From: serialhex <serialhex@...>
Date: 2011-05-13T03:46:26+09:00
Subject: Re: Math cube root

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On Thu, May 12, 2011 at 2:30 PM, Josh Cheek <josh.cheek@gmail.com> wrote:

> Which we don't here, given that the cube root of -1 isn't imaginary :/


there are actually 3 cube roots of -1 (and 4 quad roots of -1, and 5 penta
roots of -1 and...)
behold the code:

ruby-1.9.2-p180 :004 > a = (-1)**(1/3)
 => (0.5000000000000001+0.8660254037844386i)
ruby-1.9.2-p180 :005 > a.conj
 => (0.5000000000000001-0.8660254037844386i)
ruby-1.9.2-p180 :006 > a.conj**3
 => (-1.0-3.885780586188048e-16i)
ruby-1.9.2-p180 :007 > a**3
 => (-1.0+3.885780586188048e-16i)
ruby-1.9.2-p180 :008 > (-1)**3
 => -1

(conjugation simply flips the sign of the imaginary number, which is helpful
for a bunch of things)  so while there may be 1 _real_ root for a given
number, there are n roots for any number taken to the (1/n)th power.  and
while most people will want the (simple) (-1)**(1/3) == (-1) answer, they
are all right (minus the floating-point rounding errors... which is kind of
odd cause i would think that i would be getting the rational representation
of the complex numbers instead of floats... idfk... whatev!)
hex

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