# [Maxima] (no subject)

Robert Marik marik at mendelu.cz
Wed May 28 01:05:55 CDT 2008

```Thank for explanation. Maxima manual states the following

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(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

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.
>
```