Hermite Interpolation with Triangular Fuzzy Numbers

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Vol. 14, No. 2, 2017, 305-311

ISSN: 2320 –3242 (P), 2320 –3250 (online) Published on 11 December 2017

www.researchmathsci.org

DOI: http://dx.doi.org/10.22457/ijfma.v14n2a13

305

International Journal of

Hermite Interpolation with Triangular Fuzzy Numbers

Thangaraj Beaula1 and P.Caroline Mary2

1

PG & Research Department of Mathematics T.B.M.L College, Porayar 609307, Tamilnadu, India

2

Department of Mathematics, Annai Vailankanni Arts and Science College Thanjavur-7, Tamilnadu, India

Received 12 November 2017; accepted 10 December 2017

Abstract. In this paper the polynomial interpolation of triangular fuzzy number is discussed. First general form of the polynomial with fuzzy coefficients is proposed. The hermite interpolation method is studied with triangular fuzzy number an example is provided to illustrate the algorithm.

Keywords: Interpolation, polynomial ,triangular fuzzy numbers

AMS Mathematics Subject Classification (2010): 03E72, 15B15

1. Introduction

In some problem, people have to deal with massive imprecise data usually represented them as triangular fuzzy numbers. Many researchers focus on fuzzy numbers, for the convenience in dealing with uncertainty. people call this ‘fuzzy mathematics’.

Interpolation method has become very popular method in engineering, economy, and computer science etc. Polynomial interpolation is one of the most common methods while some times we do not know the exact value of the measured quantities because of the inevitable measurement inaccuracy. We can describe the uncertainty by using fuzzy numbers. Chenyi Hu Anelina etc have discussed the interval polynomial interpolation problem and its lagrange form[6].

In this paper, we study the theory and algorithm of interpolatin problem of triangular fuzzy numbers by defining the fuzzy coefficients polynomial to obtain the general from solution. Then we study the hermite interpolation method, which is more convenient for practice .

We know that lagrange interpolation formula is not most convenient for practical purpose. But hermite interpolation formula gives more accurate result than obtained in lagrange interpolation formula Finally an example is provided to illustrate the method.

2. Preliminaries

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Thangaraj Beaula and P.Caroline Mary

306

) = 0, x< , ≤ ≤

, ≤ ≤ 0, x>

Definitions 2.2. Let = (, , ) =(, , ) be two triangular fuzzy numbers. = iff = , = , =

Definitions 2.3. The operations of triangular fuzzy numbers are defined as follows: (1) + =(+ , + , + )

(2) k = (k, , ) k≥₀ (, , k),

Definitions 2.4. Let _!, _"… … … … _$ be triangular fuzzy number. A function

%&, !, "… … … … $)denoted by %_&) is called the n-oreder polynomial with triangular fuzzy numbers coefficients, if it satisfies the following conditions

(1) %_&) is an n-order polynomial about x;

(2) %_&)is a 1-order polynomial about _!, _"… … … … _$

We get an equivalent definition of the triangular polynomial %_&).

Definitions 2.5. Let (_!), (_") … … . . , (_$) be m+1 polynomials, whose degree is not more than n, and at least one of them is a polynomial of degree n .The fuzzy triangular polynomials %_&) has the following from

%&) = !(!) + "(") + ⋯ … … … + $($) The set of all triangular fuzzy polynomials of degree m (m≤_{n) is denoted by}%_&

3. Interpolation polynomials of fuzzy numbers

In this section the interpolation problem of triangular fuzzy numbers is formulated in detail by comparing with the hermite interpolation problem. Then we obtain the existence theorem of solutions and investigate its properties.

Definition 3.1. Let _!, _",…………._& be n+ 1 distinct node . Given*_!= +_!), *_"= +_"), … … … , *_&= +_&)

*!,=+_-,_.₎,*_",=+_-,_/₎,………..,*_&,=+_-,₀₎,

The problem is to find a polynomial %_&) ∈ %_&, is called interpolating polynomials such that %_&₂) = (_&₂) + 3_&₂),

where (_&₂) = *₂3_&₂) = *₂,(i=0,1,2………n)

Theorem 3.1. There exists a unique polynomial 4_5&6") ∈ 4_5&6" such that

75&6"2)= *₂,8_5&6"₂)=*₂,for i=0,1,2….,n The Hermite form of 4_5&6") is

45&6") = ∑ :&2;! 2)*2+ ∑ <&2;! 2)*2, (1)

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307

:2,) = 0 +GH B <2,=>? = @1 B+ B = C_{0 B+ B ≠ CF}

75&6"2, *!, *", … … … … *&) = *2 B = 0,1,2, … … … J

85&6"2, *!,, *",, … … … , *&,) = *2 , B = 0,1,2 … … … … J

Theorem 3.2. Let _!, _",………_& be n+1 distinct nodes. and let *₂ = *₂,*₂, *₂), (i=0,1,2,…....n) be n+1 triangular fuzzy numbers. There exists at least one fuzzy polynomial such that

4K5&6"2) = *2 JL 85&6"2) = *2, +GH B = 0,1,2 … … … J

Proof: Suppose that (_&₂) = *₂3_&₂) = *₂, is an interpolation polynomial such that

75&6"2, *!, *", … … … … *&) = *2 B = 0,1,2, … … . J

85&6"2, *!,, *",, … … … … , *&,) = *2 , B = 0,1,2 … … J Define

7M_5&6" _{) = inf} ∀RSS∈RS 2;!,",…..&

7_5&6"_{, *}_!_{, *}_"_{, … … … … *}_&₎₍₂₎

7K5&6" ) = sup ∀RSS∈RS 2;!,",…..&7

5&6", *!, *", … … … … *&) (3)

7K5&6"W ) = ∑ :&2;! 2)*2 (4) and

85&6" ) = inf ∀R_SSX∈R_S, 2;!,",…..&

8

5&6", *!,, *",, … … … … *&,) 5)

85&6" ) = inf_∀R SSX∈RS, 2;!,",…..&

8_5&6"_{, *}_!,_{, *}_",_{, … … … … *}_&,₎₍₆₎

85&6"W ) = ∑ <&2;! 2)*2 , (7)

4K5&6" ) = 7K5&6" ) + 85&6" )

4K_5&6" _{) = 7K}

5&6" ) + 85&6" )

4K5&6" ) = 7K5&6" ) + 85&6" )

Since :₂)JL <₂)are interpolating polynomials in x of degree 2n+1 using conditions

:2=>? = @1 B+ B = C_{0 B+ B ≠ C F <}2) = 0 +GH B

:2,) = 0 +GH B <2,=>? = @1 B+ B = C_{0 B+ B ≠ CF}

Let :₂) = Z₂)[₂)\5

<2) = ]2)[2)\5 where

2) = ^ − > >− 2 &

`;! `ab

, B = 0,1, … … … . J

and Z₂) and ]₂) are linear function in x

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308 Then

2 + 2= 1

c2 + L2= 0

and :₂) = ₂ + ₂)[₂)\5 (8)

<2) = c2) + L2)[2)\5

:2) = 2 + 2)[2)\5

= ₂ + ₂) 2₂)₂,) + ₂[₂)\5 Put =₂

:2,2)=₂₂+ ₂) 2₂₂)₂,₂) + ₂[₂₂)\5 0=2₂,₂) + ₂

Therefore ₂+ 2₂,₂) = 0

<2) = c2) + L2)[2)\5

= c₂ + L₂) 2₂)₂,) + c₂[₂)\5 Put =_>

<₂,₌_>_?₌_=c₂_>_{+ L}₂_?₂₂₌_>_? 2

,₌_>_{? + c}₂_d₂₌_>_?e5 1= c₂

2 + 2 = 1

c2 + L2 = 0 and

₂+ 2₂,

2) = 0

c2= 1 Solving the above equations we get

2= −22,2), 2 = 1+2₂₂,₂)

c2= 1L2 = −2 (9) Using the above values on applying in equations (8) the equation becomes

:2) = [1 − 2 − 2)2,2)\[2)\5

<2) = − 2)[2)\5

Using the above expression for :₂) JL <₂) we obtain finally

45&6") = ^[1 − 2 − 2)2,2)\[2)\5*2 &

2;!

+ ^ − 2)[2)\5*2,

&

2;! This is the required hermite interpolation formula

Therefore

4_5&6"_{) = ^[1 − 2 −}₂₎₂,

2)\[2)\5*2 &

2;!

+ ^ − 2)[2)\5*2,

&

2;!

4_5&6"_{) = ^[1 − 2 −}₂₎

2,2)\[2)\5*2 &

2;!

+ ^ − 2)[2)\5*2,

&

2;!

4_5&6"_{) = ^[1 − 2 −}₂₎

2,2)\[2)\5*2 &

2;!

+ ^ − 2)[2)\5*2,

&

2;!

4K5&6"2) = 4K5&6" ), 4K5&6" ), 4K5&6" )

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309

As a consequence the interpolating polynomial of triangular fuzzy number exists. More specifically, by theorem 3.2 4M_5&6") contains all the interpolating polynomials of

*2,which are arbitrarily chosen

Theorem 3.3. The hermite interpolating polynomials of fuzzy numbers is

4K5&6") = ^ :2 &

2;!

)*2+ ^ <2)*2, &

2;! where

4K5&6") = 7K5&6") + 85&6") Obviously

75&6"2) = *2 B = 0,1, … … … J

:2=>? = @1 B+ B = C_{0 B+ B ≠ C F <}2) = 0 +GH B

:2,) = 0 +GH B <2,=>? = @1 B+ B = C_{0 B+ B ≠ CF} Theorem 3.4.

Suppose that 4K_5&6") = d4_5&6" ), 4_5&6" )e given by

45&6" ) = inf_∀R_SS_∈R_S 2;!,",…..&

7_5&6"_{, *}_!_{, *}_"_{, … … … … *}_&₎

+ inf

∀R_SSX_∈R S, 2;!,",…..&

8

5&6", *!,, *",, … … … … *&,)

45&6" ) = sup_∀R SS∈RS 2;!,",…..&

7

5&6", *!, *", … … … … *&)

+ sup

∀R_SSX∈R f_S, 2;!,",…..&

8_5&6"_{, *}_!,_{, *}_",_{, … … … … *}_&,₎

such that 7K_5&6") = ∑ :&_2;! ₂)*₂

85&6") = ^ <2)*2, &

2;! Then we have 4K_5&6") = 7K_5&6") + 8_5&6")

Hence the hermite interpolation formula holds to replace *₂ by *₂, in (1).

We know that Lagrange interpolation formula is not convenient for a practical purpose. But hermite interpolation formula given more accurate result than obtained in lagrange interpolation formula.

4. Numerical example

1. Determine the Hermite polynomial which fits the following data and hence find an approximate value of In 4 _!= 2.0 *_!= 0.67,0.69315,0.71)*_!, = 0.4,0.5,0.6)

"= 2.5 *"= 0.90,0.0.91629,0.92)*", = 0.3,0.4,0.5)

5= 3.0 *5= 1.08,1.09861,1.1) *5, = 0.32,0.3333,0.35) Solution:

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310

2) = ^ − >

>− 2 &

`;! `ab

, B = 0,1, … … … . J

!) = − 2.5) − 3.0)_{2 − 2.5)2 − 3.0)}!) = 25− 11 + 15

") =_{2.5 − 2)2.5 − 3.0)} − 2) − 3.0) ") = −45− 20 + 24)

5) =_{3 − 2)3.0 − 2.5)} − 2) − 2.5) 5) = 25− 9 + 10

!,) = 4 − 11 ",) = −8 − 20)5,) = 4 − 9

!,!) = −3 ",") = 0, 5,5) = 3

4

5&6") = ^[1 − 2 − 2)2,2)\[2)\5*2 &

2;!

+ ^ − 2)[2)\5*2,

&

2;!

4_5&6"_{) = ^[1 − 2 −}₂₎

2,2)\[2)\5*2 &

2;!

+ ^ − 2)[2)\5*2,

&

2;!

4_5&6"_{) = ^[1 − 2 −}₂₎

2,2)\[2)\5*2 &

2;!

+ ^ − 2)[2)\5*2,

&

2;!

:2) = [1 − 2 − 2)2,2)\[2)\5

:!) = [1 − 2 − !)!,!)\[!)\5

:!) = [6 − 11\[25− 11 + 5\5

:_") = [45− 20 + 24\5

:5) = [−6 + 19\[25− 9 + 10\

Consider

:!) = [6 − 11\[25− 11 + 5\5

Put = 4,:_!) =117

:!) = [6 − 11\[25− 11 + 5\5 (0.67, 0.69315, 0.71)

:!) = (78.39, 81.09855, 83.07)

:") = [45− 20 + 24\5 Put = 4,:_") = 64

:") = [45− 20 + 24\5(0.91,0.91629,0.4) = 64 ( 57.6,58.64256,58.88)

:5) = [−6 + 19\[25− 9 + 10\

Put = 4 , :₅) = -180

:₅) = [−6 + 19\[25− 9 + 10\1.1,1.09861,1.1)

:5) = −1801.1,1.09861,1.1)

:5) = −198, −197.7498, −194.4)

Consider <₂) = − ₂)[₂)\5

<!) = − 2)[25− 11 + 5\5

Put = 4 <_!) = 18

<!) = − 2)[25− 11 + 5\5(0.4,0.5,0.6)

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311

<!) = 7.2,9,10.8)

<") = − 2.5)[45− 20 + 24\5 Put = 4 <_") = 96 0.3,0.4,0.5)

<_") = 28.8,38.4,48)

<5) = − 3.0)[25− 9 + 10\

Put = 4 <₅) = 36 0.30,0.33,0.5)

<5) = 11.52,11.9988,12.6)

45&6") = (-14.49,1.39011,18.95) Hence In (4) ≅ 1.39011

5. Conclusion

In this paper ,we get the general form of the fuzzy interpolating polynomial problem. Then, the solution formula is given, including hermite form. Finally, we obtain several interesting conclusions and illustrate their meanings by examples. In future, we will focus on other interpolation problems with uncertainty.

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