# Quod Erat Demonstrandum

## 2010/08/28

### 直線方程

Filed under: Junior Form Mathematics — johnmayhk @ 8:57 下午

## 2010/08/20

### P/NP

Filed under: Report,University Mathematics — johnmayhk @ 11:28 上午

P/NP問題是在理論信息學中計算複雜度理論領域裡至今沒有解決的問題，它被「克雷數學研究所」（Clay Mathematics Institute，簡稱CMI）在千禧年大獎難題中收錄。

http://zh.wikipedia.org/zh-tw/P/NP%E9%97%AE%E9%A2%98

http://www.nytimes.com/2010/08/17/science/17proof.html?_r=1

## 2010/08/19

### 國際數學家大會 2010-08-19 至 2010-08-27

Filed under: Report — johnmayhk @ 9:34 上午

http://www.icm2010.org.in/

http://en.wikipedia.org/wiki/International_Congress_of_Mathematicians

## 2010/08/01

### 答中四同學包

Filed under: NSS — johnmayhk @ 9:22 上午

1.

Find $\frac{dy}{dx}$ of the following from first principles.

(a) $y = \tan x$
(b) $y = xe^x$

(a)

$\frac{dy}{dx}$
$= \displaystyle \lim_{\Delta x \rightarrow 0}\frac{\tan (x + \Delta x) - \tan x}{\Delta x}$
$= \displaystyle \lim_{\Delta x \rightarrow 0}\frac{1}{\Delta x}(\frac{\sin (x + \Delta x)}{\cos (x + \Delta x)} - \frac{\sin x}{\cos x})$
$= \displaystyle \lim_{\Delta x \rightarrow 0}\frac{1}{\Delta x} \frac{\sin (x + \Delta x)\cos x - \cos (x + \Delta x)\sin x}{\cos (x + \Delta x) \cos x}$
$= \displaystyle \lim_{\Delta x \rightarrow 0}\frac{1}{\Delta x} \frac{\sin (x + \Delta x - x)}{\cos (x + \Delta x) \cos x}$
$= \displaystyle \lim_{\Delta x \rightarrow 0}\frac{\sin \Delta x}{\Delta x} \frac{1}{\cos (x + \Delta x) \cos x}$
$= \displaystyle \lim_{\Delta x \rightarrow 0} \frac{1}{\cos (x + \Delta x) \cos x}$
$= \frac{1}{\cos x \cos x}$
$= \sec^2x$

(b)

$\frac{dy}{dx}$
$= \displaystyle \lim_{\Delta x \rightarrow 0}\frac{(x + \Delta x)e^{x + \Delta x} - xe^x}{\Delta x}$
$= \displaystyle \lim_{\Delta x \rightarrow 0}\frac{(x + \Delta x)e^xe^{\Delta x} - xe^x}{\Delta x}$
$= \displaystyle \lim_{\Delta x \rightarrow 0}\frac{e^x((x + \Delta x)e^{\Delta x} - x)}{\Delta x}$
$= \displaystyle \lim_{\Delta x \rightarrow 0}\frac{e^x(xe^{\Delta x} + \Delta xe^{\Delta x} - x)}{\Delta x}$
$= \displaystyle \lim_{\Delta x \rightarrow 0}\frac{e^x(x(e^{\Delta x} - 1) + \Delta xe^{\Delta x})}{\Delta x}$
$= \displaystyle \lim_{\Delta x \rightarrow 0}e^x(x(\frac{e^{\Delta x} - 1}{\Delta x}) + e^{\Delta x})$
$= e^x(x(1) + 1)$
$= (x + 1)e^x$

$(x + \Delta x)e^{x + \Delta x} - xe^x$
$= (x + \Delta x)e^{x + \Delta x} - xe^{x + \Delta x} + xe^{x + \Delta x} - xe^x$