|
|
Last edited by hbghlyj 2023-11-14 19:33下面给出一个方程 $x^5+px+q=0$ 的机械化解法.
注意系数都是非零有理数 $(p, q)\in \mathbb{Q} \backslash\{0\}$
系数不是有理数也可以有解, 比如 $x^5+10i x+8i=0$ , 但是只能手工分析 $\mathbb{Q}(\sqrt{-1})$ 求解, 暂时无法机械化求解.
这类方程五个真根能由四个伪根线性组合得到[1], 所以只要解伪根即可
\[\begin{bmatrix} x_1\\
x_2\\
x_3\\
x_4\\
x_5 \end{bmatrix}= \begin{bmatrix} 1 & 1 & 1 & 1 \\
e^{+\frac{2 i \pi }{5}} & e^{+\frac{4 i \pi }{5}} & e^{-\frac{4 i \pi}{5}} & e^{-\frac{2 i \pi}{5}} \\
e^{+\frac{4 i \pi }{5}} & e^{-\frac{2 i \pi}{5}} & e^{+\frac{2 i \pi }{5}} & e^{-\frac{4 i \pi}{5}} \\
e^{-\frac{4 i \pi}{5}} & e^{+\frac{2 i \pi }{5}} & e^{-\frac{2 i \pi}{5}} & e^{+\frac{4 i \pi }{5}} \\
e^{-\frac{2 i \pi}{5}} & e^{-\frac{4 i \pi}{5}} & e^{+\frac{4 i \pi }{5}} & e^{+\frac{2 i \pi }{5}} \\
\end{bmatrix} \cdot \begin{bmatrix} \sqrt[5]{\zeta_1}\\
\sqrt[5]{\zeta_2}\\
\sqrt[5]{\zeta_3}\\
\sqrt[5]{\zeta_4}\\
\end{bmatrix}\]
这个系数矩阵也能用根式表达
$\exp \left(\pm\frac{2 \pi i}{5}\right)=+\frac{1}{4} \left(\sqrt{5}-1 \pm i \sqrt{ 2\sqrt{5}+10}\right)$
$\exp \left(\pm\frac{4 \pi i}{5}\right)=-\frac{1}{4} \left(\sqrt{5}+1 \pm i \sqrt{ 2\sqrt{5}+10}\right)$
就是写出来比较长所以用 $e^x$ 简写.
- getRoot[pseudo_] := Table[Sum[Surd[pseudo[[i]], 5] Exp[2Pi I / 5]^(i j), {i, 1, 4}], {j, 0, 4}];
- getPseudoRoot[a_, b_] := Module[
- {v, pseudo, points},
- points = FindInstance[{a d == 5p^4(3 - 4 e q), b d == -4p^5(11e + 2 q), e^2 == 1, d == q^2 + 1, q >= 0, p != 0}, {p, e, q, d}, Rationals];
- If[Length@points == 0, Return[Failure["NoSolution", <||>]]];
- Subscript[v, 1] = + Sqrt[d] + Sqrt[d - e Sqrt[d]];
- Subscript[v, 2] = -Sqrt[d] - Sqrt[d + e Sqrt[d]];
- Subscript[v, 3] = -Sqrt[d] + Sqrt[d + e Sqrt[d]];
- Subscript[v, 4] = + Sqrt[d] - Sqrt[d - e Sqrt[d]];
- pseudo = {
- Subscript[v, 1]^2 * Subscript[v, 3],
- Subscript[v, 3]^2 * Subscript[v, 4],
- Subscript[v, 2]^2 * Subscript[v, 1],
- Subscript[v, 4]^2 * Subscript[v, 2]
- };
- FullSimplify[
- pseudo * p^5 / d^2 /. First@points,
- ComplexityFunction -> (100 Count[#, Root[__], All] + LeafCount[#]&)
- ]
- ];
getPseudoRoot 可以解伪根, 用法是 getPseudoRoot[15, 44], 获取方程 $x^5+15x+44=0$ 的伪根.
getRoot 用于把伪根转为真根, getRoot@getPseudoRoot[15, 44] , 获得五个真根.
你说是解就是解了, 怎么验证呢?
数值的话只能作参考, 毕竟可能有数值误差, 你还是不信.
当然乘回去化简是可以的, 但是这一大坨左一个开五次方右一个开五次方, 搞不好没法化简.
我们可以看看这一堆代数式的最小生成多项式是不是 $x^5+15x+44=0$ 来验证, 这个复杂度比化简简单多了.
roots=getRoot@getPseudoRoot[15,44];
RootApproximant[roots]
SortBy[roots//N,Arg]
SortBy[x/.NSolve[x^5+15x+44==0,x],Arg]
我们来看看一些书上常见的五次方程的伪根.
$x^5+15x+44=0$
Galois 群是 $F_{20}$ 因此可解, 解得
\[\begin{aligned} \zeta_{1}&=1-\sqrt{2}\\
\zeta_{2}&=-3+2 \sqrt{2}\\
\zeta_{3}&=-3-2 \sqrt{2}\\
\zeta_{4}&=1+\sqrt{2}\\
\end{aligned}\]
$x^5 + 11 x + 44=0$
Galois 群是 $D_{5}$ 因此可解, 解得
\[\begin{aligned} \zeta_{1}&=\frac{11}{625} \left(-75+50 \sqrt{5}-\sqrt{18125-6169 \sqrt{5}}\right)\\
\zeta_{2}&=-\frac{11}{625} \left(75+50 \sqrt{5}-\sqrt{18125+6169 \sqrt{5}}\right)\\
\zeta_{3}&=-\frac{11}{625} \left(75+50 \sqrt{5}+\sqrt{18125+6169 \sqrt{5}}\right)\\
\zeta_{4}&=\frac{11}{625} \left(-75+50 \sqrt{5}+\sqrt{18125-6169 \sqrt{5}}\right)\\
\end{aligned}\]
$x^5 + 20 x + 32=0$
Galois 群是 $D_{5}$ 因此可解, 解得
\[\begin{aligned} \zeta_{1}&=-\frac{4 }{5 \sqrt{5}}\left(5+\sqrt{25-2 \sqrt{5}}\right)\\
\zeta_{2}&=-\frac{4 }{5 \sqrt{5}}\left(-5+\sqrt{25+2 \sqrt{5}}\right)\\
\zeta_{3}&=\frac{4 }{5 \sqrt{5}}\left(5+\sqrt{25+2 \sqrt{5}}\right)\\
\zeta_{4}&=\frac{4 }{5 \sqrt{5}}\left(-5+\sqrt{25-2 \sqrt{5}}\right)\\
\end{aligned}\]
$5x^5+6x+6=0$
Galois 群是 $F_{20}$ 因此可解, 解得
\[\begin{aligned} \zeta_{1}&=-\frac{432}{3125}\\
\zeta_{2}&=-\frac{24}{3125}\\
\zeta_{3}&=\frac{324}{3125}\\
\zeta_{4}&=-\frac{18}{3125}\\
\end{aligned}\]
这个方程有点意思哦
$2x^5-5x-4=0$
Galois 群是 $F_{20}$ 因此可解, 解得
\[\begin{aligned} \zeta_{1}&=\frac{1}{500} \left(150-100 \sqrt{2}+\sqrt{42500-29990 \sqrt{2}}\right)\\
\zeta_{2}&=\frac{1}{500} \left(150+100 \sqrt{2}-\sqrt{42500+29990 \sqrt{2}}\right)\\
\zeta_{3}&=\frac{1}{500} \left(150+100 \sqrt{2}+\sqrt{42500+29990 \sqrt{2}}\right)\\
\zeta_{4}&=\frac{1}{500} \left(150-100 \sqrt{2}-\sqrt{42500-29990 \sqrt{2}}\right)\\
\end{aligned}\]
参考
Characterization of Solvable Quintics tandfonline.com/doi/abs/10.1080/00029890.1994 … 8?journalCode=uamm20
Solution of Solvable Irreducible Quintic Equations jstor.org/stable/2369449?seq=1^pixiv.net/artworks/98111089 |
|
