---
title: 微積分(一)
tags: 微積分(一), 共同筆記
---

[TOC]

# 微積分(一)
>原文書:
>Calculus: Early Transcendentals (Metric Version) 9th edition
>
>>Author: James Stewart, Daniel K. Clegg, Saleem Watson
## 教授相關資訊
- 教授的名字 : 高橋亮甫(a.k.a 楊劼之)
- Office
數學系館 320
## 考試日程及配分
- 考試日期
Mid. Exam I 2020-10-14
Mid. Exam II 2020-11-25
Final Exam 2021-01-06
- 配分
HW=30%
MEI=20%
MEII=20%
FE=30%
## 教學內容

### 9/9

#### Calculus = Differentiation + Integration
Calculus is about functions.
- Differentiation 微分 $\rightarrow$ 瞬間變化率
- Integration 積分 $\rightarrow$ 函式圖形下面積

#### Fundamental Theorem of Calculus
Differentiation $\overset{反操作}\longleftrightarrow$ Integration

i.e.
$$
F \xrightarrow{differentiation} f \xrightarrow{Integration} F
$$

#### Function
*Definition*

Let $X$, $Y$ be 2 sets.
A function $f: X\rightarrow Y$ is a collection of pairs $(x,y) \in X\times Y$, denoted by $\Gamma(f)$ such that
1. For any $x\in X$, there's a $y$ such that $(x,y)\in\Gamma(f)$
2. For any $(x,y)\in\Gamma(f)$ and $(x,y')\in\Gamma(f)$, we have that $y = y'$

*Examples*
- $f(x) = x^2, \;f:\mathbb{R}\rightarrow\mathbb{R}$ is a fuction.
- $f(x) = \frac{1}{x(x-1)}, \;f:\mathbb{R}-\{0,1\}\rightarrow \mathbb{R}$ is a function.
- $f(x) = \sqrt{x}, \;f:\mathbb{R^+}\rightarrow\mathbb{R}$ is a function.
- $f(x) = sin(x), \;f:R\rightarrow R$ is a function.

#### Rational Function
> A rational function $f$ is a ratio of two polynomials:
> $$
> f(x) = \frac{P(x)}{Q(x)}
> $$
> where $P$ and $Q$ are polynomials. The domain consists of all values of $x$ such that $Q(x) \neq 0$.
> [name=Calculus: 7/e]

:::warning
老師上課時講的是沒法被準確算出為Rational Function
上面是同作者書中7th edition的說法
:::

*Examples*
- $f(x) = x^2, \;f:\mathbb{R}\rightarrow\mathbb{R}$ is a rational fuction.
- $f(x) = \frac{1}{x(x-1)}, \;f:\mathbb{R}-\{0,1\}\rightarrow \mathbb{R}$ is a rational function.

#### Irrational Function
相對於Rational Function

*Examples*
- $f(x) = \sqrt{x}, \;f:\mathbb{R^+}\rightarrow\mathbb{R}$ is irrational a function.
- $f(x) = sin(x), \;f:R\rightarrow R$ is a irrational function.

#### How to calculate $sin(x)$ ?
*Step1*
$$
\begin{aligned}
sin(2x) &= 2sin(x)cos(x) \\
&= 2sin(x)(1-2sin^2(\frac{x}{2})) \\
&= 2(2sin(\frac{x}{2})cos(\frac{x}{2}))(1-2sin^2(\frac{x}{2})) \\
&= 2[2sin(\frac{x}{2})(1-2sin^2(\frac{x}{4}))](1-2sin^2(\frac{x}{2})) \\
&= .......
\end{aligned}
$$
Value in $sin$ is getting smaller and smaller.

*Step2*

Let $X > 0$, $X$ is a small number.


We can see that
$$
sin(x) < x \tag{1}
$$


Since two triangles are similar triangles (AA),
we know that
$$
\frac{r}{1} = \frac{sin(x)}{\sqrt{1-sin^2(x)}}
$$

The area of the circular sector is
$$
\frac{1^2x}{2} = \frac{x}{2}
$$
and it should be less than the area of the triangle $\frac{r}{2}$, so we have
$$
\begin{aligned}
\frac{x}{2} &< \frac{r}{2} = \frac{sin(x)}{2\sqrt{1-sin^2(x)}} \\
&< \frac{sin(x)}{2\sqrt{1-x^2}} \\
&< \frac{sin(x)}{2(1-x^2)}
\end{aligned}
$$

$$
\begin{aligned}
&\frac{x}{2} < \frac{sin(x)}{2(1-x^2)} \\
&\implies x < \frac{sin(x)}{1-x^2} \\
&\implies x-x^3 < sin(x)
\end{aligned} \tag{2}
$$

By combining (1) and (2)
$$
x-x^3 < sin(x) < x
$$
which infers that
$sin(x) \approx x$ when $x$ is small.

With this idea and the formula in step 1, we can estimate the value of $sin(x)$ by applying the formula to it, and when the variable of $sin$ gets small enough, replace it with the variable itself.

#### Properties of functions

$X,\;Y \subset R$
Let $f: \;X \rightarrow Y$ be a function.

*Injective function* (1對1)
If $f(x) \neq f(x')$ for all $x \neq x'$, we say $f$ is injective (one to one).

*Surjective function* (映成)
For any $y \in Y$, there exists $x \in X$ such that $f(x) = y$, we say that $f$ is surjective.

*Bijective function*
If $f$ in injective and surjective, we say that $f$ is bijective (one-to-one).

*Inverse function*
For any bijective function $f: \;X \rightarrow Y$, there exists an inverse function $f^{-1}: \;Y \rightarrow X$, such that
$$
(f \circ f^{-1}): \;Y \rightarrow Y \\
(f \circ f^{-1})(y) = y \\
(f^{-1} \circ f): \;X \rightarrow X \\
(f^{-1} \circ f)(x) = x
$$

### 9/11

#### Rational Function
*Definition*
$$
\frac{P(x)}{Q(x)}, \;where \;P,Q \;are \;polynomials, \; Q(x) \neq 0
$$

#### Irrational Function
Examples: $sin(x)$, $\sqrt{x}$, $\log x$, $a^x \;(a>0)$

#### How to solve $a^x$
`TODO`

#### Interval notation
Let $a, \;b$ be 2 real numbers, $a<b$, we define that
$$
(a,b) = \{x \in \mathbb{R} \mid a\lt x\lt b\} \;\text{(open interval)} \\
[a,b] = \{x \in \mathbb{R} \mid a\le x \le b\} \;\text{(close interval)} \\
(a,b] = \{x \in \mathbb{R} \mid a\lt x\le b\} \\
[a, b) = \{x \in \mathbb{R} \mid a\le x \lt b\}
$$

#### Neighborhood
*Definition*
Let $a \in \mathbb{R}$, a neighborhood of a is $(a-c,a+c)$ for some $c \gt 0$


> The limit of a point is decided by its neighborhood.

#### Limit of a function
*Explanation*
*(nbd.=neighborhood)*
Let$f:\mathbb{R} \to\mathbb{R}$ be a function,
we call $\lim\limits_{x\to a}f(x)=L\iff$For any small nbd. of $L$, $f(x)$ is in this nbd. if $x$ is in another small nbd. of $a(x\ne a)$
*Q:為什麼$x\ne a$？*
*師說：驗證某人是否為好人，可訪問其本人及鄰居，但本人給出的評價必定有所偏頗，故此採樣不計入統計數據。*

*Definition*
Let $f(x)$ be a function,
$a\in\mathbb{R}$, $\lim\limits_{x\to a}f(x)=L\iff$For any $\varepsilon > 0$,
there exists a $\delta > 0$,
such that $\overset{\text{f(x) is in a small nbd. of L}}{|f(x)-L| < \varepsilon}$
*i.e. $f(x)\in(L-\varepsilon, L+\varepsilon)$*,

when $\overset{\text{x is in a small nbd. of a}}{0<|x-a|<\delta}$
*i.e. $x\in(a-\delta, a+\delta)$*,

:::info
$\varepsilon$: Epsilon
$\delta$: Delta
:::

### 9/16
#### Examples of limit
1. $\lim\limits_{x \to 5}1 = 1$
***analysis***
We have to find a $\delta \gt 0$ such that for every $\varepsilon \gt 0$, $|1-1| \lt \varepsilon$ when $0 \lt |x-5| \lt \delta$
It is clear that we can choose any number greater than 0 as $\delta$ because $|1-1| \lt \varepsilon$ is always true.
***proof***
Given $\varepsilon > 0$, choose $\delta = 1$.
If $0 \lt |x-5| \lt \delta$, then
$$
|1-1| = 0 \lt \varepsilon
$$
Thus, if $0 \lt |x-5| \lt \delta$ then $|1-1| < \varepsilon$
Therefore, by the definition of a limit,
$$
\lim\limits_{x \to 5}1 = 1
$$
2. $\lim\limits_{x \to 3}x = 3$
***analysis***
We have to find a $\delta \gt 0$ such that for every $\varepsilon \gt 0$, $|x-3| \lt \varepsilon$ when $0 \lt |x-3| \lt \delta$
This suggest that we can choose $\delta = \varepsilon$ so that
$$
|x-3| \lt \delta = \varepsilon
$$
is true.
***proof***
Given $\varepsilon > 0$, choose $\delta = \varepsilon$.
If $0 \lt |x - 3| \lt \delta$, then
$$
|x - 3| < \delta = \varepsilon
$$
Thus, if $0 \lt |x - 3| \lt \delta$ then $|x - 3| \lt \varepsilon$
Therefore, by the definition of a limit,
$$
\lim\limits_{x \to 3}x = 3
$$
3. $\lim\limits_{x \to 2}(3x^2+1) = 13$
***analysis***
We have to find a $\delta \gt 0$ such that for every $\varepsilon \gt 0$, $|3x^2+1-13| \lt \varepsilon$ when $0 \lt |x-2| \lt \delta$
We know that
$$
\begin{aligned}
|3x^2+1-13|
&= |3x^2-12| \\
&= 3|x^2-4| \\
&= 3|x+2||x-2|
\end{aligned}
$$
Therefore, we want a $\delta$ such that $3|x-2||x+2| \lt \varepsilon$ when $0 \lt |x-2| \lt \delta$
Since we only care about x values that are close to 2, we can assume that x is within a distance 1 from 2, that is, $|x-2| \lt 1$.
So $1 \lt x \lt 3$, then $3 \lt x+2 \lt 5$
And we can further say that
$$
3|x+2||x-2| \lt 15|x-2|
$$
So we can make $15|x-2| \lt \varepsilon$ by taking $|x-2| \lt \dfrac{\varepsilon}{15}$, and choose $\delta = \dfrac{\varepsilon}{15}$
But remember that $|x-2|$ now has 2 restrictions:
$|x-2| \lt 1$ and $|x-2| \lt \dfrac{\varepsilon}{15}$
In order to satisfy both of the inequalities, we choose $\delta = min\left\{1, \dfrac{\varepsilon}{15}\right\}$
***proof***
Given $\varepsilon$, choose $\delta = min\left\{1, \dfrac{\varepsilon}{15}\right\}$
If $0 \lt |x-2| \lt \delta$, then $|x-2| \lt 1\implies 1 \lt x \lt 3\implies |x+2| < 5$
so
$$
\begin{aligned}
|3x^2+1-13|
&= |3x^2-12| \\
&= 3|x^2-4| \\
&= 3|x+2||x-2| \\
&\lt 15\delta \\
&\lt 15\left(\dfrac{\varepsilon}{15}\right) = \varepsilon
\end{aligned}
$$
Thus, $|3x^2+1-13| \lt \varepsilon$ when $0 \lt |x-2| \lt \delta$
Therefore, by the definition of a limit
$$
\lim\limits_{x \to 2}(3x^2+1) = 13
$$

#### Basic rules of Limit
Let $\lim\limits_{x \to a}f(x) = L$, $\lim\limits_{x \to a}g(x) = M$, $c \in \mathbb{R}$, then
1. $\lim\limits_{x \to a}(f(x)+g(x)) = L+M$
2. $\lim\limits_{x \to a}cf(x) = cL$
3. $\lim\limits_{x \to a}(f(x)g(x)) = LM$
4. $\lim\limits_{x \to a}\dfrac{f(x)}{g(x)} = \dfrac{L}{M}, \;M \neq 0$