# 数学代写|实分析作业代写Real analysis代考|MATH2350

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## 数学代写|实分析作业代写Real analysis代考|Approximations Using Riemann Sums

Since upper and lower sums are in general difficult to evaluate, Darboux’s definition of the Riemann integral is not particularly useful in obtaining numerical approximations. The most elementary numerical method is to use Riemann sums. One of the first mathematicians to use numerical methods was Euler, who considered sums of the form
$$\sum_{k=1}^n f\left(x_{k-1}\right)\left(x_k-x_{k-1}\right)$$
as an approximation to the integral. This is nothing but the Riemann sum for the partition $\mathcal{P}=\left{x_0, x_1, \ldots, x_n\right}$ of $[a, b]$ with $t_k=x_{k-1}$ for all $k$.

In using Riemann sums to approximate the integral of $f$ it is convenient to take equally spaced partitions. Let $n \in \mathbb{N}$, and set $h=(b-a) / n$. Define
$$x_0=a, \quad x_1=a+h, \quad x_2=a+2 h, \cdots, \quad x_n=a+n h=b .$$
Thus if $\mathcal{P}n=\left{x_0, x_1, \ldots, x_n\right}$ with $x_i$ as defined, we always have $\Delta x_i=h$ for all $i=1, \ldots, n$, and $$\mathcal{S}\left(\mathcal{P}_n, f\right)=\sum{i=1}^n f\left(t_i\right) \Delta x_i=h \sum_{i=1}^n f\left(t_i\right)$$
where for each $i=1, \ldots, n, t_i \in\left[x_{i-1}, x_i\right]$. If we take $t_i=x_{i-1}$ for all $i$, then we obtain the above formula of Euler. Similarly, we could take $t_i$ to be the right endpoint $x_i$ of the interval $\left[x_{i-1}, x_i\right]$. Another choice of $t_i$ would be the midpoint; i.e., $t_i=\left(x_{i-1}+x_i\right) / 2$. For monotone functions, it is intuitively obvious that the midpoint gives a better approximation than either the right or left endpoint to the integral of $f$ over the interval $\left[x_{i-1}, x_i\right]$ (see Figure 6.8). With $x_i$ as defined by (17) and $t_i=\left(x_{i-1}+x_i\right) / 2$ the above formula becomes
$$M_n(f)=h \sum_{i=1}^n f\left(a+\left(i-\frac{1}{2}\right) h\right)$$

## 数学代写|实分析作业代写Real analysis代考|The Trapezoidal Rule

We now consider another common second order approximation method known as the trapezoidal rule. In using Riemann sums, regardless of the choice of the points $t_i$, we used rectangles to approximate the integral of the function $f$. In our second numerical method we will replace rectangles by trapezoids. Let $f$ be a Riemann integrable function on $[a, b]$ and let $\mathcal{P}=\left{x_0, x_1, \ldots, x_n\right}$ be a partition of $[a, b]$. As in Figure 6.9 , for each interval $\left[x_{i-1}, x_i\right]$, the area of the trapezoid formed by the points $\left(x_{i-1}, 0\right),\left(x_{i-1}, f\left(x_{i-1}\right)\right),\left(x_i, f\left(x_i\right)\right),\left(x_i, 0\right)$ is given by $\frac{1}{2}\left[f\left(x_{i-1}\right)+f\left(x_i\right)\right] \Delta x_i$. Summing these up gives
$$\frac{1}{2} \sum_{i=1}^n\left[f\left(x_{i-1}\right)+f\left(x_i\right)\right] \Delta x_i$$
as an approximation to the integral of $f$ on $[a, b]$. If as previously we set $h=(b-a) / n$ and $x_i=a+i h, i=0, \ldots, n$, then the above sum becomes
$$\frac{h}{2} \sum_{i=1}^n[f(a+(i-1) h)+f(a+i h)]=\frac{h}{2}\left[f(a)+2 \sum_{i=1}^{n-1} f(a+i h)+f(b)\right]$$
If $f$ is Riemann integrable on $[a, b]$ and $n \in \mathbb{N}$, the quantity $T_n(f)$ defined by
$$T_n(f)=\frac{h}{2}\left[f(a)+2 \sum_{i=1}^{n-1} f(a+i h)+f(b)\right]$$
where $h=(b-a) / n$, is called the $n$th trapezoidal approximation to the integral of $f$ on $[a, b]$. If we set $y_i=f(a+i h)$, then $T_n(f)$ can be expressed as
$$T_n(f)=\frac{h}{2}\left[y_0+2 \sum_{i=1}^{n-1} y_i+y_n\right]$$

# 实分析代写

## 数学代写|实分析作业代写Real analysis代考|Approximations Using Riemann Sums

$$\sum_{k=1}^n f\left(x_{k-1}\right)\left(x_k-x_{k-1}\right)$$

$$x_0=a, \quad x_1=a+h, \quad x_2=a+2 h, \cdots, \quad x_n$$

$$\mathcal{S}\left(\mathcal{P}n, f\right)=\sum i=1^n f\left(t_i\right) \Delta x_i=h \sum{i=1}^n f\left(t_i\right)$$

$$M_n(f)=h \sum_{i=1}^n f\left(a+\left(i-\frac{1}{2}\right) h\right)$$

## 数学代写|实分析作业代写Real analysis代考|The Trapezoidal Rule

$$\frac{1}{2} \sum_{i=1}^n\left[f\left(x_{i-1}\right)+f\left(x_i\right)\right] \Delta x_i$$

$$\frac{h}{2} \sum_{i=1}^n[f(a+(i-1) h)+f(a+i h)]=\frac{h}{2}[f(a)$$

$$T_n(f)=\frac{h}{2}\left[f(a)+2 \sum_{i=1}^{n-1} f(a+i h)+f(b)\right]$$

$$T_n(f)=\frac{h}{2}\left[y_0+2 \sum_{i=1}^{n-1} y_i+y_n\right]$$

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## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

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