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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|实变函数作业代写Real analysis代考|Sequences with an Infinite Range

We are now ready for our most abstract result so far.
Theorem 4.2.1 Bolzano – Weierstrass Theorem
Every bounded sequence has at least one convergent subsequence.
Proof 4.2.1
As discussed, we have already shown a sequence with a bounded finite range always has convergent subsequences. Now we prove the case where the range of the sequence of values $\left{a_{1}, a_{2} \ldots,\right$,$} has$ infinitely many distinct values. We assume the sequences start at $n=k$ and by assumption, there is a positive number $B$ so that $-B \leq a_{n} \leq B$ for all $n \geq k$. Define the interval $J_{0}=\left[\alpha_{0}, \beta_{0}\right]$ where $\alpha_{0}=-B$ and $\beta_{0}=B$. Thus at this starting step, $J_{0}=[-B, B]$. Note the length of $J_{0}$, denoted by $\ell_{0}$ is $2 B$.

Let $\mathcal{S}$ be the range of the sequence which has infinitely many points and for convenience, we will let the phrase infinitely many points be abbreviated to IMPs.
Step 1:
Bisect $\left[\alpha_{0}, \beta_{0}\right]$ into two pieces $u_{0}$ and $u_{1}$. That is the interval $J_{0}$ is the union of the two sets $u_{0}$ and $u_{1}$ and $J_{0}=u_{0} \cup u_{1}$. Now at least one of the intervals $u_{0}$ and $u_{1}$ contains IMPs of $\mathcal{S}$ as otherwise each piece has only finitely many points and that contradicts our assumption that $\mathcal{S}$ has IMPS. Now both may contain IMPS so select one such interval containing IMPS and call it $J_{1}$. Label the endpoints of $J_{1}$ as $\alpha_{1}$ and $\beta_{1}$; hence, $J_{1}=\left[\alpha_{1}, \beta_{1}\right]$. Note $\ell_{1}=\beta_{1}-\alpha_{1}=\frac{1}{2} \ell_{0}=B$ We see $J_{1} \subseteq J_{0}$ and
$$-B=\alpha_{0} \leq \alpha_{1} \leq \beta_{1} \leq \beta_{0}=B$$
Since $J_{1}$ contains IMPS, we can select a sequence value $a_{n_{1}}$ from $J_{1}$.
Step 2:
Now bisect $J_{1}$ into subintervals $u_{0}$ and $u_{1}$ just as before so that $J_{1}=u_{0} \cup u_{1}$. At least one of $u_{0}$ and $u_{1}$ contain IMPS of $\mathcal{S}$.

Choose one such interval and call it $J_{2}$. Label the endpoints of $J_{2}$ as $\alpha_{2}$ and $\beta_{2}$; hence, $J_{2}=$ $\left[\alpha_{2}, \beta_{2}\right]$. Note $\ell_{2}=\beta_{2}-\alpha_{2}=\frac{1}{2} \ell_{1}$ or $\ell_{2}=(1 / 4) \ell_{0}=\left(1 / 2^{2}\right) \ell_{0}=(1 / 2) B$. We see $J_{2} \subseteq J_{1} \subseteq J_{0}$ and
$$-B=\alpha_{0} \leq \alpha_{1} \leq \alpha_{2} \leq \beta_{2} \leq \beta_{1} \leq \beta_{0}=B$$

## 数学代写|实变函数作业代写Real analysis代考|Bounded Infinite Sets Have at Least One Accumulation Point

We can extend the Bolzano – Weierstrass Theorem for sequences to bounded sets having an infinite number of points.
Theorem 4.3.1 Bolzano – Weierstrass Theorem for Sets: Accumulation Point Version
Every bounded infinite set of real numbers has at least one accumulation point.
Proof 4.3.1
We let the bounded infinite set of real numbers be $S$. We know there is a positive number $B$ so that $-B \leq x \leq B$ for all $x$ in $S$ because $S$ is bounded.
Step 1:
By a process essentially the same as the subdivision process in the proof of the Bolzano-Weierstrass Theorem for Sequences, we can find a sequence of sets
$$J_{0}, J_{1}, \ldots, J_{p}, \ldots$$
These sets have many properties.

• $\ell_{p}=B / 2^{p-1}$ for $p \geq 1$ ( remember $B$ is the bound for $S$ )
• $J_{p} \subseteq J_{p-1}$,
• $J_{p}=\left[\alpha_{p}, \beta_{p}\right]$ and $J_{p}$ contains IMPs of $S .$
• $-B=\alpha_{0} \leq \ldots \leq \alpha_{p}<\beta_{p} \leq \ldots \leq \beta_{0}=B$
Since $J_{0}$ contains IMPs of $S$ choose $x_{0}$ in $S$ from $J_{0}$. Next, since $J_{1}$ contains IMPs of $S$, choose $x_{1}$ different from $x_{0}$ from $S$. Continuing ( note we are using what we called our relaxed use of POMI here!) we see by an induction argument we can choose $x_{p}$ in $S$, different from the previous ones, with $x_{p}$ in $J_{p}$.

We also know the sequence $\left(\alpha_{p}\right)$ has a finite supremum $\alpha$ and the sequence $\left(\beta_{p}\right)$ has a finite infimum $\beta$. Further, we know $\alpha_{p} \leq \alpha, \beta \geq \beta_{p}$ for all $p$. Since the lengths of the intervals $J_{p}$ go to 0 as $p \rightarrow \infty$, we also know $\alpha=\beta$ and the sequence $\left(x_{p}\right)$ converges to $\alpha$. Thus, the ball $B_{r}(\alpha)=(\alpha-r, \alpha+r)$ contains $J_{p}$ for $p>P$. So in particular, $\alpha \in J_{p}$ and $x_{p} \in J_{p}=\left[\alpha_{P}, \beta_{P}\right]$ for any $p \geq P$. It remains to show that $\alpha$ is an accumulation point of $S$. Choose any $r>0$. Since $\ell_{p}=B / 2^{p-1}$, we can find an integer $P$ so that $B / 2^{P-1}0$ was arbitrary, this tells us $\alpha$ is an accumulation point of $S$.

## 数学代写|实变函数作业代写Real analysis代考|Sequences with an Infinite Range

\left 的分隔符蚗失或无法识别 无限多个不同的值。我们假设序列开始于 $n=k$ 并且通过假设，有一个正数 $B$ 以便 $-B \leq a_{n} \leq B$ 对所有人 $n \geq k$. 定义间隔 $J_{0}=\left[\alpha_{0}, \beta_{0}\right]$ 在哪里 $\alpha_{0}=-B$ 和 $\beta_{0}=B$. 因此，在这个起始步骤， $J_{0}=[-B, B]$. 注意长度 $J_{0}$ ， 表示为 $\ell_{0}$ 是 $2 B$.

$$-B=\alpha_{0} \leq \alpha_{1} \leq \beta_{1} \leq \beta_{0}=B$$

$\ell_{2}=(1 / 4) \ell_{0}=\left(1 / 2^{2}\right) \ell_{0}=(1 / 2) B$. 我们看 $J_{2} \subseteq J_{1} \subseteq J_{0}$ 和
$$-B=\alpha_{0} \leq \alpha_{1} \leq \alpha_{2} \leq \beta_{2} \leq \beta_{1} \leq \beta_{0}=B$$

## 数学代写|实变函数作业代写Real analysis代考|Bounded Infinite Sets Have at Least One Accumulation Point

$$J_{0}, J_{1}, \ldots, J_{p}, \ldots$$

• $\ell_{p}=B / 2^{p-1}$ 为了 $p \geq 1$ ( 记住 $B$ 是有界的 $S$ )
• $J_{p} \subseteq J_{p-1}$,
• $J_{p}=\left[\alpha_{p}, \beta_{p}\right]$ 和 $J_{p}$ 包含 IMP $S .$
• $-B=\alpha_{0} \leq \ldots \leq \alpha_{p}<\beta_{p} \leq \ldots \leq \beta_{0}=B$ 自从 $J_{0}$ 包含 IMP $S$ 选择 $x_{0}$ 在 $S$ 从 $J_{0}$. 接下来，由于 $J_{1}$ 包含 IMP $S$ ，选择 $x_{1}$ 不同于 $x_{0}$ 从 $S$. 继续（请注意，我们在这里使用我们所谓的轻 松使用 POMI！) 我们可以通过可以选择的归纳论证看到 $x_{p}$ 在 $S$, 与前面的不同, 与 $x_{p}$ 在 $J_{p}$. 我们也知道序列 $\left(\alpha_{p}\right)$ 有一个有限的决赛 $\alpha$ 和序列 $\left(\beta_{p}\right)$ 有一个有限的最低 $\beta$. 此外，我们知道 $\alpha_{p} \leq \alpha, \beta \geq \beta_{p}$ 对所有人 $p$. 由于区间的长度 $J_{p}$ 去 0 作为 $p \rightarrow \infty$ ，我们也知道 $\alpha=\beta$ 和序列 $\left(x_{p}\right)$ 收敛到 $\alpha$. 于是，球 $B_{r}(\alpha)=(\alpha-r, \alpha+r)$ 包含 $J_{p}$ 为了 $p>P$. 所以特别是， $\alpha \in J_{p}$ 和 $x_{p} \in J_{p}=\left[\alpha_{P}, \beta_{P}\right]$ 对于任何 $p \geq P$. 仍有待证明 $\alpha$ 是一个积累点 $S$. 选择任何 $r>0$. 自从 $\ell_{p}=B / 2^{p-1}$ ，我们可以找到一个整数 $P$ 以便 $B / 2^{P-1} 0$ 是任意的，这告诉我们 $\alpha$ 是一个积累点 $S$.

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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。

assignmentutor™您的专属作业导师
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