> "Tiago Katcipis" <[EMAIL PROTECTED]> wrote
>
> > def newton_divergente(x):
> > return math.pow(x, 1.0/3.0)
> >
> > but when x = -20 it returns this error
> >
> > return math.pow(x, 1.0/3.0)
> > ValueError: math domain error
> >
> > but why is that? is it impossible to calculate -20 ^ (1/3) ?
Can I chime in late on this one? I'm catching up on some old email.
It's already been pointed out that a negative number cannot be raised to a
fractional power. But as long as you're only looking to calculate roots
you should be able to calculate them unless the number is negative *and*
the root is even. For example, you cannot (without getting into complex
numbers) take the square root (an even root) of -8. But the cube root (an
odd root) of -8 is -2.
N**1/x is equal to -(-N**1/x), as long as the constraints above are
met. Relying on this, you can easily write an "nthroot" routine that
calculates the non-complex nth root of a number, unless the number is
negative and the root-number is even:
def nthroot(x, n):
"""returns the non-complex nth root of x"""
if x > 0:
return x**(1.0/n)
elif n%2 == 1:
return -(-x)**(1.0/n)
else:
raise ValueError, "even root cannot be calculated for negative number"
For example:
>>> nthroot(100, 2)
10.0
>>> nthroot(8, 3)
2.0
>>> nthroot(-8, 3)
-2.0
>>> nthroot(-8, 2) # this is a no-no; negative number & even root
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "nthroots.py", line 8, in nthroot
raise ValueError, "even root cannot be calculated for negative number"
ValueError: even root cannot be calculated for negative number
>>>
Now, if you *are* willing to get into complex numbers, i.e., numbers with
an imaginary component, you should be aware that there's no such thing as
*the* Nth root of a number. Instead there are N Nth roots, evenly spaced
in a circle centered on 0+0i.
For example, the number 16 has two square roots, at 4 and -4; or more
completely 4+0i and -4+0i.
But it has 4 4th roots. Obviously it has 2 and -2 (2+0i and -2+0i); but
it also has 0+2i and 0-2i:
(0+2i)*(0+2i)*(0+2i)*(0+2i) = -4 * -4 = 16
(0-2i)*(0-2i)*(0-2i)*(0-2i) = 4 * 4 = 16
The following (barely-tested) routine should calculate all the Nth roots
of a given x, even when x is negative and N is even:
def nthroots(x, n):
"""returns a list of all nth roots of x"""
from math import pi, cos, sin
roots = []
if x >=1:
startangle = 0
else:
startangle = pi/2
firstroot = abs(x)**(1.0/n)
for i in range(0, n):
angle = (i*(2*pi)/n)+startangle
root = complex(firstroot*cos(angle), firstroot*sin(angle))
roots.append(root)
return roots
Example:
>>> nthroots(16,4)
[(2+0j), (1.2246063538223773e-016+2j), (-2+2.4492127076447545e-016j),
(-3.6738190614671318e-016-2j)]
Apart from precision issues, those are 2+0j, 0+2j, -2+0j and 0-2j.
_______________________________________________
Tutor maillist - Tutor@python.org
http://mail.python.org/mailman/listinfo/tutor
