factoring constants over algebraic number fields that are not
particularly evident
is a problem I didn't know how to solve in 1969. I'm not sure anyone knows
how to do this properly in general; the sqrt denesting is a special
case, and
maybe someone who has studied the papers by zippel and susan landau can
enhance radcan. Maybe Bill Gosper could say how to do it.
RJF
On 4/7/13 1:37 PM, Stavros Macrakis wrote:
> 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
> <mailto: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
>
>