# Quod Erat Demonstrandum

## 2018/02/22

### 懷古-開方2

Filed under: Fun,mathematics — johnmayhk @ 6:08 下午
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$\displaystyle \sqrt{2} \simeq 1+\frac{1}{3}+\frac{1}{3\times 4}-\frac{1}{3\times 4\times 34}$

$\displaystyle (1+\frac{1}{3}+\frac{1}{3\times 4})^2 > 2\Rightarrow 1+\frac{1}{3}+\frac{1}{3\times 4} > \sqrt{2}$

$\displaystyle w=\frac{1}{144} \div (2\times (1+\frac{1}{3}+\frac{1}{3\times 4}))=\frac{1}{3\times 4\times 34}$

$\displaystyle 1+\frac{1}{3}+\frac{1}{3\times 4}-\frac{1}{3\times 4\times 34}$

$\displaystyle (1+\frac{1}{3}+\frac{1}{3\times 4}-\frac{1}{3\times 4\times 34})^2$

$\displaystyle (1+\frac{1}{3}+\frac{1}{3\times 4}-\frac{1}{3\times 4\times 34})^2=2+\frac{1}{144}-2\times (1+\frac{1}{3}+\frac{1}{3\times 4})w+w^2=2+w^2$

$\displaystyle (1+\frac{1}{3}+\frac{1}{3\times 4}-\frac{1}{3\times 4\times 34})^2 > 2$

$\displaystyle 1+\frac{1}{3}+\frac{1}{3\times 4}-\frac{1}{3\times 4\times 34} > \sqrt{2}$

$\displaystyle w^2$ 視為很小的話，就是 Baudhayana 的

$\displaystyle \sqrt{2} \simeq 1+\frac{1}{3}+\frac{1}{3\times 4}-\frac{1}{3\times 4\times 34}$

$\displaystyle x^2=2$

$\displaystyle \Rightarrow x=\frac{2}{x}$

$\displaystyle x_0=\frac{2}{x_0}$

$\displaystyle x_0 \neq \frac{2}{x_0}$

$\displaystyle x_1=\frac{1}{2}(x_0+\frac{2}{x_0})$

$\displaystyle x_1=\frac{2}{x_1}$ 否？

$\displaystyle x_2=\frac{1}{2}(x_1+\frac{2}{x_1})$

$\displaystyle x_{n+1}=\frac{1}{2}(x_n+\frac{2}{x_n})$

$\displaystyle x_0,x_1,x_2,\dots$

（注：修應用數學的同學，相信認得上述不過是定點疊代的最簡單題目，也記得如何測試收斂性吧～）

$\displaystyle x_{n+1}-x_n=\frac{1}{2}(x_n+\frac{2}{x_n})-x_n=\frac{1}{2}(\frac{2}{x_n}-x_n)$

$\displaystyle x_1-x_0=\frac{1}{2}(\frac{2}{4/3}-\frac{4}{3})=\frac{1}{3\times 4}$

$\displaystyle x_1=1+\frac{1}{3}+\frac{1}{3\times 4}$

$\displaystyle x_2-x_1=\frac{1}{2}(\frac{2}{17/12}-\frac{17}{12})=-\frac{1}{3\times 4\times 34}$

$\displaystyle x_2=1+\frac{1}{3}+\frac{1}{3\times 4}-\frac{1}{3\times 4\times 34}$

$\displaystyle x_3=1+\frac{1}{3}+\frac{1}{3\times 4}-\frac{1}{3\times 4\times 34}-\frac{1}{3\times 4\times 34\times 1154}$

$\displaystyle x_3=$1.4142135623746899…

（有沒有發覺疊代公式右邊計出的結果形如 $\displaystyle \frac{\pm 1}{n}$，這是必然的嗎？）

(圖片來源：https://en.wikipedia.org/wiki/YBC_7289)

1;24,51,10 和 ;42,25,35

$\displaystyle 1+\frac{24}{60}+\frac{51}{60^2}+\frac{10}{60^2}$

$\displaystyle \frac{42}{60}+\frac{25}{60^2}+\frac{35}{60^2}$

1.41421296296

0.70710648148

$\displaystyle x_0=1+\frac{24}{60}=\frac{7}{5}$

$\displaystyle x_1=1+\frac{24}{60}+\frac{1}{70}$

$\displaystyle x_2=1+\frac{24}{60}+\frac{1}{70}-\frac{1}{13860}$

$\displaystyle \frac{1}{70}$$\displaystyle \frac{1}{13860}$ 寫成 60 進制（如何？其實唔難。），得

$\displaystyle \frac{1}{70}= \frac{51}{60^2}+\frac{25}{60^3}+\frac{42}{60^4}+\dots$

$\displaystyle \frac{1}{13860}= \frac{15}{60^3}+\frac{35}{60^4}+\dots$

$\displaystyle x_2=1+\frac{24}{60}+\frac{51}{60^2}+\frac{25-15}{60^3}+\frac{42-35}{60^4}+\dots\simeq 1+\frac{24}{60}+\frac{51}{60^2}+\frac{10}{60^3}$