Weighted Optimal Formulas for Approximate Integration

Abstract

Solutions to problems arising from much scientific and applied research conducted at the world level lead to integral and differential equations. They are approximately solved, mainly using quadrature, cubature, and difference formulas. Therefore, in the current work, we consider a discrete analogue of the differential operator ${\left[1-\frac{1}{{\left(2\pi \right)}^2}\frac{d^2}{d{x}^2}\right]}^{\left[m\right]}$ in the Hilbert space $H_2^{\mu }\left(R\right)$, called $D_m\left[\beta \right]$. We modify the Sobolev algorithm to construct optimal quadrature formulas using a discrete operator. We provide a weighted optimal quadrature formula, using this algorithm for the case where $m=1$. Finally, we construct an optimal quadrature formula in the Hilbert space $H_2^{\mu }\left(R\right)$ for the weight functions $p\left(x\right)=1$ and $p\left(x\right)=e^{2\pi i\omega x}$ when $m=1$.

1. Introduction and Problem Statement

Numerical integration formulas, also known as quadrature and cubature formulas, are used to calculate definite integrals approximately. They are handy for integrals where the antiderivative of the integrand cannot be expressed through elementary functions or when the integrand is only available at discrete points, such as from experimental data. Furthermore, quadrature formulas are a fundamental tool for numerically solving differential and integral equations.

In the field of quadrature theory, several methods are available to calculate integrals roughly using a finite number of integrand values. These methods include the spline method [1,2,3], the $\phi$-function method (as seen in [4,5,6,7,8,9]) and the Sobolev method [10,11,12,13]. Many authors have studied the Sard problem using these methods in different spaces (refer to [14,15,16,17,18,19,20,21,22] and the related literature).

It is worth mentioning that S.L. Sobolev studied the problem of minimizing the norm of the error functional of cubature formulas with regards to coefficients in space $L_2^{\left(m\right)}\left(R^n\right)$. He solved this problem by reducing it to a system of difference equations of the Wiener–Hopf type. As a result, he proved the existence and uniqueness of a solution to this system (see [10,11,12,13]). In addition, he described a specific analytical algorithm for finding optimal coefficients in [10]. To complete this, he defined and studied a discrete analogue $D_{hH}^{\left(m\right)}\left[\beta \right]$ of the polyharmonic operator ${\Delta }^m.$

The problem of the construction of the discrete operator $D_{hH}^{\left(m\right)}\left[\beta \right]$ in an n-dimensional case is complicated and remains an open problem. In the one-dimensional case, Z. Z. Zhamalov and K. M. Shadimetov [23] considered a discrete analogue $D_h^{\left(m\right)}\left[\beta \right]$ of the differential operator $\frac{d^{2m}}{d{E}^{2m}}$. In the work [24], it was possible to construct a discrete analogue of the differential operator $\frac{d^{2m}}{d{E}^{2m}}+2{\omega }^2\frac{d^{2m-2}}{d{E}^{2m-2}}+{\omega }^4\frac{d^{2m-4}}{d{E}^{2m-4}}$. Furthermore, in the works [25,26], discrete analogues of differential operators $\frac{d^{2m}}{d{x}^{2m}}-1$ (for odd m) and $\frac{d^{2m}}{d{x}^{2m}}+2\frac{d^{2m-2}}{d{x}^{2m-2}}+1$ (if even m) were created, and their properties were studied.

M.D. Ramazanov [27] constructed optimal cubature formulas in their work. The author studied the spaces of functions $W_2^{\mu }$, which are obtained by completing the finite Fourier series

$$f\left(x\right)=\sum _k{f}_k{e}^{2\pi ikx}$$

in the following norm:

$$∥\left.f\right|W_2^{\mu }∥={\left|\sum _k{\left|f_k\mu \left(2\pi ik\right)\right|}^2\right|}^{1/2}.$$

In the work of M.D. Ramazanov and Kh.M. Shadimetov [28], optimal cubature formulas were constructed in the space $\widetilde{L_2^{\left(m\right)}}\left(H\right)$.

In the current work, we study Sard’s problem of constructing optimal quadrature formulas in a Hilbert space.

Definition 1. 

Space $H_2^{\mu }\left(R\right)$ is defined as the closure of infinitely differentiable functions defined in R and decreasing to infinity faster than any negative degree in the

$${∥f\left(x\right)∥}_{H_2^{\mu }\left(R\right)}={\left\{{\int }_{-\infty }^{\infty }{\left|F^{-1}\left[{(1+y^2)}^{m/2}·F\left[f\left(x\right)\right]\left(y\right)\right]\right|}^{2}dx\right\}}^{1/2}.$$

(1)

Here, F and $F^{-1}$ are direct and inverse Fourier transforms

$$F\left[f\left(x\right)\right]\left(y\right)={\int }_{-\infty }^{\infty }f\left(x\right)e^{2\pi iyx}dx$$

and

$$F^{-1}\left[f\left(x\right)\right]\left(y\right)={\int }_{-\infty }^{\infty }f\left(x\right)e^{-2\pi iyx}dx.$$

When condition ${\nu }_{m/2}\left(x\right)=F^{-1}\left[{(1+y^2)}^{-m/2}\right]\left(x\right)\in L_2\left(R\right)$ is satisfied, space $H_2^{\mu }\left(R\right)$ is embedded in the space of continuous functions $C\left(R\right)$.

The inner product in $H_2^{\mu }\left(R\right)$ is defined as follows:

$$〈\phi \left(x\right),\psi \left(x\right)〉={\int }_{-\infty }^{\infty }{F}^{-1}\left[{(1+y^2)}^{m/2}·F\left[\phi \left(x\right)\right]\left(y\right)\right]·F^{-1}\left[{(1+y^2)}^{m/2}·F\left[\psi \left(x\right)\right]\left(y\right)\right]dx.$$

We consider a quadrature formula of the form

$${\int }_0^{1}p\left(x\right)f\left(x\right)dx\approx \sum _{\beta =0}^N{C}_{\beta }f\left(h\beta \right),$$

(2)

where $C_{\beta }$ and $h\beta \left(h\beta \in [0,1]\right)$ are coefficients and nodes, respectively, and $p\left(x\right)$ is a weighted function, and $f\left(x\right)$ is an element of the Hilbert space $H_2^{\mu }\left(R\right)$.

The error of the quadrature formula is the difference

$$\left(\ell ,f\right)={\int }_0^{1}p\left(x\right)f\left(x\right)dx-\sum _{\beta =0}^N{C}_{\beta }f\left(h\beta \right),$$

(3)

where

$${\ell }_N\left(x\right)={\epsilon }_{[0,1]}\left(x\right)p\left(x\right)-\sum _{\beta =0}^N{C}_{\beta }\delta (x-h\beta ).$$

(4)

The error functional of the quadrature Formula (2) is denoted by ${\ell }_N\left(x\right)$. Here, ${ϵ}_{[0,1]}\left(x\right)$ is the characteristic function of the interval $[0,1]$, and $\delta \left(x\right)$ is the Dirac’s delta-function.

The error of the quadrature formula will be a linear and continuous functional from space $H_2^{{\mu }^{*}}\left(R\right)$ dual to space $H_2^{\mu }\left(R\right)$, i.e., ${\ell }_N\left(x\right)\in H_2^{{\mu }^{*}}\left(R\right)$.

The quality of the quadrature formula is assessed using the norm of the error functional as follows:

$${∥{\ell }_N\left(x\right)∥}_{H_2^{{\mu }^{*}}\left(R\right)}=\underset{\Vert f\Vert \ne 0}{sup}\frac{\left|\ell \left(f\right)\right|}{{∥f\left(x\right)∥}_{H_2^{\mu }\left(R\right)}}.$$

The norm of the error functional ${\ell }_N\left(x\right)$ depends on the coefficients $C_{\beta }$. If

$${∥\stackrel{\circ }{{\ell }_N}\left(x\right)∥}_{H_2^{{\mu }^{*}}\left(R\right)}=\underset{C_{\beta }}{inf}{∥{\ell }_N\left(x\right)∥}_{H_2^{{\mu }^{*}}\left(R\right)},$$

(5)

then they say that the functional $\stackrel{\circ }{{\ell }_N}$ corresponds to the optimal quadrature formula in $H_2^{{\mu }^{*}}\left(R\right)$.

The focus of this work is to investigate Sard’s problem and to construct optimal quadrature formulas using the Sobolev method in the $H_2^{\mu }\left(R\right)$ space. The objective is to create a quadrature formula that accurately represents the basis functions of the norm kernel (1). However, it should be noted that this optimal quadrature formula will not be exact for algebraic polynomial, exponential or trigonometric functions.

The following article is divided into eight sections. Section 2 discusses the extremal function of the error functional of the quadrature formula and its norm. It calculates the norm of the error functional using this extremal function. Section 3 studies the existence and uniqueness of the optimal quadrature formula. Section 4 explains the Sobolev algorithm for determining the optimal coefficients of a quadrature formula of the form (2). Meanwhile, Section 5 modifies the Sobolev algorithm to find optimal coefficients. Section 6 provides an algorithm for constructing the Fourier transform of function ${\widehat{\nu }}_m\left(x\right)$. In Section 7, a discrete analogue $D_m,\left[\beta \right]$ of the differential operator ${\left(1-\frac{d^2}{{\left(2\pi \right)}^{2}d{x}^2}\right)}^{\left[m\right]}$ is constructed. Section 8 discusses the Sobolev method for constructing optimal quadrature formulas of the form (2) in space $H_2^{\mu }\left(R\right)$. Finally, under $m=1$, optimal quadrature formulas are obtained for the cases of weight functions $p\left(x\right)=1$ and $p\left(x\right)=e^{2\pi i\omega x}$.

2. Extremal Function of the Error Functional of the Quadrature Formula and Its Norm

To find the norm of the error functional (4) in space $H_2^{{\mu }^{*}}\left(R\right)$, its extremal function is used.

Function ${\psi }_{\ell }\left(x\right)$ is called an extremal function of functional ${\ell }_N\left(x\right),$ if

$$({\ell }_N\left(x\right),{\psi }_{\ell }\left(x\right))={∥{\ell }_N∥}_{H_2^{{\mu }^{*}}\left(R\right)}·{∥{\psi }_{\ell }∥}_{H_2^{\mu }\left(R\right)}.$$

(6)

Based on the Riesz theorem on the general form of a linear continuous functional in a Hilbert space for any $f\left(x\right)\in H_2^{\mu }\left(R\right)$, we have

$$\begin{array}{c}\hfill ({\ell }_N\left(x\right),f\left(x\right))=〈{\psi }_{\ell }\left(x\right),f\left(x\right)〉\\ \hfill \mathrm{and}\\ \hfill {∥{\ell }_N\left(x\right)∥}_{H_2^{{\mu }^{*}}}={∥{\psi }_{\ell }\left(x\right)∥}_{H_2^{\mu }},\end{array}$$

(7)

where ${\psi }_{\ell }\left(x\right)$ is a function from space $H_2^{\mu }\left(R\right),$ which corresponds to the functional ${\ell }_N\left(x\right)$. The following theorem is true.

Theorem 1. 

The extremal function of the error functional ${\ell }_N\left(x\right)$ has the form

$${\psi }_{\ell }\left(x\right)={\int }_0^{1}p\left(y\right){\nu }_m(x-y)dy-\sum _{\beta =0}^N{C}_{\beta }{\nu }_m(x-h\beta ),$$

where $p\left(x\right)$ is the weight function, ${\nu }_m\left(x\right)=F^{-1}\left[{(1+y^2)}^{-m}\right]\left(x\right).$

Proof. 

To prove the theorem, we use generalized function theorems and the Fourier transform. By virtue of the Fourier transform of generalized functions we have

$$({\ell }_N\left(x\right),f\left(x\right))=\left(F\left[{\ell }_N\left(x\right)\right]\left(y\right),F\left[f\left(x\right)\right]\left(y\right)\right)=$$

$$=\left({(1+y^2)}^{-m/2}F\left[{\ell }_N\left(x\right)\right]\left(y\right),{(1+y^2)}^{m/2}F\left[f\left(x\right)\right]\left(y\right)\right)=$$

$$=\left(F^{-1}\left\{{(1+y^2)}^{-m/2}F\left[{\ell }_N\left(x\right)\right]\left(y\right)\right\},F^{-1}\left\{{(1+y^2)}^{m/2}F\left[f\left(x\right)\right]\left(y\right)\right\}\right).$$

(8)

If in this equality, we assume

$$F^{-1}\left\{{(1+y^2)}^{m/2}F\left[{\psi }_{\ell }\left(x\right)\right]\left(y\right)\right\}=F^{-1}\left\{{(1+y^2)}^{-m/2}F\left[{\ell }_N\left(x\right)\right]\left(y\right)\right\},$$

i.e., if we assume

$$f\left(x\right)={\psi }_{\ell }\left(x\right)=F^{-1}\left\{{(1+y^2)}^{-m}·F\left[{\ell }_N\left(x\right)\right]\left(y\right)\right\}\left(x\right),$$

(9)

then we will have

$$({\ell }_N\left(x\right),{\psi }_{\ell }\left(x\right))=<{\psi }_{\ell }\left(x\right),{\psi }_{\ell }\left(x\right)>={∥{\ell }_N∥}_{{\hspace{1em}}_{H_2^{{\mu }^{*}}\left(R\right)}}^2.$$

(10)

It follows, firstly, that

$${\psi }_{\ell }\left(x\right)=F^{-1}\left\{{(1+y^2)}^{-m}·F\left[{\ell }_N\left(x\right)\right]\left(y\right)\right\}\left(x\right)=$$

$$=[p\left(x\right)·{\epsilon }_{[0,1]}\left(x\right)]\ast {\nu }_m\left(x\right)-\sum _{\beta =0}^N{C}_{\beta }{\nu }_m(x-h\beta )={\int }_0^{1}p\left(y\right){\nu }_m(x-y)dy-\sum _{\beta =0}^N{C}_{\beta }{\nu }_m(x-h\beta ),$$

(11)

where

$${\nu }_m\left(x\right)=F^{-1}\left[{(1+y^2)}^{-m}\right]\left(x\right).$$

(12)

Function ${\psi }_{\ell }\left(x\right)$ is an extremal function of the error functional (4). In this case, the squared norm of the error functional is calculated by the formula

$${∥{\ell }_N\left(x\right)∥}_{H_2^{\mu *}\left(R\right)}^2=〈{\psi }_{\ell }\left(x\right),{\psi }_{\ell }\left(x\right)〉={\int }_{-\infty }^{\infty }{\left|F^{-1}\left\{{(1+y^2)}^{-m/2}F\left[{\ell }_N\left(x\right)\right]\left(y\right)\right\}\left(x\right)\right|}^{2}dx.$$

(13)

The theorem is completely proven. □

3. Research on the Existence and Uniqueness of an Optimal Quadrature Formula

From (13), it is clear that the norm of the error functional (4) is a function of the coefficients $C_{\beta },\beta =0,1,2,...,N$.

The problem of minimizing the norm of the error functional for fixed nodes $h\beta$ is to determine such coefficients $C_{\beta },\beta =0,1,2,...,N$ for which

$$\underset{C_{\beta }}{inf}{∥{\ell }_N∥}_{H_2^{{\mu }^{*}}\left(R\right)}.$$

To find the coefficients $C_{\beta },\beta =0,1,2,...,N$, we rewrite equality (13) in a slightly different form

$${∥{\ell }_N∥}_{H_2^{{\mu }^{*}}\left(R\right)}^2={\int }_{-\infty }^{\infty }{\left|F^{-1}\left\{{(1+y^2)}^{-m/2}F\left[{\ell }_N\left(x\right)\right]\left(y\right)\right\}\left(x\right)\right|}^{2}dx=$$

$$={\int }_{-\infty }^{\infty }{\left|F^{-1}\left[{(1+y^2)}^{-m/2}\right]\left(x\right)\ast {\ell }_N\left(x\right)\right|}^{2}dx={\int }_{-\infty }^{\infty }{\left|{\nu }_{m/2}\left(x\right)\ast {\ell }_N\left(x\right)\right|}^{2}dx=$$

$$={\int }_{-\infty }^{\infty }{\left|{\nu }_{m/2}\left(x\right)\ast \left(p\left(x\right)·{\epsilon }_{[0,1]}\left(x\right)\right)-\sum _{\beta =0}^N{C}_{\beta }{\nu }_m(x-h\beta )\right|}^{2}dx.$$

(14)

Equality (14) connects the problem of constructing the quadrature Formula (2) with the problem of approximating functions $L_2$ in $p\left(x\right)\ast {\nu }_m\left(x\right)$ by a linear combination of functions $x_{\beta }$ shifted to ${\nu }_m\left(x\right)$.

Indeed, from (14), it is clear that finding the smallest value of the norm of the error functional by coefficients at fixed nodes $x_{\beta }$ is equivalent to the best approximation of function ${\nu }_{m/2}\left(x\right)\ast [p\left(x\right)·{\epsilon }_{[0,1]}\left(x\right)]$ by a linear combination of function ${\nu }_{m/2}\left(x\right)$, and it shifts by $x_{\beta }=\beta h,\beta =0,1,2,...,N$ in the norm of space $L_2\left(R\right)$.

Let us first prove the following lemmas.

Lemma 1. 

The following equality is true

$${∥\sum _{\beta =0}^N{C}_{\beta }\delta (x-h\beta )∥}_{H_2^{{\mu }^{*}}}^2={\int }_{-\infty }^{\infty }{\left|\sum _{\beta =0}^N{C}_{\beta }{\nu }_{\frac{m}{2}}(x-h\beta )\right|}^{2}dx.$$

Proof. 

An extremal function of functional ${\ell }_1\left(x\right)={\sum }_{\beta =0}^N{C}_{\beta }\delta (x-h\beta )$ is a function of ${\psi }_{{\ell }_1}\left(x\right)={\sum }_{\beta =0}^N{C}_{\beta }{\nu }_m(x-h\beta ).$

Indeed, taking (11) into account, we have

$${\psi }_{{\ell }_1}\left(x\right)=F^{-1}\left\{{(1+y^2)}^{-m}F\left[{\ell }_1\left(x\right)\right]\left(y\right)\right\}\left(x\right)=F^{-1}\left\{{(1+y^2)}^{-m}\right\}\ast {\ell }_1\left(x\right)=$$

$$={\nu }_m\left(x\right)\ast {\ell }_1\left(x\right)={\int }_{\infty }^{\infty }{\nu }_m(x-y)\sum _{\beta =0}^N{C}_{\beta }\delta (y-h\beta )dy=\sum _{\beta =0}^N{C}_{\beta }{\nu }_m(x-h\beta ).$$

By the Riesz theorem on the general form of a linear continuous functional on a Hilbert space, we have

$${∥\sum _{\beta =0}^N{C}_{\beta }\delta (x-h\beta )∥}_{H_2^{{\mu }^{*}}}^2={∥{\psi }_{{\ell }_1}\left(x\right)∥}_{H_2^{\mu }}^2=$$

$$={\int }_{-\infty }^{\infty }{\left|F^{-1}\left\{{(1+y^2)}^{\frac{m}{2}}F\left[{\psi }_{{\ell }_1}\left(x\right)\right]\left(y\right)\right\}\left(x\right)\right|}^{2}dx=$$

$$={\int }_{-\infty }^{\infty }{\left|F^{-1}\left\{{(1+y^2)}^{\frac{m}{2}}F\left[\sum _{\beta =0}^N{C}_{\beta }{\nu }_m(x-h\beta )\right]\left(y\right)\right\}\left(x\right)\right|}^{2}dx=$$

$$={\int }_{-\infty }^{\infty }{\left|F^{-1}\left\{{(1+y^2)}^{\frac{m}{2}}F[{\nu }_m\left(x\right)\ast \sum _{\beta =0}^N{C}_{\beta }\delta (x-h\beta )]\left(y\right)\right\}\left(x\right)\right|}^{2}dx=$$

$$={\int }_{-\infty }^{\infty }{\left|F^{-1}\{{(1+y^2)}^{\frac{m}{2}}F\left[{\nu }_m\left(x\right)\right]·F\left[\sum _{\beta =0}^N{C}_{\beta }\delta (x-h\beta )\right]\left(y\right)\}\left(x\right)\right|}^{2}dx=$$

$$={\int }_{-\infty }^{\infty }{\left|F^{-1}\{{(1+y^2)}^{\frac{m}{2}}\frac{1}{{(1+y^2)}^m}·F\left[\sum _{\beta =0}^N{C}_{\beta }\delta (x-h\beta )\right]\left(y\right)\}\left(x\right)\right|}^{2}dx=$$

$$={\int }_{-\infty }^{\infty }{\left|F^{-1}\{\frac{1}{{(1+y^2)}^{\frac{m}{2}}}·F\left[\sum _{\beta =0}^N{C}_{\beta }\delta (x-h\beta )\right]\left(y\right)\}\left(x\right)\right|}^{2}dx=$$

$$={\int }_{-\infty }^{\infty }{\left|F^{-1}\{F\left[{\nu }_{\frac{m}{2}}\left(x\right)\right]\left(y\right)·F\left[\sum _{\beta =0}^N{C}_{\beta }\delta (x-h\beta )\right]\left(y\right)\}\left(x\right)\right|}^{2}dx=$$

$$={\int }_{-\infty }^{\infty }{\left|F^{-1}\left\{F[{\nu }_{\frac{m}{2}}\left(x\right)\ast \sum _{\beta =0}^N{C}_{\beta }\delta (x-h\beta )]\left(y\right)\right\}\left(x\right)\right|}^{2}dx=$$

$$={\int }_{-\infty }^{\infty }{\left|\sum _{\beta =0}^N{C}_{\beta }{\nu }_{\frac{m}{2}}(x-h\beta )\right|}^{2}dx$$

Lemma 1 is proven. □

Lemma 2. 

The system

$${\left\{{\nu }_{m/2}(x-\beta h)\right\}}_{\beta =0}^N$$

(15)

is a linearly independent system.

Proof. 

Consider the linear combination

$$\sum _{\beta =0}^N{C}_{\beta }\delta (x-h\beta )\in H_2^{{\mu }^{*}}\left(R\right).$$

We have

$$\left|\left(\sum _{\beta =0}^N{C}_{\beta }\delta (x-h\beta ),\phi \left(x\right)\right)\right|\le {∥\sum _{\beta =0}^N{C}_{\beta }\delta (x-h\beta )∥}_{H_2^{{\mu }^{*}}\left(R\right)}·{∥\phi \left(x\right)∥}_{H_2^{\mu }\left(R\right)}.$$

Let system (15) be linearly dependent, i.e., ${\sum }_{\beta =0}^N{C}_{\beta }{\nu }_{m/2}(x-h\beta )=0$ and ${\sum }_{\beta =0}^N{C}_{\beta }^2\ne 0$. Let $C_{{\beta }^{\prime }}=max\{\left.C_{\beta }\right|C_{\beta }\ne 0,\beta =0,1,2,...,N\}.$

Hence and from Lemma 1, it follows that

$${∥\sum _{\beta =0}^N{C}_{\beta }\delta (x-h\beta )∥}_{H_2^{*}\left(R\right)}^2={\int }_{-\infty }^{\infty }{\left|\sum _{\beta =0}^N{C}_{\beta }{\nu }_{m/2}(x-h\beta )\right|}^{2}dx=0.$$

(16)

Let us take the function

$${\omega }_{{\beta }^{\prime }}\left(x\right)=C_{{\beta }^{\prime }}{e}^{-{(x-h{\beta }^{\prime })}^2}·\prod _{\beta \ne {\beta }^{\prime }}\frac{(x-h{\beta }^{\prime })}{(h\beta -h{\beta }^{\prime })}.$$

For this function, we have

$$\left(\sum _{\beta =0}^N{C}_{\beta }\delta (x-h\beta ),{\omega }_{{\beta }^{\prime }}\left(x\right)\right)=C_{{\beta }^{\prime }}^2.$$

On the other hand, we have

$$0<C_{{\beta }^{\prime }}^2=\left|\left(\sum _{\beta =0}^N{C}_{\beta }\delta (x-h\beta ),{\omega }_{{\beta }^{\prime }}\left(x\right)\right)\right|\le {∥\sum _{\beta =0}^N{C}_{\beta }\delta (x-h\beta )∥}_{H_2^{{\mu }^{*}}\left(R\right)}·{∥{\omega }_{{\beta }^{\prime }}\left(x\right)∥}_{H_2^{\mu }\left(R\right)}.$$

This inequality contradicts equality (16).

The linear independence of system (15) is proven by this contradiction.

Lemma 2 is proven. □

This also implies the linear independence of $\delta (x-h\beta ),\beta =0,1,2,...,N$ in $H_2^{\mu }\left(R\right)$. Thus, the linear span of functions ${\nu }_{m/2}(x-h\beta ),\beta =0,1,2,...,N$ is a $(N+1)$-dimensional subspace in $L_2\left(R\right)$.

As is known from the theory of Hilbert spaces, an element ${\sum }_{\beta =0}^N{C}_{\beta }{\nu }_{m/2}(x-h\beta )$ is closest to an element $p\left(x\right)·{\epsilon }_{[0,1]}\left(x\right)\ast {\nu }_{m/2}\left(x\right)$ only if the difference $[p\left(x\right)·{\epsilon }_{[0,1]}\left(x\right)]\ast {\nu }_{m/2}\left(x\right)-{\sum }_{\beta =0}^N{C}_{\beta }{\nu }_{m/2}(x-h\beta )$ is orthogonal to each element ${\nu }_{m/2}(x-h\beta ),\beta =0,1,2,...,N$ in $L_2\left(R\right)$, i.e.,

$$\left({\ell }_N\left(x\right)\ast {\nu }_{\frac{m}{2}}\left(x\right),{\nu }_{\frac{m}{2}}(x-h\alpha )\right)=0,\alpha =0,1,2\dots N.$$

(17)

Using the Fourier transform and the left side of equality (17), we reduce it to

$$\left({\ell }_N\left(x\right)\ast {\nu }_{\frac{m}{2}}\left(x\right),{\nu }_{\frac{m}{2}}(x-h\alpha )\right)$$

$$=\left({\nu }_{m/2}\left(x\right)\ast [p\left(x\right)·{\epsilon }_{[0,1]}\left(x\right)]-\sum _{\beta =0}^N{C}_{\beta }{\nu }_{m/2}(x-h\beta ),{\nu }_{m/2}(x-h\alpha )\right)=$$

$$=\left({\nu }_{m/2}\left(x\right)\ast [p\left(x\right)·{\epsilon }_{[0,1]}\left(x\right)]-{\nu }_{m/2}\left(x\right)\ast \sum _{\beta =0}^N{C}_{\beta }\delta (x-h\beta ),{\nu }_{m/2}\left(x\right)\ast \delta (x-h\alpha )\right)=$$

$$=(F[{\nu }_{m/2}\left(x\right)\ast [p\left(x\right)·{\epsilon }_{[0,1]}\left(x\right)]]\left(y\right)$$

$$-F[{\nu }_{m/2}\left(x\right)\ast \sum _{\beta =0}^N{C}_{\beta }\delta (x-h\beta )],F[{\nu }_{m/2}\left(x\right)\ast \delta (x-h\alpha )])=$$

$$=(\frac{1}{{(1+y^2)}^{m/2}}·F\left[[p\left(x\right)·{\epsilon }_{[0,1]}\left(x\right)]\right]\left(y\right)$$

$$-\frac{1}{{(1+y^2)}^{m/2}}·F\left[\sum _{\beta =0}^N{C}_{\beta }\delta (x-h\beta )\right],\frac{1}{{(1+y^2)}^{m/2}}·F\left[\delta (x-h\alpha )\right])$$

$$=\left(\frac{1}{{(1+y^2)}^m}·F\left[[p\left(x\right)·{\epsilon }_{[0,1]}\left(x\right)]\right]\left(y\right)-\frac{1}{{(1+y^2)}^m}·F\left[\sum _{\beta =0}^N{C}_{\beta }\delta (x-h\beta )\right],F\left[\delta (x-h\alpha )\right]\right).$$

From here, applying the inverse Fourier transform, we obtain

$$\left({\ell }_N\left(x\right)\ast {\nu }_{\frac{m}{2}}\left(x\right),{\nu }_{\frac{m}{2}}(x-h\alpha )\right)=$$

$$=\left({\nu }_m\left(x\right)\ast [p\left(x\right)·{\epsilon }_{[0,1]}\left(x\right)]-{\nu }_m\left(x\right)\ast \sum _{\beta =0}^N{C}_{\beta }\delta (x-h\beta ),\delta (x-h\alpha )\right)=$$

$$=\left({\nu }_m\left(x\right)\ast [p\left(x\right)·{\epsilon }_{[0,1]}\left(x\right)]-\sum _{\beta =0}^N{C}_{\beta }{\nu }_m(x-h\beta ),\delta (x-h\alpha )\right)=$$

$$={\left.{\nu }_m\left(x\right)\ast [p\left(x\right)·{\epsilon }_{[0,1]}\left(x\right)]-\sum _{\beta =0}^N{C}_{\beta }{\nu }_m(x-h\beta )\right|}_{x=h\alpha }.$$

(18)

Hence, taking (17) into account, from (18), we obtain

$${\psi }_{\ell }\left(h\alpha \right)=0,\alpha =0,1,2,...,N.$$

(19)

This theorem, known as Babushka’s theorem (see [10]), states that the error functional’s extremal function of the optimal quadrature formula becomes zero at the formula’s nodes.

System (19) represents a set of linear algebraic equations for the coefficients $C_{\beta }$. As such, it can be expressed in the following form:

$$\sum _{\beta =0}^N{C}_{\beta }{\nu }_m(h\alpha -h\beta )={\int }_{-\infty }^{\infty }p\left(y\right)·{\epsilon }_{[0,1]}\left(y\right)]·{\nu }_m(h\alpha -y)dy,\alpha =0,1,2,...,N.$$

(20)

The solution to this system is the optimal coefficients of the quadrature formula. From these lemmas and from the theory of existence and uniqueness of the best approximation under space, the existence and uniqueness of the optimal quadrature formula follows.

Thus, the following is true.

Theorem 2. 

The coefficients of the optimal quadrature formula in space $H_2^{\mu }\left(R\right)$ are a solution to the system of linear Equation (20), which exists and is unique.

4. Algorithm S.L. Sobolev to Determine Optimal Coefficients

Sometimes, it is impossible to solve the system (20) by using known methods because the determinant of the system is too small. To address this issue, S.L. Sobolev [10] proposed a method that enables us to identify the optimal coefficients of the quadrature formula.

We can make some changes to the variables in system (20) to simplify it. Let us redefine $C_{\beta }$ as $C\left[\beta \right]$ and ${\nu }_m\left(h\beta \right)$ as ${\nu }_h^{\left(m\right)}\left[\beta \right]$. We will also assume that $C\left[\beta \right]$ is defined everywhere and is equal to zero when $h\beta \notin [0,1]$. After these changes, we can express system (20) as a convolution of functions with a discrete argument as follows:

$$C\left[\beta \right]\ast {\nu }_h^{\left(m\right)}\left[\beta \right]=f_m\left[\beta \right],\beta =0,1,...,N,$$

(21)

$$C\left[\beta \right]=0,h\beta \notin [0,1],$$

(22)

where

$$f_m\left[\beta \right]={\int }_0^{1}p\left(y\right)·{\nu }_m(h\beta -y)dy.$$

(23)

The system of Equations (21) and (22) will be denoted by system B.

Let us consider the corresponding problem.

Problem B: Find a discrete function $C\left[\beta \right]$ that satisfies system B for given $f_m\left[\beta \right].$ Let us now turn to the solution of system B.

The main idea of this solution is to replace an unknown function.

Namely, instead of $C\left[\beta \right]$ we introduce the function

$$u_h^{\left(m\right)}\left[\beta \right]={\nu }_h^{\left(m\right)}\left[\beta \right]\ast C\left[\beta \right].$$

(24)

In this formulation, we only need to express $C\left[\beta \right]$ through $u_h^{\left(m\right)}\left[\beta \right]$, i.e., find an operator $D_m\left[\beta \right]$ that satisfies the equality

$$D_m\left[\beta \right]\ast {\nu }_h^{\left(m\right)}\left[\beta \right]=\delta \left[\beta \right],$$

(25)

where $\delta \left[\beta \right]=\left\{\begin{array}{c}1,\beta =0,\hfill \\ 0,\beta \ne 0.\hfill \end{array}\right.$

This will allow us to express $C\left[\beta \right]$ in turn as

$$C\left[\beta \right]=D_m\left[\beta \right]\ast u_h^{\left(m\right)}\left[\beta \right].$$

(26)

5. Modification of the Algorithm by S.L. Sobolev for Finding Optimal Coefficients

Some properties of function ${\nu }_m\left(x\right)$ are given below.

• The parity of ${\mu }^{-1}\left(x\right)={(1+x^2)}^{-m}$ implies the parity of ${\nu }_m\left(\xi \right)=F^{-1}\left[{\mu }^{-1}\left(x\right)\right]\left(\xi \right)$. This is obvious.
• The function ${\nu }_m\left(x\right)$ decreases to infinity at a faster rate than any negative power of $\left|x\right|$. Moreover, both $\mu \left(x\right)$ and ${\mu }^{-1}\left(x\right)$ are infinitely differentiable functions. Due to the infinite differentiability and summability of ${\mu }^{-1}\left(x\right)$ and its derivatives, it can be concluded that

  $${\nu }_m\left(\xi \right)={\int }_{-\infty }^{\infty }{(-2\pi i\xi )}^{-\alpha }·{\left[{(1+x^2)}^{-m}\right]}^{\alpha }·e^{-2\pi i\xi x}dx$$

  for each $\alpha \ge 0$.
  It follows that $\left|{\nu }_m\left(\xi \right){\left(2\pi i\xi \right)}^{\alpha }\right|<\infty$.
• The explicit form of ${\nu }_m\left(x\right)$. Due to the parity of ${\nu }_m\left(x\right)$, we have

  $${\nu }_m\left(x\right)={\int }_{-\infty }^{\infty }\frac{e^{2\pi iyx}}{{(1+y^2)}^m}dy=2{\int }_0^{\infty }\frac{cos2\pi yx}{{(1+y^2)}^m}dy.$$
  Further, we know that the equality is [10]

  $${\int }_0^{\infty }\frac{cos\left(ay\right)dy}{{(b^2+y^2)}^n}=\frac{\pi ·e^{-ab}}{{(2·b)}^{2n-1}(n-1)!}\sum _{\hspace{1em}}^{\hspace{1em}}\frac{(2n-k-2)!}{k!(n-k-1)!}{\left(2ab\right)}^k$$

  (27)
  We can obtain an explicit expression for the Fourier transform of the function ${\mu }^{-1}\left(x\right)$ in elementary functions by substituting $a=2\pi x,b=1$ and $n=m$ in the given equation. It is worth noting that the left side of (27) does not change when we replace x with $-x$ because of the parity of the integrand. However, the right side is valid only for $x>0$. Therefore, we replace the equality x with $\left|x\right|$ on the right side of the equation.

Then, we will have

$${\nu }_m\left(x\right)={\int }_{-\infty }^{\infty }{e}^{2\pi iyx}{(1+y^2)}^{-m}dy=\frac{\pi e^{-2\pi \left|x\right|}}{2^{2m-2}·(m-1)!}\sum _{k=o}^{m-1}\frac{(2m-k-2)!·{\left(4\pi \right)}^k}{k!·(m-k-1)!}{\left|x\right|}^k.$$

(28)

Using (24) and (28), we find the form of function $u_h^{\left(m\right)}\left[\beta \right]$ at $\beta <0$ and $\beta >N$.

Let $\beta <0$, then

$$u_h^{\left(m\right)}\left[\beta \right]=e^{2\pi h\beta }\sum _{k=0}^{m-1}{a}_k^{-}{\left[\beta \right]}^k$$

Let $\beta >N$, then

$$u_h^{\left(m\right)}\left[\beta \right]=e^{-2\pi h\beta }\sum _{k=0}^{m-1}{a}_k^{+}{\left[\beta \right]}^k$$

Here, the coefficients $a_k^{-}$ and $a_k^{+}$, $k=0,1,...,m-1$ are unknown. So, we obtain the following task to find function $u_h^{\left(m\right)}\left[\beta \right]$ having the form

$$u_h^{\left(m\right)}\left[\beta \right]=\left\{\begin{array}{c}{e}^{2\pi h\beta }\sum _{k=0}^{m-1}{a}_k^{-}{\left[\beta \right]}^k,\beta <0,\hfill \\ f_m\left[\beta \right],\beta =0,1,...,N,\hfill \\ e^{-2\pi h\beta }\sum _{k=0}^{m-1}{a}_k^{+}{\left[\beta \right]}^k,\beta >N.\hfill \end{array}\right.$$

Since $C\left[\beta \right]=0$ with $\beta <0$ and $\beta >N$ from (26), we have

$$D_m\left[\beta \right]\ast u_h^{\left(m\right)}\left[\beta \right]=0,h\beta \notin \left[0,1\right].$$

(29)

If we construct a discrete operator $D_m\left[\beta \right]$, then the unknown coefficients are determined from (29).

In the following paragraphs, we will deal with the construction of a discrete operator $D_m\left[\beta \right]$.

We will now find a solution to Equation (25). The theory of periodic generalized functions and the Fourier transform suggest that it is more convenient to search for a harrow-shaped function instead of a discrete function $D_m\left[\beta \right]$.

$${\widehat{D}}_m\left(x\right)=\sum _{\beta =-\infty }^{\infty }{D}_m\left[\beta \right]\delta (x-h\beta ).$$

Then, Equation (25) in the class of harrow-shaped functions becomes the equation

$${\widehat{D}}_m\left(x\right)\ast {\widehat{\nu }}_m\left(x\right)=\delta \left(x\right),$$

(30)

where

$${\widehat{\nu }}_m\left(x\right)=\sum _{\beta =-\infty }^{\infty }{\nu }_m\left(h\beta \right)\delta (x-h\beta ).$$

(31)

According to [10], harrow-shaped functions and discrete argument functions are isomorphic. This means that instead of studying $D_m\left[\beta \right]$, we can focus on the ${\widehat{D}}_m\left(x\right)$ function. To complete this, we can apply the Fourier transform to both sides of Equation (30). It is important to keep in mind that

$$F\left[\phi \right(x)\ast \psi (x\left)\right]=F\left[\phi \right]·F\left[\psi \right]\mathrm{and}F\left[\delta \right(x\left)\right]=1,$$

and we get

$$F\left[{\widehat{D}}_m\left(x\right)\right]·F\left[{\widehat{\nu }}_m\left(x\right)\right]=1.$$

From here, we have

$$F\left[{\widehat{D}}_m\left(x\right)\right]={\left\{F\left[{\widehat{\nu }}_m\left(x\right)\right]\right\}}^{-1}.$$

(32)

Now, let us calculate the Fourier transform $F\left[{\widehat{\nu }}_m\left(x\right)\right]$ of the harrow-shaped function ${\widehat{\nu }}_m\left(x\right)$. It is known (see [10]) that

${\varphi }_0\left(x\right)={\sum }_{\beta =-\infty }^{\infty }\delta (x-\beta )$, $\delta \left(hx\right)=h^{-1}\delta \left(x\right)$ and $F\left[{\sum }_{\beta =-\infty }^{\infty }{e}^{2\pi ix\beta }\right]={\sum }_{\beta =-\infty }^{\infty }\delta (p-\beta ).$

Keeping these equalities in mind, we get

$${\widehat{\nu }}_m\left(x\right)=\sum _{\beta =-\infty }^{\infty }{\nu }_m\left(h\beta \right)\delta (x-h\beta )={\nu }_m\left(x\right)·\sum _{\beta =-\infty }^{\infty }\delta (x-h\beta )=$$

$$=h^{-1}{\nu }_m\left(x\right)·\sum _{\beta =-\infty }^{\infty }\delta (x{h}^{-1}-\beta )=h^{-1}{\nu }_m\left(x\right)·{\varphi }_0\left(h^{-1}x\right).$$

(33)

It is also known (see [23]) that

$$F\left[{\varphi }_0\left(h^{-1}x\right)\right]=\sum _{\beta =-\infty }^{\infty }F\left[\delta (h^{-1}x-\beta )\right]=h\sum _{\beta =-\infty }^{\infty }F\left[\delta (x-h\beta )\right]=$$

$$=h\sum _{\beta =-\infty }^{\infty }{e}^{2\pi iph\beta }=h\sum _{\beta =-\infty }^{\infty }\delta (hp-\beta )=h{\varphi }_0\left(hp\right),$$

i.e.,

$$F\left[{\varphi }_0\left(h^{-1}x\right)\right]=h{\varphi }_0\left(hp\right).$$

(34)

Therefore, taking into account (33) and (34), we have

$$F\left[{\widehat{\nu }}_m\left(x\right)\right]=F[h^{-1}{\nu }_m\left(x\right)·{\varphi }_0\left(h^{-1}x\right)]=$$

$$=h^{-1}F\left[{\nu }_m\left(x\right)\right]\ast F\left[{\varphi }_0\left(h^{-1}x\right)\right]=h^{-1}F\left[{\nu }_m\left(x\right)\right]\ast {\varphi }_0\left(hp\right)h.$$

(35)

Since $F\left[{\nu }_m\left(x\right)\right]\left(p\right)={(1+p^2)}^{-m}$, then from (35), we have

$$F\left[{\nu }_m\left(x\right)\right]\left(p\right)={(1+p^2)}^{-m}\ast {\varphi }_0\left(hp\right)=$$

$$=\sum _{\beta =-\infty }^{\infty }{(1+p^2)}^{-m}\ast \delta (hp-\beta )=h^{-1}\sum _{\beta =-\infty }^{\infty }\frac{1}{{\left[1+{(p-\beta h^{-1})}^2\right]}^m}.$$

(36)

Function $F\left[{\nu }_m\left(x\right)\right]\left(p\right)$ is a periodic function with period $h^{-1}=N$. Using equality (36), we must first determine ${\left\{F\left[{\nu }_m\left(x\right)\right]\left(p\right)\right\}}^{-1}$, which will also be an N-periodic function, and then expand it into a Fourier series. Then, based on (32), we will have

$$F\left[{\widehat{D}}_m\left(x\right)\right]\left(p\right)=\sum _{\beta =-\infty }^{\infty }{\widehat{D}}_{\beta }{e}^{2\pi iph\beta },$$

(37)

where ${\widehat{D}}_{\beta }$ are the Fourier coefficients of function $F\left[{\widehat{D}}_m\left(x\right)\right]\left(p\right)$ as follows:

$${\widehat{D}}_{\beta }={\int }_0^{N}F\left[{\widehat{D}}_m\left(x\right)\right]\left(p\right)e^{2\pi iph\beta }dp.$$

(38)

Applying the Fourier inversion formula to equality (37), we arrive at the harrow-shaped function

$${\widehat{D}}_m\left(x\right)=\sum _{\beta =-\infty }^{\infty }{\widehat{D}}_{\beta }\delta (x-h\beta ).$$

To find the Fourier coefficients $F\left[{\widehat{D}}_m\left(x\right)\right]$, we need to apply the definition of a harrow-shaped function. This will give us the set of Fourier coefficients that we require. First, we need to determine the Fourier coefficients of the function $F\left[{\widehat{\nu }}_m\left(x\right)\right]\left(p\right)$.

Lemma 3. 

The following equality is true

$$\sum _{\beta =-\infty }^{\infty }{h}^{-1}{[1+{(p-\beta h^{-1})}^2]}^m=\sum _{\beta =-\infty }^{\infty }{\nu }_m\left(h\beta \right)e^{2\pi i\beta hp}.$$

(39)

Proof. 

Let $\phi \left(x\right)$ be the main function.

Then,

$$\left(\sum _{\beta =-\infty }^{\infty }{h}^{-1}{[1+{(p-h^{-1}\beta )}^2]}^m,\phi \left(p\right)\right)$$

$$=\left(F^{-1}\left[\sum _{\beta =-\infty }^{\infty }{h}^{-1}{[1+{(p-h^{-1}\beta )}^2]}^m\right],F^{-1}\left[\phi \left(p\right)\right]\right)=$$

$$=\left(F^{-1}[1/{(1+p^2)}^m\ast {\varphi }_0\left(hp\right)]\left(x\right),F^{-1}\left[\phi \left(p\right)\right]\left(x\right)\right)=$$

$$=\left(F^{-1}[1/{(1+p^2)}^m]\left(x\right)·F^{-1}\left[{\varphi }_0\left(hp\right)\right]\left(x\right),F^{-1}\left[\phi \left(p\right)\right]\left(x\right)\right)=$$

$$=\left({\nu }_m\left(x\right)·h^{-1}{\varphi }_0\left(h^{-1}x\right),F^{-1}\left[\phi \left(p\right)\right]\left(x\right)\right)=\left(\sum _{\beta =-\infty }^{\infty }{\nu }_m\left(h\beta \right)e^{2\pi ih\beta x},\phi \left(x\right)\right).$$

From the definition of equality of two generalized functions, it follows (39), which proves Lemma 3. □

From (36) and (39), it follows that

$$F\left[{\widehat{\nu }}_m\left(x\right)\right]\left(p\right)=\sum _{\beta =-\infty }^{\infty }{\nu }_m\left(h\beta \right)e^{2\pi ihp\beta }.$$

(40)

From (36), it is clear that $F\left[{\widehat{\nu }}_m\left(x\right)\right]\left(p\right)>0$ is the sum of positive functions. Therefore, ${\left\{F\left[{\widehat{\nu }}_m\left(x\right)\right]\left(p\right)\right\}}^{-1}>0.$

From (32), it is clear that to determine the Fourier coefficients of function $F\left[{\widehat{D}}_m\left(hk\right)\right]\left(p\right)$, we need to find the Fourier transform of function ${\widehat{\nu }}_m\left(x\right)$.

6. Algorithm for Constructing the Fourier Transform of Function ${\widehat{\mathbf{\nu }}}_{\mathbf{m}}\left(\mathbf{x}\right)$

We are currently studying the process of constructing the Fourier transform of the function ${\widehat{\nu }}_m\left(x\right)$ to determine the discrete analogue $D_m\left[\beta \right]$ of the differential operator ${(1-\frac{d^2}{{\left(2\pi \right)}^{2}d{x}^2})}^{\left[m\right]}$. This discrete analogue is used to construct optimal quadrature formulas in the space $H_2^{\mu }\left(R\right)$.

The following is true.

Theorem 3. 

The Fourier transform of function ${\widehat{\nu }}_m\left(x\right)$ is used to determine the discrete analogue of the differential operator ${\left[1-\frac{1}{{\left(2\pi \right)}^2}\frac{d^2}{d{x}^2}\right]}^{\left[m\right]}$, satisfying the equality (32). This analogue takes the form

$$F\left[{\widehat{\nu }}_m\left(x\right)\right]\left(p\right)=\frac{\pi \left(2m-2\right)!}{2^{2m-2}{\left[(m-1)!\right]}^2}\left[\frac{\lambda \left(e^{4\pi h}-1\right)}{-{\lambda }^2{e}^{2\pi h}+\lambda \left(e^{4\pi h}+1\right)-e^{2\pi h}}\right]+$$

$$\begin{array}{c}+\frac{\pi }{2^{2m-2}(m-1)!}\sum _{k=0}^{m-1}\frac{(2m-k-2)!{\left(4\pi h\right)}^k}{k!(m-k-1)!}\left[\frac{e^{2\pi h}}{e^{2\pi h}-\lambda }\right.\sum _{i=0}^k{\left(\frac{\lambda }{e^{2\pi h}-\lambda }\right)}^i\sum _{\rho =0}^i{\left(-1\right)}^{i-\rho }\times \hfill \\ \times \left.\frac{i!}{\rho !\left(i-\rho \right)!}{\rho }^k+\frac{\lambda e^{2\pi h}}{\lambda e^{2\pi h}-1}\sum _{i=0}^k{\left(\frac{1}{\lambda e^{2\pi h}-1}\right)}^i\sum _{\rho =0}^i{\left(-1\right)}^{i-\rho }\frac{i!}{\rho !\left(i-\rho \right)!}{\rho }^k\right].\hfill \end{array}$$

Proof. 

Using formulas (40), we calculate the Fourier transform of function ${\widehat{\nu }}_m\left(x\right).$ We calculate equality (40) for $x=h\beta$ as follows:

$$F\left[{\widehat{\nu }}_m\left(x\right)\right]\left(p\right)=\sum _{\beta =-\infty }^{\infty }\frac{\pi ·e^{-2\pi h\left|\beta \right|}}{2^{2m-2}(m-1)!}\sum _{k=0}^{m-1}\frac{(2m-k-2)!{\left(4\pi \right)}^k}{k!(m-k-1)!}{\left|h\beta \right|}^k{e}^{2\pi ih\beta p}.$$

Thus, we have

$$F\left[{\widehat{\nu }}_m\left(x\right)\right]\left(p\right)=$$

$$=\sum _{\beta =-\infty }^{\infty }{\nu }_m\left(h\beta \right)e^{2\pi ih\beta p}\sum _{\beta =-\infty }^{\infty }\frac{\pi ·e^{-2\pi h\left|\beta \right|}}{2^{2m-2}(m-1)!}\sum _{k=0}^{m-1}\frac{(2m-k-2)!{\left(4\pi \right)}^k}{k!(m-k-1)!}{\left|h\beta \right|}^k{e}^{2\pi ih\beta p}=$$

$$=\frac{\pi }{2^{2m-2}(m-1)!}\sum _{k=0}^{m-1}\frac{(2m-k-2)!{\left(4\pi h\right)}^k}{k!(m-k-1)!}\sum _{\beta =-\infty }^{\infty }{e}^{-2\pi h\left|\beta \right|}{e}^{2\pi ihp\beta }{\left|\beta \right|}^k.$$

Hence, denoting $e^{2\pi ihp}=\lambda$, we have

$$F\left[{\widehat{\nu }}_m\left(x\right)\right]\left(p\right)=\frac{\pi }{2^{2m-2}(m-1)!}\sum _{k=0}^{m-1}\frac{(2m-k-2)!{\left(4\pi h\right)}^k}{k!(m-k-1)!}\sum _{\beta =-\infty }^{\infty }{e}^{-2\pi h\left|\beta \right|}{\lambda }^{\beta }{\left|\beta \right|}^k=$$

$$=\frac{\pi }{2^{2m-2}(m-1)!}\frac{(2m-2)!}{(m-1)!}\sum _{\beta =-\infty }^{\infty }{e}^{-2\pi h\left|\beta \right|}{\lambda }^{\beta }+$$

$$+\frac{\pi }{2^{2m-2}(m-1)!}\sum _{k=1}^{m-1}\frac{(2m-k-2)!{\left(4\pi h\right)}^k}{k!(m-k-1)!}\sum _{\beta =-\infty }^{\infty }{e}^{-2\pi h\left|\beta \right|}{\lambda }^{\beta }{\left|\beta \right|}^k=$$

$$=\frac{\pi \left(2m-2\right)!}{2^{2m-2}{\left[(m-1)!\right]}^2}\left[1+\sum _{\beta =1}^{\infty }{\left(\frac{\lambda }{e^{2\pi h}}\right)}^{\beta }+\sum _{\beta =1}^{\infty }{e}^{-2\pi h\beta }{\lambda }^{-\beta }\right]+$$

$$+\frac{\pi }{2^{2m-2}(m-1)!}\sum _{k=1}^{m-1}\frac{(2m-k-2)!{\left(4\pi h\right)}^k}{k!(m-k-1)!}\left[\sum _{\beta =1}^{\infty }{\left(\frac{\lambda }{e^{2\pi h}}\right)}^{\beta }{\beta }^k+\sum _{\beta =1}^{\infty }\frac{1}{{\left(\lambda e^{2\pi h}\right)}^{\beta }}{\beta }^k\right]=$$

$$=\frac{\pi \left(2m-2\right)!}{2^{2m-2}{\left[(m-1)!\right]}^2}\left[1+\sum _{\beta =1}^{\infty }{\left(\frac{\lambda }{e^{2\pi h}}\right)}^{\beta }+\sum _{\beta =1}^{\infty }\frac{1}{{\left(\lambda e^{2\pi h}\right)}^{\beta }}\right]+$$

$$+\frac{\pi }{2^{2m-2}(m-1)!}\sum _{k=1}^{m-1}\frac{(2m-k-2)!{\left(4\pi h\right)}^k}{k!(m-k-1)!}\left[\sum _{\beta =0}^{\infty }{\left(\frac{\lambda }{e^{2\pi h}}\right)}^{\beta }{\beta }^k+\sum _{\beta =0}^{\infty }\frac{1}{{\left(\lambda e^{2\pi h}\right)}^{\beta }}{\beta }^k\right]$$

Applying the formulas for the sum of an infinitely decreasing geometric progression, we obtain

$$F\left[{\widehat{\nu }}_m\left(x\right)\right]\left(p\right)=\frac{\pi \left(2m-2\right)!}{2^{2m-2}{\left[(m-1)!\right]}^2}\left[1+\frac{\lambda }{e^{2\pi h}-1}+\frac{1}{\lambda e^{2\pi h}-1}\right]+$$

$$+\frac{\pi }{2^{2m-2}(m-1)!}\sum _{k=0}^{m-1}\frac{(2m-k-2)!{\left(4\pi h\right)}^k}{k!(m-k-1)!}\left[\sum _{\beta =0}^{\infty }{\left(\frac{\lambda }{e^{2\pi h}}\right)}^{\beta }{\beta }^k+\sum _{\beta =0}^{\infty }\frac{1}{{\left(\lambda e^{2\pi h}\right)}^{\beta }}{\beta }^k\right].$$

The following formula is valid

$$\sum _{\gamma =0}^{n-1}{q}^{\gamma }{\gamma }^k=\frac{1}{1-q}\sum _{i=0}^k{\left(\frac{q}{1-q}\right)}^i{\Delta }^i{0}^k-\frac{q^n}{1-q}\sum _{i=0}^k{\left(\frac{q}{1-q}\right)}^i{\Delta }^i{\gamma }^k\left|{\hspace{0.17em}}_{\gamma =n}\right.,$$

(41)

where ${\Delta }^i{\gamma }^k$ is a finite difference of order i from ${\gamma }^k$, ${\Delta }^i{0}^k={\Delta }^i{\gamma }^k\left|{\hspace{0.17em}}_{\gamma =n}\right.$. For $\left|q\right|<1$ from (41), we have

$$\sum _{\gamma =0}^{\infty }{q}^{\gamma }{\gamma }^k=\frac{1}{1-q}\sum _{i=0}^k{\left(\frac{q}{1-q}\right)}^i{\Delta }^i{0}^k.$$

(42)

Thus, using formulas (42), we have

$$F\left[{\widehat{\nu }}_m\left(x\right)\right]\left(p\right)=\frac{\pi \left(2m-2\right)!}{2^{2m-2}{\left[(m-1)!\right]}^2}\left[\frac{\lambda \left(e^{4\pi h}-1\right)}{-{\lambda }^2{e}^{2\pi h}+\lambda \left(e^{4\pi h}+1\right)-e^{2\pi h}}\right]+$$

$$\begin{array}{c}\frac{\pi }{2^{2m-2}(m-1)!}\sum _{k=0}^{m-1}\frac{(2m-k-2)!{\left(4\pi h\right)}^k}{k!(m-k-1)!}\left[\frac{e^{2\pi h}}{e^{2\pi h}-\lambda }\right.\sum _{i=0}^k{\left(\frac{\lambda }{e^{2\pi h}-\lambda }\right)}^i{\Delta }^i{0}^k+\hfill \\ +\frac{\lambda e^{2\pi h}}{\lambda e^{2\pi h}-1}\sum _{i=0}^k{\left(\frac{1}{\lambda e^{2\pi h}-1}\right)}^i{\Delta }^i{0}^k,\hfill \end{array}$$

(43)

where ${\Delta }^i{\gamma }^k$ is a finite difference of the order of i from ${\gamma }^k$, ${\Delta }^i{0}^k={\Delta }^i{\gamma }^k\left|{\hspace{0.17em}}_{\gamma =n}\right.$.

Using for calculation, a finite difference of order i, type ${\Delta }^i{0}^k$, the results of work [23] from equality (43), we have

$$F\left[{\widehat{\nu }}_m\left(x\right)\right]\left(p\right)=\frac{\pi \left(2m-2\right)!}{2^{2m-2}{\left[(m-1)!\right]}^2}\left[\frac{\lambda \left(e^{4\pi h}-1\right)}{-{\lambda }^2{e}^{2\pi h}+\lambda \left(e^{4\pi h}+1\right)-e^{2\pi h}}\right]+$$

$$+\frac{\pi }{2^{2m-2}(m-1)!}\sum _{k=0}^{m-1}\frac{(2m-k-2)!{\left(4\pi h\right)}^k}{k!(m-k-1)!}\left[\frac{e^{2\pi h}}{e^{2\pi h}-\lambda }\right.\sum _{i=0}^k{\left(\frac{\lambda }{e^{2\pi h}-\lambda }\right)}^i{\Delta }^i{0}^k+$$

$$+\frac{\lambda e^{2\pi h}}{\lambda e^{2\pi h}-1}\sum _{i=0}^k{\left(\frac{1}{\lambda e^{2\pi h}-1}\right)}^i{\Delta }^i{0}^k]=\frac{\pi \left(2m-2\right)!}{2^{2m-2}{\left[(m-1)!\right]}^2}\left[\frac{\lambda \left(e^{4\pi h}-1\right)}{-{\lambda }^2{e}^{2\pi h}+\lambda \left(e^{4\pi h}+1\right)-e^{2\pi h}}\right]+$$

$$\frac{\pi }{2^{2m-2}(m-1)!}\sum _{k=0}^{m-1}\frac{(2m-k-2)!{\left(4\pi h\right)}^k}{k!(m-k-1)!}\left[\frac{e^{2\pi h}}{e^{2\pi h}-\lambda }\right.\sum _{i=0}^k{\left(\frac{\lambda }{e^{2\pi h}-\lambda }\right)}^i\sum _{\rho =0}^i{\left(-1\right)}^{i-\rho }\frac{i!}{\rho !\left(i-\rho \right)!}{\rho }^k+$$

$$+\frac{\lambda e^{2\pi h}}{\lambda e^{2\pi h}-1}\sum _{i=0}^k{\left(\frac{1}{\lambda e^{2\pi h}-1}\right)}^i\sum _{\rho =0}^i{\left(-1\right)}^{i-\rho }\frac{i!}{\rho !\left(i-\rho \right)!}{\rho }^k],$$

(44)

and it follows that

$$\begin{array}{c}F\left[{\widehat{\nu }}_m\left(x\right)\right]\left(p\right)=\frac{\pi \left(2m-2\right)!}{2^{2m-2}{\left[(m-1)!\right]}^2}\left[\frac{\lambda \left(e^{4\pi h}-1\right)}{-{\lambda }^2{e}^{2\pi h}+\lambda \left(e^{4\pi h}+1\right)-e^{2\pi h}}\right]+\hfill \\ \frac{\pi }{2^{2m-2}(m-1)!}\sum _{k=0}^{m-1}\frac{(2m-k-2)!{\left(4\pi h\right)}^k}{k!(m-k-1)!}\left[\frac{e^{2\pi h}}{e^{2\pi h}-\lambda }\right.\sum _{i=0}^k{\left(\frac{\lambda }{e^{2\pi h}-\lambda }\right)}^i\sum _{\rho =0}^i{\left(-1\right)}^{i-\rho }\frac{i!}{\rho !\left(i-\rho \right)!}{\rho }^k+\hfill \end{array}$$

$$\left.+\frac{\lambda e^{2\pi h}}{\lambda e^{2\pi h}-1}\sum _{i=0}^k{\left(\frac{1}{\lambda e^{2\pi h}-1}\right)}^i\sum _{\rho =0}^i{\left(-1\right)}^{i-\rho }\frac{i!}{\rho !\left(i-\rho \right)!}{\rho }^k\right],$$

(45)

which was what was required to be proven. □

To determine the Fourier coefficients of function $F\left[{\widehat{D}}_m\left(h\beta \right)\right]\left(p\right)$, i.e., $D_m\left[\beta \right]$ applying Theorem 3, we will deal with the expansion of the function ${\left\{F\left[{\widehat{\nu }}_m\left(x\right)\right]\left(p\right)\right\}}^{-1}$ into a Fourier‘series.

7. Discrete Analogue of one Differential Operator of the 2mth Order

In this section, we will discuss the creation of a discrete analogue of the differential operator ${\left(1-\frac{1}{{\left(2\pi \right)}^2}\frac{d^2}{d{x}^2}\right)}^{\left[m\right]}$, which involves solving an equation using convolutions of a discrete argument: $D_m\left[\beta \right]\ast {\nu }_m\left[\beta \right]=\delta \left[\beta \right]$.

To complete this, we present some well-known results from [23].

If we denote the Euler polynomial by $E_k\left(x\right)={\sum }_{s=0}^k{a}_s^{\left(k\right)}{x}^s$, then the coefficient $a_s^{\left(k\right)}$ of the Euler polynomial, as was shown by Euler himself, is expressed by the formula

$$a_s^{\left(k\right)}=\sum _{j=0}^s{\left(-1\right)}^j\left({\hspace{0.17em}}_j^{k+2}\right){\left(s+1-j\right)}^{k+1}$$

The following theorem from [23] is valid.

Theorem 4. 

Polynomial

$$P_k\left(x\right)={\left(x-1\right)}^{k+1}\sum _{i=1}^{k+1}\frac{{\Delta }^i{0}^{k+1}}{{\left(x-1\right)}^i}$$

(46)

is an Euler polynomial of degree k, and in addition,

$$E_k\left(x\right)=x^k{E}_k\left(\frac{1}{x}\right)$$

(47)

or else

$$a_s^{\left(k\right)}=a_{k-s}^{\left(k\right)},s=0,1,2,k.$$

Now, let us prove the following theorem.

Theorem 5. 

The discrete operator $D_m\left[\beta \right]$ is defined by the formula

$$D_m\left[\beta \right]=\frac{2^{2m-2}{\left[\left(m-1\right)!\right]}^2}{\pi \left(2m-2\right)!}\left\{\begin{array}{c}{\sum }_{k=1}^{m-1}\frac{V^m\left({\lambda }_{1,k}\right){\lambda }_{1,k}^{\left|\beta \right|}}{{\lambda }_{1,k}^2{T}_{2m-2}^{\prime }\left({\lambda }_{1,k}\right)},\left|\beta \right|\ge 2;\hfill \\ \frac{1}{d_{2m-2}}+{\sum }_{k=1}^{m-1}\frac{V^m\left({\lambda }_{1,k}\right)}{{\lambda }_{1,k}{T}_{2m-2}^{\prime }\left({\lambda }_{1,k}\right)},\left|\beta \right|=1;\hfill \\ {\sum }_{k=1}^{m-1}\frac{V^m\left({\lambda }_{1,k}\right)}{{\lambda }_{1,k}^2{T}_{2m-2}^{\prime }\left({\lambda }_{1,k}\right)}-\frac{2mcosh\left(2\pi h\right)·d_{2m-2}+d_{2m-3}}{d_{2m-2}^2},\beta =0.\hfill \end{array}\right.$$

Proof. 

From (46), we have

$$\begin{array}{c}\frac{e^{2\pi h}}{e^{2\pi h}-\lambda }\sum _{i=0}^k{\left(\frac{\lambda }{e^{2\pi h}-\lambda }\right)}^i{\Delta }^i{0}^k=\frac{e^{2\pi h}}{e^{2\pi h}-\lambda }\sum _{i=0}^k{\left(\frac{1}{{\lambda }^{-1}{e}^{2\pi h}-1}\right)}^i{\Delta }^i{0}^k=\hfill \\ =\frac{e^{2\pi h}}{e^{2\pi h}-\lambda }·\frac{E_{k-1}\left({\lambda }^{-1}{e}^{2\pi h}\right)}{{\left({\lambda }^{-1}{e}^{2\pi h}-1\right)}^k}=\frac{{\lambda }^{-1}{e}^{2\pi h}{E}_{k-1}\left({\lambda }^{-1}{e}^{2\pi h}\right)}{{\left({\lambda }^{-1}{e}^{2\pi h}-1\right)}^{k+1}}.\hfill \end{array}$$

(48)

Also,

$$\frac{\lambda e^{2\pi h}}{\lambda e^{2\pi h}-1}\sum _{i=0}^k{\left(\frac{1}{\lambda e^{2\pi h}-1}\right)}^i{\Delta }^i{0}^k=\frac{\lambda e^{2\pi h}{E}_{k-1}\left(\lambda e^{2\pi h}\right)}{{\left(\lambda e^{2\pi h}-1\right)}^{k+1}}.$$

(49)

Substituting the found expressions (48) and (49) into (45), we obtain

$$\begin{array}{c}F\left[\widehat{\nu }\left(x\right)\right]\left(p\right)=\frac{\pi \left(2m-2\right)!}{2^{2m-2}{\left[\left(m-1\right)!\right]}^2}\left[\frac{\lambda \left(e^{4\pi h}-1\right)}{-{\lambda }^2{e}^{2\pi h}+\lambda \left(e^{4\pi h}+1\right)-e^{2\pi h}}+\right.\hfill \\ \left.+\frac{\left(m-1\right)!}{\left(2m-2\right)!}\sum _{k=1}^{m-1}\frac{\left(2m-2-k\right)!{\left(4\pi h\right)}^k}{k!\left(m-k-1\right)!}\left(\frac{{\lambda }^{-1}{e}^{2\pi h}{E}_{k-1}\left({\lambda }^{-1}{e}^{2\pi h}\right)}{{\left({\lambda }^{-1}{e}^{2\pi h}-1\right)}^{k+1}}+\frac{\lambda e^{2\pi h}{E}_{k-1}\left(\lambda e^{2\pi h}\right)}{{\left(\lambda e^{2\pi h}-1\right)}^{k+1}}\right)\right].\hfill \end{array}$$

(50)

Now, using Formula (75), we obtain

$$\frac{{\lambda }^{-1}{e}^{2\pi h}{E}_{k-1}\left({\lambda }^{-1}{e}^{2\pi h}\right)}{{\left({\lambda }^{-1}{e}^{2\pi h}-1\right)}^{k+1}}=\frac{{\lambda }^{-1}{e}^{2\pi h}{\left({\lambda }^{-1}{e}^{2\pi h}\right)}^{k-1}{E}_{k-1}\left(\lambda e^{-2\pi h}\right)}{{\left({\lambda }^{-1}{e}^{2\pi h}-1\right)}^{k+1}}=$$

$$=\frac{\lambda e^{-2\pi h}{\left({\lambda }^{-1}{e}^{2\pi h}\right)}^{k+1}{E}_{k-1}\left(\lambda e^{-2\pi h}\right)}{{\left({\lambda }^{-1}{e}^{2\pi h}-1\right)}^{k+1}}=\frac{\lambda e^{-2\pi h}{E}_{k-1}\left(\lambda e^{-2\pi h}\right)}{{\left(1-\lambda e^{-2\pi h}\right)}^{k+1}}.$$

Then, equality (50) takes the following form

$$\begin{array}{c}F\left[\widehat{\nu }\left(x\right)\right]\left(p\right)=\frac{\pi \left(2m-2\right)!}{2^{2m-2}{\left[\left(m-1\right)!\right]}^2}\left[\frac{\lambda \left(e^{4\pi h}-1\right)}{-{\lambda }^2{e}^{2\pi h}+\lambda \left(e^{4\pi h}+1\right)-e^{2\pi h}}+\right.\hfill \\ \left.+\frac{\left(m-1\right)!}{\left(2m-2\right)!}\sum _{k=1}^{m-1}\frac{\left(2m-2-k\right)!{\left(4\pi h\right)}^k}{k!\left(m-k-1\right)!}\left(\frac{\lambda e^{-2\pi h}{E}_{k-1}\left(\lambda e^{-2\pi h}\right)}{{\left(1-\lambda e^{-2\pi h}\right)}^{k+1}}+\frac{\lambda e^{2\pi h}{E}_{k-1}\left(\lambda e^{2\pi h}\right)}{{\left(\lambda e^{2\pi h}-1\right)}^{k+1}}\right)\right].\hfill \end{array}$$

(51)

It is easy to see that

$$\frac{\lambda \left(e^{4\pi h}-1\right)}{-{\lambda }^2{e}^{2\pi h}+\lambda \left(e^{4\pi h}+1\right)-e^{2\pi h}}=\frac{-2sinh\left(2\pi h\right)}{{\lambda }^2-2\lambda cosh\left(2\pi h\right)+1}$$

(52)

and

$$\begin{array}{c}\frac{\lambda e^{-2\pi h}{E}_{k-1}\left(\lambda e^{-2\pi h}\right)}{{\left(1-\lambda e^{-2\pi h}\right)}^{k+1}}+\frac{\lambda e^{2\pi h}{E}_{k-1}\left(\lambda e^{2\pi h}\right)}{{\left(\lambda e^{2\pi h}-1\right)}^{k+1}}=\hfill \\ =\frac{{\left(-1\right)}^{k+1}\lambda \left[e^{-2\pi h}{\left(\lambda e^{2\pi h}-1\right)}^{k+1}{E}_{k-1}\left(\lambda e^{-2\pi h}\right)+e^{2\pi h}{\left(1-\lambda e^{-2\pi h}\right)}^{k+1}{E}_{k-1}\left(\lambda e^{2\pi h}\right)\right]}{{\left({\lambda }^2-2cosh\left(2\pi h\right)\lambda +1\right)}^{k+1}}.\hfill \end{array}$$

(53)

Using equalities (52) and (53), we rewrite expression (51) in the form

$$\begin{array}{c}F\left[\widehat{\nu }\left(x\right)\right]\left(p\right)=\frac{\pi \left(2m-2\right)!}{2^{2m-2}{\left[\left(m-1\right)!\right]}^2}[\frac{-2sinh\left(2\pi h\right)}{{\lambda }^2-2\lambda cosh\left(2\pi h\right)+1}+\hfill \\ +\frac{\left(m-1\right)!}{\left(2m-2\right)!}\sum _{k=1}^{m-1}\frac{\left(2m-2-k\right)!{\left(4\pi h\right)}^k}{k!\left(m-k-1\right)!}\times \hfill \\ \times \frac{{\left(-1\right)}^{k+1}\lambda \left[e^{-2\pi h}{\left(\lambda e^{2\pi h}-1\right)}^{k+1}{E}_{k-1}\left(\lambda e^{-2\pi h}\right)+e^{2\pi h}{\left(1-\lambda e^{-2\pi h}\right)}^{k+1}{E}_{k-1}\left(\lambda e^{2\pi h}\right)\right]}{{\left({\lambda }^2-2\lambda cosh\left(2\pi h\right)+1\right)}^{k+1}}].\hfill \end{array}$$

(54)

Let us denote by $P_{2k}$ a polynomial of degree $2k$ in $\lambda$, i.e.,

$${\mathsf{\Psi }}_{2k}\left(\lambda \right)=e^{-2\pi h}{\left(\lambda e^{2\pi h}-1\right)}^{k+1}{E}_{k-1}\left(\lambda e^{-2\pi h}\right)+e^{2\pi h}{\left(1-\lambda e^{-2\pi h}\right)}^{k+1}{E}_{k-1}\left(\lambda e^{2\pi h}\right),$$

(55)

and through $V\left(\lambda \right)$, it is a polynomial of the second degree in $\lambda$, i.e.,

$$V\left(\lambda \right)={\lambda }^2-2\lambda cosh\left(2\pi h\right)+1,$$

(56)

and the following coefficients through $a_k$, i.e.,

$$d_k=\frac{{\left(-1\right)}^{k+1}{\left(4\pi h\right)}^k\left(2m-2-k\right)!\left(m-1\right)!}{\left(2m-2\right)!k!\left(m-k-1\right)!}.$$

(57)

Taking into account notations (55), (56) and (57), we write equality (54) in the form

$$\begin{array}{c}F\left[\widehat{\nu }\left(x\right)\right]\left(p\right)=\frac{\pi \left(2m-2\right)!}{2^{2m-2}{\left[\left(m-1\right)!\right]}^2}\left[\frac{-2\lambda sinh\left(2\pi h\right)}{V\left(\lambda \right)}+\lambda \sum _{k=1}^{m-1}\frac{d_k{\mathsf{\Psi }}_{2k}\left(\lambda \right)}{V^{k+1}\left(\lambda \right)}\right]=\hfill \\ =\frac{\pi \left(2m-2\right)!\lambda }{2^{2m-2}{\left[\left(m-1\right)!\right]}^2}\left[\frac{-2sinh\left(2\pi h\right)V^{m-1}\left(\lambda \right)+{\sum }_{k=1}^{m-1}{d}_k{\mathsf{\Psi }}_{2k}\left(\lambda \right)V^{m-k-1}\left(\lambda \right)}{V^m\left(\lambda \right)}\right]=\hfill \end{array}$$

$$=\frac{\pi \left(2m-2\right)!\lambda }{2^{2m-2}{\left[\left(m-1\right)!\right]}^2{V}^m\left(\lambda \right)}\left[-2sinh\left(2\pi h\right)V^{m-1}\left(\lambda \right)+\sum _{k=1}^{m-1}{d}_k{\mathsf{\Psi }}_{2k}\left(\lambda \right)V^{m-k-1}\left(\lambda \right)\right].$$

(58)

Let us denote by $T_{2m-2}\left(\lambda \right)$ the next polynomial of degree $2m-2$ in $\lambda$, i.e.,

$$T_{2m-2}\left(\lambda \right)=\sum _{k=1}^{m-1}{d}_k{\mathsf{\Psi }}_{2k}\left(\lambda \right)V^{m-k-1}\left(\lambda \right)-2sinh\left(2\pi h\right)V^{m-1}\left(\lambda \right).$$

(59)

From the notation (55) and (56), it follows that $T_{2m-2}\left(\lambda \right)={\lambda }^{2m-2}{T}_{2m-2}\left(\frac{1}{\lambda }\right)$ so the coefficients of the polynomial $T_{2m-2}\left(\lambda \right)$ are symmetrical. If the ${\lambda }_k$ roots of the polynomial $T_{2m-2}\left(\lambda \right)$, then ${\lambda }_k^{-1}\left(k=\overline{1,m-1}\right)$ is also a root of this polynomial. Hence, we have $\left|{\lambda }_{1,k}\right|<1,\left|{\lambda }_{2,k}\right|>1,$

$${\lambda }_{1,k}·{\lambda }_{2,k}=1,k=\overline{1,m-1}.$$

We write the polynomial $T_{2m-2}\left(\lambda \right)$ in the form

$$T_{2m-2}\left(\lambda \right)=\sum _{n=0}^{2m-2}{d}_n{\lambda }^n,a_n=a_{2m-2-n}.$$

After these notations, we write (58) in the form

$$F\left[\widehat{\nu }\left(x\right)\right]\left(p\right)=\frac{\pi \left(2m-2\right)!\lambda T_{2m-2}\left(\lambda \right)}{2^{2m-2}{\left[\left(m-1\right)!\right]}^2{V}^m\left(\lambda \right)},\lambda =e^{2\pi iph}.$$

(60)

To construct this discrete operator, we use Formulas (32) and (60), i.e.,

$$F\left[\widehat{D}\left(x\right)\right]\left(p\right)=\frac{1}{F\left[\widehat{\nu }\left(x\right)\right]\left(p\right)}=\frac{2^{2m-2}{\left[\left(m-1\right)!\right]}^2{V}^m\left(\lambda \right)}{\pi \left(2m-2\right)!\lambda T_{2m-2}\left(\lambda \right)}.$$

From here,

$$F\left[\widehat{D}\left(x\right)\right]\left(p\right)=\frac{2^{2m-2}{\left[\left(m-1\right)!\right]}^2{V}^m\left(\lambda \right)}{\pi \left(2m-2\right)!\lambda T_{2m-2}\left(\lambda \right)}.$$

(61)

To find the Fourier transform of functions $\widehat{D}\left(x\right)$, we expand the functions A into a Fourier series

$$\frac{2^{2m-2}{\left[\left(m-1\right)!\right]}^2{V}^m\left(\lambda \right)}{\pi \left(2m-2\right)!\lambda T_{2m-2}\left(\lambda \right)}.$$

(62)

In order to obtain the expansion of this function into a Fourier series, we divided the polynomial $V^m\left(\lambda \right)$ of degree $2m$ from $\lambda$ by the polynomial $\lambda T_{2m-2}\left(\lambda \right)$ of degree $2m-1$ from $\lambda$, i.e.,

$$\frac{V^m\left(\lambda \right)}{\lambda T_{2m-2}\left(\lambda \right)}=\frac{\lambda }{d_{2m-2}}+\left(-2mcosh\left(2\pi h\right)-\frac{d_{2m-3}}{d_{2m-2}}\right)\frac{1}{d_{2m-2}}+\frac{R_{2m-2}\left(\lambda \right)}{\lambda T_{2m-2}\left(\lambda \right)}.$$

(63)

Here, $d_{2m-2},d_{2m-3}$ are the coefficients of the polynomial $T_{2m-2}\left(\lambda \right)$, and $R_{2m-2}\left(\lambda \right)$ is a polynomial of degree $2m-2$ then $\lambda$. Now, we decompose the rational fraction $\frac{R_{2m-2}\left(\lambda \right)}{\lambda T_{2m-2}\left(\lambda \right)}$ into elementary fractions

$$\frac{R_{2m-2}\left(\lambda \right)}{\lambda T_{2m-2}\left(\lambda \right)}=\frac{M_0}{\lambda }+\sum _{k=1}^{m-1}\frac{M_{1,k}}{\lambda -{\lambda }_{1,k}}+\sum _{k=1}^{m-1}\frac{M_{2,k}}{\lambda -{\lambda }_{2,k}}.$$

(64)

Here, $M_0,M_{1,k},M_{2,k}$ are unknown coefficients, ${\lambda }_{1,k}$ are roots of the polynomial $T_{2m-2}\left(\lambda \right)$ with an absolute value less than one.

Taking (64) into account, equality (63) takes the following form

$$\frac{V^m\left(\lambda \right)}{\lambda T_{2m-2}\left(\lambda \right)}=\frac{\lambda }{d_{2m-2}}+\frac{-2mcosh\left(2\pi h\right)·d_{2m-2}-d_{2m-3}}{d_{2m-2}^2}+\frac{M_0}{\lambda }+\sum _{k=1}^{m-1}\frac{M_{1,k}}{\lambda -{\lambda }_{1,k}}+\sum _{k=1}^{m-1}\frac{M_{2,k}}{\lambda -{\lambda }_{2,k}}.$$

(65)

From here, we find $V^m\left(\lambda \right)$, i.e.,

$$\begin{array}{c}{V}^m\left(\lambda \right)=\frac{\lambda T_{2m-2}\left(\lambda \right)}{d_{2m-2}}-\frac{\left(2mcosh\left(2\pi h\right)d_{2m-2}+d_{2m-3}\right)\lambda T_{2m-2}\left(\lambda \right)}{d_{2m-2}^2}+\hfill \\ +M_0{T}_{2m-2}\left(\lambda \right)+\sum _{k=1}^{m-1}\frac{M_{1,k}{T}_{2m-2}\left(\lambda \right)}{\lambda -{\lambda }_{1,k}}+\sum _{k=1}^{m-1}\frac{M_{2,k}{T}_{2m-2}\left(\lambda \right)}{\lambda -{\lambda }_{2,k}}.\hfill \end{array}$$

(66)

Substituting in equality (66) instead of $\lambda :\lambda =0,\lambda ={\lambda }_{1,k},\lambda ={\lambda }_{2,k}$, we obtain $M_0{d}_0=1,$

$$\begin{array}{c}{V}^m\left({\lambda }_{1,k}\right)=M_{1,k}{\lambda }_{1,k}{T}_{2m-2}^{\prime }\left({\lambda }_{1,k}\right),\hfill \\ V^m\left({\lambda }_{2,k}\right)=M_{2,k}{\lambda }_{2,k}{T}_{2m-2}^{\prime }\left({\lambda }_{2,k}\right).\hfill \end{array}$$

From here, we consistently find

$$M_0=\frac{1}{d_0},M_{1,k}=\frac{V^m\left({\lambda }_{1,k}\right)}{{\lambda }_{1,k}{T}_{2m-2}^{\prime }\left({\lambda }_{1,k}\right)},M_{2,k}=\frac{V^m\left({\lambda }_{2,k}\right)}{{\lambda }_{2,k}{T}_{2m-2}^{\prime }\left({\lambda }_{2,k}\right)}.$$

(67)

Since ${\lambda }_{1,k}·{\lambda }_{2,k}=1$ and $V^m\left(\lambda \right)={\lambda }^{2m}{V}^m\left(\frac{1}{\lambda }\right),$ then after some calculations, we obtain

$$M_{1,k}=\frac{-M_{2,k}}{{\lambda }_{2,k}^2}.$$

(68)

Considering that $\left|{\lambda }_{1,k}\right|<1$ and $\left|{\lambda }_{2,k}\right|>1$, then the sums ${\sum }_{k=1}^{m-1}\frac{M_{1,k}}{\lambda -{\lambda }_{1,k}}$ and ${\sum }_{k=1}^{m-1}\frac{M_{2,k}}{\lambda -{\lambda }_{2,k}}$ can be represented by a Laurent series on circle $\left|{\lambda }^2\right|=1$ as follows:

$$\sum _{k=1}^{m-1}\frac{M_{1,k}}{\lambda -{\lambda }_{1,k}}=\frac{1}{\lambda }\sum _{k=1}^{m-1}\frac{M_{1,k}}{1-\frac{{\lambda }_{1,k}}{\lambda }}=\frac{1}{\lambda }\sum _{k=1}^{m-1}{M}_{1,k}\sum _{\beta =0}^{\infty }{\left(\frac{{\lambda }_{1,k}}{\lambda }\right)}^{\beta },$$

(69)

$$\sum _{k=1}^{m-1}\frac{M_{2,k}}{\lambda -{\lambda }_{2,k}}=-\sum _{k=1}^{m-1}\frac{M_{2,k}}{{\lambda }_{2,k}\left(1-\frac{{\lambda }_{2,k}}{\lambda }\right)}=-\sum _{k=1}^{m-1}\frac{M_{2,k}}{{\lambda }_{2,k}}\sum _{\beta =0}^{\infty }{\left(\frac{\lambda }{{\lambda }_{2,k}}\right)}^{\beta }.$$

(70)

Substituting expressions (69) and (70) into (65), we have

$$\begin{array}{c}\frac{V^m\left(\lambda \right)}{\lambda T_{2m-2}\left(\lambda \right)}=\frac{\lambda }{d_{2m-2}}-\frac{1}{d_{2m-2}^2}\left(2mcosh\left(2\pi h\right)·d_{2m-2}+d_{2m-3}\right)+\hfill \\ +\frac{M_0}{\lambda }+\frac{1}{\lambda }\sum _{k=1}^{m-1}{M}_{1,k}\sum _{\beta =0}^{\infty }{\left(\frac{{\lambda }_{1,k}}{\lambda }\right)}^{\beta }-\sum _{k=1}^{m-1}\frac{M_{2,k}}{{\lambda }_{2,k}}\sum _{\beta =0}^{\infty }{\left(\frac{\lambda }{{\lambda }_{2,k}}\right)}^{\beta }.\hfill \end{array}$$

(71)

From equality (71) for the Fourier transform of $\widehat{D}\left(x\right)$ given by Formula (61), we have

$$\begin{array}{c}F\left[\widehat{D}\left(x\right)\right]\left(p\right)=\frac{2^{2m-2}{\left[\left(m-1\right)!\right]}^2{V}^m\left(\lambda \right)}{\pi \left(2m-2\right)!\lambda T_{2m-2}\left(\lambda \right)}=\frac{2^{2m-2}{\left[\left(m-1\right)!\right]}^2}{\pi \left(2m-2\right)!}[\frac{\lambda }{d_{2m-2}}-\hfill \\ -\frac{2mcosh\left(2\pi h\right)·d_{2m-2}+d_{2m-3}}{d_{2m-2}^2}+\frac{M_0}{\lambda }+\frac{1}{\lambda }\sum _{k=1}^{m-1}{M}_{1,k}\sum _{\beta =0}^{\infty }{\left(\frac{{\lambda }_{1,k}}{\lambda }\right)}^{\beta }-\sum _{k=1}^{m-1}\frac{M_{2,k}}{{\lambda }_{2,k}}\sum _{\beta =0}^{\infty }{\left(\frac{\lambda }{{\lambda }_{2,k}}\right)}^{\beta }].\hfill \end{array}$$

(72)

Hence, taking into account that $\lambda =e^{2\pi iph}$ and $d_0=d_{2m-2}$, then using Formulas (67), (68) after some simplifications we obtain

$$\begin{array}{c}F\left[\widehat{D}\left(x\right)\right]\left(p\right)=\frac{2^{2m-2}{\left[\left(m-1\right)!\right]}^2}{\pi \left(2m-2\right)!}[\left(\frac{1}{d_{2m-2}}+\sum _{k=1}^{m-1}{M}_{1,k}\right)e^{2\pi iph}-\hfill \\ -\frac{2mcosh\left(2\pi h\right)·d_{2m-2}+d_{2m-3}}{d_{2m-2}^2}+\sum _{k=1}^{m-1}\frac{M_{1,k}}{{\lambda }_{1,k}}+\left(\frac{1}{d_{2m-2}}+\sum _{k=1}^{m-1}{M}_{1,k}\right)e^{-2\pi iph}+\hfill \\ +\sum _{k=1}^{m-1}{M}_{1,k}\sum _{\beta =2}^{\infty }{\lambda }_{1,k}^{\beta -1}{e}^{-2\pi iph\beta }+\sum _{k=1}^{m-1}{M}_{1,k}\sum _{\beta =2}^{\infty }{\lambda }_{1,k}^{\beta -1}{e}^{2\pi iph\beta }].\hfill \end{array}$$

(73)

This is the Fourier series for function $\widehat{D}\left(x\right)$, i.e.,

$$F\left[\widehat{D}\left(x\right)\right]\left(p\right)=\sum _{\beta =-\infty }^{\infty }{D}_m\left[\beta \right]e^{2\pi iph\beta },$$

where $D_m\left[\beta \right]$ is the desired discrete operator. Then from the last two equalities, we obtain the statement of the theorem. Theorem 5 is proven. □

8. Calculation of Coefficients of Optimal Lattice Quadrature Formulas

We can obtain $C_{\beta }$ from equality (26) as follows:

$$C_{\beta }=D_1\left[\beta \right]\ast u_h^1\left[\beta \right].$$

(74)

The value of $D_1\left[\beta \right]$ is given by Formula (45) when $m=1$. The value of $u_h^1\left[\beta \right]$ is given by formula

$$u_h^1\left[\beta \right]=f_1\left[\beta \right],$$

(75)

when $\beta =0,1,...,N$. Here, $f_1\left[\beta \right]={\int }_0^{1}p\left(y\right){\nu }_1(h\beta -y)dy$. However, we have not yet determined the value of the function $u_h^1\left[\beta \right]$ for $\beta <0$ and $\beta >N$.

Using Formula (17), we can determine the specific values of $u_h^1\left[\beta \right]$ for $\beta <0$ and $\beta >N$ by using the form of the function $u_h^1\left[\beta \right]$ for $\beta <0$ and $\beta >N$. According to Formula (17),

$$u_h^1\left(h\gamma \right)=\sum _{\beta =0}^N{C}_{\beta }{\nu }_1(h\gamma -h\beta ),$$

(76)

where ${\nu }_1\left(x\right)=e^{-2\pi \left|x\right|}$. Using the form of ${\nu }_1\left(x\right)$, we can show that

$u_h^1(-h)=e^{-2\pi h}u\left(0\right)$ and $u_h^1\left((N+1)h\right)=e^{-2\pi h}{u}_h^1\left(Nh\right)$.

From the general form $u_h^1\left(h\gamma \right)$, we have

$$u_h^1\left(0\right)=C_0{\nu }_1\left(0\right)+C_1{\nu }_1(-h)+C_2{\nu }_1(-2h)+...+C_N{\nu }_1(-Nh),$$

$$u_h^1(-h)=C_0{\nu }_1(-h)+C_1{\nu }_1(-2h)+C_2{\nu }_1(-3h)+...+C_N{\nu }_1(-(N-1)h)=$$

$$=e^{-2\pi h}\left(C_0+C_1{e}^{-h}+C_2{e}^{-2h}+...+C_N{e}^{-Nh}\right)=$$

$$=e^{-2\pi h}{u}_h^1\left(0\right)$$

(77)

and

$$u_h^1\left(Nh\right)=\left(C_0{e}^{-2\pi Nh}+C_1{e}^{-2\pi (N-1)h}+...+C_N^{-2\pi 0}\right),$$

$$u_h^1\left((N+1)h\right)=e^{-2\pi h}\left(C_0{e}^{-2\pi Nh}+C_1{e}^{-2\pi (N-1)h}+...+C_N\right)=$$

$$=e^{-2\pi h}{u}_h^1\left(Nh\right).$$

(78)

These values of function $u_h^1\left[\beta \right]$ are sufficient for us to determine the optimal coefficients. From (74),

$$C\left[\beta \right]=D_h^1\left[\beta \right]\ast u_h^1\left[\beta \right]=\sum _{\alpha =-1}^1{D}_h^1\left(h\alpha \right)u_h^1(h\alpha -h\beta ).$$

Hence, by virtue of formulas (74), (75) and (76), for the optimal coefficients $\stackrel{\circ }{C}\left[\beta \right]$, we have

$$\stackrel{\circ }{C}\left[\beta \right]=\frac{1}{\pi }coth\left(2\pi h\right)·u_h^1\left[\beta \right]-\frac{1}{2\pi sinh\left(2\pi h\right)}{u}_h^1[\beta -1]-\frac{1}{2\pi sinh\left(2\pi h\right)}{u}_1[\beta +1]=$$

$$=\frac{1}{\pi }coth\left(2\pi h\right){\int }_0^{1}p\left(y\right){\nu }_1(h\beta -y)dy-\frac{1}{2\pi sinh\left(2\pi h\right)}{\int }_0^{1}p\left(y\right){\nu }_1(h(\beta -1)-y)dy-$$

$$-\frac{1}{2\pi sinh\left(2\pi h\right)}{\int }_0^{1}p\left(y\right){\nu }_1(h(\beta +1)-y)dy,$$

at $\beta =1,2,...,N-1$. From the general form of function ${\nu }_m\left(x\right)$ from Formula (12), it follows

$${\nu }_1\left(x\right)=\pi e^{-2\pi \left|x\right|}.$$

Keeping this in mind, we obtain

$$\stackrel{\circ }{C}\left[\beta \right]=coth\left(2\pi h\right)·{\int }_0^{1}p\left(y\right)e^{-2\pi \left|h\beta -y\right|}dy-\frac{1}{2sinh\left(2\pi h\right)}{\int }_0^{1}p\left(y\right)e^{-2\pi \left|h(\beta -1)-y\right|}dy-$$

$$-\frac{1}{2sinh\left(2\pi h\right)}{\int }_0^{1}p\left(y\right)e^{-2\pi \left|h(\beta +1)-y\right|}dy,\beta =1,2,...,N-1.$$

(79)

From (74), (75), (76) and (79), we obtain

$$\stackrel{\circ }{C}\left[0\right]=\pi coth\left(2\pi h\right)u_h^1\left[0\right]-\frac{1}{2\pi sinh\left(2\pi h\right)}{u}_h^1[-h]-\frac{1}{2\pi sinh\left(2\pi h\right)}{u}_h^1\left[h\right]=$$

$$=coth\left(2\pi h\right){\int }_0^{1}p\left(y\right)e^{-2\pi y}dy-\frac{e^{-2\pi h}}{2sinh\left(2\pi h\right)}{\int }_0^{1}p\left(y\right)e^{-2\pi y}dy-$$

$$-\frac{1}{2sinh\left(2\pi h\right)}{\int }_0^{1}p\left(y\right)e^{-2\pi \left|h-y\right|}dy.$$

(80)

In the same way, taking into account (77), (78), (79) and (74), we have

$$C\left[N\right]=coth\left(2\pi h\right)·{\int }_0^{1}p\left(y\right)e^{-2\pi \left|Nh-y\right|}dy-\frac{1}{2sinh\left(2\pi h\right)}{\int }_0^{1}p\left(y\right)e^{-2\pi \left|h\left(N-1\right)-y\right|}dy-$$

$$-\frac{1}{2sinh\left(2\pi h\right)}{\int }_0^{1}p\left(y\right)e^{-2\pi \left|h(N+1)-y\right|}dy.$$

(81)

So, we have proven the following theorem:

Theorem 6. 

Among all quadrature formulas of the form

$${\int }_0^{1}p\left(x\right)f\left(x\right)dx\approx \sum _{\beta =0}^N\stackrel{\circ }{C}\left[\beta \right]f\left[\beta \right],$$

there is a unique optimal quadrature formula in space $H_2^{\mu }\left(R\right)\left(0,1\right)$, the coefficients of which $H_2^{\mu }\left(R\right)\left(0,1\right)$,

$$\stackrel{\circ }{C}\left[0\right]=\frac{1}{2sinh\left(2\pi h\right)}{\int }_0^{1}p\left(x\right)\left[\left(2cosh\left(2\pi h\right)-e^{-2\pi h}\right)e^{-2\pi x}-e^{-2\pi \left|x-h\right|}\right]dx,$$

(82)

$$\stackrel{\circ }{C}\left[\beta \right]=\frac{1}{2sinh\left(2\pi h\right)}{\int }_0^{1}p\left(x\right)[2cosh\left(2\pi h\right)e^{-2\pi \left|x-h\beta \right|}-$$

$$-e^{-2\pi \left|x-h\left(\beta +1\right)\right|}-e^{-2\pi \left|x-h\left(\beta -1\right)\right|}]dx,\beta =\overline{1,N-1},$$

(83)

$$\stackrel{\circ }{C}\left[N\right]=\frac{1}{2sinh\left(2\pi h\right)}{\int }_0^{1}p\left(x\right)\left[\left(2cosh\left(2\pi h\right)-e^{-2\pi h}\right)e^{-2\pi (x+1)}-e^{-2\pi \left|x-1+h\right|}\right]dx,$$

(84)

This leads to the following corollaries.

Corollary 1. 

Among all quadrature formulas of the form (2) with optimal coefficients, which minimizes the norm of the error functional (4) with equally spaced nodes in the $H_2^{\mu }\left(R\right)$ space for $m=1$ and $p\left(x\right)=1$, the optimal one has the following form

$${\int }_0^{1}f\left(x\right)dx\cong \frac{cosh\left(2\pi h\right)-1}{\pi sinh\left(2\pi h\right)}\left[\frac{1}{2}\left(f\left(0\right)+f\left(1\right)\right)+\sum _{\beta =1}^{N-1}f\left(h\beta \right)\right],$$

here,

$$\stackrel{\circ }{C}\left[0\right]=\frac{cosh\left(2\pi h\right)-1}{2\pi sinh\left(2\pi h\right)},$$

$\stackrel{\circ }{C}\left[\beta \right]=\frac{cosh\left(2\pi h\right)-1}{\pi sinh\left(2\pi h\right)}$ for $\beta =\overline{1,N-1}$,

$$\stackrel{\circ }{C}\left[N\right]=\frac{cosh\left(2\pi h\right)-1}{2\pi sinh\left(2\pi h\right)},$$

where h is small parameter.

Corollary 2. 

Among all quadrature formulas of the form (2) with optimal coefficients, which minimizes the norm of the error functional (4) with equally spaced nodes in the $H_2^{\mu }\left(R\right)$ space for $m=1$ and $p\left(x\right)=e^{2\pi i\omega x}$, the optimal one has the following form:

$${\int }_0^1{e}^{2\pi i\omega x}f\left(x\right)dx\approx \sum _{\beta =0}^N\stackrel{\circ }{C}\left[\beta \right]f\left[\beta \right],$$

here,

$$\stackrel{\circ }{C}\left[0\right]=\frac{1}{2\pi \left(1+{\omega }^2\right)sinh\left(2\pi h\right)}\left[e^{2\pi h}-e^{2\pi hi\omega }+i\omega sinh\left(2\pi h\right)\right],$$

$$\stackrel{\circ }{C}\left[\beta \right]=\frac{e^{2\pi i\omega h\beta }}{\pi \left({\omega }^2+1\right)sinh\left(2\pi h\right)}\left[cosh\left(2\pi h\right)-cos\left(2\pi \omega h\right)\right],\beta =\overline{1,N-1},$$

$$\stackrel{\circ }{C}\left[N\right]=\frac{e^{2\pi i\omega }}{2\pi \left(1+{\omega }^2\right)sinh\left(2\pi h\right)}\left[cosh\left(2\pi h\right)-e^{-2\pi i\omega h}-i\omega sin\left(2\pi h\right)\right],$$

where h is a small parameter.

9. Conclusions

Thus, in this work, we used the Sobolev method to solve a system of algebraic equations that determines the coefficients of quadrature formulas of the form (2). To achieve this, we created a discrete analogue $D_m\left[\beta \right]$ of the differential operator ${\left[1-\frac{1}{{\left(2\pi \right)}^2}\frac{d^2}{d{x}^2}\right]}^{\left[m\right]}$ that we used to solve the system of Equations (21) and (22). With this, we obtained explicit expressions for the optimal coefficients $\stackrel{\circ }{C}\left[\beta \right]$, which we used to construct weighted optimal quadrature formulas in space $H_2^{\mu }\left(R\right)$ in the form of Equation (2). Finally, we constructed an optimal quadrature formula in Hilbert space $H_2^{\mu }\left(R\right)$ for the case $m=1$, using weight functions $p\left(x\right)=1$ and $p\left(x\right)=e^{2\pi i\omega x}$.