5.1 Chebyshev's orthogonal polynomials

Chebyshev's orthogonal polynomials in general case have the following form [5]:

$$F\_{\nu}(\xi) = T\_{\nu}(\xi) = \cos\left(\nu \cdot \arccos\xi\right), \quad -1 \le \xi \le 1\tag{16}$$

These polynomials have the following orthogonality property:

$$\int\_{-1}^{1} \frac{T\_{\mu}(\xi)T\_{\nu}(\xi)d\xi}{\sqrt{1-\xi^{2}}} = \begin{cases} 0 \text{ if } \mu \neq \xi; \\ \frac{\pi}{2} \text{ if } \mu = \xi \neq 0; \\ \pi \text{ if } \mu = \xi = 0. \end{cases} \tag{17}$$

where ffiffiffiffiffiffiffiffiffiffiffiffi <sup>1</sup> � <sup>ξ</sup><sup>2</sup> <sup>p</sup> is a weighting coefficient ω ξð Þ in the Eq. (17).

The approximating Chebyshev's orthogonal polynomial for y is obtained on the base of function S minimization:

$$\mathcal{S} = \int\_{-1}^{1} \boldsymbol{\rho}(\boldsymbol{\xi}) \left( \boldsymbol{\mathcal{y}}(\boldsymbol{\xi}) - \sum\_{i=0}^{m} \boldsymbol{b}\_{i} T\_{i}(\boldsymbol{\xi}) \right)^{2} d\boldsymbol{\xi} \tag{18}$$

where from (18) we obtain the following expression for coefficients:

$$b\_k = \begin{cases} \frac{1}{\pi} \int\_{-1}^{1} \frac{\mathcal{Y}(\xi)}{\sqrt{1-\xi^2}} d\xi, & k=0\\ \frac{1}{\pi} \int\_{-1}^{1} \frac{\mathcal{Y}(\xi)T\_k(\xi)}{\sqrt{1-\xi^2}} d\xi, & k \neq 0 \end{cases} \tag{19}$$

Hence, the approximating equation takes the form

$$\overline{\mathcal{Y}}(\xi) = \sum\_{k=0}^{m} b\_k T\_k(\xi) \tag{20}$$

As it may be readily seen from the presented expressions, coefficient bk in Eq. (19) doesn't depend on the choice of degree m. Thus, the variable m doesn't demand recalculation of bj, ∀j ≤ m, while such recalculation is necessary for nonorthogonal approximation.

The best degree m<sup>∗</sup> of approximating may be obtained on the base of hypothesis that investigation results y ið Þ, i ¼ 1, 2, …, r have independent Gaussian distribution in the bounds of some polynomial function y of definite degree, for example, <sup>m</sup><sup>∗</sup> <sup>þ</sup> <sup>μ</sup>, where

$$\overline{\mathcal{Y}}\_{m^\*+\mu}(\mathbf{x}\_i) = \sum\_{j=0}^{m^\*+\mu} b\_j \mathbf{x}\_i^j \tag{21}$$

and a dispersion <sup>σ</sup><sup>2</sup> of distribution y‐<sup>y</sup> don't depend on <sup>μ</sup>.

It's clear that for very small m (m =0,1,2,…), σ<sup>2</sup> <sup>m</sup> decreases as m grows. As in accordance with previously formulated hypothesis, dispersion doesn't depend on <sup>μ</sup>; therefore, the best degree <sup>m</sup><sup>∗</sup> is a minimal <sup>m</sup>, for which <sup>σ</sup><sup>m</sup> ffi <sup>σ</sup><sup>m</sup>þ1.

For determining m<sup>∗</sup> it's necessary to calculate the approximating polynomials of various degrees. As coefficients bj in Eq. (20) don't depend on μ, the determination of the best degree of polynomial is accelerated.

Let us have the forecasted variable Y and input variables x1, x2, …xn. Let's search the relation between them in the following form:

$$Y = A\_{\mathfrak{z}} f\_{\mathfrak{z}}(\mathfrak{x}\_1) + A\_{\mathfrak{z}} f\_{\mathfrak{z}}(\mathfrak{x}\_2) + \dots + A\_{\mathfrak{z}} f\_{\mathfrak{z}}(\mathfrak{x}\_n) \tag{22}$$

where Ai is a fuzzy number of triangular type given as Ai ¼ α<sup>i</sup> ð Þ ;ci , functions fi are determined so [5, 6]

$$f\_i(\mathbf{x}\_i) = \sum\_{j=0}^{m\_i} b\_{ij} T\_j(\mathbf{x}\_i) \tag{23}$$

The degree mi of function fi is determined using hypothesis defined in the preceding section. So if we denote zi ¼ fi ð Þ xi , we'll get linear interval model in its classical form.

#### 5.2 Investigation of trigonometric polynomials as partial descriptions

Let function f xð Þ be periodic with period 2π defined at the interval ½ � �π; π , and its derivative f 0 ð Þ x is also defined at the interval ½ � �π; π . Then the following equality holds

$$S(\boldsymbol{\pi}) = f(\boldsymbol{\pi}); \forall \boldsymbol{\pi} \in [-\boldsymbol{\pi}, \boldsymbol{\pi}] \tag{24}$$

where

$$S(\mathbf{x}) = \frac{a\_0}{2} + \sum\_{j=1} \left( a\_j \cos \left( j\mathbf{x} \right) + b\_j \sin \left( j\mathbf{x} \right) \right) \tag{25}$$

Coefficients aj and bj are calculated by Euler formulas:

$$\begin{aligned} a\_j &= \frac{1}{\pi} \int\_{-\pi}^{\pi} f(\mathbf{x}) \cos \left( j \mathbf{x} \right) d\mathbf{x}; \\\ b\_j &= \frac{1}{\pi} \int\_{-\pi}^{\pi} f(\mathbf{x}) \sin \left( j \mathbf{x} \right) d\mathbf{x}; \end{aligned} \tag{26}$$

#### 5.3 Definition

A trigonometric polynomial of the degree M is called the following polynomial:

$$T\_M(\mathbf{x}) = \frac{a\_0}{2} + \sum\_{j=1}^{M} \left( a\_j \cos \left( j \mathbf{x} \right) + b\_j \sin \left( j \mathbf{x} \right) \right) \tag{27}$$

The following theorem is true stating that exists such M, where 2M < N, which minimizes the following criterion:

$$\sum\_{j=1}^{N} \left( f(\mathbf{x}\_i) - T\_M(\mathbf{x}\_i) \right)^2 \tag{28}$$

Investigation of Fuzzy Inductive Modeling Method in Forecasting Problems DOI: http://dx.doi.org/10.5772/intechopen.86348

Hence the coefficients of corresponding trigonometric polynomial are determined by formulas

$$\begin{aligned} a\_j &= \frac{2}{N} \sum\_{i=1}^N f(\mathbf{x}\_i) \cos \left( j \mathbf{x}\_i \right); \\ b\_j &= \frac{2}{N} \sum\_{i=1}^N f(\mathbf{x}\_i) \sin \left( j \mathbf{x}\_i \right); \end{aligned} \tag{29}$$

Let it be the forecasted variable Y and input variables x1, x2, …xn. Let's search the dependence among them in the form

$$Y = A\_1 f\_1(\mathbf{x}\_1) + A\_2 f\_2(\mathbf{x}\_2) + \dots + A\_n f\_n(\mathbf{x}\_n) \tag{30}$$

where Ai is a fuzzy number of triangular type given as Ai ¼ α<sup>i</sup> ð Þ ;ci , functions fi are determined in such a way:

$$f\_i(\mathbf{x}\_i) = T\_{\mathcal{M}\_i}(\mathbf{x}\_i) \tag{31}$$

The degree Mi of function fi is determined by the theorem described in the preceding section. Therefore, if we assign zi ¼ fi ð Þ xi , the linear interval model will be obtained in its classical form.
