五ご次じ方かた程ほど

五ご次じ方かた程ほど是ぜ一いち種しゅ最高さいこう次數じすう為ため五ご次じ的てき多項式たこうしき方かた程ほど。本條ほんじょう目め專せん指ゆび只ただ含一いち個こ未知數みちすう的てき五ご次じ方かた程ほど（一元いちげん五ご次じ方かた程ほど），即そく方かた程ほど形かたち如

$y=x^{5}+13x^{4}+35x^{3}-85x^{2}-216x+252$的てき圖形ずけい

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

其中，a、b、c、d、e和わf为复数域いき内的ないてき数すう，且a不ふ为零。例れい如：

$x^{5}-4x^{4}+2x^{3}-3x+7=0$

二に次じ方かた程ほど很早就找到了りょう公式こうしき解かい。經過けいか數學すうがく家か們的不斷ふだん努力どりょく，三さん次じ方かた程ほど及四よん次じ方かた程ほど在ざい16世紀せいき中有ちゅうう了解りょうかい答こたえ，但ただし是ぜ之の后きさき很长的てき一段时间里沒有人知道五次方程是否存在公式解。直ちょく到いた1824年ねん，保ほ羅ら·魯菲尼あま和わ尼あま爾なんじ斯·阿おもね貝かい爾なんじ證明しょうめい了りょう一般いっぱん的てき五ご次じ方かた程ほど，不ふ存在そんざい統一とういつ的てき根ね式しき解かい（即そく由よし方かた程ほど的てき係數けいすう通過つうか有限ゆうげん次じ的てき四則しそく運算うんざん及根號こんごう組合くみあい而成的てき公式こうしき解かい）。認みとめ為ため一般的五次方程沒有公式解存在的看法其实是不正確的。事實じじつ上じょう，利用りよう一いち些超越ちょうえつ函數かんすう，如Θしーた函数かんすう或ある戴德金きんηいーた函數かんすう即そく可か構造こうぞう出で五ご次じ方かた程ほど的てき公式こうしき解かい。另外，若わか只ただ需求得とく數すう值解，可か以利用りよう數すう值方法ほう（如牛うし頓ひたぶる法ほう）得とく到いた相當そうとう理想りそう的てき解答かいとう。

證明しょうめい一般五次及其以上的一元多项式方程無根式解的人是埃ほこり瓦かわら里さと斯特·伽羅きゃら瓦かわら，他た巧妙こうみょう地ち利用りよう群論ぐんろん處理しょり了りょう上述じょうじゅつ的てき問題もんだい。

布ぬの靈れい·傑すぐる拉ひしげ德とく正規せいき式しき

對たい於一般いっぱん的てき五ご次じ方程式ほうていしき

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

可か以藉由よし以下いか的てき多た项式变换

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

得え到いた一いち個こ$y\,$的てき五ご次じ多項式たこうしき，上述じょうじゅつ的てき轉換てんかん稱たたえ為ため契ちぎり爾なんじ恩おん豪ごう森もり轉換てんかん（Tschirnhaus transformation），藉由特別とくべつ選擇せんたく的てき係數けいすう$b_{i}\,$，可か以使$y^{4}\,$,$y^{3}\,$,$y^{2}\,$ 的てき係數けいすう為ため$0\,$，從したがえ而得到いた如下的てき方程式ほうていしき：

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

以上いじょう的てき化か簡方法ほう是ぜ由ゆかり厄やく蘭らん·塞ふさが缪爾·布ぬの靈れい所ところ發現はつげん，後來こうらい喬たかし治ち·傑すぐる拉ひしげ德とく也獨立どくりつ發現はつげん了りょう此法，因いん此上式しき稱たたえ為ため布ぬの靈れい·傑すぐる拉ひしげ德とく正規せいき式しき（Bring-Jerrard normal form）。 其步驟如下か： 首くび先さき令れい

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

可か消去しょうきょ四よん次じ方かた項こう，得とく到いた

$y^{5}+ay^{3}+by^{2}+cy+d=0\,$；

其中，

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

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

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

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

接せっ下か來らい，令れい$z=y^{2}+py+q\,$， 得え到いた

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

再さい令れい$P=Q=0\,$， 求もとめ得とく

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

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

第だい三さん步ほ，利用りよう契ちぎり爾なんじ恩おん豪ごう森もり想到そうとう的てき方法ほうほう，令れい：

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

代入だいにゅう

$z^{5}+Az^{2}+Bz+C=0\,$，

得え到いた

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

再さい令れい$R=S=T=0\,$， 則のり得とく$b_{4}={\frac {3Ab_{1}+4B}{5}}\,$， 若わか令れい$b_{2}=-{\frac {4Bb_{1}+5C}{3A}}\,$， 則のり$b_{1}\,$，$b_{3}\,$可か由よし以下いか兩個りゃんこ方かた程ほど解かい得どく：

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

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

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

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

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

若わか以函數すう的てき觀點かんてん來らい看み，方ぽう程ほど

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

的てき解かい有ゆう兩個りゃんこ自じ變數へんすう $U\,$, 和わ $V\,$。

若わか再さい令れい

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

則のり方程式ほうていしき可か以進一步化簡為如下形式：

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

它的解かい $\xi \,$ 是ぜ單一たんいつ變數へんすう $t\,$ 的てき函數かんすう。

特殊とくしゅ五ご次じ方かた程ほど的てき求根きゅうこん公式こうしき

雖然一般的五次方程不存在根式解，但ただし是ぜ對たい於某些特殊こと的てき五ご次じ方かた程ほど，滿足まんぞく某ぼう些條件じょうけん後ご還かえ是ぜ有ゆう根ね式しき解かい的てき。

型式けいしき1

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

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

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

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

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

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

型式けいしき2

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

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

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

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

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

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

其中

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

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

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

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

型式けいしき3

${a^{2}x^{5}+5abx^{3}+5b^{2}x+ac=0}$，当とう$a\neq 0$时，

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

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

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

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

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

通過つうか模も橢圓だえん函數かんすう求もとめ解かい

在ざい Tschirnhaus 變換へんかん的てき幫助下か，所有しょゆう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 形式けいしき。 Bring-Jerrard 形式けいしき包含ほうがん五ご次項じこう、線せん性せい項こう和わ絕對ぜったい項こう。 但ただし是ぜ四よん次じ、三次和二次項在這種形式的方程中根本不存在。 Bring-Jerrard 形式けいしき的てき廣義こうぎ橢圓だえん解かい將はた在ざい以下いか段落だんらく中ちゅう討論とうろん。根據こんきょ數學すうがく家か Glashan、Young 和わ Runge 發現はつげん的てき參さん數すう化か公式こうしき，可か以從方かた程ほど和實かずみ解かい中ちゅう導出どうしゅつ以下いか一いち對たい公式こうしき:

$x^{5}+x={\frac {2}{5}}y^{-5/4}{\frac {(1+y-y^{2}){\sqrt {2+2y^{2}}}}{\sqrt[{4}]{10+15y-10y^{2}}}}$

$x={\frac {2}{5}}y^{-1/4}{\sqrt[{4}]{10+15y-10y^{2}}}\cosh {\biggl \{}{\frac {1}{5}}{\text{arcosh}}{\biggl [}{\frac {5{\sqrt {5+5y^{2}}}}{(1+2y){\sqrt {4+6y-4y^{2}}}}}{\biggr ]}{\biggr \}}-$

$-{\frac {2}{5}}y^{-1/4}{\sqrt[{4}]{10+15y-10y^{2}}}\sinh {\biggl \{}{\frac {1}{5}}{\text{arsinh}}{\biggl [}{\frac {5y{\sqrt {5+5y^{2}}}}{(2-y){\sqrt {4+6y-4y^{2}}}}}{\biggr ]}{\biggr \}}$

這對公式こうしき對たい所有しょゆう值 0 < y < 2 都と有效ゆうこう。對たい於要用よう這種方法ほうほう求もとめ解かい的てき Bring-Jerrard 的てき一般いっぱん形式けいしき，需要じゅよう一いち個こ橢圓だえん鍵かぎ。 這個橢圓だえん密みつ鑰可以根據こんきょ 卡爾·雅みやび可か比ひ (Carl Gustav Jakob Jacobi) 使用しよう Θしーた函數かんすう 生成せいせい:

$x^{5}+x=w$

$x={\frac {2}{5}}y^{-1/4}{\sqrt[{4}]{10+15y-10y^{2}}}\cosh {\biggl \{}{\frac {1}{5}}{\text{arcosh}}{\biggl [}{\frac {5{\sqrt {5+5y^{2}}}}{(1+2y){\sqrt {4+6y-4y^{2}}}}}{\biggr ]}{\biggr \}}-$

$-{\frac {2}{5}}y^{-1/4}{\sqrt[{4}]{10+15y-10y^{2}}}\sinh {\biggl \{}{\frac {1}{5}}{\text{arsinh}}{\biggl [}{\frac {5y{\sqrt {5+5y^{2}}}}{(2-y){\sqrt {4+6y-4y^{2}}}}}{\biggr ]}{\biggr \}}$

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

現在げんざい在ざい下面かめん精確せいかく地ち解釋かいしゃく這個解決かいけつ過程かてい。 本ほん段上だんじょう式しき的てき等式とうしき刻こく度ど的てき右側みぎがわ取と值 w:

$w={\frac {2}{5}}y^{-5/4}{\frac {(1+y-y^{2}){\sqrt {2+2y^{2}}}}{\sqrt[{4}]{10+15y-10y^{2}}}}$

必須ひっす為ため值 y 求もとめ解かい該方程ほど。 這需要じゅよう一個橢圓模函數表達式，在ざい這種情況じょうきょう下か包括ほうかつ Jacobi theta 函數かんすう:

$y={\frac {5\,\vartheta _{00}{\bigl \{}q{\bigl [}{\bigl (}50{\sqrt {5}}\,w^{2}+32+2{\sqrt {3125w^{4}+256}}{\bigr )}^{-1/2}{\bigl (}{\sqrt {{\sqrt {3125w^{4}+256}}+16}}+5{\sqrt[{4}]{5}}\,w{\bigr )}{\bigr ]}^{5}{\bigr \}}^{2}}{2\,\vartheta _{00}{\bigl \{}q{\bigl [}{\bigl (}50{\sqrt {5}}\,w^{2}+32+2{\sqrt {3125w^{4}+256}}{\bigr )}^{-1/2}{\bigl (}{\sqrt {{\sqrt {3125w^{4}+256}}+16}}+5{\sqrt[{4}]{5}}\,w{\bigr )}{\bigr ]}{\bigr \}}^{2}}}-{\frac {1}{2}}$

此解表ひょう達たち式しき與あずか以下いか表ひょう達たち式しき一致いっち:

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

橢圓だえん函數かんすう的てき定義ていぎ和わ恆等こうとう式しき

現在げんざい必須ひっす定義ていぎ此表達たち式しき中ちゅう指定してい的てき函數かんすう。 所ところ示しめせ的てき主要しゅよう theta 函數かんすう具有ぐゆう以下いか總和そうわ定義ていぎ和わ以下いか等とう效こう乘じょう積せき定義ていぎ:

$\vartheta _{00}(z)=1+2\sum _{k=1}^{\infty }z^{k^{2}}$

$\vartheta _{00}(z)=\prod _{k=1}^{\infty }(1-z^{2k})(1+z^{2k-1})^{2}$

字母じぼ q 描述了りょう數學すうがく橢圓だえん nome 函數かんすう:

$q(\varepsilon )=\exp[-\pi K({\sqrt {1-\varepsilon ^{2}}})K(\varepsilon )^{-1}]$

內商中ちゅう顯示けんじ的てき字母じぼ K 表示ひょうじ完かん整せい的てき第だい一類いちるい 椭圆积分:

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

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

縮寫しゅくしゃ ctlh 表示ひょうじ函數かんすう 双そう曲きょく双そう扭线餘あまり切きり函数かんすう (Hyperbolic lemniscate cotangent)。 而縮寫しゅくしゃ aclh 表示ひょうじ函數かんすう 双そう曲きょく双そう扭线 面積めんせき餘弦よげん函数かんすう (Hyperbolic lemniscate Areacosine)。 這些函數かんすう與あずか 卡爾·弗どる里さと德とく里さと希まれ·高だか斯(Carl Friedrich Gauss) 建立こんりゅう的てき 双そう扭线函数かんすう (Lemniscate elliptic functions) sl 和わ cl 在ざい代數だいすう上じょう相關そうかん，並なみ且可以使用しよう這兩個りゃんこ函數かんすう來らい定義ていぎ：

$\mathrm {sl} (\varphi )=\tan {\biggl \langle }2\arctan {\biggl \{}{\frac {4}{G}}\sin {\bigl (}{\frac {\varphi }{G}}{\bigr )}\sum _{k=1}^{\infty }{\frac {\cosh[(2k-1)\pi ]}{\cosh[(2k-1)\pi ]^{2}-\cos(\varphi /G)^{2}}}{\biggr \}}{\biggr \rangle }$

$\mathrm {cl} (\varphi )=\tan {\biggl \langle }2\arctan {\biggl \{}{\frac {4}{G}}\cos {\bigl (}{\frac {\varphi }{G}}{\bigr )}\sum _{k=1}^{\infty }{\frac {\cosh[(2k-1)\pi ]}{\cosh[(2k-1)\pi ]^{2}-\sin(\varphi /G)^{2}}}{\biggr \}}{\biggr \rangle }$

$[{\text{sl}}(\varphi )^{2}+1][{\text{cl}}(\varphi )^{2}+1]=2$

${\text{ctlh}}(\varrho )=\operatorname {cl} ({\tfrac {1}{2}}{\sqrt {2}}\varrho ){\biggl [}{\frac {\operatorname {sl} ({\tfrac {1}{2}}{\sqrt {2}}\varrho )^{2}+1}{\operatorname {sl} ({\tfrac {1}{2}}{\sqrt {2}}\varrho )^{2}+\operatorname {cl} ({\tfrac {1}{2}}{\sqrt {2}}\varrho )^{2}}}{\biggr ]}^{1/2}$

${\text{ctlh}}(\varrho )={\frac {{\text{cd}}(\varrho ;{\tfrac {1}{2}}{\sqrt {2}})}{\sqrt[{4}]{{\text{cd}}(\varrho ;{\tfrac {1}{2}}{\sqrt {2}})^{4}+{\text{sn}}(\varrho ;{\tfrac {1}{2}}{\sqrt {2}})^{4}}}}$

$\mathrm {aclh} (s)={\tfrac {1}{2}}F[2\operatorname {arccot}(s);{\tfrac {1}{2}}{\sqrt {2}}]$

${\text{aclh}}(s)={\frac {1}{2}}{\sqrt {2}}\,\pi \,G-\int _{0}^{1}{\frac {s}{\sqrt {s^{4}t^{4}+1}}}\,\mathrm {d} t$

$G={\tfrac {1}{2}}{\sqrt {2\pi }}\,\Gamma ({\tfrac {3}{4}})^{-2}$

${\text{ctlh}}{\bigl [}{\tfrac {1}{2}}\mathrm {aclh} (s){\bigr ]}^{2}=(2s^{2}+2+2{\sqrt {s^{4}+1}})^{-1/2}({\sqrt {{\sqrt {s^{4}+1}}+1}}+s)$

${\text{sl}}{\bigl [}{\tfrac {1}{2}}{\sqrt {2}}\,\mathrm {aclh} (s){\bigr ]}={\sqrt {{\sqrt {s^{4}+1}}-s^{2}}}$

字母じぼG代表だいひょう高こう斯常數すう，可か以用伽とぎ馬ば函數かんすう用よう剛つよし才ざい所しょ示しめせ的てき方式ほうしき表示ひょうじ。

连分数すう

连分数すう是ぜ拉ひしげ马努金きん (Rogers-Ramanujan continued fraction) 允許いんきょ以 Bring-Jerrard 形式けいしき對たい廣義こうぎ五次方程進行非常緊湊的解。 這個連分數れんぶんすう函數かんすう和わ交替こうたい連分數れんぶんすう可か以定義ていぎ如下：

$R(z)=z^{1/5}{\frac {(z;z^{5})_{\infty }(z^{4};z^{5})_{\infty }}{(z^{2};z^{5})_{\infty }(z^{3};z^{5})_{\infty }}}$

$R(z)=\tan {\biggl \{}{\frac {1}{2}}\arctan {\biggl [}{\frac {\vartheta _{00}(z^{1/2})^{2}}{2\vartheta _{00}(z^{5/2})^{2}}}-{\frac {1}{2}}{\biggr ]}{\biggr \}}^{2/5}\tan {\biggl \{}{\frac {1}{2}}\operatorname {arccot} {\biggl [}{\frac {\vartheta _{00}(z^{1/2})^{2}}{2\vartheta _{00}(z^{5/2})^{2}}}-{\frac {1}{2}}{\biggr ]}{\biggr \}}^{1/5}$

$R(z^{2})=\tan {\biggl \{}{\frac {1}{2}}\arctan {\biggl [}{\frac {\vartheta _{00}(z)^{2}}{2\vartheta _{00}(z^{5})^{2}}}-{\frac {1}{2}}{\biggr ]}{\biggr \}}^{2/5}\tan {\biggl \{}{\frac {1}{2}}\operatorname {arccot} {\biggl [}{\frac {\vartheta _{00}(z)^{2}}{2\vartheta _{00}(z^{5})^{2}}}-{\frac {1}{2}}{\biggr ]}{\biggr \}}^{1/5}$

$S(z)=\tan {\biggl \{}{\frac {1}{2}}\arctan {\biggl [}{\frac {\vartheta _{00}(z)^{2}}{2\vartheta _{00}(z^{5})^{2}}}-{\frac {1}{2}}{\biggr ]}{\biggr \}}^{1/5}\cot {\biggl \{}{\frac {1}{2}}\operatorname {arccot} {\biggl [}{\frac {\vartheta _{00}(z)^{2}}{2\vartheta _{00}(z^{5})^{2}}}-{\frac {1}{2}}{\biggr ]}{\biggr \}}^{2/5}$

$S(z)={\frac {R(z^{4})}{R(z^{2})R(z)}}$

括くく號ごう，每まい個こ都みやこ有ゆう兩りょう個條かじょう目め，形成けいせい所謂いわゆる的てき Pochhammer-符號ふごう (Pochhammer symbol) 並なみ因よし此代このしろ表ひょう產品さんぴん系列けいれつ。 基もと於這些定義ていぎ，可か以為實際じっさい解かい建立こんりゅう以下いか壓縮あっしゅく精確せいかく解かい公式こうしき:

$x^{5}+x=w$

$x={\frac {S{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {5}{4}}{\sqrt[{4}]{5}}\,w)]^{2}\}{\bigr \rangle }^{2}-R{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {5}{4}}{\sqrt[{4}]{5}}\,w)]^{2}\}^{2}{\bigr \rangle }}{S{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {5}{4}}{\sqrt[{4}]{5}}\,w)]^{2}\}{\bigr \rangle }^{2}}}\times$

$\times {\frac {1-R{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {5}{4}}{\sqrt[{4}]{5}}\,w)]^{2}\}^{2}{\bigr \rangle }\,S{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {5}{4}}{\sqrt[{4}]{5}}\,w)]^{2}\}{\bigr \rangle }}{R{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {5}{4}}{\sqrt[{4}]{5}}\,w)]^{2}\}^{2}{\bigr \rangle }^{2}}}\times$

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

準じゅん確かく的てき例れい子こ

分配ぶんぱい给非初等しょとう数学すうがく实解的てき第だい一いち个自然しぜん数すう w 是ぜ数字すうじ w = 3:

$x^{5}+x=3$

$x={\frac {S{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {15}{4}}{\sqrt[{4}]{5}})]^{2}\}{\bigr \rangle }^{2}-R{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {15}{4}}{\sqrt[{4}]{5}})]^{2}\}^{2}{\bigr \rangle }}{S{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {15}{4}}{\sqrt[{4}]{5}})]^{2}\}{\bigr \rangle }^{2}}}\times$

$\times {\frac {1-R{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {15}{4}}{\sqrt[{4}]{5}})]^{2}\}^{2}{\bigr \rangle }\,S{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {15}{4}}{\sqrt[{4}]{5}})]^{2}\}{\bigr \rangle }}{R{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {15}{4}}{\sqrt[{4}]{5}})]^{2}\}^{2}{\bigr \rangle }^{2}}}\times$

$\times {\frac {\vartheta _{00}{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {15}{4}}{\sqrt[{4}]{5}})]^{2}\}^{5}{\bigr \rangle }\,\vartheta _{00}{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {15}{4}}{\sqrt[{4}]{5}})]^{2}\}^{1/5}{\bigr \rangle }^{2}-5\,\vartheta _{00}{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {15}{4}}{\sqrt[{4}]{5}})]^{2}\}^{5}{\bigr \rangle }^{3}}{2{\sqrt[{4}]{20}}\,{\text{sl}}[{\tfrac {1}{2}}{\sqrt {2}}\,{\text{aclh}}({\tfrac {15}{4}}{\sqrt[{4}]{5}})]\,\vartheta _{00}{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {15}{4}}{\sqrt[{4}]{5}})]^{2}\}{\bigr \rangle }^{3}}}$

$q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {15}{4}}{\sqrt[{4}]{5}})]^{2}\}\approx 0.452374059450344348576600264284387826377845763909$

$x\approx 1.132997565885065266721141634288532379816526027727$

與あずか此類似るいじ，數字すうじ w = 7 僅分配給はいきゅう非ひ基本きほん解かい:

$x^{5}+x=7$

$x={\frac {S{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {35}{4}}{\sqrt[{4}]{5}})]^{2}\}{\bigr \rangle }^{2}-R{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {35}{4}}{\sqrt[{4}]{5}})]^{2}\}^{2}{\bigr \rangle }}{S{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {35}{4}}{\sqrt[{4}]{5}})]^{2}\}{\bigr \rangle }^{2}}}\times$

$\times {\frac {1-R{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {35}{4}}{\sqrt[{4}]{5}})]^{2}\}^{2}{\bigr \rangle }\,S{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {35}{4}}{\sqrt[{4}]{5}})]^{2}\}{\bigr \rangle }}{R{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {35}{4}}{\sqrt[{4}]{5}})]^{2}\}^{2}{\bigr \rangle }^{2}}}\times$

$\times {\frac {\vartheta _{00}{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {35}{4}}{\sqrt[{4}]{5}})]^{2}\}^{5}{\bigr \rangle }\,\vartheta _{00}{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {35}{4}}{\sqrt[{4}]{5}})]^{2}\}^{1/5}{\bigr \rangle }^{2}-5\,\vartheta _{00}{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {35}{4}}{\sqrt[{4}]{5}})]^{2}\}^{5}{\bigr \rangle }^{3}}{2{\sqrt[{4}]{20}}\,{\text{sl}}[{\tfrac {1}{2}}{\sqrt {2}}\,{\text{aclh}}({\tfrac {35}{4}}{\sqrt[{4}]{5}})]\,\vartheta _{00}{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {35}{4}}{\sqrt[{4}]{5}})]^{2}\}{\bigr \rangle }^{3}}}$

$q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {35}{4}}{\sqrt[{4}]{5}})]^{2}\}\approx 0.53609630892200161460073096549143569900990236$

$x\approx 1.4108138510595771319852918753499397839215989$

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