Thank for explanation. Maxima manual states the following
---------------------
When radexpand is false, certain transformations are inhibited. radcan
(sqrt (1-x)) remains sqrt (1-x) and is not simplified to %i sqrt (x-1).
radcan (sqrt (x^2 - 2*x + 11)) remains sqrt (x^2 - 2*x + 1) and is not
simplified to x - 1.
---------------------
Neglecting the typo in radcan (sqrt (x^2 - 2*x + 11)) which should be
probably radcan (sqrt (x^2 - 2*x + 1)), I get the answer x-1 also with
radexpand:false. Moreover, sqrt(1-x) remains sqrt(1-x) even with
radexpand:true.
Is the documentation of radcan obsolete?
Robert Marik
On Tue, 27 May 2008, Richard Fateman wrote:
> Well, it was deliberate when I wrote radcan.
>
> it is perhaps inconsistent with this:
> z:(x-1)^2;
> sqrt(z) ==> abs(x-1)
>
> though sqrt(expand(z)); does not do that simplification.
>
> the commercial macsyma returns abs(x-1), so someone decided differently
> there.
>
> algebraically speaking, sqrt((x-1)^2) has TWO algebraic values, x-1 and
> 1-x.
> note that neither one of them is abs(x-1).
> radcan chooses one of the values based on which goes to +inf as x goes to
> +inf.
>