Re: What does power operator ^ returns?

[Guido Walter Pettinari]

Thank you for all the answers! I will stick to the sign-abs trick for  
the time being, even though the 'root' solution Doug pointed out is  
really cool :)

@Jaroslav
Thank you for explaining this strange behaviour. I tried to use the  
nthroot function instead of the hat operator but, as you pointed out,  
the calculations were much slower.

Cheers,

Guido

On Oct 19, 2009, at 19:17 , <address@hidden> wrote:

> > From: address@hidden
> > Subject: What does power operator ^ returns?
> > Date: Mon, 19 Oct 2009 16:42:06 +0000
> > To: address@hidden
> >
> > Hi all!
> >
> > I need to compute the cubic root of many negative numbers. My  
> problem
> > is real-valued, hence I am not interested in the imaginary  
> solutions.
> > However, if I use the power operator '^', Octave returns just the
> > first imaginary solution. The same applies when doing any odd root.
> >
> > Example:
> > octave> a=-8
> > a = -8
> > octave> a^(1/3)
> > ans = 1.000000000000000e+00 + 1.732050807568877e+00i
> >
> > How can I make octave to return the real solution, which in my  
> example
> > is just -2? I do not want just the absolute value (i.e. abs(a^ 
> (1/3)),
> > since it does not preserve the sign. I could do with a check on the
> > sign, but it would be inefficient.
> >
> > I am using Octave 3.2.3 on Mac Os X 10.6.
> >
> > Thank you very much,
> >
> > Guido
> > _______________________________________________
> > Help-octave mailing list
> > address@hidden
> > https://www-old.cae.wisc.edu/mailman/listinfo/help-octave
>
>
>
> if you want to see all the roots try
>
> roots([1 0 0 8])
>
> that will give you all the cube roots of -8
>
> and roots([1 0 0 0 0 8])
> will give you all 5,   fifth roots of -8
> :-)
>
> Doug