load(sqdnst)$
sqrtdenest( sqrt((1+sqrt(2))^2)-sqrt(1+2*sqrt(2)+2) ) => 0
This does make me wonder why radcan doesn't always use sqrtdenest in
relevant cases. Perhaps at the time they were written (1970's), that would
have made radcan too slow.
By the way, this is a good example of why you should always tell us about
the problem you're *really* trying to solve, not a problem you're having
with one particular approach to solving it.
-s
On Sat, Nov 24, 2012 at 11:45 AM, Ren? K?lin <renekaelin at gmx.ch> wrote:
> The problem came from this:
>
> I'd like to have one unique routine that simplifies both of the following
> expressions to zero.
>
> sqrt((1+sqrt(2))^2)-(1+sqrt(2));
> sqrt((1+sqrt(2))^2)-sqrt(1+2*sqrt(2)+2);
>
> I tried
> sqrt((1+sqrt(2))^2)-(1+sqrt(2)); --> output: 0
> but
> sqrt((1+sqrt(2))^2)-sqrt(1+2*sqrt(2)+2); --> output
> -sqrt(2^(3/2)+3)+sqrt(2)+1
>
> Compared with this:
> sqrt(expand((1+sqrt(2))^2))-(1+sqrt(2)); --> output:
> sqrt(2^(3/2)+3)-sqrt(2)-1
> and
> sqrt(expand((1+sqrt(2))^2))-sqrt(1+2*sqrt(2)+2); --> output: 0
>
> Thanks for your help!
>
>
> Am 23.11.2012 um 22:33 schrieb Stavros Macrakis <macrakis at alum.mit.edu>:
>
> I believe that the only factorization that Maxima does over quadratic
> integers is over Z[sqrt(-1)], the Gaussian integers, using the function
> gcfactor. In general (as you probably know), quadratic integers don't have
> unique factorizations.
>
> If you're interested in this topic, you might want to look at:
> http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/quadraticgrad.pdf .
> It would be great if you could implement the relevant algorithms for
> Maxima!
>
> -s
>
> On Fri, Nov 23, 2012 at 3:39 PM, Ren? K?lin <renekaelin at gmx.ch> wrote:
>
>> Hello
>>
>> Is it possible to factor 1+2*sqrt(2)+2? I'd like to get (1+sqrt(2))^2.
>> _______________________________________________
>> Maxima mailing list
>> Maxima at math.utexas.edu
>> http://www.math.utexas.edu/mailman/listinfo/maxima
>>
>
>
>
>