Asymptotics of Riemann Sums for Uniform Partitions of Intervals of Integration

Abstract

Our goal is to prove the Euler–Maclaurin summation formula using only the Taylor formula. Furthermore, the Euler–Maclaurin summation formula will be considered for the case of functions of the class $C^{2k}\left([a,b]\right)$. A stronger version of the estimation of the rest for functions of class $C^n\left([a,b]\right)$ is given. We will also present the application of the obtained Euler–Maclaurin summation formulas to determine the asymptotics of Riemann sums for uniform partitions of intervals of integration. This paper also introduces the concepts of the asymptotic smoothing of the graph of a given function, as well as the asymptotic uniform distribution of the positive and negative values of the given function (generalizing the concept of the symmetric distribution of these values with respect to the x-axis, as in the case of the Peano–Jordan measure).

1. Introduction

The standard form of the Euler–Maclaurin summation formula (in short, the E-M formula) is proven by the authors of many monographs and papers using Bernoulli numbers and polynomials (see [1,2,3,4,5,6,7,8]). The remainder in this formula is obtained using the integral, which also includes the rescaled Bernoulli polynomial. In this paper, to prove the Euler–Maclaurin summation formula, we use only the Taylor formula with the Lagrange form of the remainder, without Bernoulli numbers and Bernoulli polynomials. We note that, for the sake of clarity of the proof, this is performed only for the first eight cases with respect to the order of differentiation and the continuity of the respective derivatives. Furthermore, the Euler–Maclaurin formula will be considered also for the case of functions of the class $C^{2k}\left([a,b]\right)$. Some stronger version of the estimation of the rest for functions of class $C^n\left([a,b]\right)$ is given. We will also present the application of the obtained Euler–Maclaurin summation formula to determine the asymptotics of Riemann sums of Riemann-integrable functions for uniform partitions of intervals of integration, referred to for brevity as uniform Riemann sums. The Euler–Maclaurin formula will be considered in this work from the perspective of the theory of real functions. We are particularly concerned with the behavior of remainders in the corresponding asymptotic expansion and the reduction of lower-order terms. We will also present a variant of the Euler–Maclaurin formula for multiple integrals. Additionally, we will compare the Euler–Maclaurin formula with the asymptotic expansion for the Bernstein operator. In this paper, we also introduce the concepts of the asymptotic smoothing of the graph of a given function, as well as the asymptotic uniform distribution of the positive and negative values of the given function (generalizing the concept of the symmetric distribution of these values with respect to the x-axis, as in the case of the Peano–Jordan measure). The latter concept is often associated by early-stage mathematics students with the symmetric distribution of the positive and negative values of the given function $f:[0,1]\to \mathbb{R}$ with respect to the x-axis. It is only advanced mathematics students, familiar with measure theory and especially the concept of the Lebesgue measure, who connect the above-mentioned asymptotic concepts with the uniform distribution of the positive and negative values of the given function. This approach offers a mathematically more rigorous perspective on the problem of the asymptotic distribution of the function’s values within its domain of definition.

2. When $\mathit{f}\in {\mathit{C}}^{\mathbf{1}}\left([\mathbf{0},\mathbf{1}]\right)$

Theorem 1.

Let $f\in C^1\left([0,1]\right)$. Then, the following asymptotic formula holds:

$$\underset{0}{\overset{1}{\int }}f\left(x\right)dx-{\displaystyle \frac{1}{n}}\sum _{k=1}^{n-1}f\left({\displaystyle \frac{k}{n}}\right)={\displaystyle \frac{1}{2n}}\left(f\left(0\right)+f\left(1\right)\right)+o\left({\displaystyle \frac{1}{n}}\right).$$

(1)

Proof. 

Based on Lagrange’s mean value theorem [9,10,11], it follows that, for each $k\in \mathbb{N}$, $k\le n$, there is a function ${\mathrm{\Theta }}_{k,n}:\left[{\displaystyle \frac{k-1}{n}},{\displaystyle \frac{k}{n}}\right]\to \left[{\displaystyle \frac{k-1}{n}},{\displaystyle \frac{k}{n}}\right]$, such that $f^{\prime }\left({\mathrm{\Theta }}_{k,n}(·)\right)\left(·-{\displaystyle \frac{k-1}{n}}\right)\in C^1\left(\left[{\displaystyle \frac{k-1}{n}},{\displaystyle \frac{k}{n}}\right]\right)$ and

$$f\left(x\right)=f\left({\displaystyle \frac{k-1}{n}}\right)+f^{\prime }\left({\mathrm{\Theta }}_{k,n}\left(x\right)\right)\left(x-{\displaystyle \frac{k-1}{n}}\right)$$

for all $x\in \left[{\displaystyle \frac{k-1}{n}},{\displaystyle \frac{k}{n}}\right]$. By summing the integrals of functions from the above formula over the intervals $\left[{\displaystyle \frac{k-1}{n}},{\displaystyle \frac{k}{n}}\right]$, $k=1,\dots ,n$, we obtain

$$\begin{array}{cc}\hfill \underset{0}{\overset{1}{\int }}f\left(x\right)dx& =\sum _{k=1}^n\underset{{\displaystyle \frac{k-1}{n}}}{\overset{{\displaystyle \frac{k}{n}}}{\int }}f\left(x\right)dx=\sum _{k=1}^n\underset{{\displaystyle \frac{k-1}{n}}}{\overset{{\displaystyle \frac{k}{n}}}{\int }}\left[f\left({\displaystyle \frac{k-1}{n}}\right)+f^{\prime }\left({\mathrm{\Theta }}_{k,n}\left(x\right)\right)\left(x-{\displaystyle \frac{k-1}{n}}\right)\right]dx\hfill \\ \hfill & =\sum _{k=1}^n{\displaystyle \frac{1}{n}}f\left({\displaystyle \frac{k-1}{n}}\right)+\sum _{k=1}^n\underset{{\displaystyle \frac{k-1}{n}}}{\overset{{\displaystyle \frac{k}{n}}}{\int }}{f}^{\prime }\left({\mathrm{\Theta }}_{k,n}\left(x\right)\right)\left(x-{\displaystyle \frac{k-1}{n}}\right)dx\hfill \end{array}$$

(2)

and

$$\begin{array}{cc}\hfill & \sum _{k=1}^n\underset{{\displaystyle \frac{k-1}{n}}}{\overset{{\displaystyle \frac{k}{n}}}{\int }}{f}^{\prime }\left({\mathrm{\Theta }}_{k,n}\left(x\right)\right)\left(x-{\displaystyle \frac{k-1}{n}}\right)dx\hfill \\ \hfill & =\sum _{k=1}^n\underset{{\displaystyle \frac{k-1}{n}}}{\overset{{\displaystyle \frac{k}{n}}}{\int }}\left[f^{\prime }\left({\mathrm{\Theta }}_{k,n}\left(x\right)\right)-f^{\prime }\left(x\right)\right]\left(x-{\displaystyle \frac{k-1}{n}}\right)dx+\sum _{k=1}^n\underset{{\displaystyle \frac{k-1}{n}}}{\overset{{\displaystyle \frac{k}{n}}}{\int }}{f}^{\prime }\left(x\right)\left(x-{\displaystyle \frac{k-1}{n}}\right)dx.\hfill \end{array}$$

(3)

For each $k\in \{1,2,\dots ,n\}$, we obtain, by integration by parts,

$$\begin{array}{cc}\hfill \underset{{\displaystyle \frac{k-1}{n}}}{\overset{{\displaystyle \frac{k}{n}}}{\int }}{f}^{\prime }\left(x\right)\left(x-{\displaystyle \frac{k-1}{n}}\right)dx& =f\left(x\right)\left(x-{\displaystyle \frac{k-1}{n}}\right){|}_{{\displaystyle \frac{k-1}{n}}}^{{\displaystyle \frac{k}{n}}}-\underset{{\displaystyle \frac{k-1}{n}}}{\overset{{\displaystyle \frac{k}{n}}}{\int }}f\left(x\right)dx\hfill \\ \hfill & ={\displaystyle \frac{1}{n}}f\left({\displaystyle \frac{k}{n}}\right)-\underset{{\displaystyle \frac{k-1}{n}}}{\overset{{\displaystyle \frac{k}{n}}}{\int }}f\left(x\right)dx.\hfill \end{array}$$

(4)

Consequently, after taking into account (3) and (4), we obtain from (2)

$$\begin{array}{cc}\hfill \underset{0}{\overset{1}{\int }}f\left(x\right)dx& =\sum _{k=1}^n{\displaystyle \frac{1}{n}}f\left({\displaystyle \frac{k-1}{n}}\right)+\sum _{k=1}^n\underset{{\displaystyle \frac{k-1}{n}}}{\overset{{\displaystyle \frac{k}{n}}}{\int }}\left[f^{\prime }\left({\mathrm{\Theta }}_{k,n}\left(x\right)\right)-f^{\prime }\left(x\right)\right]\left(x-{\displaystyle \frac{k-1}{n}}\right)dx\hfill \\ \hfill & +\sum _{k=1}^n{\displaystyle \frac{1}{n}}f\left({\displaystyle \frac{k}{n}}\right)-\sum _{k=1}^n\underset{{\displaystyle \frac{k-1}{n}}}{\overset{{\displaystyle \frac{k}{n}}}{\int }}f\left(x\right)dx,\hfill \end{array}$$

which implies, after some algebraic manipulations,

$$2\underset{0}{\overset{1}{\int }}f\left(x\right)dx=2\sum _{k=1}^{n-1}{\displaystyle \frac{1}{n}}f\left({\displaystyle \frac{k}{n}}\right)+{\displaystyle \frac{1}{n}}\left(f\left(0\right)+f\left(1\right)\right)+\phi \left(n\right),$$

where

$$\phi \left(n\right):=\sum _{k=1}^n\underset{{\displaystyle \frac{k-1}{n}}}{\overset{{\displaystyle \frac{k}{n}}}{\int }}\left[f^{\prime }\left({\mathrm{\Theta }}_{k,n}\left(x\right)\right)-f^{\prime }\left(x\right)\right]\left(x-{\displaystyle \frac{k-1}{n}}\right)dx.$$

(5)

From the uniform continuity of the function $f^{\prime }$ in $[0,1]$, it follows that, for a given arbitrarily $\epsilon >0$, there exists $N_{\epsilon }\in \mathbb{N}$ such that, for every $n\in \mathbb{N}$, if $n\ge N_{\epsilon }$, then, for every $k\in \{1,2,\dots ,n\}$ and $x\in \left[{\displaystyle \frac{k-1}{n}},{\displaystyle \frac{k}{n}}\right]$, the following holds:

$$\left|f^{\prime }\left({\mathrm{\Theta }}_{k,n}\left(x\right)\right)-f^{\prime }\left(x\right)\right|\le \epsilon ,$$

which implies

$$\begin{array}{cc}\hfill \left|\phi \right(n\left)\right|& \le \sum _{k=1}^n\underset{{\displaystyle \frac{k-1}{n}}}{\overset{{\displaystyle \frac{k}{n}}}{\int }}\left|f^{\prime }\left({\mathrm{\Theta }}_{k,n}\left(x\right)\right)-f^{\prime }\left(x\right)\right|\left(x-{\displaystyle \frac{k-1}{n}}\right)dx\hfill \\ \hfill & \le \sum _{k=1}^n\epsilon \underset{{\displaystyle \frac{k-1}{n}}}{\overset{{\displaystyle \frac{k}{n}}}{\int }}\left(x-{\displaystyle \frac{k-1}{n}}\right)dx=\sum _{k=1}^n{\displaystyle \frac{\epsilon }{2}}{\left(x-{\displaystyle \frac{k-1}{n}}\right)}^2{|}_{{\displaystyle \frac{k-1}{n}}}^{{\displaystyle \frac{k}{n}}}={\displaystyle \frac{\epsilon }{2n}}.\hfill \end{array}$$

Therefore,

$$\underset{0}{\overset{1}{\int }}f\left(x\right)dx-{\displaystyle \frac{1}{n}}\sum _{k=1}^{n}f\left({\displaystyle \frac{k}{n}}\right)={\displaystyle \frac{1}{2n}}\left(f\left(0\right)-f\left(1\right)\right)+o\left({\displaystyle \frac{1}{n}}\right),$$

and, from this, we conclude (1). □

Corollary 1.

Let $f\in C^1\left([0,1]\right)$. Then, we have

$$\underset{n\to \infty }{lim}\left(n\underset{0}{\overset{1}{\int }}f\left(x\right)dx-\sum _{k=1}^{n-1}f\left({\displaystyle \frac{k}{n}}\right)-{\displaystyle \frac{1}{2}}\left(f\left(0\right)+f\left(1\right)\right)\right)=0.$$

Furthermore,

$$\underset{n\to \infty }{lim}\left(n\underset{0}{\overset{1}{\int }}g\left(x\right)dx-\sum _{k=1}^{n-1}g\left({\displaystyle \frac{k}{n}}\right)\right)=0$$

(6)

whenever $g\left(x\right):=4x(1-x)f\left(x\right)$, $x\in [0,1]$. Thus, the graph of the function $g\left(x\right)$ asymptotically smooths the graph of the function f. The condition (6) is satisfied by any function $g\in C^1\left([0,1]\right)$ such that

$$g\left(0\right)+g\left(1\right)=0.$$

This means—in some sense—an asymptotically uniform distribution of the positive and negative values of the function g.

Corollary 2.

Let $f\in C^m\left([0,1]\right)$, $m\in \mathbb{N}$, $m\ge 2$. Then, for every $n\in \mathbb{N}$, $n\ge 2$, we obtain

$${\displaystyle \frac{1}{n}}\sum _{k=1}^{n-1}{f}^{(m-1)}\left({\displaystyle \frac{k}{n}}\right)=f^{(m-2)}\left(1\right)-f^{(m-2)}\left(0\right)-{\displaystyle \frac{f^{(m-1)}\left(0\right)+f^{(m-1)}\left(1\right)}{2n}}+o\left({\displaystyle \frac{1}{n}}\right).$$

Remark 1.

The relation (1) appears to be original. In textbooks and cited articles, we usually find the remainder with the Landau symbol capital O.

Remark 2.

If the assumption about the function f in Theorem 1 is weakened, and, instead of $f\in C^1\left([0,1]\right)$, it is assumed that f is differentiable in $[0,1]$ and $f^{\prime }$ is a Riemann-integrable function in $[0,1]$ (see [12]), due to the fact that a family of such functions forms an algebra with respect to the point-wise addition and multiplication of functions guarantees the correctness of (4), then we will obtain an estimation

$$\left|\phi \left(n\right)\right|\le 2\underset{x\in [0,1]}{sup}\left\{\right|f^{\prime }\left(x\right)\left|\right\}\sum _{k=1}^n{\displaystyle \frac{1}{2}}{\left(x-{\displaystyle \frac{k-1}{n}}\right)}^2{|}_{{\displaystyle \frac{k-1}{n}}}^{{\displaystyle \frac{k}{n}}}={\displaystyle \frac{1}{n}}\underset{x\in [0,1]}{sup}\left\{\right|f^{\prime }\left(x\right)\left|\right\},$$

that is, in (2), we replace the symbol $o\left({\displaystyle \frac{1}{n}}\right)$ with $O\left({\displaystyle \frac{1}{n}}\right)$, for which

$$n\left|O\left({\displaystyle \frac{1}{n}}\right)\right|\le \underset{x\in [0,1]}{sup}\left\{\right|f^{\prime }\left(x\right)\left|\right\}foranyn\in \mathbb{N}.$$

Remark 3.

Similarly to formula (1), it can be proven that, if $f\in C^1\left({[0,1]}^r\right)$, where $r\in \mathbb{N}$, then the following asymptotic formula for the r-dimensional Riemann integral holds:

$$\begin{array}{cc}\hfill & \underset{{[0,1]}^r}{\int }fd\mathbf{x}-\sum _{k_1=1}^{n_1}\sum _{k_2=1}^{n_2}\dots \sum _{k_r=1}^{n_r}f\left({\displaystyle \frac{k_1}{n_1}},{\displaystyle \frac{k_2}{n_2}},\dots ,{\displaystyle \frac{k_r}{n_r}}\right){\displaystyle \frac{1}{n_1{n}_2\dots n_r}}\hfill \\ \hfill & =O\left({\displaystyle \frac{1}{n_1}}\right)+O\left({\displaystyle \frac{1}{n_2}}\right)+\dots +O\left({\displaystyle \frac{1}{n_r}}\right)=O\left(\sum _{k=1}^r{\displaystyle \frac{1}{n_k}}\right)\hfill \end{array}$$

(7)

as $min\{n_1,n_2,\dots ,n_r\}\to \infty$, where we additionally have

$$n_k\left|O\left({\displaystyle \frac{1}{n_k}}\right)\right|\le {\omega }_k:=max\left\{\left|{\displaystyle \frac{\partial f}{\partial x_k}}\left(\mathbf{x}\right)\right|:\mathbf{x}\in {[0,1]}^r\right\}$$

for all $k\in \{1,2,\dots ,r\}$. We note that the asymptotic formula for a 2-dimensional Riemann integral when $f\in C^3\left({[0,1]}^2\right)$ and $f\in C^4\left({[0,1]}^2\right)$ is presented in Remark 5.

We will only demonstrate the proof of Formula (7) in the case of $r=2$.

Proof. 

So, for $k\in \{1,2,\dots ,n\}$, $l\in \{1,2,\dots ,m\}$, there are functions ${\mathrm{\Theta }}_{k,l}:\left[{\displaystyle \frac{k-1}{n}},{\displaystyle \frac{k}{n}}\right]\to \left[{\displaystyle \frac{k-1}{n}},{\displaystyle \frac{k}{n}}\right]$ and ${\mathsf{\Psi }}_{k,l}:\left[{\displaystyle \frac{l-1}{m}},{\displaystyle \frac{l}{m}}\right]\to \left[{\displaystyle \frac{l-1}{m}},{\displaystyle \frac{l}{m}}\right]$ such that, for all $(x,y)\in \left[{\displaystyle \frac{k-1}{n}},{\displaystyle \frac{k}{n}}\right]\times \left[{\displaystyle \frac{l-1}{m}},{\displaystyle \frac{l}{m}}\right]$,

$$\begin{array}{cc}\hfill f(x,y)& =f\left({\displaystyle \frac{k-1}{n}},{\displaystyle \frac{l-1}{m}}\right)\hfill \\ \hfill & +f_x^{\prime }\left({\mathrm{\Theta }}_{k,l}\left(x\right),{\displaystyle \frac{l-1}{m}}\right)\left(x-{\displaystyle \frac{k-1}{n}}\right)+f_y^{\prime }\left({\displaystyle \frac{k-1}{n}},{\mathsf{\Psi }}_{k,l}\left(y\right)\right)\left(y-{\displaystyle \frac{l-1}{m}}\right)\hfill \end{array}$$

and $\left(f_x^{\prime }\left({\mathrm{\Theta }}_{k,l}(·),{\displaystyle \frac{l-1}{m}}\right)\left(·-{\displaystyle \frac{k-1}{n}}\right)+f_y^{\prime }\left({\displaystyle \frac{k-1}{n}},{\mathsf{\Psi }}_{k,l}(•)\right)\left(•-{\displaystyle \frac{l-1}{m}}\right)\right)\in C\left(\left[{\displaystyle \frac{k-1}{n}},{\displaystyle \frac{k}{n}}\right]\times \left[{\displaystyle \frac{l-1}{m}},{\displaystyle \frac{l}{m}}\right]\right)$ (see [13]). Hence,

$$\begin{array}{cc}\hfill & \underset{\left[{\displaystyle \frac{k-1}{n}},{\displaystyle \frac{k}{n}}\right]\times \left[{\displaystyle \frac{l-1}{m}},{\displaystyle \frac{l}{m}}\right]}{\int \int }f(x,y)dxdy\hfill \\ \hfill & =\underset{\left[{\displaystyle \frac{k-1}{n}},{\displaystyle \frac{k}{n}}\right]\times \left[{\displaystyle \frac{l-1}{m}},{\displaystyle \frac{l}{m}}\right]}{\int \int }\left[f\left({\displaystyle \frac{k-1}{n}},{\displaystyle \frac{l-1}{m}}\right)\right.\hfill \\ \hfill & +f_x^{\prime }\left({\mathrm{\Theta }}_{k,l}\left(x\right),{\displaystyle \frac{l-1}{m}}\right)\left(x-{\displaystyle \frac{k-1}{n}}\right)\left.+f_y^{\prime }\left({\displaystyle \frac{k-1}{n}},{\mathsf{\Psi }}_{k,l}\left(y\right)\right)\left(y-{\displaystyle \frac{l-1}{m}}\right)\right]dxdy\hfill \\ \hfill & =f\left({\displaystyle \frac{k-1}{n}},{\displaystyle \frac{l-1}{m}}\right){\displaystyle \frac{1}{n\times m}}\hfill \\ \hfill & +\underset{\left[{\displaystyle \frac{k-1}{n}},{\displaystyle \frac{k}{n}}\right]\times \left[{\displaystyle \frac{l-1}{m}},{\displaystyle \frac{l}{m}}\right]}{\int \int }\left[f_x^{\prime }\left({\mathrm{\Theta }}_{k,l}\left(x\right),{\displaystyle \frac{l-1}{m}}\right)\left(x-{\displaystyle \frac{k-1}{n}}\right)+f_y^{\prime }\left({\displaystyle \frac{k-1}{n}},{\mathsf{\Psi }}_{k,l}\left(y\right)\right)\left(y-{\displaystyle \frac{l-1}{m}}\right)\right]dxdy,\hfill \end{array}$$

which gives, after summing over $1\le k\le n$ and $1\le l\le m$,

$$\begin{array}{cc}\hfill & \sum _{k=1}^n\sum _{l=1}^m\underset{\left[{\displaystyle \frac{k-1}{n}},{\displaystyle \frac{k}{n}}\right]\times \left[{\displaystyle \frac{l-1}{m}},{\displaystyle \frac{l}{m}}\right]}{\int \int }f(x,y)dxdy=\underset{{[0,1]}^2}{\int \int }f(x,y)dxdy=\sum _{k=1}^n\sum _{l=1}^{m}f\left({\displaystyle \frac{k-1}{n}},{\displaystyle \frac{l-1}{m}}\right){\displaystyle \frac{1}{n\times m}}\hfill \\ & +\sum _{k=1}^n\sum _{l=1}^m\underset{\left[{\displaystyle \frac{k-1}{n}},{\displaystyle \frac{k}{n}}\right]\times \left[{\displaystyle \frac{l-1}{m}},{\displaystyle \frac{l}{m}}\right]}{\int \int }\left[f_x^{\prime }\left({\mathrm{\Theta }}_{k,l}\left(x\right),{\displaystyle \frac{l-1}{m}}\right)\left(x-{\displaystyle \frac{k-1}{n}}\right)+f_y^{\prime }\left({\displaystyle \frac{k-1}{n}},{\mathsf{\Psi }}_{k,l}\left(y\right)\right)\left(y-{\displaystyle \frac{l-1}{m}}\right)\right]dxdy,\hfill \end{array}$$

where

$$\begin{array}{cc}\hfill & \left|\sum _{k=1}^n\sum _{l=1}^m\underset{\left[{\displaystyle \frac{k-1}{n}},{\displaystyle \frac{k}{n}}\right]\times \left[{\displaystyle \frac{l-1}{m}},{\displaystyle \frac{l}{m}}\right]}{\int \int }\left[f_x^{\prime }\left({\mathrm{\Theta }}_{k,l}\left(x\right),{\displaystyle \frac{l-1}{m}}\right)\left(x-{\displaystyle \frac{k-1}{n}}\right)+f_y^{\prime }\left({\displaystyle \frac{k-1}{n}},{\mathsf{\Psi }}_{k,l}\left(y\right)\right)\left(y-{\displaystyle \frac{l-1}{m}}\right)\right]dxdy\right|\hfill \\ \hfill & \le \sum _{k=1}^n\sum _{l=1}^m\underset{{\displaystyle \frac{l-1}{m}}}{\overset{{\displaystyle \frac{l}{m}}}{\int }}dy\underset{{\displaystyle \frac{k-1}{n}}}{\overset{{\displaystyle \frac{k}{n}}}{\int }}{\omega }_x\left(x-{\displaystyle \frac{k-1}{n}}\right)dx+\sum _{k=1}^n\sum _{l=1}^m\underset{{\displaystyle \frac{k-1}{n}}}{\overset{{\displaystyle \frac{k}{n}}}{\int }}dx\underset{{\displaystyle \frac{l-1}{m}}}{\overset{{\displaystyle \frac{l}{m}}}{\int }}{\omega }_y\left(y-{\displaystyle \frac{l-1}{m}}\right)dy\hfill \\ \hfill & =\sum _{k=1}^n\sum _{l=1}^m\left({\displaystyle \frac{{\omega }_x}{2n^{2}m}}+{\displaystyle \frac{{\omega }_y}{2n{m}^2}}\right)={\displaystyle \frac{{\omega }_x}{2n}}+{\displaystyle \frac{{\omega }_y}{2m}}.\hfill \end{array}$$

Here, ${\omega }_x:={\omega }_1$ and ${\omega }_y:={\omega }_2$. □

Remark 4.

It is worth emphasizing that the asymptotic relation with the same form as in Theorem 1 and Remark 3 holds also for Bernstein operators. Indeed, let $f\in C^2\left([0,1]\right)$, and the respective Bernstein operators are

$$B_n(f;x):=\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)x^k{(1-x)}^{n-k}f\left({\displaystyle \frac{k}{n}}\right)$$

for $x\in [0,1]$ and $n\in \mathbb{N}$ (see [14,15,16]). Then, the Voronovskaya Theorem (1932) holds ([15], Theorem 3.1):

$$B_n(f;x)-f\left(x\right)={\displaystyle \frac{x(1-x)}{2n}}{f}^{\prime \prime }\left(x\right)+o\left({\displaystyle \frac{1}{n}}\right)$$

or in the form

$$\underset{n\to \infty }{lim}n\left(B_n(f;x)-f\left(x\right)\right)={\displaystyle \frac{x(1-x)}{2}}{f}^{\prime \prime }\left(x\right)$$

uniformly on $[0,1]$ (which indicates weak asymptotic approximation; of course, this became a stimulus in seeking modifications of Bernstein operators that provide the better asymptotic approximation of higher orders; see, for example, those for the Bernstein–Kantorovich-type operators [17,18,19,20], for the Bernstein–Stancu-type operators [21,22], and for the Lototsky–Bernstein operators [23,24]). Stancu [25,26] generalized this result for two-dimensional Bernstein operators

$$B_n(f;x,y):=\sum _{k=0}^n\sum _{l=0}^{n-k}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{n-k}{l}}\right)f\left({\displaystyle \frac{k}{n}},{\displaystyle \frac{l}{n}}\right)x^k{y}^l{(1-x-y)}^{n-k-l},$$

defined for arbitrary $x,y\in [0,1]$, $x+y\le 1$ and $n\in \mathbb{N}$. In this case, the asymptotic relation holds:

$$B_n(f;x,y)-f(x,y)={\displaystyle \frac{x(1-x)}{2n}}{f}_{xx}^{\prime \prime }(x,y)-{\displaystyle \frac{xy}{n}}{f}_{xy}^{\prime \prime }(x,y)+{\displaystyle \frac{y(1-y)}{2n}}{f}_{yy}^{\prime \prime }(x,y)+o\left({\displaystyle \frac{1}{n}}\right).$$

A variant of (7) for Bernstein polynomials was considered by Z.V. Zaritskaya in [27], deriving the asymptotic formula

$$|B_{n_1{n}_2\dots n_r}(f;x_1,x_2,\dots ,x_r)-f(x_1,x_2,\dots ,x_r)|=O\left(\sum _{k=1}^r{\displaystyle \frac{1}{n_k}}\right)$$

as $min\{n_1,n_2,\dots ,n_r\}\to \infty$, in every ${[a,b]}^r\subset {(0,1)}^r$, where

$$\begin{array}{cc}\hfill & B_{n_1{n}_2\dots n_r}(f;x_1,x_2,\dots ,x_r)=\sum _{k_1=0}^{n_1}\sum _{k_2=0}^{n_2}\dots \sum _{k_r=0}^{n_r}f\left({\displaystyle \frac{k_1}{n_1}},{\displaystyle \frac{k_2}{n_2}},\dots ,{\displaystyle \frac{k_r}{n_r}}\right)\hfill \\ \hfill & \times \left({\displaystyle \genfrac{}{}{0pt}{}{n_1}{k_1}}\right)x_1^{k_1}{(1-x_1)}^{n_1-k_1}\left({\displaystyle \genfrac{}{}{0pt}{}{n_2}{k_2}}\right)x_2^{k_2}{(1-x_2)}^{n_2-k_2}\dots \left({\displaystyle \genfrac{}{}{0pt}{}{n_r}{k_r}}\right)x_r^{k_r}{(1-x_r)}^{n_r-k_r},\hfill \end{array}$$

$f\in C\left({[0,1]}^r\right)$, and $f\in C^{(1,1)}\left({[a,b]}^r\right)$ for every ${[a,b]}^r\subset {(0,1)}^r$, i.e., f is differentiable in ${[a,b]}^r$ and satisfies the Lipschitz condition with constant one in ${[a,b]}^r$.

3. When $\mathit{f}\in {\mathit{C}}^{\mathit{m}}\left([\mathbf{0},\mathbf{1}]\right)$, $\mathit{m}\ge \mathbf{2}$

Theorem 2.

Let f be m-times differentiable in $[0,1]$ and $f^{\left(m\right)}$ be a Riemann-integrable function in $[0,1]$, where $m\in \mathbb{N}$, $m\ge 2$, and let

$$M_m:=\underset{x\in [0,1]}{max}\left\{\right|f^{\left(m\right)}\left(x\right)\left|\right\},$$

and $I_m\left(n\right)$ be a function of $n\in \mathbb{N}$ and index m such that

$$I_m\left(n\right)={\displaystyle \frac{1}{n}}\sum _{k=0}^{n-1}f\left({\displaystyle \frac{k}{n}}\right)-\underset{0}{\overset{1}{\int }}f\left(t\right)dt.$$

Index m could be treated only in terms of its auxiliary role (parameter)—its sole purpose is to point towards the differentiability class currently considered.

(1) 

If $m=2$, then

$$I_2\left(n\right)=-{\displaystyle \frac{f\left(1\right)-f\left(0\right)}{2n}}+O\left({\displaystyle \frac{1}{n^2}}\right),asn\to \infty ,$$

(8)

where

$$n^2\left|O\left({\displaystyle \frac{1}{n^2}}\right)\right|\le {\displaystyle \frac{5}{12}}{M}_2.$$

(9)

(2) 

If $m=3$, then

$$I_3\left(n\right)=-{\displaystyle \frac{f\left(1\right)-f\left(0\right)}{2n}}+{\displaystyle \frac{f^{\prime }\left(1\right)-f^{\prime }\left(0\right)}{12n^2}}+O\left({\displaystyle \frac{1}{n^3}}\right),asn\to \infty ,$$

(10)

where $n^3\left|O\left({\displaystyle \frac{1}{n^3}}\right)\right|\le {\displaystyle \frac{1}{6}}{M}_3$.

(3) 

If $m=4$, then

$$I_4\left(n\right)=-{\displaystyle \frac{f\left(1\right)-f\left(0\right)}{2n}}+{\displaystyle \frac{f^{\prime }\left(1\right)-f^{\prime }\left(0\right)}{12n^2}}+O\left({\displaystyle \frac{1}{n^4}}\right),asn\to \infty ,$$

(11)

where $n^4\left|O\left({\displaystyle \frac{1}{n^4}}\right)\right|\le {\displaystyle \frac{9}{125}}{M}_4$.

If, in cases (1), (2), and (3), we assume more strongly that $f\in C^m\left([0,1]\right)$, then, in Formulas (8), (10), and (11), we can replace the Landau big O symbol with the Landau small o symbol (see, for example, [28,29]).

The proof of this fact proceeds similarly to the proof of Theorem 1, where integrals of the form

$$\underset{{\displaystyle \frac{k-1}{n}}}{\overset{{\displaystyle \frac{k}{n}}}{\int }}{f}^{\left(r\right)}\left({\mathrm{\Theta }}_{k,n}\left(t\right)\right){\left(t-{\displaystyle \frac{k-1}{n}}\right)}^{r}dt,$$

where $r\in \{2,3,4\}$, respectively, are replaced by the sum of two integrals:

$$\underset{{\displaystyle \frac{k-1}{n}}}{\overset{{\displaystyle \frac{k}{n}}}{\int }}\left(f^{\left(r\right)}\left({\mathrm{\Theta }}_{k,n}\left(t\right)\right)-f^{\left(r\right)}\left(t\right)\right){\left(t-{\displaystyle \frac{k-1}{n}}\right)}^{r}dt+\underset{{\displaystyle \frac{k-1}{n}}}{\overset{{\displaystyle \frac{k}{n}}}{\int }}{f}^{\left(r\right)}\left(t\right){\left(t-{\displaystyle \frac{k-1}{n}}\right)}^{r}dt.$$

The estimation of the first integral uses the uniform continuity of the function $f^{\left(r\right)}$ on the interval $[0,1]$ and yields the asymptotic relation $o\left({\displaystyle \frac{1}{n^{r+1}}}\right)$ as $n\to \infty$. The second integral, in turn, is simplified using r-times computation by parts,

$$\begin{array}{cc}\hfill & \underset{{\displaystyle \frac{k-1}{n}}}{\overset{{\displaystyle \frac{k}{n}}}{\int }}{f}^{\left(r\right)}\left(t\right){\left(t-{\displaystyle \frac{k-1}{n}}\right)}^{r}dt=f^{(r-1)}\left(t\right){\left(t-{\displaystyle \frac{k-1}{n}}\right)}^r{|}_{{\displaystyle \frac{k-1}{n}}}^{{\displaystyle \frac{k}{n}}}-r\underset{{\displaystyle \frac{k-1}{n}}}{\overset{{\displaystyle \frac{k}{n}}}{\int }}{f}^{(r-1)}\left(t\right){\left(t-{\displaystyle \frac{k-1}{n}}\right)}^{r-1}dt\hfill \\ \hfill & ={\displaystyle \frac{1}{n^r}}{f}^{(r-1)}\left({\displaystyle \frac{k}{n}}\right)-r\left[{\displaystyle \frac{1}{n^{r-1}}}{f}^{(r-2)}\left({\displaystyle \frac{k}{n}}\right)-(r-1)\underset{{\displaystyle \frac{k-1}{n}}}{\overset{{\displaystyle \frac{k}{n}}}{\int }}{f}^{(r-2)}\left(t\right){\left(t-{\displaystyle \frac{k-1}{n}}\right)}^{r-2}dt\right]\hfill \\ \hfill & ={\displaystyle \frac{1}{n^r}}{f}^{(r-1)}\left({\displaystyle \frac{k}{n}}\right)-{\displaystyle \frac{r}{n^{r-1}}}{f}^{(r-2)}\left({\displaystyle \frac{k}{n}}\right)+{\displaystyle \frac{r(r-1)}{n^{r-2}}}{f}^{(r-3)}\left({\displaystyle \frac{k}{n}}\right)-\dots +{(-1)}^{r}r!\underset{{\displaystyle \frac{k-1}{n}}}{\overset{{\displaystyle \frac{k}{n}}}{\int }}f\left(t\right)dt\hfill \end{array}$$

and the obtained components, after summing over $1\le k\le n$, are added to the existing sums, and an appropriate reduction is performed.

 Proof

.

(1) From Taylor’s formula with the Lagrange form of the remainder [10,11], we find that, for every $k=1,2,\dots ,n$, there exists ${\mathrm{\Theta }}_k\in \left({\displaystyle \frac{k-1}{n}},{\displaystyle \frac{k}{n}}\right)$ such that

$$f\left({\displaystyle \frac{k}{n}}\right)=f\left({\displaystyle \frac{k-1}{n}}\right)+f^{\prime }\left({\displaystyle \frac{k-1}{n}}\right){\displaystyle \frac{1}{n}}+f^{\prime \prime }\left({\mathrm{\Theta }}_k\right){\displaystyle \frac{1}{2n^2}},$$

hence,

$$f^{\prime }\left({\displaystyle \frac{k-1}{n}}\right)=n\left[f\left({\displaystyle \frac{k}{n}}\right)-f\left({\displaystyle \frac{k-1}{n}}\right)\right]-f^{\prime \prime }\left({\mathrm{\Theta }}_k\right){\displaystyle \frac{1}{2n}}.$$

(12)

Moreover, there exists a function ${\phi }_{k,n}\in C\left(\left[{\displaystyle \frac{k-1}{n}},{\displaystyle \frac{k}{n}}\right]\right)$ such that

$$f\left(t\right)=f\left({\displaystyle \frac{k-1}{n}}\right)+f^{\prime }\left({\displaystyle \frac{k-1}{n}}\right)\left(t-{\displaystyle \frac{k-1}{n}}\right)+{\phi }_{k,n}\left(t\right)$$

and

$$\mathsf{|}{\phi }_{k,n}\left(t\right)|\le {\displaystyle \frac{M_2}{2}}{\left(t-{\displaystyle \frac{k-1}{n}}\right)}^2$$

for every $t\in [{\displaystyle \frac{k-1}{n}},{\displaystyle \frac{k}{n}}]$. Therefore, on account of (12), we obtain

$$\begin{array}{cc}\hfill f\left(t\right)& =f\left({\displaystyle \frac{k-1}{n}}\right)+n\left[f\left({\displaystyle \frac{k}{n}}\right)-f\left({\displaystyle \frac{k-1}{n}}\right)\right]\left(t-{\displaystyle \frac{k-1}{n}}\right)\hfill \\ \hfill & -f^{\prime \prime }\left({\mathrm{\Theta }}_k\right){\displaystyle \frac{1}{2n}}\left(t-{\displaystyle \frac{k-1}{n}}\right)+{\phi }_{k,n}\left(t\right)\hfill \end{array}$$

for every $t\in [{\displaystyle \frac{k-1}{n}},{\displaystyle \frac{k}{n}}]$. Integrating the above equality with respect to t over the interval $\left[{\displaystyle \frac{k-1}{n}},{\displaystyle \frac{k}{n}}\right]$, we obtain

$$\begin{array}{cc}\hfill \underset{{\displaystyle \frac{k-1}{n}}}{\overset{{\displaystyle \frac{k}{n}}}{\int }}f\left(t\right)dt& ={\displaystyle \frac{1}{n}}f\left({\displaystyle \frac{k-1}{n}}\right)+n\left[f\left({\displaystyle \frac{k}{n}}\right)-f\left({\displaystyle \frac{k-1}{n}}\right)\right]{\displaystyle \frac{1}{2}}{\left(t-{\displaystyle \frac{k-1}{n}}\right)}^2{|}_{{\displaystyle \frac{k-1}{n}}}^{{\displaystyle \frac{k}{n}}}+{\epsilon }_{k,n}\hfill \\ \hfill & ={\displaystyle \frac{1}{2n}}\left[f\left({\displaystyle \frac{k-1}{n}}\right)+f\left({\displaystyle \frac{k}{n}}\right)\right]+{\epsilon }_{k,n},\hfill \end{array}$$

where

$$\mathsf{|}{\epsilon }_{k,n}|\le {\displaystyle \frac{M_2}{4n^3}}+{\displaystyle \frac{M_2}{6n^3}}={\displaystyle \frac{5M_2}{12n^3}}.$$

(13)

Therefore,

$$\sum _{k=1}^n\underset{{\displaystyle \frac{k-1}{n}}}{\overset{{\displaystyle \frac{k}{n}}}{\int }}f\left(t\right)dt=\underset{0}{\overset{1}{\int }}f\left(t\right)dt={\displaystyle \frac{1}{n}}\sum _{k=1}^n{\displaystyle \frac{f\left(\frac{k-1}{n}\right)+f\left(\frac{k}{n}\right)}{2}}+\sum _{k=1}^n{\epsilon }_{k,n},$$

and, hence,

$$-I_2\left(n\right)={\displaystyle \frac{1}{n}}\sum _{k=1}^n{\displaystyle \frac{f\left(\frac{k-1}{n}\right)+f\left(\frac{k}{n}\right)}{2}}-{\displaystyle \frac{1}{n}}\sum _{k=0}^{n-1}f\left({\displaystyle \frac{k}{n}}\right)+\sum _{k=1}^n{\epsilon }_{k,n}\sim {\displaystyle \frac{f\left(1\right)-f\left(0\right)}{2n}}+O\left({\displaystyle \frac{1}{n^2}}\right),$$

where, from (13), we obtain $n^2\left|O\left({\displaystyle \frac{1}{n^2}}\right)\right|\le {\displaystyle \frac{5}{12}}{M}_2$, which implies (8).

(2) From Taylor’s formula with the Lagrange form of the remainder, it follows that, for every $k=1,2,\dots ,n$, there exist number ${\mathrm{\Theta }}_{k,n}\in \left({\displaystyle \frac{k-1}{n}},{\displaystyle \frac{k}{n}}\right)$ and function ${\mathrm{\Theta }}_{k,n}(·):\left[{\displaystyle \frac{k-1}{n}},{\displaystyle \frac{k}{n}}\right]\to \mathbb{R}$, ${\mathrm{\Theta }}_{k,n}\left(t\right)\in \left[{\displaystyle \frac{k-1}{n}},t\right]$ for every $t\in \left[{\displaystyle \frac{k-1}{n}},{\displaystyle \frac{k}{n}}\right]$ and $f^{\prime \prime \prime }\left({\mathrm{\Theta }}_{k,n}(·)\right)\left(·-{\displaystyle \frac{k-1}{n}}\right)\in C\left(\left[{\displaystyle \frac{k-1}{n}},{\displaystyle \frac{k}{n}}\right]\right)$ such that

$$f\left({\displaystyle \frac{k}{n}}\right)=f\left({\displaystyle \frac{k-1}{n}}\right)+f^{\prime }\left({\displaystyle \frac{k-1}{n}}\right){\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{2}}{f}^{\prime \prime }\left({\displaystyle \frac{k-1}{n}}\right){\displaystyle \frac{1}{n^2}}+{\displaystyle \frac{1}{6}}{f}^{\prime \prime \prime }\left({\mathrm{\Theta }}_{k,n}\right){\displaystyle \frac{1}{n^3}},$$

hence,

$${\displaystyle \frac{1}{2n^2}}{f}^{\prime \prime }\left({\displaystyle \frac{k-1}{n}}\right)=\left[f\left({\displaystyle \frac{k}{n}}\right)-f\left({\displaystyle \frac{k-1}{n}}\right)\right]-f^{\prime }\left({\displaystyle \frac{k-1}{n}}\right){\displaystyle \frac{1}{n}}-{\displaystyle \frac{1}{6n^3}}{f}^{\prime \prime \prime }\left({\mathrm{\Theta }}_{k,n}\right)$$

(14)

and

$$\begin{array}{cc}\hfill \underset{{\displaystyle \frac{k-1}{n}}}{\overset{{\displaystyle \frac{k}{n}}}{\int }}f\left(t\right)dt& =\underset{{\displaystyle \frac{k-1}{n}}}{\overset{{\displaystyle \frac{k}{n}}}{\int }}\left[f\left({\displaystyle \frac{k-1}{n}}\right)+f^{\prime }\left({\displaystyle \frac{k-1}{n}}\right)\left(t-{\displaystyle \frac{k-1}{n}}\right)+{\displaystyle \frac{1}{2}}{f}^{\prime \prime }\left({\displaystyle \frac{k-1}{n}}\right){\left(t-{\displaystyle \frac{k-1}{n}}\right)}^2\right.\hfill \\ \hfill & \left.+{\displaystyle \frac{1}{6}}{f}^{\prime \prime \prime }\left({\mathrm{\Theta }}_{k,n}\left(t\right)\right){\left(t-{\displaystyle \frac{k-1}{n}}\right)}^3\right]dt=f\left({\displaystyle \frac{k-1}{n}}\right){\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{2}}{f}^{\prime }\left({\displaystyle \frac{k-1}{n}}\right){\displaystyle \frac{1}{n^2}}\hfill \\ \hfill & +{\displaystyle \frac{1}{6}}{f}^{\prime \prime }\left({\displaystyle \frac{k-1}{n}}\right){\displaystyle \frac{1}{n^3}}+{\mathsf{\Psi }}_{k,n}\stackrel{\left(14\right)}{=}f\left({\displaystyle \frac{k-1}{n}}\right){\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{2}}{f}^{\prime }\left({\displaystyle \frac{k-1}{n}}\right){\displaystyle \frac{1}{n^2}}\hfill \\ \hfill & +{\displaystyle \frac{1}{3n}}\left[f({\displaystyle \frac{k}{n}}\right)-f\left({\displaystyle \frac{k-1}{n}}\right)-f^{\prime }\left({\displaystyle \frac{k-1}{n}}\right){\displaystyle \frac{1}{n}}-{\displaystyle \frac{1}{6n^3}}{f}^{\prime \prime \prime }\left({\mathrm{\Theta }}_{k,n}\right)]+{\mathsf{\Psi }}_{k,n}\hfill \\ \hfill & =f\left({\displaystyle \frac{k-1}{n}}\right){\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{6n^2}}{f}^{\prime }\left({\displaystyle \frac{k-1}{n}}\right)+{\displaystyle \frac{1}{3n}}\left[f\left({\displaystyle \frac{k}{n}}\right)-f\left({\displaystyle \frac{k-1}{n}}\right)\right]\hfill \\ \hfill & -{\displaystyle \frac{1}{18n^4}}{f}^{\prime \prime \prime }\left({\mathrm{\Theta }}_{k,n}\right)+{\mathsf{\Psi }}_{k,n},\hfill \end{array}$$

where $|{\mathsf{\Psi }}_{k,n}|\le {\displaystyle \frac{M_3}{24n^4}}$. To summarize, we obtain

$$\begin{array}{cc}\hfill -I_3\left(n\right)& =\sum _{k=1}^n\left[\underset{{\displaystyle \frac{k-1}{n}}}{\overset{{\displaystyle \frac{k}{n}}}{\int }}f\left(t\right)dt-f\left({\displaystyle \frac{k-1}{n}}\right){\displaystyle \frac{1}{n}}\right]\hfill \\ \hfill & =\sum _{k=1}^n\left[{\displaystyle \frac{1}{6n^2}}{f}^{\prime }\left({\displaystyle \frac{k-1}{n}}\right)-{\displaystyle \frac{1}{18n^4}}{f}^{\prime \prime \prime }\left({\mathrm{\Theta }}_{k,n}\right)+{\mathsf{\Psi }}_{k,n}\right]+{\displaystyle \frac{1}{3n}}\sum _{k=1}^n\left[f\left({\displaystyle \frac{k}{n}}\right)-f\left({\displaystyle \frac{k-1}{n}}\right)\right]\hfill \\ \hfill & \stackrel{\left(8\right)}{=}{\displaystyle \frac{1}{6n}}\left[\underset{0}{\overset{1}{\int }}{f}^{\prime }\left(t\right)dt-{\displaystyle \frac{f^{\prime }\left(1\right)-f^{\prime }\left(0\right)}{2n}}\right]+{\mathsf{\Phi }}_n+\sum _{k=1}^n\left[{\mathsf{\Psi }}_{k,n}-{\displaystyle \frac{1}{18n^4}}{f}^{\prime \prime \prime }\left({\mathrm{\Theta }}_{k,n}\right)\right]\hfill \\ \hfill & +{\displaystyle \frac{1}{3n}}\left(f\left(1\right)-f\left(0\right)\right)\hfill \\ & ={\displaystyle \frac{1}{2n}}\left(f\left(1\right)-f\left(0\right)\right)-{\displaystyle \frac{1}{12n^2}}\left(f^{\prime }\left(1\right)-f^{\prime }\left(0\right)\right)+{\mathsf{\Phi }}_n+\sum _{k=1}^n\left[{\mathsf{\Psi }}_{k,n}-{\displaystyle \frac{1}{18n^4}}{f}^{\prime \prime \prime }\left({\mathrm{\Theta }}_{k,n}\right)\right],\hfill \end{array}$$

where, by (9), we have $|{\mathsf{\Phi }}_n|\le {\displaystyle \frac{5}{72n^3}}{M}_3$. Finally, we obtain the estimate

$$\begin{array}{cc}\hfill \left|{\mathsf{\Phi }}_n+\sum _{k=1}^n\left[{\mathsf{\Psi }}_{k,n}-{\displaystyle \frac{1}{18n^4}}{f}^{\prime \prime \prime }\left({\mathrm{\Theta }}_{k,n}\right)\right]\right|& \le |{\mathsf{\Phi }}_n|+n\times max\{|{\mathsf{\Psi }}_{k,n}|:k=1,\dots ,n\}\hfill \\ \hfill & +n\times {\displaystyle \frac{1}{18n^4}}\times max\left\{\right|f^{\prime \prime \prime }\left({\theta }_{k,n}\right)|:k=1,\dots ,n\}\hfill \\ \hfill & \le {\displaystyle \frac{5}{72n^3}}{M}_3+{\displaystyle \frac{1}{24n^3}}{M}_3+{\displaystyle \frac{1}{18n^3}}{M}_3={\displaystyle \frac{1}{6n^3}}{M}_3,\hfill \end{array}$$

which implies relation (10).

(3) Similarly to the above, from Taylor’s formula with the Lagrange form of the remainder, it follows that, for every $k=1,2,\dots ,n$, there exist ${\mathrm{\Theta }}_{k,n}\in \left({\displaystyle \frac{k-1}{n}},{\displaystyle \frac{k}{n}}\right)$ and function ${\mathrm{\Theta }}_{k,n}(·):\left[{\displaystyle \frac{k-1}{n}},{\displaystyle \frac{k}{n}}\right]\to \mathbb{R}$, ${\mathrm{\Theta }}_{k,n}\left(t\right)\in \left[{\displaystyle \frac{k-1}{n}},t\right]$ for any $t\in \left[{\displaystyle \frac{k-1}{n}},{\displaystyle \frac{k}{n}}\right]$ and $f^{IV}\left({\mathrm{\Theta }}_{k,n}(·)\right)\left(·-{\displaystyle \frac{k-1}{n}}\right)\in C\left(\left[{\displaystyle \frac{k-1}{n}},{\displaystyle \frac{k}{n}}\right]\right)$ such that

$$\begin{array}{cc}\hfill {\displaystyle \frac{1}{6n^3}}{f}^{\prime \prime \prime }\left({\displaystyle \frac{k-1}{n}}\right)& =\left[f\left({\displaystyle \frac{k}{n}}\right)-f\left({\displaystyle \frac{k-1}{n}}\right)\right]-f^{\prime }\left({\displaystyle \frac{k-1}{n}}\right){\displaystyle \frac{1}{n}}\hfill \\ \hfill & -f^{\prime \prime }\left({\displaystyle \frac{k-1}{n}}\right){\displaystyle \frac{1}{2n^2}}-{\displaystyle \frac{1}{24}}{f}^{IV}\left({\mathrm{\Theta }}_{k,n}\right){\displaystyle \frac{1}{n^4}}\hfill \end{array}$$

(15)

and

$$\begin{array}{cc}\hfill \underset{{\displaystyle \frac{k-1}{n}}}{\overset{{\displaystyle \frac{k}{n}}}{\int }}f\left(t\right)dt& =\underset{{\displaystyle \frac{k-1}{n}}}{\overset{{\displaystyle \frac{k}{n}}}{\int }}\left[f\left({\displaystyle \frac{k-1}{n}}\right)+f^{\prime }\left({\displaystyle \frac{k-1}{n}}\right)\left(t-{\displaystyle \frac{k-1}{n}}\right)+{\displaystyle \frac{1}{2}}{f}^{\prime \prime }\left({\displaystyle \frac{k-1}{n}}\right){\left(t-{\displaystyle \frac{k-1}{n}}\right)}^2\right.\hfill \\ \hfill & \left.+{\displaystyle \frac{1}{6}}{f}^{\prime \prime \prime }\left({\displaystyle \frac{k-1}{n}}\right){\left(t-{\displaystyle \frac{k-1}{n}}\right)}^3+{\displaystyle \frac{1}{24}}{f}^{IV}\left({\mathrm{\Theta }}_{k,n}\left(t\right)\right){\left(t-{\displaystyle \frac{k-1}{n}}\right)}^4\right]dt\hfill \\ & ={\displaystyle \frac{1}{n}}f\left({\displaystyle \frac{k-1}{n}}\right)+{\displaystyle \frac{1}{2n^2}}{f}^{\prime }\left({\displaystyle \frac{k-1}{n}}\right)+{\displaystyle \frac{1}{6n^3}}{f}^{\prime \prime }\left({\displaystyle \frac{k-1}{n}}\right)+{\displaystyle \frac{1}{24n^4}}{f}^{\prime \prime \prime }\left({\displaystyle \frac{k-1}{n}}\right)+{\mathsf{\Psi }}_{k,n}\hfill \\ & \left(\mathrm{where}{n}^5\left|{\mathsf{\Psi }}_{k,n}\right|\le {\displaystyle \frac{1}{125}}{M}_4\right)\hfill \\ \hfill & \stackrel{\left(15\right)}{=}{\displaystyle \frac{1}{n}}f\left({\displaystyle \frac{k-1}{n}}\right)+{\displaystyle \frac{1}{2n^2}}{f}^{\prime }\left({\displaystyle \frac{k-1}{n}}\right)+{\displaystyle \frac{1}{6n^3}}{f}^{\prime \prime }\left({\displaystyle \frac{k-1}{n}}\right)+{\displaystyle \frac{1}{4n}}\left\{\left[f\left({\displaystyle \frac{k}{n}}\right)-f\left({\displaystyle \frac{k-1}{n}}\right)\right]\right.\hfill \\ \hfill & \left.-f^{\prime }\left({\displaystyle \frac{k-1}{n}}\right){\displaystyle \frac{1}{n}}-f^{\prime \prime }\left({\displaystyle \frac{k-1}{n}}\right){\displaystyle \frac{1}{2n^2}}-{\displaystyle \frac{1}{24}}{f}^{IV}\left({\mathrm{\Theta }}_{k,n}\right){\displaystyle \frac{1}{n^4}}\right\}+{\mathsf{\Psi }}_{k,n}\hfill \\ \hfill & ={\displaystyle \frac{1}{n}}f\left({\displaystyle \frac{k-1}{n}}\right)+{\displaystyle \frac{1}{4n}}\left[f\left({\displaystyle \frac{k}{n}}\right)-f\left({\displaystyle \frac{k-1}{n}}\right)\right]+{\displaystyle \frac{1}{4n^2}}{f}^{\prime }\left({\displaystyle \frac{k-1}{n}}\right)+{\displaystyle \frac{1}{24n^3}}{f}^{\prime \prime }\left({\displaystyle \frac{k-1}{n}}\right)\hfill \\ \hfill & +{\mathsf{\Psi }}_{k,n}-{\displaystyle \frac{1}{96n^5}}{f}^{IV}\left({\mathrm{\Theta }}_{k,n}\right)\hfill \end{array}$$

where $|{\mathsf{\Psi }}_{k,n}|\le {\displaystyle \frac{M_4}{120n^5}}$. Based on this relation, we obtain

$$\begin{array}{cc}\hfill -I_4\left(n\right)& =\sum _{k=1}^n\left[\underset{{\displaystyle \frac{k-1}{n}}}{\overset{{\displaystyle \frac{k}{n}}}{\int }}f\left(t\right)dt-{\displaystyle \frac{1}{n}}f\left({\displaystyle \frac{k-1}{n}}\right)\right]={\displaystyle \frac{1}{4n}}(f\left(1\right)-f\left(0\right))+{\displaystyle \frac{1}{4n^2}}\sum _{k=1}^n{f}^{\prime }\left({\displaystyle \frac{k-1}{n}}\right)\hfill \\ \hfill & +{\displaystyle \frac{1}{24n^3}}\sum _{k=1}^n{f}^{\prime \prime }\left({\displaystyle \frac{k-1}{n}}\right)+\sum _{k=1}^n{\mathsf{\Psi }}_{k,n}-{\displaystyle \frac{1}{96n^5}}\sum _{k=1}^n{f}^{IV}\left({\mathrm{\Theta }}_{k,n}\right)\hfill \\ \hfill & \stackrel{\left(10\right)}{=}{\displaystyle \frac{1}{4n}}\left(f\left(1\right)-f\left(0\right)\right)\hfill \\ \hfill & +{\displaystyle \frac{1}{4n}}\left[\underset{0}{\overset{1}{\int }}{f}^{\prime }\left(t\right)dt-{\displaystyle \frac{f^{\prime }\left(1\right)-f^{\prime }\left(0\right)}{2n}}+{\displaystyle \frac{f^{\prime \prime }\left(1\right)-f^{\prime \prime }\left(0\right)}{12n^2}}+{\mathsf{\Omega }}_2\left({\displaystyle \frac{1}{n^3}}\right)\right]\hfill \\ \hfill & +{\displaystyle \frac{1}{24n^3}}\sum _{k=1}^n{f}^{\prime \prime }\left({\displaystyle \frac{k-1}{n}}\right)+{\mathsf{\Phi }}_n\hfill \\ \hfill & \left(\mathrm{where}{n}^3\left|{\mathsf{\Omega }}_2\left({\displaystyle \frac{1}{n^3}}\right)\right|\le {\displaystyle \frac{1}{6}}{M}_4\mathrm{and}{\mathsf{\Phi }}_n:=\sum _{k=1}^n{\mathsf{\Psi }}_{k,n}-{\displaystyle \frac{1}{96n^5}}\sum _{k=1}^n{f}^{IV}\left({\mathrm{\Theta }}_{k,n}\right)\right)\hfill \\ \hfill & \stackrel{\left(8\right)}{=}{\displaystyle \frac{1}{2n}}\left(f\left(1\right)-f\left(0\right)\right)-{\displaystyle \frac{f^{\prime }\left(1\right)-f^{\prime }\left(0\right)}{8n^2}}+{\displaystyle \frac{1}{4n}}{\mathsf{\Omega }}_2\left({\displaystyle \frac{1}{n^3}}\right)\hfill \\ \hfill & +{\mathsf{\Phi }}_n+{\displaystyle \frac{f^{\prime \prime }\left(1\right)-f^{\prime \prime }\left(0\right)}{48n^3}}+{\displaystyle \frac{1}{24n^2}}\left[\underset{0}{\overset{1}{\int }}{f}^{\prime \prime }\left(t\right)dt-{\displaystyle \frac{f^{\prime \prime }\left(1\right)-f^{\prime \prime }\left(0\right)}{2n}}+{\mathsf{\Omega }}_1\left({\displaystyle \frac{1}{n^2}}\right)\right]\hfill \\ \hfill & \left(\mathrm{where}{n}^2\left|{\mathsf{\Omega }}_1\left({\displaystyle \frac{1}{n^2}}\right)\right|\le {\displaystyle \frac{5}{12}}{M}_4\right)\hfill \\ \hfill & ={\displaystyle \frac{1}{2n}}\left(f\left(1\right)-f\left(0\right)\right)-{\displaystyle \frac{1}{12n^2}}\left(f^{\prime }\left(1\right)-f^{\prime }\left(0\right)\right)+{\displaystyle \frac{1}{4n}}{\mathsf{\Omega }}_2\left({\displaystyle \frac{1}{n^3}}\right)+{\mathsf{\Phi }}_n+{\displaystyle \frac{1}{24n^2}}{\mathsf{\Omega }}_1\left({\displaystyle \frac{1}{n^2}}\right),\hfill \end{array}$$

hence,

$$\left|{\displaystyle \frac{1}{4n}}{\mathsf{\Omega }}_2\left({\displaystyle \frac{1}{n^3}}\right)+{\mathsf{\Phi }}_n+{\displaystyle \frac{1}{24n^2}}{\mathsf{\Omega }}_1\left({\displaystyle \frac{1}{n^2}}\right)\right|\le {\displaystyle \frac{M_4}{125n^4}}+{\displaystyle \frac{M_4}{24n^4}}+{\displaystyle \frac{5M_4}{288n^4}}<{\displaystyle \frac{9}{125}}{M}_4.$$

□

Corollary 3.

If $f\in C^2[0,1]$, then

$$\underset{n\to \infty }{lim}n\left[n\underset{0}{\overset{1}{\int }}f\left(t\right)dt-\sum _{k=1}^{n-1}f\left({\displaystyle \frac{k}{n}}\right)-{\displaystyle \frac{f\left(1\right)+f\left(0\right)}{2}}\right]=0.$$

If $f\in C^3[0,1]$, then

$$\underset{n\to \infty }{lim}n\left(n\left[n\underset{0}{\overset{1}{\int }}f\left(t\right)dt-\sum _{k=1}^{n-1}f\left({\displaystyle \frac{k}{n}}\right)-{\displaystyle \frac{f\left(0\right)+f\left(1\right)}{2}}\right]+{\displaystyle \frac{f^{\prime }\left(0\right)-f^{\prime }\left(1\right)}{12}}\right)=0.$$

If $f\in C^4[0,1]$, then

$$\underset{n\to \infty }{lim}{n}^2\left(n\left[n\underset{0}{\overset{1}{\int }}f\left(t\right)dt-\sum _{k=1}^{n-1}f\left({\displaystyle \frac{k}{n}}\right)-{\displaystyle \frac{f\left(0\right)+f\left(1\right)}{2}}\right]+{\displaystyle \frac{f^{\prime }\left(0\right)-f^{\prime }\left(1\right)}{12}}\right)=0.$$

Moreover, if we define (the condition of the asymptotic smoothing of the graph of the given function f)

$$g\left(x\right):=4x(1-x)f\left(x\right),$$

then,

(1) 

${\displaystyle \underset{n\to \infty }{lim}n\left[n\underset{0}{\overset{1}{\int }}g\left(x\right)dx-\sum _{k=1}^{n-1}g\left({\displaystyle \frac{k}{n}}\right)\right]=0}$ whenever $f\in C^2\left([0,1]\right)$,

(2) 

${\displaystyle \underset{n\to \infty }{lim}n\left(n\left[n\underset{0}{\overset{1}{\int }}g\left(x\right)dx-\sum _{k=1}^{n-1}g\left({\displaystyle \frac{k}{n}}\right)\right]+{\displaystyle \frac{f\left(0\right)-f\left(1\right)}{3}}\right)=0}$ whenever $f\in C^3\left([0,1]\right)$,

(3) 

${\displaystyle \underset{n\to \infty }{lim}{n}^2\left(n\left[n\underset{0}{\overset{1}{\int }}g\left(x\right)dx-\sum _{k=1}^{n-1}g\left({\displaystyle \frac{k}{n}}\right)\right]+{\displaystyle \frac{f\left(0\right)-f\left(1\right)}{3}}\right)=0}$ whenever $f\in C^4\left([0,1]\right)$.

In particular, when $f\left(0\right)=f\left(1\right)$, then conditions (2) and (3) reduce to the form

$$\underset{n\to \infty }{lim}{n}^2\left[n\underset{0}{\overset{1}{\int }}g\left(x\right)dx-\sum _{k=1}^{n-1}g\left({\displaystyle \frac{k}{n}}\right)\right]=0$$

whenever $f\in C^3\left([0,1]\right)$ and

$$\underset{n\to \infty }{lim}{n}^3\left[n\underset{0}{\overset{1}{\int }}g\left(x\right)dx-\sum _{k=1}^{n-1}g\left({\displaystyle \frac{k}{n}}\right)\right]=0$$

whenever $f\in C^4\left([0,1]\right)$.

Corollary 4.

Let $f\in C^3\left([0,1]\right)$. Then, for any $n,\rho ,r\in \mathbb{N}$, $\rho \ne r$, we obtain the relation

$$\begin{array}{cc}\hfill {\displaystyle \frac{1}{\rho n}}\sum _{k=0}^{\rho n-1}f\left({\displaystyle \frac{k}{\rho n}}\right)-{\displaystyle \frac{1}{rn}}\sum _{\tau =0}^{rn-1}f\left({\displaystyle \frac{\tau }{rn}}\right)& =-{\displaystyle \frac{r-\rho }{\rho rn}}\times {\displaystyle \frac{f\left(1\right)-f\left(0\right)}{2}}+{\displaystyle \frac{r^2-{\rho }^2}{{\rho }^2{r}^2{n}^2}}\times {\displaystyle \frac{f^{\prime }\left(1\right)-f^{\prime }\left(0\right)}{12}}\hfill \\ \hfill & +\left\{\begin{array}{cc}O\left({\displaystyle \frac{1}{n^3}}\right),\hfill & iff\in C^3\left([0,1]\right),\hfill \\ O\left({\displaystyle \frac{1}{n^4}}\right),\hfill & iff\in C^4\left([0,1]\right),\hfill \end{array}\right.\hfill \end{array}$$

as $n\to \infty$.

Corollary 5.

Let $f\in C^m\left([0,1]\right)$, $m\in \mathbb{N}$, $m\ge 3$. Then, we have

$$\begin{array}{cc}\hfill {\displaystyle \frac{1}{n}}\sum _{k=1}^{n-1}{f}^{(m-3)}\left({\displaystyle \frac{k}{n}}\right)& =f^{(m-2)}\left(1\right)-f^{(m-2)}\left(0\right)-{\displaystyle \frac{f^{(m-3)}\left(0\right)+f^{(m-3)}\left(1\right)}{2n}}\hfill \\ \hfill & +{\displaystyle \frac{f^{(m-2)}\left(1\right)-f^{(m-2)}\left(0\right)}{12n^2}}+o\left({\displaystyle \frac{1}{n^3}}\right)as\mathbb{N}\ni n\to \infty .\hfill \end{array}$$

If $m\ge 4$, then we have the asymptotic relation

$$\begin{array}{cc}\hfill {\displaystyle \frac{1}{n}}\sum _{k=1}^{n-1}{f}^{(m-3)}\left({\displaystyle \frac{k}{n}}\right)& =f^{(m-2)}\left(1\right)-f^{(m-2)}\left(0\right)-{\displaystyle \frac{f^{(m-3)}\left(0\right)+f^{(m-3)}\left(1\right)}{2n}}\hfill \\ \hfill & +{\displaystyle \frac{f^{(m-2)}\left(1\right)-f^{(m-2)}\left(0\right)}{12n^2}}+o\left({\displaystyle \frac{1}{n^4}}\right)as\mathbb{N}\ni n\to \infty .\hfill \end{array}$$

Example 1.

Let $N,x\in \mathbb{R}$, $N>0$, $1+{\displaystyle \frac{x}{N}}>0$. Then, we have

$$\begin{array}{cc}\hfill & \underset{0}{\overset{1}{\int }}{\left(1+{\displaystyle \frac{x}{t+N}}\right)}^{t+N}dt-{\displaystyle \frac{1}{n}}\sum _{k=1}^{n-1}{\left(1+{\displaystyle \frac{nx}{k+nN}}\right)}^{{\displaystyle \frac{k+nN}{n}}}={\displaystyle \frac{1}{2n}}\left({\left(1+{\displaystyle \frac{x}{N}}\right)}^N+{\left(1+{\displaystyle \frac{x}{N+1}}\right)}^{N+1}\right)\hfill \\ \hfill & -{\displaystyle \frac{1}{12n^2}}\left({\left(1+{\displaystyle \frac{x}{N+1}}\right)}^{N+1}\left[ln\left(1+{\displaystyle \frac{x}{N+1}}\right)-{\displaystyle \frac{x}{x+N+1}}\right]-{\left(1+{\displaystyle \frac{x}{N}}\right)}^N\left[ln\left(1+{\displaystyle \frac{x}{N}}\right)-{\displaystyle \frac{x}{x+N}}\right]\right)\hfill \\ \hfill & +o\left({\displaystyle \frac{1}{n^4}}\right),\hfill \end{array}$$

since ${\left(1+{\displaystyle \frac{x}{•+N}}\right)}^{•+N}\in C^{\infty }\left([0,1]\right)$. Hence, it follows that

$$\underset{N\to \infty }{lim}\left[\underset{n\to \infty }{lim}n\left(\underset{0}{\overset{1}{\int }}{\left(1+{\displaystyle \frac{x}{t+N}}\right)}^{t+N}dt-{\displaystyle \frac{1}{n}}\sum _{k=1}^{n-1}{\left(1+{\displaystyle \frac{nx}{k+nN}}\right)}^{{\displaystyle \frac{k+nN}{n}}}\right)\right]=e^x,$$

and, if $x>0$, then, for sufficiently large $n\in \mathbb{N}$ (see [30]),

$${\left(1+{\displaystyle \frac{x}{N}}\right)}^N\le n\left(\underset{0}{\overset{1}{\int }}{\left(1+{\displaystyle \frac{x}{t+N}}\right)}^{t+N}dt-{\displaystyle \frac{1}{n}}\sum _{k=1}^{n-1}{\left(1+{\displaystyle \frac{nx}{k+nN}}\right)}^{{\displaystyle \frac{k+nN}{n}}}\right)\le {\left(1+{\displaystyle \frac{x}{N+1}}\right)}^{N+1}.$$

Remark 5.

If $f\in C^3\left({[0,1]}^2\right)$, then the following asymptotic formula for a 2-dimensional Riemann integral holds

$$\begin{array}{cc}\hfill & \underset{{[0,1]}^2}{\int }fd\mathbf{x}-\sum _{k=1}^n\sum _{l=1}^{m}f\left({\displaystyle \frac{k-1}{n}},{\displaystyle \frac{l-1}{m}}\right){\displaystyle \frac{1}{n\times m}}\hfill \\ \hfill & -{\displaystyle \frac{1}{2}}\sum _{k=1}^n\sum _{l=1}^m\left[f_x^{\prime }\left({\displaystyle \frac{k-1}{n}},{\displaystyle \frac{l-1}{m}}\right){\displaystyle \frac{1}{n^2\times m}}+f_y^{\prime }\left({\displaystyle \frac{k-1}{n}},{\displaystyle \frac{l-1}{m}}\right){\displaystyle \frac{1}{n\times m^2}}\right]\hfill \\ \hfill & =o\left(1\right)\left({\displaystyle \frac{1}{n^3}}+{\displaystyle \frac{1}{m^3}}\right)\hfill \end{array}$$

(16)

for all $n,m\in \mathbb{N}$, as $min\{n,m\}\to \infty$, and we suppose that there exists $r>0$ such that $max\{n,m\}\le r\times {(min\{n,m\})}^2$. On the other hand, if $f\in C^4\left({[0,1]}^2\right)$, then the following asymptotic formula holds

$$\begin{array}{cc}\hfill & \underset{{[0,1]}^2}{\int }fd\mathbf{x}-\sum _{k=1}^n\sum _{l=1}^{m}f\left({\displaystyle \frac{k-1}{n}},{\displaystyle \frac{l-1}{m}}\right){\displaystyle \frac{1}{n\times m}}\hfill \\ \hfill & -{\displaystyle \frac{1}{2}}\sum _{k=1}^n\sum _{l=1}^m\left[f_x^{\prime }\left({\displaystyle \frac{k-1}{n}},{\displaystyle \frac{l-1}{m}}\right){\displaystyle \frac{1}{n^2\times m}}+f_y^{\prime }\left({\displaystyle \frac{k-1}{n}},{\displaystyle \frac{l-1}{m}}\right){\displaystyle \frac{1}{n\times m^2}}\right]\hfill \\ \hfill & -\sum _{k=1}^n\sum _{l=1}^m\left[{\displaystyle \frac{1}{6}}{f}_{xx}^{\prime \prime }\left({\displaystyle \frac{k-1}{n}},{\displaystyle \frac{l-1}{m}}\right){\displaystyle \frac{1}{n^3\times m}}+{\displaystyle \frac{1}{6}}{f}_{yy}^{\prime \prime }\left({\displaystyle \frac{k-1}{n}},{\displaystyle \frac{l-1}{m}}\right){\displaystyle \frac{1}{n\times m^3}}\right.\hfill \\ \hfill & \left.+{\displaystyle \frac{1}{4}}{f}_{xy}^{\prime \prime }\left({\displaystyle \frac{k-1}{n}},{\displaystyle \frac{l-1}{m}}\right){\displaystyle \frac{1}{n^2\times m^2}}\right]=o\left(1\right)\left({\displaystyle \frac{1}{n^4}}+{\displaystyle \frac{1}{m^4}}\right)\hfill \end{array}$$

for all $n,m\in \mathbb{N}$, as $min\{n,m\}\to \infty$, and we suppose that there exists $r>0$ such that $max\{n,m\}\le r\times {(min\{n,m\})}^2$.

Sketch of the proof of 

(16). If $f\in C^3\left({[0,1]}^2\right)$, then the following decomposition holds (see [13])

$$\begin{array}{cc}\hfill f(x,y)& =f\left({\displaystyle \frac{k-1}{n}},{\displaystyle \frac{l-1}{m}}\right)\hfill \\ \hfill & +f_x^{\prime }\left({\displaystyle \frac{k-1}{n}},{\displaystyle \frac{l-1}{m}}\right)\left(x-{\displaystyle \frac{k-1}{n}}\right)+f_y^{\prime }\left({\displaystyle \frac{k-1}{n}},{\displaystyle \frac{l-1}{m}}\right)\left(y-{\displaystyle \frac{l-1}{m}}\right)\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}{f}_{xx}^{\prime \prime }\left({\displaystyle \frac{k-1}{n}},{\displaystyle \frac{l-1}{m}}\right){\left(x-{\displaystyle \frac{k-1}{n}}\right)}^2+{\displaystyle \frac{1}{2}}{f}_{yy}^{\prime \prime }\left({\displaystyle \frac{k-1}{n}},{\displaystyle \frac{l-1}{m}}\right){\left(y-{\displaystyle \frac{l-1}{m}}\right)}^2\hfill \\ \hfill & +f_{xy}^{\prime \prime }\left({\displaystyle \frac{k-1}{n}},{\displaystyle \frac{l-1}{m}}\right)\left(x-{\displaystyle \frac{k-1}{n}}\right)\left(y-{\displaystyle \frac{l-1}{m}}\right)+o\left(d{\mathbf{x}}^3\right)\hfill \end{array}$$

for any $(x,y)\in \left[{\displaystyle \frac{k-1}{n}},{\displaystyle \frac{k}{n}}\right]\times \left[{\displaystyle \frac{l-1}{m}},{\displaystyle \frac{l}{m}}\right]$ and $k,l\in \mathbb{N}$, $k\le n$, $l\le m$, where

$$d{\mathbf{x}}^3={\left(x-{\displaystyle \frac{k-1}{n}}\right)}^3+{\left(y-{\displaystyle \frac{l-1}{m}}\right)}^3.$$

Hence,

$$\begin{array}{cc}\hfill & \underset{{[0,1]}^2}{\int \int }f(x,y)dxdy=\sum _{k=1}^n\sum _{l=1}^m\underset{\left[{\displaystyle \frac{k-1}{n}},{\displaystyle \frac{k}{n}}\right]\times \left[{\displaystyle \frac{l-1}{m}},{\displaystyle \frac{l}{m}}\right]}{\int \int }\left[f\left({\displaystyle \frac{k-1}{n}},{\displaystyle \frac{l-1}{m}}\right)\right.\hfill \\ \hfill & \left.+f_x^{\prime }\left({\displaystyle \frac{k-1}{n}},{\displaystyle \frac{l-1}{m}}\right)\left(x-{\displaystyle \frac{k-1}{n}}\right)+f_y^{\prime }\left({\displaystyle \frac{k-1}{n}},{\displaystyle \frac{l-1}{m}}\right)\left(y-{\displaystyle \frac{l-1}{m}}\right)+\dots +o\left(d{\mathbf{x}}^3\right)\right]dxdy\hfill \\ \hfill & =\sum _{k=1}^n\sum _{l=1}^{m}f\left({\displaystyle \frac{k-1}{n}},{\displaystyle \frac{l-1}{m}}\right){\displaystyle \frac{1}{n\times m}}\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}\sum _{k=1}^n\sum _{l=1}^m\left[f_x^{\prime }\left({\displaystyle \frac{k-1}{n}},{\displaystyle \frac{l-1}{m}}\right){\displaystyle \frac{1}{n^2\times m}}+f_y^{\prime }\left({\displaystyle \frac{k-1}{n}},{\displaystyle \frac{l-1}{m}}\right){\displaystyle \frac{1}{n\times m^2}}\right]+o\left(1\right)\left({\displaystyle \frac{1}{n^3}}+{\displaystyle \frac{1}{m^3}}\right)\hfill \end{array}$$

as $min\{n,m\}\to \infty$. □

Remark 6.

Let $f:[0,1]\to \mathbb{R}$, $n\in \mathbb{N}$, $n\ge 2$. Setting $f\left(x\right)=ng\left(nx\right)$, $x\in [0,1]$ and then substituting $t=nx$, $x\in [0,1]$, we obtain

$${\displaystyle \frac{1}{n}}\sum _{k=0}^{n-1}f\left({\displaystyle \frac{k}{n}}\right)-\underset{0}{\overset{1}{\int }}f\left(x\right)dx=\sum _{k=0}^{n-1}g\left(k\right)-\underset{0}{\overset{1}{\int }}g\left(nx\right)d\left(nx\right)=\sum _{k=0}^{n-1}g\left(k\right)-\underset{0}{\overset{n}{\int }}g\left(t\right)dt.$$

Additionally, suppose that g is m-times differentiable in $[0,n]$ and $g^{\left(m\right)}$ is a Riemann-integrable function in $[0,n]$, $m,n\in \mathbb{N}$, $m,n\ge 2$ and $M_m\left(g\right):=\underset{x\in [0,n]}{sup}\left\{\left|g^{\left(m\right)}\left(x\right)\right|\right\}$. Hence, using items (1), (2), and (3) of Theorem 2, we obtain, respectively, the following.

(1) 

If $m=2$, then

$$\sum _{k=0}^{n-1}g\left(k\right)-\underset{0}{\overset{n}{\int }}g\left(x\right)dx={\displaystyle \frac{1}{2}}\left(g\left(0\right)-g\left(n\right)\right)+O\left({\displaystyle \frac{1}{n}}\right),asn\to \infty ,$$

where $n\left|O\left({\displaystyle \frac{1}{n}}\right)\right|\le {\displaystyle \frac{5}{12}}{M}_2\left(g\right)$.

(2) 

If $m=3$, then

$$\sum _{k=0}^{n-1}g\left(k\right)-\underset{0}{\overset{n}{\int }}g\left(x\right)dx={\displaystyle \frac{1}{2}}(g\left(0\right)-g\left(n\right))+{\displaystyle \frac{1}{12}}(g^{\prime }\left(n\right)-g^{\prime }\left(0\right))+O\left({\displaystyle \frac{1}{n^2}}\right),asn\to \infty ,$$

where $n^2\left|O\left({\displaystyle \frac{1}{n^2}}\right)\right|\le {\displaystyle \frac{1}{6}}{M}_3\left(g\right)$.

(3) 

If $m=4$, then

$$\sum _{k=0}^{n-1}g\left(k\right)-\underset{0}{\overset{n}{\int }}g\left(x\right)dx={\displaystyle \frac{1}{2}}(g\left(0\right)-g\left(n\right))+{\displaystyle \frac{1}{12}}(g^{\prime }\left(n\right)-g^{\prime }\left(0\right))+O\left({\displaystyle \frac{1}{n^3}}\right),asn\to \infty ,$$

where $n^3\left|O\left({\displaystyle \frac{1}{n^3}}\right)\right|\le {\displaystyle \frac{9}{125}}{M}_4\left(g\right)$.

If we suppose that $g\in C^m\left({[0,1]}^r\right)$, then all O asymptotic forms could be replaced by small o asymptotic forms.

Remark 7.

The formulas from Remark 6 can be generalized to the form (see Theorem 10.2.3 in [5]), the general form of the Euler–Maclaurin summation formula, and the respective form for an even-order derivative:

$$\begin{array}{cc}\hfill \sum _{k=m}^{n}g\left(k\right)-\underset{m}{\overset{n}{\int }}g\left(x\right)dx& ={\displaystyle \frac{g\left(m\right)+g\left(n\right)}{2}}+\sum _{k=1}^r{\displaystyle \frac{B_{2k}}{\left(2k\right)!}}\left(g^{(2k-1)}\left(n\right)-g^{(2k-1)}\left(m\right)\right)\hfill \\ \hfill & +\left\{\begin{array}{cc}O\left({\displaystyle \frac{1}{n^{2r-1}}}\right),\hfill & ifg\in C^{2r+1}\left([m,n]\right),\hfill \\ O\left({\displaystyle \frac{1}{n^{2r}}}\right),\hfill & ifg\in C^{2r+2}\left([m,n]\right)\hfill \end{array}\right.\hfill \end{array}$$

as $n\to \infty$, which holds for all $n,m\in {\mathbb{N}}_0,n>m$ and all functions $g\in C^{2r+1}\left([m,n]\right)$, where $B_{2k},k\in \mathbb{N}$ denote Bernoulli numbers [31,32].

Remark 8.

In particular (see Theorem (300) in Knopp’s monograph [33] and Theorem in Chapter 4.4 of Varadarajan’s monograph [6]), if $g\in C^{\infty }\left([m,\infty )\right)$, the function $sgng$ is constant on $(m,\infty )$ and all $g^{\left(k\right)}$, $k\in {\mathbb{N}}_0$, are monotonic on $(m,\infty )$ with

$$\underset{x\to \infty }{lim}{g}^{\left(k\right)}\left(x\right)=0$$

for all $k\in \mathbb{N}$, then, for every $n\in (m,\infty )$ and for every $r\in \mathbb{N}$, there exists $\theta =\theta (m,n;r)\in [0,1]$ such that the following asymptotic relation holds with the rest of the Lagrange form:

$$\begin{array}{cc}\hfill \sum _{k=m}^{n}g\left(k\right)-\underset{m}{\overset{n}{\int }}g\left(x\right)dx& ={\displaystyle \frac{g\left(m\right)+g\left(n\right)}{2}}+\sum _{k=1}^{r-1}{\displaystyle \frac{B_{2k}}{\left(2k\right)!}}\left(g^{(2k-1)}\left(n\right)-g^{(2k-1)}\left(m\right)\right)\hfill \\ \hfill & +\theta \times {\displaystyle \frac{B_{2r}}{\left(2r\right)!}}\left(g^{(2r-1)}\left(n\right)-g^{(2r-1)}\left(m\right)\right),\hfill \end{array}$$

where we have (see [31] and Section 4.1 in [6])

$$1<{\displaystyle \frac{2^{2k-1}{\pi }^{2k}\left|B_{2k}\right|}{\left(2k\right)!}}<2$$

for all $k\in \mathbb{N}$, $k>1$.

4. Conclusions

The aim of this paper was to consider the asymptotics of Riemann sums for uniform partitions of the interval of integration. Although the mentioned asymptotic relations are closely related to the Euler–Maclaurin summation formula, in this paper, only the Taylor formula with the remainder in Lagrange form was used to derive and analyze them. In this work, proofs are provided only for the first eight initial cases, while the proof of the general case is omitted for the sake of clarity, as it would require the introduction of Bernoulli numbers and polynomials. The authors aimed to make the idea behind such a proof more accessible by avoiding additional technical details. The authors hope that they have successfully met these expectations. This study focused significantly on discussing the remainders in the obtained asymptotic relations. Notably, this paper also derived asymptotic formulas for the Riemann sums of n-dimensional Riemann integrals for uniform partitions of n-dimensional intervals of integration. The obtained results were compared with approximation formulas for Bernstein operators (specifically, with Voronovskaya-type theorems). It is also worth noting that Formula (16) was quite controversial in its reception by our students; its particular variant of the symmetric division of the square ${[0,1]}^2$ for $m=n$ was more acceptable.