# Fuzzy Least Squares Approximation Using Fuzzy Polynomial

Authors: Kun Liu; Xiaobin Guo
Journal: Mathematical Problems in Engineering
 (2021)

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## Abstract

In this paper, the fuzzy polynomial is introduced and applied to investigate the least squares approximation problem based on LR fuzzy numbers. A new and simple approach to solve the original problem is constructed by using approximate fuzzy polynomial. Two numerical examples are given to illustrate the proposed method. Since a large number of data exist as an uncertain property and need a function relation to reflect the laws between different variables, our results enrich fuzzy numerical approximation theory.


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## Body

## 1. Introduction

There are many systems in which part or all parameters may be uncertain and can be represented and computed by the fuzzy numbers in real world. So, both theory and applications of fuzzy mathematics have been paid more attention by many scholars in the past decades, see [1–24].In production practice, it is often necessary to conduct statistics and analysis on a large number of data. Due to equipment or technology and other reasons, many measurement or experimental data are uncertain. Sometimes, this property can be expressed and processed by means of fuzzy numbers. Therefore, the problem of numerical approximation based on the uncertainty of fuzzy numbers has attracted more and more attention of scholars in the field of fuzzy mathematics in recent years. Since a large number of data exist as an uncertain property and need a function relation to reflect the laws between different variables, the fuzzy interpolation and approximation problem has been getting more and more important. However, the research work to fuzzy interpolation and approximation was little in the past two decades. In 1990, Lowen [25] firstly introduced the fuzzy Lagrange interpolation theorem. In 1994, fuzzy interpolation was investigated by Kaleva [18] again. Later, a few researchers proposed other ways to construct fuzzy approximation functions. For instance, in 2005, Allahviranloo et al. [5] studied the method of fuzzy number approximation. In 2006, Allahviranloo et al. [1] researched numerical approximation of fuzzy function by using the fuzzy polynomials.In this paper, we propose the fuzzy least squares problem and find a polynomial function to approximate the given data. The structure of this paper is organized as follows.In Section2, we introduce the basic concepts of LR fuzzy numbers. In Section 3, we propose fuzzy least squares approximation problem and present a model to solve fuzzy least squares approximate. In Section 4, we obtain a method which is made of three steps to approximate a fuzzy function. Two numerical examples are given in Section 5.


## 2. Preliminaries

There are several definitions for the concept of fuzzy numbers (see [10, 25–28]).Definition 1.
A fuzzy number is a fuzzy set likeu:R⟶I=0,1 which satisfies(1)
u is upper semicontinuous(2)
u is fuzzy convex, i.e., uλx+1−λy≥minux,uy, for all x,y∈R,λ∈0,1(3)
u is normal, i.e., there exists x0∈R such that ux0=1(4)
suppu=x∈R|ux>0 is the support of u, and its closure cl(suppu) is compact
LetE1 be the set of all fuzzy numbers on R.Definition 2.
A fuzzy numberM˜ is said to be a LR fuzzy number if(1)μM˜x=Lm−xα,x≤m,α>0,Rx−mβ,x≥m,β>0,where m and α and β are called the mean value and left and right spreads of M˜, respectively. The function L⋅, which is called left shape function, satisfies(1)
Lx=L−x(2)
L0=1 and L1=0(3)
Lx is nonincreasing on 0,∞
The definition of a right shape functionR⋅ is similar to that of L⋅.
Clearly, two LR fuzzy numbers,M˜=m,α,βLR and N˜=n,γ,δLR, are said to be equal if and only if m=n,α=γ, and β=δ. Also, M˜=m,α,βLR is positive (negative) if and only if m−α>0m+β<0.Definition 3.
For arbitrary LR fuzzy numbers,M˜=m,α,βLR and N˜=n,γ,δLR, we have the following.(1)
Addition:(2)M˜+N˜=m,α,βLR+n,γ,δLR=m+n,α+γ,β+δLR.(2)
Subtraction:(3)M˜−N˜=m,α,βLR−n,γ,δLR=m−n,α+δ,β+γLR.(3)
Scalar multiplication:(4)λM˜=λm,α,βLR≅λm,λα,λβ,λ≥0,λm,−λβ,−λαRL,λ<0.


## 3. Fuzzy Least Squares Approximation

Definition 4.
A fuzzy number value functionfx˜=fx,flx,frx is said to be a LR fuzzy number value function if(5)μf˜fx=Lm−fxflx,fx≤m,flx>0,Rfx−mfrx,fx≥m,frx>0,where m and flx and frx are called the mean value and left and right spreads of fx˜, respectively. The function L⋅, which is called left shape function, satisfies(1)
Lfx=L−fxx(2)
L0=1 and L1=0(3)
Lfx is nonincreasing on 0,∞
The definition of a right shape functionR⋅ is similar to that of L⋅.Lemma 1 (see [25, 29–31]).
Least squares principle.
Least squares method is using curvey=φx to approximate data xi,yii=1,2,…,m. The standard is that you want to choose φx, and it makes the difference between the function φxi at xi and the measured data xi are very small. If you want to do that, you have to minimize the sum of squares of deviation, which is ∑i=1mφxi−yi2 a minimum.
Minimum square error:(6)δ1=∑i=1mwiφxi−yi21/2.
Maximum deviation:(7)δ2=max1≤i≤mφxi−yi.


### 3.1. LR Fuzzy Least Squares Approximation Problem

Consider the following set of data forxi and f˜xi, where xi∈R, x1<x2<⋯<xm, f˜xi are LR numbers, and f˜xi=fxi,flxi,frxi.Select a function class inCa,b, i.e.,(8)S=Spanφ0x,φ1x,…,φnx.Meanwhile, letX=x1,x2,…,xm be a set of m points in R, and m>n. We suppose(9)p˜x=px,plx,prx=∑j=0najφjx,∑j=0nbjφjx,∑j=0ncjφjx.We seekpx,plx,prx, making(10)∑i=1mwiδi2=∑i=1mwif˜xi−p˜xi2=minpx,plx,prx∈S∑i=1mwif˜xi−p˜xi2,where wii=1,2,…,m are the weight function, and wi>0.Therefore,p˜x=px,plx,prx is the solution of fuzzy least squares approximation.Lemma 2 (see [25]).
The canonical systemGa=d has an unique solution a0,a1,…,an, and the corresponding function px=∑j=0najφjx is satisfied with equation (11); therefore, px is the least squares solution of the data set xi,fxi.Theorem 1.
The three canonical systemsGa=d,Gb=e,andGc=f have a unique solution a0,a1,…,an,b0,b1,…,bn,c0,c1,…,cn, separately, and the corresponding function px=∑j=0najφjx,plx=∑j=0nbjφjx,andprx=∑j=0ncjφjx are satisfied with equations (11)–(13); therefore, p˜x are the least squares solution of the data set xi,f˜xi.Proof.
Supposep˜x=px,plx,prx=a0φ0x+a1φ1x+⋯+anφnx,b0φ0x+b1φ1x+⋯+bnφnx,c0φ0x+c1φ1x+⋯+cnφnx; then,(11)px=a0φ0x+a1φx+⋯+anφnx,(12)plx=b0φ0x+b1φ1x+⋯+bnφnx,(13)prx=c0φ0x+c1φ1x+⋯+cnφnx.
Based on least squares solution,px=a0φ0x+a1φ1x+⋯+anφnx is satisfied with equation (11). If we want to get the least squares solution, we must resolve the coefficient a0,a1,…,an.
In fact, we will solve the coefficienta0,a1,…,an, and it can transform into solve the minimum points a0,a1,…,an of the multivariate functions Ga0,a1,…,an:(14)Ga0,a1,…,an=∑i=1mwifx−∑j=0najφjxi2,and we can obtain ∂G/∂ak=0,k=0,1,…,n:(15)∂G∂ak=2∑i=1mwifx−∑j=0najφjxiφkxi=0,(16)∑j=0naj∑i=1mwiφjxiφkxi=∑i=1mwifxiφkxi,k=0,1,…,n.
According to the inner product symbol of the discrete point set, equation (16) can be written as(17)∑j=0nφj,φkaj=f,φk,k=0,1,…,n.
The matrix form can be expressed asGa=d, i.e.,(18)φ0,φ0φ0,φ1…φ0,φnφ1,φ0φ1,φ1…φ1,φn⋮⋮⋱⋮φn,φ0φn,φ1…φn,φna0a1⋮an=f,φ0f,φ1⋮f,φn,when φ0x,φ1x,…,φnx are linearly independent, due to the Gram matrix, detGn+1≠0; therefore, there is unique solution a0,a1,…,an. It can be seen the least squares solution exists. So, a0,a1,…,an are the unique solution with equation (18). It proved that a unique function can be obtained px=a0φ0x+a1φ1x+⋯+anφnx.
Let us prove the latter part. For anyp∗x=∑j=0naj∗φjx∈S, then(19)∑i=1mwifxi−p∗xi2=∑i=1mwifxi−pxi+pxi−p∗xi2=∑i=1mwifxi−pxi2+2∑i=1mwifxi−pxipxi−p∗xi+∑i=1mwifxi−p∗xi2.
From equation (16), we know(20)∑i=1mwifxi−pxipxi−p∗xi=∑i=1mwifxi−pxi∑j=0naj−aj∗φjxi=∑j=0naj−aj∗∑i=1mwifxi−pxiφjxi=0,and thus,(21)∑i=1mwifxi−p∗xi2=∑i=1mwifxi−pxi2+∑i=1mwipxi−p∗xi2≥∑i=1mwifxi−pxi2,i.e.,(22)∑i=1mwifxi−pxi2=minpx∈S∑i=1mwifxi−p∗xi2.
Therefore,px is the least squares solution of data set xi,fxii=1,2,…,m, and it is satisfied with equation (10).
In the same way,Gb=e,Gc=g can solve the b0,b1,…,bn,c0,c1,…,cn which are the unique solutions of plx=b0φ0x+b1φ1x+⋯+bnφnx,prx=c0φ0x+c1φ1x+⋯+cnφnx, and plx,prx is the least squares solution of data set xi,flxi,xi,frxii=1,2,…,m, and they are also satisfied with equation (10).
At last, we can get the fuzzy least squares approximation:(23)p˜x=px,plx,prx.
Minimum square error:(24)δ1=∑i=1mwifxi−pxi21/2,δ1l=∑i=1mwiflxi−plxi21/2,δ1r=∑i=1mwifrxi−prxi21/2.
Maximum deviation:(25)δ2=max1≤i≤mfxi−pxi,δ2l=max1≤i≤mflxi−plxi,δ2r=max1≤i≤mfrxi−prxi.


## 3.1. LR Fuzzy Least Squares Approximation Problem

Consider the following set of data forxi and f˜xi, where xi∈R, x1<x2<⋯<xm, f˜xi are LR numbers, and f˜xi=fxi,flxi,frxi.Select a function class inCa,b, i.e.,(8)S=Spanφ0x,φ1x,…,φnx.Meanwhile, letX=x1,x2,…,xm be a set of m points in R, and m>n. We suppose(9)p˜x=px,plx,prx=∑j=0najφjx,∑j=0nbjφjx,∑j=0ncjφjx.We seekpx,plx,prx, making(10)∑i=1mwiδi2=∑i=1mwif˜xi−p˜xi2=minpx,plx,prx∈S∑i=1mwif˜xi−p˜xi2,where wii=1,2,…,m are the weight function, and wi>0.Therefore,p˜x=px,plx,prx is the solution of fuzzy least squares approximation.Lemma 2 (see [25]).
The canonical systemGa=d has an unique solution a0,a1,…,an, and the corresponding function px=∑j=0najφjx is satisfied with equation (11); therefore, px is the least squares solution of the data set xi,fxi.Theorem 1.
The three canonical systemsGa=d,Gb=e,andGc=f have a unique solution a0,a1,…,an,b0,b1,…,bn,c0,c1,…,cn, separately, and the corresponding function px=∑j=0najφjx,plx=∑j=0nbjφjx,andprx=∑j=0ncjφjx are satisfied with equations (11)–(13); therefore, p˜x are the least squares solution of the data set xi,f˜xi.Proof.
Supposep˜x=px,plx,prx=a0φ0x+a1φ1x+⋯+anφnx,b0φ0x+b1φ1x+⋯+bnφnx,c0φ0x+c1φ1x+⋯+cnφnx; then,(11)px=a0φ0x+a1φx+⋯+anφnx,(12)plx=b0φ0x+b1φ1x+⋯+bnφnx,(13)prx=c0φ0x+c1φ1x+⋯+cnφnx.
Based on least squares solution,px=a0φ0x+a1φ1x+⋯+anφnx is satisfied with equation (11). If we want to get the least squares solution, we must resolve the coefficient a0,a1,…,an.
In fact, we will solve the coefficienta0,a1,…,an, and it can transform into solve the minimum points a0,a1,…,an of the multivariate functions Ga0,a1,…,an:(14)Ga0,a1,…,an=∑i=1mwifx−∑j=0najφjxi2,and we can obtain ∂G/∂ak=0,k=0,1,…,n:(15)∂G∂ak=2∑i=1mwifx−∑j=0najφjxiφkxi=0,(16)∑j=0naj∑i=1mwiφjxiφkxi=∑i=1mwifxiφkxi,k=0,1,…,n.
According to the inner product symbol of the discrete point set, equation (16) can be written as(17)∑j=0nφj,φkaj=f,φk,k=0,1,…,n.
The matrix form can be expressed asGa=d, i.e.,(18)φ0,φ0φ0,φ1…φ0,φnφ1,φ0φ1,φ1…φ1,φn⋮⋮⋱⋮φn,φ0φn,φ1…φn,φna0a1⋮an=f,φ0f,φ1⋮f,φn,when φ0x,φ1x,…,φnx are linearly independent, due to the Gram matrix, detGn+1≠0; therefore, there is unique solution a0,a1,…,an. It can be seen the least squares solution exists. So, a0,a1,…,an are the unique solution with equation (18). It proved that a unique function can be obtained px=a0φ0x+a1φ1x+⋯+anφnx.
Let us prove the latter part. For anyp∗x=∑j=0naj∗φjx∈S, then(19)∑i=1mwifxi−p∗xi2=∑i=1mwifxi−pxi+pxi−p∗xi2=∑i=1mwifxi−pxi2+2∑i=1mwifxi−pxipxi−p∗xi+∑i=1mwifxi−p∗xi2.
From equation (16), we know(20)∑i=1mwifxi−pxipxi−p∗xi=∑i=1mwifxi−pxi∑j=0naj−aj∗φjxi=∑j=0naj−aj∗∑i=1mwifxi−pxiφjxi=0,and thus,(21)∑i=1mwifxi−p∗xi2=∑i=1mwifxi−pxi2+∑i=1mwipxi−p∗xi2≥∑i=1mwifxi−pxi2,i.e.,(22)∑i=1mwifxi−pxi2=minpx∈S∑i=1mwifxi−p∗xi2.
Therefore,px is the least squares solution of data set xi,fxii=1,2,…,m, and it is satisfied with equation (10).
In the same way,Gb=e,Gc=g can solve the b0,b1,…,bn,c0,c1,…,cn which are the unique solutions of plx=b0φ0x+b1φ1x+⋯+bnφnx,prx=c0φ0x+c1φ1x+⋯+cnφnx, and plx,prx is the least squares solution of data set xi,flxi,xi,frxii=1,2,…,m, and they are also satisfied with equation (10).
At last, we can get the fuzzy least squares approximation:(23)p˜x=px,plx,prx.
Minimum square error:(24)δ1=∑i=1mwifxi−pxi21/2,δ1l=∑i=1mwiflxi−plxi21/2,δ1r=∑i=1mwifrxi−prxi21/2.
Maximum deviation:(25)δ2=max1≤i≤mfxi−pxi,δ2l=max1≤i≤mflxi−plxi,δ2r=max1≤i≤mfrxi−prxi.


## 4. Solving Fuzzy Least Squares Approximation

In order to solve the fuzzy least squares approximation polynomials, we use the least squares principle to approximate each polynomial, wheref˜xi=fxi,flxi,frxi.In particular, we generally choose the basis inS to be 1,x,x2,…,xn, and the weight coefficient is wi=1i=1,2,…,m.Solve the fuzzy least squares approximation polynomialp˜x, which satisfies(26)p˜xi=f˜xi,i=0,1,…,m.Because ofp˜x=f˜x, p˜x can be converted to solve three polynomials by least squares approximation, that is, px,plx,prx. We can divide it into three steps:Step 1: solvepx.At first, let(27)px=a0+a1x+a2x2+⋯+anxn.Then, calculate(28)∑i=1mwi,∑i=1mxi,∑i=1mxi2,…,∑i=1mxin,∑i=1mxin+1,…,∑i=1mxi2n,∑i=1mfxi,∑i=1mxifxi,∑i=1mxi2fxi,…,∑i=1mxinfxi.Last, by solving normal equation,(29)∑i=1m1∑i=1mxi…∑i=1mxin∑i=1mxi∑i=1mxi2…∑i=1mxin+1⋮⋮⋱⋮∑i=1mxin∑i=1mxin+1…∑i=1mxi2na0a1⋮an=∑i=1mfxi∑i=1mxifxi⋮∑i=1mxinfxi,we get the solutiona0,a1,…,an, i.e., px=a0+a1x+a2x2+⋯+anxn.Step 2: solveplx.At first, let(30)plx=b0+b1x+b2x2+⋯+bnxn.Then, calculate(31)∑i=1mflxi,∑i=1mxiflxi,∑i=1mxi2flxi,…,∑i=1mxinflxi.Last, by solving the normal equation,(32)∑i=1m1∑i=1mxi…∑i=1mxin∑i=1mxi∑i=1mxi2…∑i=1mxin+1⋮⋮⋱⋮∑i=1mxin∑i=1mxin+1…∑i=1mxi2nb0b1⋮bn=∑i=1mflxi∑i=1mxiflxi⋮∑i=1mxinflxi,we get the solutionb0,b1,…,bn, i.e., plx=b0+b1x+b2x2+⋯+bnxn.Step 3: solveprx.At first, let(33)prx=c0+c1x+c2x2+⋯+cnxn.Then, calculate(34)∑i=1mfrxi,∑i=1mxifrxi,∑i=1mxi2frxi,…,∑i=1mxinfrxi.Last, by solving normal equation,(35)∑i=1m1∑i=1mxi…∑i=1mxin∑i=1mxi∑i=1mxi2…∑i=1mxin+1⋮⋮⋱⋮∑i=1mxin∑i=1mxin+1…∑i=1mxi2nc0c1⋮cn=∑i=1mfrxi∑i=1mxifrxi⋮∑i=1mxinfrxi,we get the solutionc0,c1,…,cn, i.e., prx=c0+c1x+c2x2+⋯+cnxn.From the above steps, we can get the fuzzy least squares approximation polynomial:(36)f˜x≈p˜x=px,plx,prx.Remark 1.
Sometimes the fitting function can be of hyperbolic,(37)px=aeb/t,or exponential type,(38)px=tat+b.
By means of variable substitution,(39)Y=1y,X=1t,or(40)Y=Lny,X=1t,we transform the fitting problem into a polynomial fitting of low order,(41)SX=A+BX,to complete it.


## 5. Numerical Examples

Example 1.
Consider the following data.
Step 1: solvepx.Letpx=a0+a1x+a2x2:(42)∑i=131=3,∑i=13xi=3,∑i=1mxi2=5,∑i=13xi3=9,∑i=13xi4=17,∑i=13fxi=8,∑i=13xifxi=12,∑i=13xi2fxi=22.Then,(43)3353595917a0a1a2=8125.We get the solutiona0=1,a1=0,a2=1, i.e., a0,a1,…,an,i.e.,px=a0+a1x+a2x2+⋯+anxn.Step 2: solveplx.Letplx=b0+b1x+b2x2:(44)∑i=13flxi=5,∑i=13xiflxi=7,∑i=13xi2flxi=13.Then,(45)3353595917b0b1b2=5713.We get the solutionb0=1,b1=−1,b2=1, i.e., plx=1−x+x2.Step 3: solveprx.Letprx=c0+c1x+c2x2:(46)∑i=13frxi=21,∑i=13xifrxi=31,∑i=13xi2frxi=57.Then,(47)3353595917c0c1c2=213157.We get the solutionc0=3,c1=−1,c2=3, i.e., prx=3−x+3x2.
Therefore, the fuzzy least squares approximation polynomial is(48)f˜x=p˜x=px,plx,prx=1+x2,1−x+x2,3−x+3x2.
The allowable minimum square error and maximum deviation:(49)δ1=0,δ1l=0,δ1r=0,δ2=0,δ2l=0,δ2r=0.Example 2.
Consider the following data.
Letpx=a0+a1x:(50)∑i=131=3,∑i=13xi=7,∑i=1mxi2=21,∑i=13fxi=13.7,∑i=13xifxi=42,and then,(51)37721a0a1=13.742.
We get the solutiona0=−0.45,a1=2.15, i.e., px=−0.45+2.15x.Step 1: solveplx.Letplx=b0+b1x:(52)∑i=13flxi=6.8,∑i=13xiflxi=20.3.Then,(53)37721b0b1=6.820.3.We get the solutionb0=0.05,b1=0.95, i.e., plx=0.05+0.95x.Step 2: solveprx.Letprx=c0+c1x:(54)∑i=13frxi=20.5,∑i=13xifrxi=61.2.Then,(55)37721c0c1=20.561.2.We get the solutionb0=0.15,b1=2.86, i.e., prx=0.15+2.86x.
Therefore, the fuzzy least squares approximation polynomial is(56)f˜x=p˜x=px,plx,prx=−0.45+2.15x,0.05+0.95x,0.15+2.86x.
The allowable minimum square error and maximum deviation:(57)δ1=0.0641,δ1l=0.1539,δ1r=0.0131,δ2=0.1,δ2l=0.1,δ2r=0.21.Remark 2.
Sometimes, we can choose different types of fitting functions for the same set of data. There is only a criterion to decide which function has the best fitting effect, that is, whose error is the minimum.


## 6. Conclusion

In this work, we investigated the fuzzy least squares approximation problem based on LR fuzzy numbers and proposed an approach to solve the fuzzy least squares approximation by using fuzzy polynomial approximation. The total deviation between the obtained approximation function and the known function is minimized. Numerical examples show that our method is feasible. This idea and method can be applied to the case in which the fitting function is of other types such as hyperbolic or exponential form.



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