[Maxima-bugs] [ maxima-Bugs-3577666 ] radcan seems to give non-equvalent function

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Bugs item #3577666, was opened at 2012-10-16 08:03
Message generated for change (Comment added) made by rtoy
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Category: Lisp Core - Simplification
Group: None
>Status: Pending
>Resolution: Invalid
Priority: 5
Private: No
Submitted By: Jean-Yves (jyoberle)
Assigned to: Nobody/Anonymous (nobody)
Summary: radcan seems to give non-equvalent function

Initial Comment:
Hi,

When doing radcan(-(2*x)/(sqrt(1-(x^2-1)^2)*sqrt(1-asin(x^2-1)^2))), the result is -2/(sqrt(2-x^2)*sqrt(1-asin(x^2-1))*sqrt(asin(x^2-1)+1)). But these two expressions are not equivalent. For x = -0.4, -(2*x)/(sqrt(1-(x^2-1)^2)*sqrt(1-asin(x^2-1)^2)) gives 20.01585798944382 wheras -2/(sqrt(2-x^2)*sqrt(1-asin(x^2-1))*sqrt(asin(x^2-1)+1)) gives -20.01585798944383.
-(2*x)/(sqrt(1-(x^2-1)^2)*sqrt(1-asin(x^2-1)^2)) is the result of diff(acos(asin(x^2-1)),x) if this can help.
Build info is build_info("5.28.0-2","2012-08-27 23:16:48","i686-pc-mingw32","GNU Common Lisp (GCL)","GCL 2.6.8").

Best regards,

Jean-Yves

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>Comment By: Raymond Toy (rtoy)
Date: 2012-10-30 18:21

Message:
Although the documentation of radcan isn't very clear on this, radcan is
expected to produce results like this.  I think the idea is if x is very
large, then both expressions are equivalent.  I think that's true for your
expressions.

If this is not what you want, use ratsimp or some other combination of
expand and factor.

Marking as pending/invalid.

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Comment By: Valery Lovchikov (lvch)
Date: 2012-10-18 23:19

Message:
use function factor

%i1 factor(sqrt(x^2-u*x^4));
%o1 sqrt(1-u*x^2)*abs(x)

but

%i1 radcan(sqrt(x^2-u*x^4));
%o1 x*sqrt(1-u*x^2)


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