Parametrized Half-Hyperbolic Tangent Function-Activated Complex-Valued Neural Network Approximation

Abstract

In this paper, we create a family of neural network (NN) operators employing a parametrized and deformed half-hyperbolic tangent function as an activation function and a density function produced by the same activation function. Moreover, we consider the univariate quantitative approximations by complex-valued neural network (NN) operators of complex-valued functions on a compact domain. Pointwise and uniform convergence results on Banach spaces are acquired through trigonometric, hyperbolic, and hybrid-type hyperbolic–trigonometric approaches.

1. Motivation

Under the umbrella of Artificial Intelligence (AI), neural network (NN) operators supply a vigorous structure for solving complex real-life problems in various fields of science such as cybersecurity, healthcare, economics, medicine, psychology, sociology, etc. Thanks to their differentiability properties, they can optimize parameters directly. Moreover, NN operators need correctly selected activation functions to work effectively. When we look at the literature, we encounter a very wide range of activation functions [1]. In this study, we choose ${\widehat{h}}_{\alpha }$—a parametrized and deformed half-hyperbolic tangent function—as the activation function. $tanh$-like functions can be more effective in reaching the optimum solution due to their trainability [2]. One of the interesting aspects of this work is that all higher-order approximations are based on trigonometric and hyperbolic-type Taylor’s formulae inequalities (see [3,4,5]). Next, we emphasize the following: as is well known, the human brain has been proven medically to be a non-symmetrical human organ. As a result, NNs try to imitate its operation and are not symmetrical mathematical structures. But in our paper, we build an approximation apparatus that is as close as possible to symmetry. Namely the activation function we initially use is described as reciprocal anti-symmetrical (see Proposition 1). This is the building block for our density function which is used in our approximation NN operators. We prove that this density function is a reciprocal symmetric function (see Equation (3)). This represents our study’s interesting connection to the general “symmetry” phenomenon.

The first construction of approximation by NN operators in the sense of the “Cardaliaguet Euvrard” and “Squashing” types was made by G.A. Anastassiou in 1997. As a fruit of this construction, he also brought to the literature the ability to calculate the convergence speed with the help of convergence rates using the modulus of continuity [6].

The mathematical expression of the one-hidden-layer neural network (NN) architecture is presented as

$$N_{\psi ,n}\left(x\right)=\sum _{i=0}^n{k}_i\psi \left(〈{\tilde{\omega }}_i·x〉+{\varpi }_i\right),x\in {\mathbb{R}}^s,s\in \mathbb{N},$$

for $0\le i\le n$, where ${\tilde{\omega }}_i=\left({\tilde{\omega }}_{i,1},{\tilde{\omega }}_{i,2},\dots ,{\tilde{\omega }}_{i,s}\right)\in {\mathbb{R}}^s$ are the connection weights, $k_i$ is the coefficient, $〈{\tilde{\omega }}_i·x〉$ is the inner product of ${\tilde{\omega }}_i$, $x;$ ${\varpi }_i\in \mathbb{R}$ is the threshold, and $\psi$ is the activation function. For more knowledge of NNs, readers are recommended to read [7,8,9].

This paper is organized as follows: in Section 1, we present our motivation to the readers. In Section 2, step by step, we build up our activation function ${\widehat{h}}_{\alpha }$. We also include our basic estimations, which form the basis for the main results. We devote Section 3 and Section 4 to the construction of density function ${\mathit{A}}_{\alpha }$, and the creation of the $\mathbb{C}-$valued linear NN operators, respectively. In Section 5, the authors perform deformed and parametrized half-hyperbolic tangent function-activated high-order neural network approximations on continuous functions over compact intervals of a real line with complex values. All convergences have rates expressed via the modulus of continuity of the involved functions’ high-order derivatives, derived from very tight Jackson-type inequalities. We conclude with Section 6.

2. Activation Function: Parametrized Half-Hyperbolic Tangent Function ${\widehat{\mathit{h}}}_{\mathit{\alpha }}$

We consider the following function inspired by [10]

$${\widehat{h}}_{\alpha }\left(x\right):=\frac{1-\alpha e^{-\gamma x}}{1+\alpha e^{-\gamma x}},\forall x\in \mathbb{R},$$

(1)

to be a parametrized half-hyperbolic tangent function for $\alpha ,\gamma >0.$ Also, one has ${\widehat{h}}_{\alpha }\left(0\right)=\frac{1-\alpha }{1+\alpha }.$

Proposition 1.

We observe that

$${\widehat{h}}_{\alpha }\left(-x\right)=-{\widehat{h}}_{\frac{1}{\alpha }}\left(x\right)$$

for $x\in \mathbb{R};\alpha >0.$

Proposition 2.

The function ${\widehat{h}}_{\alpha }$ is strictly increasing on $\mathbb{R}.$ Because

$${\widehat{h}}_{\alpha }^{\prime }\left(x\right)={\left(\frac{1-\alpha e^{-\gamma x}}{1+\alpha e^{-\gamma x}}\right)}^{\prime }=\frac{2\alpha \gamma e^{\gamma x}}{{\left(e^{\gamma x}+\alpha \right)}^2}>0,$$

for every $x\in \mathbb{R};\alpha ,\gamma >0.$

Proposition 3.

When we take the second-order derivative of ${\widehat{h}}_{\alpha },$ we have

$${\widehat{h}}_{\alpha }^{″}\left(x\right)={\left(\frac{2\alpha \gamma e^{\gamma x}}{{\left(e^{\gamma x}+\alpha \right)}^2}\right)}^{\prime }=2\alpha {\gamma }^2{e}^{\gamma x}\frac{\left(\alpha -e^{\gamma x}\right)}{{\left(\alpha +e^{\gamma x}\right)}^3}\in C\left(\mathbb{R}\right),\forall x\in \mathbb{R};\alpha ,\gamma >0.$$

(2)

According to the above calculation, the following are true:

Case 1:

if $x<\frac{ln\alpha }{\gamma }$, then ${\widehat{h}}_{\alpha }$ is strictly concave upward, with ${\widehat{h}}_{\alpha }^{″}\left(\frac{ln\alpha }{\gamma }\right)=0.$

Case 2:

if $x>\frac{ln\alpha }{\gamma }$, then ${\widehat{h}}_{\alpha }$ is strictly concave downward.

3. Construction of Density Function ${\mathit{A}}_{\mathbf{\alpha }}$

In this section, we aim to create the density function ${\mathit{A}}_{\alpha }$, and we also present its basic properties, which will be used throughout the paper. It is clear that $1>-1⟹x+1>x-1.$ So, for every $x\in \mathbb{R};\alpha ,\gamma >0,$ let us consider the following density function:

$${\mathit{A}}_{\alpha }\left(x\right):=\frac{1}{4}\left({\widehat{h}}_{\alpha }\left(x+1\right)-{\widehat{h}}_{\alpha }\left(x-1\right)\right)>0.$$

Furthermore, note that

$$\underset{x\to +\infty }{lim}{\mathit{A}}_{\alpha }\left(x\right)=\underset{x\to -\infty }{lim}{\mathit{A}}_{\alpha }\left(x\right)=0,$$

so the x-axis is a horizontal asymptote. One can write that

$$\begin{array}{ccc}\hfill {\mathit{A}}_{\alpha }\left(-x\right)& =& \frac{1}{4}\left({\widehat{h}}_{\alpha }\left(-x+1\right)-{\widehat{h}}_{\alpha }\left(-x-1\right)\right)\hfill \\ & =& \frac{1}{4}\left(-\frac{\frac{1}{\alpha }{e}^{-\gamma \left(x+1\right)}-1}{\frac{1}{\alpha }{e}^{-\gamma \left(x+1\right)}+1}+\frac{\frac{1}{\alpha }{e}^{-\gamma \left(x-1\right)}-1}{\frac{1}{\alpha }{e}^{-\gamma \left(x-1\right)}+1}\right)\hfill \\ & =& \frac{1}{4}\left({\widehat{h}}_{\frac{1}{\alpha }}\left(x+1\right)-{\widehat{h}}_{\frac{1}{\alpha }}\left(x-1\right)\right)\hfill \\ & =& {\mathit{A}}_{\frac{1}{\alpha }}\left(x\right),x\in \mathbb{R};\alpha >0.\hfill \end{array}$$

Thus,

$${\mathit{A}}_{\alpha }\left(-x\right)={\mathit{A}}_{\frac{1}{\alpha }}\left(x\right),\forall x\in \mathbb{R};\alpha >0.$$

(3)

Remark 1.

We have

$${\mathit{A}}_{\alpha }^{\prime }\left(x\right):=\frac{1}{4}\left({\widehat{h}}_{\alpha }^{\prime }\left(x+1\right)-{\widehat{h}}_{\alpha }^{\prime }\left(x-1\right)\right).$$

Then,

(i) 

Let $x<\frac{ln\alpha }{\gamma }-1,$ then $x-1<x+1<\frac{ln\alpha }{\gamma },$ and ${\widehat{h}}_{\alpha }^{\prime }\left(x+1\right)>{\widehat{h}}_{\alpha }^{\prime }\left(x-1\right)$, that is, ${\mathit{A}}_{\alpha }^{\prime }\left(x\right)>0$. Thus, ${\mathit{A}}_{\alpha }$ is strictly increasing on $\left(-\infty ,\frac{ln\alpha }{\gamma }-1\right)$.

(ii) 

Let $x-1>\frac{ln\alpha }{\gamma },$ then $x+1>x-1>\frac{ln\alpha }{\gamma },$ and ${\widehat{h}}_{\alpha }^{\prime }\left(x+1\right)<{\widehat{h}}_{\alpha }^{\prime }\left(x-1\right)$, namely, ${\mathit{A}}_{\alpha }^{\prime }\left(x\right)<0$. Therefore, ${\mathit{A}}_{\alpha }$ is strictly decreasing on $\left(\frac{ln\alpha }{\gamma }+1,+\infty \right)$.

Remark 2.

Let $\frac{ln\alpha }{\gamma }-1\le x\le \frac{ln\alpha }{\gamma }+1$. Then,

$$\begin{array}{ccc}\hfill {\mathit{A}}_{\alpha }^{″}\left(x\right)& =& \frac{1}{4}\left({\widehat{h}}_{\alpha }^{″}\left(x+1\right)-{\widehat{h}}_{\alpha }^{″}\left(x-1\right)\right)\hfill \\ & =& \frac{\alpha {\gamma }^2}{2}\left(e^{\gamma \left(x+1\right)}\frac{\left(\alpha -e^{\gamma \left(x+1\right)}\right)}{{\left(\alpha +e^{\gamma \left(x+1\right)}\right)}^3}-e^{\gamma \left(x-1\right)}\frac{\left(\alpha -e^{\gamma \left(x-1\right)}\right)}{{\left(\alpha +e^{\gamma \left(x-1\right)}\right)}^3}\right)\hfill \end{array}$$

(4)

Explicitly, according to $\left(4\right);$ one determines that ${\mathit{A}}_{\alpha }^{″}\left(x\right)\le 0$ for $x\in \left[\frac{ln\alpha }{\gamma }-1,\frac{ln\alpha }{\gamma }+1\right]$, and is strictly concave downward on $\left(\frac{ln\alpha }{\gamma }-1,\frac{ln\alpha }{\gamma }+1\right).$ Therefore, ${\mathit{A}}_{\alpha }$ is a bell-shaped function on $\mathbb{R}.$ Moreover, ${\mathit{A}}_{\alpha }^{″}\left(\frac{ln\alpha }{\gamma }\right)<0$ is satisfied.

Remark 3.

The maximum value of ${\mathit{A}}_{\alpha }$ is

$${\mathit{A}}_{\alpha }\left(\frac{ln\alpha }{\gamma }\right)=\frac{{\widehat{h}}_1\left(1\right)}{2}.$$

Theorem 1

([10]). We determine that

$$\sum _{j=-\infty }^{\infty }{\mathit{A}}_{\alpha }\left(x-j\right)=1,\forall x\in \mathbb{R},\forall \alpha >0.$$

Thus

$$\sum _{j=-\infty }^{\infty }{\mathit{A}}_{\alpha }\left(nx-j\right)=1,\forall n\in \mathbb{N},\forall x\in \mathbb{R}.$$

In this manner, it holds that

$$\sum _{j=-\infty }^{\infty }{\mathit{A}}_{\frac{1}{\alpha }}\left(x-j\right)=1,\forall x\in \mathbb{R}.$$

However,

$${\mathit{A}}_{\frac{1}{\alpha }}\left(x-j\right)\stackrel{\left(3\right)}{=}{\mathit{A}}_{\alpha }\left(j-x\right).$$

So

$$\sum _{j=-\infty }^{\infty }{\mathit{A}}_{\alpha }\left(j-x\right)=1,\forall x\in \mathbb{R},\alpha >0,$$

and

$$\sum _{j=-\infty }^{\infty }{\mathit{A}}_{\alpha }\left(j+x\right)=1,\forall x\in \mathbb{R},\alpha >0.$$

Theorem 2

([10]). It holds that

$${\int }_{-\infty }^{\infty }{\mathit{A}}_{\alpha }\left(x\right)dx=1;\alpha >0.$$

Thus, this means that ${\mathit{A}}_{\alpha }$ is a density function over $\mathbb{R}$ such that $\alpha >0.$

Next, the following result is needed.

Theorem 3

([10]). Let $0<\lambda <1$, and $n\in \mathbb{N}$ with $n^{1-\lambda }>2$; $\alpha ,\gamma >0$. Then,

$$\sum _{\left\{\begin{array}{c}k=-\infty \\ :\left|nx-k\right|\ge n^{1-\lambda }\end{array}\right.}^{\infty }{\mathit{A}}_{\alpha }\left(nx-k\right)<2max\left\{\alpha ,\frac{1}{\alpha }\right\}{e}^{2\gamma }{e}^{-\gamma n^{\left(1-\lambda \right)}}=:C{e}^{-\gamma n^{\left(1-\lambda \right)}},$$

(5)

where $C:=2max\left\{\alpha ,\frac{1}{\alpha }\right\}{e}^{2\gamma }.$

Let $⌈·⌉$ and $⌊·⌋$ be the ceiling of the number and the integral part of the number, respectively.

Let us continue with the following conclusion:

Theorem 4

([10]). Let $x\in \left[a,b\right]\subset \mathbb{R}$ and $n\in \mathbb{N}$ so that $⌈na⌉\le ⌊nb⌋$. For $\alpha ,\gamma >0$, we consider the number ${\gamma }_{\alpha }>{\rho }_0>0$ with ${\mathit{A}}_{\alpha }\left({\rho }_0\right)={\mathit{A}}_{\alpha }\left(0\right),$ and ${\gamma }_{\alpha }>1$. Then,

$$\frac{1}{{\displaystyle \sum _{k=⌈na⌉}^{⌊nb⌋}}{\mathit{A}}_{\alpha }\left(nx-k\right)}<\mathit{max}\left\{\frac{1}{{\mathit{A}}_{\alpha }\left({\gamma }_{\alpha }\right)},\frac{1}{\mathit{A}\frac{1}{\alpha }\left(\gamma \frac{1}{\alpha }\right)}\right\}=:\nabla \left(\alpha \right).$$

(6)

We also mention the following:

Remark 4

([10]). (i) We also notice that

$$\underset{n\to +\infty }{lim}\sum _{k=⌈na⌉}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right)\ne 1,\mathit{for}\mathit{at}\mathit{least}\mathit{some}x\in \left[a,b\right],$$

where $\alpha >0.$

(ii) Let $\left[a,b\right]\subset \mathbb{R}$. For large $n,$ we always have $⌈na⌉\le ⌊nb⌋$. Also $a\le \frac{k}{n}\le b$, iff $⌈na⌉\le k\le ⌊nb⌋$. In general, it holds that

$$\sum _{k=⌈na⌉}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right)\le 1.$$

(7)

4. Generation of $\mathbb{C}$-Valued Linear NN Operators

Let $\left(\mathbb{C},\left|·\right|\right)$ be the Banach space of the complex numbers on the reals.

Definition 1.

Let $f\in C\left(\left[a,b\right],\mathbb{C}\right)$ and $n\in \mathbb{N}:⌈na⌉\le ⌊nb⌋$. We introduce and define the $\mathbb{C}$-valued linear neural network operators as follows:

$${\mathsf{\Theta }}_n\left(f,x\right):=\frac{{\displaystyle \sum _{k=⌈na⌉}^{⌊nb⌋}}f\left(\frac{k}{n}\right){\mathit{A}}_{\alpha }\left(nx-k\right)}{{\displaystyle \sum _{k=⌈na⌉}^{⌊nb⌋}}{\mathit{A}}_{\alpha }\left(nx-k\right)},x\in \left[a,b\right];\alpha >0,\alpha \ne 1.$$

(8)

For large enough n, we always obtain $⌈na⌉\le ⌊nb⌋.$ Also, $a\le \frac{k}{n}\le b,$ iff $⌈na⌉\le k\le ⌊nb⌋.$ The same ${\mathsf{\Theta }}_n$ may be used for real-valued functions. Here, we study the pointwise and uniform convergence of ${\mathsf{\Theta }}_n\left(f,x\right)$ to $f\left(x\right)$ with rates.

Clearly, here, ${\mathsf{\Theta }}_n\left(f\right)\in C\left(\left[a,b\right],\mathbb{C}\right).$

For the sake of usefulness, we can follow the following:

$${\mathsf{\Theta }}_n^{*}\left(f,x\right):=\sum _{k=⌈na⌉}^{⌊nb⌋}f\left(\frac{k}{n}\right){\mathit{A}}_{\alpha }\left(nx-k\right),$$

(9)

that is,

$${\mathsf{\Theta }}_n\left(f,x\right):=\frac{{\mathsf{\Theta }}_n^{*}\left(f,x\right)}{{\displaystyle \sum _{k=⌈na⌉}^{⌊nb⌋}}{\mathit{A}}_{\alpha }\left(nx-k\right)}.$$

(10)

so that

$${\mathsf{\Theta }}_n\left(f,x\right)-f\left(x\right)=\frac{{\mathsf{\Theta }}_n^{*}\left(f,x\right)-f\left(x\right)\left({\displaystyle \sum _{k=⌈na⌉}^{⌊nb⌋}}{\mathit{A}}_{\alpha }\left(nx-k\right)\right)}{{\displaystyle \sum _{k=⌈na⌉}^{⌊nb⌋}}{\mathit{A}}_{\alpha }\left(nx-k\right)}.$$

(11)

Therefore, we state that

$$\left|{\mathsf{\Theta }}_n\left(f,x\right)-f\left(x\right)\right|\le \nabla \left(\alpha \right)\left|{\mathsf{\Theta }}_n^{*}\left(f,x\right)-f\left(x\right)\left(\sum _{k=⌈na⌉}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right)\right)\right|$$

$$=\nabla \left(\alpha \right)\left|\sum _{k=⌈na⌉}^{⌊nb⌋}\left(f\left(\frac{k}{n}\right)-f\left(x\right)\right){\mathit{A}}_{\alpha }\left(nx-k\right)\right|,$$

(12)

where $\nabla \left(\alpha \right)$ is as in (6).

We will calculate (12) by virtue of the classical first modulus of continuity for $f\in C\left(\left[a,b\right],\mathbb{C}\right)$ defined below:

$${\omega }_1\left(f,\delta \right):=\underset{\begin{array}{c}x,y\in \left[a,b\right]\\ \left|x-y\right|\le \delta \end{array}}{sup}\left|f\left(x\right)-f\left(y\right)\right|,\delta >0.$$

Moreover, $f\in C\left(\left[a,b\right],\mathbb{C}\right)$ is equivalent to ${\displaystyle \underset{\delta \to 0}{lim}}{\omega }_1\left(f,\delta \right)=0$ (see [6]).

5. Approximation Results

Now, we are ready to perform $\mathbb{C}$-valued neural network high-order approximations to a function f given with rates. Let us start with a trigonometric approach.

Theorem 5.

Let $f\in C^2\left(\left[a,b\right],\mathbb{C}\right)$, $0<\lambda <1$, $n\in \mathbb{N}:n^{1-\lambda }>2$, $x\in \left[a,b\right].$ Then,

(1) 

$$\left|{\mathsf{\Theta }}_n\left(f,x\right)-f\left(x\right)\right|\le \nabla \left(\alpha \right)\left[\left|f^{\prime }\left(x\right)\right|\left(\frac{1}{n^{\lambda }}+\left(b-a\right)C{e}^{-\gamma n^{\left(1-\lambda \right)}}\right)\right.$$

$$+\frac{\left|f^{\prime \prime }\left(x\right)\right|}{2}\left(\frac{1}{n^{2\lambda }}+{\left(b-a\right)}^{2}C{e}^{-\gamma n^{\left(1-\lambda \right)}}\right)+$$

$$\left.\left(\frac{{\omega }_1\left(f^{\prime \prime }+f,\frac{1}{n^{\lambda }}\right)}{2n^{2\lambda }}+{∥f^{\prime \prime }+f∥}_{\infty }{\left(b-a\right)}^{2}C{e}^{-\gamma n^{\left(1-\lambda \right)}}\right)\right],$$

(2) 

If $f^{\prime }\left(x\right)=f^{\prime \prime }\left(x\right)=0,$ we obtain

$$\left|{\mathsf{\Theta }}_n\left(f,x\right)-f\left(x\right)\right|\le \nabla \left(\alpha \right)\left[\frac{{\omega }_1\left(f^{\prime \prime }+f,\frac{1}{n^{\lambda }}\right)}{2n^{2\lambda }}+{∥f^{\prime \prime }+f∥}_{\infty }{\left(b-a\right)}^{2}C{e}^{-\gamma n^{\left(1-\lambda \right)}}\right],$$

Note here that there is a high rate of convergence at $n^{-3\lambda }.$

(3) 

In addition, we obtain

$$\begin{array}{ccc}\hfill {∥{\mathsf{\Theta }}_{n}f-f∥}_{\infty }& \le & \nabla \left(\alpha \right)\left[{∥f^{\prime }∥}_{\infty }\left(\frac{1}{n^{\lambda }}+\left(b-a\right)C{e}^{-\gamma n^{\left(1-\lambda \right)}}\right)\right.\hfill \\ & & +\frac{{∥f^{\prime \prime }∥}_{\infty }}{2}\left(\frac{1}{n^{2\lambda }}+{\left(b-a\right)}^{2}C{e}^{-\gamma n^{\left(1-\lambda \right)}}\right)\hfill \\ & & \left.+\left(\frac{{\omega }_1\left(f^{\prime \prime }+f,\frac{1}{n^{\lambda }}\right)}{2n^{2\lambda }}+{∥f^{\prime \prime }+f∥}_{\infty }\left(b-a\right)C{e}^{-\gamma n^{\left(1-\lambda \right)}}\right)\right],\hfill \end{array}$$

namely, ${\displaystyle \underset{n\to +\infty }{lim}}{\mathsf{\Theta }}_n\left(f\right)=f$, pointwise and uniformly,

(4) 

Eventually, it holds that

$$\begin{array}{ccc}& & \left|{\mathsf{\Theta }}_n\left(f,x\right)-f^{\prime }\left(x\right){\mathsf{\Theta }}_n\left(sin\left(·-x\right),x\right)-2f^{\prime \prime }\left(x\right){\mathsf{\Theta }}_n\left({sin}^2\left(\frac{·-x}{2}\right),x\right)-f\left(x\right)\right|\hfill \\ & \le & \nabla \left(\alpha \right)\left[\frac{{\omega }_1\left(f^{\prime \prime }+f,\frac{1}{n^{\lambda }}\right)}{2n^{2\lambda }}+{∥f^{\prime \prime }+f∥}_{\infty }{\left(b-a\right)}^{2}C{e}^{-\gamma n^{\left(1-\lambda \right)}}\right],\hfill \end{array}$$

and a high speed of convergence at $n^{-3\lambda }$ is gained.

Proof.  

Inspired by [3], and applying the trigonometric Taylor’s formula for $f\in C^2\left(\left[a,b\right],\mathbb{C}\right)$, let $\frac{k}{n},x\in \left[a,b\right]$; then,

$$f\left(\frac{k}{n}\right)=f\left(x\right)+f^{\prime }\left(x\right)sin\left(\frac{k}{n}-x\right)+2f^{\prime \prime }\left(x\right){sin}^2\left(\frac{\frac{k}{n}-x}{2}\right)$$

$$+{\int }_x^{\frac{k}{n}}\left[\left(f^{\prime \prime }\left(t\right)+f\left(t\right)\right)-\left(f^{\prime \prime }\left(x\right)+f\left(x\right)\right)\right]sin\left(\frac{k}{n}-t\right)dt.$$

Furthermore, it holds that

$$f\left(\frac{k}{n}\right){\mathit{A}}_{\alpha }\left(nx-k\right)=f\left(x\right){\mathit{A}}_{\alpha }\left(nx-k\right)$$

$$+f^{\prime }\left(x\right)sin\left(\frac{k}{n}-x\right){\mathit{A}}_{\alpha }\left(nx-k\right)+2f^{\prime \prime }\left(x\right){sin}^2\left(\frac{\frac{k}{n}-x}{2}\right){\mathit{A}}_{\alpha }\left(nx-k\right)$$

$$+{\mathit{A}}_{\alpha }\left(nx-k\right)\left({\int }_x^{\frac{k}{n}}\left[\left(f^{\prime \prime }\left(t\right)+f\left(t\right)\right)-\left(f^{\prime \prime }\left(x\right)+f\left(x\right)\right)\right]sin\left(\frac{k}{n}-t\right)dt\right).$$

So, we have

$$\begin{array}{ccc}& & \sum _{k=⌈na⌉}^{⌊nb⌋}f\left(\frac{k}{n}\right){\mathit{A}}_{\alpha }\left(nx-k\right)-f\left(x\right)\sum _{k=⌈na⌉}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right)\hfill \\ & =& f^{\prime }\left(x\right)\sum _{k=⌈na⌉}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right)sin\left(\frac{k}{n}-x\right)+2f^{\prime \prime }\left(x\right)\sum _{k=⌈na⌉}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right){sin}^2\left(\frac{\frac{k}{n}-x}{2}\right)\hfill \\ & & +\sum _{k=⌈na⌉}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right)\left({\int }_x^{\frac{k}{n}}\left[\left(f^{\prime \prime }\left(t\right)+f\left(t\right)\right)-\left(f^{\prime \prime }\left(x\right)+f\left(x\right)\right)\right]sin\left(\frac{k}{n}-t\right)dt\right).\hfill \end{array}$$

Hence, we gain

$$\begin{array}{ccc}& & {\mathsf{\Theta }}_n^{*}\left(f,x\right)-f\left(x\right)\sum _{k=⌈na⌉}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right)\hfill \\ & =& f^{\prime }\left(x\right){\mathsf{\Theta }}_n^{*}\left(sin\left(·-x\right),x\right)+2f^{\prime \prime }\left(x\right){\mathsf{\Theta }}_n^{*}\left({sin}^2\left(\frac{·-x}{2}\right),x\right)+{\mathsf{\Gamma }}_n\left(x\right),\hfill \end{array}$$

where

$${\mathsf{\Gamma }}_n\left(x\right):=\sum _{k=⌈na⌉}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right)\left({\int }_x^{\frac{k}{n}}\left[\left(f^{\prime \prime }\left(t\right)+f\left(t\right)\right)-\left(f^{\prime \prime }\left(x\right)+f\left(x\right)\right)\right]sin\left(\frac{k}{n}-t\right)dt\right).$$

We assume that

$$\widehat{\mathsf{\Gamma }}\left(n\right):={\int }_x^{\frac{k}{n}}\left[\left(f^{\prime \prime }\left(t\right)+f\left(t\right)\right)-\left(f^{\prime \prime }\left(x\right)+f\left(x\right)\right)\right]sin\left(\frac{k}{n}-t\right)dt.$$

when $n>⌈{\left(b-a\right)}^{-\frac{1}{\lambda }}⌉,$ in other words, when large enough $n\in \mathbb{N}$, $b-a>\frac{1}{n^{\lambda }}$ is assumed. Thus, this yields $\left|\frac{k}{n}-x\right|\le \frac{1}{n^{\lambda }}$ or $\left|\frac{k}{n}-x\right|>\frac{1}{n^{\lambda }}$.

For the case $\left|\frac{k}{n}-x\right|\le \frac{1}{n^{\lambda }}$, the following are obtained:

(i)

If $\frac{k}{n}\ge x$, then

$$\begin{array}{ccc}\hfill \left|\widehat{\mathsf{\Gamma }}\left(n\right)\right|& =& \left|{\int }_x^{\frac{k}{n}}\left[\left(f^{\prime \prime }\left(t\right)+f\left(t\right)\right)-\left(f^{\prime \prime }\left(x\right)+f\left(x\right)\right)\right]sin\left(\frac{k}{n}-t\right)dt\right|\hfill \\ & \le & {\int }_x^{\frac{k}{n}}{\omega }_1\left(f^{\prime \prime }+f,t-x\right)\left|sin\left(\frac{k}{n}-t\right)\right|dt\hfill \end{array}$$

(employing $\left|sinx\right|\le \left|x\right|$, ∀$x\in \mathbb{R}$)

$$\begin{array}{ccc}& \le & {\int }_x^{\frac{k}{n}}{\omega }_1\left(f^{\prime \prime }+f,t-x\right)\left(\frac{k}{n}-t\right)dt\le {\omega }_1\left(f^{\prime \prime }+f,\frac{k}{n}-x\right)\frac{{\left(\frac{k}{n}-x\right)}^2}{2}\hfill \\ & \le & \frac{{\omega }_1\left(f^{\prime \prime }+f,\frac{1}{n^{\lambda }}\right)}{2n^{2\lambda }},\hfill \end{array}$$

namely,

$$\left|\widehat{\mathsf{\Gamma }}\left(n\right)\right|\le \frac{{\omega }_1\left(f^{\prime \prime }+f,\frac{1}{n^{\lambda }}\right)}{2n^{2\lambda }}.$$

(ii)

If $\frac{k}{n}<x$, then

$$\begin{array}{ccc}\hfill \left|\widehat{\mathsf{\Gamma }}\left(n\right)\right|& =& \left|{\int }_x^{\frac{k}{n}}\left[\left(f^{\prime \prime }\left(t\right)+f\left(t\right)\right)-\left(f^{\prime \prime }\left(x\right)+f\left(x\right)\right)\right]sin\left(\frac{k}{n}-t\right)dt\right|\hfill \\ & =& \left|{\int }_{\frac{k}{n}}^x\left[\left(f^{\prime \prime }\left(t\right)+f\left(t\right)\right)-\left(f^{\prime \prime }\left(x\right)+f\left(x\right)\right)\right]sin\left(\frac{k}{n}-t\right)dt\right|\hfill \\ & \le & {\int }_{\frac{k}{n}}^x\left|\left(f^{\prime \prime }\left(t\right)+f\left(t\right)\right)-\left(f^{\prime \prime }\left(x\right)+f\left(x\right)\right)\right|\left|sin\left(\frac{k}{n}-t\right)\right|dt\hfill \\ & \le & {\int }_{\frac{k}{n}}^x{\omega }_1\left(f^{\prime \prime }+f,x-\frac{k}{n}\right)\left(t-\frac{k}{n}\right)dt\hfill \\ & \le & {\omega }_1\left(f^{\prime \prime }+f,x-\frac{k}{n}\right)\frac{{\left(x-\frac{k}{n}\right)}^2}{2}\hfill \\ & \le & \frac{{\omega }_1\left(f^{\prime \prime }+f,\frac{1}{n^{\lambda }}\right)}{2n^{2\lambda }}.\hfill \end{array}$$

Therefore,

$$\left|\widehat{\mathsf{\Gamma }}\left(n\right)\right|\le \frac{{\omega }_1\left(f^{\prime \prime }+f,\frac{1}{n^{\lambda }}\right)}{2n^{2\lambda }}.$$

So, we have proven that when $\left|\frac{k}{n}-x\right|\le \frac{1}{n^{\lambda }}$, it is always true that

$$\left|\widehat{\mathsf{\Gamma }}\left(n\right)\right|\le \frac{{\omega }_1\left(f^{\prime \prime }+f,\frac{1}{n^{\lambda }}\right)}{2n^{2\lambda }}.$$

(a)

Again let $\frac{k}{n}\ge x$:

$$\begin{array}{ccc}\hfill \left|\widehat{\mathsf{\Gamma }}\left(n\right)\right|& =& \left|{\int }_x^{\frac{k}{n}}\left[\left(f^{\prime \prime }\left(t\right)+f\left(t\right)\right)-\left(f^{\prime \prime }\left(x\right)+f\left(x\right)\right)\right]sin\left(\frac{k}{n}-t\right)dt\right|\hfill \\ & \le & {\int }_x^{\frac{k}{n}}\left|\left(f^{\prime \prime }\left(t\right)+f\left(t\right)\right)-\left(f^{\prime \prime }\left(x\right)+f\left(x\right)\right)\right|\left|sin\left(\frac{k}{n}-t\right)\right|dt\hfill \end{array}$$

(employing $\left|sinx\right|\le \left|x\right|$, ∀ $x\in \mathbb{R}$)

$$\begin{array}{ccc}& \le & 2{∥f^{\prime \prime }+f∥}_{\infty }\left({\int }_x^{\frac{k}{n}}\left(\frac{k}{n}-t\right)\right)dt\hfill \\ & =& 2{∥f^{\prime \prime }+f∥}_{\infty }\frac{{\left(\frac{k}{n}-x\right)}^2}{2}\le {∥f^{\prime \prime }+f∥}_{\infty }{\left(b-a\right)}^2.\hfill \end{array}$$

Thus, we obtain

$$\left|\widehat{\mathsf{\Gamma }}\left(n\right)\right|\le {∥f^{\prime \prime }+f∥}_{\infty }{\left(b-a\right)}^2.$$

(b)

One more time, let $\frac{k}{n}<x,$

$$\begin{array}{ccc}\hfill \left|\widehat{\mathsf{\Gamma }}\left(n\right)\right|& =& \left|{\int }_x^{\frac{k}{n}}\left[\left(f^{\prime \prime }\left(t\right)+f\left(t\right)\right)-\left(f^{\prime \prime }\left(x\right)+f\left(x\right)\right)\right]sin\left(\frac{k}{n}-t\right)dt\right|\hfill \\ & =& \left|{\int }_{\frac{k}{n}}^x\left[\left(f^{\prime \prime }\left(x\right)+f\left(x\right)\right)-\left(f^{\prime \prime }\left(t\right)+f\left(t\right)\right)\right]sin\left(\frac{k}{n}-t\right)dt\right|\hfill \\ & \le & {\int }_{\frac{k}{n}}^x\left|\left(f^{\prime \prime }\left(x\right)+f\left(x\right)\right)-\left(f^{\prime \prime }\left(t\right)+f\left(t\right)\right)\right|\left|sin\left(\frac{k}{n}-t\right)\right|dt\hfill \\ & \le & 2{∥f^{\prime \prime }+f∥}_{\infty }{\int }_{\frac{k}{n}}^x\left|sin\left(\frac{k}{n}-t\right)\right|dt\hfill \\ & \le & 2{∥f^{\prime \prime }+f∥}_{\infty }{\int }_{\frac{k}{n}}^x\left(t-\frac{k}{n}\right)dt={∥f^{\prime \prime }+f∥}_{\infty }{\left(x-\frac{k}{n}\right)}^2\hfill \\ & \le & {∥f^{\prime \prime }+f∥}_{\infty }{\left(b-a\right)}^2.\hfill \end{array}$$

So, we gain

$$\left|\widehat{\mathsf{\Gamma }}\left(n\right)\right|\le {∥f^{\prime \prime }+f∥}_{\infty }{\left(b-a\right)}^2.$$

Also, we have

$${\mathsf{\Gamma }}_n\left(x\right)=\sum _{\left\{\begin{array}{c}k=⌈na⌉\\ :\left|\frac{k}{n}-x\right|\le \frac{1}{n^{\lambda }}\end{array}\right.}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right)\widehat{\mathsf{\Gamma }}\left(n\right)$$

$$+\sum _{\left\{\begin{array}{c}k=⌈na⌉\\ :\left|\frac{k}{n}-x\right|>\frac{1}{n^{\lambda }}\end{array}\right.}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right)\widehat{\mathsf{\Gamma }}\left(n\right).$$

Thus, it follows that

$$\left|{\mathsf{\Gamma }}_n\left(x\right)\right|\le \sum _{\left\{\begin{array}{c}k=⌈na⌉\\ :\left|\frac{k}{n}-x\right|\le \frac{1}{n^{\lambda }}\end{array}\right.}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right)\left|\widehat{\mathsf{\Gamma }}\left(n\right)\right|$$

$$+\sum _{\left\{\begin{array}{c}k=⌈na⌉\\ :\left|\frac{k}{n}-x\right|>\frac{1}{n^{\lambda }}\end{array}\right.}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right)\left|\widehat{\mathsf{\Gamma }}\left(n\right)\right|$$

$$\le \left(\sum _{\left\{\begin{array}{c}k=⌈na⌉\\ :\left|\frac{k}{n}-x\right|\le \frac{1}{n^{\lambda }}\end{array}\right.}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right)\right)\frac{{\omega }_1\left(f^{\prime \prime }+f,\frac{1}{n^{\lambda }}\right)}{2n^{2\lambda }}$$

$$+\sum _{\left\{\begin{array}{c}k=⌈na⌉\\ :\left|\frac{k}{n}-x\right|>\frac{1}{n^{\lambda }}\end{array}\right.}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right){∥f^{\prime \prime }+f∥}_{\infty }{\left(b-a\right)}^2\stackrel{(\mathrm{by}(7\left)\right)}{\le }$$

$$\frac{{\omega }_1\left(f^{\prime \prime }+f,\frac{1}{n^{\lambda }}\right)}{2n^{2\lambda }}+{∥f^{\prime \prime }+f∥}_{\infty }{\left(b-a\right)}^2$$

$$\left(\sum _{\left\{\begin{array}{c}k=⌈na⌉\\ :\left|\frac{k}{n}-x\right|>\frac{1}{n^{\lambda }}\end{array}\right.}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right)\right)\stackrel{(\mathrm{by}\mathrm{Theorem}3)}{\le }$$

$$\frac{{\omega }_1\left(f^{\prime \prime }+f,\frac{1}{n^{\lambda }}\right)}{2n^{2\lambda }}+{∥f^{\prime \prime }+f∥}_{\infty }{\left(b-a\right)}^{2}C{e}^{-\gamma n^{\left(1-\lambda \right)}}.$$

As a result, we derive

$$\left|{\mathsf{\Gamma }}_n\left(x\right)\right|\le \frac{{\omega }_1\left(f^{\prime \prime }+f,\frac{1}{n^{\lambda }}\right)}{2n^{2\lambda }}+{∥f^{\prime \prime }+f∥}_{\infty }{\left(b-a\right)}^{2}C{e}^{-\gamma n^{\left(1-\lambda \right)}}.$$

(13)

Again, we apply $\left|sinx\right|\le \left|x\right|$, ∀$x\in \mathbb{R}.$

We obtain

$${\mathsf{\Theta }}_n^{*}\left(sin\left(·-x\right),x\right)=\sum _{k=⌈na⌉}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right)sin\left(\frac{k}{n}-x\right),$$

and

$$\left|{\mathsf{\Theta }}_n^{*}\left(sin\left(·-x\right),x\right)\right|\le \sum _{k=⌈na⌉}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right)\left|sin\left(\frac{k}{n}-x\right)\right|$$

$$=\sum _{\left\{\begin{array}{c}k=⌈na⌉\\ :\left|\frac{k}{n}-x\right|\le \frac{1}{n^{\lambda }}\end{array}\right.}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right)\left|sin\left(\frac{k}{n}-x\right)\right|$$

$$+\sum _{\left\{\begin{array}{c}k=⌈na⌉\\ :\left|\frac{k}{n}-x\right|>\frac{1}{n^{\lambda }}\end{array}\right.}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right)\left|sin\left(\frac{k}{n}-x\right)\right|$$

$$\le \sum _{\left\{\begin{array}{c}k=⌈na⌉\\ :\left|\frac{k}{n}-x\right|\le \frac{1}{n^{\alpha }}\end{array}\right.}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right)\left|\frac{k}{n}-x\right|$$

$$+\sum _{\left\{\begin{array}{c}k=⌈na⌉\\ :\left|\frac{k}{n}-x\right|>\frac{1}{n^{\lambda }}\end{array}\right.}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right)\left|\frac{k}{n}-x\right|$$

$$\le \frac{1}{n^{\lambda }}+\left(b-a\right)\left(\sum _{\left\{\begin{array}{c}k=⌈na⌉\\ :\left|\frac{k}{n}-x\right|>\frac{1}{n^{\lambda }}\end{array}\right.}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right)\right)$$

$$\stackrel{(\mathrm{by}(7\left)\right)}{\le }\frac{1}{n^{\lambda }}+\left(b-a\right)C{e}^{-\gamma n^{\left(1-\lambda \right)}}.$$

We determine that

$$\left|{\mathsf{\Theta }}_n^{*}\left(sin\left(·-x\right),x\right)\right|\le \frac{1}{n^{\lambda }}+\left(b-a\right)C{e}^{-\gamma n^{\left(1-\lambda \right)}}.$$

Then, we calculate

$${\mathsf{\Theta }}_n^{*}\left({sin}^2\left(\frac{·-x}{2}\right),x\right)=\sum _{k=⌈na⌉}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right){sin}^2\left(\frac{\frac{k}{n}-x}{2}\right),$$

using $\left|sinx\right|\le \left|x\right|$, ∀$x\in \mathbb{R}$, and calculate

$${\mathsf{\Theta }}_n^{*}\left({sin}^2\left(\frac{·-x}{2}\right),x\right)=\sum _{k=⌈na⌉}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right){\left|sin\left(\frac{\frac{k}{n}-x}{2}\right)\right|}^2$$

$$\le \frac{1}{4}\sum _{k=⌈na⌉}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right){\left|\frac{k}{n}-x\right|}^2$$

$$=\frac{1}{4}\left[\sum _{\left\{\begin{array}{c}k=⌈na⌉\\ :\left|\frac{k}{n}-x\right|\le \frac{1}{n^{\lambda }}\end{array}\right.}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right){\left|\frac{k}{n}-x\right|}^2\right.$$

$$+\left.\sum _{\left\{\begin{array}{c}k=⌈na⌉\\ :\left|\frac{k}{n}-x\right|>\frac{1}{n^{\lambda }}\end{array}\right.}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right){\left|\frac{k}{n}-x\right|}^2\right]$$

$$\le \frac{1}{4}\left[\frac{1}{n^{2\lambda }}+{\left(b-a\right)}^{2}C{e}^{-\gamma n^{\left(1-\lambda \right)}}\right],$$

i.e.,

$${\mathsf{\Theta }}_n^{*}\left({sin}^2\left(\frac{·-x}{2}\right),x\right)\le \frac{1}{4}\left[\frac{1}{n^{2\lambda }}+{\left(b-a\right)}^{2}C{e}^{-\gamma n^{\left(1-\lambda \right)}}\right].$$

As a consequence, we obtain the following:

(1)

$$\begin{array}{ccc}\hfill \left|{\mathsf{\Theta }}_n\left(f,x\right)-f\left(x\right)\right|& \le & \nabla \left(\alpha \right)\left[\left|f^{\prime }\left(x\right)\right|\left(\frac{1}{n^{\lambda }}+\left(b-a\right)C{e}^{-\gamma n^{\left(1-\lambda \right)}}\right)\right.\hfill \\ & & +\frac{\left|f^{\prime \prime }\left(x\right)\right|}{2}\left(\frac{1}{n^{2\lambda }}+{\left(b-a\right)}^{2}C{e}^{-\gamma n^{\left(1-\lambda \right)}}\right)\hfill \end{array}$$

$$+\left.\left(\frac{{\omega }_1\left(f^{\prime \prime }+f,\frac{1}{n^{\lambda }}\right)}{2n^{2\lambda }}+{∥f^{\prime \prime }+f∥}_{\infty }{\left(b-a\right)}^{2}C{e}^{-\gamma n^{\left(1-\lambda \right)}}\right)\right].$$

(14)

(2)

If $f^{\prime }\left(x\right)=f^{\prime \prime }\left(x\right)=0,$ according to (14), we have

$$\left|{\mathsf{\Theta }}_n\left(f,x\right)-f\left(x\right)\right|\le \nabla \left(\alpha \right)\left[\frac{{\omega }_1\left(f^{\prime \prime }+f,\frac{1}{n^{\lambda }}\right)}{2n^{2\lambda }}+{∥f^{\prime \prime }+f∥}_{\infty }{\left(b-a\right)}^{2}C{e}^{-\gamma n^{\left(1-\lambda \right)}}\right].$$

Here, we keep in mind that $n^{-3\lambda }$ has a high convergence rate.

(3)

In addition, according to (14), we have

$$\begin{array}{ccc}\hfill {∥{\mathsf{\Theta }}_{n}f-f∥}_{\infty }& \le & \nabla \left(\alpha \right)\left[{∥f^{\prime }∥}_{\infty }\left(\frac{1}{n^{\lambda }}+\left(b-a\right)C{e}^{-\gamma n^{\left(1-\lambda \right)}}\right)\right.\hfill \\ & & +\frac{{∥f^{\prime \prime }∥}_{\infty }}{2}\left(\frac{1}{n^{2\lambda }}+{\left(b-a\right)}^{2}C{e}^{-\gamma n^{\left(1-\lambda \right)}}\right)\hfill \\ & & +\left.\left(\frac{{\omega }_1\left(f^{\prime \prime }+f,\frac{1}{n^{\lambda }}\right)}{2n^{2\lambda }}+{∥f^{\prime \prime }+f∥}_{\infty }{\left(b-a\right)}^{2}C{e}^{-\gamma n^{\left(1-\lambda \right)}}\right)\right].\hfill \end{array}$$

We state that convergence is pointwise and uniform such that ${\displaystyle \underset{n\to +\infty }{lim}}{\mathsf{\Theta }}_n\left(f\right)=f$.

We consider that

$${\mathsf{\Theta }}_n\left(f,x\right)-f^{\prime }\left(x\right){\mathsf{\Theta }}_n\left(sin\left(·-x\right),x\right)-2f^{\prime \prime }\left(x\right){\mathsf{\Theta }}_n\left({sin}^2\left(\frac{·-x}{2}\right),x\right)-f\left(x\right)$$

$$=\frac{{\mathsf{\Theta }}_n^{*}\left(f,x\right)}{{\sum }_{k=⌈na⌉}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right)}-f^{\prime }\left(x\right)\frac{{\mathsf{\Theta }}_n^{*}\left(sin\left(·-x\right),x\right)}{{\sum }_{k=⌈na⌉}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right)}-$$

$$2f^{\prime \prime }\left(x\right)\frac{{\mathsf{\Theta }}_n^{*}\left({sin}^2\left(\frac{·-x}{2}\right),x\right)}{{\sum }_{k=⌈na⌉}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right)}-f\left(x\right)\left(\frac{{\sum }_{k=⌈na⌉}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right)}{{\sum }_{k=⌈na⌉}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right)}\right)$$

$$=\frac{1}{{\sum }_{k=⌈na⌉}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right)}\left[{\mathsf{\Theta }}_n^{*}\left(f,x\right)-f^{\prime }\left(x\right){\mathsf{\Theta }}_n^{*}\left(sin\left(·-x\right),x\right)\right.$$

$$-\left.2f^{\prime \prime }\left(x\right){\mathsf{\Theta }}_n^{*}\left({sin}^2\left(\frac{·-x}{2}\right),x\right)-f\left(x\right)\sum _{k=⌈na⌉}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right)\right]$$

$$=\frac{1}{{\sum }_{k=⌈na⌉}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right)}\left({\mathsf{\Gamma }}_n\left(x\right)\right).$$

Lastly, we obtain (∀$x\in \left[a,b\right]$, $n\in \mathbb{N}$):

(4)

$$\begin{array}{ccc}& & \left|{\mathsf{\Theta }}_n\left(f,x\right)-f^{\prime }\left(x\right){\mathsf{\Theta }}_n\left(sin\left(·-x\right),x\right)-2f^{\prime \prime }\left(x\right){\mathsf{\Theta }}_n\left({sin}^2\left(\frac{·-x}{2}\right),x\right)-f\left(x\right)\right|\hfill \\ & & \stackrel{\left(6\right)}{\le }\nabla \left(\alpha \right)\left|{\mathsf{\Gamma }}_n\left(x\right)\right|\hfill \\ & & \stackrel{\left(13\right)}{\le }\nabla \left(\alpha \right)\left[\frac{{\omega }_1\left(f^{\prime \prime }+f,\frac{1}{n^{\lambda }}\right)}{2n^{2\lambda }}+{∥f^{\prime \prime }+f∥}_{\infty }{\left(b-a\right)}^{2}C{e}^{-\gamma n^{\left(1-\lambda \right)}}\right].\hfill \end{array}$$

The theorem is proved. □

We move ahead with a hyperbolic high-order neural network approximation.

Theorem 6.

Let $f\in C^2\left(\left[a,b\right],\mathbb{C}\right)$, $0<\lambda <1$, $n\in \mathbb{N}:n^{1-\lambda }>2$, $x\in \left[a,b\right].$ Then,

(1) 

$$\begin{array}{ccc}\hfill \left|{\mathsf{\Theta }}_n\left(f,x\right)-f\left(x\right)\right|& \le & \nabla \left(\alpha \right)cosh\left(b-a\right)\left[\left|f^{\prime }\left(x\right)\right|\left(\frac{1}{n^{\lambda }}+\left(b-a\right)C{e}^{-\gamma n^{\left(1-\lambda \right)}}\right)\right.\hfill \\ & & \frac{\left|f^{\prime \prime }\left(x\right)\right|}{2}\left(\frac{1}{n^{2\lambda }}+{\left(b-a\right)}^{2}C{e}^{-\gamma n^{\left(1-\lambda \right)}}\right)\hfill \\ & & +\left.\left(\frac{{\omega }_1\left(f^{\prime \prime }-f,\frac{1}{n^{\lambda }}\right)}{2n^{2\lambda }}+{∥f^{\prime \prime }-f∥}_{\infty }{\left(b-a\right)}^{2}C{e}^{-\gamma n^{\left(1-\lambda \right)}}\right)\right],\hfill \end{array}$$

(2) 

If $f^{\prime }\left(x\right)=f^{\prime \prime }\left(x\right)=0,$ we obtain

$$\begin{array}{ccc}\hfill \left|{\mathsf{\Theta }}_n\left(f,x\right)-f\left(x\right)\right|& \le & \nabla \left(\alpha \right)cosh\left(b-a\right)\hfill \\ & & \left[\frac{{\omega }_1\left(f^{\prime \prime }-f,\frac{1}{n^{\lambda }}\right)}{2n^{2\lambda }}+{∥f^{\prime \prime }-f∥}_{\infty }{\left(b-a\right)}^{2}C{e}^{-\gamma n^{\left(1-\lambda \right)}}\right],\hfill \end{array}$$

considering the high rate of convergence at $n^{-3\lambda }$.

(3) 

In addition, we obtain

$$\begin{array}{ccc}\hfill {∥{\mathsf{\Theta }}_{n}f-f∥}_{\infty }& \le & \nabla \left(\alpha \right)cosh\left(b-a\right)\left[{∥f^{\prime }∥}_{\infty }\left(\frac{1}{n^{\lambda }}+\left(b-a\right)C{e}^{-\gamma n^{\left(1-\lambda \right)}}\right)\right.\hfill \\ & & \frac{{∥f^{\prime \prime }∥}_{\infty }}{2}\left(\frac{1}{n^{2\lambda }}+{\left(b-a\right)}^{2}C{e}^{-\gamma n^{\left(1-\lambda \right)}}\right)\hfill \\ & & +\left.\left(\frac{{\omega }_1\left(f^{\prime \prime }-f,\frac{1}{n^{\lambda }}\right)}{2n^{2\lambda }}+{∥f^{\prime \prime }-f∥}_{\infty }{\left(b-a\right)}^{2}C{e}^{-\gamma n^{\left(1-\lambda \right)}}\right)\right],\hfill \end{array}$$

it yields that ${\displaystyle \underset{n\to +\infty }{lim}}{\mathsf{\Theta }}_n\left(f\right)=f$, pointwise and uniformly, and

(4) 

$$\left|{\mathsf{\Theta }}_n\left(f,x\right)-f^{\prime }\left(x\right){\mathsf{\Theta }}_n\left(sinh\left(·-x\right),x\right)-2f^{\prime \prime }\left(x\right){\mathsf{\Theta }}_n\left({sinh}^2\left(\frac{·-x}{2}\right),x\right)-f\left(x\right)\right|$$

$$\le \nabla \left(\alpha \right)cosh\left(b-a\right)\left[\frac{{\omega }_1\left(f^{\prime \prime }-f,\frac{1}{n^{\lambda }}\right)}{2n^{2\lambda }}+{∥f^{\prime \prime }-f∥}_{\infty }{\left(b-a\right)}^{2}C{e}^{-\gamma n^{\left(1-\lambda \right)}}\right],$$

Here, again, $n^{-3\lambda }$ is our high convergence speed.

Proof. 

Using the mean value theorem, we write

$$sinhx=sinhx-sinh0=\left(cosh\xi \right)\left(x-0\right),$$

for some $\xi$ in $\left(0,x\right)$, for any $x\in \mathbb{R}$.

Thus,

$$\left|sinhx\right|\le {∥cosh∥}_{\infty ,\left[-\left(b-a\right),b-a\right]}\left|x\right|,\forall x\in \left[-\left(b-a\right),b-a\right].$$

In other words, there exists $M\ge 1$ such that

$$\left|sinhx\right|\le M\left|x\right|,\forall x\in \left[-\left(b-a\right),b-a\right],$$

(15)

where $M:={∥cosh∥}_{\infty ,\left[-\left(b-a\right),b-a\right]}=cosh\left(b-a\right).$

Inspired by [3,4], and applying the hyperbolic Taylor’s formula for $f\in C^2\left(\left[a,b\right],\mathbb{C}\right)$, when $\frac{k}{n},x\in \left[a,b\right]$, then

$$\begin{array}{ccc}\hfill f\left(\frac{k}{n}\right)& =& f\left(x\right)+f^{\prime }\left(x\right)sinh\left(\frac{k}{n}-x\right)+2f^{\prime \prime }\left(x\right){sinh}^2\left(\frac{\frac{k}{n}-x}{2}\right)\hfill \\ & & +{\int }_x^{\frac{k}{n}}\left[\left(f^{\prime \prime }\left(t\right)-f\left(t\right)\right)-\left(f^{\prime \prime }\left(x\right)-f\left(x\right)\right)\right]sinh\left(\frac{k}{n}-t\right)dt.\hfill \end{array}$$

So, it holds that

$$\begin{array}{ccc}\hfill f\left(\frac{k}{n}\right){\mathit{A}}_{\alpha }\left(nx-k\right)& =& f\left(x\right){\mathit{A}}_{\alpha }\left(nx-k\right)\hfill \\ & & +f^{\prime }\left(x\right)sinh\left(\frac{k}{n}-x\right){\mathit{A}}_{\alpha }\left(nx-k\right)+2f^{\prime \prime }\left(x\right){sinh}^2\left(\frac{\frac{k}{n}-x}{2}\right){\mathit{A}}_{\alpha }\left(nx-k\right)\hfill \\ & & +{\mathit{A}}_{\alpha }\left(nx-k\right)\left({\int }_x^{\frac{k}{n}}\left[\left(f^{\prime \prime }\left(t\right)-f\left(t\right)\right)-\left(f^{\prime \prime }\left(x\right)-f\left(x\right)\right)\right]sinh\left(\frac{k}{n}-t\right)dt\right).\hfill \end{array}$$

Accordingly, we obtain

$$\begin{array}{ccc}& & \sum _{k=⌈na⌉}^{⌊nb⌋}f\left(\frac{k}{n}\right){\mathit{A}}_{\alpha }\left(nx-k\right)-f\left(x\right)\sum _{k=⌈na⌉}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right)\hfill \\ & =& f^{\prime }\left(x\right)\sum _{k=⌈na⌉}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right)sinh\left(\frac{k}{n}-x\right)+2f^{\prime \prime }\left(x\right)\sum _{k=⌈na⌉}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right){sinh}^2\left(\frac{\frac{k}{n}-x}{2}\right)\hfill \\ & & +\sum _{k=⌈na⌉}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right)\left({\int }_x^{\frac{k}{n}}\left[\left(f^{\prime \prime }\left(t\right)-f\left(t\right)\right)-\left(f^{\prime \prime }\left(x\right)-f\left(x\right)\right)\right]sinh\left(\frac{k}{n}-t\right)dt\right).\hfill \end{array}$$

So, we obtain

$${\mathsf{\Theta }}_n^{*}\left(f,x\right)-f\left(x\right)\sum _{k=⌈na⌉}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right)$$

$$=f^{\prime }\left(x\right){\mathsf{\Theta }}_n^{*}\left(sinh\left(·-x\right),x\right)+2f^{\prime \prime }\left(x\right){\mathsf{\Theta }}_n^{*}\left({sinh}^2\left(\frac{·-x}{2}\right),x\right)+{\mathsf{\Gamma }}_n\left(x\right),$$

(16)

where

$${\mathsf{\Gamma }}_n\left(x\right):=\sum _{k=⌈na⌉}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right)\left({\int }_x^{\frac{k}{n}}\left[\left(f^{\prime \prime }\left(t\right)-f\left(t\right)\right)-\left(f^{\prime \prime }\left(x\right)-f\left(x\right)\right)\right]sinh\left(\frac{k}{n}-t\right)dt\right).$$

We assume that

$$\widehat{\mathsf{\Gamma }}\left(n\right):={\int }_x^{\frac{k}{n}}\left[\left(f^{\prime \prime }\left(t\right)-f\left(t\right)\right)-\left(f^{\prime \prime }\left(x\right)-f\left(x\right)\right)\right]sinh\left(\frac{k}{n}-t\right)dt.$$

For large enough $n\in \mathbb{N}$, let $b-a>\frac{1}{n^{\lambda }}$, that is, when $n>⌈{\left(b-a\right)}^{-\frac{1}{\lambda }}⌉$.

Hence, $\left|\frac{k}{n}-x\right|\le \frac{1}{n^{\lambda }}$ or $\left|\frac{k}{n}-x\right|>\frac{1}{n^{\lambda }}$.

For $\left|\frac{k}{n}-x\right|\le \frac{1}{n^{\lambda }}$, we have the following:

(i)

If $\frac{k}{n}\ge x$, then

$$\begin{array}{ccc}\hfill \left|\widehat{\mathsf{\Gamma }}\left(n\right)\right|& =& \left|{\int }_x^{\frac{k}{n}}\left[\left(f^{\prime \prime }\left(t\right)-f\left(t\right)\right)-\left(f^{\prime \prime }\left(x\right)-f\left(x\right)\right)\right]sinh\left(\frac{k}{n}-t\right)dt\right|\hfill \\ & \le & {\int }_x^{\frac{k}{n}}{\omega }_1\left(f^{\prime \prime }-f,t-x\right)\left|sinh\left(\frac{k}{n}-t\right)\right|dt\hfill \end{array}$$

$$\stackrel{\left(15\right)}{\le }{\int }_x^{\frac{k}{n}}{\omega }_1\left(f^{\prime \prime }-f,t-x\right)M\left(\frac{k}{n}-t\right)dt\le M{\omega }_1\left(f^{\prime \prime }-f,\frac{k}{n}-x\right){\int }_x^{\frac{k}{n}}\left(\frac{k}{n}-t\right)dt$$

$$=M{\omega }_1\left(f^{\prime \prime }-f,\frac{k}{n}-x\right)\frac{{\left(\frac{k}{n}-x\right)}^2}{2}\le \frac{M{\omega }_1\left(f^{\prime \prime }-f,\frac{1}{n^{\lambda }}\right)}{2n^{2\lambda }},$$

that is,

$$\left|\widehat{\mathsf{\Gamma }}\left(n\right)\right|\le \frac{M{\omega }_1\left(f^{\prime \prime }-f,\frac{1}{n^{\lambda }}\right)}{2n^{2\lambda }}.$$

(ii)

If $\frac{k}{n}<x$, then

$$\begin{array}{ccc}\hfill \left|\widehat{\mathsf{\Gamma }}\left(n\right)\right|& =& \left|{\int }_x^{\frac{k}{n}}\left[\left(f^{\prime \prime }\left(t\right)-f\left(t\right)\right)-\left(f^{\prime \prime }\left(x\right)-f\left(x\right)\right)\right]sinh\left(\frac{k}{n}-t\right)dt\right|\hfill \\ & =& \left|{\int }_{\frac{k}{n}}^x\left[\left(f^{\prime \prime }\left(t\right)-f\left(t\right)\right)-\left(f^{\prime \prime }\left(x\right)-f\left(x\right)\right)\right]sinh\left(\frac{k}{n}-t\right)dt\right|\hfill \\ & \le & {\int }_{\frac{k}{n}}^x\left|\left(f^{\prime \prime }\left(t\right)-f\left(t\right)\right)-\left(f^{\prime \prime }\left(x\right)-f\left(x\right)\right)\right|\left|sinh\left(\frac{k}{n}-t\right)\right|dt\hfill \\ & \le & M{\omega }_1\left(f^{\prime \prime }-f,x-\frac{k}{n}\right){\int }_{\frac{k}{n}}^x\left(t-\frac{k}{n}\right)dt=M{\omega }_1\left(f^{\prime \prime }-f,x-\frac{k}{n}\right)\frac{{\left(x-\frac{k}{n}\right)}^2}{2}\hfill \\ & \le & \frac{M{\omega }_1\left(f^{\prime \prime }-f,\frac{1}{n^{\lambda }}\right)}{2n^{2\lambda }},\hfill \end{array}$$

i.e.,

$$\left|\widehat{\mathsf{\Gamma }}\left(n\right)\right|\le \frac{M{\omega }_1\left(f^{\prime \prime }-f,\frac{1}{n^{\lambda }}\right)}{2n^{2\lambda }}.$$

So, we have verified that when $\left|\frac{k}{n}-x\right|\le \frac{1}{n^{\alpha }}$, it gives

$$\left|\widehat{\mathsf{\Gamma }}\left(n\right)\right|\le \frac{M{\omega }_1\left(f^{\prime \prime }-f,\frac{1}{n^{\lambda }}\right)}{2n^{2\lambda }}.$$

Now, let us again assume that $\frac{k}{n}\ge x$; then,

$$\begin{array}{ccc}\hfill \left|\widehat{\mathsf{\Gamma }}\left(n\right)\right|& =& \left|{\int }_x^{\frac{k}{n}}\left[\left(f^{\prime \prime }\left(t\right)-f\left(t\right)\right)-\left(f^{\prime \prime }\left(x\right)-f\left(x\right)\right)\right]sinh\left(\frac{k}{n}-t\right)dt\right|\hfill \\ & \le & {\int }_x^{\frac{k}{n}}\left|\left(f^{\prime \prime }\left(t\right)-f\left(t\right)\right)-\left(f^{\prime \prime }\left(x\right)-f\left(x\right)\right)\right|\left|sinh\left(\frac{k}{n}-t\right)\right|dt\hfill \\ & \le & 2M{∥f^{\prime \prime }-f∥}_{\infty }{\int }_x^{\frac{k}{n}}\left(\frac{k}{n}-t\right)dt\hfill \\ & =& 2M{∥f^{\prime \prime }-f∥}_{\infty }\frac{{\left(\frac{k}{n}-x\right)}^2}{2}\le M{∥f^{\prime \prime }-f∥}_{\infty }{\left(b-a\right)}^2.\hfill \end{array}$$

Thus,

$$\left|\widehat{\mathsf{\Gamma }}\left(n\right)\right|\le M{∥f^{\prime \prime }-f∥}_{\infty }{\left(b-a\right)}^2.$$

If $\frac{k}{n}<x$, then

$$\begin{array}{ccc}\hfill \left|\widehat{\mathsf{\Gamma }}\left(n\right)\right|& =& \left|{\int }_{\frac{k}{n}}^x\left[\left(f^{\prime \prime }\left(t\right)-f\left(t\right)\right)-\left(f^{\prime \prime }\left(x\right)-f\left(x\right)\right)\right]sinh\left(\frac{k}{n}-t\right)dt\right|\hfill \\ & \le & {\int }_{\frac{k}{n}}^x\left|\left(f^{\prime \prime }\left(t\right)-f\left(t\right)\right)-\left(f^{\prime \prime }\left(x\right)-f\left(x\right)\right)\right|\left|sinh\left(\frac{k}{n}-t\right)\right|dt\hfill \\ & \le & 2M{∥f^{\prime \prime }-f∥}_{\infty }{\int }_{\frac{k}{n}}^x\left(t-\frac{k}{n}\right)dt\hfill \\ & =& 2M{∥f^{\prime \prime }-f∥}_{\infty }\frac{{\left(x-\frac{k}{n}\right)}^2}{2}\le M{∥f^{\prime \prime }-f∥}_{\infty }{\left(b-a\right)}^2.\hfill \end{array}$$

Hence, it verifies that

$$\left|\widehat{\mathsf{\Gamma }}\left(n\right)\right|\le M{∥f^{\prime \prime }-f∥}_{\infty }{\left(b-a\right)}^2.$$

Also, we have

$${\mathsf{\Gamma }}_n\left(x\right)=\sum _{\left\{\begin{array}{c}k=⌈na⌉\\ :\left|\frac{k}{n}-x\right|\le \frac{1}{n^{\lambda }}\end{array}\right.}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right)\widehat{\mathsf{\Gamma }}\left(n\right)$$

$$+\sum _{\left\{\begin{array}{c}k=⌈na⌉\\ :\left|\frac{k}{n}-x\right|>\frac{1}{n^{\lambda }}\end{array}\right.}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right)\widehat{\mathsf{\Gamma }}\left(n\right).$$

Thus, it yields that

$$\left|{\mathsf{\Gamma }}_n\left(x\right)\right|\le \sum _{\left\{\begin{array}{c}k=⌈na⌉\\ :\left|\frac{k}{n}-x\right|\le \frac{1}{n^{\lambda }}\end{array}\right.}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right)\left|\widehat{\mathsf{\Gamma }}\left(n\right)\right|$$

$$+\sum _{\left\{\begin{array}{c}k=⌈na⌉\\ :\left|\frac{k}{n}-x\right|>\frac{1}{n^{\lambda }}\end{array}\right.}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right)\left|\widehat{\mathsf{\Gamma }}\left(n\right)\right|$$

$$\le \left(\sum _{\left\{\begin{array}{c}k=⌈na⌉\\ :\left|\frac{k}{n}-x\right|\le \frac{1}{n^{\lambda }}\end{array}\right.}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right)\right)\frac{{\omega }_1\left(f^{\prime \prime }-f,\frac{1}{n^{\lambda }}\right)M}{2n^{2\lambda }}$$

$$+\left(\sum _{\left\{\begin{array}{c}k=⌈na⌉\\ :\left|\frac{k}{n}-x\right|>\frac{1}{n^{\lambda }}\end{array}\right.}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right)\right)M{∥f^{\prime \prime }-f∥}_{\infty }{\left(b-a\right)}^2$$

$$\stackrel{(\mathrm{by}\left(5\right))}{\le }\frac{M{\omega }_1\left(f^{\prime \prime }-f,\frac{1}{n^{\lambda }}\right)}{2n^{2\lambda }}+M{∥f^{\prime \prime }-f∥}_{\infty }{\left(b-a\right)}^2$$

$$\left(\sum _{\left\{\begin{array}{c}k=⌈na⌉\\ :\left|\frac{k}{n}-x\right|>\frac{1}{n^{\lambda }}\end{array}\right.}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right)\right)$$

$$\stackrel{(\mathrm{by}\mathrm{Theorem}3)}{\le }M\frac{{\omega }_1\left(f^{\prime \prime }-f,\frac{1}{n^{\lambda }}\right)}{2n^{2\lambda }}+M{∥f^{\prime \prime }-f∥}_{\infty }{\left(b-a\right)}^{2}C{e}^{-\gamma n^{\left(1-\lambda \right)}}.$$

Therefore, we determine that

$$\left|{\mathsf{\Gamma }}_n\left(x\right)\right|\le M\left[\frac{{\omega }_1\left(f^{\prime \prime }-f,\frac{1}{n^{\lambda }}\right)}{2n^{2\lambda }}+{∥f^{\prime \prime }-f∥}_{\infty }{\left(b-a\right)}^{2}C{e}^{-\gamma n^{\left(1-\lambda \right)}}\right].$$

(17)

Also, we have that

$${\mathsf{\Theta }}_n^{*}\left(sinh\left(·-x\right),x\right)=\sum _{k=⌈na⌉}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right)sinh\left(\frac{k}{n}-x\right),$$

and

$$\left|{\mathsf{\Theta }}_n^{*}\left(sinh\left(·-x\right),x\right)\right|\le \sum _{k=⌈na⌉}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right)\left|sinh\left(\frac{k}{n}-x\right)\right|$$

$$=\sum _{\left\{\begin{array}{c}k=⌈na⌉\\ :\left|\frac{k}{n}-x\right|\le \frac{1}{n^{\lambda }}\end{array}\right.}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right)\left|sinh\left(\frac{k}{n}-x\right)\right|$$

$$+\sum _{\left\{\begin{array}{c}k=⌈na⌉\\ :\left|\frac{k}{n}-x\right|>\frac{1}{n^{\lambda }}\end{array}\right.}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right)\left|sinh\left(\frac{k}{n}-x\right)\right|$$

$$\le M\left[\sum _{\left\{\begin{array}{c}k=⌈na⌉\\ :\left|\frac{k}{n}-x\right|\le \frac{1}{n^{\lambda }}\end{array}\right.}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right)\left|\frac{k}{n}-x\right|\right.$$

$$+\left.\sum _{\left\{\begin{array}{c}k=⌈na⌉\\ :\left|\frac{k}{n}-x\right|>\frac{1}{n^{\lambda }}\end{array}\right.}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right)\left|\frac{k}{n}-x\right|\right]$$

$$\le M\left[\frac{1}{n^{\lambda }}+\left(b-a\right)\left(\sum _{\left\{\begin{array}{c}k=⌈na⌉\\ :\left|\frac{k}{n}-x\right|>\frac{1}{n^{\lambda }}\end{array}\right.}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right)\right)\right]\stackrel{(\mathrm{by}\left(5\right))}{\le }$$

$$M\left[\frac{1}{n^{\alpha }}+\left(b-a\right)C{e}^{-\gamma n^{\left(1-\lambda \right)}}\right].$$

We obtain that

$$\left|{\mathsf{\Theta }}_n^{*}\left(sinh\left(·-x\right),x\right)\right|\le M\left[\frac{1}{n^{\lambda }}+\left(b-a\right)C{e}^{-\gamma n^{\left(1-\lambda \right)}}\right].$$

(18)

Then, we calculate

$${\mathsf{\Theta }}_n^{*}\left({sinh}^2\left(\frac{·-x}{2}\right),x\right)=\sum _{k=⌈na⌉}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right){sinh}^2\left(\frac{\frac{k}{n}-x}{2}\right).$$

One determines that

$${\mathsf{\Theta }}_n^{*}\left({sinh}^2\left(\frac{·-x}{2}\right),x\right)=\sum _{k=⌈na⌉}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right){\left(sinh\left(\frac{\frac{k}{n}-x}{2}\right)\right)}^2$$

$$\le \frac{M}{4}\sum _{k=⌈na⌉}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right){\left(\frac{k}{n}-x\right)}^2$$

$$=\frac{M}{4}\left[\sum _{\left\{\begin{array}{c}k=⌈na⌉\\ :\left|\frac{k}{n}-x\right|\le \frac{1}{n^{\lambda }}\end{array}\right.}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right){\left(\frac{k}{n}-x\right)}^2\right.$$

$$+\left.\sum _{\left\{\begin{array}{c}k=⌈na⌉\\ :\left|\frac{k}{n}-x\right|>\frac{1}{n^{\lambda }}\end{array}\right.}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right){\left(\frac{k}{n}-x\right)}^2\right]$$

$$\le \frac{M}{4}\left[\frac{1}{n^{2\lambda }}+{\left(b-a\right)}^{2}C{e}^{-\gamma n^{\left(1-\lambda \right)}}\right].$$

That is,

$${\mathsf{\Theta }}_n^{*}\left({sinh}^2\left(\frac{·-x}{2}\right),x\right)\le \frac{M}{4}\left[\frac{1}{n^{2\lambda }}+{\left(b-a\right)}^{2}C{e}^{-\gamma n^{\left(1-\lambda \right)}}\right].$$

(19)

According to (12), and based on (16)–(19), we acquire the following:

(1)

$$\left|{\mathsf{\Theta }}_n\left(f,x\right)-f\left(x\right)\right|\le \nabla \left(\alpha \right)M\left[\left|f^{\prime }\left(x\right)\right|\left(\frac{1}{n^{\lambda }}+\left(b-a\right)C{e}^{-\gamma n^{\left(1-\lambda \right)}}\right)\right.$$

$$+\frac{\left|f^{\prime \prime }\left(x\right)\right|}{2}\left(\frac{1}{n^{2\lambda }}+{\left(b-a\right)}^{2}C{e}^{-\gamma n^{\left(1-\lambda \right)}}\right)$$

$$+\left.\left(\frac{{\omega }_1\left(f^{\prime \prime }-f,\frac{1}{n^{\lambda }}\right)}{2n^{2\lambda }}+{∥f^{\prime \prime }-f∥}_{\infty }{\left(b-a\right)}^{2}C{e}^{-\gamma n^{\left(1-\lambda \right)}}\right)\right].$$

(20)

(2)

If $f^{\prime }\left(x\right)=f^{\prime \prime }\left(x\right)=0,$ according to (20), we achieve that

$$\left|{\mathsf{\Theta }}_n\left(f,x\right)-f\left(x\right)\right|$$

$$\le \nabla \left(\alpha \right)M\left[\frac{{\omega }_1\left(f^{\prime \prime }-f,\frac{1}{n^{\lambda }}\right)}{2n^{2\lambda }}+{∥f^{\prime \prime }-f∥}_{\infty }{\left(b-a\right)}^{2}C{e}^{-\gamma n^{\left(1-\lambda \right)}}\right],$$

with the high rate of convergence at $n^{-3\lambda }.$

(3)

Moreover, according to (20), we gain

$$\begin{array}{ccc}\hfill {∥{\mathsf{\Theta }}_{n}f-f∥}_{\infty }& \le & \nabla \left(\alpha \right)M\left[{∥f^{\prime }∥}_{\infty }\left(\frac{1}{n^{\lambda }}+\left(b-a\right)C{e}^{-\gamma n^{\left(1-\lambda \right)}}\right)\right.\hfill \\ & & +\frac{{∥f^{\prime \prime }∥}_{\infty }}{2}\left(\frac{1}{n^{2\lambda }}+{\left(b-a\right)}^{2}C{e}^{-\gamma n^{\left(1-\lambda \right)}}\right)\hfill \\ & & +\left.\left(\frac{{\omega }_1\left(f^{\prime \prime }-f,\frac{1}{n^{\lambda }}\right)}{2n^{2\lambda }}+{∥f^{\prime \prime }-f∥}_{\infty }{\left(b-a\right)}^{2}C{e}^{-\gamma n^{\left(1-\lambda \right)}}\right)\right].\hfill \end{array}$$

It yields a pointwise and uniform convergence such that ${\displaystyle \underset{n\to +\infty }{lim}}{\mathsf{\Theta }}_n\left(f\right)=f$.

We note that

$${\mathsf{\Theta }}_n\left(f,x\right)-f^{\prime }\left(x\right){\mathsf{\Theta }}_n\left(sinh\left(·-x\right),x\right)-2f^{\prime \prime }\left(x\right){\mathsf{\Theta }}_n\left({sinh}^2\left(\frac{·-x}{2}\right),x\right)-f\left(x\right)$$

$$=\frac{{\mathsf{\Theta }}_n^{*}\left(f,x\right)}{{\sum }_{k=⌈na⌉}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right)}-f^{\prime }\left(x\right)\frac{{\mathsf{\Theta }}_n^{*}\left(sinh\left(·-x\right),x\right)}{{\sum }_{k=⌈na⌉}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right)}$$

$$-2f^{\prime \prime }\left(x\right)\frac{{\mathsf{\Theta }}_n^{*}\left({sinh}^2\left(\frac{·-x}{2}\right),x\right)}{{\sum }_{k=⌈na⌉}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right)}-f\left(x\right)\left(\frac{{\sum }_{k=⌈na⌉}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right)}{{\sum }_{k=⌈na⌉}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right)}\right)$$

$$=\frac{1}{{\sum }_{k=⌈na⌉}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right)}\left[{\mathsf{\Theta }}_n^{*}\left(f,x\right)-f^{\prime }\left(x\right){\mathsf{\Theta }}_n^{*}\left(sinh\left(·-x\right),x\right)\right.$$

$$-\left.2f^{\prime \prime }\left(x\right){\mathsf{\Theta }}_n^{*}\left({sinh}^2\left(\frac{·-x}{2}\right),x\right)-f\left(x\right)\sum _{k=⌈na⌉}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right)\right]$$

$$=\frac{1}{{\sum }_{k=⌈na⌉}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right)}\left({\mathsf{\Gamma }}_n\left(x\right)\right).$$

Consequently, we gain (∀$x\in \left[a,b\right]$, $n\in \mathbb{N}$):

(4)

$$\begin{array}{ccc}& & \left|{\mathsf{\Theta }}_n\left(f,x\right)-f^{\prime }\left(x\right){\mathsf{\Theta }}_n\left(sinh\left(·-x\right),x\right)-2f^{\prime \prime }\left(x\right){\mathsf{\Theta }}_n\left({sinh}^2\left(\frac{·-x}{2}\right),x\right)-f\left(x\right)\right|\hfill \\ & & \stackrel{\left(6\right)}{\le }\nabla \left(\alpha \right)\left|{\mathsf{\Gamma }}_n\left(x\right)\right|\hfill \\ & & \stackrel{\left(17\right)}{\le }\nabla \left(\alpha \right)M\left[\frac{{\omega }_1\left(f^{\prime \prime }-f,\frac{1}{n^{\lambda }}\right)}{2n^{2\lambda }}+{∥f^{\prime \prime }-f∥}_{\infty }{\left(b-a\right)}^{2}C{e}^{-\gamma n^{\left(1-\lambda \right)}}\right].\hfill \end{array}$$

The theorem is accomplished. □

Now, we go further with a hybrid-type, i.e., hyperbolic–trigonometric high-order NN approximation.

Theorem 7.

Let $f\in C^4\left(\left[a,b\right],\mathbb{C}\right)$, $0<\lambda <1$, $n\in \mathbb{N}:n^{1-\lambda }>2$, $x\in \left[a,b\right].$ Then,

(1) 

$$\begin{array}{ccc}\hfill & & \left|{\mathsf{\Theta }}_n\left(f,x\right)-f\left(x\right)-\frac{f^{\prime }\left(x\right)}{2}{\mathsf{\Theta }}_n\left(\left(sinh\left(·-x\right)+sin\left(·-x\right)\right),x\right)\right.\hfill \\ & & -\frac{f^{\prime \prime }\left(x\right)}{2}{\mathsf{\Theta }}_n\left(\left(cosh\left(·-x\right)-cos\left(·-x\right)\right),x\right)\hfill \\ & & -\frac{f^{\left(3\right)}\left(x\right)}{2}{\mathsf{\Theta }}_n\left(\left(sinh\left(·-x\right)-sin\left(·-x\right)\right),x\right)\hfill \\ & & \left.-f^{\left(4\right)}\left(x\right){\mathsf{\Theta }}_n\left(\left({sinh}^2\left(\frac{·-x}{2}\right)-{sin}^2\left(\frac{·-x}{2}\right)\right),x\right)\right|\hfill \\ & \le & \frac{\nabla \left(\alpha \right)\left(cosh\left(b-a\right)+1\right)}{2}\hfill \\ & & \left[\frac{{\omega }_1\left(f^{\left(4\right)}-f,\frac{1}{n^{\lambda }}\right)}{2n^{2\lambda }}+{∥f^{\left(4\right)}-f∥}_{\infty }{\left(b-a\right)}^{2}C{e}^{-\gamma n^{\left(1-\lambda \right)}}\right],\hfill \end{array}$$

(2) 

If $f^{\left(i\right)}\left(x\right)=0,$ $i=1,2,3,4,$ we obtain

$$\left|{\mathsf{\Theta }}_n\left(f,x\right)-f\left(x\right)\right|\le \frac{\nabla \left(\alpha \right)\left(cosh\left(b-a\right)+1\right)}{2}$$

$$\left[\frac{{\omega }_1\left(f^{\left(4\right)}-f,\frac{1}{n^{\lambda }}\right)}{2n^{2\lambda }}+{∥f^{\left(4\right)}-f∥}_{\infty }{\left(b-a\right)}^{2}C{e}^{-\gamma n^{\left(1-\lambda \right)}}\right],$$

(21)

and in (21), $n^{-3\lambda }$ appears as the highest speed.

Proof. 

Inspired by [4], we employ the hyperbolic–trigonometric Taylor’s formula for $f\in C^4\left(\left[a,b\right],\mathbb{C}\right)$:

When $\frac{k}{n},x\in \left[a,b\right]$, then

$$\begin{array}{ccc}& & f\left(\frac{k}{n}\right)-f\left(x\right)-f^{\prime }\left(x\right)\left(\frac{sinh\left(\frac{k}{n}-x\right)+sin\left(\frac{k}{n}-x\right)}{2}\right)\hfill \\ & & -f^{\prime \prime }\left(x\right)\left(\frac{cosh\left(\frac{k}{n}-x\right)-cos\left(\frac{k}{n}-x\right)}{2}\right)\hfill \\ & & -f^{\left(3\right)}\left(x\right)\left(\frac{sinh\left(\frac{k}{n}-x\right)-sin\left(\frac{k}{n}-x\right)}{2}\right)\hfill \\ & & -f^{\left(4\right)}\left(x\right)\left({sinh}^2\left(\frac{\frac{k}{n}-x}{2}\right)-{sin}^2\left(\frac{\frac{k}{n}-x}{2}\right)\right)\hfill \end{array}$$

$$\begin{array}{ccc}& =& {\int }_x^{\frac{k}{n}}\left[\left(f^{\left(4\right)}\left(t\right)-f\left(t\right)\right)-\left(f^{\left(4\right)}\left(x\right)-f\left(x\right)\right)\right]\left(\frac{sinh\left(\frac{k}{n}-t\right)-sin\left(\frac{k}{n}-t\right)}{2}\right)dt\hfill \\ & =& :\mathsf{\Phi }\left(\frac{k}{n},x\right).\hfill \end{array}$$

From Theorems 5 and 6, we determine that

$$\begin{array}{ccc}& & {\mathsf{\Theta }}_n^{*}\left(f,x\right)-f\left(x\right)\sum _{k=⌈na⌉}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right)\hfill \\ & & -\frac{f^{\prime }\left(x\right)}{2}{\mathsf{\Theta }}_n^{*}\left(\left(sinh\left(·-x\right)+sin\left(·-x\right)\right),x\right)\hfill \\ & & -\frac{f^{\prime \prime }\left(x\right)}{2}{\mathsf{\Theta }}_n^{*}\left(\left(cosh\left(·-x\right)-cos\left(·-x\right)\right),x\right)\hfill \\ & & -\frac{f^{\left(3\right)}\left(x\right)}{2}{\mathsf{\Theta }}_n^{*}\left(\left(sinh\left(·-x\right)-sin\left(·-x\right)\right),x\right)\hfill \end{array}$$

$$-\frac{f^{\left(4\right)}\left(x\right)}{2}{\mathsf{\Theta }}_n^{*}\left(\left({sinh}^2\left(\frac{·-x}{2}\right)-{sin}^2\left(\frac{·-x}{2}\right)\right),x\right)={\widehat{\mathsf{\Phi }}}_n\left(x\right),$$

where

$${\widehat{\mathsf{\Phi }}}_n\left(x\right):=\sum _{k=⌈na⌉}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right){\mathsf{\Phi }}_n\left(\frac{k}{n},x\right).$$

Without loss of generality, let us consider that $n>⌈{\left(b-a\right)}^{-\frac{1}{\lambda }}⌉$.

Hence, $\left|\frac{k}{n}-x\right|\le \frac{1}{n^{\lambda }}$ or $\left|\frac{k}{n}-x\right|>\frac{1}{n^{\lambda }}$.

For $\left|\frac{k}{n}-x\right|\le \frac{1}{n^{\lambda }}$, we gain the following cases:

(i)

If $\frac{k}{n}\ge x$, then

$$\begin{array}{ccc}& & \left|\mathsf{\Phi }\left(\frac{k}{n},x\right)\right|\hfill \\ & =& \left|\frac{1}{2}{\int }_x^{\frac{k}{n}}\left[\left(f^{\left(4\right)}\left(t\right)-f\left(t\right)\right)-\left(f^{\left(4\right)}\left(x\right)-f\left(x\right)\right)\right]\left(sinh\left(\frac{k}{n}-t\right)-sin\left(\frac{k}{n}-t\right)\right)dt\right|\hfill \end{array}$$

$$\begin{array}{ccc}& \le & \frac{1}{2}{\int }_x^{\frac{k}{n}}\left|\left(f^{\left(4\right)}\left(t\right)-f\left(t\right)\right)-\left(f^{\left(4\right)}\left(x\right)-f\left(x\right)\right)\right|\left|sinh\left(\frac{k}{n}-t\right)-sin\left(\frac{k}{n}-t\right)\right|dt\hfill \\ & \le & \frac{1}{2}{\int }_x^{\frac{k}{n}}{\omega }_1\left(f^{\left(4\right)}-f,t-x\right)\left(\left|sinh\left(\frac{k}{n}-t\right)\right|+\left|sin\left(\frac{k}{n}-t\right)\right|\right)dt\hfill \\ & \le & \frac{{\omega }_1\left(f^{\left(4\right)}-f,\frac{k}{n}-x\right)}{2}{\int }_x^{\frac{k}{n}}\left(cosh\left(b-a\right)\left(\frac{k}{n}-t\right)+\left(\frac{k}{n}-t\right)\right)dt\hfill \\ & =& \frac{\left(cosh\left(b-a\right)+1\right){\omega }_1\left(f^{\left(4\right)}-f,\frac{k}{n}-x\right)}{2}{\int }_x^{\frac{k}{n}}\left(\frac{k}{n}-t\right)dt\hfill \\ & =& \frac{\left(cosh\left(b-a\right)+1\right){\omega }_1\left(f^{\left(4\right)}-f,\frac{k}{n}-x\right)}{4}{\left(\frac{k}{n}-x\right)}^2\hfill \\ & \le & \frac{\left(cosh\left(b-a\right)+1\right){\omega }_1\left(f^{\left(4\right)}-f,\frac{1}{n^{\lambda }}\right)}{4n^{2\lambda }}.\hfill \end{array}$$

Namely, if $\frac{k}{n}\ge x$, then

$$\left|\mathsf{\Phi }\left(\frac{k}{n},x\right)\right|\le \frac{\left(cosh\left(b-a\right)+1\right){\omega }_1\left(f^{\left(4\right)}-f,\frac{1}{n^{\lambda }}\right)}{4n^{2\lambda }}.$$

(ii)

When $\frac{k}{n}<x$, then

$$\begin{array}{ccc}& & \left|\mathsf{\Phi }\left(\frac{k}{n},x\right)\right|\hfill \\ & =& \left|\frac{1}{2}{\int }_{\frac{k}{n}}^x\left[\left(f^{\left(4\right)}\left(t\right)-f\left(t\right)\right)-\left(f^{\left(4\right)}\left(x\right)-f\left(x\right)\right)\right]\left(sinh\left(\frac{k}{n}-t\right)-sin\left(\frac{k}{n}-t\right)\right)dt\right|\hfill \\ & \le & \frac{{\omega }_1\left(f^{\left(4\right)}-f,x-\frac{k}{n}\right)}{2}{\int }_{\frac{k}{n}}^x\left(cosh\left(b-a\right)\left(t-\frac{k}{n}\right)+\left(t-\frac{k}{n}\right)\right)dt\hfill \\ & =& \frac{\left(cosh\left(b-a\right)+1\right){\omega }_1\left(f^{\left(4\right)}-f,x-\frac{k}{n}\right)}{2}{\int }_{\frac{k}{n}}^x\left(t-\frac{k}{n}\right)dt\hfill \\ & =& \frac{\left(cosh\left(b-a\right)+1\right){\omega }_1\left(f^{\left(4\right)}-f,x-\frac{k}{n}\right)}{4}{\left(x-\frac{k}{n}\right)}^2\hfill \\ & \le & \frac{\left(cosh\left(b-a\right)+1\right){\omega }_1\left(f^{\left(4\right)}-f,\frac{1}{n^{\lambda }}\right)}{4n^{2\lambda }}.\hfill \end{array}$$

Finally, when $\left|\frac{k}{n}-x\right|\le \frac{1}{n^{\lambda }}$, we always determine that

$$\left|\mathsf{\Phi }\left(\frac{k}{n},x\right)\right|\le \frac{\left(cosh\left(b-a\right)+1\right){\omega }_1\left(f^{\left(4\right)}-f,\frac{1}{n^{\lambda }}\right)}{4n^{2\lambda }}.$$

Again, let $\frac{k}{n}\ge x$; then,

$$\begin{array}{ccc}& & \left|\mathsf{\Phi }\left(\frac{k}{n},x\right)\right|\hfill \\ & \le & \frac{1}{2}{\int }_x^{\frac{k}{n}}\left|\left(f^{\left(4\right)}\left(t\right)-f\left(t\right)\right)-\left(f^{\left(4\right)}\left(x\right)-f\left(x\right)\right)\right|\left|sinh\left(\frac{k}{n}-t\right)-sin\left(\frac{k}{n}-t\right)\right|dt\hfill \\ & \le & {∥f^{\left(4\right)}-f∥}_{\infty }{\int }_x^{\frac{k}{n}}\left[cosh\left(b-a\right)\left(\frac{k}{n}-t\right)+\left(\frac{k}{n}-t\right)\right]dt\hfill \\ & =& {∥f^{\left(4\right)}-f∥}_{\infty }\left(cosh\left(b-a\right)+1\right){\int }_x^{\frac{k}{n}}\left(\frac{k}{n}-t\right)dt\hfill \\ & =& \frac{{∥f^{\left(4\right)}-f∥}_{\infty }\left(cosh\left(b-a\right)+1\right)}{2}{\left(\frac{k}{n}-x\right)}^2\hfill \\ & \le & \frac{{∥f^{\left(4\right)}-f∥}_{\infty }\left(cosh\left(b-a\right)+1\right){\left(b-a\right)}^2}{2}.\hfill \end{array}$$

Thus,

$$\left|\mathsf{\Phi }\left(\frac{k}{n},x\right)\right|\le \frac{{∥f^{\left(4\right)}-f∥}_{\infty }\left(cosh\left(b-a\right)+1\right){\left(b-a\right)}^2}{2}.$$

If we let $\frac{k}{n}<x$, we obtain

$$\begin{array}{ccc}& & \left|\mathsf{\Phi }\left(\frac{k}{n},x\right)\right|\hfill \\ & \le & \frac{1}{2}{\int }_{\frac{k}{n}}^x\left|\left(f^{\left(4\right)}\left(t\right)-f\left(t\right)\right)-\left(f^{\left(4\right)}\left(x\right)-f\left(x\right)\right)\right|\left|sinh\left(\frac{k}{n}-t\right)-sin\left(\frac{k}{n}-t\right)\right|dt\hfill \\ & \le & {∥f^{\left(4\right)}-f∥}_{\infty }{\int }_{\frac{k}{n}}^x\left[cosh\left(b-a\right)\left(t-\frac{k}{n}\right)+\left(t-\frac{k}{n}\right)\right]dt\hfill \\ & =& {∥f^{\left(4\right)}-f∥}_{\infty }\left(cosh\left(b-a\right)+1\right){\int }_{\frac{k}{n}}^x\left(t-\frac{k}{n}\right)dt\hfill \\ & =& \frac{{∥f^{\left(4\right)}-f∥}_{\infty }\left(cosh\left(b-a\right)+1\right)}{2}{\left(x-\frac{k}{n}\right)}^2\hfill \\ & \le & \frac{{∥f^{\left(4\right)}-f∥}_{\infty }\left(cosh\left(b-a\right)+1\right){\left(b-a\right)}^2}{2}.\hfill \end{array}$$

which is why there exists that

$$\left|\mathsf{\Phi }\left(\frac{k}{n},x\right)\right|\le \frac{{∥f^{\left(4\right)}-f∥}_{\infty }\left(cosh\left(b-a\right)+1\right){\left(b-a\right)}^2}{2}.$$

So,

$$\left|{\widehat{\mathsf{\Phi }}}_n\left(x\right)\right|\le \sum _{k=⌈na⌉}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right)\left|\mathsf{\Phi }\left(\frac{k}{n},x\right)\right|$$

$$=\sum _{\left\{\begin{array}{c}k=⌈na⌉\\ :\left|\frac{k}{n}-x\right|\le \frac{1}{n^{\lambda }}\end{array}\right.}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right)\left|\mathsf{\Phi }\left(\frac{k}{n},x\right)\right|$$

$$+\sum _{\left\{\begin{array}{c}k=⌈na⌉\\ :\left|\frac{k}{n}-x\right|>\frac{1}{n^{\lambda }}\end{array}\right.}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right)\left|\mathsf{\Phi }\left(\frac{k}{n},x\right)\right|$$

$$\le \left(\sum _{\left\{\begin{array}{c}k=⌈na⌉\\ :\left|\frac{k}{n}-x\right|\le \frac{1}{n^{\lambda }}\end{array}\right.}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right)\right)\frac{\left(cosh\left(b-a\right)+1\right){\omega }_1\left(f^{\left(4\right)}-f,\frac{1}{n^{\lambda }}\right)}{4n^{2\lambda }}$$

$$+\left(\sum _{\left\{\begin{array}{c}k=⌈na⌉\\ :\left|\frac{k}{n}-x\right|>\frac{1}{n^{\lambda }}\end{array}\right.}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right)\right)$$

$$\left(\frac{{∥f^{\left(4\right)}-f∥}_{\infty }\left(cosh\left(b-a\right)+1\right){\left(b-a\right)}^2}{2}\right)$$

$$\le \frac{\left(cosh\left(b-a\right)+1\right){\omega }_1\left(f^{\left(4\right)}-f,\frac{1}{n^{\lambda }}\right)}{4n^{2\lambda }}$$

(22)

$$+\frac{{∥f^{\left(4\right)}-f∥}_{\infty }\left(cosh\left(b-a\right)+1\right){\left(b-a\right)}^2}{2}C{e}^{-\gamma n^{\left(1-\lambda \right)}}.$$

We try to prove that

$$\left|{\widehat{\mathsf{\Phi }}}_n\left(x\right)\right|\le \frac{\left(cosh\left(b-a\right)+1\right)}{2}$$

$$\left[\frac{{\omega }_1\left(f^{\left(4\right)}-f,\frac{1}{n^{\lambda }}\right)}{2n^{2\lambda }}+{∥f^{\left(4\right)}-f∥}_{\infty }{\left(b-a\right)}^{2}C{e}^{-\gamma n^{\left(1-\lambda \right)}}\right].$$

(23)

We examine that

$$\begin{array}{ccc}& & {\mathsf{\Theta }}_n\left(f,x\right)-f\left(x\right)-\frac{f^{\prime }\left(x\right)}{2}{\mathsf{\Theta }}_n\left(\left(sinh\left(·-x\right)+sin\left(·-x\right)\right),x\right)\hfill \\ & & -\frac{f^{\prime \prime }\left(x\right)}{2}{\mathsf{\Theta }}_n\left(\left(cosh\left(·-x\right)-cos\left(·-x\right)\right),x\right)\hfill \\ & & -\frac{f^{\left(3\right)}\left(x\right)}{2}{\mathsf{\Theta }}_n\left(\left(sinh\left(·-x\right)-sin\left(·-x\right)\right),x\right)\hfill \\ & & -f^{\left(4\right)}\left(x\right){\mathsf{\Theta }}_n\left(\left({sinh}^2\left(\frac{·-x}{2}\right)-{sin}^2\left(\frac{·-x}{2}\right)\right),x\right)\hfill \\ & =& \left[{\mathsf{\Theta }}_n^{*}\left(f,x\right)-f\left(x\right)\sum _{k=⌈na⌉}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right)-\frac{f^{\prime }\left(x\right)}{2}{\mathsf{\Theta }}_n^{*}\left(\left(sinh\left(·-x\right)+sin\left(·-x\right)\right),x\right)\right.\hfill \\ & & -\frac{f^{\prime \prime }\left(x\right)}{2}{\mathsf{\Theta }}_n^{*}\left(\left(cosh\left(·-x\right)-cos\left(·-x\right)\right),x\right)\hfill \\ & & -\frac{f^{\left(3\right)}\left(x\right)}{2}{\mathsf{\Theta }}_n^{*}\left(\left(sinh\left(·-x\right)-sin\left(·-x\right)\right),x\right)\hfill \end{array}$$

$$\begin{array}{ccc}& & \left.-f^{\left(4\right)}\left(x\right){\mathsf{\Theta }}_n^{*}\left(\left({sinh}^2\left(\frac{·-x}{2}\right)-{sin}^2\left(\frac{·-x}{2}\right)\right),x\right)\right]\frac{1}{{\sum }_{k=⌈na⌉}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right)}\hfill \\ & =& \frac{{\widehat{\mathsf{\Phi }}}_n\left(x\right)}{{\sum }_{k=⌈na⌉}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right)}.\hfill \end{array}$$

As a result, we obtain (∀$x\in \left[a,b\right]$, $n\in \mathbb{N}$):

$$\left|{\mathsf{\Theta }}_n\left(f,x\right)-f\left(x\right)-\frac{f^{\prime }\left(x\right)}{2}{\mathsf{\Theta }}_n\left(\left(sinh\left(·-x\right)+sin\left(·-x\right)\right),x\right)\right.$$

$$-\frac{f^{\prime \prime }\left(x\right)}{2}{\mathsf{\Theta }}_n\left(\left(cosh\left(·-x\right)-cos\left(·-x\right)\right),x\right)$$

$$-\frac{f^{\left(3\right)}\left(x\right)}{2}{\mathsf{\Theta }}_n\left(\left(sinh\left(·-x\right)-sin\left(·-x\right)\right),x\right)$$

$$\left.-f^{\left(4\right)}\left(x\right){\mathsf{\Theta }}_n\left(\left({sinh}^2\left(\frac{·-x}{2}\right)-{sin}^2\left(\frac{·-x}{2}\right)\right),x\right)\right|$$

$$=\frac{\left|{\widehat{\mathsf{\Phi }}}_n\left(x\right)\right|}{{\sum }_{k=⌈na⌉}^{⌊nb⌋}{\mathit{A}}_{\alpha }\left(nx-k\right)}\le \nabla \left(\alpha \right)\left|{\widehat{\mathsf{\Phi }}}_n\left(x\right)\right|\stackrel{(\mathrm{by}(23\left)\right)}{\le }\frac{\nabla \left(\alpha \right)\left(cosh\left(b-a\right)+1\right)}{2}$$

$$\left[\frac{{\omega }_1\left(f^{\left(4\right)}-f,\frac{1}{n^{\lambda }}\right)}{2n^{2\lambda }}+{∥f^{\left(4\right)}-f∥}_{\infty }{\left(b-a\right)}^{2}C{e}^{-\gamma n^{\left(1-\lambda \right)}}\right].$$

The theorem is established. □

Now, a general trigonometric result will be considered.

Theorem 8.

Let $f\in C^4\left(\left[a,b\right],\mathbb{C}\right)$, $0<\lambda <1$, $n\in \mathbb{N}:n^{1-\lambda }>2$, $x\in \left[a,b\right],$ and $\tilde{\xi },\overline{\xi }\in \mathbb{R}$ such that $\tilde{\xi }\overline{\xi }\left({\tilde{\xi }}^2-{\overline{\xi }}^2\right)\ne 0.$ Then,

(1) 

$$\begin{array}{ccc}& & \left|{\mathsf{\Theta }}_n\left(f,x\right)-f\left(x\right)-\frac{f^{\prime }\left(x\right)}{\tilde{\xi }\overline{\xi }\left({\overline{\xi }}^2-{\tilde{\xi }}^2\right)}{\mathsf{\Theta }}_n\left(\left({\overline{\xi }}^{3}sin\left(\tilde{\xi }\left(·-x\right)\right)-{\tilde{\xi }}^{3}sin\left(\overline{\xi }\left(·-x\right)\right)\right),x\right)\right.\hfill \\ & & -\frac{f^{\prime \prime }\left(x\right)}{\left({\overline{\xi }}^2-{\tilde{\xi }}^2\right)}{\mathsf{\Theta }}_n\left(\left(cos\left(\tilde{\xi }\left(·-x\right)\right)-cos\left(\overline{\xi }\left(·-x\right)\right)\right),x\right)\hfill \\ & & -\frac{f^{\left(3\right)}\left(x\right)}{\tilde{\xi }\overline{\xi }\left({\overline{\xi }}^2-{\tilde{\xi }}^2\right)}{\mathsf{\Theta }}_n\left(\left(\overline{\xi }sin\left(\tilde{\xi }\left(·-x\right)\right)-\tilde{\xi }sin\left(\overline{\xi }\left(·-x\right)\right)\right),x\right)\hfill \\ & & -\left(\frac{2f^{\left(4\right)}\left(x\right)+\left({\tilde{\xi }}^2+{\overline{\xi }}^2\right)f^{\prime \prime }\left(x\right)}{{\left(\tilde{\xi }\overline{\xi }\right)}^2\left({\overline{\xi }}^2-{\tilde{\xi }}^2\right)}\right)\hfill \\ & & \left.{\mathsf{\Theta }}_n\left(\left({\overline{\xi }}^2{sin}^2\left(\frac{\tilde{\xi }\left(·-x\right)}{2}\right)-{\tilde{\xi }}^2{sin}^2\left(\frac{\overline{\xi }\left(·-x\right)}{2}\right)\right),x\right)\right|\hfill \\ & \le & \frac{\nabla \left(\alpha \right)}{\left|{\overline{\xi }}^2-{\tilde{\xi }}^2\right|}\left[\frac{{\omega }_1\left(\left(f^{\left(4\right)}+\left({\tilde{\xi }}^2+{\overline{\xi }}^2\right)f^{\prime \prime }+{\tilde{\xi }}^2{\overline{\xi }}^{2}f\right),\frac{1}{n^{\lambda }}\right)}{n^{2\lambda }}\right.\hfill \\ & & +\left.2{∥f^{\left(4\right)}+\left({\tilde{\xi }}^2+{\overline{\xi }}^2\right)f^{\prime \prime }+{\tilde{\xi }}^2{\overline{\xi }}^{2}f∥}_{\infty }{\left(b-a\right)}^{2}C{e}^{-\gamma n^{\left(1-\lambda \right)}}\right],\hfill \end{array}$$

(2) 

If $f^{\left(i\right)}\left(x\right)=0,$ $i=1,2,3,4,$ we obtain

$$\begin{array}{ccc}\hfill \left|{\mathsf{\Theta }}_n\left(f,x\right)-f\left(x\right)\right|& \le & \frac{\nabla \left(\alpha \right)}{\left|{\overline{\xi }}^2-{\tilde{\xi }}^2\right|}\hfill \\ & & \left[\frac{{\omega }_1\left(\left(f^{\left(4\right)}+\left({\tilde{\xi }}^2+{\overline{\xi }}^2\right)f^{\prime \prime }+{\tilde{\xi }}^2{\overline{\xi }}^{2}f\right),\frac{1}{n^{\lambda }}\right)}{n^{2\lambda }}\right.\hfill \\ & & +\left.2{∥f^{\left(4\right)}+\left({\tilde{\xi }}^2+{\overline{\xi }}^2\right)f^{\prime \prime }+{\tilde{\xi }}^2{\overline{\xi }}^{2}f∥}_{\infty }{\left(b-a\right)}^{2}C{e}^{-\gamma n^{\left(1-\lambda \right)}}\right].\hfill \end{array}$$

$n^{-3\lambda }$ is the high convergence speed in both (1) and (2).

Proof. 

This proof is inspired by [4] (Chapter 3, Theorem 3.13, pp. 84–89), and also the proof of Theorem 7. □

We finalize with a general hyperbolic result.

Theorem 9.

Let $f\in C^4\left(\left[a,b\right],\mathbb{C}\right)$, $0<\lambda <1$, $n\in \mathbb{N}:n^{1-\lambda }>2$, $x\in \left[a,b\right]$, and let $\tilde{\xi },\overline{\xi }\in \mathbb{R}$ with $\tilde{\xi }\overline{\xi }\left({\tilde{\xi }}^2-{\overline{\xi }}^2\right)\ne 0.$ Then,

(1) 

$$\begin{array}{ccc}& & \left|{\mathsf{\Theta }}_n\left(f,x\right)-f\left(x\right)-\frac{f^{\prime }\left(x\right)}{\tilde{\xi }\overline{\xi }\left({\overline{\xi }}^2-{\tilde{\xi }}^2\right)}{\mathsf{\Theta }}_n\left(\left({\overline{\xi }}^{3}sinh\left(\tilde{\xi }\left(·-x\right)\right)-{\tilde{\xi }}^{3}sinh\left(\overline{\xi }\left(·-x\right)\right)\right),x\right)\right.\hfill \\ & & -\frac{f^{\prime \prime }\left(x\right)}{{\overline{\xi }}^2-{\tilde{\xi }}^2}{\mathsf{\Theta }}_n\left(\left(cosh\left(\overline{\xi }\left(·-x\right)\right)-cosh\left(\tilde{\xi }\left(·-x\right)\right)\right),x\right)\hfill \\ & & -\frac{f^{\prime \prime \prime }\left(x\right)}{\tilde{\xi }\overline{\xi }\left({\overline{\xi }}^2-{\tilde{\xi }}^2\right)}{\mathsf{\Theta }}_n\left(\left(\tilde{\xi }sinh\left(\overline{\xi }\left(·-x\right)\right)-\overline{\xi }sinh\left(\tilde{\xi }\left(·-x\right)\right)\right),x\right)\hfill \\ & & -\left(\frac{2(f^{\left(4\right)}\left(x\right)-\left({\tilde{\xi }}^2+{\overline{\xi }}^2\right)f^{\prime \prime }\left(x\right))}{{\left(\tilde{\xi }\overline{\xi }\right)}^2\left({\overline{\xi }}^2-{\tilde{\xi }}^2\right)}\right)\hfill \\ & & \left.{\mathsf{\Theta }}_n\left(\left({\tilde{\xi }}^2{sinh}^2\left(\frac{\overline{\xi }\left(·-x\right)}{2}\right)-{\overline{\xi }}^2{sinh}^2\left(\frac{\tilde{\xi }\left(·-x\right)}{2}\right)\right),x\right)\right|\hfill \\ & \le & \frac{\nabla \left(\alpha \right)cosh\left(b-a\right)}{\left|{\overline{\xi }}^2-{\tilde{\xi }}^2\right|}\left[\frac{{\omega }_1\left(\left(f^{\left(4\right)}-\left({\tilde{\xi }}^2+{\overline{\xi }}^2\right)f^{\prime \prime }+{\tilde{\xi }}^2{\overline{\xi }}^{2}f\right),\frac{1}{n^{\lambda }}\right)}{n^{2\lambda }}\right.\hfill \\ & & +\left.2{∥f^{\left(4\right)}-\left({\tilde{\xi }}^2+{\overline{\xi }}^2\right)f^{\prime \prime }+{\tilde{\xi }}^2{\overline{\xi }}^{2}f∥}_{\infty }{\left(b-a\right)}^{2}C{e}^{-\gamma n^{\left(1-\lambda \right)}}\right],\hfill \end{array}$$

(2) 

If $f^{\left(i\right)}\left(x\right)=0,$ $i=1,2,3,4,$ we obtain

$$\left|{\mathsf{\Theta }}_n\left(f,x\right)-f\left(x\right)\right|\le \frac{\nabla \left(\alpha \right)cosh\left(b-a\right)}{\left|{\overline{\xi }}^2-{\tilde{\xi }}^2\right|}$$

$$\left[\frac{{\omega }_1\left(\left(f^{\left(4\right)}-\left({\tilde{\xi }}^2+{\overline{\xi }}^2\right)f^{\prime \prime }+{\tilde{\xi }}^2{\overline{\xi }}^{2}f\right),\frac{1}{n^{\lambda }}\right)}{n^{2\lambda }}\right.$$

$$+\left.2{∥f^{\left(4\right)}-\left({\tilde{\xi }}^2+{\overline{\xi }}^2\right)f^{\prime \prime }+{\tilde{\xi }}^2{\overline{\xi }}^{2}f∥}_{\infty }{\left(b-a\right)}^{2}C{e}^{-\gamma n^{\left(1-\lambda \right)}}\right].$$

Again, $n^{-3\lambda }$ is the high convergence speed in both (1) and (2).

Proof. 

This proof is inspired by [4], and also the proof of Theorem 7. □

6. Conclusions

As we have seen, the foundations of neural operators are based on a rich tapestry of branches such as approximation theory and computational analysis. The sophisticated architecture of single-layer neural networks offers effective and faster solutions to many problems in engineering and science. Our study stands out in terms of revealing complex-valued hyperbolic, trigonometric, and hybrid convergence cases with the help of an activation function that is relatively easy to train. Moreover, the types of convergence that reach a promising speed of approximation are our key findings. Finally, we would like to point out that we worked with Ostrowski- and Opial-type inequalities, norms, and also trigonometric- and hyperbolic-type Taylor formulae to reach this convergence rate result.