求解一个三次方程 [英] Solving a cubic equation

问题描述

作为一个程序我写的一部分，我需要解决一个三次方程的准确位置（而不是使用数值求根）：

As part of a program I'm writing, I need to solve a cubic equation exactly (rather than using a numerical root finder):

a*x**3 + b*x**2 + c*x + d = 0.

我想这里使用来自方程。然而，考虑以下code（这是Python的，但它是pretty的通用code）：

I'm trying to use the equations from here. However, consider the following code (this is Python but it's pretty generic code):

a =  1.0
b =  0.0
c =  0.2 - 1.0
d = -0.7 * 0.2

q = (3*a*c - b**2) / (9 * a**2)
r = (9*a*b*c - 27*a**2*d - 2*b**3) / (54*a**3)

print "q = ",q
print "r = ",r

delta = q**3 + r**2

print "delta = ",delta

# here delta is less than zero so we use the second set of equations from the article:

rho = (-q**3)**0.5

# For x1 the imaginary part is unimportant since it cancels out
s_real = rho**(1./3.)
t_real = rho**(1./3.)

print "s [real] = ",s_real
print "t [real] = ",t_real

x1 = s_real + t_real - b / (3. * a)

print "x1 = ", x1

print "should be zero: ",a*x1**3+b*x1**2+c*x1+d

但输出是：

q =  -0.266666666667
r =  0.07
delta =  -0.014062962963
s [real] =  0.516397779494
t [real] =  0.516397779494
x1 =  1.03279555899
should be zero:  0.135412149064

所以输出不为零，所以x1为不实际的解决方案。有维基百科的文章中的错误？

so the output is not zero, and so x1 isn't actually a solution. Is there a mistake in the Wikipedia article?

PS：我知道numpy.roots会解决这类方程，但我必须这样做以百万计的方程，所以我需要实现这个工作对系数数组

ps: I know that numpy.roots will solve this kind of equation but I need to do this for millions of equations and so I need to implement this to work on arrays of coefficients.

推荐答案

维基百科的形式（RHO ^（1/3），THETA / 3）不等于 RHO ^（1/3）的实部和 THETA / 3 是虚部。相反，这是极坐标。因此，如果你想真正的一部分，你会采取 RHO ^（1/3）* COS（THETA / 3）。

Wikipedia's notation (rho^(1/3), theta/3) does not mean that rho^(1/3) is the real part and theta/3 is the imaginary part. Rather, this is in polar coordinates. Thus, if you want the real part, you would take rho^(1/3) * cos(theta/3).

我做了这些改变到code和它的工作对我来说：

I made these changes to your code and it worked for me:

theta = arccos(r/rho)
s_real = rho**(1./3.) * cos( theta/3)
t_real = rho**(1./3.) * cos(-theta/3)

（当然， s_real = t_real 在这里，因为 COS 是偶数。）

(Of course, s_real = t_real here because cos is even.)

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