On Dec 18, 8:26=A0pm, jzakiya <jzak... / mail.com> wrote:
> On Dec 18, 8:02=A0pm, jzakiya <jzak... / mail.com> wrote:
>
>
>
> > On Dec 18, 5:24=A0pm, jzakiya <jzak... / mail.com> wrote:
>
> > > On Dec 18, 2:13=A0pm, "William James" <> wrote:
>
> > > > Fleck Jean-Julien wrote:
> > > > > > That is WRONG, you cannot do that.
>
> > > > > Well, I never said that you should do that, I just explained how =
Ruby
> > > > > interpreted it...
>
> > > > > > That only works for odd roots of negative numbers.
> > > > > > The even root of negative numbers are imaginary.
>
> > > > > > -27**3**-1 =3D> -3 **correct
> > > > > > -27**2**-1 =3D> -5.19615242270663 **WRONG, its 5.196152i
>
> > > > > Sure. That's quite a hint why ** does not accept a negative numbe=
r
> > > > > with a non integer exponent. To take into account all the special
> > > > > cases, you should first see if your exponent is a rational and in=
 that
> > > > > case, see if the denominator is odd (after all due simplification=
s of
> > > > > course). In this case (and only this case), you could try to deci=
pher
> > > > > a root for this negative number.
>
> > > > > Cheers,
>
> > > > def root base, n
> > > > =A0 exp =3D 1.0/n
> > > > =A0 return base ** exp if base >=3D 0 or n.even?
> > > > =A0 -( base.abs ** exp )
> > > > end
>
> > > > --
>
> > > Remember
>
> > > i =3D (-1)^(1/2)
>
> > > i^1 =3D =A0i
> > > i^2 =3D -1
> > > i^3 =3D -i
> > > i^4 =3D =A01
> > > Then it repeats, for example: i^5 =3D i*(i^4) =3D i
>
> > > For negative real value roots:
>
> > > x =3D (-a)^(1/n) where n is odd integer =3D> x =3D -[a^(1/n)]
>
> > > But for negative real value roots where n is even:
>
> > > x =3D (-a)^(1/n) where n is even gives
>
> > > x =3D |a^(1/n)|*(-1)^(1/n)
> > > x =3D |a^(1/n)|*(i^2)^(1/n)
> > > x =3D |a^(1/n)|*(i)^(2/n)
> > > from e^(i*x) =3D cos(x) + i*sin(x) where x =3D PI/2
> > > x =3D |a^{1/n)|*e^(PI*i/2)^(2/n)
> > > x =3D |a^(1/n)|*e^(PI*i/n)
> > > x =3D |a^(1/n)|*(cos(PI/n) + i*sin(PI/n)) for n even
>
> > > (-256)^(1/2) =3D |256^(1/2)|*(cos(PI/2) + i*sin(PI/2))
> > > =A0 =A0 =A0 =A0 =A0 =A0 =A0=3D (16)(0 + i) =3D 16i
>
> > > (-256)^(/4) =3D |256^(1/4)|*(cos(PI/4) + i*sin(PI/4))
> > > =A0 =A0 =A0 =A0 =A0 =A0 =3D (4)*(0.707 + 0.707*i)
> > > =A0 =A0 =A0 =A0 =A0 =A0 =3D 2.828 + i*2.828
> > > =A0 =A0 =A0 =A0 =A0 =A0 =3D 2.828*(1+i)
>
> > > Check in irb
>
> > > > require 'complex'
> > > > include Math
>
> > > x =3D Complex(-256,0)
>
> > > x**(1/2.0)
> > > =3D> (9.79685083057902e-16+16.0i)
>
> > > X**(1/4.0)
> > > =3D> (2.82842712474619+2.82842712474619i)
>
> > BTW there is an error (sort of) in 'complex' too
>
> > >require 'complex'
> > >include Math
> > > x =3D Complex(-27,0)
>
> > =3D> (-27+0i)
>
> > >y =3D x**(1/3.0) # or x**3**-1
>
> > =3D> (1.5+2.59807621135332i) =A0# should be (-3+0i)
>
> > >y**3
>
> > =3D> (-27.0+1.24344978758018e-14i)
>
> > > Complex(-3,0)**3
>
> > =3D> -27
>
> > Whenever you take the root n of a number you actually
> > get n values. If the value is positive you get n copies
> > of the same positive real value.
>
> > When you take the root of a negative real value you
> > get n roots too, for n even and odd.
>
> > For even odd, you get one real root and n/2 Complex Conjugate Pairs
> > (CCP).
>
> > Thus, for n=3D3 for (-27)^(1/3) the real root is x1=3D-3
> > and x2 is y above and x3 is the CCP of y.
>
> > For n=3D5, you get one real root and 2 pairs of CCPs, etc.
>
> > For n even, you get n/2 CCPs only.
> > So, for n=3D2 there is one pair of CCP roots.
> > For n=3D4 you get 2 different CCP roots, etc,
> > Thus for n even there are no real roots.
>
> > So, I think it's more intuitive (for most people)
> > to expect Complex(-27,0)**(1/n-odd) to return the real
> > root x1 only (i.e. (-3)*(-3)*(-3) =3D -27), so have it
> > act as Complex(-27,0).real (for n odd) be the default.
>
> > I guess complex variables aren't called complex for nothing. :-)
>
> Oooh, I just noticed:
> Complex(-27**(1/3.0),0) =3D> (-3.0+0i).
>
> So that acts I expected/want.
>
> So maybe it would be nice to be able to do:
>
> Complex(-a**(1/n),0).roots or
> Complex(a**(1/n)).roots where a is already complex
> and have this return all the roots in an array so you
> can see/pick which one(s) you want.

So if you have these methods:

Complex(a).root(n).roots # all the roots of a
Complex(a).root(n).real  # the real root(s) if exists

Then for any value a (real or complex) you will get
what you want under the different conditions.