There are some notes I took while learning math from this online course. I am still young and ambitious to learn math again.

Week 1

Properties of Operations

\[\begin{align} \bf{x + y} &= \bf{y + x}\\ \bf{(x + y) + z} &= \bf{x + (y + z)}\\ \bf{x + 0} &= \bf{0}\\ \bf{(-x) + x} &= \bf{0}\\ \lambda\bf{(x+y)} &= \lambda \mathbf{x} + \lambda\mathbf{y}\\ (\lambda + \mu)\bf{x} &=\lambda\mathbf{x} + \mu \mathbf{x}\\ 1 \cdot \mathbf{x} &= \mathbf{x}\\ \lambda (\mu \mathbf{x}) &= (\lambda \mu) \mathbf{x} \end{align}\]

Inner Product in $\mathbb{R}^n$

\[\begin{aligned} \langle \mathbf{x}, \mathbf{y}\rangle &= \sum_{i=1}^n{x_i y_i} &(\bf{x, y} \in \mathbb{R}^n)\\ \sqrt{\langle \mathbf{x}, \mathbf{x} \rangle}&= \|\mathbf{x}\| & \text{The norm}\\ & \|\bf{x - y}\| & \text{Distance} \end{aligned}\]

Open Ball

The definition of an open ball in $\mathbb{R}^n$ is

\[\begin{aligned} B_\varepsilon\left(\mathbf{x}^{(0)}\right) &= \left\{ \mathbf{x} \in \mathbb{R}^n | \left\| \mathbf{x} - \mathbf{x}^{(0)} \right\| < \varepsilon \right\}.\\ \end{aligned}\]

It is open in a sense that we can always find open balls for every point inside one open ball.

Open Set

The set $S$ is open if $\forall \mathbf{x} \in S$ there exist a open ball $B_\varepsilon(\mathbf{x}) \in S$

Example:

  1. The first quadrant \(\mathbb{R}_+^2 = \left\{ \mathbf{x} \in \mathbb{R}^2 | x_1, x_2 > 0 \right\}\) is an open set.
  2. The open balls are open sets.
  3. The union set of all open sets is a open set. ( \(\cup_{\alpha \in \mathbb{I}} \mathcal{A}_\alpha\) is open set if $\mathcal{A}_\alpha$ are open sets.)

Intersection of Open Sets

Suppose all the following finite number of sets are open

\[\mathcal{A}_1,\ \mathcal{A}_2,\ \mathcal{A}_3,\ \dots\ \mathcal{A}_m\]

Their intersection,

\[\mathcal{A} = \bigcap_{i=1}^m \mathcal{A}_i\]

is an open set. (This is not true for infinite amount of open sets, ie. intersection up to $\mathcal{A}_\infty$ )

Example:

\[\begin{aligned} \mathcal{A}_i &= \left( -\frac1i, \frac1i \right)\\[0.5em] \mathcal{A} &= \bigcap_{i=1}^{m}\mathcal{A}_i = \mathcal{A}_m & \textrm{open set}\\[1em] \tilde{\mathcal{A}} &= \bigcap_{i=1}^{\infty}\mathcal{A}_i = \{0\} & \textrm{not an open set}\\ \end{aligned}\]

Sequences in $\mathbb{R}^n$

For \(\mathbf{x} \in \mathbb{R}^n\) to be one element of a sequence, we need to label $\mathbf{x}$, we note the label as,

\[\mathbf{x}^{(m)} \; (m \in \mathbb{N})\]

and the sequance is therefore written as,

\[\left\{ \mathbf{x}^{(m)} \right\}_{m=1}^{\infty}.\]

Concretely, $\mathbf{x}^{(3)}$ is the 3rd term of the sequence.

Convergence of the Sequence

Being convergent means there is a limiting value when $m \rightarrow \infty$. Suppose this value is \(\mathbf{x}^{(0)}\), we write

\[\lim_{m\rightarrow\infty} \mathbf{x}^{(m)} = \mathbf{x}^{(0)}.\]

We define convergence as the following,

\[\forall \varepsilon>0,\ \exists\ m_\varepsilon \textrm{ such that } \forall\ m \ge m_\varepsilon,\ \| \mathbf{x}^{(m)} - \mathbf{x}^{(0)} \| \le \varepsilon.\]

(The symbol $m_\varepsilon$ means a typical value of $m$ that depends on $\varepsilon$.)

Closed Set

Intuition: for a convergent sequence, whose term belongs to a set, we want to make sure its limit also belongs to the same set.

The set $S \in \mathbb{R}^n$ is closed, if any convergent sequence

\[\left\{ \mathbf{x}^{(m)} \right\}_{m=1}^{\infty}, \; \mathbf{x}^{(m)} \in S​\]

has the limit that belongs to $S$.

Example:

$\mathbb{R}^2_+$ is not closed because we can find the following sequence that contradicts the definition.

\[\mathbf{x}^{(m)} = \left(\begin{array}{c} 1 \\ 1/m \end{array}\right) \rightarrow \left(\begin{array}{c} 1 \\ 0 \end{array}\right) \notin \mathbb{R}^2_+\]

The set

\[S = \left\{ x \in \mathbb{R}^2 | \underbrace{x_1 > 0}_{\text{not closed}},\; \underbrace{x_2 \ge 0}_{\text{not open}} \right\}\]

is not open amd not closed at the same time.

Operations on Closed Sets

For all the $\mathcal{A}_\alpha$ that are closed set, we have the follwoing conclusions

\[\begin{aligned} \bigcap_{\alpha \in \mathbb{I}}{\mathcal{A}_\alpha} &- \text{closed set} \\[2em] \bigcup_{i = 1}^{m}{\mathcal{A}_i} &- \text{closed set} \\[2em] \bigcup_{\alpha \in \mathbb{I}}{\mathcal{A}_\alpha} &- \text{may not be closed set} \\[2em] \end{aligned}\]

Bounded and Compact

A set $S \in \mathbb{R}^n$ is called bounded if there exists a ball $B_R(\mathbb{0})$ such that $S \in B_R(\mathbf{0})$.

A set $S \in \mathbb{R}^n$ which is closed and bounded is called compact

Example:

A closed ball, expressed as the following function

\[\begin{aligned} \mathrm{cl}\ B_\varepsilon\left(\mathbf{x}^{(0)}\right) &= \left\{ \mathbf{x} \in \mathbb{R}^n | \left\| \mathbf{x} - \mathbf{x}^{(0)} \right\| \le \varepsilon \right\} & (\textrm{cl means closure})\\ \end{aligned}\]

is a compact.

Functions in $\mathbb{R}^n$

The normal functions is something like

\[y = f(x); x \in \mathcal{D}\\ \mathcal{D} \subset \mathbb{R}\]

The functions in $\mathbb{R}^n$ is something like,

\[y = f(\mathbf{x}) = f(x_1, \dots, x_n); \mathbf{x} \in \mathcal{D}\\ \mathcal{D} \subset \mathbb{R}^n\]

Example,

The function is $y = \sqrt{1 - x_1^2 - x_2^2}$, and its natural domain is

\[\mathcal{D} = \left\{ x \in \mathbb{R}^2 | x_1^2 + x_2^2 \le 1 \right\}.\]

Domain of a function

$\mathcal{D} \subset \mathbb{R}^n$ is the domain of a function, it consists of all permissable values of $\mathbf{x}$, where the funciton looks like the following.

\[\mathbf{x} \in \mathcal{D}\\ y = f(\mathbf{x}) \in \mathbb{R}\]

we can give $f(\mathbf{x})$ different values, forming the level sets

\[\left\{ x \in \mathcal{D} | f(\mathbf{x}) = y \right\}\]

Continuity

For function $f(\mathbf{x})$ where $\mathcal{D}$ is the domain; for $\mathbf{x}^{(0)} \in \mathcal{D}$, the function is continous if the following expression is true.

\[\lim_{\mathbf{x}\rightarrow\mathbf{x}^{(0)}} f(\mathbf{x}) = f(\mathbf{x}^{(0)})\]

If a function is continues on any point in $\mathcal{D}$ we call the function continous on $\mathcal{D}$

Th. Weierstrass

Let $y=f(\mathbf{x})$ be continues on interval $[a, b]$. Then there exist $m, M$ and $\mathbf{x}_1, \mathbf{x}_2 \in [a, b]$ such that \(m \le f(\mathbf{x}) \le M\) and \(f(\mathbf{x}_1) = m; f(\mathbf{x}_2) = M\)

Composite Function

For function $f(\mathbf{x}), g(\mathbf{x})$ that are continous at $\mathbf{x}^{(0)} \in \mathcal{D}$ ($\mathcal{D}$ is called the common domain), the following composite functions are also continous at $\mathbf{x}^{(0)}$.

\[f(\mathbf{x}) \pm g(\mathbf{x}) \\[1em] f(\mathbf{x}) g(\mathbf{x}) \\[1em] \frac{f(\mathbf{x})} {g(\mathbf{x})} \; \left(g(\mathbf{x}^{(0)}) \ne 0\right)\]

A “difficult” example for composite function is the following, let

\[\begin{aligned} g(x, y) &= xy \\ f(x, y) &= \sin(xy) \\ \rightarrow f(t) &= \sin(t); t = g(x, y) \end{aligned}\]

But what can we say about the domains of these functions?

A generalised example would be,

\[\begin{aligned} t &= g(x, y) \\ f(x, y) &= h(t) \end{aligned}\]

we have to make sure the image of $g$ matches the domain of $h$. Like the follwoing graph,

                ┌──────────────┐
                │   D(h)       │
                │              │
┌───────┐       │  ┌───────┐   │
│       │       │  │       │   │
│ D(g)  │───────┼─▶│ Im(g) │   │
│       │       │  │       │   │
└───────┘       │  └───────┘   │
                └──────────────┘