# Quod Erat Demonstrandum

## 2015/09/13

### 一式過

Filed under: Pure Mathematics — johnmayhk @ 5:19 下午
Tags: , ,

$-1,1,-1,1,\dots$

$(-1)^n$

$\cos(180^on)$

$1,2,3,1,2,3,\dots$

$T_n = \left \{ \begin{array}{ll} 1 \hspace{10 mm} \mbox{for}\hspace{5 mm} n = 1,4,7,\dots\\2 \hspace{10 mm} \mbox{for}\hspace{5 mm} n=2,5,8,\dots \\3 \hspace{10 mm} \mbox{for}\hspace{5 mm} n = 3,6,9,\dots \end{array}\right.$

$T_n=\frac{1}{3}((1+2\cos(120^o(n-1)))+2(1+2\cos(120^o(n+1)))+3(1+2\cos(120^on)))$

$T_n=2+\frac{2}{3}(\cos 120^o(n-1)+2\cos120^o(n+1)+3\cos120^on)$　for $n=1,2,3,\dots$

$f(x) = \left \{ \begin{array}{ll} 1 \hspace{10 mm} \mbox{for}\hspace{5 mm} x = rational\\0 \hspace{10 mm} \mbox{for}\hspace{5 mm} x = irrational\end{array}\right.$

$sgn(x) = \left \{ \begin{array}{ll} 1 \hspace{10 mm} \mbox{for}\hspace{5 mm} x > 0\\0 \hspace{10 mm} \mbox{for}\hspace{5 mm} x = 0\\-1 \hspace{8 mm} \mbox{for}\hspace{5 mm} x < 0\end{array}\right.$

$sgn(x)=\displaystyle \lim_{n\rightarrow \infty}\frac{2}{\pi}\tan^{-1}(nx)$

$x > 0$
$x = 0$
$x < 0$

$\displaystyle \lim_{y\rightarrow +\infty}\tan^{-1}y=\frac{\pi}{2}$$\displaystyle \lim_{y\rightarrow -\infty}\tan^{-1}y=-\frac{\pi}{2}$

$sgn(\sin^2(m!\pi x))$

$x$ 是有理數，設 $x=\frac{p}{q}$，只要有足夠大的 $m$，比如 $m > q$，那麼 $\sin^2(m!\pi x)=0$，即

$y=\displaystyle \lim_{m\rightarrow +\infty}sgn(\sin^2(m!\pi x))=0$

$x$ 是無理數，$\sin^2(m!\pi x) > 0$，那麼 $sgn(\sin^2(m!\pi x))=1$，即

$y=\displaystyle \lim_{m\rightarrow +\infty}sgn(\sin^2(m!\pi x))=1$

$f(x) = \left \{ \begin{array}{ll} 1 \hspace{10 mm} \mbox{for}\hspace{5 mm} x = rational\\0 \hspace{10 mm} \mbox{for}\hspace{5 mm} x = irrational\end{array}\right.$

$f(x)=1-\displaystyle \lim_{m\rightarrow +\infty}sgn(\sin^2(m!\pi x))$

$g(x) = \left \{ \begin{array}{ll} a \hspace{10 mm} \mbox{for}\hspace{5 mm} x = rational\\b \hspace{10 mm} \mbox{for}\hspace{5 mm} x = irrational\end{array}\right.$

$g(x)=a+(b-a)\displaystyle \lim_{m\rightarrow +\infty}sgn(\sin^2(m!\pi x))$

$\displaystyle \lim_{m\rightarrow \infty}(\displaystyle \lim_{n\rightarrow \infty}(\cos(m!\pi x))^{2n})$

$x$ 是有理數，設 $x=\frac{p}{q}$，只要有足夠大的 $m$，比如 $m > q$，那麼

$(\cos(m!\pi x))^2=1\Rightarrow (\cos(m!\pi x))^{2n}=1$

$\displaystyle \lim_{m\rightarrow \infty}(\displaystyle \lim_{n\rightarrow \infty}(\cos(m!\pi x))^{2n})=1$

$x$ 是無理數，$m!x$ 也必是無理數，故 $\displaystyle \lim_{n\rightarrow \infty}(\cos(m!\pi x))^{2n}=0$

$\displaystyle \lim_{m\rightarrow \infty}(\displaystyle \lim_{n\rightarrow \infty}(\cos(m!\pi x))^{2n})=0$

(a) Show that if $k\le x < k+1$，（$k=0,1,2,\dots$），$\int [x]|\sin \pi x|dx=(-1)^{[x]+1}\frac{[x]}{\pi}\cos \pi x+C_k$

(b) Let $F(x)=(-1)^{[x]+1}\frac{[x]}{\pi}\cos \pi x+C_k, k \le x < k+1$$k=0,1,2,\dots$）be a continuous function for all $x\ge 0$, show that $C_k=\frac{1}{\pi}(2k-1)+C_{k-1}$.

(c) If $C_0=0$, find $\int [x]|\sin \pi x|dx$　（$x\ge 0$） （答案在這裡

$\cos k\pi=(-1)^k$

http://www.gutenberg.org/files/38769/38769-pdf.pdf

or

https://dl.dropboxusercontent.com/u/19150457/SFXC/38769-pdf.pdf

## 1 則迴響 »

1. 以前番中學上堂講sin cos function果陣
諗過d類似嘅野
想用sin and limit 黎 “扮" square wave:
f(x)= limit n->inf (sin(x))^( 1 /( 2n+1))
當然果陣唔知可以用fourier series表達
要set好佢take邊個root。
同發現最後都係要定義野, 要定義n係integer先 lol

迴響 由 Yik Fung Luk — 2015/09/13 @ 11:42 下午 | 回應