Trigonometric $F^m$-transform and its approximative properties

Abstract

The paper is devoted to generalizing the F-transform with constant components to the trigonometric $F^m$-transform (or $^{t}F^{m}$-transform for short) where $^{t}F^{m}$-transform components are trigonometric polynomials up to m degree, $m\ge 0$. The involving basic functions for $^{t}F^{m}$-transform are sinusoidal shaped functions which are smooth functions. Applying the Gram–Schmidt procedure in order to achieve an orthogonal system leads us to simple calculations of trigonometric basis functions, and we obtain an explicit representation for them for arbitrary m. The approximation and convergence properties of the direct and inverse $^{t}F^{m}$-transforms are discussed, and the applicability of $^{t}F^{m}$-transforms is illustrated by some examples.

1 Introduction

The direct and inverse fuzzy transforms (or F-transform for short) introduced in Perfilieva (2006) have been developed extensively in recent years. The main idea of the F-transforms is based on using the fuzzy partition $A_k, k=1,\ldots ,n$ of [a, b]. Applications of the F-transform are wide and can be used in the construction of approximate models, approximation of functions, filtering, solution of differential equations, and data compression (Perfilieva 2006; Bede et al. 2011; Perfilieva et al. 2008a, b; Novák et al. 2008; Di Martino et al. 2008). In Perfilieva et al. (2011), F-transform has been generalized to $F^m$-transform with polynomial components of arbitrary degree $m\ge 0$. In fact, in Perfilieva et al. (2011), the Gram–Schmidt process has been applied to the system $\{1, x, x^2,\ldots , x^m\}$ in order to convert it to an orthogonal system in a linear space of functions, $L_2(A_k)$, equipped by a weighted inner product with the weight function $A_k$.

In this paper, we study another generalization of the F-transform which has trigonometric components of the degree m, $m\ge 0$. We call them the trigonometric $F^m$-transform ($^{t}F^{m}$-transform for short). In this study, the involving basic functions, $A_k$, for $^{t}F^{m}$-transform are sinusoidal shaped functions with smoothness property. The idea of $^{t}F^{m}$-transform is that, in some applications, if we approximate the function f on each element $A_k$ not by a polynomial functions but by a trigonometric function, we get an even better approximation to f. In fact, we consider $\{1, \cos k\pi x, \sin k\pi x \}$ for $k=1,2,\ldots m$ as a linearly independent system of trigonometric functions and we apply the Gram–Schmidt procedure in order to convert it to an orthogonal system in $L_2(A_k)$. Therefore, we calculate trigonometric basis functions explicitly and we obtain a representation for them with arbitrary m. A linear subspace of $L_2(A_k )$ spanned by these trigonometric basis is denoted by $L^m_2 (A_k )$. Then, we construct trigonometric components, $^{t}F_k^m$, as orthogonal projections of an original function on $L^m_2 (A_k )$. The decisive advantage of ${^{t}}F^{m}$-transform is the simple calculations and the obtained explicit formulas as m is arbitrary.

The structure of this paper is as follows. In Sect. 2, we focus on definition of the F-transform and some of its properties. Moreover, we present some theorems and definitions related to $L_2(A_k)$. In Sect. 3, we construct the trigonometric basis functions by applying Gram–Schmidt procedure to the linearly independent system of trigonometric functions. In Sect. 4, we define the direct and inverse $^{t}F^m$-transforms of a function f with trigonometric components of degree m which are generalization of F-transform. In Sect. 5, some useful approximation properties of $^{t}F^{m}$-transform components of an original function have been investigated. We prove that every trigonometric component approximates an original function on $A_k$ and the accuracy of the resulting approximation using $^{t}F^{m}$-transform increases with an increase in degree m. Also, the accuracy of approximation of the inverse $^{t}F^{m}$-transform will be shown in the space of continuous functions and the space $L^1$. It is worth noting that the inverse $^{t}F^{m}$-transform approximates the original function on the whole domain. Moreover, uniform convergence property of inverse $^{t}F^{m}$-transform is established for arbitrary m. In the last section, in order to illustrate the established theory, we report plots of the error function for some test functions.

2 Preliminary

Generalizing the F-transform is our main propose in this paper; therefore, in the first part of this section, we focus on the definition of the F-transform and some of its properties (Perfilieva 2006). Moreover, the space $L_2(A_k)$ plays crucial role in our study, and so we review some results on this topic in the last part of this section.

2.1 F-transform

The fuzzy partition is the basic concept of F-transform, and so we state it first from Perfilieva (2006). We take an interval [a, b] of real numbers as a universal set. Fuzzy sets on [a, b] are determined by the membership functions from [a, b] into [0, 1]. A fuzzy partition of [a, b] is given by the set of fuzzy subsets of [a, b] which is defined as follows.

Definition 1

(Perfilieva 2006) Let $x_0=x_1<\cdots < x_n=x_{n+1}$ be fixed nodes within [a, b] such that $x_1 = a, x_n = b$ and $n\ge 3$. We say that fuzzy sets $A_1,\ldots , A_n :$ [a, b] $\rightarrow [0, 1]$, which are identified with their membership functions, constitute a fuzzy partition of [a, b] if for $k = 1,\ldots , n$, they fulfill the following conditions:

• (1) $A_k:$ [a, b] $\rightarrow [0,1]$, $A_k(x_k)=1$,

• (2) $A_1 (x) = 0$ if $x\notin [x_1,x_2)$, $A_n(x)=0$ if $x \notin (x_{n-1},x_n]$, $A_k (x) = 0$ if $x\notin (x_{k-1},x_{k+1})$,

• (3) $A_k (x$) is continuous,

• (4) For each $k = 2, \ldots , n$, $A_k (x)$ strictly increases on $[x_{k-1}, x_{k}]$ and strictly decreases on $[x_k , x_{k+1}]$,

• (5) for all $x\in [a,b]$, $\sum _{k=1}^{k=n}A_k(x)=1$.

The membership functions $A_1, \ldots , A_n$ are called basic functions.

Fig. 1

An example of a uniform fuzzy partition of [0, 2] by sinusoidal membership functions

We say that the fuzzy partition $A_1, \ldots , A_n$, $n\ge 3$, is h-uniform if nodes $x_1, \ldots , x_{n}$ are h-equidistant, i.e., for all $k = 2, \ldots , n$, $x_k = x_{k-1}+h$, where $h = (b-a)/(n-1)$ and the following two additional properties are fulfilled:

• (6) for all $k = 2, \ldots , n-1$, and for all $x \in [0, h]$, $A_k (x_k - x) = A_k (x_k + x)$,

• (7) $A_k(x) = A_{k-1}(x- h)$ and $A_{k+1}(x) = A_k(x- h)$ for all $k = 2,\ldots ,n- 1$ and $x\in [x_k,x_{k+1}]$.

The following property of a uniform partition shows that the definite integral of a basic function does not depend on its shape and it will be used throughout this study.

Lemma 1

(Perfilieva 2006) Let a h-uniform partition of [a, b] be given by basic functions $A_1,\ldots ,A_n$, $n\ge 3$. Then

$$\begin{aligned} \int _{x_1}^{x_2}A_1(x) \mathrm{d}x=\int _{x_{n-1}}^{x_n} A_n(x)\mathrm{d}x=\frac{h}{2}, \end{aligned}$$

and for $k = 2,\ldots ,n-1$

$$\begin{aligned} \int _{x_{k-1}}^{x_{k+1}}A_k(x) \mathrm{d}x=h. \end{aligned}$$

The following example regarding the basic functions is essential in the next section (see, e.g., Perfilieva 2006).

Example 1

We consider $A_1,\ldots , A_n$ as a h-uniform partition by sinusoidal shaped basic functions with the following analytic representation which is illustrated in Fig. 1,

$$\begin{aligned} A_1(x)=\left\{ \begin{array}{ll}\frac{1}{2}(\cos \frac{\pi }{h} (x-x_1) +1),&{}\quad x\in [x_1,x_2],\\ 0, &{}\quad o.w \end{array}\right. \end{aligned}$$

(2.1)

and for $k=2,\ldots ,n-1$

$$\begin{aligned} A_k(x)=\left\{ \begin{array}{ll}\frac{1}{2}(\cos \frac{\pi }{h} (x-x_k) +1),&{}\quad x\in [x_{k-1},x_{k+1}],\\ 0, &{}\quad o.w \end{array}\right. \nonumber \\ \end{aligned}$$

(2.2)

and

$$\begin{aligned} A_n(x)=\left\{ \begin{array}{ll}\frac{1}{2}(\cos \frac{\pi }{h} (x-x_n) +1),&{}\quad x\in [x_{n-1},x_n].\\ 0, &{}\quad o.w \end{array}\right. \end{aligned}$$

(2.3)

Now, we recall the definitions of direct and inverse F-transforms with constant components (Perfilieva 2006).

Definition 2

Let $A_1$,...,$A_n$ be a fuzzy partition of [a, b]. We define a (direct) F-transform of a continuous function f with respect to $A_1,\ldots ,A_n$ as a vector $\mathbf F _n[ f] = [F_1,\ldots , F_n]$, where the k-th component $F_k$ is given by

$$\begin{aligned} F_k =\frac{\int _a^b f(x)A_k(x) \mathrm{d}x}{\int _a^b A_k(x) \mathrm{d}x},\quad k= 1,\ldots , n.\end{aligned}$$

Definition 3

Let $f\in C[a, b]$ and $A_1,\ldots ,A_n$ be basic functions which form a fuzzy partition of [a, b]. Let $F_n[f]$ $= [F_1, \ldots , F_n]$ be the F-transform of f with respect to $A_1,\ldots ,A_n$. Then the function

$$\begin{aligned}f_{F,n}(x) = \sum ^n_{ k=1} F_kA_k(x),\end{aligned}$$

is called the inverse F-transform.

2.2 Space $L_2(A_k)$

Let $A_1,\ldots ,A_n$ be a fuzzy partition of [a, b] and $k\in \{1,\ldots , n\}$. The space $L_2(A_k)$ is a set of square-integrable functions $f : [x_{k-1}, x_{k+1}]\rightarrow \mathbb R$ and $L_2(A_1,\ldots , A_n)$ is a set of functions $f:[a, b] \rightarrow \mathbb R$ such that for all $k = 1,\ldots , n$, $f\mid [x_{k-1}, x_{k+1}] \in L_2(A_k )$.

The inner product of $f, g \in L_2(A_k )$ is defined by

$$\begin{aligned} \langle f, g \rangle _k = \frac{\int ^{x_{k+1}}_{ x_{k-1}} f(x)g(x)A_k(x) \mathrm{d}x}{\int ^{x_{k+1}}_{ x_{k-1}}A_k(x) \mathrm{d}x}. \end{aligned}$$

$L_2(A_k)$ is a Hilbert space equipped with the norm $\Vert \cdot \Vert _k$ defined by

$$\begin{aligned} \Vert f\Vert _k=\sqrt{\langle f,f\rangle _k}. \end{aligned}$$

The functions $f, g \in L_2(A_k)$ are orthogonal in $L_2(A_k)$ if $\langle f, g\rangle _ k =0$. The function $f\in L_2(A_k )$ is orthogonal to a subspace B of $L_2(A_k)$ if $\langle f, g\rangle _ k = 0$ for all $g \in B$ and it is denoted by $f \bot B$ or $f \in B^\bot$.

Theorem 1

(see DeVore et al. 1993) Let H be a Hilbert space with the norm $\Vert \cdot \Vert$, and let B be its closed linear subspace. Then, for every element $f\in H$, there exists a unique best approximation $g_0\in B$ in the sense that $g_0$ fulfills

$$\begin{aligned} \Vert f-g_0\Vert =\inf \{\Vert f-g\Vert | g\in B\}. \end{aligned}$$

Moreover, $f- g_0 \in B^\bot$ and $g_0$ is called an orthogonal projection of f on B.

3 Subspace $L_2^m (A_k)$ with sinusoidal shaped basic functions

In this section, we consider $A_1,\ldots , A_n$ as a h-uniform partition by sinusoidal shaped basic functions as in Example 1. For each $k=1,\ldots ,n,$ the set $\{1, \cos \frac{l\pi }{h} (x-x_k), \sin \frac{l\pi }{h} (x-x_k) \}_{l=1}^m$ is a linearly independent system of trigonometric functions restricted to the interval $[x_{k-1},x_{k+1}]$. In what follows, we apply the Gram–Schmidt process to this system to obtain an orthogonal basis for $L_2(A_k)$.

Remark 1

Here, we give inner product of two trigonometric functions of the above-mentioned system in $L_2(A_k)$ that will be useful in sequence. Let $\alpha , \beta \in \mathbb {N}$, for $k=1,\ldots ,n$, we have

$$\begin{aligned} \langle 1,1\rangle _k=1, \left\langle 1,\cos \frac{\pi }{h} (x-x_k)\right\rangle _k=\frac{1}{2}, \end{aligned}$$

and

$$\begin{aligned} \left\langle 1,\cos \frac{\alpha \pi }{h} (x-x_k)\right\rangle _k=0, \end{aligned}$$

where $\alpha \ne 1$. Also

$$\begin{aligned} \left\langle \cos \frac{\alpha \pi }{h} (x-x_k), \cos \frac{\beta \pi }{h} (x-x_k)\right\rangle _k=\left\{ \begin{array}{cc} \frac{1}{2} &{}\quad \alpha =\beta , \\ \frac{1}{4} &{}\quad |\alpha -\beta |=1, \\ 0 &{} |\alpha -\beta |\ne 0,1, \end{array}\right. \end{aligned}$$

and

$$\begin{aligned}&\left\langle \sin \frac{\alpha \pi }{h} (x-x_k), \sin \frac{\beta \pi }{h} (x-x_k)\right\rangle _k\\&\quad =\left\{ \begin{array}{cc} \frac{1}{2} &{}\quad \alpha =\beta , \\ \frac{1}{4} &{}\quad |\alpha -\beta |=1, \\ 0 &{}\quad |\alpha -\beta |\ne 0,1. \end{array}\right. \end{aligned}$$

Moreover, we have

$$\begin{aligned}\left\langle 1, \sin \frac{\beta \pi }{h} (x-x_k)\right\rangle _k=0,\end{aligned}$$

and

$$\begin{aligned} \left\langle \cos \frac{\alpha \pi }{h} (x-x_k), \sin \frac{\beta \pi }{h} (x-x_k)\right\rangle _k=0.\end{aligned}$$

Lemma 2

Let [a, b] be an interval, $n\ge 3$, and $x_0, x_1,\ldots , x_n, x_{n+1}$ be nodes such that $x_0= x_1=a< \cdots < x_n$ $=x_{n+1}=b$. Let $A_1,\ldots , A_n$ be as a h-uniform partition by sinusoidal shaped basic functions as in Example 1. For every k=1,...,n, consider the set $\{1, \cos \frac{l\pi }{h} (x-x_k), \sin \frac{l\pi }{h} (x-x_k) \}_{l=1}^m$ of trigonometric functions restricted to the interval $[x_{k-1},x_{k+1}]$. Then, there exist the orthogonal trigonometric functions $u_k^0, u_k^l, v_k^l$ $(l=1,\ldots ,m)$ in $L_2(A_k)$ as follows:

$$\begin{aligned} u_k^0(x)=1, \end{aligned}$$

$$\begin{aligned} u_k^l(x)=\sum _{i=1}^l (-1)^{l+i}\cos \frac{i\pi }{h}(x-x_k)+(-1)^l\frac{1}{2}, \end{aligned}$$

(3.1)

and

$$\begin{aligned} v_k^l(x)=\sum _{i=1}^l (-1)^{l+i}\frac{i}{l}\sin \frac{i\pi }{h}(x-x_k). \end{aligned}$$

(3.2)

The orthogonal trigonometric functions $u^0_k$, $u^1_k,\ldots , u^m_k, v^1_k,\ldots , v^m_k$ are linearly independent. Moreover, $\langle u^0_k, u^0_k\rangle _k$ $=\langle 1, 1\rangle _k$ $=1$, $\langle u^i_k, u^i_k\rangle _k=\frac{1}{4}$ and $\langle v^i_k, v^i_k\rangle _k=\frac{i+1}{4i}$ for $i=1,\ldots ,m$ and $k=1,\ldots ,n$.

Proof

Applying the Gram–Schmidt process to the system $\{1,\cos \frac{l\pi }{h} (x-x_k), \sin \frac{l\pi }{h} (x-x_k) \}_{l=1}^m$, we gain the orthogonal trigonometric functions $u^0_k, u^1_k, \ldots ,u^m_k, v^1_k, \ldots , v^m_k$ by the following recursive equations:

$$\begin{aligned}u^0_k= 1,\end{aligned}$$

and for $l = 0,\ldots ,m - 1,$

$$\begin{aligned} \ u^{l+1}_k = \cos \frac{(l+1)\pi }{h} (x-x_k) -\sum _ {i=0}^ l \lambda _i^l u^i_k. \ \end{aligned}$$

(3.3)

Also

$$\begin{aligned}v^1_k=\sin \frac{\pi }{h} (x-x_k),\end{aligned}$$

and for $l = 1,\ldots ,m - 1,$

$$\begin{aligned} v^{l+1}_k = \sin \frac{(l+1)\pi }{h} (x-x_k) -\sum _ {i=1}^ {l} \mu _i^l v^i_k,\ \end{aligned}$$

(3.4)

where

$$\begin{aligned}\lambda _i^l = \frac{\left\langle \cos \frac{(l+1)\pi }{h} (x-x_k) , u^i_k \right\rangle _k}{\left\langle u^i_k , u^i_k \right\rangle _k},\ i = 0,\ldots , l\end{aligned}$$

and

$$\begin{aligned}\mu _i^l = \frac{\left\langle \sin \frac{(l+1)\pi }{h} (x-x_k) , v^i_k \right\rangle _k }{ \left\langle v^i_k , v^i_k \right\rangle _k},\ i = 1,\ldots , l.\end{aligned}$$

By (3.3) and Remark 1, we have $u_k^0(x)=1,$

$$\begin{aligned} u_k^1(x)= & {} \cos \frac{\pi }{h} (x-x_k)-\frac{\left\langle \cos \frac{\pi }{h} (x-x_k),1\right\rangle _k}{\left\langle 1,1\right\rangle _k}\\= & {} \cos \frac{\pi }{h} (x-x_k)-\frac{1}{2}. \end{aligned}$$

Now we use the mathematics induction in order to prove (3.1). Therefore, we assume formula (3.1) is true for $n=l-1$; that is, we assume

$$\begin{aligned} u_k^{l-1}(x)=\sum _{i=1}^{l-1} (-1)^{l+i-1}\cos \frac{i\pi }{h}(x-x_k)+(-1)^{l-1}\frac{1}{2}. \end{aligned}$$

We claim that the formula is true for $n=l$. Utilizing the Gram–Schmidt process and recalling Remark 1, we have

$$\begin{aligned} u^{l}_k(x)= & {} \cos \frac{l\pi }{h} (x-x_k) -\sum _ {i=0}^ {l-1} \lambda _i^{l-1} u^i_k(x),\\= & {} \cos \frac{l\pi }{h} (x-x_k)\\&\quad -\sum _ {i=0}^ {l-1} \frac{\left\langle \cos \frac{l\pi }{h} (x-x_k) , u^i_k (x)\right\rangle _k}{ \left\langle u^i_k(x) , u^i_k(x) \right\rangle _k}u^i_k(x)\\= & {} \cos \frac{l\pi }{h} (x-x_k)\\&\quad -\frac{\left\langle \cos \frac{l\pi }{h} (x-x_k) , u^{l-1}_k(x) \right\rangle _k}{ \left\langle u^{l-1}_k(x) , u^{l-1}_k(x) \right\rangle _k}u^{l-1}_k(x)\\= & {} \cos \frac{l\pi }{h} (x-x_k)\\&\quad -\frac{\left\langle \cos \frac{l\pi }{h} (x-x_k) , \cos \frac{(l-1)\pi }{h} (x-x_k) \right\rangle _k}{ \left\langle u^{l-1}_k(x) , u^{l-1}_k(x) \right\rangle _k}u^{l-1}_k(x)\\= & {} \cos \frac{l\pi }{h} (x-x_k) -u^{l-1}_k(x)\\= & {} \cos \frac{l\pi }{h} (x-x_k) -\sum _{i=1}^{l-1} (-1)^{l+i-1}\cos \frac{i\pi }{h}(x-x_k)\\&\quad -\,(-1)^{l-1}\frac{1}{2}\\= & {} \sum _{i=1}^{l} (-1)^{l+i}\cos \frac{i\pi }{h}(x-x_k)+(-1)^{l}\frac{1}{2}. \end{aligned}$$

From (3.4), we have

$$\begin{aligned}v_k^1(x)=\sin \frac{\pi }{h} (x-x_k). \end{aligned}$$

We now employ the mathematics induction in order to prove (3.2). Therefore, we assume formula (3.2) is true for $n=l-1$, i.e.,

$$\begin{aligned} v_k^{l-1}(x)=\sum _{i=1}^{l-1} (-1)^{l+i-1}\frac{i}{l-1}\sin \frac{i\pi }{h}(x-x_k). \end{aligned}$$

We claim that the formula is true for $n=l$. Utilizing the Gram–Schmidt process (3.4) and applying Remark 1, we have

$$\begin{aligned} v^{l}_k(x)= & {} \sin \frac{l\pi }{h} (x-x_k) -\sum _ {i=0}^ {l-1} \lambda _i^{l-1} v^i_k(x),\\= & {} \sin \frac{l\pi }{h} (x-x_k) \\&\quad -\sum _ {i=0}^ {l-1} \frac{\left\langle \sin \frac{l\pi }{h} (x-x_k) , v^i_k (x)\right\rangle _k}{ \left\langle v^i_k(x) , v^i_k(x) \right\rangle _k}v^i_k(x)\\= & {} \sin \frac{l\pi }{h} (x-x_k)\\&\quad -\frac{\left\langle \sin \frac{l\pi }{h} (x-x_k) , v^{l-1}_k(x) \right\rangle _k}{ \left\langle v^{l-1}_k(x) , v^{l-1}_k(x) \right\rangle _k}v^{l-1}_k(x)\\= & {} \sin \frac{l\pi }{h} (x-x_k) \\&\quad -\frac{\left\langle \sin \frac{l\pi }{h} (x-x_k) , \sin \frac{(l-1)\pi }{h} (x-x_k) \right\rangle _k}{ \left\langle v^{l-1}_k(x) , v^{l-1}_k(x) \right\rangle _k}v^{l-1}_k(x)\\= & {} \sin \frac{l\pi }{h} (x-x_k) -v^{l-1}_k(x)\\= & {} \sin \frac{l\pi }{h} (x-x_k)\\&\quad -\sum _{i=1}^{l-1} (-1)^{l+i-1}\frac{i}{l-1}\sin \frac{i\pi }{h}(x-x_k)\\= & {} \sum _{i=1}^l (-1)^{l+i}\frac{i}{l}\sin \frac{i\pi }{h}(x-x_k). \end{aligned}$$

Remark 2

For $l=2,\ldots ,m,$ we have

$$\begin{aligned}u_k^l(x)= \cos \frac{l\pi }{h}(x-x_k)-u_k^{l-1}(x), \end{aligned}$$

$$\begin{aligned} v_k^l(x)=\sin \frac{l\pi }{h}(x-x_k)-\frac{l-1}{l}v_k^{l-1}(x). \end{aligned}$$

Let $m\ge 0$, the linear subspace of $L_2(A_k )$ spanned by the trigonometric basis $u^0_k$, $u^1_k,\ldots , u^m_k,$ $v^1_k,\ldots , v^m_k$ is denoted by $L^m_2 (A_k )$. It is clear that

$$\begin{aligned} L^0_2 (A_k )\subset L^1_2 (A_k ) \subset \cdots L^m_2 (A_k)\subset \cdots .\end{aligned}$$

In the following lemmas, we state some explicit relations in order to simplify the calculations throughout this study and get an explicit representation for coefficients $c_{k,i}$ involving in components of the trigonometric $F^{m}$-transform.

Lemma 3

Let $A_1,\ldots , A_n$ be sinusoidal shaped basic functions and let $u^0_k, u^i_k, v^i_k$ $(i=1,\ldots ,m)$, be the trigonometric basis. Then for $k=1,\ldots ,n$ and $x\in [x_{k-1},x_{k+1}]$, we have

$$\begin{aligned} u_k^iA_k=\frac{1}{4}\left( \cos \frac{(i+1)\pi }{h}(x-x_k)+\cos \frac{i\pi }{h}(x-x_k)\right) ,\nonumber \\ \end{aligned}$$

(3.5)

and

$$\begin{aligned} v_k^iA_k=\frac{1}{4}\left( \sin \frac{(i+1)\pi }{h}(x-x_k)+\frac{i+1}{i}\sin \frac{i\pi }{h}(x-x_k)\right) . \end{aligned}$$

(3.6)

Proof

Let $x\in [x_{k-1},x_{k+1}]$. For $i=1$, we have

$$\begin{aligned} u_k^1A_k= & {} \frac{1}{2}\left( \cos \frac{\pi }{h}(x-x_k)-\frac{1}{2})(\cos \frac{\pi }{h}(x-x_k)+1\right) \\= & {} \frac{1}{2}\left( \cos ^2\frac{\pi }{h}(x-x_k)+\frac{1}{2}\cos \frac{\pi }{h}(x-x_k)-\frac{1}{2}\right) \\= & {} \frac{1}{4}\left( \cos \frac{2\pi }{h}(x-x_k)+\cos \frac{\pi }{h}(x-x_k)\right) . \end{aligned}$$

By Lemma 2, together with applying induction, we get

$$\begin{aligned} u_k^m A_k= & {} \frac{1}{2}\left( \cos \frac{m\pi }{h}(x-x_k)-u_k^{m-1}(x))(\cos \frac{\pi }{h}(x-x_k)+1\right) \\= & {} \frac{1}{2}\cos \frac{m\pi }{h}(x-x_k)\left( \cos \frac{\pi }{h}(x-x_k)+1\right) \\&\quad -u_k^{m-1}(x)A_k(x)\\= & {} \frac{1}{4}\left( \cos \frac{(m+1)\pi }{h}(x-x_k)+\cos \frac{(m-1)\pi }{h}(x-x_k)\right) \\&\quad +\frac{1}{2}\cos \frac{m\pi }{h}(x-x_k)) -\frac{1}{4}\cos \frac{m\pi }{h}(x-x_k)\\&\quad - \frac{1}{4}\cos \frac{(m-1)\pi }{h}(x-x_k)\\= & {} \frac{1}{4}\cos \frac{(m+1)\pi }{h}(x-x_k)+ \frac{1}{4}\cos \frac{m\pi }{h}(x-x_k). \end{aligned}$$

This prove identity (3.5). We can show (3.6) in an analogous manner.

An orthogonal projection of a function $f\in L_2(A_k )$ on the subspace $L^m_2 (A_k)$ has been given in the following lemma. It is worth noting that in the lemma below, we have obtained the explicit formulas for the involving coefficients.

Lemma 4

Let trigonometric functions ${}^{t}F^{m}_k$ be the orthogonal projection of $f\in L_2(A_k )$ on $L^m_2 (A_k )$. Then,

$$\begin{aligned} ^{t}F^{m}_k= & {} c_{k,0}u^0_k + c_{k,1}u^1_k +\cdots +c_{k,m} u^m_k+ d_{k,1}v^1_k +\cdots \\&+\,d_{k,m} v^m_k, \end{aligned}$$

where

$$\begin{aligned} c_{k,0}= & {} \frac{\langle f, 1\rangle _k}{\langle 1, 1\rangle _k}\nonumber \\= & {} \left\{ \begin{array}{cc} \frac{1}{h} \displaystyle \int _{x_{k-1}}^{x_{k+1}}f(x)(\cos \frac{\pi }{h}(x-x_k)+1)\mathrm{d}x&{}\quad k=1,n\\ \displaystyle \frac{1}{2h} \int _{x_{k-1}}^{x_{k+1}}f(x)(\cos \frac{\pi }{h}(x-x_k)+1)\mathrm{d}x &{}\quad o.w, \end{array}\right. \nonumber \\ \end{aligned}$$

(3.7)

and for all $i = 1, \ldots ,m$,

$$\begin{aligned} c_{k,i}= & {} \frac{\left\langle f, u_k^i\right\rangle _k}{\left\langle u_k^i, u_k^i\right\rangle _k} =\left\{ \begin{array}{cc} \frac{2}{h} \displaystyle \int _{x_{k-1}}^{x_{k+1}}f(x)\left( \cos \frac{(i+1)\pi }{h}(x-x_k)+\cos \frac{i\pi }{h}(x-x_k)\right) \mathrm{d}x&{}\quad k=1,n\\ \displaystyle \frac{1}{h} \int _{x_{k-1}}^{x_{k+1}}f(x)\left( \cos \frac{(i+1)\pi }{h}(x-x_k)+\cos \frac{i\pi }{h}(x-x_k)\right) \mathrm{d}x&{}\quad o.w, \end{array}\right. \end{aligned}$$

(3.8)

$$\begin{aligned} d_{k,i}= & {} \frac{\left\langle f, v_k^i\right\rangle _k}{\left\langle v_k^i, v_k^i \right\rangle _k} =\left\{ \begin{array}{cc}\frac{2}{h} \displaystyle \int _{x_{k-1}}^{x_{k+1}}f(x)\left( \frac{i}{i+1}\sin \frac{(i+1)\pi }{h}(x-x_k)+\sin \frac{i\pi }{h}(x-x_k)\right) \mathrm{d}x&{}\quad k=1,n\\ \displaystyle \frac{1}{h} \int _{x_{k-1}}^{x_{k+1}}f(x)\left( \frac{i}{i+1}\sin \frac{(i+1)\pi }{h}(x-x_k)+\sin \frac{i\pi }{h}(x-x_k)\right) \mathrm{d}x,&{}\quad o.w. \end{array}\right. \end{aligned}$$

(3.9)

Proof

Since $F^m_k$ is the orthogonal projection of $f\in L_2(A_k )$ on $L^m_2 (A_k )$, therefore $f-F_k^m \bot L_2^m(A_k)$ and we have for $i=0,1,\ldots ,m$, $j=1,\ldots ,m$

$$\begin{aligned} \langle f-F_k^m, u_k^i\rangle _k=0,\ \ \ \langle f-F_k^m, v_k^j\rangle _k=0. \end{aligned}$$

Since

$$\begin{aligned} F_k^m= & {} c_{k,0}u_k^0-c_{k,1}u_k^1-\cdots -c_{k,m}u_k^m-d_{k,1}v_k^1-\cdots \\&-\,d_{k,m}v_k^m, \end{aligned}$$

therefore

$$\begin{aligned}\left\langle f,u_k^i\right\rangle _k -c_{k,i}\left\langle u_k^i,u_k^i\right\rangle _k= 0 \Rightarrow c_{k,i}=\frac{\langle f, u_k^i\rangle _k}{\langle u_k^i, u_k^i\rangle _k},\end{aligned}$$

and

$$\begin{aligned} \langle f,v_k^j\rangle _k -d_{k,j}\langle v_k^j,v_k^j\rangle _k= 0 \Rightarrow d_{k,j}=\frac{\langle f, v_k^j\rangle _k}{\langle v_k^j, v_k^j\rangle _k}. \end{aligned}$$

By Lemma 2, together with the definition of inner product, we have

$$\begin{aligned} c_{k,0}= & {} \frac{\langle f, 1\rangle _k}{\langle 1, 1\rangle _k}\\= & {} \left\{ \begin{array}{cc} \frac{1}{h} \displaystyle \int _{x_{k-1}}^{x_{k+1}}f(x)\left( \cos \frac{\pi }{h}(x-x_k)+1\right) \mathrm{d}x&{}\quad k=1,n\\ \displaystyle \frac{1}{2h} \int _{x_{k-1}}^{x_{k+1}}f(x)\left( \cos \frac{\pi }{h}(x-x_k)+1\right) \mathrm{d}x &{}\quad o.w, \end{array}\right. \end{aligned}$$

and for all $i = 1, \ldots ,m$,

$$\begin{aligned} c_{k,i}= & {} \frac{\langle f, u_k^i\rangle _k}{\langle u_k^i, u_k^i\rangle _k}\\= & {} \left\{ \begin{array}{cc} \displaystyle \frac{4}{h} \int _{x_{k-1}}^{x_{k+1}}f(x)u_k^i(x)\left( \cos \frac{\pi }{h}(x-x_k)+1\right) \mathrm{d}x&{}\quad k=1,n\\ \displaystyle \frac{2}{h} \int _{x_{k-1}}^{x_{k+1}}f(x)u_k^i(x)\left( \cos \frac{\pi }{h}(x-x_k)+1\right) \mathrm{d}x &{}\quad o.w, \end{array}\right. \end{aligned}$$

$$\begin{aligned} d_{k,i}= & {} \frac{\langle f, v_k^i\rangle _k}{\langle v_k^i, v_k^i\rangle _k}\\= & {} \left\{ \begin{array}{cc} \frac{4i}{h(i+1)} \displaystyle \int _{x_{k-1}}^{x_{k+1}}f(x)v_k^i(x)\left( \cos \frac{\pi }{h}(x-x_k)+1\right) \mathrm{d}x&{}\quad k=1,n\\ \displaystyle \frac{2i}{h(i+1)} \int _{x_{k-1}}^{x_{k+1}}f(x)v_k^i(x)\left( \cos \frac{\pi }{h}(x-x_k)+1\right) \mathrm{d}x &{}\quad o.w. \end{array}\right. \end{aligned}$$

Recalling Lemma 3, we conclude the result.

4 Direct and inverse trigonometric $F^m$-transforms

In this section, we define the direct and inverse $^{t}F^m$- transforms, $m\ge 0$, of a function f with trigonometric components of degree m which is a generalization of F-transform.

Definition 4

Let $A_1,\ldots , A_n$ be sinusoidal shaped basic functions and $u^0_k, u^i_k, v^i_k$ $(i=1,\ldots ,m)$, be the trigonometric basis of $L_2^m(A_k)$. We define a direct $^{t}F^m$-transform of a function $f\in L_2(A_1, \ldots , A_n)$ with respect to $A_1,\ldots ,A_n$ as a vector $^{t}F^{m}[ f ] = (^{t}F^{m} _1 , \ldots , tF^{m}_ n )$ where the k-th component $^{t}F^{m}_k$ is given by

$$\begin{aligned}^{t}F^{m}_k= & {} c_{k,0}u^0_k + c_{k,1}u^1_k +\cdots +c_{k,m} u^m_k+ d_{k,1}v^1_k +\cdots \\&+\,d_{k,m} v^m_k,\end{aligned}$$

where $c_{k,0}, c_{k,i}$ and $d_{k,i}$ are as the relations (3.7–3.9) respectively.

The following lemma shows that $^{t}F^{m}$-transform is a generalization of F-transform, i.e., for $m = 0$, the $^{t}F^{0}$-transform and F-transform are the same.

Lemma 5

Let $m=0$ and $(^{t}F^{0} _1 ,\ldots , ^{t}F^{0}_ n )$ be the $^{t}F^{m}$- transform of f with respect to $A_1,\ldots , A_n$. Then,

$$\begin{aligned}(^{t}F^{0} _1 ,\ldots , ^{t}F^{0}_ n ) =(c_{1,0},\ldots , c_{n,0}),\end{aligned}$$

is the F-transform of f with respect to $A_1,\ldots , A_n$.

Proof

For every $k=1,\ldots ,n$, we consider only coefficients $c_{k,0}$ that corresponds to the basis function $u_k^0$. Because $u_k^0=1$ we have

$$\begin{aligned}c_{k,0}=\frac{\langle f, u_k^0\rangle _k}{\langle u_k^0, u_k^0\rangle _k}=\frac{\frac{1}{2} \int _{x_{k-1}}^{x_{k+1}}f(x)\left( \cos \frac{\pi }{h}(x-x_k)+1\right) \mathrm{d}x}{\frac{1}{2}\int _{x_{k-1}}^{x_{k+1}}\left( \cos \frac{\pi }{h}(x-x_k)+1\right) \mathrm{d}x},\end{aligned}$$

which coincides with the k-th component $F_k$ with respect to trigonometric fuzzy partition.

In the following, analogous to some properties of $F^m$-transform in Bede et al. (2011), we point out some main properties of the $^{t}F^{m}$-transform.

• (A) Because $^tF^{m}_k$ is an orthogonal projection of a function $f\in L_2(A_k )$ on the subspace $L^m_2 (A_k)$, it is the best approximation of f in $L^m_2 (A_k )$.

• (B) Every ${^{t}}F^{m}$-transform component $^tF^{m}_k$, $k = 1,\ldots , n$, satisfies the following equation:

  $$\begin{aligned}^tF^{m}_ k = {^{t}}F^{m-1}_ k + c_{k,m} u^m_ k+ d_{k,m} v^m_ k,\ \ \ m = 1, 2,\ldots .\end{aligned}$$

• (C) The ${^{t}}F^m$-transform of f is an image of a mapping ${^{t}}\mathcal {F}^m : L_2(A_1, \ldots , A_n) \rightarrow L^m_2(A_1)\times \cdots \times L^m_2(A_n)$ such that ${^{t}}\mathcal {F}^m(f)=({^{t}}F_1^m,\ldots ,{^{t}}F_n^m)$. Moreover, ${^{t}}\mathcal {F}^m$ is linear, i.e., for all $f, g \in L_2(A_1,\ldots , A_n)$, and for all $\alpha ,\beta \in \mathbb {R}$, ${^{t}}\mathcal {F}^m(\alpha f+\beta g) = \alpha \ {^{t}}\mathcal {F}^m( f ) + \beta \ {^{t}}\mathcal {F}^m( g),$

• (D) Every function that can be spanned by trigonometric basis functions from $L_2(A_k)$ coincides with its trigonometric ${^{t}}F^m$ transform components on the corresponding interval.

In what follow, we present the definition of the inverse $^tF^{m}$-transform similar to inverse F-transform.

Definition 5

Let $A_1,\ldots , A_n$ be sinusoidal shaped basic functions. Let $f:[a,b]\rightarrow \mathbb R$ be a given function, $m\ge 0$, and let $(^tF_1^m,\ldots ,^tF_n^m)$ be the $^tF^{m}$-transform of f. Then, the following function $^tf_{F,n}^m:[a,b]\rightarrow \mathbb R$

$$\begin{aligned} { ^{t}}f_{F,n}^m=\sum _{k=1}^n {^tF_k^m} A_k \end{aligned}$$

is called the inverse $^tF^{m}$-transform of f with respect to $(^tF_1^m,\ldots ,^tF_n^m)$.

5 Approximation properties

In this section, we apply the truncated Taylor expansion of f to derive the estimation error under sufficient smoothness assumption on f.

5.1 Approximation using the direct $^tF^{m}$-transform

In this subsection, some useful approximation properties of $^tF^{m}$-transform components of an original function have been investigated.

Lemma 6

Let $A_1,\ldots , A_n$ be sinusoidal shaped basic functions. Moreover, let trigonometric functions $^tF^{m}_ k$, $^tF^{m+1}_ k$ ($m \ge 0$) be orthogonal projections of $f \in L_2(A_k )$ on $L^m_2 (A_k )$ and $L^{m+1}_ 2 (A_k )$, respectively. Then,

$$\begin{aligned} \parallel f |_[x_{k-1},x_{k+1}] - ^tF^{m+1}_ k \parallel _k \le \parallel f |_[x_{k-1},x_{k+1}] - ^tF^{m}_ k \parallel _k. \end{aligned}$$

Proof

Our proof is similar to Lemma 2 in Perfilieva et al. (2011).

Theorem 2

Let trigonometric functions $^tF^{m}_k$ be the $^tF^{m}$-transform component of $f\in L_2(A_k )$ and let f be four times continuously differentiable in [a, b]. Then for every $k=2,\ldots ,n-1$

$$\begin{aligned} ^tF_{k}^{m}(x)=f(x)+O(h), \,\,\, x\in [x_{k-1}, x_{k+1}]. \end{aligned}$$

Proof

By Definition 4, we have

$$\begin{aligned} ^tF_k^m(x)=c_{k0}+\sum _{i=1}^m c_{ki} u_k^i(x)+d_{ki}v_k^i(x). \end{aligned}$$

Applying truncated Taylor series for f, we obtain for $k=2,\ldots ,n-1$

$$\begin{aligned}&c_{k,0}= \frac{1}{2h}\int _{x_{k-1}}^{x_{k+1}}f(x)\left( (\cos \frac{\pi }{h}(x-x_k)+1\right) \mathrm{d}x \\&\qquad = \frac{1}{2h}\int _{x_{k-1}}^{x_{k+1}}(f(x_{k})+(x-x_{k})f'(\zeta _k)\\&\qquad \quad \,\times \left( (\cos \frac{\pi }{h}(x-x_k)+1\right) \mathrm{d}x \\&\qquad = f(x_k)+O(h), \end{aligned}$$

$$\begin{aligned}c_{k,i}= & {} \frac{1}{h} \int _{x_{k-1}}^{x_{k+1}}f(x)\left( \cos \frac{(i+1)\pi }{h}(x-x_k)+\cos \frac{i\pi }{h}(x-x_k)\right) \mathrm{d}x\\= & {} \frac{1}{h}\int _{x_{k-1}}^{x_{k+1}}(f(x_{k})+(x-x_{k})f'(\zeta _k))\\&\left( \cos \frac{(i+1)\pi }{h}(x-x_k)+\cos \frac{i\pi }{h}(x-x_k)\right) \mathrm{d}x=O(h), \end{aligned}$$

and

$$\begin{aligned} d_{ki}= & {} \frac{1}{h} \int _{x_{k-1}}^{x_{k+1}}f(x)\left( \frac{i}{i+1}\sin \frac{(i+1)\pi }{h}(x-x_k) +\sin \frac{i\pi }{h}(x-x_k)\right) \mathrm{d}x\\= & {} \frac{1}{h}\int _{x_{k-1}}^{x_{k+1}}(f(x_{k})+(x-x_{k})f'(\zeta _k))\\&\left( \frac{i}{i+1}\sin \frac{(i+1)\pi }{h}(x-x_k)+\sin \frac{i\pi }{h}(x-x_k)\right) \mathrm{d}x =O(h). \end{aligned}$$

By estimations given above, we obtain

$$\begin{aligned}&^tF_k^m(x)\nonumber \\&\quad =c_{k0}+\sum _{i=1}^m c_{ki} u_k^i(x)+d_{ki}v_k^i(x)\\&\quad = f(x_k)+O(h)\nonumber \\&\qquad +\sum _{i=1}^m O(h)\left( \sum _{j=1}^i (-1)^{i+j}\cos \frac{j\pi }{h}(x-x_k)+(-1)^i\frac{1}{2}\right) \\&\qquad + \sum _{i=1}^m O(h)\left( \sum _{j=1}^i (-1)^{i+j}\frac{j}{i}\sin \frac{j\pi }{h}(x-x_k)\right) \\&\quad = f(x_k)+O(h) \\&\qquad +\sum _{i=1}^m \sum _{j=1}^i O(h) \left( \sum _{n=0}^{\infty }(-1)^{n}\left( \frac{j\pi }{h}\right) ^{2n}\frac{(x-x_k)^{2n}}{(2n)!}+(-1)^i\frac{1}{2}\right) \\&\qquad + \sum _{i=1}^m \sum _{j=1}^i O(h)\left( \frac{j}{i}\sum _{n=0}^{\infty }(-1)^{n}\left( \frac{j\pi }{h}\right) ^{2n+1}\frac{(x-x_k)^{2n+1}}{(2n+1)!}\right) \\&\quad = f(x_k)+ O(h)= f(x)+O(h)\nonumber . \end{aligned}$$

5.2 Approximation using the inverse trigonometric $F^m$-transform

In this section, the accuracy of approximation of the inverse $^tF^{m}$-transform will be shown in the space of continuous functions and the space $L^1$. In this study, uniformly convergence property of inverse $^tF^{m}$-transforms has been established for arbitrary m.

Theorem 3

Let $A_1,\ldots , A_n$ be sinusoidal shaped basic functions and the function f be once continuously differentiable on [a, b]. Let $^tf^{ m}_{ F,n}$ be the inverse $^tF^{m}$-transform of f. Then, for all $x\in [a+h, b-h]$, the following estimation holds true:

$$\begin{aligned}f(x) - ^tf^{m}_ {F,n}(x) = O(h).\end{aligned}$$

Proof

Firstly, we claim that

$$\begin{aligned} \sum _{i=1}^{n}A_i(x)=1,\quad \forall x\in [a,b]. \end{aligned}$$

(5.1)

Indeed, if $x\in [a,b]$, there exists $k\in \{2,\ldots ,n\}$ so that $x\in [x_{k-1},x_{k}]$. By Definition 1, we have

$$\begin{aligned} \sum _{i=1}^{n}A_i(x)= & {} \frac{1}{2}\left( \cos \frac{\pi }{h}(x-x_{k-1})+1\right) \\&+\,\frac{1}{2}\left( \cos \frac{\pi }{h}(x-x_{k})+1\right) \\= & {} 1+ \frac{1}{2}\left( \cos \frac{\pi }{h}(x-x_{k-1})+\cos \frac{\pi }{h}(x-x_k)\right) \\= & {} 1+ \frac{1}{2}\left( \cos \frac{\pi }{h}(x-x_k+x_k-x_{k-1})\right. \\&\left. +\,\cos \frac{\pi }{h}(x-x_k)\right) \\= & {} 1. \end{aligned}$$

Now, let $x\in [a+h , b-h ]$ so that $x \in [x_k , x_{k+1}]$ for some $k = 2, \ldots , n-2$. By Theorem 2, we have

$$\begin{aligned}&f(x) - {^tf^{m}}_{ F,n}(x)\\&\quad = f (x)-\sum ^n_{ i=1} {^{t}}F^{m}_ i A_i(x)\\&\quad = f(x)\sum ^n_{i=1} A_i(x)- \sum ^n_ {i=1} {^{t}}F^{m}_i A_i(x)\\&\quad = \sum ^n_{i=1} \frac{1}{2}\left( \cos \frac{\pi }{h}(x-x_{i})+1)( f (x)- ^tF^{m}_i \right) \\&\quad =O(h) \end{aligned}$$

Corollary 1

Let the function f be once continuously differentiable on [a, b]. Then the sequence $\{ ^tf^{m}_ {F,n} \}$ of the inverse $^tF^{m}$-transforms of f uniformly converges to f in the metric space of continuous functions on $[a + h, b - h]$, i.e., as $n\rightarrow \infty$,

$$\begin{aligned} \max _{ [a+h,b-h]} | f (x) - ^tf^{m}_ {F,n}(x)|\rightarrow 0. \end{aligned}$$

Theorem 4

Let $A_1,\ldots , A_n$ be sinusoidal shaped basic functions and the function f be ones continuously differentiable on [a, b]. Let $^tf^ {m}_ {F,n}$ be the inverse $^tF^{m}$-transform of f, where $m \ge 1$. Then

$$\begin{aligned}\int _{a+h}^{b-h} |f(x)-^tf_{F,n}^m|\mathrm{d}x\le O(h).\end{aligned}$$

Proof

For fixed $m \ge 1$, by Theorem 3, we deduce

$$\begin{aligned} \int _{a+h}^{b-h} |f(x)-^tf_{F,n}^m|\mathrm{d}x\le \int _{a}^{b} |f(x)-^tf_{F,n}^m|\mathrm{d}x= O(h). \end{aligned}$$

Corollary 2

Let the function f be once continuously differentiable on [a, b]. Then the sequence $\{ ^tf^{m}_ {F,n} \}_{n-1}^{n=\infty }$ of the inverse $^tF^{m}$-transforms of f, converges uniformity to f in the norm space $L_1$ on $[a + h, b - h]$, i.e.,

$$\begin{aligned}\ \parallel f (x) - ^tf^{m}_ {F,n}(x)\parallel _{L^1}\rightarrow 0,\end{aligned}$$

as $n\rightarrow \infty$.

Fig. 2

Plot of absolute error for the function $f(x)=\sqrt{(x)}+\sqrt{(1-x)}, \ \ x\in [0, 1]$. a $h = .1, m = 5$, b $h = .05, m = 5$, c $h = .1, m = 10$

Fig. 3

Plot of absolute error for the function $\mathrm{cos}(10\pi x), \ \ x\in [0, 1], \ \ h=.1, \ \ \ m=5$. a $h = .1, m = 5$, b $h = .05, m = 5$, c $h = .025, m = 5$

Fig. 4

Plot of absolute error for the function $f(x)=\mathrm{sin}(1/x), \ \ x\in [0, 1]$. a $h = .05, m = 10$, b $h = .02, m = 10$, c $h = .02, m = 20$

6 Illustrative examples

Example 2

In order to illustrate the established theory, we now report in Figs. 2, 3 and 4 plots of the error functions

$$\begin{aligned} E(x)=|f(x)-^tf^{m}_{ F,n}(x)| \end{aligned}$$

which are resulted by applying the $^tF^{m}$-transform to approximate the following test functions for various amounts of h and m.

• a) $\sqrt{x}+\sqrt{1-x}, \ \ \ x\in [0, 1]$.

• b) $\cos (10\pi x), \ \ \ \ x\in [0, 1]$.

• c) $\sin (1/x), \ \ \ x\in (0, 1].$

Functions 1 and 3 have nonsmooth behavior at the end points of [0, 1]. Recalling the error estimation given in Sect. 5, we expect the large error at the boundary points. To overcome this disadvantage, we applied linear extrapolation at boundary points by adding two extra points $x_0=a-h, x_{n+1}=b+h$ and extending the existing fuzzy partition to these points. These results are based on applying extrapolation at boundary points which reduced the error. It is worthy noting that for $\cos (10\pi x)$, the best approximation happens in the case $h=0.1$ at the middle points of [0, 1], with arbitrary $m>1$. This is because that the set of basis functions includes

$$\begin{aligned} u_k^1=\cos \frac{\pi }{h}(x-x_k)-\frac{1}{2}=\cos {10\pi }(x-x_k)-\frac{1}{2}, \end{aligned}$$

which is matched with $\cos {10\pi }(x-x_k)$ for $h=0.1$. The convergence of approximations is illustrated in all figures.

Example 3

In this example, we report in Figs. 5 and 6, plots of the behavior of an average error for various h

$$\begin{aligned} E(x)=\parallel f (x) - ^tf^{m}_ {F,n}(x)\parallel _{L^1},\end{aligned}$$

which are resulted by applying the $^tF^{m}$-transform to approximate the following test functions.

• 1. $\sqrt{x}+\sqrt{1-x}, \ \ \ x\in [0, 1]$.

• 2. $\sin (1/x), \ \ \ x\in (0, 1].$

Fig. 5

Plot of average error for the function $f(x)=\sqrt{x}+\sqrt{1-x}, \ \ x\in [0, 1], \ \ 0\le h\le 0.1, \ \ m=10$.

Fig. 6

Plot of average error for the function $f(x)=\sin (1/x), \ \ x\in [0, 1], \ \ 0\le h \le 0.1, \ \ m=10$.