[Maxima] (no subject)
marik at mendelu.cz
Wed May 28 01:05:55 CDT 2008
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
Is the documentation of radcan obsolete?
On Tue, 27 May 2008, Richard Fateman wrote:
> Well, it was deliberate when I wrote radcan.
> it is perhaps inconsistent with this:
> 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
> algebraically speaking, sqrt((x-1)^2) has TWO algebraic values, x-1 and
> 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
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