astrofrog - 7 months ago 66

Python Question

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.`

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

But the output is:

`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

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

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.

Answer

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)`

.

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)
```

(Of course, `s_real = t_real`

here because `cos`

is even.)

Source (Stackoverflow)