三次方程求根公式的三角函数形式

hbghlyj

TrigSolve=((q Sqrt[-4/(3q)]Cos[1/3 ArcCos[-3r/(2q)Sqrt[3/-q]]+2Pi/3{0,1,2}]-b/(3a))/.{q->-(b^2/(3 a^2))+c/a,r->-(b^3/(27 a^3))+b^3/(9 a)-(b c)/3+d/a})/.Thread[{d,c,b,a}->CoefficientList[#1,#2]]&;

示例:
g[x_] = x^3 - 3 x + 1; TrigSolve[g[x], x]
输出\[\left\{-2 \cos \left(\frac{\pi }{9}\right),2 \cos \left(\frac{2 \pi }{9}\right),2 \sin \left(\frac{\pi }{18}\right)\right\}\]验证:
Simplify[g/@%]
输出
{0,0,0}

hbghlyj

这个东西应该只适用于方程有三个实根的情况.
比如
g[x_] = 2 - 9 x - 4 x^2 + 12 x^3; Chop[N[g /@ TrigSolve[g[x], x]]]
输出
{150.062, 150.062, 150.062}
都不是0,说明公式失效
为何三个数值相等呢?

hbghlyj

改用维基教科书上的公式写成代码如下:
TrigSolve =
Module[{L, Q, R, \[CapitalDelta], \[Theta]},
   L = CoefficientList[#1, #2]; {c, b, a} = Delete[L/L[[4]], 4];
   Q = (3 b - a^2)/9;
   R = (9 a b - 27 c - 2 a^3)/54; \[CapitalDelta] =
    Q^3 + R^2; \[Theta] = ArcCos[R/Sqrt[-Q^3]];
   2 Sqrt[-Q] Cos[\[Theta]/3 + 2 Pi/3 {0, 1, 2}] - a/3] &;
再测试2#的例子,这回都是0,说明公式在方程有两个虚根时仍然适用.

hbghlyj

发现ComplexExpand可以把根式化成三角函数形式:
In:=
       ComplexExpand[ToRadicals[Root[#^3 - 3 # + 1 &, 1]]]
Out:=
       $-2 \cos \left(\frac{\pi }{9}\right)$
但是有的时候需要手动用辅助角继续化简:
In:=
       ComplexExpand[ToRadicals[Root[-1+21 #1-35 #1^2+7 #1^3&,1]]]
Out:=
       $\frac{5}{3}-\frac{4}{3}\cos \left(\frac{1}{3} \tan ^{-1}\left(\frac{3 \sqrt{3}}{13}\right)\right)-\frac{4 \sin \left(\frac{1}{3} \tan ^{-1}\left(\frac{3 \sqrt{3}}{13}\right)\right)}{\sqrt{3}}$
In:=
       %/. (A_ : 1) Cos[t_] + (B_ : 1) Sin[t_] :> Sqrt[A^2 + B^2] Cos[t - ArcTan[A, B]]
Out:=
       $\frac{5}{3}-\frac{8}{3} \sin \left(\frac{\pi }{6}+\frac{1}{3} \tan ^{-1}\left(\frac{3 \sqrt{3}}{13}\right)\right)$
更有甚者,需要先用TrigReduce,再手动用辅助角化简:
In:=
       ComplexExpand[ToRadicals[Root[#^3-6 #+1&,1]]]
Out:=
       $-\frac{\text{Cos}\left[\frac{1}{3} \left(\pi
   -\text{ArcTan}\left[\sqrt{31}\right]\right)\right]}{\sqrt{2}}-\frac{\text{Cos}\left[\frac{1}{3} \left(-\pi
   +\text{ArcTan}\left[\sqrt{31}\right]\right)\right]}{\sqrt{2}}-\sqrt{\frac{3}{2}} \text{Sin}\left[\frac{1}{3}
   \left(\pi -\text{ArcTan}\left[\sqrt{31}\right]\right)\right]+\sqrt{\frac{3}{2}} \text{Sin}\left[\frac{1}{3}
   \left(-\pi +\text{ArcTan}\left[\sqrt{31}\right]\right)\right]$
       $+i \left(\sqrt{\frac{3}{2}} \text{Cos}\left[\frac{1}{3}
   \left(\pi -\text{ArcTan}\left[\sqrt{31}\right]\right)\right]-\sqrt{\frac{3}{2}} \text{Cos}\left[\frac{1}{3}
   \left(-\pi +\text{ArcTan}\left[\sqrt{31}\right]\right)\right]-\frac{\text{Sin}\left[\frac{1}{3} \left(\pi
   -\text{ArcTan}\left[\sqrt{31}\right]\right)\right]}{\sqrt{2}}-\frac{\text{Sin}\left[\frac{1}{3} \left(-\pi
   +\text{ArcTan}\left[\sqrt{31}\right]\right)\right]}{\sqrt{2}}\right)$
In:=
       % // TrigReduce
Out:=
       $-\sqrt{6} \text{Cos}\left[\frac{\pi
   }{6}+\frac{\text{ArcTan}\left[\sqrt{31}\right]}{3}\right]-\sqrt{2} \text{Sin}\left[\frac{\pi
   }{6}+\frac{\text{ArcTan}\left[\sqrt{31}\right]}{3}\right]$
In:=
       %/. (A_ : 1) Cos[t_] + (B_ : 1) Sin[t_] :> Sqrt[A^2 + B^2] Cos[t - ArcTan[A, B]]
Out:=
       $-2 \sqrt{2}
   \text{Cos}\left[\frac{\text{ArcTan}\left[\sqrt{31}\right]}{3}\right]$

hbghlyj

Trigonometric solution in terms of real quantities
While casus irreducibilis cannot be solved in radicals in terms of real quantities, it can be solved trigonometrically in terms of real quantities.^[7] Specifically, the depressed monic cubic equation $t^{3}+pt+q=0$ is solved by
$$ t_{k}=2{\sqrt {-{\frac {p}{3}}}}\cos \left[{\frac {1}{3}}\arccos \left({\frac {3q}{2p}}{\sqrt {\frac {-3}{p}}}\right)-k{\frac {2\pi }{3}}\right]\quad {\text{for}}\quad k=0,1,2\,.$$
These solutions are in terms of real quantities if and only if ${q^{{2}} \over 4}+{p^{{3}} \over 27}<0$ — i.e., if and only if there are three real roots. The formula involves starting with an angle whose cosine is known, trisecting the angle by multiplying it by 1/3, and taking the cosine of the resulting angle and adjusting for scale.

Although cosine and its inverse function (arccosine) are transcendental functions, this solution is algebraic in the sense that ${\displaystyle \cos \left[\arccos \left(x\right)/3\right]}$ is an algebraic function, equivalent to angle trisection.

   REFERENCE   
[7] Cox, David A. (2012), Galois Theory, Pure and Applied Mathematics (2nd ed.), John Wiley & Sons,
Section 1.3B Trigonometric Solution of the Cubic, pp. 18–19.

hbghlyj

ursoswald.ch/download/CUBIC.pdf
Theorem 10
(i) $w_{1}=\sinh \left(\frac{1}{3} \operatorname{arsinh} C\right)$ if $p>0$,
(ii) $w_{1}=\left\{\begin{array}{ll}\cosh \left(\frac{1}{3} \operatorname{arcosh} C\right) & \text { if } p<0 \text { and } C \geq 1 \\ -\cosh \left(\frac{1}{3} \operatorname{arcosh}(-C)\right) & \text { if } p<0 \text { and } C \leq-1\end{array}\right.$,
(iii) $w_{k}=\cos \left(\frac{1}{3} \arccos C+(k-1) \cdot \frac{2 \pi}{3}\right)$, where $k=1,2,3$ if $p<0$ and $|C| \leq 1$.

青青子衿

hbghlyj 发表于 2022-8-16 01:37
ursoswald.ch/download/CUBIC.pdf

下面的这个结论正确吗？
{AA -> b^2 - 3*a*c,
BB -> b*c - 9*a*d,
CC -> c^2 - 3*b*d};
Delta = BB^2-4*AA*CC;

当Delta>0
三次方程a*t^3+b*t^2+c*t+d=0有一个实根和两个共轭复根

当Delta=0
当A≠B时，
三次方程a*t^3+b*t^2+c*t+d=0有一个单实根和一对二重实根
当A=B=0时，
三次方程a*t^3+b*t^2+c*t+d=0有三重实根
当A=B≠0时，
三次方程a*t^3+b*t^2+c*t+d=0有一个单实根和一对二重实根

当Delta<0
三次方程a*t^3+b*t^2+c*t+d=0有三个相异的单实根