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Abstract

This is the place where the subjects that most undergraduates majoring Mathematics would learn. Nothing special, just going write what I want.

학부생으로서 배웠던 수학을 총망라하여 두서없이 작성하는 곳. 아래 글 내용은 각 과목에서 배운 것 중에서 하나씩 빼와서 작성한 것이다.

Discrete Mathematics

(23-1)제대로 안한건지, 얕게 한건지는 몰라도 수학들 중에서 제일 신선하고 재미있었다. 좀 더 수학을 공부해야겠다는 생각이 든 과목이다.

$$F_{n+2} = F_{n+1}+F_{n}, n \in \mathbb{Z}_{\geq 0}$$ $$F_{0} = 0, F_{1} = 1$$

Put the non-negative interger $n$:

Here's one way to derive it. \begin{align*} \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^n \begin{bmatrix} F_1 \\ F_0 \end{bmatrix} &= \begin{bmatrix} F_{n+1} \\ F_n \end{bmatrix} \end{align*}

Differential Equation

(23-1)미방은 고등수학의 연장선 느낌이 든다. 공대애들이 훨씬 더 잘하는 것 같다. 문제는 공대생들보다도 못푸니 정리들이라도 증명해야겠다.

(Laplace Transform)
Let $f$ be a peicewise continuous function on $[0,\infty)$. We define $L[f](s) = \int_{0}^{\infty}e^{-st}f(t)dt$ provided that the integral converges.

For example, $L[t^{n}](s) = \frac{n!}{s^{n+1}}(s>0)$.
Know that $L$ is a linear operation.

Topology

(24-1)수학과 해석학을 듣고 바로 들어보는 입장에서, 뭔가 실수에서만 사용되는 개념을 일반화시키고 정리로 사용된 open, closed를 정의로 사용하면서 시작하니 뭔가 신기하긴 하다.

In Topology, we use an open set. That is, 'metric' is not needed. We defined and open sets in $\mathbb{R}$. Let's discuss about properties of open sets in $\mathbb{R}$.
1. $\emptyset, \mathbb{R}$ are open.
2. If $u_{\alpha}$ is open $\forall \alpha\in J$, then arbitrary union $\bigcup\limits_{\alpha\in J}u_{\alpha}$ is open.
3. If $u_{i}$ is open $\forall i = 1,2,...n$, then finite intersection $\bigcap\limits_{i=1}^{n}u_{i}$ is open.
In Topology, we use these properties as a definition.

$\mathcal{T}$, topology on $X$, is a collection of subsets of $X$ such that
1. $\mathcal{T} \ni \emptyset, X$
2. $\mathcal{T} \ni$arbitrary union of elements of $\mathcal{T}$
3. $\mathcal{T} \ni$finite intersection of elements of $\mathcal{T}$.
We say 'an element of $\mathcal{T}$ is an open set'.

(Urysohn Metriation Theorem)
Every regular space $X$ with a countable basis is metrizable.

(Tietze Extension)
Let $X$ be a normal space, and $A$ be a closed subspace of $X$. Then,

Any continuous map of $A$ into $[a,b]\subset \mathbb{R}$ may be extended to a continious map of all $X$ into $[a,b]$.
Any continuous map of $A$ into $\mathbb{R}$ may be extended to a continious map of all $X$ into $\mathbb{R}$.

Number Theory

(23-2)정수론이지만, integer에 그치지 않는, 영문 그대로 수에 대해 배웠던 과목이다. 데데킨트 컷이 젤 인상깊고, 데데킨트 컷밖에 생각이 나지 않았던..

A subset $L\subset \mathbb{Q}$ is called a Dedekind Cut if :

$L$ is proper;
$L$ has no maximal element;
For all elements $a, b \in \mathbb{Q}$ with $a < b$, if $b\in L$ then $a\in L$.

$\mathbb{R}$ is complete, $i.e.$ Every cauchy sequence is convergent.

Complex Analysis

(24-1)기억이 잘..

(Cauchy's Theorem)
If $f(z)$ is analytic in a simply connected domain $D$ and $C$ is a closed cantour lying inside $D$, then $\int\limits_{C}f(z)dz = 0.$

(Taylor's Theorem)
Let $f(z)$ be analytic in a domain $D$ with boundary $\partial D$. If $z_{0}\in D$, then $f(z)$ may be expressed as a Taylor series about $z_{0}$.
$f(z) = \sum\limits_{n=0}^{\infty}\frac{f^{(n)}(z_{0})}{n!}(z-z_{0})^{n}$ for $|z-z_{0}|<\delta = dist(z_{0},\partial D)$.

Abstract Algebra

(25-1)

Differential Geometry

(25-1)

Show that for $\beta(t) = \alpha\circ \varphi(t)$, $\kappa[\beta](t) = \kappa[\alpha](\varphi(t)), \tau[\beta](t) = \tau[\alpha](\varphi(t))$

Know that \begin{align*} \dot{\beta}(t) &= \dot{\alpha}(\varphi(t))\dot{\varphi}(t), \ddot{\beta}(t) = \ddot{\alpha}(\varphi(t))\dot{\varphi}^{2}(t)+\dot{\alpha}(\varphi(t))\ddot{\varphi}(t)\\ \dddot{\beta}(t) &= \dddot{\alpha}(\varphi(t))\dot{\varphi}^{3}(t)+2\ddot{\alpha}(\varphi(t))\ddot{\varphi}(t)+\ddot{\alpha}(\varphi(t))\dot{\varphi}(t)\ddot{\varphi}(t) +\dot{\alpha}(\varphi(t))\dddot{\varphi}(t)\\ \dot{\beta}(t)\times\ddot{\beta}(t) &= \dot{\varphi}^{3}(t)(\dot{\alpha}(\varphi(t))\times\ddot{\alpha}(\varphi(t)))\\ (\dot{\beta}(t)\times\ddot{\beta}(t))\cdot\dddot{\beta}(t) &= \dot{\varphi}^{6}(t)(\dot{\alpha}(\varphi(t))\times\ddot{\alpha}(\varphi(t)))\cdot\dddot{\alpha}(\varphi(t))\\ \text{Then, we have}\\ \kappa[\beta](t) &=\frac{|\dot{\beta}(t)\times\ddot{\beta}(t)|}{|\dot{\beta}(t)|^{3}}=\frac{|\dot{\varphi}^{3}(t)||\dot{\alpha}(\varphi(t))\times\ddot{\alpha}(\varphi(t))|}{|\dot{\varphi}^{3}(t)||\dot{\alpha}(\varphi(t))|} = \kappa[\alpha](\varphi(t))\\ \tau[\beta](t) &= \frac{(\dot{\beta}(t)\times\ddot{\beta}(t))\cdot\dddot{\beta}(t)}{|\dot{\beta}(t)\times\ddot{\beta}(t)|^{2}} =\frac{\dot{\varphi}^{6}(t)(\dot{\alpha}(\varphi(t))\times\ddot{\alpha}(\varphi(t)))\cdot\dddot{\alpha}(\varphi(t))}{\dot{\varphi}^{6}(t)|\dot{\alpha}(\varphi(t))\times\ddot{\alpha}(\varphi(t))|^{2}}= \tau[\alpha](\varphi(t)) \end{align*}

Auction Theory

(25-1)

(Power distribution) Suppose there are two bidders with private values that are distributed independently according to the distribution $F(x)=x^{a}$ over $[0,1]$, where $a>0$. Then, the symmetric equilibrium bidding strategies in a first-price auction is \begin{align*} \Pi_{1}(b,x) &= (x-b)P(\beta^{\mathrm{I}}(X_{2})\leq b) = (x-\beta^{\mathrm{I}}(z))z^{a}\\ \frac{\partial}{\partial z}\Pi_{1}\bigg|_{z=x} &= axx^{a-1}-\frac{\partial}{\partial z}(\beta^{\mathrm{I}}(z)z^{a})\bigg|_{z=x} = 0, \beta^{\mathrm{I}}(x) = \frac{a}{a+1}x. \end{align*}

Partial Differential Equation

(24-2 Exchange Student, GSU) Wave Equation이랑 Heat Equation 을 $(-\infty,\infty)$, $[0,\infty)$, $[0,l]$에서 푸는 방법들을 배웠다.

(Wave Eq) $u_{tt}-c^{2}u_{xx} = 0$
(Heat Eq) $u_{t}-ku_{xx} = 0$

Applied Numerical Methods

(24-2 Exchange Student, GSU)수치해석학의 기본을 배운 것 같다. 함수들의 근부터, 미분계수, 특정한 적분값, n차연립방정식, 심지어는 고유값과 고유벡터까지.. 컴퓨터의 대단함을 느끼는 기분이다.

Environment

Theorem and Proofs(Temporary placement, 철거예정)

(Schwartz inequality)Let $a_{1}, \ldots , a_{n}$ and $b_{1}, \ldots , b_{n}$ be complex numbers. $$|\sum_{k=1}^{n}a_{k}\bar{b_{k}}|^{2}\leq (\sum_{k=1}^{n}|a_{k}|^{2})(\sum_{k=1}^{n}|b_{k}|^{2})$$

Know that $\langle \vec{x},\vec{x} \rangle \geq 0 \textbf{ } \forall \vec{x}$. Also, since we defined $e^{i\theta} = \frac{\langle \vec{x},\vec{y} \rangle}{|\langle \vec{x},\vec{y} \rangle|}$, we know that $e^{-i\theta} = \frac{|\langle \vec{x},\vec{y} \rangle|}{\langle \vec{x},\vec{y} \rangle}$. By the definition of inner product, $\langle c\vec{x},\vec{y} \rangle = c\langle \vec{x},\vec{y} \rangle$. Then, we also get $\langle \vec{x},c\vec{y} \rangle = \bar{c}\langle \vec{x},\vec{y} \rangle$ begin{align*} 0\leq f(t) = \langle \vec{x}+te^{i\theta}\vec{y}, \vec{x}+te^{i\theta}\vec{y}\rangle &= \langle \vec{x},\vec{x} \rangle + \langle \vec{x},te^{i\theta}\vec{y} \rangle + \langle te^{i\theta}\vec{y},\vec{x} \rangle + \langle te^{i\theta}\vec{y},te^{i\theta}\vec{y} \rangle\\ &= \langle \vec{x},\vec{x} \rangle + te^{-i\theta}\langle \vec{x},\vec{y} \rangle + te^{i\theta}\langle \vec{y},\vec{x} \rangle + te^{i\theta}te^{-i\theta}\langle \vec{y},\vec{y} \rangle\\ &= \langle \vec{x},\vec{x} \rangle + t|\langle \vec{x},\vec{y} \rangle| + t\frac{\langle \vec{x},\vec{y} \rangle \langle \vec{y},\vec{x} \rangle}{|\langle \vec{x},\vec{y} \rangle|} + t^{2}\langle \vec{y},\vec{y} \rangle\\ &= \langle \vec{x},\vec{x} \rangle + 2t|\langle \vec{x},\vec{y} \rangle| + t^{2}\langle \vec{y},\vec{y} \rangle \text{(Quadratic inequality of t)}\\ \end{align*} Since this holds for every $t\in \mathbb{R}$, $$D = |\langle \vec{x},\vec{y} \rangle|^{2} - \langle \vec{x},\vec{x} \rangle \langle \vec{y},\vec{y} \rangle \leq 0.$$ $LHS$ of Schwartz inequality is $|\langle \vec{x},\vec{y} \rangle|^{2}$ and $RHS$ is $\langle \vec{x},\vec{x} \rangle \langle \vec{y},\vec{y} \rangle = |\vec{x}|^{2}|\vec{y}|^{2}$, so it's showed by $D$.

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