Posted on Categories:Real analysis, 实分析, 数学代写

# 数学代写|实分析代写Real Analysis代考|Alternative proof of Intermediate value theorem

avatest™

## avatest™帮您通过考试

avatest™的各个学科专家已帮了学生顺利通过达上千场考试。我们保证您快速准时完成各时长和类型的考试，包括in class、take home、online、proctor。写手整理各样的资源来或按照您学校的资料教您，创造模拟试题，提供所有的问题例子，以保证您在真实考试中取得的通过率是85%以上。如果您有即将到来的每周、季考、期中或期末考试，我们都能帮助您！

•最快12小时交付

•200+ 英语母语导师

•70分以下全额退款

## 数学代写|实分析代写Real Analysis代考|Alternative proof of Intermediate value theorem

Let $[a, b]$ be a closed and bounded interval and $f:[a, b] \rightarrow \mathbb{R}$ is continuous on $[a, b]$. If $f(a) \neq f(b)$ then for every real number $r$ lying between $f(a)$ and $f(b)$ there is a point $c$ in $(a, b)$ such that $f(c)=r$.
Proof. Suppose on the contrary, there does not exist a point $c$ in $(a, b)$ such that $f(c)=r$.
Let us define a function $g: \mathbb{R} \rightarrow \mathbb{R}$ by $g(x)=f(a), x \in(-\infty, a)$
\begin{aligned} & =f(x), x \in[a, b]^{\circ} \ & =f(b), x \in(b, \infty) . \end{aligned}
Then $g$ is continuous on $\mathbb{R}$ and $g=f$ on $[a, b]$.
Let $G_1=(-\infty, r), G_2=(r, \infty)$. Then $\mathbb{R}=G_1 \cup{r} \cup G_2$.

$G_1$ and $G_2$ are open sets in $\mathbb{R}$. Since $g$ is continuous on $\mathbb{R}, g^{-1}\left(G_1\right)$ and $g^{-1}\left(G_2\right)$ are both open sets in $\mathbb{R}$.

Since $g^{-1}(r)=\phi, g^{-1}\left(G_1\right)=\mathbb{R}-g^{-1}\left(G_2\right)$. Since $g^{-1}\left(G_2\right)$ is open, $g^{-1}\left(G_1\right)$ is closed. Thus $g^{-1}\left(G_1\right)$ is both open and closed in $\mathbb{R}$.

But $g^{-1}\left(G_1\right)$ is non-empty, since $a \in g^{-1}\left(G_1\right)$ and $g^{-1}\left(G_1\right) \neq \mathbb{R}$, since $b \notin g^{-1}\left(G_1\right)$.

So $g^{-1}\left(G_1\right)$ is neither $\mathbb{R}$ nor $\phi$ and at the same time $g^{-1}\left(G_1\right)$ is both open and closed. This is a contradiction, since the only subsets in $\mathbb{R}$ which are both open and closed are $\phi$ and $\mathbb{R}$.
Hence our assumption is wrong and the theorem is proved.

## 数学代写|实分析代写Real Analysis代考|Monotone functions and continuity

Theorem 8.6.1. Let $I=(a, b)$ be an interval. Let $f: I \rightarrow \mathbb{R}$ be monotone increasing on $I$. Then at any point $c \in I$,
(i) $f(c-0)=\sup {x \in(a, c)} f(x)$, (ii) $f(c+0)=\inf {x \in(c, b)} f(x)$,
(iii) $f(c-0) \leq f(c) \leq f(c+0)$.
Proof. (i) If $x \in I$ and $x{x \rightarrow c-} f(x)=u$, i.e., $f(c-0)=u=\sup {x \in(a, c)} f(x)$ (ii) If $x \in I$ and $x>c$, then $f(x) \geq f(c)$.
Hence the set ${f(x): c<x<b}$ is bounded below, $f(c)$ being a lower bound. The set being non-empty, has a greatest lower bound, say $l$.

Then $l \geq f(c)$, and for a pre-assigned positive $\epsilon$, there exists a point $x_1$ in $(c, b)$ such that $l \leq f\left(x_1\right)<l+\epsilon$.
Let $x_1=c+\delta, 0<\bar{\delta}<b-c$.
Since $f$ is monotone increasing on $(c, b)$, for all $x$ in $c<x<x_1$ $l-\epsilon<l \leq f(x) \leq f\left(x_1\right)<l+\epsilon$ for all $x$ in $c<x<x_1$ Consequently, $|f(x)-l|<\epsilon$ for all $x$ in $c<x<x_1+\delta$ This implies that $\lim {x \rightarrow c+} f(x)=l$, i.e., $f(c+0)=l=\inf {x \in(c, b)} f(x)$
(iii) We have $f(c-0)=u \leq f(c)$ and $f(c+0)=l \geq f(c)$. Therefore $f(c-0) \leq f(c) \leq f(c+0)$.
Note. If $f: I \rightarrow \mathbb{R}$ be monotone decreasing on $I=(a, b)$ then at any point $c \in I$,
(i) $f(c-0)=\inf {x \in(a, c)} f(x)$, (ii) $f(c+0)=\sup {x \in(c, b)} f(x)$
(iii) $f(c-0) \geq f(c) \geq f(c+0)$.

## 数学代写|实分析代写Real Analysis代考|Alternative proof of Intermediate value theorem

\begin{aligned} & =f(x), x \in[a, b]^{\circ} \ & =f(b), x \in(b, \infty) . \end{aligned}

$G_1$$G_2$是$\mathbb{R}$中的开放集。因为$g$在$\mathbb{R}, g^{-1}\left(G_1\right)$上是连续的，而$g^{-1}\left(G_2\right)$在$\mathbb{R}$上都是开集。

## 数学代写|实分析代写Real Analysis代考|Monotone functions and continuity

(i) $f(c-0)=\sup {x \in(a, c)} f(x)$， (ii) $f(c+0)=\inf {x \in(c, b)} f(x)$，
(iii) $f(c-0) \leq f(c) \leq f(c+0)$。

(三)我们已经 $f(c-0)=u \leq f(c)$ 和 $f(c+0)=l \geq f(c)$． 因此 $f(c-0) \leq f(c) \leq f(c+0)$．

(i) $f(c-0)=\inf {x \in(a, c)} f(x)$(ii) $f(c+0)=\sup {x \in(c, b)} f(x)$
(iii) $f(c-0) \geq f(c) \geq f(c+0)$．

## 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。