if 119899 is odd for 119898 = 0 119877

0119899(119909 119910) = 119880

0(119910) =

11986401198650 by Lemma 10 we find that it is of type [119899119899 minus 1]

For 119898 = 1 because deg1198640 = deg119864

1 = 119899 deg119865

0 =

deg1198651 = 119899 minus 1

1198771119899

(119909 119910) =1198731

1198631

= 1198800(119910) + 119880

1(119910) (119909 minus 119909

0)

=1198640

1198650

+1198641

1198651

(119909 minus 1199090)

=11986401198651+ 11986411198650(119909 minus 119909

0)

11986501198651

(38)

is of type [21198992119899 minus 2]

Journal of Applied Mathematics 7

For119898 = 2 from the formula above we know that deg1198731

= 2119899 deg1198631 = 2119899 minus 2 deg119864

2 = 119899 and deg119865

2 = 119899 minus 1

one has

1198772119899

(119909 119910) =1198732

1198632

=1198731

1198631

+1198642

1198652

(119909 minus 1199090) (119909 minus 119909

1)

=11987311198652+ 11986421198631(119909 minus 119909

0) (119909 minus 119909

1)

11986311198652

(39)

We get that 1198772119899

(119909 119910) is of type [31198993119899 minus 3]For 119898 = 119896 let

119877119896119899

(119909 119910) =119873119896

119863119896

(40)

be of type [(119896 + 1)119899(119896 + 1)(119899 minus 1)]For 119898 = 119896 + 1 we consider that

119877119896+1119899

(119909 119910) =119873119896

119863119896

+119864119896+1

119865119896+1

(119909 minus 1199090) sdot sdot sdot (119909 minus 119909

119896)

=119873119896119865119896+1

+ 119864119896+1

119863119896(119909 minus 119909

0) sdot sdot sdot (119909 minus 119909

119896)

119863119896119865119896

=119873119896+1

119863119896+1

(41)

where

deg 119873119896+1

= (119896 + 1) (119899 minus 1) + 119896 + 1 = (119896 + 2) 119899

deg 119863119896+1

= (119896 + 1) (119899 minus 1) + (119899 minus 1) = (119896 + 2) (119899 minus 1)

(42)

Therefore when 119899 is odd119877 is of type [(119898+1)119899(119898+1)(119899minus

1)]

Definition 13 A matrix-valued Newton-Thiele type ratio-nal fraction 119877

119898119899(119909 119910) = 119873(119909 119910)119863(119909 119910) is defined to

be a bivariate generalized inverse and rational interpolant(BGIRINT) on the rectangular grid Λ

119898119899if

(i) 119877119898119899

(119909119897 119910119896) = 119884119897119896 (119909119897 119910119896) isin Λ

119898119899

(ii) 119863(119909119897 119910119896) = 0 (119909

119897 119910119896) isin Λ

119898119899

(iii) (a) if 119899 is even

deg 119873 (119909 119910) = 119898119899+ 119898+119899 deg 119863 (119909 119910) = 119898119899+119899

(43)

(b) if n is odd

deg 119873 (119909 119910) = 119898119899+119899 deg 119863 (119909 119910) = 119898119899+119899minus119898minus1

(44)

(iv) 119863(119909 119910) | 119873(119909 119910)2

(v) 119863(119909 119910) is real and 119863(119909 119910) ge 0

Now let us turn to the error estimate of the BGIRINTFirstly we give Lemma 14

Lemma 14 Let

119877lowast

119898119899(119909 119910) = 119880

lowast

0(119910) + 119880

lowast

1(119910) (119909 minus 119909

0)

+ sdot sdot sdot + 119880lowast

119898(119910) (119909minus119909

0) (119909minus119909

1) sdot sdot sdot (119909minus119909

119898minus1)

(45)

where for 119897 = 0 1 119898

119880lowast

119897(119910) = 120593

1198970(1199090 119909

119897 1199100) +

119910 minus 1199100

1205931198971

(1199090 119909

119897 1199100 1199101)

+ sdot sdot sdot +119910 minus 119910119899minus1

|

| 120593119897119899

(1199090 119909

119897 1199100 119910

119899)

+119910 minus 119910119899|

| 120593119897119899+1

(1199090 119909

119897 y0 119910

119899 119910)

(46)

then 119877lowast

119898119899(119909 119910) satisfies

119877lowast

119898119899(119909119894 119910) = 119891 (119909

119894 119910) 119894 = 0 1 119898

119877lowast

119898119899(119909 119910119895) = 119877119898119899

(119909 119910119895) 119895 = 0 1 119899

(47)

Remark 15 We delete the proof of the above lemma since itcan be easily generalized from [3]

Now the error estimate Theorem 16 will be put which isalso motivated by [3] We let

119877119898119899

(119909 119910) =119873 (119909 119910)

119863 (119909 119910) 119877

lowast

119898119899(119909 119910) =

119873lowast

(119909 119910)

119863lowast (119909 119910) (48)

Theorem 16 Let a matrix 119891(119909 119910) bemax(119898 + 1 119899 + 1) timescontinuously differentiable on a set119863 = 119886 le 119909 le 119887 119888 le 119910 le 119889

containing the points Λ119898119899

= (119909119897 119910119896) 119897 = 0 1 119898 119896 =

0 1 119899 (119909119897 119910119896) isin R2 then for any point (119909 119910) isin 119863 there

exists a point (120585 120578) isin 119863 such that

119891 (119909 119910) minus 119877119898119899

(119909 119910)

=1

119863 (119909 119910)119863lowast (119909 119910)

times 120596119898+1

(119909)

(119898 + 1)

120597119898+1

120597119909119898+11198641(120585 119910) +

120603119899+1

(119910)

(119899 + 1)

120597119899+1

120597119910119899+11198642(119909 120578)

(49)

where

120596119898+1

(119909) = (119909 minus 1199090) (119909 minus 119909

1) sdot sdot sdot (119909 minus 119909

119898)

120603119899+1

(119910) = (119910 minus 1199100) (119910 minus 119910

1) sdot sdot sdot (119910 minus 119910

119899)

1198641(119909 119910) = 119863 (119909 119910)119863

lowast

(119909 119910) [119891 (119909 119910) minus 119877lowast

119898119899(119909 119910)]

1198642(119909 119910) = 119863 (119909 119910)119863

lowast

(119909 119910) [119877lowast

119898119899(119909 119910) minus 119877

119898119899(119909 119910)]

(50)

8 Journal of Applied Mathematics

Proof Let

1198641(119909 119910) = 119863 (119909 119910)119863

lowast

(119909 119910) [119891 (119909 119910) minus 119877lowast

119898119899(119909 119910)]

1198642(119909 119910) = 119863 (119909 119910)119863

lowast

(119909 119910) [119877lowast

119898119899(119909 119910) minus 119877

119898119899(119909 119910)]

119864 (119909 119910) = 1198641(119909 119910) + 119864

2(119909 119910)

(51)

From Lemma 14 we know

1198641(119909119894 119910) = 0 119894 = 0 1 119898 (52)

which results in

1198641(119909 119910) =

120596119898+1

(119909)

(119898 + 1)

120597119898+1

120597119909119898+11198641(119909 119910)

100381610038161003816100381610038161003816100381610038161003816119909=120585

(53)

where 120596119898+1

(119909) = (119909 minus 1199090)(119909 minus 119909

1) sdot sdot sdot (119909 minus 119909

119898) and 120585 is a

number contained in the interval (119886 119887) which may dependon 119910 Similarly from

1198642(119909 119910119895) = 0 119895 = 0 1 119899 (54)

we can get that

1198642(119909 119910) =

120603119899+1

(119910)

(119899 + 1)

120597119899+1

120597119910119899+11198642(119909 119910)

100381610038161003816100381610038161003816100381610038161003816119910=120578

(55)

where120603119899+1

(119910) = (119910minus1199100)(119910minus119910

1) sdot sdot sdot (119910minus119910

119899) and 120578 is a number

contained in the interval (119888 119889) which may depend on 119909Therefore

119864 (119909 119910) = 1198641(119909 119910) + 119864

2(119909 119910)

=120596119898+1

(119909)

(119898 + 1)

120597119898+1

120597119909119898+11198641(119909 119910)

100381610038161003816100381610038161003816100381610038161003816119909=120585

+120603119899+1

(119910)

(119899 + 1)

120597119899+1

120597119910119899+11198642(119909 119910)

100381610038161003816100381610038161003816100381610038161003816119910=120578

(56)

The proof is thus completed

4 Numerical Examples

Example 17 Let (119909119894 119910119895) and 119884

119894119895 119894 119895 = 0 1 2 be given in

Table 1

11987722

(119909 119910) = 1198800(119910) + 119880

1(119910) 119909 + 119880

2(119910) 119909 (119909 minus 1) (57)

Using Algorithm 7 we can get

1198800(119910) = (

1 0

0 0) +

119910

( 0 00 1

)+

119910 minus 1 |

| (0 0

minus25 minus15)

1198801(119910) = (

0 minus1

0 0) +

119910

(0 12

minus12 0)

+119910 minus 1 |

| ( 0 11 0

)

1198802(119910) = (

0 1

1

20) +

119910

(0 minus35

15 0)

+119910 minus 1 |

| (0 minus1514

minus514 minus57)

(58)

Based on generalized inverse (4) we obtain

11987722

(119909 119910)

= (11988611

11988612

11988621

11988622

) times (701199106

minus 3681199105

+ 8921199104

minus12401199103

+10401199102

minus512119910+128)minus1

=119873 (119909 119910)

119863 (119909 119910)

(59)

where

11988611

= 701199106

minus 3681199105

+ 8921199104

minus 12401199103

+ 10401199102

minus 512119910 + 128

11988612

= 1781199105

119909 minus 8581199104

119909 + 16881199103

119909

minus 17761199091199102

+ 1024119909119910 minus 256119909 minus 381199105

1199092

+ 2621199104

1199092

minus 6401199103

1199092

minus 5121199092

119910

minus 51199106

1199092

+ 51199106

119909 + 8001199092

1199102

+ 1281199092

11988621

= 1081199105

minus 2001199104

+ 2001199103

minus 1121199102

+ 32119910

minus 281199106

+ 601199106

119909 minus 3441199105

119909 + 7361199104

119909 minus 7841199103

119909

+ 4081199091199102

minus 32119909119910 minus 64119909 minus 241199105

1199092

+ 161199104

1199092

minus 961199092

119910 + 101199106

1199092

+ 401199092

1199102

+ 641199092

11988622

= 681199105

minus 1401199104

+ 1601199103

minus 961199102

+ 32119910 minus 141199106

minus 501199106

1199092

+ 2701199105

1199092

minus 6401199104

1199092

+ 8201199103

1199092

minus 5601199092

1199102

+ 1601199092

119910 + 501199106

119909 minus 2701199105

119909

+ 6401199104

119909 minus 8201199103

119909 + 5601199091199102

minus 160119909119910

(60)

FromDefinition 13 we know that 11987722

(119909 119910) is a BGIRINTsince 119877

22(119909119894 119910119895) = 119884119894119895and deg 119873(119909 119910) = 8 deg 119863(119909 119910) =

6 and 119863(119909 119910) | 119873(119909 119910)2

In paper [9] a bivariateThieletypematrix-valued rationalinterpolant 119877

119879(119909 119910) = 119866

00(119910) + (119909 minus 119909

0)11986610

(119910) + sdot sdot sdot + (119909 minus

119909119898minus1

)1198661198990

(119910) (page 73) where for 119897 = 0 1 119899

1198661198970

(119910) = 1198611198970

(1199090 119909

119897 1199100)

+119910 minus 1199100

1198611198971

(1199090 119909

119897 1199100 1199101)

+ sdot sdot sdot +119910 minus 119910119898minus1

|

| 119861119897119898

(1199090 119909

119897 1199100 119910

119898)

(61)

Here 119861119897119896(1199090 119909

119897 1199100 119910

119896) 119897 = 0 1 119899 119896 = 0 1 119898

is defined different from that of this paper (see [9]) We

Journal of Applied Mathematics 9

Table 1

119910119895119909119894

1199090= 0 119909

1= 1 119909

2= 2

1199100= 0 (

1 0

0 0) (

1 minus1

0 0) (

1 0

1 0)

1199101= 1 (

1 0

0 1) (

1 0

minus1 1) (

1 minus1

0 1)

1199102= 2 (

1 0

minus1 0) (

1 1

minus1 0) (

1 1

0 minus1)

Table 2

119884(13 13) 119877NT(13 13) 119877NT minus 119884119865

(minus0856888 05155013 3669296

130000 366929 2600) (

minus0856888 05155010 3669297

130000 366929 2600) 109119890 minus 006

119877119879(13 13) 119877

119879minus 119884119865

(minus098999 014111 547394

129999 547394 2999) 2613697

119884(15 15) 119877NT(15 15) 119877NT minus 119884119865

(minus0989999 01411200 4481689

150000 4481689 300) (

minus0989992 01411203 4481688

150000 4481688 300) 915119890 minus 007

119877119879(15 15) 119877

119879minus 119884119865

(minus099829372 minus005837 547394

149999 547394 31999) 143144

now give another numerical example to compare the twoalgorithms which shows that the method of this paper isbetter than the one of [9]

Example 18 Let

119884 (119909 119910) = (cos (119909 + 119910) sin (119909 + 119910) 119890

119910

119909 119890119910

119909 + 119910) (62)

and we suppose 1199090 1199091 1199092 1199093 1199094 1199095 = 119910

0 1199101 1199102 1199103 1199104 1199105

= 10 12 14 16 18 20 The numerical results 119877NT(119909 119910)

in Step 9 of Algorithm 7 and 119877119879(119909 119910) in [9] are given in

Table 2

Remark 19 InTable 2 from the F-normof119877NTminus119884 and119877119879minus119884

we can see that the error using Newton-Thiele type formulais much less than the one usingThieletype formula

5 Conclusion

In this paper the bivariate generalized inverse Newton-Thieletype matrix interpolation to approximate a matrix function119891(119909 119910) is given and we also give a recursive algorithmaccompanied with some other important conclusions such asdivisibility characterization and some numerical examplesFrom the second example it is easy to find that the approxi-mant using Newton-Thiele type formula is much better thanthe one usingThieletype formula

Acknowledgments

The work is supported by Shanghai Natural Science Foun-dation (10ZR1410900) and by Key Disciplines of ShanghaiMunicipality (S30104)

References

[1] G A Baker and P R Graves-Morris Pade Approximation PartII Addison-Wesley Publishing Company Reading Mass USA1981

[2] I Kh Kuchminska OM Sus and SMVozna ldquoApproximationproperties of two-dimensional continued fractionsrdquo UkrainianMathematical Journal vol 55 pp 36ndash54 2003

[3] J Tan and Y Fang ldquoNewton-Thielersquos rational interpolantsrdquoNumerical Algorithms vol 24 no 1-2 pp 141ndash157 2000

[4] P R Graves-Morris ldquoVector valued rational interpolants IrdquoNumerische Mathematik vol 42 no 3 pp 331ndash348 1983

[5] G Chuan-qing and C Zhibing ldquoMatrix valued rational inter-polants and its error formulardquo Mathematica Numerica Sinicavol 17 pp 73ndash77 1995

[6] G Chuan-qing ldquoMatrix Pade type approximat and directionalmatrix Pade approximat in the inner product spacerdquo Journal ofComputational and Applied Mathematics vol 164-165 pp 365ndash385 2004

[7] C Gu ldquoThiele-type and Lagrange-type generalized inverserational interpolation for rectangular complexmatricesrdquo LinearAlgebra and Its Applications vol 295 no 1ndash3 pp 7ndash30 1999

[8] C Gu ldquoA practical two-dimensional Thiele-type matrix Padeapproximationrdquo IEEE Transactions on Automatic Control vol48 no 12 pp 2259ndash2263 2003

10 Journal of Applied Mathematics

[9] C Gu ldquoBivariate Thiele-type matrix-valued rational inter-polantsrdquo Journal of Computational and Applied Mathematicsvol 80 no 1 pp 71ndash82 1997

[10] N K Bose and S Basu ldquoTwo-dimensional matrix Pade approx-imants existence nonuniqueness and recursive computationrdquoIEEE Transactions on Automatic Control vol 25 no 3 pp 509ndash514 1980

[11] A Cuyt and B Verdonk ldquoMultivariate reciprocal differencesfor branched Thiele continued fraction expansionsrdquo Journal ofComputational and Applied Mathematics vol 21 no 2 pp 145ndash160 1988

[12] W Siemaszko ldquoThiele-tyoe branched continued fractions fortwo-variable functionsrdquo Journal of Computational and AppliedMathematics vol 9 no 2 pp 137ndash153 1983

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

6 Journal of Applied Mathematics

Then we can get

119878(119895+1)

(119909 119910) = 119878(119895)

(119909 119910) + 119880119895+1

(119909 minus 1199090) sdot sdot sdot (119909 minus 119909

119895)

=119873119895(119909 119910)

119863119895(119909 119910)

+119864119895+1

(119910)

119865119895+1

(119910)(119909 minus 119909

0) sdot sdot sdot (119909 minus 119909

119895)

= (119873119895(119909 119910) 119865

119895+1(119910) + 119863

119895(119909 119910) 119864

119895+1(119910)

times (119909 minus 1199090) sdot sdot sdot (119909 minus 119909

119895))

times (119863119895(119909 119910) 119865

119895+1(119910))minus1

=119873119895+1

(119909 119910)

119863119895+1

(119909 119910)

(29)

where

119873119895+1

(119909 119910) = 119873119895(119909 119910) 119865

119895+1(119910)

+ 119863119895(119909 119910) 119864

119895+1(119910) (119909 minus 119909

0)

sdot sdot sdot (119909 minus 119909119895)

119863119895+1

(119909 119910) = 119863119895(119909 119910) 119865

119895+1(119910)

(30)

Obviously

10038171003817100381710038171003817119873119895+1

10038171003817100381710038171003817

2

=10038171003817100381710038171003817119873119895(119909 119910)

10038171003817100381710038171003817

2

1198652

(119895+1)(119910) + 119863

2

119895(119909 119910)

times10038171003817100381710038171003817119864(119895+1)

(119910)10038171003817100381710038171003817

2

(119909 minus 1199090)2

sdot sdot sdot (119909 minus 119909119895)2

+ 119873119895(119909 119910) sdot 119864

119895+1(119910) 119865119895+1

(119910)119863119895(119909 119910)

times (119909 minus 1199090) sdot sdot sdot (119909 minus 119909

119895)

+ 119873119895(119909 119910) sdot 119864

119895+1(119910) 119865119895+1

(119910)119863119895(119909 119910)

times (119909 minus 1199090) sdot sdot sdot (119909 minus 119909

119895)

(31)

since

119863119895(119909 119910) |

10038171003817100381710038171003817119873119895(119909 119910)

10038171003817100381710038171003817

2

119865119895+1

(119910) |10038171003817100381710038171003817119864119895+1

(119910)10038171003817100381710038171003817

2

119863119895+1

(119909 119910) |10038171003817100381710038171003817119873119895+1

(119909 119910)10038171003817100381710038171003817

2

(32)

Termination 119878(119898)(119909 119910) = 119877119898119899

(119909 119910)

Theorem 12 (Characterization) Let 119877119898119899

(119909 119910) = 119873(119909 119910)

119863(119909 119910) be expressed as

119877 = 119877119898119899

(119909 119910) = 1198800(119910) + 119880

1(119910) (119909 minus 119909

0)

+ sdot sdot sdot + 119880119898

(119910) (119909 minus 1199090) sdot sdot sdot (119909 minus 119909

119898minus1)

(33)

where 119880119897(119910) = 119864

119894119865119894 as in (12) we assume 119865

119894 119865119895have no

common factor 119894 119895 = 0 1 119898 119894 = 119895 then

(i) if n is even 119877 is of type [119898119899 + 119898 + 119899119898119899 + 119899](ii) if n is odd 119877 is of type [119898119899 + 119899119898119899 + 119899 minus 119898 minus 1]

Proof The proof is recursive Let 119899 be evenFor 119898 = 0 119877

0119899(119909 119910) = 119880

0(119910) = 119864

01198650 by Lemma 10 we

find that it is of type [119899119899]For 119898 = 1 because deg119864

0 = deg119865

0 = 119899 deg119864

1 =

deg1198651 = 119899

1198771119899

(119909 119910) =1198731

1198631

= 1198800(119910) + 119880

1(119910) (119909 minus 119909

0)

=1198640

1198650

+1198641

1198651

(119909 minus 1199090)

=11986401198651+ 11986411198650(119909 minus 119909

0)

11986501198651

(34)

is of type [2119899 + 12119899]For 119898 = 2 from the formula above we know that

deg1198731 = 2119899 + 1 deg119863

1 = 2119899 deg119864

2 = 119899 and deg119865

2 =

119899 one has1198772119899

(119909 119910) =1198732

1198632

=1198731

1198631

+1198642

1198652

(119909 minus 1199090) (119909 minus 119909

1)

=11987311198652+ 11986421198631(119909 minus 119909

0) (119909 minus 119909

1)

11986311198652

(35)

It is easy to find that 1198772119899

(119909 119910) is of type [3119899 + 23119899]For 119898 = 119896 let 119877

119896119899(119909 119910) = 119873

119896119863119896be of type [(119896 + 1)119899 +

119896(119896 + 1)119899]For 119898 = 119896 + 1 we consider that

119877119896+1119899

(119909 119910) =119873119896

119863119896

+119864119896+1

119865119896+1

(119909 minus 1199090) sdot sdot sdot (119909 minus 119909

119896)

=119873119896119865119896+1

+ 119864119896+1

119863119896(119909 minus 119909

0) sdot sdot sdot (119909 minus 119909

119896)

119863119896119865119896

=119873119896+1

119863119896+1

(36)

wheredeg 119873

119896+1 = (119896 + 1) (119899 + 1) + 119899

= (119896 + 2) 119899 + (119896 + 1)

deg 119863119896+1

= (119896 + 1) 119899 + 119899 = (119896 + 2) 119899

(37)

Therefore when 119899 is even 119877 is of type [119898119899 + 119898 + 119899119898119899 + 119899]Similarly if 119899 is odd for 119898 = 0 119877

0119899(119909 119910) = 119880

0(119910) =

11986401198650 by Lemma 10 we find that it is of type [119899119899 minus 1]

For 119898 = 1 because deg1198640 = deg119864

1 = 119899 deg119865

0 =

deg1198651 = 119899 minus 1

1198771119899

(119909 119910) =1198731

1198631

= 1198800(119910) + 119880

1(119910) (119909 minus 119909

0)

=1198640

1198650

+1198641

1198651

(119909 minus 1199090)

=11986401198651+ 11986411198650(119909 minus 119909

0)

11986501198651

(38)

is of type [21198992119899 minus 2]

Journal of Applied Mathematics 7

For119898 = 2 from the formula above we know that deg1198731

= 2119899 deg1198631 = 2119899 minus 2 deg119864

2 = 119899 and deg119865

2 = 119899 minus 1

one has

1198772119899

(119909 119910) =1198732

1198632

=1198731

1198631

+1198642

1198652

(119909 minus 1199090) (119909 minus 119909

1)

=11987311198652+ 11986421198631(119909 minus 119909

0) (119909 minus 119909

1)

11986311198652

(39)

We get that 1198772119899

(119909 119910) is of type [31198993119899 minus 3]For 119898 = 119896 let

119877119896119899

(119909 119910) =119873119896

119863119896

(40)

be of type [(119896 + 1)119899(119896 + 1)(119899 minus 1)]For 119898 = 119896 + 1 we consider that

119877119896+1119899

(119909 119910) =119873119896

119863119896

+119864119896+1

119865119896+1

(119909 minus 1199090) sdot sdot sdot (119909 minus 119909

119896)

=119873119896119865119896+1

+ 119864119896+1

119863119896(119909 minus 119909

0) sdot sdot sdot (119909 minus 119909

119896)

119863119896119865119896

=119873119896+1

119863119896+1

(41)

where

deg 119873119896+1

= (119896 + 1) (119899 minus 1) + 119896 + 1 = (119896 + 2) 119899

deg 119863119896+1

= (119896 + 1) (119899 minus 1) + (119899 minus 1) = (119896 + 2) (119899 minus 1)

(42)

Therefore when 119899 is odd119877 is of type [(119898+1)119899(119898+1)(119899minus

1)]

Definition 13 A matrix-valued Newton-Thiele type ratio-nal fraction 119877

119898119899(119909 119910) = 119873(119909 119910)119863(119909 119910) is defined to

be a bivariate generalized inverse and rational interpolant(BGIRINT) on the rectangular grid Λ

119898119899if

(i) 119877119898119899

(119909119897 119910119896) = 119884119897119896 (119909119897 119910119896) isin Λ

119898119899

(ii) 119863(119909119897 119910119896) = 0 (119909

119897 119910119896) isin Λ

119898119899

(iii) (a) if 119899 is even

deg 119873 (119909 119910) = 119898119899+ 119898+119899 deg 119863 (119909 119910) = 119898119899+119899

(43)

(b) if n is odd

deg 119873 (119909 119910) = 119898119899+119899 deg 119863 (119909 119910) = 119898119899+119899minus119898minus1

(44)

(iv) 119863(119909 119910) | 119873(119909 119910)2

(v) 119863(119909 119910) is real and 119863(119909 119910) ge 0

Now let us turn to the error estimate of the BGIRINTFirstly we give Lemma 14

Lemma 14 Let

119877lowast

119898119899(119909 119910) = 119880

lowast

0(119910) + 119880

lowast

1(119910) (119909 minus 119909

0)

+ sdot sdot sdot + 119880lowast

119898(119910) (119909minus119909

0) (119909minus119909

1) sdot sdot sdot (119909minus119909

119898minus1)

(45)

where for 119897 = 0 1 119898

119880lowast

119897(119910) = 120593

1198970(1199090 119909

119897 1199100) +

119910 minus 1199100

1205931198971

(1199090 119909

119897 1199100 1199101)

+ sdot sdot sdot +119910 minus 119910119899minus1

|

| 120593119897119899

(1199090 119909

119897 1199100 119910

119899)

+119910 minus 119910119899|

| 120593119897119899+1

(1199090 119909

119897 y0 119910

119899 119910)

(46)

then 119877lowast

119898119899(119909 119910) satisfies

119877lowast

119898119899(119909119894 119910) = 119891 (119909

119894 119910) 119894 = 0 1 119898

119877lowast

119898119899(119909 119910119895) = 119877119898119899

(119909 119910119895) 119895 = 0 1 119899

(47)

Remark 15 We delete the proof of the above lemma since itcan be easily generalized from [3]

Now the error estimate Theorem 16 will be put which isalso motivated by [3] We let

119877119898119899

(119909 119910) =119873 (119909 119910)

119863 (119909 119910) 119877

lowast

119898119899(119909 119910) =

119873lowast

(119909 119910)

119863lowast (119909 119910) (48)

Theorem 16 Let a matrix 119891(119909 119910) bemax(119898 + 1 119899 + 1) timescontinuously differentiable on a set119863 = 119886 le 119909 le 119887 119888 le 119910 le 119889

containing the points Λ119898119899

= (119909119897 119910119896) 119897 = 0 1 119898 119896 =

0 1 119899 (119909119897 119910119896) isin R2 then for any point (119909 119910) isin 119863 there

exists a point (120585 120578) isin 119863 such that

119891 (119909 119910) minus 119877119898119899

(119909 119910)

=1

119863 (119909 119910)119863lowast (119909 119910)

times 120596119898+1

(119909)

(119898 + 1)

120597119898+1

120597119909119898+11198641(120585 119910) +

120603119899+1

(119910)

(119899 + 1)

120597119899+1

120597119910119899+11198642(119909 120578)

(49)

where

120596119898+1

(119909) = (119909 minus 1199090) (119909 minus 119909

1) sdot sdot sdot (119909 minus 119909

119898)

120603119899+1

(119910) = (119910 minus 1199100) (119910 minus 119910

1) sdot sdot sdot (119910 minus 119910

119899)

1198641(119909 119910) = 119863 (119909 119910)119863

lowast

(119909 119910) [119891 (119909 119910) minus 119877lowast

119898119899(119909 119910)]

1198642(119909 119910) = 119863 (119909 119910)119863

lowast

(119909 119910) [119877lowast

119898119899(119909 119910) minus 119877

119898119899(119909 119910)]

(50)

8 Journal of Applied Mathematics

Proof Let

1198641(119909 119910) = 119863 (119909 119910)119863

lowast

(119909 119910) [119891 (119909 119910) minus 119877lowast

119898119899(119909 119910)]

1198642(119909 119910) = 119863 (119909 119910)119863

lowast

(119909 119910) [119877lowast

119898119899(119909 119910) minus 119877

119898119899(119909 119910)]

119864 (119909 119910) = 1198641(119909 119910) + 119864

2(119909 119910)

(51)

From Lemma 14 we know

1198641(119909119894 119910) = 0 119894 = 0 1 119898 (52)

which results in

1198641(119909 119910) =

120596119898+1

(119909)

(119898 + 1)

120597119898+1

120597119909119898+11198641(119909 119910)

100381610038161003816100381610038161003816100381610038161003816119909=120585

(53)

where 120596119898+1

(119909) = (119909 minus 1199090)(119909 minus 119909

1) sdot sdot sdot (119909 minus 119909

119898) and 120585 is a

number contained in the interval (119886 119887) which may dependon 119910 Similarly from

1198642(119909 119910119895) = 0 119895 = 0 1 119899 (54)

we can get that

1198642(119909 119910) =

120603119899+1

(119910)

(119899 + 1)

120597119899+1

120597119910119899+11198642(119909 119910)

100381610038161003816100381610038161003816100381610038161003816119910=120578

(55)

where120603119899+1

(119910) = (119910minus1199100)(119910minus119910

1) sdot sdot sdot (119910minus119910

119899) and 120578 is a number

contained in the interval (119888 119889) which may depend on 119909Therefore

119864 (119909 119910) = 1198641(119909 119910) + 119864

2(119909 119910)

=120596119898+1

(119909)

(119898 + 1)

120597119898+1

120597119909119898+11198641(119909 119910)

100381610038161003816100381610038161003816100381610038161003816119909=120585

+120603119899+1

(119910)

(119899 + 1)

120597119899+1

120597119910119899+11198642(119909 119910)

100381610038161003816100381610038161003816100381610038161003816119910=120578

(56)

The proof is thus completed

4 Numerical Examples

Example 17 Let (119909119894 119910119895) and 119884

119894119895 119894 119895 = 0 1 2 be given in

Table 1

11987722

(119909 119910) = 1198800(119910) + 119880

1(119910) 119909 + 119880

2(119910) 119909 (119909 minus 1) (57)

Using Algorithm 7 we can get

1198800(119910) = (

1 0

0 0) +

119910

( 0 00 1

)+

119910 minus 1 |

| (0 0

minus25 minus15)

1198801(119910) = (

0 minus1

0 0) +

119910

(0 12

minus12 0)

+119910 minus 1 |

| ( 0 11 0

)

1198802(119910) = (

0 1

1

20) +

119910

(0 minus35

15 0)

+119910 minus 1 |

| (0 minus1514

minus514 minus57)

(58)

Based on generalized inverse (4) we obtain

11987722

(119909 119910)

= (11988611

11988612

11988621

11988622

) times (701199106

minus 3681199105

+ 8921199104

minus12401199103

+10401199102

minus512119910+128)minus1

=119873 (119909 119910)

119863 (119909 119910)

(59)

where

11988611

= 701199106

minus 3681199105

+ 8921199104

minus 12401199103

+ 10401199102

minus 512119910 + 128

11988612

= 1781199105

119909 minus 8581199104

119909 + 16881199103

119909

minus 17761199091199102

+ 1024119909119910 minus 256119909 minus 381199105

1199092

+ 2621199104

1199092

minus 6401199103

1199092

minus 5121199092

119910

minus 51199106

1199092

+ 51199106

119909 + 8001199092

1199102

+ 1281199092

11988621

= 1081199105

minus 2001199104

+ 2001199103

minus 1121199102

+ 32119910

minus 281199106

+ 601199106

119909 minus 3441199105

119909 + 7361199104

119909 minus 7841199103

119909

+ 4081199091199102

minus 32119909119910 minus 64119909 minus 241199105

1199092

+ 161199104

1199092

minus 961199092

119910 + 101199106

1199092

+ 401199092

1199102

+ 641199092

11988622

= 681199105

minus 1401199104

+ 1601199103

minus 961199102

+ 32119910 minus 141199106

minus 501199106

1199092

+ 2701199105

1199092

minus 6401199104

1199092

+ 8201199103

1199092

minus 5601199092

1199102

+ 1601199092

119910 + 501199106

119909 minus 2701199105

119909

+ 6401199104

119909 minus 8201199103

119909 + 5601199091199102

minus 160119909119910

(60)

FromDefinition 13 we know that 11987722

(119909 119910) is a BGIRINTsince 119877

22(119909119894 119910119895) = 119884119894119895and deg 119873(119909 119910) = 8 deg 119863(119909 119910) =

6 and 119863(119909 119910) | 119873(119909 119910)2

In paper [9] a bivariateThieletypematrix-valued rationalinterpolant 119877

119879(119909 119910) = 119866

00(119910) + (119909 minus 119909

0)11986610

(119910) + sdot sdot sdot + (119909 minus

119909119898minus1

)1198661198990

(119910) (page 73) where for 119897 = 0 1 119899

1198661198970

(119910) = 1198611198970

(1199090 119909

119897 1199100)

+119910 minus 1199100

1198611198971

(1199090 119909

119897 1199100 1199101)

+ sdot sdot sdot +119910 minus 119910119898minus1

|

| 119861119897119898

(1199090 119909

119897 1199100 119910

119898)

(61)

Here 119861119897119896(1199090 119909

119897 1199100 119910

119896) 119897 = 0 1 119899 119896 = 0 1 119898

is defined different from that of this paper (see [9]) We

Journal of Applied Mathematics 9

Table 1

119910119895119909119894

1199090= 0 119909

1= 1 119909

2= 2

1199100= 0 (

1 0

0 0) (

1 minus1

0 0) (

1 0

1 0)

1199101= 1 (

1 0

0 1) (

1 0

minus1 1) (

1 minus1

0 1)

1199102= 2 (

1 0

minus1 0) (

1 1

minus1 0) (

1 1

0 minus1)

Table 2

119884(13 13) 119877NT(13 13) 119877NT minus 119884119865

(minus0856888 05155013 3669296

130000 366929 2600) (

minus0856888 05155010 3669297

130000 366929 2600) 109119890 minus 006

119877119879(13 13) 119877

119879minus 119884119865

(minus098999 014111 547394

129999 547394 2999) 2613697

119884(15 15) 119877NT(15 15) 119877NT minus 119884119865

(minus0989999 01411200 4481689

150000 4481689 300) (

minus0989992 01411203 4481688

150000 4481688 300) 915119890 minus 007

119877119879(15 15) 119877

119879minus 119884119865

(minus099829372 minus005837 547394

149999 547394 31999) 143144

now give another numerical example to compare the twoalgorithms which shows that the method of this paper isbetter than the one of [9]

Example 18 Let

119884 (119909 119910) = (cos (119909 + 119910) sin (119909 + 119910) 119890

119910

119909 119890119910

119909 + 119910) (62)

and we suppose 1199090 1199091 1199092 1199093 1199094 1199095 = 119910

0 1199101 1199102 1199103 1199104 1199105

= 10 12 14 16 18 20 The numerical results 119877NT(119909 119910)

in Step 9 of Algorithm 7 and 119877119879(119909 119910) in [9] are given in

Table 2

Remark 19 InTable 2 from the F-normof119877NTminus119884 and119877119879minus119884

we can see that the error using Newton-Thiele type formulais much less than the one usingThieletype formula

5 Conclusion

In this paper the bivariate generalized inverse Newton-Thieletype matrix interpolation to approximate a matrix function119891(119909 119910) is given and we also give a recursive algorithmaccompanied with some other important conclusions such asdivisibility characterization and some numerical examplesFrom the second example it is easy to find that the approxi-mant using Newton-Thiele type formula is much better thanthe one usingThieletype formula

Acknowledgments

The work is supported by Shanghai Natural Science Foun-dation (10ZR1410900) and by Key Disciplines of ShanghaiMunicipality (S30104)

References

[1] G A Baker and P R Graves-Morris Pade Approximation PartII Addison-Wesley Publishing Company Reading Mass USA1981

[2] I Kh Kuchminska OM Sus and SMVozna ldquoApproximationproperties of two-dimensional continued fractionsrdquo UkrainianMathematical Journal vol 55 pp 36ndash54 2003

[3] J Tan and Y Fang ldquoNewton-Thielersquos rational interpolantsrdquoNumerical Algorithms vol 24 no 1-2 pp 141ndash157 2000

[4] P R Graves-Morris ldquoVector valued rational interpolants IrdquoNumerische Mathematik vol 42 no 3 pp 331ndash348 1983

[5] G Chuan-qing and C Zhibing ldquoMatrix valued rational inter-polants and its error formulardquo Mathematica Numerica Sinicavol 17 pp 73ndash77 1995

[6] G Chuan-qing ldquoMatrix Pade type approximat and directionalmatrix Pade approximat in the inner product spacerdquo Journal ofComputational and Applied Mathematics vol 164-165 pp 365ndash385 2004

[7] C Gu ldquoThiele-type and Lagrange-type generalized inverserational interpolation for rectangular complexmatricesrdquo LinearAlgebra and Its Applications vol 295 no 1ndash3 pp 7ndash30 1999

[8] C Gu ldquoA practical two-dimensional Thiele-type matrix Padeapproximationrdquo IEEE Transactions on Automatic Control vol48 no 12 pp 2259ndash2263 2003

10 Journal of Applied Mathematics

[9] C Gu ldquoBivariate Thiele-type matrix-valued rational inter-polantsrdquo Journal of Computational and Applied Mathematicsvol 80 no 1 pp 71ndash82 1997

[10] N K Bose and S Basu ldquoTwo-dimensional matrix Pade approx-imants existence nonuniqueness and recursive computationrdquoIEEE Transactions on Automatic Control vol 25 no 3 pp 509ndash514 1980

[11] A Cuyt and B Verdonk ldquoMultivariate reciprocal differencesfor branched Thiele continued fraction expansionsrdquo Journal ofComputational and Applied Mathematics vol 21 no 2 pp 145ndash160 1988

[12] W Siemaszko ldquoThiele-tyoe branched continued fractions fortwo-variable functionsrdquo Journal of Computational and AppliedMathematics vol 9 no 2 pp 137ndash153 1983

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Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Journal of Applied Mathematics 7

For119898 = 2 from the formula above we know that deg1198731

= 2119899 deg1198631 = 2119899 minus 2 deg119864

2 = 119899 and deg119865

2 = 119899 minus 1

one has

1198772119899

(119909 119910) =1198732

1198632

=1198731

1198631

+1198642

1198652

(119909 minus 1199090) (119909 minus 119909

1)

=11987311198652+ 11986421198631(119909 minus 119909

0) (119909 minus 119909

1)

11986311198652

(39)

We get that 1198772119899

(119909 119910) is of type [31198993119899 minus 3]For 119898 = 119896 let

119877119896119899

(119909 119910) =119873119896

119863119896

(40)

be of type [(119896 + 1)119899(119896 + 1)(119899 minus 1)]For 119898 = 119896 + 1 we consider that

119877119896+1119899

(119909 119910) =119873119896

119863119896

+119864119896+1

119865119896+1

(119909 minus 1199090) sdot sdot sdot (119909 minus 119909

119896)

=119873119896119865119896+1

+ 119864119896+1

119863119896(119909 minus 119909

0) sdot sdot sdot (119909 minus 119909

119896)

119863119896119865119896

=119873119896+1

119863119896+1

(41)

where

deg 119873119896+1

= (119896 + 1) (119899 minus 1) + 119896 + 1 = (119896 + 2) 119899

deg 119863119896+1

= (119896 + 1) (119899 minus 1) + (119899 minus 1) = (119896 + 2) (119899 minus 1)

(42)

Therefore when 119899 is odd119877 is of type [(119898+1)119899(119898+1)(119899minus

1)]

Definition 13 A matrix-valued Newton-Thiele type ratio-nal fraction 119877

119898119899(119909 119910) = 119873(119909 119910)119863(119909 119910) is defined to

be a bivariate generalized inverse and rational interpolant(BGIRINT) on the rectangular grid Λ

119898119899if

(i) 119877119898119899

(119909119897 119910119896) = 119884119897119896 (119909119897 119910119896) isin Λ

119898119899

(ii) 119863(119909119897 119910119896) = 0 (119909

119897 119910119896) isin Λ

119898119899

(iii) (a) if 119899 is even

deg 119873 (119909 119910) = 119898119899+ 119898+119899 deg 119863 (119909 119910) = 119898119899+119899

(43)

(b) if n is odd

deg 119873 (119909 119910) = 119898119899+119899 deg 119863 (119909 119910) = 119898119899+119899minus119898minus1

(44)

(iv) 119863(119909 119910) | 119873(119909 119910)2

(v) 119863(119909 119910) is real and 119863(119909 119910) ge 0

Now let us turn to the error estimate of the BGIRINTFirstly we give Lemma 14

Lemma 14 Let

119877lowast

119898119899(119909 119910) = 119880

lowast

0(119910) + 119880

lowast

1(119910) (119909 minus 119909

0)

+ sdot sdot sdot + 119880lowast

119898(119910) (119909minus119909

0) (119909minus119909

1) sdot sdot sdot (119909minus119909

119898minus1)

(45)

where for 119897 = 0 1 119898

119880lowast

119897(119910) = 120593

1198970(1199090 119909

119897 1199100) +

119910 minus 1199100

1205931198971

(1199090 119909

119897 1199100 1199101)

+ sdot sdot sdot +119910 minus 119910119899minus1

|

| 120593119897119899

(1199090 119909

119897 1199100 119910

119899)

+119910 minus 119910119899|

| 120593119897119899+1

(1199090 119909

119897 y0 119910

119899 119910)

(46)

then 119877lowast

119898119899(119909 119910) satisfies

119877lowast

119898119899(119909119894 119910) = 119891 (119909

119894 119910) 119894 = 0 1 119898

119877lowast

119898119899(119909 119910119895) = 119877119898119899

(119909 119910119895) 119895 = 0 1 119899

(47)

Remark 15 We delete the proof of the above lemma since itcan be easily generalized from [3]

Now the error estimate Theorem 16 will be put which isalso motivated by [3] We let

119877119898119899

(119909 119910) =119873 (119909 119910)

119863 (119909 119910) 119877

lowast

119898119899(119909 119910) =

119873lowast

(119909 119910)

119863lowast (119909 119910) (48)

Theorem 16 Let a matrix 119891(119909 119910) bemax(119898 + 1 119899 + 1) timescontinuously differentiable on a set119863 = 119886 le 119909 le 119887 119888 le 119910 le 119889

containing the points Λ119898119899

= (119909119897 119910119896) 119897 = 0 1 119898 119896 =

0 1 119899 (119909119897 119910119896) isin R2 then for any point (119909 119910) isin 119863 there

exists a point (120585 120578) isin 119863 such that

119891 (119909 119910) minus 119877119898119899

(119909 119910)

=1

119863 (119909 119910)119863lowast (119909 119910)

times 120596119898+1

(119909)

(119898 + 1)

120597119898+1

120597119909119898+11198641(120585 119910) +

120603119899+1

(119910)

(119899 + 1)

120597119899+1

120597119910119899+11198642(119909 120578)

(49)

where

120596119898+1

(119909) = (119909 minus 1199090) (119909 minus 119909

1) sdot sdot sdot (119909 minus 119909

119898)

120603119899+1

(119910) = (119910 minus 1199100) (119910 minus 119910

1) sdot sdot sdot (119910 minus 119910

119899)

1198641(119909 119910) = 119863 (119909 119910)119863

lowast

(119909 119910) [119891 (119909 119910) minus 119877lowast

119898119899(119909 119910)]

1198642(119909 119910) = 119863 (119909 119910)119863

lowast

(119909 119910) [119877lowast

119898119899(119909 119910) minus 119877

119898119899(119909 119910)]

(50)

8 Journal of Applied Mathematics

Proof Let

1198641(119909 119910) = 119863 (119909 119910)119863

lowast

(119909 119910) [119891 (119909 119910) minus 119877lowast

119898119899(119909 119910)]

1198642(119909 119910) = 119863 (119909 119910)119863

lowast

(119909 119910) [119877lowast

119898119899(119909 119910) minus 119877

119898119899(119909 119910)]

119864 (119909 119910) = 1198641(119909 119910) + 119864

2(119909 119910)

(51)

From Lemma 14 we know

1198641(119909119894 119910) = 0 119894 = 0 1 119898 (52)

which results in

1198641(119909 119910) =

120596119898+1

(119909)

(119898 + 1)

120597119898+1

120597119909119898+11198641(119909 119910)

100381610038161003816100381610038161003816100381610038161003816119909=120585

(53)

where 120596119898+1

(119909) = (119909 minus 1199090)(119909 minus 119909

1) sdot sdot sdot (119909 minus 119909

119898) and 120585 is a

number contained in the interval (119886 119887) which may dependon 119910 Similarly from

1198642(119909 119910119895) = 0 119895 = 0 1 119899 (54)

we can get that

1198642(119909 119910) =

120603119899+1

(119910)

(119899 + 1)

120597119899+1

120597119910119899+11198642(119909 119910)

100381610038161003816100381610038161003816100381610038161003816119910=120578

(55)

where120603119899+1

(119910) = (119910minus1199100)(119910minus119910

1) sdot sdot sdot (119910minus119910

119899) and 120578 is a number

contained in the interval (119888 119889) which may depend on 119909Therefore

119864 (119909 119910) = 1198641(119909 119910) + 119864

2(119909 119910)

=120596119898+1

(119909)

(119898 + 1)

120597119898+1

120597119909119898+11198641(119909 119910)

100381610038161003816100381610038161003816100381610038161003816119909=120585

+120603119899+1

(119910)

(119899 + 1)

120597119899+1

120597119910119899+11198642(119909 119910)

100381610038161003816100381610038161003816100381610038161003816119910=120578

(56)

The proof is thus completed

4 Numerical Examples

Example 17 Let (119909119894 119910119895) and 119884

119894119895 119894 119895 = 0 1 2 be given in

Table 1

11987722

(119909 119910) = 1198800(119910) + 119880

1(119910) 119909 + 119880

2(119910) 119909 (119909 minus 1) (57)

Using Algorithm 7 we can get

1198800(119910) = (

1 0

0 0) +

119910

( 0 00 1

)+

119910 minus 1 |

| (0 0

minus25 minus15)

1198801(119910) = (

0 minus1

0 0) +

119910

(0 12

minus12 0)

+119910 minus 1 |

| ( 0 11 0

)

1198802(119910) = (

0 1

1

20) +

119910

(0 minus35

15 0)

+119910 minus 1 |

| (0 minus1514

minus514 minus57)

(58)

Based on generalized inverse (4) we obtain

11987722

(119909 119910)

= (11988611

11988612

11988621

11988622

) times (701199106

minus 3681199105

+ 8921199104

minus12401199103

+10401199102

minus512119910+128)minus1

=119873 (119909 119910)

119863 (119909 119910)

(59)

where

11988611

= 701199106

minus 3681199105

+ 8921199104

minus 12401199103

+ 10401199102

minus 512119910 + 128

11988612

= 1781199105

119909 minus 8581199104

119909 + 16881199103

119909

minus 17761199091199102

+ 1024119909119910 minus 256119909 minus 381199105

1199092

+ 2621199104

1199092

minus 6401199103

1199092

minus 5121199092

119910

minus 51199106

1199092

+ 51199106

119909 + 8001199092

1199102

+ 1281199092

11988621

= 1081199105

minus 2001199104

+ 2001199103

minus 1121199102

+ 32119910

minus 281199106

+ 601199106

119909 minus 3441199105

119909 + 7361199104

119909 minus 7841199103

119909

+ 4081199091199102

minus 32119909119910 minus 64119909 minus 241199105

1199092

+ 161199104

1199092

minus 961199092

119910 + 101199106

1199092

+ 401199092

1199102

+ 641199092

11988622

= 681199105

minus 1401199104

+ 1601199103

minus 961199102

+ 32119910 minus 141199106

minus 501199106

1199092

+ 2701199105

1199092

minus 6401199104

1199092

+ 8201199103

1199092

minus 5601199092

1199102

+ 1601199092

119910 + 501199106

119909 minus 2701199105

119909

+ 6401199104

119909 minus 8201199103

119909 + 5601199091199102

minus 160119909119910

(60)

FromDefinition 13 we know that 11987722

(119909 119910) is a BGIRINTsince 119877

22(119909119894 119910119895) = 119884119894119895and deg 119873(119909 119910) = 8 deg 119863(119909 119910) =

6 and 119863(119909 119910) | 119873(119909 119910)2

In paper [9] a bivariateThieletypematrix-valued rationalinterpolant 119877

119879(119909 119910) = 119866

00(119910) + (119909 minus 119909

0)11986610

(119910) + sdot sdot sdot + (119909 minus

119909119898minus1

)1198661198990

(119910) (page 73) where for 119897 = 0 1 119899

1198661198970

(119910) = 1198611198970

(1199090 119909

119897 1199100)

+119910 minus 1199100

1198611198971

(1199090 119909

119897 1199100 1199101)

+ sdot sdot sdot +119910 minus 119910119898minus1

|

| 119861119897119898

(1199090 119909

119897 1199100 119910

119898)

(61)

Here 119861119897119896(1199090 119909

119897 1199100 119910

119896) 119897 = 0 1 119899 119896 = 0 1 119898

is defined different from that of this paper (see [9]) We

Journal of Applied Mathematics 9

Table 1

119910119895119909119894

1199090= 0 119909

1= 1 119909

2= 2

1199100= 0 (

1 0

0 0) (

1 minus1

0 0) (

1 0

1 0)

1199101= 1 (

1 0

0 1) (

1 0

minus1 1) (

1 minus1

0 1)

1199102= 2 (

1 0

minus1 0) (

1 1

minus1 0) (

1 1

0 minus1)

Table 2

119884(13 13) 119877NT(13 13) 119877NT minus 119884119865

(minus0856888 05155013 3669296

130000 366929 2600) (

minus0856888 05155010 3669297

130000 366929 2600) 109119890 minus 006

119877119879(13 13) 119877

119879minus 119884119865

(minus098999 014111 547394

129999 547394 2999) 2613697

119884(15 15) 119877NT(15 15) 119877NT minus 119884119865

(minus0989999 01411200 4481689

150000 4481689 300) (

minus0989992 01411203 4481688

150000 4481688 300) 915119890 minus 007

119877119879(15 15) 119877

119879minus 119884119865

(minus099829372 minus005837 547394

149999 547394 31999) 143144

now give another numerical example to compare the twoalgorithms which shows that the method of this paper isbetter than the one of [9]

Example 18 Let

119884 (119909 119910) = (cos (119909 + 119910) sin (119909 + 119910) 119890

119910

119909 119890119910

119909 + 119910) (62)

and we suppose 1199090 1199091 1199092 1199093 1199094 1199095 = 119910

0 1199101 1199102 1199103 1199104 1199105

= 10 12 14 16 18 20 The numerical results 119877NT(119909 119910)

in Step 9 of Algorithm 7 and 119877119879(119909 119910) in [9] are given in

Table 2

Remark 19 InTable 2 from the F-normof119877NTminus119884 and119877119879minus119884

we can see that the error using Newton-Thiele type formulais much less than the one usingThieletype formula

5 Conclusion

In this paper the bivariate generalized inverse Newton-Thieletype matrix interpolation to approximate a matrix function119891(119909 119910) is given and we also give a recursive algorithmaccompanied with some other important conclusions such asdivisibility characterization and some numerical examplesFrom the second example it is easy to find that the approxi-mant using Newton-Thiele type formula is much better thanthe one usingThieletype formula

Acknowledgments

The work is supported by Shanghai Natural Science Foun-dation (10ZR1410900) and by Key Disciplines of ShanghaiMunicipality (S30104)

References

[1] G A Baker and P R Graves-Morris Pade Approximation PartII Addison-Wesley Publishing Company Reading Mass USA1981

[2] I Kh Kuchminska OM Sus and SMVozna ldquoApproximationproperties of two-dimensional continued fractionsrdquo UkrainianMathematical Journal vol 55 pp 36ndash54 2003

[3] J Tan and Y Fang ldquoNewton-Thielersquos rational interpolantsrdquoNumerical Algorithms vol 24 no 1-2 pp 141ndash157 2000

[4] P R Graves-Morris ldquoVector valued rational interpolants IrdquoNumerische Mathematik vol 42 no 3 pp 331ndash348 1983

[5] G Chuan-qing and C Zhibing ldquoMatrix valued rational inter-polants and its error formulardquo Mathematica Numerica Sinicavol 17 pp 73ndash77 1995

[6] G Chuan-qing ldquoMatrix Pade type approximat and directionalmatrix Pade approximat in the inner product spacerdquo Journal ofComputational and Applied Mathematics vol 164-165 pp 365ndash385 2004

[7] C Gu ldquoThiele-type and Lagrange-type generalized inverserational interpolation for rectangular complexmatricesrdquo LinearAlgebra and Its Applications vol 295 no 1ndash3 pp 7ndash30 1999

[8] C Gu ldquoA practical two-dimensional Thiele-type matrix Padeapproximationrdquo IEEE Transactions on Automatic Control vol48 no 12 pp 2259ndash2263 2003

10 Journal of Applied Mathematics

[9] C Gu ldquoBivariate Thiele-type matrix-valued rational inter-polantsrdquo Journal of Computational and Applied Mathematicsvol 80 no 1 pp 71ndash82 1997

[10] N K Bose and S Basu ldquoTwo-dimensional matrix Pade approx-imants existence nonuniqueness and recursive computationrdquoIEEE Transactions on Automatic Control vol 25 no 3 pp 509ndash514 1980

[11] A Cuyt and B Verdonk ldquoMultivariate reciprocal differencesfor branched Thiele continued fraction expansionsrdquo Journal ofComputational and Applied Mathematics vol 21 no 2 pp 145ndash160 1988

[12] W Siemaszko ldquoThiele-tyoe branched continued fractions fortwo-variable functionsrdquo Journal of Computational and AppliedMathematics vol 9 no 2 pp 137ndash153 1983

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

8 Journal of Applied Mathematics

Proof Let

1198641(119909 119910) = 119863 (119909 119910)119863

lowast

(119909 119910) [119891 (119909 119910) minus 119877lowast

119898119899(119909 119910)]

1198642(119909 119910) = 119863 (119909 119910)119863

lowast

(119909 119910) [119877lowast

119898119899(119909 119910) minus 119877

119898119899(119909 119910)]

119864 (119909 119910) = 1198641(119909 119910) + 119864

2(119909 119910)

(51)

From Lemma 14 we know

1198641(119909119894 119910) = 0 119894 = 0 1 119898 (52)

which results in

1198641(119909 119910) =

120596119898+1

(119909)

(119898 + 1)

120597119898+1

120597119909119898+11198641(119909 119910)

100381610038161003816100381610038161003816100381610038161003816119909=120585

(53)

where 120596119898+1

(119909) = (119909 minus 1199090)(119909 minus 119909

1) sdot sdot sdot (119909 minus 119909

119898) and 120585 is a

number contained in the interval (119886 119887) which may dependon 119910 Similarly from

1198642(119909 119910119895) = 0 119895 = 0 1 119899 (54)

we can get that

1198642(119909 119910) =

120603119899+1

(119910)

(119899 + 1)

120597119899+1

120597119910119899+11198642(119909 119910)

100381610038161003816100381610038161003816100381610038161003816119910=120578

(55)

where120603119899+1

(119910) = (119910minus1199100)(119910minus119910

1) sdot sdot sdot (119910minus119910

119899) and 120578 is a number

contained in the interval (119888 119889) which may depend on 119909Therefore

119864 (119909 119910) = 1198641(119909 119910) + 119864

2(119909 119910)

=120596119898+1

(119909)

(119898 + 1)

120597119898+1

120597119909119898+11198641(119909 119910)

100381610038161003816100381610038161003816100381610038161003816119909=120585

+120603119899+1

(119910)

(119899 + 1)

120597119899+1

120597119910119899+11198642(119909 119910)

100381610038161003816100381610038161003816100381610038161003816119910=120578

(56)

The proof is thus completed

4 Numerical Examples

Example 17 Let (119909119894 119910119895) and 119884

119894119895 119894 119895 = 0 1 2 be given in

Table 1

11987722

(119909 119910) = 1198800(119910) + 119880

1(119910) 119909 + 119880

2(119910) 119909 (119909 minus 1) (57)

Using Algorithm 7 we can get

1198800(119910) = (

1 0

0 0) +

119910

( 0 00 1

)+

119910 minus 1 |

| (0 0

minus25 minus15)

1198801(119910) = (

0 minus1

0 0) +

119910

(0 12

minus12 0)

+119910 minus 1 |

| ( 0 11 0

)

1198802(119910) = (

0 1

1

20) +

119910

(0 minus35

15 0)

+119910 minus 1 |

| (0 minus1514

minus514 minus57)

(58)

Based on generalized inverse (4) we obtain

11987722

(119909 119910)

= (11988611

11988612

11988621

11988622

) times (701199106

minus 3681199105

+ 8921199104

minus12401199103

+10401199102

minus512119910+128)minus1

=119873 (119909 119910)

119863 (119909 119910)

(59)

where

11988611

= 701199106

minus 3681199105

+ 8921199104

minus 12401199103

+ 10401199102

minus 512119910 + 128

11988612

= 1781199105

119909 minus 8581199104

119909 + 16881199103

119909

minus 17761199091199102

+ 1024119909119910 minus 256119909 minus 381199105

1199092

+ 2621199104

1199092

minus 6401199103

1199092

minus 5121199092

119910

minus 51199106

1199092

+ 51199106

119909 + 8001199092

1199102

+ 1281199092

11988621

= 1081199105

minus 2001199104

+ 2001199103

minus 1121199102

+ 32119910

minus 281199106

+ 601199106

119909 minus 3441199105

119909 + 7361199104

119909 minus 7841199103

119909

+ 4081199091199102

minus 32119909119910 minus 64119909 minus 241199105

1199092

+ 161199104

1199092

minus 961199092

119910 + 101199106

1199092

+ 401199092

1199102

+ 641199092

11988622

= 681199105

minus 1401199104

+ 1601199103

minus 961199102

+ 32119910 minus 141199106

minus 501199106

1199092

+ 2701199105

1199092

minus 6401199104

1199092

+ 8201199103

1199092

minus 5601199092

1199102

+ 1601199092

119910 + 501199106

119909 minus 2701199105

119909

+ 6401199104

119909 minus 8201199103

119909 + 5601199091199102

minus 160119909119910

(60)

FromDefinition 13 we know that 11987722

(119909 119910) is a BGIRINTsince 119877

22(119909119894 119910119895) = 119884119894119895and deg 119873(119909 119910) = 8 deg 119863(119909 119910) =

6 and 119863(119909 119910) | 119873(119909 119910)2

In paper [9] a bivariateThieletypematrix-valued rationalinterpolant 119877

119879(119909 119910) = 119866

00(119910) + (119909 minus 119909

0)11986610

(119910) + sdot sdot sdot + (119909 minus

119909119898minus1

)1198661198990

(119910) (page 73) where for 119897 = 0 1 119899

1198661198970

(119910) = 1198611198970

(1199090 119909

119897 1199100)

+119910 minus 1199100

1198611198971

(1199090 119909

119897 1199100 1199101)

+ sdot sdot sdot +119910 minus 119910119898minus1

|

| 119861119897119898

(1199090 119909

119897 1199100 119910

119898)

(61)

Here 119861119897119896(1199090 119909

119897 1199100 119910

119896) 119897 = 0 1 119899 119896 = 0 1 119898

is defined different from that of this paper (see [9]) We

Journal of Applied Mathematics 9

Table 1

119910119895119909119894

1199090= 0 119909

1= 1 119909

2= 2

1199100= 0 (

1 0

0 0) (

1 minus1

0 0) (

1 0

1 0)

1199101= 1 (

1 0

0 1) (

1 0

minus1 1) (

1 minus1

0 1)

1199102= 2 (

1 0

minus1 0) (

1 1

minus1 0) (

1 1

0 minus1)

Table 2

119884(13 13) 119877NT(13 13) 119877NT minus 119884119865

(minus0856888 05155013 3669296

130000 366929 2600) (

minus0856888 05155010 3669297

130000 366929 2600) 109119890 minus 006

119877119879(13 13) 119877

119879minus 119884119865

(minus098999 014111 547394

129999 547394 2999) 2613697

119884(15 15) 119877NT(15 15) 119877NT minus 119884119865

(minus0989999 01411200 4481689

150000 4481689 300) (

minus0989992 01411203 4481688

150000 4481688 300) 915119890 minus 007

119877119879(15 15) 119877

119879minus 119884119865

(minus099829372 minus005837 547394

149999 547394 31999) 143144

now give another numerical example to compare the twoalgorithms which shows that the method of this paper isbetter than the one of [9]

Example 18 Let

119884 (119909 119910) = (cos (119909 + 119910) sin (119909 + 119910) 119890

119910

119909 119890119910

119909 + 119910) (62)

and we suppose 1199090 1199091 1199092 1199093 1199094 1199095 = 119910

0 1199101 1199102 1199103 1199104 1199105

= 10 12 14 16 18 20 The numerical results 119877NT(119909 119910)

in Step 9 of Algorithm 7 and 119877119879(119909 119910) in [9] are given in

Table 2

Remark 19 InTable 2 from the F-normof119877NTminus119884 and119877119879minus119884

we can see that the error using Newton-Thiele type formulais much less than the one usingThieletype formula

5 Conclusion

In this paper the bivariate generalized inverse Newton-Thieletype matrix interpolation to approximate a matrix function119891(119909 119910) is given and we also give a recursive algorithmaccompanied with some other important conclusions such asdivisibility characterization and some numerical examplesFrom the second example it is easy to find that the approxi-mant using Newton-Thiele type formula is much better thanthe one usingThieletype formula

Acknowledgments

The work is supported by Shanghai Natural Science Foun-dation (10ZR1410900) and by Key Disciplines of ShanghaiMunicipality (S30104)

References

[1] G A Baker and P R Graves-Morris Pade Approximation PartII Addison-Wesley Publishing Company Reading Mass USA1981

[2] I Kh Kuchminska OM Sus and SMVozna ldquoApproximationproperties of two-dimensional continued fractionsrdquo UkrainianMathematical Journal vol 55 pp 36ndash54 2003

[3] J Tan and Y Fang ldquoNewton-Thielersquos rational interpolantsrdquoNumerical Algorithms vol 24 no 1-2 pp 141ndash157 2000

[4] P R Graves-Morris ldquoVector valued rational interpolants IrdquoNumerische Mathematik vol 42 no 3 pp 331ndash348 1983

[5] G Chuan-qing and C Zhibing ldquoMatrix valued rational inter-polants and its error formulardquo Mathematica Numerica Sinicavol 17 pp 73ndash77 1995

[6] G Chuan-qing ldquoMatrix Pade type approximat and directionalmatrix Pade approximat in the inner product spacerdquo Journal ofComputational and Applied Mathematics vol 164-165 pp 365ndash385 2004

[7] C Gu ldquoThiele-type and Lagrange-type generalized inverserational interpolation for rectangular complexmatricesrdquo LinearAlgebra and Its Applications vol 295 no 1ndash3 pp 7ndash30 1999

[8] C Gu ldquoA practical two-dimensional Thiele-type matrix Padeapproximationrdquo IEEE Transactions on Automatic Control vol48 no 12 pp 2259ndash2263 2003

10 Journal of Applied Mathematics

[9] C Gu ldquoBivariate Thiele-type matrix-valued rational inter-polantsrdquo Journal of Computational and Applied Mathematicsvol 80 no 1 pp 71ndash82 1997

[10] N K Bose and S Basu ldquoTwo-dimensional matrix Pade approx-imants existence nonuniqueness and recursive computationrdquoIEEE Transactions on Automatic Control vol 25 no 3 pp 509ndash514 1980

[11] A Cuyt and B Verdonk ldquoMultivariate reciprocal differencesfor branched Thiele continued fraction expansionsrdquo Journal ofComputational and Applied Mathematics vol 21 no 2 pp 145ndash160 1988

[12] W Siemaszko ldquoThiele-tyoe branched continued fractions fortwo-variable functionsrdquo Journal of Computational and AppliedMathematics vol 9 no 2 pp 137ndash153 1983

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Journal of Applied Mathematics 9

Table 1

119910119895119909119894

1199090= 0 119909

1= 1 119909

2= 2

1199100= 0 (

1 0

0 0) (

1 minus1

0 0) (

1 0

1 0)

1199101= 1 (

1 0

0 1) (

1 0

minus1 1) (

1 minus1

0 1)

1199102= 2 (

1 0

minus1 0) (

1 1

minus1 0) (

1 1

0 minus1)

Table 2

119884(13 13) 119877NT(13 13) 119877NT minus 119884119865

(minus0856888 05155013 3669296

130000 366929 2600) (

minus0856888 05155010 3669297

130000 366929 2600) 109119890 minus 006

119877119879(13 13) 119877

119879minus 119884119865

(minus098999 014111 547394

129999 547394 2999) 2613697

119884(15 15) 119877NT(15 15) 119877NT minus 119884119865

(minus0989999 01411200 4481689

150000 4481689 300) (

minus0989992 01411203 4481688

150000 4481688 300) 915119890 minus 007

119877119879(15 15) 119877

119879minus 119884119865

(minus099829372 minus005837 547394

149999 547394 31999) 143144

now give another numerical example to compare the twoalgorithms which shows that the method of this paper isbetter than the one of [9]

Example 18 Let

119884 (119909 119910) = (cos (119909 + 119910) sin (119909 + 119910) 119890

119910

119909 119890119910

119909 + 119910) (62)

and we suppose 1199090 1199091 1199092 1199093 1199094 1199095 = 119910

0 1199101 1199102 1199103 1199104 1199105

= 10 12 14 16 18 20 The numerical results 119877NT(119909 119910)

in Step 9 of Algorithm 7 and 119877119879(119909 119910) in [9] are given in

Table 2

Remark 19 InTable 2 from the F-normof119877NTminus119884 and119877119879minus119884

we can see that the error using Newton-Thiele type formulais much less than the one usingThieletype formula

5 Conclusion

In this paper the bivariate generalized inverse Newton-Thieletype matrix interpolation to approximate a matrix function119891(119909 119910) is given and we also give a recursive algorithmaccompanied with some other important conclusions such asdivisibility characterization and some numerical examplesFrom the second example it is easy to find that the approxi-mant using Newton-Thiele type formula is much better thanthe one usingThieletype formula

Acknowledgments

The work is supported by Shanghai Natural Science Foun-dation (10ZR1410900) and by Key Disciplines of ShanghaiMunicipality (S30104)

References

[1] G A Baker and P R Graves-Morris Pade Approximation PartII Addison-Wesley Publishing Company Reading Mass USA1981

[2] I Kh Kuchminska OM Sus and SMVozna ldquoApproximationproperties of two-dimensional continued fractionsrdquo UkrainianMathematical Journal vol 55 pp 36ndash54 2003

[3] J Tan and Y Fang ldquoNewton-Thielersquos rational interpolantsrdquoNumerical Algorithms vol 24 no 1-2 pp 141ndash157 2000

[4] P R Graves-Morris ldquoVector valued rational interpolants IrdquoNumerische Mathematik vol 42 no 3 pp 331ndash348 1983

[5] G Chuan-qing and C Zhibing ldquoMatrix valued rational inter-polants and its error formulardquo Mathematica Numerica Sinicavol 17 pp 73ndash77 1995

[6] G Chuan-qing ldquoMatrix Pade type approximat and directionalmatrix Pade approximat in the inner product spacerdquo Journal ofComputational and Applied Mathematics vol 164-165 pp 365ndash385 2004

[7] C Gu ldquoThiele-type and Lagrange-type generalized inverserational interpolation for rectangular complexmatricesrdquo LinearAlgebra and Its Applications vol 295 no 1ndash3 pp 7ndash30 1999

[8] C Gu ldquoA practical two-dimensional Thiele-type matrix Padeapproximationrdquo IEEE Transactions on Automatic Control vol48 no 12 pp 2259ndash2263 2003

10 Journal of Applied Mathematics

[9] C Gu ldquoBivariate Thiele-type matrix-valued rational inter-polantsrdquo Journal of Computational and Applied Mathematicsvol 80 no 1 pp 71ndash82 1997

[10] N K Bose and S Basu ldquoTwo-dimensional matrix Pade approx-imants existence nonuniqueness and recursive computationrdquoIEEE Transactions on Automatic Control vol 25 no 3 pp 509ndash514 1980

[11] A Cuyt and B Verdonk ldquoMultivariate reciprocal differencesfor branched Thiele continued fraction expansionsrdquo Journal ofComputational and Applied Mathematics vol 21 no 2 pp 145ndash160 1988

[12] W Siemaszko ldquoThiele-tyoe branched continued fractions fortwo-variable functionsrdquo Journal of Computational and AppliedMathematics vol 9 no 2 pp 137ndash153 1983

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

10 Journal of Applied Mathematics

[9] C Gu ldquoBivariate Thiele-type matrix-valued rational inter-polantsrdquo Journal of Computational and Applied Mathematicsvol 80 no 1 pp 71ndash82 1997

[10] N K Bose and S Basu ldquoTwo-dimensional matrix Pade approx-imants existence nonuniqueness and recursive computationrdquoIEEE Transactions on Automatic Control vol 25 no 3 pp 509ndash514 1980

[11] A Cuyt and B Verdonk ldquoMultivariate reciprocal differencesfor branched Thiele continued fraction expansionsrdquo Journal ofComputational and Applied Mathematics vol 21 no 2 pp 145ndash160 1988

[12] W Siemaszko ldquoThiele-tyoe branched continued fractions fortwo-variable functionsrdquo Journal of Computational and AppliedMathematics vol 9 no 2 pp 137ndash153 1983

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of