# 方根

## 基本運算

$\sqrt[n]{ab} = \sqrt[n]{a} \sqrt[n]{b} \qquad a \ge 0, b \ge 0$
$\sqrt[n]{\frac{a}{b}} = \frac{\sqrt[n]{a}}{\sqrt[n]{b}} \qquad a \ge 0, b > 0$
$\sqrt[n]{a^m} = \left(\sqrt[n]{a}\right)^m = \left(a^{\frac{1}{n}}\right)^m = a^{\frac{m}{n}},$

$a^m a^n = a^{m+n} \,$
$({\frac{a}{b}})^m = \frac{a^m}{b^m}$
$(a^m)^n = a^{mn} \,$

$\sqrt[3]{a^5}\sqrt[5]{a^4} = a^{5/3} a^{4/5} = a^{5/3 + 4/5} = a^{37/15}$

$\sqrt[3]{a^5} = \sqrt[3]{aaaaa} = \sqrt[3]{a^3a^2} = a\sqrt[3]{a^2}$

$\sqrt[3]{a^5}+\sqrt[3]{a^8}$
$=\sqrt[3]{a^3a^2}+\sqrt[3]{a^6 a^2}$
$=a\sqrt[3]{a^2}+a^2\sqrt[3]{a^2}$
$=({a+a^2})\sqrt[3]{a^2}$

## 不盡根數

• $\sqrt{a^2 b} = a \sqrt{b}$
• $\sqrt[n]{a^m b} = a^{\frac{m}{n}}\sqrt[n]{b}$
• $\sqrt{a} \sqrt{b} = \sqrt{ab}$
• $(\sqrt{a}+\sqrt{b})^{-1} = \frac{1}{(\sqrt{a}+\sqrt{b})} = \frac{\sqrt{a}-\sqrt{b}}{(\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})} = \frac{\sqrt{a}- \sqrt{b}} {a - b}$

## 無窮級數

\begin{align} &(1+x)^\frac{s}{t} = \sum_{n=0}^\infty \frac{\displaystyle\prod_{k=0}^n (s+t-kt)}{(s+t)n!t^n}x^n\\ &(|x|<1) \end{align}

## 找到所有的方根

$e^{(\frac{\varphi+2k\pi}{n})i} \times \sqrt[n]{a}$

### 正實數

$e^{2\pi i \frac{k}{n}} \times \sqrt[n]{a}$

## 解多項式

$\ x^5=x+1$

## 算法

1. 猜一個$\sqrt[n]{A}$的近似值，將其作為初始值$x_0$
2. $x_{k+1} = \frac{1}{n} \left[{(n-1)x_k +\frac{A}{x_k^{n-1}}}\right]$。記誤差為$\Delta x_k = \frac{1}{n} \left[{\frac{A}{x_k^{n-1}}} - x_k\right]$，即$x_{k+1} = x_{k} + \Delta x_k$
3. 重複步驟2，直至絕對誤差足夠小，即：$| \Delta x_k | < \epsilon$

### 從牛頓法導出

$\sqrt[n]{A}$之值，亦即求方程$x^n-A=0$的根。

$f(x)=x^n-A$，其導函數$f'(x)=nx^{n-1}$

$x_{k+1} = x_k - \frac{f(x_k)}{f'(x_k)}$
$= x_k - \frac{x_k^n - A}{n x_k^{n-1}}$
$= x_k - \frac{x_k}{n}+\frac{A}{n x_k^{n-1}}$
$= \frac{1}{n} \left[{(n-1)x_k +\frac{A}{x_k^{n-1}}}\right]$

### 從牛頓二項式定理導出

$x_k$為疊代值，$y$為誤差值。

$A=(x_k-y)^n$（*），作牛頓二項式展開，取首兩項：$A\approx x_k^n-n x^{n-1}_k y$