Multivariate Smooth Symmetrized and Perturbed Hyperbolic Tangent Neural Network Approximation over Infinite Domains

Abstract

In this article, we study the multivariate quantitative smooth approximation under differentiation of functions. The approximators here are multivariate neural network operators activated by the symmetrized and perturbed hyperbolic tangent activation function. All domains used here are infinite. The multivariate neural network operators are of quasi-interpolation type: the basic type, the Kantorovich type, and the quadrature type. We give pointwise and uniform multivariate approximations with rates. We finish with illustrations.

1. Introduction

The author of [1,2] (see Chapters 2–5) was the first to establish neural network approximation to continuous functions with rates by very specific neural network operators of Cardaliaguet–Euvrard and “squashing” types, by using the modulus of continuity of the engaged function or its high order derivative, and producing very tight Jackson type inequalities. He treats both the univariate and multivariate cases. These operators “bell-shaped” and “squashing” functions are assumed to be of compact support.

Further, the author, inspired by [3], continued his studies on neural network approximation by introducing and using the proper quasi-interpolation operators of sigmoidal and hyperbolic tangent types which resulted in the complete monographs [4,5], by treating both the univariate and multivariate cases. He included also the corresponding fractional cases.

The author here performs symmetrized and perturbed hyperbolic tangent activated multivariate neural network approximation to differentiated functions from ${\mathbb{R}}^N$ into $\mathbb{R}$, $N\in \mathbb{N}$.

We present real multivariate quasi-interpolation quantitative approximations. We derive very tight Jackson type multivariate inequalities.

Real feed-forward neural networks (FNNs) with one hidden layer, the ones we use here are mathematically expressed by

$$N_n\left(x\right)=\sum _{j=0}^n{c}_j\sigma \left(〈{\alpha }_j·x〉+b_j\right),x\in {\mathbb{R}}^s,s\in \mathbb{N},$$

where for $0\le j\le n$, $b_j\in \mathbb{R}$ are the thresholds, ${\alpha }_j\in {\mathbb{R}}^s$ are the connection weights, $c_j\in \mathbb{R}$ are the coefficients, $〈{\alpha }_j·x〉$ is the inner product of ${\alpha }_j$ and x, and $\sigma$ is the activation function of the network. About neural networks in general read [6,7,8].

Recent developments in neural network approximation are [9,10,11,12,13,14,15,16,17,18].

2. Basics

Initially we follow [5], pp. 455–460.

Our perturbed hyperbolic tangent activation function is

$$g_{q,\lambda }\left(x\right):={\displaystyle \frac{e^{\lambda x}-q{e}^{-\lambda x}}{e^{\lambda x}+q{e}^{-\lambda x}}},\lambda ,q>0,x\in \mathbb{R}.$$

(1)

Above $\lambda$ is the parameter and q is the deformation coefficient.

For more read Chapter 18 of [5]: “q-deformed and $\lambda$-Parametrized Hyperbolic Tangent based Banach space Valued Ordinary and Fractional Neural Network Approximation”.

Chapters 17 and 18 of [5] motivate our current work.

The proposed “symmetrization method” aims to use half data feed to our multivariate neural networks.

We will employ the following density function

$$M_{q.\lambda }\left(x\right):={\displaystyle \frac{1}{4}}\left(g_{q,\lambda }\left(x+1\right)-g_{q,\lambda }\left(x-1\right)\right)>0,$$

(2)

$\forall x\in \mathbb{R}$; $q,\lambda >0$.

We have that

$$M_{q,\lambda }\left(-x\right)=M_{{\displaystyle \frac{1}{q}},\lambda }\left(x\right),\forall x\in \mathbb{R};q,\lambda >0,$$

(3)

and

$$M_{{\displaystyle \frac{1}{q}},\lambda }\left(-x\right)=M_{q,\lambda }\left(x\right),\forall x\in \mathbb{R};q,\lambda >0.$$

(4)

Adding (3) and (4) we obtain

$$M_{q,\lambda }\left(-x\right)+M_{{\displaystyle \frac{1}{q}},\lambda }\left(-x\right)=M_{q,\lambda }\left(x\right)+M_{{\displaystyle \frac{1}{q}},\lambda }\left(x\right),$$

(5)

a key to this work.

So that

$$\mathsf{\Phi }\left(x\right):={\displaystyle \frac{M_{q,\lambda }\left(x\right)+M_{\frac{1}{q},\lambda }\left(x\right)}{2}}$$

(6)

is an even function, symmetric with respect to the y-axis.

By (18.18) of [5], we have

$$\begin{array}{c}{M}_{q,\lambda }\left({\displaystyle \frac{lnq}{2\lambda }}\right)={\displaystyle \frac{tanh\left(\lambda \right)}{2}},\hfill \\ \mathrm{and}\hfill \\ M_{{\displaystyle \frac{1}{q}},\lambda }\left(-{\displaystyle \frac{lnq}{2\lambda }}\right)={\displaystyle \frac{tanh\left(\lambda \right)}{2}},\lambda >0.\hfill \end{array}$$

(7)

sharing the same maximum at symmetric points.

By Theorem 18.1, p. 458 of [5], we have that

$$\begin{array}{c}{\displaystyle \sum _{i=-\infty }^{\infty }}{M}_{q,\lambda }\left(x-i\right)=1,\forall x\in \mathbb{R},\lambda ,q>0,\hfill \\ \mathrm{and}\hfill \\ {\displaystyle \sum _{i=-\infty }^{\infty }}{M}_{{\displaystyle \frac{1}{q}},\lambda }\left(x-i\right)=1,\forall x\in \mathbb{R},\lambda ,q>0.\hfill \end{array}$$

(8)

Consequently, we derive that

$${\displaystyle \sum _{i=-\infty }^{\infty }}\mathsf{\Phi }\left(x-i\right)=1,\forall x\in \mathbb{R}.$$

(9)

By Theorem 18.2, p. 459 of [5], we have that

$$\begin{array}{c}{\int }_{-\infty }^{\infty }{M}_{q,\lambda }\left(x\right)dx=1,\hfill \\ \mathrm{and}\hfill \\ {\int }_{-\infty }^{\infty }{M}_{{\displaystyle \frac{1}{q}},\lambda }\left(x\right)dx=1,\hfill \end{array}$$

(10)

so that

$${\int }_{-\infty }^{\infty }\mathsf{\Phi }\left(x\right)dx=1,$$

(11)

therefore $\mathsf{\Phi }$ is a density function.

By Theorem 18.3, p. 459 of [5], we have:

Let $0<\alpha <1$, and $n\in \mathbb{N}$ with $n^{1-\alpha }>2$; $q,\lambda >0$. Then

$${\displaystyle \sum _{\left\{\begin{array}{c}k=-\infty \\ :\left|nx-k\right|\ge n^{1-\alpha }\end{array}\right.}^{\infty }}{M}_{q,\lambda }\left(nx-k\right)<2max\left\{q,{\displaystyle \frac{1}{q}}\right\}{e}^{4\lambda }{e}^{-2\lambda n^{\left(1-\alpha \right)}}=T{e}^{-2\lambda n^{\left(1-\alpha \right)}},$$

(12)

where $T:=2max\left\{q,{\displaystyle \frac{1}{q}}\right\}{e}^{4\lambda }.$

Similarly, we get that

$$\sum _{k=-\infty }^{\infty }{M}_{{\displaystyle \frac{1}{q}},\lambda }\left(nx-k\right)<T{e}^{-2\lambda n^{\left(1-\alpha \right)}}.$$

(13)

Consequently we obtain that

$$\sum _{\left\{\begin{array}{c}k=-\infty \\ :\left|nx-k\right|\ge n^{1-\alpha }\end{array}\right.}^{\infty }\mathsf{\Phi }\left(nx-k\right)<T{e}^{-2\lambda n^{\left(1-\alpha \right)}},$$

(14)

where $T:=2max\left\{q,{\displaystyle \frac{1}{q}}\right\}{e}^{4\lambda }.$

Remark 1. 

We introduce

$$Z\left(x_1,\dots ,x_N\right):=Z\left(x\right):=\prod _{i=1}^N\mathsf{\Phi }\left(x_i\right),x=\left(x_1,\dots ,x_N\right)\in {\mathbb{R}}^N,N\in \mathbb{N}.$$

(15)

It has the properties:

(i)

$$Z\left(x\right)>0,\forall x\in {\mathbb{R}}^N,$$

(ii)

$$\sum _{k=-\infty }^{\infty }Z\left(x-k\right):=\sum _{k_1=-\infty }^{\infty }\sum _{k_2=-\infty }^{\infty }\dots \sum _{k_N=-\infty }^{\infty }Z\left(x_1-k_1,\dots ,x_N-k_N\right)=1,$$

(16)

where $k:=\left(k_1,\dots ,k_n\right)\in {\mathbb{Z}}^N$, ∀$x\in {\mathbb{R}}^N,$ hence

(iii)

$$\sum _{k=-\infty }^{\infty }Z\left(nx-k\right)=1,$$

(17)

∀$x\in {\mathbb{R}}^N;$ $n\in \mathbb{N}$, and

(iv)

$${\int }_{{\mathbb{R}}^N}Z\left(x\right)dx=1,$$

(18)

that is Z is a multivariate density function.

Here, denote ${∥x∥}_{\infty }:=max\left\{\left|x_1\right|,\dots ,\left|x_N\right|\right\}$, $x\in {\mathbb{R}}^N$, also set $\infty :=\left(\infty ,\dots ,\infty \right)$, $-\infty :=\left(-\infty ,\dots ,-\infty \right)$ upon the multivariate context,

(v)

$$\sum _{\left\{\begin{array}{c}k=-\infty \\ {∥{\displaystyle \frac{k}{n}}-x∥}_{\infty }>{\displaystyle \frac{1}{n^{\beta }}}\end{array}\right.}^{\infty }Z\left(nx-k\right)\stackrel{\left(14\right)}{<}T{e}^{-2\lambda n^{\left(1-\beta \right)}},$$

(19)

where $T:=2max\left\{q,{\displaystyle \frac{1}{q}}\right\}{e}^{4\lambda },$ $0<\beta <1,$ $n\in \mathbb{N}:n^{1-\beta }>2$, $x\in {\mathbb{R}}^N.$

Theorem 1. 

Let $0<\beta <1$, and $n\in \mathbb{N}$ with $n^{1-\beta }>2$. It holds

$$\sum _{\left\{\begin{array}{c}k=-\infty \\ \left\{k\in {\mathbb{Z}}^N:{∥nx-k∥}_{\infty }\ge n^{1-\beta }\right\}\end{array}\right.}^{\infty }{∥nx-k∥}_{\infty }^{m}Z\left(nx-k\right)<$$

(20)

$${\displaystyle \frac{2m!}{{\lambda }^m}}\left(q+{\displaystyle \frac{1}{q}}\right)e^{2\lambda }{e}^{-\lambda \left(n^{1-\beta }-1\right)}.$$

Proof. 

The condition $k\in {\mathbb{Z}}^N:{∥nx-k∥}_{\infty }\ge n^{1-\beta }$, implies that there exists at least one $k_r\in \mathbb{Z}:\left|n{x}_r-k_r\right|\ge n^{1-\beta }$, where $r\in \left\{1,\dots ,N\right\}.$

Indeed, it is

$$\left\{k\in {\mathbb{Z}}^N:{∥nx-k∥}_{\infty }\ge n^{1-\beta }\right\}\subset$$

$${\cup }_{r=1}^N\left\{{\mathbb{Z}}^{N-1}\cup \left\{k_r\in \mathbb{Z}:\left|n{x}_r-k_r\right|\ge n^{1-\beta }\right\}\right\}\subset$$

$${\mathbb{Z}}^{N-1}\cup \left\{k_{r^{*}}\in \mathbb{Z}:\left|n{x}_{r^{*}}-k_{r^{*}}\right|\ge n^{1-\beta }\right\},$$

(21)

for some $r^{*}\in \left\{1,\dots ,N\right\}.$

Let $x\ge 1$, that is $0\le x-1<x+1$. Using the mean value theorem we have that

$$M_{q.\lambda }\left(x\right)={\displaystyle \frac{1}{4}}\left(g_{q,\lambda }\left(x+1\right)-g_{q,\lambda }\left(x-1\right)\right)={\displaystyle \frac{2q\lambda e^{2\lambda \xi }}{{\left(e^{2\lambda \xi }+q\right)}^2}}=:\left(\psi \right),$$

(22)

for some $0\le x-1<\xi <x+1.$

Hence

$$\left(\psi \right)<{\displaystyle \frac{2q\lambda \left(e^{2\lambda \xi }+q\right)}{{\left(e^{2\lambda \xi }+q\right)}^2}}={\displaystyle \frac{2q\lambda }{\left(e^{2\lambda \xi }+q\right)}}<{\displaystyle \frac{2q\lambda }{\left(e^{2\lambda \left(x-1\right)}+q\right)}}<{\displaystyle \frac{2q\lambda }{e^{2\lambda \left(x-1\right)}}},x\ge 1.$$

(23)

That is

$$M_{q.\lambda }\left(x\right)<2q\lambda e^{2\lambda }{e}^{-2\lambda x},\forall x\ge 1.$$

(24)

Similarly, it holds

$$M_{{\displaystyle \frac{1}{q}}.\lambda }\left(x\right)<{\displaystyle \frac{2}{q}}\lambda e^{2\lambda }{e}^{-2\lambda x},\forall x\ge 1,$$

(25)

and hence

$$\mathsf{\Phi }\left(x\right)<\left(q+{\displaystyle \frac{1}{q}}\right)\lambda e^{2\lambda }{e}^{-2\lambda x},\forall x\ge 1.$$

(26)

We also have that ($x\ge 0$)

$$e^x=\sum _{k=0}^{\infty }{\displaystyle \frac{x^k}{k!}}>{\displaystyle \frac{x^m}{m!}},m\in \mathbb{N}.$$

(27)

Hence it is $x^m<e^{x}m!$, so that ${\displaystyle \frac{x^m}{2^m}}<e^{{\displaystyle \frac{x}{2}}}m!$ and

$$x^m<2^m{e}^{{\displaystyle \frac{x}{2}}}m!,x\ge 0,m\in \mathbb{N}.$$

(28)

Next, we observe that

$$\underset{\left\{\begin{array}{c}k=-\infty \\ :\left\{k\in {\mathbb{Z}}^N:{∥nx-k∥}_{\infty }\ge n^{1-\beta }\right\}\end{array}\right.}{\sum ^{\infty }}{∥nx-k∥}_{\infty }^{m}Z\left(nx-k\right)\le$$

$$\underset{\left\{\begin{array}{c}{k}_{r^{*}}=-\infty \\ :\left\{k_{r^{*}}\in \mathbb{Z}:\left|n{x}_{r^{*}}-k_{r^{*}}\right|\ge n^{1-\beta }\right\}\end{array}\right.}{\sum ^{\infty }}{\left|n{x}_{r^{*}}-k_{r^{*}}\right|}^m\mathsf{\Phi }\left(\left|n{x}_{r^{*}}-k_{r^{*}}\right|\right)\stackrel{\left(26\right)}{<}$$

$$(\mathrm{set}\mu :=\left(q+{\displaystyle \frac{1}{q}}\right)\lambda e^{2\lambda })$$

(29)

$${\displaystyle \frac{\mu }{{\left(2\lambda \right)}^m}}\underset{\left\{\begin{array}{c}{k}_{r^{*}}=-\infty \\ :\left\{k_{r^{*}}\in \mathbb{Z}:\left|n{x}_{r^{*}}-k_{r^{*}}\right|\ge n^{1-\beta }\right\}\end{array}\right.}{\sum ^{\infty }}{\left(2\lambda \left|n{x}_{r^{*}}-k_{r^{*}}\right|\right)}^m{e}^{-2\lambda \left|n{x}_{r^{*}}-k_{r^{*}}\right|}\stackrel{\left(28\right)}{<}$$

$${\displaystyle \frac{2^{m}m!\mu }{{\left(2\lambda \right)}^m}}\underset{\left\{\begin{array}{c}{k}_{r^{*}}=-\infty \\ :\left\{k_{r^{*}}\in \mathbb{Z}:\left|n{x}_{r^{*}}-k_{r^{*}}\right|\ge n^{1-\beta }\right\}\end{array}\right.}{\sum ^{\infty }}{e}^{\lambda \left|n{x}_{r^{*}}-k_{r^{*}}\right|}{e}^{-2\lambda \left|n{x}_{r^{*}}-k_{r^{*}}\right|}=$$

$$2^{m}m!{\displaystyle \frac{\mu }{{\left(2\lambda \right)}^m}}\underset{\left\{\begin{array}{c}{k}_{r^{*}}=-\infty \\ :\left\{k_{r^{*}}\in \mathbb{Z}:\left|n{x}_{r^{*}}-k_{r^{*}}\right|\ge n^{1-\beta }\right\}\end{array}\right.}{\sum ^{\infty }}{e}^{-\lambda \left|n{x}_{r^{*}}-k_{r^{*}}\right|}\le$$

$${\displaystyle \frac{2^{m+1}}{\lambda }}m!{\displaystyle \frac{\mu }{{\left(2\lambda \right)}^m}}{\int }_{n^{1-\beta }-1}^{\infty }{e}^{-\lambda x}d\lambda x\stackrel{\left(y:=\lambda x\right)}{=}{\displaystyle \frac{2^{m+1}}{\lambda }}m!{\displaystyle \frac{\mu }{{\left(2\lambda \right)}^m}}{\int }_{n^{1-\beta }-1}^{\infty }{e}^{-y}dy=$$

(30)

$${\displaystyle \frac{2^{m+1}}{\lambda }}m!{\displaystyle \frac{\mu }{{\left(2\lambda \right)}^m}}{e}^{-\lambda \left(n^{1-\beta }-1\right)}={\displaystyle \frac{2^{m+1}m!}{\lambda 2^m{\lambda }^m}}\left(q+{\displaystyle \frac{1}{q}}\right)\lambda e^{2\lambda }{e}^{-\lambda \left(n^{1-\beta }-1\right)}=$$

$${\displaystyle \frac{2m!}{{\lambda }^m}}\left(q+{\displaystyle \frac{1}{q}}\right)e^{2\lambda }{e}^{-\lambda \left(n^{1-\beta }-1\right)}.$$

(31)

We have proved that

$$\underset{\left\{\begin{array}{c}k=-\infty \\ :\left\{k\in {\mathbb{Z}}^N:{∥nx-k∥}_{\infty }\ge n^{1-\beta }\right\}\end{array}\right.}{\sum ^{\infty }}{∥nx-k∥}_{\infty }^{m}Z\left(nx-k\right)\le$$

(32)

$${\displaystyle \frac{2m!}{{\lambda }^m}}\left(q+{\displaystyle \frac{1}{q}}\right)e^{2\lambda }{e}^{-\lambda \left(n^{1-\beta }-1\right)}.$$

□

We need

Definition 1. 

The modulus of continuity here is defined by

$${\omega }_1\left(f,\delta \right):=\underset{{∥x-y∥}_{\infty }<\delta }{\underset{x,y\in {\mathbb{R}}^N:}{sup}}\left|f\left(x\right)-f\left(y\right)\right|,\delta >0,$$

(33)

where $f:{\mathbb{R}}^N\to \mathbb{R}$ is a bounded and continuous function, denoted by $f\in C_B\left({\mathbb{R}}^N\right)$, $N\in \mathbb{N}$. Similarly ${\omega }_1$ is defined for $f\in C_U\left({\mathbb{R}}^N\right)$ (uniformly continuous functions). We have that $f\in C_U\left({\mathbb{R}}^N\right)$, iff ${\omega }_1\left(f,\delta \right)\to 0$ as $\delta \downarrow 0.$

Notation 1. 

Let $f\in C^m\left({\mathbb{R}}^N\right)$, $m,N\in \mathbb{N}$. Here, $f_{\alpha }$ denotes a partial derivative of f, $\alpha :=\left({\alpha }_1,\dots ,{\alpha }_N\right)$, ${\alpha }_i\in {\mathbb{Z}}_{+}$, $i=1,\dots ,N,$ and $\left|\alpha \right|={\displaystyle \sum _{i=1}^N}{\alpha }_i=l$, where $l=0,1,\dots ,m$. We write also $f_{\alpha }:={\displaystyle \frac{{\partial }^{\alpha }f}{\partial x^{\alpha }}}$ and we say it is of order l.

We denote

$${\omega }_{1,m}^{max}\left(f_{\alpha },h\right):=\underset{\alpha :\left|\alpha \right|=m}{max}{\omega }_1\left(f_{\alpha },h\right),h>0.$$

(34)

Call also

$${∥f_{\alpha }∥}_{\infty ,m}^{max}:=\underset{\left|\alpha \right|=m}{max}\left\{{∥f_{\alpha }∥}_{\infty }\right\},$$

(35)

where ${∥·∥}_{\infty }$ is the supremum norm.

Under differentiation the speed of convergence of our neural network operators improves a lot, that is the aim of this work.

Next, we describe our neural network operators.

Definition 2. 

Let $f\in C_B\left({\mathbb{R}}^N\right)$, we define

$$A_n\left(f,x\right):=A_n\left(f,x_1,\dots ,x_N\right):=\sum _{k=-\infty }^{\infty }f\left({\displaystyle \frac{k}{n}}\right)Z\left(nx-k\right):=$$

(36)

$$\sum _{k_1=-\infty }^{\infty }\sum _{k_2=-\infty }^{\infty }\dots \sum _{k_N=-\infty }^{\infty }f\left({\displaystyle \frac{k_1}{n}},{\displaystyle \frac{k_2}{n}},\dots ,{\displaystyle \frac{k_N}{n}}\right)\left(\prod _{i=1}^N\mathsf{\Phi }\left(n{x}_i-k_i\right)\right),$$

$n\in \mathbb{N}$, ∀$x\in {\mathbb{R}}^N$, $N\in \mathbb{N}$, the multivariate quasi-interpolation neural network operator.

Also for $f\in C_B\left({\mathbb{R}}^N\right)$ we define the multivariate Kantorovich type neural network operator

$$K_n\left(f,x\right):=K_n\left(f,x_1,\dots ,x_N\right):=\sum _{k=-\infty }^{\infty }\left(n^N{\int }_{{\displaystyle \frac{k}{n}}}^{{\displaystyle \frac{k+1}{n}}}f\left(t\right)dt\right)Z\left(nx-k\right):=$$

(37)

$$\sum _{k_1=-\infty }^{\infty }\sum _{k_2=-\infty }^{\infty }\dots \sum _{k_N=-\infty }^{\infty }\left(n^N{\int }_{{\displaystyle \frac{k_1}{n}}}^{{\displaystyle \frac{k_1+1}{n}}}{\int }_{{\displaystyle \frac{k_2}{n}}}^{{\displaystyle \frac{k_2+1}{n}}}\dots {\int }_{{\displaystyle \frac{k_N}{n}}}^{{\displaystyle \frac{k_N+1}{n}}}f\left(t_1,\dots ,t_N\right)d{t}_1\dots d{t}_N\right)$$

$$\left(\prod _{i=1}^N\mathsf{\Phi }\left(n{x}_i-k_i\right)\right),$$

$n\in \mathbb{N}$, ∀$x\in {\mathbb{R}}^N.$

Again for $f\in C_B\left({\mathbb{R}}^N\right),$ $N\in \mathbb{N},$ we define the multivariate neural network operator of quadrature type $Q_n\left(f,x\right)$, $n\in \mathbb{N}$, as follows. Let $\theta =\left({\theta }_1,\dots ,{\theta }_N\right)\in {\mathbb{N}}^N$, $r=\left(r_1,\dots ,r_N\right)\in {\mathbb{Z}}_{+}^N$, $w_r=w_{r_1{r}_2\dots r_N}\ge 0$, such that

$$\sum _{r=0}^{\theta }{w}_r=\sum _{r_1=0}^{{\theta }_1}\sum _{r_2=0}^{{\theta }_2}\dots \sum _{r_N=0}^{{\theta }_N}{w}_{r_1{r}_2\dots r_N}=1;k\in Z^N,$$

and

$${\delta }_{nk}\left(f\right):={\delta }_{n,k_1,k_2,\dots ,k_N}\left(f\right):=\sum _{r=0}^{\theta }{w}_{r}f\left({\displaystyle \frac{k}{n}}+{\displaystyle \frac{r}{n\theta }}\right):=$$

$$\sum _{r_1=0}^{{\theta }_1}\sum _{r_2=0}^{{\theta }_2}\dots \sum _{r_N=0}^{{\theta }_N}{w}_{r_1{r}_2\dots r_N}f\left({\displaystyle \frac{k_1}{n}}+{\displaystyle \frac{r_1}{n{\theta }_1}},{\displaystyle \frac{k_2}{n}}+{\displaystyle \frac{r_2}{n{\theta }_2}},\dots ,{\displaystyle \frac{k_N}{n}}+{\displaystyle \frac{r_N}{n{\theta }_N}}\right),$$

(38)

where ${\displaystyle \frac{r}{\theta }}:=\left({\displaystyle \frac{r_1}{{\theta }_1}},{\displaystyle \frac{r_2}{{\theta }_2}},\dots ,{\displaystyle \frac{r_N}{{\theta }_N}}\right).$

We put

$$Q_n\left(f,x\right):=Q_n\left(f,x_1,\dots ,x_N\right):=\sum _{k=-\infty }^{\infty }{\delta }_{nk}\left(f\right)Z\left(nx-k\right):=$$

(39)

$$\sum _{k_1=-\infty }^{\infty }\sum _{k_2=-\infty }^{\infty }\dots \sum _{k_N=-\infty }^{\infty }{\delta }_{n,k_1,k_2,\dots ,k_N}\left(f\right)\left(\prod _{i=1}^N\mathsf{\Phi }\left(n{x}_i-k_i\right)\right),\forall x\in {\mathbb{R}}^N.$$

Remark 2. 

We notice that

$${\int }_{{\displaystyle \frac{k}{n}}}^{{\displaystyle \frac{k+1}{n}}}f\left(t\right)dt={\int }_{{\displaystyle \frac{k_1}{n}}}^{{\displaystyle \frac{k_1+1}{n}}}{\int }_{{\displaystyle \frac{k_2}{n}}}^{{\displaystyle \frac{k_2+1}{n}}}\dots {\int }_{{\displaystyle \frac{k_N}{n}}}^{{\displaystyle \frac{k_N+1}{n}}}f\left(t_1,t_2,\dots ,t_N\right)d{t}_{1}d{t}_2\dots d{t}_N=$$

$${\int }_0^{{\displaystyle \frac{1}{n}}}{\int }_0^{{\displaystyle \frac{1}{n}}}\dots {\int }_0^{{\displaystyle \frac{1}{n}}}f\left(t_1+{\displaystyle \frac{k_1}{n}},t_2+{\displaystyle \frac{k_2}{n}},\dots ,t_N+{\displaystyle \frac{k_N}{n}}\right)d{t}_1\dots d{t}_N$$

$$={\int }_0^{{\displaystyle \frac{1}{n}}}f\left(t+{\displaystyle \frac{k}{n}}\right)dt.$$

(40)

Thus, it holds

$$K_n\left(f,x\right)=\sum _{k=-\infty }^{\infty }\left(n^N{\int }_0^{{\displaystyle \frac{1}{n}}}f\left(t+{\displaystyle \frac{k}{n}}\right)dt\right)Z\left(nx-k\right).$$

(41)

Motivation here comes from the following.

Here, $\left(X,{∥·∥}_{\gamma }\right)$ is an arbitrary Banach space.

Theorem 2.

([5], p. 58). Let $f\in C_B\left({\mathbb{R}}^N,X\right)$ (continuous and bounded X-valued functions from ${\mathbb{R}}^N$), $0<\beta <1$, $\mu >0$, $x\in {\mathbb{R}}^N$, $N,n\in \mathbb{N}$ with $n^{1-\beta }>2$, ${\omega }_1$ is defined similarly as in Definition 1. Then

(i)

$${∥A_n\left(f,x\right)-f\left(x\right)∥}_{\gamma }\le {\omega }_1\left(f,{\displaystyle \frac{1}{n^{\beta }}}\right)+{\displaystyle \frac{4{∥{∥f∥}_{\gamma }∥}_{\infty }}{e^{\mu \left(n^{1-\beta }-2\right)}}}=:{\lambda }_1^{*}\left(n\right),$$

(42)

(ii)

$${∥{∥A_n\left(f\right)-f∥}_{\gamma }∥}_{\infty }\le {\lambda }_1^{*}\left(n\right).$$

(43)

Given that $f\in \left(C_U\left({\mathbb{R}}^N,X\right)\cap C_B\left({\mathbb{R}}^N,X\right)\right)$, we obtain ${\displaystyle \underset{n\to \infty }{lim}}{A}_n\left(f\right)=f$, uniformly.

The speed of convergence above is ${\displaystyle \frac{1}{n^{\beta }}}.$

The space $C_U\left({\mathbb{R}}^N,X\right)$ denotes the uniformly continuous functions. The operator $A_n\left(f,x\right)$ is defined similarly to (36).

3. Main Results

Next, we study the approximation properties of $A_n$, $K_n$, and $Q_n$ neural network operators.

Theorem 3. 

Let $0<\beta <1$, $n\in \mathbb{N}:n^{1-\beta }>2$, $x\in {\mathbb{R}}^N$, $f\in C^m\left({\mathbb{R}}^N\right)$, $m,N\in \mathbb{N},$ with $f_{\alpha }\in C_B\left({\mathbb{R}}^N\right)$, for all $\alpha :=\left({\alpha }_1,\dots ,{\alpha }_N\right)$, ${\alpha }_i\in {\mathbb{Z}}_{+},$ $i=1,\dots ,N,$ and $\left|\alpha \right|={\displaystyle \sum _{i=1}^N}{\alpha }_i=l$, where $l=0,1,\dots ,m$. Then

(1)

$$\left|A_n\left(f,x\right)-f\left(x\right)-\sum _{j=1}^m\sum _{\left\{\begin{array}{c}\alpha :\\ \left|\alpha \right|=j\end{array}\right\}}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)f_{\alpha }\left(x\right)A_n\left(\left(\prod _{i=1}^N{\left(·-x_i\right)}^{{\alpha }_i}\right)\right)\left(x\right)\right|$$

$$\le \left({\displaystyle \frac{N^m}{m!n^{m\beta }}}\right){\omega }_{1,m}^{max}\left(f_{\alpha },{\displaystyle \frac{1}{n^{\beta }}}\right)+$$

(44)

$${\displaystyle \frac{4{∥f∥}_{\infty ,m}^{max}{N}^m}{n^m{\lambda }^m}}\left(q+{\displaystyle \frac{1}{q}}\right)e^{2\lambda }{e}^{-\lambda \left(n^{1-\beta }-1\right)}=:{\mathsf{\Phi }}_1^{*},$$

(2) assume that $f_{\alpha }\left(x\right)=0$, for all $\alpha :\left|\alpha \right|=j$, $j=1,\dots ,m$, we have

$$\left|A_n\left(f,x\right)-f\left(x\right)\right|\le {\mathsf{\Phi }}_1^{*},$$

(45)

with the high speed of convergence $n^{-\beta \left(m+1\right)},$

(3)

$$\left|A_n\left(f,x\right)-f\left(x\right)\right|\le$$

$$\sum _{j=1}^N\sum _{\left\{\begin{array}{c}\alpha :\\ \left|\alpha \right|=j\end{array}\right.}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)\left|f_{\alpha }\left(x\right)\right|$$

$$\left\{{\displaystyle \frac{1}{n^{\beta j}}}+{\displaystyle \frac{1}{n^j}}{\displaystyle \frac{2j!}{{\lambda }^j}}\left(q+{\displaystyle \frac{1}{q}}\right)e^{2\lambda }{e}^{-\lambda \left(n^{1-\beta }-1\right)}\right\}+{\mathsf{\Phi }}_1^{*},$$

(46)

and

(4)

$${∥A_n\left(f\right)-f∥}_{\infty }\le$$

$$\sum _{j=1}^N\sum _{\left\{\begin{array}{c}\alpha :\\ \left|\alpha \right|=j\end{array}\right.}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right){∥f_{\alpha }∥}_{\infty }\left\{{\displaystyle \frac{1}{n^{\beta j}}}+{\displaystyle \frac{1}{n^j}}{\displaystyle \frac{2j!}{{\lambda }^j}}\left(q+{\displaystyle \frac{1}{q}}\right)e^{2\lambda }{e}^{-\lambda \left(n^{1-\beta }-1\right)}\right\}+{\mathsf{\Phi }}_1^{*}.$$

(47)

We have that $A_n\left(f\right)\to f$, as $n\to \infty$, pointwise and uniformly.

Proof. 

Consider $g_z\left(t\right):=f\left(x_0+t\left(z-x_0\right)\right)$, $t\ge 0$; $x_0,z\in {\mathbb{R}}^N$. Then

$$g_z^{\left(j\right)}\left(t\right)=\left[{\left(\sum _{i=1}^N\left(z_i-x_{0i}\right){\displaystyle \frac{\partial }{\partial x_i}}\right)}^{j}f\right]\left(x_{01}+t\left(z_1-x_{01}\right),\dots ,x_{0N}+t\left(z_N-x_{0N}\right)\right),$$

(48)

for all $j=0,1,\dots ,m.$

We have the multivariate Taylor’s formula

$$f\left(z_1,\dots ,z_N\right)=g_z\left(1\right)=\sum _{j=0}^m{\displaystyle \frac{g_z^{\left(j\right)}\left(0\right)}{j!}}+$$

(49)

$${\displaystyle \frac{1}{\left(m-1\right)!}}{\int }_0^1{\left(1-\theta \right)}^{m-1}\left(g_z^{\left(m\right)}\left(\theta \right)-g_z^{\left(m\right)}\left(0\right)\right)d\theta .$$

Notice $g_z\left(0\right)=f\left(z_0\right)$. Also for $j=0,1,\dots ,m,$ we have

$$g_z^{\left(j\right)}\left(0\right)=\sum _{\left\{\begin{array}{c}\alpha :=\left({\alpha }_1,\dots ,{\alpha }_N\right),{\alpha }_i\in {\mathbb{Z}}_{+}\\ i=1,\dots ,N,\left|\alpha \right|:={\displaystyle \sum _{i=1}^N}{\alpha }_i=j\end{array}\right\}}\left({\displaystyle \frac{j!}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)$$

$$\left({\displaystyle \prod _{i=1}^N}{\left(z_i-x_{0i}\right)}^{{\alpha }_i}\right)f_{\alpha }\left(x_0\right).$$

(50)

Furthermore

$$g_z^{\left(m\right)}\left(\theta \right)=\sum _{\left\{\begin{array}{c}\alpha :=\left({\alpha }_1,\dots ,{\alpha }_N\right),{\alpha }_i\in {\mathbb{Z}}_{+}\\ i=1,\dots ,N,\left|\alpha \right|:={\displaystyle \sum _{i=1}^N}{\alpha }_i=m\end{array}\right\}}\left({\displaystyle \frac{m!}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)$$

$$\left({\displaystyle \prod _{i=1}^N}{\left(z_i-x_{0i}\right)}^{{\alpha }_i}\right)f_{\alpha }\left(x_0+\theta \left(z-x_0\right)\right),$$

(51)

$0\le \theta \le 1.$

So, we treat $f\in C^m\left({\mathbb{R}}^N\right).$

Thus, we have for $x\in {\mathbb{R}}^N,$ $x=\left(x_1,\dots ,x_N\right)$, $k_i\in \mathbb{Z}$, $i=1,\dots ,N,$ $k:=\left(k_1,\dots ,k_N\right);$ $n\in \mathbb{N},$ that

$$f\left({\displaystyle \frac{k_1}{n}},\dots ,{\displaystyle \frac{k_N}{n}}\right)-f\left(x\right)=$$

$${\displaystyle \sum _{j=1}^m}\sum _{\left\{\begin{array}{c}\alpha :=\left({\alpha }_1,\dots ,{\alpha }_N\right),{\alpha }_i\in {\mathbb{Z}}_{+}\\ i=1,\dots ,N,\left|\alpha \right|:={\displaystyle \sum _{i=1}^N}{\alpha }_i=j\end{array}\right\}}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)$$

$$\left({\displaystyle \prod _{i=1}^N}{\left({\displaystyle \frac{k_i}{n}}-x_i\right)}^{{\alpha }_i}\right)f_{\alpha }\left(x\right)+R,$$

(52)

where

$$R=m{\int }_0^1{\left(1-\theta \right)}^{m-1}\sum _{\left\{\begin{array}{c}\alpha :=\left({\alpha }_1,\dots ,{\alpha }_N\right),{\alpha }_i\in {\mathbb{Z}}_{+}\\ i=1,\dots ,N,\left|\alpha \right|:={\displaystyle \sum _{i=1}^N}{\alpha }_i=m\end{array}\right\}}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)$$

(53)

$$\left({\displaystyle \prod _{i=1}^N}{\left({\displaystyle \frac{k_i}{n}}-x_i\right)}^{{\alpha }_i}\right)\left[f_{\alpha }\left(x+\theta \left({\displaystyle \frac{k}{n}}-x\right)\right)-f_{\alpha }\left(x\right)\right]d\theta .$$

We see that

$$\left|R\right|\le m{\int }_0^1{\left(1-\theta \right)}^{m-1}\sum _{\left|\alpha \right|=m}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)$$

$$\left({\displaystyle \prod _{i=1}^N}{\left|{\displaystyle \frac{k_i}{n}}-x_i\right|}^{{\alpha }_i}\right)\left|f_{\alpha }\left(x+\theta \left({\displaystyle \frac{k}{n}}-x\right)\right)-f_{\alpha }\left(x\right)\right|d\theta \le$$

$$m{\int }_0^1{\left(1-\theta \right)}^{m-1}\left(\sum _{\left|\alpha \right|=m}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)\left({\displaystyle \prod _{i=1}^N}{\left|{\displaystyle \frac{k_i}{n}}-x_i\right|}^{{\alpha }_i}\right)\right.$$

(54)

$$\left.{\omega }_1\left(f_{\alpha },\theta {∥{\displaystyle \frac{k}{n}}-x∥}_{\infty }\right)\right)d\theta \le \left(\ast \right).$$

Notice here that

$${∥{\displaystyle \frac{k}{n}}-x∥}_{\infty }\le {\displaystyle \frac{1}{n^{\beta }}},\mathrm{iff}\left|{\displaystyle \frac{k_i}{n}}-x_i\right|\le {\displaystyle \frac{1}{n^{\beta }}},i=1,\dots ,N.$$

(55)

We assume (55) and we further see that

$$\left(\ast \right)\le m{\omega }_{1,m}^{max}\left(f_{\alpha },{\displaystyle \frac{1}{n^{\beta }}}\right){\int }_0^1{\left(1-\theta \right)}^{m-1}\left(\sum _{\left|\alpha \right|=m}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)\left({\displaystyle \prod _{i=1}^N}{\left({\displaystyle \frac{1}{n^{\beta }}}\right)}^{{\alpha }_i}\right)\right)d\theta$$

$$=\left({\displaystyle \frac{{\omega }_{1,m}^{max}\left(f_{\alpha },\frac{1}{n^{\beta }}\right)}{\left(m!\right)n^{m\beta }}}\right)\left(\sum _{\left|\alpha \right|=m}{\displaystyle \frac{m!}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)=$$

$$\left({\displaystyle \frac{{\omega }_{1,m}^{max}\left(f_{\alpha },\frac{1}{n^{\beta }}\right)}{\left(m!\right)n^{m\beta }}}\right)N^m.$$

(56)

Conclusion: When ${∥{\displaystyle \frac{k}{n}}-x∥}_{\infty }\le {\displaystyle \frac{1}{n^{\beta }}}$, we proved that

$$\left|R\right|\le \left({\displaystyle \frac{N^m}{m!n^{m\beta }}}\right){\omega }_{1,m}^{max}\left(f_{\alpha },{\displaystyle \frac{1}{n^{\beta }}}\right).$$

(57)

By (53) we have that

$$\left|R\right|\le m\left({\int }_0^1{\left(1-\theta \right)}^{m-1}\sum _{\left\{\begin{array}{c}\alpha :=\left({\alpha }_1,\dots ,{\alpha }_N\right),{\alpha }_i\in {\mathbb{Z}}_{+}\\ i=1,\dots ,N,\left|\alpha \right|=m\end{array}\right\}}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)\right.$$

$$\left.\left({\displaystyle \prod _{i=1}^N}\left({\left|{\displaystyle \frac{k_i}{n}}-x_i\right|}^{{\alpha }_i}\right)2{∥f_{\alpha }∥}_{\infty }\right)d\theta \right)=$$

$$2\sum _{\left|\alpha \right|=m}{\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\left({\displaystyle \prod _{i=1}^N}{\left|{\displaystyle \frac{k_i}{n}}-x_i\right|}^{{\alpha }_i}\right){∥f_{\alpha }∥}_{\infty }\le$$

$$\left({\displaystyle \frac{2{∥\frac{k}{n}-x∥}_{\infty }^m{∥f_{\alpha }∥}_{\infty ,m}^{max}}{m!}}\right)\left(\sum _{\left|\alpha \right|=m}{\displaystyle \frac{m!}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)=$$

$${\displaystyle \frac{2{∥\frac{k}{n}-x∥}_{\infty }^m{∥f_{\alpha }∥}_{\infty ,m}^{max}{N}^m}{m!}}.$$

(58)

We proved in general that

$$\left|R\right|\le {\displaystyle \frac{2{∥\frac{k}{n}-x∥}_{\infty }^m{∥f_{\alpha }∥}_{\infty ,m}^{max}{N}^m}{m!}}.$$

(59)

Next we see that, let

$${\theta }_n:={\displaystyle \sum _{k=-\infty }^{\infty }}RZ\left(nx-k\right),$$

(60)

then

$$\left|{\theta }_n\right|\le {\displaystyle \sum _{k=-\infty }^{\infty }}\left|R\right|Z\left(nx-k\right)=$$

$${\displaystyle \sum _{\begin{array}{c}k=-\infty \\ :\left\{k\in {\mathbb{Z}}^N:{∥nx-k∥}_{\infty }<n^{1-\beta }\right\}\end{array}}^{\infty }}\left|R\right|Z\left(nx-k\right)+{\displaystyle \sum _{\begin{array}{c}k=-\infty \\ :\left\{k\in {\mathbb{Z}}^N:{∥nx-k∥}_{\infty }\ge n^{1-\beta }\right\}\end{array}}^{\infty }}\left|R\right|Z\left(nx-k\right)$$

$$\stackrel{\mathrm{by}\left\{\right(57), (\mathrm{59}\left)\right\}}{\le }\left({\displaystyle \frac{N^m}{m!n^{m\beta }}}\right){\omega }_{1,m}^{max}\left(f_{\alpha },{\displaystyle \frac{1}{n^{\beta }}}\right)+$$

(61)

$$\left({\displaystyle \frac{2{∥f_{\alpha }∥}_{\infty ,m}^{max}{N}^m}{m!}}\right)\left(\sum _{\begin{array}{c}k=-\infty \\ :\left\{k\in {\mathbb{Z}}^N:{∥nx-k∥}_{\infty }\ge n^{1-\beta }\right\}\end{array}}^{\infty }{∥{\displaystyle \frac{k}{n}}-x∥}_{\infty }^{m}Z\left(nx-k\right)\right)\le$$

$$\left({\displaystyle \frac{N^m}{m!n^{m\beta }}}\right){\omega }_{1,m}^{max}\left(f_{\alpha },{\displaystyle \frac{1}{n^{\beta }}}\right)+$$

$$\left({\displaystyle \frac{2{∥f_{\alpha }∥}_{\infty ,m}^{max}{N}^m}{n^{m}m!}}\right)\sum _{\begin{array}{c}k=-\infty \\ :\left\{k\in {\mathbb{Z}}^N:{∥nx-k∥}_{\infty }\ge n^{1-\beta }\right\}\end{array}}^{\infty }{∥nx-k∥}_{\infty }^{m}Z\left(nx-k\right)\stackrel{\left(20\right)}{<}$$

$$\left({\displaystyle \frac{N^m}{m!n^{m\beta }}}\right){\omega }_{1,m}^{max}\left(f_{\alpha },{\displaystyle \frac{1}{n^{\beta }}}\right)+$$

$$\left({\displaystyle \frac{2{∥f_{\alpha }∥}_{\infty ,m}^{max}{N}^m}{n^{m}m!}}\right){\displaystyle \frac{2m!}{{\lambda }^m}}\left(q+{\displaystyle \frac{1}{q}}\right)e^{2\lambda }{e}^{-\lambda \left(n^{1-\beta }-1\right)}=$$

$$\left({\displaystyle \frac{N^m}{m!n^{m\beta }}}\right){\omega }_{1,m}^{max}\left(f_{\alpha },{\displaystyle \frac{1}{n^{\beta }}}\right)+$$

$${\displaystyle \frac{4{∥f_{\alpha }∥}_{\infty ,m}^{max}{N}^m}{n^m{\lambda }^m}}\left(q+{\displaystyle \frac{1}{q}}\right)e^{2\lambda }{e}^{-\lambda \left(n^{1-\beta }-1\right)}.$$

(62)

We have proved that

$$\left|{\theta }_n\right|\le \left({\displaystyle \frac{N^m}{m!n^{m\beta }}}\right){\omega }_{1,m}^{max}\left(f_{\alpha },{\displaystyle \frac{1}{n^{\beta }}}\right)+$$

$${\displaystyle \frac{4{∥f_{\alpha }∥}_{\infty ,m}^{max}{N}^m}{n^m{\lambda }^m}}\left(q+{\displaystyle \frac{1}{q}}\right)e^{2\lambda }{e}^{-\lambda \left(n^{1-\beta }-1\right)}.$$

(63)

By (17) and (52), we have

$$A_n\left(f,x\right)-f\left(x\right)=\sum _{k=-\infty }^{\infty }f\left({\displaystyle \frac{k}{n}}\right)Z\left(nx-k\right)-f\left(x\right)=$$

$$\sum _{j=1}^m\sum _{\left\{\begin{array}{c}\alpha :\\ \left|\alpha \right|=j\end{array}\right\}}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)f_{\alpha }\left(x\right)A_n\left(\left({\displaystyle \prod _{i=1}^N}{\left(·-x_i\right)}^{{\alpha }_i}\right)\right)\left(x\right)+{\theta }_n.$$

(64)

Consequently, it holds

$$\left|A_n\left(f,x\right)-f\left(x\right)-{\displaystyle \sum _{j=1}^m}\sum _{\left\{\begin{array}{c}\alpha :\\ \left|\alpha \right|=j\end{array}\right\}}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)f_{\alpha }\left(x\right)A_n\left(\left({\displaystyle \prod _{i=1}^N}{\left(·-x_i\right)}^{{\alpha }_i}\right)\right)\left(x\right)\right|$$

$$\le \left({\displaystyle \frac{N^m}{m!n^{m\beta }}}\right){\omega }_{1,m}^{max}\left(f_{\alpha },{\displaystyle \frac{1}{n^{\beta }}}\right)+$$

(65)

$${\displaystyle \frac{4{∥f_{\alpha }∥}_{\infty ,m}^{max}{N}^m}{n^m{\lambda }^m}}\left(q+{\displaystyle \frac{1}{q}}\right)e^{2\lambda }{e}^{-\lambda \left(n^{1-\beta }-1\right)}.$$

Next, we estimate

$$\left|A_n\left(\left(\prod _{i=1}^N{\left(·-x_i\right)}^{{\alpha }_i}\right)\right)\left(x\right)\right|=$$

$$\left|\sum _{k=-\infty }^{\infty }Z\left(nx-k\right)\left(\prod _{i=1}^N{\left({\displaystyle \frac{k_i}{n}}-x_i\right)}^{{\alpha }_i}\right)\right|\le$$

$$\sum _{k=-\infty }^{\infty }Z\left(nx-k\right)\prod _{i=1}^N{\left|{\displaystyle \frac{k_i}{n}}-x_i\right|}^{{\alpha }_i}=$$

$$\sum _{\left\{\begin{array}{c}k=-\infty \\ :{∥{\displaystyle \frac{k}{n}}-x∥}_{\infty }<{\displaystyle \frac{1}{n^{\beta }}}\end{array}\right.}^{\infty }Z\left(nx-k\right)\prod _{i=1}^N{\left|{\displaystyle \frac{k_i}{n}}-x_i\right|}^{{\alpha }_i}+$$

$$\sum _{\left\{\begin{array}{c}k=-\infty \\ :{∥{\displaystyle \frac{k}{n}}-x∥}_{\infty }\ge {\displaystyle \frac{1}{n^{\beta }}}\end{array}\right.}^{\infty }Z\left(nx-k\right)\prod _{i=1}^N{\left|{\displaystyle \frac{k_i}{n}}-x_i\right|}^{{\alpha }_i}\le$$

$${\displaystyle \frac{1}{n^{\beta j}}}+{\displaystyle \frac{1}{n^j}}\left(\sum _{\left\{\begin{array}{c}k=-\infty \\ :{∥{\displaystyle \frac{k}{n}}-x∥}_{\infty }\ge {\displaystyle \frac{1}{n^{\beta }}}\end{array}\right.}^{\infty }Z\left(\left|nx-k\right|\right)\prod _{i=1}^N{\left|n{x}_i-k_i\right|}^{{\alpha }_i}\right)\le$$

(66)

$${\displaystyle \frac{1}{n^{\beta j}}}+{\displaystyle \frac{1}{n^j}}\left(\sum _{\left\{\begin{array}{c}k=-\infty \\ :{∥nx-k∥}_{\infty }\ge n^{1-\beta }\end{array}\right.}^{\infty }{∥nx-k∥}_{\infty }^{j}Z\left(\left|nx-k\right|\right)\right)\stackrel{\left(20\right)}{\le }$$

$${\displaystyle \frac{1}{n^{\beta j}}}+{\displaystyle \frac{1}{n^j}}{\displaystyle \frac{2j!}{{\lambda }^j}}\left(q+{\displaystyle \frac{1}{q}}\right)e^{2\lambda }{e}^{-\lambda \left(n^{1-\beta }-1\right)}.$$

We have proved that

$$\left|A_n\left(\left(\prod _{i=1}^N{\left(·-x_i\right)}^{{\alpha }_i}\right)\right)\left(x\right)\right|\le$$

$${\displaystyle \frac{1}{n^{\beta j}}}+{\displaystyle \frac{1}{n^j}}{\displaystyle \frac{2j!}{{\lambda }^j}}\left(q+{\displaystyle \frac{1}{q}}\right)e^{2\lambda }{e}^{-\lambda \left(n^{1-\beta }-1\right)},$$

(67)

for $j=1,\dots ,m.$

The proof of the theorem is now completed. □

We continue with the following result.

Theorem 4. 

Assume all the conditions are the same as in Theorem 3. Then

(1)

$$\left|K_n\left(f,x\right)-f\left(x\right)-{\displaystyle \sum _{j=1}^m}\sum _{\left\{\begin{array}{c}\alpha :\\ \left|\alpha \right|=j\end{array}\right\}}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)f_{\alpha }\left(x\right)K_n\left(\left({\displaystyle \prod _{i=1}^N}{\left(·-x_i\right)}^{{\alpha }_i}\right)\right)\left(x\right)\right|$$

$$\le {\displaystyle \frac{N^m}{m!}}{\left({\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)}^m{\omega }_{1,m}^{max}\left(f_{\alpha },{\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)+$$

(68)

$$\left({\displaystyle \frac{{\left(2N\right)}^m{∥f_{\alpha }∥}_{\infty ,m}^{max}}{n^{m}m!}}\right)\left[T{e}^{-2\lambda n^{\left(1-\beta \right)}}+{\displaystyle \frac{2m!}{{\lambda }^m}}\left(q+{\displaystyle \frac{1}{q}}\right)e^{2\lambda }{e}^{-\lambda \left(n^{1-\beta }-1\right)}\right]=:{\mathsf{\Phi }}_2^{*},$$

(2) assume that $f_{\alpha }\left(x\right)=0$, for all $\alpha :\left|\alpha \right|=j$, $j=1,\dots ,m$, we have

$$\left|K_n\left(f,x\right)-f\left(x\right)\right|\le {\mathsf{\Phi }}_2^{*},$$

(69)

with the high speed of convergence ${\left({\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)}^{m+1},$

(3)

$$\left|K_n\left(f,x\right)-f\left(x\right)\right|\le \sum _{j=1}^N\sum _{\left\{\begin{array}{c}\alpha :\\ \left|\alpha \right|=j\end{array}\right.}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)\left|f_{\alpha }\left(x\right)\right|$$

(70)

$$\left\{{\left({\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)}^j+{\displaystyle \frac{2^{j-1}}{n^j}}\left[T{e}^{-2\lambda n^{\left(1-\beta \right)}}+{\displaystyle \frac{2j!}{{\lambda }^j}}\left(q+{\displaystyle \frac{1}{q}}\right)e^{2\lambda }{e}^{-\lambda \left(n^{1-\beta }-1\right)}\right]\right\}+{\mathsf{\Phi }}_2^{*},$$

and

(4)

$${∥K_n\left(f\right)-f∥}_{\infty }\le \sum _{j=1}^N\sum _{\left\{\begin{array}{c}\alpha :\\ \left|\alpha \right|=j\end{array}\right.}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right){∥f_{\alpha }∥}_{\infty }$$

$$\left\{{\left({\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)}^j+{\displaystyle \frac{2^{j-1}}{n^j}}\left[T{e}^{-2\lambda n^{\left(1-\beta \right)}}+{\displaystyle \frac{2j!}{{\lambda }^j}}\left(q+{\displaystyle \frac{1}{q}}\right)e^{2\lambda }{e}^{-\lambda \left(n^{1-\beta }-1\right)}\right]\right\}+{\mathsf{\Phi }}_2^{*}.$$

(71)

We have that $K_n\left(f\right)\to f$, as $n\to \infty$, pointwise and uniformly.

Proof. 

It holds that

$$f\left(t+{\displaystyle \frac{k}{n}}\right)-f\left(x\right)-\sum _{j=1}^m\sum _{\left\{\begin{array}{c}\alpha :\\ \left|\alpha \right|=j\end{array}\right.}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)\left({\displaystyle \prod _{i=1}^N}{\left(t_i+{\displaystyle \frac{k_i}{n}}-x_i\right)}^{{\alpha }_i}\right)f_{\alpha }\left(x\right)=R,$$

(72)

where

$$R:=m{\int }_0^1{\left(1-\theta \right)}^{m-1}\sum _{\left\{\begin{array}{c}\alpha :\\ \left|\alpha \right|=m\end{array}\right.}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)$$

(73)

$$\left({\displaystyle \prod _{i=1}^N}{\left(t_i+{\displaystyle \frac{k_i}{n}}-x_i\right)}^{{\alpha }_i}\right)\left[f_{\alpha }\left(x+\theta \left(t+{\displaystyle \frac{k}{n}}-x\right)\right)-f_{\alpha }\left(x\right)\right]d\theta .$$

We see that

$$\left|R\right|\le m{\int }_0^1{\left(1-\theta \right)}^{m-1}\sum _{\left|\alpha \right|=m}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)$$

$$\left({\displaystyle \prod _{i=1}^N}{\left|t_i+{\displaystyle \frac{k_i}{n}}-x_i\right|}^{{\alpha }_i}\right)\left|f_{\alpha }\left(x+\theta \left(t+{\displaystyle \frac{k}{n}}-x\right)\right)-f_{\alpha }\left(x\right)\right|d\theta \le$$

$$m{\int }_0^1{\left(1-\theta \right)}^{m-1}\left(\sum _{\left|\alpha \right|=m}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)\left({\displaystyle \prod _{i=1}^N}{\left|t_i+{\displaystyle \frac{k_i}{n}}-x_i\right|}^{{\alpha }_i}\right)\right.$$

$$\left.{\omega }_1\left(f_{\alpha },\theta {∥t+{\displaystyle \frac{k}{n}}-x∥}_{\infty }\right)\right)d\theta \le \left(\ast \right).$$

(74)

Notice that

$${∥{\displaystyle \frac{k}{n}}-x∥}_{\infty }\le {\displaystyle \frac{1}{n^{\beta }}},\mathrm{iff}\left|{\displaystyle \frac{k_i}{n}}-x_i\right|\le {\displaystyle \frac{1}{n^{\beta }}},i=1,\dots ,N.$$

(75)

Let ${∥{\displaystyle \frac{k}{n}}-x∥}_{\infty }\le {\displaystyle \frac{1}{n^{\beta }}}.$

Here, we consider $0\le t_i\le {\displaystyle \frac{1}{n}}$, $i=1,\dots ,N.$

We further see that

$$\left(\ast \right)\le m{\omega }_{1,m}^{max}\left(f_{\alpha },{\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right){\int }_0^1{\left(1-\theta \right)}^{m-1}$$

$$\left(\sum _{\left|\alpha \right|=m}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)\left({\displaystyle \prod _{i=1}^N}{\left({\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)}^{{\alpha }_i}\right)\right)d\theta =$$

$${\displaystyle \frac{{\omega }_{1,m}^{max}\left(f_{\alpha },\frac{1}{n}+\frac{1}{n^{\beta }}\right)}{m!}}{\left({\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)}^m\left(\sum _{\left|\alpha \right|=m}{\displaystyle \frac{m!}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)=$$

$${\displaystyle \frac{{\omega }_{1,m}^{max}\left(f_{\alpha },\frac{1}{n}+\frac{1}{n^{\beta }}\right)}{m!}}{\left({\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)}^m{N}^m.$$

(76)

Conclusion: When ${∥{\displaystyle \frac{k}{n}}-x∥}_{\infty }\le {\displaystyle \frac{1}{n^{\beta }}}$, we proved that

$$\left|R\right|\le {\displaystyle \frac{N^m}{m!}}{\left({\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)}^m{\omega }_{1,m}^{max}\left(f_{\alpha },{\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right).$$

(77)

By (73) we have that

$$\left|R\right|\le m{\int }_0^1{\left(1-\theta \right)}^{m-1}\sum _{\begin{array}{c}\alpha :\\ \left|\alpha \right|=m\end{array}}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)$$

$$\left({\displaystyle \prod _{i=1}^N}{\left(\left|t_i\right|+\left|{\displaystyle \frac{k_i}{n}}-x_i\right|\right)}^{{\alpha }_i}2{∥f_{\alpha }∥}_{\infty }\right)d\theta \le$$

$$m{\int }_0^1{\left(1-\theta \right)}^{m-1}\sum _{\begin{array}{c}\alpha :\\ \left|\alpha \right|=m\end{array}}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)$$

$$\left({\displaystyle \prod _{i=1}^N}{\left({\displaystyle \frac{1}{n}}+\left|{\displaystyle \frac{k_i}{n}}-x_i\right|\right)}^{{\alpha }_i}2{∥f_{\alpha }∥}_{\infty }\right)d\theta =$$

(78)

$$\sum _{\begin{array}{c}\alpha :\\ \left|\alpha \right|=m\end{array}}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)\left({\displaystyle \prod _{i=1}^N}{\left({\displaystyle \frac{1}{n}}+\left|{\displaystyle \frac{k_i}{n}}-x_i\right|\right)}^{{\alpha }_i}\right)2{∥f_{\alpha }∥}_{\infty }\le$$

$$\left({\displaystyle \frac{2{\left({∥\frac{k}{n}-x∥}_{\infty }+\frac{1}{n}\right)}^m{∥f_{\alpha }∥}_{\infty ,m}^{max}}{m!}}\right)N^m.$$

We proved in general that

$$\left|R\right|\le {\displaystyle \frac{2{\left({∥\frac{k}{n}-x∥}_{\infty }+\frac{1}{n}\right)}^m{∥f_{\alpha }∥}_{\infty ,m}^{max}{N}^m}{m!}}.$$

(79)

Next, we see that

$$n^N{\int }_{{\left[0,{\displaystyle \frac{1}{n}}\right]}^N}f\left(t+{\displaystyle \frac{k}{n}}\right)dt-f\left(x\right)-\sum _{j=1}^m\sum _{\begin{array}{c}\alpha :\\ \left|\alpha \right|=j\end{array}}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)f_{\alpha }\left(x\right)$$

(80)

$$n^N{\int }_{{\left[0,{\displaystyle \frac{1}{n}}\right]}^N}\left({\displaystyle \prod _{i=1}^N}{\left(t_i+{\displaystyle \frac{k_i}{n}}-x_i\right)}^{{\alpha }_i}\right)dt=n^N{\int }_{{\left[0,{\displaystyle \frac{1}{n}}\right]}^N}Rdt.$$

So, when ${∥{\displaystyle \frac{k}{n}}-x∥}_{\infty }\le {\displaystyle \frac{1}{n^{\beta }}}$, we get

$$n^N{\int }_{{\left[0,{\displaystyle \frac{1}{n}}\right]}^N}\left|R\right|dt\le {\displaystyle \frac{N^m}{m!}}{\left({\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)}^m{\omega }_{1,m}^{max}\left(f_{\alpha },{\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right).$$

(81)

And, in general it holds

$$n^N{\int }_{{\left[0,{\displaystyle \frac{1}{n}}\right]}^N}\left|R\right|dt\le$$

(82)

$${\displaystyle \frac{2{\left({∥\frac{k}{n}-x∥}_{\infty }+\frac{1}{n}\right)}^m{∥f_{\alpha }∥}_{\infty ,m}^{max}{N}^m}{m!}}.$$

Furthermore, it holds

$$K_n\left(f,x\right)-f\left(x\right)-\sum _{j=1}^N\sum _{\begin{array}{c}\alpha :\\ \left|\alpha \right|=j\end{array}}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)f_{\alpha }\left(x\right)K_n\left(\left({\displaystyle \prod _{i=1}^N}{\left(·-x_i\right)}^{{\alpha }_i}\right)\right)\left(x\right)=$$

(83)

$$\sum _{k=-\infty }^{\infty }\left(n^N{\int }_{{\left[0,{\displaystyle \frac{1}{n}}\right]}^N}Rdt\right)Z\left(nx-k\right).$$

Call

$$U_n^{*}:=\sum _{k=-\infty }^{\infty }\left(n^N{\int }_{{\left[0,{\displaystyle \frac{1}{n}}\right]}^N}Rdt\right)Z\left(nx-k\right).$$

(84)

When ${∥{\displaystyle \frac{k}{n}}-x∥}_{\infty }<{\displaystyle \frac{1}{n^{\beta }}}$, we clearly obtain that

$$\left|U_n^{*}\right|{|}_{{∥{\displaystyle \frac{k}{n}}-x∥}_{\infty }<{\displaystyle \frac{1}{n^{\beta }}}}\le {\displaystyle \frac{N^m}{m!}}{\left({\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)}^m{\omega }_{1,m}^{max}\left(f_{\alpha },{\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right).$$

(85)

And, it is

$$\left|U_n^{*}\right|{|}_{{∥{\displaystyle \frac{k}{n}}-x∥}_{\infty }\ge {\displaystyle \frac{1}{n^{\beta }}}}\le$$

$$\left({\displaystyle \frac{2{∥f_{\alpha }∥}_{\infty ,m}^{max}{N}^m}{m!}}\right)\sum _{\begin{array}{c}k=-\infty \\ :{∥{\displaystyle \frac{k}{n}}-x∥}_{\infty }\ge {\displaystyle \frac{1}{n^{\beta }}}\end{array}}^{\infty }{\left({∥{\displaystyle \frac{k}{n}}-x∥}_{\infty }+{\displaystyle \frac{1}{n}}\right)}^{m}Z\left(nx-k\right)=$$

$${\displaystyle \frac{2{∥f_{\alpha }∥}_{\infty ,m}^{max}{N}^m}{n^{m}m!}}\sum _{\begin{array}{c}k=-\infty \\ :{∥nx-k∥}_{\infty }\ge n^{1-\beta }\end{array}}^{\infty }{\left(1+{∥nx-k∥}_{\infty }\right)}^{m}Z\left(nx-k\right)\le$$

(86)

$${\displaystyle \frac{2^m{∥f_{\alpha }∥}_{\infty ,m}^{max}{N}^m}{n^{m}m!}}\sum _{\begin{array}{c}k=-\infty \\ :{∥nx-k∥}_{\infty }\ge n^{1-\beta }\end{array}}^{\infty }\left(1+{∥nx-k∥}_{\infty }^m\right)Z\left(nx-k\right)=$$

$$\left({\displaystyle \frac{2^m{∥f_{\alpha }∥}_{\infty ,m}^{max}{N}^m}{n^{m}m!}}\right)$$

$$\left[\left(\sum _{\begin{array}{c}k=-\infty \\ :{∥nx-k∥}_{\infty }\ge n^{1-\beta }\end{array}}^{\infty }Z\left(nx-k\right)\right)+\left(\sum _{\begin{array}{c}k=-\infty \\ :{∥nx-k∥}_{\infty }\ge n^{1-\beta }\end{array}}^{\infty }{∥nx-k∥}_{\infty }^{m}Z\left(nx-k\right)\right)\right]\le$$

(87)

(by (19) and (20))

$$\left({\displaystyle \frac{2^m{∥f_{\alpha }∥}_{\infty ,m}^{max}{N}^m}{n^{m}m!}}\right)\left[T{e}^{-2\lambda n^{\left(1-\beta \right)}}+{\displaystyle \frac{2m!}{{\lambda }^m}}\left(q+{\displaystyle \frac{1}{q}}\right)e^{2\lambda }{e}^{-\lambda \left(n^{1-\beta }-1\right)}\right].$$

(88)

That is

$$\left|U_n^{*}\right|{|}_{{∥{\displaystyle \frac{k}{n}}-x∥}_{\infty }\ge {\displaystyle \frac{1}{n^{\beta }}}}\le \left({\displaystyle \frac{2^m{∥f_{\alpha }∥}_{\infty ,m}^{max}{N}^m}{n^{m}m!}}\right)$$

$$\left[T{e}^{-2\lambda n^{\left(1-\beta \right)}}+{\displaystyle \frac{2m!}{{\lambda }^m}}\left(q+{\displaystyle \frac{1}{q}}\right)e^{2\lambda }{e}^{-\lambda \left(n^{1-\beta }-1\right)}\right].$$

(89)

Consequently it holds

$$\left|U_n^{*}\right|\le \left|U_n^{*}\right|{{|}_{{∥{\displaystyle \frac{k}{n}}-x∥}_{\infty }<{\displaystyle \frac{1}{n^{\beta }}}}+\left|U_n^{*}\right||}_{{∥{\displaystyle \frac{k}{n}}-x∥}_{\infty }\ge {\displaystyle \frac{1}{n^{\beta }}}}\le$$

$${\displaystyle \frac{N^m}{m!}}{\left({\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)}^m{\omega }_{1,m}^{max}\left(f_{\alpha },{\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)+$$

$$\left({\displaystyle \frac{2^m{∥f_{\alpha }∥}_{\infty ,m}^{max}{N}^m}{n^{m}m!}}\right)\left[T{e}^{-2\lambda n^{\left(1-\beta \right)}}+{\displaystyle \frac{2m!}{{\lambda }^m}}\left(q+{\displaystyle \frac{1}{q}}\right)e^{2\lambda }{e}^{-\lambda \left(n^{1-\beta }-1\right)}\right].$$

(90)

Finally, we estimate

$$\left|K_n\left(\left(\prod _{i=1}^N{\left(·-x_i\right)}^{{\alpha }_i}\right)\right)\left(x\right)\right|=$$

$$\left|\sum _{k=-\infty }^{\infty }\left(n^N{\int }_{{\left[0,{\displaystyle \frac{1}{n}}\right]}^N}\left(\prod _{i=1}^N{\left(t_i+{\displaystyle \frac{k_i}{n}}-x_i\right)}^{{\alpha }_i}\right)dt\right)Z\left(nx-k\right)\right|\le$$

$$\sum _{k=-\infty }^{\infty }\left(n^N{\int }_{{\left[0,{\displaystyle \frac{1}{n}}\right]}^N}\left(\prod _{i=1}^N{\left(\left|t_i\right|+\left|{\displaystyle \frac{k_i}{n}}-x_i\right|\right)}^{{\alpha }_i}\right)dt\right)Z\left(nx-k\right)=$$

(91)

$$\sum _{\left\{\begin{array}{c}k=-\infty \\ :{∥x-{\displaystyle \frac{k}{n}}∥}_{\infty }<{\displaystyle \frac{1}{n^{\beta }}}\end{array}\right.}^{\infty }\left(n^N{\int }_{{\left[0,{\displaystyle \frac{1}{n}}\right]}^N}\left(\prod _{i=1}^N{\left(\left|t_i\right|+\left|{\displaystyle \frac{k_i}{n}}-x_i\right|\right)}^{{\alpha }_i}\right)dt\right)Z\left(nx-k\right)+$$

$$\sum _{\left\{\begin{array}{c}k=-\infty \\ :{∥x-{\displaystyle \frac{k}{n}}∥}_{\infty }\ge {\displaystyle \frac{1}{n^{\beta }}}\end{array}\right.}^{\infty }\left(n^N{\int }_{{\left[0,{\displaystyle \frac{1}{n}}\right]}^N}\left(\prod _{i=1}^N{\left(\left|t_i\right|+\left|{\displaystyle \frac{k_i}{n}}-x_i\right|\right)}^{{\alpha }_i}\right)dt\right)Z\left(nx-k\right)\le$$

$$\sum _{\left\{\begin{array}{c}k=-\infty \\ :{∥x-{\displaystyle \frac{k}{n}}∥}_{\infty }<{\displaystyle \frac{1}{n^{\beta }}}\end{array}\right.}^{\infty }\left(\prod _{i=1}^N{\left({\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)}^{{\alpha }_i}\right)Z\left(nx-k\right)+$$

$${\displaystyle \frac{1}{n^j}}\sum _{\left\{\begin{array}{c}k=-\infty \\ :{∥nx-k∥}_{\infty }\ge n^{1-\beta }\end{array}\right.}^{\infty }\left(\prod _{i=1}^N{\left(1+\left|n{x}_i-k_i\right|\right)}^{{\alpha }_i}\right)Z\left(nx-k\right)\le$$

$${\left({\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)}^j+{\displaystyle \frac{1}{n^j}}\sum _{\left\{\begin{array}{c}k=-\infty \\ :{∥nx-k∥}_{\infty }\ge n^{1-\beta }\end{array}\right.}^{\infty }{\left(1+{∥nx-k∥}_{\infty }\right)}^{j}Z\left(nx-k\right)\le$$

$${\left({\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)}^j+{\displaystyle \frac{2^{j-1}}{n^j}}\sum _{\left\{\begin{array}{c}k=-\infty \\ :{∥nx-k∥}_{\infty }\ge n^{1-\beta }\end{array}\right.}^{\infty }\left(1+{∥nx-k∥}_{\infty }^j\right)Z\left(nx-k\right)=$$

$${\left({\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)}^j+{\displaystyle \frac{2^{j-1}}{n^j}}\left[\left(\sum _{\left\{\begin{array}{c}k=-\infty \\ :{∥nx-k∥}_{\infty }\ge n^{1-\beta }\end{array}\right.}^{\infty }Z\left(nx-k\right)\right)+\right.$$

$$\left.\sum _{\left\{\begin{array}{c}k=-\infty \\ :{∥nx-k∥}_{\infty }\ge n^{1-\beta }\end{array}\right.}^{\infty }{∥nx-k∥}_{\infty }^{j}Z\left(nx-k\right)\right]\stackrel{(\mathrm{by} \left(19\right), \left(20\right))}{\le }$$

$${\left({\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)}^j+{\displaystyle \frac{2^{j-1}}{n^j}}\left[T{e}^{-2\lambda n^{\left(1-\beta \right)}}+{\displaystyle \frac{2j!}{{\lambda }^j}}\left(q+{\displaystyle \frac{1}{q}}\right)e^{2\lambda }{e}^{-\lambda \left(n^{1-\beta }-1\right)}\right].$$

(92)

We have found that

$$\left|K_n\left(\left(\prod _{i=1}^N{\left(·-x_i\right)}^{{\alpha }_i}\right)\right)\left(x\right)\right|\le$$

$${\left({\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)}^j+{\displaystyle \frac{2^{j-1}}{n^j}}\left[T{e}^{-2\lambda n^{\left(1-\beta \right)}}+{\displaystyle \frac{2j!}{{\lambda }^j}}\left(q+{\displaystyle \frac{1}{q}}\right)e^{2\lambda }{e}^{-\lambda \left(n^{1-\beta }-1\right)}\right],$$

(93)

for $j=1,\dots ,m.$

The proof of the theorem now it is completed. □

Theorem 5. 

Assume all the conditions are the same as in Theorem 3. Then

(1)

$$\left|Q_n\left(f,x\right)-f\left(x\right)-{\displaystyle \sum _{j=1}^m}\sum _{\left\{\begin{array}{c}\alpha :\\ \left|\alpha \right|=j\end{array}\right\}}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)f_{\alpha }\left(x\right)Q_n\left(\left({\displaystyle \prod _{i=1}^N}{\left(·-x_i\right)}^{{\alpha }_i}\right)\right)\left(x\right)\right|$$

$$\le {\displaystyle \frac{N^m}{m!}}{\left({\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)}^m{\omega }_{1,m}^{max}\left(f_{\alpha },{\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)+$$

(94)

$$\left({\displaystyle \frac{{\left(2N\right)}^m{∥f_{\alpha }∥}_{\infty ,m}^{max}}{n^{m}m!}}\right)\left[T{e}^{-2\lambda n^{\left(1-\beta \right)}}+{\displaystyle \frac{2m!}{{\lambda }^m}}\left(q+{\displaystyle \frac{1}{q}}\right)e^{2\lambda }{e}^{-\lambda \left(n^{1-\beta }-1\right)}\right]=:{\mathsf{\Phi }}_2^{*},$$

(2) assume that $f_{\alpha }\left(x\right)=0$, for all $\alpha :\left|\alpha \right|=j$, $j=1,\dots ,m$, we have

$$\left|Q_n\left(f,x\right)-f\left(x\right)\right|\le {\mathsf{\Phi }}_2^{*},$$

(95)

with the high speed of convergence ${\left({\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)}^{m+1},$

(3)

$$\left|Q_n\left(f,x\right)-f\left(x\right)\right|\le \sum _{j=1}^N\sum _{\left\{\begin{array}{c}\alpha :\\ \left|\alpha \right|=j\end{array}\right.}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)\left|f_{\alpha }\left(x\right)\right|$$

(96)

$$\left\{{\left({\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)}^j+{\displaystyle \frac{2^{j-1}}{n^j}}\left[T{e}^{-2\lambda n^{\left(1-\beta \right)}}+{\displaystyle \frac{2j!}{{\lambda }^j}}\left(q+{\displaystyle \frac{1}{q}}\right)e^{2\lambda }{e}^{-\lambda \left(n^{1-\beta }-1\right)}\right]\right\}+{\mathsf{\Phi }}_2^{*},$$

and

(4)

$${∥Q_n\left(f\right)-f∥}_{\infty }\le \sum _{j=1}^N\sum _{\left\{\begin{array}{c}\alpha :\\ \left|\alpha \right|=j\end{array}\right.}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right){∥f_{\alpha }∥}_{\infty }$$

$$\left\{{\left({\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)}^j+{\displaystyle \frac{2^{j-1}}{n^j}}\left[T{e}^{-2\lambda n^{\left(1-\beta \right)}}+{\displaystyle \frac{2j!}{{\lambda }^j}}\left(q+{\displaystyle \frac{1}{q}}\right)e^{2\lambda }{e}^{-\lambda \left(n^{1-\beta }-1\right)}\right]\right\}+{\mathsf{\Phi }}_2^{*}.$$

(97)

We have that $Q_n\left(f\right)\to f$, as $n\to \infty$, pointwise and uniformly.

Proof. 

We have that

$$f\left({\displaystyle \frac{k}{n}}+{\displaystyle \frac{r}{n\theta }}\right)-f\left(x\right)-\sum _{j=1}^m\sum _{\begin{array}{c}\alpha :\\ \left|\alpha \right|=j\end{array}}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)\left(\prod _{i=1}^N{\left({\displaystyle \frac{k_i}{n}}+{\displaystyle \frac{r_i}{n{\theta }_i}}-x_i\right)}^{{\alpha }_i}\right)f_{\alpha }\left(x\right)=R,$$

(98)

where

$$R:=m{\int }_0^1{\left(1-\theta \right)}^{m-1}\sum _{\begin{array}{c}\alpha :\\ \left|\alpha \right|=m\end{array}}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)$$

(99)

$$\left({\displaystyle \prod _{i=1}^N}{\left({\displaystyle \frac{k_i}{n}}+{\displaystyle \frac{r_i}{n{\theta }_i}}-x_i\right)}^{{\alpha }_i}\right)\left[f_{\alpha }\left(x+\theta \left({\displaystyle \frac{k}{n}}+{\displaystyle \frac{r}{n\theta }}-x\right)\right)-f_{\alpha }\left(x\right)\right]d\theta .$$

We see that

$$\left|R\right|\le m{\int }_0^1{\left(1-\theta \right)}^{m-1}\sum _{\left|\alpha \right|=m}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)$$

$$\left({\displaystyle \prod _{i=1}^N}{\left|{\displaystyle \frac{k_i}{n}}+{\displaystyle \frac{r_i}{n{\theta }_i}}-x_i\right|}^{{\alpha }_i}\right)\left|f_{\alpha }\left(x+\theta \left({\displaystyle \frac{k}{n}}+{\displaystyle \frac{r}{n\theta }}-x\right)\right)-f_{\alpha }\left(x\right)\right|d\theta \le$$

(100)

$$m{\int }_0^1{\left(1-\theta \right)}^{m-1}\left(\sum _{\left|\alpha \right|=m}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)\left({\displaystyle \prod _{i=1}^N}{\left|{\displaystyle \frac{k_i}{n}}+{\displaystyle \frac{r_i}{n{\theta }_i}}-x_i\right|}^{{\alpha }_i}\right)\right.$$

$$\left.{\omega }_1\left(f_{\alpha },\theta {∥{\displaystyle \frac{k}{n}}+{\displaystyle \frac{r}{n\theta }}-x∥}_{\infty }\right)\right)d\theta \le \left(\ast \right).$$

Notice that

$${∥{\displaystyle \frac{k}{n}}-x∥}_{\infty }\le {\displaystyle \frac{1}{n^{\beta }}},\mathrm{iff}\left|{\displaystyle \frac{k_i}{n}}-x_i\right|\le {\displaystyle \frac{1}{n^{\beta }}},i=1,\dots ,N.$$

We further see that

$$\left(\ast \right)\le m{\omega }_{1,m}^{max}\left(f_{\alpha },{\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right){\int }_0^1{\left(1-\theta \right)}^{m-1}$$

$$\left(\sum _{\left|\alpha \right|=m}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)\left({\displaystyle \prod _{i=1}^N}{\left({\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)}^{{\alpha }_i}\right)\right)d\theta =$$

$${\displaystyle \frac{{\omega }_{1,m}^{max}\left(f_{\alpha },\frac{1}{n}+\frac{1}{n^{\beta }}\right)}{m!}}{\left({\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)}^m\left(\sum _{\left|\alpha \right|=m}{\displaystyle \frac{m!}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)=$$

(101)

$${\displaystyle \frac{{\omega }_{1,m}^{max}\left(f_{\alpha },\frac{1}{n}+\frac{1}{n^{\beta }}\right)}{m!}}{\left({\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)}^m{N}^m.$$

Conclusion: When ${∥{\displaystyle \frac{k}{n}}-x∥}_{\infty }\le {\displaystyle \frac{1}{n^{\beta }}}$, we proved that

$$\left|R\right|\le {\displaystyle \frac{N^m}{m!}}{\left({\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)}^m{\omega }_{1,m}^{max}\left(f_{\alpha },{\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right).$$

(102)

By (99), we obtain that

$$\left|R\right|\le m{\int }_0^1{\left(1-\theta \right)}^{m-1}\sum _{\begin{array}{c}\alpha :\\ \left|\alpha \right|=m\end{array}}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)$$

$$\left(\prod _{i=1}^N{\left(\left|{\displaystyle \frac{k}{n}}-x_i\right|+{\displaystyle \frac{1}{n}}\right)}^{{\alpha }_i}2{∥f_{\alpha }∥}_{\infty }\right)d\theta =$$

(103)

$$\sum _{\begin{array}{c}\alpha :\\ \left|\alpha \right|=m\end{array}}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)\left({\displaystyle \prod _{i=1}^N}{\left(\left|{\displaystyle \frac{k_i}{n}}-x_i\right|+{\displaystyle \frac{1}{n}}\right)}^{{\alpha }_i}\right)2{∥f_{\alpha }∥}_{\infty }\le$$

$$\left({\displaystyle \frac{2{\left({∥\frac{k}{n}-x∥}_{\infty }+\frac{1}{n}\right)}^m{∥f_{\alpha }∥}_{\infty ,m}^{max}}{m!}}\right)\left(\sum _{\left|\alpha \right|=m}{\displaystyle \frac{m!}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)=$$

$$\left({\displaystyle \frac{2{\left({∥\frac{k}{n}-x∥}_{\infty }+\frac{1}{n}\right)}^m{∥f_{\alpha }∥}_{\infty ,m}^{max}}{m!}}\right)N^m.$$

We have extablished in general

$$\left|R\right|\le {\displaystyle \frac{2{\left({∥\frac{k}{n}-x∥}_{\infty }+\frac{1}{n}\right)}^m{∥f_{\alpha }∥}_{\infty ,m}^{max}{N}^m}{m!}}.$$

(104)

Next, we observe that

$$\sum _{r=0}^{\theta }{w}_{r}f\left({\displaystyle \frac{k}{n}}+{\displaystyle \frac{r}{n\theta }}\right)-f\left(x\right)-{\displaystyle \sum _{j=1}^m}\sum _{\begin{array}{c}\alpha :\\ \left|\alpha \right|=j\end{array}}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)f_{\alpha }\left(x\right)$$

$$\left(\sum _{r=0}^{\theta }{w}_r\left(\prod _{i=1}^N{\left({\displaystyle \frac{k_i}{n}}+{\displaystyle \frac{r_i}{n{\theta }_i}}-x_i\right)}^{{\alpha }_i}\right)\right)=\sum _{r=0}^{\theta }{w}_{r}R.$$

(105)

So, when ${∥{\displaystyle \frac{k}{n}}-x∥}_{\infty }\le {\displaystyle \frac{1}{n^{\beta }}}$, we get

$$\left|\sum _{r=0}^{\theta }{w}_{r}R\right|\le \sum _{r=0}^{\theta }{w}_r\left|R\right|\le {\displaystyle \frac{N^m}{m!}}{\left({\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)}^m{\omega }_{1,m}^{max}\left(f_{\alpha },{\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right).$$

(106)

And, it holds in general

$$\left|\sum _{r=0}^{\theta }{w}_{r}R\right|\le {\displaystyle \frac{2{\left({∥\frac{k}{n}-x∥}_{\infty }+\frac{1}{n}\right)}^m{∥f_{\alpha }∥}_{\infty ,m}^{max}{N}^m}{m!}}.$$

(107)

Furthermore, it holds

$$Q_n\left(f,x\right)-f\left(x\right)-{\displaystyle \sum _{j=1}^N}\sum _{\begin{array}{c}\alpha :\\ \left|\alpha \right|=j\end{array}}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)f_{\alpha }\left(x\right)Q_n\left(\left({\displaystyle \prod _{i=1}^N}{\left(·-x_i\right)}^{{\alpha }_i}\right)\right)\left(x\right)=$$

$$\sum _{k=-\infty }^{\infty }\left(\sum _{r=0}^{\theta }{w}_{r}R\right)Z\left(nx-k\right).$$

(108)

Call

$$E_n^{*}:=\sum _{k=-\infty }^{\infty }\left(\sum _{r=0}^{\theta }{w}_{r}R\right)Z\left(nx-k\right).$$

(109)

We derive

$$\left|E_n^{*}\right|{|}_{{∥{\displaystyle \frac{k}{n}}-x∥}_{\infty }<{\displaystyle \frac{1}{n^{\beta }}}}\le {\displaystyle \frac{N^m}{m!}}{\left({\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)}^m{\omega }_{1,m}^{max}\left(f_{\alpha },{\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right).$$

(110)

And, furthermore we get that

$$\left|E_n^{*}\right|{|}_{{∥{\displaystyle \frac{k}{n}}-x∥}_{\infty }\ge {\displaystyle \frac{1}{n^{\beta }}}}\le \left({\displaystyle \frac{2{∥f_{\alpha }∥}_{\infty ,m}^{max}{N}^m}{m!}}\right)\sum _{k=-\infty }^{\infty }{\left({∥{\displaystyle \frac{k}{n}}-x∥}_{\infty }+{\displaystyle \frac{1}{n}}\right)}^{m}Z\left(nx-u\right).$$

(111)

As in the proof of Theorem 4, we obtain

$$\left|E_n^{*}\right|{|}_{{∥{\displaystyle \frac{k}{n}}-x∥}_{\infty }\ge {\displaystyle \frac{1}{n^{\beta }}}}\le \left({\displaystyle \frac{2^m{∥f_{\alpha }∥}_{\infty ,m}^{max}{N}^m}{n^{m}m!}}\right)$$

$$\left[T{e}^{-2\lambda n^{\left(1-\beta \right)}}+{\displaystyle \frac{2m!}{{\lambda }^m}}\left(q+{\displaystyle \frac{1}{q}}\right)e^{2\lambda }{e}^{-\lambda \left(n^{1-\beta }-1\right)}\right].$$

(112)

Consequently it holds

$$\left|E_n^{*}\right|\le \left|E_n^{*}\right|{{|}_{{∥{\displaystyle \frac{k}{n}}-x∥}_{\infty }<{\displaystyle \frac{1}{n^{\beta }}}}+\left|E_n^{*}\right||}_{{∥{\displaystyle \frac{k}{n}}-x∥}_{\infty }\ge {\displaystyle \frac{1}{n^{\beta }}}}\le$$

$${\displaystyle \frac{N^m}{m!}}{\left({\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)}^m{\omega }_{1,m}^{max}\left(f_{\alpha },{\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)+$$

$$\left({\displaystyle \frac{2^m{∥f_{\alpha }∥}_{\infty ,m}^{max}{N}^m}{n^{m}m!}}\right)\left[T{e}^{-2\lambda n^{\left(1-\beta \right)}}+{\displaystyle \frac{2m!}{{\lambda }^m}}\left(q+{\displaystyle \frac{1}{q}}\right)e^{2\lambda }{e}^{-\lambda \left(n^{1-\beta }-1\right)}\right].$$

(113)

Finally, we estimate

$$\left|Q_n\left(\left(\prod _{i=1}^N{\left(·-x_i\right)}^{{\alpha }_i}\right)\right)\left(x\right)\right|=$$

$$\left|\sum _{k=-\infty }^{\infty }\left(\sum _{r=0}^{\theta }{w}_r\left(\prod _{i=1}^N{\left({\displaystyle \frac{k_i}{n}}+{\displaystyle \frac{r_i}{n{\theta }_i}}-x_i\right)}^{{\alpha }_i}\right)\right)Z\left(nx-k\right)\right|\le$$

$$\sum _{k=-\infty }^{\infty }\left(\sum _{r=0}^{\theta }{w}_r\left(\prod _{i=1}^N{\left(\left|{\displaystyle \frac{k_i}{n}}-x_i\right|+{\displaystyle \frac{1}{n}}\right)}^{{\alpha }_i}\right)\right)Z\left(nx-k\right)=$$

$$\sum _{k=-\infty }^{\infty }\left(\prod _{i=1}^N{\left(\left|{\displaystyle \frac{k_i}{n}}-x_i\right|+{\displaystyle \frac{1}{n}}\right)}^{{\alpha }_i}\right)Z\left(nx-k\right)\le$$

(114)

$$\sum _{\left\{\begin{array}{c}k=-\infty \\ :{∥x-{\displaystyle \frac{k}{n}}∥}_{\infty }<{\displaystyle \frac{1}{n^{\beta }}}\end{array}\right.}^{\infty }\left(\prod _{i=1}^N{\left(\left|{\displaystyle \frac{k_i}{n}}-x_i\right|+{\displaystyle \frac{1}{n}}\right)}^{{\alpha }_i}\right)Z\left(nx-k\right)+$$

$$\sum _{\left\{\begin{array}{c}k=-\infty \\ :{∥x-{\displaystyle \frac{k}{n}}∥}_{\infty }\ge {\displaystyle \frac{1}{n^{\beta }}}\end{array}\right.}^{\infty }\left(\prod _{i=1}^N{\left(\left|{\displaystyle \frac{k_i}{n}}-x_i\right|+{\displaystyle \frac{1}{n}}\right)}^{{\alpha }_i}\right)Z\left(nx-k\right)\le$$

$$\sum _{\left\{\begin{array}{c}k=-\infty \\ :{∥x-{\displaystyle \frac{k}{n}}∥}_{\infty }<{\displaystyle \frac{1}{n^{\beta }}}\end{array}\right.}^{\infty }\left(\prod _{i=1}^N{\left({\displaystyle \frac{1}{n^{\beta }}}+{\displaystyle \frac{1}{n}}\right)}^{{\alpha }_i}\right)Z\left(nx-k\right)+$$

$${\displaystyle \frac{1}{n^j}}\sum _{\left\{\begin{array}{c}k=-\infty \\ :{∥nx-k∥}_{\infty }\ge n^{1-\beta }\end{array}\right.}^{\infty }\left({\displaystyle \prod _{i=1}^N}{\left(1+\left|n{x}_i-k_i\right|\right)}^{{\alpha }_i}\right)Z\left(nx-k\right)$$

$$\le \dots \left(\mathrm{as}\mathrm{in}\mathrm{Theorem}4\right)\le$$

$${\left({\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)}^j+{\displaystyle \frac{2^{j-1}}{n^j}}\left[T{e}^{-2\lambda n^{\left(1-\beta \right)}}+{\displaystyle \frac{2j!}{{\lambda }^j}}\left(q+{\displaystyle \frac{1}{q}}\right)e^{2\lambda }{e}^{-\lambda \left(n^{1-\beta }-1\right)}\right],$$

(115)

for $j=1,\dots ,m.$

The theorem is proved. □

4. Illustrations for m = 1

We present

Corollary 1. 

Let $0<\beta <1$, $n\in \mathbb{N}:n^{1-\beta }>2$, $x\in {\mathbb{R}}^N$, $f\in C^1\left({\mathbb{R}}^N\right)$, $N\in \mathbb{N},$ with $f_{\alpha }\in C_B\left({\mathbb{R}}^N\right)$, for all $\alpha :=\left({\alpha }_1,\dots ,{\alpha }_N\right)$, ${\alpha }_i\in {\mathbb{Z}}_{+},$ $i=1,\dots ,N,$ and $\left|\alpha \right|={\displaystyle \sum _{i=1}^N}{\alpha }_i=l$, where $l=0,1$. Then

(1)

$$\left|A_n\left(f,x\right)-f\left(x\right)-\sum _{\left\{\begin{array}{c}\alpha :\\ \left|\alpha \right|=1\end{array}\right\}}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)f_{\alpha }\left(x\right)A_n\left(\left({\displaystyle \prod _{i=1}^N}{\left(·-x_i\right)}^{{\alpha }_i}\right)\right)\left(x\right)\right|$$

$$\le \left({\displaystyle \frac{N}{n^{\beta }}}\right){\omega }_{1,1}^{max}\left(f_{\alpha },{\displaystyle \frac{1}{n^{\beta }}}\right)+{\displaystyle \frac{4{∥f_{\alpha }∥}_{\infty ,1}^{max}N}{n\lambda }}\left(q+{\displaystyle \frac{1}{q}}\right)e^{2\lambda }{e}^{-\lambda \left(n^{1-\beta }-1\right)}=:\overline{{\mathsf{\Phi }}_1^{*}},$$

(116)

(2) assume that $f_{\alpha }\left(x\right)=0$, for all $\alpha :\left|\alpha \right|=1$, we have

$$\left|A_n\left(f,x\right)-f\left(x\right)\right|\le \overline{{\mathsf{\Phi }}_1^{*}},$$

(117)

at the high speed of convergence $n^{-2\beta },$

(3)

$$\left|A_n\left(f,x\right)-f\left(x\right)\right|\le \sum _{\left\{\begin{array}{c}\alpha :\\ \left|\alpha \right|=1\end{array}\right\}}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)\left|f_{\alpha }\left(x\right)\right|$$

$$\left\{{\displaystyle \frac{1}{n^{\beta }}}+{\displaystyle \frac{2}{n\lambda }}\left(q+{\displaystyle \frac{1}{q}}\right)e^{2\lambda }{e}^{-\lambda \left(n^{1-\beta }-1\right)}\right\}+\overline{{\mathsf{\Phi }}_1^{*}},$$

(118)

(4)

$${∥A_n\left(f\right)-f∥}_{\infty }\le$$

$$\sum _{\left\{\begin{array}{c}\alpha :\\ \left|\alpha \right|=1\end{array}\right\}}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right){∥f_{\alpha }∥}_{\infty }\left\{{\displaystyle \frac{1}{n^{\beta }}}+{\displaystyle \frac{2}{n\lambda }}\left(q+{\displaystyle \frac{1}{q}}\right)e^{2\lambda }{e}^{-\lambda \left(n^{1-\beta }-1\right)}\right\}+\overline{{\mathsf{\Phi }}_1^{*}}.$$

(119)

We have that $A_n\left(f\right)\to f$, as $n\to \infty$, pointwise and uniformly.

Proof. 

By Theorem 3. □

We finish with

Corollary 2. 

Assume all the conditions are the same as in Corollary 1. Then

(1)

$$\left\{\begin{array}{c}\left|K_n\left(f,x\right)-f\left(x\right)-{\displaystyle \sum _{\left\{\begin{array}{c}\alpha :\\ \left|\alpha \right|=1\end{array}\right\}}}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)f_{\alpha }\left(x\right)K_n\left(\left({\displaystyle \prod _{i=1}^N}{\left(·-x_i\right)}^{{\alpha }_i}\right)\right)\left(x\right)\right|\\ \left|Q_n\left(f,x\right)-f\left(x\right)-{\displaystyle \sum _{\left\{\begin{array}{c}\alpha :\\ \left|\alpha \right|=1\end{array}\right\}}}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)f_{\alpha }\left(x\right)Q_n\left(\left({\displaystyle \prod _{i=1}^N}{\left(·-x_i\right)}^{{\alpha }_i}\right)\right)\left(x\right)\right|\end{array}\right\}$$

$$\le N\left({\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right){\omega }_{1,1}^{max}\left(f_{\alpha },{\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)+$$

(120)

$$\left({\displaystyle \frac{\left(2N\right){∥f_{\alpha }∥}_{\infty ,1}^{max}}{n}}\right)\left[T{e}^{-2\lambda n^{\left(1-\beta \right)}}+{\displaystyle \frac{2}{\lambda }}\left(q+{\displaystyle \frac{1}{q}}\right)e^{2\lambda }{e}^{-\lambda \left(n^{1-\beta }-1\right)}\right]=:\overline{{\mathsf{\Phi }}_2^{*}},$$

(2) assume that $f_{\alpha }\left(x\right)=0$, for all $\alpha :\left|\alpha \right|=1$, we have

$$\left\{\begin{array}{c}\left|K_n\left(f,x\right)-f\left(x\right)\right|\\ \left|Q_n\left(f,x\right)-f\left(x\right)\right|\end{array}\right\}\le \overline{{\mathsf{\Phi }}_2^{*}},$$

(121)

with the high speed of convergence ${\left({\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)}^2,$

(3)

$$\left\{\begin{array}{c}\left|K_n\left(f,x\right)-f\left(x\right)\right|\\ \left|Q_n\left(f,x\right)-f\left(x\right)\right|\end{array}\right\}\le {\displaystyle \sum _{\left\{\begin{array}{c}\alpha :\\ \left|\alpha \right|=1\end{array}\right.}}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right)\left|f_{\alpha }\left(x\right)\right|$$

$$\left\{\left({\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)+{\displaystyle \frac{1}{n}}\left[T{e}^{-2\lambda n^{\left(1-\beta \right)}}+{\displaystyle \frac{2}{\lambda }}\left(q+{\displaystyle \frac{1}{q}}\right)e^{2\lambda }{e}^{-\lambda \left(n^{1-\beta }-1\right)}\right]\right\}+\overline{{\mathsf{\Phi }}_2^{*}},$$

(122)

and

(4)

$$\left\{\begin{array}{c}{∥K_n\left(f\right)-f∥}_{\infty }\\ {∥Q_n\left(f\right)-f∥}_{\infty }\end{array}\right\}\le {\displaystyle \sum _{\left\{\begin{array}{c}\alpha :\\ \left|\alpha \right|=1\end{array}\right.}}\left({\displaystyle \frac{1}{{\displaystyle \prod _{i=1}^N}{\alpha }_i!}}\right){∥f_{\alpha }∥}_{\infty }$$

$$\left\{\left({\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^{\beta }}}\right)+{\displaystyle \frac{1}{n}}\left[T{e}^{-2\lambda n^{\left(1-\beta \right)}}+{\displaystyle \frac{2}{\lambda }}\left(q+{\displaystyle \frac{1}{q}}\right)e^{2\lambda }{e}^{-\lambda \left(n^{1-\beta }-1\right)}\right]\right\}+\overline{{\mathsf{\Phi }}_2^{*}}.$$

(123)

We have that $K_n\left(f\right)\to f$, $Q_n\left(f\right)\to f$, as $n\to \infty$, pointwise and uniformly.

Proof. 

By Theorems 4 and 5. □

5. Conclusions

Here, we presented multivariate neural network approximation over infinite domains under differentiation of functions. The activation function was the symmetrized and perturbed hyperbolic tangent function. The rates of convergence were higher and the data needed to feed the neural network were half due to symmetry.