![](//images.weserv.nl/?url=//upload.wikimedia.org/wikipedia/commons/thumb/f/f8/Quintic_function.png/250px-Quintic_function.png)
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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e979e3e7158d50859b2bcfba1e44144f233f2a20)
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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e007690b4449cc2e02ef7b19f4f2c77067a4f831)
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)[1]。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
![{\displaystyle x^{5}+a_{1}x^{4}+a_{2}x^{3}+a_{3}x^{2}+a_{4}x+a_{5}=0\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/338f428ab0553eebb86941c59185d289f2f40799)
可以藉由以下的多项式变换
![{\displaystyle y=x^{4}+b_{1}x^{3}+b_{2}x^{2}+b_{3}x+b_{4}\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dcb7ac587f2dfc6dab2bd18fddc60cfd58260916)
得到一個
的五次多項式,上述的轉換稱為契爾恩豪森轉換(Tschirnhaus transformation),藉由特別選擇的係數
,可以使
,
,
的係數為
,從而得到如下的方程式:
![{\displaystyle y^{5}+my+n=0\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a44cd4c04da764b470c34a47e8cf70c97cd4ea6a)
以上的化簡方法是由厄蘭·塞缪爾·布靈所發現,後來喬治·傑拉德也獨立發現了此法,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}}\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cc36007a1f6a593e4f57cc5028827b487b281f78)
可消去四次方項,得到
;
其中,
![{\displaystyle a={\frac {5a_{2}-2a_{1}^{2}}{5}}\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6df99900639ca182f4e9f41cfaef5e2bdc006ea5)
![{\displaystyle b={\frac {25a_{3}-15a_{1}a_{2}+4a_{1}^{3}}{25}}\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1fe294c9c03f9bae3fc67aee78589c324ac52be7)
![{\displaystyle c={\frac {125a_{4}-50a_{1}a_{3}+15a_{1}^{2}a_{2}-3a_{1}^{4}}{125}}\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4d81e13879209a8d9590244ddd5cf520295a68f4)
![{\displaystyle d={\frac {3125a_{5}-625a_{1}a_{4}+125a_{1}^{2}a_{3}-25a_{1}^{3}a_{2}+4a_{1}^{5}}{3125}}\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5c7c1db7a213ee2ae1d2b6a03c36ce19d06e2b9f)
接下來,令
,
得到
,
再令
,
求得
;
![{\displaystyle q={2a \over 5}.\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/249abe21940537e2ff6d75afcb914fd452af1eda)
第三步,利用契爾恩豪森想到的方法,令:
,
代入
,
得到
,
再令
,
則得
,
若令
,
則
,
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;\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/be7509d7f02529f32d1c9cd7eca2288060fda3e7)
![{\displaystyle 675A^{3}b_{3}^{3}+(3375A^{2}Cb_{1}-3600AB^{2}b_{1}-2025A^{4}\,-4500ABC)b_{3}^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2a7a9cde287fb17f744c20fce6f5725a806d8e89)
![{\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}+\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/312e8372c28b0e0be255e81019adc5a1c9781251)
![{\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}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d2ddd0659284b63041cb90eb80345af8cc9dc084)
![{\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.\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/37008c6918dadcf1dab3b103ac595a76d6a6e658)
若以函數的觀點來看,方程
![{\displaystyle X^{5}+UX+V=0\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1c185a68175907b4f2ddd1f0505ce8b8692d3ef7)
的解有兩個自變數
, 和
。
若再令
![{\displaystyle X={\sqrt[{4}]{-U}}\xi \,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6bd43d9f7585c795d5e8d8cc31c800afc62030a6)
則方程式可以進一步化簡為如下形式:
![{\displaystyle \xi ^{5}-\xi +t=0\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/88abe21fe16b5829dae86037242aa75c18ee4ba0)
它的解
是單一變數
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
,当
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}}}\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/41534601cefc864bd44175223fef11fe8f103504)
![{\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}}}}}}\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/edb76f2d579c3751bd87df28fc26f2a5a22f29e1)
![{\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}}}}}}\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/439bb34b4fbd0a517aef6bd0fa12e9f3f8b10c5e)
![{\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}}}}}}\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1e750282109340dc1847e6aa55068da0d14fe1ce)
![{\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}}}}}}\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/efeec4551d62e5f387d5cb8caf742aeb6b1c3ff1)
型式2
,当
时,
![{\displaystyle x_{1}=d\left[A+B+C+D\right]\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cb3ceb4d8c8af292245476b032c23686d84cd8e1)
![{\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]\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/969b5c4103a99ca143dfed982f95ae09a62340c0)
![{\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]\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f9dd190318aa0f2861ff774a9318b1e5a2b40d3c)
![{\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]\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/55b805972973f9400647541ecf972f99c3800b43)
![{\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]\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4f31d39b75547e5ec87efd9b09607175722f13b4)
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}}}}\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ff76407b0d088cfb30a4e5119f7c368e43f5951b)
![{\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}}}}\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/70b1739322d635e8e35a48f40c3131dcb5033f82)
![{\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}}}}\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e11a61c3a500a1413df8e25af673b794c6cf039a)
![{\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}}}}\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c2ee86c99a482aeaf16ee35794d9caedb3a13448)
型式3
![{\displaystyle {a^{2}x^{5}+5abx^{3}+5b^{2}x+ac=0}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ceb50ce652fd4de4c3b01e5331a1255c09156ca0)
,当
![{\displaystyle a\neq 0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f455a7f96d74aa94573d8e32da3b240ab0aa294f)
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}}}}}}\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9f0fcd42827a462df25fd7a36a4b2469066a9cb1)
![{\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}}}}}}\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/529b3dcbccfd1ea105640642c8852804b3edf657)
![{\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}}}}}}\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/74d870c45f0ca05e98bdceb93e0a36258920c76f)
![{\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}}}}}}\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3c122abf4e6b34fa0559df179e552b2db930a5ed)
![{\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}}}}}}\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/29657589f119014ed854fbe93e9ad19d2b82ff77)
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}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/87861bb26eaf7d3781068c9a08c6caf9f5d7ff09)
![{\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 \}}-}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9c520010591fbdcf2f5f61bdb7454116b63c2072)
![{\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 \}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/01f503a8d0944a1d12408b892d7c2cc400abab0d)
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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f17c55e95771c38bb4adcc79798c07dccd059de3)
![{\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 \}}-}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9c520010591fbdcf2f5f61bdb7454116b63c2072)
![{\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 \}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/01f503a8d0944a1d12408b892d7c2cc400abab0d)
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7e4d1ae884002e262f9b25bae9607cd82412f402)
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}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7d82652e8e635936157861c2d45cba23453ce663)
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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ac512dbe529ffe8e5d415cb8040aa9f92452d11c)
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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7e4d1ae884002e262f9b25bae9607cd82412f402)
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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/48303127dcb00f886fe38e381650bff156a7c7dd)
![{\displaystyle \vartheta _{00}(z)=\prod _{k=1}^{\infty }(1-z^{2k})(1+z^{2k-1})^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/465159171cd1a4e0b0338ed49f1c025d8adf7ac0)
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}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6c64c06dc297c68502bb06e6c07e36bbccd609e2)
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 }](https://wikimedia.org/api/rest_v1/media/math/render/svg/c3f049dee2ea013d3cfceb8f726fc8ed08698f5c)
![{\displaystyle K(r)=2\int _{0}^{1}{\frac {1}{\sqrt {(u^{2}+1)^{2}-4r^{2}u^{2}}}}\mathrm {d} u}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7a6572f0d8477dfe0ffc8adf30cec7e530d3cdc4)
縮寫 ctlh 表示函數 双曲双扭线餘切函数[3] (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[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 }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1791550a2f621d4e3af70d6d487ed8d762a4a462)
![{\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 }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/814a459471e82cb54c3bc0988069b83be88ce2c4)
![{\displaystyle [{\text{sl}}(\varphi )^{2}+1][{\text{cl}}(\varphi )^{2}+1]=2}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5010c59ea047a03787ec123d3cddf7707106be7c)
![{\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}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/835ac16e842bf820bd2afd699320c4bb8483be82)
![{\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}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e1441608657939547bd8e37ef4a89f3bc9213c89)
![{\displaystyle \mathrm {aclh} (s)={\tfrac {1}{2}}F[2\operatorname {arccot}(s);{\tfrac {1}{2}}{\sqrt {2}}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e497893056711eac6ba37933bdfd418cdcfde8f3)
![{\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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8327b0e6b0bacd9fa599d338d41d870285ca94c2)
![{\displaystyle G={\tfrac {1}{2}}{\sqrt {2\pi }}\,\Gamma ({\tfrac {3}{4}})^{-2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f154593d437465ed468352364b5bf3b4d701736a)
![{\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)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fb0e85a53ae3dee4d6260bf34574bdda2677ab5e)
![{\displaystyle {\text{sl}}{\bigl [}{\tfrac {1}{2}}{\sqrt {2}}\,\mathrm {aclh} (s){\bigr ]}={\sqrt {{\sqrt {s^{4}+1}}-s^{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6417f3c2a1d5e63bededbc392ece42de7e2cf02d)
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
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 }}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/18547c53c229e61f5879f040633acb00a74f63a6)
![{\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}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e05fd6a136f16b01c04ca730bf084d9004b4581b)
![{\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}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fa7ecda83ca3a83d770dca4b783ba22086abd63a)
![{\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}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c0283e6b959ed0567349c3d9a92d9f5b73238462)
![{\displaystyle S(z)={\frac {R(z^{4})}{R(z^{2})R(z)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/75136febb4badd46fbbc350fbf6169e39942d00d)
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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f17c55e95771c38bb4adcc79798c07dccd059de3)
![{\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 }](https://wikimedia.org/api/rest_v1/media/math/render/svg/8c7e6cbec3ecdeb892514072f67a3c67e970d6fb)
![{\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 }](https://wikimedia.org/api/rest_v1/media/math/render/svg/9e2665ac8c95edccf1f8be194a49db90e21553ac)
![{\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}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/be4e06136ee2d75943c483914dcdf36f1489d0e3)
準確的例子
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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f72e41b09d5a1851073a15a92f47000d8947107f)
![{\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 }](https://wikimedia.org/api/rest_v1/media/math/render/svg/f0c89500c3bb1767fb3430d839d84e3ae480af8f)
![{\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 }](https://wikimedia.org/api/rest_v1/media/math/render/svg/781028992714c0946b51d0f4beb4404be08a6957)
![{\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}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8d8c01516b64dd0311e47b51ba3d0d855ef50d55)
![{\displaystyle q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {15}{4}}{\sqrt[{4}]{5}})]^{2}\}\approx 0.452374059450344348576600264284387826377845763909}](https://wikimedia.org/api/rest_v1/media/math/render/svg/177ba0691acd01a59eda4ab442a6dd6b3995110c)
![{\displaystyle x\approx 1.132997565885065266721141634288532379816526027727}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6256c17d05f7d4f1a329eb780bef90e69d356313)
與此類似,數字 w = 7 僅分配給非基本解:
![{\displaystyle x^{5}+x=7}](https://wikimedia.org/api/rest_v1/media/math/render/svg/90b0d7f5a79c052b36d64661628dbb6082f96b12)
![{\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 }](https://wikimedia.org/api/rest_v1/media/math/render/svg/07fff627922c2d8c366dc8825dc5126cecc8a01b)
![{\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 }](https://wikimedia.org/api/rest_v1/media/math/render/svg/3e95f9ebcc86c1af3e2a91f7c05b13d3249dfeba)
![{\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}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7bb028d4ca204cfbcf87201b4b37c676ac96e017)
![{\displaystyle q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {35}{4}}{\sqrt[{4}]{5}})]^{2}\}\approx 0.53609630892200161460073096549143569900990236}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0eddcb1393869a86603d07d9ba98c79ccd0fd20a)
![{\displaystyle x\approx 1.4108138510595771319852918753499397839215989}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0e7223e74e8293feb1d5c8f80b782d727f7a050c)
參見
引文