Just to be clear, none of this is specific to radcan. You'll get the same
results with ratsimp or expand.
-s
On Mon, Feb 6, 2012 at 10:59, zxl <zhaxiaolei at gmail.com> wrote:
> some fix.
>
> 1?if B=5^k C, where 5 not divide C, then B 5^(1/2) will be represented
> as C 5^(k+1/2);
>
> 2?radcan can expand (a+b sqrt(5))^n to A+B 5^m where 5 not divide B.
>
> On Mon, Feb 6, 2012 at 11:42 PM, zxl <zhaxiaolei at gmail.com> wrote:
> > thanks Eric Reyssat and Stavros Macrakis. Now I know that
> >
> > 1?5^(3/2) is somewhat "atom" in maxima;
> >
> > 2?radcan can expand (a+b sqrt(5))^n to A+B (5)^(1/2) or A+B (5)^(3/2).
> >
> > thanks
> >
> > On Sat, Feb 4, 2012 at 11:41 PM, zxl <zhaxiaolei at gmail.com> wrote:
> >>
> >> hello:
> >>
> >> I hope expand (a+b sqrt(5))^n to the form A+B sqrt(5), but expand
> ((1+sqrt(5))) give me 5^(3/2)+ 3 sqrt(5) + 16. How can I convert 5^(3/2) to
> 5 5^(1/2) ?
> >>
> >> thanks
> >>
> >> z.x.l
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