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assignmentutor-lab™ 为您的留学生涯保驾护航 在代写实变函数Real analysis方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写实变函数Real analysis代写方面经验极为丰富，各种代写实变函数Real analysis相关的作业也就用不着说。

• 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|Limit Theorems

There are many results we can prove about convergence sequences. Here are a few.
Theorem 3.4.1 A Convergent Sequence is Bounded
If a sequence $\left(a_{n}\right)$ converges it must be bounded; i.e. $\exists D>0 \ni\left|a_{n}\right| \leq D \forall n$. Further, if the sequence limit $a$ is not zero, $\exists N \ni\left|a_{n}\right|>|a| / 2 \forall n>N$.

Proof 3.4.1
Let $\left(a_{n}\right)$ be a sequence which converges to $a$. Pick $\epsilon=1$. Then there is an $N$ so that $n>N \Rightarrow$ $\left|a_{n}-a\right|<1$. Use the backwards triangle inequality to write $n>N \Rightarrow\left|a_{n}\right|-|a|<\left|a_{n}-a\right|<1$. The first part of this is what we need: we have $\left|a_{n}\right|<1+|a|$ when $n>N$ ( we ignore the middle part and only use the far left and the far right $)$. Let $B=\max \left{\left|a_{1}\right|, \ldots,\left|a_{N}\right|\right}$. Then we see $\left|a_{n}\right|<\max {B, 1+|a|}=D$ for all $n$. This shows the sequence is bounded.

For the next part, use $\epsilon=|a| / 2>0$. Then $\exists N$ so that $n>N$ implies $\left|a_{n}-a\right|<\left|a_{n}\right| / 2$. Using the backwards triangle inequality again, we have $|a|-\left|a_{n}\right|<\left|a_{n}-a\right|<|a| / 2$. Now use the far left and far right of this to see $\Rightarrow\left|a_{n}\right|>|a| / 2$ when $n>N$.
A very important fact is called the squeezing lemma which we use a lot.
Lemma 3.4.2 The Squeezing Lemma
The Squeezing Lemma
Let $\left(a_{n}\right)$ and $\left(b_{n}\right)$ be two sequences that converge to $L$. If $\left(c_{n}\right)$ is another sequence satisfying there is an $N$ so that $a_{n} \leq c_{n} \leq b_{n}$ for $n>N$, then $\left(c_{n}\right)$ converges to $L$ also.

## 数学代写|实变函数作业代写Real analysis代考|Bounded Sequences with a Finite Range

We have already looked at sequences with finite ranges. Since their range is finite, they are bounded sequences. We also know they have subsequences that converge which we have explicitly calculated. If the range of the sequence is a single value, then we know the sequence will converge to that value and we now know how to prove convergence of a sequence. Let’s formalize this into a theorem. But this time, we will argue more abstractly. Note how the argument is still essentially the same.
Theorem 4.1.1 A Sequence with a Finite Range Diverges Unless the Range is One Value
Let the sequence $\left(a_{n}\right)$ have a finite range $\left{y_{1}, \ldots, y_{p}\right}$ for some positive integer $p \geq 1$. If $p=1$, the sequence converges to $y_{1}$ and if $p>1$, the sequence does not converge but there is a subsequence $\left(a_{n_{k}^{i}}\right)$ which converges to $y_{i}$ for each $y_{i}$ in the range of the sequence.
Proof 4.1.1
If the range of the sequence consists of just one point, then $a_{n}=y_{1}$ for all $n$ and it is easy to see $a_{n} \rightarrow y_{1}$ as given $\epsilon>0,\left|a_{n}-y_{1}\right|=\left|y_{1}-y_{1}\right|=0<\epsilon$ for all $n$ which shows convergence.

If the range has $p>1$, let $a$ be any number not in the range and calculate $d_{i}=\left|a-y_{i}\right|$, the distance from a to each point $y_{i}$ in the range. Let $d=(1 / 2) \min \left{d_{1}, \ldots, d_{p}\right}$ and choose $\epsilon=d$. Then $\left|a_{n}-a\right|$ takes on $p$ values, $\left|y_{i}-a\right|=d_{i}$ for all $n$. But $d_{i}>d$ for all $i$ which shows us that $\left|a_{n}-a\right|>\epsilon$ for all $n$. A little thought then shows us this is precisely the definition of the sequence $\left(a_{n}\right)$ not converging to $a$.

If $a$ is one of the range values, say $a=y_{i}$, then the distances we defined above are positive except $d_{i}$ which is zero. So $\left|a_{n}-y_{i}\right|$ is zero for all indices $n$ which give range value $y_{i}$ but positive for all other range values. Let $\epsilon=d=(1 / 2) \min {j \neq i}\left|d{i}-d_{j}\right|$. Then, for any index $n$ with $a_{n} \neq y_{i}$, we have $\left|a_{n}-y_{i}\right|=\left|y_{j}-y_{i}\right|=d_{j}>$ d for some $j$. Thus, no matter what $N$ we pick, we can always find $n>N$ giving $\left|a_{n}-y_{i}\right|>\epsilon$. Hence, the limit can not be $y_{i}$. Since this argument works for any range value $y_{i}$, we see the limit value can not be any range value.

## 数学代写|实变函数作业代写Real analysis代考|Bounded Sequences with a Finite Range

$\backslash 1$ left 的分隔符缺失或无法识别 并选择 $\epsilon=d$. 然后 $\left|a_{n}-a\right|$ 承担 $p$ 价值观， $\left|y_{i}-a\right|=d_{i}$ 对所有人 $n$. 但 $d_{i}>d$ 对所有人 $i$ 这向我们展示了 $\left|a_{n}-a\right|>\epsilon$ 对所有人 $n$. 稍加思考，我们就知道这正是序列的定义 $\left(a_{n}\right)$ 不收敛到 $a$.

$\left|a_{n}-y_{i}\right|=\left|y_{j}-y_{i}\right|=d_{j}>\mathrm{d}$ 对于 些 $j$. 因此，无论怎样 $N$ 我们挑选，我们总能找到 $n>N$ 给予 $\left|a_{n}-y_{i}\right|>\epsilon$. 因此，极限不能 $y_{i}$. 由 于此参数适用于任何范围值 $y_{i}$ ，我们看到限制值不能是任何范围值。

## 有限元方法代写

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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
assignmentutor™您的专属作业导师