### 数学代写|实变函数作业代写Real analysis代考|Proving Propositions

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• 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|Understanding Smoothness

This text gathers material from many other sets of notes that we have written over the years since our first senior analysis course taught in 1982 . From all of our experiences, we have selected the following material to cover.

Part One: Introduction These are our beginning remarks you are reading now which are in Chapter $1 .$

Part Two: Understanding Smoothness Here we are concerned with developing continuity and the idea of differentiation for both calculus on the real line and calculus in $\Re^{2}$.

• We think learning abstraction is a hard thing, so we deliberately start slow. You have probably seen treatments of the real line and mathematical induction already, but in Chapter 2, we start at that basic level. We go over induction carefully and explain how to organize your proofs. We also work out the triangle inequality for numbers and introduce the important concept of the infimum and supremum of sets of numbers.
• In Chapter 3, we introduce sequences of real numbers and lay out a lot of notational machinery we will use later.
• In Chapter 4, we prove the important Bolzano – Weierstrass Theorem for bounded sequences and bounded sets and introduce sequential compactness. We also discuss carefully the limit inferior and superior of sequences in two equivalent ways.
• In Chapter 5, we go over more set theory and introduce topological compactness and finally prove the full equivalence that sequential and topological compactness on the real line are equivalent to the set being closed and bounded.
• In Chapter 6, we define limits of functions and explore limit inferior and limit superiors carefully.
• In Chapter 7, we talk about continuity.

## 数学代写|实变函数作业代写Real analysis代考|Mathematical Induction

We begin our study of analysis by looking at a powerful tool for proving certain types of propositions: the Principle of Mathematical Induction;
Theorem 2.1.1 The Principle of Mathematical Induction
For each natural number $n$, let $P(n)$ be a statement or proposition about the numbers $n$.

• If $P(1)$ is true: This is called the BASIS STEP
• If $P(k+1)$ is true when $P(k)$ is true: This is called the INDUCTIVE STEP then we can conclude $P(n)$ is true for all natural numbers $n$.
A proof using the POMI is organized as follows:
Proof 2.1.1
State the Proposition Here
Proof:
BASIS
Verify $P(1)$ is true
INDUCTIVE
Assume $P(k)$ is true for arbitrary $k>1$ and use that information to prove $P(k+1)$ is true.
We have verified the inductive step. Hence, by the POMI, $P(n)$ holds for all $n$.
QED
You must include this finishing statement as part of your proof and show the QED as above. Here QED is an abbreviation for the Latin Quod Erat Demonstratum or that which was to be shown. We often use the symbol $\mathbf{\square}$ instead of QED.

Note, the natural numbers or counting numbers are usually denoted by the symbol $\mathbb{N}$. The set of all integers, positive, negative and zero is denoted by $\mathbb{Z}$ and the real numbers is denoted by $\Re$ or $\mathbb{R}$. There are many alternative versions of this. One useful one is this.

## 数学代写|实变函数作业代写Real analysis代考|More Examples

INDUCTIVE. We assume $P(k)$ is true for an arbitrary $k>1$. Hence, we know
$$1+2+3+\cdots+k=\frac{1}{2} k(k+1)$$
Now look at $P(k+1)$. We note
$$1+2+3+\cdots+(k+1)={1+2+3+\cdots+k}+(k+1)$$
Now apply the induction hypothesis and let $1+2+3+\cdots+k=\frac{1}{2} k(k+1)$ We find
$$1+2+3+\cdots+(k+1)=\frac{1}{2} k(k+1)+(k+1)=(k+1)\left{\frac{1}{2} k+1\right}=\frac{1}{2}(k+1)(k+2)$$
This is precisely the statement $P(k+1)$. Thus $P(k+1)$ is true and we have verified the inductive step. Hence, by the POMI, $P(n)$ holds for all $n$.

Recall when you first encountered Riemann integration, you probably looked at taking the limit of Riemann sums using right side partitions. So for example, for $f(x)=2+x$ on the interval $[0,1]$ using a partition width of $\frac{1}{n}$, the Riemann sum is
$$\sum_{i=1}^{n} f\left(0+\frac{i}{n}\right) \frac{1}{n}=\sum_{i=1}^{n}\left(2+\frac{i}{n}\right) \frac{1}{n}=\frac{2}{n} \sum_{i=1}^{n} 1+\frac{1}{n^{2}} \sum_{i=1}^{n} i$$
The first sum, $\sum_{i=1}^{n} 1=n$ and so the first term is $2 \frac{n}{n}=2$. To evaluate the second term, we use our formula from above: $\sum_{i=1}^{n} i=\frac{1}{2} n(n+1)$ and so the second term becomes $\frac{n(n+1)}{2 n n}$ which simplifies to $\frac{1}{2}\left(1+\frac{1}{n}\right)$. So the Riemann sum here is $2+\frac{1}{2}\left(1+\frac{1}{n}\right)$ which as $n$ gets large clearly approaches the value $2.5$. The terms $2.5+\frac{1}{2 n}$ form what is called a sequence and the limit of this sequence is 2.5. We will talk about this a lot more later. From your earlier calculus courses, you know
$$\int_{0}^{1}(2+x) d x=\left.\left(2 x+\frac{1}{2} x^{2}\right)\right|_{0} ^{1}=2+\frac{1}{2}$$
which matches what we found with the Riemann sum limit. In later chapters, we discuss the theory of Riemann integration much more carefully, so consider this just a taste of that kind of theory!
Theorem 2.2.2
$$1^{2}+2^{2}+3^{2}+\cdots+n^{2}=\frac{1}{6} n(n+1)(2 n+1) \forall n \geq 1$$
Proof 2.2.2
BASIS $P(1)$ is the statement $1=\frac{1}{6}(1)(2)(3)=1$ which is true. So the basis step is verified. INDUCTIVE. We assume $P(k)$ is true for an arbitrary $k>1$. Hence, we know
$$1^{2}+2^{2}+3^{2}+\cdots+k^{2}=\frac{1}{6} k(k+1)(2 k+1)$$
Now look at $P(k+1)$. We note
$$1^{2}+2^{2}+3^{2}+\cdots+(k+1)^{2}=\left{1^{2}+2^{2}+3^{2}+\cdots+k^{2}\right}+(k+1)^{2}$$

## 数学代写|实变函数作业代写Real analysis代考|Understanding Smoothness

• 我们认为学习抽象是一件很难的事情，所以我们故意慢慢开始。您可能已经看过实数线和数学归纳法的处理方法，但在第 2 章中，我们将从这个基本层次开始。我们仔细检查归纳并解释如何组织你的证明。我们还解决了数的三角不等式，并引入了数集的下确界和上确界的重要概念。
• 在第 3 章中，我们介绍了实数序列并列出了许多我们稍后将使用的符号机制。
• 在第 4 章中，我们证明了有界序列和有界集的重要 Bolzano-Weierstrass 定理，并引入了序列紧致性。我们还以两种等效的方式仔细讨论了序列的下限和上限。
• 在第 5 章中，我们将讨论更多的集合论并介绍拓扑紧致性，并最终证明实线上的序列紧致性和拓扑紧致性等价于封闭和有界的集合。
• 在第 6 章中，我们定义了函数的极限，并仔细探讨了下限和上限。
• 在第 7 章中，我们讨论了连续性。

## 数学代写|实变函数作业代写Real analysis代考|Mathematical Induction

n， 让磷(n)是关于数字的陈述或命题n.

• 如果磷(1)是真的：这被称为基础步骤
• 如果磷(ķ+1)是真的磷(ķ)是真的：这被称为归纳步骤然后我们可以得出结论磷(n)对所有自然数都成立n.
使用 POMI 的证明组织如下：
证明 2.1.1
在此处陈述命题
证明：
基础
验证磷(1)是真的
归纳
假设磷(ķ)任意为真ķ>1并使用该信息来证明磷(ķ+1)是真的。
我们已经验证了归纳步骤。因此，通过 POMI，磷(n)适用于所有人n.
QED
您必须将此完成声明作为证明的一部分，并按上述方式出示 QED。这里的 QED 是拉丁语 Quod Erat Demonstratum 或将要显示的内容的缩写。我们经常使用符号而不是 QED。

## 数学代写|实变函数作业代写Real analysis代考|More Examples

1+2+3+⋯+ķ=12ķ(ķ+1)

1+2+3+⋯+(ķ+1)=1+2+3+⋯+ķ+(ķ+1)

1+2+3+\cdots+(k+1)=\frac{1}{2} k(k+1)+(k+1)=(k+1)\left{\frac{1}{2 } k+1\right}=\frac{1}{2}(k+1)(k+2)1+2+3+\cdots+(k+1)=\frac{1}{2} k(k+1)+(k+1)=(k+1)\left{\frac{1}{2 } k+1\right}=\frac{1}{2}(k+1)(k+2)

∑一世=1nF(0+一世n)1n=∑一世=1n(2+一世n)1n=2n∑一世=1n1+1n2∑一世=1n一世

∫01(2+X)dX=(2X+12X2)|01=2+12

12+22+32+⋯+n2=16n(n+1)(2n+1)∀n≥1

12+22+32+⋯+ķ2=16ķ(ķ+1)(2ķ+1)

1^{2}+2^{2}+3^{2}+\cdots+(k+1)^{2}=\left{1^{2}+2^{2}+3^{2} +\cdots+k^{2}\right}+(k+1)^{2}

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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。