Re: [Maxima-discuss] is(somefunction((a*b)^k=a^k*b^k)) -> true

[Eduardo O. <edu...@gm...>]

Hi all,
thanks a lot - radcan is exactly what I needed! =)
In my case neither is(a=b) nor is(equal(a,b)) would work.
Look:

o1 : a^k*b^k = (a*b)^k;       /* a^k*b^k = (a*b)^k */
o2 : radcan(o1);              /* a^k*b^k = a^k*b^k */
is(o1);                       /* false */
is(lhs(o1) = rhs(o1));        /* false */
is(equal(lhs(o1), rhs(o1)));  /* unknown */
is(radcan(o1));               /* true */
is(o2);                       /* true */

Cheers =),
  Eduardo Ochs

On Sat, 28 Oct 2023 at 22:13, Robert Dodier <rob...@gm...> wrote:

> On Sat, Oct 28, 2023 at 12:36 PM Eduardo Ochs <edu...@gm...>
> wrote:
>
> >   o1 : a^k*b^k = (a*b)^k;
> >   o2 : a^k*b^k + a^k*c^k = a^k*(b^k+c^k);
> >   is(o1);             /* false */
> >   is(o2);             /* false */
> >   is(ratsimp(o1));    /* false */
> >   is(ratsimp(o2));    /* true */
> >   is(ratexpand(o1));  /* false */
> >   is(ratexpand(o2));  /* true */
>
> Bear in mind that is(foo = bar) only looks at whether foo and bar are
> just copies of the same thing; that is, whether foo and bar are
> identical, and not whether foo and bar are equivalent. On the other
> hand, is(equal(foo, bar)) looks at whether foo and bar are equivalent,
> specifically in the sense that ratsimp(foo - bar) is 0.
>
> E.g. is(x*(x + 1) = x^2 + x) yields false, while is(equal(x*(x + 1),
> x^2 + x)) yields true.
>
> For what it's worth,
>
> Robert
>