Thank you for the hint, stavros.
Well, I tried a more general case:
sqrt((x+y)^2)-abs(x+y); --> output: 0
sqrt((x+y)^2)-sqrt(x^2+2*x*y+y^2); --> output: not 0
Here, the sqrtdenest-function doesn't simplify both expressions to 0, either.
Am 24.11.2012 um 17:52 schrieb Stavros Macrakis <macrakis at alum.mit.edu>:
> 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.
>> _______________________________________________
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>>
>>
>
>