In the first case, the general simplifier has already simplified the
expression by the time radcan sees it:
(%i5) sqrt((x+1)^2);
(%o5) abs(x + 1)
radcan intentionally treats sqrt(x) as meaning *either* the positive or
negative square root, so doesn't return abs. But in this case, the abs has
already been introduced by the general simplifier.
-s
On Sat, Feb 26, 2011 at 05:51, Lo?c <xlogo at free.fr> wrote:
> Hi list
>
> I don't understand some simplifications with radcan:
>
> (%i5) radcan(sqrt((x+1)^2));
> (%o5) abs(x + 1)
> (%i6) radcan(sqrt(x^2+2*x+1));
> (%o6) x + 1
>
> Seems to be a bug, no?
>
> Best
>
>
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