y=x^{5}+13x^{4}+35x^{3}-85x^{2}-216x+252的圖形
**五次方程**是一種最高次數為五次的多項式方程。本條目專指只含一個未知數的五次方程（一元五次方程），即方程形如

{\displaystyle ax^{5}+bx^{4}+cx^{3}+dx^{2}+ex+f=0}
其中，*a*、*b*、*c*、*d*、*e*和*f*为复数域内的数，且*a*不为零。例如：

{\displaystyle x^{5}-4x^{4}+2x^{3}-3x+7=0}
二次方程很早就找到了公式解。經過數學家們的不斷努力，三次方程及四次方程在16世紀中有了解答，但是之后很长的一段时间里沒有人知道五次方程是否存在公式解。直到1824年，保羅·魯菲尼和尼爾斯·阿貝爾證明了一般的五次方程，不存在統一的根式解（即由方程的係數通過有限次的四則運算及根號組合而成的公式解）。認為一般的五次方程沒有公式解存在的看法其实是不正確的。事實上，利用一些超越函數，如Θ函数或戴德金η函數即可構造出五次方程的公式解。另外，若只需求得數值解，可以利用數值方法（如牛頓法）得到相當理想的解答。

證明一般五次及其以上的一元多项式方程無根式解的人是埃瓦里斯特·伽羅瓦，他巧妙地利用群論處理了上述的問題。

布靈·傑拉德正規式
---------

對於一般的五次方程式

x^{5}+a\_{1}x^{4}+a\_{2}x^{3}+a\_{3}x^{2}+a\_{4}x+a\_{5}=0\,
可以藉由以下的多项式变换

y=x^{4}+b\_{1}x^{3}+b\_{2}x^{2}+b\_{3}x+b\_{4}\,
得到一個y\,的五次多項式，上述的轉換稱為契爾恩豪森轉換（英语：）（Tschirnhaus transformation），藉由特別選擇的係數b\_{i}\,，可以使y^{4}\,,y^{3}\,,y^{2}\, 的係數為0\,，從而得到如下的方程式： 

y^{5}+my+n=0\,
以上的化簡方法是由厄蘭·塞缪爾·布靈（英语：）所發現，後來喬治·傑拉德（英语：）也獨立發現了此法，因此上式稱為**布靈·傑拉德正規式**（**Bring-Jerrard normal form**）。 
其步驟如下： 
首先令 

x=y-{\frac {a\_{1}}{5}}\,
可消去四次方項，得到 

y^{5}+ay^{3}+by^{2}+cy+d=0\,；
其中，

a={\frac {5a\_{2}-2a\_{1}^{2}}{5}}\,
b={\frac {25a\_{3}-15a\_{1}a\_{2}+4a\_{1}^{3}}{25}}\,
c={\frac {125a\_{4}-50a\_{1}a\_{3}+15a\_{1}^{2}a\_{2}-3a\_{1}^{4}}{125}}\,
d={\frac {3125a\_{5}-625a\_{1}a\_{4}+125a\_{1}^{2}a\_{3}-25a\_{1}^{3}a\_{2}+4a\_{1}^{5}}{3125}}\,
接下來，令z=y^{2}+py+q\,， 
得到

z^{5}+Pz^{4}+Qz^{3}+Az^{2}+Bz+C=0\,，
再令P=Q=0\,， 
求得 

p={\frac {-15b+{\sqrt {60a^{3}+225b^{2}-200ac}}}{10a}}\,；
q={2a \over 5}.\,
第三步，利用契爾恩豪森想到的方法，令： 

X=z^{4}+b\_{1}z^{3}+b\_{2}z^{2}+b\_{3}z+b\_{4}\,，
代入 

z^{5}+Az^{2}+Bz+C=0\,，
得到 

X^{5}+RX^{4}+SX^{3}+TX^{2}+UX+V=0\,，
再令R=S=T=0\,， 
則得b\_{4}={\frac {3Ab\_{1}+4B}{5}}\,， 
若令b\_{2}=-{\frac {4Bb\_{1}+5C}{3A}}\,， 
則b\_{1}\,，b\_{3}\,可由以下兩個方程解得： 

(27A^{4}-160B^{3}+300ABC)b\_{1}^{2}+(27A^{3}B-400B^{2}C+375C^{2}A)b\_{1}\,+(18A^{2}B^{2}-45A^{3}C-250BC^{2})=0;\,

  

675A^{3}b\_{3}^{3}+(3375A^{2}Cb\_{1}-3600AB^{2}b\_{1}-2025A^{4}\,-4500ABC)b\_{3}^{2}
+(675A^{3}Bb\_{1}^{2}+6000B^{2}Cb\_{1}^{2}\,+7200A^{2}B^{2}b\_{1}-4050A^{3}Cb\_{1}+15000C^{2}Bb\_{1}\,\!+9375C^{3}+9675A^{2}BC+2025A^{5})b\_{3}+\,
(-4770A^{3}BC-1125A^{2}C^{2}b\_{1}^{2}-1500B^{2}C^{2}\,-320AB^{3}b\_{1}^{3}-960B^{4}b\_{1}^{2}-3843A^{3}B^{2}b\_{1}+1485A^{4}Cb\_{1}-54A^{5}b\_{1}^{3}
-6250AC^{3}-2400B^{3}Cb\_{1}-108A^{2}B^{3}-675A^{6}\,-756A^{4}Bb\_{1}^{2}-9375AC^{2}Bb\_{1}-3900AB^{2}Cb\_{1}^{2}-225A^{2}BCb\_{1}^{3})=0.\,
若以函數的觀點來看，方程

X^{5}+UX+V=0\,
的解有兩個自變數 U\,, 和 V\,。

若再令

X={\sqrt[ {4}]{-U}}\xi \,
則方程式可以進一步化簡為如下形式：

\xi ^{5}-\xi +t=0\,
它的解 \xi \, 是單一變數 t\, 的函數。

特殊五次方程的求根公式
-----------

雖然一般的五次方程不存在根式解，但是對於某些特殊的五次方程，滿足某些條件後還是有根式解的。

### 型式1

{\displaystyle {ax^{5}+bx^{4}+cx^{3}+{\frac {15abc-4b^{3}}{25a^{2}}}x^{2}+{\frac {25a^{2}c^{2}-5ab^{2}c-b^{4}}{125a^{3}}}x+f=0}}，当a\ne0时，

{\displaystyle {x\_{1}={\frac {-2b+{\sqrt[{5}]{176b^{5}-1200ab^{3}c+2000a^{2}bc^{2}-50000a^{4}f+16{\sqrt {\left(11b^{5}-75ab^{3}c+125a^{2}bc^{2}-3125a^{4}f\right)^{2}-4\left(2b^{2}-5ac\right)^{5}}}}}+{\sqrt[{5}]{176b^{5}-1200ab^{3}c+2000a^{2}bc^{2}-50000a^{4}f-16{\sqrt {\left(11b^{5}-75ab^{3}c+125a^{2}bc^{2}-3125a^{4}f\right)^{2}-4\left(2b^{2}-5ac\right)^{5}}}}}}{10a}}}\,}

{\displaystyle {x\_{2}=-{\frac {b}{5a}}+{\frac {-1+{\sqrt {5}}+{\sqrt {10+2{\sqrt {5}}}}{\rm {i}}}{40a}}{\sqrt[{5}]{176b^{5}-1200ab^{3}c+2000a^{2}bc^{2}-50000a^{4}f+16{\sqrt {\left(11b^{5}-75ab^{3}c+125a^{2}bc^{2}-3125a^{4}f\right)^{2}-4\left(2b^{2}-5ac\right)^{5}}}}}+{\frac {-1+{\sqrt {5}}-{\sqrt {10+2{\sqrt {5}}}}{\rm {i}}}{40a}}{\sqrt[{5}]{176b^{5}-1200ab^{3}c+2000a^{2}bc^{2}-50000a^{4}f-16{\sqrt {\left(11b^{5}-75ab^{3}c+125a^{2}bc^{2}-3125a^{4}f\right)^{2}-4\left(2b^{2}-5ac\right)^{5}}}}}}\,}

{\displaystyle {x\_{3}=-{\frac {b}{5a}}+{\frac {-1-{\sqrt {5}}+{\sqrt {10-2{\sqrt {5}}}}{\rm {i}}}{40a}}{\sqrt[{5}]{176b^{5}-1200ab^{3}c+2000a^{2}bc^{2}-50000a^{4}f+16{\sqrt {\left(11b^{5}-75ab^{3}c+125a^{2}bc^{2}-3125a^{4}f\right)^{2}-4\left(2b^{2}-5ac\right)^{5}}}}}+{\frac {-1-{\sqrt {5}}-{\sqrt {10-2{\sqrt {5}}}}{\rm {i}}}{40a}}{\sqrt[{5}]{176b^{5}-1200ab^{3}c+2000a^{2}bc^{2}-50000a^{4}f-16{\sqrt {\left(11b^{5}-75ab^{3}c+125a^{2}bc^{2}-3125a^{4}f\right)^{2}-4\left(2b^{2}-5ac\right)^{5}}}}}}\,}

{\displaystyle {x\_{4}=-{\frac {b}{5a}}+{\frac {-1-{\sqrt {5}}-{\sqrt {10-2{\sqrt {5}}}}{\rm {i}}}{40a}}{\sqrt[{5}]{176b^{5}-1200ab^{3}c+2000a^{2}bc^{2}-50000a^{4}f+16{\sqrt {\left(11b^{5}-75ab^{3}c+125a^{2}bc^{2}-3125a^{4}f\right)^{2}-4\left(2b^{2}-5ac\right)^{5}}}}}+{\frac {-1-{\sqrt {5}}+{\sqrt {10-2{\sqrt {5}}}}{\rm {i}}}{40a}}{\sqrt[{5}]{176b^{5}-1200ab^{3}c+2000a^{2}bc^{2}-50000a^{4}f-16{\sqrt {\left(11b^{5}-75ab^{3}c+125a^{2}bc^{2}-3125a^{4}f\right)^{2}-4\left(2b^{2}-5ac\right)^{5}}}}}}\,}

{\displaystyle {x\_{5}=-{\frac {b}{5a}}+{\frac {-1+{\sqrt {5}}-{\sqrt {10+2{\sqrt {5}}}}{\rm {i}}}{40a}}{\sqrt[{5}]{176b^{5}-1200ab^{3}c+2000a^{2}bc^{2}-50000a^{4}f+16{\sqrt {\left(11b^{5}-75ab^{3}c+125a^{2}bc^{2}-3125a^{4}f\right)^{2}-4\left(2b^{2}-5ac\right)^{5}}}}}+{\frac {-1+{\sqrt {5}}+{\sqrt {10+2{\sqrt {5}}}}{\rm {i}}}{40a}}{\sqrt[{5}]{176b^{5}-1200ab^{3}c+2000a^{2}bc^{2}-50000a^{4}f-16{\sqrt {\left(11b^{5}-75ab^{3}c+125a^{2}bc^{2}-3125a^{4}f\right)^{2}-4\left(2b^{2}-5ac\right)^{5}}}}}}\,}
### 型式2

{\displaystyle {(c^{2}+1)x^{5}+5d^{4}(3\mp c)x-4d^{5}(\pm 11+2c)=0}}，当{\displaystyle c\neq {\pm {i}}}时，

  

x\_{1}=d\left[A+B+C+D\right]\,

  

{\displaystyle x\_{2}=d\left[{\frac {(-1+{\sqrt {5}})+{\sqrt {10+2{\sqrt {5}}}}{\rm {i}}}{4}}A+{\frac {(-1-{\sqrt {5}})+{\sqrt {10-2{\sqrt {5}}}}{\mathrm {i} }}{4}}B\right]+d\left[{\frac {(-1-{\sqrt {5}})-{\sqrt {10-2{\sqrt {5}}}}{\mathrm {i} }}{4}}C+{\frac {(-1+{\sqrt {5}})-{\sqrt {10+2{\sqrt {5}}}}{\mathrm {i} }}{4}}D\right]\,}

{\displaystyle x\_{3}=d\left[{\frac {(-1-{\sqrt {5}})+{\sqrt {10-2{\sqrt {5}}}}{\mathrm {i} }}{4}}A+{\frac {(-1-{\sqrt {5}})-{\sqrt {10-2{\sqrt {5}}}}{\mathrm {i} }}{4}}B\right]+d\left[{\frac {(-1+{\sqrt {5}})-{\sqrt {10+2{\sqrt {5}}}}{\mathrm {i} }}{4}}C+{\frac {(-1+{\sqrt {5}})+{\sqrt {10+2{\sqrt {5}}}}{\mathrm {i} }}{4}}D\right]\,}

{\displaystyle x\_{4}=d\left[{\frac {(-1-{\sqrt {5}})-{\sqrt {10-2{\sqrt {5}}}}{\mathrm {i} }}{4}}A+{\frac {(-1+{\sqrt {5}})-{\sqrt {10+2{\sqrt {5}}}}{\mathrm {i} }}{4}}B\right]+d\left[{\frac {(-1+{\sqrt {5}})+{\sqrt {10+2{\sqrt {5}}}}{\mathrm {i} }}{4}}C+{\frac {(-1-{\sqrt {5}})+{\sqrt {10-2{\sqrt {5}}}}{\mathrm {i} }}{4}}D\right]\,}

{\displaystyle x\_{5}=d\left[{\frac {(-1+{\sqrt {5}})-{\sqrt {10+2{\sqrt {5}}}}{\rm {i}}}{4}}A+{\frac {(-1+{\sqrt {5}})+{\sqrt {10+2{\sqrt {5}}}}{\mathrm {i} }}{4}}B\right]+d\left[{\frac {(-1-{\sqrt {5}})+{\sqrt {10-2{\sqrt {5}}}}{\mathrm {i} }}{4}}C+{\frac {(-1-{\sqrt {5}})-{\sqrt {10-2{\sqrt {5}}}}{\mathrm {i} }}{4}}D\right]\,}
其中

  

A={\sqrt[ {5}]{{\frac {\left({\sqrt {c^{2}+1}}+{\sqrt {c^{2}+1\mp {\sqrt {c^{2}+1}}}}\right)^{2}\left(-{\sqrt {c^{2}+1}}+{\sqrt {c^{2}+1\pm {\sqrt {c^{2}+1}}}}\right)}{(c^{2}+1)^{2}}}}}\,
B={\sqrt[ {5}]{{\frac {\left({\sqrt {c^{2}+1}}+{\sqrt {c^{2}+1\mp {\sqrt {c^{2}+1}}}}\right)^{2}\left(-{\sqrt {c^{2}+1}}+{\sqrt {c^{2}+1\pm {\sqrt {c^{2}+1}}}}\right)}{(c^{2}+1)^{2}}}}}\,
C={\sqrt[ {5}]{{\frac {\left({\sqrt {c^{2}+1}}+{\sqrt {c^{2}+1\pm {\sqrt {c^{2}+1}}}}\right)^{2}\left({\sqrt {c^{2}+1}}-{\sqrt {c^{2}+1\mp {\sqrt {c^{2}+1}}}}\right)}{(c^{2}+1)^{2}}}}}\,
D=-{\sqrt[ {5}]{{\frac {\left({\sqrt {c^{2}+1}}-{\sqrt {c^{2}+1\mp {\sqrt {c^{2}+1}}}}\right)^{2}\left(-{\sqrt {c^{2}+1}}+{\sqrt {c^{2}+1\pm {\sqrt {c^{2}+1}}}}\right)}{(c^{2}+1)^{2}}}}}\,
### 型式3

{\displaystyle {a^{2}x^{5}+5abx^{3}+5b^{2}x+ac=0}}，当a\ne0时，

{\displaystyle {x\_{1}={\sqrt[{5}]{-{\frac {c}{2a}}+{\sqrt {\left({\frac {c}{2a}}\right)^{2}+\left({\frac {b}{a}}\right)^{5}}}}}+{\sqrt[{5}]{-{\frac {c}{2a}}-{\sqrt {\left({\frac {c}{2a}}\right)^{2}+\left({\frac {b}{a}}\right)^{5}}}}}}\,}

{\displaystyle {x\_{2}={\frac {-1+{\sqrt {5}}+{\sqrt {10+2{\sqrt {5}}}}{\rm {i}}}{4}}{\sqrt[{5}]{-{\frac {c}{2a}}+{\sqrt {\left({\frac {c}{2a}}\right)^{2}+\left({\frac {b}{a}}\right)^{5}}}}}+{\frac {-1+{\sqrt {5}}-{\sqrt {10+2{\sqrt {5}}}}{\rm {i}}}{4}}{\sqrt[{5}]{-{\frac {c}{2a}}-{\sqrt {\left({\frac {c}{2a}}\right)^{2}+\left({\frac {b}{a}}\right)^{5}}}}}}\,}

{\displaystyle {x\_{3}={\frac {-1-{\sqrt {5}}+{\sqrt {10-2{\sqrt {5}}}}{\rm {i}}}{4}}{\sqrt[{5}]{-{\frac {c}{2a}}+{\sqrt {\left({\frac {c}{2a}}\right)^{2}+\left({\frac {b}{a}}\right)^{5}}}}}+{\frac {-1-{\sqrt {5}}-{\sqrt {10-2{\sqrt {5}}}}{\rm {i}}}{4}}{\sqrt[{5}]{-{\frac {c}{2a}}-{\sqrt {\left({\frac {c}{2a}}\right)^{2}+\left({\frac {b}{a}}\right)^{5}}}}}}\,}

{\displaystyle {x\_{4}={\frac {-1-{\sqrt {5}}-{\sqrt {10-2{\sqrt {5}}}}{\rm {i}}}{4}}{\sqrt[{5}]{-{\frac {c}{2a}}+{\sqrt {\left({\frac {c}{2a}}\right)^{2}+\left({\frac {b}{a}}\right)^{5}}}}}+{\frac {-1-{\sqrt {5}}+{\sqrt {10-2{\sqrt {5}}}}{\rm {i}}}{4}}{\sqrt[{5}]{-{\frac {c}{2a}}-{\sqrt {\left({\frac {c}{2a}}\right)^{2}+\left({\frac {b}{a}}\right)^{5}}}}}}\,}

{\displaystyle {x\_{5}={\frac {-1+{\sqrt {5}}-{\sqrt {10+2{\sqrt {5}}}}{\rm {i}}}{4}}{\sqrt[{5}]{-{\frac {c}{2a}}+{\sqrt {\left({\frac {c}{2a}}\right)^{2}+\left({\frac {b}{a}}\right)^{5}}}}}+{\frac {-1+{\sqrt {5}}+{\sqrt {10+2{\sqrt {5}}}}{\rm {i}}}{4}}{\sqrt[{5}]{-{\frac {c}{2a}}-{\sqrt {\left({\frac {c}{2a}}\right)^{2}+\left({\frac {b}{a}}\right)^{5}}}}}}\,}

通過模橢圓函數求解
---------

在 Tschirnhaus 變換的幫助下，所有五次方程都可以在初等數學函數表達式的幫助下轉換為 Bring-Jerrard 形式。 Bring-Jerrard 形式包含五次項、線性項和絕對項。 但是四次、三次和二次項在這種形式的方程中根本不存在。 Bring-Jerrard 形式的廣義橢圓解將在以下段落中討論。根據數學家 Glashan、Young 和 Runge 發現的參數化公式，可以從方程和實解中導出以下一對公式:

{\displaystyle x^{5}+x={\frac {2}{5}}y^{-5/4}{\frac {(1+y-y^{2}){\sqrt {2+2y^{2}}}}{\sqrt[{4}]{10+15y-10y^{2}}}}}
{\displaystyle x={\frac {2}{5}}y^{-1/4}{\sqrt[{4}]{10+15y-10y^{2}}}\cosh {\biggl \{}{\frac {1}{5}}{\text{arcosh}}{\biggl [}{\frac {5{\sqrt {5+5y^{2}}}}{(1+2y){\sqrt {4+6y-4y^{2}}}}}{\biggr ]}{\biggr \}}-}
{\displaystyle -{\frac {2}{5}}y^{-1/4}{\sqrt[{4}]{10+15y-10y^{2}}}\sinh {\biggl \{}{\frac {1}{5}}{\text{arsinh}}{\biggl [}{\frac {5y{\sqrt {5+5y^{2}}}}{(2-y){\sqrt {4+6y-4y^{2}}}}}{\biggr ]}{\biggr \}}}
這對公式對所有值 0 < y < 2 都有效。對於要用這種方法求解的 Bring-Jerrard 的一般形式，需要一個橢圓鍵。 這個橢圓密鑰可以根據 卡爾·雅可比 (Carl Gustav Jakob Jacobi) 使用 Θ函數 生成:

{\displaystyle x^{5}+x=w}
{\displaystyle x={\frac {2}{5}}y^{-1/4}{\sqrt[{4}]{10+15y-10y^{2}}}\cosh {\biggl \{}{\frac {1}{5}}{\text{arcosh}}{\biggl [}{\frac {5{\sqrt {5+5y^{2}}}}{(1+2y){\sqrt {4+6y-4y^{2}}}}}{\biggr ]}{\biggr \}}-}
{\displaystyle -{\frac {2}{5}}y^{-1/4}{\sqrt[{4}]{10+15y-10y^{2}}}\sinh {\biggl \{}{\frac {1}{5}}{\text{arsinh}}{\biggl [}{\frac {5y{\sqrt {5+5y^{2}}}}{(2-y){\sqrt {4+6y-4y^{2}}}}}{\biggr ]}{\biggr \}}}
{\displaystyle y={\frac {5\,\vartheta \_{00}{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {5}{4}}{\sqrt[{4}]{5}}\,w)]^{2}\}^{5}{\bigr \rangle }^{2}}{2\,\vartheta \_{00}{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {5}{4}}{\sqrt[{4}]{5}}\,w)]^{2}\}{\bigr \rangle }^{2}}}-{\frac {1}{2}}}
現在在下面精確地解釋這個解決過程。 本段上式的等式刻度的右側取值 w:

{\displaystyle w={\frac {2}{5}}y^{-5/4}{\frac {(1+y-y^{2}){\sqrt {2+2y^{2}}}}{\sqrt[{4}]{10+15y-10y^{2}}}}}
必須為值 y 求解該方程。 這需要一個橢圓模函數表達式，在這種情況下包括 Jacobi theta 函數:

{\displaystyle y={\frac {5\,\vartheta \_{00}{\bigl \{}q{\bigl [}{\bigl (}50{\sqrt {5}}\,w^{2}+32+2{\sqrt {3125w^{4}+256}}{\bigr )}^{-1/2}{\bigl (}{\sqrt {{\sqrt {3125w^{4}+256}}+16}}+5{\sqrt[{4}]{5}}\,w{\bigr )}{\bigr ]}^{5}{\bigr \}}^{2}}{2\,\vartheta \_{00}{\bigl \{}q{\bigl [}{\bigl (}50{\sqrt {5}}\,w^{2}+32+2{\sqrt {3125w^{4}+256}}{\bigr )}^{-1/2}{\bigl (}{\sqrt {{\sqrt {3125w^{4}+256}}+16}}+5{\sqrt[{4}]{5}}\,w{\bigr )}{\bigr ]}{\bigr \}}^{2}}}-{\frac {1}{2}}}
此解表達式與以下表達式一致:

{\displaystyle y={\frac {5\,\vartheta \_{00}{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {5}{4}}{\sqrt[{4}]{5}}\,w)]^{2}\}^{5}{\bigr \rangle }^{2}}{2\,\vartheta \_{00}{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {5}{4}}{\sqrt[{4}]{5}}\,w)]^{2}\}{\bigr \rangle }^{2}}}-{\frac {1}{2}}}

橢圓函數的定義和恆等式
-----------

現在必須定義此表達式中指定的函數。 所示的主要 theta 函數具有以下總和定義和以下等效乘積定義:

{\displaystyle \vartheta \_{00}(z)=1+2\sum \_{k=1}^{\infty }z^{k^{2}}}
{\displaystyle \vartheta \_{00}(z)=\prod \_{k=1}^{\infty }(1-z^{2k})(1+z^{2k-1})^{2}}
字母 q 描述了數學橢圓 nome 函數:

{\displaystyle q(\varepsilon )=\exp[-\pi K({\sqrt {1-\varepsilon ^{2}}})K(\varepsilon )^{-1}]}
內商中顯示的字母 K 表示完整的第一類 椭圆积分:

{\displaystyle K(r)=\int \_{0}^{\pi /2}{\frac {1}{\sqrt {1-r^{2}\sin(\varphi )^{2}}}}\mathrm {d} \varphi }
{\displaystyle K(r)=2\int \_{0}^{1}{\frac {1}{\sqrt {(u^{2}+1)^{2}-4r^{2}u^{2}}}}\mathrm {d} u}
縮寫 **ctlh** 表示函數 *双曲双扭线餘切函数* (Hyperbolic lemniscate cotangent)。 而縮寫 **aclh** 表示函數 *双曲双扭线 面積餘弦函数* (Hyperbolic lemniscate Areacosine)。 這些函數與 卡爾·弗里德里希·高斯(Carl Friedrich Gauss) 建立的 双扭线函数 (Lemniscate elliptic functions) sl 和 cl 在代數上相關，並且可以使用這兩個函數來定義： 

{\displaystyle \mathrm {sl} (\varphi )=\tan {\biggl \langle }2\arctan {\biggl \{}{\frac {4}{G}}\sin {\bigl (}{\frac {\varphi }{G}}{\bigr )}\sum \_{k=1}^{\infty }{\frac {\cosh[(2k-1)\pi ]}{\cosh[(2k-1)\pi ]^{2}-\cos(\varphi /G)^{2}}}{\biggr \}}{\biggr \rangle }}
{\displaystyle \mathrm {cl} (\varphi )=\tan {\biggl \langle }2\arctan {\biggl \{}{\frac {4}{G}}\cos {\bigl (}{\frac {\varphi }{G}}{\bigr )}\sum \_{k=1}^{\infty }{\frac {\cosh[(2k-1)\pi ]}{\cosh[(2k-1)\pi ]^{2}-\sin(\varphi /G)^{2}}}{\biggr \}}{\biggr \rangle }}
{\displaystyle [{\text{sl}}(\varphi )^{2}+1][{\text{cl}}(\varphi )^{2}+1]=2}
{\displaystyle {\text{ctlh}}(\varrho )=\operatorname {cl} ({\tfrac {1}{2}}{\sqrt {2}}\varrho ){\biggl [}{\frac {\operatorname {sl} ({\tfrac {1}{2}}{\sqrt {2}}\varrho )^{2}+1}{\operatorname {sl} ({\tfrac {1}{2}}{\sqrt {2}}\varrho )^{2}+\operatorname {cl} ({\tfrac {1}{2}}{\sqrt {2}}\varrho )^{2}}}{\biggr ]}^{1/2}}
{\displaystyle {\text{ctlh}}(\varrho )={\frac {{\text{cd}}(\varrho ;{\tfrac {1}{2}}{\sqrt {2}})}{\sqrt[{4}]{{\text{cd}}(\varrho ;{\tfrac {1}{2}}{\sqrt {2}})^{4}+{\text{sn}}(\varrho ;{\tfrac {1}{2}}{\sqrt {2}})^{4}}}}}
{\displaystyle \mathrm {aclh} (s)={\tfrac {1}{2}}F[2\operatorname {arccot}(s);{\tfrac {1}{2}}{\sqrt {2}}]}
{\displaystyle {\text{aclh}}(s)={\frac {1}{2}}{\sqrt {2}}\,\pi \,G-\int \_{0}^{1}{\frac {s}{\sqrt {s^{4}t^{4}+1}}}\,\mathrm {d} t}
{\displaystyle G={\tfrac {1}{2}}{\sqrt {2\pi }}\,\Gamma ({\tfrac {3}{4}})^{-2}}
{\displaystyle {\text{ctlh}}{\bigl [}{\tfrac {1}{2}}\mathrm {aclh} (s){\bigr ]}^{2}=(2s^{2}+2+2{\sqrt {s^{4}+1}})^{-1/2}({\sqrt {{\sqrt {s^{4}+1}}+1}}+s)}
{\displaystyle {\text{sl}}{\bigl [}{\tfrac {1}{2}}{\sqrt {2}}\,\mathrm {aclh} (s){\bigr ]}={\sqrt {{\sqrt {s^{4}+1}}-s^{2}}}}
字母G代表高斯常數，可以用伽馬函數用剛才所示的方式表示。

连分数
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连分数是拉马努金 (Rogers-Ramanujan continued fraction) 允許以 Bring-Jerrard 形式對廣義五次方程進行非常緊湊的解。 這個連分數函數和交替連分數可以定義如下：

{\displaystyle R(z)=z^{1/5}{\frac {(z;z^{5})\_{\infty }(z^{4};z^{5})\_{\infty }}{(z^{2};z^{5})\_{\infty }(z^{3};z^{5})\_{\infty }}}}
{\displaystyle R(z)=\tan {\biggl \{}{\frac {1}{2}}\arctan {\biggl [}{\frac {\vartheta \_{00}(z^{1/2})^{2}}{2\vartheta \_{00}(z^{5/2})^{2}}}-{\frac {1}{2}}{\biggr ]}{\biggr \}}^{2/5}\tan {\biggl \{}{\frac {1}{2}}\operatorname {arccot} {\biggl [}{\frac {\vartheta \_{00}(z^{1/2})^{2}}{2\vartheta \_{00}(z^{5/2})^{2}}}-{\frac {1}{2}}{\biggr ]}{\biggr \}}^{1/5}}
{\displaystyle R(z^{2})=\tan {\biggl \{}{\frac {1}{2}}\arctan {\biggl [}{\frac {\vartheta \_{00}(z)^{2}}{2\vartheta \_{00}(z^{5})^{2}}}-{\frac {1}{2}}{\biggr ]}{\biggr \}}^{2/5}\tan {\biggl \{}{\frac {1}{2}}\operatorname {arccot} {\biggl [}{\frac {\vartheta \_{00}(z)^{2}}{2\vartheta \_{00}(z^{5})^{2}}}-{\frac {1}{2}}{\biggr ]}{\biggr \}}^{1/5}}
{\displaystyle S(z)=\tan {\biggl \{}{\frac {1}{2}}\arctan {\biggl [}{\frac {\vartheta \_{00}(z)^{2}}{2\vartheta \_{00}(z^{5})^{2}}}-{\frac {1}{2}}{\biggr ]}{\biggr \}}^{1/5}\cot {\biggl \{}{\frac {1}{2}}\operatorname {arccot} {\biggl [}{\frac {\vartheta \_{00}(z)^{2}}{2\vartheta \_{00}(z^{5})^{2}}}-{\frac {1}{2}}{\biggr ]}{\biggr \}}^{2/5}}
{\displaystyle S(z)={\frac {R(z^{4})}{R(z^{2})R(z)}}}
括號，每個都有兩個條目，形成所謂的 Pochhammer-符號 (Pochhammer symbol) 並因此代表產品系列。 基於這些定義，可以為實際解建立以下壓縮精確解公式:

{\displaystyle x^{5}+x=w}
{\displaystyle x={\frac {S{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {5}{4}}{\sqrt[{4}]{5}}\,w)]^{2}\}{\bigr \rangle }^{2}-R{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {5}{4}}{\sqrt[{4}]{5}}\,w)]^{2}\}^{2}{\bigr \rangle }}{S{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {5}{4}}{\sqrt[{4}]{5}}\,w)]^{2}\}{\bigr \rangle }^{2}}}\times }
{\displaystyle \times {\frac {1-R{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {5}{4}}{\sqrt[{4}]{5}}\,w)]^{2}\}^{2}{\bigr \rangle }\,S{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {5}{4}}{\sqrt[{4}]{5}}\,w)]^{2}\}{\bigr \rangle }}{R{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {5}{4}}{\sqrt[{4}]{5}}\,w)]^{2}\}^{2}{\bigr \rangle }^{2}}}\times }
{\displaystyle \times {\frac {\vartheta \_{00}{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {5}{4}}{\sqrt[{4}]{5}}\,w)]^{2}\}^{5}{\bigr \rangle }\,\vartheta \_{00}{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {5}{4}}{\sqrt[{4}]{5}}\,w)]^{2}\}^{1/5}{\bigr \rangle }^{2}-5\,\vartheta \_{00}{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {5}{4}}{\sqrt[{4}]{5}}\,w)]^{2}\}^{5}{\bigr \rangle }^{3}}{2{\sqrt[{4}]{20}}\,{\text{sl}}[{\tfrac {1}{2}}{\sqrt {2}}\,{\text{aclh}}({\tfrac {5}{4}}{\sqrt[{4}]{5}}\,w)]\,\vartheta \_{00}{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {5}{4}}{\sqrt[{4}]{5}}\,w)]^{2}\}{\bigr \rangle }^{3}}}}

準確的例子
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分配给非初等数学实解的第一个自然数 w 是数字 w = 3:

{\displaystyle x^{5}+x=3}
{\displaystyle x={\frac {S{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {15}{4}}{\sqrt[{4}]{5}})]^{2}\}{\bigr \rangle }^{2}-R{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {15}{4}}{\sqrt[{4}]{5}})]^{2}\}^{2}{\bigr \rangle }}{S{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {15}{4}}{\sqrt[{4}]{5}})]^{2}\}{\bigr \rangle }^{2}}}\times }
{\displaystyle \times {\frac {1-R{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {15}{4}}{\sqrt[{4}]{5}})]^{2}\}^{2}{\bigr \rangle }\,S{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {15}{4}}{\sqrt[{4}]{5}})]^{2}\}{\bigr \rangle }}{R{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {15}{4}}{\sqrt[{4}]{5}})]^{2}\}^{2}{\bigr \rangle }^{2}}}\times }
{\displaystyle \times {\frac {\vartheta \_{00}{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {15}{4}}{\sqrt[{4}]{5}})]^{2}\}^{5}{\bigr \rangle }\,\vartheta \_{00}{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {15}{4}}{\sqrt[{4}]{5}})]^{2}\}^{1/5}{\bigr \rangle }^{2}-5\,\vartheta \_{00}{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {15}{4}}{\sqrt[{4}]{5}})]^{2}\}^{5}{\bigr \rangle }^{3}}{2{\sqrt[{4}]{20}}\,{\text{sl}}[{\tfrac {1}{2}}{\sqrt {2}}\,{\text{aclh}}({\tfrac {15}{4}}{\sqrt[{4}]{5}})]\,\vartheta \_{00}{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {15}{4}}{\sqrt[{4}]{5}})]^{2}\}{\bigr \rangle }^{3}}}}
{\displaystyle q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {15}{4}}{\sqrt[{4}]{5}})]^{2}\}\approx 0.452374059450344348576600264284387826377845763909}
{\displaystyle x\approx 1.132997565885065266721141634288532379816526027727}
與此類似，數字 w = 7 僅分配給非基本解:

{\displaystyle x^{5}+x=7}
{\displaystyle x={\frac {S{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {35}{4}}{\sqrt[{4}]{5}})]^{2}\}{\bigr \rangle }^{2}-R{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {35}{4}}{\sqrt[{4}]{5}})]^{2}\}^{2}{\bigr \rangle }}{S{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {35}{4}}{\sqrt[{4}]{5}})]^{2}\}{\bigr \rangle }^{2}}}\times }
{\displaystyle \times {\frac {1-R{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {35}{4}}{\sqrt[{4}]{5}})]^{2}\}^{2}{\bigr \rangle }\,S{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {35}{4}}{\sqrt[{4}]{5}})]^{2}\}{\bigr \rangle }}{R{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {35}{4}}{\sqrt[{4}]{5}})]^{2}\}^{2}{\bigr \rangle }^{2}}}\times }
{\displaystyle \times {\frac {\vartheta \_{00}{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {35}{4}}{\sqrt[{4}]{5}})]^{2}\}^{5}{\bigr \rangle }\,\vartheta \_{00}{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {35}{4}}{\sqrt[{4}]{5}})]^{2}\}^{1/5}{\bigr \rangle }^{2}-5\,\vartheta \_{00}{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {35}{4}}{\sqrt[{4}]{5}})]^{2}\}^{5}{\bigr \rangle }^{3}}{2{\sqrt[{4}]{20}}\,{\text{sl}}[{\tfrac {1}{2}}{\sqrt {2}}\,{\text{aclh}}({\tfrac {35}{4}}{\sqrt[{4}]{5}})]\,\vartheta \_{00}{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {35}{4}}{\sqrt[{4}]{5}})]^{2}\}{\bigr \rangle }^{3}}}}
{\displaystyle q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {35}{4}}{\sqrt[{4}]{5}})]^{2}\}\approx 0.53609630892200161460073096549143569900990236}
{\displaystyle x\approx 1.4108138510595771319852918753499397839215989}

參見
--

* 方程式論
* 布靈根式

引文
--

1. ↑  阿米爾·艾克塞爾（Amir D. Aezel）.. 台北: 時報出版. 1998: p.87. ISBN 957-13-2648-8.  引文格式1维护：冗余文本 (link)
2. ↑ .  [2022-02-18]. （原始内容存档于2022-02-18）.
3. ↑ .  [2022-02-18]. （原始内容存档于2022-02-18）.
4. ↑ .  [2022-02-19]. （原始内容存档于2022-02-19）.

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| * 
* 论
* 编
多項式 |
| --- |
|  |
| 函數 | 
* 零次函數(常數函數)
* 一次函數
* 二次函数
* 三次函數
* 四次函數
* 五次函數

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|  |
| 方程 | 
* 一次方程
* 二次方程
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| 算法 | 
* 多项式除法
* 因式
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* 秦九韶算法
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