2013年5月19日 星期日

[微分拓樸] 淺論 Manifold (2) - Manifold 的 Boundary 與 Regular point

回憶前篇,我們說一個 Manifold with boundary 定義如下:

==========================
Definition: Manifold with Boundary
集合 $M \subset \mathbb{R}^n$ 為 $k$-dimensional manifold of class $C^r$ 若下列條件成立:
對任意點 $p \in M$,存在鄰域 $U_p \subset \mathbb{R}^k$ open 或者 $U_p \subset \mathbb{H}^k$ open in $\mathbb{H}^k$ 且 $V_p \subset M$;與 coordinate patch $\alpha: U_p \to V_p$ 滿足
 (1) $\alpha \in C^r$
 (2) $\alpha^{-1} \in C^0$
 (3) $D \alpha$ 有 rank $k$
==========================


接著我們可以介紹 Manifold 的 Interior Point 與 Boundary point:

==========================
Definition: Interior Point and Boundary Point of a Manifold
令 $M \subset \mathbb{R}^n$ 為 $k$-manifold 且 $p \in M$
我們說 $p$ 為 Manifold $M$ 中的 interior point 若下列條件成立:
對上述的 $p$ 而言,存在 coordinate patch $\alpha : U_p \to V_p$ 使得 $U_p$ 為 open in $\mathbb{R}^k$

反之,我們則稱此點 $p$ 為 boundary point
==========================
Comments:
上述的定義的 interior/boundary point 與 一般的 topology 中定義的 interior/ boundary point 不盡相同! 讀者須小心分辨


以下我們有個更好的判斷法來辨別是否為 interior point 或者 boundary point :
==========================
Lemma: 令 $M$ 為 $k$-manifold in $\mathbb{R}^n$ 且定義 $\alpha: U_p \to V_p$ 為 coordinate patch about $p \in M$
1. 若 $U_p$ 為 open in $\mathbb{R}^k$ 則 $p$ 為 interior point of $M$
2. 若 $U_p$ 為 open in  $\mathbb{H}^k$ 且存在  $x_0 \in \mathbb{R}^{k-1} \times (0,\infty)$  使得 $p = \alpha(x_0)$ 則 $p$ 為 interior point of $M$
3. 若 $U_p$ 為 open in  $\mathbb{H}^k$ 且存在 $x_0 \in \mathbb{R}^{k-1} \times \{0\}$ 使得 $p = \alpha(x_0)$  則 $p$ 為 boundary point of $M$
==========================


$k$-Manifold 的非空邊界 為 $k-1$-Manifold。
==========================
Theorem: 令 $M$ 為 $k$-manifold of class $C^r$ 。令 $\partial M$ 為 $M$ 所有的 Boundary point 所形成的集合,若 $\partial M \neq \emptyset $ 則 $\partial M$ 為 $k-1$ manifold (without boundary)。
==========================


==========================
Definition: Critical Point, Regular Point, and its Values
令 $f : U \subset \mathbb{R}^m \to \mathbb{R}^n$ 為 $C^1$ mapping 且 令 $D f(x)$ 為 $f$ 的導數,則
1. 我們說 點 $p \in U$ 為 critical point of $f$ 若 $Df(p)$ 不為 full rank。
2. 我們說  $q \in \mathbb{R}^n$ 為 critical value of $f$ 若 存在 $p$ 為 critical point of  $f$ 使得 $f(p) = q$。
3. 我們說 $q \in \mathbb{R}^n$ 為 regular value of $f$ 若其並非 critical value
==========================

Comment:
在一維空間時候,可知所謂的 critical point 即為 一階導數為零的點。 


Graph 為 manifold。
==========================
FACT 1: Graph of a function is a manifold
令 $\beta: U \subset \mathbb{R}^n \to \mathbb{R}^k$ 為 $C^r$-mapping。則 graph of $\beta$
\[
\{(x, \beta(x)): x \in \mathbb{R}^n\} \subset \mathbb{R}^{n+k}
\] 為 $n$-manifold in $\mathbb{R}^{n+k}$
==========================


==========================
FACT 2: 任意 $M \subset \mathbb{R}^m$ 為 $k$-manifold 為一個 Local property:亦即
考慮 $M \subset \mathbb{R}^m$ 為子集合, 若 對任意 $p \in M$ 存在 open neighborhood $M'$ of $p$ 使得 $M'$ 為 $k$-manifold,則 $M$ 為 $k$-manifold。
==========================

Proof:
考慮 $M \subset \mathbb{R}^m$ 為子集合, 且給定任意 $p \in M$,可知 存在一個 open neighborhood $M'$ of $p$ 使得 $M'$ 為 $k$-manifold,

若 $p \in M'$ 為 $k$-manifold 則由 manifold 定義 存在 coordinate patch $\alpha :U \to V$ 其中 $U \subset \mathbb{H}^k$ 或者 $U \subset \mathbb{R}^k$ open  且 $V \subset M'$ open。

由於 $M'$ 為 open 故 $V \subset M$ 必為 open in $M$,因此 $\alpha$ 為 coordinate patch about $p$ into $M$ 且滿足所有 coordinate patch 所需的性質。故 $M$ 為 $k$-manifold;且  $k$-manifold 為一個 Local property。$\square$


上述兩個 FACT 可推得下面的重要結果:

==========================
Theorem: Regular Value Theorem
若 $f: U \subset \mathbb{R}^{n+k} \to \mathbb{R}^n$ 為 $C^r$-mapping 且 點 $p \in \mathbb{R}^n$ 為 regular value,則 對此點 $p$ 的 inverse image  $f^{-1}(p)$ 為 $k$-manifold。
==========================
comments:
注意到上述定理中 $p$ 為 regular value 故只為單值,不可放置多值 e.g., $f^{-1}[0,1]$。

現在我們給出上述 Regular Value Theorem 的證明:

Proof of Regular Value Theorem by using Rank Theorem
我們要證明其 inverse image $f^{-1}(p)$ 為 $k$-manifold。亦即給定 對任意 $p \in f^{-1}(p)$,要證  存在兩 open 鄰域 $U_p \subset \mathbb{R}^{k}, V_p \subset \mathbb{R}^n$ 且 有 coordinate patch $\alpha: U_p \to V_p$ 滿足三個條件。

由假設可知  $f: U \subset \mathbb{R}^{n+k} \to \mathbb{R}^n$ 為 $C^r$-mapping 且 點 $p \in \mathbb{R}^n$ 為 regular value, 故可知 $Df$ 具有 常數 rank $n$ 對任意 $q \in f^{-1}(p)$

現在回憶 Rank theorem:若 $f: U \subset \mathbb{R}^{n+k}$ 且導數 $Df(x)$ 具有 常數 rank 值,則:對點 $p_1 \in U$ 存在 $U_1, V_1 \subset \mathbb{R}^{n+k}$ 且有函數 $H: V_1 \to U_1$為 1-1 & onto 使得 對任意 $x \in V_1$,
\[
(f \circ H)(x) = f'(p_1) x + \varphi(f'(p_1))
\]且有 projection $P: \mathbb{R}^n \to \mathbb{R}(f(p_1))$ satisfying $P(\varphi(f(p_1)))x = 0$

令 $p \in \mathbb{R}^n$為 regular value, 我們考慮 $x$ 使得
\[
(f \circ H)(x) =p
\]但 $\{x \in V_1: (f \circ H)(x) =p\} = \{x \in V_1: f'(q)x = Pp\}:=L $ 其中 $f(q) = p$

注意到 $H^{-1}(q) \in L$,故 若$x_0 \in L, $ 則
\[
f'(q)x_0 = Pp
\]因此
\[
f'(q) (x_0 - H^{-1}(q)) = f'(q)x_0 - f'(q) H^{-1}(q) = Pp - Pp= 0
\]故 $x_0 = H^{-1}(q) + y$ 其中$y \in \mathcal{N}(f'(q))$. 現在觀察 $\mathcal{N}(f'(q))$ is a vector space with dimension $n+k-n = k$. (因為 range space 為 dimension $k$ )

令 $Q$ 為 composition of translation by $H^{-1}(q)$ and the linear map taking $\mathcal{N}(f'(q)) \to \mathbb{R}^k$

定義 $U_p := Q(L)$ 與 coordinate patch $\alpha := H \circ Q^{-1}: U_0 \to V_1$ 即為 所需的 coordinate patch. $\square$


看個例子:

Example 
考慮 $S:=\{(x,y): x^2 + y^2 = r^2\}$ 若我們選 $f(x,y) := x^2 + y^2$ 則 $f: \mathbb{R}^{1+1} \to \mathbb{R}^1$ 且為 $C^2$ mapping。 故現在我們檢驗何時會是 regular value,利用 critical value 的定義:檢驗其一階導數
 \[Df\left( x \right) = \left[ {\begin{array}{*{20}{c}}
{2x}\\
{2y}
\end{array}} \right]\]故可發現若 $(x,y) = (0,0)$ 時 其 一階導數無法維持 full rank。故除了 $(0,0)$ 以外的點皆為 regular point,其餘任意一點 $(x,y)$ 所對應的值 ( regular value ) 我們記做 $r$ 。由上述 Regular Value Theorem 可知 其 inverse image $f^{-1}(\{r\})$ 為 1-manifold。

注意到
\[{f^{ - 1}}(\{ 1\} ): = \{ (x,y):{x^2} + {y^2} = 1\}  = {S^1} = \{ (x,y):{x^2} + {y^2} = 1\} \]

2013年5月15日 星期三

[微分拓樸] 淺論 Manifold (1)

基本想法:
Manifold 一般譯為 "流形" ,本質上是作為  $\mathbb{R}^n$ 空間中 曲線 or 曲面 的進一步推廣。故我們可將 Manifold 視為 $\mathbb{R}^n$ 空間中的子集,這樣一來,整個 $\mathbb{R}^n$ 之上定義的概念 (極限、微分、積分) 都可以在 manifold上面做處理。

==============
Definition: k-dimensional manifold of class $C^r$ without Boundary
一個子集合 $M \subset \mathbb{R}^n$ 為 k-dimensional manifold of class $C^r$ without Boundary 若下列條件成立:對任意 $p \in M$ 存在兩個 open sets: $U_p \subset \mathbb{R}^k$ 與 $V_p \subset M$ ($V_p$ open in $M$) 且 存在 連續 bijection 函數 $\alpha:U_p \to V_p$ 滿足
 (i) $\alpha \in C^1$
 (ii) $\alpha^{-1}$ 為 連續
 (iii) $D \alpha$ 具備 rank $k$
==============

Comment: 
$\alpha: U_p \to V_p$ 一般稱為 coordinate patch。

以下我們看幾個例子:
-----------
Example 1: 1-manifold example
考慮 $M:=\{(x,y) \in \mathbb{R}^2: x^2 + y^2 = 1 \}$,試問此集合是否為 1-dimensional manifold of class $C^1$?
-----------

Proof:
要證明上述集合為 manifold 我們需要證明:
對任意 $p \in M$ 存在兩個 open sets $U_p \subset \mathbb{R}^k$ 與 $V_p \subset M$ open in $M$ 且 存在 一連續 bijection 函數 $\alpha:U_p \to V_p$ 且
 (i) $\alpha \in C^1$
 (ii) $\alpha^{-1}$ 為 連續
 (iii) $D \alpha$ 具備 rank $k$

注意到 $S$ 為 $\mathbb{R}^2$ 中的圓,故我們可以利用 參數表示法 將其用單變數表示:

現在給定 $p \in M \setminus (0,-1)$ ,考慮 $\alpha: (-\pi, \pi) \to V \subset M$,且滿足 $\alpha(x) := (\cos(x), \sin(x))$ 則可得
\[
\alpha(-\pi, \pi) \to M \setminus (0,-1)
\]現在我們宣稱此 參數表示 $\alpha$ 為 coordinate patch:故逐步檢驗三個性質:
(1) 注意到 $\alpha(x) := (\cos(x), \sin(x))$ ,由於 $(cos(x), sin(x)) $ 微分後仍為連續函數故滿足 $C^1$。
(2) 其反函數 $\alpha^{-1}(x,y)$

-----------
Example 2: 1-manifold example
考慮 $M:=\mathbb{R}$,試問此集合是否為 1-dimensional manifold of class $C^1$?
-----------
Proof:
令 $p \in M$,我們要證明存在兩個 open sets: $U_p \subset \mathbb{R}^1$ 與 $V_p \subset M$ ($V_p$ open in $M$) 且 存在 連續 bijection 函數 $\alpha:U_p \to V_p$ 滿足
 (i) $\alpha \in C^1$
 (ii) $\alpha^{-1}$ 為 連續
 (iii) $D \alpha$ 具備 rank $1$

故令 $U_p := \mathbb{R}$ 且取 coordinate patch $\alpha:U_p \to V_p:=\alpha(U_p)$ 滿足 $\alpha(x) = x$ ,接著我們逐步檢驗 $\alpha$ 所需的條件如下
(i) $\alpha (x) = x$ 為 $C^1$ (因為 $x$ 為 $C^1$ 函數),

(ii) 檢驗 $\alpha^{-1}$ 為 連續,故檢驗 $\alpha^{-1}$ 是否存在 (亦即 檢驗是否為 one-to-one:) 觀察
\[
\alpha(x) = \alpha(y) \Rightarrow x = y
\]故 $\alpha$ 為 one-to-one 且 onto (因為 $V_p := \alpha(U_p)$;故可推知 $\alpha^{-1}$ 存在,且 $\alpha^{-1}(x) = x$ 亦為 連續函數 (因為 $x$ 為連續函數)。

(iii) 檢驗 $D \alpha$ 具備 rank $1$,觀察 $D \alpha = 1 \neq 0$ (full rank) 故 $M:=\mathbb{R}$ 為 1-dimensional manifold of class $C^1$?

Exercise:
同上例,試證 $M:=[0,1]$ 為 1-manifold;
同上例,試證 $M:=\mathbb{R}^n$ 為 n-manifold;


Example: NOT a 2-manifolds
令 $M:= \mathbb{R}^3$ 試問 $M$ 是否為 2-manifolds?

Proof:
令 $p:=(p_1,p_2,p_3) \in M$,選 $U_p:=\mathbb{R}^2$,$V_p:=\alpha(\mathbb{R}^2)$  且 $\alpha: \mathbb{R}^2 \to \alpha(\mathbb{R}^2)$ 滿足
\[\begin{array}{l}
\alpha (x,y): = p + (x,y,0)\\
\begin{array}{*{20}{c}}
{}&{}&{}&{}
\end{array} = ({p_1},{p_2},{p_3}) + (x,y,0) = ({p_1} + x,{p_2} + y,{p_3})
\end{array}\]但注意到 $V_p:=\alpha(\mathbb{R}^2)$ 照上面的定法只能映射出平面! NEVER open in $\mathbb{R}^3$


Example 
令 $\alpha: \mathbb{R} \to \mathbb{R}^2$ 且 $\alpha(x) := (x,x^2)$ 且 $M:=\alpha(\mathbb{R})$ 。試證明  $M$ 為 1-manifold 且  $M$ 僅被一個 coordinate patch 蓋住。

Proof:
要證 $M:=\alpha(\mathbb{R})$ 為 1-manifold。 令 $p \in M$ 且 取 鄰域 $U_p:=\mathbb{R}$ 為 open,定義  coordinate patch $\alpha: U_p \to V_p:=\alpha(\mathbb{R}) = M$ 滿足 $\alpha(x) = (x,x^2)$。

接著我們逐步檢驗 $\alpha$ 滿足三項條件:
(i)  $\alpha$ 為 $C^1$ (因為 $x,x^2$ 皆為 $C^1$),
(故此 $V_p:=\alpha(\mathbb{R}) \subset M$ $V_p := \alpha(\mathbb{R})$ 為 open 。)

(ii) 若 $\alpha(x) = \alpha(y) \Rightarrow (x-y,x^2-y^2)=(0,0)$,此暗示了 $x=y$ 。故可推知 $\alpha$ 為 one-to-one 因此 $\alpha^{-1}:V_p \to U_p$ 存在 且 $x = \alpha^{-1}(x,y)$ 為連續函數;

(iii) 最後我們檢驗 $D \alpha = [1, \; 2x]^T$ 為 full rank (對任意 $x$ 而言)。

總結以上 $M= \alpha(\mathbb{R})$ 為 1-manifold 且此表明 $\alpha(\mathbb{R}) \supset M$ 亦即 $M$僅被 一個 coordinate patch 蓋住 $\square$


現在我們可介紹 有邊界的 Manifold。
令 $\mathbb{H}^k := \mathbb{R}^{k-1} \times [0, \infty)$ 為 upper-half plane。現在我們可以定義有邊界的 Manifold:

==========================
Definition: Manifold with Boundary
我們稱 集合 $M \subset \mathbb{R}^n$ 為 $k$-dimensional manifold of class $C^r$  (with boundary) 若下列條件成立:對任意點 $p \in M$,存在鄰域 $U_p \subset \mathbb{R}^k$ open 或者 $U_p \subset \mathbb{H}^k$ open in $\mathbb{H}^k$ 且 $V_p \subset M$ open,且有 coordinate patch $\alpha: U_p \to V_p$ 滿足
 (1) $\alpha \in C^r$
 (2) $\alpha^{-1} \in C^0$
 (3) $D \alpha$ 有 rank $k$
==========================

上述的定義有些基本問題需要解決:
  1. 何謂 $C^r$ 函數?
  2. 任意集合上的 $C^r$ 函數該如何界定?
  3. $D \alpha$ on $\mathbb{H}^k$ 該如何界定?

ANS1: $\alpha$ 為 $C^r$ in $U \subset \mathbb{R}^k$ 若 任意 $r$ 階 偏導數 存在且連續。
ANS2: 我們引入 extension function:

====================
Definition: $C^r$-Extension
令 $f : S \subset \mathbb{R}^k \to \mathbb{R}^n$ ,我們稱 $g$ 為 $f$ 的 extension 若下列條件成立: 存在 $U \supset S $ open set 與 $g: U \to \mathbb{R}^n$ 使得 對任意 $x \in S$, $g(x) = f(x)$ 且 $g$ 為 $C^r$ on $U$。
====================

若 extension 函數 $g$ 為 $C^r$ 則原函數 $f$ 必為 $C^r$
====================
FACT:
令 $f : S \subset \mathbb{R}^k \to \mathbb{R}^n$ 若存在 extension $g$ of $f$ ,則 $f \in C^r$ on $S$。====================

事實上 class $C^r$ 是一個 local property! 亦即我們可以透過 extension function $g$ 來說函數 $f$ 在某點的鄰域附近仍為 $C^r$ 函數。
====================
Lemma: 
令 $f: S \subset \mathbb{R}^k \to \mathbb{R}^n$ 若 對任意 $x \in S$ 存在鄰域 $U_x$ 使得 $x \in U_x$ 且 $g_x: U_x \to \mathbb{R}^n$ 為 $C^r$-extension of $f$ on $U_x \cap S$ 則 $f$ 亦為 $C^r $ on S。
====================

Proof: omitted.

現在我們可以回答 問題3: $D \alpha$ 在 $\mathbb{H}^k$ 如何界定? 我們可以 extend $\alpha$ 到 $\mathbb{R}^k$ 但此 extension 是否會影響 $D \alpha$? 以下我們將證明上述 manifold 定義中,$D \alpha $ 與 我們引入的 $C^r$-extension 無關。

現在考慮 coordinate patch: $\alpha: U \subset \mathbb{H}^k \to \mathbb{R}^n$ 且令 $\beta$ 為其 $C^r$-extension of $\alpha$,現在注意到由於 $\beta$ 為 $C^r$ 故一階導數存在 (且 $h \uparrow 0$ 或者 $h \downarrow 0$ 都成立),故我們可只寫導數的單邊極限:
\[{D_j}\beta \left( x \right): = \mathop {\lim }\limits_{h \to 0} \frac{{\beta \left( {x + h{e_j}} \right) - \beta \left( x \right)}}{h} = \mathop {\lim }\limits_{h \searrow 0} \frac{{\beta \left( {x + h{e_j}} \right) - \beta \left( x \right)}}{h}\]注意到 $x \in U \subset \mathbb{H}^k$ 則其附近 $x + h e_j \in U$ 故
\[\begin{array}{l}
{D_j}\beta \left( x \right) = \mathop {\lim }\limits_{h \searrow 0} \frac{{\beta \left( {x + h{e_j}} \right) - \beta \left( x \right)}}{h}\\
\begin{array}{*{20}{c}}
{}&{}&{}&{}
\end{array} = \mathop {\lim }\limits_{h \searrow 0} \frac{{\alpha \left( {x + h{e_j}} \right) - \alpha \left( x \right)}}{h} = {D_j}\alpha \left( x \right)
\end{array}\]因此 $D_j \alpha$ 與 extension 無關,且 $D \alpha$ 在 $U \subset \mathbb{H}^{k}$ 定義亦可被接受。

2013年5月3日 星期五

[整理] 多元智能理論(英)

最近涉獵了一些 多元智能理論 (Multiple Intelligence theory),個人覺得頗為有趣。 整理的資料如下 

Multiple Intelligence theory, Edited by Chung-Han Hsieh


The Multiple Intelligence theory, or so called MI theory, which is proposed by Professor Howard Gardner. He believed that the human beings can have a number of relatively different but interacted intelligence not just only one general intelligence.
(from Harvard Graduate school of Education)

Two Definitions of Intelligence
  • 1. From Merriam-Webster Dictionary
    Definition: the ability to learn or understand or to deal with new or trying (difficult) situations. (Merriam-Webster Dict.)
  • 2. From Professor Howard Gardner:
    Howard Gardner viewed intelligence as 'the capacity to solve problems or to fashion products that are valued in one or more cultural setting' (Gardner & Hatch, 1989)

The main idea of Multiple Intelligence theory, MI theory

According to this MI theory, human beings have a number of relatively different but interacted intelligences. This theory is about understanding the intellect and the cognitive aspects of the human mind. It also criticized the traditional intelligence standards for the narrow psychological aspect; in other words, MI theory is against the traditional single intelligence standard such as the general IQ test.

Why is it so influential?
Many educators and teachers, based on their teaching experiences, tend to apply the MI theory’s idea to teaching their students because this theory successfully provides a guideline to find out the intelligence of different students in different ways. These educators believe that everyone can be educated to achieve certain level of intelligence by using an appropriate way.

 Professor Gardner's perspectives:
  • the intelligence of human beings is domain-specific, which means each person has their own special intelligence on certain subjects
  • Every people have at least 8 or 9 different type intelligence and it provides in more details than just one single IQ score.
  • People can be educated to achieve certain level intelligence.
  • Intelligence has multiple display ways. EX: Even the twins does not hold the same intelligence
  • The Multiple intelligence theory is not used for determining the type of intelligence of people, but for recognizing the type of intelligence of people.
Merit of MI theory (educations & some examples)
  • Students are likely to become more engaged in learning as they match their intelligence strengths.
  • Every learner is unique. Each student is seen as an individual with his or her own strengths and weaknesses.
  • A good possibility for teacher to guide, identify, celebrate their students intelligence.
  • Students may be motivated and confident when using an intelligence they know is one of their strengths.
How to apply the MI theory-One Example
In 1998 Nicholsen–Nelson suggested one approach to applying MI theory in language teaching:
  1. Play to your student's strengths.If your student is good a specific intelligence then you should structure the learning material to this strength.
  2.  Variety is the spice of life. Try to let every student participate in as many different intelligences during the lesson as possible.
  3. Pick a tool suited for the job.Language has many different dimensions, aspects or functions. These different facets could be linked to the most appropriate intelligence.
  4. One size fits all.Everyone has to participate in all the exercises to make sure that their use all their Intelligences. The multiple intelligences approach aims to develop the whole person and not just enhance already high intelligences.
  5. Me and my people.Be aware that different cultures value different intelligences. Western culture with its IQ test has a biased view on the intelligences. Language learning in particular needs to increase more than the students' IQ and is also a progress of understanding, communication and culture.

Criticism of MI theory
  1. lack of evidence or empirical support with certain systematic researches
  2. It's NOT new. Critics of MI theory maintain that Gardner's work isn't groundbreaking -- that what he calls "intelligences" are primary abilities that educators and cognitive psychologists have always acknowledged.
  3. It isn't well defined. Some critics wonder if the number of "intelligences" will continue to increase. These opposing theorists believe that notions such as bodily-kinesthetic or musical ability represent individual aptitude or talent rather than intelligence. Critics also believe that M.I. theory lacks the rigor and precision of a real science. Gardner claims that it would be impossible to guarantee a definitive list of intelligences.
  4. It's culturally embedded. M.I. theory states that one's culture plays an important role in determining the strengths and weaknesses of one's intelligences. Critics counter that intelligence is revealed when an individual must confront an unfamiliar task in an unfamiliar environment.
  5. It defeats National Standards. Widespread adoption of multiple intelligence pedagogy would make it difficult to compare and classify students' skills and abilities across classrooms.
  6. It is idealized and impractical. Educators faced with overcrowded classrooms and lack of resources see multiple intelligence theory as utopian.
Potential Abused part
  • Using MI theory as the only one instruction rule
  • Using MI theory to tag student.
  References


  1. “FAQ – Multiple Intelligences and Related Educational Topics”–howardgardner.com
  2.  “Theory of Multiple Intelligences” – Wikipedia.
  3.  Howard Gardner, "Multiple Intelligences and education” – infed.org
  4. “Education World:Multiple Intelligences: A Theory for Everyone” – educationworld.com
  5. “21 years later, Multiple Intelligences Still Debated” – Washington Post
  6. “Summary of the Criticism of Multiple Intelligences” – mdecgateway.org
  7.  Lynn Waterhouse, “Multiple Intelligences, the Mozart Effect, and Emotional Intelligence : A Critical Review” (2006) 
  8.  “Multiple Intelligences”, –The Wiki of English Teaching.
  9.  “Are there Multiple intelligence” ?

2013年5月2日 星期四

[隨機分析] Ito Isometry Property in H^2 Space

已知 $f \in \cal{H}^2[0,t]$,我們有 Ito Isometry 如下:
\[
E\left[ \left( \int_0^t f(s) dB_s \right)^2 \right] = E \left [ \int_0^t f(s)^2 ds \right]
\]
現在我們看看 cross term 會怎麼樣?

考慮 $f,g \in \cal{H}^2[0,t]$
\[E\left[ {\int_0^t f (s)d{B_s} \cdot \int_0^t g (s)d{B_s}} \right] = E\left[ {\int_0^t f (s)g\left( s \right)ds} \right]
\]
Proof:
觀察下式:
\[\begin{array}{l}
E\left[ {{{\left( {\int_0^t {f(s)} d{B_s} + \int_0^t {g\left( s \right)} d{B_s}} \right)}^2}} \right]\\
\begin{array}{*{20}{c}}
{}&{}
\end{array} = E\left[ {{{\left( {\int_0^t {f(s)} d{B_s}} \right)}^2} + 2\int_0^t {f(s)} d{B_s}\int_0^t {g\left( s \right)} d{B_s} + {{\left( {\int_0^t {g(s)} d{B_s}} \right)}^2}} \right]\\
\begin{array}{*{20}{c}}
{}&{}
\end{array} = E\left[ {{{\left( {\int_0^t {f(s)} d{B_s}} \right)}^2}} \right] + E\left[ {2\int_0^t {f(s)} d{B_s}\int_0^t {g\left( s \right)} d{B_s}} \right]\\
\begin{array}{*{20}{c}}
{}&{}&{}&{}&{}&{}&{}&{}
\end{array} + E\left[ {{{\left( {\int_0^t {g(s)} d{B_s}} \right)}^2}} \right]\\
\begin{array}{*{20}{c}}
{}&{}
\end{array} = E\left[ {\int_0^t {{f^2}(s)} ds} \right] + E\left[ {2\int_0^t {f(s)} d{B_s}\int_0^t {g\left( s \right)} d{B_s}} \right]\\
\begin{array}{*{20}{c}}
{}&{}&{}&{}&{}&{}&{}&{}
\end{array} + E\left[ {\int_0^t {{g^2}(s)} ds} \right] \ \ \ \  (*)
\end{array}
\]但注意到我們所觀察的式子亦可寫成
\[\begin{array}{l}
E\left[ {{{\left( {\int_0^t {f(s)} d{B_s} + \int_0^t {g\left( s \right)} d{B_s}} \right)}^2}} \right] = E\left[ {{{\left( {\int_0^t {\left( {f(s) + g\left( s \right)} \right)} d{B_s}} \right)}^2}} \right]\\
\begin{array}{*{20}{c}}
{}&{}
\end{array} = E\left[ {\int_0^t {{{\left( {f(s) + g\left( s \right)} \right)}^2}} ds} \right]\\
\begin{array}{*{20}{c}}
{}&{}
\end{array} = E\left[ {\int_0^t {{f^2}(s)} ds} \right] + 2E\left[ {\int_0^t {f(s)g\left( s \right)} ds} \right] + E\left[ {\int_0^t {{g^2}\left( s \right)} ds} \right] \ \ \ \ \ (\star)
\end{array}
\]比較 $(*)$ 與 $(\star)$,可得
\[E\left[ {\int_0^t {f(s)} d{B_s}\int_0^t {g\left( s \right)} d{B_s}} \right] = E\left[ {\int_0^t {f(s)g\left( s \right)} ds} \right]  \]