> To: maxima at math.utexas.edu
> From: toy.raymond at gmail.com
> Date: Wed, 9 Jun 2010 17:01:13 -0400
> Subject: Re: [Maxima] bfloat(sqrt(%i))?
>
> On 6/9/10 10:02 AM, Richard Fateman wrote:
> > Raymond Toy wrote:
> >
> > Going back to the original question, if we cannot choose the "positive"
> > s, can we choose a "principal" root? Sometimes, but you
> > do not have any assurance that this is the one of interest.
>
> I think principal root is nice. If it's not the root of interest, you
> should do something else, like solve(x^2=%i,x).
>
> >
> > Here is another reason, or at least food for thought.
> > Mathematica 7.0 returns (-1)^(1/4).
> >
> > evaluating it numerically however gives 0.707107 + 0.707107 I.
>
> What exactly does this mean? If (-1)^(1/4) is meant to be any (or all)
> of the 4 roots of -1, then it seems that 0.7+0.7I is the wrong answer.
> But if it is the right answer, then (-1)^(1/4) really means the
> principal root.
>
> Unless numerical evaluation is meant to produce the principal root if
> there is ambiguity.
>
> How is this meant to interact with say, solve(x^2+3*x+1,x) which returns
>
> [x = -(sqrt(5)-3)/2, x = (sqrt(5)+3)/2].
>
> Does that mean sqrt(5) is +/- 2.23 so there are 4 possible roots to this
> quadratic? (I didn't actually check that this produces 4 values).
>
>
> Ray
>
No Ray you missed that It is already +- sqrt(5)
x= (3 -sqrt(5))/2 and x=(3+sqrt(5))/2DougHTH