五次方程

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

${\displaystyle ax^{5}+bx^{4}+cx^{3}+dx^{2}+ex+f=0}$

${\displaystyle y=x^{5}+13x^{4}+35x^{3}-85x^{2}-216x+252}$的圖形

其中，a、b、c、d、e和f為複數域內的數，且a不為零。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，但是之後很長的一段時間裏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（即由方程的系數通過有限次的四則運算及根號組合而成的公式解）^[1]。認為一般的五次方程沒有公式解存在的看法其實是不正確的。事實上，利用一些超越函數，如Θ函數或戴德金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項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

${\displaystyle x^{5}+a_{1}x^{4}+a_{2}x^{3}+a_{3}x^{2}+a_{4}x+a_{5}=0\,}$

可以藉由以下的多項式轉換

${\displaystyle y=x^{4}+b_{1}x^{3}+b_{2}x^{2}+b_{3}x+b_{4}\,}$

得到一個${\displaystyle y\,}$的五次多項式，上述的轉換稱為契爾恩豪森轉換（英語：Tschirnhaus transformation）（Tschirnhaus transformation），藉由特別選擇的系數${\displaystyle b_{i}\,}$，可以使${\displaystyle y^{4}\,}$,${\displaystyle y^{3}\,}$,${\displaystyle y^{2}\,}$ 的系數為${\displaystyle 0\,}$，從而得到如下的方程式：

${\displaystyle y^{5}+my+n=0\,}$

以上的化簡方法是由厄蘭·塞繆爾·布靈（英語：Erland Samuel Bring）所發現，後來喬治·傑拉德（英語：George Jerrard）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

${\displaystyle x=y-{\frac {a_{1}}{5}}\,}$

可消去四次方項，得到

${\displaystyle y^{5}+ay^{3}+by^{2}+cy+d=0\,}$；

其中，

${\displaystyle a={\frac {5a_{2}-2a_{1}^{2}}{5}}\,}$

${\displaystyle b={\frac {25a_{3}-15a_{1}a_{2}+4a_{1}^{3}}{25}}\,}$

${\displaystyle c={\frac {125a_{4}-50a_{1}a_{3}+15a_{1}^{2}a_{2}-3a_{1}^{4}}{125}}\,}$

${\displaystyle d={\frac {3125a_{5}-625a_{1}a_{4}+125a_{1}^{2}a_{3}-25a_{1}^{3}a_{2}+4a_{1}^{5}}{3125}}\,}$

接下來，令${\displaystyle z=y^{2}+py+q\,}$， 得到

${\displaystyle z^{5}+Pz^{4}+Qz^{3}+Az^{2}+Bz+C=0\,}$，

再令${\displaystyle P=Q=0\,}$， 求得

${\displaystyle p={\frac {-15b+{\sqrt {60a^{3}+225b^{2}-200ac}}}{10a}}\,}$；

${\displaystyle q={2a \over 5}.\,}$

第三步，利用契爾恩豪森想到的方法，令：

${\displaystyle X=z^{4}+b_{1}z^{3}+b_{2}z^{2}+b_{3}z+b_{4}\,}$，

代入

${\displaystyle z^{5}+Az^{2}+Bz+C=0\,}$，

得到

${\displaystyle X^{5}+RX^{4}+SX^{3}+TX^{2}+UX+V=0\,}$，

再令${\displaystyle R=S=T=0\,}$， 則得${\displaystyle b_{4}={\frac {3Ab_{1}+4B}{5}}\,}$， 若令${\displaystyle b_{2}=-{\frac {4Bb_{1}+5C}{3A}}\,}$， 則${\displaystyle b_{1}\,}$，${\displaystyle b_{3}\,}$可由以下兩個方程解得：

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

${\displaystyle 675A^{3}b_{3}^{3}+(3375A^{2}Cb_{1}-3600AB^{2}b_{1}-2025A^{4}\,-4500ABC)b_{3}^{2}}$

${\displaystyle +(675A^{3}Bb_{1}^{2}+6000B^{2}Cb_{1}^{2}\,+7200A^{2}B^{2}b_{1}-4050A^{3}Cb_{1}+15000C^{2}Bb_{1}\,\!+9375C^{3}+9675A^{2}BC+2025A^{5})b_{3}+\,}$

${\displaystyle (-4770A^{3}BC-1125A^{2}C^{2}b_{1}^{2}-1500B^{2}C^{2}\,-320AB^{3}b_{1}^{3}-960B^{4}b_{1}^{2}-3843A^{3}B^{2}b_{1}+1485A^{4}Cb_{1}-54A^{5}b_{1}^{3}}$

${\displaystyle -6250AC^{3}-2400B^{3}Cb_{1}-108A^{2}B^{3}-675A^{6}\,-756A^{4}Bb_{1}^{2}-9375AC^{2}Bb_{1}-3900AB^{2}Cb_{1}^{2}-225A^{2}BCb_{1}^{3})=0.\,}$

若以函數的觀點來看，方程

${\displaystyle X^{5}+UX+V=0\,}$

的解有兩個自變量 ${\displaystyle U\,}$, 和 ${\displaystyle V\,}$。

若再令

${\displaystyle X={\sqrt[{4}]{-U}}\xi \,}$

則方程式可以進一步化簡為如下形式：

${\displaystyle \xi ^{5}-\xi +t=0\,}$

它的解 ${\displaystyle \xi \,}$ 是單一變數 ${\displaystyle t\,}$ 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

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

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

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

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

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

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

型式2

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

${\displaystyle x_{1}=d\left[A+B+C+D\right]\,}$

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

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

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

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

其中

${\displaystyle A={\sqrt[{5}]{\frac {\left({\sqrt {c^{2}+1}}+{\sqrt {c^{2}+1\mp {\sqrt {c^{2}+1}}}}\right)^{2}\left(-{\sqrt {c^{2}+1}}+{\sqrt {c^{2}+1\pm {\sqrt {c^{2}+1}}}}\right)}{(c^{2}+1)^{2}}}}\,}$

${\displaystyle B={\sqrt[{5}]{\frac {\left({\sqrt {c^{2}+1}}+{\sqrt {c^{2}+1\mp {\sqrt {c^{2}+1}}}}\right)^{2}\left(-{\sqrt {c^{2}+1}}+{\sqrt {c^{2}+1\pm {\sqrt {c^{2}+1}}}}\right)}{(c^{2}+1)^{2}}}}\,}$

${\displaystyle C={\sqrt[{5}]{\frac {\left({\sqrt {c^{2}+1}}+{\sqrt {c^{2}+1\pm {\sqrt {c^{2}+1}}}}\right)^{2}\left({\sqrt {c^{2}+1}}-{\sqrt {c^{2}+1\mp {\sqrt {c^{2}+1}}}}\right)}{(c^{2}+1)^{2}}}}\,}$

${\displaystyle D=-{\sqrt[{5}]{\frac {\left({\sqrt {c^{2}+1}}-{\sqrt {c^{2}+1\mp {\sqrt {c^{2}+1}}}}\right)^{2}\left(-{\sqrt {c^{2}+1}}+{\sqrt {c^{2}+1\pm {\sqrt {c^{2}+1}}}}\right)}{(c^{2}+1)^{2}}}}\,}$

型式3

${\displaystyle {a^{2}x^{5}+5abx^{3}+5b^{2}x+ac=0}}$，當${\displaystyle a\neq 0}$時，

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

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

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

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

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

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^[2] 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 橢圓積分:

${\displaystyle K(r)=\int _{0}^{\pi /2}{\frac {1}{\sqrt {1-r^{2}\sin(\varphi )^{2}}}}\mathrm {d} \varphi }$

${\displaystyle K(r)=2\int _{0}^{1}{\frac {1}{\sqrt {(u^{2}+1)^{2}-4r^{2}u^{2}}}}\mathrm {d} u}$

縮寫 ctlh 表示函數 雙曲雙扭線餘切函數^[3] (Hyperbolic lemniscate cotangent)。 而縮寫 aclh 表示函數 雙曲雙扭線 面積餘弦函數 (Hyperbolic lemniscate Areacosine)。 這些函數與 卡爾·弗里德里希·高斯(Carl Friedrich Gauss) 建立的 雙扭線函數^[4] (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。

連分數

連分數是拉馬努金 (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

分配給非初等數學實解的第一個自然數 w 是數字 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）. 費馬最後定理. 台北: 時報出版. 1998: p.87. ISBN 957-13-2648-8. 引文格式1維護：冗餘文本 (link)
2. 存档副本. [2022-02-18]. （原始內容存檔於2022-02-18）.
3. 存档副本. [2022-02-18]. （原始內容存檔於2022-02-18）.
4. 存档副本. [2022-02-19]. （原始內容存檔於2022-02-19）.