### 数学代写|实变函数作业代写Real analysis代考|Sequences of Real Numbers

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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|Basic Definitions

To define sequences and subsequences of a sequence, we must have a way of stating precisely what kinds of subsets of the integers we want to focus on.
Definition 3.1.1 Right Increasing Subsets of Integers
We know $\mathbb{Z}={\ldots,-2,-1,0,1,2,3, \ldots}$ is the set of integers. Let $\mathbb{Z}{\geq k}$ denote the set of all integers $\geq k$ Thus, $\mathbb{Z}{\geq-3}={-3,-2,-1,0,1,2,3, \ldots}$. Let $T$ be a subset of $\mathbb{Z}$. We say $T$ is a Right Increasing Infinite Subset or $\mathbf{R I I}$ of $\mathbb{Z}$ if $T$ is not bounded above and $T$ is bounded below and the entries in $T$ are always increasing.
Example 3.1.1
\begin{aligned} &T={2,4,6, \ldots, 2 n, \ldots}={2 k}_{k=1}^{\infty}=(2 k){k=1}^{\infty}=(2 k) \ &T={-17,-5,7,19, \ldots}={-17+12 k}{k=0}^{\infty}=(-17+12 k){k=0}^{\infty} \end{aligned} In general, a RII subset $T$ of $\mathbb{Z}$ can be characterized by $$T=\left{n{0}, n_{1}, n_{2}, \ldots, n_{k}, \ldots\right}$$
where $n_{0}$ is the starting integer or index and
$$n_{0}<n_{1}<n_{2}<\ldots<n_{k}<\ldots$$
with $n_{k} \rightarrow \infty$ as $k \rightarrow \infty$.
Note $T=\mathbb{Z}_{\geq 1}={1,2,3, \ldots}$ is our usual set of counting numbers $\mathbb{N}$. We can use this ideas to define sequences and subsequences carefully.

## 数学代写|实变函数作业代写Real analysis代考| The Definition of a Sequence

A sequence of real numbers is simply a real valued function whose domain is the set $\mathbb{Z}{\geq k}$ for some $k$. A subset of the set $f\left(\mathbb{Z}{\geq k}\right)$ defined by $f(T)$ for any $R I I T$ of $\mathbb{Z}{\geq k}$ is called a subsequence of the sequence. To help you see what sequences are all about, you need to look at examples! Example 3.2.1 Consider $f: \mathbb{N} \rightarrow \Re$ defined by $f(n)=a{n}$ where each $a_{n}$ is a number. There are many notations for this sequence:
\begin{aligned} {f(n): n \in \mathbb{N}} &={f(n)}_{n=1}^{\infty}=(f(n)){n=1}^{\infty}=(f(n)) \ &={f(1), f(2), \ldots, f(n), \ldots}=\left{a{1}, a_{2}, \ldots, a_{n}, \ldots\right} \end{aligned}
Example 3.2.2 Consider $f: \mathbb{N} \rightarrow \Re$ defined by $f(n)=\sin (n \pi / 4)$. Then
\begin{aligned} {f(n)}_{n=1}^{\infty}=&{\sin (\pi / 4), \sin (2 \pi / 4), \sin (3 \pi / 4), \sin (4 \pi / 4),\ &\sin (5 \pi / 4), \sin (6 \pi / 4), \sin (7 \pi / 4), \sin (8 \pi / 4), \ldots} \ =&{\sqrt{2} / 2,1, \sqrt{2} / 2,0,-\sqrt{2} / 2,-1,-\sqrt{2} / 2,0, \ldots} \end{aligned}
Let the numbers ${\sqrt{2} / 2,1, \sqrt{2} / 2,0,-\sqrt{2} / 2,-1,-\sqrt{2} / 2,0}$ be called block $B$. Then we see the values of this sequence consist of the infinitely repeating blocks $B$ as follows:
$$(\sin (n \pi / 4))={B, B, \ldots, B, \ldots}$$
where each block $B$ consists of 8 numbers.
So the range of this function, i.e. the range of this sequence, is the set ${-1,-\sqrt{2} / 2,0, \sqrt{2} / 2,1}$ which is just 5 values.
Now define RII sets as follows:
\begin{aligned} &T_{1}={1+8 k}_{k \geq 0}={1,9,17, \ldots} \ &T_{2}={2+8 k}_{k \geq 0}={2,10,18, \ldots} \ &T_{3}={4+8 k}_{k \geq 0}={4,12,20, \ldots} \ &T_{4}={5+8 k}_{k \geq 0}={5,13,21, \ldots} \ &T_{5}={6+8 k}_{k \geq 0}={6,14,22, \ldots} \end{aligned}
Define new functions $f_{i}: T_{i} \rightarrow \Re$ by
\begin{aligned} &f_{1}(1+8 k)=f(1+8 k)=\sin ((1+8 k) \pi / 4)=\sin (\pi / 4)=\sqrt{2} / 2 \ &f_{2}(2+8 k)=f(2+8 k)=\sin ((2+8 k) \pi / 4)=\sin (2 \pi / 4)=1 \ &f_{3}(4+8 k)=f(4+8 k)=\sin ((4+8 k) \pi / 4)=\sin (4 \pi / 4)=0 \ &f_{4}(5+8 k)=f(5+8 k)=\sin ((5+8 k) \pi / 4)=\sin (5 \pi / 4)=-\sqrt{2} / 2 \ &f_{5}(6+8 k)=f(6+8 k)=\sin ((6+8 k) \pi / 4)=\sin (6 \pi / 4)=-1 \end{aligned}
Each of these functions extracts a subset of the original set $(f(n))=(\sin (n \pi / 4))$ where we did not explicitly indicate the subscripts $n \geq 1$ as it is understood at this point. These five functions give five subsequences of the original sequence.

We usually don’t go through all this trouble to find subsequences. Instead of the $f_{i}$ notation, we define these subsequences by another notation. The original sequence is $f(n)=\sin (n \pi / 4)$ so the values of the sequence are $a_{n}=\sin (n \pi / 4)$. These five subsequences are then defined by

$a_{1+8 k}=f(1+8 k)=\sin ((1+8 k) \pi / 4)$
$a_{2+8 k}=f(2+8 k)=\sin ((2+8 k) \pi / 4)=\sin (2 \pi / 4)=1$
$a_{4+8 k}=f(4+8 k)=\sin ((4+8 k) \pi / 4)=\sin (4 \pi / 4)=0$
$a_{5+8 k}=f(5+8 k)=\sin ((5+8 k) \pi / 4)=\sin (5 \pi / 4)=-\sqrt{2} / 2$
$a_{6+8 k}=f(6+8 k)=\sin ((6+8 k) \pi / 4)=\sin (6 \pi / 4)=-1$

## 数学代写|实变函数作业代写Real analysis代考|The Convergence of a Sequence

What do we mean by the phrase a sequence converges? For a simple sequence like $\left(a_{n}=1 / n\right)$ it is easy to see that as $n$ gets large the value of the sequence gets closer and closer to 0 . It is also easy to see for sequences with finite blocks that repeat in their range, like $\left((-1)^{n}\right)$ or $(\sin (n \pi / 3))$, the values of the sequences bounce around among the finite possibilities in the block. But we need a careful way to say when the sequence gets closer to some fixed value and a careful way to say the sequence can not do that. This language is handled in the definition below:

Theorem 3.3.1 $\left((-1)^{n}\right)$ Diverges
The sequence $\left((-1)^{n}\right){n \geq 1}$ does not converge. Proof 3.3.1 The range of this sequence is the set ${-1,1}$. We will show there is no number a so that $(-1)^{n} \rightarrow a$ for this sequence. Case 1: Let $a=1$. Note the difference between $-1$ and $+1$ is 2 which suggests we pick a tolerance $\epsilon<2$ to show the sequence does not converge to 1 as we want to isolate which value we are trying to get close to. Let $\epsilon=1$. Then $$\left|(-1)^{n}-1\right|=\left{\begin{array}{c} |1-1|=0, \text { if } n \text { is even. } \ |-1-1|=2, \text { if } n \text { is odd. } \end{array}\right.$$ Now pick an $N$. Then there is an odd integer larger than $N$, say $2 N+1$, for which $\left|(-1)^{2 N+1}-1\right|=$ $2>\epsilon$. Since we can do this for all $N$, we see the sequence can not converge to $1 .$ Case-1: We can repeat this argument for this case. Let $\epsilon=1$. Then $$\left|(-1)^{n}-(-1)\right|=\left{\begin{array}{l} |1+1|=2, \text { if } n \text { is even. } \ |-1+1|=0, \text { if } n \text { is odd } \end{array}\right.$$ Now pick an $N$. Then there is an even integer larger than $N$, say $2 N$, for which $\left|(-1)^{2 N}-(-1)\right|=$ $2>\epsilon$. Since we can do this for all $N$, we see the sequence can not converge to $-1$. Case $a \neq 1,-1$ : If $a$ is not $-1$ or 1 , let $d{1}$ be the distance for a to 1 which is $|a-1|$ and let $d_{2}$ be the distance to $-1$ which is $|a-(-1)|$. Let $d$ be the minimum of these two distances, $d=\min \left(d_{1}, d_{2}\right)$. Then, we have
$$\left|(-1)^{n}-a\right|=\left{\begin{array}{l} |1-a|=d_{1}, \text { if } n \text { is even. } \ |-1-a|=d_{2}, \text { if } n \text { is odd } . \end{array}\right.$$

## 数学代写|实变函数作业代写Real analysis代考|Basic Definitions

T=\left{n{0}, n_{1}, n_{2}, \ldots, n_{k}, \ldots\right}T=\left{n{0}, n_{1}, n_{2}, \ldots, n_{k}, \ldots\right}

n0<n1<n2<…<nķ<…

## 数学代写|实变函数作业代写Real analysis代考| The Definition of a Sequence

\begin{对齐} {f(n): n \in \mathbb{N}} &={f(n)}_{n=1}^{\infty}=(f(n)){n=1 }^{\infty}=(f(n)) \ &={f(1), f(2), \ldots, f(n), \ldots}=\left{a{1}, a_{2 }, \ldots, a_{n}, \ldots\right} \end{对齐}\begin{对齐} {f(n): n \in \mathbb{N}} &={f(n)}_{n=1}^{\infty}=(f(n)){n=1 }^{\infty}=(f(n)) \ &={f(1), f(2), \ldots, f(n), \ldots}=\left{a{1}, a_{2 }, \ldots, a_{n}, \ldots\right} \end{对齐}

\begin{对齐} {f(n)}_{n=1}^{\infty}=&{\sin (\pi / 4), \sin (2 \pi / 4), \sin (3 \pi / 4), \sin (4 \pi / 4),\ &\sin (5 \pi / 4), \sin (6 \pi / 4), \sin (7 \pi / 4), \sin (8 \pi / 4), \ldots} \ =&{\sqrt{2} / 2,1, \sqrt{2} / 2,0,-\sqrt{2} / 2,-1,-\sqrt{2 } / 2,0, \ldots} \end{对齐}\begin{对齐} {f(n)}_{n=1}^{\infty}=&{\sin (\pi / 4), \sin (2 \pi / 4), \sin (3 \pi / 4), \sin (4 \pi / 4),\ &\sin (5 \pi / 4), \sin (6 \pi / 4), \sin (7 \pi / 4), \sin (8 \pi / 4), \ldots} \ =&{\sqrt{2} / 2,1, \sqrt{2} / 2,0,-\sqrt{2} / 2,-1,-\sqrt{2 } / 2,0, \ldots} \end{对齐}

(罪⁡(n圆周率/4))=乙,乙,…,乙,…

F1(1+8ķ)=F(1+8ķ)=罪⁡((1+8ķ)圆周率/4)=罪⁡(圆周率/4)=2/2 F2(2+8ķ)=F(2+8ķ)=罪⁡((2+8ķ)圆周率/4)=罪⁡(2圆周率/4)=1 F3(4+8ķ)=F(4+8ķ)=罪⁡((4+8ķ)圆周率/4)=罪⁡(4圆周率/4)=0 F4(5+8ķ)=F(5+8ķ)=罪⁡((5+8ķ)圆周率/4)=罪⁡(5圆周率/4)=−2/2 F5(6+8ķ)=F(6+8ķ)=罪⁡((6+8ķ)圆周率/4)=罪⁡(6圆周率/4)=−1

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