---
tags: course_note
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# Calculus A 1

[TOC]

## 09/12

### Syllabus

[hackmd link](/@teshenglin/ByzbCMses)

### intro

Math is a language for scientists to communicate with other scientints.

We care about the final answer and how you come up with the answer

### ch 1

A function $f$ is a rule that assigns to each element $x$ in a set $D$ exactly one element, call $f(x)$, in a set $E$.

$D$ : domain, what can be feed into the function

if we define $y=f(x)$
$y$ : dependent variable

range is the set of all possible output values of the function

---

for now on we consider $f:\mathbb{R}\rightarrow \mathbb{R}$

$\mathbb{R}$ : the set of real number

$[0,1]=\{y \mid 0\leq y\leq 1\}$
*notice that $y$ is default in $\mathbb{R}$ because the number has been compared to other real numbers*

---

Common notation in Calculus I

$x$ : independent variable
$f$ : function
$y$ : dependent variable

### Image classification

MNIST data set.

$f(some\ handwriting\ image) = {0,1,2,...,9}$

convolutional neural network (CNN)

D = specific resolution image
range = {0,1,2,...,9}

### Ch 1.1

Find the domain of the function

#39 $f(x)=\frac{x+4}{x^2-9}$

$x^2-9\neq0$
$\Rightarrow x\neq\pm3$

answer: $\{x\mid x\in\mathbb{R},x\neq\pm3\}$

#42 $g(t)=\sqrt{3-t}-\sqrt{2+t}$

$2+t\geq0$
$\Rightarrow t\geq-2$
$3-t\geq0$
$\Rightarrow t\leq 3$
answer: $\{t\mid-2\leq t\leq 3\}$

#45 $F(p)=\sqrt{2-\sqrt{p}}$

$p\geq0$
$2-\sqrt{p}\geq0$
$\Rightarrow 2\geq\sqrt{p}$
$\Rightarrow [0,4]$
answer: $\{p\mid p\in[0,4]\}$

---

composite function

$(f\circ g)(x)=f(g(x))$

e.g.

$f(x)=\sqrt{x}$ , $g(x)=\sqrt{2-x}$

what is the domain?

(a) $(f\circ g)(x)$

$(f\circ g)(x)=\sqrt{\sqrt{2-x}}$
$\Rightarrow \sqrt{2-x}\geq0$
$\Rightarrow 2-x\geq0$
$\Rightarrow x\leq 2$
answer: $\{x\mid x\leq 2\}$

(b) $(g\circ f)(x)$

$(g\circ f)(x)=\sqrt{2-\sqrt{x}}$
$\Rightarrow 2-\sqrt{x}\geq0$

answer: $\{x\mid 0\leq x\leq 4\}$

(<!---->c) $(f\circ f)(x)$

answer: $\{x\mid -2\leq x\leq 2\}$

---

e.g.

$f(x)=\begin{cases}x&x<0\\x^2&x\geq0\end{cases}$

---

e.g.

Heaviside function

$H(t)=\begin{cases}0&t<0\\1&t\geq0\end{cases}$


---

**polynomial function**

$p(t)=p_0+p_1t+p_2t^2+...+p_nt^n$

e.g.

$p(t)=1+3t+t^4+t^{10}$

**Power function**

$f(x)=x^\alpha, \alpha\in\mathbb{R}$

e.g.

$f(x)=x^\frac{1}{2}=\sqrt{x}, x\geq0$

$p(t)=t^{-\frac{1}{4}}=(t^{-1})^\frac{1}{4}=(\frac{1}{t})^\frac{1}{4}=\frac{1}{\sqrt[4]{t}}, t>0$

---

**sinusoidal function**

$\sin(x),\cos(x),\tan(x),\cot(x),\sec(x),\csc(x)$

## 09/15

Machine learning: To learn a function

MNIST, Yolo v4, AlphaGo, Language Translation, Natural language Programming...

**piecewise function**

e.g.

$f(x)=\begin{cases}\sin(x)&x>0\\\cos(x)&x\leq 0\end{cases}$

### sec 1.4

**Exponential functions**

e.g.

$f(x)=b^x$

$for x\in \mathbb{N}$

$\mathbb{N}$ : natural number = $\{1,2,3...\}$

e.g.
$f(3)=b^3=b\times b\times b$

$f(-3)=(b^3)^{-1}=\frac{1}{b^3}=\frac{1}{b\times b\times b}$

$f(0)=b^0=b^{3-3}=b^3\times\frac{1}{b^3}=1$

$\Rightarrow b^0=1\ for\ b\neq 0$
because $0^0$ if not defined

For $x\in \mathbb{Q}$

$x=\frac{g}{p}$
$b^x=b^{\frac{g}{p}}=(b^g)^\frac{1}{p}$

**law of exponents**

$b^x\cdot b^y=b^{x+y}$
$(b^x)^y=b^{xy}$

---

e.g.

$f(x)=2^x$

|$x=1$|$x=2$|$x=10$|$x=100$|
|-|-|-|-|
|$2$|$4$|$1024$|$\approx10^{30}$|

:::info
Exponential functions grow much faster than any polynomial fucntions
:::

e.g.

$f(x)=x^2$

|$x=1$|$x=2$|$x=10$|$x=100$|
|-|-|-|-|
|$1$|$4$|$100$|$10^4$|

$f(x)=x^10$

|$x=1$|$x=2$|$x=10$|$x=100$|
|-|-|-|-|
|$1$|$1024$|$10^{10}$|$10^{20}$|

**in small scale**

**in bigger scale**

$f(x)=2^x$ is going to catch up $x^10$

**There are 3 intersections between $y=2^x$ and $y=x^{10}$**

---

exercise

$8^{\frac{4}{3}}=2^{?}$

$(2^3)^\frac{4}{3}=2^?$

$2^{3\times \frac{4}{3}}=2^?$

$?=3\times\frac{4}{3}=4$

---

$\frac{\sqrt{a\sqrt{b}}}{\sqrt[3]{ab}}=a^?b^?$

$= \frac{(ab^{\frac{1}{2}})^\frac{1}{2}}{(ab)^\frac{1}{3}}$
$\Rightarrow a^\frac{1}{6}b^{11}{12}$

---

graph $y=4^x-1$


---

### sec 1.4 Natural exponential function

$e$ : natural exponent

$f(x)=e^x$ : natural exponential function

:::info
**def:** $e$ is such that the slope of the tangent line of $y=f(x)=e^x$ at $(0,1)$ is one
:::


:::info
suppose you have 1 dollar with 100% interest rate. At the end if the first year you get

if interest paid annually
$1+1=2$

if paid monthly
$(1+\frac{1}{12})^{12}$

if paid daily
$(1+\frac{1}{365})^{365}$

if

$(1+\frac{1}{n})^n$ , n is huge

as $n\rightarrow\infty$

$e\approx 2.xxx$

:::

---

### sec 1.5 inverse function

:::info
Def: A function $f:D\rightarrow E$ is a rule that assigns an element $x\in D$ exactly one element in $E$.

Def A function is called one-to-one if $f(x)=f(y)$ implies $x=y$
:::

e.g.

$f(x)=x^2$


To show that $f(x)=x^2$ is not 1-1

$f(1)=1=f(-1)$

but $1\neq -1$ so $x^2$ is not 1-1

**however**

$f(x)=x^2,\ x\geq 0$

then $f$ is a 1-1 function

---


in this case, $f$ is a 1-1 function, but $g$ is not.

:::info
if $f$ is 1-1 and $f(x)=y$ then it's inverse fucntion is $f^{-1}(y)=x$
:::

:::danger
$f^{-1}(x)$ and $(f(x))^{-1}$ are not the same
:::

---

if $A\rightarrow B$ is 1-1

then $f^{-1}(f(x))=x,x\in A$
$f(f^{-1}(x))=x,x\in B$

---

:::info
**how to find the inverse funciton**

1. $f(x)=y$
2. $f^{-1}(y)=x$
3. change the symbol, $f^{-1}(x)$
:::

e.g.

$f(x)=x^2,\ x\geq 0$

1. $x^2=y$
2. $x=(x^2)^\frac{1}{2}=y^\frac{1}{2}=f^{-1}(y)$
3. $f^{-1}(x)=x^{\frac{1}{2}}=\sqrt{x}$

---

### Logarithmic function

1. $f(x)=b^x=y$
2. $\log_{b}b^{x}=\log_{b}y$
$\Rightarrow x=\log_{b}y=f^{-1}(y)$
3. $f^{-1}(x)=\log_{b}x$

---

:::info
**some logarithmic formula**

$\log_b(xy)=\log_bx+\log_by$
$\log_b(\frac{x}{y})=\log_bx-\log_by$
:::

---

### Natural logorithm

$\log_ex=\ln x$

in general

$\ln x=\log_e x$
$\log x=\log_{10} x$

---

### Inverse Trigonometric function

e.g. $\sin(x)$

if we restrict $x\in[-\frac{\pi}{2},\frac{\pi}{2}]$


for $\sin(x)$
domain: $[-\frac{\pi}{2},\frac{\pi}{2}]$
range: $[-1,1]$

for $\sin^{-1}(x)$
domain: $[-1,1]$
range: $[-\frac{\pi}{2},\frac{\pi}{2}]$

$\sin^{-1}(x)=asin(x)=\arcsin(x)$

---

similar to $\cos$

$[0,\pi]\rightarrow [-1,1]$

$\cos^{-1}(x)$ : $[-1,1]\rightarrow[0,\pi]$

e.g.

$\cos^{-1}=0$ , not $2\pi$
because of the range we define

---

e.g. $e^{\ln(\ln e^3)}$

$y=e^{\ln(\ln e^3)}$
$\begin{align}\ln y=&\ln e^{\ln(\ln e^3)}\\=&\ln(\ln e^3)\\=&\ln(3)\end{align}$

$\Rightarrow y=3$

e.g. $sin^{-1}(sin\frac{5}{4}\pi)$

$\begin{align}\sin^{-1}(\sin\frac{5}{4}\pi)=\sin^{-1}(\sin(\frac{-\pi}{4}))=-\frac{\pi}{4}\end{align}$

e.g. $\cos(\sin^{-1}(x))$

assume $y=\sin^{-1}x$
$\Rightarrow\sin y=x$


$\cos(\sin^{-1}(x))=\cos(y)=\sqrt{1-x^2}$

---

e.g. $\sin(\tan^{-1}x)$

$y=\tan^{-1}x\Rightarrow \tan y=x$


$\sin(\tan^{-1}x)=\sin(y)=\frac{x}{\sqrt{1+x^2}}$

---

## 09/19

### Ch 2 Limits and Derivatives

### Ch 2.1 tangent line

tangent: touching ~~at only one point~~
$\rightarrow$ but also move in the same direction

### Ch 2.2 Limits of a function

$\lim_{x\rightarrow a}f(x)=L$

the limit of $f(x)$ as $x$ approaches $a$ equals to $L$

e.g. $f(x)=\frac{x}{x}$

domain: $\{x\mid x\in\mathbb{R}\setminus\{0\}\}$


|$x$|0.1|0.01|0.001|...|$\lim_{x\rightarrow 1}$|
|-|-|-|-|-|-|
|$f(x)$|1|1|1|...|1|

:::info
$\lim_{x\rightarrow a}f(x)$ has nothing to do with the value at $x=a$
:::

e.g.

$f(x)=\begin{cases}x&x\neq0\\a&x=0\end{cases}$

$\lim_{x\rightarrow a}f(x)=0$

e.g. $f(x)=\frac{x^2-2x+1}{x-1}, x\neq 1$

$\lim_{x\rightarrow 1}f(x)$

$f(x)=\frac{(x-1)^2}{x-1}=x-1,x\neq1$

$\Rightarrow\lim_{x\rightarrow 1}f(x)= \lim_{x\rightarrow 1}(x-1)=0$

:::info
Def: One-sided limit

$\lim_{x\rightarrow a^+}f(x)=L$
or
$\lim_{x\rightarrow a^-}f(x)=L$
:::

e.g.

$H(x)=\begin{cases}1&x\geq0\\-1&x<0\end{cases}$

$\lim_{x\rightarrow0^+}H(x)=1$
$\lim_{x\rightarrow0^-}H(x)=-1$

$\Rightarrow\lim_{x\rightarrow0}H(x)$ does not exist

:::info
$\lim_{x\rightarrow0}f(x)$
if and only if (iff)
$\lim_{x\rightarrow0^+}f(x)=L=\lim_{x\rightarrow0^-}f(x)$
:::

e.g. $f(x)=\frac{1}{x^2}$


$\lim_{x\rightarrow 0}f(x)=\infty$

:::danger
Note: the limit equals to infinity means that the liimit **does not exist**

because
$\lim_{x\rightarrow a}f(x)=L$
where $L\in\mathbb{R}$
:::

e.g. $f(x)=\frac{1}{x},x\neq 0$


$\lim_{x\rightarrow 0^+}f(x)=\infty$
$\lim_{x\rightarrow 0^-}f(x)=-\infty$

$\Rightarrow \lim_{x\rightarrow 0}f(x)$ does not exist

:::info

If the limits is infinity at $x=a$

$\Rightarrow$ there exists a vertical asymptote at $x=a$

vertical asymptote: 垂直漸進線

:::

sec 2.2 #38

$\lim_{x\rightarrow 3^-}\frac{x^2+4x}{x^2-2x-3}$

$\lim_{x\rightarrow 3}x^2+4x=21$
$\lim_{x\rightarrow 3}x^2-2x-3=0$

$x^2-2x-3=(x-3)(x+1)$

for $x$ approaches $3$ from the left
$\Rightarrow x^2-2x-3<0$

sec 2.2 #34

$\lim_{x\rightarrow 0^+}\ln(\sin(x))$

for $x>0$ and close to $0$ , $\sin(x)>0$ and close to $0$.

sec 2.2 #32

$\lim_{x\rightarrow 3^-}\frac{\sqrt{x}}{(x-3)^5}$

because
$\lim_{x\rightarrow 3^-}\sqrt{x}=\sqrt{3}$
$\lim_{x\rightarrow 3^-}(x-3)^5=0^-$
$\Rightarrow \lim_{x\rightarrow 3^-}\frac{\sqrt{x}}{(x-3)^5}=-\infty$

e.g.

$\lim_{x\rightarrow 0}\frac{\sqrt{9+x}-3}{x}$

for $x\neq0$

$\begin{align}\lim_{x\rightarrow 0}\frac{\sqrt{9+x}-3}{x}=&\frac{\sqrt{9+x}-3}{x}\cdot\frac{\sqrt{9+x}+3}{\sqrt{9+x}+3}\\=&\frac{9+x-9}{x(\sqrt{9+x}+3)}\\=&\frac{x}{x(\sqrt{9+x}+3)}=\frac{1}{\sqrt{9+x}+3}\\=&\frac{1}{3+3}=\frac{1}{6}\end{align}$

### sec 2.3 Limit Laws

If the limit exists, then

$\lim(f\pm g)=\lim f\pm \lim g$

$\lim(f\cdot g)=(\lim f)(\lim g)$

$\lim(\frac{f}{g})=\frac{\lim f}{\lim g}$

### squeezing theorem

if $f(x)\leq g(x)\leq h(x)$
and $\lim_{x\rightarrow a}f(x)=L=\lim_{x\rightarrow a}h(x)$
then $\lim_{x\rightarrow a} g(x)=L$

e.g. $f(x)=\sin\frac{1}{x}$


$\lim_{x\rightarrow 0} f(x)$ does not exist

e.g. $f(x)=x\sin\frac{1}{x}$


$\lim_{x\rightarrow 0}f(x)=0$

e.g. $f(x)=x^2\sin\frac{1}{x}$


$-1\leq \sin\frac{1}{x}\leq 1, x\neq 0$

$-x^2\leq x^2\sin\frac{1}{x}\leq x^2, \forall x\in \mathbb{R}$

$\Rightarrow\lim_{x\rightarrow 0}-x^2=0=\lim_{x\rightarrow 0}x^2$

by squeezing thm.

$\Rightarrow \lim_{x\rightarrow 0}f(x)=0$

#63

$f(x)=\begin{cases}x^2&x\ is\ rational\\0&x\ is\ irrational\end{cases}$


$0\leq f(x)\leq x^2$

by squeezing thm.

$\Rightarrow \lim_{x\rightarrow 0}f(x)=0$

## 10/03

### Chain rule

consider $(f\cdot g)(x)=f(g(x))=F(x)$

given $x$, evaluate $g(x)$, then evaluate $f(g(x))$

Q: $\frac{d}{dx}F(x)=\frac{d}{dx}f(g(x))$

Let $z=g(x)$ then $F(x)=f(z)$

$[\frac{d}{dx}F]=\frac{[F]}{[x]}=\frac{[f]}{[x]}$

$([\frac{d}{dz}f]=\frac{[f]}{[z]})\times \frac{[z]}{[x]}=\frac{[f]}{[x]}$

:::info
**Chain rule**

$\frac{dF}{dx}=f'(g(x))\times g'(x)$

:::

---

$\begin{align}\frac{d}{dx}F(x)=&\frac{d}{dx}f(g(x))\\=&\lim_{h\rightarrow 0}\frac{f(g(x+h))-f(g(x))}{h}\\=&\lim_{h\rightarrow 0}\frac{f(g(x+h))-f(g(x))}{g(x+h)-g(x)}\times \frac{g(x+h)-g(x)}{h}\end{align}$

Assume $f$ and $g$ are both differentiable

$=\lim_{h\rightarrow 0}\frac{f(g(x+h))-f(g(x))}{g(x+h)-g(x)}\times \lim_{h\rightarrow 0}\frac{g(x+h)-g(x)}{h}=f'(g(x))\times g'(x)$

**Remark:**
$f'=\frac{df}{dz}$
$g'=\frac{dg}{dx}$

e.g.

$\frac{d}{dx}(\sin(x^2))$

$f=\sin(x),\ g=x^2$

$=\cos(x^2)\times 2x$

e.g.

$\frac{d}{dx}(\sin^2(x))$

$f=x^2,\ g=\sin(x)$

$=2\sin(x)\times \cos(x)$

e.g.

$\frac{d}{dx}\sqrt{1+x^2}$

$f=\sqrt{x},\ g=1+x^2$

$=\frac{1}{2}(1+x^2)^\frac{-1}{2}\times 2x$
$=\frac{x}{\sqrt{1+x^2}}$

e.g.

$\frac{d}{dx}(e^{x^2})$

$=e^{x^2}\times 2x$

e.g.

$\frac{d}{dx}(2^x)$

$2^x=e^{x\ln2}$

$=e^{x\ln2}\times \ln 2$
$=2^x\times \ln 2$

e.g.

$\frac{d}{dt}(2^{t^3})=\frac{d}{dt}(e^{t^3\ln 2})$

$=e^{t^3\ln 2}\times 3t^2\ln 2$
$=2^{t^3}\times 3t^2 \ln 2$

e.g.

$\frac{d}{dx}(x^2)=\frac{d}{dx}(e^{x\ln x})$

$=e^{x\ln x}\times (1\times \ln x+x\times \frac{1}{x})$

$=x^x(1+\ln x)$