On Jan 24, 2008 10:03 AM, Carl Witty <[EMAIL PROTECTED]> wrote: > > On Jan 23, 11:41 pm, Paul Zimmermann <[EMAIL PROTECTED]> wrote: > > Thus you have constructed a nice expression for 1: > > > > sage: sol[2].subs(a=1).right() > > (2/(3*sqrt(3)) + 10/27)^(1/3) - 2/(9*(2/(3*sqrt(3)) + 10/27)^(1/3)) + 1/3 > > > > Quiz: how to simplify that expression to 1 within SAGE? I've tried simplify, > > and radical_simplify, but neither succeeds... > > The Sage rings AA and QQbar can decide equalities between radical > expressions (over the reals and complex numbers respectively): > > sage: a = AA((2/(3*sqrt(3)) + 10/27)^(1/3) - 2/(9*(2/(3*sqrt(3)) + > 10/27)^(1/3)) + 1/3) > sage: a > [0.99999999999999988 .. 1.0000000000000003] > sage: a == 1 > True > > By the way, when I implemented AA/QQbar, I decided on the following > behavior for exponentiation: > > sage: AA(-1)^(1/3) > -1 > sage: QQbar(-1)^(1/3) > [0.49999999999999994 .. 0.50000000000000012] + [0.86602540378443859 .. > 0.86602540378443871]*I > > So when taking roots, for AA we prefer real roots if they exist, but > for QQbar we take the principal root.
Wow! That's all very impressive actually. It would be great if you could add the above examples to the tutorials for AA and QQbar (i.e., the tops of the relevant code files). Nice. -- William --~--~---------~--~----~------------~-------~--~----~ To post to this group, send email to sage-support@googlegroups.com To unsubscribe from this group, send email to [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/sage-support URLs: http://www.sagemath.org -~----------~----~----~----~------~----~------~--~---
