粉筆灰的數學小站

Direct Limit Part 1: Direct Limit Definition 1.1. (Direct Limit) A partially ordered set $I$ is said to be a direct set  if for all pairs $i , j \in I$, there exist $k \in I$ such that $i \leqslant k$ and $j \leqslant k$. Let $\mathfrak{C}$ be a category. A direct system  is a pair $(X_i, \mu_{ij})$, where $\{X_i\}_{i \in I}$ is a family of objects in $\mathfrak{C}$ indexed by a direct set $I$, and for all $i \leqslant j$, the morphism $\mu_{ij} \colon X_i \to X_j$ satisfies the property: $\mu_{ii} = \text{Id}_{X_i}$. For $i \leqslant j \leqslant k$, we have $\mu_{ik} = \mu_{jk} \circ \mu_{ij}$. Given an direct system $(X_i, \mu_{ij})$ in $\mathfrak{C}$, a direct limit  of the direct system is a pair $(X, \mu_i)$, where $X \in \text{Ob}(\mathfrak{C})$ and morphism $\mu_i \colon X_i \to X$ satisfying $\mu_i = \mu_j \circ \mu_{ij}$ for all $i \leqslant j$. $X$ satisfies the following universal property: Given an object $Y$ and morphisms $\alpha_i \colon X_i...

Definition 1.1. (Presheaf) Let $X$ be a topological space, where we denote $\mathfrak{Top}(X)$ be the category with: $\text{Ob}(\mathfrak{Top}(X)) = \{\text{open sets of }X\}$. $\text{Mor}_{\mathfrak{Top}(X)}(U,V) = \begin{cases} \text{inclusion} &,U \subset V\\ \varnothing &, \text{otherwise}\end{cases}$ A presheaf of abelien group on $X$ is a contravariant functor $\mathscr{F} \colon \mathfrak{Top}(X) \to \mathfrak{Ab}$, where $\mathfrak{Ab}$ is the category of abelien groups. Precisely, that is: $\mathscr{F}$ assigns an abelien group $\mathscr{F}(U)$ to every open set $U \in \mathfrak{Top}(X)$. $\mathscr{F}(\varnothing) = \text{Trivial group}$. For each $U \subset V$, $\mathscr{F}$ assigns a group homomorphism $\rho_{VU} \colon \mathscr{F}(V) \to \mathscr{F}(U)$. $\rho_{UU} = \text{Id}_{U}$. For each $U \subset V \subset W$, we have $\rho_{WU} = \rho_{WV} \circ \rho_{VU}$ We called $\mathscr{F}(U) = \Gamma(U, \mathscr{F})$ the sections of the presheaf ...

Note 1. Varieties (1) Part 1: Definitions Definition 1.1. (Variety) Let $k$ be an algebraically closed field, then an affine $n-$space  is $\mathbb{A}_k^n = k^n$. For an ideal $I \triangleleft R = k[x_i \mid 1 \leqslant i \leqslant n]$, an affine algebraic set  is defined by: $$V(I) = \{a \in \mathbb{A}_k^n \mid f(a) = 0 \text{ for all } f \in I\}$$ Proposition 1.2. Affine algebraic set satisfy the following: $V(R) = \varnothing$, $V((0)) = \mathbb{A}_k^n$. $V(I) \cup V(J) = V(IJ)$ For aribitrary index set $A$, we have $\displaystyle \bigcap_{\alpha \in A}V(I_\alpha) = V \left(\sum_{\alpha \in A}I_{\alpha}\right)$ These fact shows that by defining $\{V(I) \mid I \triangleleft R\}$ be exactly the closed set of $\mathbb{A}_k^n$, we may induce a topological space, namely, Zariski topology . Definition 1.2. A toplogical space $X$ is said to be irreducible if for any proper pair of closed set $C_1$, $C_2$, $X \neq C_1 \cup C_2$. ($X$ is not a union o...

Homework 6. Date: 108/10/31 Place: 共同教學館 102 [J] 代表微積分用書 Calculus: Early Transcendetals (8E), James Stewart. Problem 1. WeBWorK: homework 4.3, 4.4, 4.5, 4.7 Problem 2. ([J], Section 4.3, Problem 33., Page 301) Suppose $f$ is a continuous function where $f(x)>0$ for all $x$, $f(0) = 4$, $f'(x)>0$ if $x<0$ or $x>2$, $f'(x)<0$ is $0<x<2$, $f''(-1) = f''(1) = 0$, $f''(x)>0$ if $x<-1$ or $x>1$, $f''(x)<0$ if $-1<x<1$. Can $f$ have an absolute maximum? If so, sketch a possible graph of $f$. If not, explain why. Can $f$ have an absolute minimum? If so, sketch a possible graph of $f$. If not, explain why. Sketch a possible graph for $f$ that does not achieve an absolute minimum. Solution. Impossible, since $f'(x)>0$, $f''(x)>0$ for $x>>0$. (In particular, this is true for $x>2$), so $\displaystyle \lim_{x \to \infty }f(x) = \infty$, there is no absolute ma...

Homework 4. Date: 108/10/14 (Week 06) Place: 新生教學館 301 Problem 1. Find the value of $a$ that makes the following function differentiable everywhere. $$f(x) = \begin{cases} ax &,\text{ if } x<0\\ x^2-3x &, \text{ if } x \geqslant 0\end{cases}$$ Solution. Obviously, $f(x)$ is differentiable anywhere except $x = 0$. Note that we have $$\lim_{x \to 0^-}\dfrac{f(x)-f(0)}{x-0} = a$$, and, $$\lim_{x \to 0^+}\dfrac{f(x)-f(0)}{x-0} = -3$$ In order to make $f(x)$ differentiable at $x = 0$, we have $a = -3$. Problem 2. Let $f(x)$ be a strictly increasing, continuous function on $(a,b)$. Then there exists an inverse function $g(y)$ defined on $(f(a), f(b))$. Suppose furthermore that $f(x)$ is differentiable on $(a,b)$. Prove that $g(y)$ is differentiable on $(f(a), f(b))$ and $g'(y_0) = \dfrac{1}{f'(x_0)}$ if $y_0 = f(x_0)$. Solution. We assume the following fact: Since $f \colon (a, b) \to (f(a), f(b))$ is strictly increasing, there is an inverse fu...

Homework 3. Date: 108/10/03 (Week 04) Place: 新生教學館 301 Problem 1. Let $f(x)$ be a strictly increasing continuous function on $[a, b]$. That is, for any $x_1 < x_2 \in [a, b]$ we have $f(x_1) < f(x_2)$. Show that $f(b)$ is the maximum and $f(a)$ is the minimum. Show that for any $y_0 \in [f(a), f(b)]$, there exists $x_0$ such that $f(x_0) = y_0$. Show that the inverse function $f^{−1}$ is defined everywhere on $[f(a), f(b)]$. Show that $f^{-1}$ is continuous on $[f(a), f(b)]$. Solution. Since $x < b$ for all $x \in [a,b)$, we have $f(x) < f(b)$ for all $x \in [a,b)$, so $f(b)$ is the maximum, similarly, $f(a)$ is the minimum. Since $y_0 \in [f(a),f(b)]$ and $f(x)$ is continuous, by Intermediate value theorem, there exist $x_0 \in [a,b]$ such that $f(x_0) = y_0$. The previous problem gives that $f \colon [a,b] \to [f(a),f(b)]$ is surjective, and it is injective by the property strictly increasing (Why?). Hence, the function $f \colon [a,b] \to [f(a),f(b...

Note 4: Review Date: 108/10/28 (Week 08) Place: 新生教學館 301 Part 1: Graphing Problem Reference. 微積分用書 Calculus: Early Transcendetals (8E), James Stewart , Section 2.6, Limits at infinity; Horizontal Asymptotes (P.126 - P.140) 微積分用書 Calculus: Early Transcendetals (8E), James Stewart , Section 4.5, Summary of Curve Sketching (P.315 - P.323) Problem 1.1. (95.上微積分甲統一教學一組.第5題) Study the function $f(x) = \dfrac{(x-2)^2}{x+1}$, and answer the following questions. The domain of $y = f(x)$. $f'(x)$. $y = f(x)$ has critical point(s) at $x = $? $y = f(x)$ is increasing / decreasing on intervals? $f''(x)$. $y = f(x)$ is concave upward / down on intervals? Find the $(x,y)$ coordinates of the following points if exist. local maximum point(s) local minimum point(s) inflection point(s) Find the asymptotes of the graph $y = f(x)$ if exist. Vertical asymptotes(s) Horizontal asymptotes(s) Slanted asymptotes(s) Sketch the graph of $y = f(x)$ below. Solut...

Note 3: Concave and Strictly concave Date: 108/10/21 (Weel 07) Place: 新生教學館 301 Part 1: Concave v.s. Strictly Concave Problem 1.1 Suppose that $f(x)$ and $f'(x)$ are differentiable on $(a, b)$. If $f''(x) \geqslant 0$, for all $x \in (a, b)$, then $f\left(\dfrac{x_1+x_2}{2}\right) \leqslant \dfrac{f(x_1)+f(x_2)}{2}$, for all $x_1 < x_2 \in (a,b)$. If $f''(x) > 0$, for all $x \in (a, b)$, then $f\left(\dfrac{x_1+x_2}{2}\right) < \dfrac{f(x_1)+f(x_2)}{2}$, for all $x_1 < x_2 \in (a,b)$. Is it true that if $f\left(\dfrac{x_1+x_2}{2}\right) \leqslant \dfrac{f(x_1)+f(x_2)}{2}$ for all $x_1 < x_2 \in (a,b)$ then $f''(x) \geqslant 0$ ? Is it true that if $f\left(\dfrac{x_1+x_2}{2}\right) < \dfrac{f(x_1)+f(x_2)}{2}$ for all $x_1 < x_2 \in (a,b)$ then $f''(x) > 0$? Solution. By direct computation:\begin{align*} & \dfrac{f(x_1)+f(x_2)}{2} - f\left( \dfrac{x_1 + x_2}{2} \right) \\ = \hspace{0.2cm}& \dfrac...

Note 2. Completeness of $\mathbb{R}$ & IVT / IFT Date: 108/10/14 (Week 06) Place: 新生教學館 301 Part 1: Completeness of Real number $\mathbb{R}$ There are various way to construct real number system. Depending on the construction, we may regard some of the statement in Proposition 1.1  as axioms$^1$. This relate to some tedious theory and mathematical proof (though, essential), which is not what we are discussing here. ($^1$: Axioms (公設/公理)：指的是數學上定義性的命題，常常用於建構特定數學領域(算數、代數、分析等)。比方說，在代數上我們定義一個 原群(Magma) 是一個有二元運算的集合(可以先把二元運算理解為像是加法的東西)。在這個例子中，「有二元運算」就是原群的公設。) In short, we will not "prove" if any statement in Proposition 1.1  holds in $\mathbb{R}$, but showing they are logically equivalent. How will this even benefit us? Well, if the equivalence is proved, then by admitting one of the statement, we will immediately have the other three property. Proposition 1.1 The following are equivalent. Completeness. Monotone convergence theorem. Least upper bound prop...