# 方根

（重定向自开方术

## 基本运算

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

${\displaystyle a^{m}a^{n}=a^{m+n}}$
${\displaystyle \left({\frac {a}{b}}\right)^{m}={\frac {a^{m}}{b^{m}}}}$
${\displaystyle \left(a^{m}\right)^{n}=a^{mn}}$

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

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

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

An unresolved root, especially one using the radical symbol, is sometimes referred to as a surd[1] or a radical.[2] Any expression containing a radical, whether it is a square root, a cube root, or a higher root, is called a radical expression, and if it contains no transcendental functions or transcendental numbers it is called an algebraic expression.

## 不尽根数

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

## 无穷级数

{\displaystyle {\begin{aligned}&(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{aligned}}}

## 找到所有的方根

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

### 正实数

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

## 解多项式

${\displaystyle \ x^{5}=x+1}$

## 算法

1. 猜一個${\displaystyle {\sqrt[{n}]{A}}}$的近似值，將其作為初始值${\displaystyle x_{0}}$
2. ${\displaystyle x_{k+1}={\frac {1}{n}}\left[{(n-1)x_{k}+{\frac {A}{x_{k}^{n-1}}}}\right]}$。記誤差為${\displaystyle \Delta x_{k}={\frac {1}{n}}\left[{\frac {A}{x_{k}^{n-1}}}-x_{k}\right]}$，即${\displaystyle x_{k+1}=x_{k}+\Delta x_{k}}$
3. 重複步驟2，直至絕對誤差足夠小，即：${\displaystyle |\Delta x_{k}|<\epsilon }$

### 從牛頓法導出

${\displaystyle {\sqrt[{n}]{A}}}$之值，亦即求方程${\displaystyle x^{n}-A=0}$的根。

${\displaystyle f(x)=x^{n}-A}$，其導函數${\displaystyle f'(x)=nx^{n-1}}$

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

### 從牛頓二項式定理導出

${\displaystyle x_{k}}$為迭代值，${\displaystyle y}$為誤差值。

${\displaystyle A=(x_{k}-y)^{n}}$（*），作牛頓二項式展開，取首兩項：${\displaystyle A\approx x_{k}^{n}-nx_{k}^{n-1}y}$