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