### 数学代写|微积分代写Calculus代写|MTH2010

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• 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|微积分代写Calculus代写|Some More Examples

In the following examples a particular procedure is adopted to show continuity or discontinuity of a function. The reader may adopt any other alternate procedure, for instance, any one of the characterizations in Theorem 2.2.1.

Example 2.2.4 For given $x_{0} \in \mathbb{R}$, let $f(x)=\left|x-x_{0}\right|, x \in \mathbb{R}$. Then $f$ is continuous on $\mathbb{R}$. To see this, note that, for $a \in \mathbb{R}$,
$$|f(x)-f(a)|=|| x-x_{0}|-| a-x_{0}|| \leq\left|\left(x-x_{0}\right)-\left(a-x_{0}\right)\right|=|x-a| .$$
Hence, for every $\varepsilon>0$, we have
$$|x-a|<\varepsilon \Rightarrow|f(x)-f(a)|<\varepsilon$$
Example 2.2.5 Recall from Example 2.2.9 that
$$\lim _{x \rightarrow 2} \frac{x^{2}-4}{x-2}=4$$
Hence, $f$ defined by
$$f(x):= \begin{cases}\frac{x^{2}-4}{x-2}, & x \neq 2 \ 4, & x=2\end{cases}$$

is continuous at 2 . However, the function
$$g(x):= \begin{cases}\frac{x^{2}-4}{x-2}, & x \neq 2 \ \alpha, & x=2\end{cases}$$
is not continuous at 2 for any $\alpha \neq 4$.
Also, if $a \neq 2$, then $x-2$ is nonzero in a neighbourhood of $a$, and the functions $x-2$ and $x^{2}-4$ are continuous on $\mathbb{R}$. Hence, by Theorem $2.2 .6, f$ is continuous at every $a \neq 2$ as well. By similar arguments, $g$ is continuous at every point $a \neq 2$. $\diamond$
Example 2.2.6 We already observed in Example 2.2.3 that the functions $f, g, h$ defined by
$$f(x)=\sin x, \quad g(x)=\cos x, \quad h(x)= \begin{cases}\frac{\sin x}{x}, & x \neq 0 \ 1, & x=0\end{cases}$$
are continuous at 0 . Now, we show that they are continuous at every point in $\mathbb{R}$.
Note that for $x, y \in \mathbb{R}$,
$$\sin x-\sin y=2 \sin \left(\frac{x-y}{2}\right) \cos \left(\frac{x+y}{2}\right)$$
so that
$$|\sin x-\sin y| \leq|x-y| \quad \forall x, y \in \mathbb{R}$$

## 数学代写|微积分代写Calculus代写|Some Properties of Continuous Functions

Recall that a subset $S$ of $\mathbb{R}$ is said to be bounded if there exists $M>0$ such that $|s| \leq M$ for all $s \in S$, and a set which is not bounded is called an unbounded set. Recall also that if $S$ is a bounded subset of $\mathbb{R}$, then $S$ has the infimum and the supremum, not necessarily in $S$.
For $S \subseteq \mathbb{R}$, we have the following:

1. Suppose $S$ is bounded, and say $\alpha:=\inf S$ and $\beta:=\sup S$. Then there exist sequences $\left(s_{n}\right)$ and $\left(t_{n}\right)$ in $S$ such that $s_{n} \rightarrow \alpha$ and $t_{n} \rightarrow \beta$.
2. $S$ is unbounded if and only if there exists a sequence $\left(s_{n}\right)$ in $S$ which is unbounded.
3. $S$ is unbounded if and only if there exists a sequence $\left(s_{n}\right)$ in $S$ such that $\left|s_{n}\right| \rightarrow \infty$ as $n \rightarrow \infty$
4. If $\left(s_{n}\right)$ is a sequence in $S$ which is unbounded, then there exists a subsequence $\left(s_{k_{n}}\right)$ of $\left(s_{n}\right)$ such that $\left|s_{k_{n}}\right| \rightarrow \infty$ as $n \rightarrow \infty$
5. If $\left(s_{n}\right)$ is a sequence in $S$ such that $\left|s_{n}\right| \rightarrow \infty$ as $n \rightarrow \infty$, and if $\left(s_{k_{n}}\right)$ is a subsequence of $\left(s_{n}\right)$, then $\left|s_{k_{n}}\right| \rightarrow \infty$ as $n \rightarrow \infty$.
Exercise 2.2.4 Prove the above statements.
Theorem 2.2.7 Suppose $f$ is a real valued continuous function defined on a closed and bounded interval $[a, b]$. Then $f$ is a bounded function.

Proof Assume for a moment that $f$ is not a bounded function. Then, there exists a sequence $\left(x_{n}\right)$ in $[a, b]$ such that $\left|f\left(x_{n}\right)\right| \rightarrow \infty$. Since $\left(x_{n}\right)$ is a bounded sequence, by Bolzano-Weierstrass theorem (Theorem 1.1.13), there exists a subsequence $\left(x_{k_{n}}\right)$ of $\left(x_{n}\right)$ such that $x_{k_{n}} \rightarrow x$ for some $x \in[a, b]$. Therefore, by the continuity of $f$, $f\left(x_{k_{n}}\right) \rightarrow f(x)$. In particular, $\left(f\left(x_{k_{n}}\right)\right.$ ) is a bounded sequence. This is a contradiction to the fact that $\left|f\left(x_{n}\right)\right| \rightarrow \infty$. Thus, we have proved that $f$ cannot be unbounded.

## 数学代写|微积分代写Calculus代写|Continuity of the Inverse of a Function

Suppose $f$ is defined on a set $D \subseteq \mathbb{R}$. We may recall the following from elementary set theory:

If $f$ is injective, i.e., one-one, then we know that a function $g$ can be defined on the range $E:=f(D)$ of $f$ by $g(y)=x$ for $y \in E$, where $x \in D$ is the unique element in $D$ such that $f(x)=y$. The above function $g$ is called the inverse of $f$. Note that the domain of the inverse of $f$ is the range of $f$.

By Corollary 2.2.10, we know that range of a continuous function defined on an interval $I$ is also an interval. Suppose $f$ is also injective. Then a natural question one would like to ask is whether its inverse is also continuous. First we answer this question affirmatively by assuming that the domain of the function is a closed and bounded interval.

Theorem 2.2.13 (Inverse function theorem (IFT)) Let $f$ be a continuous injective function defined on a closed and bounded interval I. Then its inverse from its range is continuous.

Proof Suppose $J=f(I)$, the range of $f$. Let $y_{0} \in J$ and $\left(y_{n}\right)$ be an arbitrary sequence in $J$ which converges to $y_{0}$. Let $x_{n}=f^{-1}\left(y_{n}\right), n \in \mathbb{N}$ and $x_{0}=f^{-1}\left(y_{0}\right)$. We have to show that $x_{n} \rightarrow x_{0}$.

Suppose, on the contrary, that $x_{n} \nrightarrow x_{0}$. Then there exists $\varepsilon_{0}>0$ and a subsequence $\left(u_{n}\right)$ of $\left(x_{n}\right)$ such that $u_{n} \notin\left(x_{0}-\varepsilon_{0}, x_{0}+\varepsilon_{0}\right)$ for all $n \in \mathbb{N}$. Since $I$ is a bounded interval, $\left(u_{n}\right)$ is a bounded sequence. Hence, $\left(u_{n}\right)$ has a subsequence $\left(v_{n}\right)$

which converges to some $v \in \mathbb{R}$. Since $I$ is a closed interval, $v \in I$. Now, continuity of $f$ implies that $f\left(v_{n}\right) \rightarrow f(v)$. But, since $\left(f\left(v_{n}\right)\right)$ is a subsequence of $\left(y_{n}\right)$, and since $y_{n} \rightarrow y_{0}$, we have $f(v)=y_{0}=f\left(x_{0}\right)$. Now, since $f$ is injective, $v=x_{0}$. Thus we have proved that $v_{n} \rightarrow x_{0}$. This is a contradiction to the fact that $v_{n} \notin\left(x_{0}-\varepsilon_{0}, x_{0}+\varepsilon_{0}\right)$ for all $n \in \mathbb{N}$.

Next we shall prove the conclusion in the last theorem by dropping the condition that $I$ is closed and bounded, but assuming an additional condition on $f$, namely that it is strictly monotonic.

## 数学代写|微积分代写Calculus代写|Some More Examples

|F(X)−F(一个)|=||X−X0|−|一个−X0||≤|(X−X0)−(一个−X0)|=|X−一个|.

|X−一个|<e⇒|F(X)−F(一个)|<e

F(X):={X2−4X−2,X≠2 4,X=2

G(X):={X2−4X−2,X≠2 一个,X=2

F(X)=罪⁡X,G(X)=因⁡X,H(X)={罪⁡XX,X≠0 1,X=0

|罪⁡X−罪⁡是|≤|X−是|∀X,是∈R

## 数学代写|微积分代写Calculus代写|Some Properties of Continuous Functions

1. 认为小号是有界的，说一个:=信息小号和b:=支持小号. 那么存在序列(sn)和(吨n)在小号这样sn→一个和吨n→b.
2. 小号无界当且仅当存在一个序列(sn)在小号这是无界的。
3. 小号无界当且仅当存在一个序列(sn)在小号这样|sn|→∞作为n→∞
4. 如果(sn)是一个序列小号无界，则存在子序列(sķn)的(sn)这样|sķn|→∞作为n→∞
5. 如果(sn)是一个序列小号这样|sn|→∞作为n→∞， 而如果(sķn)是一个子序列(sn)， 然后|sķn|→∞作为n→∞.
练习 2.2.4 证明上述陈述。
定理 2.2.7 假设F是定义在闭有界区间上的实值连续函数[一个,b]. 然后F是有界函数。

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