Hello! On Wed, Jun 11, 2008 at 04:32:22PM +0200, Timo Schlüßler wrote:
is there a way to compute a sqrt() algebraic? E.g. that sqrt(2) * sqrt(2) gives exact 2 and not 2.0000001 or whatever else.
sqrt(2) is a bit ambigous. I.e. there are GiNaC::sqrt(const GiNaC::numeric&), GiNaC::sqrt(const GiNaC::ex&), and also std::sqrt(double) std::sqrt(float). So, you need to specify the type of the argument expicitly: ex e = sqrt(ex(2)); cout << e << endl; // prints sqrt(2) e = e*e; cout << e << endl; // prints 2 ex a = sqrt(ex(45))*sqrt(ex(3)); cout << a << endl; // prints sqrt(3)*sqrt(45)
Or that sqrt(45) * sqrt(3) gives sqrt(135) and not the rounded value.
You'll need to do the transformation yourself. It's not quite clear what is the canonical form of the expression sqrt(45)*sqrt(3) (and in general, any non-rational expression), so GiNaC won't bother to rewrite it. Best regards, Alexei -- All science is either physics or stamp collecting.