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