Disclaimer: I do not know the SAGE code here, just general theory. As BFJ pointed out, there is no 'canonical' form for such expressions, where 'canonical' means (see 2.3.1, p.79 of Davenport, Siret & Tournier) that there is a unique representation for every expression. But there may well be a 'normal' form, in the sense that zero only has one representation, so you could try full_simplify(A-B)==0 rather than full_simplify(A)==full_simplify(B).
On Nov 20, 3:57 pm, pevzi <[email protected]> wrote: > Thank you for your reply. > But so is there any way to check if the expressions are equivalent? > > On 20 ноя, 05:05, BFJ <[email protected]> wrote: > > > The two expressions you give may be algebraically equivalent, but > > they're not identical. There is no canonical "fully simplified" form > > for a general algebraic expression, so you can't expect > > full_simplify() to output this non-existant form. If the expressions > > are simple enough, like polynomials, you might, but not in general. > > > On Nov 19, 3:31 pm, pevzi <[email protected]> wrote: > > > > I have two expressions: > > > > (x/(2*sqrt(x+1)) + 1/(2*sqrt(x+1)*(sqrt(x+1)+1))) > > > ((x*(sqrt(x+1)+1)+1)/(2*sqrt(x+1)*(sqrt(x+1)+1))) > > > > As you see, they are identical, so full_simplify() method should > > > return the same result for both expressions. But: > > > > sage: (x/(2*sqrt(x+1)) + 1/(2*sqrt(x+1)*(sqrt(x > > > +1)+1))).full_simplify() > > > 1/2*(x + sqrt(x + 1))/(sqrt(x + 1) + 1) > > > sage: ((x*(sqrt(x+1)+1)+1)/(2*sqrt(x+1)*(sqrt(x > > > +1)+1))).full_simplify() > > > 1/2*(sqrt(x + 1)*x + x + 1)/(x + sqrt(x + 1) + 1) > > > > Although > > > > sage: ((1/2*(x + sqrt(x + 1))/(sqrt(x + 1) + 1))/(1/2*(sqrt(x + 1)*x + > > > x + 1)/(x + sqrt(x + 1) + 1))).full_simplify() > > > 1 > > > > Is this really a bug or I misunderstand something? -- To post to this group, send email to [email protected] To unsubscribe from this group, send email to [email protected] For more options, visit this group at http://groups.google.com/group/sage-support URL: http://www.sagemath.org
