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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|微积分代写Calculus代写|Continuous Functions

Let $f$ be defined on $(a, b)$, and choose $a<c<b$. We say that $f$ is continuous at $c$ if
$$\lim _{x \rightarrow c} f(x)=f(c) .$$
If $f$ is continuous at every real $c$ in $(a, b)$, then, we say that $f$ is continuous on $(a, b)$ or, if $(a, b)$ is understood from the context, $f$ is continuous.

Recalling the definition of $\lim _{x \rightarrow c}$, we see that $f$ is continuous at $c$ iff, for all sequences $\left(x_n\right)$ satisfying $x_n \rightarrow c$ and $x_n \neq c, n \geq 1, f\left(x_n\right) \rightarrow f(c)$. In fact, $f$ is continuous at $c$ iff $x_n \rightarrow c$ implies $f\left(x_n\right) \rightarrow f(c)$, i.e., the condition $x_n \neq c, n \geq$ 1 , is superfluous. To see this, suppose that $f$ is continuous at $c$, and suppose that $x_n \rightarrow c$, but $f\left(x_n\right) \nrightarrow f(c)$. Since $f\left(x_n\right) \nrightarrow f(c)$, by Exercise $1.5 .8$, there is an $\epsilon>0$ and a subsequence $\left(x_n^{\prime}\right)$, such that $\left|f\left(x_n^{\prime}\right)-f(c)\right| \geq \epsilon$ and $x_n^{\prime} \rightarrow c$, for $n \geq 1$. But, then, $f\left(x_n^{\prime}\right) \neq f(c)$ for all $n \geq 1$, hence, $x_n^{\prime} \neq c$ for all $n \geq 1$. Since $x_n^{\prime} \rightarrow c$, by the continuity at $c$, we obtain $f\left(x_n^{\prime}\right) \rightarrow f(c)$, contradicting $\left|f\left(x_n^{\prime}\right)-f(c)\right| \geq \epsilon$. Thus, $f$ is continuous at $c$ iff $x_n \rightarrow c$ implies $f\left(x_n\right) \rightarrow f(c)$. In the previous section we saw that $f(x)=x^2$ is continuous at $c$. Since this works for any $c, f$ is continuous. Repeating this argument, one can show that $f(x)=x^4$ is continuous, since $x^4=x^2 x^2$. A simpler example is to choose a real $k$ and to set $f(x)=k$ for all $x$. Here, $f\left(x_n\right)=k$, and $f(c)=k$ for all sequences $\left(x_n\right)$ and all $c$, so, $f$ is continuous. Another example is $f:(0, \infty) \rightarrow \mathbf{R}$ given by $f(x)=1 / x$. By the division property of sequences, $x_n \rightarrow c$ implies $1 / x_n \rightarrow 1 / c$ for $c>0$, so, $f$ is continuous.

Functions can be continuous at various points and not continuous at other points. For example, the function $f$ in Exercise 2.2.1 is continuous at every irrational $c$ and not continuous at every rational $c$. On the other hand, the function $f: \mathbf{R} \rightarrow \mathbf{R}$, given by $(\S 2.2)$
$$f(x)= \begin{cases}1, & x \in \mathbf{Q} \ 0, & x \notin \mathbf{Q}\end{cases}$$
is continuous at no point.

数学代写|微积分代写Calculus代写|Derivatives

Let $f$ be defined on $(a, b)$, and choose $c \in(a, b)$. We say that $f$ is differentiable at $c$ if
$$\lim _{x \rightarrow c} \frac{f(x)-f(c)}{x-c}$$
exists as a real, i.e., exists and is not $\pm \infty$. If it exists, we denote this limit $f^{\prime}(c)$ or $\frac{d f}{d x}(c)$, and we say that $f^{\prime}(c)$ is the derivative of $f$ at $c$. If $f$ is differentiable at $c$ for all $a<c<b$, we say that $f$ is differentiable on $(a, b)$ or, if it is clear from the context, differentiable. In this case, the derivative $f^{\prime}:(a, b) \rightarrow \mathbf{R}$ is a function defined on all of $(a, b)$.

For example, the function $f(x)=m x+b$ is differentiable on $\mathbf{R}$ with derivative $f^{\prime}(c)=m$ for all $c$ since
$$\lim {x \rightarrow c} \frac{(m x+b)-(m c+b)}{x-c}=\lim {x \rightarrow c} m=m .$$
Since its graph is a line, the derivative of $f(x)=m x+b$ (at any real) is the slope of its graph. In particular, the derivative of a constant function $f(x)=b$ for all $x$ is zero.
If $f(x)=x^2$, then, $f$ is differentiable with derivative
$$f^{\prime}(c)=\lim {x \rightarrow c} \frac{x^2-c^2}{x-c}=\lim {x \rightarrow c} \frac{(x-c)(x+c)}{x-c}=\lim {x \rightarrow c}(x+c)=2 c .$$ If $f$ is differentiable at $c$, then, \begin{aligned} \lim {x \rightarrow c} f(x) &=\lim {x \rightarrow c}\left[\left(\frac{f(x)-f(c)}{x-c}\right)(x-c)+f(c)\right] \ &=\lim {x \rightarrow c}\left(\frac{f(x)-f(c)}{x-c}\right) \lim _{x \rightarrow c}(x-c)+f(c) \end{aligned}

$$=f^{\prime}(c) \cdot 0+f(c)=f(c) .$$
So, $f$ is continuous at $c$. Hence, a differentiable function is continuous. However, $f(x)=|x|$ is continuous at 0 but not differentiable there since
$$\lim {x \rightarrow 0+} \frac{|x|-|0|}{x-0}=1,$$ whereas $$\lim {x \rightarrow 0-} \frac{|x|-|0|}{x-0}=-1 .$$
However,
$$(|x|)^{\prime}=\frac{x}{|x|}, \quad x \neq 0,$$
since $|x|=x$, hence, $(|x|)^{\prime}=1$ on $(0, \infty)$, and $|x|=-x$, hence, $(|x|)^{\prime}=-1$ on $(-\infty, 0)$
Derivatives are computed using their arithmetic properties.

微积分代考

数学代写|微积分代写微积分代写|连续函数

$$\lim _{x \rightarrow c} f(x)=f(c) .$$

$$f(x)= \begin{cases}1, & x \in \mathbf{Q} \ 0, & x \notin \mathbf{Q}\end{cases}$$

数学代写|微积分代写calculus代写|Derivatives

.

$$\lim _{x \rightarrow c} \frac{f(x)-f(c)}{x-c}$$

$$\lim {x \rightarrow c} \frac{(m x+b)-(m c+b)}{x-c}=\lim {x \rightarrow c} m=m .$$

$$f^{\prime}(c)=\lim {x \rightarrow c} \frac{x^2-c^2}{x-c}=\lim {x \rightarrow c} \frac{(x-c)(x+c)}{x-c}=\lim {x \rightarrow c}(x+c)=2 c .$$如果$f$在$c$可微，则\begin{aligned} \lim {x \rightarrow c} f(x) &=\lim {x \rightarrow c}\left[\left(\frac{f(x)-f(c)}{x-c}\right)(x-c)+f(c)\right] \ &=\lim {x \rightarrow c}\left(\frac{f(x)-f(c)}{x-c}\right) \lim _{x \rightarrow c}(x-c)+f(c) \end{aligned}

$$=f^{\prime}(c) \cdot 0+f(c)=f(c) .$$

$$\lim {x \rightarrow 0+} \frac{|x|-|0|}{x-0}=1,$$而$$\lim {x \rightarrow 0-} \frac{|x|-|0|}{x-0}=-1 .$$

$$(|x|)^{\prime}=\frac{x}{|x|}, \quad x \neq 0,$$

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MATLAB代写

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