RJF,
I wonder why radcan doesn't handle simple cases of sqrt denesting, like:
sqrt(3+2*sqrt(2)) => sqrt(2)+1 (handled by sqrtdenest)
sqrt(a+2*sqrt(a)+1) => sqrt(a)+1 (not handled by sqrtdenest)
Should it be extended to cover them, or is there some good reason it should
not?
More generally, are there ways of quasi-factoring expressions with radicals
in some useful way? I realize that in general factorization is not unique
in these cases.
-s
On Sat, Apr 6, 2013 at 9:29 PM, Barton Willis <willisb at unk.edu> wrote:
> > (%05)sqrt(a^2+(2-2*a)*sqrt(a)-1)
>
>>
> > Yes, as I pointed out in my email, that is a difficult case.
>
> Hardly a method, but ...
>
> (%i1) load(to_poly_solver)$
>
> (%i2) ((sqrt(a)+1)*(sqrt(a)-1)^3)^(1/2)-w$
>
> (%i3) subst(a=b^2,%)$
>
> (%i4) to_poly(%,[w,b])$
>
> (%i5) elim_allbut(first(%),[b,w])$
>
> (%i6) first(first(%))$
>
> (%i7) solve(%,w)$
>
> (%i8) subst(b=sqrt(a),%)$
>
> A Taylor expansion* indicates* (doesn't prove) that the last solution is
> correct:
>
> (%i9) facsum(%,sqrt(a-1));
> (%o9)
> [w=-%i*(a-2*sqrt(a)+1),w=%i*(a-2*sqrt(a)+1),w=-%i*(a+2*sqrt(a)+1),w=%i*(a+2*sqrt(a)+1),w=-(sqrt(a)+1)*sqrt(a-1),
> w=(sqrt(a)+1)*sqrt(a-1),w=-(sqrt(a)-1)*sqrt(a-
> 1),w=(sqrt(a)-1)*sqrt(a-1)]
>
> --Barton
>
>