## 2008年7月29日 星期二

### 循環論證 (II) : 圓周公式

Continue from 循環論證 (I) : 圓周公式

There should be no controversy before the integral arises,

let's verify each steps carefully starting from the arc length.

$\int_{0}^{r}\sqrt{1+\frac{dy}{dx}^2}\,dx$

The arc length formula itself is correct.

Its proof is based on Mean Value Theorem and Riemann Sum.

Here, what we need to verify is the differentiability of $\inline y=\sqrt{r^2-x^2}$ on the interval $\inline \left(0,r\right)$ and the continuity on $\inline \left[0,r\right]$.

Both of them are clear except the continuity at the end-point 0.

But, since we are considering the first quadrant only, the existence of one hand limit is enough.

For the symmetry part, there should be no problem.

{If you really think about it, just consider the properties of the odd function and even function}

Afterwards, the simplification of integral is obviously correct.

Then, we investigate the computation of integral.

$L=4r\left[\sin^{-1} \frac{x}{r} \right]_{0}^{r} \cdots (1)$

$L=4r\left[\frac{\pi}{2}-0\right]=2\pi r \cdots (2)$

As I mentioned in the first part, there must be a flaw in the proof, so the flaw should arise in the computation part.

Before we proceed, let's recall the following high-school definitions:

$\inline \pi$ is defined as the common ratio of circumference to diameter of any circle.

Radian is defined as the ratio of arc length to the radius of a circle, in particular, it is the arc length of the unit circle.

$\inline \left(\sin x,\cos x\right)$ is defined as the point lying on the unit circle making an angle $x$ (in radian) with the x-axis.

$\inline \sin^{-1} x$ is then defined as the inverse mapping on the right half of the unit circle

Let's consider the second equation first, as this is easier to explain.

Here, we use the common sense that $\inline \sin^{-1}1=\tfrac{\pi}{2}$ and $\inline\sin^{-1}0=0$ , but why!?

somehow we know $\inline \sin\tfrac{\pi}{2}=1$

(we must rely on the fact that $\inline \pi$ is the ratio)

So, we found that the second one contains flaw.

However, this does not mean that the first one contains no circular reasoning.
For those who knows a little bit of analysis,

For the first equation, we used Fundamental theorem of Calculus, which itself require the function $\inline \sin^{-1}x$ to be integrable, being the inverse function of $\inline \sin x$, it is continuous, and hence integrable.

To show that $\inline \sin x$ is a continuous function, from epsilon-delta definition, the inequality $\inline \left|\sin x - \sin y\right| \leq \left|x - y\right|$ must be involved, which is often proved using geometric argument

## 2008年7月28日 星期一

### The Game Theory in Batman

The following is an article from Mingpao, which analyses the recent Batman movie with game theory.

【咫尺地球】奧運前夕昆明巴士連環爆炸，手法似仿效前期蓋達，卻又不像國際組織所為。真正專業的，要麼在遙遠的省份輪番襲擊，要麼在同一地方真正real-time放炸彈，似不會產生1小時真空。假如策劃者不具備規模，卻希望製造恐慌來誘惑國內同道中人仿效，也就是在設計一場博弈。這教人想起上映中的《蝙蝠俠黑夜之神》的天才罪犯「小丑」。

## 2008年7月24日 星期四

### 平方根的疑惑

平方根這個概念看似簡單，可是很多中學生都無法清楚了解它。以下是兩個最常見的問題：

它們的答案都可以由定義直接回答。

平方根的定義：若某數 xy 符合 x2 = y，我們稱 xy 的平方根。因為 32 = (-3)2 = 9，所以 3 和 -3 都是 9 的平方根。那麼 $\sqrt{x}$ 是甚麼？原來這是 x 的非負平方根。我們不難證明任何不小於 0 的實數都剛好有一個非負平方根。換言之，當 x 不是負數時，$\sqrt{x}$ 只有一個值。由此可知，$\sqrt{9}$ 的值是 3。

## 2008年7月19日 星期六

### 立法會選舉

• 李華明、計明華、黃啟明、黃偉達
• 胡志偉
• 陳鑑林、黎榮浩、陳曼琪、洪錦鉉

## 2008年7月10日 星期四

### 拓撲學(一)

1)
V E + F = 2 這條公式，應該在小學或初中的數學課本出現過。如果你數出以下圖案的角(vertex)的總數V，邊(edge)的總數E及面(face)的總數F，再計算它們V E + F 的值，答案永遠是2

2) 一個拓撲學學者不懂得分辨一個甜圈(冬甩doughnut)及一隻(有手抦的)咖啡杯。原因是，它們都共同有一個孔。