### 数学代写|实变函数作业代写Real analysis代考|Uniqueness of Limits and So Forth

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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|Convergent Sequences have Unique Limits

First, we need to show the limit of a convergent sequence is unique.
Theorem 3.3.3 Convergent Sequences have Unique Limits
If the sequence $\left(x_{n}\right)$ converges, then the limit is unique.
Proof 3.3.3
Assume $x_{n} \rightarrow a$ and $x_{n} \rightarrow b$. Then for an arbitrary $\epsilon>0$, there is an $N_{1}$ and an $N_{2}$ so that
\begin{aligned} &n>N_{1} \Rightarrow\left|x_{n}-a\right|<\epsilon / 2 \ &n>N_{2} \Rightarrow\left|x_{n}-b\right|<\epsilon / 2 \end{aligned} Now pick a $P>\max \left{N_{1}, N_{2}\right}$. Then, we have
$$|a-b|=\left|a-x_{P}+x_{P}-b\right| \leq\left|a-x_{P}\right|+\left|x_{P}-b\right|<\epsilon / 2+\epsilon / 2=\epsilon$$ Since $\epsilon>0$ is arbitrary and $|a-b|<\epsilon$ for all $\epsilon>0$, this shows $a=b$.
Next we need to look at subsequences of convergent sequences.
Theorem 3.3.4 Subsequences of Convergent Sequences have the Same Limit
Assume $\left(x_{n}\right)$ converges and $\left(x_{n_{k}}\right)$ is a subsequence. Then $\left(x_{n_{k}}\right)$ converges to the same limiting value.
Proof 3.3.4
Since $x_{n} \rightarrow$ a for some a, for arbitrary $\epsilon>0$, there is an $N$ so that $n>N \Rightarrow\left|x_{n}-a\right|<\epsilon$. In particular, $n_{k}>N \Rightarrow\left|x_{n_{k}}-a\right|<\epsilon$. This says $x_{n_{k}} \rightarrow$ a also.

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

The range of this sequence is not finite and indeed does not repeat any values. However, we can guess that as $n$ gets large, the values in the sequence are closer and closer to 1 . So let’s assume $a=1$ as our guess for the limit. Pick an arbitrary $\epsilon>0$ and consider
$$\left|a_{n}-a\right|=|(1+(4 / n))-1|=|4 / n|=4|1 / n|$$
Pick any $N$ so that $4 / N<\epsilon$ or $N>4 / \epsilon$. Then for any $n>N, n>4 / \epsilon$ or $4 / n<\epsilon$ So we have shown $$n>N>4 / \epsilon \Rightarrow|(1+(4 / n))-1|<\epsilon$$ Since $\in$ was chosen arbitrarily, we know the sequence converges to $1 .$ The convergence proofs follows a nice template which we show next. Theorem 3.3.6 $a_{n} \rightarrow a$ Template $$a_{n} \rightarrow a \text {. }$$ Proof 3.3.6 Step 1: Identify what $a$ is. Step 2: Choose $\epsilon>0$ arbitrarily.
Step 3: Now follow this argument:
\begin{aligned} \mid \text { original sequence – proposed limit } \mid &=\mid \text { simplify using algebra etc to get a new expression } \mid \ & \leq \mid \text { use triangle inequality etc to get a new expression again } \mid \ &=\text { call this last step the overestimate. We have now } \ &=\mid \text { overestimate } \mid<\epsilon \end{aligned} Step 4: Solve for $n$ in terms of $\epsilon$ to give a simple equation. Step 5: Choose $N$ to satisfy the inequality you get from Step $4 .$ Step 6: Then for any $n>N$, |overestimate $\mid<\epsilon$ and we have $\mid$ (original ) $-$ (proposed limit) $\mid<\epsilon$ proving $a_{n} \rightarrow a$. Theorem 3.3.7 $\left(\frac{1+4 n}{5+6 n}\right)$ Converges $\left(\frac{1+4 n}{5+6 n}\right)_{n \geq 1}$ converges. Proof 3.3.7 We can guess the value of the limit is $a=2 / 3$ so pick $\epsilon>0$ arbitrarily. Consider
$$\left|\frac{1+4 n}{5+6 n}-\frac{2}{3}\right|=\left|\frac{3(1+4 n)-2(5+6 n)}{(3)(5+6 n)}\right|=\left|\frac{-7}{(3)(5+6 n)}\right|$$

## 数学代写|实变函数作业代写Real analysis代考|Sequence Spaces

Let $S$ denote the set of all sequences of real numbers. To make it easy to write them down, let’s assume all these sequences start at the same integer, say $k$. So $S$ is the set of all objects $x$ where $x$ is a sequence of the form $\left(a_{n}\right){n \geq k}$. Thus, $S=\left{x: x=\left(a{n}\right)_{n \geq k}\right}$.

• We define addition operation + like this: $x+y$ is the new sequence $\left(a_{n}+b_{n}\right){n \geq k}$ when $x=\left(a{n}\right){n \geq k}$ and $y=\left(b{n}\right)_{n \geq k}$. So if $x=\left(3+4 / n+5 / n^{2}\right)$ and $y=\left(-7+\sin (n+2)+3 / n^{3}\right)$ the sequence $x+y=\left(-4+4 / n+5 / n^{2}+3 / n^{3}+\sin (n+2)\right)$.
• We can do a similar thing with the subtraction operation, $-$.
• We can scale a sequence with any number $\alpha$ by defining the sequence $\alpha x$ to be $\alpha x=$ $\left(\alpha a_{n}\right)_{n \geq k}$. Thus, the sequence $2\left(13+5 / n^{4}\right)=\left(26+10 / n^{4}\right)$.
• With these operations, $S$ is a vector space over the Real numbers. This is an idea you probably heard about in your Linear Algebra course.

The set of all sequences that converge is also a vector space which we denote by $c$ but we have to prove that the new sequence $\alpha\left(a_{n}\right)+\beta\left(b_{n}\right)$ also converges when we know $\left(a_{n}\right)$ and $\left(b_{n}\right)$ converge.
The set of all sequences that converge to zero is also a vector space which we denote by $c_{0}$ but we have to prove that the new sequence $\alpha\left(a_{n}\right)+\beta\left(b_{n}\right)$ also converges to 0 when we know $\left(a_{n}\right)$ and $\left(b_{n}\right.$ ) converge to 0 .

## 数学代写|实变函数作业代写Real analysis代考|Convergent Sequences have Unique Limits

n>ñ1⇒|Xn−一种|<ε/2 n>ñ2⇒|Xn−b|<ε/2现在选一个P>\max \left{N_{1}, N_{2}\right}P>\max \left{N_{1}, N_{2}\right}. 那么，我们有

|一种−b|=|一种−X磷+X磷−b|≤|一种−X磷|+|X磷−b|<ε/2+ε/2=ε自从ε>0是任意的并且|一种−b|<ε对全部ε>0，这表明一种=b.

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

|一种n−一种|=|(1+(4/n))−1|=|4/n|=4|1/n|

n>ñ>4/ε⇒|(1+(4/n))−1|<ε自从∈任意选择，我们知道序列收敛到1.收敛证明遵循我们接下来展示的一个很好的模板。定理 3.3.6一种n→一种模板

∣ 原始序列 – 建议的限制 ∣=∣ 使用代数等简化以获得新表达式 ∣ ≤∣ 使用三角不等式等再次获得新的表达式 ∣ = 称这最后一步为高估。我们现在有  =∣ 高估 ∣<ε第 4 步：求解n按照ε给出一个简单的方程。第 5 步：选择ñ满足你从 Step 得到的不等式4.第6步：然后对于任何n>ñ, |高估∣<ε我们有∣（原来的 ）−（建议限制）∣<ε证明一种n→一种. 定理 3.3.7(1+4n5+6n)收敛(1+4n5+6n)n≥1收敛。证明 3.3.7 我们可以猜测极限的值是一种=2/3所以选择ε>0任意。考虑

|1+4n5+6n−23|=|3(1+4n)−2(5+6n)(3)(5+6n)|=|−7(3)(5+6n)|

## 数学代写|实变函数作业代写Real analysis代考|Sequence Spaces

• 我们这样定义加法运算+：X+是是新的序列(一种n+bn)n≥ķ什么时候X=(一种n)n≥ķ和是=(bn)n≥ķ. 因此，如果X=(3+4/n+5/n2)和是=(−7+罪⁡(n+2)+3/n3)序列X+是=(−4+4/n+5/n2+3/n3+罪⁡(n+2)).
• 我们可以用减法运算做类似的事情，−.
• 我们可以用任意数字缩放一个序列一种通过定义序列一种X成为一种X= (一种一种n)n≥ķ. 因此，序列2(13+5/n4)=(26+10/n4).
• 通过这些操作，小号是实数上的向量空间。这是您可能在线性代数课程中听说过的一个想法。

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## MATLAB代写

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